Monomial Algebras, 2nd edition

Mathematics MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS Monomial Algebras, Second Edition presents algebraic, combina...

Author: Rafael H. Villarreal

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Mathematics

MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS

Monomial Algebras, Second Edition presents algebraic, combinatorial, and computational methods for studying monomial algebras and their ideals, including Stanley–Reisner rings, monomial subrings, Ehrhart rings, and blowup algebras. It emphasizes square-free monomials and the corresponding graphs, clutters, or hypergraphs.

Monomial Algebras

Bringing together several areas of pure and applied mathematics, this book shows how monomial algebras are related to polyhedral geometry, combinatorial optimization, and combinatorics of hypergraphs. It directly links the algebraic properties of monomial algebras to combinatorial structures (such as simplicial complexes, posets, digraphs, graphs, and clutters) and linear optimization problems.

Villarreal

Features • Presents computational and combinatorial methods in commutative algebra • Shows how to solve a variety of problems of monomial algebras • Covers various affine and graded rings, including Cohen–Macaulay, complete intersection, and normal • Examines their basic algebraic invariants, such as multiplicity, Betti numbers, projective dimension, and Hilbert polynomial • Contains more than 550 exercises and over 50 examples, many of which illustrate the use of computer algebra systems

Second Edition

New to the Second Edition • Four new chapters that focus on the algebraic properties of blowup algebras in combinatorial optimization problems of clutters and hypergraphs • Two new chapters that explore the algebraic and combinatorial properties of the edge ideal of clutters and hypergraphs • Full revisions of existing chapters to provide an up-to-date account of the subject

Monomial Algebras

MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS

Second Edition

Rafael H. Villarreal

K23008

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2/13/15 12:23 PM

MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS

Monomial Algebras Second Edition

Rafael H. Villarreal Centro de Investigación y de Estudios Avanzados del IPN Mexico City, Mexico

MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS

Series Editors John A. Burns Thomas J. Tucker Miklos Bona Michael Ruzhansky Chi-Kwong Li

Published Titles Application of Fuzzy Logic to Social Choice Theory, John N. Mordeson, Davender S. Malik and Terry D. Clark Blow-up Patterns for Higher-Order: Nonlinear Parabolic, Hyperbolic Dispersion and Schrödinger Equations, Victor A. Galaktionov, Enzo L. Mitidieri, and Stanislav Pohozaev Difference Equations: Theory, Applications and Advanced Topics, Third Edition, Ronald E. Mickens Dictionary of Inequalities, Second Edition, Peter Bullen Iterative Optimization in Inverse Problems, Charles L. Byrne Modeling and Inverse Problems in the Presence of Uncertainty, H. T. Banks, Shuhua Hu, and W. Clayton Thompson Monomial Algebras, Second Edition, Rafael H. Villarreal Set Theoretical Aspects of Real Analysis, Alexander B. Kharazishvili Signal Processing: A Mathematical Approach, Second Edition, Charles L. Byrne Sinusoids: Theory and Technological Applications, Prem K. Kythe Special Integrals of Gradshetyn and Ryzhik: the Proofs – Volume l, Victor H. Moll

Forthcoming Titles Actions and Invariants of Algebraic Groups, Second Edition, Walter Ferrer Santos and Alvaro Rittatore Analytical Methods for Kolmogorov Equations, Second Edition, Luca Lorenzi Complex Analysis: Conformal Inequalities and the Bierbach Conjecture, Prem K. Kythe Computational Aspects of Polynomial Identities: Volume l, Kemer’s Theorems, 2nd Edition Belov Alexey, Yaakov Karasik, Louis Halle Rowen Cremona Groups and Icosahedron, Ivan Cheltsov and Constantin Shramov Geometric Modeling and Mesh Generation from Scanned Images, Yongjie Zhang Groups, Designs, and Linear Algebra, Donald L. Kreher Handbook of the Tutte Polynomial, Joanna Anthony Ellis-Monaghan and Iain Moffat Lineability: The Search for Linearity in Mathematics, Juan B. Seoane Sepulveda, Richard W. Aron, Luis Bernal-Gonzalez, and Daniel M. Pellegrinao Line Integral Methods and Their Applications, Luigi Brugnano and Felice Iaverno Microlocal Analysis on Rˆn and on NonCompact Manifolds, Sandro Coriasco

Forthcoming Titles (continued) Nonlinear Functional Analysis in Banach Spaces and Banach Algebras: Fixed Point Theory Under Weak Topology for Nonlinear Operators and Block Operators with Applications Aref Jeribi and Bilel Krichen Partial Differential Equations with Variable Exponents: Variational Methods and Quantitative Analysis, Vicentiu Radulescu and Dusan Repovs Practical Guide to Geometric Regulation for Distributed Parameter Systems, Eugenio Aulisa and David S. Gilliam Reconstructions from the Data of Integrals, Victor Palamodov Special Integrals of Gradshetyn and Ryzhik: the Proofs – Volume ll, Victor H. Moll Stochastic Cauchy Problems in Infinite Dimensions: Generalized and Regularized Solutions, Irina V. Melnikova and Alexei Filinkov Symmetry and Quantum Mechanics, Scott Corry

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2015 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Version Date: 20150202 International Standard Book Number-13: 978-1-4822-3470-1 (eBook - PDF) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

Contents Preface

ix

Preface to First Edition

xv

1 Polyhedral Geometry and Linear Optimization 1.1 Polyhedral sets and cones . . . . . . . . . . . . . 1.2 Relative volumes of lattice polytopes . . . . . . . 1.3 Hilbert bases and TDI systems . . . . . . . . . . 1.4 Rees cones and clutters . . . . . . . . . . . . . . 1.5 The integral closure of a semigroup . . . . . . . . 1.6 Unimodularity of matrices and normality . . . . 1.7 Normaliz, a computer program . . . . . . . . . . 1.8 Cut-incidence matrices and integrality . . . . . . 1.9 Elementary vectors and matroids . . . . . . . . .

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1 1 20 28 39 46 48 50 51 55

2 Commutative Algebra 2.1 Module theory . . . . . . . . . . . . . . . 2.2 Graded modules and Hilbert polynomials 2.3 Cohen–Macaulay modules . . . . . . . . . 2.4 Normal rings . . . . . . . . . . . . . . . . 2.5 Valuation rings . . . . . . . . . . . . . . . 2.6 Krull rings . . . . . . . . . . . . . . . . . . 2.7 Koszul homology . . . . . . . . . . . . . . 2.8 A vanishing theorem of Grothendieck . . .

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61 61 76 79 86 92 94 104 107

3 Aﬃne and Graded Algebras 3.1 Cohen–Macaulay graded algebras 3.2 Hilbert Nullstellensatz . . . . . . 3.3 Gr¨obner bases . . . . . . . . . . . 3.4 Projective closure . . . . . . . . . 3.5 Minimal resolutions . . . . . . .

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111 111 123 126 132 134

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vi 4 Rees Algebras and Normality 4.1 Symmetric algebras . . . . . . . . . . . 4.2 Rees algebras and syzygetic ideals . . 4.3 Complete and normal ideals . . . . . . 4.4 Multiplicities and a criterion of Herzog 4.5 Jacobian criterion . . . . . . . . . . . .

CONTENTS

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141 141 142 145 159 165

5 Hilbert Series 5.1 Hilbert–Serre Theorem . . . . . . . . . . 5.2 a-invariants and h-vectors . . . . . . . . 5.3 Extremal algebras . . . . . . . . . . . . 5.4 Initial degrees of Gorenstein ideals . . . 5.5 Koszul homology and Hilbert functions . 5.6 Hilbert functions of some graded ideals .

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171 171 177 182 189 196 199

6 Stanley–Reisner Rings and Edge Ideals of Clutters 6.1 Primary decomposition . . . . . . . . . . . . . . . . . 6.2 Simplicial complexes and homology . . . . . . . . . . 6.3 Stanley–Reisner rings . . . . . . . . . . . . . . . . . 6.4 Regularity and projective dimension . . . . . . . . . 6.5 Unmixed and shellable clutters . . . . . . . . . . . . 6.6 Admissible clutters . . . . . . . . . . . . . . . . . . . 6.7 Hilbert series of face rings . . . . . . . . . . . . . . . 6.8 Simplicial spheres . . . . . . . . . . . . . . . . . . . . 6.9 The upper bound conjectures . . . . . . . . . . . . .

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201 201 209 212 226 234 246 250 255 258

7 Edge Ideals of Graphs 7.1 Graph theory . . . . . . . . . . . . . . 7.2 Edge ideals and B-graphs . . . . . . . 7.3 Cohen–Macaulay and chordal graphs . 7.4 Shellable and sequentially C–M graphs 7.5 Regularity, depth, arithmetic degree . 7.6 Betti numbers of edge ideals . . . . . . 7.7 Associated primes of powers of ideals .

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261 261 268 274 282 293 298 303

8 Toric Ideals and Aﬃne Varieties 8.1 Binomial ideals and their radicals . . . . . 8.2 Lattice ideals . . . . . . . . . . . . . . . . 8.3 Monomial subrings and toric ideals . . . . 8.4 Toric varieties . . . . . . . . . . . . . . . . 8.5 Aﬃne Hilbert functions . . . . . . . . . . 8.6 Vanishing ideals over ﬁnite ﬁelds . . . . . 8.7 Semigroup rings of numerical semigroups . 8.8 Toric ideals of monomial curves . . . . . .

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311 311 316 326 335 342 345 347 352

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9 Monomial Subrings 9.1 Integral closure of monomial subrings 9.2 Homogeneous monomial subrings . . . 9.3 Ehrhart rings . . . . . . . . . . . . . . 9.4 The degree of lattice and toric ideals . 9.5 Laplacian matrices and ideals . . . . . 9.6 Gr¨obner bases and normal subrings . . 9.7 Toric ideals generated by circuits . . . 9.8 Divisor class groups of semigroup rings

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365 366 372 380 392 396 403 410 416

10 Monomial Subrings of Graphs 10.1 Edge subrings and ring graphs . . . . 10.2 Incidence matrices and circuits . . . . 10.3 The integral closure of an edge subring 10.4 Ehrhart rings of edge polytopes . . . . 10.5 Integral closure of Rees algebras . . . 10.6 Edge subrings of complete graphs . . . 10.7 Edge cones of graphs . . . . . . . . . . 10.8 Monomial birational extensions . . . .

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423 423 440 448 454 457 461 467 477

11 Edge Subrings and Combinatorial Optimization 11.1 The canonical module of an edge subring . . . . 11.2 Integrality of the shift polyhedron . . . . . . . . 11.3 Generators for the canonical module . . . . . . . 11.4 Computing the a-invariant . . . . . . . . . . . . . 11.5 Algebraic invariants of edge subrings . . . . . . .

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483 483 484 487 489 493

12 Normality of Rees Algebras of Monomial Ideals 12.1 Integral closure of monomial ideals . . . . . . . . 12.2 Normality criteria . . . . . . . . . . . . . . . . . 12.3 Rees cones and polymatroidal ideals . . . . . . . 12.4 Veronese subrings and the a-invariant . . . . . . 12.5 Normalizations of Rees algebras . . . . . . . . . . 12.6 Rees algebras of Veronese ideals . . . . . . . . . . 12.7 Divisor class group of a Rees algebra . . . . . . . 12.8 Stochastic matrices and Cremona maps . . . . .

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499 499 505 508 513 519 524 529 531

13 Combinatorics of Symbolic Rees Algebras of Edge Ideals of Clutters 13.1 Vertex covers of clutters . . . . . . . . . . . 13.2 Symbolic Rees algebras of edge ideals . . . 13.3 Blowup algebras in perfect graphs . . . . . 13.4 Algebras of vertex covers of graphs . . . . . 13.5 Edge subrings in perfect matchings . . . . .

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537 537 540 551 555 558

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viii

CONTENTS

13.6 Rees cones and perfect graphs . . . . . . . . . . . . . . . . . . 561 13.7 Perfect graphs and algebras of covers . . . . . . . . . . . . . . 564 14 Combinatorial Optimization and Blowup Algebras 14.1 Blowup algebras of edge ideals . . . . . . . . . . . . 14.2 Rees algebras and polyhedral geometry . . . . . . . . 14.3 Packing problems and blowup algebras . . . . . . . . 14.4 Uniform ideal clutters . . . . . . . . . . . . . . . . . 14.5 Clique clutters of comparability graphs . . . . . . . . 14.6 Duality and integer rounding problems . . . . . . . . 14.7 Canonical modules and integer rounding . . . . . . . 14.8 Clique clutters of Meyniel graphs . . . . . . . . . . .

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567 568 570 583 599 610 615 628 633

Appendix Graph Diagrams 635 A.1 Cohen–Macaulay graphs . . . . . . . . . . . . . . . . . . . . . 635 A.2 Unmixed graphs . . . . . . . . . . . . . . . . . . . . . . . . . 638 Bibliography

639

Notation Index

669

Index

673

Preface The main purpose of this book is to introduce algebraic, combinatorial, and computational methods to study monomial algebras and their presentation ideals, including Stanley–Reisner rings, monomial subrings, Ehrhart rings, blowup algebras, emphasizing square-free monomials and its corresponding graphs, clutters, or hypergraphs. Monomial algebras are related to various ﬁelds, for instance to numerical semigroups, semigroup rings, algebraic geometry, commutative algebra and combinatorics, integer programming and polyhedral geometry, graph theory and combinatorial optimization. We develop links between the areas shown as the vertices of the following graph.

Polyhedral Geometry A A AA AA AA AA Monomial AA Algebras AA AA , l , AA l Combinatorial Combinatorics Optimization of Hypergraphs

This allows us to solve a variety of problems of monomial algebras using the methods of the other areas and vice versa. An eﬀort has been made to give a unifying presentation of the techniques and notions of these areas. In this book we are interested in the algebraic properties of monomial algebras that can be directly linked to combinatorial structures—such as simplicial

x

Preface

complexes, posets, digraphs, graphs, and clutters—and to linear optimization problems. We study various types of aﬃne and graded rings (Cohen– Macaulay, sequentially Cohen–Macaulay, Gorenstein, Artinian, complete intersection, unmixed, normal, reduced) and examine their basic algebraic invariants (type, multiplicity, a-invariant, regularity, Betti numbers, Krull dimension, projective dimension, h-vector, Hilbert polynomial). In recent years the algebraic properties of blowup algebras have been linked to combinatorial optimization problems of clutters and hypergraphs. In this edition, we include four new chapters (Chapters 1, 11, 13, and 14) to introduce this area of research and to present some of the main advances. The study of algebraic and combinatorial properties of the edge ideal of clutters and hypergraphs has attracted a great deal of interest in the last two decades [163, 189, 224, 326, 408]. In the present edition we include two chapters with some of the advances in the area (Chapters 6 and 7). In order to present an up-to-date account of the subject, we have made a full revision of all chapters in the ﬁrst edition. The chapters have been reorganized and arranged in a diﬀerent order. In particular Chapters 9 and 12 were originally a single chapter, while Chapters 8 and 10 were originally divided into two chapters for each of them. This book brings together several areas of pure and applied mathematics. It contains over 550 exercises and over 50 examples, many of them illustrating the use of computer algebra systems. It has extensive indices of terminology and notation. The contents of this book are as follows. In Chapter 1 we begin by introducing several notions and results coming from polyhedral geometry, combinatorial optimization, and linear programming. Relative volumes of lattice polytope are studied here. We present relations between Hilbert bases, TDI linear systems of inequalities, max-ﬂow min-cut properties of clutters, and normality of aﬃne and Rees semigroups. For instance the max-ﬂow min-cut property is classiﬁed in terms of Hilbert bases and the integrality of certain polyhedra. This is used in Chapter 14 to prove that the edge ideal of a clutter is normally torsion-free if and only if the clutter has the max-ﬂow min-cut property. We present a slight generalization of a theorem of Lucchesi and Younger [298] that is useful to detect TDI systems arising from incidence matrices of digraphs. This is used in Chapter 11 to express the a-invariant of the edge subring of a bipartite graph in terms of directed cuts. Elementary integral vectors and matroids are introduced at the end of the chapter. The notion of a matroid will appear in several places in this book in connection with monomial rings and ideals. The computer program we have used to study linear systems of inequalities and to compute Hilbert bases of rational cones is Normaliz [68]. The prerequisite for this chapter is a course on linear algebra and familiarity with point set topology. Chapter 2 discusses certain topics and results on commutative algebra

Preface

xi

(module theory, normal and graded rings). It includes some detailed proofs, and points the reader to the appropriate references when proofs are omitted. However we make free use of the standard terminology and notation of homological algebra (including Tor and Ext) as described in [363] and [428]. This edition has three new sections on valuation rings, Krull rings, and a vanishing theorem of Grothendieck (Sections 2.5, 2.6, and 2.8). A number of topics connected to aﬃne and graded algebras are studied in Chapter 3, e.g., Gr¨obner bases, Hilbert Nullstellensatz and aﬃne varieties, projective closure, minimal resolutions, and Betti numbers. We present two versions of the Noether normalization lemma and some of its applications to aﬃne algebras and to Cohen–Macaulay graded algebras. In Chapter 4 a thorough presentation of complete and normal ideals is given. Here the systematic use of blowup algebras makes clear their importance for the area. In this chapter we introduce symbolic powers of ideals and normally torsion-free ideals. Then we present a few cases where equality of symbolic and ordinary powers can be described in terms of properties of the associated graded ring. An elegant and useful Cohen– Macaulay criterion due to Herzog is included here. Chapter 5 deals with the Hilbert series of graded modules and algebras, a topic that is quite useful in Stanley’s proof of the upper bound conjecture for simplicial spheres. The h-vector and a-invariant of a graded algebra are deﬁned through the Hilbert–Serre theorem. For Cohen–Macaulay graded algebras we present the main properties of their h-vectors and a-invariants. Some optimal upper bounds for the number of generators in the least degree of Gorenstein and Cohen–Macaulay graded ideals are given, which naturally leads to the notion of an extremal algebra. As an application the Koszul homology of Cohen–Macaulay ideals with pure resolutions is studied using Hilbert function techniques. This edition has a new section on Hilbert functions of a certain type of graded ideals that occur in algebraic coding theory (Section 5.6). Chapter 6 is an introduction to monomial ideals, Stanley–Reisner rings, and edge ideals of clutters. An understated goal here is to highlight some of the works of T. Hibi, J. Herzog, M. Hochster, G. Reisner, and R. Stanley. We study algebraic and combinatorial properties of edge ideals and simplicial complexes. In particular we examine shellable, unmixed, and sequentially Cohen–Macaulay simplicial complexes and their corresponding edge ideals and invariants (projective dimension, regularity, depth, Hilbert series). As for applications, the proofs of the upper bound conjectures for polytopes are discussed to give a ﬂavor to some of the methods and ideas of the area. Monomial ideals can also be used to solve certain problems of general polynomial ideals using the theory of Gr¨ obner bases. In Chapter 7 we give an introduction to graph theory and study algebraic properties and invariants of edge ideals using the combinatorial structure of

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graphs. The study of edge ideals of graphs has been a very active area of research in the last decade because of its connections to graph theory. We present classiﬁcations of the following families: unmixed bipartite graphs, Cohen–Macaulay bipartite graphs, Cohen–Macaulay trees, shellable bipartite graphs, sequentially Cohen–Macaulay bipartite graphs. The invariants examined here include regularity, depth, Krull dimension, and multiplicity. Edge ideals of graphs are shown to have the persistence property. In Chapter 8 we study algebraic and geometric aspects of three special types of polynomial ideals and their quotient rings: toric ideals → lattice ideals → binomial ideals. These ideals are interesting from a computational point of view and are related to diverse ﬁelds, such as combinatorics, algebraic geometry, integer programming, semigroup rings, coding theory, and algebraic statistics. Chapter 9 deals with point conﬁgurations and their lattice polytopes. We consider monomial subrings and binomial ideals associated with them, e.g., Ehrhart rings, Rees algebras, homogeneous subrings, lattice ideals, and toric ideals. Using Hilbert functions, polyhedral geometry, and Gr¨ obner bases, we study normalizations of monomial subrings, initial ideals of toric ideals, normal monomial subrings, primary decompositions and multiplicities of lattice ideals. Algebraic graph theory is used to study matrix ideals of Laplacian matrices. The reciprocity law of Ehrhart for integral polytopes and the Danilov–Stanley formula for canonical modules of monomial subrings will be introduced here. Applications of these results will be given. In Chapter 10 we study monomial subrings associated with graphs and their toric ideals. We relate the even closed walks and circuits of the vector matroid of a graph with Gr¨obner bases theory. A description of the integral closure of the edge subring of a multigraph will be presented along with a description of the circuits of its toric ideal. We study the family of graphs whose number of primitive cycles equals its cycle rank. It is shown that this family is precisely the family of ring graphs. These graphs are characterized in algebraic and combinatorial terms. We classify edge subrings of bipartite graphs which are complete intersections. Then we present sharp upper bounds for the multiplicity of edge subrings. Several connections between monomial subrings, graph theory, and polyhedral geometry will occur in this chapter. We study in detail the irreducible representation of an edge cone of a graph and show some applications to graph theory (for instance we show the marriage theorem). Chapter 11 focuses on edge subrings of connected bipartite graphs and their algebraic invariants. We show how to compute the canonical module and the a-invariant of an edge subring using linear programming techniques and introduce an integral polyhedron whose vertices correspond to minimal generators of the canonical module. The a-invariant of an edge subring will

Preface

xiii

be interpreted in combinatorial optimization terms as the maximum number of edge disjoint directed cuts. We study the Gorenstein property and the type of an edge subring. Chapter 12 is about Rees algebras of monomial ideals. We study the integral closure of a monomial ideal, and the normality and invariants of its Rees algebra. The normalization of a Rees algebra is examined using the Danilov–Stanley formula, Carath´eodory’s theorem, and Hilbert bases of Rees cones. Interesting classes of normal ideals, such as ideals of Veronese type and polymatroidal ideals are introduced in the chapter. The divisor class group of a normal Rees algebra is computed using polyhedral geometry. Chapter 13 shows the interaction between graph theory, combinatorial optimization, commutative algebra, and the theory of blowup algebras. In this chapter, we give a description—using notions from combinatorial optimization—of the minimal generators of the symbolic Rees algebra of the edge ideal of a clutter and show a graph theoretical description of the minimal generators of the symbolic Rees algebra of the ideal of covers of a graph. The minimal sets of generators of symbolic Rees algebras of edge ideals are studied using polyhedral geometry. Indecomposable graphs are related to the strong perfect graph theorem. We give a description—in terms of cliques—of the symbolic Rees algebra and the Simis cone of the edge ideal of a perfect graph. In Chapter 14 we relate combinatorial optimization with commutative algebra, and present applications to both areas. We establish some links between the algebraic properties of blowup algebras of edge ideals and the combinatorial optimization properties of clutters and polyhedra. A longstanding conjecture of Conforti and Cornu´ejols about packing problems is examined from an algebraic point of view. We study max-ﬂow min-cut problems of clutters, packing problems, and integer rounding properties of various systems of linear inequalities to gain insight about the algebraic properties of blowup algebras. Systems with integer rounding properties and clutters with the max-ﬂow min-cut property come from linear optimization problems. The equality between the Rees algebra and the symbolic Rees algebra of an edge ideal is characterized in combinatorial and algebraic terms. A number of properties of clutters and edge ideals are shown to be closed (under taking) minors, Alexander duals, and parallelizations. The max-ﬂow min-cut property of a clutter is characterized in algebraic and combinatorial terms. The structure of ideal uniform clutters is presented here. If a clutter satisﬁes the max-ﬂow min-cut property, we prove that all invariant factors of its incidence matrix are equal to 1 and that the columns of this matrix form a Hilbert basis. It is shown that the clique clutter of a comparability graph satisﬁes the max-ﬂow min-cut property. The normality of an ideal is described in terms of the integer rounding property of a linear system and we establish a duality theorem for monomial subrings. We show

xiv

Preface

that the Rees algebra of the ideal of covers of a perfect graph is normal and that clique clutters of Meynel graphs are Ehrhart clutters. This book stresses the use of computational and combinatorial methods in commutative algebra because they have been a major factor in discovering new and interesting results. The main computer algebra programs that we use in this book are Normaliz [68] and Macaulay2 [199]. These programs provide an invaluable tool to study monomial algebras and their algebraic invariants. As a handy reference we include a section summarizing the type of computations that can be done using Normaliz (see Section 1.7). We also occasionally use other computer algebra systems, such as CoCoA [88], TM R [431]. Singular [200] is another system Maple [80], and Mathematica that can be used for computations in commutative algebra with Gr¨ obner bases. Porta [84] and polymake [177] are two systems that can be used for computations in convex polyhedra and ﬁnite simplicial complexes. Combinatorial commutative algebra is an extensive area of mathematics. The main references for the area are the books of Bruns and Herzog [65], Hibi [240], Miller and Sturmfels [317], and Stanley [395]. This book emphasizes the use of discrete mathematics and combinatorial optimization methods in combinatorial commutative algebra. There are a number of excellent recent books that oﬀer a complementary view of the subject, namely, Beck and Robins [21], Bruns and Gubelazde [61], De Loera, Hemmecke and K¨oppe [105], Ene and Herzog [142], Herzog and Hibi [224], Huneke and Swanson [259], and Vasconcelos [414]. Outstanding references for computational and combinatorial aspects that complement and—in some cases—extend some of the material included here are the books of Berge [25, 27], Brøndsted [57], Chv´ atal [87], Cornu´ejols [93], De Loera, Rambau and Santos [106], Diestel [111], Cox, Little and O’Shea [99], Eisenbud [128], Ewald [151], Gitler and Villarreal [189], Godsil and Royle [190], Golumbic [191], Greuel and Pﬁster [200], Harary [208], Kreuzer and Robbiano [282], Schenck [369], Schrijver [372, 373], Stanley [394, 396], Sturmfels [400], Vasconcelos [413], and Ziegler [438]. We would like to thank Winfried Bruns, Enrique Reyes, and Aron Simis for their helpful comments and corrections. Thanks to executive editor Robert B. Stern for suggesting we prepare an up-to-date second edition of Monomial Algebras for the new series in mathematics Monographs and Research Notes in Mathematics, Taylor & Francis (Chapman and Hall/CRC Press Group). The support of Marsha Pronin and Samantha White at Taylor & Francis is much appreciated. Finally, we are grateful to Karen Simon for her careful editorial work on the manuscript. Rafael H. Villarreal Cinvestav-IPN Mexico City, D.F.

Preface to First Edition Let R = K[x] = K[x1 , . . . , xn ] be a polynomial ring in the indeterminates x1 , . . . , xn , over the ﬁeld K. Let fi = xvi = xv1i1 · · · xvnin , i = 1, . . . , q, be a ﬁnite set of monomials of R. We are interested in studying several algebras and ideals associated with these monomials. Some of these are: • the monomial subring: K[f1 , . . . , fq ] ⊂ K[x], • the Rees algebra: K[x, f1 t, . . . , fq t] ⊂ K[x, t], which is also a monomial subring, • the face ring or Stanley–Reisner ring: K[x]/(f1 , . . . , fq ), if the monomials are square-free, and • the toric ideal: the ideal of relations of a monomial subring. In the following diagram we stress the most relevant relations between the properties of those algebras that will take place in this text. Rees algebra @ I @ @ @ @ R @ Face ring Monomial subring Toric ideal If such monomials are square-free they are indexed by a hypergraph built on the set of indeterminates, which provides a second combinatorial structure in addition to the associated Stanley–Reisner simplicial complex.

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Preface to First Edition

This book was written with the aim of providing an introduction to the methods that can be used to study monomial algebras and their presentation ideals, with emphasis on square-free monomials. We have striven to provide methods that are eﬀective for computations. A substantial part of this volume is dedicated to the case of monomial algebras associated to graphs, that is, those deﬁned by square-free quadratic monomials deﬁning a simple graph. We will systematically use graph theory to study those algebras. Such a systematic treatment is a gap in the literature that we intend to ﬁll. Two outstanding references for graph theory are [48] and [208]. In the text, special attention is paid to providing means to determine whether a given monomial algebra or ideal is Cohen–Macaulay or normal. Those means include diverse characterizations and qualities of those two properties. Throughout this work base rings are assumed to be Noetherian and modules ﬁnitely generated. An eﬀort has been made to make the book self-contained by including a chapter on commutative algebra (Chapter 2) that includes some detailed proofs and often points the reader to the appropriate references when proofs are omitted. However, we make free use of the standard terminology and notation of homological algebra (including Tor and Ext) as described in [363] and [428]. The ﬁrst goal is to present basic properties of monomial algebras. For this purpose in Chapter 3 we study aﬃne and graded algebras. The topics include Noether normalizations and their applications, diverse attributes of Cohen–Macaulay graded algebras, Hilbert Nullstellensatz and aﬃne varieties, some Gr¨obner bases theory, and minimal resolutions. In Chapter 4 a thorough presentation of complete and normal ideals is given. Here the systematic use of Rees algebras and associated graded rings makes clear their importance for the area. Chapter 5 deals with the Hilbert series of graded modules and algebras, a topic that is quite useful in Stanley’s proof of the upper bound conjecture for simplicial spheres. Here we introduce the h-vector and a-invariant of graded algebras and give several interpretations of the a-invariant when the algebra is Cohen–Macaulay. Some optimal upper bounds for the number of generators in the least degree of Gorenstein and Cohen–Macaulay ideals are presented, which naturally leads to the notion of an extremal algebra. As an application the Koszul homology of Cohen–Macaulay ideals with pure resolutions is studied using Hilbert function techniques. General monomial ideals and Stanley–Reisner rings are examined in Chapter 6. The ﬁrst version of this chapter was some notes originally prepared to teach a short course during the XXVII Congreso Nacional de la Sociedad Matem´ atica Mexicana in October of 1994. In this course we pre-

Preface to First Edition

xvii

sented some applications of commutative algebra to combinatorics. We have expanded these notes to include a more complete treatment of shellable and Cohen–Macaulay complexes. The presentation of the last three sections of this chapter, discussing the Hilbert series of face rings and the upper bound conjectures, was inspired by [41, 65] and [393]. Since monomial algebras deﬁned by square-free monomials of degree two have an underlying graph theoretical structure it is natural that some interaction will occur between monomial algebras, graph theory, and polyhedral geometry. We have included three chapters that focus on monomial algebras associated to graphs. One of them is Chapter 7, where we present connections between graphs and ideals and study the Cohen–Macaulay property of the face ring. Another is Chapter 10, where we present a combinatorial description of the integral closure of the corresponding monomial subring and give some applications to graph theory. In Chapter 10 we consider monomial subrings and toric ideals of complete graphs with the aim of computing their Hilbert series, Noether normalizations, and Gr¨obner bases. The central topic of Chapters 9 and 12 is the normality of monomial subrings and ideals; some features of toric ideals are presented here. Chapter 8 is devoted to the study of monomial curves and their toric ideals, where the focus of our attention will be on monomial space curves and monomial curves in four variables. Aﬃne toric varieties and their toric ideals are studied in Chapter 8. Most of the material in this textbook has been written keeping in mind a typical graduate student with a basic knowledge of abstract algebra and a non-expert who wishes to learn the subject. We hope that this book can be read by people from diverse subjects and ﬁelds, such as combinatorics, graph theory, and computer algebra. Various units are accessible to upper undergraduates. In the last ﬁfteen years a dramatic increase in the number of research articles and books in commutative algebra that stress its connections with computational issues in algebraic geometry and combinatorics has taken place. Excellent references for computational and combinatorial aspects that complement some of the material included here are [99, 128, 413], [65, 395, 400] and [57, 438]. A constant concern during the writing of this text was to give appropriate credits for the proofs and results that were adapted from printed material or communicated to us. We apologize for any involuntary omission and would appreciate any comments and suggestions in this regard. During the fall of 1999 a course on monomial algebras associated to graphs was given at the University of Messina covering Chapter 7 to Chapter 10 with the support of the Istituto Nazionale Di Alta Matematica Francesco Severi. It is a pleasure to thank Vittoria Bonanzinga, Marilena Crupi, Gaetana Restuccia, Rossana Utano, Maurizio Imbesi, Giancarlo Rinaldo,

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Fabio Ciolli, and Giovanni Molica for the opportunity to improve those chapters and for their hospitality. We thank Wolmer V. Vasconcelos for his comments and encouragement to write this book. A number of colleagues and students provided helpful annotations to some early drafts. We are especially grateful to Adri´an Alc´ antar, Joe Brennan, Alberto Corso, Jos´e Mart´ınez-Bernal, Susan Morey, Carlos Renter´ıa, Enrique Reyes, and Aron Simis. We are also grateful to Laura Valencia for her competent secretarial assistance. The Consejo Nacional de Ciencia y Tecnolog´ıa (CONACyT) and the Sistema Nacional de Investigadores (SNI) deserve special acknowledgment for their generous support. It should be mentioned that the development of this book was included in the project Estudios sobre Algebras Monomiales, which was supported by the CONACyT grant 27931E. In the homepage “http://www.math.cinvestav.mx/∼vila” we will maintain an updated list of corrections. Rafael H. Villarreal Cinvestav-IPN Mexico City, D.F.

Chapter 1

Polyhedral Geometry and Linear Optimization In this chapter we introduce several notions and results from polyhedral geometry, combinatorial optimization and linear programming. Excellent general references for these areas are [35, 87, 281, 372, 373, 438]. Then we study relative volumes of lattice polytopes. We present various relations between Hilbert bases, TDI linear systems of inequalities, the max-ﬂow mincut property of clutters, integral linear systems, and the normality of aﬃne semigroups and Rees semigroups. Elementary integral vectors and matroids are introduced at the end of the chapter.

1.1

Polyhedral sets and cones

An aﬃne space or linear variety in Rn is by deﬁnition a translation of a linear subspace of Rn . Let A be a subset of Rn . The aﬃne space generated by A, denoted by aﬀ(A), is the set of all aﬃne combinations of points in A: aﬀ(A) = {a1 v1 + · · · + ar vr | vi ∈ A, ai ∈ R, a1 + · · · + ar = 1}. There is a unique linear subspace V of Rn such that aﬀ(A) = x0 + V , for some x0 ∈ Rn . The dimension of A is deﬁned as dim A = dim R (V ). Deﬁnition 1.1.1 An aﬃne map is a function between two aﬃne spaces that preserves aﬃne combinations. A point x ∈ Rn is called a convex combination of v1 , . . . , vr ∈ Rn if there are a1 , . . . , ar in R such that ai ≥ 0 for all i, x = i ai vi and i ai = 1. Let A be a subset of Rn . The convex hull of A, denoted by conv(A), is the set of all convex combinations of points in A. If A = conv(A) we say that A is a convex set.

2

Chapter 1

Deﬁnition 1.1.2 A convex polytope P ⊂ Rn is the convex hull of a ﬁnite set of points v1 , . . . , vr in Rn , that is, P = conv(v1 , . . . , vr ). The inner product of two vector x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ) in Rn is deﬁned by x, y = x · y = x1 y1 + · · · + xn yn . Deﬁnition 1.1.3 Let x = (x1 , . . . , xn ) be a point in Rn . The Euclidean norm of x is deﬁned as x = x, x. We shall always assume that a subset of Rn has the topology induced by the usual topology of Rn . Deﬁnition 1.1.4 A point a of a set A in Rn is said to be a relative interior point of A if there exists r > 0 such that Br (a) ∩ aﬀ(A) ⊂ A, where Br (a) = {x| x − a < r}. Let A ⊂ Rn . The set of relative interior points of A, denoted by ri(A), is called the relative interior of A. We denote the closure of A by A. The set A \ ri(A) is called the relative boundary of A and is denoted by rb(A), points in rb(A) are called the relative boundary points of A. If dim(A) = n, the relative interior of A is sometimes denoted by Ao . An aﬃne space of Rn of dimension n − 1 is called an aﬃne hyperplane. Given a ∈ Rn \ {0} and c ∈ R, deﬁne the aﬃne hyperplane H(a, c) = {x ∈ Rn | x, a = c}. Notice that any aﬃne hyperplane of Rn has this form. The two closed halfspaces bounded by H(a, c) are H + (a, c) = {x ∈ Rn | x, a ≥ c} and H − (a, c) = H + (−a, −c). If a is a rational vector and c is a rational number, the aﬃne hyperplane H(a, c) (resp. the halfspace H + (a, c)) is called a rational hyperplane (resp. rational halfspace). If c = 0, for simplicity the set Ha will denote the hyperplane of Rn through the origin with normal vector a, that is, Ha := H(a, 0) = {x ∈ Rn | x, a = 0}. The two closed halfspaces bounded by Ha are denoted by Ha+ = {x ∈ Rn | x, a ≥ 0} and Ha− = {x ∈ Rn | x, a ≤ 0}.

Polyhedral Geometry and Linear Optimization

3

Deﬁnition 1.1.5 A polyhedral set or convex polyhedron is a subset of Rn which is the intersection of a ﬁnite number of closed halfspaces of Rn . The set Rn is considered a polyhedron. The transpose of a matrix A (resp. vector x) will be denoted by At or A (resp. xt or x ). Often a vector x will denote a column vector or a row vector, from the context the meaning should be clear. Thus, a polyhedral set Q can be represented as

Q = {x ∈ Rn | Ax ≤ b} for some matrix A and for some vector b. As usual, if a = (ai ) and c = (ci ) are vectors in Rq , then a ≤ c means ai ≤ ci for all i. If A and b have rational entries, Q is called a rational polyhedron. Since the intersection of an arbitrary family of convex sets in Rn is convex, one derives that any polyhedral set is convex and closed. Deﬁnition 1.1.6 Let Q be a closed convex set in Rn . A hyperplane H of Rn is called a supporting hyperplane of Q if Q is contained in one of the two closed halfspaces bounded by H and Q ∩ H = ∅. Deﬁnition 1.1.7 A proper face of a polyhedral set Q is a set F ⊂ Q such that there is a supporting hyperplane H(a, c) satisfying the conditions: (a) F = Q ∩ H(a, c) = ∅, (b) Q ⊂ H(a, c), and Q ⊂ H + (a, c) or Q ⊂ H − (a, c). The improper faces of a polyhedral set Q are Q itself and ∅. Deﬁnition 1.1.8 A proper face F of a polyhedral set Q ⊂ Rn is called a facet of Q if dim(F ) = dim(Q) − 1. Proposition 1.1.9 If Q is a polyhedral set in Rn and F1 , F2 are faces of Q, then their intersection F = F1 ∩ F2 is a face of Q. Proof. Let Hi = H(ai , ci ) be a supporting hyperplane of Q, where ci ∈ R and 0 = ai ∈ Rn , such that Fi = Q ∩ Hi and Q ⊂ Hi+ for i = 1, 2. Next we prove the equality F = Q ∩ H(a1 + a2 , c1 + c2 ). The left-hand side is clearly contained in the right-hand side. On the other hand if x ∈ Q ∩ H(a1 + a2 , c1 + c2 ), then using x, ai ≥ ci one has: c1 + c2 = x, a1 + a2 = x, a1 + x, a2 ≥ c1 + c2 , hence x, ai = ci for i = 1, 2 and x ∈ F . As Q ⊂ H + (a1 + a2 , c1 + c2 ), the set F is a face of Q. 2

4

Chapter 1

Deﬁnition 1.1.10 A partially ordered set or poset is a pair P = (V, ), where V is a ﬁnite set of vertices and is a binary relation on V satisfying: (a) u u, ∀u ∈ V (reﬂexivity). (b) u v and v u, imply u = v (antisymmetry). (c) u v and v w, imply u w (transitivity). A poset P = (V, ) with vertex set V = {x1 , . . . , xn } can be displayed by an inclusion diagram , where xi is joined to xj by raising a line if xi ≺ xj and there is no other vertex in between. Example 1.1.11 The inclusion diagram of the divisors of 12 is: r12 6r @ H @r 4 HHr 3r 2 @ @r 1 A poset in which any two elements x, y have a greatest lower bound inf{x, y} and a lowest upper bound sup{x, y} is called a lattice. A lattice is called complete if inf S and sup S exist for any subset S. A mapping ϕ from one lattice L1 = (V1 , ) onto another lattice L2 = (V2 , ) is called an isomorphism when it is one-to-one and we have x y if and only if ϕ(x) ϕ(y) for all x, y ∈ V1 . Corollary 1.1.12 Let Q be a polyhedral set in Rn and let F be the set of all faces of Q. The partially ordered set (F , ⊂) is a complete lattice with the lattice operations inf G = ∩{F | F ∈ G} and sup G = ∩{G ∈ F | ∀F ∈ G; F ⊂ G}. Proof. It follows from Proposition 1.1.9 and from the fact that a convex polyhedron has only ﬁnitely many faces [427, Theorem 3.2.1(v)]. 2 Deﬁnition 1.1.13 The lattice (F , ⊂) is called the face-lattice of Q. Deﬁnition 1.1.14 A polytopal complex C is a ﬁnite collection of polytopes in Rn such that (i) ∅ ∈ C, (ii) if P ∈ C, then all faces of P are in C, and (iii) the intersection P ∩ Q of two polytopes P, Q ∈ C is a face of both P and Q. Similarly, one can deﬁne the notion of a polyhedral complex replacing polytope by polyhedron. The dimension of a polytopal complex C, denoted by dim(C), is the largest dimension of a polytope in C. Deﬁnition 1.1.15 Let P be a convex polytope. All proper faces of P form the boundary complex C(∂P), whose facets are the facets of P.

Polyhedral Geometry and Linear Optimization

5

The boundary complex C(∂P) of a polytope P of dimension d + 1 is a pure polytopal complex of dimension d [57, Chapter 2]. Deﬁnition 1.1.16 Let P be a convex polytope of dimension d + 1. The boundary complex C(∂P) is shellable if there is a linear ordering F1 , . . . , Fs of the facets of P such that for every 1 ≤ i < j ≤ s, there is 1 ≤ < j satisfying Fi ∩ Fj ⊂ F ∩ Fj and F ∩ Fj is a facet of Fj . Theorem 1.1.17 [58] The boundary complex of a polytope is shellable. The set of nonnegative real numbers and the set of nonnegative integers are denoted by: R+ = {x ∈ R| x ≥ 0} and N = {0, 1, 2, . . . } respectively, R+ is also denoted by R≥0 . If A is a set of points in Rn , the cone generated by A, denoted by R+ A or cone(A), is deﬁned as q R+ A = ai βi ai ∈ R+ , βi ∈ A for all i . i=1

The vector space spanned by A is denoted by RA. Given a vector v ∈ Rn , we deﬁne: R+ v := {λv| λ ∈ R+ }. Theorem 1.1.18 (Carath´eodory’s theorem) Let v1 , . . . , vq be a sequence of vectors in Rn not all of them zero. If x ∈ R+ v1 + · · · + R+ vq , then there is a linearly independent set V ⊂ {v1 , . . . , vq } such that x ∈ R+ V. Proof. By induction on q. The case q = 1 is clear. If q ≥ 2, we can write x = a1 v1 + · · · + aq vq , ai ≥ 0 for i = 1, . . . , q. One may assume that v1 , . . . , vq are linearly dependent and ai > 0 for all i, otherwise the result follows by induction. There are real numbers b1 , . . . , bq such that at least q one bi is positive and i=1 bi vi = 0. Setting 0 < c = min{ai /bi | bi > 0} = aj /bj , q we get x = x − c(b1 v1 + · · · + bq vq ) = i=1 (ai − cbi )vi , with ai − cbi ≥ 0 for all i and aj − cbj = 0. Hence, by induction, the result follows. 2 Deﬁnition 1.1.19 A set of vectors α1 , . . . , αq ∈ Rn is aﬃnely independent if forevery sequence λ1 , . . . , λq of real numbers satisfying qi=1 λi αi = 0 q and i=1 λi = 0, one has λi = 0 for all i. Corollary 1.1.20 Let v1 , . . . , vq be a set of vectors in Rn and let x be in its convex hull. Then there exists an aﬃnely independent set V ⊂ {v1 , . . . , vq } such that x ∈ conv(V).

6

Chapter 1

Proof. Consider B = {(v1 , 1), . . . , (vq , 1)}. Since (x, 1) belongs to R+ B, the result follows from Carath´eodory’s theorem and Exercise 1.1.67. 2 Deﬁnition 1.1.21 If C ⊂ Rn is closed under linear combinations with nonnegative real coeﬃcients, we say that C is a convex cone. A polyhedral cone is a convex cone which is also a polyhedral set. Proposition 1.1.22 If C is a closed convex cone and H is a supporting hyperplane of C, then H is a hyperplane passing through the origin. Proof. Let H = H(a, c). Since 0 ∈ C ⊂ H + , one has c ≤ 0. If C ∩H = {0}, then c = 0 as required. Assume C ∩ H = {0} and pick 0 = z ∈ C such that z, a = c. Using that tz ∈ C ⊂ H + (a, c) for all t ≥ 0, one derives tc = tz, a = tz, a ≥ c ∀ t ≥ 0 ⇒ c = 0.

2

An aﬃne space of dimension 1 is called a line. The following result is quite useful in determining the facets of a polyhedral cone. Proposition 1.1.23 Let A be a ﬁnite set of points in Zn and let F be a face of R+ A. The following hold. (a) If F = {0}, then F = R+ V for some V ⊂ A. (b) If dim(F ) = 1 and R+ A contains no lines, then F = R+ α with α ∈ A. (c) If dim(R+ A) = n and F is a facet deﬁned by the supporting hyperplane Ha , then Ha is generated by a linearly independent subset of A. Proof. Let F = R+ A ∩ Ha with R+ A ⊂ Ha− . Then F is equal to the cone generated by the set V = {α ∈ A| α, a = 0}. This proves (a). Parts (b) and (c) follow from (a). 2 Most of the notions and results considered thus far make sense if we replace R by an intermediate ﬁeld Q ⊂ K ⊂ R, i.e., we can work in the aﬃne space Kn . However with very few exceptions, like for instance Theorem 1.1.24, we will always work in the Euclidean space Rn or Qn . For convenience we state the fundamental theorem of linear inequalities: Theorem 1.1.24 [372, Theorem 7.1] Let Q ⊂ K ⊂ R be an intermediate ﬁeld and let C ⊂ Kn be a cone generated by A = {α1 , . . . , αq }. If α ∈ Kn \C and t = rank{α1 , . . . , αq , α}, then there exists a hyperplane Ha containing t−1 linearly independent vectors from A such that a, α > 0 and a, αi ≤ 0 for i = 1, . . . , q. Theorem 1.1.25 (Farkas’s Lemma) Let A be an n × q matrix with entries in a ﬁeld K and let α ∈ Kn . Assume Q ⊂ K ⊂ R. Then either there exists x ∈ Kq with Ax = α and x ≥ 0, or there exists v ∈ Kn with vA ≥ 0 and v, α < 0, but not both.

Polyhedral Geometry and Linear Optimization

7

Proof. Let A = {α1 , . . . , αq } be the set of column vectors of A. Assume that there is no x ∈ Kq with Ax = α and x ≥ 0, i.e., α is not in R+ A. By Theorem 1.1.24 there is a hyperplane Hv such that v, α < 0 and v, αi ≥ 0 for all i. Hence vA ≥ 0, as required. If both conditions hold, then 0 > v, α = v, Ax = vA, x ≥ 0, a contradiction. 2 The reader is referred to [438] for other versions of Farkas’s lemma. The next result tells us how to separate a point from a cone. Corollary 1.1.26 Let C ⊂ Kn be a cone generated by A = {a1 , . . . , am }. / C, then there is a hyperplane H through the origin such If γ ∈ Kn and γ ∈ that γ ∈ H − \H and C ⊂ H + . Proof. Let A be the matrix with column vectors a1 , . . . , am . By Farkas’s lemma (see Theorem 1.1.25) there exists μ ∈ Kn such that μA ≥ 0 and γ, μ < 0. Thus μ, ai ≥ 0 for all i. If H is the hyperplane through the 2 origin with normal vector μ we get C ⊂ H + , as required. Corollary 1.1.27 Let A be a ﬁnite set in Zn , then ZA ∩ R+ A = ZA ∩ Q+ A

and

Zn ∩ R+ A = Zn ∩ Q+ A,

where ZA is the subgroup of Zn spanned by A and Q+ = {x ∈ Q| x ≥ 0}. Proof. It follows at once from Theorem 1.1.25.

2

Deﬁnition 1.1.28 We say that C is a ﬁnitely generated cone if C = R+ A, for some ﬁnite set A = {v1 , . . . , vq }. The proof of the next result yields the duality theorem for cones. Theorem 1.1.29 If C ⊂ Rn , then C is a ﬁnitely generated cone (resp. ﬁnitely generated cone by rational vectors) if and only if C is a polyhedral cone (resp. rational polyhedral cone) in Rn . Proof. ⇒) Assume that C = (0) is a cone generated by A = {α1 , . . . , αm }. We set r = dim(R+ A). Notice that aﬀ(C) is the real vector space generated by A, because 0 ∈ C. If aﬀ(C) = C, then C is a polyhedral cone, because C is the intersection of n − r hyperplanes of Rn through the origin. Now assume that C aﬀ(C). Consider the family F = {F | F = Ha ∩ C; dim(F ) = r − 1; C ⊂ Ha− }. By Theorem 1.1.24 the family F is non-empty. Notice that F is a ﬁnite set because each F in F is a cone generated by a subset of A; see Proposition 1.1.23. Assume that F = {F1 , . . . , Fs }, where Fi = Hai ∩ C. We claim that the following equality holds C = Ha−1 ∩ · · · ∩ Ha−s ∩ aﬀ(C).

(1.1)

8

Chapter 1

The inclusion “⊂” is clear. To show the inclusion “⊃”, we proceed by contradiction. Assume that there exists α ∈ C such that α belongs to the right-hand side of Eq. (1.1). By Theorem 1.1.24 and by reordering the elements of A if necessary, there is a hyperplane Ha containing linearly independent vectors α1 , . . . , αr−1 such that (i) a, α > 0, and (ii) a, αi ≤ 0 for i = 1, . . . , m. Thus F = Ha ∩ C = Hak ∩ C for some 1 ≤ k ≤ s. We may assume that α1 , . . . , αr form a basis of aﬀ(C). Since α ∈ aﬀ(C) there are scalars λ1 , . . . , λr with α = λ1 α1 + · · · + λr αr . Using (ii) we get a, α = λr αr , a > 0. Hence λr < 0. On the other hand ak , αr < 0 because C ⊂ Ha−k and dim(Fk ) = r − 1. Therefore ak , α = λr ak , αr > 0, <0

<0

a contradiction to the fact that α ∈ Ha−k . ⇐) Assume that C = Hb−1 ∩ · · · ∩ Hb−r , where b1 , . . . , br ∈ Rn . Consider the cone C generated by b1 , . . . , br . From the ﬁrst part of the proof we can write C = R+ {b1 , . . . , br } = Hα−1 ∩ · · · ∩ Hα−m , (∗) for some set of vectors A = {α1 , . . . , αm } in Rn . Next we show the equality C = R+ A. Notice that bi , αj ≤ 0 for all i, j, because bi ∈ C for all i. Thus R+ A ⊂ C. Assume that there is α ∈ C \ R+ A. By Corollary 1.1.26, there exists a hyperplane Ha such that R+ A ⊂ Ha− and a, α > 0. Hence αi , a ≤ 0 for all i, and by Eq. (∗) we conclude that a ∈ R+ {b1 , . . . , br }. Therefore, we can write a = λ1 b1 + · · · + λr br , λi ≥ 0 for all i. Since α ∈ C, we have α, a = λ1 α, b1 + · · · + λr α, br ≤ 0, contradicting a, α > 0. Thus C = R+ A. The respective statement about the rationality character of the representations is left as an exercise. 2 Corollary 1.1.30 (Duality theorem for cones) Let B = {β1 , . . . , βr } be a subset of Rn , and let {α1 , . . . , αm } and {c1 , . . . , cs } be subsets of Rn \ {0}. − − − (a) If R+ B = ∩m i=1 Hαi , then Hβ1 ∩ · · · ∩ Hβr = R+ α1 + · · · + R+ αm .

(b) If R+ B = ∩si=1 Hc+i , then Hβ+1 ∩ · · · ∩ Hβ+r = R+ c1 + · · · + R+ cs . Proof. Part (a) follows from the second part of the proof of Theorem 1.1.29. − Part (b) follows using the equality Hc+i = H−c and using part (a). 2 i By Theorem 1.1.29 a polyhedral cone C Rn has two representations: Minkowski representation C = R+ B with B = {β1 , . . . , βr } a ﬁnite set, and Implicit representation C = Hc+1 ∩ · · · ∩ Hc+s for some c1 , . . . , cs ∈ Rn \ {0}.

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From the duality theorem for cones these two representations satisfy: Hβ+1 ∩ · · · ∩ Hβ+r = R+ c1 + · · · + R+ cs .

(1.2)

The dual cone of C is deﬁned as C ∗ := Hc+ = Ha+ . c∈C

a∈B ∗∗

By the duality theorem one has C = C. An implicit representation of C is called irredundant or irreducible if none of the closed half-spaces Hc+1 , . . . , Hc+s can be omitted from the intersection. Remark 1.1.31 The left-hand side of Eq. (1.2) is an irredundant representation of C ∗ if and only if no proper subset of B generates C. Corollary 1.1.32 Let C ⊂ Rn be a ﬁnitely generated cone and let F = {F | F = Ha ∩ C; dim(F ) = r − 1; C ⊂ Ha− } = {F1 , . . . , Fs } = ∅, where Fi = C ∩ Hai , dim(C) = r. Then C = aﬀ(C) ∩ Ha−1 ∩ · · · ∩ Ha−s . Proof. It follows from the ﬁrst part of the proof of Theorem 1.1.29.

2

One of the fundamental results in polyhedral geometry is the following remarkable decomposition theorem for polyhedra. See [372, Corollary 7.1b] and [427, Theorem 4.1.3] for historical comments and for more information about this result. Theorem 1.1.33 (Finite basis theorem) If Q is a set in Rn , then Q is a polyhedron (resp. rational polyhedron) if and only if Q can be expressed as Q = P + C, where P is a convex polytope (resp. rational polytope) and C is a ﬁnitely generated cone (resp. ﬁnitely generated rational cone). Proof. ⇒) Let Q = {x| Ax ≤ b} be a polyhedron in Rn , where A is a matrix and b is a vector. Consider the set

x n C = x ∈ R ; λ ∈ R+ ; Ax − λb ≤ 0 . λ Notice that C can be written as

x A −b x n ; λ ∈ R; ≤ 0 , C = x ∈ R λ 0 −1 λ where −b is a column vector. Thus C is a polyhedral cone in Rn+1 . Using Corollary 1.1.29 we get that C can be expressed as

x1 x C = R+ ,..., m (λi ≥ 0; xi ∈ Rn ). λ1 λm

10

Chapter 1

We may assume that λi ∈ {0, 1} for all i. Set A = {xi | λi = 1} = {x1 , . . . , xr }, B = {xi | λi = 0} = {xr+1 , . . . , xm }, P = conv(A), and C = R+ B. Notice that x ∈ Q if and only if (x, 1) ∈ C . Thus x ∈ Q if and only if (x, 1) can be written as x1 xr xr+1 x x + · · · + μr + μr+1 + · · · + μm m = μ1 1 1 0 0 1 with μi ≥ 0 for all i. Consequently x ∈ Q if and only if x ∈ P +C. Therefore we obtain Q = P + C. ⇐) Assume that Q is equal to P + C with P = conv(x1 , . . . , xr ) and C = R+ (xr+1 , . . . , xm ). Consider the following ﬁnitely generated cone

xr xr+1 x x1 C = R+ ,..., , ,..., m . 1 1 0 0 By Corollary 1.1.29 the cone C is a polyhedron. Thus there exists a matrix A and a vector b such that C can be written as

x n x ∈ R C = ; λ ∈ R; Ax + λb ≤ 0 . λ Since x ∈ Q if and only if (x, 1) ∈ C we conclude that Q = {x| Ax ≤ −b}, that is Q is a polyhedron. This proof is due to Schrijver [372]. 2 Thus, by the ﬁnite basis theorem, a polyhedron has two representations. The computer program PORTA [84] can be used to switch between these two representations. This program is available for Unix and Windows systems. For polyhedral cones one can use Normaliz [68]. Corollary 1.1.34 A set Q ⊂ Rn is a convex polytope if and only if Q is a bounded polyhedral set. Proof. If Q = conv(α1 , . . . , αm ) is a polytope, then by the triangle inequality for all x ∈ Q we have x ≤ α1 + · · · + αm . Thus Q is bounded. Conversely if Q is a bounded polyhedron, then by Theorem 1.1.33 we can decompose Q as Q = P + C, with P a polytope and C a ﬁnitely generated cone. Notice that C = {0}, otherwise ﬁxing p0 ∈ P and c0 ∈ C \ {0} we get p0 + λc0 ∈ Q for all λ > 0, a contradiction because Q is bounded. Thus P = Q, as required. 2 Deﬁnition 1.1.35 Let Q be a polyhedral set and x0 ∈ Q. The point x0 is called a vertex or an extreme point of Q if {x0 } is a proper face of Q. Proposition 1.1.36 [57, Theorem 7.2] If P = conv(α1 , . . . , αr ) ⊂ Rn and V is the set of vertices of P, then V ⊂ {α1 , . . . , αr } and P = conv(V ).

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Lemma 1.1.37 [57, Theorem 5.6] If Q is a polyhedral set in Rn and v ∈ Q is not a vertex of Q, then there is a face F of Q such that v ∈ ri(F ). Lemma 1.1.38 Let Q be a convex polyhedron in Rn and x0 ∈ Q. Then x0 is a vertex of Q if and only if Q \ {x0 } is a convex set. Proof. ⇒) Let H(a, c) be a proper supporting hyperplane of Q such that {x0 } = Q ∩ H(a, c) and Q ⊂ H + (a, c). Take x, y ∈ Q \ {x0 } and consider z = ty + (1 − t)x with 0 < t < 1. Note that x, y are not in H(a, c), thus z, a = ty, a + (1 − t)x, a > tc + (1 − t)c = c. Hence z, a > c, that is, z = x0 and z ∈ Q, as required. ⇐) Let Q = ∩ri=1 H + (ai , ci ) be a decomposition of Q as an intersection of closed halfspaces, where 0 = ai ∈ Rn and ci ∈ R for all i. First observe that x0 is in H(ai , ci ) for some i, otherwise if x0 , ai > ci for all i, there is an open ball Bδ (x0 ) in Rn , of radius δ centered at x0 , whose closure lies in Q. Hence taking two antipodal points x1 , x2 in the boundary of Bδ (x0 ) one obtains x0 ∈ Q \ {x0 }, a contradiction. One may now assume there is k ≥ 1 such that x0 ∈ H(ai , ci ) for i ≤ k and x0 , ai > ci for i > k. Set A = H(a1 , c1 ) ∩ · · · ∩ H(ak , ck ). We claim that A = {x0 }. If there is x1 ∈ A \ {x0 }, pick Bδ (x0 ) whose closure is contained in H + (ai , ci ) for all i > k. Since z = tx0 + (1 − t)x1 is in A for all t ∈ R, making t = 1 + δ/ x1 − x0 one derives z ∈ A and z − x0 = δ, thus z ∈ Q. Note that if k = r, then x1 is already in Q and in this case we set z = x1 . Making t = −1 in z1 = tz + (1 − t)x0 one concludes z1 ∈ A and z1 − x0 = δ. Altogether z, z1 are in Q \ {x0 } and x0 = (z + z1 )/2, a contradiction because Q \ {x0 } is a convex set. From the equality {x0 } = A = A∩Q one has that {x0 } is an intersection of faces. Using Proposition 1.1.9 yields that {x0 } is a face, as required. 2 Proposition 1.1.39 If Q = C + P ⊂ Rn with C a polyhedral cone and P a polytope, then every vertex of Q is a vertex of P. Proof. Let x be a vertex of Q and write x = c + p for some c ∈ C and p ∈ P. We claim that p is a vertex of P. If p is not a vertex of P, then by Lemma 1.1.37 there is a face F of P such that p ∈ ri(F ). Pick p = v ∈ F . By [57, Theorem 3.5], there is y ∈ F with p ∈]y, v[. Hence p = λy + (1 − λ)v with 0 < λ < 1. Note that x = y + c ∈ Q, x = v + c ∈ Q, and x = λ(y + c) + (1 − λ)(v + c)

(0 < λ < 1),

thus Q \ {x} is not a convex set, a contradiction to Lemma 1.1.38. This proves that p is a vertex of P. Assume c = 0, then p + λc = x = p + c = p + 2c

(0 < λ < 1).

12

Chapter 1

Notice that x = p+c = λ0 (p+λc)+(1−λ0 )(p+2c) is a convex combination, where 0 < λ0 = 1/(2 − λ) < 1. Since p + λc and p + 2c are in Q \ {x}, this shows that Q \ {x} is not a convex set, a contradiction. Thus c = 0 and x is a vertex of P, as required. 2 Theorem 1.1.40 [372, Theorem 8.4] Let Q ⊂ Rn be a polyhedral set. Then Q has at least one vertex if and only if Q does not contain any lines in Rn . Proposition 1.1.41 Let Q be a polyhedron containing no lines and let f (x) = x, a. If max{f (x)| x ∈ Q} < ∞, then the maximum is attained at a vertex of Q. Proof. Let x0 ∈ Q be an optimal solution and let λ = f (x0 ) be the optimal value. Note that F = Q ∩ H(a, λ) is a face of Q because Q ⊂ H − (a, λ) and x0 ∈ F . Since F contains no lines, by Theorem 1.1.40, the face F contains at least one vertex z0 , which is also a vertex of Q by transitivity. Thus the 2 optimal value λ is attained at the vertex z0 , as required. Theorem 1.1.42 If Q ⊂ Rn is a polyhedral set containing no lines and α1 , . . . , αr are the vertices of Q, then there are β1 , . . . , βs ∈ Rn such that Q = conv(α1 , . . . , αr ) + (R+ β1 + · · · + R+ βs ). 2

Proof. It follows from the proof of [427, Theorem 4.1.3]. Deﬁnition 1.1.43 If a polyhedron Q in Rn is represented as Q = aﬀ(Q) ∩ H + (a1 , c1 ) ∩ · · · ∩ H + (ar , cr )

(1.3)

with ai ∈ Rn \ {0}, ci ∈ R for all i, and none of the closed halfspaces H + (a1 , c1 ), . . . , H + (ar , cr ) can be omitted from the intersection, we say that Eq. (1.3) is an irreducible representation of Q. Theorem 1.1.44 [427, Theorem 3.2.1] Let Q be a polyhedral set in Rn which is not an aﬃne space and let Q = aﬀ(Q) ∩ H + (a1 , c1 ) ∩ · · · ∩ H + (ar , cr ) be an irreducible representation. If Fi = Q ∩ H(ai , ci ), i = 1, . . . , r, then (a) ri(Q) = {x ∈ Q| x, a1 > c1 , . . . , x, ar > cr }; (b) rb(Q) = F1 ∪ · · · ∪ Fr ; (c) the facets of Q are precisely the sets F1 , . . . , Fr ; (d) each face F Q is the intersection of all Fi such that F ⊂ Fi .

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Corollary 1.1.45 Let Q be a polyhedron and let β be a vector in ri(Q). If F is a proper face of Q, then β ∈ F . r Proposition 1.1.46 Let Q = i=1 H + (ai , ci ) be a polyhedral set in Rn which is not an aﬃne space, where 0 = ai ∈ Rn and ci ∈ R for all i. If x0 ∈ Rn , then x0 is a vertex of Q if and only if (a) {x0 } = ∩i∈I H(ai , ci ) for some I ⊂ {1, . . . , r}, and (b) x0 , ai ≥ ci for all i = 1, . . . , r. Proof. ⇒) This direction follows at once from the proof of Lemma 1.1.38. ⇐) Note x0 ∈ Q. Since the intersection of faces of Q is a face (see Proposition 1.1.9), one has that {x0 } is a face. 2 Corollary 1.1.47 Let A = (aij ) be an r×n matrix, let c = (ci ) be a column vector and let a1 , . . . , ar be the rows of A. If Q = {x ∈ Rn | Ax ≥ c} and x0 ∈ Q, then x0 is a vertex of Q if and only if there is J ⊂ {1, . . . , r} with |J| = n such that the set {ai | i ∈ J} is linearly independent and {x0 } = {(xi ) ∈ Rn | ai1 x1 + · · · + ain xn = ci for all i ∈ J}. Proof. Assume x0 is a vertex of Q. Hence, by Proposition 1.1.46, x0 is the unique solution of a system of linear equations: ai1 x1 + · · · + ain xn = ci (i ∈ I) for some I ⊂ {1, . . . , r}. Let [A |c ] be the augmented matrix of this system, by Gaussian elimination one obtains that this matrix reduces to In | c . 0 | 0 Hence the rank of A is equal to n. Thus there are n linearly independent rows of A and x0 is the unique solution of the system ai1 x1 + · · · + ain xn = ci (i ∈ J ⊂ I) for some J ⊂ I with |J| = n. The converse is clear.

2

Let Q be a polyhedron in Rn represented by a system of linear constraints (i = 1, . . . , r), ai , x ≥ bi where ai ∈ Rn and bi ∈ R for all i. With a slight abuse of language, we will say that the constraints are linearly independent if the corresponding ai are linearly independent.

14

Chapter 1

Deﬁnition 1.1.48 A vector x0 in Rn is called a basic feasible solution of a system of linear constraints ai , x ≥ bi , i = 1, . . . , r, if (a) x0 satisﬁes all linear constraints, and (b) out of the constraints that satisfy ai , x0 = bi , there are n of them that are linearly independent. The next result is a restatement of the corollary above. Corollary 1.1.49 Let Q be a polyhedron in Rn . A vector x0 in Rn is a vertex of Q if and only if x0 is a basic feasible solution of any system of linear constraints that represents Q. This result yields a method to ﬁnd the vertices of a polyhedron which is by no means the best in practice. See [10, 11] for a thorough discussion on how to ﬁnd all the vertices of a polyhedron. Corollary 1.1.50 If C = Rn is a polyhedral cone of dimension n, then there is a unique irreducible representation C = Ha+1 ∩ · · · ∩ Ha+r , where ai ∈ Rn \ {0}. Proof. According to Proposition 1.1.22 a set F is a proper face of C if there is a supporting hyperplane Ha of C such that F = C ∩ Ha = ∅ and C ⊂ Ha . In particular the facets of C are deﬁned by hyperplanes through the origin. Therefore by Theorem 1.1.44 the irreducible representation of C has the required form and is unique. 2 Proposition 1.1.51 If Q = Rn is a rational polyhedral cone of dimension n in Rn , then there are unique (up to sign) a1 , . . . , ar in Zn with relatively prime entries such that the irreducible representation of Q is Q = Ha+1 ∩ Ha+2 ∩ · · · ∩ Ha+r Proof. By the ﬁnite basis theorem, there are α1 , . . . , αq in Qn such that Q = R+ α1 + · · · + R+ αq . First note that if Hb is a supporting hyperplane of Q generated by a set of n − 1 linearly independent vectors in {α1 , . . . , αq }, then by the Gram–Schmidt process Hb has an orthogonal basis of vectors in Qn , and consequently there is a normal vector a to Hb such that a ∈ Qn and Ha = Hb . Hence multiplying a by a suitable integer and then dividing by the greatest common divisor of the entries, one may assume Ha = Hb , where a is in Zn and has relative prime entries. Observe that a, b are linearly dependent because the orthogonal complement of Ha is one dimensional. It is readily seen that a is uniquely determined up to sign. To complete the proof use Proposition 1.1.23 (c) to see that any facet 2 of Q is deﬁned by a supporting hyperplane Hb as above.

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Deﬁnition 1.1.52 A polyhedron containing no lines is called pointed. If F is a face of dimension 1 of a pointed polyhedral cone, then F = R+ v for some v (see Proposition 1.1.23). The face F is called an extremal ray. Lemma 1.1.53 [372, Section 8.8] If C is a pointed polyhedral cone in Rn , then C is generated by non-zero representatives of its extremal rays. Theorem 1.1.54 Let A be a ﬁnite set of non-zero points in Zn and let V be the set of all α ∈ A such that R+ α is a face of R+ A of dimension 1. If R+ A is a pointed cone, then R+ A = R+ V. Proof. We set A = {α1 , . . . , αq }. Let F be a face of R+ A of dimension 1. Then, by Proposition 1.1.23, F = R+ αi for some i. Thus, the result follows from Lemma 1.1.53. 2 Deﬁnition 1.1.55 Let A be an n × q real matrix and let b, c be two real vectors of sizes q and n, respectively. The primal problem is deﬁned as n Maximize f (x) = i=1 ci xi (∗) Subject to xA ≤ b and x ≥ 0. Its dual problem is deﬁned as Minimize g(y) = qi=1 bi yi

(∗∗)

Subject to Ay ≥ c and y ≥ 0. A solution x0 ∈ Rn of the primal (resp. dual) problem that maximizes the objective function f is called and optimal solution, the corresponding value f (x0 ) of the objective function is called an optimal value. The most important theorem in Linear Programming (LP) theory is: Theorem 1.1.56 (LP duality theorem, [372, Corollary 7.1.g]) If the primal problem (∗) has an optimal solution (x1 , . . . , xn ), then the dual q problem (∗∗) n has an optimal solution (y1 , . . . , yq ) such that i=1 ci xi = i=1 bi yi . Theorem 1.1.57 [372, Corollary 7.1g, p. 92] Let A be a matrix and b, c vectors. Then max{c, x | xA ≤ b} = min{b, y | y ≥ 0, Ay = c} provided that at least one of the sets in (1.4) is non-empty. An immediate consequence of the duality theorem is:

(1.4)

16

Chapter 1

Corollary 1.1.58 Consider the LP primal-dual pair max{x, c| xA ≤ b} = min{y, b| y ≥ 0; Ay = c}. Let x and y be feasible solutions, i.e., xA ≤ b, y ≥ 0 and Ay = c. Then the following conditions are equivalent : (a) x and y are both optimum solutions. (b) x, c = y, b. (c) (b − xA)y = 0. Condition (c) is called complementary slackness. See Exercise 1.1.81 for another form of complementary slackness Proposition 1.1.59 [281, Proposition 3.3] Let Q = {x ∈ Rn | xA ≤ b} = ∅ be a polyhedral set and let F = ∅ be a proper subset of Q. Then the following conditions are equivalent : (a) F = {x ∈ Q | A x = b } for some subsystem xA ≤ b of xA ≤ b. (b) F is a proper face of Q. (c) F = {x ∈ Q | x, c = δ} for some vector 0 = c ∈ Rn such that the maximum value δ = max{x, c | x ∈ Q} is ﬁnite. Deﬁnition 1.1.60 A minimal face of a polyhedron is a face not containing any other face. Lemma 1.1.61 [372, p. 104] Let Q = {x ∈ Rn | xA ≤ b} = ∅ be a convex polyhedron. A set F is a minimal face of Q if and only if ∅ = F Q and F is equal to F = {x ∈ Rn | A x = b } for some subsystem xA ≤ b of xA ≤ b. Deﬁnition 1.1.62 A rational polyhedron Q ⊂ Rn is called integral if Q is equal to conv(Zn ∩ Q). If Q is a pointed rational polyhedron, then Q is integral if and only if Q has only integral vertices. This is shown below. Theorem 1.1.63 [281, Theorem 5.12] If Q is a rational polyhedron, then Q is integral if and only if any of the following statements hold: (a) Each face of Q contains integral vectors. (b) Each minimal face of Q contains integral vectors. (c) max{c, x| x ∈ Q} is attained by an integral vector for each c for which the maximum is ﬁnite. (d) max{c, x| x ∈ Q} is an integer for each integral vector c for which the maximum is ﬁnite.

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Corollary 1.1.64 If Q is a pointed polyhedron, then Q is integral if and only if all vertices of Q are integral. Proof. Let x0 be a vertex of Q. There is a supporting hyperplane H = H(a, c) such that {x0 } = H ∩ Q and Q ⊂ H + . Since x0 is a convex combination of points in Q ∩ Zn , it is seen that x0 is integral. The converse follows from Theorem 1.1.63 and Proposition 1.1.41. 2 Proposition 1.1.65 Let Q ⊂ Rn be a rational polyhedron and let QI = conv(Q ∩ Zn ) be its integral hull. Then (a) QI is a polyhedron. (b) If Q is a pointed polyhedron, then QI is integral. Proof. We may assume that QI is non-empty. By the ﬁnite basis theorem we can write Q = P + C, where C is a cone generated by integral vectors α1 , . . . , αs and P is a polytope. Consider the linear map T : Rs → Rn induced by T (ei ) = αi . Notice that B := T ([0, 1]s ) is a polytope whose elements have the form λ1 α1 + · · · + λs αs , 0 ≤ λi ≤ 1 for all i. It is not hard to see that QI = conv((P +B)∩Zn )+C. Since P +B is bounded, we get that QI is a polyhedron by the ﬁnite basis theorem. This proves (a). To prove part (b) observe that the vertices of QI are contained in (P + B) ∩ Zn , this follows from Proposition 1.1.39. Hence by Corollary 1.1.64, QI is integral. This proof was adapted from [372]. 2

Exercises 1.1.66 If f : Rq → Rn is an aﬃne map, then f (x) = Ax + b, for some n × q matrix A and some vector b. 1.1.67 Let A = {α0 , . . . , αr } be a sequence of distinct vectors in Rn . Prove: (a) A is aﬃnely independent ⇔ α1 − α0 , . . . , αr − α0 are linearly independent ⇔ (α0 , 1), . . . , (αr , 1) are linearly independent. (b) If A is contained in an aﬃne hyperplane not containing the origin, then A is aﬃnely independent if and only if it is linearly independent. 1.1.68 If V ⊂ Rn , prove that V is an aﬃne space in Rn if and only if λ1 V + λ2 V ⊂ V for all λ1 , λ2 in R such that λ1 + λ2 = 1. 1.1.69 If C ⊂ Rn , prove that C is a convex cone if and only if λx + μy ∈ C for all x, y ∈ C and λ, μ ≥ 0. 1.1.70 Let π1 : R2 → R be the projection π1 (x1 , x2 ) = x1 . Prove that π1 is not a closed linear map. Hint Consider the closed set {(x, 1/x)| x > 0}.

18

Chapter 1

1.1.71 Let Q be a polyhedral set in Rn and let f : Rn → Rm be a linear function. Prove that f (Q) is a polyhedral set. 1.1.72 Let Q be a polyhedron in Rn and let f : Q → R be a linear function. If f is bounded from above and Q contains no lines, prove that f attains its maximum at a vertex of Q. Notice that f (Q) is a closed convex set. 1.1.73 Determine the facets of the convex polytope P = conv(±e1 , . . . , ±en ) ⊂ Rn . 1.1.74 Let C be a polyhedral cone in Rn and let C ∗ be its dual cone. Then C is a pointed cone if and only if dim(C ∗ ) = n. 1.1.75 If C = Rn and C ∗ is its dual cone, then C ∗ = (0). 1.1.76 If C = He+2 ⊂ R2 , then C = R+ e1 + R+ (−e1 ) + R+ e2 . Use the duality theorem to show that C ∗ = R+ e2 . + + 1.1.77 If C = R+ (1, 1), prove that C = H(1,−1) ∩ H(−1,1) ∩ He+1 and that + ∗ the dual cone is given by C = H(1,1) .

1.1.78 If C is a rational cone and CI := conv(C ∩ Zn ), then CI = C. 1.1.79 Let Q = {x| Ax ≤ b} = ∅ be a polyhedron. If Q = P + C, where P is a polytope and C is a polyhedral cone, prove that {y|Ay ≤ 0} = {y|x + y ∈ Q, ∀ x ∈ Q}. The cone C = {y|Ay ≤ 0} is called the characteristic cone of Q. 1.1.80 Let Q = {x| Ax ≤ b} be a polyhedron. The lineality space of Q is the linear space lin.space(Q) = {x| Ax = 0}. Prove that Q is pointed if and only if lin.space(Q) = (0). 1.1.81 (Complementary slackness) Consider the LP primal-dual pair min{x, c| x ≥ 0; xA ≥ b} = max{y, b| y ≥ 0; Ay ≤ c}. Let x and y be feasible solutions, i.e., xA ≥ b, Ay ≤ c and x, y ≥ 0. Use Theorem 1.1.56 to show that the following conditions are equivalent: (a) x and y are both optimum solutions. (b) x, c = y, b. (c) (b − xA)y = 0 and x(c − Ay) = 0.

Polyhedral Geometry and Linear Optimization

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1.1.82 (P. Gordan [194]) The system Ax < 0 is unsolvable if and only if the system yA = 0, y ≥ 0, y = 0 is solvable. The vector inequality (ai ) < (bi ) means that ai < bi for all i. In the following exercises K will denote an intermediate ﬁeld Q ⊂ K ⊂ R. 1.1.83 Let H ⊂ Kn be a linear subspace and {v1 , v2 , . . . , vm } a basis of H. Prove that H has an orthogonal basis using the Gram–Schmidt process: 1. Set u1 := v1 and u2 := v2 − ((v2 · u1 )/(u1 · u1 ))u1 . Verify that {u1 , u2 } is an orthogonal set and Ku1 + Ku2 = Kv1 + Kv2 . 2. If u1 , u2 , . . . , uk (k < m) have been constructed so that they form an orthogonal basis for the linear space generated by v1 , . . . , vk , setting vk+1 · u1 vk+1 · ui vk+1 · uk uk+1 = vk+1 − u1 + · · · + ui + · · · + uk , u1 · u1 ui · ui uk · uk verify that {u1 , u2 , . . . , uk , uk+1 } is an orthogonal set and that this set generates Kv1 + Kv2 + · · · + Kvk + Kvk+1 . 1.1.84 If A = (aij ) is a matrix with entries in K, then dim(RA ) is equal to dim(im(A)), where RA is the row space of A and im(A) is the image of the linear map deﬁned by A. 1.1.85 If H is a linear subspace of Kn , then H ⊕ H ⊥ = Kn , where H ⊥ = {x ∈ Kn | x · h = 0 for all h ∈ H}. 1.1.86 If A ∈ Mn×q (K), then (im(A))⊥ = ker(At ), im(A) ⊕ ker(At ) = Kn , and (im(At ))⊥ = ker(A). 1.1.87 Let α, β, γ be three vectors in Kn such that α, β ≤ 0 and β, γ > 0. n Prove that the vector α = α − α,β

γ,β γ is in K , α , β = 0, and α belongs to the cone generated by α and γ. 1.1.88 Using the previous exercises show that the proof of Farkas’s lemma given in [304, p. 86] is valid for any intermediate ﬁeld Q ⊂ K ⊂ R. 1.1.89 Show that the proof of the fundamental theorem of linear inequalities given in [372, Theorem 7.1] works for any intermediate ﬁeld Q ⊂ K ⊂ R. 1.1.90 If A ∈ Mn×q (K) and γ ∈ Kn , then either there exists a vector x ∈ Kq with Ax ≤ γ, or there exists a vector α ∈ Kn with α ≥ 0, αA = 0 and α, γ < 0 , but not both.

20

1.2

Chapter 1

Relative volumes of lattice polytopes

In this section we introduce relative volumes and unimodular coverings. Two main references for relative volumes are [21, 146]. Let A = {v1 , . . . , vq } be a set of distinct vectors in Zn and let P = conv(v1 , . . . , vq ) be the convex hull of A. A point in Zn is called a lattice point and P is called a lattice polytope. Here a lattice point (resp. lattice polytope) is used as a synonym of integral point (resp. integral polytope). In the sequel to simplify the exposition and the proofs we set m = q − 1 and α0 = v1 , . . . , αm = vq . If V = {0, α1 − α0 , . . . , αm − α0 } is the image of A under the translation f : Rn −→ Rn ,

f

x −→ x − α0 ,

then aﬀ(A) = α0 + RV. In particular, from this expression we get d = dim(P) = dimR (RV), where RV is the linear space spanned by V. Theorem 1.2.1 [261, Theorem 3.7] Let Zn be the free Z-module of rank n. Then any submodule of Zn is free of rank at most n. Using this result, and the inclusions ZV ⊂ Zn ∩ RV ⊂ RV, it follows that Zn ∩ RV is a free abelian group of rank d. A constructive proof of this fact, which is useful for computing relative volumes, is included below in the proof of Lemma 1.2.3. Theorem 1.2.2 [332, Theorem II.9, pp. 26-27] Let A be an integral matrix of rank d. Then there are invertible integral matrices U and Q such that U −1 AQ = diag{s1 , . . . , sd , 0, . . . , 0}, si > 0 for 1 ≤ i ≤ d and si divides si+1 for 1 ≤ i ≤ d − 1. The matrix D = diag{s1 , . . . , sd , 0, . . . , 0} is called the Smith normal form of A and s1 , . . . , sd are called the invariant factors of A. Lemma 1.2.3 RV ∩ Zn = Zγ1 ⊕ · · · ⊕ Zγd for some γ1 , . . . , γd in Zn . Proof. Consider the matrix A of size n × m whose column vectors correspond to the non-zero vectors in V. By Theorem 1.2.2, there are unimodular integral matrices U and Q, of orders n and m, respectively, such that U −1 AQ = D = diag{s1 , . . . , sd , 0, . . . , 0}

(si = 0; si |si+1 ∀ i).

Polyhedral Geometry and Linear Optimization

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Let u1 , . . . , un be the columns of U . We claim that the leftmost d columns of U can serve as γ1 , . . . , γd . We regard some vectors as column vectors. Take α ∈ RV ∩ Zn , thus α = Aβ for some β ∈ Rm . Set β = Q−1 β and denote the ith entry of β by βi . Notice the equalities α = U (Dβ ) and U −1 α = Dβ = (s1 β1 , . . . , sd βd , 0, . . . , 0)t . Hence si βi ∈ Z for all i, and consequently α is in Zu1 + · · · + Zud . For the reverse inclusion it suﬃces to prove that ui belongs to RV for i = 1, . . . , d. Using the equality (AQ)ei = (U D)ei (ei =ith unit vector) we derive (1 ≤ i ≤ d),

Aqi = U (si ei ) = si (U ei ) = si ui

where qi = Qei is the ith column of Q. Hence ui ∈ RV for i = 1, . . . , d. 2 Lemma 1.2.4 There is an aﬃne isomorphism φ : aﬀ(A) → Rd such that φ(aﬀ(A) ∩ Zn ) = Zd , where d = dim(P). Proof. By Lemma 1.2.3 one has RV ∩ Zn = Zγ1 ⊕ · · · ⊕ Zγd for some γ1 , . . . , γd in Zn . Therefore there is a linear map ψ : RV −→ Rd ,

ψ

γi −→ ei .

Hence φ = ψf : aﬀ(A) −→ Rd satisﬁes the required condition.

2

Observe that φ(P) is an integral polytope of dimension d with a positive Lebesgue measure denoted by m(φ(P)). Deﬁnition 1.2.5 The relative volume of the integral polytope P is: vol(P) := m(φ(P)). If P has dimension n, then the relative volume of P agrees with its usual volume. To see that the relative volume is independent of φ we need the following fact about the standard volume. Theorem 1.2.6 [427, Theorem 6.2.14] Let T : Rn → Rn be an aﬃne map given by T (x) = Ax + x0 , where A is a real matrix of order n and x0 ∈ Rn . If P is a bounded convex set in Rn , then T (P) has volume |det(A)|vol(P). Lemma 1.2.7 Let T : Zn → Zn be the aﬃne map given by T (x) = Ax + β, where A is an integral matrix of order n and β ∈ Zn . If T is bijective, then det(A) = ±1. Proof. Note that the matrix A determines a bijective linear map, thus A has an inverse with integral entries. Indeed, for each i there is a column vector βi in Zn such that Aβi = ei . Therefore A (β1 · · · βn ) = I, and consequently det(A) = ±1. 2

22

Chapter 1

Proposition 1.2.8 The relative volume of P is independent of φ. Proof. If φ1 is another aﬃne isomorphism φ1 : aﬀ(A) → Rd such that φ1 (aﬀ(A) ∩ Zn ) = Zd , then the aﬃne map φ1 φ−1 : Zd −→ Zd is bijective. Using Theorem 1.2.6 and Lemma 1.2.7 one obtains vol(φ(P)) = vol((φ1 φ−1 )(φ(P))) = vol(φ1 (P)).

2

In practice one can compute volumes of lattice polytopes using Normaliz [68] and Polyprob [104]. Example 1.2.9 We give an illustration in R3 of the procedure outlined above to compute relative volumes. We begin by setting A = {v1 , v2 , v3 , v4 } = {(1, 3, 1), (1, 5, 3), (4, 9, 4), (2, 8, 5)} and P = conv(A). Then V = {(0, 0, 0), (0, 2, 2), (3, 6, 3), (1, 5, 4)} and f : R3 −→ R3 ,

f

x −→ x − α0 ,

where α0 = (1, 3, 1). Consider the matrix ⎡ ⎤ 0 3 1 A = ⎣ 2 6 5 ⎦. 2 3 4 Next using Maple [80] to compute the Smith normal form of A we obtain invertible integral matrices U and Q such that U AQ = D, where ⎤ ⎤ ⎡ ⎤ ⎡ ⎡ −2 1 0 1 0 0 0 1 0 0 0 ⎦; U −1 = ⎣ 1 2 0 ⎦. U =⎣ 1 D = ⎣ 0 1 0 ⎦; 1 −1 1 0 0 0 1 1 1 By the proof of Lemma 1.2.3, the ﬁrst two columns of U −1 form a Z-basis for RV ∩Z3 . Using the aﬃne map ψ : RV → R3 induced by ψ(0, 1, 1) = (1, 0) and ψ(1, 2, 1) = (0, 1), and the map φ = ψf : aﬀ(A) → R, we get φ(P) = conv((0, 0), (2, 0), (0, 3), (3, 1)) and vol(P) = m(φ(P)) = 11/2. Proposition 1.2.10 ([176], [394, Proposition 4.6.30]) The relative volume of P is given by: |Zn ∩ iP| . vol(P) = lim i→∞ id Deﬁnition 1.2.11 For an abelian group (M, +) its torsion subgroup T (M ) is the set of all x in M such that px = 0 for some 0 = p ∈ N. The group M is torsion-free if T (M ) = (0).

Polyhedral Geometry and Linear Optimization

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Lemma 1.2.12 If B = {β1 , . . . , βs } ⊂ Zn , then (a) (RB ∩ Zn )/ZB = T (Zn /ZB), and

(∗)

(b) RB ∩ Zn = ZB if and only if Zn /ZB is torsion-free. Proof. (a): If z ∈ T (Zn /ZB) ⊂ Zn /ZB, then kz ∈ ZB for some 0 = k ∈ N, so z ∈ QB ⊂ RB. Since also z ∈ Zn , z is in the left-hand side of Eq. (∗). This shows the inclusion “⊃”. The other inclusion is left as an exercise. (b): This follows from (a). 2 Lemma 1.2.13 [397, pp. 32-33] If H ⊂ G are free abelian groups of the same rank d with Z-bases δ1 , . . . , δd and γ1 , . . . , γd related by δi = j gij γj , where gij ∈ Z for all i, j, then |G/H| = | det(gij )|. Theorem 1.2.14 [146] If A = {v1 , . . . , vq } ⊂ Zn is a set of vectors lying on an aﬃne hyperplane not containing the origin and P = conv(A) has dimension d, then vol(P) = |T (Zn /(v2 − v1 , . . . , vq − v1 ))| lim

i→∞

|ZA ∩ iP| . id

Proof. As in the beginning of this section for convenience of notation we set m = q − 1 and α0 = v1 , . . . , αm = vq . Let x0 ∈ Qn be such that αi , x0 = 1 for all i. As αi − α0 , x0 = 0, there is a decomposition ZA = Zα0 ⊕ ZV with V = {α1 − α0 , . . . , αm − α0 }. Pick δ1 , . . . , δd ∈ Zn such that ZV = Zδ1 ⊕ · · · ⊕ Zδd .

(1.5)

Therefore one can write αi = α0 + fi1 δ1 + · · · + fid δd

(fij ∈ Z).

(1.6)

Consider the lattice polytope P1 = conv((1, 0, . . . , 0), (1, f11 , . . . , f1d ), . . . , (1, fm1 , . . . , fmd )) ⊂ Rd+1 . ϕ

There is a bijective map ZA∩iP −→ Zd+1 ∩iP1 , namely ϕ is the restriction of the linear map from ZA to Zd+1 that maps each vector into its coordinate vector with respect to the basis {α0 , δ1 , . . . , δd }. Therefore |ZA ∩ iP| |Zd+1 ∩ iP1 | = lim = vol(P1 ). i→∞ i→∞ id id lim

To estimate vol(P1 ) consider the lattice polytope P2 = conv((0, . . . , 0), (f11 , . . . , f1d ), . . . , (fm1 , . . . , fmd )) ⊂ Rd

24

Chapter 1

and note that applying the translation α → α − e1 to P1 gives: vol(P1 ) =

vol(conv((0, . . . , 0), (0, f11 , . . . , f1d ), . . . , (0, fm1 , . . . , fmd ))).

Thus vol(P1 ) = vol(P2 ). The next step is to relate vol(P2 ) with vol(P) by obtaining a second expression for vol(P2 ). According to Lemma 1.2.3, there are γ1 , . . . , γd ∈ Zn such that RV ∩ Zn = Zγ1 ⊕ · · · ⊕ Zγd .

(1.7)

Writing δi = gi1 γ1 + · · · + gid γd

(gij ∈ Z; i, j = 1, . . . , d),

(1.8)

from Eqs. (1.6) and (1.8) one derives the matrix equality (fij )(gij ) = (cij ),

(1.9)

where αi − α0 = ci1 γ1 + · · · + cid γd , 1 ≤ i ≤ m. By deﬁnition vol(P) = vol(conv((0, . . . , 0), (c11 , . . . , c1d ), . . . , (cm1 , . . . , cmd ))). By Eq. (1.9), the linear transformation σ : Zd −→ Zd ,

σ

x −→ x(gij ),

satisﬁes σ(fi1 , . . . , fid ) = (ci1 , . . . , cid ). Hence by Theorem 1.2.6 we get vol(P) = vol(σ(P2 )) = | det(gij )|vol(P2 ). To ﬁnish the proof it suﬃces to show that | det(gij )| is the order of the torsion subgroup of Zn /ZV. Now we have (a)

(b)

T (Zn /ZV) = (RV ∩ Zn )/ZV = (Zγ1 ⊕ · · · ⊕ Zγd )/(Zδ1 ⊕ · · · ⊕ Zδd ), where in (a) we use Lemma 1.2.12, and in (b) the identities (1.5) and (1.7). Hence, from the identity (1.8) and Lemma 1.2.13, we get |T (Zn /ZV)| = | det(gij )|.

2

Proposition 1.2.15 Let α0 , . . . , αn be a set of aﬃnely independent points in Rn and Δ = conv(α0 , . . . , αn ). Then the volume of the simplex Δ is ⎞ ⎞ ⎛ ⎛ α0 1 α1 − α0 ⎟ ⎜ . ⎜ .. ⎟ .. det ⎝ .. det ⎝ ⎠ . ⎠ . αn 1 αn − α0 vol(Δ) = = . n! n!

Polyhedral Geometry and Linear Optimization

25

Proof. The result follows using linear algebra or applying the change of variables formula. See Theorem 1.2.6. 2 Deﬁnition 1.2.16 A set Δ in Rn is called a lattice d-simplex if Δ is the convex hull of a set of d + 1 aﬃnely independent points in Zn . Let Δ be a lattice d-simplex in Rn . The normalized volume of Δ is deﬁned as d!vol(Δ). If d = n, then from Proposition 1.2.15 one has vol(Δ) ≥

1 . n!

Deﬁnition 1.2.17 A lattice d-simplex Δ in Rn is called unimodular if d!vol(Δ) = 1. Proposition 1.2.18 Let Δ = conv(α0 , . . . , αd ) be a lattice d-simplex. If R(α1 − α0 , . . . , αd − α0 ) ∩ Zn = Zγ1 ⊕ · · · ⊕ Zγd for some γ1 , . . . , γd ∈ Zn and αi − α0 = j cij γj , then vol(Δ) =

| det(cij )| . d!

Proof. Note vol(Δ) = vol(conv(0, c1 , . . . , cd )), where ci = (ci1 , . . . , cin ). Hence the formula follows from Proposition 1.2.15. 2 Lemma 1.2.19 If β1 , . . . , βm ∈ Zn , then R(β1 , . . . βm ) ∩ Zn = Zβ1 + · · · + Zβm + Zδ1 + · · · + Zδs , where δ1 , . . . , δs generate the torsion subgroup of Zn /(β1 , . . . , βm ). Proof. It is straightforward and will be left as an exercise.

2

Lemma 1.2.20 If α0 , . . . , αm ∈ Zn , then there is an isomorphism of groups ϕ : T (Zn /(α1 − α0 , . . . , αm − α0 )) −→ T Zn+1 /((α0 , 1), . . . , (αm , 1)) , given by ϕ(α) = (α, 0). Here T (M ) denotes the torsion subgroup of M . Proof. The map ϕ is clearly well-deﬁned and injective. To prove that ϕ is onto consider an equation s(α, b) = λ0 (α0 , 1) + · · · + λm (αm , 1), where λi ∈ Z, 0 = s ∈ N, α ∈ Zn and b ∈ Z. Then sα

= λ0 α0 + · · · + λm αm ,

sb

= λ0 + · · · + λm .

Hence s(α − bα0 ) = λ1 (α1 − α0 ) + · · · + λm (αm − α0 ). It follows readily 2 that ϕ(α − bα0 ) = (α, b), as required.

26

Chapter 1

Proposition 1.2.21 Let α = α0 , . . . , αd be a set of aﬃnely independent vectors in Zn deﬁning a simplex Δ. Then the following are equivalent: (a) d!vol(Δ) = 1. (b) R(α1 − α0 , . . . , αd − α0 ) ∩ Zn = Z(α1 − α0 ) + · · · + Z(αd − α0 ). (c) Zn /Z{α1 − α0 , . . . , αd − α0 } is torsion-free. (d) Zn+1 /Z{(α0 , 1), . . . , (αd , 1)} is torsion-free. Proof. (a) ⇔ (b) follows from Proposition 1.2.18 and (b) ⇔ (c) ⇔ (d) follows from Lemmas 1.2.19 and 1.2.20. 2 Corollary 1.2.22 If Δ = conv(α0 , α1 , . . . , αd ) is a unimodular lattice dsimplex in Rn , then Δ ∩ Zn = {α0 , α1 , . . . , αd }. Proof. Let α ∈ Δ ∩ Zn . Then α = λ0 α0 + · · · + λd αd , where λi ∈ Q+ for all i and λ0 + · · · + λd = 1. Using the equality α − α0 = λ1 (α1 − α0 ) + · · · + λd (αd − α0 ) and Proposition 1.2.21(c) we get α − α0

= =

η1 (α1 − α0 ) + · · · + ηd (αd − α0 ) λ1 (α1 − α0 ) + · · · + λd (αd − α0 ),

where ηi ∈ Z for all i. Since α1 − α0 , . . . , αd − α0 are linearly independent we obtain λi = ηi for all i. Hence exactly one of the λi ’s is equal to 1 and the others are equal to zero, that is, α ∈ {α0 , α1 , . . . , αd }. 2 Deﬁnition 1.2.23 A lattice polytope P = conv(A) of dimension d is said to have a unimodular covering with support in A if there are simplices Δ1 , . . . , Δm of dimension d such that the following two conditions hold (i) P = ∪m i=1 Δi and the vertices of Δi are contained in A for all i. (ii) d!vol(Δi ) = 1. If condition (ii) is replaced by (ii)’ [ZA : ZAi ] = 1 for all i, where Ai denotes the vertex set of Δi , then we say that Δ1 , . . . , Δm is a weakly unimodular covering of P with support in A. For a discussion on the existence of unimodular covers of rational cones see [60] and the references therein.

Polyhedral Geometry and Linear Optimization

27

Proposition 1.2.24 If A = {v1 , . . . , vq } ⊂ Zn and the vectors in A lie in a hyperplane of Rn not containing the origin, then any unimodular covering of P = conv(A) with support in A is weakly unimodular. Proof. Let Δi = conv(Ai ) be any unimodular simplex of dimension d with vertex set Ai ⊂ A and d = dim(P). For convenience of notation assume that Ai = {v1 , . . . , vd+1 }. To show the equality ZA = ZAi it suﬃces to prove ZA ⊂ ZAi . Take an arbitrary vector α in ZA and write: α = (α, η) =

η1 v1 + · · · + ηq vq η1 (v1 , 1) + · · · + ηq (vq , 1)

(ηi ∈ Z), (η = η1 + · · · + ηq ).

Since A lies in an aﬃne hyperplane not containing the origin and Ai is aﬃnely independent, the set Ai is seen to be linearly independent. Therefore d + 1 = dim(R{(v1 , 1), . . . , (vd+1 , 1)}) = dim(R{(v1 , 1), . . . , (vq , 1)}). Thus one can write (α, η) = λ1 (v1 , 1) + · · · + λd+1 (vd+1 , 1), λi ∈ Q. By Proposition 1.2.21 the group Zn+1 /Z{(v1 , 1), . . . , (vd+1 , 1)} is torsion-free. Hence it follows from Lemma 1.2.12(b) that λi ∈ Z for all i, and consequently α is in ZAi . 2 Example 1.2.25 Let A = {v1 , . . . , v5 } be the set of vectors in Z6 given by v1 = (1, 0, 1, 0, 0, 1), v2 = (1, 0, 0, 1, 1, 0), v3 = (0, 1, 1, 0, 1, 0), v4 = (0, 1, 0, 1, 0, 1), v5 = (1, 0, 0, 1, 0, 1). Consider A1 = {v2 , v3 , v4 , v5 }, A2 = {v1 , v3 , v4 , v5 }, A3 = {v1 , v2 , v3 , v5 }, Δ1 = conv(A1 ),

Δ2 = conv(A2 ),

Δ3 = conv(A3 ),

and P = conv(A). The groups Z6 /ZA and Z6 /ZAi are torsion-free for all i. This readily implies that ZA = ZAi , i = 1, 2, 3. Using the relation v1 + v2 + v4 = v3 + 2v5 it is not hard to see that P = Δ1 ∪ Δ2 ∪ Δ3 . Thus Δ1 , Δ2 , Δ3 is a weakly unimodular covering. The relative volume of P can be computed using the procedure described before. It is left to the reader to verify that dim(P) = 3 and vol(P) = 1/2.

Exercises 1.2.26 Let Δ be an n-dimensional simplex in Rn with vertices α0 , . . . , αn in Zn . Prove that Δ is unimodular if and only if any of the following two equivalent conditions hold (a) Zn+1 = Z(α0 , 1) + · · · + Z(αn , 1). (b) Zn = Z(α1 − α0 ) + · · · + Z(αn − α0 ).

28

Chapter 1

1.2.27 Construct a lattice simplex of normalized volume greater than 1. Then prove that the converse of Proposition 1.2.24 fails. 1.2.28 If P is the integral polytope with vertices (0, 0) and (1, 3): y

6

3 2 1

0

s

s

1

2

x

Prove that the relative volume of P is equal to 1. 1.2.29 If P = conv((3, 1), (1, 3)), then vol(P) = 2. 1.2.30 Consider the set A = {e1 + e2 , e2 + e3 , e3 + e4 , e1 + e4 } and the polytope P = conv(A). Prove that dim(P) = 2 and vol(P) = 1. 1.2.31 Let P be the tetrahedron in R3 whose vertices are v1 = (1, 1, 0), v2 = (2, 1, −1), v3 = (2, −5, 0), v4 = (7, 1, −8). Prove that vol(P) = 2. Then verify this using Maple [80]. 1.2.32 If A = {v1 , . . . , vq } ⊂ Zn , prove that the rank of RA ∩ Zn , as a free abelian group, is equal to the dimension of RA as a real vector space.

1.3

Hilbert bases and TDI systems

The set of integral points of a rational polyhedral cone form a semigroup that arises in many branches of mathematics, such as combinatorial commutative algebra [65], toric varieties [176], and integer programming [372]. A ﬁnite set H ⊂ Rn is called a Hilbert basis if Zn ∩ R+ H = NH, where NH is the semigroup spanned by H consisting of all linear combinations of H with coeﬃcients in N. Notice that all vectors in a Hilbert basis are integral. A nice introduction to Hilbert bases can be found in [215]. Let C ⊂ Rn be a rational polyhedral cone. A ﬁnite set H is called a Hilbert basis of C if C = R+ H and H is a Hilbert basis. A Hilbert basis of C is minimal if it does not strictly contain any other Hilbert basis of C.

Polyhedral Geometry and Linear Optimization

29

Hilbert bases of rational polyhedral cones always exist. For a pointed cone there is only one minimal Hilbert basis. To prove these two facts, we need Gordan’s lemma. Below we give two versions of this lemma. Deﬁnition 1.3.1 Given a = (ai ) ∈ Rn , we deﬁne the value of a as: |a| = a1 + · · · + an . Lemma 1.3.2 (Gordan, version 1) If A = {v1 , . . . , vq } ⊂ Zn and ZA is the subgroup generated by A, then there exist γ1 , . . . , γm in Zn such that ZA ∩ R+ A = Nγ1 + · · · + Nγm and γi ∈ [N, M ]n for all i, where N = −q max |vi− | and M = q max |vi+ |. 1≤i≤q

1≤i≤q

Proof. Recall that C = ZA ∩ R+ A = ZA ∩ Q+ A; see Corollary 1.1.27. Let β ∈ C. Then one can write β=

q xi i=1

yi

vi ,

where xi ∈ N and 0 = yi ∈ N. By the division algorithm there are ri , ni in N such that xi = ni yi + ri and 0 ≤ ri < yi . Therefore one can write β=

q i=1

ni vi +

q

ai vi ,

i=1

q with ai ∈ [0, 1] ∩ Q. As i=1 ai vi ∈ C ∩ [N, M ]n , the set A ∪ (C ∩ [N, M ]n ) is a generating set for C with the required property. Our argument was based on the proof of Gordan’s lemma given in [65]. 2 Deﬁnition 1.3.3 A semigroup (S, +, 0) of Zn is said to be ﬁnitely generated if there exists a ﬁnite set Γ = {γ1 , . . . , γr } ⊂ S such that: S = NΓ := Nγ1 + · · · + Nγr . A set of generators Γ of S is called minimal if γi = 0 ∀i and none of its elements is a linear combination with coeﬃcients in N of the others. Remark 1.3.4 There are examples of subsemigroups of Nn , with n ≥ 2, which are not ﬁnitely generated; see Exercise 1.3.34. Lemma 1.3.5 (Gordan, version 2) If R+ A is a cone in Rn generated by a ﬁnite set A ⊂ Zn , then the semigroup Zn ∩ R+ A is ﬁnitely generated.

30

Chapter 1

Proof. If A = {v1 , . . . , vq }, consider the ﬁnite set of integral points: {a1 v1 + · · · + aq vq | 0 ≤ ai ≤ 1} ∩ Zn = {γ1 , . . . , γr }. It is left to the reader to prove that γ1 , . . . , γr is the required set of generators for Zn ∩ R+ A. See [372, p. 233]. 2 Let A be a ﬁnite set in Zn and let G = Zn or G = ZA. Then, by Gordan’s lemma (versions 1 and 2) there exists γ1 , . . . , γr ∈ Zn such that: G ∩ R+ A = Nγ1 + · · · + Nγr . Computing the γi ’s is in general diﬃcult [69]. Fortunately, the γi ’s can be computed using Normaliz [68]. Therefore Hilbert bases of rational polyhedral cones always exist: Proposition 1.3.6 Let A be a ﬁnite set in Zn . Then there exist γ1 , . . . , γr such that R+ A ∩ Zn = Nγ1 + · · · + Nγr , and H = {γ1 , . . . , γr } is a Hilbert basis of R+ A. Proof. The existence follows from Lemma 1.3.5. That H is a Hilbert basis of R+ A follows from the equality R+ A = R+ γ1 + · · · + R+ γr . 2 In Deﬁnition 1.1.52 we considered pointed polyhedra. For cones we have the following equivalent deﬁnition (see Exercise 1.3.36). Deﬁnition 1.3.7 A polyhedral cone C = {x| Ax ≤ 0} is called pointed if the lineality space {x| Ax = 0} is equal to {0}. Lemma 1.3.8 Let C be a rational polyhedral cone. If C is pointed, then there exists an integral vector b such that b, x > 0 for all 0 = x ∈ C. Proof. There is a rational matrix A such that C = {x| Ax ≤ 0}; see Theorem 1.1.29. We may assume that A is integral. If u1 , . . . , ut are the 2 rows of A, then b = −u1 − · · · − ut satisﬁes the required condition. Theorem 1.3.9 [371] Let A be a ﬁnite set in Zn and let C = R+ A. If C is pointed, then there exists a unique minimal Hilbert basis of C given by H = {x ∈ C ∩ Zn | 0 = x ∈ / Ny1 + Ny2 ; ∀y1 , y2 ∈ (C \ {0}) ∩ Zn }. Proof. Let Γ = {γ1 , . . . , γr } be an arbitrary Hilbert basis of C. We claim r that H ⊂ Γ. Let x ∈ H. Since Zn ∩ C = NΓ we can write x = i=1 ai γi , ai ∈ N. Thus, by construction of H, we get x = γi for some i, which proves the claim. To ﬁnish the proof it suﬃces to prove NH = NΓ, because this

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equality implies R+ H = R+ Γ and consequently Zn ∩ R+ H = NH. We proceed by contradiction by assuming that the set: V = {x| x ∈ NΓ = C ∩ Zn ; x ∈ / NH} is not empty. By Lemma 1.3.8 there exists b ∈ Zn such that b, x > 0 for all 0 = x ∈ C. Let x0 ∈ V such that x0 , b = min{x, b| x ∈ V}. Since x0 ∈ / H we can write x0 = x1 + x2 with x1 , x2 in (C \ {0}) ∩ Zn . Thus, since xi , b < x0 , b for i = 1, 2, we have x1 , x2 ∈ NH and x1 + x2 ∈ NH, a contradiction. 2 Deﬁnition 1.3.10 (Edmonds and Giles [125]) A rational system xA ≤ b of linear inequalities is called totally dual integral, abbreviated TDI, if the minimum in the LP-duality equation max{x, c| xA ≤ b} = min{y, b| y ≥ 0; Ay = c}

(1.10)

has an integral optimum solution y for each integral vector c for which the minimum is ﬁnite. Note that there are rational systems of linear inequalities which deﬁne the same polyhedron and such that one system is totally dual integral while the other is not (see Exercise 1.3.39). TDI systems occur in the theory of Gr¨ obner bases of toric ideals [252], in the theory of perfect graphs [296, 86], and in combinatorial commutative algebra [147, 185, 188]. See Theorems 9.6.21, 13.6.8, and 14.3.6. Proposition 1.3.11 [372, Corollary 22.1c] If xA ≤ b is a TDI-system and b is integral, then the polyhedron {x | xA ≤ b} is integral. Proof. By hypothesis min{y, b| y ≥ 0; Ay = c} has an integral optimum solution y for each integral vector c for which the minimum is ﬁnite. Then by the LP duality theorem (see Theorem 1.1.57), we get that max{x, c| xA ≤ b} is an integer for each integral vector c for which the maximum is ﬁnite. 2 Thus the polyhedron {x | xA ≤ b} is integral by Theorem 1.1.63. Deﬁnition 1.3.12 A set P in Rn is called a parallelotope if it is the image of [0, 1]n under a non-singular linear transformation, i.e., P has the form P = {λ1 v1 + · · · + λn vn | 0 ≤ λi ≤ 1} for some linearly independent vectors v1 , . . . , vn in Rn .

32

Chapter 1

Lemma 1.3.13 Let v1 , . . . , vn be a basis of Rn and let P = {λ1 v1 + · · · + λn vn | 0 ≤ λi ≤ 1} be a parallelotope. Then vol(P) = | det[v1 , . . . , vn ]|. Proof. Since P is the image of [0, 1]n under the linear map T induced by ei → vi , the formula follows from Theorem 1.2.6. 2 Proposition 1.3.14 Let A be an integral matrix of order n × n and let A = {v1 , . . . , vn } be the set of columns of A. If det(A) = 0, then A is a Hilbert basis if and only if | det(A)| = 1. Proof. ⇒) Let Q = [0, 1]n be the unit cube and let P be the parallelotope P = {λ1 v1 + · · · + λn vn | 0 ≤ λi ≤ 1}. By Lemma 1.3.13 one has vol(P) = | det(A)|. As A is a Hilbert basis and A is linearly independent, we have (k + 1)n = |kQ ∩ Zn | = |kP ∩ Zn |,

∀ k ∈ N.

Therefore by Proposition 1.2.10 and the fact that [0, 1]n has volume 1, we conclude that 1 = vol(P) = | det(A)|, as required. ⇐) Let α be a vector in R+ A ∩ Zn . As | det(A)| = 1 and since A is linearly independent, by Cramer’s rule, it follows that α ∈ NA. Thus A is a Hilbert basis. 2 Notation Let A = (0) be an integral matrix. The greatest common divisor of all the non-zero r × r sub determinants of A will be denoted by Δr (A). Theorem 1.3.15 [261, Theorem 3.9] Let A be an integral matrix of rank r and let d1 , . . . , dr be the invariant factors of A. Then d1 = Δ1 (A), d2 = Δ2 (A)Δ1 (A)−1 , . . . , dr = Δr (A)Δr−1 (A)−1 . The next result is called the fundamental structure theorem for ﬁnitely generated abelian groups. Theorem 1.3.16 [261, pp. 187-188] Let M be a ﬁnitely generated Z-module with a presentation M Zn /(a1 , . . . , aq ). If A is the matrix of rank r with columns a1 , . . . , aq and d1 , . . . , dr are the invariant factors of A, then M Z/(d1 ) ⊕ Z/(d2 ) ⊕ · · · ⊕ Z/(dr ) ⊕ Zn−r . Lemma 1.3.17 If A is an integral matrix of size n × q and rank r, then Δr (A) = |T (Zn /(a1 , . . . , aq ))|, where ai is the ith column of A. In particular Δr (A) = 1 if and only if the quotient group Zn /(a1 , . . . , aq ) is torsion-free.

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Proof. Let d1 , . . . , dr be the invariant factors of the matrix A. On one hand using Theorem 1.3.15, we get the equality d1 d2 · · · dr = Δr (A). On the other hand, by Theorem 1.3.16, we obtain an isomorphism T (Zn /(a1 , . . . , aq )) Z/(d1 ) ⊕ · · · ⊕ Z/(dr ). 2

Therefore the order of the torsion subgroup is Δr (A), as required.

Lemma 1.3.18 Let A = {v1 , . . . , vd } ⊂ Zn be a set of linearly independent vectors and let A be the matrix with column vectors v1 , . . . , vd . Then d!vol(conv(0, v1 , . . . , vd )) = Δd (A). Proof. Let Γ = {γ1 , . . . , γd } be a set of vectors such that RA ∩ Zn = Zγ1 ⊕ · · · ⊕ Zγd ,

(1.11)

see Lemma 1.2.3. Then we can write vi = ci1 γ1 + · · · + cid γd

(i = 1, . . . , d)

(1.12)

where C = (cij ) is an integral matrix. By Proposition 1.2.18, we have d!vol(conv(0, v1 , . . . , vd )) = | det(C)|. Using Eqs. (1.11) and (1.12), we get that the map T (Zn /ZA) −→ T (Zd /Z{c1 , . . . , cd }), α → [α]Γ , is an isomorphism of groups, where ci is the ith row of C and [α]Γ is the coordinate vector of α in the basis Γ. Hence, by Lemma 1.3.17, we get Δd (A) = |T (Zn /ZA)| = |T (Zd /Z{c1 , . . . , cd })| = | det(C)|.

2

Proposition 1.3.19 Let A = {v1 , . . . , vd } ⊂ Zn be a linearly independent set and let A be the matrix with column vectors v1 , . . . , vd . Then A is a Hilbert basis if and only if Δd (A) = 1. Proof. ⇒) Let C = (cij ) be as in the proof of Lemma 1.3.18. As A is a Hilbert basis, it is seen that the rows of C form a Hilbert basis. Then | det(C)| = 1 by Proposition 1.3.14. Therefore Δd (A) = 1 by Lemma 1.3.18. ⇐) If Δd (A) = 1, then Zn /ZA is torsion-free. Since the vi ’s are linearly independent, it follows readily that A is a Hilbert basis. 2 Theorem 1.3.20 [179] Let A = {v1 , . . . , vq } ⊂ Zn be a Hilbert basis and let r = rank (ZA). If R+ A is pointed, then there exist H ⊂ A such that H is linearly independent, H is a Hilbert basis and |H| = r.

34

Chapter 1

Corollary 1.3.21 If A = {v1 , . . . , vq } ⊂ Zn and R+ A is a pointed cone, then A is a Hilbert basis if and only if R+ A ∩ ZA = NA and Zn /ZA is a torsion-free group. Proof. Assume that A is a Hilbert basis. Let A be the matrix with column vectors v1 , . . . , vq and let r = rank (A). Then clearly R+ A ∩ ZA = NA. By Theorem 1.3.20 and Proposition 1.3.19, we get Δr (A) = 1. Thus Zn /ZA is a torsion-free group. The converse follows readily. Notice that for the 2 converse the hypothesis that the cone R+ A is pointed is not needed. Let P = {x | xA ≤ b} be a rational polyhedron and let F be a face of P. A column of A is active in F if the corresponding inequality in xA ≤ b is satisﬁed with equality for all vectors x in F . A minimal face of P is a face not containing any other face. A face F of P is minimal if and only if F is an aﬃne subspace [372]. Theorem 1.3.22 [372, Theorem 22.5] A rational system xA ≤ b is TDI if and only if for each minimal face F of the polyhedron P = {x | xA ≤ b}, the columns of A which are active in F form a Hilbert basis. Proof. ⇒) Let v1 , . . . , vq be the column vectors of A and let b = (bi ). Assume that v1 , . . . , vr are the columns of A which are active in F . Take an integral vector c in R+ {v1 , . . . , vr }. Then the maximum in the LP-duality equation max{x, c| xA ≤ b} = min{y, b| y ≥ 0; Ay = c} (1.13) is attained by each vector of F . Indeed if x0 ∈ F and c = λ1 v1 + · · · + λr vr , with λi ≥ 0 for all i, then x0 , c = λ1 b1 + · · · + λr br and min{y, b| y ≥ 0; Ay = c} ≤ λ1 b1 + · · · + λr br ≤ max{x, c| xA ≤ b}. Thus we have equality everywhere and the maximum is attained at any vector of F . The minimum has an integral optimum solution y. Then by complementary slackness (see Corollary 1.1.58), we get (b − xA)y = 0 for any x ∈ F . Therefore (bi − x, vi )yi = 0 for i = 1, . . . , q. If vi is inactive in F , then x, vi < bi for some x ∈ F . Then yi = 0. This proves that yi = 0 for i > r, i.e., c ∈ N{v1 , . . . , vr }. Altogether v1 , . . . , vr is a Hilbert basis. ⇐) Let c be an integral vector for which the optimum values of Eq. (1.13) are ﬁnite. Let F be a face of P such that each vector in F attains the maximum in Eq. (1.13). We may assume that F is a minimal face of P. By the minimality of F if a vector x0 in F satisﬁes x0 , vi = bi for some i, then any other vector x in F satisﬁes x0 , vi = bi , i.e., vi is active in F . Thus we may assume that there is 1 ≤ r ≤ q such that x, vi = bi for i ≤ r and x ∈ F , and x, vi < bi for i > r and x ∈ F . Thus by hypothesis the set A1 = {v1 , . . . , vr } is a Hilbert basis. Let x and y = (yi ) be optimum solutions of Eq. (1.13) with x ∈ F . Then by complementary slackness (see

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Corollary 1.1.58), we get (b − xA)y = 0, Ay = c and y ≥ 0. If yi > 0, then . Hence c is in NA1 because x, vi = bi , i.e., i ≤ r. Therefore c is in R+ A1 r A1 is a Hilbert basis. Then we can write c = i=1 ηi vi , ηi ∈ N. Consider the vector y0 = (η1 , . . . , ηr , 0 . . . , 0) in Nq whose last q − r entries are equal to zero. Thus c = Ay0 . Since bi −x, vi = 0 for i ≤ r, we get (b−xA)y0 = 0. Thus the minimum in Eq. (1.13) is attained in y0 , as required. 2 Notice that we have in fact shown that Theorem 1.3.22 is valid if we replace “each minimal face F ” by “each face F ”. Deﬁnition 1.3.23 Let A be an integral matrix and let w be an integral vector. The system xA ≤ w is said to have the integer rounding property if min{y, w| y ≥ 0; Ay = a} = min{y, w| Ay = a; y ≥ 0 ; y integral} for each integral vector a for which min{y, w| y ≥ 0; Ay = a} is ﬁnite. Theorem 1.3.24 [372, Theorem 22.18] Let A be an integral matrix of size n × q with column vectors v1 , . . . , vq and w = (wi ) ∈ Zq . Then the system xA ≤ w has the integer rounding property if and only if the set H = {(v1 , w1 ), . . . , (vq , wq ), en+1 } ⊂ Zn+1 is a Hilbert basis. Corollary 1.3.25 Let A be an integral matrix and let w = (wi ) be an integral vector. The system xA ≤ w is TDI if and only if {x| xA ≤ w} is an integral polyhedron and the set H ⊂ Zn+1 is a Hilbert basis. Proof. It follows from Proposition 1.3.11, Theorem 1.3.24, and using the deﬁnition of a TDI system. 2 Theorem 1.3.26 Let A be an integer matrix with column vectors v1 , . . . , vq and let w = (wi ) be an integral vector. If the polyhedron P = {x| xA ≤ w} is integral and H = {(v1 , w1 ), . . . , (vq , wq )} is a Hilbert basis, then the system xA ≤ w is TDI. Proof. Let F be a minimal face of P. Recall that a column of A is active in F if the corresponding inequality in xA ≤ w is satisﬁed with equality for all vectors in F . We may assume that v1 , . . . , vr are the columns of A which are active in F . Then x, vi = wi for x ∈ F and 1 ≤ i ≤ r. If y, vi < wi for some y ∈ F , then x, vi < wi for any other x ∈ F . Indeed if x, vi = wi for some x ∈ F , consider the supporting hyperplane of P given by H = {x|x, vi = wi }, then x ∈ F ∩ H F because y ∈ F and y ∈ / F ∩ H, a contradiction to the minimality of the face F . Thus we may also assume that x, vi < wi for x ∈ F and i > r.

36

Chapter 1

Since P is integral, by Theorem 1.1.63, each face of P contains integral vectors. Pick an integral vector x0 ∈ F . By Theorem 1.3.22, it suﬃces to prove that B = {v1 , . . . , vr } is a Hilbert basis. Take a ∈ R+ B ∩ Zn . Then a = λ1 v1 + · · · + λr vr with λi ≥ 0 for all i. Thus we have b = a, x0 = =

λ1 v1 , x0 + · · · + λr vr , x0 λ1 w1 + · · · + λr wr . r Hence b is an integer and q (a, b) = i=1 λi (vi , wi ). As H is a Hilbert basis, we can write (a, b) = i=1 ηi (vi , wi ), ηi ∈ N for all i. Therefore on one hand 0 = (a, b), (x0 , −1). On the other hand (a, b), (x0 , −1) is equal to r i=1

ηi (vi , wi ), (x0 , −1) +

=0

q i=r+1

ηi (vi , wi ), (x0 , −1) .

<0

Hence ηi = 0 for i > r and a = η1 v1 + · · · + ηr vr . Thus a ∈ NB.

2

The converse is not true in general. However there are some interesting linear systems where the converse holds. Deﬁnition 1.3.27 Let A be an integral matrix and w an integral vector. The system x ≥ 0; xA ≤ w is TDI if the minimum in the LP-duality equation max{a, x| x ≥ 0; xA ≤ w} = min{y, w| y ≥ 0; Ay ≥ a}

(1.14)

has an integral optimum solution y for each integral vector a with ﬁnite minimum. Proposition 1.3.28 Let A be a nonnegative integral matrix of size n × q with column vectors v1 , . . . , vq and let w = (wi ) ∈ Nq . Then, the system x ≥ 0; xA ≤ w is TDI if and only if the polyhedron P = {x| x ≥ 0; xA ≤ w} is integral and H = {(v1 , w1 ), . . . , (vq , wq ), −e1 , . . . , −en } is a Hilbert basis. Proof. Assume the system x ≥ 0; xA ≤ w is TDI. By Proposition 1.3.11, we get that P is integral. Next we prove that H is a Hilbert basis. Take (a, b) ∈ R+ H ∩Zn+1 , where a ∈ Zn and b ∈ Z. By hypothesis, the minimum in Eq. (1.14) has an integral optimum solution y = (yi ) such that y, w ≤ b. Since y ≥ 0 and a ≤ Ay, we can write a = y1 v1 + · · · + yq vq − δ1 e1 − · · · − δn en (δi ∈ N) =⇒ (a, b) = y1 (v1 , w1 ) + · · · + yq−1 (vq−1 , wq−1 ) +(yq + b − y, w)(vq , wq ) − (b − y, w)vq − δ, where δ = (δi ). As the entries of A are in N, the vector −vq can be written as a nonnegative integer combination of −e1 , . . . , −en . Thus (a, b) ∈ NH. This proves that H is a Hilbert basis. The converse follows readily from Theorem 1.3.26. 2

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37

Lemma 1.3.29 Let A be an integral matrix of size n×q with column vectors v1 , . . . , vq and w = (wi ) ∈ Zq . If P = {x| xA ≤ w} is pointed, then α is a − vertex of P if and only if H(α,−1) is a closed halfspace deﬁning a facet of the cone C = R+ {(v1 , w1 ), . . . , (vq , wq ), en+1 }. Proof. Since P is pointed, the rank of A is equal to n and C is a cone of full dimension n + 1. Using that α is a vertex of P if and only if α is a basic feasible solution of the system xA ≤ w (see Corollary 1.1.49), the proof follows from Theorem 1.1.44. 2 Procedure 1.3.30 If P = {x| xA ≤ w} is pointed, using Corollary 1.3.25 and Lemma 1.3.29 we can check whether a system xA ≤ w is TDI. Using Normaliz [68] this can be achieved using a single input ﬁle of the form: q+1 n+1 v1,w1 ... vq,wq 0 1 0 The ﬁrst and third block of the output ﬁle contain the Hilbert basis and the support hyperplanes. To verify if P is integral, by Lemma 1.3.29, it suﬃces to verify that all rows of the support hyperplanes having its last entry positive are integral and its last entry is equal to 1. Example 1.3.31 To illustrate Procedure 1.3.30 consider the integer vector w = (3, 0, 0, 1, 0) and the matrix ⎛ ⎞ 1 1 1 1 1 A=⎝ 0 1 0 1 2 ⎠ 0 0 1 1 2 We now verify that the system xA ≤ w is not TDI. Using the following input ﬁle for Normaliz 6 4 1 1 1 1 1 0 0

0 1 0 1 2 0

0 0 1 1 2 0

3 0 0 1 0 1

38

Chapter 1

We obtain the following output ﬁle (we only show the ﬁrst and third block): 6 Hilbert basis elements: 1 0 0 3 1 1 0 0 1 0 1 0 0 0 0 1 1 2 2 0 1 1 1 0

6 support hyperplanes: -3 3 3 1 0 0 0 1 0 0 1 0 0 1 0 0 2 -2 1 0 2 1 -2 0

Thus the polyhedron {x| xA ≤ w} is integral and the rows in the input ﬁle do not form a Hilbert basis. For instance the vector (1, 1, 1, 0) is in the cone generated by the rows of the input ﬁle but it is not an N-linear combination of the rows. Corollary 1.3.32 Let xA ≤ w be a system satisfying the integer rounding property and let A = {v1 , . . . , vq } be the set of column vectors of A. Then (a) A is a Hilbert basis. (b) Zn /ZA is a torsion-free group provided that R+ A is a pointed cone. q Proof. (a): Take a ∈ R+ A ∩ Zn , then we can write a = i=1 λi vi , for some λ1 , . . . , λq in R+ . Hence (a, i λi wi ) = λ1 (v1 , w1 ) + · · · + λq (vq , wq ) + δ(0, 1), where δ ≥ 0. Therefore, by Theorem 1.3.24, there are λ1 , . . . λq ∈ N and δ ∈ N such that (a, i wi λi ) = λ1 (v1 , w1 ) + · · · + λq (vq , wq ) + δ (0, 1), Thus a ∈ NA. (b): It follows at once from (a) and Corollary 1.3.21.

2

Exercises 1.3.33 Prove that any subsemigroup of N is ﬁnitely generated. 1.3.34 Prove that the subsemigroup of N2 given by S = (N+ × N) ∪ {(0, 0)} is not ﬁnitely generated, where N+ = {1, 2, . . .}. 1.3.35 (Gordan’s lemma) Let L be a lattice in Rn , i.e., L is an additive subgroup of Zn , and let C be a rational polyhedral cone in Rn . Prove that L ∩ C is a ﬁnitely generated semigroup. Hint Adapt the proof of Lemma 1.3.2. 1.3.36 Let C = {x| Ax ≤ 0} be a rational polyhedral cone. Prove that C contains no lines if and only if {x| Ax = 0} = (0).

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1.3.37 Let A be an integral matrix. If C = {x| x ≥ 0; Ax = 0}, prove that C is a rational polyhedral cone. 1.3.38 If w = (3, 0, 0, −1, 0) and A is ⎛ 1 1 A=⎝ 0 1 0 0

the matrix ⎞ 1 1 1 0 1 1 ⎠, 1 1 2

use Normaliz to verify that the system xA ≤ w is TDI. 1.3.39 Consider the two systems ⎛

1 (x1 , x2 , x3 ) ⎝ 1 0

0 1 1

1 0 1

⎛ ⎞ ⎞ ⎛ 2 1 1 ⎜ 2 ⎟ ⎟ ; (x1 , x2 , x3 ) ⎝ 1 1 ⎠≤⎜ ⎝ 2 ⎠ 1 0 3

0 1 1

⎞ ⎛ ⎞ 1 2 0 ⎠≤⎝ 2 ⎠ 1 2

Prove that they deﬁne the same polyhedron. Then prove that the ﬁrst system is TDI but the second is not. Use Normaliz to verify these assertions. 1.3.40 Let A be an n × d integral matrix of rank d whose set of column vectors A = {v1 , . . . , vd } form a Hilbert basis. Prove that the system xA ≤ b is TDI for each integral vector b.

1.4

Rees cones and clutters

In this section we characterize the max-ﬂow min-cut property of clutters in terms of Hilbert bases of Rees cones and the integrality of polyhedra. Let A be an n × q matrix with entries in N, let A = {v1 , . . . , vq } be the set of columns of A, and let A := {e1 , . . . , en , (v1 , 1), . . . , (vq , 1)} ⊂ Rn+1 , where ei is the ith unit vector. The cone R+ A , generated by A , is called the Rees cone of A or the Rees cone of A. The term Rees cone was coined in [147] because this cone encodes some information about the Rees algebra of the monomial ideal associated to A; see Chapters 12–14. The ﬁrst aim of this section is to study the irreducible representation, as an intersection of closed halfspaces, of a Rees cone. We shall always assume that all rows and columns of A are non-zero. The set covering polyhedron (in Proposition 13.1.2 we clarify this terminology) is by deﬁnition the rational polyhedron: Q(A) := {x ∈ Rn | x ≥ 0, xA ≥ 1},

40

Chapter 1

where 0 and 1 are vectors whose entries are equal to 0 and 1, respectively. Often we denote the vectors 0, 1 simply by 0, 1. As is seen below, one can express the irreducible representation of R+ A in terms of Q(A). Some of the facets of a Rees cone are easy to identify. Consider the index set J = {i | 1 ≤ i ≤ n and ei , vj = 0 for some j} ∪ {n + 1}. Notice that R+ A has dimension n + 1. It is not hard to see that the set F = R+ A ∩ Hei

(1 ≤ i ≤ n + 1)

deﬁnes a facet of R+ A if and only i ∈ J . Therefore, by Theorem 1.1.44 and Proposition 1.1.51, the Rees cone has the following unique irreducible representation ! ! r + + R+ A = Hei Hαi (1.15) i∈J

i=1

such that 0 = αi ∈ Q and αi , en+1 = −1 for all i. In many interesting cases, from the viewpoint of commutative algebra, one has the equality J = {1, . . . , n}; see Section 14.2 and Exercise 14.2.34. n+1

Lemma 1.4.1 Let A be a matrix with entries in N and let a = (ai1 , . . . , aiq ) be its ith row. Set k = min{aij | 1 ≤ j ≤ q}. If aij > 0 for all j, then ei /k is a vertex of Q(A). Proof. Set x0 = ei /k. Clearly x0 ∈ Q(A), x0 , vj = 1 for some j, and 2 x0 , e = 0 for = i. It is seen that x0 is a vertex of Q(A). The irreducible representation of a Rees cone can be expressed in terms of the vertices of the set covering polyhedron. Theorem 1.4.2 Let V be the vertex set of Q(A). Then the irreducible representation of the Rees cone is given by ! ! + + Hei H(u,−1) . R+ A = i∈J

u∈V

Proof. We set B = {ei | i ∈ J } ∪ {(u, −1)| u ∈ V } and V = {u1 , . . . , up }. First we dualize Eq. (1.15) and use the duality theorem for cones to obtain (R+ A )∗

= = =

{y ∈ Rn+1 | y, x ≥ 0, ∀ x ∈ R+ A } + + ∩ · · · ∩ H(v He+1 ∩ · · · ∩ He+n ∩ H(v 1 ,1) q ,1) R+ ei + R+ α1 + · · · + R+ αr . i∈J

(1.16)

Polyhedral Geometry and Linear Optimization

41

Next we show the equality (R+ A )∗ = R+ B. The right-hand side is clearly contained in the left-hand side because a vector α belongs to Q(A) if and only if (α, −1) is in (R+ A )∗ . To prove the reverse inclusion observe that by Eq. (1.16) it suﬃces to show that αk ∈ R+ B for all k. Writing αk = (ck , −1) and using αk ∈ (R+ A )∗ gives ck ∈ Q(A). The set covering polyhedron can be written as Q(A) = R+ e1 + · · · + R+ en + conv(V ), where conv(V ) denotes the convex hull of V, this follows from the structure of polyhedra (see Theorem 1.1.33) by noticing that the characteristic cone of Q(A) is precisely Rn+ . Thus we can write ck = λ1 e1 + · · · + λn en + μ1 u1 + · · · + μp up , where λi ≥ 0, μj ≥ 0 for all i, j and μ1 +· · ·+μp = 1. If 1 ≤ i ≤ n and i ∈ / J, then the ith row of A has all its entries positive. Thus by Lemma 1.4.1 we get that ei /ki is a vertex of Q(A) for some ki > 0. To avoid cumbersome notation we denote ei and (ei , 0) simply by ei , from the context the meaning of ei should be clear. Therefore from the equalities ! ei ei λi ei = λi ki λi ki , −1 + λi ki en+1 = ki ki i∈J /

i∈J /

we conclude that αk

i∈J /

i∈J /

i∈J /

λi ei is in R+ B. From the identities

= (ck , −1) = λ1 e1 + · · · + λn en + μ1 (u1 , −1) + · · · + μp (up , −1) p = λi ei + λi ei + μi (ui , −1) i∈J /

i∈J \{n+1}

i=1

we obtain αk ∈ R+ B, as required. Taking duals in (R+ A )∗ = R+ B yields R+ A = Ha+ . (1.17) a∈B

Thus, by Remark 1.1.31, the proof reduces to showing that β ∈ / R+ (B \ {β}) for all β ∈ B. To show this we will assume that β ∈ R+ (B \ {β}) for some β ∈ B and derive a contradiction. Case (I): β = (uj , −1). For simplicity assume β = (up , −1). Then (up , −1) =

λi ei +

=

λi ei +

i∈J \{n+1}

−1

μj (uj , −1),

(λi ≥ 0; μj ≥ 0) ⇒

j=1

i∈J

up

p−1

p−1

μj u j

(1.18)

j=1

= λn+1 − (μ1 + · · · + μp−1 ).

(1.19)

42

Chapter 1

To derive a contradiction we claim that Q(A) = Rn+ + conv(u1 , . . . , up−1 ), which is impossible because by Proposition 1.1.39 the vertices of Q(A) would be contained in {u1 , . . . , up−1 }. To prove the claim note that the right-hand side is clearly contained in the left-hand side. For the other inclusion take γ ∈ Q(A) and write γ

=

n

bi ei +

i=1 (1.18)

=

δ+

p

(bi , ci ≥ 0;

ci u i

i=1 p−1

p

ci = 1)

i=1

(δ ∈ Rn+ ).

(ci + cp μi )ui

i=1

Therefore using the inequality p−1

(ci + cp μi ) =

i=1

p−1

ci + cp

i=1

p−1

! μi

(1.19)

= (1 − cp ) + cp (1 + λn+1 ) ≥ 1

i=1

we get γ ∈ Rn+ + conv(u1 , . . . , up−1 ). This proves the claim. Case (II): β = ek for some k ∈ J . We only consider the subcase k ≤ n. The subcase k = n + 1 can be treated similarly. We can write ek =

p λi ei + μi (ui , −1), i∈J \{k}

(λi ≥ 0; μi ≥ 0).

i=1

p p From this equality we get ek = i=1 μi ui . Hence ek A ≥ ( i=1 μi )1 > 0, a 2 contradiction because k ∈ J and ek , vj = 0 for some j. As a consequence we obtain: Theorem 1.4.3 [188, Theorem 3.2] The mapping ϕ : Qn → Qn+1 given by ϕ(α) = (α, −1) induces a bijective mapping ϕ : V −→ {α1 , . . . , αr } between the set V of vertices of Q(A) and the set {α1 , . . . , αr } of vectors that occur in the irreducible representation of R+ A given in Eq. (1.15). Deﬁnition 1.4.4 Let A be a matrix with entries in {0, 1}. The system x ≥ 0; xA ≥ 1 is called totally dual integral (TDI) if the maximum in the LP-duality equation min{α, x| x ≥ 0; xA ≥ 1} = max{y, 1| y ≥ 0; Ay ≤ α} has an integral optimum solution y for each integral vector α with ﬁnite maximum.

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43

Proposition 1.4.5 If the system x ≥ 0; xA ≥ 1 is TDI, then Q(A) has only integral vertices. Proof. It follows from Theorem 1.1.63.

2

Deﬁnition 1.4.6 A clutter C with vertex set X = {x1 , . . . , xn } is a family of subsets of X, called edges, none of which is included in another. The set of vertices and edges of C are denoted by V (C) and E(C), respectively. Let C be a clutter with vertex set X = {x1 , . . . , xn } and let f1 , . . . , fq be the edges of C. The incidence matrix of C is the n × q matrix A = (aij ) given by aij = 1 if xi ∈ fj and aij = 0 otherwise. Deﬁnition 1.4.7 A clutter C, with incidence matrix A, has the max-ﬂow min-cut (MFMC) property if both sides of the LP-duality equation min{α, x| x ≥ 0; xA ≥ 1} = max{y, 1| y ≥ 0; Ay ≤ α}

(1.20)

have integral optimum solutions x, y for each nonnegative integral vector α. It turns out that the max-ﬂow min-cut property is equivalent to require that a certain blowup ring has no nilpotent elements; see Theorem 14.3.6. Proposition 1.4.8 Let C be a clutter and let A be its incidence matrix. The system x ≥ 0; xA ≥ 1 is TDI if and only if C has the max-ﬂow min-cut property. Proof. If follows from Propositions 1.4.5 and 1.1.41.

2

Let C be a clutter with n vertices and let A = {v1 , . . . , vq } be the set of columns of its incidence matrix. For use below we denote by A the Rees conﬁguration of C: A = {e1 , . . . , en , (v1 , 1), . . . , (vq , 1)} ⊂ Nn+1 , where ei is the ith unit vector of Rn Theorem 1.4.9 [188, Theorem 3.4] If C is a clutter and A is its incidence matrix, then C has the max-ﬂow min-cut property if and only if Q(A) is an integral polyhedron and A is a Hilbert basis of R+ A . Proof. ⇒) By Proposition 1.4.5 the polyhedron Q(A) is integral. Next we show that A is an integral Hilbert basis. Take (α, αn+1 ) ∈ Zn+1 ∩ R+ A . Then Ay ≤ α and y, 1 = αn+1 for some vector y ≥ 0. Therefore one concludes that the optimal value of the linear program max{y, 1| y ≥ 0; Ay ≤ α}

44

Chapter 1

is greater than or equal to αn+1 . Since C has the MFMC property, this linear program has an optimal integral solution y0 . By Exercise 1.4.12, there exists an integral vector y0 such that 0 ≤ y0 ≤ y0 and |y0 | = αn+1 . Therefore

α

αn+1

A α A A (y0 − y0 ) + − y0 = y0 + 0 0 0 1

and (α, αn+1 ) ∈ NA , as required. ⇐) Assume that C does not satisfy the MFMC property. Then there is an α0 ∈ Nn such that if y0 is an optimal solution of the linear program: max{y, 1| y ≥ 0; Ay ≤ α0 },

(∗)

then y0 is not integral. We claim that also the optimal value |y0 | = y0 , 1 of this linear program is not integral. If |y0 | is integral, then (α0 , |y0 |) is in Zn+1 ∩ R+ A . As A is a Hilbert basis we get that (α0 , |y0 |) is in NA , but this readily yields that the linear program (∗) has an integral optimal solution, a contradiction. This completes the proof of the claim. Now, consider the dual linear program: min{x, α0 | x ≥ 0, xA ≥ 1}. By Proposition 1.1.41 the optimal value of this linear program is attained at a vertex x0 of Q(A). Then by the duality theorem (Theorem 1.1.56) we get x0 , α0 = |y0 | ∈ / Z. Hence x0 is not integral, a contradiction to the integrality of the set covering polyhedron Q(A). 2 Remark 1.4.10 Normaliz [68] computes the irreducible representation and the minimal integral Hilbert basis of a Rees cone. Thus we can eﬀectively use Theorems 1.4.2 and 1.4.9 to determine whether a given clutter has the max-ﬂow min-cut property; see example below. Example 1.4.11 Let A be the ⎡ 1 ⎢ 0 ⎢ ⎣ 0 1

transpose of the matrix: ⎤ 0 0 0 1 1 0 1 0 ⎥ ⎥ 0 1 1 1 ⎦ 1 1 0 0

and let C be the clutter with incidence matrix A. Consider the following input ﬁle for Normaliz that we call “example.in”:

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45

4 5 1 0 0 1

0 1 0 1

0 0 1 1

0 1 1 0

1 0 1 0

3

Applying Normaliz , i.e., typing “normaliz example” in the directory where one has the ﬁle normaliz.exe we get the output ﬁle “example.out”: 9 generators of integral closure of Rees algebra: 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 1 1 0 1 0 1 0 1 0 0 1 1 1 1 1 1 1 0 0 1 10 support hyperplanes: 0 0 1 1 1 -1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 -1 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 1 -1 1 1 1 0 0 -1 semigroup is not homogeneous

The ﬁrst block shows that A is a Hilbert basis for the Rees cone. The second block shows the irreducible representation of the Rees cone of A, thus using Theorem 1.4.2 we obtain that Q(A) is integral. Altogether Theorem 1.4.9 proves that the clutter C has the max-ﬂow min-cut property.

Exercises 1.4.12 Let b, η1 , . . . , ηq be a sequence in N. If η1 + · · · + ηq ≥ b, then there are 1 , . . . , q ∈ N such that 0 ≤ i ≤ ηi for all i and 1 + · · · + q = b. 1.4.13 Let A be the incidence matrix of a clutter C and let v1 , . . . , vq be its column vectors. The system x ≥ 0; xA ≥ 1 is TDI if and only if the set Au = {vi | vi , u = 1} ∪ {ei | ei , u = 0} is a Hilbert basis for any vertex u of Q(A).

46

1.5

Chapter 1

The integral closure of a semigroup

Let A = {v1 , . . . , vq } be a set of vectors in Nn \ {0}. The integral closure or normalization of the aﬃne semigroup NA := Nv1 + · · · + Nvq ⊂ Nn , is deﬁned as NA := ZA ∩ R+ A, where ZA is the subgroup of Zn generated by A. The semigroup NA is called normal or integrally closed if NA = NA. The Rees semigroup of A is by deﬁnition NA ⊂ Nn+1 , where A = {(v1 , 1), . . . , (vq , 1), e1 , . . . , en }. Notice that en+1 = (v1 , 1) − (v1 , 1), e1 e1 − · · · − (v1 , 1), en en . Hence ZA = Zn+1 . As a consequence we get NA = R+ A ∩ Zn+1 . From this equality we obtain a characterization of the normality of a Rees semigroup in terms of Hilbert bases: Proposition 1.5.1 NA is normal if and only if A is a Hilbert basis. The algebraic invariants of semigroup rings of semigroups generated by a Hilbert basis are easier to understand; see [405, Section 6]. This family of semigroups and their cones are studied throughout this book. Proposition 1.5.2 If A is a Hilbert basis, then NA is normal. Proof. Since NA ⊂ R+ A ∩ ZA ⊂ R+ A ∩ Zn = NA we obtain the equality NA = R+ A ∩ ZA. 2 Example 1.5.3 If A = {(1, 1, 0), (0, 1, 1), (1, 0, 1)}, the aﬃne semigroup NA is normal because A is linearly independent, but A is not a Hilbert basis because NA Z3 ∩ R+ A (see Exercise 1.5.8). Notation Given β = (βi ) ∈ Rn+1 , we deﬁne degn+1 (β) = βn+1 . Proposition 1.5.4 [147] Let R+ A be the Rees cone of A and let B be its minimal integral Hilbert basis. If β ∈ B, then degn+1 (β) < n. Proof. We set β = (α, b), with α ∈ Nn and b = degn+1 (β) ∈ N. Note that R+ A ∩ Zn+1 = Q+ A ∩ Zn+1 and dim(R+ A ) = n + 1. Thus, since (α, b) ∈ R+ A , by Carath´eodory’s theorem (Theorem 1.1.18) we can write (α, b) = λ1 (vi1 , 1) + · · ·+ λr (vir , 1) + μ1 ej1 + · · · + μs ejs (λi , μk ∈ Q+ ), (∗)

Polyhedral Geometry and Linear Optimization

47

where {(vi1 , 1), . . . , (vir , 1), ej1 , . . . , ejs } is a linearly independent subset of A . By the minimality of B it is seen that 0 ≤ λi < 1 and 0 ≤ μk < 1 for all i, k. From Eq. (∗) we get b = λ1 + · · · + λr < r. Thus b ≤ r − 1. If r ≤ n, then b ≤ n − 1. Thus from now on we may assume r = n + 1 and Eq. (∗) takes the simpler form (α, b) = λ1 (vi1 , 1) + · · · + λn+1 (vin+1 , 1). Consider the cone C generated by A = {(vi1 , 1), . . . , (vin+1 , 1)}. Since −e1 is not in C, using Exercise 1.5.9, we obtain a point x0 = (1 − λ0 )(α, b) + λ0 (−e1 ) (0 ≤ λ0 < 1) in the relative boundary of C. By Theorem 1.1.44 the relative boundary of C is the union of its facets. Hence using that any facet of C is an n-dimensional cone generated by a subset of A (see Proposition 1.1.23) together with the minimality of B it follows that we can write (α, b) as: (α, b) = ρ0 e1 + ρ1 (vj1 , 1) + · · · + ρn (vjn , 1) and consequently b ≤ n − 1, as required.

(0 ≤ ρi < 1 ∀i), 2

Deﬁnition 1.5.5 A matrix A is called totally unimodular if each i × i subdeterminant of A is 0 or ±1 for all i ≥ 1. Examples of totally unimodular matrices include incidence matrices of clutters without odd cycles [27, Chapter 5], incidence matrices of bipartite graphs and digraphs [372, p. 274], and network matrices [372, Chapter 19]. Theorem 1.5.6 (Hoﬀman, Kruskal) If A is a totally unimodular matrix, then Q = {x | x ≥ 0; Ax ≤ b} is integral for each integral vector b. Proof. Assume that A has size n × q. Notice that Q contains no lines, thus it suﬃces to show that Q has only integral vertices. Let x0 be a vertex of Q. By Corollary 1.1.47, A x0 = b for some subsystem A x ≤ b of b A ≤ 0 −Iq such that A is a square matrix of order q with linearly independent rows. By hypothesis det(A ) = ±1, hence x0 is integral because the inverse of A is an integral matrix by Cramer’s rule. 2 Corollary 1.5.7 [49] If A = (aij ) is a totally unimodular {0, 1}-matrix of size n × q with column vectors v1 , . . . , vq , then R+ A is normal. Proof. By the Hoﬀman–Kruskal theorem, the system x ≥ 0; xA ≥ 1 is TDI. Thus, by Theorem 1.4.9, the set A is a Hilbert basis, i.e., the Rees 2 semigroup R+ A is normal.

48

Chapter 1

Exercises 1.5.8 If A = {(1, 1, 0), (0, 1, 1), (1, 0, 1)}, use Normaliz , with option 0, to verify that the minimal integral Hilbert basis of R+ A is A ∪ {(1, 1, 1)}. 1.5.9 Let a be a point of a set V in Rn and let x be a point in aﬀ(V) not in V. If λ0 = sup{λ ∈ [0, 1]| (1 − λ)a + λx ∈ V}, then x0 = (1 − λ0 )a + λ0 x is a relative boundary point of V lying between a and x. 1.5.10 If A is a totally unimodular matrix and A is obtained from A by adding columns of unit vectors, then the matrix A is totally unimodular.

1.6

Unimodularity of matrices and normality

In this section, we characterize when a vector belongs to a subgroup of Zn and prove Heger’s theorem for the existence of solutions of an integer linear system. Then we show the normality of semigroups arising from unimodular matrices and unimodular coverings. A minor of order r (r-minor for short) of a matrix A is deﬁned as the determinant of a square submatrix of A of order r. Deﬁnition 1.6.1 An integral matrix A = (0) is t-unimodular if all the nonzero r-minors of A have absolute value equal to t, where r is the rank of A. If A is 1-unimodular, we say that A is a unimodular matrix . Theorem 1.6.2 [372, Theorem 19.2] Let A be an integral matrix of full row rank. Then the polyhedron {x| x ≥ 0; Ax = b} is integral for each integral vector b if and only if A is unimodular. Theorem 1.6.3 Let A be an n × q integral matrix whose set of column vectors is A = {v1 , . . . , vq } and let b ∈ Zn be a column vector such that rank(A) = rank([A b]). The following conditions are equivalent: (a) b ∈ ZA. (b) Zn /ZA and Zn /Z(A ∪ {b}) have the same invariant factors. (c) The matrices [A 0] and [A b] have the same Smith normal form. (d) Δr (A) = Δr ([A b]), where r = rank(A) and Δr (A) is the gcd of all the non-zero r-minors of A. q Proof. There is k ∈ N \ {0} such that kb = i=1 λi vi , λi ∈ Z. Therefore, there is a canonical epimorphism of ﬁnite groups ϕ : T (Zn /ZA) −→ T (Zn /ZB)

ϕ

(α + ZA −→ α + ZB),

where B = A ∪ {b}. Notice that ϕ is injective if and only if (a) holds.

Polyhedral Geometry and Linear Optimization

49

The implication (a) ⇒ (b) is straightforward. There are invertible integral matrices Pi and Qi such that D1 = Q1 [A 0]P1 = diag(d1 , . . . , dr , 0, . . . , 0), D2 = Q2 [A b]P2 = diag(e1 , . . . , er , 0, . . . , 0), are the Smith normal forms of [A 0] and [A b], respectively; that is, di , ei are positive integers satisfying that di divides di+1 and ei divides ei+1 for all i. By the fundamental structure theorem for ﬁnitely generated abelian groups (see Theorem 1.3.16) there are isomorphisms: T (Zn /ZA) (Z/d1 Z) × · · · × (Z/dr Z), T (Zn /ZB) (Z/e1 Z) × · · · × (Z/er Z). Thus (b) ⇔ (c). Note Δi (A) = d1 · · · di and Δi ([A b]) = e1 · · · ei for all i. Hence (c) ⇒ (d). To prove (d) ⇒ (i) observe |T (Zn /ZA)| = |T (Zn /ZB)| and consequently ϕ must be injective. 2 As an immediate consequence of Theorem 1.6.3 we get: Theorem 1.6.4 (I. Heger, [372, p. 51]) Let A be an n × q integral matrix and let b ∈ Zn be a column vector. If rank(A) = rank([A b]), then the system Ax = b has an integral solution if and only if Δr (A) = Δr ([A b]). Remark 1.6.5 Let A ⊂ Zn be a ﬁnite set. Note that the ﬁnite basis theorem together with Theorem 1.6.3 yield a membership test to decide when a given vector b in Zn belongs to NA = ZA ∩ R+ A. Proposition 1.6.6 If A is a t-unimodular matrix with columns v1 , . . . , vq and vi1 , . . . , vir is a Q-basis for the column space of A, then ZA = Zvi1 ⊕ · · · ⊕ Zvir Proof. For each j one has Δr (vi1 · · · vir ) = Δr (vi1 · · · vir vj ) = t. Then by Theorem 1.6.3, we get vj ∈ Zvi1 ⊕ · · · ⊕ Zvir , as required. 2 Theorem 1.6.7 If A is a t-unimodular matrix whose set of columns is A, then the aﬃne semigroup NA is normal. Proof. If b ∈ ZA ∩ R+ A, by Theorem 1.1.18 and Proposition 1.6.6 there are vi1 , . . . , vir linearly independent vectors in A, r = rank(A), such that b ∈ R+ vi1 + · · · + R+ vir and b ∈ Zvi1 + · · · + Zvir .

(1.21)

By comparing the coeﬃcients of b with respect to the two representations obtained from Eq. (1.21), one derives b ∈ NA. 2

50

Chapter 1

Corollary 1.6.8 If A is a totally unimodular matrix whose set of columns is A, then NA = Zn ∩ R+ A; that is, A is a Hilbert basis. Proof. Let A = {v1 , . . . , vq } be the set of column vectors of A. Take b ∈ Zn ∩ R+ A, then by Carath´eodory’s theorem r(see Theorem 1.1.18) and after permutation of the vi ’s we can write b = i=1 ηi vi with ηi ≥ 0 for all i, where r is the rank of A and v1 , . . . , vr are linearly independent. The submatrix A = (v1 · · · vr ) is totally unimodular. Therefore, by Theorem 1.6.3, the system of equations A x = b has an integral solution. Thus b is a linear combination of v1 , . . . , vr with coeﬃcients in Z. It follows 2 that ηi ∈ N for all i, that is, b ∈ NA. The other inclusion is clear. Proposition 1.6.9 If A = {v1 , . . . , vq } ⊂ Zn lie in a hyperplane of Rn not containing the origin and P = conv(A) has a weakly unimodular covering with support in A, then NA is a normal semigroup. Proof. Take 0 = β ∈ R+ A ∩ ZA. There exists λ1 . . . , λq ∈ Q+ such that β = λ1 v1 +· · ·+λq vq . We set |λ| = λ1 +· · ·+λq . Since β/|λ| is in P and using that P has a weakly unimodular covering, there is an aﬃnely independent set A1 ⊂ A deﬁning a lattice simplex Δ = conv(A1 ) such that β/|λ| ∈ Δ and ZA1 = ZA. Altogether β belongs to ZA1 ∩ R+ A1 . Note that A1 is a linearly independent set because A1 lie in a hyperplane not containing the 2 origin. Therefore it follows rapidly that β ∈ NA1 , as required.

Exercises 1.6.10 If A = {v1 , . . . , vq } ⊂ Zn and P = conv(A) has a unimodular covering with support in A, then N(v1 , 1) + · · · + N(vq , 1) is normal.

1.7

Normaliz, a computer program

Throughout this book we will frequently use Normaliz [68], a program that provides an invaluable eﬀective tool to study monomial subrings and their algebraic invariants. This program computes the following: • normalization (or integral closure) of an aﬃne semigroup; • Hilbert basis of a pointed rational cone; • lattice points of an integral polytope; • support hyperplanes of a Rees cone; • generators of the integral closure of the Rees algebra of a monomial ideal I = (xv1 , . . . , xvq ) ⊂ R = K[x1 , . . . , xn ]; • generators of the integral closure of I;

Polyhedral Geometry and Linear Optimization

51

• Hilbert series and Hilbert polynomial of a homogeneous aﬃne semigroup; • Ehrhart ring (Theorem 9.3.6) and relative volume (Proposition 1.2.10) of a lattice polytope. It can be used to check the following integer programming properties: • integer rounding property of any of the linear systems x ≥ 0; xA ≥ 1,

x ≥ 0; xA ≤ 1,

xA ≤ 1,

where A is the matrix with column vectors v1 , . . . , vq (Corollary 14.6.9, Theorem 14.6.16, and Proposition 14.6.30); • integer rounding property and total dual integrality (TDI) of a linear system xA ≤ w (Theorem 1.3.24, Procedure 1.3.30); • perfection of a graph (Remark 13.6.4). If I is square-free, it can also be used to determine the following: • minimal primes of I (Remark 13.2.23); • reducedness of the associated graded ring grI (R) (Example 14.2.20); • the max-ﬂow min-cut property of a clutter (Example 1.4.11); • integrality of the set covering polyhedron Q(A) = {x| xA ≥ 1; x ≥ 0}, where A = (v1 , . . . , vq ); • total dual integrality of the system x ≥ 0; xA ≥ 1; • the generators of the symbolic Rees algebra Rs (I) (Example 13.2.22).

1.8

Cut-incidence matrices and integrality

In this section we introduce incidence matrices of digraphs, and present a generalization of a theorem of Lucchesi and Younger [298] which is very useful to detect TDI systems arising from incidence matrices of digraphs. This theorem will be used in Theorem 11.3.2 to express the a-invariant of the edge subring of a bipartite graph in terms of directed cuts. This illustrates that some deep results in combinatorial optimization can be used to study algebraic invariants of rings. Next we introduce some more TDI systems which are suited for our purposes (see [281, Chapter 5] and [372, p. 311]).

52

Chapter 1

Deﬁnition 1.8.1 Let B be an integral matrix. A system Bx ≥ b, x ≥ 0 is totally dual integral (TDI) if the maximum in min{c, x| x ≥ 0; Bx ≥ b} = max{y, b| y ≥ 0; yB ≤ c} has an integral optimum solution y for each integral vector c with ﬁnite maximum. Proposition 1.8.2 If the matrix B is totally unimodular, then the system Bx ≥ b; x ≥ 0 is TDI. Proof. It follows from Theorem 1.5.6.

2

Proposition 1.8.3 Let B be an integral matrix and let b be an integral vector. If the system Bx ≥ b; x ≥ 0 is TDI, then Q = {x| Bx ≥ b; x ≥ 0} has only integral vertices. Proof. It follows from Theorem 1.1.63.

2

Deﬁnition 1.8.4 A digraph G consists of a ﬁnite vertex set V (G) and a family E(G) of ordered pairs of elements of V (G). The pairs (v1 , v2 ) ∈ E(G) are called directed edges or arrows. Deﬁnition 1.8.5 Let G be a digraph with vertex set V (G) and edge set E(G). Given a family F of subsets of V (G), the one-way cut-incidence matrix of F is the matrix B = (bX,e )X∈F ,e∈E(G) , where

1 if e ∈ δ + (X) bX,e = 0 otherwise. The two-way cut-incidence matrix is ⎧ ⎨ 1 −1 bX,e = ⎩ 0

B = (bX,e )X∈F ,e∈E(G) , where if e ∈ δ + (X) if e ∈ δ − (X) otherwise.

Here δ + (X) = {e = (z, w) ∈ E(G)| z ∈ X, w ∈ / X} is the set of edges leaving the vertex set X and δ − (X) is the set of edges entering the vertex set X. If F = {{v} | v ∈ V (G)}, the matrix B is called the incidence matrix of G. Remark 1.8.6 An interesting case occurs when the one-way and the twoway cut-incidence matrix of the family F coincide. In this case the rows of the matrix B correspond to directed cuts only. Recall that δ + (X) is a directed cut of a digraph G if ∅ = X ⊂ V (G) and δ − (X) = ∅. It was pointed out to us by Jens Vygen that the next result is a slight generalization of the Lucchesi, Younger theorem [298]. It follows using the technique of proof of [281, Theorem 19.10].

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53

Theorem 1.8.7 [405] Let G be a digraph and F a family of subsets of V (G) such that the one-way cut incidence matrix B of F is equal to the two-way cut incidence matrix of F . If F satisﬁes the following three conditions: (a) the rows of B are non-zero, (b) if X, Y ∈ F and X ∪ Y = V (G), then X ∪ Y ∈ F , and (c) if X, Y ∈ F and X ∩ Y = ∅, then X ∩ Y ∈ F , then the system Bx ≥ 1, x ≥ 0 is totally dual integral and the rational polyhedron {x| Bx ≥ 1; x ≥ 0} is integral. Proof. Let c be an integral vector such that the maximum in min{c, x| x ≥ 0; Bx ≥ 1} = max{y, 1| y ≥ 0; yB ≤ c}

(1.22)

is attained at y0 = (yF1 , . . . , yFr ), where F = {F1 , . . . , Fr }. Note c ≥ 0. Since the set of vectors y for which the maximum in Eq. (1.22) is attained is a face of the polytope P = {y ∈ Rr | y ≥ 0; yB ≤ c} and since any face of P is a compact set, one may assume r r yFi |Fi |2 = max yi |Fi |2 y, 1 = y0 , 1; y ≥ 0; yB ≤ c . (1.23) i=1

i=1

We claim that the family G = {Fi | yFi > 0} is cross-free, that is, for any Fi , Fj ∈ G, at least one of the four sets Fi \ Fj , Fj \ Fi , Fi ∩ Fj , V (G)\ Fi ∪Fj is empty. To prove the claim suppose F1 \ F2 = ∅, F2 \ F1 = ∅, F1 ∩ F2 = ∅, F1 ∪ F2 = V (G) for some F1 , F2 ∈ G (for simplicity we are assuming i = 1, j = 2). Set = min{yF1 , yF2 } and consider the vector y0 in Rr whose entries are given by yF 1 := yF1 − , yF 2 := yF2 − , yF 1 ∩F2 := yF1 ∩F2 + , yF 1 ∪F2 := yF1 ∪F2 + , and yS := yS for all other S ∈ F . Let B = (bFi ,αj ) be the one-way cutincidence matrix of F , where α1 , . . . , αq are the edges of G. Consider any edge αj = (z, w). For each 1 ≤ j ≤ q one has the following inequality bF1 ,j (yF1 − ) + bF2 ,j (yF2 − ) + bF1 ∩F2 ,j (yF1 ∩F2 + ) + bF1 ∪F2 ,j (yF1 ∪F2 + ) ≤ bF1 ,j yF1 + bF2 ,j yF2 + bF1 ∩F2 ,j yF1 ∩F2 + bF1 ∪F2 ,j yF1 ∪F2 . To show the inequality consider two cases. Case (I): bF1 ,j = bF2 ,j = 0. If i = 1 or i = 2 and z ∈ Fi , then w ∈ Fi and bF1 ∪F2 ,j = 0. If i = 1 or i = 2 and z ∈ / Fi , then w ∈ / Fi and bF1 ∩F2 ,j = 0. Hence in this case yF1 ∩F2 ,j = 0 and yF1 ∪F2 ,j = 0. Case (II): bF1 ,j = 0 and bF2 ,j = 1. By the previous considerations one has that either yF1 ∩F2 ,j = 0 or yF1 ∪F2 ,j = 0.

54

Chapter 1

From the inequality above we obtain y0 B ≤ y0 B. Since y0 , 1 = y0 , 1, we obtain a contradiction with the choice of y0 because r i=1

yFi |Fi |2 <

r

yF i |Fi |2 .

i=1

Note that for any numbers a > b ≥ c > d > 0 with a + d = b + c one has the inequality a2 + d2 > b2 + c2 . In our case to show the strict inequality we take a = |F1 ∪ F2 |, b = |F1 | ≥ c = |F2 | and d = |F1 ∩ F2 |. Thus we have shown that the family G is cross-free. Let B be the submatrix of B whose rows correspond to the elements of the family G. Then max{y , 1| y ≥ 0; y B ≤ c} = max{y, 1| y ≥ 0; yB ≤ c}.

(1.24)

The inequality ≤ is clear because B is a submatrix of B. As the maximum in the right-hand side is attained at y0 , where y0 has zero entries in positions corresponding to rows outside B , the reverse inequality follows. Now the matrix B , being the two-way cut-incidence matrix of a cross-free family G, is a network matrix, hence it is totally unimodular; see [281, Theorem 5.27]. Hence by the Hoﬀman–Kruskal theorem the polytope {y | y ≥ 0; y B ≤ c} is integral, and so the maximum in the left-hand side of Eq. (1.24) is attained at an integral vector y . To complete the proof that the system Bx ≥ 1, x ≥ 0 is TDI observe that y can be extended (by adding zero entries) to an integral optimum solution y of the right-hand side of Eq. (1.24). Since the system Bx ≥ 1, x ≥ 0 is totally dual integral it follows from Proposition 1.8.3 that the polyhedron {x| Bx ≥ 1; x ≥ 0} is integral. 2 Corollary 1.8.8 Let G be a connected digraph and let F = {X| ∅ = X V (G); δ − (X) = ∅}. If B is the one-way cut-incidence matrix of F , then {x| Bx ≥ 1; x ≥ 0} is a non-empty integral polyhedron. Proof. It suﬃces to verify the hypothesis of Theorem 1.8.7. First we prove that δ + (X) = ∅ for X ∈ F . Assume δ + (X) = ∅. Pick z ∈ V (G) \ X and x ∈ X. If (z, w) ∈ E(G) or (w, z) ∈ E(G), then w ∈ / X. There is an / X for undirected path {z0 = z, z1 , . . . , zr = x}, a contradiction since zi ∈ all i. Thus (a) is satisﬁed. Next take X, Y ∈ F . From the inequality |δ − (X)| + |δ − (Y )| ≥ |δ − (X ∩ Y )| + |δ − (X ∪ Y )|,

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see [281, Lemma 2.1(b)], we get δ − (X ∩ Y ) = ∅ and δ − (X ∪ Y ) = ∅. Thus if X ∩ Y = ∅ (resp. X ∪ Y = V (G)), then X ∩ Y ∈ F (resp. X ∪ Y ∈ F ). Thus conditions (b) and (c) are satisﬁed. On the other hand by construction of F the matrix B is also the two-way cut-incidence matrix of F . 2 Lemma 1.8.9 Let G be a connected bipartite graph with bipartition (V1 , V2 ) and let F be the family F = {A ∪ A | ∅ = A V1 ; N (A) ⊂ A ⊂ V2 } ∪ {A | ∅ = A ⊂ V2 }. If G is regarded as the digraph with all its arrows leaving the vertex set V2 , then the following equality holds F = {X| ∅ = X V (G); δ − (X) = ∅}. Proof. It follows readily from the deﬁnitions.

2

Exercises 1.8.10 If A is the incidence matrix of a digraph G, prove that A is totally unimodular. 1.8.11 Let T : Rq → Rn be a linear map and let Q ⊂ Rq be a rational polyhedron. If T (Zq ) ⊂ Zn , then T (Q ) is a rational polyhedron.

1.9

Elementary vectors and matroids

In this section, we give formulae to compute the circuits of the kernel of an integral matrix and show that the circuits generate the kernel. We introduce the notion of a matroid and explain the relation with vector matroids. The notion of an elementary integral vector or circuit occurs in convex analysis [357], in the theory of toric ideals of graphs [34, 400, 417], and in matroid theory [338]. Given α = (α1 , . . . , αq ) ∈ Rq , its support is supp(α) = {i | αi = 0}. Note that α can be written uniquely as α = α+ − α− , where α+ and α− are two nonnegative vectors with disjoint support which are called the positive part and the negative part of α, respectively. Sometimes we denote the positive and negative part of α by α+ and α− , respectively. Remark 1.9.1 If α is a linear combination of β1 , . . . , βr ∈ Rq , then one has the inclusion supp(α) ⊂ supp(β1 ) ∪ · · · ∪ supp(βr ). Deﬁnition 1.9.2 Let V be a linear subspace of Qq . An elementary vector of V is a non-zero vector α in V whose support is minimal with respect to inclusion, i.e., supp(α) does not properly contain the support of any other non-zero vector in V .

56

Chapter 1

The concept of an elementary vector arises in graph theory when V is the kernel of the incidence matrix of a graph G [358, Section 22]. Lemma 1.9.3 If V is a linear subspace of Qq and α, β are two elementary vectors of V with the same support, then α = λβ for some λ ∈ Q. Proof. If i ∈ supp(α), then one can write αi = λβi for some scalar λ. Since supp(α − λβ) supp(α), one concludes α − λβ = 0, as required. 2 Deﬁnition 1.9.4 Two vectors α = (αi ) and β = (βi ) in Qq are in harmony if αi βi ≥ 0 for every i. Lemma 1.9.5 [357] Let V be a linear subspace of Qq . If 0 = α ∈ V , then there is an elementary vector γ ∈ V in harmony with α such that supp(γ) ⊂ supp(α). Proof. Let β = (β1 , . . . , βq ) be an elementary vector of V whose support is contained in supp(α). By replacing β by −β one may assume αi βi > 0 for some i. Consider

αi αj α = min β > 0 . λj = i i 1≤i≤q βj βi Note that zi = (αi − λj βi )αi ≥ 0 for all i. Indeed if αi βi ≤ 0, then clearly zi ≥ 0, and if αi βi > 0 by the minimality of λj one has zi ≥ 0. Thus the vector α − λj β is in harmony with α and its support is strictly contained in the support of α. If α − λj β = 0 we are done, otherwise we apply the same argument with α − λj β playing the role of α. Since being in harmony is an equivalence relation, the result follows by a recursive application of this procedure. 2 Theorem 1.9.6 [357] If V is a vector subspace of Qq and α ∈ V \ {0}, then α can be written as α = ri=1 βi for some elementary vectors β1 , . . . , βr of V with r ≤ dim V such that (i)

β1 , . . . , βr are in harmony with α,

(ii) supp(βi ) ⊂ supp(α) for all i, and (iii) supp(βi ) is not contained in the union of the supports of β1 , . . . , βi−1 for all i ≥ 2. Proof. By induction on the number of elements in the support of α. If α is not an elementary vector of V , then by Lemma 1.9.5 there is an elementary vector α1 ∈ V in harmony with α such that supp(α1 ) ⊂ supp(α). Using the proof of Lemma 1.9.5 it follows that there is a positive scalar λ1 > 0 such that α − λ1 α1 is in harmony with α and the support of α − λ1 α1 is strictly contained in supp(α). Therefore the result follows applying induction. This proof uses the original argument given in [357]. 2

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Deﬁnition 1.9.7 Let V be a linear subspace of Qq . An elementary integral vector or circuit of V is an elementary vector α of V such that α ∈ Zq and the non-zero entries of α are relatively prime. Corollary 1.9.8 If V is a linear subspace of Qq , then there are only a ﬁnite number of circuits of N and they generate V as a Q-vector space. Proof. It follows from Lemma 1.9.3 and Theorem 1.9.6, respectively.

2

Proposition 1.9.9 [134] Let A be an n × q integral matrix and let α be a non-zero vector in ker(A). If A has rank n, then α is an elementary vector of ker(A) if and only if there is 0 = λ ∈ Q such that α=λ

n+1

(−1)k det[vi1 , . . . , vik−1 , vik+1 , . . . , vin+1 ]eik ,

(1.25)

k=1

for some column vectors vij of A. Here ek is the kth unit vector in Rq . Proof. ⇐) Let A be the submatrix of A consisting of the column vectors vi1 , . . . , vin+1 . Since A has size n × (n + 1) and rank n, ker(A ) is generated by a non-zero vector. It follows readily that the vector α given by Eq. (1.25) is an elementary vector. ⇒) Let supp(α) = {j1 , . . . , jr } and let v1 , . . . , vq be the column vectors of A. Since α ∈ ker(A) one has α = λj1 ej1 + · · · + λjr ejr and λj1 vj1 + · · · + λjr vjr = 0,

(1.26)

where 0 = λjk ∈ Q and vjk are columns of A for all k. By the minimality of supp(α) we may assume that {vj1 , . . . , vjr−1 } is linearly independent. Thus, using that rank(A) = n, there are column vectors vjr+1 , . . . , vjn+1 such that the set A = {vj1 , . . . , vjr−1 , vjr+1 , . . . , vjn+1 } is linearly independent. After permuting j1 , . . . , jn+1 , one can write A ∪ {vjr } = {vi1 , . . . , vin+1 } with 1 ≤ i1 < · · · < in+1 ≤ q. Consider the vector β=

n+1

(−1)k det[vi1 , . . . , vik−1 , vik+1 , . . . , vin+1 ]eik .

(1.27)

k=1

Note that in Eq. (1.26) and Eq. (1.27) the coeﬃcient of ejr is non-zero, hence there are scalars 0 = λ and cik in Q such that cik vik . α − λβ = ik =jr

Since the support of α−λβ is strictly contained in the support of β and since β is an elementary vector of ker(A), one obtains α − λβ = 0, as required. 2

58

Chapter 1

Theorem 1.9.10 Let A = (0) be an n × q integral matrix and let ψ be the Z-linear homomorphism ψ : Zq −→ Zn given by ψ(α) = A(α). Then ker(ψ) is generated as a Z-module by the circuits of ker(ψ). Proof. By Theorem 1.2.2, there are invertible integral matrices P and Q such that P AQ = D, where D is an r × q matrix with r = rank(A). Since A and DQ−1 have the same kernel and DQ−1 has size r × q and rank r, we may assume that n = rank(A). We set N = ker(ψ). Let α1 , . . . , αq be the column vectors of A and let ei be the ith unit vector of Zq . By Proposition 1.9.9, it follows that the set of circuits of N is the set of vectors of the form n+1 ±1 ±1 · v, (−1)j · det(αi1 , . . . , αij−1 , αij+1 , . . . , αin+1 ) · eij = h j=1 h where 1 ≤ i1 < · · · < in+1 ≤ q, and h is the greatest common divisor of the entries of the vector v, if v is non-zero. For convenience we introduce the notation v = v(i1 , . . . , in+1 ) and [i1 , . . . , 'ij , . . . , in+1 ] = det(αi1 , . . . , αij−1 , αij+1 , . . . , αin+1 ). Set d equal to the greatest common divisor of all the non-zero integers having the special form [i1 , . . . , 'ij , . . . , in+1 ] and

v(i1 , . . . , in+1 ) = 0 1 ≤ i1 < · · · < in+1 ≤ q . B= d Note the inclusions ZB ⊂ L ⊂ N , where L is the Z-module generated by the circuits of N . We claim that ZB = N . First observe that L and N have both rank q − n; see Corollary 1.9.8. Since ZB and L have equal rank, one concludes that N/ZB is a ﬁnite group. Thus to show ZB = N it suﬃces to prove that N/ZB is torsion-free or equivalently that Zq /ZB is torsion-free of rank n. Consider the matrix M whose rows are the vectors of B. By the fundamental theorem for ﬁnitely generated abelian groups (see Theorem 1.3.16) the proof reduces to showing that the ideal Iq−n (M ) of Z generated by the (q − n)-minors of M is equal to Z. Fix Δ = [i1 , . . . , 'ij , . . . , in+1 ]/d. For simplicity of notation one may consider the case Δ = [1, . . . , n]/d. Note that Δ occurs in v(1, . . . , n, i)/d with a ± sign for i = n + 1, . . . , q. Thus the matrix ⎛ ⎞ ··· m1n Δ 0 0 ··· 0 m11 ⎜ m21 ··· m2n 0 Δ 0 ··· 0 ⎟ ⎜ ⎟ M = ⎜ .. .. .. .. .. .. ⎟ ⎝ . . . . . . ⎠ m(q−n)1 · · · m(q−n)n 0 0 0 · · · Δ

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59

whose ith row is the vector v(1, . . . , n, i)/d is a submatrix of M . Therefore Δq−n belongs to Iq−n (M ). If Iq−n (M ) = Z, there is a prime number p dividing Δq−n for all Δ. Thus p divides Δ for all Δ, and this implies that pd divides d, a contradiction. 2 Corollary 1.9.11 If A is at-unimodular matrix of size n × q and α is a q circuit of ker(A), then α = i=1 i ei for some 1 , . . . , q in {0, ±1}. Proof. It follows readily from the explicit description of the circuits given in Proposition 1.9.9 and from Exercise 1.9.17. 2 Matroids Let M = (E, I) be a matroid on E, i.e., there is a collection I of subsets of E satisfying the following three conditions: (i1 ) ∅ ∈ I. (i2 ) If I ∈ I and I ⊂ I, then I ∈ I. (i3 ) If I1 and I2 are in I and |I1 | < |I2 |, then there is an element e of I2 \ I1 such that I1 ∪ {e} ∈ I. The members of I are the independent sets of M . It is convenient to write I(M ) for I and E(M ) for E, particularly when several matroids are being considered. A subset of E that is not in I is called dependent . A maximal independent set of M with respect to inclusion is called a basis. A minimal dependent set in M is called a circuit . The reader is referred to [338] for the general theory of matroids. Proposition 1.9.12 [338, Proposition 1.1.1, p. 8] Let E be the set of column labels of an m × n matrix A over a ﬁeld K, and let I be the set of subsets B of E for which the multiset of columns labeled by B is linearly independent in K m . Then M [A] = (E, I) is a matroid. The matroid M [A] obtained as above from the matrix A is called the vector matroid of A. Any matroid M arising from a matrix A is called representable over the ﬁeld K. We are especially interested in vector matroid arising from a nonnegative integral matrix A with column vectors v1 , . . . , vq . By Proposition 1.9.12 there is a matroid M [A] on A = {v1 , . . . , vq } over the ﬁeld Q of rational numbers, whose independent sets are the independent subsets of A. Deﬁnition 1.9.13 A minimal dependent set or circuit of M [A] is a dependent set all of whose proper subsets are independent. A subset B of A is called a basis of M [A] if B is a maximal independent set.

60

Chapter 1

There is a correspondence: Circuits of ker(A) −→ Circuits of M [A] given by α = (α1 , . . . , αq ) −→ C(α) = {vi | i ∈ supp(α)}. Thus the set of circuits of the kernel of A is the algebraic realization of the set of circuits of the vector matroid M [A]. The circuits of ker(A) can be used to study the normality of monomial subrings (see Theorem 9.7.1). Theorem 1.9.14 [338, Corollary 1.2.5, p. 18] Let B be a non-empty family of subsets of X. Then B is the collection of bases of a matroid on X if and only if the following exchange property is satisﬁed: If B1 and B2 are members of B and b1 ∈ B1 \ B2 , then there is an element b2 ∈ B2 \ B1 such that (B1 \ {b1 }) ∪ {b2 } is in B. The family of bases of any matroid satisﬁes the following symmetric exchange property (see [286]): Theorem 1.9.15 If B1 and B2 are bases of a matroid M and b1 ∈ B1 \ B2 , then there is an element b2 ∈ B2 \ B1 such that both (B1 \ {b1 }) ∪ {b2 } and (B2 \ {b2 }) ∪ {b1 } are bases of M . It is well known [338] that all bases of a matroid M have the same number of elements. This common number is called the rank of the matroid.

Exercises 1.9.16 Prove that the circuits of the kernel of the matrix M are precisely the row vectors of the matrix A with a ± sign ⎛ ⎞ 2 −3 1 0 ⎜ 3 −4 4 3 1 0 0 1 ⎟ ⎟. M= A=⎜ ⎝ 0 1 3 4 0 1 −3 2 ⎠ 1 0 −4 3 1.9.17 Let A be an n × q matrix with entries in an integral domain R and let F1 , . . . , Fn be the rows of A. Assume that Fi = j=i λj Fj , for some λj ’s in the ﬁeld of fractions of R. Consider the R-linear maps ϕ

ϕ : Rq −→ Rn

α −→ Aα

ϕ : Rq −→ Rn−1

α −→ A α,

ϕ

where A is the matrix obtained from A by removing its ith row, prove that ker(ϕ) = ker(ϕ ). 1.9.18 If X = {x1 , . . . , x4 } and B = {{x1 , x2 }, {x2 , x3 }, {x3 , x4 }, {x1 , x4 }}, prove that B satisﬁes the bases exchange property of matroids.

Chapter 2

Commutative Algebra In this chapter some basic notions and results from commutative algebra will be introduced. All rings considered in this book are commutative and Noetherian and modules are ﬁnitely generated. Our main references are the books of Bruns and Herzog [65], Eisenbud [128], Matsumura [309, 310], and Vasconcelos [413]. Some of the results presented below are just stated without giving proofs, if need be, the reader may locate the missing proofs in those references.

2.1

Module theory

Noetherian modules and localizations Let R be a commutative ring with unit and let M be an R-module. Recall that M is called Noetherian if every submodule N of M is ﬁnitely generated, that is, N = Rf1 + · · ·+ Rfq , for some f1 , . . . , fq in N . Theorem 2.1.1 The following conditions are equivalent: (a) M is Noetherian. (b) M satisﬁes the ascending chain condition for submodules; that is, for every ascending chain of submodules of M N0 ⊂ N1 ⊂ · · · ⊂ Nn ⊂ Nn+1 ⊂ · · · ⊂ M there exists an integer k such that Ni = Nk for every i ≥ k. (c) Any family F of submodules of M partially ordered by inclusion has a maximal element, i.e., there is N ∈ F such that if N ⊂ Ni and Ni ∈ F , then N = Ni .

62

Chapter 2

Proof. (a)⇒(b): Consider the submodule N = ∪i≥0 Ni . By hypothesis there are m1 , . . . , mr such that N = Rm1 + · · · + Rmr . Then, there is k such that mi ∈ Nk for all i. It follows that Ni = Nk for all i ≥ k. (b)⇒(c): Let N1 ∈ F . If N1 is not maximal, there is N2 ∈ F such that N1 N2 . If N2 is not maximal, there is N3 ∈ F such that N2 N3 . Applying this argument repeatedly we get that F has a maximal element. (c)⇒(a): Let N be a submodule of M and let F be the family of submodules of N that are ﬁnitely generated. By hypothesis F has a maximal element N . It follows that N = N . 2 In particular a Noetherian ring R is a commutative ring with unit with the property that every ideal of I is ﬁnitely generated; that is, given an ideal I of R there exists a ﬁnite number of generators f1 , . . . , fq such that I = {a1 f1 + · · · + aq fq | ai ∈ R, ∀ i} . As usual, if I is generated by f1 , . . . , fq , we write I = (f1 , . . . , fq ). Proposition 2.1.2 If M is a ﬁnitely generated R-module over a Noetherian ring R, then M is a Noetherian module. Corollary 2.1.3 If R is a Noetherian ring and I is an ideal of R, then R/I and Rn are Noetherian R-modules. In particular any submodule of Rn is ﬁnitely generated. Theorem 2.1.4 (Hilbert’s basis theorem [9, Theorem 7.5]) A polynomial ring R[x] over a Noetherian ring R is Noetherian. One of the important examples of a Noetherian ring is a polynomial ring over a ﬁeld k. Often we will denote a polynomial ring in several variables by k[x] and a polynomial ring in one variable by k[x]. The letters k and K will always denote ﬁelds. In this book, unless otherwise stated, by a ring (resp. module) we shall always mean a Noetherian ring (resp. ﬁnitely generated module). The prime spectrum of a ring R, denoted by Spec(R), is the set of prime ideals of R. The minimal primes of R are the minimal elements of Spec(R) with respect to inclusion and the maximal ideals of R are the maximal elements of the set of proper ideals of R with respect to inclusion. Let R be a ring and let X = Spec(R) be its prime spectrum. Given an ideal I of R, the set of all prime ideals of R containing I will be denoted by V (I). The minimal primes of I are the minimal elements of V (I) with respect to inclusion. It is not hard to verify that the pair (X, Z) is a topological space, where Z is the family of open sets of X, and where U is in Z iﬀ U = X \ V (I), for some ideal I. This topology is called the Zariski topology of the prime spectrum of R.

Commutative Algebra

63

A local ring (R, m, k) is a Noetherian ring R with exactly one maximal ideal m, the ﬁeld k = R/m is called the residue ﬁeld of R. A homomorphism of rings is a map ϕ : R → S such that: (i)

ϕ(a + b) = ϕ(a) + ϕ(b), ∀ a, b ∈ A,

(ii) ϕ(ab) = ϕ(a)ϕ(b), ∀ a, b ∈ A, and (iii) ϕ(1) = 1. Let R be a ring and let ϕ : Z → R be the canonical homomorphism ϕ(a) = a · 1R , then ker(ϕ) = nZ, for some n ≥ 0. The integer n is called the characteristic of R and is denoted by char(R). Proposition 2.1.5 Let (R, m) be a local ring, then either char(R) = 0 or char(R) = pk , for some prime number p and some integer k ≥ 1. Proof. Let n = char(R) ≥ 2. If n = pqm, where p, q are prime numbers with q = p, then p · 1R and q · 1R are both in m because they are non invertible, but this is impossible because 1 = ap + bq and from the equality 1R = (a · 1R )(p · 1R ) + (b · 1R )(q · 1R ), one derives 1R ∈ m, a contradiction.

2

Modules of fractions and localizations Let R be a ring, M an R-module, and S a multiplicatively closed subset of R so that 1 ∈ S. Then the module of fractions of M with respect to S, or the localization of M with respect to S, is deﬁned by S −1 (M ) = {m/s| m ∈ M, s ∈ S} , where m/s = m1 /s1 if and only if t(s1 m − sm1 ) = 0 for some t ∈ S. In particular S −1 R has a ring structure given by the usual rules of addition and multiplication, and S −1 M is a module over the ring S −1 R with the operations: m1 /s1 + m2 /s2

=

(a/s) · (m1 /s1 ) =

(s2 m1 + s1 m2 )/s1 s2

(si ∈ S, mi ∈ M ),

am1 /ss1

(a ∈ R, s ∈ S).

There is a map ϕ : M → S −1 M , given by ϕ(m) = m/1. If f : M → N is a homomorphism of R-modules, then there is an induced homomorphism S −1 f : S −1 M −→ S −1 N of S −1 R-modules given by f (m/s) = f (m)/s.

64

Chapter 2

Example 2.1.6 If f is in a ring R and S = {f i | i ∈ N}, then S −1 R is usually written Rf . For instance if R = C [x] is a polynomial ring in one variable over the ﬁeld C of complex numbers, then Rx = C [x, x−1 ] is the ring of Laurent polynomials. Deﬁnition 2.1.7 Let p be a prime ideal of a ring R and S = R \ p. In this case S −1 R is written Rp and is called the localization of R at p. Example 2.1.8 Let p be a prime ideal of a ring R. The local ring (Rp , pRp , k(p)) is the prototype of a local ring, where k(p) = Rp /pRp denotes the residue ﬁeld of Rp . Krull dimension and height By a chain of prime ideals of a ring R we mean a ﬁnite strictly increasing sequence of prime ideals p0 ⊂ p1 ⊂ · · · ⊂ pn , the integer n is called the length of the chain. The Krull dimension of R, denoted by dim(R), is the supremum of the lengths of all chains of prime ideals in R. Let p be a prime ideal of R, the height of p, denoted by ht (p) is the supremum of the lengths of all chains of prime ideals p0 ⊂ p1 ⊂ · · · ⊂ pn = p which end at p. Note dim(Rp ) = ht (p). If I is an ideal of R, then ht (I), the height of I, is deﬁned as ht(I) = min{ht(p)| I ⊂ p and p ∈ Spec(R)}. In general dim(R/I) + ht (I) ≤ dim(R). The diﬀerence dim(R) − dim(R/I) is called the codimension of I and dim(R/I) is called the dimension of I. Let M be an R-module. The annihilator of M is given by annR (M ) = {x ∈ R| xM = 0}, if m ∈ M the annihilator of m is ann (m) = ann (Rm). It is convenient to generalize the notion of annihilator to ideals and submodules. Let N1 and N2 be submodules of M , their ideal quotient or colon ideal is deﬁned as (N1 : R N2 ) = {x ∈ R| xN2 ⊂ N1 }. Let us recall that the dimension of an R-module M is dim(M ) = dim(R/ann (M )) and the codimension of M is codim(M ) = dim(R) − dim(M ). Theorem 2.1.9 If R[x] is a polynomial ring over a Noetherian ring R, then dim R[x] = dim(R) + 1.

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Primary decomposition of modules Let I be an ideal of a ring R. The radical of I is rad (I) = {x ∈ R| xn ∈ I for some n > 0}, √ the radical is also denoted by I. In particular (0), denoted by NR or nil(R), is the set of nilpotent elements of R and is called the nilradical of R. A ring is reduced if its nilradical is zero. The Jacobson radical of R is the intersection of all the maximal ideals of R. Proposition 2.1.10 If I is a proper ideal of a ring R, then rad (I) is the intersection of all prime ideals containing I. Deﬁnition 2.1.11 Let M be a module over a ring R. The set of associated primes of M , denoted by AssR (M ), is the set of all prime ideals p of R such that there is a monomorphism φ of R-modules: φ

R/p → M. Note that p = ann (φ(1)). Lemma 2.1.12 If M = 0 is an R-module, then AssR (M ) = ∅. Proof. Consider the family of ideals F = {ann(m)| 0 = m ∈ M } ordered by inclusion. By Theorem 2.1.1 this family has a maximal element that we denote by ann(m). It suﬃces to show that ann(m) is prime. Let x, y be two elements of R such that xy ∈ ann(m). Assume that x ∈ / ann(m). Then ann(m) ⊂ ann(xm) and consequently ann(m) = ann(xm). This shows that y ∈ ann(m), as required. 2 If M = R/I, it is usual to say that an associated prime ideal of R/I is an associated prime ideal of I and to set Ass(I) = Ass(R/I). Proposition 2.1.13 Let R be a ring and let S be a multiplicatively closed subset of R. If M is an R-module and p is a prime ideal of R with S ∩p = ∅, then p is an associated prime of M if and only if S −1 p is an associated prime of S −1 M . Proof. If p is in Ass(M ), then R/p → M . Hence S −1 R/S −1 p → S −1 M . Thus, S −1 p is an associated prime of S −1 M . For the converse assume S −1 p = ann(m/1). Since R is Noetherian, p is generated by a ﬁnite set a1 , . . . , an . Hence for each i there is si ∈ S such that si ai m = 0. Set s = s1 · · · sn . We claim p = ann(sm). Clearly one has p ⊂ ann(sm). To show the other containment take x ∈ ann(sm), then 2 xsm = 0 and x/1 ∈ ann(m/1) = S −1 p. Hence x ∈ p.

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Deﬁnition 2.1.14 Let M be an R-module, the support of M , denoted by Supp(M ), is the set of all prime ideals p of R such that Mp = 0. f

g

A sequence 0 → M → M → M → 0 of R-modules is called a short exact sequence if f is a monomorphism, g is an epimorphism, and im(f ) = ker(g). A sequence of R modules and homomorphisms: fi−1

fi

fi+1

· · · −→ Mi−1 −→ Mi −→ Mi+1 −→ · · · is said to be exact at Mi if im(fi−1 ) = ker(fi ). If the sequence is exact at each Mi it is called an exact sequence. Lemma 2.1.15 If 0 → M → M → M → 0 is a short exact sequence of modules over a ring R, then Supp(M ) = Supp(M ) ∪ Supp(M ). Proof. Let p be a prime ideal of R. It suﬃces to observe that from the exact sequence 0 → Mp → Mp → Mp → 0, we get Mp = 0 if and only if Mp = 0 or Mp = 0.

2

Theorem 2.1.16 If M is an R-module, then there is a ﬁltration of submodules (0) = M0 ⊂ M1 ⊂ · · · ⊂ Mn = M and prime ideals p1 , . . . , pn of R such that Mi /Mi−1 R/pi for all i. Proof. By Lemma 2.1.12 there is a prime ideal p1 and a submodule M1 of M such that R/p1 M1 . If M1 M , then there is an associated prime ideal p2 of M/M1 such that R/p2 is isomorphic to a submodule of M/M1 , i.e., R/p2 M2 /M1 , where M2 is a submodule of M containing M1 . If M2 M , we pick an associated prime p3 of M/M2 and repeat the argument. Since M is Noetherian a repeated use of this procedure yields the required ﬁltration. 2 In general the primes p1 , . . . , pn that occur in a ﬁltration of the type described in the previous result are not associated primes of the module M ; see [115] for a careful discussion of ﬁltrations and for some of their applications to combinatorics. f

g

Lemma 2.1.17 If 0 → M → M → M → 0 is a short exact sequence of modules over a ring R, then Ass(M ) ⊂ Ass(M ) ∪ Ass(M ). Proof. We may assume (without loss of generality) that f is the identity map by replacing M by f (M ). Let p be an associated prime of M . There is 0 = m ∈ M such that p = ann(m).

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Case (I): Rm ∩ M = (0). There is r ∈ R such that 0 = rm ∈ M . From the equality p = ann(m) we get p = ann(rm). Thus p ∈ Ass(M ). Case (II): Rm ∩ M = (0). Notice that g(m) = 0. From the equality p = ann(g(m)) we get p ∈ Ass(M ). 2 Corollary 2.1.18 If M is an R-module, then AssR (M ) is a ﬁnite set. Proof. Let p1 , . . . , pn be prime ideals as in Theorem 2.1.16. By a repeated 2 use of Lemma 2.1.17 one has AssR (M ) ⊂ {p1 , . . . , pn } ⊂ Supp(M ). Let M be an R-module. An element x ∈ R is a zero divisor of M if there is 0 = m ∈ M such that xm = 0. The set of zero divisors of M is denoted by Z(M ). If x is not a zero divisor on M , we say that x is a regular element of M . Lemma 2.1.19 If M is an R-module, then ( p. Z(M ) = p∈AssR (M)

Proof. The right-hand side is clearly contained in the left-hand side by deﬁnition of an associated prime. Let r be a zero divisor of M and consider the family F = {ann(m)| 0 = m ∈ M ; rm = 0}. Notice that any maximal element of this family is a prime ideal. 2 Proposition 2.1.20 If M is an R-module, then Ass(M ) ⊂ Supp(M ) = V (ann (M )), and any minimal element of Supp(M ) is in Ass(M ). Proof. If p is an associated prime of M , then there is a monomorphism R/p→M and thus 0 = (R/p)p →Mp . Hence p is in the support of M , this shows the ﬁrst containment. Next, we show Supp(M ) = V (ann (M )). Let p ∈ Supp(M ) and let x ∈ ann (M ). If x ∈ p, then xm = 0 for all m ∈ M and Mp = (0), which is absurd. Therefore p is in V (ann (M )). Conversely let p be in V (ann (M )) and let m1 , . . . , mr be a ﬁnite set of generators of M . If Mp = (0), then for each i there is si ∈ p so that si mi = 0, therefore s1 · · · sr is in ann (M ) ⊂ p, which is impossible. Hence Mp = 0 and p is in the support of M . To prove the last part take a minimal prime p in the support of M . As Mp = (0) there is an associated prime p1 Rp of Mp , where p1 is a prime ideal of R contained in p. Since Mp1 (Mp )p1 = (0), we get that p1 is in the support of M and p = p1 . Therefore using Proposition 2.1.13 one concludes p ∈ Ass(M ). 2

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Let M be an R-module, the minimal primes of M are deﬁned to be the minimal elements of Supp(M ) with respect to inclusion. A minimal prime of M is called an isolated associated prime of M . An associated prime of M which is not isolated is called an embedded prime. If R is a ring and I is an ideal, note that the minimal primes of I are precisely the minimal primes of AssR (R/I). In particular the minimal primes of R are precisely the minimal primes of AssR (R). Deﬁnition 2.1.21 Let M be an R-module. A submodule N of M is said to be a p-primary submodule if AssR (M/N ) = {p}. An ideal q of a ring R called a p-primary ideal if AssR (R/q) = {p}. Proposition 2.1.22 An ideal q = R of a ring R is a primary ideal if and only if xy ∈ q and x ∈ / q implies y n ∈ q for some n ≥ 1. Proof. Assume q is a primary ideal. Let x, y ∈ R such that xy ∈ q and x∈ / q. Hence y is a zero divisor of R/q because yx = 0 and x = 0. Since Z(R/q) = {p} and rad(q) = p, we get y n ∈ q for some positive integer n. The converse is left as an exercise. 2 Deﬁnition 2.1.23 Let M be an R-module. A submodule N of M is said to be irreducible if N cannot be written as an intersection of two submodules of M that properly contain N . Proposition 2.1.24 Let M be an R-module. If Q = M is an irreducible submodule of M , then Q is a primary submodule. Proof. Assume there are p1 and p2 distinct associated prime ideals of M/Q and pick r0 ∈ p1 \ p2 (or vice versa). There is xi in M \ Q such that pi = ann(xi ), where xi = xi + Q. We claim that (Rx1 + Q) ∩ (Rx2 + Q) = Q. If z is in the intersection, then z = λ1 x1 + q1 = λ2 x2 + q2 , for some λi ∈ R and qi ∈ Q. Note that r0 z ∈ Q, hence r0 λ2 x2 is in Q and consequently r0 λ2 ∈ p2 . Thus λ2 ∈ p2 and we get λ2 x2 ∈ Q. This shows z ∈ Q and completes the proof of the claim. As Q is irreducible one has Q = Rx1 + Q or Q = Rx2 + Q, which is a contradiction because xi ∈ Q for i = 1, 2. 2 Lemma 2.1.25 If N1 , N2 are p-primary submodules of M , then N1 ∩ N2 is a p-primary submodule. Proof. Set N = N1 ∩ N2 . There is an inclusion M/N → M/N1 ⊕ M/N2 . Hence using Exercise 2.1.53 we get Ass(M/N ) ⊂ Ass(M/N1 ⊕ M/N2 ) = Ass(M/N1 ) ∪ Ass(M/N2 ) = {p}. Therefore Ass(M/N ) = {p}.

2

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Deﬁnition 2.1.26 Let M be an R-module and let N M be a proper submodule. An irredundant primary decomposition of N is an expression of N as an intersection of submodules, say N = N1 ∩ · · · ∩ Nr , such that: (a) (Submodules are primary) AssR (M/Ni ) = {pi } for all i. (b) (Irredundancy) N = N1 ∩ · · · ∩ Ni−1 ∩ Ni+1 ∩ · · · ∩ Nr for all i. (c) (Minimality) pi = pj if Ni = Nj . Theorem 2.1.27 Let M be an R-module. If N M is a proper submodule of M , then N has an irredundant primary decomposition Proof. First we claim that N is an intersection of a ﬁnite number of irreducible submodules of M . Let F be the family of submodules of M such that the claim is false. Assume F = ∅. By Theorem 2.1.1, F has a maximal element that we denote by N . Since N is not irreducible, we can write N = N2 ∩ N3 , for some submodules N2 , N3 strictly containing N . By the maximality of N we get that N1 and N2 can be written as an intersection of irreducible submodules, a contradiction. Thus F = ∅. Hence the result follows from Proposition 2.1.24 and Lemma 2.1.25. 2 Corollary 2.1.28 If R is a Noetherian ring and I a proper ideal of R, then I has an irredundant primary decomposition I = q1 ∩ · · · ∩ qr such that qi is a pi -primary ideal and AssR (R/I) = {p1 , . . . , pr }. Proof. Let (0) = I/I = (q1 /I) ∩ · · · ∩ (qr /I) be an irredundant decomposition of the zero ideal of R/I. Then I = q1 ∩ · · · ∩ qr and qi /I is pi -primary; that is, AssR ((R/I)/(qi /I)) = Ass(R/qi ) = {pi }. Let us show that qi is a primary ideal. If xy ∈ qi and x ∈ / qi , then y is a zero-divisor of R/qi , but Z(R/qi ) = pi , hence y ∈ pi = rad (ann (R/qi )) = rad (qi ) and y n is in qi for some n > 0. 2 Corollary 2.1.29 If M is an R-module, then rad(ann(M )) =

p.

p∈Ass(M)

Proof. Let p1 , . . . , pr be the minimal elements of Supp(M ). By Proposition 2.1.20 we have Supp(M ) = V (ann (M )) and pi is in Ass(M ) for all i. 2 By Corollary 2.1.28 we have that rad(ann(M )) = p1 ∩ · · · ∩ pr . Corollary 2.1.30 If N M and N = N1 ∩ · · · ∩ Nr is an irredundant primary decomposition of N with AssR (M/Ni ) = {pi }, then AssR (M/N ) = {p1 , . . . , pr }, and ann (M/Ni ) is a pi -primary ideal for all i.

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Proof. There is a natural monomorphism M/(N1 ∩ · · · ∩ Nr ) → (M/N1 ) ⊕ · · · ⊕ (M/Nr ). Hence Ass(M/N ) ⊂ {p1 , . . . , pr }. There are natural monomorphisms (N2 ∩ · · · ∩ Nr )/N → M/N1 ;

(N2 ∩ · · · ∩ Nr )/N → M/N.

Since Ass(M/N1 ) = {p1 } we get p1 ∈ Ass(N2 ∩ · · · ∩ Nr /N ), consequently p1 ∈ Ass(M/N ). Similarly one can show that any other pi is an associated prime of M/N . This proves the asserted equality. By Corollary 2.1.29 we have rad(ann(M/Ni )) = pi . Thus it suﬃces to show that I = ann(M/Ni ) is a primary ideal. Assume that xy ∈ I for some x, y ∈ R. If x is not in I, then xM is not contained in Ni . Pick m ∈ M such that xm ∈ / Ni . Since y(xm) ⊂ Ni , we get that y is a zero divisor of M/Ni , but the zero divisors of this module are precisely the elements of pi according to Lemma 2.1.19. Hence y r ∈ I for some r. 2 Deﬁnition 2.1.31 Let R be a ring and let S be the set of nonzero divisors of R. The ring S −1 R is called the total ring of fractions of R. If R is a domain, S −1 R is the ﬁeld of fractions of R. Proposition 2.1.32 Let R be a ring and let K be the total ring of fractions of R. If R is reduced, then K is a direct product of ﬁelds. Proof. Let p1 , . . . , pr be the minimal primes of R and S = R \ ∪ri=1 pi . Since R is reduced one has (0) = p1 ∩ · · · ∩ pr and K = S −1 R. Deﬁne φ : K −→ S −1 R/S −1 p1 × · · · × S −1 R/S −1 pr by φ(x) = (x + S −1 p1 , . . . , x + S −1 pr ). As S −1 p1 , . . . , S −1 pr are maximal ideals and its intersection is zero, it follows from the Chinese remainder theorem that φ is an isomorphism (see Exercise 2.1.46). 2 Lemma 2.1.33 Let M be an R-module and L an ideal of R. If LM = M , then there is x ∈ R such that x ≡ 1 (mod L) and xM = (0). Proof. Let M = Rα 1 + · · · + Rαn , αi ∈ M . As LM = M , there are bij n in L such that αi = j=1 bij αi . Set α = (α1 , . . . , αn ) and H = (bij ) − I, where I is the identity matrix. Since Hαt = 0 and Hadj(H) = det(H)I, one concludes det(H)αi = 0 for all i. Hence xM = (0) and x ≡ 1(mod L), where x = det(H). 2 Lemma 2.1.34 (Nakayama) Let R be a ring and let N be a submodule of an R-module M . If I is an ideal of R contained in the intersection of all the maximal ideals of R such that M = IM + N , then M = N .

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Proof. Note I(M/N ) = M/N . By Lemma 2.1.33 there exists an element x ≡ 1(mod I) such that x(M/N ) = (0). To ﬁnish the proof note that x is a unit; otherwise x belongs to some maximal ideal m and this yields a contradiction because I ⊂ m. 2 Let M be an R-module. The minimum number of generators of M will be denoted by μ(M ). A consequence of Nakayama’s lemma is an expression for μ(M ), when the ring R is local (cf. Corollary 3.5.2). Corollary 2.1.35 If M is a module over a local ring (R, m), then μ(M ) = dimk (M/mM ), where k = R/m. Proof. Let α1 , . . . , αq be a minimal generating set for M and αi = αi +mM . After a permutation of the αi one may assume that α1 , . . . , αr is a basis for M/mM as a k-vector space, for some r ≤ q. Set N = Rα1 + · · · + Rαr . Note the equality M = N + mM , then by Nakayama’s Lemma N = M . Therefore r = q, as required. 2 Modules of ﬁnite length An R-module M has ﬁnite length if there is a composition series (0) = M0 ⊂ M1 ⊂ · · · ⊂ Mn = M,

(2.1)

where Mi /Mi−1 is a non-zero simple module (that is, Mi /Mi−1 has no proper submodules other than (0)) for all i. Note that Mi /Mi−1 must be cyclic and thus isomorphic to R/m, for some maximal ideal m. The number n is independent of the composition series and is called the length of M ; it is usually denoted by R (M ) or simply (M ). Proposition 2.1.36 [9, Proposition 6.9] If 0 → M → M → M → 0 is an exact sequence of R-modules of ﬁnite length, then R (M ) = R (M ) + R (M ). An R-module M is called Artinian if M satisﬁes the descending chain condition for submodules; that is, for every chain of submodules of M · · · ⊂ Nn+1 ⊂ Nn ⊂ · · · ⊂ N2 ⊂ N1 ⊂ N0 = M there exists an integer k such that Ni = Nk for every i ≥ k. It is easy to verify that M is Artinian if and only if any family F of submodules of M partially ordered by inclusion has a minimal element, i.e., there is N ∈ F such that if Ni ⊂ N and Ni ∈ F , then N = Ni . Proposition 2.1.37 Let M be an R-module. Then R (M ) < ∞ if and only if M is Noetherian and Artinian.

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Proof. ⇒) Set n = R (M ). If M is not Noetherian or Artinian, pick submodules of M such that (0) = N0 N1 · · · Nn Nn+1 = M, a contradiction because this chain can be reﬁned to a composition series of length n. ⇐) We construct a ﬁnite composition series as follows. Set M0 = M . Consider the family F1 of proper submodules of M and pick a maximal element M1 , which exists because M is Noetherian. By induction consider the family Fi of proper submodules of Mi and pick a maximal element Mi+1 . Notice that this process must stop at the zero module because M is Artinian. 2 Lemma 2.1.38 Let M be a k-vector space. Then M is Artinian if and only if dimk (M ) = k (M ) < ∞. Proof. Assume that M is Artinian and dimk (M ) = ∞. Let B ⊂ M be an inﬁnite linearly independent set, say B = {α1 , . . . , αi , . . . , }. Then · · · k(B \ {α1 , . . . , αi }) · · · k(B \ {α1 , α2 }) k(B \ {α1 }) M, a contradiction. The converse is also easy to show.

2

Proposition 2.1.39 If 0 → M → M → M → 0 is an exact sequence of modules over a ring R, then dim(M ) = max{dim(M ), dim(M )}. Proof. Set d = dim(M ), d = dim(M ) and d = dim(M ). First note that d = dim(R/p) for some prime p containing ann (M ), by Proposition 2.1.20 we obtain Mp = (0). Therefore using Lemma 2.1.15 one has Mp = (0) or Mp = (0), thus either p contains ann (M ) or p contains ann (M ). This proves d ≤ max{d , d }. On the other hand ann (M ) is contained in ann (M ) ∩ ann (M ) and consequently max{d , d } ≤ d. 2 Proposition 2.1.40 If M is an R-module, then M has ﬁnite length if and only if every prime ideal in Supp(M ) is a maximal ideal. Proof. ⇒) Let {Mi }ni=0 be a composition series as in Eq. (2.1). Notice that pi are maximal ideals for all i. Indeed if N is a simple R-module and R/p N for some prime ideal p, then p is a maximal ideal. Hence from Lemma 2.1.15 we obtain that Supp(M ) is equal to {p1 , . . . , pn }. ⇐) There is a ﬁltration (0) = M0 ⊂ M1 ⊂ · · · ⊂ Mn = M of submodules and prime ideals p1 , . . . , pn of R such that Mi /Mi−1 R/pi for all i (see Theorem 2.1.16). By the proof of Corollary 2.1.18 one has Ass(M ) ⊂ {p1 , . . . , pn } ⊂ Supp(M ).

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Hence by Proposition 2.1.20 we get Ass(M ) = {p1 , . . . , pn } = Supp(M ). Thus the ﬁltration above is a composition series because pi is a maximal ideal for all i. 2 Theorem 2.1.41 Let R be a ring. Then R is Artinian if and only if (a) R is Noetherian, and (b) every prime ideal of R is maximal. Proof. ⇒) First we show that R (R) < ∞. Let F = {m1 m2 · · · ms | mi ∈ Max(R), ∀ i}, we do not require m1 , . . . , ms to be distinct. The family F has a minimal element that we denote by I = m1 · · · mr . We claim that I = (0). Assume that I = (0). Clearly I 2 ∈ F and I 2 ⊂ I, thus I = I 2 . Consider the family G = {J| J is an ideal; JI = (0)}. This family is non-empty because I ∈ G. Since R is Artinian, there exists a minimal element J of G. Then (0) = JI = JI 2 = (JI)I ⇒ JI ∈ G. Since IJ ⊂ J, we get J = IJ. Notice that J is a principal ideal. Indeed since IJ = (0), there is x ∈ J such that xI = (0). Hence (x) ∈ G, and consequently J = (x). If m is a maximal ideal of R, then mI ∈ F and mI ⊂ I. Hence mI = I and I ⊂ m. Thus I is contained in the Jacobson radical of I. From the equality (x) = J = IJ = Ix, we obtain that x = λx for some λ ∈ I. As x(1 − λ) = 0, with 1 − λ a unit of R, we get x = 0. A contradiction to J = (0). This proves that I = (0). Set Ik = m1 · · · mk for 1 ≤ k ≤ r and consider the ﬁltration (0) = I = Ir ⊂ Ir−1 ⊂ Ir−2 ⊂ · · · ⊂ I1 ⊂ I0 = R. Observe that Ii /Ii+1 is an R/mi+1 vector space. The vector space Ii /Ii+1 is Artinian because the subspaces of Ii /Ii+1 correspond bijectively to the intermediate ideals of Ii+1 ⊂ Ii . Therefore by Lemma 2.1.38 we get that Ii /Ii+1 has ﬁnite length as an R/mi+1 vector space and consequently as an R-module. Hence the ﬁltration can be reﬁned to a ﬁnite composition series, i.e., R (R) < ∞. Therefore R is Noetherian by Proposition 2.1.37 and (a) holds. To prove (b) take a minimal prime p of R. Since p contains I = (0) it follows that p = mi for some 1 ≤ i ≤ r. ⇐) Let (0) = q1 ∩ · · · ∩ qr be an irredundant primary decomposition. Then all the associated primes are maximal. By Proposition 2.1.20 we obtain AssR (R) = Supp(R). Hence R is Artinian by Proposition 2.1.40. 2

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Deﬁnition 2.1.42 A ring is called semilocal if it has only ﬁnitely many maximal ideals. Using the Chinese reminder theorem (see Exercise 2.1.46) and the proof of Theorem 2.1.41 we obtain: Proposition 2.1.43 If R is an Artinian ring, then R has only a ﬁnite number of maximal ideals m1 , . . . , ms and R R/ma1 1 × · · · × R/mas s . Deﬁnition 2.1.44 Let R be a ring with total ring of fractions Q. An Rmodule M has rank r if M ⊗R Q is a free Q-module of rank equal to r. Lemma 2.1.45 If M is an R-module of positive rank r, then dim(M ) = dim(R). Proof. Let p1 , . . . , pn be the associated primes of R and S = R \ ∪ni=1 pi . By hypothesis S −1 M (S −1 R)r . Since Mpi (S −1 M )pi [(S −1 R)r ]pi [(S −1 R)pi ]r (Rpi )r = (0), we obtain that all the minimal primes of R are in the support of M and consequently one has dim(M ) = dim(R). 2

Exercises 2.1.46 (Chinese remainder theorem) Let I1 , . . . , Ir be ideals of a ring R. If Ii + Ij = R for i = j, prove: (a) I1 ∩ · · · ∩ Ir = I1 · · · Ir , (b) The rings R/I1 ∩ · · · ∩ Ir and R/I1 × · · · × R/Ir are isomorphic. 2.1.47 Let I be an ideal of a ring R. Prove that rad(I) = I if and only if for any x ∈ R such that x2 ∈ I one has x ∈ I. 2.1.48 Let I1 , . . . , In be ideals of a ring R and let p be a prime ideal of R. (a) If I1 · · · In ⊂ p, prove that Ii ⊂ p for some i. (b) If ∩ni=1 Ii ⊂ p, then Ii ⊂ p for some i. (c) If ∩ni=1 Ii = p, then p = Ii for some i. 2.1.49 Let I, p1 , . . . , pr be ideals of a ring R. If I is contained in ∪ri=1 pi and pi is a prime ideal for i ≥ 3, prove that I ⊂ pi for some i (cf. Lemma 3.1.31). 2.1.50 Let R be a ring and let I be an ideal. If p is a prime ideal such that I ⊂ p, prove that ht(I) ≤ ht(Ip ). Give an example where the strict inequality holds.

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2.1.51 Let R be a ring and let I be an ideal of R. If all the minimal primes of I have the same height and p is a prime ideal such that I ⊂ p, prove that ht(I) = ht(Ip ). 2.1.52 Let I be an ideal of a ring R. Then a prime ideal p of R is an associated prime of R/I if and only if p = (I : x) for some x ∈ R. 2.1.53 If M , N are R-modules, prove Ass(M ⊕ N ) = Ass(M ) ∪ Ass(N ). 2.1.54 Let M be an R-module and I an ideal of R contained in annR (M ). Note that M inherits a structure of R/I-module. Prove that p ∈ AssR (M ) if and only if p/I ∈ AssR/I (M ). 2.1.55 Let M be an R-module and let p be a prime ideal of R. Prove that p ∈ AssR (M ) if and only if pRp ∈ AssRp (Mp ). 2.1.56 Let A be a ring and q a proper ideal of A. Prove that q is a primary ideal if and only if Z(A/q) ⊂ NA/q , where NA/q is the nilradical of A/q. 2.1.57 Let A be a ring and q a proper ideal of A. If every element in A/q is either nilpotent or invertible, prove that q is a primary ideal. 2.1.58 Let A be a ring and q an ideal of A such that rad (q) = m is a maximal ideal. Show that q is a primary ideal. 2.1.59 If B = A[x] is a polynomial ring over a ring A and q a primary ideal of A, then qB is a primary ideal of B. 2.1.60 Let M be an R-module and S ⊂ T two multiplicatively closed subsets of R. Prove T −1 M T −1 (S −1 M ) as T −1 R modules. 2.1.61 Let M be an R-module and S a multiplicatively closed subset of R. If p is a prime ideal such that S ∩ p = ∅, prove Mp (S −1 M )p . 2.1.62 Let R be a ring and I an ideal. If x ∈ R \ I, then there is an exact sequence of R-modules: ψ

φ

0 −→ R/(I : x) −→ R/I −→ R/(I, x) −→ 0, where ψ(r) = xr is multiplication by x and φ(r) = r. f

g

2.1.63 If 0 → M → M → M → 0 is an exact sequence of R-modules, prove that the following sequence is also exact S −1 f

0 −→ S −1 M −→ S −1 M

S −1 g

−→ S −1 M −→ 0.

76

Chapter 2

2.1.64 Let N1 and N2 be submodules of an R-module M . If S is a multiplicatively closed subset of R, then (a) S −1 (N1 + N2 ) = S −1 (N1 ) + S −1 (N2 ), (b) S −1 (N1 ∩ N2 ) = S −1 (N1 ) ∩ S −1 (N2 ), (c) S −1 (N1 /N2 ) S −1 (N1 )/S −1 (N2 ), as S −1 (R)-modules, (d) S −1 (ann (N2 )) = ann (S −1 (N2 )), and (e) S −1 (N1 : N2 ) = (S −1 (N1 ) : S −1 (N2 )). Hint Use that N2 is a ﬁnitely generated R-module. 2.1.65 If I and J are ideals of a ring R and S is a multiplicatively closed subset of R, then √ (a) S −1 (I) = S −1 ( I), and (b) S −1 (IJ) = S −1 (I)S −1 (J).

2.2

Graded modules and Hilbert polynomials

Let (H, +) be an abelian semigroup. An H-graded ring is a ring R together with a decomposition ) Ra (as a Z-module), R= a∈H

such that Ra Rb ⊂ Ra+b for all a, b ∈ H. A graded ring is by deﬁnition a Z-graded ring. If R is an H-graded ring and M is an R-module with a decomposition ) M= Ma , a∈H

such that Ra Mb ⊂ Ma+b for all a, b ∈ H, we say that M is an H-graded module. An element 0 = f ∈ M is said to be homogeneous of degree a if f ∈ Ma , in this case we set deg(f ) = a. The non-zero elements in Ra are also called forms of degree a. Any element f ∈ M can be written uniquely as f = a∈H fa with only ﬁnitely many fa = 0. A map ϕ : M → N between H-graded modules is graded if ϕ(Ma ) ⊂ Na for all a ∈ H. Let M = ⊕a∈H Ma be an H-graded module and N a graded submodule; that is, N is graded with the induced grading N = ⊕a∈H N ∩Ma . Then M/N is an H-graded R-module with (M/N )a = Ma /N ∩ Ma for a ∈ H, R0 ⊂ R is a subring and Ma is an R0 -module for a ∈ H.

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Proposition 2.2.1 [310, p. 92] Let M = ⊕a∈H Ma be an H-graded module and N ⊂ M a submodule. Then the following conditions are equivalent: (g1 ) N is generated over R by homogeneous elements. (g2 ) If f = a∈H fa is in N , fa ∈ Ma for all a, then each fa is in N . (g3 ) N is a graded submodule of M . Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and let d1 , . . . , dn be a sequence in N+ . For a = (ai ) in Nn we set xa = xa1 1 · · · xann n and |a| = i=1 ai di . The induced N-grading on R is given by: R=

∞ )

Ri , where Ri =

i=0

)

Kxa .

|a|=i

Notice that deg(xi ) = di for all i. The induced grading extends to a Zgrading by setting Ri = 0 for i < 0. The homogeneous elements of R are called quasi-homogeneous polynomials. Let I be a homogeneous ideal of R generated by a set f1 , . . . , fr of homogeneous polynomials. Setting deg(fi ) = δi , I becomes a graded ideal with the grading Ii = I ∩ Ri = f1 Ri−δ1 + · · · + fr Ri−δr . Hence R/I is an N-graded R-module graded by (R/I)i = Ri /Ii . Deﬁnition 2.2.2 The standard grading or usual grading of a polynomial ring K[x1 , . . . , xn ] is the N-grading induced by setting deg(xi ) = 1 for all i. Hilbert polynomial and multiplicity Let R = ⊕∞ i=0 Ri be an N-graded ring. We recall that R is a Noetherian ring if and only if R0 is a Noetherian ring and R = R0 [x1 , . . . , xn ] for some x1 , . . . , xn in R. If M is a ﬁnitely generated N-graded R-module and R0 is an Artinian local ring, deﬁne the Hilbert function of M as H(M, i) = R0 (Mi )

(R0 = length w.r.t R0 ).

Deﬁnition 2.2.3 An N-graded ring R = ⊕∞ i=0 Ri is called a homogeneous ring if R = R0 [x1 , . . . , xn ], where deg(xi ) = 1 for all i. For use below, by convention the zero polynomial has degree −1. Theorem 2.2.4 (Hilbert [65, Theorem 4.1.3]) Let R = ⊕∞ i=0 Ri be a homogeneous ring and let M be a ﬁnitely generated N-graded R-module with d = dim(M ). If R0 is an Artinian local ring, then there is a unique polynomial ϕM (t) ∈ Q[t] of degree d − 1 such that ϕM (i) = H(M, i) for i 0. Deﬁnition 2.2.5 The polynomial ϕM (t) is called the Hilbert polynomial of M . If ϕM (t) = ad−1 td−1 + · · · + a0 , the multiplicity or degree of M , denoted by e(M ) or deg(M ), is (d − 1)!ad−1 if d ≥ 1 and R0 (M ) if d = 0.

78

Chapter 2

Graded primary decomposition Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K endowed with a positive grading induced by setting deg(xi ) = di for all i, where di is a positive integer for i = 1, . . . , n. Lemma 2.2.6 [65, Lemma 1.5.6] If M is an N-graded R-module and p is in Ass(M ), then p is a graded ideal and there is m ∈ M homogeneous such that p = ann(m). Proposition 2.2.7 [309, p. 63] Let M be an N-graded R-module and let Q be a p-primary submodule of M . If p is graded and Q∗ is the submodule of M generated by the homogeneous elements in Q, then Q∗ is again a p-primary submodule. Theorem 2.2.8 Let M be an N-graded R-module and let N be a proper graded submodule of M . Then N has an irredundant primary decomposition N = N1 ∩ · · · ∩ Nr such that Ni is a graded submodule for all i. Proof. Use Lemma 2.2.6, Proposition 2.2.7, and Theorem 2.1.27.

2

Finding primary decompositions of graded ideals in polynomial rings over ﬁelds is a diﬃcult task. For a treatment of the principles of primary decomposition consult the book of Vasconcelos [413, Chapter 3]. Notation If R = ⊕∞ i=0 Ri is an N-graded ring, we set R+ = ⊕i≥1 Ri . Lemma 2.2.9 (Graded Nakayama lemma) Let R be an N-graded ring and let M be an N-graded R-module. If N is a graded submodule of M and I ⊂ R+ is a graded ideal of R such that M = N + IM , then N = M . Proof. Since M/N = I(M/N ), one may assume N = (0). If x ∈ M is a homogeneous element of degree r, then a recursive use of the equality 2 M = IM yields that x ∈ I r+1 M and x must be zero.

Exercises 2.2.10 Let I be a graded ideal of a polynomial ring R over a ﬁeld K. Prove that the radical of I is also graded. 2.2.11 Let I be a graded ideal of a polynomial ring R = K[x1 , . . . , xn ] over a ﬁeld K. If m = (x1 , . . . , xn ) is the irrelevant maximal ideal of R, prove that m ∈ Ass(R/I) if and only if depth(R/I) = 0. Hint Use Lemma 2.1.19. 2.2.12 Let R = ⊕i∈Z Ri be a graded ring. Prove that R0 is a subring of R with 1 ∈ R0 . A graded ring R with Ri = 0 for i < 0 is called a graded R0 -algebra. 2.2.13 Let R = K[x1 , x2 ] be a polynomial ring over a ﬁeld K with the grading induced by deg(xi ) = (−1)i . Determine the subring R0 .

Commutative Algebra

2.3

79

Cohen–Macaulay modules

Here we introduce some special types of rings and modules and present the following fundamental result of dimension theory. Theorem 2.3.1 (Dimension theorem [310, Theorem 13.4]) Let (R, m) be a local ring and let M be an R-module. Set δ(M ) = min{r| there are x1 , . . . , xr ∈ m with R (M/(x1 , . . . xr )M ) < ∞}, then dim(M ) = δ(M ). Deﬁnition 2.3.2 Let (R, m) be a local ring and let M be an R-module of dimension d. A system of parameters (s.o.p for short) of M is a set of elements θ1 , . . . , θd in m such that R (M/(θ1 , . . . , θd )M ) < ∞. Corollary 2.3.3 Let (R, m) be a local ring and let M be an R-module of dimension d. If h1 , . . . , hd ∈ m is a system of parameters of M , then dim M/(h1 , . . . , hi )M = d − i for 1 ≤ i ≤ d. Proof. Set M = M/(h1 , . . . , hi )M . First note that hi+1 , . . . , hd is a s.o.p for M , hence dim(M ) ≤ d − i by Theorem 2.3.1. On the other hand if θ1 , . . . , θr is a s.o.p for M , then θ1 , . . . , θr , h1 , . . . , hi is a s.o.p for M and again by the dimension theorem one has d ≤ r + i = dim(M ) + i. 2 Deﬁnition 2.3.4 Let M be an R-module. A sequence θ = θ1 , . . . , θn in R is called a regular sequence of M or an M -regular sequence if (θ)M = M / Z(M/(θ1 , . . . , θi−1 )M ) for all i. and θi ∈ Proposition 2.3.5 Let M be an R-module and let I be an ideal of R such that IM = M . If θ = θ1 , . . . , θr is an M -regular sequence in I, then θ can be extended to a maximal M -regular sequence in I. Proof. By induction assume there is an M -regular sequence θ1 , . . . , θi in I for some i ≥ r. Set M = M/(θ1 , . . . , θi )M . If I ⊂ Z(M ), pick θi+1 in I which is regular on M . Since (θ1 ) ⊂ (θ1 , θ2 ) ⊂ · · · ⊂ (θ1 , . . . , θi ) ⊂ (θ1 , . . . , θi+1 ) ⊂ R is an increasing sequence of ideals in a Noetherian ring R, this inductive construction must stop at a maximal M -regular sequence in I. 2 Lemma 2.3.6 Let M be a module over a local ring (R, m). If θ1 , . . . , θr is an M -regular sequence in m, then r ≤ dim(M ).

80

Chapter 2

Proof. By induction on dim(M ). If dim(M ) = 0, then m is an associated prime of M and every element of m is a zero divisor of M . We claim that dim(M/θ1 M ) < dim(M ). If this equality does not hold, there is a saturated chain of prime ideals ann (M ) ⊂ ann (M/θ1 M ) ⊂ p0 ⊂ · · · ⊂ pd , where d is the dimension of M and p0 is minimal over ann (M ). According to Proposition 2.1.20 the ideal p0 consists of zero divisors, a contradiction since θ1 ∈ ann (M/θ1 M ) ⊂ p0 . This proves the claim. Since θ2 , . . . , θr is a 2 regular sequence on M/θ1 M by induction one derives r ≤ dim(M ). Proposition 2.3.7 Let M be an R-module and let I be an ideal of R. (a) HomR (R/I, M ) = (0) iﬀ there is x ∈ I which is regular on M . (b) ExtrR (R/I, M ) HomR (R/I, M/θM ), where θ = θ1 , . . . , θr is any M -regular sequence in I. Proof. (a) ⇒) Assume I ⊂ Z(M ). Using Lemma 2.1.19 one has I ⊂ p for some p ∈ AssR (M ). Hence there is a monomorphism ψ : R/p → M . To derive a contradiction note that the composition ϕ

ψ

R/I −→ R/p −→ M is a non-zero map, where ϕ is the canonical map from R/I to R/p. The converse is left as an exercise. (b) Consider the exact sequence θ

1 M −→ M = M/θ1 M −→ 0. 0 −→ M −→

According to [363, Theorem 7.3] there is a long exact sequence with natural connecting homomorphisms θ

∂

1 r−1 r−1 · · · −→ Extr−1 R (R/I, M ) −→ ExtR (R/I, M ) −→ ExtR (R/I, M ) −→

θ

1 ExtrR (R/I, M ) −→ ExtrR (R/I, M ) −→ · · ·

Since θ1 is in I, using [363, Theorem 7.16] it follows that in the last exact sequence the maps given by multiplication by θ1 are zero. Hence r Extr−1 R (R/I, M ) ExtR (R/I, M ),

and the proof follows by induction on r.

2

Let M = (0) be a module over a local ring (R, m). The depth of M , denoted by depth(M ), is the length of any maximal regular sequence on M which is contained in m. From Proposition 2.3.7 one derives depth(M ) = inf{r| ExtrR (R/m, M ) = (0)}. In general by Lemma 2.3.6 we have depth(M ) ≤ dim(M ).

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Deﬁnition 2.3.8 An R-module M is called Cohen–Macaulay (C–M for short) if depth(M ) = dim(M ), or if M = (0). Lemma 2.3.9 (Depth lemma [413, p. 305]) If 0 → N → M → L → 0 is a short exact sequence of modules over a local ring R, then (a) If depth(M ) < depth(L), then depth(N ) = depth(M ). (b) If depth(M ) = depth(L), then depth(N ) ≥ depth(M ). (c) If depth(M ) > depth(L), then depth(N ) = depth(L) + 1. Lemma 2.3.10 If M is a module over a local ring (R, m) and z ∈ m is a regular element of M , then (a) depth(M/zM ) = depth(M ) − 1, and (b) dim(M/zM ) = dim(M ) − 1. Proof. As depth M > depth M/zM applying the depth lemma to the exact sequence z 0 −→ M −→ M −→ M/zM −→ 0 yields depth(M ) = depth(M/zM )+1. To prove the second equality ﬁrst observe the inequality dim(M ) > dim(M/zM ), which follows from the proof Lemma 2.3.6 by making z = θ1 . On the other hand the reverse inequality dim(M ) ≤ dim(M/zM ) + 1 follows from Theorem 2.3.1. 2 Proposition 2.3.11 [65, p. 58] If M is a Cohen–Macaulay R-module, then S −1 M is Cohen–Macaulay for every multiplicatively closed set S of R. Proposition 2.3.12 [65, p. 58] Let M be an R-module and let x be an M -regular sequence. If M is Cohen–Macaulay, then M/xM is Cohen– Macaulay (over R or R/(x)). The converse holds if R is local. Proposition 2.3.13 [309, Theorem 30] If M = (0) is a Cohen–Macaulay R-module over a local ring R and p ∈ Ass(M ), then dim(R/p) = depth(M ). Deﬁnition 2.3.14 Let (R, m) be a local ring of dimension d. A system of parameters (s.o.p) of R is a set θ1 , . . . , θd generating an m-primary ideal. Theorem 2.3.15 Let (R, m, k) be a local ring. If δ(R) := min{μ(I) = dimk (I/mI) | I is an ideal of R with rad (I) = m}, then dim(R) = δ(R). Proof. This result follows readily from Theorem 2.3.1.

2

82

Chapter 2

Theorem 2.3.16 (Krull principal ideal theorem) Let I be an ideal of a ring R generated by a sequence h1 , . . . , hr . Then (a) ht (p) ≤ r for any minimal prime p of I. (b) If h1 , . . . , hr is a regular sequence, then ht (p) = r for any minimal prime p of I. Proof. (a): Since IRp = pRp from Theorem 2.3.15 one has the inequality ht (p) = dim(Rp ) ≤ r. (b): Set J = (h1 , . . . , hi−1 ) and L = (h1 , . . . , hi ). Assume ht (P ) = i − 1 for any minimal prime P over J. Since hi is regular on R/J and L/J is a principal ideal one has ht (L/J) = 1, thus there is a prime ideal p0 minimal over J such that J ⊂ p0 ⊂ p and consequently ht (p) ≥ i. Using (a) one gets ht (p) = i. 2 Corollary 2.3.17 If (R, m, k) is a local ring, then dim(R) ≤ dimk (m/m2 ) and R has ﬁnite Krull dimension. Proof. Let x1 , . . . , xq be a set of elements in m whose images in m/m2 form a basis of this vector space. Then m = (x1 , . . . , xq ) + m2 . Hence by Nakayama’s lemma we get m = (x1 , . . . , xq ). Hence dim R ≤ q by Theorem 2.3.15. 2 Remark If R is a Noetherian ring, then dim Rp < ∞ for all p ∈ Spec(R). There are examples of Noetherian rings of inﬁnite Krull dimension [9]. Deﬁnition 2.3.18 A local ring (R, m, k) is called regular if dim(R) = dimk (m/m2 ). A ring R is regular if Rp is a regular local ring for every p ∈ Spec(R). Cohen–Macaulay rings A local ring (R, m) is called Cohen–Macaulay if R is Cohen–Macaulay as an R-module. If R is non local and Rp is a C–M local ring for all p ∈ Spec(R), then we say that R is a Cohen–Macaulay ring. An ideal I of R is Cohen–Macaulay if R/I is a Cohen–Macaulay R-module. If R is a Cohen–Macaulay ring and S is a multiplicatively closed subset of R, then S −1 (R) is a Cohen–Macaulay ring (see [65, Theorem 2.1.3]). Proposition 2.3.19 Let M be a module of dimension d over a local ring (R, m) and let θ = θ1 , . . . , θd be a system of parameters of M . Then M is Cohen–Macaulay if and only if θ is an M -regular sequence. Proof. ⇒) Let p be an associated prime of M . By Proposition 2.3.13 one has dim(R/p) = d. We claim that θ1 is not in p. If θ1 ∈ p, then by Nakayama’s lemma one has (M/θ1 M )p = (0). Hence p is in the support

Commutative Algebra

83

of M/θ1 M , a contradiction because by Corollary 2.3.3 one has the equality dim(M/θ1 M ) = d − 1. Therefore θ1 ∈ p and θ1 is regular on M . Thus the proof follows by induction, because according to Lemma 2.3.10 one has that M/θ1 M is a C–M module of dimension d − 1 and the images of θ2 , . . . , θd in R/θ1 form a system of parameters of M/(θ1 )M . ⇐) By Corollary 2.3.3 one has dim(M/θM ) = 0 and M/(θ)M is C–M. Therefore M is Cohen–Macaulay by Lemma 2.3.10. 2 Lemma 2.3.20 Let (R, m) be a local ring and let (f1 , . . . , fr ) be an ideal of height equal to r. Then there are fr+1 , . . . , fd in m such that f1 , . . . , fd is a system of parameters of R. Proof. Set d = dim(R) and I = (f1 , . . . , fr ). One may assume r < d, otherwise there is nothing to prove. Let p1 , . . . , ps be the minimal primes of I. Note that ht (pi ) = r for all i by Theorem 2.3.16. Hence if we pick fr+1 in m \ ∪si=1 pi , one has the equality ht (I, fr+1 ) = r + 1, and the result follows by induction. 2 Deﬁnition 2.3.21 Let R be a ring and let I be an ideal of R. If I is generated by a regular sequence we say that I is a complete intersection. Deﬁnition 2.3.22 An ideal I of a ring R is called a set-theoretic complete intersection if there are f1 , . . . , fr in I such that rad (I) = rad (f1 , . . . , fr ), where r = ht (I). Deﬁnition 2.3.23 An ideal I of a ring R is height unmixed or unmixed if ht (I) = ht (p) for all p in AssR (R/I). Proposition 2.3.24 Let (R, m) be a Cohen–Macaulay local ring and let I be an ideal of R. If I is a complete intersection, then R/I is Cohen– Macaulay and I is unmixed. Proof. Set d = dim(R) and r = ht (I). By Lemma 2.3.10 R/I is Cohen– Macaulay and dim(R/I) = d − r. Let p be an associated prime of R/I, then using Proposition 2.3.13 yields dim(R) − ht (p) ≥ dim(R/p) = depth(R/I) = d − r. As a consequence ht (p) ≤ r and thus ht (p) = r. Hence I is unmixed.

2

Theorem 2.3.25 Let R be a Cohen–Macaulay ring and I a proper ideal of R of height r. If I is generated by r elements f1 , . . . , fr , then I is unmixed. Proof. It is enough to prove that I has no embedded primes, because by Krull’s theorem all minimal primes of I have height r. Let p and q be two associated primes of R/I and assume p ⊂ q, then IRq ⊂ pRq ⊂ qRq .

84

Chapter 2

Since IRq has height r and is generated by r elements, one obtains (using Lemma 2.3.20) that f1 /1, . . . , fr /1 is part of a system of parameters of Rq . Therefore by Proposition 2.3.19 the sequence f1 /1, . . . , fr /1 is a regular sequence and thus Proposition 2.3.24 proves that IRq is unmixed. Noticing that pRq and qRq are both associated primes of IRq , one derives pRq = qRq and hence p = q. 2 Remark 2.3.26 The case r = 0 in the statement of Theorem 2.3.25 means that (0) is unmixed. Theorem 2.3.27 (Unmixedness theorem [310, Theorem 17.6]) A ring R is Cohen–Macaulay if and only if every proper ideal I of R of height r generated by r elements is unmixed. Proposition 2.3.28 If I is an ideal of height r in a Cohen–Macaulay ring R, then there is a regular sequence f1 , . . . , fr in I. Proof. By induction assume that f1 , . . . , fs is a regular sequence in I such that s < r. Note that (f1 , . . . , fs ) is unmixed by Theorem 2.3.25. If I ⊂ Z(R/(f1 , . . . , fs )), then ht (I) = s, which is a contradiction. Therefore there is an element fs+1 in I which is regular modulo (f1 , . . . , fs ). 2 Let A be a (Noetherian) ring one says that A is a catenary ring if for every pair p ⊂ q of prime ideals ht (q/p) is equal to the length of any maximal chain of prime ideals between p and q. If A is a domain, then A is catenary if and only if ht (q/p) = ht (q) − ht (p) for every pair of prime ideals p ⊂ q. Theorem 2.3.29 [65, Theorem 2.1.12] If R is a Cohen–Macaulay ring, then R is catenary. Proposition 2.3.30 Let A and B be two rings and let C = A × B. If C is a Cohen–Macaulay ring, then A is Cohen–Macaulay. Proof. From the ring homomorphism A × B → A, (a, b) → a, we get A × B/{0} × B A. It follows that P is a prime ideal of C containing the ideal {0} × B if and only if P = p × B for some prime ideal p of A. Let I A be an ideal and let I = I × B. We are going to establish a correspondence between the associated primes of I and I . If p is an associated prime ideal of I, there is z ∈ A such that p = (I : z). Notice that p × B = (I : (z, 1)), thus p × B is an associated prime of I . Conversely if P is an associated prime of I , then there is a prime ideal p containing I and a pair (z1 , z2 ) in C such that P = p × B = (I : (z1 , z2 )). Thus p is an associated prime of I because p = (I : z1 ). Assume that I = (x1 , . . . , xr ) is an ideal of A of height r generated by r elements. By Theorem 2.3.27 it suﬃces to prove that I is unmixed.

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Case (I): Assume r = 0, that is I = (0). Any associated prime ideal of I consists of zero divisors only. Since C is C–M, the ideal I has no embedded primes. Hence I has no embedded primes by the correspondence between associated primes of I and I established above. Case (II): Assume r ≥ 1 and set s = ht(I ). The ideal I is generated by r elements. Indeed if (x, y) ∈ I , then (x, y) = (a1 , y)(x1 , 1) + (a2 , 0)(x2 , 0) + · · · + (ar , 0)(xr , 0)

(ai ∈ A).

This means that I = ((x1 , 1), (x2 , 0), . . . , (xr , 0)) and by Krull theorem we obtain s ≤ r. Pick a minimal prime ideal P of I such that s = ht(P ). There is a prime ideal p such that P = p × B. It is seen that p is a minimal prime of I and by Krull theorem ht(p) = r. Thus there is a strictly descending chain of prime ideals pr · · · p0 = p =⇒ pr × B · · · p0 × B = p × B. Therefore s = ht(P ) ≥ r, and we obtain that ht(I ) = r. Thus since C is Cohen–Macaulay, the ideal I has no embedded primes and consequently I has no embedded primes either; by the correspondence between associated 2 primes of I and I . Gorenstein rings Let M = (0) be a module over a local ring (R, m) and let k = R/m be the residue ﬁeld of R. The socle of M is deﬁned as Soc(M ) = (0 :

M m)

= {z ∈ M | mz = (0)},

and the type of M is deﬁned as type(M ) = dimk Soc(M/xM ), where x is a maximal M -sequence in m. Observe that the type of M is well-deﬁned because by Proposition 2.3.7 one has: ExtrR (k, M ) HomR (k, M/xM ) Soc(M/xM ), where r = depth(M ). The ring R is said to be Gorenstein if R is a Cohen– Macaulay ring of type 1. An ideal I ⊂ R is called Gorenstein if R/I is a Gorenstein ring. For a thorough study of Gorenstein rings see [18, 65].

Exercises 2.3.31 Let φ : A → B be an epimorphism of rings. If dim(A) = dim(B) < ∞ and A is a domain, prove that φ is injective. 2.3.32 Let R be a catenary ring. If S ⊂ R is a multiplicatively closed set and I is an ideal of R, prove that R/I and S −1 R are both catenary rings. 2.3.33 Let I be an ideal of a ring R and let x ∈ R \ I. If I is unmixed and I = rad (I), then (I : x) is unmixed and ht(I) = ht(I : x).

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2.3.34 Let M = Zr × Zpα1 1 × · · · × Zpαmm be a ﬁnitely generated Z-module, where p1 , . . . , pm are prime numbers. Prove the following: (a) M has ﬁnite length iﬀ r = 0. (b) (M ) = α1 + · · · + αm , if (M ) < ∞. (c) Ass(M ) = Supp(M ) if r = 0. (d) Z(M ) = ∪m i=1 (pi ). (e) Ass(M ) = {(p1 ), . . . , (pm ), (0)} if r = 0. (f) Ass(M ) = {(p1 ), . . . , (pm )} if r = 0. 2.3.35 (Cayley–Hamilton) Let ϕ : M → M be an endomorphism of an Rmodule M . If I is an ideal of R such that ϕ(M ) ⊂ IM , then ϕ satisﬁes an equation of the form ϕn + an−1 ϕn−1 + · · · + a1 ϕ + a0 = 0 (ai ∈ I). 2.3.36 If R is a principal ideal domain, prove that R is a regular ring. 2.3.37 Let M be a module over a local ring (R, m). If θ = θ1 , . . . , θr is an M -regular sequence in m, prove that θ can be extended to a system of parameters of M . 2.3.38 Let M be an R-module and let I be an ideal of R. If there is x ∈ I which is regular on M , show that HomR (R/I, M ) = (0). 2.3.39 Let M be an R-module and let I be an ideal of R. If J ⊂ I, then: HomR (R/I, M/JM ) (JM :

M I)/JM.

2.3.40 If A and B are two Cohen–Macaulay rings, prove that A × B is a Cohen–Macaulay ring.

2.4

Normal rings

Let A and B be two rings. One says that B is an A-algebra if there is a homomorphism of rings ϕ : A −→ B. Note that B has an A-module structure (compatible with its ring structure) given by a · b = ϕ(a)b for all a ∈ A and b ∈ B. Thus if A ⊂ B is a ring extension, that is, A is a subring of B, then B has in a natural way an A-algebra structure induced by the inclusion map. If A = K is a ﬁeld and B is a K-algebra, then ϕ is injective; in this case one may always assume that K ⊂ B is a ring extension. Given F = {f1 , . . . , fq } a ﬁnite subset of B, we denote the subring of B generated by F and ϕ(A) by A[F ] = A[f1 , . . . , fq ]. Sometimes A[F ] is

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called the A-subalgebra or A-subring of B spanned by F . If B = A[F ], one says that B is an A-algebra of ﬁnite type or a ﬁnitely generated A-algebra; on the other hand one says that ϕ is ﬁnite or that B is a ﬁnite A-algebra if B is a ﬁnitely generated A-module. A homomorphism φ : B → C of A-algebras is a map φ which is both a homomorphism of rings and a homomorphism of A-modules. Note that φ(a · 1) = a · 1 for all a ∈ A. Deﬁnition 2.4.1 Let A be a subring of B. An element b ∈ B is integral over A, if there is a monic polynomial 0 = f (x) ∈ A[x] such that f (b) = 0. The set A of all elements b ∈ B which are integral over A is called the integral closure of A in B. Let A be a subring of B, one says that A ⊂ B is an integral extension of rings or that B is integral over A, if b is integral over A for all b ∈ B. A homomorphism of rings ϕ : A → B is called integral if B is integral over ϕ(A); in this case one also says that B is integral over A. Proposition 2.4.2 Let A be a subring of B. If B is a ﬁnitely generated A-module, then B is integral over A. B = Aα1 + · · · + Aαn . Let Proof. There are α1 , . . . , αn ∈ B such that n β ∈ B. Then, one can write βαi = m ij αj , where mij ∈ A. Set j=1 M = (mij ), N = M − βI and α = (α1 , . . . , αn ). Here I denotes the identity matrix. Since N α = 0, one can use the formula N adj(N ) = det(N )I to conclude αi det(N ) = 0 for all i. Hence det(N ) = 0. To complete the proof note that f (x) = (−1)n det(M − xI) is a monic polynomial in A[x] and f (β) = (−1)n det(N ) = 0.

2

Corollary 2.4.3 If A is a subring of B, then the integral closure A of A in B is a subring of B. Proof. Let α, β ∈ A. If α and β satisfy monic polynomials with coeﬃcients in A of degree m and n, respectively, then A[α, β] is a ﬁnitely generated Amodule with basis {αi β j | 0 ≤ i ≤ m − 1 and 0 ≤ j ≤ n − 1}. Hence, by Proposition 2.4.2, A[α, β] is integral over A. In particular α ± β and αβ are integral over A. 2 Corollary 2.4.4 If B is an A-algebra of ﬁnite type, then B is integral over A if and only if B is ﬁnite over A. Proof. ⇒) It follows by the arguments given in the proof of Corollary 2.4.3. ⇐) It follows from Proposition 2.4.2. 2

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Lemma 2.4.5 Let A ⊂ B be a ring extension. If B is a domain and b ∈ B is integral over A, then A[b] is a ﬁeld if and only if A is a ﬁeld. Proof. ⇒) Let a ∈ A \ {0} and c = a−1 its inverse in A[b]. Since A[b] is integral over A, there is an equation cn + an−1 cn−1 + · · · + a1 c + a0 = 0 (ai ∈ A), multiplying by an it follows rapidly that c ∈ A. ⇐) Let A[x] be a polynomial ring and ψ : A[x] → A[b] the epimorphism given by ψ(f (x)) = f (b). As ker(ψ) is a non-zero prime ideal and A[x] is a principal ideal domain, ker(ψ) is a maximal ideal and A[b] is a ﬁeld. 2 Proposition 2.4.6 Let A ⊂ B be an integral extension of rings. If B is a domain, then A is a ﬁeld if and only if B is a ﬁeld. Proof. It follows from Lemma 2.4.5.

2

Corollary 2.4.7 Let A ⊂ B be an integral extension of rings. If P is a prime ideal of B and p = P ∩ A, then P is a maximal ideal of B if and only if p is a maximal ideal of A. Proof. The result is a direct consequence of Proposition 2.4.6, because A/p ⊂ B/P is an integral extension of rings. 2 Corollary 2.4.8 Let A ⊂ B be an integral extension of rings. If P ⊂ Q are two prime ideals of B such that P ∩ A = Q ∩ A, then P = Q. Proof. Set p = P ∩ A. As P Bp ∩ Ap = pAp and Bp is integral over Ap , 2 using Corollary 2.4.7 we get that P Bp is maximal and hence P = Q. Lemma 2.4.9 If A ⊂ B is an integral extension of rings and p is a maximal ideal of A, then pB = B and p = P ∩ A for any maximal ideal P of B containing pB. Proof. If pB = B, one can write 1 = a1 b1 + · · · + aq bq with ai ∈ p and bi ∈ B. Set S = A[b1 , . . . , bq ], the subring of B generated by b1 , . . . , bq . Since S is a ﬁnitely generated A-module and pS = S, then by Lemma 2.1.33 there is x ≡ 1(mod p) such that xS = (0). As 1 ∈ B, one derives x = 0 and 1 ∈ p, which is impossible. Hence pB = B. Note that this part of the proof holds if p is a proper ideal of A. If pB ⊂ P is a maximal ideal of B, then p ⊂ P ∩ A and thus p = A ∩ P by the maximality of p. 2 Proposition 2.4.10 (Lying over) If A ⊂ B is an integral extension and p is a prime ideal of A, then there is a prime ideal P of B such that p = P ∩A.

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Proof. Since Ap ⊂ Bp is an integral extension, then by Lemma 2.4.9 pBp = Bp . Note that P ∩ A = p for any prime ideal P of B such that P Bp is prime and contains pBp (see Exercise 2.4.22). 2 Proposition 2.4.11 If A ⊂ B is an integral extension of rings and A is local, then B is semilocal. Proof. Let m be the maximal ideal of A. By Corollary 2.4.7 every prime ideal of B/mB is maximal. Hence, B/mB is an Artinian semilocal ring by Proposition 2.1.43. Since the maximal ideals of B are in one to one correspondence with the maximal ideals of B/mB, the result follows. 2 Theorem 2.4.12 (Going-up) Let A ⊂ B be an integral ring extension. If p ⊂ q are two prime ideals of A and p = P ∩ A for some P in Spec(B), then there is Q ∈ Spec(B) such that q = Q ∩ A and P ⊂ Q. Proof. Since A/p ⊂ B/P is an integral extension, by Proposition 2.4.10 one has q/p = (Q/P ) ∩ (A/p) for some prime ideal Q of B containing P . It follows readily that q = A ∩ Q. 2 Proposition 2.4.13 Let A ⊂ B be an integral extension of rings. If B is integral over A, then dim(A) = dim(B). Proof. The formula follows using Theorem 2.4.12, Corollary 2.4.8, and Proposition 2.4.10. 2 Let A be an integral domain and let KA be its ﬁeld of fractions. The integral closure or normalization of A, denoted A, is the set of all f ∈ KA satisfying an equation of the form f n + an−1 f n−1 + · · · + a1 f + a0 = 0 (ai ∈ A and n ≥ 1). By Corollary 2.4.3 the integral closure A of A is a subring of KA . If A = A we say that A is integrally closed or normal . If A is not a domain we say that A is normal if Ap is a normal domain for every prime ideal p of A. Any normal ring is a direct product of ﬁnitely many normal domains [128]. Proposition 2.4.14 Let A ⊂ B be an integral extension of rings. If B is a domain and A is a normal domain, then (a) (going-down) if p ⊂ q are two prime ideals of A and q = Q ∩ A for some Q in Spec(B), then there is P ∈ Spec(B) such that p = P ∩ A and P ⊂ Q. (b) ht (I) = ht (I ∩ A) for any ideal I of B. (c) ht (I) = ht (IB) for any ideal I of A.

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Proof. For parts (a) and (b) see [309, Theorems 5 and 20]. Notice that, by part (a) and Proposition 2.4.10, A ⊂ B satisﬁes the going down and the lying over conditions. Hence, by [259, Proposition B.2.4], ht (I) = ht (IB) for any ideal I of A. This proves (c). 2 Theorem 2.4.15 (Serre’s criterion [310, Theorem 23.8]) A ring A is normal if and only if (S2 ) depth(Ap ) ≥ inf{2, ht (p)} for all p ∈ Spec(A), and (R1 ) Ap is regular for all p ∈ Spec(A) with ht (p) ≤ 1. Theorem 2.4.16 (Auslander–Buchsbaum [310, Theorem 20.3]) If (A, m) is a regular local ring, then A is a unique factorization domain. Deﬁnition 2.4.17 Let R be a ring and let G ≤ Aut(R) be a group of ring automorphisms of R. The invariant ring of G is: RG = {r ∈ R| φ(r) = r, ∀ φ ∈ G}. Proposition 2.4.18 If R is a normal domain and G is a group of ring automorphisms of R, then RG is a normal domain. Proof. Let f be an element in the ﬁeld of fractions of RG which is integral over RG . Since R is normal f ∈ R and r2 f = r1 for some r1 , r2 ∈ RG . Thus r2 φ(f ) = φ(r2 )φ(f ) = φ(r1 ) = r1 for all φ ∈ R . Hence f ∈ RG , as required.

2

G

Example 2.4.19 If R = K[x1 , . . . , xn ] is a polynomial ring over a ﬁeld K and G is a subgroup of GLn (K), then for every A in G there is a ring automorphism φA : R → R, f →f (Ax), where x = (x1 , . . . , xn )t . Thus G is a group of ring automorphisms of R and RG = {f ∈ R| f (Ax) = f, ∀ A ∈ G}. Flat and faithfully ﬂat algebras Let φ : A → B be a homomorphism of rings and consider an exact sequence of A-modules fi−1

fi

fi+1

· · · −→ Ni−1 −→ Ni −→ Ni+1 −→ · · ·

(N )

If the sequence 1⊗fi−1

1⊗fi

1⊗fi+1

· · · −→ B ⊗A Ni−1 −→ B ⊗A Ni −→ B ⊗A Ni+1 −→ · · · (B ⊗A N ) is exact for any N , B is called a ﬂat A-algebra and φ is said to be ﬂat . The homomorphism φ is said to be faithfully ﬂat if any sequence N of A-modules is exact if and only if B ⊗A N is exact. In a similar way one can deﬁne ﬂat and faithfully ﬂat A-modules. A basic property is that any free A-module (not necessarily ﬁnitely generated) is faithfully ﬂat. In particular a polynomial ring A[x] in several variables with coeﬃcients in A is a faithfully ﬂat A-algebra.

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Theorem 2.4.20 [309, Sections 3.H, 5.D and 9.C] Let φ : A → B be a ﬂat homomorphism of rings. The following hold: (a) (I1 ∩ I2 )B = I1 B ∩ I2 B, for any two ideals I1 , I2 of A. (b) The going down theorem holds for φ. (c) If q is a p-primary ideal of A such that pB is prime, then qB is a pB-primary ideal. Theorem 2.4.21 [309, Sections 4.C and 13.B] Let φ : A → B be a faithfully ﬂat homomorphism of rings. The following hold: (a) The map φ∗ : Spec(B) → Spec(A), φ∗ (P ) = φ−1 (P ) is surjective. (b) IB ∩ A = I and ht (I) = ht (IB) for any ideal I of A.

Exercises 2.4.22 Let A ⊂ B be a ring extension and p a prime ideal of A. If S = A \ p and Bp = S −1 B, prove that the map P −→ P Bp gives a bijection between the set of prime ideals of P of B such that P ∩ A = p and the set of prime ideals of Bp containing pBp . 2.4.23 Let K be a ﬁeld and ϕ : A → B an isomorphism of K-algebras, note ϕ(r) = r for all r ∈ K. If A = ⊕i≥0 Ai is a graded K-algebra, then B is also a graded K-algebra graded by B = ⊕i≥0 ϕ(Ai ). 2.4.24 Let ϕ : A → B be an isomorphism between two integral domains. If KA (resp. KB ) is the ﬁeld of fractions of A (resp. B), then ϕ extends to an isomorphism ϕ from KA to KB such that ϕ(A) = B. 2.4.25 If A is a unique factorization domain, show that A is normal. 2.4.26 Show an example of an integral extension of rings A ⊂ B, such that A is a ﬁeld and B is not an integral domain (cf. Lemma 2.4.5). 2.4.27 Let B = A[x] be a polynomial ring over a ring A and let I be an ideal of A. Prove (A/I)[x] B/IB, where the left-hand side is a polynomial ring with coeﬃcients in A/I. 2.4.28 If B = A[x] is a polynomial ring over a ring A and q is a p-primary ideal of A, then qB is a pB-primary ideal of B. 2.4.29 Let B = A[x] be a polynomial ring over a ring A and let I be an ideal of A. If I = ∩ri=1 qi is a primary decomposition of I, then IB = ∩ri=1 qi B is a primary decomposition of IB. 2.4.30 If A ⊂ B is an integral extension of rings and A is semilocal, then B is semilocal. 2.4.31 Let A be a domain and let x be an indeterminate over A. Then A is normal if and only if A[x] is normal.

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2.5

Chapter 2

Valuation rings

In this part we introduce valuation rings. Let A be an integral domain and let K be its ﬁeld of fractions. Deﬁnition 2.5.1 A is a valuation ring of K if for each 0 = x ∈ K either x ∈ A or x−1 ∈ A. Proposition 2.5.2 Let A be a valuation ring of K. The following hold. (a) If I, J are two ideals of A, then I ⊂ J or J ⊂ I. (b) A is a local ring. (c) If A is an intermediate ring A ⊂ A ⊂ K, then A is a valuation ring. (d) A is integrally closed in K. Proof. (a) Assume I ⊂ J and pick x ∈ I \ J. Let y ∈ J. One has y −1 x ∈ / A, because if y −1 x ∈ A, then x = (xy −1 )y ∈ J, a contradiction. Thus yx−1 ∈ A and consequently y = x(x−1 y) ∈ I. This proves J ⊂ I, as required. Part (b) follows from (a). (c) Let x ∈ K and assume that x ∈ / A . Then x ∈ / A and consequently −1 −1 x ∈ A. Thus x ∈ A . Hence, A is a valuation ring of K. (d) Let x ∈ K be an integral element over A. There are a0 , a1 , . . . , an−1 in A such that xn + an−1 xn−1 + · · · + a1 x + a0 = 0. Assume x ∈ / A. Then x−1 ∈ A. Multiplying this equation by x1−n we get x = −(an−1 + an−2 x−1 + · · · + a1 x2−n + a0 x1−n ). Thus x ∈ A, a contradiction. Therefore x ∈ A.

2

Deﬁnition 2.5.3 An abelian group H with a total order ≺ is called an ordered group if x ≺ y and u ≺ v implies x + u ≺ y + v. Example 2.5.4 Let G1 , . . . , Gn be abelian ordered groups and let G be its cartesian product. To make G and ordered group for a = (ai ), b = (bi ) in G and a = b we deﬁne a ≺ b if the ﬁrst non-zero entry of b − a is positive. This deﬁnes a total order on G called the lexicographic order. Let H be an abelian ordered group. We order H ∪ {∞} by adding an element ∞ bigger than any element in H and declaring ∞ + x = ∞ for x ∈ H and ∞ + ∞ = ∞. Deﬁnition 2.5.5 Let H be an ordered group. A map ν : K → H ∪ {∞} is called a valuation of K if ν satisﬁes: (1) ν(xy) = ν(x) + ν(y),

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(2) ν(x + y) ≥ min{ν(x), ν(y)}, (3) ν(x) = ∞ if and only if x = 0. The valuation ring of a valuation ν, denoted by Aν , is the set of all a ∈ K with ν(a) ≥ 0. The value group of ν is the subgroup ν(K \ {0}). If ν is a valuation of a ﬁeld K, then Aν is a valuation ring of K and the ideal mν = {x ∈ K|ν(x) > 0} is its maximal ideal. Example 2.5.6 Let p ≥ 2 be a prime number and let ν : Q → Z ∪ {∞} be the map deﬁned by ⎧ a a1 * a + ⎨ n if 0 = = pn and (a1 b1 , p) = 1, b b1 ν = a ⎩ ∞ if b = 0. b It is easy to see that ν is a valuation of Q whose valuation ring is Z(p) , the localization of Z at the prime ideal p = (p). Deﬁnition 2.5.7 Let G be an ordered group. A subgroup H of G is called isolated if for each α ∈ H such that 0 ≺ β ≺ α, one has β ∈ H. If H is an isolated subgroup of G and H = G we say that H is proper. The rank of G is the number of isolated proper subgroups of G. Deﬁnition 2.5.8 Let ν be a valuation of a ﬁeld K. The rank of the valuation ring Aν is the rank of the value group of ν. Example 2.5.9 The group G = Zn with the lexicographical order has rank n because the proper non-zero isolated subgroups of G have the form: Hi = {(0, . . . , 0, xi , . . . , xn )|xi , . . . , xn ∈ Z}

(i = 2, . . . , n).

Deﬁnition 2.5.10 A valuation ring whose value group is isomorphic to Z is called a discrete valuation ring (DVR for short). Theorem 2.5.11 [310, Chapter 4] A is a discrete valuation ring if and only if any of the following equivalent conditions hold. (a) A is a Noetherian valuation ring. (b) A is a normal Noetherian local ring of dimension 1. (c) A is a Noetherian local ring, dim(A) ≥ 1 and the maximal ideal of A is principal. Theorem 2.5.12 [310, p. 82] Let A be a Noetherian domain. Then, the following three conditions are equivalent: (a) A is normal. (b) Ap is a discrete valuation ring for each prime ideal p of A of height 1. (c) All associated primes of non-zero principal ideals of A have height 1.

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Exercises 2.5.13 Let A be a valuation ring of K. Prove that there is an ordered group G and a valuation ν : K → G ∪ {∞} such that the valuation ring of ν is A and G is its value group. 2.5.14 Let K(x) be the ﬁeld of rational functions in one variable over a ﬁeld K. For f (x) ∈ K[x], write f (x) = xm g(x), where m ∈ N and g(x) is a polynomial with g(0) = 0 and deﬁne ν(f ) = m. Prove that extend ν to a discrete valuation of K(x) by setting ν(f /g) = ν(f ) − ν(g). 2.5.15 Let G be an ordered abelian group and let K[{xi |i ∈ G}] be a polynomial ring over a ﬁeld K with quotient ﬁeld Q. Deﬁne ν : Q → G∪{∞} as follows ⎧ k1 kn ⎪ ⎨ k1 g1 + · · · + kn gn if f = axg1 · · · xgn is a monomial, if f = f1 + · · · + fn is a sum of monomials, min{ν(fi )} ν(f ) = ⎪ ⎩ ∞i if f = 0. Prove that ν is a valuation of Q whose value group is G. 2.5.16 If (A, m) is a discrete valuation ring, prove that m = (x) for some x ∈ A and that every non-zero ideal of A is a power of m.

2.6

Krull rings

In this section we study the notion of a Krull ring and introduce its divisor class group. The main references for this section are [161, 309, 310]. Let A be an integral domain with quotient ﬁeld K and let Z = X (1) (A) be the set of prime ideals of A of height 1. Deﬁnition 2.6.1 An integral domain A is called a Krull ring if the following three conditions are satisﬁed: (a) Ap is a discrete valuation ring for p ∈ Z. (b) A = ∩p∈Z Ap . (c) Each 0 = x ∈ A belongs to at most a ﬁnite number of ideals in Z. Theorem 2.6.2 (Mori–Nagata [161], [309]) Let A be an integral domain with quotient ﬁeld K and let L be a ﬁnite algebraic ﬁeld extension of K. If A is Noetherian, then the integral closure A of A in L is a Krull ring (not necessarily Noetherian). If dim(A) = 2, then A is Noetherian. Corollary 2.6.3 If A is Noetherian and normal, then A is a Krull ring.

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Theorem 2.6.4 (Krull–Akizuki [309, p. 297]) Let A be a domain with quotient ﬁeld K and let L be a ﬁnite ﬁeld extension of K. If A is Noetherian and dim(A) = 1, then every subring between A and L is Noetherian. Theorem 2.6.5 [309, Theorem 38] If A is a Noetherian normal domain, then A = ∩p∈Z Ap . Theorem 2.6.6 [310, Theorem 12.4(iii)] If A is a Krull ring, then A[x] and A[[x]] are Krull rings. Proposition 2.6.7 Let A be a valuation ring. If dim(A) ≥ 2, then the formal power series ring A[[x]] is not normal. Proof. Let (0) p1 p2 be a chain of prime ideals of A. Pick 0 = b ∈ p1 and a ∈ p2 \ p1 . Then ba−n ∈ A for n > 0. Indeed if ba−n ∈ / A, then an ∈ bA ⊂ p and a ∈ p , a contradiction. It is not hard to see that there 1 1 ∞ is f = i=1 ui xi that satisﬁes the equation z 2 + az + x = 0 and such that u1 = a−1 and ui ∈ a−2i+1 A for i ≥ 2. Thus bf ∈ A[[x]] and consequently / A we get that f ∈ / A[[x]]. f is in the ﬁeld of fractions of A[[x]]. Since a−1 ∈ Thus A[[x]] is not normal. 2 Corollary 2.6.8 Let A be a valuation ring. If dim(A) ≥ 2, then A is not a Krull ring. Proof. If A is a Krull ring, then A[[x]] is a Krull ring by Theorem 2.6.6. In particular A[[x]] is normal, a contradiction to Proposition 2.6.7. 2 Divisor class groups of Krull rings Let p ∈ Z such that Ap is a discrete valuation ring. The maximal ideal pAp of the local ring Ap is principal and i ∩∞ i=1 p = (0) because Ap is a Noetherian domain. Let 0 = x ∈ A. Since any non-zero ideal of Ap is a power of pAp , there is a unique nonnegative integer np satisfying (x)Ap = pnp Ap and (x)Ap = pnp +1 Ap . This deﬁnes a valuation vp : K ∗ −→ Z of K given by

np if 0 = x ∈ A, vp (x) = vp (a) − vp (b) if x = a/b; a, b ∈ A \ {0}. The valuation vp is called the associated valuation of p. Lemma 2.6.9 Let A be a Krull ring and let 0 = a ∈ A. Then aA = p(np ) , p∈Z

where np = vp (a) and p(n) := pn Ap ∩ A is the symbolic power of order n.

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Proof. From the equality A = ∩p∈Z Ap we get aA = ∩p∈Z (aAp ∩ A). As 2 aAp = pnp Ap , we conclude that p(np ) = aAp ∩ A as desired. Deﬁnition 2.6.10 Let A be a Krull ring and let D(A) be the free abelian group on Z. That is D(A) consists of the formal sums p∈Z np · p (with np ∈ Z and np = 0 except for a ﬁnite number) with the sum deﬁned by * + * + (np + np ) · p. np · p + np · p = For each 0 = x in K, we set div(x) :=

vp (x) · p.

p∈Z

The function div gives a group homomorphism from K ∗ to D(A), because div(xy) = div(x) + div(y). The divisor class group of A is deﬁned as Cl(A) := D(A)/F (A), where F (A) is the image of K ∗ under the homomorphism div. Lemma 2.6.11 If A is a unique factorization domain and 0 = p ∈ A is a prime element, then ht(p) = 1. Proof. Assume there is a prime ideal q such that (0) q (p). Take 0 = x ∈ q. There is a prime divisor q of x that belongs to q. Thus q ∈ (p) and p = uq for some unit u ∈ A, a contradiction. 2 The next result is a measure, in terms of Cl(A), of how far is A from being a unique factorization domain. Theorem 2.6.12 An integral domain A is a unique factorization domain if and only if A is a Krull ring and Cl(A) = (0). Proof. ⇒) As A is a unique factorization domain the notions of prime element and irreducible element coincide. Hence every prime ideal of A of height 1 is principal. To prove this assertion take p ∈ Z. Let 0 = x ∈ p. Notice that some prime factor x1 of x belongs to p. Thus (x1 ) = p. Next we show that A is a Krull ring. Let p ∈ Z and let x ∈ A such that p = (x). Clearly Ap is normal and dim(Ap ) = 1. Let 0 J Ap be an ideal of Ap . Set n equal to the smallest integer n > 0 such that xn ∈ J, it follows readily that J = (xn ). This proves that Ap is a discrete valuation ring. Now we show the equality Ap . A= p∈Z

Commutative Algebra

97

Let 0 = x = a/b with a, b ∈ A. We may assume that a and b are relatively prime. Assume that x belongs to the right-hand side of the equality above. To prove that x ∈ A it suﬃces to show that b is a unit of A. If b is not a unit pick a prime divisor p of b and set p = (p), then b ∈ p and by Lemma 2.6.11 one has p ∈ Z. Since a/b ∈ Ap we get a ∈ p, a contradiction because a and b are relatively prime. Thus b is a unit and x ∈ A. Since A is a UFD every 0 = x ∈ A is contained in at most a ﬁnite number of principal prime ideals. Altogether we have shown that A is a Krull ring. It remains to show that Cl(A) = (0). This follows immediately by observing that if p0 = (x0 ) ∈ Z, then as an element of D(A) one has p0 = div(x0 ). ⇐) Let p0 ∈ Z. We claim that p0 is a principal ideal. As Cl(A) = (0), there exists 0 = x0 ∈ K such that vp (x0 ) · p = p0 . div(x0 ) = p∈Z

Hence vp0 (x0 ) = 1 and vp (x0 ) = 0 for p = p0 . Writing x0 = a/b, with p0 . By a, b ∈ A∗ we get vp0 (a) = vp0 (b) + 1 and vp (a) = vp (b) for p = Lemma 2.6.9, we have the equalities aA = p(vp (a)) and bA = p(vp (b)) . p∈Z

p∈Z

Consequently, using p(vp0 (b)+1) ⊂ p(vp0 (b)) , one has aA ⊂ bA. This proves (1) x0 ∈ A. Again by Lemma 2.6.9 we get x0 A = p0 = p0 . This proves that p0 is principal. Let 0 = x ∈ A be a non-unit of A. Notice there exists at least one ideal in Z that contains x. Otherwise consider x−1 , the inverse of x in K, and observe that x−1 ∈ ∩p∈Z Ap = A, a contradiction because x is not a unit of A. Let p1 = (x1 ), . . . , ps = (xs ) be the ideals in Z that contain x. There are elements λ1 , . . . , λs in A such that x = λ1 x1 = · · · = λs xs . Notice that λ1 x1 ∈ (x2 ). Since (x2 ) is a prime ideal of height 1 we get λ1 ∈ (x2 ). Using this argument we readily obtain that x = λxa1 1 · · · xas s with λ ∈ A, in ak +1 i does not addition since ∩∞ i=1 pk = (0) for all k we may assume that xk divide x for all k. It follows that λ is a unit of A, because if λ is not a unit it must belong to one of the ideals p1 , . . . , ps which contradicts the choice of a1 , . . . , as . Clearly the elements x1 , . . . , xs are irreducible and we have proved that x is a product of irreducible elements. Finally assume that x has two representations x = y1 · · · yr = z1 · · · zm with yi and zj irreducible elements for all i, j. Notice that every irreducible element z ∈ A is prime. Indeed since z is contained in at least one prime ideal p = (w) of height 1, we get z = λw. As z is irreducible it follows that

98

Chapter 2

λ is a unit and z is prime. Therefore it follows readily that r = m and there 2 is a permutation σ such that yi is associated of zσ(i) for all i. Corollary 2.6.13 If K is a ﬁeld and K[X] is a ring of polynomials in any number of variables (ﬁnite or inﬁnite), then K[X] is a Krull ring. Deﬁnition 2.6.14 An A-submodule I of K is called a fractional ideal of A if I = (0) and xI ⊂ A for some 0 = x ∈ A (or equivalently 0 = x ∈ K). If I is a fractional ideal of A deﬁne I −1 = (A : K I) = {x ∈ K| xI ⊂ A}. We say that a fractional ideal I is divisorial if (I −1 )−1 = I. Remark 2.6.15 An A-submodule I = (0) of K is a fractional ideal of A if and only if there is an ideal J of A and z ∈ K such that I = zJ. This follows readily using that if xI ⊂ A for some x ∈ K ∗ , then J = xI is an ideal of A and I = x−1 J. Lemma 2.6.16 If I is a fractional ideal of A, then I −1 is a fractional ideal of A and II −1 ⊂ A. Deﬁnition 2.6.17 A fractional ideal I of A is invertible if II −1 = A. A fractional ideal of the form xA with x ∈ K ∗ is called principal . Every principal fractional ideal is invertible and divisorial. Lemma 2.6.18 Let I be an invertible fractional ideal of A. If IJ = A for some fractional ideal J of A, then J = I −1 = (A :K I). In particular if I, J are invertible fractional ideals, then the usual product IJ is invertible and (IJ)−1 = I −1 J −1 . Deﬁnition 2.6.19 The Picard group of A is deﬁned as

- invertible fractional principal fractional Pic(A) = . ideals ideals Theorem 2.6.20 [310, Theorem 11.3, p. 80] Let I be a fractional ideal of A. The following conditions are equivalent : (a) I is invertible. (b) I is a projective A-module. (c) I is ﬁnitely generated as an A-module and for every maximal ideal m of A, the fractional ideal Im = IAm of Am is principal. Theorem 2.6.21 Let p = (0) be a prime ideal of A. If A is Noetherian and p is invertible, then ht(p) = 1 and Ap is a discrete valuation ring.

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Lemma 2.6.22 Let I be a fractional ideal of A. Then I −1 is a fractional ideal and yA. (I −1 )−1 = (A : K (A : K I)) = I⊂yA

In particular I is divisorial if and only if I is an intersection of principal fractional ideals. Lemma 2.6.23 Let I = x0 J be a fractional ideal of A such that 0 = x0 ∈ K and J is a fractional ideal of A. Then J is divisorial if and only if I is divisorial. Proof. Assume J is divisorial. By Lemma 2.6.22 we have J = (J −1 )−1 = (A : K (A : K J)) = xA. J⊂xA

Multiplying by x0 and making the change of variable z = x0 x we obtain I = x0 J = zA = zA = (I −1 )−1 . J⊂zx−1 0 A

x0 J⊂zA

2

Therefore I is divisorial. The converse follows similarly.

Proposition 2.6.24 Let I, J be two fractional ideals of A. Then every homomorphism of A-modules f : J → I is of the form f (x) = αx for some α ∈ (I : K J). In particular there is an isomorphism of A-modules given by HomA (J, I) −→ (I :

K J);

f −→ α.

Proof. Let 0 = x, y ∈ J. There are 0 = a, b ∈ A such that ax = by. Therefore the element af (x) bf (y) f (y) f (x) = = = =α x ax by y is independent of x. Hence f (x) = αx for x ∈ J and α ∈ (I :

K J).

2

Corollary 2.6.25 Let I, J be two fractional ideals of A. If I J as Amodules, then I is divisorial if and only if J is divisorial. Proof. By Proposition 2.6.24 we have I = αJ for some α ∈ K. Thus the result follows from Lemma 2.6.23. 2 Deﬁnition 2.6.26 Let M be an A-module and let M ∗ = HomA (M, A) be its dual. We say that M is reﬂexive if the canonical homomorphism ϕ : M → M ∗∗ = HomA (M ∗ , A); is an isomorphism.

ϕ(m)(f ) = f (m); ∀ m ∈ M, f ∈ M ∗

100

Chapter 2

Lemma 2.6.27 A direct summand of a reﬂexive module is reﬂexive. 2

Proof. It is left as an exercise.

Proposition 2.6.28 If F is a free A-module of ﬁnite rank, then F is a reﬂexive module. Corollary 2.6.29 Every ﬁnitely generated projective module is reﬂexive. Proposition 2.6.30 Let I be a fractional ideal of A. If I is reﬂexive (as an A-module), then I is divisorial. Proof. As I is always contained in (I −1 )−1 , it is enough to show that (I −1 )−1 is contained in I. Let x ∈ (I −1 )−1 . The idea is to show that x deﬁnes an element ψ of I ∗∗ and to use that ϕ is onto to obtain that x belongs to I. Let f ∈ I ∗ . By Proposition 2.6.24 there exists a unique element αf in (A : K I) such that f (z) = zαf for all z ∈ I. Consider the map ψ : I ∗ −→ A;

f −→ xαf .

First let us show that ψ is well-deﬁned, that is we must show xαf ∈ A. We may assume αf = 0. Since αf I ⊂ A we obtain I ⊂ α−1 f A. Hence, using (I −1 )−1 =

yA,

I⊂yA

we conclude that x ∈ α−1 f A, this proves that xαf ∈ A. It is not hard to see that ψ is a homomorphism. Hence since ϕ is onto there exists x0 ∈ I such that ϕ(x0 ) = ψ. Therefore αf x = ψ(f ) = ϕ(x0 )(f ) = f (x0 ) = x0 αf ∀ f ∈ I ∗ ⇒ x = x0 ∈ I.

2

Proposition 2.6.31 [161, Proposition 5.2, Corollary 5.5.] Let I be a fractional ideal of a Krull ring A. Then I is reﬂexive if and only if I is divisorial. Proposition 2.6.32 Let A be a Krull ring and let I be an ideal of A. Then I is divisorial if and only if I = q1 ∩ · · · ∩ qs , where qi is a primary ideal of A of height 1 for all i. Example 2.6.33 Let R = K[x1 , x2 ] be a polynomial ring over a ﬁeld K and let A = K[F ], the subring of R generated by F = {x21 , x1 x2 , x22 }. Let p = x1 R ∩ A = (x21 , x1 x2 )A. We claim that p is a divisorial prime ideal such that p2 is not divisorial. The ring A is a normal Noetherian domain, thus it is a Krull ring. Since A ⊂ R is an integral extension, by the going down theorem (see Proposition 2.4.14) we have that p is a prime ideal of A of height 1. Hence p is divisorial.

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Now we show that p2 is not divisorial. By Proposition 2.6.32 it suﬃces to observe that from the equality (x21 , x1 x2 , x22 )A = (p2 : A x1 x2 ) it follows that the maximal ideal m = (x21 , x1 x2 , x22 )A of A is an embedded associated prime of p2 . Deﬁnition 2.6.34 The set of divisorial ideals of A is denoted by Div(A) and the set of principal fractional ideals by Prin(A). Proposition 2.6.35 The set Div(A) with the binary operation Div(A) × Div(A) −→ Div(A);

(I, J) −→ I ∗ J := ((IJ)−1 )−1

is a monoid whose identity is A. Deﬁnition 2.6.36 Let A be an integral domain and K its ﬁeld of fractions. An element x ∈ K is almost integral over A if there is 0 = a ∈ A such that axn ∈ A for all n ≥ 0. Proposition 2.6.37 Let A be an integral domain and let K be its ﬁeld of fractions. Then an element x ∈ K is integral over A if and only if x is almost integral over A. Proof. Let x = c/d, where c, d are in A and d = 0. If x is integral over A, then there is an equation xm + b1 xm−1 + · · · + bm−1 x + bm = 0, bi ∈ A. Setting a = dm one has axn ∈ A for all n > 0. Conversely assume there is 0 = a ∈ A such that axn ∈ A for all n > 0. Since a−1 A is a Noetherian A-module and A[x] ⊂ a−1 A one has that A[x] is a ﬁnitely generated A-module. As A is a subring of A[x] by Proposition 2.4.2 one derives that x is integral over A. 2 Deﬁnition 2.6.38 The set of elements x ∈ K that are almost integral over . and this set is called the almost integral closure of A. If A is denoted by A . A = A it is said that A is completely normal . Lemma 2.6.39 [161] An element x ∈ K is almost integral over A if and only if there exists a fractional ideal I of A such that xI ⊂ I. . is a subring of K and Corollary 2.6.40 The set A . ⊂ K. A⊂A⊂A . holds if A is Noetherian. In particular if A is completely The equality A = A normal, then A is normal.

102

Chapter 2

Proof. Taking into account Proposition 2.6.37 it suﬃces to show that A is . By Lemma 2.6.39 there are fractional ideals a subring of K. Let x, y ∈ A. I, J of A such that x ∈ (I :K I), y ∈ (J :K J). Observe that xy ∈ (IJ :K IJ);

x + y ∈ (I ∩ J :K I ∩ J).

Since IJ, I ∩ J are fractional ideals of A, by Lemma 2.6.39 we get xy ∈ A and x + y ∈ A. 2 Corollary 2.6.41 If A is a Krull ring, then A is completely normal. Proof. A is an intersection of discrete valuation rings and these rings are normal and Noetherian, hence completely normal. 2 Theorem 2.6.42 [289, Theorem 5.19, p. 113] If A is a valuation ring with A = K, then A is completely normal if and only if the value group of A has rank 1. Remark 2.6.43 By Exercise 2.5.15 there exists a ﬁeld Q and a valuation ν of Q such that its value group is Z × Z. Thus Aν is not completely normal because the rank of Z×Z is 2. Any valuation ring of Krull dimension greater than or equal to two is normal but not completely normal. See [436]. Theorem 2.6.44 The set Div(A) with the binary operation Div(A) × Div(A) −→ Div(A);

(I, J) −→ I ∗ J := ((IJ)−1 )−1

is a group if and only if A is completely normal. Proposition 2.6.45 Let A be a Krull ring and let I A be an ideal of A. Then I is divisorial if and only if I can be written as I = p1 ∗ · · · ∗ pr = (A :

K (A : K p1

· · · pr ))

for some pi ∈ Z and this “factorization” is unique. Corollary 2.6.46 If A is a Krull ring, then there is an isomorphism from the group (D(A), +, 0) to the group (Div(A), ∗, A) given by: ρ : D(A) −→ Div(A);

a1 p1 + · · · + ar pr −→ pa1 1 ∗ · · · ∗ par r ,

where pa1 1 means pa1 1 = p1 ∗ · · · ∗ p1 (a1 times) if a1 > 0. Proof. Clearly ρ is a homomorphism. To prove that ρ is onto take a divisorial fractional ideal I of A. There is 0 = a ∈ A such that aI ⊂ A. Notice that aI is a divisorial ideal of A because xA =⇒ aI = zA. I= I⊂xA

aI⊂zA

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Thus (aA)I = aI = ((aI)−1 )−1 = (aA) ∗ I. By Proposition 2.6.45 we have (aA) ∗ I = pa1 1 ∗ · · · ∗ par r

and

aA = pb11 ∗ · · · ∗ pbrr (ai , bi ≥ 0).

Therefore solving for I we get that I is in the image of ρ. To ﬁnish the proof notice that from the uniqueness of the factorization in Proposition 2.6.45 we readily get that ρ is injective. 2 Another way of representing the divisor class group of a Krull ring is: Theorem 2.6.47 If A is a Krull ring, then Cl(A) Div(A)/Prin(A), where Prin(A) is the group of principal fractional ideals of A. Proposition 2.6.48 If I is an invertible fractional ideal of A, then I is divisorial. Proof. Let x ∈ (I −1 )−1 . Then xI −1 ⊂ A. Now since I is invertible one has II −1 = A, then there are ai ∈ I and bi ∈ I −1 such that 1 = i ai bi . Hence x = i ai (xbi ). Notice that xbi ∈ A because bi ∈ I −1 , hence x ∈ I. Thus we have shown (I −1 )−1 ⊂ I. To prove the reverse inclusion take x ∈ I. Notice that xy ∈ A for all y ∈ I −1 , thus x ∈ (I −1 )−1 . 2 Lemma 2.6.49 If I, J are two invertible fractional ideals, then IJ = I ∗ J = ((IJ)−1 )−1 . Proof. By Lemma 2.6.18 one has (IJ)−1 = I −1 J −1 . Therefore taking the inverse on both sides gives the required equality. 2 As a consequence of the last two results we obtain: Corollary 2.6.50 The Picard group of A is a subgroup of the semigroup Div(A)/Prin(A) of divisor classes. Example 2.6.51 Let R = K[x1 , x2 , . . . , xn , . . .] be a polynomial ring in an inﬁnite number of variables over a ﬁeld K and let A = K[F ] be the subring of R generated by F = {xi xj | 1 ≤ i ≤ j}. Consider the ideal p = x1 R ∩ A = (x21 , x1 x2 , x1 x3 , . . .)A. We claim that p is a divisorial prime ideal of A and that p is not invertible. The ring A is a Krull ring [161]. Since A ⊂ R is an integral extension by the going down theorem one has that p is a height 1 prime ideal of A. Therefore p is divisorial. Since p is not ﬁnitely generated it cannot be invertible. Deﬁnition 2.6.52 A is a Dedekind ring if every fractional ideal of A is invertible.

104

Chapter 2

Notice that by Remark 2.6.15 A is a Dedekind ring if and only if every ideal (0) = I ⊂ A of A is invertible. Principal ideal domains are Dedekind rings. If S is a multiplicatively closed subset of A and A is a Dedekind ring, then S −1 A is a Dedekind ring. If dim(A) = 1, then A is a Dedekind ring if and only if A is a Krull ring. Dedekind rings occur in algebraic number theory because the ring of algebraic integers in a ﬁnite extension of Q is a Dedekind ring. By the Krull–Akizuki theorem if A = Z, K = Q and L is a ﬁnite extension of Q, then the integral closure of A in L is a Dedekind ring. More generally if A is a Dedekind ring and L is a ﬁnite extension of K, then the integral closure A of A in L is a Dedekind ring. If A = Z, then Pic(A) is ﬁnite. See [329]. Theorem 2.6.53 [310, Theorem 11.6] An integral domain A is a Dedekind ring if and only if any of the following equivalent conditions hold: (a) A is a ﬁeld or A is a normal Noetherian domain of dimension 1. (b) Each non-zero ideal I of A has a unique factorization I = p1 p2 · · · ps into prime ideals. Corollary 2.6.54 If A is a Dedekind ring, then Div(A) is a free abelian group generated by the non-zero prime ideals of A and Pic(A) = Cl(A). Proof. It follows from Corollary 2.6.46 and Lemma 2.6.49.

2

Exercises 2.6.55 Let A be a domain with quotient ﬁeld K. Prove that the set given by G = {xA| x ∈ K ∗ } is an ordered abelian group with the usual multiplication of fractional ideals and where G is ordered by xA ≺ yA if and only if xA ⊃ yA. 2.6.56 Let K = Z2 be the ﬁeld with two elements and let F = ⊕i∈N Fi , where Fi = K for all i. Prove that F ∗ = HomK (F, K) is uncountable and that F is not a reﬂexive K-module.

2.7

Koszul homology

Let a be an element of the ring R and let K (a) be the complex deﬁned as

R for i = 0, 1, Ki = 0 otherwise, with d1 : K1 (a) → K0 (a) being multiplication by a.

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Let I be an ideal of R generated by the sequence x = {x1 , . . . , xn }. The ordinary Koszul complex associated to x is deﬁned as K (x; R) = K (x1 ) ⊗ · · · ⊗ K (xn ). For an R-module M we shall write K (x; M ) for K (x; R) ⊗ M . The Koszul complex K (x; R) is then the exterior algebra complex associated to E = Rn and the map θ : E −→ R, deﬁned as θ(z1 , . . . , zn ) = z1 x1 + /· · · + zn xn . That is, θ deﬁnes a diﬀerential ∂ = dθ on the exterior algebra (E) of E given in degree r by ∂(e1 ∧ · · · ∧ er ) =

r

(−1)i−1 θ(ei )e1 ∧ · · · ∧ e'i ∧ · · · ∧ er .

i=1

From the deﬁnition of the diﬀerential of K (x; R), we get that if w and w / are homogeneous elements of (E), of degrees p and q, respectively, then ∂(w ∧ w ) = (−1)p w ∧ ∂(w ) + ∂(w) ∧ w .

/ This implies that the cycles Z(K ) form a subalgebra of (E), and that the boundaries B(K ) form a two-sided ideal of Z(K ). As a consequence the homology of the Koszul complex, H (x), inherits a skew commutative R-algebra structure. One can also see that H (x) is annihilated by I = (x). Indeed, if e ∈ E and w ∈ Zr (K ), we have from the last formula ∂(e ∧ w) = θ(e)w. The ordinary Koszul complex K (x) = K (x; R) is simply the complex of free modules /n n /n−1 n /1 n /0 n K (x) : 0→ R → R → ··· → R → R → 0, where ∧k Rn is the kth exterior power of Rn ; thus ∧k Rn is a free R-module of rank nk with basis {ei1 ∧ · · · ∧ eik |1 ≤ i1 < · · · < ik ≤ n}. Proposition 2.7.1 [410, Theorem 2.3] If x is a regular sequence in R, then the Koszul complex is acyclic; that is, the complex K (x) is exact. Sliding depth Let (R, m) be a Cohen–Macaulay local ring and let I be an ideal of R generated by x1 , . . . , xn , denote by H (x) the homology of the ordinary Koszul complex built on the sequence x = {x1 , . . . , xn }. Deﬁnition 2.7.2 (i) (SD ) I satisﬁes sliding depth if depth Hi (x) ≥ dim(R) − n + i, ∀ i ≥ 0. (ii) (SCM ) I is strongly Cohen–Macaulay if Hi (x) are C–M, ∀ i ≥ 0. (Depths are computed with respect to maximal ideals. It is usual to set depth(0) equal to ∞.)

106

Chapter 2

Remark 2.7.3 (a) The (SD) condition localizes [234], (b) If I satisﬁes (SD) with respect to some generating set, then it will satisfy (SD) with respect to any other generating set of I. This follows from the isomorphisms: Hi ({x1 , . . . , xn , 0}) Hi (x) ⊕ Hi−1 (x), and Hi ({x1 , . . . , xn , y}) Hi ({x1 , . . . , xn , 0}), where y ∈ (x). Linkage Let I and J be two ideals in a Cohen–Macaulay local ring R. The ideals I and J are said to be (algebraically) linked if there is an R-sequence x = {x1 , . . . , xn } in I ∩ J such that I = ((x) : J) and J = ((x) : I), if in addition I and J are unmixed ideals of the same height n without common components and such that I ∩ J = (x), then I and J are said to be geometrically linked . When I and J are linked we shall write I ∼ J. We say that J is in the linkage class of I if there are ideals I1 , . . . , Im such that I ∼ I1 ∼ · · · ∼ Im ∼ J. The ideal J is said to be in the even linkage class of I if m is odd. Let R be a Gorenstein ring and let I be a Cohen–Macaulay ideal of R. If J is linked to I, then Peskine and Szpiro [339] showed that J is Cohen–Macaulay. An interesting result of Huneke [253] proves that (SCM) is preserved under even linkage. His method can be adapted to prove the following result. Proposition 2.7.4 Let I and J be two ideals in a Gorenstein local ring R of dimension d, and let x = {x1 , . . . , xn } be a generating set for I. Assume that J is evenly linked to I. If I satisﬁes the condition (SDk )

depth Hi (x; R) ≥ d − n + i, 0 ≤ i ≤ k,

then J satisﬁes the (SDk ) condition as well.

Exercises 2.7.5 Let R = K[x1 , . . . , x6 ] be a polynomial ring over a ﬁeld K and let I = I2 (X) be the ideal of 2 × 2-minors of the symmetric matrix: ⎤ ⎡ x1 x2 x3 X = ⎣ x2 x4 x5 ⎦ . x3 x5 x6

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Prove that I is linked to I2 (X ), where X is the symmetric matrix ⎤ ⎡ x1 x2 −x3 x5 ⎦ . X = ⎣ x2 x4 −x3 x5 x6 In particular if char(K) = 2 the ideal I is self-linked. For other characteristics this is an open question.

2.8

A vanishing theorem of Grothendieck

Let R be a ring and let I be an R-module. We say that I is injective if the functor HomR ( · , I) is exact. Note that this functor is always left exact. Deﬁnition 2.8.1 Let R be a ring and let M be an R-module. A complex ∂

∂

∂

0 1 2 I 1 −→ I 2 −→ ··· I : 0 −→ I 0 −→

of injective R-modules is an injective resolution of M if Hi (I ) = 0 for i > 0 and H0 (I ) = ker(∂0 ) ∼ = M. The injective dimension of M , denoted inj dim M , is the smallest integer n for which there exist an injective resolution I of M with I m = 0 for m > n. If there is no such n, the injective dimension of M is inﬁnite. For the proofs of the next three results and for additional information on Gorenstein rings and injective resolutions see [18, 65, 255, 363]. Theorem 2.8.2 Let (R, m) be a local ring and let M be an R-module of ﬁnite injective dimension. Then dim M ≤ inj dim M = depth R. Deﬁnition 2.8.3 A local ring R is Gorenstein if inj dim R < ∞. A ring R is Gorenstein if Rm is a Gorenstein ring for all m maximal ideal of R. Deﬁnition 2.8.4 Let R be a ring and let N ⊂ M be R-modules. M is an essential extension of N if for any non-zero R-submodule U of M one has U ∩ N = 0. An essential extension M of N is called proper if N = M . Proposition 2.8.5 Let R be a ring. An R-module N is injective if and only if it has no proper essential extensions. Deﬁnition 2.8.6 Let R be a ring and M an R-module. An injective module E such that M ⊂ E is an essential extension is called an injective hull of M and is denoted by E = E(M ) or E = ER (M ).

108

Chapter 2

Let M be an R-module. Then M admits an injective hull and M has a minimal injective resolution E (M ): ∂−1

∂

0 → M → E0 (M ) →0 E1 (M ) → · · · ∂i−1

∂

→ Ei−1 (M ) → Ei (M ) →i Ei+1 (M ) → · · · where E0 (M ) = E(M ) and Ei+1 = E(coker∂i−1 ). Here ∂−1 denotes the embedding M → E(M ), and ∂i is deﬁned in a natural way. Any two minimal injective resolutions of M are isomorphic. If I is an injective resolution of M , then E (M ) is isomorphic to a direct summand of I . Deﬁnition 2.8.7 Let (R, m, K) be a local ring R and let M = 0 be an R-module with depth r. The type of M is the number type(M ) = dimK ExtrR (K, M ). Theorem 2.8.8 Let R be a local ring. Then R is a Gorenstein ring if and only if R is a Cohen–Macaulay ring of type 1. Local cohomology Here we present a vanishing theorem of Grothendieck (Theorem 2.8.12) which is useful to prove Reisner criterion (see Theorem 6.3.12) for Cohen–Macaulay complexes. Our main references for homological algebra and local cohomology are [65, 202, 245, 363]. Let (R, m) be a local ring and let M be an R-module. Denote by Γm (M ) the submodule of M of all the elements with support in {m}, that is, Γm (M ) = {x ∈ M | mk x = 0 for some k > 0}. Let x = {x1 , . . . , xn } be a sequence of elements in R generating an mprimary ideal. Set xk = {xk1 , . . . , xkn }. The family xk gives the m-adic topology on R, hence Γm (M ) = {z ∈ M | (x)k z = 0 for some k ≥ 0}. Since HomR (R/I, M ) = {x ∈ M | Ix = 0} for any ideal I of R, we obtain a natural isomorphism Γm (M ) ∼ = lim HomR (R/mk , M ) ∼ = lim HomR (R/(xk ), M ). −→

−→

(2.2)

Proposition 2.8.9 Γm ( · ) is a left exact additive functor. i ( · ) are Deﬁnition 2.8.10 The local cohomology functors, denoted by Hm the right derived functors of Γm ( · ).

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109

Remark Let M and I be R-modules. (a) If I is an injective resolution i 0 (M ) = H i (Γm (I )) for i ≥ 0, (b) Hm (M ) = Γm (M ) and of M , then Hm i i Hm (M ) = 0 for i < 0. If I is injective, then Hm (I) = 0 for i > 0. Proposition 2.8.11 If (R, m) is a local ring and M is an R-module, then i (M ) ∼ Hm = lim ExtiR (R/mk , M ) ∼ = lim ExtiR (R/(xk ), M ), −→

−→

for i ≥ 0, where x is a sequence in R generating an m-primary ideal. Proof. Recall that if P is a projective resolution of L and I is an injective resolution of M , then ExtiR (L, M ) can be computed as follows: ExtiR (L, M ) ∼ = H i (HomR (P , M )) ∼ = H i (HomR (L, I )), see [245, Proposition 8.1]. Assume I is an injective resolution of M , then i (M ) ∼ Hm = H i (Γm (I )) and Γm (I ) ∼ = lim HomR (R/mk , I ). −→

Therefore i (M ) ∼ Hm = H i (lim HomR (R/mk , I )) ∼ = lim H i (HomR (R/mk , I )) −→

∼ = lim ExtiR (R/mk , M).

−→

−→

Since Γm (M ) ∼ = lim HomR (R/mk , M ) ∼ = lim HomR (R/(xk ), M ), −→

−→

the second isomorphism follows using the same arguments.

2

Next we recall the following vanishing theorem. Theorem 2.8.12 (Grothendieck [202]) If (R, m) is a local ring and M is an R-module of depth t and dimension d, then i (a) Hm (M ) = 0 for i < t and i > d. t d (b) Hm (M ) = 0 and Hm (M ) = 0.

Local cohomology of face rings Let Δ be a simplicial complex and let R = K[Δ] = K[X1 , . . . , Xn ]/IΔ be the Stanley–Reisner ring of Δ, where IΔ is the ideal of K[X1 , . . . , Xn ] generated by all Xi1 · · · Xir such that {Xi1 , . . . , Xir } ∈ / Δ. Let m be the maximal ideal generated by the residue classes xi of the i (R) be the local cohomology modules of R. indeterminates Xi and let Hm

110

Chapter 2

Consider the complex C C :

0 −→ C 0 −→ C 1 −→ · · · −→ C n −→ 0, ) Rxi1 ...xit , Ct = 1≤i1 <···
where Ry denotes R localized at S = {y i }i≥0 and the diﬀerentiation map dt : C t → C t+1 is given on the component d

t Rxj1 ···xjt+1 Rxi1 ···xit −→

to be the natural homomorphism (−1)s−1 · η : Rxi1 ···xit −→ R(xi1 ···xit )xjs if {i1 , . . . , it } = {j1 , . . . , j's , . . . , jt+1 } and 0 otherwise. If x = xi1 · · · xir is in K[Δ], then Rx = 0 iﬀ supp x ∈ Δ. Hence H i (C ) = 0 for i > dim K[Δ] (cf. Theorem 2.8.12). Recall that there is an isomorphism i Hm (R) ∼ = H i (C ).

The reader should consult [65] for further details and results about the local cohomology of face rings.

Chapter 3

Aﬃne and Graded Algebras A few topics connected with aﬃne and graded algebras are studied in this chapter, e.g., Gr¨obner bases, Hilbert Nullstellensatz and aﬃne varieties, projective closure, minimal resolutions, and Betti numbers. We present the aﬃne and graded versions of the Noether normalization lemma and some of its applications to aﬃne algebras and Cohen–Macaulay graded algebras. As before all base rings considered here are Noetherian and modules are ﬁnitely generated.

3.1

Cohen–Macaulay graded algebras

In this section we will emphasize the relationship between graded Cohen– Macaulay rings and their homogeneous Noether normalizations. Then some useful characterizations of those rings will be given. First we introduce aﬃne algebras and aﬃne Noether normalizations. Deﬁnition 3.1.1 Let k be a ﬁeld and let S be a k-algebra. We say that S is an aﬃne k-algebra if S = k[y1 , . . . , yr ] for some y1 , . . . , yr ∈ S. Deﬁnition 3.1.2 Let α and β be in Nn . The lexicographical order in Nn is obtained by declaring β " α if the last non-zero entry of β − α is positive. Notation The set of positive integers will be denoted by N+ . If α, β ∈ Rn , here α · β will denote the usual inner product of α and β. Lemma 3.1.3 Let α1 " · · · " αm be a sequence of m distinct points in Nn ordered lexicographically. Then, there is w = (w1 , . . . , wn ) ∈ Nn+ such that w1 = 1 and w · α1 > w · αi for i ≥ 2.

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Chapter 3

Proof. Let αi = (αi1 , . . . , αin ) and βi = (αi1 , . . . , αi(n−1) ). We proceed by induction on n ≥ 2. There is k so that α1n = · · · = αkn and αkn > αin for i > k. One may assume k < m; for otherwise β1 " · · · " βm and one can use induction. Since β1 " · · · " βk by induction there is w = (1, w2 , . . . , wn−1 ) such that w · β1 > w · βi for 2 ≤ i ≤ k. On the other hand for every i > k one can choose δi ∈ N+ so that w · β1 + α1n δi > w · βi + αin δi . Setting wn = max{δi | k < i ≤ m}, w = (w , wn ), we get w · α1 > w · αi for i ≥ 2. 2 Proposition 3.1.4 Let R = k[x1 , . . . , xn ] be a polynomial ring over a ﬁeld k and let f be a polynomial in R \ k. Then, there is a change of variables i xi = xw 1 + yi for i ≥ 2 such that wn r−1 r 2 f (x1 , xw + · · · + cr−1 x1 + cr , 1 + y2 , . . . , x1 + yn ) = c0 x1 + c1 x1

where 0 = c0 ∈ k, r > 0, and ci ∈ k[y2 , . . . , yn ] for i ≥ 1. m αi Proof. The polynomial f can be written as f = i=1 bi x , where 0 = bi ∈ k for all i. One may assume that α1 " · · · " αm are ordered lexicographically. By Lemma 3.1.3 there is w ∈ Nn+ such that xi = xwi + yi satisﬁes the required properties. Note r = w · α1 . 2 Lemma 3.1.5 If R = k[x1 , . . . , xn ] is an aﬃne k-algebra of dimension n, then R is a polynomial ring. Proof. One may assume R B/I, where B is a polynomial ring in n variables with coeﬃcients in the ﬁeld k and I is an ideal of B. Let I ⊂ p0 ⊂ · · · ⊂ pn be a chain of prime ideals of B of length n. If I = (0), then adding (0) to the chain yields a chain of length n + 1, which is impossible because dim(B) = n. Hence I = (0). 2 Theorem 3.1.6 Let R = k[x1 , . . . , xn ] be a polynomial ring over a ﬁeld k and let I = R be an ideal of R. Then there are z1 , . . . , zn in R such that (a) k[x1 , . . . , xn ] is integral over k[z1 , . . . , zn ], and (b) I ∩ k[z1 , . . . , zn ] = z1 k[z1 , . . . , zn ] + · · · + zg k[z1 , . . . , zn ]. Proof. The proof is by induction on n. If n = 1 and I = (f (x1 )) = 0, then one sets z1 = f (x1 ). Assume n ≥ 2 and I = 0. One may assume that I contains a monic polynomial in x1 . Otherwise take a non-zero polynomial g in I and apply Proposition 3.1.4 to get an isomorphism of k-algebras ϕ

k[x1 , . . . , xn ] −→ k[x1 , y2 . . . , yn ]

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113

i induced by ϕ(x1 ) = x1 and ϕ(xi ) = xw 1 + yi for i ≥ 2, such that ϕ(g) is monic in x1 . Let

f = xr1 + c1 xr−1 + · · · + cr−1 x1 + cr 1 be a polynomial in I with ci ∈ k[x2 , . . . , xn ] and r > 0. Set I1 = I ∩ k[x2 , . . . , xn ]. By induction there are z2 , . . . , zn such that: (i) k[x2 , . . . , xn ] is integral over k[z2 , . . . , xn ], and (ii) I1 ∩ k[z2 , . . . , zn ] = (z2 , . . . , zn ). Set z1 = f . It is not hard to see that R is integral over k[z1 , . . . , zn ] and k[z1 , . . . , zn ] ∩ I = (z1 , . . . , zg ).

2

Corollary 3.1.7 If R = k[x1 , . . . , xn ] is a polynomial ring over a ﬁeld k and I = R is an ideal of R, then dim(R/I) = dim(R) − ht (I). Proof. One may assume that there are z1 , . . . , zn in R and an integer g such that the conditions (a) and (b) of Theorem 3.1.6 are satisﬁed. We will show that g is equal to the height of I. By Lemma 3.1.5 the zi ’s are algebraically independent. Hence Proposition 2.4.14 yields ht (I) = ht (I ∩ k[z1 , . . . , zn ]) = g. Note that there is an integral extension ϕ

k[zg+1 , . . . , zn ] k[z1 , . . . , zn ]/(z1 , . . . , zg ) → k[x1 , . . . , xn ]/I. Therefore n − g = dim(R/I).

2

Corollary 3.1.8 (Noether normalization lemma) If R = k[x1 , . . . , xn ] is a polynomial ring over a ﬁeld k and I = R is an ideal, then there is an integral extension k[h1 , . . . , hd ]→R/I, where h1 , . . . , hd are in R and d = dim(R/I). Proof. By Theorem 3.1.6 there is an integral extension ϕ

k[zg+1 , . . . , zn ] k[z1 , . . . , zn ]/(z1 , . . . , zg ) → R/I. To conclude the argument set hi = zg+i for i = 1, . . . , d.

2

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Chapter 3

Corollary 3.1.9 If R = k[x1 , . . . , xn ] is a polynomial ring over a ﬁeld k, then R is a catenary ring. Proof. Note ht (q/p) = dim(R/p) − dim(R/q) for any two prime ideals p ⊂ q in R. Hence by Corollary 3.1.7 we get ht (q/p) = ht (q) − ht (p). 2 Deﬁnition 3.1.10 Let k ⊂ L be a ﬁeld extension. A subset of L which is algebraically independent and is maximal with respect to inclusions is called a transcendence basis of L over k. Theorem 3.1.11 [262, Theorem 8.35] If k ⊂ L is a ﬁeld extension, then any two transcendence bases have the same cardinality. Deﬁnition 3.1.12 Let k ⊂ L be a ﬁeld extension. The transcendence degree of L over k, denoted trdegk (L), is the cardinality of any transcendence basis of L over k. Corollary 3.1.13 Let k be a ﬁeld and let A be a ﬁnitely generated kalgebra. If A is a domain with ﬁeld of fractions L, then dim(A) = trdegk (L). Deﬁnition 3.1.14 Let k be a ﬁeld. A standard algebra or homogeneous algebra is a ﬁnitely generated N-graded k-algebra S=

∞ )

Si = k[y1 , . . . , yr ]

i=0

such that yi ∈ S1 for all i and S0 = k. If we only require yi homogeneous and deg(yi ) > 0 for all i, we say that S is a positively graded k-algebra. The irrelevant maximal ideal m of S is deﬁned by m = S+ =

∞ )

Si .

i=1

Deﬁnition 3.1.15 Let k be a ﬁeld and S = k[y1 , . . . , yr ] a positively graded k-algebra with yi homogeneous of degree di . There is a graded epimorphism ϕ : R = k[x1 , . . . , xr ] −→ S ϕ

given by f (x1 , . . . , xr ) −→ f (y1 , . . . , yr ), where R is a polynomial ring graded by deg(xi ) = di , the presentation of S is the k-algebra R/ ker(ϕ). The graded ideal ker(ϕ) is called the ideal of relations or the presentation ideal of S.

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115

Proposition 3.1.16 If S is a Cohen–Macaulay standard algebra over a ﬁeld k and L is an ideal of S, then dim(S) = dim(S/L) + ht (L). Proof. One may assume S = R/I and L = J/I, where R is a polynomial ring over k and J is an ideal of R containing I. Set g = ht (J/I) and r = ht (I). Let p be a prime ideal such that J ⊂ p and g = ht (p/I). There is a saturated chain of prime ideals I ⊂ p0 ⊂ p1 ⊂ · · · ⊂ pg = p. Since p0 is a minimal prime ideal of I using Proposition 2.3.13 one obtains dim(R/I) = dim(R/p0 ), that is, r = ht (p0 ). Hence there is a saturated chain of prime ideals q0 = (0) ⊂ q1 ⊂ · · · ⊂ qr = p0 ⊂ p1 ⊂ · · · ⊂ pg = p. As R is a catenary domain one obtains ht (p) = r + g and consequently ht (J) ≤ r + g. Therefore, we get dim(S) ≤ ht (L) + dim(S/L). To conclude the proof note that the reverse inequality holds in general. 2 Corollary 3.1.17 Let R be a positively graded polynomial ring over a ﬁeld k and I a graded ideal of R. If R/I is Cohen–Macaulay, then I is unmixed. Proof. It follows from Propositions 2.3.13 and 3.1.16.

2

Deﬁnition 3.1.18 Let k be a ﬁeld and let S be a positively graded kalgebra. A set of homogeneous elements θ = {θ1 , . . . , θd } is called a homogeneous system of parameters (h.s.o.p for short) if d = dim(S) and rad (θ) = S+ . Corollary 3.1.19 Let S be a positively graded algebra over a ﬁeld k and let h1 , . . . , hd be a h.s.o.p for S. If 1 ≤ i ≤ dim(S), then dim S/(h1 , . . . , hi ) = dim(S) − i. Proof. Set S = S/(h1 , . . . , hi ) and d = dim(S). Note that the set of images of hi+1 , . . . , hd in S generate an m-primary ideal, where m = S+ is the irrelevant maximal ideal of S and m = mS. Hence, by the graded version of the dimension theorem (Theorem 2.3.15), we get dim(S) ≤ d − i. On the other hand if θ1 , . . . , θr is a h.s.o.p for S, then the sequence θ1 , . . . , θr , h 1 , . . . , h i

116

Chapter 3

generates an m-primary ideal of S. Applying the dimension theorem once again one has d ≤ r + i = dim(S) + i. Thus dim(S) = d − i.

2

Proposition 3.1.20 Let S be a positively graded algebra over a ﬁeld k and θ = θ1 , . . . , θd a h.s.o.p for S. Then S is Cohen–Macaulay if and only if θ is a regular sequence. Proof. By Corollary 3.1.19 dim S/(θ1 , . . . , θi ) = d − i. If S is C–M, then by Proposition 3.1.16 one has ht (θ1 , . . . , θi ) = i. Assume θ1 , . . . , θi−1 is a regular sequence. Next we show that θi is regular on A = S/(θ1 , . . . , θi−1 ). Otherwise if θi is a zero divisor of A, then θi belongs to some minimal prime p of (θ1 , . . . , θi−1 ). Since ht (p) ≤ i − 1 (see Theorem 2.3.16) we obtain ht (θ1 , . . . , θi ) ≤ i − 1, which is impossible. Conversely if θ is a regular sequence, then depth(S) = d and S is C–M. 2 Proposition 3.1.21 Let S be a positively graded algebra over a ﬁeld k. If S is Cohen–Macaulay and θ = θ1 , . . . , θq is a regular sequence of homogeneous elements in S+ , then S/(θ) is Cohen–Macaulay. Proof. It suﬃces to prove the case q = 1, which is a direct consequence of Lemma 2.3.10. 2 Lemma 3.1.22 Let V = {0} be a vector space over an inﬁnite ﬁeld K. Then V is not a ﬁnite union of proper subspaces of V . Proof. We proceed by contradiction.0Assume that there are proper subspaces V1 , . . . , Vm of V such that V = m i=1 Vi , where m is the least positive integer with this property. Let v1 ∈ V1 \ (V2 ∪ · · · ∪ Vm ) and v2 ∈ V2 \ (V1 ∪ V3 ∪ · · · ∪ Vm ). Pick m + 1 distinct non-zero scalars k0 , . . . , km in K. Consider the vectors βi = v1 − ki v2 for i = 0, . . . , m. By the pigeon-hole principle there are distinct vectors βr , βs in Vj for some j. Since βr − βs ∈ Vj we get v2 ∈ Vj . Thus j = 2 by the choice of v2 . To ﬁnish the proof observe that βr ∈ V2 imply v1 ∈ V2 , which contradicts the choice of v1 . 2 Proposition 3.1.23 Let R = k[x1 , . . . , xn ] be a polynomial ring over a ﬁeld k with a positive grading and let I be a graded ideal of R. Then there are homogeneous polynomials h1 , . . . , hd in R+ such that dim R/(I, h1 , . . . , hi ) = d − i, for i = 1, . . . , d, where dim(R/I) = d. If k is an inﬁnite ﬁeld and deg(xi ) = 1 for all i, then h1 , . . . , hd can be chosen in R1 .

Aﬃne and Graded Algebras

117

Proof. Assume d > 0. Let p1 , . . . , pr be the set of minimal primes of I of height g = ht (I). We claim that there is a homogeneous polynomial h1 not in ∪ri=1 pi . To show it we use induction on r. Since p1 · · · pr−1 ⊂ pr and because pi is graded we can pick f ∈ p1 · · · pr−1 ∩ Rd1 and f ∈ pr , d1 > 0. r By induction there is g ∈ Rd2 and g ∈ ∪r−1 i=1 pi , d2 > 0. Assume Ri ⊂ ∪i=1 pi for all i > 0, hence g ∈ pr . To complete the proof of the claim consider h = f d2 − g d1 to derive a contradiction. Note dim(R/I) > dim R/(I, h1 ), by the choice of h1 . Hence a repeated application of the claim rapidly yields a sequence h1 , . . . , hs of homogeneous polynomials in R+ with s ≤ d and such that ht (I, h1 , . . . , hs ) = dim(R). Therefore by Theorem 2.3.15 one concludes d = s, as required. If k is inﬁnite and deg(xi ) = 1 for all i, then there is h1 in R1 and not in ∪ri=1 pi ; for otherwise one has R1 = ∪ri=1 (pi )1 and since R1 cannot be a ﬁnite union of proper subspaces (see Lemma 3.1.22) we derive R+ = pi for some i, which is impossible. Thus one may proceed as above to get the required sequence. 2 Theorem 3.1.24 (Homogeneous Noether normalization lemma) Let k be a ﬁeld and let R be a polynomial ring over k. If R is positively graded and I is a graded ideal of R, then there are homogeneous polynomials h1 , . . . , hd in R+ , with d = dim(R/I), and a natural embedding ϕ

A = k[h1 , . . . , hd ] → R/I such that R/I is a ﬁnitely generated A-module. Moreover if k is inﬁnite and deg(xi ) = 1 for all i, then h1 , . . . , hd can be chosen in R1 . Proof. We set R = k[x1 , . . . , xn ] and Bi = {xa | deg(xa ) = i}. According to Proposition 3.1.23, there are homogeneous polynomials h1 , . . . , hd in m with rad (I, h1 , . . . , hd ) = m, where m = R+ . Hence, there is r such that mr ⊂ (I, h1 , . . . , hd ). Note that Bi ⊂ (I, h1 , . . . , hd ) for i ≥ s = rδ, where δ is the maximum of the degrees of x1 , . . . , xn . We set B = ∪si=1 Bi and J = I + Bk[h1 , . . . , hd ]. Using induction on i, we now show that Bi ⊂ J for i ≥ s. The inclusion is clear for i ≤ s. Assume i > s and take xa in Bi . Let f1 , . . . , fq be a homogeneous generating set for I. Since (I, h1 , . . . , hd ) is graded, we can write xa = qj=1 bj fj + dj=1 cj hj , where b1 , . . . , bq and c1 , . . . , cd are homogeneous and deg(cj hj ) = i. Hence all the monomials in the support of cj have degree less than i. Thus, by induction, one gets that cj is in J for all j. Consequently xa ∈ J. Observe that the image of B in R/I generates R/I as an A-module. Next we verify that the canonical map ϕ from A to R/I is injective. Since R/I is integral over ϕ(A), by Proposition 2.4.13, one obtains that d is equal

118

Chapter 3

to dim ϕ(A). Using A/ ker(ϕ) ϕ(A) and the fact that A has dimension at most d we get ker(ϕ) = 0. If k is inﬁnite and deg(xi ) = 1, then according to Proposition 3.1.23 one can choose h1 , . . . , hd of degree one. 2 Deﬁnition 3.1.25 Let R be a positively graded polynomial ring over a ﬁeld k and I = R a graded ideal. A homogeneous Noether normalization of S = R/I is an integral extension k[h1 , . . . , hd ] → S, where h1 , . . . , hd are homogeneous polynomials in R+ and d = dim(S). Proposition 3.1.26 (Stanley) Let R = k[x1 , . . . , xn ] be a polynomial ring over a ﬁeld k and let I be a monomial ideal with d = dim(R/I). Then A = k[σ1 , . . . , σd ]→S = R/I is a Noether normalization, where σi is the ith symmetric polynomial. Proof. It suﬃces to prove that S is integral over A. Let xi = xi + I. Since d = dim(R/rad(I)), one has σi ∈ rad(I) for i > d. Hence, from the equality (x − x1 ) · · · (x − xn ) = xn − σ1 xn−1 + · · · + (−1)n σn , + · · · + (−1)d σ d xn−d = h, where h is nilpotent in S. we get xni − σ 1 xn−1 i i m Hence if h ∈ I, raising the last equality to the mth power yields that xi is integral over A. 2 Proposition 3.1.27 Let R = k[x1 , . . . , xn ] be a positively graded polynomial ring over a ﬁeld k and let I = R be a graded ideal of R. If S = R/I is a Cohen–Macaulay ring and A = k[h1 , . . . , hd ] → S is a homogeneous Noether normalization of S, then S = Axβ1 ⊕ · · · ⊕ Axβm for any set of monomials B = {xβ1 , . . . , xβm } whose image in the Artinian ring S = R/(I, h1 , . . . , hd ) is a k-vector space basis of S. Proof. Set M i = {xα ∈ R| deg(xα ) = i}. First we show that the image of B generate S as an A-module. It suﬃces to prove M i ⊂ J = I + AB for all i ≥ 0. Let xα ∈ M i . There are homogeneous m polynomials d μ1 , . . . , μd in R and λ1 , . . . , λm in k such that xα = f + j=1 λj xβj + j=1 μj hj , where f ∈ Ri ∩ I and deg(μj hj ) = i if μj = 0. Since deg μj < i, by induction one obtains that μj ∈ J for all j. Consequently xα ∈ J. We claim that if c1 xβ1 +· · ·+cm xβm belongs to (I, h1 , . . . , hi−1 ) for some 2 ≤ i ≤ d and c1 , . . . , cm are in k[hi , . . . , hd ], then cj = 0 for all j. To prove the claim note that h1 , . . . , hd is a regular sequence (see Proposition 3.1.20) and use descending induction on i starting with i = d.

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119

show the equality S = Axβ1 ⊕ · · · ⊕ Axβm . Assume that mNext, βwe i ai x is in I for some a1 , . . . , am in A. If a = 0 for some , write i=1 si ai = j=0 aij hj1 , where aij are in k[h2 , . . . , hd ]. One may assume ar0 = 0 for some r; otherwise we may factor out h1 to some power and apply that h1 is regular on S. Hence a10 xβ1 + · · · + am0 xβm is in (I, h1 ), and applying the claim with i = 2 we derive ar0 = 0, which is a contradiction. Therefore aj = 0 for all j, as required. 2 Lemma 3.1.28 Let k be an inﬁnite ﬁeld and let S be a standard k-algebra. If S is Cohen–Macaulay, then there exists a h.s.o.p θ = {θ1 , . . . , θd } such that θ is a regular sequence and θi is homogeneous of degree 1 for all i. Proof. Let S = ⊕∞ i=0 Si and Ass(S) = {p1 , . . . , ps }. Notice that Ass(S) = Min(S), because S is Cohen–Macaulay. We may assume d > 0, otherwise there is nothing to prove. If S1 ⊂ Z(S) = ∪si=1 pi , then S1 = ∪si=1 (pi )1 . Since k is inﬁnite, by Lemma 3.1.22, we obtain that s = 1 and p1 = S+ , that is, dim(S) = 0, which is a contradiction. Hence there is θ1 ∈ S1 which is regular on S. By Proposition 3.1.16 dim S/(h1 ) = d − 1. Applying the depth lemma (see Lemma 2.3.9) to the exact sequence θ

1 S −→ S/(θ1 ) −→ 0 0 −→ S(−1) −→

yields depth S/(θ1 ) = d−1. Altogether S/(θ1 ) is C–M and the result follows by induction. 2 Lemma 3.1.29 Let R be an N-graded polynomial ring over a ﬁeld k and I a graded ideal of R of height g. If I is a complete intersection, then there are homogeneous polynomials f1 , . . . , fg such that I = (f1 , . . . , fg ). Proof. Let h1 , . . . , hg be a regular sequence generating the ideal I. Using the graded version of Corollary 2.1.35 one has r = μ(I) = dimk (I/mI) ≤ g, where m = R+ . On the other hand, since I is graded, there are homogeneous polynomials f1 , . . . , fr generating I and such that {fi + mI}ri=1 is a k-basis for I/mI. By Theorem 2.3.16 one concludes r = g. 2 Proposition 3.1.30 Let R be a positively graded polynomial ring over a ﬁeld k and let I be a graded ideal of R. If I is a complete intersection, then R/I is Cohen–Macaulay. Proof. By Lemma 3.1.29 the ideal I is generated by a sequence f1 , . . . , fg consisting of homogeneous polynomials, where g denotes the height of I.

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According to Proposition 3.1.23 there is a sequence h1 , . . . , hn−g consisting of homogeneous polynomials such that rad (f1 , . . . , fg , h1 , . . . , hn−g ) = m = R+ . Therefore {f1 , . . . , fg , h1 , . . . , hn−g } is a homogeneous system of parameters for R and by Proposition 3.1.20 we derive that f1 , . . . , fg is a regular sequence. Finally Lemma 2.3.10 yields that R/I is Cohen–Macaulay. 2 Lemma 3.1.31 Let I, p1 , . . . , pm be graded ideals of a polynomial ring R over a ﬁeld k such that p1 , . . . , pm are prime ideals. If f ∈ ∪m i=1 pi for any f ∈ I homogeneous, then I ⊂ pi for some i. Proof. By induction on m. If m = 2 and I ⊂ pi for i = 1, 2, then for each i pick a homogeneous polynomial fi ∈ I \ pi of degree ui . Since f1u2 + f2u1 is homogeneous we readily derive a contradiction. Hence I ⊂ pi for some i. Let G be the set of homogeneous elements in I and suppose G ⊂ ∪m i=1 pi . One may assume pi ⊂ pj for i = j. If I ⊂ pi for all i, then by induction there is f1 in G \ ∪m−1 i=1 pi . On the other hand since Ip1 · · · pm−1 ⊂ pm (see Exercise 2.1.48) there is a homogeneous polynomial f2 in Ip1 · · · pm−1 and f2 not in pm . Since f1u2 + f2u1 , deg(fi ) = ui , is homogeneous we rapidly derive a contradiction. Thus I ⊂ pi for some i. 2 Proposition 3.1.32 Let R be a polynomial ring over a ﬁeld k and let I be a graded ideal of height r, then there is a regular sequence f1 , . . . , fr in I of homogeneous polynomials. Proof. We will proceed by induction. Assume that f1 , . . . , fs is a regular sequence of homogeneous polynomials in I, where s < r. Note that the ideal (f1 , . . . , fs ) is Cohen–Macaulay by Proposition 3.1.30, and hence it is unmixed by Corollary 3.1.17. If all the homogeneous polynomials in I belong to Z(R/(f1 , . . . , fs )), then by Lemma 3.1.31 the ideal I must be contained in an associated prime of (f1 , . . . , fs ) and consequently ht (I) ≤ s, which is impossible. Thus there is a homogeneous polynomial fs+1 in I such that fs+1 is regular modulo (f1 , . . . , fs ). 2 Tensor product of aﬃne algebras Let A, B be two aﬃne algebras over a ﬁeld k and consider presentations A R1 /I1 , B R2 /I2 , where R1 = k[x], R2 = k[y] are polynomial rings in disjoint sets of variables and Ii is an ideal of Ri . Set R = k[x, y] and I = I1 + I2 . The map R −→ R1 /I1 ⊗k R2 /I2 ,

xi → xi ⊗ 1, yi → 1 ⊗ yi ,

induces a k-algebra homomorphism: ϕ : R/I → R1 /I1 ⊗k R2 /I2 ,

ϕ(f (x)g(y)) = f (x) ⊗ g(y).

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On the other hand there is a k-bilinear map ψ : R1 /I1 × R2 /I2 → R/I, given by multiplication ψ(f , g) = f g. By the universal property of the tensor product there is a map ψ that makes the following diagram φ - R1 /I1 ⊗k R2 /I2 R1 /I1 × R2 /I2 ψ ? R/I

) ψ

commutative, where φ is the canonical map and ψ = ψφ. As a consequence ψ is the inverse of ϕ. Thus we have proved: Proposition 3.1.33 If A and B are two aﬃne algebras over a ﬁeld k with presentations A k[x]/I1 and B k[y]/I2 , then A ⊗k B k[x, y]/(I1 + I2 ). Theorem 3.1.34 If A and B are two standard algebras over a ﬁeld k, then depth(A ⊗k B) = depth(A) + depth(B). Proof. Pick a regular sequence g = g1 , . . . , gr (resp. h = h1 , . . . , hs ) on A = k[x]/I1 (resp. B = k[y]/I2 ) with g ⊂ (x) (resp. h ⊂ (y)) and such that g (resp. h) consist of forms, where r is the depth of A and s is the depth of B. As B is a faithfully ﬂat k-algebra, applying the functor (·) ⊗k B to the injective map gi

0 −→ k[x]/(I1 , g1 , . . . , gi−1 ) −→ k[x]/(I1 , g1 , . . . , gi−1 ) gives a natural commutative diagram k[x]/(I1 , g1 , .⏐. . , gi−1 ) ⊗k B ⏐ 2

gi ⊗1

k[x, y]/(I1 , I2 , g1 , . . . , gi−1 )

−→

−→

k[x]/(I1 , g1 , .⏐. . , gi−1 ) ⊗k B ⏐ 2

gi

k[x, y]/(I1 , I2 , g1 , . . . , gi−1 )

such that the map in the ﬁrst row is injective. Since the vertical arrows are natural isomorphisms by Proposition 3.1.33, one concludes that the map in the second row is also injective. As a consequence g is a regular sequence on k[x, y]/(I1 , I2 ). Similarly if we tensor the injective map h

i 0 −→ k[y]/(I2 , h1 , . . . , hi−1 ) −→ k[y]/(I2 , h1 , . . . , hi−1 )

122

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with the faithfully ﬂat k-algebra k[x]/(I1 , g) it follows rapidly that h is a regular sequence on k[x, y]/(I1 , I2 , g). Altogether g, h is a regular sequence on k[x, y]/(I1 , I2 ). To ﬁnish the proof note that (x, y) is an associated prime ideal of k[x, y]/(I1 , I2 , g, h) by Exercise 3.1.41, thus r + s is the length of a maximal regular sequence in (x, y) and by Proposition 2.3.7 one obtains depth(A ⊗k B) = r + s. For an alternative proof see [171]. 2 Corollary 3.1.35 If A and B are two standard algebras over a ﬁeld k, then A ⊗k B is Cohen–Macaulay if and only if A and B are Cohen–Macaulay. Proof. As the dimension is always greater than or equal to the depth by Lemma 2.3.6, the result follows at once using Theorem 3.1.34 together with Exercise 3.1.36. 2 A similar statement holds if one replaces Cohen–Macaulay by Gorenstein; indeed the type of the tensor products of two Cohen–Macaulay standard algebras is the product of their types [171].

Exercises 3.1.36 If A and B are standard algebras over a ﬁeld k, then dim(A ⊗k B) = dim(A) + dim(B). 3.1.37 Let R = k[x1 , . . . , xn ] be a polynomial ring over a ﬁeldk and let n A = (aij ) be an invertible matrix with entries in k. If yi = j=1 aij xj for i = 1, . . . , n, prove that R = k[y1 , . . . , yn ]. In particular y1 , . . . , yn are algebraically independent. 3.1.38 Let V be a vector space over an inﬁnite ﬁeld k and W, V1 , . . . , Vm vector subspaces of V . If W ⊂ Vi for all i, prove that W ⊂ ∪m i=1 Vi . 3.1.39 Let I, I1 , . . . , Im be ideals of a ring R and let k be an inﬁnite ﬁeld. If I ⊂ ∪m i=1 Ii and k is a subring of R, then I ⊂ Ii for some i. 3.1.40 Let R be a polynomial ring over an inﬁnite ﬁeld k and I a graded ideal of height r. If I is minimally generated by forms of degree p ≥ 1, prove that there are forms f1 , . . . , fm of degree p in I such that f1 , . . . , fr is a regular sequence and I is minimally generated by f1 , . . . , fm . 3.1.41 Let A = k[x]/I1 and B = k[y]/I2 be aﬃne algebras over a ﬁeld k. If (x) (resp. (y)) is an associated prime of k[x]/I1 (resp. k[y]/I2 ), then (x, y) is an associated prime of k[x, y]/(I1 , I2 ).

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3.1.42 Let k[x] be a polynomial ring and I an ideal (resp. graded). If k ⊂ K is a ﬁeld extension, then there is a natural (resp. graded) isomorphism of K-algebras: K[x]/IK[x] k[x]/I ⊗k K. 3.1.43 Let A be an aﬃne algebra over a ﬁeld k and let k ⊂ K be a ﬁeld extension. Prove: (a) dim(A) = dim(A ⊗k K). (b) depth(A) = depth(A ⊗k K) if A is a standard algebra. 3.1.44 Let K be a ﬁeld and let A be an aﬃne K-algebra. Prove that A is Artinian if and only if dimK (A) < ∞.

3.2

Hilbert Nullstellensatz

Let K be a ﬁeld and let S = K[t1 , . . . , tq ] be a polynomial ring. In what follows t stands for the set of variables of S, that is, S = K[t]. We deﬁne the aﬃne space of dimension q over K, denoted by AqK , to be the cartesian product K q = K × · · · × K. Given an ideal I ⊂ S, deﬁne the zero set or aﬃne variety of I as V (I) = {α ∈ AqK | f (α) = 0, ∀f ∈ I}. By the Hilbert’s basis theorem V (I) is the zero locus of a ﬁnite collection of polynomials (see Theorem 2.1.4). Conversely, for any X ⊂ AqK deﬁne I(X), the vanishing ideal of X, as the set of polynomials of S that vanish at all points of X. An aﬃne variety is the zero set of an ideal. The dimension of a variety X is the Krull dimension of its coordinate ring S/I(X). Proposition 3.2.1 (Zariski topology [200, Lemma A.2.4]) (a) V (1) = ∅ and V (0) = AqK . (b) V (I ∩ J) = V (I) ∪ V (J) = V (IJ), for all ideals I and J of S. (c) ∩V (Iα ) = V (∪Iα ), where {Iα } is any family of ideals of S. By the previous result the sets in the family τ = {AqK \ V (I)| I is an ideal of S} are the open sets of a topology on AqK , called the Zariski topology. Deﬁnition 3.2.2 Given X ⊂ AqK , the Zariski closure of X, denoted by X, is the closure of X in the Zariski topology of AqK , i.e., X is the smallest aﬃne variety of AqK containing X.

124

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Proposition 3.2.3 If X ⊂ AqK , then X = V (I(X)). Proof. As V (I(X)) is closed in the Zariski topology and X ⊂ V (I(X)), taking closures one has X ⊂ V (I(X)). For the other inclusion note that X is closed; that is, X = V (J), where J is an ideal of S. Applying I to X ⊂ V (J), one gets I(V (J)) ⊂ I(X). Since one also has the inclusion J ⊂ I(V (J)), by transitivity J ⊂ I(X). Therefore V (I(X)) ⊂ V (J) and 2 consequently V (I(X)) ⊂ X. Lemma 3.2.4 Let X and Y be aﬃne varieties in AqK . If I(X) = I(Y ), then X = Y Proof. By symmetry it suﬃces to show the inclusion X ⊂ Y . Take α ∈ X. There are g1 , . . . , gr in S such that Y = V (g1 , . . . , gr ). Clearly gi ∈ I(Y ) for all i. Then any gi vanishes at all points of X because I(Y ) = I(X). Thus gi (α) = 0 for all i, i.e., α ∈ Y . 2 Deﬁnition 3.2.5 An aﬃne variety X ⊂ AqK is reducible if there are aﬃne varieties X1 = X and X2 = X such that X = X1 ∪ X2 ; otherwise, X is irreducible. Theorem 3.2.6 Let K be a ﬁeld and let X be an aﬃne variety of AqK , then X is irreducible if and only if I(X) is a prime ideal. Proof. ⇒) Let I be an ideal of S with X = V (I) and let f, g be polynomials of S such that f g ∈ I(X). Then V (I, f ) ∪ V (I, g) = V (I). As X is irreducible, we get that V (I, f ) = V (I) or V (I, g) = V (I). Hence f ∈ I(X) or g ∈ I(X). Therefore I(X) is a prime ideal. ⇐) Let X1 and X2 be aﬃne varieties such that X = X1 ∪ X2 . Then I(X) = I(X1 ) ∩ I(X2 ). As I(X) is prime, by Exercise 2.1.48, we get that I(X) = I(X1 ) or I(X) = I(X2 ). Hence, by Lemma 3.2.4, we have X = X1 or X = X2 . Therefore X is irreducible. 2 Proposition 3.2.7 [210, Proposition 1.5] If X ⊂ AqK is an aﬃne variety over a ﬁeld K, then there are unique irreducible aﬃne varieties X1 , . . . , Xr in AqK such that Xi ⊂ Xj for all i = j and X = ∪ri=1 Xi . Proof. The existence follows using that the Zariski topology is Artinian (Exercise 3.2.16). To show uniqueness assume that Y1 , . . . , Ys is another decomposition of X into aﬃne varieties such that Yi ⊂ Yj for all i = j. We have the equality ∩ri=1 I(Xi ) = ∩si=1 I(Yi ). Fix 1 ≤ i ≤ r. By Theorem 3.2.6, I(Xi ) is a prime ideal of S. Then, by Exercise 2.1.48, we obtain that I(Xi ) ⊃ I(Yj ) for some j. Hence, by Exercise 3.2.15, Xi ⊂ Yj . Similarly, there is k such that Yj ⊂ Xk . Altogether, we have Xi ⊂ Yj ⊂ Xk . Thus, i = k and Xi = Yj for some j. This means that {X1 , . . . , Xr } is contained 2 in {Y1 , . . . , Ys }. A symmetric argument shows the reverse inclusion.

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The irreducible aﬃne varieties X1 , . . . , Xr of Proposition 3.2.7 are called the irreducible components of X. Theorem 3.2.8 Let S = K[t1 , . . . , tq ] be a polynomial ring over a ﬁeld K and let m be a maximal ideal of S. If K is algebraically closed, then there are a1 , . . . , aq ∈ K such that m = (t1 − a1 , . . . , tq − aq ). Proof. There is an integral extension K[h1 , . . . , hd ]→S/m, where d = dim(S/m) = 0; see Corollary 3.1.8. Hence, the canonical map ϕ : K → S/m is an isomorphism because K is algebraically closed. To complete the proof 2 choose ai ∈ K so that ϕ(ai ) = ai = ti = ϕ(ti ). Proposition 3.2.9√ [200, Lemma 1.8.8] If I is an ideal of a ring S and f ∈ S, then f ∈ I if and only if (I, 1 − tf ) = S[t], where t is a new variable. Theorem 3.2.10 (Hilbert Nullstellensatz) Let S be a polynomial ring over an algebraically closed ﬁeld K and let I be an ideal of S, then √ I(V (I)) = I. √ Proof. One clearly has I ⊂ I(V (I)), hence I ⊂ I(V (I)) because I(V (I)) is a radical ideal. For the other inclusion take f ∈ I(V (I)) and consider the ideal J = (I, 1 − f t) of the ring S[t], where S = K[t1 , . . . , tq ] and t is a new variable. Next we show the equality J = S[t]. If J = S[t] by Theorem 3.2.8 one has (I, 1 − f t) ⊂ (t1 − a1 , . . . , tq − aq , t − aq+1 ), ai ∈ K. Hence β ∈ V (I), where β = (a1 , . . . , aq ). Using that f ∈ I(V (I)) gives f (β) = 0, √ but this is impossible because 1 − f (β)aq+1 = 0. Hence J = S[t]. Thus f ∈ I by Proposition 3.2.9. 2 Corollary 3.2.11 Let X = V (f1 , . . . , fr ) be an aﬃne variety, deﬁned by polynomials f1 , . . . , fr in q variables over an algebraically closed ﬁeld K. Then r ≥ q − dim(X). Proof. By the Nullstellensatz I(X) = rad (f1 , . . . , fr ). Let p be a minimal prime of (f1 , . . . , fr ) of height q − dim(X). Applying Theorem 2.3.16 we get q − dim(X) ≤ r. 2

Exercises 3.2.12 Let I and J be two ideals in a polynomial ring S over a ﬁeld K. If rad (I) = rad (J), prove that V (I) = V (J).

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3.2.13 If K is a ﬁeld and X ⊂ AqK , then X ⊂ V (I(X)) with equality if X is an aﬃne variety. 3.2.14 Let J be an ideal of a polynomial ring S over a ﬁeld K. Prove that J ⊂ I(V (J)) with equality if J is the ideal of an aﬃne variety. 3.2.15 Let X and Y be aﬃne varieties in AqK . Then X ⊂ Y if and only if I(X) ⊃ I(Y ). 3.2.16 (The Zariski topology is Artinian) For any descending chain X1 ⊃ X2 ⊃ · · · ⊃ Xi ⊃ · · · of aﬃne varieties in AqK there is k ∈ N \ {0} such that Xi = Xk for i ≥ k. 3.2.17 Let I be an ideal of a polynomial ring S = K[t] over an algebraically closed ﬁeld K and let p1 , . . . , ps be the minimal primes of I. Prove that Y is an irreducible component of V (I) if and only if Y = V (pi ) for some i. 3.2.18 Let K be an algebraically closed ﬁeld. Then there is a one to one correspondence between aﬃne varieties (resp. irreducible varieties) in AqK and radical ideals (resp. prime ideals) in the polynomial ring K[t1 , . . . , tq ]. 3.2.19 Let I be a monomial ideal in K[t1 , . . . , tq ] and X = V (I) ⊂ AqK the associated monomial variety. If K is inﬁnite, prove that X is irreducible if and only if X = V (ti1 , . . . , tir ). 3.2.20 Let K be an inﬁnite ﬁeld. Prove: (a) AqK is an irreducible variety, (b) any two non-empty open sets of AqK intersect, (c) any non-empty open set of AqK is dense, (d) (K ∗ )q is an open set of AqK , (e) (K ∗ )q is not an aﬃne variety. 3.2.21 Let S = K[t1 , . . . , tq ] be a polynomial ring over a ﬁeld K and let m be a maximal ideal of S. If K is uncountable, then S/m is a ﬁnite extension of K.

3.3

Gr¨ obner bases

In this section we review some basic facts and deﬁnitions on Gr¨obner bases. The reader may consult [1, 99, 142] for a detailed discussion of Gr¨ obner bases and for the missing proofs of this section. Let K be a ﬁeld and let R = K[x1 , . . . , xn ] be a polynomial ring. The monomials or terms of R are the power products xa = xa1 1 · · · xann ,

a = (a1 , . . . , an ) ∈ Nn .

The set of monomials of R is denoted by Mn = {xa | a ∈ Nn }.

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Deﬁnition 3.3.1 A total order " of Mn is called a monomial order or term order if (a) xa # 1 for all xa ∈ Mn , and (b) for all xa , xb , xc ∈ Mn , xa " xb implies xa xc " xb xc . Two examples of monomial orders of Mn are the lexicographical order or lex order deﬁned as xb " xa iﬀ the last non-zero entry of b − a is positive, and the reverse lexicographical order or revlex order given by xb " xa iﬀ the last non-zero entry of b − a is negative. In the sequel we assume that a monomial order ≺ for Mn has been ﬁxed. Let f be a non-zero polynomial in R. One can write f=

r

ai M i ,

i=1

with ai ∈ K ∗ = K \ {0}, Mi ∈ Mn and M1 " · · · " Mr . The leading term M1 of f is denoted by in≺ (f ) or lt≺ (f ), or simply by in(f ). The leading coeﬃcient a1 of f and a1 M1 are denoted by lc(f ) and lm(f ), respectively. Deﬁnition 3.3.2 Let I be an ideal of R. The initial ideal of I, denoted by in≺ (I) or simply by in(I), is given by in≺ (I) = ({in≺ (f )| f ∈ I}). Lemma 3.3.3 (Dickson) If {Mi }∞ i=1 is a sequence in Mn , then there is an integer k so that Mi is a multiple of some term in the set {M1 , . . . , Mk } for every i > k. Proof. Let I ⊂ K[x1 , . . . , xn ] be the ideal generated by {Mi }∞ i=1 . By the Hilbert’s basis theorem I is ﬁnitely generated (see Theorem 2.1.4). It is seen that I can be generated by a ﬁnite set of terms M1 , . . . , Mk . Hence for 2 each i > k, there is 1 ≤ j ≤ k such that Mi is a multiple of Mj . Deﬁnition 3.3.4 Let F = {f1 , . . . , fq } ⊂ R \ {0} be a set of polynomials in R. One says that f reduces to g modulo F , denoted f →F g , if g = f − (au/ lc(fi ))fi for some fi ∈ F , u ∈ Mn , a ∈ K ∗ such that a · u · in≺ (fi ) occurs in f with coeﬃcient a. Proposition 3.3.5 The reduction relation “−→F ” is Noetherian, that is, any sequence of reductions g1 −→F · · · −→F gi −→F · · · is stationary.

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Chapter 3

Proof. Notice that at the ith step of the reduction some term of gi is replaced by terms of lower degree. Therefore if the sequence above is not stationary, then there is a never ending decreasing sequence of terms in Mn , but this is impossible according to Dickson’s lemma. 2 Theorem 3.3.6 (Division algorithm [142, Theorem 2.11]) If f, f1 , . . . , fq are polynomials in R, then f can be written as f = a1 f1 + · · · + aq fq + r, where ai , r ∈ R and either r = 0 or r = 0 and no term of r is divisible by one of in(f1 ), . . . , in(fq ). Furthermore if ai fi = 0, then in(f ) ≥ in(ai fi ). Deﬁnition 3.3.7 The polynomial r in the division algorithm is called a remainder of f with respect to F = {f1 , . . . , fq }. Deﬁnition 3.3.8 Let I = (0) be an ideal of R and let F = {f1 , . . . , fr } be a subset of I. The set F is called a Gr¨ obner basis of I if in≺ (I) = (in≺ (f1 ), . . . , in≺ (fr )). Deﬁnition 3.3.9 A Gr¨obner basis F = {f1 , . . . , fr } of an ideal I is called a reduced Gr¨ obner basis for I if: (i) lc(fi ) = 1 ∀i, and (ii) none of the terms occurring in fi belongs to in≺ (F \ {fi }) ∀i. Theorem 3.3.10 [142, Theorem 2.17] Each ideal I has a unique reduced Gr¨ obner basis. Deﬁnition 3.3.11 Let f, g ∈ R and let [f, g] = lcm(f, g) be its least common multiple. The S-polynomial of f and g is given by S(f, g) =

[in(f ), in(g)] [in(f ), in(g)] f− g, lm(f ) lm(g)

Given a set of generators of a polynomial ideal one can determine a Gr¨ obner basis using the next fundamental procedure: Theorem 3.3.12 (Buchberger [74]) If F = {f1 , . . . , fq } is a set of generators of an ideal I of R, then one can construct a Gr¨ obner basis for I using the following algorithm: Input: F Output: a Gr¨ obner basis G for I Initialization: G := F , B := {{fi , fj }| fi = fj ∈ G} while B = ∅ do

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pick any {f, g} ∈ B B := B \ {{f, g}} r := remainder of S(f, g) with respect to G if r = 0 then B := B ∪ {{r, h}| h ∈ G} G := G ∪ {r} Proposition 3.3.13 Let I be an ideal of R and let F = {f1 , . . . , fq } be a Gr¨ obner basis of I. If B = {u | u ∈ Mn and u ∈ (in(f1 ), . . . , in(fq ))}, then B is a basis for the K-vector space R/I. Proof. First we show that B is a generating set for R/I.Take f ∈ R/I. q r Since “−→F ” is Noetherian, we can write f = i=1 ai fi + i=1 λi ui , where ∗ λi ∈ K and such that every ui is a term which is not a multiple of any of the terms in(fj ). Accordingly ui is in B for all i and f is a linear combination of the ui ’s. s To prove that B is linearly independent assume h = i=1 λi ui ∈ I, where ui ∈ B and λi ∈ K. We must show h = 0. If h = 0, then we can label the ui ’s so that u1 " · · · " us and λ1 = 0. Hence in(h) = u1 ∈ in(I), but this is a clear contradiction because in(I) = (in(f1 ), . . . , in(fq )). Therefore h = 0, as required. 2 Deﬁnition 3.3.14 A monomial in B is called a standard monomial with respect to f1 , . . . , fq . Corollary 3.3.15 (Macaulay) If I is a graded ideal of R, then R/I and R/ in≺ (I) have the same Hilbert function. Lemma 3.3.16 [142, Proposition 2.15] Let f , g be polynomials in R and let F = {f, g}. If in(f ) and in(g) are relatively prime, then S(f, g) →F 0. Theorem 3.3.17 [74] Let I be an ideal of R and let F = {f1 , . . . , fq } be a set of generators of I, then F is a Gr¨ obner basis for I if and only if S(fi , fj ) −→F 0 for all i = j. Deﬁnition 3.3.18 Let I be an ideal of R generated by F = {f1 , . . . , fr } and consider the homomorphism of R-modules ϕ : Rr → I, ei −→ fi , where ei is the ith unit vector. The kernel of ϕ, denoted by Z(F ), is called the syzygy module of I with respect to F .

130

Chapter 3

Theorem 3.3.19 [1, Theorem 3.4.1] Let G = {g1 , . . . , gr } be a Gr¨ obner basis and write [in(gi ), in(gj )] [in(gi ), in(gj )] · gi − · gj = aijk gk , lm(gi ) lm(gj ) r

S(gi , gj ) =

k=1

where aijk ∈ R and in(S(gi , gj )) # in(aijk gk ) for all i, j, k. Then the set r [in(gi ), in(gj )] [in(gi ), in(gj )] · ei − · ej − aijk ek lm(gi ) lm(gj ) k=1

1≤i, j≤r

generates the syzygy module of I = (g1 , . . . , gr ) with respect to G. Elimination of variables Let K[x1 , . . . , xn , t1 , . . . , tq ] be a polynomial ring over a ﬁeld K. A useful monomial order is the elimination order with respect to the variables x1 , . . . , xn . This order is given by xa tc " xb td if and only if deg(xa ) > deg(xb ), or both degrees are equal and the last nonzero entry of (a, c)− (b, d) is negative. The elimination order with respect to all variables x1 , . . . , xn , t1 , . . . , tq is deﬁned accordingly. This order is called the GRevLex order. Theorem 3.3.20 Let B = K[x1 , . . . , xn , t1 , . . . , tq ] be a polynomial ring over a ﬁeld K with a term order ≺ such that terms in the xi ’s are greater obner basis G, then than terms in the ti ’s. If I is an ideal of B with a Gr¨ G ∩ K[t1 , . . . , tq ] is a Gr¨ obner basis of I ∩ K[t1 , . . . , tq ]. Proof. Set S = K[t1 , . . . , tq ] and I c = I ∩ S. If M ∈ in(I c ), there is f ∈ I c with lt(f ) = M . Hence M = λ lt(g) for some g ∈ G, because G is a Gr¨obner basis. Since M ∈ S and xα " tβ for all α and β we obtain g ∈ G ∩ S, that 2 is, M ∈ (in(G ∩ S)). Thus in(I c ) = (in(G ∩ S)), as required. Example 3.3.21 Let ≺ be the elimination order with respect to x1 , . . . , x4 . Using Macaulay2 [199], we can compute the reduced Gr¨ obner basis of I = (t1 − x1 x2 , t2 − x1 x3 , t3 − x1 x4 , t4 − x2 x3 , t5 − x2 x4 , t6 − x3 x4 ). By Theorem 3.3.20, it follows that I ∩K[t1 , . . . , t6 ] = (t3 t4 −t1 t6 , t2 t5 −t1 t6 ). Deﬁnition 3.3.22 Let I and J be two ideals of a ring R. The ideal ( (I : R J i ) (I : J ∞ ) = i≥1

is the saturation of I w.r.t J. If f ∈ R, we set (I : (f )∞ ) = (I : f ∞ ).

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The saturation can be computed by elimination of variables using the following result. Proposition 3.3.23 Let R[t] be a polynomial ring in one variable over a ring R and let I be an ideal of R. If f ∈ R, then ( (I : R f i ) = (I, 1 − tf ) ∩ R. (I : f ∞ ) = i≥1

q Proof. Let g ∈ (I, 1 − tf ) ∩ R. Then g = i=1 ai fi + aq+1 (1 − tf ), where fi ∈ I and ai ∈ R[t]. Making t = 1/f in the last equation and multiplying by f m , with m large enough, one derives an equality gf m = b1 f1 + · · · + bq fq , where bi ∈ R. Hence gf m ∈ I and g ∈ (I : f ∞ ). Conversely let g ∈ (I : f ∞ ), hence there is m ≥ 1 such that gf m ∈ I. Since one can write g = (1 − tm f m )g + tm f m g and 1 − tm f m = (1 − tf )b, for some b ∈ R[t], one derives g ∈ (I, 1 − tf ) ∩ R.

2

Exercises 3.3.24 If I and J are two ideals of a polynomial ring R over a ﬁeld K and let t be a new variable, then I ∩ J = (t · I, (1 − t) · J) ∩ R. 3.3.25 Let I be an ideal of a ring R and f a non-zero element of R, then (f )(I : f ) = I ∩ (f ). 3.3.26 Let R = k[x] be a polynomial ring, where k is a ﬁeld. Recall that f is a binomial in R if f = xα − xβ for some α, β in Nn . Use Buchberger’s algorithm to prove that an ideal of R generated by a ﬁnite set of binomials has a Gr¨ obner basis consisting of binomials. 3.3.27 Let I be an ideal of a ring R and f ∈ R. Prove that the following statements are equivalent: (a) f is regular on R/I. (b) (I : R f ) = I. (c) I = (I, 1 − tf ) ∩ R, where t is a new variable. 3.3.28 Let Nn be the set of n-tuples of nonnegative integers endowed with the partial order given by (a1 , . . . , an ) ≥ (b1 , . . . , bn ) if and only if ai ≥ bi for all i. If A ⊂ Nn , use Dickson’s lemma to prove that A has only a ﬁnite number of minimal elements.

132

3.4

Chapter 3

Projective closure

Let S = K[t1 , . . . , tq ] be a polynomial ring over a ﬁeld K and let u be a new variable. For f ∈ S of degree e deﬁne t1 tq h e ,..., , f =u f u u that is, f h is the homogenization of the polynomial f with respect to u. The homogenization of an ideal I ⊂ S is the ideal I h of S[u] given by I h = ({f h | f ∈ I}). √ √ h Proposition 3.4.1 If I is an ideal of S, then I h = I . √ √ h Proof. To show the inclusion I ⊂ I h , we need only observe the equality√(f h )m = (f m )h for√f ∈ S and m ≥ 0. To show the reverse inclusion let g ∈ I h . As the ideal I h is a graded ideal one may assume that g is s h m h homogeneous. √ Write g = u g1 , where g1 ∈ S. Since g is in I , it follows 2 that g1 ∈ I, as required. Let " be the elimination order on the monomials of S[u] with respect to t1 , . . . , tq , u, this order extends the elimination order with respect to t1 , . . . , tq on the monomials of S. Proposition 3.4.2 Let I be an ideal of S spanned by a ﬁnite set G. Setting G h = {g h | g ∈ G} and in{G h } = {in(g h )| g ∈ G}, the following hold. (a) If in(G h ) = (in{G h }), then in(I) = ({in(g)| g ∈ G}) and I h = (G h ). obner basis of I h . (b) G is a Gr¨ obner basis of I if and only if G h is a Gr¨ Proof. (a): We set G = {g1 , . . . , gr } and in{G} = {in(g)| g ∈ G}. To show the ﬁrst equality we need only show that in(I) ⊂ (in{G}). Let m be a monomial in the ideal in(I). There is g ∈ I, of degree e, such that in(g) = m. Writing g = ri=1 fi gi for some f1 , . . . , fr in S, from the equality gh = fi ue i=1 r

t1 tq ,..., u u

gi

t1 tq ,..., u u

(3.1)

0. As in(g h ) = in(g), one has in(us g h ) = us m. we get us g h ∈ (G h ) for s s h Hence u m ∈ (in{G }). Using in(gi ) = in(gih ) yields m ∈ (in{G}). To show the second equality it is enough to show that {g h | g ∈ I} ⊂ (G h ). By the ﬁrst equality r G is a Gr¨obner basis of I. Hence any g ∈ I can be written as g = i=1 fi gi , where in(g) # in(fi gi ) for all i. Notice that e = deg(g) ≥ deg(fi gi ) = deg(fi ) + deg(gi ). Since Eq. (3.1) holds, it follows that g h ∈ (g1h , . . . , grh ).

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(b) ⇒) By part (a) we need only show that in(G h ) ⊂ (in{G h }). Let m ∈ in(G h ) be a term, then m = in(g) for some g ∈ (G h ). We may assume that g is homogeneous. We can write g = up (m1 + m2 ue2 + · · · + ms ues ), where m1 , . . . , ms are monomials in S such that m1 " m2 ue2 " · · · " ms ues . As all mi uei have the same degree we obtain 0 ≤ e2 ≤ · · · ≤ es . It follows that g = g(t1 , . . . , tq , 1) belongs to I and in(g ) = m1 . Therefore m1 belongs to in(I) = (in{G}). Since in(gi ) = in(gih ) we obtain that m ∈ (in{G h }). ⇐) This implication follows from part (a). 2 Projective closure Let K be a ﬁeld. We deﬁne the projective space of dimension q over K, denoted by PqK , to be the quotient space (K q+1 \ {0})/ ∼ where two points α, β in K q+1 \ {0} are equivalent under ∼ if α = cβ for some c ∈ K. It is usual to denote the equivalence class of α by [α]. For any set X ⊂ PqK deﬁne I(X), the vanishing ideal of X, as the ideal generated by the homogeneous polynomials in S[u] that vanish at all points of X. Conversely, given a homogeneous ideal I ⊂ S[u] deﬁne its zero set as V (I) = {[α] ∈ PqK | f (α) = 0, ∀f ∈ I homogeneous} . A projective variety is the zero set of a homogeneous ideal. It is not diﬃcult to see that the members of the family τ = {PqK \ V (I)| I is an ideal of S[u]} are the open sets of a topology on PqK , called the Zariski topology. Deﬁnition 3.4.3 Let Y ⊂ AqK . The projective closure of Y is deﬁned as Y := ϕ(Y ), where ϕ is the map ϕ : AqK → PqK , α → [(α, 1)], and ϕ(Y ) is the closure of ϕ(Y ) in the Zariski topology of PqK . Proposition 3.4.4 If Y ⊂ AqK , then Y = V (I(ϕ(Y ))). Proof. It is left as an exercise; see Proposition 3.2.3 for a similar formula for the Zariski closure in the aﬃne case. 2 Proposition 3.4.5 Let Y ⊂ AqK be a set and let Y ⊂ PqK be its projective closure. If f1 , . . . , fr is a Gr¨ obner basis of I(Y ), then I(Y ) = (f1h , . . . , frh ). Proof. Note that I(Y ) = I(Y )h and use Proposition 3.4.2.

2

Corollary 3.4.6 Let Y ⊂ AqK be a set and let Y ⊂ PqK be its projective closure. Then the height of I(Y ) in S is equal to the height of I(Y ) in S[u].

134

Chapter 3

Exercises 3.4.7 If I is a prime ideal of a polynomial ring S, prove that I h is also a prime ideal. 3.4.8 Let Y = {(x31 , x21 , x1 )| x1 ∈ K} ⊂ A3K be a monomial curve and let Y be its projective closure. If K is an inﬁnite ﬁeld prove that I(Y ) = (t23 − t2 u, t2 t3 − t1 u, t22 − t1 t3 ). 3.4.9 Let S = K[t1 , . . . , tq ] be a polynomial ring over a ﬁeld K. If I is an h ideal of S generated by f1 , . . . , fm , then ((f1h , . . . , fm ) : u∞ ) = I h . d

3.4.10 Let Y = {(xd11 , . . . , x1q )| x1 ∈ K} be a monomial curve in the aﬃne space AqK . If d1 > d2 > · · · > dq and K is the ﬁeld of complex numbers, then the projective closure Y of Y is equal to 3 4 d d −d Y = [(xd11 , xd12 ud11 −d2 , . . . , x1q u11 q , ud11 )] ∈ PqK u1 , x1 ∈ K .

3.5

Minimal resolutions

The aim of this section is to study homogeneous resolutions of positively graded modules over polynomial rings. We shall be interested in the numerical data of these resolutions and in particular in the Betti numbers of these modules. We begin with a result which is a consequence of the graded Nakayama’s lemma (see Lemma 2.2.9). Proposition 3.5.1 Let R = K[x] be a polynomial ring over a ﬁeld K and let M be an N-graded R-module. If F = {f1 , . . . , fq } is a set of homogeneous elements of M and m = (x), then F is a minimal set of generators for M if and only if the image of F in M/mM is a K-basis of M/mM . Corollary 3.5.2 Let R = K[x] be a polynomial ring over a ﬁeld K and let m = (x). If M is an N-graded R-module, then μ(M ) = dimR/m (M/mM ) = dimK (M/mM ), where μ(M ) is the minimum number of generators of M . Proof. Let f1 , . . . , fr be a minimum set of generators of M . The dimension of M/mM , as a K-vector space, is less than or equal to r = μ(M ) because the images of f1 , . . . , fr generate M/mM . Hence, by Proposition 3.5.1, we get the required equality. 2

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135

Lemma 3.5.3 Let R = K[x] be a polynomial ring over a ﬁeld K and let M be an N-graded module over R with a presentation ϕ

0 −→ ker(ϕ) −→ Rq −→ M −→ 0,

ϕ

ei −→ fi .

If m = (x) and F = {f1 , . . . , fq } is a set of homogeneous elements, then F is a minimal set of generators for M if and only if L = ker(ϕ) ⊂ mRq . Proof. ⇐) Assume L ⊂ mRq . By Proposition 3.5.1 we need only show that the image of F in M/mM is linearly independent over K. If λ1 f1 + · · · + λq fq ∈ mM, for some λi ’s in K, then (λi ) − (ai ) ∈ L ⊂ mRq , for some ai ’s in m. We are using (λi ) as a short hand for (λ1 , . . . , λq ). Hence (λi ) is in mRq . Thus, λj = 0 for all j. ⇒) If L ⊂ mRq , pick z = (z1 , . . . , zq ) ∈ L such that z ∈ / mRq . Then zi is not in m for some i. Since the fi ’s are homogeneous this readily implies that fi is a linear combination of elements in F \ {fi }, a contradiction. 2 Deﬁnition 3.5.4 Let R = ⊕∞ i=0 Ri be a positively graded ring and a ∈ N. The graded R-module obtained by a shift in the graduation of R is given by R(−a) =

∞ )

R(−a)i ,

i=0

where the ith graded component of R(−a) is R(−a)i = R−a+i . Proposition 3.5.5 Let R = ⊕∞ i=0 Ri be a positively graded polynomial ring over a ﬁeld K with maximal irrelevant ideal m = R+ and M an N-graded R-module. Then there is an exact sequence of graded free modules F.

· · · −→

bk )

ϕk

R(−dki ) −→ · · · −→

i=1

b0 )

ϕ0

R(−d0i ) −→ M −→ 0

i=1

where ϕk is a degree preserving map and im(ϕk ) ⊂ mRbk−1 for k ≥ 1. Proof. Let f1 , . . . , fq be a set of homogeneous elements that minimally generate M . Set d0i = deg(fi ) and b0 = q. There is an exact sequence i

0 −→ Z1 = ker(ϕ0 ) −→

b0 )

ϕ0

R(−d0i ) −→ M −→ 0,

i=1

where ϕ0 is a degree preserving homomorphism such that ϕ0 (ei ) = fi for all i. Note that Z1 ⊂ mRb0 by Lemma 3.5.3. Since Z1 is once again a ﬁnitely generated graded R-module one may iterate the process to obtain the required exact sequence F. 2

136

Chapter 3

Theorem 3.5.6 (Hilbert syzygy theorem [128]) Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and let M be an N-graded R-module. Then M has a graded free resolution of length at most n. Proof. Set m = R+ . First notice that TorR i (M, R/m) = 0 for i ≥ n + 1, because the ordinary Koszul complex K (x) is a graded free resolution of R/m = K of length n. On the other hand assume that ϕk

ϕ1

· · · −→ Fk −→ · · · −→ F1 −→ F0 −→ M −→ 0 is a graded free resolution of M as in Proposition 3.5.5. Applying the functor (·) ⊗R R/m yields the complex ϕk ⊗1

· · · −→ Fk ⊗ R/m −→ · · · −→ F1 ⊗ R/m ϕ1 ⊗1

−→ F0 ⊗ R/m −→ M ⊗ R/m −→ 0.

Using im(ϕk ) ⊂ mFk−1 for all k ≥ 1 one obtains that all the maps ϕk ⊗ 1 are zero. Hence TorR i (M, R/m) Fi ⊗ R/m for i ≥ 1. In particular Fi ⊗ R/m Fi /mFi = 0 for i ≥ n + 1 and by 2 Nakayama’s lemma one obtains Fi = 0 for all i ≥ n + 1. Corollary 3.5.7 Let R = ⊕∞ i=0 Ri be a polynomial ring of dimension n over a ﬁeld K and let M be an N-graded R-module. Then there is a unique (up to complex isomorphism) exact sequence of graded modules 0 −→

bg )

ϕg

R(−dgi ) −→ · · · −→

i=1

bk )

ϕk

R(−dki ) −→ · · ·

i=1

−→

b1 )

ϕ1

R(−d1i ) −→

i=1

b0 )

ϕ0

R(−d0i ) −→ M −→ 0,

i=1

such that (a) g = sup{i | TorR i (M, K) = 0} ≤ n, and (b) im(ϕk ) ⊂ mRbk−1 for all k ≥ 1, where m = R+ . Proof. Notice the following isomorphisms: TorR j (M, K)

bj )

K(−dji ) Fj /mFj ,

i=1

where Fj is the jth free module in the resolution of M . Hence the bj ’s and the dji ’s are uniquely determined and so is the length of the resolution. 2

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Remark 3.5.8 The entries of the matrices ϕk are in m. This condition is equivalent to require that at each stage we use a minimal generating set. Remark 3.5.9 If dk1 ≤ · · · ≤ dkbk for all k, then d11 < d21 < · · · < dg1 .

Indeed if ϕk (ei ) = j rij ej , where rij ∈ m and rij homogeneous, then deg(ϕk (ei )) > d(k−1)1 because rij have positive degree. Hence one obtains dk1 > d(k−1)1 . Deﬁnition 3.5.10 The integers b0 , . . . , bg are the Betti numbers of M . The dji ’s are the twists, they indicate a shift in the graduation. In general the Betti numbers and twists of M may depend on the base ﬁeld K; see for instance [149, Example 2.10] and [346, Remark 3]. Deﬁnition 3.5.11 The R-module Zk = ker(ϕk−1 ) is called the k-syzygy module of M . Deﬁnition 3.5.12 The homogeneous resolution or minimal resolution of M is the unique graded resolution of M by free R-modules described in Corollary 3.5.7. If M has a minimal free resolution as above, then note that pdR (M ), the projective dimension of M , is equal to g. In our situation the notions of “free resolution” and “projective resolution” coincide because of the theorem of Quillen-Suslin: if K is a principal ideal domain, then all ﬁnitely generated projective K[x1 , . . . , xn ]-modules are free [363]. Theorem 3.5.13 (Auslander–Buchsbaum [150, Theorem 3.1]) Let M be an R-module. If R is a regular local ring, then pdR (M ) + depth(M ) = dim(R). Corollary 3.5.14 If R is a polynomial ring over a ﬁeld K and I is a graded ideal, then pdR (R/I) ≥ ht (I), with equality if and only if R/I is a Cohen–Macaulay R-module. Proof. The result follows from an appropriate graded version of the Auslander–Buchsbaum formula. 2 Proposition 3.5.15 Let S be a positively graded algebra of dimension d over a ﬁeld K. If A = K[h1 , . . . , hd ] → S is a homogeneous Noether normalization of S, then S is a Cohen–Macaulay ring if and only if S is a free A-module.

138

Chapter 3

Proof. By Theorem 3.5.13 pdA (S) + depth(S) = dim(A), where the depth of S is taken with respect to A+ . Assume S is a free A-module, then depth(S) = d and thus the depth of S with respect to S+ is also equal d. Therefore S is Cohen–Macaulay. Conversely assume S Cohen–Macaulay. Note that A is a polynomial ring. Let p ∈ Spec S be a minimal (graded) prime over A+ S. There is an integral extension K = A/A+ → S/p, hence S/p is a zero dimensional domain and consequently p is a maximal ideal. Therefore rad (A+ S) = S+ , that is, h1 , . . . , hd is a h.s.o.p for S. Now use Proposition 3.1.20 to conclude that h1 , . . . , hd is a regular sequence in S, that is, depth S = d. Another application of the Auslander–Buchsbaum formula yields that S is a free A-module. 2 Theorem 3.5.16 (Hilbert–Burch [128, Theorem 20.15]) Let R be a polynomial ring over a ﬁeld K and let I be a graded Cohen–Macaulay ideal of height two. If ϕ

0 −→ Rq−1 −→ Rq −→ R −→ R/I −→ 0 is the minimal resolution of R/I, then I is generated by all the minors of size q − 1 of the matrix ϕ. Example Let R = Q[x, y, z, w] and let I = (f1 , f2 , f3 ), where f1 = y 2 − xz, f2 = x3 − yzw, f3 = x2 y − z 2 w. Let us construct the minimal resolution of R/I. Consider the exact sequence ϕ1

id

0 −→ ker(ϕ1 ) = Z1 −→ R(−2) ⊕ R2 (−3) −→ I −→ 0,

ϕ1

ei −→ fi .

Here Z1 is the module of syzygies of I. Using Macaulay we ﬁnd that Z1 is generated as an R-modulo by the column vectors, ψ1 , ψ2 , of the matrix: ⎤ ⎡ zw x2 z ⎦ A=⎣ y −x −y Notice that R(−2) ⊕ R2 (−3) is graded by (R(−2) ⊕ R2 (−3))i = R(−2)i ⊕ R(−3)i ⊕ R(−3)i . 5 6 Hence ψ1 , ψ2 ∈ R(−2) ⊕ R2 (−3) 4 . Next consider the exact sequence id

ϕ2

0 −→ ker(ϕ2 ) = Z2 −→ R2 (−4) −→ Z1 −→ 0,

ϕ2

ei −→ ψi .

Since ψ1 R + ψ2 R is a free R-module, we have Z2 = (0). Altogether the minimal homogeneous resolution of R/I is: ϕ2

ϕ1

0 −→ R2 (−4) −→ R(−2) ⊕ R2 (−3) −→ R −→ R/I −→ 0. In this example R/I is Cohen–Macaulay because pdR (R/I) = 2 = ht (I).

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Pure and linear resolutions Let R = ⊕∞ i=0 Ri be a polynomial ring over a ﬁeld K with its usual graduation and I a graded ideal of R. Let 0→

bg )

R(−dgj ) → · · · →

j=1

b1 )

R(−d1j ) → R → S = R/I → 0

(3.2)

j=1

be the minimal graded resolution of S by free R-modules. The ideal I (or the algebra S) has a pure resolution if there are constants d1 < d2 < · · · < dg such that d1j = d1 , . . . , dgj = dg for all j. If in addition dj = d1 + j − 1 for 2 ≤ j ≤ g the resolution is said to be d1 -linear . Theorem 3.5.17 (Herzog–K¨ uhl [230]) If S is a Cohen–Macaulay algebra with a pure resolution, then bi =

7 j=i

dj |dj − di |

(1 ≤ i ≤ g).

Theorem 3.5.18 (Huneke-Miller [256]) If S is a Cohen–Macaulay algebra with a pure resolution, then the multiplicity e(S) of S is given by e(S) = d1 · · · dg /g!. Eisenbud and Schreyer proved the following long-standing multiplicity conjecture stated in [237, Conjecture 1, p. 2880]. Theorem 3.5.19 (Huneke–Srinivasan Multiplicity Conjecture [133]) If S is a Cohen–Macaulay algebra of codimension g, then g g 1 7 1 7 mi ≤ e(S) ≤ Mi , g! i=1 g! i=1

where Mi = max{dij | j = 1, . . . , bi } and mi = min{dij | j = 1, . . . , bi }. Herzog and Srinivasan [237, Conjecture 2, p. 2881] conjectured that the second inequality holds for non-Cohen–Macaulay algebras. This conjecture was shown by Boij and S¨ oderberg [47]. Since the Betti numbers and the twists are positive integers, these formulas impose restrictions on the numbers that can occur in the resolution of S.

140

Chapter 3

Exercises 3.5.20 Let R be a polynomial ring and let I be a graded Cohen–Macaulay ideal of height 3 with a pure resolution: 0 → Rb3 (−(d + a + b)) → Rb2 (−(d + a)) → Rb1 (−d) → I → 0, where a, b ≥ 1. Assume b1 = 6 and b = 1. Show that x = a + 1 and y = d + a + 1 is a solution of the diophantine equation 6x(x − 1) = y(y − 1). Then prove that the integral solutions of this equation are given by x = (X + 1)/2 and y = (Y + 1)/2, √ √ √ where 6X + Y = ±(1 ± 6)(5 + 2 6)n , n ∈ Z. 3.5.21 Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and let I be a graded ideal, then (x1 , . . . , xn ) is an associated prime of I if and only if pdR (R/I) = n. 3.5.22 Let R = K[x, y, z] be a polynomial ring and let I be the ideal (xn + y n − z n | n ≥ 2). If char(K) = 2, then I = (x2 + y 2 − z 2 , x3 + y 3 − z 3 , x2 y 2 ). If char(K) = 2, prove that q1 ∩ q2 ∩ q3 is a primary decomposition of I, where q1 = (y 2 , x − z), q2 = (x2 , y − z), q3 = (y 3 , z 3 , x2 y 2 , x2 + y 2 − z 2 ). 3.5.23 Let R = K[x] be a polynomial ring over a ﬁeld K. If I is a homogeneous ideal of R of height r which is generated by r polynomials, then I is generated by r homogeneous polynomials.

Chapter 4

Rees Algebras and Normality In this part a detailed presentation of complete and normal ideals is given. The systematic use of Rees algebras and associated graded rings will make clear their importance for the area. Some outstanding references for blowup algebras and the normality of Rees algebras are [59, 412, 414]. A general reference for integral closure of ideals, rings, and modules is [259].

4.1

Symmetric algebras

Let M be an R-module. Given n ≥ 0, we deﬁne T n (M ) = M ⊗ · · · ⊗ M and T 0 (M ) = R.

n−times

The tensor algebra T (M ) of M is the noncommutative graded algebra T (M ) =

∞ )

T n (M ),

n=0

where the product in T (M ) is induced by juxtaposition, that is, the product of x1 ⊗ · · · ⊗ xm and y1 ⊗ · · · ⊗ yn is x1 ⊗ · · · ⊗ xm ⊗ y1 ⊗ · · · ⊗ yn . The symmetric algebra of M , denoted by SymR (M ) or simply Sym(M ), is deﬁned as the quotient algebra Sym(M ) = T (M )/J where J is the two-sided ideal generated by all xy − yx = x ⊗ y − y ⊗ x with x and y running through M . Observe that Sym(M ) is commutative.

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Since J is a graded ideal generated by homogeneous elements of degree two, the symmetric algebra is graded by Symn (M ) = T n (M )/J ∩ T n (M ) and Sym0 (M ) = R. Note that Sym2 (M ) = M ⊗M/(x⊗y −y ⊗x), with x and y running through all the elements of M .

Exercises 4.1.1 If M is a free R-module of rank n prove that the symmetric algebra of M is a polynomial ring in n variables with coeﬃcients in R.

4.2

Rees algebras and syzygetic ideals

Let I be an ideal of a ring R generated by f1 , . . . , fq . The Rees algebra of I, denoted by R[It] or R(I), is the subring of R[t] given by R[It] = R[f1 t, . . . , fq t] ⊂ R[t], where t is a new variable. Note R[It] = R ⊕ It ⊕ · · · ⊕ I n tn ⊕ · · · ⊂ R[t]. There is an epimorphism of R-algebras ϕ : B = R[t1 , . . . , tq ] −→ R[It] −→ 0,

ϕ

ti −→ fi t,

where B = R[t] is a polynomial ring over the ring R. The kernel of ϕ, denoted by J, is the presentation ideal of R[It] with respect 8∞ to f1 , . . . , fq . Notice that J is a graded ideal in the ti -variables: J = i=1 Ji , where B has the standard grading induced by setting deg(ti ) = 1. The mapping ψ : Rq −→ I given by ψ(z1 , . . . , zq ) = qi=1 zi fi induces an R-algebra epimorphism β : R[t1 , . . . , tq ] −→ SymR (I). Thus, the symmetric algebra of I is: SymR (I) R[t1 , . . . , tq ]/ker(β), where ker(β) is an ideal of R[t] generated by linear forms: q q ! ker(β) = b i ti bi fi = 0 and bi ∈ R . i=1

i=1

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On the other hand the kernel of ϕ is generated by all forms F (t1 , . . . , tq ) such that F (f1 , . . . , fq ) = 0. In particular, one may factor ϕ through SymR (I) and obtain the commutative diagram: ϕ

−→

R[t1 , . . . , tq ] ↓β

α

R[It]

%

SymR (I) We say that I is an ideal of linear type if α is an isomorphism. An important module-theoretic obstruction to “SymR (I) R[It]” is given by the following result. Proposition 4.2.1 (Herzog–Simis–Vasconcelos [233]) Let I be an ideal of a ring R. If SymR (I) R[It], then for each prime p containing I, Ip can be generated by ht (p) elements. Syzygetic ideals Let I be an ideal of a ring R and let H1 (I) be the ﬁrst homology module of the Koszul complex H (x, R) associated to a set x = x1 , . . . xn of generators of I. In [382] it is pointed out that H1 (I) is related to I/I 2 , the conormal module of I, by the following exact sequence: f

h

H1 (I) −→ Rn ⊗ (R/I) −→ I ⊗ (R/I) = I/I 2 −→ 0.

(∗)

Here f ([z]) = z ⊗ 1 and h(ei ⊗ 1) = xi ⊗ 1. Set δ(I) = ker(f ). Deﬁnition 4.2.2 The ideal I is called syzygetic if δ(I) = 0. Although H1 (I) may depend on the set of generators for I, δ(I) depends only on I. Indeed Simis and Vasconcelos [382] proved that δ(I) ker(Sym2 (I) → I 2 ), where Sym2 (I) → I 2 is the surjection induced by the multiplication map. Deﬁnition 4.2.3 An ideal I of a ring R is said to be generically a complete intersection if IRp is a complete intersection for all p ∈ AssR (R/I). Remark 4.2.4 If I is unmixed and generically a complete intersection, then the R/I-torsion of H1 (I) equals δ(I). To prove this, notice that Ip is syzygetic for all p ∈ AssR (R/I); consequently, δ(I) is a torsion module.

144

Chapter 4

Computing the presentation ideal of a Rees algebra Let I be a graded ideal of a polynomial ring R over a ﬁeld k and let f1 , . . . , fq be a generating set for I. Consider the presentation of the Rees algebra: ϕ : B = R[t1 , . . . , tq ] −→ R[It] −→ 0,

(ti −→ tfi ).

The kernel J of ϕ can be obtained as follows: J = (t1 − tf1 , . . . , tq − tfq ) ∩ B. Since J is a graded ideal in the ti -variables, J = ⊕i≥1 Ji . The relationship between J and the Koszul homology of I is very tight. The exact sequence of Eq. (∗) can be made precise: 0 −→ J2 /B1 J1 = δ(I) −→ H1 (I) −→ (R/I)q −→ I/I 2 −→ 0. In particular one can decide whether I is syzygetic – that is, J2 = B1 J1 –or of linear type–that is, J = J1 B.

Exercises 4.2.5 Let A[x] be a polynomial ring over a ring A and let f1 , . . . , fq be forms of degree d ≥ 1 in A[x]. Then there is a graded isomorphism of A-algebras ϕ : A[f1 , . . . , fq ] −→ A[tf1 , . . . , tfq ], with ϕ(fi ) = tfi , where t is a new variable and both rings have an appropriate grading such that deg(fi ) = 1 and deg(tfi ) = 1. 4.2.6 Let I = (f1 , . . . , fq ) be an ideal of a polynomial ring R = K[x] over a ﬁeld K. Prove that R[It] R[t1 , . . . , tq ]/((f2 t1 − f1 t2 , . . . , fq t1 − f1 tq ) : f1∞ ). 4.2.7 Let I = I2 (X) be the ideal of 2 × 2-minors of the symmetric matrix: ⎡

x1 X = ⎣ x2 x3

x2 x4 x5

⎤ x3 x5 ⎦ , x6

where the entries of X are indeterminates over a ﬁeld K. Prove that I is a syzygetic ideal that satisﬁes sliding depth. See [238].

Rees Algebras and Normality

4.3

145

Complete and normal ideals

Let R be a ring and let I be an ideal of R, an element z ∈ R is integral over I if z satisﬁes an equation z + a1 z −1 + · · · + a−1 z + a = 0, ai ∈ I i , the integral closure of I is the set of all elements z ∈ R which are integral over I. This set will be denoted by I or Ia . Deﬁnition 4.3.1 If I = I, I is said to be integrally closed or complete. If all the powers I k are complete, the ideal I is said to be normal. The simplest kinds of integrally closed monomial ideals are the Stanley– Reisner ideals and the powers of face ideals. Lemma 4.3.2 Let R be a ring and I an ideal of R. An element α ∈ R is in the integral closure of I if and only if there is an ideal L of R such that (i) αL ⊂ IL, and (ii) αr ann(L) = (0) for some integer r ≥ 0. Proof. ⇒) As α is in integral closure of I there is an equation: αn = a1 αn−1 + · · · + an−1 α + an , where ai ∈ I i , and n ≥ 1 is some integer. The required L is obtained by setting L = Rαn−1 + Iαn−2 + · · · + I n−2 α + I n−1 . Indeed, from the equation above Rαn ⊂ Iαn−1 + I 2 αn−2 + · · · + I n−1 α + I n = IL, and α(I i αn−i−1 ) = I i αn−i = I(I i−1 αn−i ) ⊂ IL for i ≥ 1. Hence αL ⊂ IL. To ﬁnish this part of the proof note αn−1 ann(L) = (0). ⇐) As R is Noetherian, L = (f1 , . . . , fn ). Then αfi = nj=1 bij fj , where bij ∈ I. Set B = (bij ), C = B − αIn and f = (f1 , . . . , fn ). Here In denotes the identity matrix. As Cf t = 0, one can use the formula C · adj(C) = det(C)In to conclude fi det(C) = 0 for all i. Therefore det(C) is in ann(L) and by hypothesis αr det(C) = 0 for some r. Expanding in powers of α gives that α is integral over I. 2 Proposition 4.3.3 If I is an ideal of a ring R, then I is an ideal of R.

146

Chapter 4

Proof. Let αi ∈ I, i = 1, 2. By the proof of Lemma 4.3.2, there is an ideal Li such that αi Li ⊂ ILi , αri i ann(Li ) = (0) and I ri ⊂ Li for some ri ≥ 0. Since αi is integral over I, one has αni i ∈ I for some ni ≥ 1. Therefore for n large enough β n is in L1 L2 , where β = α1 + α2 . Thus β n ann(L1 L2 ) = (0). On the other hand one clearly has the inclusion βL1 L2 ⊂ IL1 L2 . Hence, by Lemma 4.3.2, one concludes β ∈ I. To ﬁnish proving that I is an ideal note xα1 ∈ I for x ∈ R, this follows 2 at once from the deﬁnition of I. Proposition 4.3.4 If I is an ideal of a ring R and S is a multiplicatively closed subset of R, then S −1 (I) = S −1 (I). Proof. To prove S −1 (I) ⊂ S −1 (I) take f ∈ S −1 (I). One may assume f = x/1 with x ∈ R because the integral closure of an ideal is again an ideal. There is an equation fn +

a1 n−1 an−1 an 0 f + ···+ f+ = , s1 sn−1 sn 1

where ai /si ∈ S −1 (I)i = S −1 (I i ). Thus one may assume ai ∈ I i and si ∈ S for all i. Clearing denominators and multiplying by an appropriate element of S one has an equality (in the ring R) of the form sxn + t1 a1 xn−1 + · · · + tn−1 an−1 x + tn an = 0 (s, ti ∈ S), multiplying both sides of this equation by sn−1 yields sx is in I. Therefore f = (xs)/s is in S −1 (I). The other inclusion follows readily using that localizations commute with powers of ideals. 2 Lemma 4.3.5 Let A be a domain and x ∈ A \ {0} such that Ax is normal. Then A is normal if and only if (x) is a complete ideal. Proof. Assume that A is a normal domain. Take z ∈ (x). Since z satisﬁes an equation of the form z m + (a1 x)z m−1 + · · · + (am−1 xm−1 )z + (am xm ) = 0, ai ∈ A, dividing by xm yields that z/x is integral over A and z ∈ (x). Conversely assume (x) is complete. Let z be an element of the quotient ﬁeld of A which is integral over A. As A and Ax have the same ﬁeld of fractions and Ax is normal one may assume z = a/xr , for some a ∈ A and r ≥ 1. To show z ∈ A it suﬃces to verify a ∈ (x). There is an equation z m + b1 z m−1 + · · · + bm−1 z + bm = 0, bi ∈ A,

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multiplying by xrm yields the identity am + (b1 xr )am−1 + · · · + (bm−1 xr(m−1) )a + (bm xrm ) = 0, 2

hence a ∈ (x) = (x), as required. Proposition 4.3.6 If A is a domain and x ∈ A \ {0}, then ⎛ ⎞ Ap ⎠ . A = Ax ⎝ p∈Ass A/(x)

Proof. Note that the left-hand side of the equality is contained in the right-hand side because A is a domain. Conversely take an element z ∈ Ax and z ∈ Ap , for all p ∈ Ass A/(x). Write z = a/xn , a ∈ A and n ≥ 1. It is enough to prove that a ∈ (x). If a is not in (x) note that ((x) : a) ⊂ Z(A/(x)). Hence ((x) : a) ⊂ p, for some p ∈ Ass A/(x). Since z ∈ Ap one may write z = a/xn = b/s, b ∈ A and s ∈ p. Therefore s ∈ ((x) : a) ⊂ p, which yields a contradiction. 2 Proposition 4.3.7 If A is an integral domain and S is a multiplicatively closed subset of A, then S −1 (A) = S −1 (A). Proof. Note that A and S −1 (A) have the same ﬁeld of fractions K. First we prove S −1 (A) ⊂ S −1 (A). Take any x in K integral over S −1 (A). There is an equation xn +

a1 n−1 an−1 an 0 x + ···+ x+ = , s1 sn−1 sn 1

where ai ∈ A and si ∈ S for all i. Set s = s1 · · · sn . If we multiply by sn , it follows that sx ∈ A and x ∈ S −1 (A). Conversely take x ∈ S −1 (A). There is s ∈ S such that sx is integral over A. Hence sx satisﬁes (sx)n + a1 (sx)n−1 + · · · + an−1 (sx) + an = 0, for some a1 , . . . , an in A, dividing by sn immediately yields that x is integral over S −1 (A), as required. 2 Corollary 4.3.8 If A is a normal domain and S is a multiplicatively closed subset of A, then S −1 (A) is a normal domain.

148

Chapter 4

Corollary 4.3.9 Let R = k[x1 , . . . , xn ] be a polynomial ring over a ﬁeld k ±1 and let KR be its ﬁeld of fractions. If R = k[x±1 1 , . . . , xn ] ⊂ KR is the ring of Laurent polynomials, then R is a normal domain. Proof. Notice that R is the localization of R at the multiplicative set of monomials of R. Hence, by Corollary 4.3.8, we get that R is normal. 2 Corollary 4.3.10 Let A be a domain and x ∈ A \ {0}. Then A is normal if and only if Ax and Ap are normal for every p ∈ Ass A/(x). Proof. If A is normal, by Corollary 4.3.8, Ax and Ap are normal for every x ∈ A \ {0} and p ∈ Spec(A). For the converse use Proposition 4.3.6 and 2 observe that Ax and Ap have the same quotient ﬁeld as A. To state a useful criterion of normality of Rees algebras, it is convenient to introduce the extended Rees algebra of an ideal I in a ring R: A = R[It, u], u = t−1 . Part of its usefulness is derived from the equality Au = R[t, t−1 ]. Proposition 4.3.11 Let I be an ideal of a ring R and A = R[It, t−1 ] its extended Rees algebra. If R is a normal domain, then A is normal if and only if Ap is normal for each associated prime p of t−1 A. Proof. Note At−1 = R[t, t−1 ]. Hence At−1 is a normal domain and one can use Corollary 4.3.10. 2 Lemma 4.3.12 Let A be a ring and x a regular element of A. If A/(x) is reduced and p ∈ AssA A/(x), then Ap is a normal domain. Proof. Let AssA A/(x) = {p1 , . . . , pr }. Since A/(x) is reduced (x) = p1 ∩ · · · ∩ pr , hence (x)Api = pi Api for all i. Note that ht (pi ) = 1, by Krull’s principal ideal theorem. Therefore the maximal ideal of Api is generated by a system of parameters, that is, Api is a regular local ring and consequently Api is a normal domain by Theorem 2.4.16. 2 Proposition 4.3.13 Let A be a ring and x a regular element of A. If Ax is a normal domain and A/(x) is reduced, then A is a normal domain. Proof. As Ax is a domain and x is a regular element on A one obtains that A is a domain. By Lemma 4.3.12 Ap is normal for all p ∈ AssA A/(x), hence A must be normal according to Corollary 4.3.10. 2

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There is another graded algebra associated to an ideal I of a ring R whose properties are related to the normality of the Rees algebra of I, it is called the associated graded ring of I and is deﬁned as: grI (R) = R/I ⊕ I/I 2 ⊕ · · · ⊕ I i /I i+1 ⊕ · · · , with multiplication (a + I i )(b + I j ) = ab + I i+j−1

(a ∈ I i−1 , b ∈ I j−1 ).

Given a generating set f1 , . . . , fq of I, it is not diﬃcult to verify that grI (R) is a graded algebra over R/I generated by the following elements of degree one: f 1 = f1 + I 2 , . . . , f q = fq + I 2 . Thus one has grI (R) = (R/I)[ f 1 , . . . , f q ]. Lemma 4.3.14 If I is an ideal of a ring R, then R[It]/IR[It] grI (R) and A/t−1 A grI (R), where A = R[It, t−1 ] is the extended Rees algebra of I. 2

Proof. It is left as an exercise.

Theorem 4.3.15 Let I be an ideal of a ring R. If I is generated by a regular sequence f1 , . . . , fq , then the epimorphism of graded algebras: ϕ : (R/I)[t1 , . . . , tq ] −→ grI (R),

ϕ(ti ) = f i = fi + I 2 ,

is an isomorphism, where t1 , . . . , tq are indeterminates over R/I. Proof. The proof is by induction on q, the case q = 1 is easy to show. Let J be the ideal (f1 , . . . , fq−1 ) and consider the epimorphism: ϕ : (R/J)[t1 , . . . , tq−1 ] −→ grJ (R) = (R/J)[ f 1 , . . . , f q−1 ], ϕ (ti ) = f i = fi + J 2 . Note (J : fq ) = J because fq is regular on R/J; moreover, since ϕ is an isomorphism it follows (by induction on m) that one has the following equalities (J m : fq ) = J m for all m ≥ 1. Let F ∈ R[t] = R[t1 , . . . , tq ] be a homogeneous polynomial of degree d such that its image in (R/I)[t] is in the kernel of ϕ, that is, F (f ) = F (f1 , . . . , fq ) ∈ I d+1 .

150

Chapter 4

As ϕ is graded it suﬃces to prove F ∈ IR[t]. To show F ∈ IR[t] we proceed by induction on d, the case d = 0 is clear. There is W ∈ R[t] of degree q d + 1 with F (f ) = W (f ), write W = i=1 ti Wi , where Wi = 0 or Wi is a homogeneous polynomial in R[t] of degree d. There are polynomials G in R[t1 , . . . , tq−1 ] and H in R[t] of degrees d and d − 1, respectively, such that F = F −

q

fi Wi = G + tq H.

i=1

Observe that F (f ) = 0, hence H(f ) ∈ (J d : fq ) = J d ⊂ I d and by induction on d one concludes that H has coeﬃcients in I. It only remains to prove that G has coeﬃcients in I. There is a polynomial H ∈ R[t1 , . . . , tq−1 ] of degree d with H(f ) = H (f ). Set F = G + fq H ∈ R[t1 , . . . , tq−1 ], noting F (f ) = F (f ) = 0 and using that ϕ is an isomorphism one derives that F has coeﬃcients in J, which implies that G has coeﬃcients in I as required. 2 Theorem 4.3.16 Let I be an ideal of a ring R and A = R[It, t−1 ] its extended Rees algebra. If R is a normal domain and grI (R) is reduced, then A is normal. Proof. As A/t−1 A grI (R) is reduced and At−1 = R[t, t−1 ] is a normal domain, by Proposition 4.3.13, we get that A is normal. 2 Theorem 4.3.17 [236] Let I be an ideal of a normal domain R. Then the following are equivalent: (a) I is a normal ideal of R. (b) The Rees algebra R[It] is normal. (c) The ideal IR[It] ⊂ R[It] is complete. (d) The ideal (t−1 ) ⊂ R[It, t−1 ] is complete. (e) The extended Rees algebra R[It, t−1 ] is normal. Proof. (a) ⇒ (b) Set A = R[It]. Let z ∈ A ⊂ R[t] and write z = si=0 bi ti . It suﬃces to prove bs ts ∈ A. First we prove that bs ts ∈ A. As z is almost integral over A there is 0 = f ∈ A such that f z n ∈ A for all n > 0. Hence there is 0 = fm ∈ I m such that (fm tm )(bs ts )n is in A for all n > 0; that is, bs ts is almost integral over A. As R is a Noetherian integral domain using Proposition 2.6.37 one derives that bs ts is integral over A. Thus there is an equation of the form (bs ts )m + a1 (bs ts )m−1 + · · · + am−1 (bs ts ) + am = 0,

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ri where ai = j=0 aij tj and aij ∈ I j . Grouping all the terms of t-degree equal to sm one has the equation bm s +

m

ai,si bm−i = 0. s

i=1

Thus bs is integral over I s and consequently bs ∈ I s , as required. (b) ⇒ (c) Let z = b0 + b1 t + · · · + bs ts be an element of R[It] which is integral over IR[It]. Then z satisﬁes an equation of the form z m + a1 z m−1 + · · · + am−1 z + am = 0, ai ∈ I i R[It], multiplying by tm one obtains that tz is integral over R[It]. Therefore tz ∈ R[It], which proves that z ∈ IR[It]. (c) ⇒ (d) Set B = R[It, t−1 ]. Let z be an element of B integral over −1 of the Laurent expansion of z is in t−1 B, one t B. As the negative s part i may assume z = i=0 bi t , s ≥ 0 and bi ∈ I i for all i ≥ 0. By descending induction on s it suﬃces to prove that bs ∈ I s+1 . There is an equation z m + a1 z m−1 + · · · + am−1 z + am = 0, ai ∈ (t−1 )i B, hence zt is almost integral over B. It follows rapidly that bs ts+1 is almost integral over B and thus bs ts+1 is integral over B. A direct calculation yields that bs ts+1 is integral over IR[It], which shows that bs is in I s+1 . (d) ⇒ (e) Set u = t−1 and note R[It, u]u = R[t, u] is a normal domain. Therefore, by Lemma 4.3.5, one concludes that R[It, u] is normal. (e) ⇒ (a) If z ∈ I r , then z satisﬁes a polynomial equation z m + a1 z m−1 + · · · + am−1 z + am = 0, ai ∈ I ri , thus multiplying by trm we get that the element ztr is integral over the ring 2 R[It, t−1 ]. Hence z ∈ I r . Corollary 4.3.18 Let I be an ideal of a normal domain R and R[It] its Rees algebra. If grI (R) is reduced, then R[It] is normal. Proof. It follows from Theorems 4.3.16 and 4.3.17.

2

Theorem 4.3.19 (Huneke [254]) Let R be a Cohen–Macaulay ring and let I be an ideal of R containing regular elements. If R[It] is Cohen–Macaulay, then grI (R) is Cohen–Macaulay. An ideal I of a ring R is a radical ideal if I = rad (I). Note: (i) a proper ideal I is radical if and only if I is an intersection of ﬁnitely many primes, and (ii) I ⊂ rad (I) and equality occurs if I is a radical ideal.

152

Chapter 4

Corollary 4.3.20 Let I be a radical ideal of a normal domain R. If I is generated by a regular sequence, then grI (R) is reduced and R[It] is a normal domain. Proof. It follows from Theorem 4.3.15 and Corollary 4.3.18.

2

Example 4.3.21 Let R = Q[x, y] be a polynomial ring over the ﬁeld Q and let I be the ideal (x2 , y 2 ). Observe that I is equal to (x2 , y 2 , xy). Thus R[It] is not normal because I is not even complete. Deﬁnition 4.3.22 Let I be an ideal of a ring R and p1 , . . . , pr the minimal primes of I. Given an integer n ≥ 1, the nth symbolic power of I is deﬁned to be the ideal I (n) = q1 ∩ · · · ∩ qr , where qi is the primary component of I n corresponding to pi . Let (R, m) be a regular local ring and I an unmixed ideal. An interesting problem is whether I (2) is contained in mI, although the answer is negative in general, the problem remains open in characteristic zero. For some insight into this problem see [132, 257]. Proposition 4.3.23 Let I be a proper ideal of a ring R and S = R\∪ri=1 pi , where p1 , . . . , pr are the minimal primes of I. Then I (n) = S −1 I n ∩ R for n ≥ 1. Proof. Let I n = q1 ∩ · · · ∩ qr ∩ qr+1 ∩ · · · ∩ qs be a primary decomposition, where qi is the primary component of pi for i ≤ r and qi is a primary component of the embedded prime pi for i > r. Since pi ∩ S = ∅ for i > r one has S −1 qi = S −1 R. Hence ! s r −1 n −1 qi = S −1 qi . S I =S i=1

i=1

To ﬁnish the argument note that S −1 qi ∩ R = qi and intersect with R the equality above. 2 Proposition 4.3.24 Let I be a radical ideal of a ring R and p1 , . . . , pr the minimal primes of I. Then (n)

for n ≥ 1. I (n) = p1 ∩ · · · ∩ p(n) r Proof. Let I n = q1 ∩ · · · ∩ qr ∩ qr+1 ∩ · · · ∩ qs be a primary decomposition of I n , where qi is pi -primary for i ≤ r and qi is an embedded primary

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component of I n for i > r. Localizing at pi yields I n Rpi = qi Rpi and from I = p1 ∩ · · · ∩ pr one obtains: I n Rpi = (IRpi )n = (pi Rpi )n = pni Rpi . (n)

Thus pni Rpi = qi Rpi and contracting one has pi

= qi , as required.

2

This result and also Exercise 6.1.25 have interesting generalizations to monomial ideals [92, Theorem 3.7]. Proposition 4.3.25 Let R be a polynomial ring over a ﬁeld K and let I be an ideal of R generated by square-free monomials. If n ≥ 1 and p1 , . . . , pr are the minimal primes of I, then I (n) = pn1 ∩ · · · ∩ pnr . (n)

Proof. From Proposition 6.1.7 one has pni = pi . Since I is a radical ideal the result follows from Proposition 4.3.24. 2 Corollary 4.3.26 Let R be a polynomial ring over a ﬁeld K. If I is an ideal of R generated by square-free monomials, then I (n) is integrally closed for n ≥ 1. Proof. Set J = I (n) . Let p1 , . . . , pr be the associated primes of I. If f ∈ J, for all i. By Corollary 4.3.20 pni is then there is m ≥ 1 so that f m ∈ pnm i n complete, thus f ∈ pi for all i. Hence Proposition 4.3.25 shows f ∈ J. 2 Corollary 4.3.27 Let R be a polynomial ring over a ﬁeld K and let I be a monomial ideal. If I is a radical ideal, then I n ⊂ I (n) for n ≥ 1. Proof. Let x ∈ I n , there is k ≥ 1 so that xk ∈ (I n )k ⊂ (I (n) )k , thus x is in ∈ I (n) . Note that I (n) is complete by Corollary 4.3.26, hence x ∈ I (n) . 2 Deﬁnition 4.3.28 An ideal I of a ring R is called normally torsion-free if Ass(R/I i ) is contained in Ass(R/I) for all i ≥ 1 and I = R. Brodmann [56] showed that when R is a Noetherian ring and I is an ideal of R the sets Ass(R/I n ) stabilize for large n. If I is a radical ideal which is normally torsion-free, then Ass(R/I n ) = Ass(R/I) for all n ≥ 1; that is, the notion of normally torsion-free is a strong form of stability. An aspect of symbolic powers that has attracted a lot of attention is to describe when the symbolic and ordinary powers of a given ideal I coincide; see [249, 333] and [185, 188]. Later in this chapter, we present a few cases where equality of symbolic and ordinary powers can be described in terms of properties of the associated graded ring. In Chapter 14 we characterize normally torsion-free monomial ideals in algebraic and combinatorial optimization terms (see Theorem 14.3.6). For a certain type of ideals, the next result characterizes normally torsionfree ideals as those ideals whose ordinary and symbolic powers coincide.

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Chapter 4

Proposition 4.3.29 Let I be an ideal of a ring R. If I has no embedded primes, then I is normally torsion-free if and only if I n = I (n) for all n ≥ 1. Proof. ⇒) Note Ass(R/I n ) = Ass(R/I) for n ≥ 1, because any associated prime p of I is a minimal prime of I n and thus p ∈ Ass(R/I n ). As I n has no embedded primes one concludes that I n has a unique irredundant minimal primary decomposition and I n = I (n) for n ≥ 1. ⇐) Let p1 , . . . , pr be the associated primes of I. Since pi is a minimal prime of I for all i, one derives Ass(R/I n ) = Ass(R/I (n) ) = {p1 , . . . , pr }.

2

Proposition 4.3.30 Let R be a polynomial ring with coeﬃcients in a ﬁeld. If I is a radical monomial ideal of R which is normally torsion-free, then its Rees algebra R[It] is a normal domain. Proof. By Proposition 4.3.29 I n = I (n) for n ≥ 1. Hence, thanks to Corollary 4.3.27, I n is integrally closed for n ≥ 1, i.e., R[It] is normal. 2 Proposition 4.3.31 Let I be an ideal of a Cohen–Macaulay ring R. If I is generated by a regular sequence, then I n = I (n) for n ≥ 1. Proof. Let f1 , . . . , fr be an R-regular sequence that generates I. By Krull’s principal ideal theorem ht (I) = r. Hence, by Theorem 2.3.25, I is unmixed. From Theorem 4.3.15 there is an isomorphism of graded rings ϕ : B = (R/I)[t1 , . . . , tr ] −→ grI (R),

ti −→ fi + I 2 ,

where B is a polynomial ring over R/I. Hence I i /I i+1 is a free R/I-modulo because it is isomorphic to Bi , the ith graded component of B. Thus AssR (I i /I i+1 ) = AssR (R/I). Using the exact sequence 0 −→ I i /I i+1 −→ R/I i+1 −→ R/I i −→ 0, it follows by induction that AssR (R/I n ) ⊂ AssR (R/I) for n ≥ 1. To ﬁnish the proof, we apply Proposition 4.3.29 to conclude the equality between the ordinary and symbolic powers of I. 2 Deﬁnition 4.3.32 A proper ideal I of a ring R is said to be locally a complete intersection if IRp is a complete intersection for all p ∈ V (I). Proposition 4.3.33 Let R be a Cohen–Macaulay ring and I a prime ideal. If I is locally a complete intersection, then I n = I (n) for all n ≥ 1. Proof. By Proposition 4.3.29 it suﬃces to prove Ass(R/I n ) ⊂ Ass(R/I) for n ≥ 1. If p is an associated prime of R/I n , then pRp is an associated prime of Rp /Ipn . Using Proposition 4.3.31 yields that Ip is normally torsion-free, that is, the only associated prime of Rp /Ipn is Ip . Therefore pRp = Ip and p = I. 2

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Proposition 4.3.34 Let p be a prime ideal of a ring R such that pRp is a complete intersection, then p(n) = pn for all n ≥ 1 if and only if grp (R) is a domain. Proof. ⇒) Since grp (R) = R[pt]/pR[pt], it is enough to show that pR[pt] is a prime ideal of R[pt]. Let x = a0 + a1 t + · · · + ar tr be an element in R[pt] and x = (a0 /1) + (a1 /1)t + · · · + (ar /1)tr the image of x in Rp [pRp t]. Note that ai is in pi+1 if and only if (ai /1) is in pi+1 Rp , because the ordinary and symbolic powers of p coincide, thus x is in pR[pt] if and only if x is in pRp [pRp t]. As a consequence pR[pt] is prime if and only if pRp [pRp t] is prime. To ﬁnish the argument use Theorem 4.3.15 and the hypothesis that pRp is generated by a regular sequence to get that grpRp (Rp ) is a domain. ⇐) First we prove AssR (pi /pi+1 ) = {p} for i ≥ 1. Let p1 be an associated prime of pi /pi+1 , that is, p1 = ann (x + pi+1 ), for some x in pi \ pi+1 . Note a ∈ p1 iﬀ ax ∈ pi+1 , and since grp (R) is a domain, one readily concludes that a ∈ p1 iﬀ a ∈ p. Hence p1 = p. There is an exact sequence: 0 −→ pi /pi+1 −→ R/pi+1 −→ R/pi −→ 0. Hence, using induction on i ≥ 1, Lemma 2.1.17 and the exact sequence above, we get Ass(R/pi ) ⊂ Ass(R/p) = {p}. Thus p is normally torsionfree and by Proposition 4.3.29 one has pn = p(n) for all n ≥ 1. 2 Theorem 4.3.35 [383] Let R be a normal domain and I a radical ideal which is generically a complete intersection. If I is normally torsion-free, then its Rees algebra R[It] is a normal domain. Proof. By Corollary 4.3.18 we need only show that the associated graded ring grI (R) = R[It]/IR[It] is reduced. Let f = a0 + a1 t + · · · + as ts + IR[It] be a nilpotent element of grI (R). Hence (as ts )m is in IR[It] for some m ≥ 1, by induction it suﬃces to verify that as ts is in IR[It]. Let p1 , . . . , pr be the minimal primes of I. Since IRpi = pi Rpi is a complete intersection one derives that grpi Rpi (Rpi ) is reduced (see Corollary 4.3.20). Therefore the image of as ts in grpi Rpi (Rpi ) is zero for all i, and one readily concludes that as belongs to the following intersection: (s+1)

s+1 (R ∩ ps+1 Rpr ) = p1 1 Rp1 ) ∩ · · · ∩ (R ∩ pr

∩ · · · ∩ p(s+1) . r

As I is normally torsion-free one can write I n = q1 ∩ · · · ∩ qr , where qi is pi -primary, localizing yields I n Rpi = pni Rpi = qi Rpi and consequently (n) qi = pi . Making n equal to s + 1 proves that as is in I s+1 . 2

156

Chapter 4

Gorenstein Rees algebras For use below R = K[x1 , . . . , xn ] will denote a polynomial ring over a ﬁeld K and S = R[It] will denote the Rees algebra of an ideal I of R, which is assumed to be Cohen–Macaulay. Deﬁnition 4.3.36 If (1, t)m = R ⊕ Rt ⊕ · · · ⊕ Rtm ⊕ Itm+1 · · · , we say that the canonical module ωS of S has the expected form if ωS = ωR (1, t)m for some m ≥ −1. Theorem 4.3.37 [235] If ωS has the expected form, then ωS ωR (1, t)g−2 if and only if I is generically a complete intersection. Lemma 4.3.38 [412, Page 142] If S is a regular local ring and J is an ideal of S generated by a regular sequence h1 , . . . , hg , then the Rees algebra S[Jt] is determinantal: z1 · · · zg S[Jt] S[z1 , . . . , zg ]/I2 h1 · · · hg and its canonical module is ωS (1, t)g−2 . Proposition 4.3.39 If I is an ideal of height g generated by square-free monomials and S = R[It] is Gorenstein, then g = 1 or g = 2. Proof. Assume g > 1. Then ωS S = (1, t)0 = R(1, t)0 = ωR (1, t)0 and therefore ωS has the expected form with m = 0. Clearly I is generically a complete intersection because I is the intersection of prime ideals of R; see Proposition 6.1.4. Thus, by Theorem 4.3.37, we get S ωS ωR (1, t)g−2 = R ⊕ Rt ⊕ · · · ⊕ Rtg−2 ⊕ Itg−1 ⊕ · · ·

(4.1)

Take a minimal prime p of I of height g. Then Sp = Rp [Ip t] is the Rees algebra of the ideal Ip , which is generated by a regular sequence. Thus localizing the extremes of Eq. (4.1) at p and using Lemma 4.3.38 we obtain Sp = Rp [Ip t] ωRp (1, t)g−2 ωSp . Note that it is important to know a priori that the canonical module of Sp is ωRp (1, t)g−2 . Hence Sp is Gorenstein. To ﬁnish the proof note that the only Gorenstein determinantal rings that occur in Lemma 4.3.38 are those with g = 2; see [65, 71, 73]. 2 Descent of normality Let A ⊂ B be an extension of rings such that B is normal; in general the normality of A is not inherited from B. We discuss some suﬃcient conditions for A to be normal. Lemma 4.3.40 Let A ⊂ B be an extension of rings. If B = A ⊕ C (as A-modules), then IB ∩ A = I for every ideal I of A.

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q Proof. Let z ∈ IB ∩ A and write z = i=1 bi fi , where bi ∈ B and fi ∈ I. By hypothesis bi = ai + ci , ai ∈ A and ci ∈ C. Since z ∈ A it follows that q z = i=1 ai fi ∈ I. This proves the containment IB ∩ A ⊂ I, the reverse containment is clear. 2 Proposition 4.3.41 Let A ⊂ B be integral domains with ﬁeld of fractions KA and KB , respectively. If B = A ⊕ C (as A-modules), then KA ∩ B ⊂ A. In particular if B is normal, then A is normal. Proof. By Lemma 4.3.40 one has IB ∩ A = I for every ideal I of A. Let b = a/c ∈ B, a, c ∈ A, then a ∈ (c)B ∩ A = (c), hence a = λc = bc, with λ ∈ A. Therefore b = λ ∈ A. 2 Proposition 4.3.42 Let R be a polynomial ring over a ﬁeld K and I an ideal of R generated by homogeneous polynomials f1 , . . . , fq . Assume deg(fi ) = d for all i. If R[It] is normal, then K[f1 , . . . , fq ] is normal. Proof. Let m = R+ be the irrelevant maximal ideal of the polynomial ring R and A = K[tf1 , . . . , tfq ]. Observe that there is a decomposition of Amodules: R[It] = K[tf1 , . . . , tfq ] ⊕ mR[It]. As A K[f1 , . . . , fq ] (as rings), by Proposition 4.3.41, the result follows. 2

Exercises 4.3.43 Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and let m be the maximal ideal (x1 , . . . , xn ). If I = md and J = (xd1 , . . . , xdn ) for some positive integer d, then R[Jt] = R[It]. 4.3.44 Let I, J be ideals of a normal domain R. Deﬁne the multi Rees algebra of I, J as R(I ⊕ J) = R[uI, vJ], here u, v are new variables. Then R(I ⊕ J) = ⊕I n J m un v m . 4.3.45 Let R be a ring and F = {Ii }i∈N a family of ideals in R. It is said that F is a ﬁltration of R, if Ii+1 ⊂ Ii , I0 = R, and Ii Ij ⊂ Ii+j for all i, j ∈ N. If I is an ideal of R, prove that the following are ﬁltrations of R: (a) Ii = I i , the ordinary powers of I, (b) Ii = I i , the integral closure of I i , (c) Ii = I (i) , the symbolic powers of I, and (d) In = ⊕i≥n Ri , where R = ⊕i≥0 Ri is a graded ring. Hint For (b) use the method of proof of Proposition 4.3.3.

158

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4.3.46 Let R be a normal domain and F = {Ii }i∈N a ﬁltration of R. If Ii is integrally closed for all i, prove that the Rees algebra R(F ) =

∞ )

Ii ti ⊂ R[t]

i=0

of the ﬁltration F is integrally closed. See [198] for a study of the Cohen– Macaulay and Gorenstein property of R(F ). 4.3.47 Let R be an integral domain. If {Rα }α∈Λ is a collection of normal subrings of R, then ∩α∈Λ Rα is normal. 4.3.48 Let R be a ring and I, J ideals of R. If I and J are integrally closed, prove that I ∩ J is integrally closed. 8 i i 4.3.49 If R is a domain and I ⊂ R is an ideal, then ∞ i=0 I t ⊂ R[It] with equality if R is a normal domain. 4.3.50 Let R be a domain and let K be its ﬁeld of fractions. If K[x] is a polynomial ring in one variable, then R[x] ⊂ R[x] ⊂ K[x] and R[x] = R[x]. 4.3.51 Let F and G be two ﬁnite sets of monomials in a polynomial ring R over a ﬁeld K. If (F ) = (G) prove that R[F t] = R[Gt]. 4.3.52 If F = {x1 x2 , x3 x4 x5 , x1 x3 , x2 x4 , x2 x5 , x1 x5 } ⊂ K[x1 , . . . , x5 ] and I = (F ), prove that K[F ] is not normal and R(I) is normal. 4.3.53 If R = Q[x1 , . . . , x7 ] and I is the principal ideal generated by the binomial f = x1 x3 x5 x7 − x2 x24 x6 , prove that R/I is a normal domain. Give a description of the irreducible binomials f such that R/(f ) is normal. 4.3.54 Let R be a Noetherian ring and a an element in the Jacobson radical of R. If a is regular and R/(a) is an integral domain, prove that R is an integral domain. 4.3.55 Let R = K[x] be a polynomial ring over a ﬁeld K. If I is an ideal generated by square-free monomials of degree 2, then I and I 2 are complete. 4.3.56 Let R = K[x1 , . . . , xn ] be a polynomial ring and let R(I) be the Rees algebra of an ideal I = (f1 , . . . , fq ), where K is a ﬁeld. If fi is homogeneous of degree d ≥ 1 for all i. Prove (a) R(I) R(I)/mR(I) ⊕ mR(I) as R(I) modules, where m = R+ , (b) K[f1 , . . . , fq ] R(I)/mR(I) as K-algebras, (c) if F = {f1 , . . . , fq } is a set of monomials, then R(I) K[F, tx] as K-algebras, where x = {x1 , . . . , xn }.

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4.4

159

Multiplicities and a criterion of Herzog

In this section we present an elegant and useful Cohen–Macaulay criterion due to Herzog (Theorem 4.4.13). Once more we recall that all rings considered in this book are Noetherian and modules are ﬁnitely generated. Quasi regular sequences It is useful to generalize Theorem 4.3.15 to modules by introducing a polynomial with coeﬃcients in a module. Let M be an R-module and R[t1 , . . . , tq ] a polynomial ring over the ring R. Set M [t] = M [t1 , . . . , tq ] = M ⊗R R[t1 , . . . , tq ], and note that an element of M [t] can be regarded as a polynomial in the ti variables with coeﬃcients in M , thus M [t] is naturally graded. If f1 , . . . , fq is a sequence in R and I = (f1 , . . . , fq ), there is a degree preserving map of additive groups ϕ : (M/IM )[t1 , . . . , tq ] −→ grI (M ) =

∞ )

I i M/I i+1 M,

i=0

such that ϕ(F (t)) = F (f1 , . . . , fq ) + I i+1 M , for all F (t) in M [t] homogeneous of degree i, where the notation F (t) means reducing the coeﬃcients of F (t) modulo IM . It is not hard to verify the equivalence between the following two conditions: (a) ϕ is injective. (b) For every homogeneous polynomial F in M [t] of positive degree n such that F (f ) ∈ I n+1 M , one has F ∈ IM [t]. Deﬁnition 4.4.1 If the map ϕ is an isomorphism and IM = M , the sequence f1 , . . . , fq is called an M -quasi-regular sequence. Theorem 4.4.2 Let M be an R-module and f = f1 , . . . , fq a sequence in R. If f is an M -regular sequence, then f is an M -quasi-regular sequence. Proof. It follows adapting the proof of Theorem 4.3.15.

2

Theorem 4.4.3 (Krull’s intersection theorem [310, Theorem 8.9]) Let M be an R-module and let ideal of R. Then there is a ∈ R such that I∞be an i I M = 0. a ≡ 1 mod (I) and a · i=1 Lemma 4.4.4 Let (R, m) be a local ring (resp. N-graded) and M an Rmodule (resp. N-graded). If N is a submodule of M (resp. ∞graded submodule) and I ⊂ m is an ideal (resp. graded ideal), then N = i=1 (N + I i M ).

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Chapter 4

∞ Proof. Using Krull’s intersection theorem one has i=1 I i M = (0) (this equality holds also in the graded case), where M = M/N . Hence the required equality follows at once. 2 Proposition 4.4.5 Let (R, m) be a local ring (resp. N-graded ring) and M an R-module (resp. N-graded module). If f = f1 , . . . , fq is an M -quasi regular sequence of elements in m (resp. homogeneous elements in m = R+ ), then f is an M -regular sequence. Proof. Fix an integer 1 ≤ r ≤ q and set I = (f1 , . . . , fq ). To begin with we split the ideal I as I = J +L, where J = (f1 , . . . , fr−1 ) and L = (fr , . . . , fq ), it is convenient to set J = (0) if r = 1. It suﬃces to show the equality (JM : M fr ) = JM , because this is equivalent to prove that fr is not a zero divisor of M/JM . Let m ∈ (JM : M fr ), according to Lemma 4.4.4 one has ∞ n n=1 (JM + L M ) = JM, thus the proof reduces to proving by induction on n that m ∈ JM + Ln M for n ≥ 1. Since m ∈ (JM : M fr ), there are mi in M such that mfr = m1 f1 + · · · + mr−1 fr−1

(m, mi ∈ M ).

(4.2)

Consider the polynomial F = mtr − m1 t1 − · · · − mr−1 tr−1 ∈ M [t]. Note degt (F ) = 1 and F (f ) = 0. As f is an M -quasi regular sequence we get m ∈ IM = JM + LM . By induction assume m ∈ JM + Ln M , that is, there is G ∈ M [tr , . . . , tq ] homogeneous of degree n such that m = b1 f1 + · · · + br−1 fr−1 + G(fr , . . . , fq ) (bi ∈ M ).

(4.3)

From Eqs. (4.2) and (4.3) we get fr G(fr , . . . , fq ) = a1 f1 + · · · + ar−1 fr−1 , ai ∈ M . Using that f is an M -quasi regular sequence one may assume (after an induction argument) that ai ∈ I n M , and thus tr G(tr , . . . , tq ) is in IM [t], that is, G(tr , . . . , tq ) is in IM [t]. One can write G = G1 + G2 , where G1 , G2 are homogeneous polynomials in M [t] of degree n with G1 ∈ JM [t] and G2 ∈ LM [t]. Since G1 (fr , . . . , fq ) is in JM and G2 (fr , . . . , fq ) is in Ln+1 M , from Eq. (4.3) we get that m is in JM + Ln+1 M , as required. 2 Lemma 4.4.6 Let f (t) ∈ Q[t] be a polynomial of degree d − 1 such that f (n) ∈ Z for n ∈ Z, then there are unique integers a0 , . . . , ad−1 such that d−1 t+i f (t) = ai fi (t), where fi (t) = . i i=0 Proof. The polynomials fi (t), i ∈ N, are a basis for Q[t] as a Q-vector d−1 space. Hence f (t) = i=0 ai fi (t), for some ai ∈ Q. Using the Pascal triangle we get d−1 t + i t + i − 1 d−1 t + i − 1 ai ai − = , f (t) − f (t − 1) = i i i−1 i=0 i=0

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thus by induction on the degree it follows that ai ∈ Z for all i.

2

Multiplicities Let (R, m) be a local ring, M a ﬁnitely generated Rmodule and q an ideal with rad (q) = m. As grq (M ) is a ﬁnitely generated grq (R)-module and A0 = R/q is Artinian, by Theorem 2.2.4 there exists a polynomial P (t) ∈ Q[t] of degree d − 1, where d = dim(grq (M )), such that P (i) = A0 (qi M/qi+1 M ), for i ≥ n0 . By Lemma 4.4.6 there are integers a0 , . . . , ad−1 such that d−1 i + j P (i) = aj , for all i ≥ 0. j j=0 From the short exact sequences 0 −→ qi M/qi+1 M −→ M/qi+1 M −→ M/qi M −→ 0, one derives (M/qi+1 M ) − (M/qi M ) = P (i) for i ≥ n0 . Hence, using the identity of Exercise 4.4.14, we get χqM (i) := (M/qi+1 M ) = (M/qn0 M ) +

i

P (j)

j=n0

j+i+1 aj = c0 + , j+1 j=0 d−1

for i ≥ n0 . Thus the function χqM (i) = (M/qi+1 M ) is polynomial of degree d, and is called the Samuel function of q with respect to M . The integer ad−1 , is the multiplicity of q on M and is denoted by e(q, M ) or simply by e(q) if M = R. Note that e(q, M )/d! is the leading coeﬃcient of χqM (i). By [259, Proposition 11.2.1], we have e(q, M ) = e(q, M ). If R is a polynomial ring with coeﬃcients in a ﬁeld K, there are eﬃcient methods to compute the multiplicity of a zero dimensional monomial ideal; see [104, 424] and Section 12.5. We recall the following result in dimension theory that relates the degree of the Samuel function with the dimension of the module. Theorem 4.4.7 [310, Theorem 13.4] χqM is a polynomial function of degree equal to the dimension of M . Deﬁnition 4.4.8 Let (R, m) be a local ring of dimension d, M an R-module and q an ideal of R with rad (q) = m. We deﬁne

e(q, M ) if dim(M ) = d, ed (q, M ) = 0 if dim(M ) < d.

162

Chapter 4

In order to show how the “multiplicity” behaves under short exact sequences we need to recall a result of E. Artin and D. Rees. Theorem 4.4.9 (Artin–Rees lemma [310, Theorem 8.5]) Let M be a module over a ring R. If N is a submodule of M and I an ideal of R, then there is a positive integer c such that I n M ∩ N = I n−c (I c M ∩ N ), ∀ n > c. Proposition 4.4.10 Let (R, m) be a local ring of dimension d and let q be an m-primary ideal of R. If 0 −→ N −→ M −→ N −→ 0 is an exact sequence of R-modules, then ed (q, M ) = ed (q, N ) + ed (q, N ). Proof. By Proposition 2.1.39, dim(M ) = max{dim(N ), dim(N )}. Hence, one may assume d = dim(M ). Tensoring with R/qn+1 the exact sequence above yields an exact sequence 0 → (N ∩ qn+1 M )/qn+1 N → N/qn+1 N → M/qn+1 M → N /qn+1 N → 0. Taking lengths with respect to R/q gives χqM (n) = χqN (n) + χqN (n) − (N ∩ qn+1 M/qn+1 N ).

(∗)

By Theorem 4.4.9, N ∩ qn+1 M ⊂ qn+1−c N , for some integer c > 0. Hence (N ∩ qn+1 M/qn+1 N ) ≤ (qn+1−c N/qn+1 N ) = χqN (n) − χqN (n − c), as χqN (n) − χqN (n − c) is a polynomial function of degree at most d − 1, the result follows by dividing Eq. (∗) by nd and taking limits when n goes to inﬁnity. 2 Next we show that the multiplicity is additive (cf. Proposition 8.5.9). Proposition 4.4.11 Let (R, m) be a local ring of dimension d and q an m-primary ideal. If M is a ﬁnitely generated R-module and A is the set of all prime ideals p of R with dim(R/p) = d, then (Mp )ed (q, R/p). ed (q, M ) = p∈A

Proof. Set B = A ∩ Supp(M ). If B = ∅, then dim(M ) < d and ed (q, M ) is equal to 0, thus in this case the identity above holds. Hence we may assume B = ∅, this yields the equality dim(M ) = d. By Theorem 2.1.16 there are prime ideals p1 , . . . , pn of R and a ﬁltration of submodules: (0) = M0 ⊂ M1 ⊂ · · · ⊂ Mn = M,

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such that Mi /Mi−1 R/pi for all i. Note B = {pi | dim(R/pi ) = d}. To show this equality observe that Ass(M ) ⊂ {p1 , . . . , pn } ⊂ Supp(M ), see Corollary 2.1.18 and its proof, and recall that the minimal elements of Supp(M ) are in Ass(M ). Using Lemma 4.4.10 and the exact sequences 0 −→ Mi−1 −→ Mi −→ R/pi −→ 0 we get e(q, M ) = e(q, R/pi ), where the sum multiset {p1 , . . . , pn }, such that dim(R/pi ) = d.

(0) if (Mi /Mi−1 )p Rp /pRp if

(i = 1, . . . , n), is taken over all pi , in the Let p ∈ B, then p = pi p = pi .

(4.4)

Since the second module is a ﬁeld we get that the length of Mp is equal to the number of times that p occurs in the multiset {p1 , . . . , pn }. Therefore e(q, M ) = (Mpi )e(q, R/pi ), pi ∈B

2

as required. This proof was adapted from [65].

Proposition 4.4.12 Let (S, m) be a local ring and M an S-module with a positive rank r = rank(M ). If q is an m-primary ideal, then e(q, M ) = e(q, S)rank(M ). Proof. First note d = dim(M ) = dim(S) by Lemma 2.1.45. Let A be the set of all prime ideals p of S with dim(S/p) = d. Since M has rank r one has Mp (Sp )r for any associated prime p of S, thus Mp (Sp )r for any p ∈ A. Therefore using Proposition 4.4.11 one obtains (Mp )ed (q, S/p) = (Spr )ed (q, S/p) ed (q, M ) = p∈A

=

p∈A

r(Sp )ed (q, S/p) = rank(M )ed (q, S).

p∈A

As e(q, M ) = ed (q, M ) and e(q, S) = ed (q, S), the proof is complete.

2

In Chapter 5 we use the following important Cohen–Macaulay criterion to study the Koszul homology of graded ideals. Theorem 4.4.13 (Herzog [217]) Let (S, m) be a Cohen–Macaulay local ring and let M be a ﬁnitely generated S-module with a well-deﬁned and positive rank. If y = y1 , . . . , yd is a system of parameters of S, then (S/(y)) · rank(M ) ≤ (M/(y)M ). Furthermore equality holds if and only if M is Cohen–Macaulay.

164

Chapter 4

Proof. By Lemma 2.1.45 one has d = dim(M ) = dim(S). Let q be the m-primary ideal generated by y. There is a (graded) epimorphism ϕ : (M/qM )[t] = (M/qM )[t1 , . . . , td ] → grq (M ) =

∞ )

qi M/qi+1 M (4.5)

i=0

such that ϕ(F ) = F (y1 , . . . , yd ) + qi+1 M , for all F in M [t] homogeneous of degree i, where F means reducing the coeﬃcients of F modulo qM . By restriction of ϕ to the ith graded component of (M/qM )[t] gives i+d−1 (M/qM ) ≥ (qi M/qi+1 M ) for i ≥ 0, (4.6) d−1 observe that both sides of the inequality are polynomial functions of degree d − 1 and consequently (M/qM ) ≥ e(q, M ). On the other hand y is a regular sequence, because S is Cohen–Macaulay (see Proposition 2.3.19), this forces ϕ to be an isomorphism when M = S (see Theorem 4.4.2). It follows rapidly that (S/q) = e(q, S). To ﬁnish the ﬁrst part of the proof note that making use of Proposition 4.4.12 one obtains the required inequality. At this point one should observe that, by the arguments above, the proof reduces to show that (M/qM ) = e(q, M ) if and only if M is Cohen– Macaulay. Next we show both implications. ⇒) Thanks to Propositions 4.4.5 and 2.3.19 it suﬃces to prove that ϕ is injective; because this implies that y is an M -regular sequence and hence M is Cohen–Macaulay. Using Eq. (4.5) we obtain an exact sequence 0 −→ ker(ϕ)i −→ ((M/qM )[t])i −→ qi M/qi+1 M −→ 0, hence we get

i+d−1 (ker(ϕ)i ) = (M/qM ) − (qi M/qi+1 M ) for i ≥ 0. d−1

As the polynomial functions on the right-hand side have both degree d − 1 and their leading terms cancel out, one concludes that (ker(ϕ)i ) grows as a polynomialfunction of degree at most d − 2. Let F = mα tα be a homogeneous polynomial in M [t] of degree p, with mα in M , and denote mα = mα + qM . Assume F is in ker(ϕ) and F = 0. Since mk ⊂ q, there is an s ≥ 1 such that ms−1 F = 0 and ms F = 0, thus one may assume mF = 0 and F = 0. Hence there is a graded epimorphism ψ induced by multiplication by F ψ : (S/m)[t](−p) −→ (S/q)[t]F . We now show that ψ is injective. Let g = bβ tβ be an element in the kernel of ψ, where bβ ∈ S and bβ = bβ + m. Thus ( .bβ tβ )( mα tα ) = 0, where

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165

.bβ = bβ + q. One may assume that bβ tβ and mα tα are the leading terms of g and F , respectively, w.r.t the lexicographical ordering of the ti variables. Thus .bβ mα tα+β is the leading term of gF and bβ mα ∈ qM . Hence bβ + q is a zero divisor of M/qM and since AssR/q (M/qM ) = {m/q}, one has that bβ + q is in m/q, which proves bα ∈ m and bβ = 0. Altogether one derives g = 0 and ψ is injective. Using that ψ is a graded isomorphism yields i−p+d−1 = (((S/q)[t]F )i ) ≤ (ker(ϕ)i ), d−1 which is a contradiction because the left-hand side is a polynomial function of degree d − 1. Therefore F = 0 and ϕ is injective, as required. ⇐) As M is Cohen–Macaulay, the sequence y is an M -regular sequence, an application of Theorem 4.4.2 yields that ϕ is an isomorphism. Therefore Eq. (4.6) becomes an equality and we get (M/qM ) = e(q, M ). 2

Exercises 4.4.14 Let d, m ∈ N. Prove the equality m d+m j+d−1 = . d d−1 j=0 4.4.15 Let f : Z → Z be a numerical function, f is said to be a polynomial function of degree d if there is a polynomial P (t) ∈ Q[t] such that P (i) = f (i) for i 0. Prove that f is a polynomial function of degree d if and only if the numerical function g : Z → Z given by g(i) = f (i) − f (i − 1) is a polynomial function of degree d − 1.

4.5

Jacobian criterion

Here we introduce the Jacobian criterion and present some applications and examples to illustrate its use. Along the way some facts about regular local rings are discussed. This type of rings are used in algebraic geometry to show for instance that the coordinate ring of a smooth irreducible aﬃne variety is normal [100, Proposition 1.0.9]. Let (R, m) be a local ring. By Nakayama’s Lemma it follows readily that R is a regular local ring if and only if m is generated by a system of parameters of R. If R is a polynomial ring or a formal power series ring over a ﬁeld k, then R is a regular ring [310, Theorem 19.5], that is, Rp is a regular local ring for all p in Spec(R) or equivalently Rm is a regular local ring for any maximal ideal m of R (see [310, Theorem 19.3]). Proposition 4.5.1 If R is a regular local ring, then R is Cohen–Macaulay.

166

Chapter 4

Proof. Let m be the maximal ideal of R and let k = R/m be its residue ﬁeld. Assume x1 , . . . , xd is a set of generators of m, where d = dim(R). If k[t1 , . . . , td ] is a ring of polynomial over the ﬁeld k, there is an epimorphism of graded k-algebras ϕ : k[t] = k[t1 , . . . , td ] −→ grm (R)

(4.7)

induced by ϕ(ti ) = xi + m2 for all i. It suﬃces to prove that the map ϕ is injective. Indeed if ϕ is injective, then x1 , . . . , xd is a regular sequence by Proposition 4.4.5, and hence R is Cohen–Macaulay by Proposition 2.3.19. If I = ker(ϕ) = 0, pick a homogeneous polynomial f in I of degree s ≥ 1. Using the isomorphism k[t](−s) f k[t], together with Eq. (4.7), one concludes: i+d−1 i−s+d−1 i i+1 − . dimk (m /m ) = dimk k[t]i − dimk Ii ≤ d−1 d−1 By Theorem 4.4.7 the left-hand side is a polynomial function of degree d − 1 while the right-hand side is a polynomial function of degree d− 2. Therefore I must be zero and ϕ is injective. 2 Corollary 4.5.2 If (R, m) is a regular local ring, then R is a domain. Proof. By Proposition 4.5.1 m is generated by a regular system of parameters. Thus grm (R) is a domain. Let x, y ∈ R such that xy = 0. If x = 0 and y = 0, then by Theorem 4.4.3 there are r, s ∈ N with x ∈ mr \ mr+1 and y ∈ ms \ ms+1 . Multiplying the images x, y of x, y in grm (R), one obtains xy = 0. Hence x = 0 or y = 0, which is impossible. This proves R is a domain. 2 Proposition 4.5.3 Let (R, m) be a regular local ring and I = R an ideal of R. If R/I is a regular local ring, then I is a complete intersection. Proof. Let x1 , . . . , xd be a generating set of m = m/I, where xi = xi + I and d is the dimension of R/I. Note that the set of images of x1 , . . . , xd in m/m2 is a basis for m/m2 as a vector space over k = R/m. Hence x1 + m2 , . . . , xd + m2 are linearly independent in m/m2 . Set n = dim(R) and m = (x1 , . . . , xd ). As R is a regular local ring n = dimk (m/m2 ), hence using the proof of Corollary 2.1.35 and the equality m = m + I one derives m = m + J, for some ideal J = (xd+1 , . . . , xn ) contained in I. By Proposition 4.5.1 x1 , . . . , xn is a regular sequence, thus it suﬃces to show J = I, to prove this equality observe that R/J is a regular local ring of dimension d, and therefore R/J is a domain by Corollary 4.5.2. Since R/I has also dimension d, we see that the canonical homomorphism ϕ : R/J → R/I is an isomorphism, thus I = J, as asserted. 2

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167

Deﬁnition 4.5.4 Let R = k[x1 , . . . xn ] be a polynomial ring over a ﬁeld k and I = (f1 , . . . , fq ) an ideal. The Jacobian matrix of I is the matrix J = (∂fi /∂xj ). We denote the Jacobian matrix of I taken modulo an ideal P by (∂fi /∂xj )(P ). Theorem 4.5.5 (Jacobian criterion [309, p. 213]) Let B = R/I be a quotient ring, where R = k[x1 , . . . , xn ] is a polynomial ring in n variables over a ﬁeld k, and let I = (f1 , . . . , fq ) ⊂ R be an ideal. If P ⊂ R is a prime ideal containing I and p = P/I, one has: (a) rank(∂fi /∂xj )(P ) ≤ ht(IP ). (b) If rank(∂fi /∂xj )(P ) = ht(IP ), then Bp is a regular ring. (c) If k is a perfect ﬁeld and Bp is regular, then rank(∂fi /∂xj )(P ) = ht(IP ). Lemma 4.5.6 Let R be a ring and I an ideal. If I is height unmixed and P is a prime ideal such that I ⊂ P , then IP is also height unmixed and ht(I) = ht(IP ). Proof. Let I = q1 ∩· · ·∩qr be a primary decomposition of I, we may assume √ qi ⊂ P for i = 1, . . . s. Localizing at P gives a primary decomposition √ s IRP = i=1 qi RP . Hence the associated primes of IP are qi RP , where √ √ i ≤ s. To ﬁnish the proof note ht( qi RP ) = ht( qi ) = ht(I) for i ≤ s. 2 Deﬁnition 4.5.7 Let k be a ﬁeld and I = (f1 , . . . , fq ) an ideal of height g of k[x1 , . . . , xn ]. The Jacobian ideal J of I is the ideal generated by the g × g minors of the Jacobian matrix J = (∂fi /∂xj ). Corollary 4.5.8 Let R = k[x1 , . . . , xn ] be a polynomial ring over a perfect ﬁeld k and I = (f1 , . . . , fq ) ⊂ R an unmixed ideal. If P is a prime ideal of R containing I and p = P/I, then (R/I)p is regular if and only if J/I ⊂ p, where J is the Jacobian ideal of I. Proof. ⇒) By Theorem 4.5.5 and Lemma 4.5.6 one has rank(∂fi /∂xj )(P ) = ht(I), hence there is f ∈ J \ P and J/I ⊂ P/I. ⇐) If (R/I)p is not regular, then by Theorem 4.5.5 one has the inequality rank(∂fi /∂xj )(P ) < ht(I), hence J/P = (0) and J/I ⊂ P/I, which is impossible. Note that this part of the proof is valid for any ﬁeld k. 2

168

Chapter 4

Proposition 4.5.9 Let R be a polynomial ring over a ﬁeld k and I an unmixed ideal of height g. If J is the Jacobian ideal of I and ht (J, I) ≥ g+2, then (R/I)p is regular for any prime p of R/I such that ht (p) ≤ 1. Proof. Let B = R/I and p = P/I, where P is a prime ideal of R. If Bp is not regular, then according to Corollary 4.5.8 one has J/I ⊂ p and hence (J, I) ⊂ P . Since R is a catenary ring and I is unmixed one obtains ht (P ) = g if ht (p) = 0 and ht (P ) = g + 1 if ht (p) = 1. Therefore ht (J, I) ≤ ht (P ) ≤ g + 1, which is impossible. Thus Bp must be regular, as required. 2 Another consequence of the Jacobian criterion is the following radical test. Let k be a ﬁeld and I = (f1 , . . . , fq ) ⊂ R = k[x1 , . . . , xn ] an ideal of height g. Let J be the Jacobian ideal of I generated by the g × g minors of the Jacobian matrix J = (∂fi /∂xj ). Theorem 4.5.10 (Vasconcelos [411]) If k is a perfect ﬁeld and I is unmixed, then I is a radical ideal if and only if there is an element f in J \ I such that (I : f ) = I. Proof. ⇒) Let P1 , . . . , Pr be the associated primes of I. If J ⊂ Z(R/I), then J ⊂ Pi for some i because Z(R/I) = ∪ri=1 Pi . Hence by Corollary 4.5.8 one derives that (R/I)Pi is not a regular ring, which is a contradiction because (R/I)Pi = RPi /Pi RPi is a ﬁeld. ⇐) Let I = q1 ∩ · · · ∩ qr be an irredundant primary decomposition of I √ and Pi = qi an associated prime ideal of I. Note (R/I)Pi (R/I)pi and IPi = (qi )Pi , where pi = Pi /I. First we observe that (R/I)pi is a regular ring; otherwise by Corollary 4.5.8 one has J/I ⊂ Pi /I, hence f ∈ Pi , which is impossible because f is regular modulo I. Therefore (R/I)Pi is a regular local ring and thus an integral domain by Theorem 2.4.16. As IPi = (qi )Pi must be prime, one concludes that qi is also prime and consequently qi = Pi . Note that here the hypothesis k perfect is not being used. 2 Example 4.5.11 Let R = k[x1 , . . . , x7 ] and I = (x1 x7 −x2 x6 , x3 x5 −x4 x7 ), where k is a ﬁeld. We now show that B = R/I is a normal ring using Theorem 2.4.15. The Jacobian matrix of I is: x7 −x6 0 0 0 −x2 x1 J= . 0 0 x5 −x7 x3 0 −x4 Note that x27 , x5 x6 , x3 x6 , x4 x6 , x2 x5 , x1 x5 , x2 x4 are in the Jacobian ideal J of I. Hence ht (J, I) ≥ g + 2, where g = ht (I) = 2. By Proposition 4.5.9, B satisﬁes (R1 ) and, since I is a complete intersection, it follows by Proposition 3.1.30 that B satisﬁes (S2 ). Thus, by Serre’s normality criterion, the ring B is normal.

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169

Lemma 4.5.12 Let R be a ring and I a prime ideal. If f ∈ R \ I and (n) If = Ifn for some n ≥ 1, then I (n) = (I n : f ∞ ). Proof. If z ∈ (I n : f ∞ ), then zf k ∈ I n for some k ≥ 1, hence z ∈ I (n) . Conversely if z ∈ I (n) , then sz ∈ I n for some s ∈ / I. Since s/1 ∈ / If and (s/1)(z/1) ∈ Ifn , then z/1 is in the nth symbolic power of If . Thus by hypothesis z/1 ∈ Ifn and this means that z ∈ (I n : f ∞ ). 2 There are a few methods to compute symbolic powers of prime ideals in polynomial rings, see [378] and [413, Chapter 3]. The following is a subtle application of the Jacobian criterion due to Vasconcelos. Proposition 4.5.13 Let R be a polynomial ring over a ﬁeld k and I a prime ideal. If f is an element in the Jacobian ideal J of I which is not in the ideal I, then I (n) = (I n : f ∞ ) for n ≥ 1, and such element f exist if k is a perfect ﬁeld. Proof. Since the localization Rf is Cohen–Macaulay, by Proposition 4.3.33 and Lemma 4.5.12 it is enough to prove that If is locally a complete intersection. Let P be a prime ideal such that I ⊂ P and f ∈ P . Set p = P/I. One has (Rf /If )Pf (Rf )Pf /(If )Pf RP /IP (R/I)P (R/I)p . As J/I ⊂ p, using Corollary 4.5.8 we conclude that RP /IP is a regular ring and consequently (If )Pf is a complete intersection (see Proposition 4.5.3). The last assertion follows from Theorem 4.5.10. 2 Example 4.5.14 Let I = (x3 − yz, y 2 − xz, z 2 − x2 y) ⊂ Q[x, y, z]. The ideal I is prime because it is the kernel of the homomorphism: ϕ : Q[x, y, z] −→ Q[t], where ϕ(h(x, y, z)) = h(t3 , t4 , t5 ). Using CoCoA [88] and Proposition 4.5.13 we get I (2) = (I 2 : f ∞ ) = (I 2 , Δ), where f = 2y 2 + xz and Δ = −x5 + 3x2 yz − xy 3 − z 3 . See [280].

Exercises 4.5.15 Let R = Q[x, y, z] and I the prime ideal generated by f1 = x3 − yz, f2 = y 2 − xz, f3 = z 2 − x2 y. Prove that Ix = (f1 , f2 ) and I (n) = (I n : x∞ ) for n ≥ 1.

170

Chapter 4

4.5.16 Use the following procedure in CoCoA [88] to verify the formula for the second symbolic power of Example 4.5.14 F1 := x'3 − y ∗ z; - - variables must begin with capital letters F2 := y'2 − x ∗ z; F3 := z'2 − x2 ∗ y; Jac := Jacobian([F1, F2, F3]); J := Minors(2, Mat(Jac)); - - 2 × 2 minors of the Jacobian matrix I := Ideal(F1, F2, F3); H := Elim(t, I'2 + Ideal(1 − t(2 ∗ y'2 + x ∗ z))); Print(H); 4.5.17 Let p be a prime ideal of a ring R such that pRp is a complete intersection, then p(n) = pn for all n ≥ 1 if and only if grp (R) is a domain. Hint Use Theorem 4.3.15, Lemma 2.1.17, and Proposition 4.3.29. 4.5.18 Let R be a Cohen–Macaulay ring and I an ideal of R which is height unmixed. If I is a complete intersection, then I is locally a complete intersection. Hint Use Lemma 2.3.20. 4.5.19 Let R be a Cohen–Macaulay ring and I a prime ideal. If f ∈ R \ I and If is a complete intersection, then I (n) = (I n : f ∞ ) for n ≥ 1. 4.5.20 Let I = I2 (X) be the ideal of 2 × 2-minors of the symmetric matrix: ⎤ ⎡ x1 x2 x3 X = ⎣ x2 x4 x5 ⎦ , x3 x5 x6 where the entries of X are indeterminates over a ﬁeld k. Prove that Ix1 is a complete intersection. Find a set of generators for I (2) and prove that depth(H1 ) = 1, where H1 is the ﬁrst Koszul homology module of I.

Chapter 5

Hilbert Series Hilbert series of graded modules and graded algebras are introduced and studied in this chapter. The h-vector and the a-invariant of a graded algebra are deﬁned using the Hilbert–Serre theorem. For Cohen–Macaulay algebras, we present their main properties. Some features of Cohen–Macaulay and Gorenstein extremal algebras will be presented.

5.1

Hilbert–Serre Theorem

Unless otherwise stated we shall always assume that modules are ﬁnitely generated and N-graded. We refer to Section 2.2 for an introduction to graded modules, Hilbert polynomials, and multiplicities. Let R = R0 [x1 , . . . , xn ] = ⊕∞ i=0 Ri be an R0 -algebra of ﬁnite type over an Artinian local ring R0 with the grading induced by deg(xi ) = di , where di is a positive integer for i = 1, . . . , n. If M = M0 ⊕ M1 ⊕ · · · ⊕ Mi ⊕ · · · is a ﬁnitely generated N-graded module over R, its Hilbert function and Hilbert series are deﬁned by H(M, i) = (Mi ) and F (M, t) =

∞

H(M, i)ti

i=0

respectively, where (Mi ) denotes the length of Mi as an R0 -module, if R0 is a ﬁeld (Mi ) = dimR0 (Mi ). Lemma 5.1.1 If 0 → M → M → M → 0 is a degree preserving short exact sequence of N-graded R-modules, then (a) H(M, i) = H(M , i) + H(M , i) for all i, and (b) F (M, t) = F (M , t) + F (M , t).

172

Chapter 5 2

Proof. It follows Proposition 2.1.36.

If j ∈ N, then M (−j) is the regrading of M obtained by a shift of the graduation of M ; more precisely M (−j) =

∞ )

M (−j)i ,

i=0

where M (−j)i = M−j+i . Note that we are assuming Mi = 0 for i < 0. In this way M (−j) becomes an N-graded R-module. Lemma 5.1.2 F (M (−j), t) = tj F (M, t). Proof. Since M (−j)i = Mi−j one has: F (M (−j), t) = tj

∞

(Mi−j )ti−j = tj F (M, t),

i=j

where the ﬁrst equality follows using that Mi−j = 0 for i = 0, . . . , j − 1. 2 Lemma 5.1.3 If z ∈ Rj , there is a degree preserving exact sequence φ

z

0 −→ (M/(0 : z))(−j) −→ M −→ M/zM −→ 0 (φ(m) = m + zM ), where (0 : z) = {m ∈ M | zm = 0} and the ﬁrst map is multiplication by z. Proof. As the map ψ : M (−j) → M , given by ψ(m) = zm, is a degree zero homomorphism one has that (0 : z)(−j) is a graded submodule of M (−j). The exactness of the sequence above follows because ψ induces an exact sequence ψ

ı

φ

0 −→ (0 : z)(−j) −→ M (−j) −→ M −→ M/zM −→ 0, 2

where ı is an inclusion.

Theorem 5.1.4 (Hilbert–Serre Theorem) The Hilbert series F (M, t) of M is a rational function that can be written as F (M, t) =

h(t) n 7

for some h(t) ∈ Z[t].

(1 − t ) di

i=1

If di = 1 for all i, there is a unique polynomial h(t) ∈ Z[t] such that F (M, t) =

h(t) and h(1) = 0. (1 − t)d

Hilbert Series

173

Proof. If n = 0, the Hilbert function of M is zero for i 0 and F (M, t) is a polynomial. If n > 0, consider the exact sequence of graded modules x

n M −→ M/xn M −→ 0. 0 −→ (0 : xn )(−dn ) −→ M (−dn ) −→

Since the ends of this exact sequence are ﬁnitely generated modules over 2 R0 [x1 , . . . , xn−1 ] the proof follows readily by induction on n. The integer d in the Hilbert–Serre theorem is denoted by d(M ). This integer is the Krull dimension of M ; see Proposition 5.1.6 and its proof. Deﬁnition 5.1.5 The degree of F (M, t) as a rational function is denoted by a(M ); it is called the a-invariant of M . If R has the standard grading, below we see how the a-invariant measures the diﬀerence between the Hilbert polynomial and the Hilbert function. For the rest of this section we shall always assume that the ring R has the standard grading induced by setting deg(xi ) = 1 for i = 1, . . . , n. We denote the Hilbert polynomial of M by ϕM (t). Proposition 5.1.6 If M is a graded R-module of dimension d. Then, there are integers a−d , . . . , a−1 so that i+d−j−1 H(M, i) = a−(d−j) , ∀ i ≥ a(M ) + 1, d−j−1 j=0 d−1

where a(M ) is the degree of F (M, t) as a rational function. Proof. By the Hilbert–Serre Theorem there is a polynomial h(t) ∈ Z[t] such that h(t) F (M, t) = (1 − t)d and h(1) = 0. If d = 0, then a(M ) is equal to deg(h) and H(M, i) = 0 for i ≥ a(M ) + 1. Thus one may assume d > 0. Observe that by the division algorithm we can ﬁnd e(t) ∈ Z[t] so that the Laurent expansion of F (M, t) − e(t), in negative powers of (1 − t), is equal to F (M, t) − e(t) =

d−1 j=0

a−(d−j) (−1)j h(j) (1) , , where a−(d−j) = d−j (1 − t) j!

where h(j) is the jth derivative of h. Next we expand (1 − t)d−j in powers of t to obtain ⎡ ⎤ ∞ d−1 ∞ i + d − j − 1 ⎣ ⎦ ti = e(t) + a−(d−j) ϕ(i)ti , F (M, t) = e(t) + d − j − 1 i=0 j=0 i=0

174

Chapter 5

observe that ϕ(i) = H(M, i) for i ≥ deg e(t)+1, where the degree of the zero polynomial is set equal to −1. To complete the proof note that a−d , . . . , a−1 are integers by Lemma 4.4.6, and that d = d(M ) is the dimension of M by Theorem 2.2.4. 2 Remark 5.1.7 The leading coeﬃcient of the Hilbert polynomial ϕM (t) of M is equal to h(1)/(d − 1)!, where h(t) is the polynomial F (M, t)(1 − t)d . If d = 0, then h(1) = (M ) and (M ) = dimK (M ) if R is a polynomial ring over a ﬁeld K. Deﬁnition 5.1.8 The index of regularity of M is the least integer ≥ 0 such that H(M, i) = ϕM (i) for i ≥ . Corollary 5.1.9 Let M be a graded R-module. If r0 = min{r ∈ N |H(M, i) = ϕM (i), ∀ i ≥ r}, then r0 = 0 if a(M ) < 0 and r0 = a(M ) + 1 otherwise. Corollary 5.1.10 Let S be a standard graded algebra of dimension d over a ﬁeld K. If S is Cohen–Macaulay and h = {h1 , . . . , hd } is a system of parameters of S consisting of linear forms, then the multiplicity of S is: e(S) = (S/hS). Proof. By Proposition 3.1.20 {h1 , . . . , hd } is a regular sequence. Thus there are exact sequences of K-vector spaces h

1 Si −→ (S/h1 S)i −→ 0. 0 −→ (S[−1])i = Si−1 −→

Therefore one has the relation ϕS (i) = ϕS (i) − ϕS (i − 1) between Hilbert polynomials for i 0, where S = S/h1 S. Let ϕS (t) =

e(S) d−1 t + ad−2 td−2 + · · · + a0 , (d − 1)! ϕS (t) =

e(S) d−2 + bd−3 td−3 + · · · + b0 . t (d − 2)!

Now, observe that thanks to ϕS (i) = ϕS (i) − ϕS (i − 1) one derives: ϕS (t) =

e(S)(d − 1) d−2 t + lower order terms, (d − 1)!

consequently e(S) = e(S). Since S is Cohen–Macaulay of dimension d − 1 (see Lemma 2.3.10), the result follows by induction. 2

Hilbert Series

175

Proposition 5.1.11 If A, B are standard algebras over a ﬁeld K, then dim(A ⊗K B) = dim(A) + dim(B). Proof. One has the next equality between Hilbert series: 9∞ :9 ∞ : i i F (A, t)F (B, t) = (dimK Ai )t (dimK Bi )t i=0

=

∞

i=0

(dimK Ci )ti = F (A ⊗K B, t),

i=0

8∞

where A⊗K B = r=0 Cr and Cr = yields the asserted equality.

8 i+j=r

Ai ⊗K Bj . Thus, Theorem 5.1.4 2

Using the Noether normalization lemma it is not diﬃcult to prove that the result above holds for arbitrary aﬃne algebras over a ﬁeld K. Computation of Hilbert series Let R be a polynomial ring over a ﬁeld K with a monomial order ≺ and let I ⊂ R be a graded ideal. Since R/I and R/ in≺ (I) have the same Hilbert function (see Corollary 3.3.15), the actual computation of the Hilbert series of R/I is a two-step process: • ﬁrst one ﬁnds a Gr¨ obner basis of I using Buchberger’s algorithm, and • second one computes the Hilbert series of R/ in(I) using elimination of variables (see [20, 37]). Other approaches to compute Hilbert series use minimal resolutions and Stanley decompositions [398]. An ad hoc method to compute Hilbert series of Cohen–Macaulay N-graded algebras is given by Proposition 3.1.27. Some examples are given below. Example 5.1.12 Let R = K[x1 , x2 , x3 ] be a polynomial ring and let I be the ideal (x21 , x22 x3 , x32 ). Let us compute the Hilbert series of R/I using elimination. Pick any monomial involving more than one variable, say x22 x3 . The idea is to eliminate x22 from the monomials containing more than one variable. From the exact sequence of graded modules: x2

2 R/I−→R/(x21 , x22 ) −→ 0, 0 −→ R/(x21 , x3 , x2 )(−2) −→

and applying Exercise 5.1.20 to the ends of this sequence, we get (1 − t)2 (1 − t2 ) −t4 + 2t2 + 2t + 1 (1 − t2 )2 F (R/I, t) = t2 . = + 3 3 (1 − t) (1 − t) (1 − t)

176

Chapter 5

Example 5.1.13 Let R = Q[x, y, z, w] and I = (f1 , f2 , f3 ), where f1 = y 2 − xz, f2 = x3 − yzw, f3 = x2 y − z 2 w. Using Macaulay2 [199] one ﬁnds that the minimal resolution of R/I is: ϕ2

ϕ1

0 −→ R2 (−4) −→ R(−2) ⊕ R2 (−3) −→ R −→ R/I −→ 0. Let us compute the Hilbert series of R/I using its minimal resolution: F (R/I, t) = F (R, t) − F (R(−2) ⊕ R2 (−3), t) + F (R2 (−4), t) 1 t2 2t3 2t4 = − − + (1 − t)4 (1 − t)4 (1 − t)4 (1 − t)4 2 1 + 2t + 2t = . (1 − t)2 Example 5.1.14 Let R = K[x1 , . . . , x5 ] and I = (x1 x2 , x3 x4 x5 ). Note that I is a complete intersection. Hence, by Exercise 5.1.20, one has F (R/I, t) =

(1 − t2 )(1 − t3 ) 1 + 2t + 2t2 + t3 5t2 − t + 2 = = −1 + . (1 − t)5 (1 − t)3 (1 − t)3

The Laurent expansion of F (R/I, t) + 1, in negative powers of (1 − t), is 5t2 − t + 2 a−3 a−2 a−1 , = + + (1 − t)3 (1 − t)3 (1 − t)2 (1 − t) where a−1 = 5, a−2 = −9, a−3 = 6. Therefore the Hilbert polynomial of S = R/I is ϕS (t) = 3t2 + 2 because by Proposition 5.1.6 one has: i+2 i+1 i H(S, i) = 6 −9 +5 = 3i2 + 2 for i ≥ 1. i i i

Exercises 5.1.15 If n ≥ 1 is an integer, then ∞ m+n−1 m 1 = t . m (1 − t)n m=0 5.1.16 Let R be a polynomial ring in n variables over a ﬁeld K. If R has the standard grading, then m+n−1 1 and H(R, m) = . F (R, t) = n−1 (1 − t)n

Hilbert Series

177

5.1.17 Let K be a ﬁeld and R = K[x1 , . . . , xn ] a polynomial ring graded by deg(xi ) = di ∈ N+ for i = 1, . . . , n. Then the Hilbert series of R is ; F (R, t) = 1/ ni=1 (1 − tdi ). 5.1.18 Let R = K[x, y, z] be a polynomial ring over a ﬁeld K and I is the ideal (xa , y b , z c , xa1 y b1 z c1 ). If a ≥ a1 , b ≥ b1 and c ≥ c1 , then e(R/I) = dimK (R/I) = abc1 + (c − c1 )(a1 b + (a − a1 )b1 ). 5.1.19 Let R = K[x, y] be a polynomial ring over a ﬁeld K. If I is the ideal (xn , xa y b , y n ) and a < n, b < n, then e(R/I) = n(a + b) − ab. 5.1.20 Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K graded by deg(xi ) = di ∈ N+ for i = 1, . . . , n. If f1 , . . . , fr is a homogeneous regular sequence in R and ai = deg(fi ), then ; ; F (R/(f1 , . . . , fr ), t) = ri=1 (1 − tai )/ ni=1 1 − tdi . 5.1.21 Let R = K[x1 , x2 , x3 ] and let I be the ideal (x21 , x22 x3 , x32 ). Prove that the a-invariant of R/I is 3 and H(R/I, i) = 4 for i ≥ 4. 5.1.22 Let A, B be two standard algebras over a ﬁeld K and deﬁne their Segre product as the graded algebra S = A ⊗S B = (A0 ⊗K B0 ) ⊕ (A1 ⊗K B1 ) ⊕ · · · ⊂ A ⊗K B, where (A ⊗K B)p = i+j=p Ai ⊗K Bj . Use Hilbert functions to prove dim(S) = dim(A) + dim(B) − 1. Note dim(Ai ⊗K Bi ) = dim(Ai ) dim(Bi ).

5.2

a-invariants and h-vectors

Our goal here is to relate the a-invariant of a Cohen–Macaulay N-graded algebra with its minimal resolution and to prove that h-vectors of such algebras are nonnegative. We include Stanley’s characterization of Cohen– Macaulay algebras in terms of Hilbert series. a-invariants of positively graded algebras Let R = K[x1 , . . . , xn ] be a positively graded polynomial ring over a ﬁeld K, let I be a graded ideal of R, and let F : 0 →

bg ) i=1

ϕg

R(−dgi ) → · · · →

bk ) i=1

ϕk

R(−dki ) → · · · →

b1 ) i=1

ϕ1

R(−d1i ) → R

178

Chapter 5

be the minimal resolution of S = R/I (see Section 3.5). From the resolution of S, using Lemmas 5.1.1 and 5.1.2, one can write the Hilbert series as:

1+

g

(−1)

k=1 n 7

F (S, t) =

k

bk

!! t

dki

i=1

(1 − t

deg(xi )

.

(5.1)

)

i=1

We begin with a general inequality, that follows from Eq. (5.1), relating a(S), the a-invariant S, with the shifts of F , the minimal resolution of S. Proposition 5.2.1 a(S) ≤ max{dki | 1 ≤ k ≤ g, 1 ≤ i ≤ bk } + a(R). Deﬁnition 5.2.2 If I ⊂ R is a Cohen–Macaulay graded ideal of height g, the canonical module of S = R/I is deﬁned as ωS = ExtgR (R/I, ωR ), where ωR = R(−δ) and δ =

n i=1

deg(xi ).

Proposition 5.2.3 If I ⊂ R is a Cohen–Macaulay graded ideal and a(S) is the a-invariant of S = R/I, then (a) maxi {d1i } < · · · < maxi {dgi }, and (b) a(S) = max1≤i≤bg {dgi } + a(R) = −min{i | (ωS )i = 0}. Proof. (a): We may assume that dk1 ≤ · · · ≤ dkbk for k = 1, . . . , g. Since S is Cohen–Macaulay, the height of I is equal g, this follows from Corollary 3.5.14. Recall that by Proposition 3.1.32 there exists a regular sequence f1 , . . . , fg inside I. Hence, using Proposition 2.3.7, one has: ExtiR (S, ωR ) = 0 for i < g. n We set δ = i=1 deg(xi ) and ωR = R(−δ). Therefore dualizing F , the minimal resolution of S, with respect to the canonical module of R, one obtains the (exact) complex: ⎞ ⎛ bg ) R(−dgi ), ωR ⎠ → 0. Hom(F , ωR ) : 0 → Hom(R, ωR ) → · · · → Hom ⎝ i=1

Hilbert Series

179

Observe the following natural isomorphisms: ⎛ ⎞ ⎛ ⎞ bg bg ) ) Hom ⎝ R(−dgi ), R(−δ)⎠ Hom ⎝ R(−dgi ), R⎠ (−δ) i=1

i=1

) bg

R(dgi )(−δ)

i=1 bg )

R(−(δ − dgi )).

i=1

Since Hom(F , ωR ) is a minimal resolution of ExtgR (S, ωR ), by Remark 3.5.9, one has d1b1 < d2b2 < · · · < dgbg . (b): By part (a), we have d1b1 < d2b2 < · · · < dgbg . As a consequence, from the expression for F (S, t) given in Eq. (5.1), one concludes the equality a(S) = dgbg + a(R). In particular one has a surjection bg )

R(−(δ − dgi )) → ExtgR (S, R(−δ)) = ωS → 0.

i=1

To complete the proof note that since δ − dgbg ≤ · · · ≤ δ − dg1 , one has the 2 equalities min{i| (ωS )i = 0} = δ − dgbg = −a(S). h-vectors of standard algebras Next we introduce the notion of hvector of standard algebras through the Hilbert–Serre theorem for Hilbert series (see Theorem 5.1.4). Deﬁnition 5.2.4 Let S be a standard algebra and let h(t) be the (unique) polynomial with integral coeﬃcients such that h(1) = 0 and F (S, t) =

h(t) , (1 − t)d

where d = dim(S). If h(t) = h0 + h1 t + · · · + hr tr , the h-vector of S is deﬁned as h(S) = (h0 , . . . , hr ). Theorem 5.2.5 [392] Let S be a standard algebra and let θ1 , . . . , θd be a homogeneous system of parameters for S with ai = deg(θi ). Then S is Cohen–Macaulay if and only if F (S, t) =

F (S/(θ1 , . . . , θd ), t) . d 7 ai (1 − t ) i=1

(5.2)

180

Chapter 5

Proof. ⇒) Assume S is Cohen–Macaulay. Hence, by Proposition 3.1.20, θ1 , . . . , θd is a regular sequence. It is enough to prove the equality F (S/(θ1 , . . . , θs ), t) =

s 7 (1 − tai )F (S, t),

(5.3)

i=1

for 1 ≤ s ≤ d. We proceed by induction on s. Assume s = 1. From the exact sequence of graded S-modules θ

1 0 −→ S[−a1 ] −→ S −→ S/(θ1 ) −→ 0,

we obtain F (S/(θ1 ), t) = F (S, t) − F (S[−a1 ], t). Since ∞

F (S[−a1 ], t) =

S[−a1 ]i ti =

i=0

ta1

=

∞

S[−a1 ]i ti

i=a1 ∞

Si−a1 ti−a1 = ta1 F (S, t),

i=a1

we conclude F (S/(θ1 ), t) = (1 − t )F (S, t). Now assume s > 1 and that the equality (5.3) is true for s − 1. From the exact sequence a1

θ

s S/(θ1 , . . . , θs−1 ) −→ S/(θ1 , . . . , θs ) → 0 0 → (S/(θ1 , . . . , θs−1 )) [−as ] −→

we obtain F (S/(θ1 , . . . , θs ), t) = (1 − tas )F (S/(θ1 , . . . , θs−1 ), t). Then, the induction hypothesisyields the required equality. i i ⇐) Let f (t) = i≥0 ai t and g(t) = i≥0 bi t be two formal power series, we say that f (t) ≥ g(t) if ai ≥ bi for all i. From the exact sequence θ

1 0 −→ (S/ann (θ1 )) [−a1 ] −→ S −→ S/(θ1 ) −→ 0,

we obtain F (S/(θ1 ), t) − (1 − ta1 )F (S, t) = ta1 F (ann (θ1 ), t). Hence, the power series F (S/(θ1 ), t) is greater than or equal to (1 − ta1 )F (S, t), and by induction it follows that r 7 F (S/(θ1 , . . . θr ), t) ≥ (1 − tai )F (S, t) for 1 ≤ r ≤ d. i=1

Set S = S/(θ1 , . . . , θd−1 ). Using the exact sequence d 0 −→ (S /ann (θd ))[−ad ] −→ S −→ S /(θd ) = S/(θ1 , . . . , θd ) −→ 0,

θ

we get F (S /(θd ), t) = (1 − t)ad F (S , t) + tad F (ann (θd ), t). By hypothesis, ; we have F (S /(θd ), t) = di=1 (1 − tai )F (S, t). Therefore (1 − t)ad (F (S , t) − (1 − ta1 ) · · · (1 − tad−1 )F (S, t))

≥0

≥0

= −tad F (ann (θd ), t).

Hilbert Series

181

Hence F (ann (θd ), t) = 0. This shows that θd is a regular element on S . By induction it follows that θ1 , . . . , θd is a regular sequence, Then S is Cohen– Macaulay by Proposition 3.1.20. This proof was adapted from [392]. 2 Theorem 5.2.6 Let S be a Cohen–Macaulay standard algebra over a ﬁeld k and h(S) = (hi ) the h-vector of S, then hi ≥ 0 for all i. Proof. One may assume that k is inﬁnite, otherwise one may change the coeﬃcient ﬁeld k using the functor (·) ⊗k K, where K is an inﬁnite ﬁeld extension of k. There is a h.s.o.p y = {y1 , . . . , yd } for S, where each yi is a form in S of degree one. Set S = S/(y)S. From Theorem 5.2.5 we derive hi = H(S, i) for all i and consequently hi ≥ 0. 2 Theorem 5.2.7 (Stanley) Let K be a ﬁeld and let S be a positively graded K-algebra of dimension d. If S is a Cohen–Macaulay domain and F (S, t) = (−1)d ta F (S, t−1 ) for some a ∈ Z, then S is Gorenstein and a is the a-invariant of S. Proof. By Exercise 5.2.10, F (S, t) = ta F (ωS , t) = F (ωS [−a], t). Pick an element 0 = x ∈ ωS [−a] of degree zero and deﬁne ϕ : S −→ ωS [−a]

ϕ

(r −→ rx),

the map ϕ is injective by Exercise 5.2.12. Therefore the equality above gives that ϕ is an isomorphism. Thus ωS S[a], as required. 2 Proposition 5.2.8 Let S be a standard graded algebra and let M ⊂ N be ﬁnitely generated graded S-modules of the same dimension d and multiplicity e. If M and N are Cohen–Macaulay and HM (t) =

f (t) , (1 − t)d

HN (t) =

g(t) (1 − t)d

are their Hilbert series, then deg f (t) ≥ deg g(t). Proof. There is an exact sequence 0 → M → N → N/M → 0 of graded modules. If M = N , N/M is a module of dimension < d since M and N have the same multiplicity. Since M and N are Cohen–Macaulay, standard depth chasing (see Lemma 2.3.9) implies that N/M is Cohen–Macaulay of dimension d − 1. We have the equality of Hilbert series, f (t) h(t) g(t) = + , (1 − t)d (1 − t)d (1 − t)d−1 so g(t) − f (t) = (1 − t)h(t). The assertion follows since the h-vectors of these two modules are positive (see Exercise 5.2.9). 2

182

Chapter 5

Exercises 5.2.9 Let R be a polynomial ring over a ﬁeld K with the usual grading and M a ﬁnitely generated N-graded module. If M is Cohen–Macaulay and h(t) = h0 + h1 t + · · · + hr tr the polynomial in Z[t] so that h(1) = 0 and F (M, t) = h(t)/(1 − t)d , where d = dim(M ). Prove that hi ≥ 0 for all i. 5.2.10 (Stanley) Let K be a ﬁeld and let S be a positively graded Kalgebra. If S is Cohen–Macaulay of dimension d prove the equality: F (ωS , t) = (−1)d F (S, t−1 ). Hint Note S R/I, with R a polynomial ring. Compute the Hilbert series of S and ωS using the minimal resolution of R/I as an R-module. 5.2.11 (Stanley) Let K be a ﬁeld and let S be a positively graded Kalgebra. If S is Gorenstein of dimension d and a(S) is its a-invariant, prove: F (S, t) = (−1)d ta(S) F (S, t−1 ). Hint ωS S[a(S)]. 5.2.12 Let S be a Cohen–Macaulay local domain and let M be a Cohen– Macaulay S-module. If dim(S) = dim(M ), prove that Z(M ) = (0). Hint Note dim(M ) = dim(S/p) for p ∈ Ass(M ) and ann(M ) = 0. 5.2.13 Let K be a ﬁeld and let S be a positively graded K-algebra. If S is a Cohen–Macaulay domain and the h-vector of S is symmetric, prove that S is a Gorenstein ring. Hint Use Theorem 5.2.7. 5.2.14 Let R be a polynomial ring over a ﬁeld K with the standard grading, and let M1 , M2 be two ﬁnitely generated graded R-modules. Then depth(M1 ⊕ M2 ) = max{depth(M1 ), depth(M2 )}.

5.3

Extremal algebras

The concept of an extremal Cohen–Macaulay—or Gorenstein—standard algebra (resp. ideal) has its origins in the works of Sally [366] and Schenzel [370]. Those algebras (resp. ideals) have the smallest possible reduction number (resp. smallest possible number of generators of the least degree). We will study with certain detail those two classes of algebras.

Hilbert Series

183

The Cohen–Macaulay case Let 8 R be a polynomial ring over a ﬁeld k with its usual grading and let I = ∞ i=p Ii be a graded ideal with Ip = (0). The integer p is the initial degree of I. As usual μ(I) denotes the minimal number of generators of I, and μ(Ip ) stands for the minimal number of generators of I in degree p. Proposition 5.3.1 Let R be a polynomial ring over a ﬁeld k and I a graded ideal in R of height g and initial degree p. If I is Cohen–Macaulay, then p+g−1 . μ(Ip ) = H(I, p) ≤ p Proof. We may assume that k is inﬁnite, since a change of the coeﬃcient ﬁeld can be easily carried out using the functor (·) ⊗k K, where K is an inﬁnite ﬁeld extension of k. This change of the coeﬃcient ﬁeld preserves the hypothesis on I and leaves both the height of I and the dimensions of the vector spaces of forms of a given degree in I unchanged. We set S = R/I. By Lemma 3.1.28, there is a homogeneous system of parameters y = {y1 , . . . , yd } for S, where each yi is a linear form in R. If we tensor the minimal resolution of S with R = R/(y) it follows that 0 ≤ H(S, p) = H(R, p) − H(I, p), where S = S/(y)S. To get the desired inequality it suﬃces to observe that R is a polynomial ring in g variables over the ﬁeld k. 2 Deﬁnition 5.3.2 If S is an Artinian positively graded algebra the socle of S is given by Soc(S) = (0 : S S+ ). Lemma 5.3.3 [174] Let S = R/I be a quotient ring of a polynomial ring R over a ﬁeld k modulo a graded ideal I. Let F : 0 →

bg ) i=1

R(−dgi ) → · · · →

b1 )

R(−d1i ) → R → S = R/I → 0,

i=1

be the minimal resolution of S. If S is Artinian, then there is a degree zero 8bg k[g − dgi ]. isomorphism of graded k-vector spaces Soc(S) i=1 Proof. Let R = k[x1 , . . . , xg ] and x = {x1 , . . . , xg }. The ordinary Koszul complex K associated to the regular sequence x is an R-free resolution of k = R/(x) and thus TorR i (S, k) Hi (K ⊗R S).

184

Chapter 5

8bg On the other hand, we have TorR g (S, k) i=1 k[−dgi ] (see the proof of Corollary 3.5.7). To complete the argument note Hg (K ⊗R S)[g] Zg (K ⊗R S)[g] g i−1 = ze1 ∧ · · · ∧ eg (−1) zxi e1 ∧ · · · ∧ e'i ∧ · · · ∧ eg = 0 [g] i=1

Soc(S). Altogether one has Soc(S)

8bg i=1

k[g − dgi ], as required.

2

Proposition 5.3.4 If S is a Cohen–Macaulay positively graded k-algebra over a ﬁeld k with presentation R/I, then the type of S is equal to the last Betti number in the minimal free resolution of R/I as an R-module. Proof. By making a change of coeﬃcients using the functor (·)⊗k K, where K is an inﬁnite ﬁeld extension of k, one may assume that k is inﬁnite. By Theorem 3.1.24 there exists a system of parameters y = {y1 , . . . , yd } for S = R/I with each yi a form of degree one of R. Since Tori (S, R/(y)) = 0 for i ≥ 1, the minimal resolution of S/(yS) as an R/(y)-module has the same twists and Betti numbers as the minimal resolution of S over R. Since S/(yS) is Artinian the result follows from Lemma 5.3.3. 2 Corollary 5.3.5 If R is a polynomial ring over a ﬁeld k, then a graded ideal I of R is Gorenstein iﬀ R/I is Cohen–Macaulay and the last Betti number in the minimal graded resolution of R/I is equal to 1. Corollary 5.3.6 Let S be a Cohen–Macaulay positively graded k-algebra over a ﬁeld k and let ωS be its canonical module. Then type(S) = μ(ωS ), where μ(ωS ) is the minimum number of generators of ωS . Proof. It follows from the proofs of Propositions 5.2.3 and 5.3.4.

2

Theorem 5.3.7 ([79], [350]) Let R be a polynomial ring over a ﬁeld k and let I be a graded ideal in R of height g and initial degree p. If S = R/I is a Cohen–Macaulay ring, then S has a p-linear resolution if and only if the following equality holds p+g−1 . (5.4) μ(Ip ) = H(I, p) = p Proof. Let F be the minimal resolution of S as in Lemma 5.3.3. We can order the shifts so that dk1 ≤ · · · ≤ dkbk for k = 1, . . . , g; by the

Hilbert Series

185

minimality of the resolution d11 < d21 < · · · < dg1 (see Remark 3.5.9). Observe that if the resolution of S is linear, then the Herzog–K¨ uhl formulas of Theorem 3.5.17 give the required equality. Conversely assume the equality above. Since we may assume that k is inﬁnite, by Lemma 3.1.28 there is a h.s.o.p y = {y1 , . . . , yd } for S, where each yi is a form of degree one in R. Set S = S/(y)S and R = R/(y). From Theorem 5.2.5 we get that the h-vector of S satisﬁes hp = 0, and we also get hi = H(S, i) for all i. Therefore hi = 0 for all i ≥ p. Using Lemma 5.3.3 one obtains a degree zero isomorphism Soc(S)

bg )

k[g − dgi ]

i=1

of graded k-vector spaces. In particular, the socle of S can only live in degrees dgi − g, hence we conclude the inequality dgi − g ≤ p − 1. On the other hand the minimality of the resolution of S gives p + (g − 1) ≤ dgi for all i. Altogether we have dgi = p + g − 1. Because I is C–M, then by Proposition 5.2.3 we must have p ≤ d1b1 < d2b2 < · · · < dgbg = p + g − 1, which implies that the resolution is p-linear, as required. 2 Example 5.3.8 [346] Let R = k[a, . . . , f ] be a polynomial ring over a ﬁeld k and let I be the ideal I = (abc, abd, ace, adf, aef, bcf, bde, bef, cde, cdf ). Then R/I is C–M if char(k) = 2 and non C–M otherwise. Thus in the ﬁrst case R/I has a 3-linear resolution. Other nice examples of Cohen–Macaulay algebras with linear resolutions include rings of minimal multiplicity [366], the coordinate ring of a variety deﬁned by the submaximal minors of a generic symmetric matrix [264], the coordinate ring of a variety deﬁned by the maximal minors of a generic matrix [123], and some face rings [170]. Cohen–Macaulay rings with linear resolutions have been studied by Sally [366] for the case p = 2, and by Schenzel [370] for the general case; more general rings with linear resolutions have been examined in [130, 233]. The Gorenstein case To give a sharp bound for the number of generators in the least degree of a graded Gorenstein ideal is harder than the Cohen–Macaulay case. We deal with this problem in the next section. The aim here is to introduce and characterize extremal Gorenstein rings in terms of their minimal resolutions and to determine their Betti numbers. Those Betti numbers are our “natural candidates” to bound the initial Betti numbers of graded Gorenstein ideals.

186

Chapter 5

Proposition 5.3.9 If S = S0 + · · · + Ss is a positively graded Artinian Gorenstein algebra over a ﬁeld k, then the multiplication maps Si × Ss−i → Ss = Soc(S) k are a perfect pairing, that is, the homomorphisms: ϕi : Si → Homk (Ss−i , Ss ) Ss−i ,

ϕi (r)(x) = rx

are isomorphisms of k-vector spaces. Proof. To show that ϕi is one to one take r ∈ ker(ϕ) and assume r = 0 and i < s. Since r is not in the socle of S, one may pick z ∈ S homogeneous of maximal positive degree such that zr = 0. Note that 0 < deg(z) < s − i, because rx = 0 for x ∈ Ss−i . Hence zr has degree less than s, using that the socle of S lives only in degree s one obtains a homogeneous element w with deg(w) > 0 such that wzr = 0, which contradicts the choice of z. Therefore r = 0 and ϕ is injective. Using the injectivity of ϕi one has dimk Si ≤ dimk Ss−i ≤ dimk Si , and thus ϕi is an isomorphism. 2 Corollary 5.3.10 Let (h0 , . . . , hs ) be the h-vector of a standard Gorenstein k-algebra, where k is a ﬁeld. Then h is symmetric, that is, hi = hs−i for all i. Proposition 5.3.11 Let R be a polynomial ring over a ﬁeld k with the standard grading and I a graded Gorenstein ideal of initial degree p and height g. If the minimal resolution of S = R/I by free R-modules is: F : 0 →

bg )

R(−dgi ) → · · · →

i=1

then

8bg−k i=1

bk )

R(−dki ) → · · · →

i=1

R(−(dg − d(g−k)i ))

8bk i=1

b1 )

R(−d1i ) → R,

i=1

R(−dki ) for 1 ≤ k ≤ g − 1.

Proof. One may always order the shifts so that dk1 ≤ · · · ≤ dkbk for k = 1, . . . , g. As bg = 1 and dgi = dg , by Proposition 5.2.3, one has that the a-invariant of S is a(S) = dg − n, where n is the number of variables of R. Since Hom(F , ωR ) is a minimal resolution of ωS ExtgR (S, ωR ) and using ωS S(dg − n), one obtains that F is self dual.

2

With the same assumptions and notation of Proposition 5.3.11 one has:

Hilbert Series

187

Proposition 5.3.12 If (h0 , . . . , hs ) is the h-vector of S, then s ≥ 2(p − 1) with equality if and only if the minimal resolution of S is of the form 0 → R(−(2p + g − 2)) → Rbg−1 (−(p + g − 2)) → · · · → Rb2 (−(p + 1)) → Rb1 (−p) → R. Proof. As in the proof of Theorem 5.2.6, one may assume that S is Artinian and that hi is equal to H(S, i). Note that the socle of S lives in degree dg − g by Lemma 5.3.3. Hence s ≥ dg − g. By Proposition 5.3.11 and the minimality of the resolution one concludes s ≥ dg − g = d1b1 + d(g−1)1 − g ≥ p − g + d(g−1)1 ≥ 2(p − 1), where the last inequality uses dk1 ≥ p + k − 1 for 1 ≤ k ≤ g − 1. For the second assertion use the chain of inequalities above and compute the h-vector of S using the resolution of S. 2 Deﬁnition 5.3.13 If s = 2(p − 1), S is called an extremal Gorenstein ring and its resolution is called pure and almost linear . Proposition 5.3.14 [370] Let R be a polynomial ring over a ﬁeld k and I a graded ideal of initial degree p and height g. If S = R/I is an extremal Gorenstein algebra, then for 1 ≤ m < g the mth Betti number of S is: p+g−1 p+m−2 bm = p+m−1 m−1 p+g−1 p+g−m−2 g p+g−2 + − . m p−1 m p−1 Proof. As in the proof of Theorem 5.3.7, one may assume that S is Artinian and that R a polynomial ring in g variables. Set s = 2(p − 1). From the minimal resolution of S of Proposition 5.3.12, and using the additivity of Hilbert series (see Lemma 5.1.1), we get ! g−1 s i hi t (1 − t)g = 1 + (−1)i bi tp+i−1 + (−1)g bg ts+g , i=0

i=1

where hi = H(S, i) = hs−i . Therefore p+m−1 g bm = (−1)p−1−i hi , 1 ≤ m ≤ g − 1 p+m−1−i i=0 Using the identity p−1 g g−1+i p−1−i (−1) p+m−1−i i i=0 p+g−1 p+m−2 = p+m−1 m−1

(5.5)

188

Chapter 5

and the symmetry of the h-vector of S (see Corollary 5.3.10) one has bm =

p+m−1 p+g−1 p+m−2 g p−1−i + (−1) hs−i . p+m−1 m−1 p+m−1−i i=p

Note p+m−1

p−1−i

(−1)

i=p

=

g hs−i p+m−1−i

m−1

i+1

(−1)

i=0

g p+g−3−i , m−1−i g−1

since the terms in the right-hand side of the equality are zero for i > p − 1 the last summation reduces to p−2 i+1 (−1)

p+g−3−i g−1 i=0 p−1 g g−1+i g g+p−2 p−1−i (−1) − = g−m+p−1−i i m p−1 i=0 p+g−1 p+g−m−2 g g+p−2 = − , m p−1 m p−1 g m−1−i

the last equality follows from Eq. (5.5) making m equal to g − m.

2

Exercises 5.3.15 Let R be a polynomial ring over a ﬁeld k and let I be a Cohen– Macaulay graded ideal in R of height g and initial degree p. Let (h0 , . . . , hs ) be the h-vector of R/I. Prove that s ≥ p − 1, with equality if and only if the minimal resolution of R/I is p-linear. If s = p − 1, the algebra R/I is called an extremal Cohen–Macaulay algebra. 5.3.16 Let R be a polynomial ring over a ﬁeld k and let I be a C–M graded ideal in R of height g and initial degree p. Prove that the Hilbert series of S = R/I can be written as

F (S, t) =

p−1 i+g−1 i t g−1 i=0 (1 − t)d

if and only if S has a p-linear resolution.

, where d = dim S,

Hilbert Series

189

5.3.17 Let R be a polynomial ring over a ﬁeld k and I a Cohen–Macaulay graded ideal in R of height g and initial degree p. If S has a linear resolution, show that the Betti numbers of S = R/I and its multiplicity are given by p+k−2 p+g−1 p+g−1 · and e(S) = . bk = k−1 g−k g 5.3.18 Let S = R/I be a standard Gorenstein algebra so that I has initial degree p and codimension g. Show that the multiplicity of S satisﬁes: g+p−1 g+p−2 e(S) ≥ e(g, p) = + , g g with equality if and only if S is an extremal Gorenstein algebra. Hint The h-vector h(S) = (h0 , . . . , hs ) is symmetric and s ≥ 2(p − 1). Then p−2 compute explicitly hp−1 + 2 i=0 hi . 5.3.19 Let R be a polynomial ring over a ﬁeld k and I a graded Gorenstein ideal of height 4 generated by forms of degree p. Let h(S) = (h0 , . . . , hs ) be the h-vector of S = R/I. Then the minimal resolution of S has the form: 0 → R(−(s+4)) → Rb (−(s−p+4)) →

s−2p+3 )

Rai (−(p+i)) → Rb (−p) → R,

i=1

where r = s − 2p + 3 and ai = ar−i+1 for i ≥ 1. 5.3.20 If x > 1 is an integer, prove that x(x + 8) is never a perfect square. 5.3.21 Let I be a graded Cohen–Macaulay ideal of a polynomial ring R with a minimal resolution: 0 → Rb3 (−(p + a + b)) → Rb2 (−(p + a)) → Rb1 (−p) → R → R/I. If b1 = 9, use the Herzog–K¨ uhl formulas to prove that I is Gorenstein. Hint Note b3 − b2 + b1 − 1 = 0 and prove that b1 b2 b3 = 9b3 (b3 + 8) is a perfect square, then use Exercise 5.3.20.

5.4

Initial degrees of Gorenstein ideals

Assume that R is a polynomial ring over a ﬁeld and that I is a homogeneous Gorenstein ideal of codimension g ≥ 3 and initial degree p ≥ 2. We set p+g−1 p+g−3 g+p−1 g+p−2 − . e(g, p) = + and μ0 = g−1 g−1 g g

190

Chapter 5

By the symmetry of the h-vector of R/I given in Corollary 5.3.10, the multiplicity of R/I satisﬁes e(R/I) ≥ e(g, p), see Exercise 5.3.18. Given codimension g and initial degree p, a graded Gorenstein algebra R/J with multiplicity e(R/J) = e(g, p) is extremal. This has strong structural implications: the minimal free resolution of R/J must be pure and almost linear; hence all the Betti numbers bi (R/J) are determined, and J is generated by μ0 forms of degree p (see Section 5.3). Results in the literature dealing with Cohen–Macaulay ideals (such as [140, 361]) give upper bounds for the minimal number of generators μ(I) of I in terms of codimension, initial degree, and multiplicity e(R/I). One of the aims below is to elucidate the multiplicity information that is already determined by the codimension and initial degree (see Theorem 5.4.3). In general one expects no such conclusion, but for graded Gorenstein algebras the symmetry of the h-vector h(R/I) can be eﬀectively exploited. Macaulay’s Theorem First we introduce the binomial expansion, and then state a famous result of Macaulay on the growth of Hilbert functions that will be needed later. Lemma 5.4.1 Let h, j be positive integers. Then h can be written uniquely aj aj−1 ai h= + + ···+ , (5.6) j j−1 i where aj > aj−1 > · · · > ai ≥ i ≥ 1. Proof. By induction on j. If j = 1, then h = h1 . Assume j > 1, then aj +1 aj there is a unique aj such that j ≤ h < . Let r = h − ajj . By j induction aj aj−1 ai r =h− = + ···+ where aj−1 > · · · > ai ≥ i ≥ 1. j j−1 i From the inequality aj aj−1 ai aj + 1 aj aj h= + + ··· + < + = j j−1 i j j−1 j aj j−1 < j−1 and this implies aj−1 < aj . Let us show the we obtain aj−1 uniqueness. Assume h can be written as bj bj−1 bk h= + + ··· + , j j−1 k

Hilbert Series

191

where bj > bj−1 > · · · > bk ≥ k ≥ 1. By induction on j one has bj bj + 1 ≤h< . j j Since aj is unique we derive aj = bj . Hence by induction hypothesis k = i 2 and aj−1 = bj−1 , . . . , ai = bi . The unique expression for h in Eq. (5.6) is called the binomial expansion of h in base j. We set aj aj−1 ai h(j) = + + ···+ , (5.7) j+1 j i+1 aj + 1 aj−1 + 1 ai + 1 h j = + + ···+ , (5.8) j+1 j i+1 sometimes h j is called the Macaulay symbol. Theorem 5.4.2 (Macaulay [413, Appendix B]) Let h : N → N be a numerical function and k a ﬁeld. The following conditions are equivalent : (a) There exists a homogeneous k-algebra S with H(S, i) = h(i) for i ≥ 0. (b) h(0) = 1 and h(i + 1) ≤ h(i) i for all i ≥ 1. Theorem 5.4.3 [318] If I is a graded Gorenstein ideal of height g ≥ 3 and initial degree p ≥ 2, then p+g−1 p+g−3 μ(Ip ) ≤ μ0 = − , g−1 g−1 and I is itself extremal if equality holds. Proof. If either μ(Ip ) > μ0 , or μ(Ip ) = μ0 and I is not extremal, then by the symmetry of the h-vector h(R/I) there is some j ≥ p so that h(R/I) = (h0 , h1 , . . . , hp−1 , hp , . . . , hj , hp−1 , . . . , h1 , h0 ), where hj = hp ≤ p+g−3 = hp−2 . The idea of the argument is to use the g−1 j

Macaulay bound hj for hj+1 to see that such a small value of hj cannot grow to such a large value of p+g−2 . hj+1 = hp−1 = g−1 Recall that this estimate is calculated from the binomial expansion for hj : aj aj−1 ai hj = + + ··· + , (5.9) j j−1 i

192

Chapter 5

where aj > aj−1 > · · · > ai ≥ i ≥ 1. Then aj + 1 aj−1 + 1 ai + 1 j

hj+1 ≤ hj = + + ··· + . j+1 j i+1

(5.10)

We may assume that hj > j, for if not, then hj ≤ j would imply that a = for all , and hence hj ≥ hj+1 , which contradicts our assumption. Notice that by grouping the terms of (5.9) according to the value of a − the binomial expansion for hj can be written as r jn + kn jn − 1 + kn jn − in + kn hj = + + ··· + jn jn − 1 jn − in n=0 r jn + kn + 1 jn − in + kn = − , (5.11) jn − in − 1 j n n=0 where aj − j = k0 > k1 > · · · > kr ≥ 0, j = j0 > j1 > · · · > jr , jr − ir = i, and jn = jn−1 − in−1 − 1 for 1 ≤ n ≤ r. Set k = k0 . Since p ≤ j and p+g−3 j+k ≤ hj ≤ g−1 j it follows that k ≤ g − 2. From (5.9), (5.10), and (5.11), together with Pascal’s identity and a+b+1 a+1 a+b+1 = , b+1 b+1 b we have hj+1 − hj

≤ =

r jn + kn + 1

jn − in + kn jn − in jn + 1 n=0 r kn + 1 jn + kn + 1 kn + 1 jn − in + kn − . jn jn + 1 jn − in jn − in − 1 n=0 −

On the other hand from the upper bound on hj and hj+1 = hp−1 we get p+g−3 g−1 hj ≤ ≤ hj+1 − hj . g−2 p−1 Since (g − 1)/(p − 1) > (k + 1)/(j + 1) it follows from (5.11) and the last two inequalities that F0 < 0, where for 0 ≤ s ≤ r we set r ks + 1 kn + 1 jn + kn + 1 Fs = − js + 1 jn + 1 jn n=s r kn + 1 jn − in + kn ks + 1 + − . jn − in js + 1 jn − in − 1 n=s

Hilbert Series

193

To derive a contradiction we are going to show the following inequalities kr + 1 ai kr + 1 F0 > F1 > · · · > Fr = − > 0. jr − ir jr + 1 i−1 Assume 1 ≤ s + 1 ≤ r. Notice that 0 ≤ =

r−1 jn − in + kn

jr − ir + kr jn+1 + kn+1 + 1 + jn − in − 1 jr − ir − 1 jn+1 n=s r js − is + ks jn + kn + 1 jn − in + kn − − . js − is − 1 jn − in − 1 jn n=s+1 −

Therefore r r js − is + ks jn + kn + 1 jn − in + kn − ≥ . js − is − 1 jn − in − 1 jn n=s+1 n=s+1 Let As denote the ﬁrst summation and Bs the second in this last inequality; clearly As − Bs > 0. Also note that ks + 1 ks + 1 > js − is js + 1

and

ks + 1 ks+1 + 1 . > js − is js+1 + 1

Putting these together, we compute ks + 1 js − is + ks ks + 1 Fs − Fs+1 = − js − is js + 1 js − is − 1 ks + 1 ks+1 + 1 ks+1 + 1 ks + 1 − − + As + Bs js + 1 js+1 + 1 js+1 + 1 js + 1 ks + 1 ks + 1 ks+1 + 1 ks + 1 − ≥ − (As − Bs ) + As js − is js + 1 js + 1 js+1 + 1 ks+1 + 1 ks + 1 − + Bs js+1 + 1 js + 1 ks + 1 ks+1 + 1 = − (As − Bs ) > 0. js − is js+1 + 1 Hence F0 > Fr > 0, which contradicts the observation that F0 < 0.

2

One might hope for even stronger estimates than suggested by the result above, for instance that μ(I) ≤ μ0 . Next we present a counterexample showing that Theorem 5.4.3 cannot be extended to bound the number of generators in all degrees.

194

Chapter 5

Example 5.4.4 If I = ((x21 , x42 , x33 , x44 ) : (x1 x2 − x3 x4 )), using Macaulay2 [199] one readily obtains that I is given by: (x21 , x1 x2 x3 + x23 x4 , x33 , x1 x23 , x42 , x44 , x1 x34 , x1 x2 x24 + x3 x34 , x32 x23 , x32 x34 ). Then R/I is a Gorenstein Artin algebra with h-vector (1, 4, 9, 13, 13, 9, 4, 1) and Betti sequence (1, 10, 18, 10, 1), whereas the h-vector for an extremal Gorenstein algebra of codimension four and initial degree two is (1, 4, 1) and the Betti sequence is (1, 9, 16, 9, 1); notice that the multiplicity e(R/I) = 54 is far greater than the minimal value of six exhibited by an extremal algebra. In height three one can bound the Betti numbers, thanks to the structure theorem of Buchsbaum and Eisenbud. For initial degree p, the extremal Gorenstein algebra of codimension three has b1 = b2 = 2p + 1. Theorem 5.4.5 Let R be a polynomial ring over a ﬁeld k and let I be a homogeneous Gorenstein ideal of height 3 and initial degree p. Then μ(I) ≤ 2p + 1 and b2 (R/I) ≤ 2p + 1. Proof. We may assume, without loss of generality, that R is equal to k[x1 , x2 , x3 ] and A = R/I is Artinian and local with socle(A) = Aσ . Then by [75] the minimal free resolution of A has the form 0 → R(−σ − 3) →

μ )

Y

R(−nj ) −→

j=1

μ )

R(−mi ) → R,

i=1

where Y is an alternating matrix, and the generators f1 , . . . , fμ of I are the maximal pfaﬃans of Y . If the theorem fails every generator has degree at least (2p + 1)/2 > p, which is a contradiction since at least one minimal generator has degree p. 2 Conjecture 5.4.6 (Miller–Villarreal) Let I ⊂ R be a graded Gorenstein ideal of initial degree p and height g. If 1 ≤ i < g and γi is the ith initial Betti number of R/I, then p+g−1 p+i−2 γi ≤ p+i−1 i−1 p+g−1 p+g−i−2 g p+g−2 + − . i p−1 i p−1 Proposition 5.4.7 ([39], [65, Lemma 4.2.13]) Let n be a given positive integer and f, g : N → N the numerical functions given by g(x) = n x

and

f (x) = x n ,

where h j is the Macaulay symbol. Then g(x) is non increasing and f (x) is strictly increasing.

Hilbert Series

195

There is a short and elegant argument, based on the behavior of the combinatorial functions g(x) = n x and f (x) = x n , to show the inequality of Theorem 5.4.3. We include the details of this argument in the proof below. Corollary 5.4.8 If R is a polynomial ring over a ﬁeld k and I is a graded Gorenstein ideal of R of height g ≥ 3 and initial degree p ≥ 2, then p+g−1 p+g−3 μ(Ip ) ≤ μ0 = − . g−1 g−1 Proof. First we make a change of coeﬃcients using the functor (·) ⊗k K, where K is an inﬁnite ﬁeld extension of k. Hence we may assume that k is inﬁnite. Let A be an Artinian reduction of S = R/I, that is, A = S/(h1 , . . . , hd ), where h = h1 , . . . , hd is a system of parameters of S with hi a form of degree 1 for all i. Note that one can write A = R/I, where R = R/(h) is a polynomial ring in g variables and I = IR is a graded Gorenstein ideal of R of codimension g and initial degree p. By Proposition 5.3.9 the Hilbert function of A is symmetric. Since the h-vector h(S) = (hi ) satisfy hi = dimk (Ai ) there is some j ≥ p so that h(S) = (h0 , h1 , . . . , hp−1 , hp , . . . , hj , hp−1 , . . . , h1 , h0 ), thus hj+1 = hp−1

p+g−2 = and hj = hp g−1

(1)

The idea of the argument is to use the Macaulay bound j

hj+1 ≤ hj ,

(2)

see Theorem 5.4.2. From the inequalities: p−2

p+g−2 p+g−3 (1) = hp−1 = = hj+1 p−1 p−2 j

(2) p+g−1 j (1) j

≤ hj = hp = − μ(Ip ) p p−2

use g(x) p+g−1 ≤ − μ(Ip ) p we conclude p−2

p−2 p+g−1 p+g−3 ≤ − μ(Ip ) . p p−2 Using that f (x) is non decreasing yields the required inequality.

2

196

Chapter 5

Exercises 5.4.9 Let I be a graded Gorenstein ideal of height 4 and initial degree p, prove μ(Ip ) ≤ (p + 1)2 . 5.4.10 Let I be a graded Gorenstein ideal of height 4 generated by forms of degree p and let R/J be an extremal algebra of the same codimension and initial degree. Prove that their Betti numbers satisfy bi (R/I) ≤ bi (R/J) for all i.

5.5

Koszul homology and Hilbert functions

The ﬁrst Koszul homology module of a graded Cohen–Macaulay ideal is studied here using the Cohen–Macaulay criterion of Herzog introduced in Section 4.4. Our approach makes use of Hilbert functions. Lemma 5.5.1 Let p and g be positive integers and let 2 2p + g p+g−1 −2 . ψ(p) = g g−1 If p ≥ 2 and g ≥ 6 then ψ(p) > 0. Proof. Notice the equality 6ψ(2) = (g + 1)(g 3 − 3g 2 − 13g − 12), which is certainly positive for g ≥ 6. We proceed by induction on p. Assume ψ(p) > 0. It is easy to check that ψ(p + 1) is greater than 2p + g (p + 1)(4(g − 2)p2 + (3g 2 − 3g − 8)p + g 2 − 3g − 2) 2 . g (p + 1)2 (2p + 1)(2p + 2) Since the right-hand side of this inequality is positive for p ≥ 2 and g ≥ 3, the induction step is complete. 2 Theorem 5.5.2 [350] Let R = ⊕∞ i=0 Ri be a polynomial ring with its usual graduation over an inﬁnite ﬁeld k and let I be a Cohen–Macaulay graded ideal of R of height g with a p-linear resolution 0 → Rbg (−(p+g−1)) → · · · → Rbk (−(p+k−1)) → · · · → Rb1 (−p) → I → 0. If I is generically a complete intersection with p ≥ 2 and g ≥ 3, then H1 (I), the ﬁrst Koszul homology module of I, is not Cohen–Macaulay. Proof. The module H1 = H1 (I) has a well-deﬁned rank equal to b1 − g. Thus, by Theorem 4.4.13, it suﬃces to prove that for some homogeneous system of parameters y of S = R/I, one has (H1 (I)/yH1 (I)) > rank(H1 (I)) · (S/yS).

Hilbert Series

197

We start by making a specialization to the case of a polynomial ring of dimension g. Since k is inﬁnite according to Theorem 3.1.24, there exists a system of parameters y = {y1 , . . . , yd } for S with each yi a form of degree one of R. We make two observations: (i) Because Tori (S, R/(y)) = 0 for i ≥ 1, it is clear that the minimal resolution of S = S/(yS) as an R (= R/(y))-module has the same twists and Betti numbers as the minimal resolution of S over R. (ii) With H1 Cohen–Macaulay it is easy to see that the ﬁrst Koszul homology module of I ⊗ R over R is precisely H1 (I) ⊗ R. We may then in the sequel assume that S is zero dimensional. We will compare the integer r=rank(original H1 (I))·(S), with partial Hilbert sums contributing to (H1 ). First notice (see Exercise 5.3.17) that the length of S and its Betti numbers can be calculated using: p+k−2 p+g−1 p+g−1 bk = · and (S) = . (5.12) k−1 g−k g From the Koszul complex we obtain the exact sequences 0 → B1 → Z1 → H1 → 0, b1

0 → Z2 → R( 2 ) (−2p) → B1 → 0, where Zi and Bi are the modules of cycles and boundaries deﬁning Hi (I). To simplify notation we set (M )i = H(M, i), the dimension of the ith component of the graded module M . One may write b1 (H1 )i = (Z1 )i − (B1 )i = (Z1 )i − (R(−2p))i + (Z2 )i , 2 and therefore 2p b1 (H1 ) ≥ (H1 )i ≥ (Z1 )i − (R(−2p))i . 2 i=0 i=0 i=0 2p

2p

(5.13)

From the minimal resolution of S we obtain (S)i = (R)i − b1 (R(−p))i + (Z1 )i ≥ 0. Hence (Z1 )i ≥ b1 (R(−p))i − (R)i and (Z1 )i = 0 for 0 ≤ i ≤ p, which leads to 2p 2p 2p (Z1 )i ≥ b1 (R(−p))i − (R)i . (5.14) i=0

i=p+1

i=p+1

Finally by (5.13) and (5.14) we have (H1 ) ≥ b1

2p i=p+1

(R(−p))i −

2p i=p+1

(R)i −

b1 . 2

(5.15)

198

Chapter 5

The proof now reduces to showing that the right-hand side of (5.15) is greater than rank(H1 (I)) · (S) = (b1 − g) · (S); that is, we must show that the following inequality holds for p ≥ 2 and g ≥ 3 b1 (R(−p))i − (R)i − f (p, g) = b1 − (b1 − g) · (S) > 0. 2 i=p+1 i=p+1 2p

2p

It is not hard to see that f (p, g) simpliﬁes to f (p, g) =

b1 + 1 p+g−1 2p + g + (1 + g) − . 2 p−1 g

Observe that from this equality we obtain the inequality: 2f (p, g) >

2 p+g−1 2p + g −2 . p g

It is easy to check that f (p, g) > 0 for g ∈ {3, 4, 5} and p ≥ 2. The required inequality is now a direct consequence of Lemma 5.5.1. 2 The result above complements one of Ulrich [404] in which the twisted conormal module ωS ⊗R I = ωS ⊗R I/I 2 was shown not to be Cohen– Macaulay; either result implies that I is not in the linkage class of a complete intersection. Corollary 5.5.3 Let I be a graded Cohen–Macaulay ideal of height g ≥ 3 in a polynomial ring R. If I is generically a complete intersection and has a p-linear resolution with p ≥ 2, then I is not in the linkage class of a complete intersection. Proof. If I is in the linkage class of a complete intersection, then I is a strongly Cohen–Macaulay ideal, a contradiction because H1 (I) is not Cohen–Macaulay. See [253] and [412, Corollary 4.2.4]. 2

Exercises 5.5.4 Let I be a graded ideal of a polynomial ring R over a ﬁeld K with a pure resolution: 0 → R3 (−15) → R27 (−10) → R25 (−9) → I → 0. Prove that H1 (I) is not Cohen–Macaulay.

Hilbert Series

5.6

199

Hilbert functions of some graded ideals

This section is about the behavior of the Hilbert function in some special cases of interest in algebraic coding theory [349, 348] (see Section 8.6). Throughout this section R = K[x1 , . . . , xn ] = ⊕∞ i=0 Ri is a polynomial ring over a ﬁeld K with the standard grading induced by setting deg(xi ) = 1 for i = 1, . . . , n and m = (x1 , . . . , xn ) is the irrelevant maximal ideal of R. Proposition 5.6.1 Let I be a graded ideal of R. If K is inﬁnite and m is not in Ass(R/I), then there is h1 ∈ R1 such that h1 ∈ / Z(R/I). Proof. Let p1 , . . . , pm be the associated primes of R/I. As R/I is graded, p1 , . . . , pm are graded ideals by Lemma 2.2.6. We proceed by contradiction. Assume that R1 , the degree 1 part of R, is contained in Z(R/I). Thanks to Lemma 2.1.19, one has that Z(R/I) = ∪m i=1 pi . Hence R1 ⊂ (p1 )1 ∪ (p2 )1 ∪ · · · ∪ (pm )1 ⊂ R1 , where (pi )1 is the homogeneous part of degree 1 of the graded ideal pi . Since K is inﬁnite, from Lemma 3.1.22, we get R1 = (pi )1 for some i. Hence, pi = m, a contradiction. 2 The hypothesis that m ∈ / Ass (R/I) is equivalent to require that R/I has positive depth. This follows from Lemma 2.1.19 (see Exercise 2.2.11). Theorem 5.6.2 Let I be a graded ideal of R. If depth(R/I) > 0, and HI is the Hilbert function of R/I, then HI (i) ≤ HI (i + 1) for i ≥ 0. Proof. Case (I): If K is inﬁnite, by Proposition 5.6.1, there exists h ∈ R1 a non-zero divisor of R/I. The homomorphism of K-vector spaces (R/I)i −→ (R/I)i+1 ,

z → hz

is injective, therefore HI (i) = dimK (R/I)i ≤ dimK (R/I)i+1 = HI (i + 1). Case (II): If K is ﬁnite, consider the algebraic closure K of K. We set R = R ⊗K K,

I = IR.

Hence, from [392, Lemma 1.1], one has that HI (i) = HI (i). This means that the Hilbert function does not change when the base ﬁeld is extended 2 from K to K. Applying Case (I) to HI (i) we obtain the result. Lemma 5.6.3 Let I be a graded ideal of R. The following hold. (a) If Ri = Ii for some i, then R = I for all ≥ i. (b) If dim R/I ≥ 2, then dimK (R/I)i > 0 for i ≥ 0.

200

Chapter 5

Proof. (a): It suﬃces to prove the case = i + 1. As Ii+1 ⊂ Ri+1 , we need only show Ri+1 ⊂ Ii+1 . Take a non-zero monomial xa in Ri+1 . Then, xa = xa1 1 · · · xann with aj > 0 for some j. Thus, xa ∈ R1 Ri . As R1 Ii ⊂ Ii+1 , we get xa ∈ Ii+1 . (b): If dimK (R/I)i = 0 for some i, then Ri = Ii . Thus, by (a), HI (j) vanishes for j ≥ i, a contradiction because the Hilbert polynomial of R/I has degree dim(R/I) − 1 ≥ 1; see Theorem 2.2.4. 2 Theorem 5.6.4 [178] Let I be a graded ideal with depth(R/I) > 0. (i) If dim(R/I) ≥ 2, then HI (i) < HI (i + 1) for i ≥ 0. (ii) If dim(R/I) = 1, then there is an integer r and a constant c such that 1 = HI (0) < HI (1) < · · · < HI (r − 1) < HI (i) = c

for i ≥ r.

Proof. Consider the algebraic closure K of K. Notice that |K| = ∞. As in the proof of Theorem 5.6.2, we make a change of coeﬃcients using the functor (·) ⊗K K. Hence we may assume that K is inﬁnite. By Proposition 5.6.1, there is h ∈ R1 a non-zero divisor of R/I. From the exact sequence h

0 −→ (R/I)[−1] −→ R/I −→ R/(h, I) −→ 0, we get HI (i + 1) − HI (i) = HS (i + 1), where S = R/(h, I). (i): If HS (i + 1) = 0, then, by Lemma 5.6.3, dimK (S) < ∞. Hence S is Artinian (see Exercise 3.1.44). Thus dim(S) = 0, a contradiction. (ii): Let r ≥ 0 be the ﬁrst integer such that HI (r) = HI (r + 1), thus Sr+1 = (0) and Rr+1 = (h, I)r+1 . Then, by Lemma 5.6.3, Sk = (0) for k ≥ r + 1. Hence, the Hilbert function of R/I is constant for k ≥ r and strictly increasing on [0, r − 1]. 2

Exercises 5.6.5 Let Pn−1 be a projective space over a ﬁeld K and let [P1 ], . . . , [Pm ] be a set of distinct points in Pn−1 , then there exists F ∈ K[x1 , . . . , xn ] a form of degree m − 1 such that F (P1 ) = 0 and F (Pi ) = 0 for 2 ≤ i ≤ m. 5.6.6 Let X be a ﬁnite set in a projective space Pn−1 over a ﬁeld K. If I = I(X) ⊂ R is the vanishing ideal of X and HI is the Hilbert function of R/I(X), then HI (i) = |X| for all i ≥ |X| − 1.

Chapter 6

Stanley–Reisner Rings and Edge Ideals of Clutters This chapter is intended as an introduction to the study of combinatorial commutative algebra, some of the topics were chosen in order to motivate the subject. Part of our exposition was inspired by [41, 393]. An understated goal here is to highlight some of the works of T. Hibi, J. Herzog, M. Hochster, G. Reisner, and R. Stanley. This chapter deals with monomial ideals, Stanley–Reisner rings and edge ideals of clutters. We study algebraic and combinatorial properties of edge ideals of clutters and simplicial complexes. In particular we examine shellable, unmixed, and sequentially Cohen–Macaulay simplicial complexes and their corresponding edge ideals and invariants (projective dimension, regularity, depth, Hilbert series). As applications, the proofs of the upper bound conjectures for convex polytopes and simplicial spheres are discussed in order to give a ﬂavor of some of the methods and ideas of the area.

6.1

Primary decomposition

General monomial ideals will be examined ﬁrst, subsequently we specialize to the case of square-free monomial ideals and edge ideals of clutters, and present some of their relevant properties. Let R = K[x] = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K. To make notation simpler, we will use the following multi-index notation: xa := xa1 1 · · · xann for a = (a1 , . . . , an ) ∈ Nn .

202

Chapter 6

Deﬁnition 6.1.1 An ideal I of R is called a monomial ideal if there is A ⊂ Nn+ such that I is generated by {xa |a ∈ A}. If I is a monomial ideal the quotient ring R/I is called a monomial ring. Note that, by Dickson’s lemma (see Lemma 3.3.3), a monomial ideal I is always minimally generated by a unique ﬁnite set of monomials. This unique set of generators of I is denoted by G(I). Deﬁnition 6.1.2 A face ideal is an ideal p of R generated by a subset of the set of variables, that is, p = (xi1 , . . . , xik ) for some variables xij . Deﬁnition 6.1.3 A monomial f in R is called square-free if f = xi1 . . . xir for some 1 ≤ i1 < · · · < ir ≤ n. A square-free monomial ideal is an ideal generated by square-free monomials. Any square-free monomial ideal is a ﬁnite intersection of face ideals: Theorem 6.1.4 Let I ⊂ R be a monomial ideal. The following hold. (i) Every associated prime of I is a face ideal. (ii) If I is square-free, then I = ∩si=1 pi , where p1 , . . . , ps are the associated primes of I. In particular pi is a minimal prime of I for all i. Proof. (i): By induction on the number of variables that occur in G(I). Set m = (x1 , . . . , xn ). Let p be an associated prime of I. If rad(I) = m, then p = m. Hence we may assume rad(I) = m. Pick a variable x1 not in rad(I) and consider the ascending chain of ideals I0 = I and Ii+1 = (Ii : x1 )

(i ≥ 0).

Since R is Noetherian, one has Ik = (Ik : x1 ) for some k. There are two cases to consider. If p is an associated prime of (Ii , x1 ) for some i, then by induction p is a face ideal because one can write (Ii , x1 ) = (Ii , x1 ), where Ii is an ideal minimally generated by a ﬁnite set of monomials in the variables x2 , . . . , xn (cf. Exercise 6.1.26). Assume we are in the opposite case. By Lemma 5.1.3 for each i there is an exact sequence x

1 0 −→ R/(Ii : x1 ) −→ R/Ii −→ R/(Ii , x1 ) −→ 0,

hence making a recursive application of Lemma 2.1.17 one obtains that p is an associated prime of Ii for all i. Since x1 is regular on R/Ik one concludes that Ik is an ideal minimally generated by monomials in the variables x2 , . . . , xn , thus by induction p is a face ideal. (ii): We only have to check ∩si=1 pi ⊂ I because I is contained in any of its associated primes. Take a monomial f in ∩si=1 pi and write f = xai11 · · · xairr , where i1 < · · · < ir and ai > 0 for all i. By Corollary 2.1.29, f k ∈ I for

Stanley–Reisner Rings and Edge Ideals of Clutters

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some k ≥ 1. Then, using that I is generated by square-free monomials, we obtain xi1 · · · xir ∈ I. Hence f ∈ I. To ﬁnish the proof observe that, by part (i), ∩si=1 pi is a monomial ideal because the intersection of monomial ideals is again a monomial ideal (see Exercise 6.1.22). 2 Deﬁnition 6.1.5 Let xa = xa1 1 · · · xann be a monomial in R. The support of xa is given by supp(xa ) := {xi | ai > 0}. Proposition 6.1.6 [169] Let I ⊂ R be a monomial ideal. If S = R/I, then there is a polynomial ring R and a square-free monomial ideal I of R such that S = S /(h), where S = R /I and h is a regular sequence on S of forms of degree one. Proof. Let G(I) = {f1 , . . . , fr } be the set of monomials that minimally generate I. Assume that one of the variables, say x1 , occurs in at least one of the monomials in G(I) with multiplicity greater than 1. We may assume that the elements of G(I) can be written as f1 = xa1 1 g1 , . . . , fs = xa1 s gs , / supp(gi ) for all i ≤ s and x1 ∈ / supp(fi ) where ai ≥ 1 for all i, a1 ≥ 2, x1 ∈ for i > s. We set I = (x0 xa1 1 −1 g1 , . . . , x0 xa1 s −1 gs , fs+1 , . . . , fr ) ⊂ R = R[x0 ], where x0 is a new variable. We claim that x0 − x1 is a non-zero divisor of S = R /I . On the contrary assume that x0 − x1 belongs to an associated prime p of I and write p = (I : h), for some monomial h. Since p is / I gives a face ideal, one has xi h ∈ I for i = 0, 1. Hence using h ∈ ai −1 ai −2 x1 h = x0 x1 gi h1 for some i and consequently h = x0 x1 gi h1 . Hence x0 h = x20 xa1 i −2 gi h1 =

a −1

x0 x1 j fj M

gj M

or, for some j > s.

In both cases one obtains x0 ∈ supp(M ) and h ∈ I , a contradiction. Since S /(x0 − x1 ) = S one can repeat the construction to obtain the asserted monomial ideal I . This proof is due to R. Fr¨oberg. 2 The ideal I constructed above is called the polarization of I. Thus any monomial ring is a deformation by linear forms of a monomial ring with square-free relations. Note that I is Cohen–Macaulay (resp. Gorenstein) iﬀ I is Cohen–Macaulay (resp. Gorenstein).

204

Chapter 6

Proposition 6.1.7 A monomial ideal q ⊂ R is primary if and only if, after permutation of the variables, q has the form: q = (xa1 1 , . . . , xar r , xb1 , . . . , xbs ), where ai ≥ 1 and ∪si=1 supp(xbi ) ⊂ {x1 , . . . , xr }. Proof. If Ass(R/q) = {p}, then by permuting the variables x1 , . . . , xn and using Theorem 6.1.4 one may assume that p is equal to (x1 , . . . , xr ). Since rad (q) = p, the ideal q is minimally generated by a set of the form: {xa1 1 , . . . , xar r , xb1 , . . . , xbs }. Let xj ∈ supp(xbi ), then xbi = xj xc , where xc is a monomial not in q. Since q is primary, a power of xj is in q. Thus xj ∈ (x1 , . . . , xr ) and consequently 1 ≤ j ≤ r, as required. For the converse note that any associated prime p of R/q can be written as p = (q : f ), for some monomial f . It follows readily that (x1 , . . . , xr ) is the only associated prime of q. 2 Corollary 6.1.8 If p is a face ideal, then pn is a primary ideal for all n. Proposition 6.1.9 If I ⊂ R is a monomial ideal, then I has an irredundant primary decomposition I = q1 ∩ · · · ∩ qr , where qi is a primary monomial ideal for all i and rad (qi ) = rad (qj ) if i = j. Proof. Let G(I) = {f1 , . . . , fq } be the set of monomials that minimally generate I. We proceed by induction on the number of variables that occur in the union of the supports of f1 , . . . , fq . One may assume that one of the variables in ∪qi=1 supp(fi ), say xn , satisfy i xn ∈ / I for all i, otherwise I is a primary ideal and there is nothing to prove. Next we permute the fi in order to ﬁnd integers 0 ≤ a1 ≤ · · · ≤ aq , with aq ≥ 1, such that fi is divisible by xani but by not higher power of xn . If we a apply this procedure to (I, xnq ), instead of I, note that one must choose a variable diﬀerent from xn . a Since (I : xnq ) is generated by monomials in less than n variables and because of the equality I = (I, xanq ) ∩ (I : xanq ) one may apply the argument above recursively to the two monomial ideals occurring in the intersection— and use induction—to obtain a decomposition of I into primary monomial ideals q1 , . . . , qs of R. Finally we remove redundant primary ideals from q1 , . . . , qs and group those primary ideals with the same radical. 2 Even for monomial ideals a minimal irredundant primary decomposition is not unique. What is unique is the number of terms in such a decomposition and the primary components that correspond to minimal primes [9].

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Example 6.1.10 If I = (x2 , xy) ⊂ K[x, y], then I = (x) ∩ (x2 , xy, y 2 ) = (x) ∩ (x2 , y), are two minimal irredundant primary decompositions of I. The computation of a primary decomposition of a monomial ideal can be carried out by successive elimination of powers of variables, as described in the proof of Proposition 6.1.9. Now we illustrate this procedure with a speciﬁc ideal. Example 6.1.11 If R = K[x, y, z] and I = (yz 2 , x2 z, x3 y 2 ). 3 2 * (I : x ) = (z, y ) I HH j J = (I, x3 ) = (x3 , zx2 , z 2 y) H

2 2 * (J : z ) = (x , y) H H j (J, z 2 ) = (z 2 , x3 , x2 z). H

Thus I = (z, y 2 ) ∩ (x2 , y) ∩ (z 2 , x3 , x2 z). Corollary 6.1.12 If I ⊂ R is a monomial ideal, then there is a primary decomposition I = q1 ∩ · · · ∩ qm , such that qi is generated by powers of variables for all i. Proof. Let q be a primary ideal. Then, by Proposition 6.1.7, q is minimally generated by a set of monomials xa1 1 , . . . , xar r , f1 , . . . , fs such that s (

supp(fi ) ⊂ {x1 , . . . , xr },

i=1

where ai > 0 for all i. Note that if f1 = xb11 · · · xbrr and b1 > 0, then a1 > b1 and one has a decomposition: q = (xb11 , xa2 2 , . . . , xar r , f2 , . . . , fs ) ∩ (xa1 1 , xa2 2 , . . . , xar r , xb22 · · · xbrr , f2 , . . . , fs ), where in the ﬁrst ideal of the intersection we have lower the degree of xa1 1 and have eliminated f1 , while in the second ideal we have eliminated the variable x1 from f1 . Applying the same argument repeatedly it follows that one can write q as an intersection of primary monomial ideals such that for each of those ideals the only minimal generators that contain x1 are pure powers of x1 . Therefore, by induction, q is the intersection of ideals generated by powers of variables. Hence the result follows from Proposition 6.1.9. 2 Proposition 6.1.13 Let I be an ideal of R = K[x1 , . . . , xn ] generated by monomials in the variables x1 , . . . , xr with r < n. If I = I1 ∩ · · · ∩ Is is an irredundant decomposition of I into monomial ideals, then none of the ideals Ii can contain a monomial in K[xr+1 , . . . , xn ].

206

Chapter 6

Proof. Set X = {x1 , . . . , xn } and X = X \ {x1 , . . . , xr }. Assume some of the Ii ’s contain monomials in K[X ]. One may split the Ii ’s into two sets so that I1 , . . . , Im do not contain monomials in K[X ] (note m could be zero), while Im+1 , . . . , Is contain monomials in the set of variables X . For i ≥ m + 1 pick a monomial gi in Ii whose support is contained in X . Since the decomposition of I is irredundant there is a monomial f ∈ ∩m i=1 Ii and f ∈ ∩si=m+1 Ii , where we set f = 1 if m = 0. To derive a contradiction consider f1 = f gm+1 · · · gs . As f1 ∈ I, we get f ∈ I which is absurd. 2 Corollary 6.1.14 Let I ⊂ R = K[x1 , . . . , xn ] be an ideal generated by monomials in the variables x1 , . . . , xr with r < n. If I = q1 ∩ · · · ∩ qs is an irredundant primary decomposition into monomial ideals, then qi is generated by monomials in K[x1 , . . . , xr ]. Proof. It follows from Propositions 6.1.7 and 6.1.13.

2

Deﬁnition 6.1.15 An ideal I of a ring R is called irreducible if I cannot be written as an intersection of two ideals of R that properly contain I. Proposition 6.1.16 If I ⊂ R = K[x1 , . . . , xn ] is a monomial ideal, then I is irreducible if and only if up to permutation of the variables I = (xa1 1 , . . . , xar r ), where ai > 0 for all i. Proof. ⇒) Since I must be primary (see Proposition 2.1.24) from the proof of Corollary 6.1.12 one derives that I is generated by powers of variables. ⇐) If I is reducible, then I = I1 ∩ I2 for some ideals I1 and I2 that properly contain I. Pick f ∈ I1 \ I (resp. g ∈ I2 \ I) with the smallest possible number of terms. We can write f = λ1 xγ1 + · · · + λs xγs ;

0 = λi ∈ K for all i.

Let 1 ≤ k ≤ r be an integer and let xbki be the maximum power of xk that divides xγi , i.e., we can write xγi = xbki xδi , where xk does not divide xδi . After permuting terms we may assume that b1 ≥ · · · ≥ bs . Note that xakk does not divide xγi for all i, k, otherwise we can ﬁnd a polynomial in I1 \ I, namely f − λi xγi , with less than s terms. Thus bi < ak for all i. We claim that b1 = · · · = bs . We proceed by contradiction assuming that bp > bp+1 for some p. From the equality a −bp

xkk

a −bp

f = xkk

a −bp

(λ1 xγ1 + · · · + λp xγp ) + xkk

(λp+1 xγp+1 + · · · + λs xγs )

Stanley–Reisner Rings and Edge Ideals of Clutters

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a −b

we obtain that the polynomial xkk p (λp+1 xγp+1 + · · · + λs xγs ) is in I1 \ I and has fewer terms than f , a contradiction. This completes the proof of the claim. Therefore we can write f = xc11 · · · xcrr f1 and g = xd11 · · · xdr r g1 , where f1 and f2 are in R = K[xr+1 , . . . , xn ]. Setting ei = max{ci , di } we get that h = xe11 · · · xerr f1 g1 is in I1 ∩ I2 , i.e., h ∈ I, a contradiction because 2 ei < ai for all i and f1 g1 ∈ R . Thus I is irreducible. Theorem 6.1.17 If I ⊂ R is a monomial ideal, then there is a unique irredundant decomposition I = q1 ∩ · · · ∩ qr such that qi is an irreducible monomial ideal. Proof. The existence follows from Corollary 6.1.12. For the uniqueness assume one has two irredundant decompositions: q1 ∩ · · · ∩ qr = q1 ∩ · · · ∩ qs , where qi and qj are irreducible for all i, j. Using the arguments given in the proof of Proposition 6.1.16 one concludes that for each i, there is σi such that qσi ⊂ qi and vice versa for each j there is πj such that qπj ⊂ qj . 2 Therefore r = s and qi = qρi for some permutation ρ. For further information on the primary decomposition of more general monomial ideals, e.g., monomial ideals obtained from a regular sequence or with coeﬃcients in a ring other than a ﬁeld, consult [122, 213, 214]. Example 6.1.18 Let I = (xy 2 , x2 y) ⊂ K[x, y] and V (I) the aﬃne variety deﬁned by I. From the primary decomposition I = (x) ∩ (y) ∩ (x2 , y 2 ) one has V (I) = {(0, 0)} ∪ V (x) ∪ V (y) = V (x) ∪ V (y) ⊂ A2K , where (x2 , y 2 ) corresponds to {(0, 0)} which is embedded in V (x) ∪ V (y). For this reason (x2 , y 2 ) is said to be an embedded primary component. If K is inﬁnite, then the coordinate axes V (x) and V (y) are the irreducible components of V (I). Monomial ideals form a lattice Let L be a lattice, that is, L is a poset in which any two elements x, y have a greatest lower bound or a meet x ∧ y and a lowest upper bound or a join x ∨ y. The lattice L is distributive if (a) x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z), and (b) x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z), ∀ x, y, z in L. Proposition 6.1.19 If L is the family of monomial ideals of R, order by inclusion, then L is a distributive lattice under the operations I ∧ J = I ∩ J and I ∨ J = I + J. Proof. It follows from Exercises 6.1.22 and 6.1.23.

2

208

Chapter 6

Exercises 6.1.20 If I is a monomial ideal of R, then any associated prime p of R/I can be written as p = (I : f ), for some monomial f . 6.1.21 If L is a lattice, then the following conditions are equivalent: (a) x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z), ∀ x, y, z in L. (b) x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z), ∀ x, y, z in L. 6.1.22 Let I and J be two ideals generated by ﬁnite sets of monomials F and G, respectively, prove that the intersection I ∩ J is generated by the set {lcm(f, g)| f ∈ F and g ∈ G}. 6.1.23 If I, J, L are monomial ideals, prove the equalities I ∩ (J + L) =

(I ∩ J) + (I ∩ L),

I + (J ∩ L) =

(I + J) ∩ (I + L).

6.1.24 If I and J are two monomial ideals, then (I : J) is a monomial ideal. 6.1.25 If q1 , . . . , qr are primary monomial ideals of R with non-comparable radicals and I is an ideal such that I = q1 ∩ · · · ∩ qr , then I (n) = qn1 ∩ · · · ∩ qnr . Hint Proceed as in the proof of Proposition 4.3.24 (cf. [92, Theorem 3.7]). 6.1.26 Let R = k[x2 , . . . , xn ] and R = R [x1 ] be polynomial rings over a ﬁeld k. If I is an ideal of R and p ∈ AssR (R/(I , x1 )), then (a) p = x1 R + p R, where p is a prime ideal of R , and (b) p is an associated prime of R /I . 6.1.27 If I = (x31 , x22 x23 , x1 x22 x3 ) ⊂ K[x1 , x2 , x3 ], show that I = q1 ∩ q2 is a primary decomposition of I, where q1 = (x31 , x22 ) and q2 = (x31 , x1 x3 , x23 ). Prove that I 2 = I (2) and I 3 = I (3) . Hint If z = x31 x42 x33 and a = x1 + x22 + x3 , then az ∈ I 3 and z ∈ I 3 . 6.1.28 If I = (x1 x22 , x21 x2 , x1 x23 , x21 x3 , x2 x23 , x22 x3 , x1 x2 x3 ), ﬁnd a primary decomposition of I. Notice that (x1 , x2 , x3 ) is an associated prime of I. 6.1.29 Show that I = (x21 , x1 x2 ) is a non-primary ideal with a prime radical. 6.1.30 If n ≥ 5 prove that the number of monomials of degree n in n variables whose support is at least three is given by 2n − 1 n μ(n) = − n − (n − 1) . n−1 2

Stanley–Reisner Rings and Edge Ideals of Clutters

6.2

209

Simplicial complexes and homology

A ﬁnite simplicial complex consists of a ﬁnite set V of vertices and a collection Δ of subsets of V called faces or simplices such that (i) If v ∈ V , then {v} ∈ Δ. (ii) If F ∈ Δ and G ⊂ F , then G ∈ Δ. Let Δ be a simplicial complex and F a face of Δ. Deﬁne the dimensions of F and Δ by dim(F ) = |F | − 1 and dim(Δ) = sup{dim(F )| F ∈ Δ} respectively. A face of dimension q is sometimes referred to as a q-face or as a q-simplex. Example 6.2.1 A triangulation of a disk with a whisker attached. x1 s A

x3 s A A A A A A A As As x4 x2

0-simplices: x1 , x2 , x3 , x4 1-simplices: x1 x2 , x2 x3 , x3 x4 , x2 x4 2-simplices: x2 x3 x4

Let F be a q-simplex in Δ with vertices v0 , v1 , . . . , vq . We say that two total orderings of the vertices vi0 < · · · < viq and vj0 < · · · < vjq are equivalent if (i0 , . . . , iq ) is an even permutation of (j0 , . . . , jq ). This is an equivalence relation because the set of even permutations forms a group, and for q > 1, it partitions the total orderings of v0 , . . . , vq into two equivalence classes. An oriented q-simplex of Δ is a q-simplex F with a choice of one of these equivalence classes. If v0 , . . . , vq are the vertices of F , the oriented simplex determined by the ordering v0 < · · · < vq will be denoted by [v0 , . . . , vq ]. Suppose A is a commutative ring with unit. Let Cq (Δ) be the free Amodule with basis consisting of the oriented q-simplices in Δ, modulo the relations [v0 , v1 , v2 , . . . , vq ] + [v1 , v0 , v2 , . . . , vq ]. In particular Cq (Δ) is deﬁned for any ﬁeld K and dimK Cq (Δ) equals the number of q-simplices of Δ. For q ≥ 1 we deﬁne the homomorphism ∂q : Cq (Δ) −→ Cq−1 (Δ)

210

Chapter 6

induced by ∂q ([v0 , v1 , . . . , vq ]) =

q

(−1)i [v0 , v1 , . . . , vi−1 , ' vi , vi+1 , . . . , vq ],

i=0

where v'i means that the symbol vi is to be deleted. Since ∂q ∂q+1 = 0 we obtain the chain complex C(Δ) = {Cq (Δ), ∂q }, which is called the oriented chain complex of Δ. The augmented oriented chain complex of Δ is the complex ∂

d 0 −→ Cd (Δ) −→ Cd−1 (Δ) −→ · · · −→ C0 (Δ) −→ A −→ 0,

where d = dim(Δ) and (v) = 1 for every vertex v of Δ. This chain complex is denoted by C (Δ) and we set ∂0 = and C−1 (Δ) = A. Setting Zq (Δ, A) = ker(∂q ), Bq = im(∂q+1 ) and . q (Δ; A) = Zq (Δ, A)/Bq (Δ, A), for q ≥ 0, H the elements of Zq (Δ, A) and Bq (Δ, A) are called cycles and boundaries, . q (Δ; A) is the qth reduced simplicial homology group of respectively, and H Δ with coeﬃcients in A. . i (Δ; A) = 0 for i < 0. Remark 6.2.2 Note that if Δ = ∅, then H Let C (Δ) be the augmented chain complex of Δ over the ring A. The q-reduced singular cohomology group with coeﬃcients in A is deﬁned as . q (Δ; A) = H . q (HomA (C (Δ), A)). H If K is a ﬁeld, there are canonical isomorphisms . q (Δ; K) ∼ . q (Δ, K), K) and H = HomK (H . q (Δ; K) ∼ . q (Δ, K), K). H = HomK (H . q (Δ; K) ∼ . q (Δ; K). Thus, in particular, we have H =H . 0 (Δ; A) Proposition 6.2.3 If Δ is a non-empty simplicial complex, then H is a free A-module of rank r − 1, where r is the number of connected components of Δ. Proof. Let V = {x1 , . . . , xn } be the vertex set of Δ and let Δ1 , . . . , Δr be its connected components. Denote the vertex set of Δi by Vi and pick a vertex in each one of the Vi ; one may assume xi ∈ Vi for i = 1, . . . , r. Set βi = xi − xr and β i = βi + im(∂1 ) for i = 1, . . . , r − 1. We claim that . 0 (Δ; A) = ker(∂0 )/im(∂1 ). the set B = {β 1 , . . . , β r−1 } is an A-basis for H

Stanley–Reisner Rings and Edge Ideals of Clutters

211

r−1 First we prove that B is linearly independent. Assume i=1 ai βi is in im(∂1 ) for some ai in A, by making an appropriate grouping of terms one can write r−1

a i βi =

i=1

bij (xi − xj ) + · · · +

xi ,xj ∈V1

bij (xi − xj ),

xi ,xj ∈Vr

for some bij in A. Since C0 (Δ) is a free A-module with basis V and the Vi are mutually disjoint one has ak xk = bij (xi − xj ) for 1 ≤ k ≤ r − 1, xi ,xj ∈Vk

applying ∂0 to both sides of this equality we obtain ak = 0 for all k. Next we show that B is a set of generators. Let xi1 and xj1 be two vertices of Δ, it suﬃces to prove that xi1 − xj1 + im(∂1 ) is in the submodule generated by B because ker(∂0 ) is generated by the elements of the form xi − xj . Assume xi1 ∈ Vm and xj1 ∈ Vs . There is a path xi1 , . . . , xit , xm in Δm and a path xj1 , . . . , xjp , xs in Δs such that every two consecutive vertices of those paths form a 1-face of Δ. From the equalities (xi1 − xi2 ) + (xi2 − xi3 ) + · · · + (xit − xm ) = (xj1 − xj2 ) + (xj2 − xj3 ) + · · · + (xjp − xs ) =

xi1 − xm xj1 − xs ,

one concludes xi1 − xj1 + im(∂1 ) = (xm − xr ) − (xs − xr ) + im(∂1 ).

2

Deﬁnition 6.2.4 The reduced Euler characteristic χ .(Δ) of a simplicial complex Δ of dimension d is given by χ .(Δ) = −1 +

d

(−1)i fi = −1 + χ(Δ),

i=0

where fi is the number of i-faces in Δ and χ(Δ) is its Euler characteristic.

Exercises 6.2.5 Let Δ be a simplicial complex of dimension d and fi the number of i-faces in Δ. If K is a ﬁeld, then d

. i (Δ; K) = −1 + (−1)i rank H

i=−1

d

(−1)i fi .

i=0

6.2.6 Let Δ be a simplicial complex, prove that there is an exact sequence . 0 (Δ; A) −→ H0 (Δ; A) −→ A −→ 0. 0 −→ H . 0 (Δ, A) ⊕ A. Then, show H0 (Δ; A) H

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6.2.7 Let Δ be a simplicial complex with vertex set V = {x1 , . . . , xn } and ∂

∂

1 0 · · · −→ C0 (Δ) −→ C−1 (Δ) = A −→ 0,

the right end of the augmented chain complex over a ring A. Prove that ker(∂0 ) is generated by the cycles of the form xi − xj . 6.2.8 Let Δ be a simplicial complex and fi the number of i-faces of Δ. If ∂

∂

1 0 C0 (Δ) −→ C−1 (Δ) = K −→ 0, · · · −→ C1 (Δ) −→

is the augmented chain complex of Δ over a ﬁeld K, then dimK (Z1 ) is equal to f1 − f0 + r; where r is the number of connected components of Δ and Z1 = ker(∂1 ).

6.3

Stanley–Reisner rings

In this section we study algebraic and combinatorial properties of edge ideals of clutters and simplicial complexes. In particular we examine shellable, unmixed, and sequentially Cohen–Macaulay simplicial complexes and their corresponding edge ideals. Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K. If I is an ideal of R generated by square-free monomials, the Stanley–Reisner simplicial / I} and its faces complex ΔI associated to I has vertex set V = {xi | xi ∈ are deﬁned by ΔI = {{xi1 , . . . , xik }| i1 < · · · < ik , xi1 · · · xik ∈ I}. Conversely if Δ is a simplicial complex with vertex set V contained in {x1 , . . . , xn }, the Stanley–Reisner ideal IΔ is deﬁned as IΔ = ({xi1 · · · xir | i1 < · · · < ir , {xi1 , . . . , xir } ∈ / Δ}) , and its Stanley–Reisner ring K[Δ] is deﬁned as the quotient ring R/IΔ . Deﬁnition 6.3.1 Let I ⊂ R be a square-free monomial ideal. The quotient ring R/I is called a face ring. Example 6.3.2 A simplicial complex Δ of dimension 1 and its Stanley– Reisner ideal: sx2 @ @ @sx3 x1 s IΔ = (x1 x3 , x1 x2 x4 , x2 x3 x4 ) @ @ @s x4

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Deﬁnition 6.3.3 A face F of a simplicial complex Δ is said to be a facet if F is not properly contained in any other face of Δ. Proposition 6.3.4 If Δ is a simplicial complex with vertices x1 , . . . , xn , then the primary decomposition of the Stanley–Reisner ideal of Δ is: pF , IΔ = F

where the intersection is taken over all facets F of Δ, and pF denotes the / F. face ideal generated by all xi such that xi ∈ Proof. Let F be a face of Δ and pF the face ideal generated by the xi such that xi ∈ / F . Note that pF is a minimal prime of IΔ if and only if F is a facet. Therefore the result follows from Theorem 6.1.4. 2 Corollary 6.3.5 Let Δ be a simplicial complex of dimension d on the vertex set V = {x1 , . . . , xn } and let K be a ﬁeld. Then dim K[Δ] = d + 1 = max{s | xi1 · · · xis ∈ / IΔ and i1 < · · · < is }. Proof. Let F be a facet of Δ with d + 1 vertices. By Proposition 6.3.4 the face ideal pF generated by the variables not in F has height equal to the height of IΔ . Hence ht(IΔ ) = n − d − 1, and thanks to Proposition 3.1.16 one has dim K[Δ] = d + 1. 2 Deﬁnition 6.3.6 Let K be a ﬁeld. A simplicial complex Δ is said to be Cohen–Macaulay over K if the Stanley–Reisner ring K[Δ] is a Cohen– Macaulay ring. Corollary 6.3.7 A Cohen–Macaulay simplicial complex Δ is pure; that is, all its maximal faces have the same dimension. Proof. It follows from Proposition 6.3.4 and Corollary 3.1.17.

2

Deﬁnition 6.3.8 A pure d-dimensional complex Δ is called strongly connected or connected in codimension 1 if each pair of facets F, G can be connected by a sequence of facets F = F0 , F1 , . . . , Fs = G, such that dim(Fi−1 ∩ Fi ) = d − 1 for 1 ≤ i ≤ s. Theorem 6.3.9 [43, Proposition 11.7] Every Cohen–Macaulay complex is strongly connected. Deﬁnition 6.3.10 Let Δ be a simplicial complex. For F ∈ Δ, deﬁne lk(F ), the link of F , as lk(F ) := {H ∈ Δ| H ∩ F = ∅ and H ∪ F ∈ Δ}.

214

Chapter 6

Theorem 6.3.11 (Hochster [65, Theorem 5.3.8]) Let Δ be a simplicial complex. If K is a ﬁeld, then the Hilbert series of the local cohomology modules of K[Δ] with respect to the ﬁne grading is given by p (K[Δ]), t) = F (Hm

. p−|F |−1 (lk F ; K) dimK H

F ∈Δ

7

t−1 j

xj ∈F

1 − t−1 j

.

Theorem 6.3.12 (Reisner [346]) Let Δ be a simplicial complex. If K is a ﬁeld, then the following conditions are equivalent: (a) Δ is Cohen–Macaulay over K. . i (lk F ; K) = 0 for F ∈ Δ and i < dim lk F . (b) H i (K[Δ]) be the local cohomology modules of K[Δ]. Recall Proof. Let Hm i that Hm (K[Δ]) ∼ = H i (C ); see Section 2.8 for the deﬁnition of C . By Theorem 2.8.12, Δ is C–M if and only if H i (C ) = 0 for i < d + 1, where d = dim(Δ). . i−|F |−1 (lk(F ); K) = 0 ⇒) If Δ is C–M, then by Hochster’s theorem H for F ∈ Δ and i < d + 1. By Corollary 6.3.7 we get that Δ is pure. If F ∈ Δ, there is a face F1 of dimension d containing F , since F1 \ F ∈ lk(F ) . i (lk(F ); K) = 0 it follows that dim lk(F ) = |F1 \ F | − 1 = d − |F |. Hence H for all F ∈ Δ and all i < dim lk(F ). . i (lk(F ); K) = 0 for ⇐) Using dim lk(F ) ≤ d − |F |, the hypothesis H . i−|F |−1 (lk(F ); K) = 0 for F ∈ Δ and F ∈ Δ and i < dim lk(F ) implies H i < d + 1. Hence by Hochster’s theorem H i (C ) = 0 for i < d + 1. Thus Δ is Cohen–Macaulay. This proof was adapted from [65]. The original proof can be found in [346]. 2

In particular a Cohen–Macaulay complex Δ, being the link of its empty . i (Δ; K) = 0 for i < dim Δ. face, must satisfy H In general the Cohen–Macaulay property of a face ring may depend on the characteristic of the base ﬁeld; classical examples of this dependence are triangulations of the projective plane [346]. See the exercises at the end of this section. Example 6.3.13 If Δ is a discrete set with n vertices x6 s x7 s xn s

sx5

sx1

sx4 sx3 s x2

then Δ is Cohen–Macaulay and IΔ = (xi xj |1 ≤ i < j ≤ n).

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Corollary 6.3.14 If Δ is a 1-dimensional simplicial complex, then Δ is connected if and only if Δ is Cohen–Macaulay. . 0 (Δ; K) = 0, Proof. By Reisner’s theorem Δ is Cohen–Macaulay iﬀ H because the link of a vertex of Δ consists of a discrete set of vertices. To complete the argument apply Proposition 6.2.3. 2 Proposition 6.3.15 [250] If K is a ﬁeld, then a simplicial complex Δ is Cohen–Macaulay over K if and only if lk(F ) is Cohen–Macaulay over K for all F ∈ Δ. Proof. ⇒) Let F be a ﬁxed face in Δ and Δ = lk(F ). If G ∈ Δ , using the equality lkΔ (G) = lkΔ (G ∪ F ) . i (lkΔ (G), K) = 0 for i < dim lkΔ (G), thus and Reisner’s theorem we get H Δ is Cohen–Macaulay. ⇐) Since all the links are Cohen–Macaulay, it suﬃces to take F = ∅ and observe Δ = lk(∅) to conclude that Δ is Cohen–Macaulay. 2 Deﬁnition 6.3.16 The q-skeleton of a simplicial complex Δ is the simplicial complex Δq consisting of all p-simplices of Δ with p ≤ q. Proposition 6.3.17 Let Δ be a simplicial complex of dimension d and Δq its q-skeleton. If Δ is Cohen–Macaulay over a ﬁeld K, then Δq is Cohen– Macaulay for q ≤ d. Proof. Set Δ = Δq and take F ∈ Δ with dim(F ) < q ≤ d. Note lkΔ (F ) = (lkΔ (F ))q−|F | . Since Δ is pure there is a facet F of Δ of dimension d containing F . Hence F \ F is a face of lkΔ (F ) of dimension d − |F |. It follows that the dimension of lkΔ (F ) is q − |F |. Using Proposition 6.3.15 we get . i (lkΔ (F ); K) = 0 for i < dim(lkΔ (F )) = d − |F |, H and consequently . i ((lkΔ (F ))q−|F | ; K) = H . i (lkΔ (F ); K) = 0 for i < q − |F | ≤ d − |F |. H Observe that the equality between the homology modules holds because the augmented oriented chain complexes of (lkΔ (F ))q−|F | and lkΔ (F ) are equal up to degree q − |F | and thus their homologies have to agree up to degree 2 one less. Hence by Reisner’s theorem Δq is Cohen–Macaulay.

216

Chapter 6

Example 6.3.18 [170] Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and Δ the (p − 2)-skeleton of a simplex of dimension n − 1, where p ≥ 2. Then IΔ is a Cohen–Macaulay ideal and the face ring K[Δ] = R/({xi1 · · · xip | 1 ≤ i1 < · · · < ip ≤ n}) has a p-linear resolution by Theorem 5.3.7. Face rings with pure and linear resolutions have been studied in [66] and [170, 172]; special results have been obtained when the corresponding Stanley–Reisner ideal is generated in degree two. Deﬁnition 6.3.19 Let Δ1 , Δ2 be two simplicial complexes with vertex sets V1 , V2 . The join Δ1 ∗ Δ2 is the complex on the vertex set V1 ∪ V2 with faces F1 ∪ F2 , where Fi ∈ Δi for i = 1, 2. Proposition 6.3.20 If Δ1 and Δ2 are simplicial complexes, then their join Δ1 ∗Δ2 is Cohen–Macaulay if and only if Δi is Cohen–Macaulay for i = 1, 2. Proof. Since one has the equality IΔ1 ∗Δ2 = IΔ1 + IΔ2 , there is a graded isomorphism of K-algebras K[Δ1 ∗ Δ2 ] K[Δ1 ] ⊗K K[Δ2 ], see Proposition 3.1.33. To complete the proof use Corollary 3.1.35.

2

Given a simplicial complex Δ and Δ1 , . . . , Δs sub complexes of Δ, we deﬁne their union s ( Δi i=1

as the simplicial complex with vertex set ∪si=1 Vi and such that F is a face of ∪si=1 Δi if and only if F is a face of Δi for some i. The intersection can be deﬁned similarly. Deﬁnition 6.3.21 A simplicial complex Δ of dimension d is called pure shellable if Δ is pure and the facets (maximal faces) of Δ can be order F1 , . . . , Fs such that ⎛ ⎞ i−1 ( Fi ⎝ F j⎠ j=1

is pure of dimension d − 1 for all i ≥ 2. Here F i = {σ ∈ Δ| σ ⊂ Fi }. If Δ is pure shellable, F1 , . . . , Fs is called a shelling. The next deﬁnition of shellable is due to Bj¨orner and Wachs [44] and is usually referred to as non-pure shellable, although in this book we will drop the adjective “non-pure.”

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Deﬁnition 6.3.22 A simplicial complex Δ is shellable if the facets (maximal faces) of Δ can be ordered F1 , . . . , Fs such that for all 1 ≤ i < j ≤ s, there exists some v ∈ Fj \ Fi and some ∈ {1, . . . , j − 1} with Fj \ F = {v}. We call F1 , . . . , Fs a shelling of Δ. For a ﬁxed shelling of Δ, if F, F ∈ Δ then we write F < F to mean that F appears before F in the ordering. If Δ is a pure simplicial complex, then Δ is shellable if and only if Δ is pure shellable (Exercise 6.3.60). Theorem 6.3.23 Let Δ be a simplicial complex. If Δ is pure shellable, then Δ is Cohen–Macaulay over any ﬁeld K. Proof. Let V = {x1 , . . . , xn } be the vertex set of Δ and let R = K[V ] be a polynomial ring over a ﬁeld K. Assume that F1 , . . . , Fs is a shelling of Δ such that F i ∩ (∪i−1 j=1 F j ) is pure of dimension d − 1 for all i ≥ 2, where d is the dimension of Δ. Set ⎧ ⎛ ⎞⎫ ⎬ ⎨ s−1 ( ⎝ F j⎠ , σ = xi ∈ Fs Fs \ {xi } ∈ F s ⎭ ⎩ j=1 one may assume σ = {x1 , . . . , xr }. There is a short exact sequence f

0 −→ R/(I : f ) −→ R/I −→ R/(I, f ) −→ 0,

(∗)

where f = x1 · · · xr and I = IΔ . First we show the equality R/(I : f ) = K[F s ], note that K[F s ] is a polynomial ring in d+1 variables. By Proposition 6.3.4 one can write I = ∩si=1 pi , where pi is generated by V \ Fi for all i. Hence (I : f ) =

s

(pi : f ) =

i=1

(pi : f ).

σ⊂Fi

Observe that if σ ⊂ Fi for some i, then i = s; otherwise if i < s, then σ belongs to F s ∩ (∪s−1 j=1 F j ), but since this simplicial complex is pure of dimension d − 1, there is v ∈ σ such that σ ⊂ Fs \ {v}, a contradiction. Hence i = s and (I : f ) = (ps : f ) = ps , as asserted. Next, we show R/(I, f ) = K[F 1 ∪ · · · ∪ F s−1 ]. Note that f ∈ pi for i = 1, . . . , s − 1; otherwise if f ∈ / pi for some i < s, then σ ⊂ Fi and the previous argument yields a contradiction. If p is a minimal prime of (I, f ) diﬀerent from p1 , . . . , ps−1 , then ps ⊂ p and xi ∈ p for some xi ∈ σ; thus by construction of σ one has Fs \ {xi } ⊂ Fk for some k < s.

218

Chapter 6

Therefore V \Fk ⊂ {xi }∪(V \Fs ) ⊂ p, and by the minimality of p one derives conclude that p1 , . . . , ps−1 are pk = ps , which is impossible. Altogether we s−1 the minimal primes of (I, f ) and (I, f ) = i=1 pi , as required. Note that the two ends of Eq. (∗) are Cohen–Macaulay R-modules of dimension d + 1. Hence, using induction on s and Eq. (∗), by the depth lemma it follows that R/I is Cohen–Macaulay. Our proof follows ideas of Hibi [239]. 2 Deﬁnition 6.3.24 A simplicial complex Δ is constructible if Δ can be obtained by the following recursive procedure: (i) any simplex is constructible, (ii) if Δ1 , Δ2 are constructible complexes of dimension d, and if Δ1 ∩ Δ2 is constructible of dimension d − 1, then Δ1 ∪ Δ2 is constructible. Proposition 6.3.25 If Δ is a pure shellable complex of dimension d, then Δ is constructible. Proof. We proceed by induction on d. If d = 0, then Δ is a discrete set of vertices, which is constructible. Assume d > 0 and that Δ has at least two facets. Let F1 , . . . , Fr be a shelling of Δ. Thus, the complex Δm = (F 1 ∪ · · · ∪ F m ) ∩ F m+1 is pure of dimension d − 1 for 1 ≤ m ≤ r − 1. Next we prove that Δm is pure shellable. Consider the set F

= {Fi ∩ Fm+1 | dim(Fi ∩ Fm+1 ) = d − 1; i ≤ m} = {F1 ∩ Fm+1 , . . . , Fs ∩ Fm+1 }

where 1 < · · · < s < m + 1 and |F | = s. We claim that a face of the form F = Fi ∩ Fj ∩ Fm+1 has dimension d − 2 for j < i < m + 1. Clearly dim(F ) ≤ d − 2, otherwise one has Fi ∩ Fm+1 = Fj ∩ Fm+1 , a contradiction because |F | = s. By Exercise 6.3.59, there are vi ∈ Fm+1 \ Fi and vj ∈ Fm+1 \ Fj such that Fm+1 ∩ Fi Fm+1 ∩ Fj

⊂ ⊂

Fm+1 ∩ Fk1 = Fm+1 \ {vi }; Fm+1 ∩ Fk2 = Fm+1 \ {vj };

By dimension considerations we get Fm+1 ∩ Fi Fm+1 ∩ Fj

= =

Fm+1 \ {vi }, Fm+1 \ {vj }.

(k1 < m + 1), (k2 < m + 1).

Stanley–Reisner Rings and Edge Ideals of Clutters

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Hence F = Fm+1 \{vi , vj }, consequently dim(F ) is equal to d−2, as claimed. It follows readily that F gives a shelling of Δm because the complex (F 1 ∩ F m+1 ) ∪ · · · ∪ (F k ∩ F m+1 ) ∩ (F k+1 ∩ F m+1 ). is pure of dimension d − 2 for 1 ≤ k ≤ s − 1. Therefore by induction hypothesis Δm is constructible for 1 ≤ m ≤ r − 1. Hence, using induction on m, it follows that the complex F 1 ∪ · · · ∪ F m ∪ F m+1 is constructible for 1 ≤ m ≤ r − 1, as required.

2

Deﬁnition 6.3.26 Let R = K[x1 , . . . , xn ]. A graded R-module M is called sequentially Cohen–Macaulay (over K) if there exists a ﬁnite ﬁltration of graded R-modules 0 = M0 ⊂ M1 ⊂ · · · ⊂ Mr = M such that each Mi /Mi−1 is Cohen–Macaulay, and the Krull dimensions of the quotients are increasing: dim(M1 /M0 ) < dim(M2 /M1 ) < · · · < dim(Mr /Mr−1 ). A ﬁltration with these properties is called a C–M ﬁltration of M . As ﬁrst observed by Stanley [395, p. 87], shellable implies sequentially Cohen–Macaulay. In [207] Haghighi, Terai, Yassemi, and Zaare-Nahandi introduce and study the notion of a sequentially Sr module which is very reminiscent of the deﬁnition of a sequentially Cohen–Macaulay module. Theorem 6.3.27 If Δ is a shellable simplicial complex, then its associated Stanley–Reisner ring R/IΔ is sequentially Cohen–Macaulay. If Δ is a simplicial complex and v is a vertex of Δ, the deletion of v, denoted by delΔ (v), is the subcomplex consisting of the faces of Δ that do not contain v. The deletion of a face is deﬁned similarly. Deﬁnition 6.3.28 [43] A simplicial complex Δ is vertex decomposable if either Δ is a simplex, or Δ = ∅, or Δ contains a vertex v, called a shedding vertex, such that both the link lkΔ (v) and the deletion delΔ (v) are vertexdecomposable, and such that every facet of delΔ (v) is a facet of Δ. Deﬁnition 6.3.29 Let Δ be a simplicial complex. The pure i-skeleton of Δ is deﬁned as: Δ[i]

= {F ∈ Δ| dim(F ) = i}; −1 ≤ i ≤ dim(Δ),

where F denotes the subcomplex generated by F . Notice that Δ[i] is always pure of dimension i.

220

Chapter 6

We say that a simplicial complex Δ is sequentially Cohen–Macaulay if its Stanley–Reisner ring has this property. Theorem 6.3.30 [121, Theorem 3.3] A simplicial complex Δ is sequentially Cohen–Macaulay if and only if the pure i-skeleton Δ[i] is Cohen–Macaulay for −1 ≤ i ≤ dim(Δ). Corollary 6.3.31 A simplicial complex Δ is Cohen–Macaulay if and only if Δ is sequentially Cohen–Macaulay and Δ is pure. Proposition 6.3.32 The following hold for simplicial complexes: (I) pure shellable ⇒ constructible ⇒ Cohen–Macaulay ⇒ pure. (II) vertex decomposable ⇒ shellable ⇒ sequentially Cohen–Macaulay. Proof. (I): The ﬁrst implication is Proposition 6.3.25. The second implication follows using Exercise 6.3.62 together with the depth lemma. The third implication follows at once from Corollary 3.1.17. (II): The ﬁrst implication is shown in [45, Theorem 11.3]. The second implication is Theorem 6.3.27. 2 Deﬁnition 6.3.33 A clutter C is a ﬁnite ground set X together with a family E of subsets of X such that if f1 , f2 ∈ E, then f1 ⊂ f2 . The ground set X is called the vertex set of C and E is called the edge set of C; they are denoted by V (C) and E(C), respectively. Clutters are simple hypergraphs (see below) and they are called Sperner families in the literature. One example of a clutter is a graph with the vertices and edges deﬁned in the usual way. Deﬁnition 6.3.34 A hypergraph H is a pair (V, E) such that V is a ﬁnite set and E is a subset of the set of all subsets of V . The elements of E are called edges and the elements of V are called vertices. A hypergraph is simple if f1 ⊂ f2 for any two edges f1 , f2 . An excellent reference for hypergraph theory is the 3-volume book of Schrijver on combinatorial optimization [373]. Deﬁnition 6.3.35 Let C be a clutter with vertex set X = {x1 , . . . , xn }. The edge ideal of;C, denoted by I(C), is the ideal of R generated by all monomials xe = xi ∈e xi such that e ∈ E(C). Edge ideals of graphs, clutters, and hypergraphs were introduced in [416], [188], and [206], respectively. The assignment C −→ I(C)

Stanley–Reisner Rings and Edge Ideals of Clutters

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establishes a one-to-one correspondence between the family of clutters and the family of square-free monomial ideals. Edge ideals of clutters are in oneto-one correspondence with simplicial complexes via the Stanley–Reisner correspondence I(C) −→ ΔI(C) . We shall be interested in studying the relationships between the algebraic and combinatorial properties of C, ΔI(C) and R/I(C). Let C be a clutter with vertex set X = {x1 , . . . , xn }. A subset F of X is called independent or stable if e ⊂ F for any e ∈ E(C). The dual concept of a stable vertex set is a vertex cover , i.e., a subset C of X is a vertex cover if and only if X \ C is a stable vertex set. A minimal vertex cover is a vertex cover which is minimal with respect to inclusion. The number of vertices in any smallest vertex cover, denoted by α0 (C), is called the vertex covering number. The vertex independence number, denoted by β0 (C), is the number of vertices in any largest independent set of vertices. Notice that the Stanley–Reisner complex of I(C) is given by ΔI(C) = ΔC , where ΔC is the simplicial complex whose faces are the independent vertex sets of C. Thus K[ΔC ] = R/I(C), where K[ΔC ] is the Stanley–Reisner ring of ΔC . Deﬁnition 6.3.36 The simplicial complex ΔC whose faces are the independent vertex sets of C is called the independence complex of C. Lemma 6.3.37 Let C be a set of vertices of a clutter C and let p be the face ideal of R generated by C. The following are equivalent: (a) C is a minimal vertex cover of C. (b) p is a minimal prime of I(C). (c) V (C) \ C is a maximal face of ΔC . Proof. Note that I(C) ⊂ p if and only if C is a vertex cover of C. Taking this into account, the proof follows readily from the fact that any minimal prime of I(C) is a face ideal; see Theorem 6.1.4. 2 Deﬁnition 6.3.38 The clutter of minimal vertex covers of C, denoted by C ∨ or b(C) is called the Alexander dual clutter or blocker of C. The edge ideal of C ∨ , denoted by Ic (C), is called the ideal of covers of C. The ideal Ic (C) is also called the Alexander dual of I(C) and is also denoted by I(C)∨ .

222

Chapter 6

Theorem 6.3.39 Let e1 , . . . , eq and c1 , . . . , cs be the edges and minimal vertex covers of a clutter C. Then, there is a duality given by: = (c1 ) ∩ (c2 ) ∩ · · · ∩ (cs ) (xe1 , xe2 , . . . , xeq ) ' ' (xc1 , xc2 , . . . , xcs ), Ic (C) = (e1 ) ∩ (e2 ) ∩ · · · ∩ (eq ) = ; ; where xek = xi ∈ek xi and xck = xi ∈ck xi for 1 ≤ k ≤ s. In particular the heights of I(C) and Ic (C) are mini {|ci |} and mini {|ei |}, respectively. I(C)

=

Proof. If pi is the prime ideal generated by ci , then p1 , . . . , ps are the minimal primes of I(C) (see Lemma 6.3.37). Thus, by Theorem 6.1.4, we get I(C) = ∩si=1 pi . As C is a clutter, we have (C ∨ )∨ = C and Ic (C ∨ ) = I(C). Hence, the minimal vertex covers of C ∨ are e1 , . . . , eq . Thus, by the previous argument, we have I(C ∨ ) = ∩qi=1 (ei ). 2 The survey article [218] explains the role of Alexander duality to prove combinatorial and algebraic theorems. Deﬁnition 6.3.40 Let Δ be a simplicial complex on the vertex set V , the Alexander dual Δ∨ of Δ is the simplicial complex given by / Δ}. Δ∨ = {G ⊂ V | V \ G ∈ Theorem 6.3.41 (Eagon–Reiner [124]) If Δ is a simplicial complex, then the Alexander dual Δ∨ is Cohen–Macaulay if and only if the Stanley–Reisner ideal IΔ has a linear resolution. This result can be rephrased as: Theorem 6.3.42 Let C be a clutter. Then, R/I(C) is Cohen–Macaulay if and only if Ic (C) has a linear resolution. This result has been generalized [220] replacing linear resolution by the notion of componentwise linear ideal , and Cohen–Macaulay by sequentially Cohen–Macaulay. Next, we introduce the result of Herzog and Hibi that link these two notions. Deﬁnition 6.3.43 Let (Id ) denote the ideal generated by all degree d elements of a homogeneous ideal I. Then I is called componentwise linear if (Id ) has a linear resolution for all d. If I is a square-free monomial ideal we write I[d] for the ideal generated by all the square-free monomial ideals of degree d in I. Theorem 6.3.44 [220] Let I be a square-free monomial ideal of R. Then (a) R/I is sequentially Cohen–Macaulay if and only if I ∨ is componentwise linear. (b) I is componentwise linear if and only if I[d] has a linear resolution for all d ≥ 0.

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Deﬁnition 6.3.45 A monomial ideal I has linear quotients if the monomials that minimally generate I can be ordered g1 , . . . , gq such that for all 1 ≤ i ≤ q − 1, ((g1 , . . . , gi ) : gi+1 ) is generated by linear forms xi1 , . . . , xit . Recall that a clutter is called uniform if all its edges have the same cardinality. Proposition 6.3.46 [155, Lemma 5.2] If I is an edge ideal of a uniform clutter and I has linear quotients, then I has a linear resolution. Theorem 6.3.47 [228, Theorem 1.4(c)] If C is a uniform clutter, then ΔC is shellable if and only if Ic (C) has linear quotients.

Exercises 6.3.48 Let R be a polynomial ring over a ﬁeld K. Prove that the family of ideals of R generated by square-free monomials is a sublattice of the lattice of monomial ideals (cf. Proposition 6.1.19). 6.3.49 A monomial ideal I is square-free monomials iﬀ any of the following conditions hold: (a) I is an intersection of prime ideals. (b) rad (I) = I. (c) A monomial f is in I iﬀ x1 · · · xr ∈ I, where supp(f ) = {xi }ri=1 . 6.3.50 If Δ is the simplicial complex: sx2 @ @ @sx3 x1 s @ @ @s x4 then IΔ = (x1 , x2 ) ∩ (x1 , x3 ) ∩ (x1 , x4 ) ∩ (x2 , x3 ) ∩ (x3 , x4 ). 6.3.51 Let Δ be a Cohen–Macaulay simplicial complex and F ∈ Δ. If lk(F ) is not a discrete set of vertices, prove that lk(F ) is connected. 6.3.52 Let Δ be a simplicial complex on the vertex set V and I its Stanley– Reisner ideal. Take x ∈ V and set J = (I : x). (a) Prove that J = ∩x∈p / i pi , where Ass(I) = {p1 , . . . , pr }. (b) Find a relation between Δ and ΔJ . 6.3.53 Let R be a polynomial ring over a ﬁeld K and I an ideal minimally generated by square-free monomials f1 , . . . , fq . If R/I is Cohen–Macaulay and x is a variable not in I such that x ∈ ∪qi=1 supp(fi ), then R/(I : x) is Cohen–Macaulay.

224

Chapter 6

6.3.54 Let Δ be a pure simplicial complex on the vertex set V and I = IΔ . Assume that x is a variable in ∪qi=1 supp(fi ), where the fi ’s are monomials that minimally generate I. If (I : x) and (I, x) are Cohen–Macaulay, then I is Cohen–Macaulay. Hint Show ht(I) = ht(I : x) = ht(I, x). 6.3.55 Let Δ be a simplicial complex and let σ ∈ Δ, deﬁne the star of the face σ as star(σ) = {G ∈ Δ| σ ∪ G ∈ Δ}. Prove that the facets of star(σ) are the facets of Δ that contain σ. 6.3.56 Let R = K[x] be a polynomial ring over a ﬁeld K and Δ a simplicial complex with vertex set x. If σ = {x1 , . . . , xr } ∈ Δ, prove the equality

where f =

;r i=1

K[star(σ)] = R/(IΔ : f ), xi and star(σ) is the star of σ.

6.3.57 If Δ is a Cohen–Macaulay complex and F a face of Δ, then star(F ) is a Cohen–Macaulay complex. Hint star(F ) = F ∗ lk(F ). 6.3.58 A simplicial complex Δ of dimension d is pure shellable if and only if Δ is pure and the facets of Δ can be listed F1 , . . . , Fs such that for every 1 ≤ i < j ≤ s, there is 1 ≤ < j with Fi ∩ Fj ⊂ F ∩ Fj and dim(F ∩ Fj ) = d − 1. 6.3.59 A simplicial complex Δ is pure shellable if and only if Δ is pure and the facets of Δ can be ordered F1 , . . . , Fs such that for all 1 ≤ j < i ≤ s, there is v ∈ Fi \ Fj and k < i with Fi ∩ Fj ⊂ Fi ∩ Fk = Fi \ {v}. 6.3.60 A simplicial complex Δ is pure shellable if and only if Δ is pure and the facets of Δ can be listed F1 , . . . , Fs such that for all 1 ≤ j < i ≤ s, there exist some v ∈ Fi \ Fj and some k ∈ {1, . . . , i − 1} with Fi \ Fk = {v}. 6.3.61 Let R be a ring and I1 , I2 ideals of R. Prove that there is an exact sequence of R-modules: ϕ

φ

0 −→ R/(I1 ∩ I2 ) −→ R/I1 ⊕ R/I2 −→ R/(I1 + I2 ) −→ 0, where ϕ(r) = (r, −r) and φ(r 1 , r2 ) = r1 + r2 . 6.3.62 Let R = K[x] be a polynomial ring over a ﬁeld K and Δ1 , Δ2 simplicial complexes whose vertex sets are contained in x. Prove that there is an exact sequence of R-modules: 0 −→ K[Δ1 ∪ Δ2 ] −→ K[Δ1 ] ⊕ K[Δ2 ] −→ K[Δ1 ∩ Δ2 ] −→ 0.

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6.3.63 Prove that any constructible simplicial complex is Cohen–Macaulay. 6.3.64 (Reisner) Let R = K[a, . . . , f ] be a polynomial ring over a ﬁeld K and let Δ be the following triangulation of the real projective plane P2 : b s sa L A @ A L@ f A L @ @s ` A ` L ` ``` AA c X s As c L A XXXXLs A s A d e

JJ A J A

Js b a A s Show that IΔ = (abc, abd, ace, adf, aef, bcf, bde, bef, cde, cdf ). The ring K[Δ] is Cohen–Macaulay and has a 3-linear resolution if K has characteristic other than 2, and is not Cohen–Macaulay and has a non-linear resolution otherwise. Prove that Δ is not pure shellable and dim Δ = 2. The ﬁgure below gives a subdivision of the minimal triangulation of the projective plane given above (cf. Reisner [346, Remark 3]). 6.3.65 (N. Terai) If K is a ﬁeld and Δ is the following triangulation of the real projective plane P2 x5 x sP1 s A PPPssx2 x4 s A J

B x6As B J

@ @ B J

Js x3 s sx11 @

s x7 Bsx3 H x10 A A J B HH B HA A J H As Asx xH 8 B 9 PP J s AA PP x2 B Jssx4 HH s s H A x1 x5 then the link of any vertex is a cycle and K[Δ] is Cohen–Macaulay if and only if char(K) = 2. . 1 (Δ; K) H . 1 (P2 ; K) = K/2K and use Theorem 6.3.12. Hint H 6.3.66 [228] Let Δ be a simplicial complex with vertex set X and facets c c F1 , . . . , Fs . Then, Δ is shellable if and only if the ; ideal (xF1 , . . . , xFs ) has c c linear quotients, where Fi = X \ Fi and xF = xi ∈F c xi .

226

Chapter 6

6.3.67 Let Δ be a simplicial complex and let v be a vertex. The following are equivalent: (i) v is a shedding vertex, i.e., every facet of delΔ (v) is a facet of Δ, (ii) No face of lkΔ (v) is a facet of delΔ (v). 6.3.68 Let Δ be a simplicial complex on the vertex set V = {x1 , . . . , xn } and Δ∨ its Alexander dual. Prove that the Stanley–Reisner ideal of Δ∨ is equal to the ideal of minimal covers of I = IΔ , that is: IΔ∨ = ({xi1 · · · xir | (xi1 , . . . , xir ) is a minimal prime of I}) = I ∨ .

6.4

Regularity and projective dimension

Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K with the standard grading and let C be a clutter with vertex set X = {x1 , . . . , xn }. In this section we study the regularity, depth, and projective dimension of R/I(C), where I = I(C) is the edge ideal of C. There are several well-known results relating these invariants. We collect some of them here for ease of reference. The ﬁrst result is a basic relation between the dimension and the depth (see Lemma 2.3.6): depth R/I(C) ≤ dim R/I(C).

(6.1)

The deviation from equality in the above relationship can be quantiﬁed using the projective dimension, as is seen in a formula discovered by Auslander and Buchsbaum (see Theorem 3.5.13): pdR (R/I(C)) + depth R/I(C) = dim(R).

(6.2)

Another invariant of interest also follows from a closer inspection of a minimal projective resolution of R/I. Consider the minimal graded free resolution of M = R/I as an R-module: F :

0→

) j

R(−j)bgj → · · · →

)

R(−j)b1j → R → R/I → 0.

j

The Castelnuovo–Mumford regularity of M (regularity of M for short) is deﬁned as reg(M ) = max{j − i| bij = 0}. The a-invariant, the regularity, and the depth of M are closely related. Theorem 6.4.1 [413, Corollary B.4.1] a(M ) ≤ reg(M ) − depth(M ), with equality if M is Cohen–Macaulay.

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An excellent reference for the regularity of graded ideals is the book of Eisenbud [129]. There are methods to compute the regularity of R/I avoiding the construction of a minimal graded free resolution; see [31, 32] and [200, p. 614]. These methods work for any homogeneous ideal over an arbitrary ﬁeld. Theorem 6.4.2 [401] Let C be a clutter. If ht(I(C)) ≥ 2, then reg I(C) = 1 + reg R/I(C) = pd R/Ic (C). This formula also holds for edge ideals of height one: Corollary 6.4.3 If ht(I(C)) = 1, then reg R/I(C) = pd R/Ic (C) − 1. Proof. We set I = I(C). The formula clearly holds if I is a principal ideal. Assume that I is not principal. By permuting variables, we may assume that the primary decomposition of I has the form I = (x1 ) ∩ · · · ∩ (xr ) ∩ p1 ∩ · · · ∩ pm , where L = p1 ∩ · · · ∩ pm is an edge ideal of height at least 2. Notice that I = f L, where f = x1 · · · xr . Then, the Alexander dual of I is I ∨ = (x1 , . . . , xr ) + L∨ . f

The multiplication map L[−r] → f L induces an isomorphism of graded R-modules. Thus reg(L[−r]) = r + reg(L) = reg(I). By the Auslander– Buchsbaum formula, one has the equality pd(R/I ∨ ) = r + pd(R/L∨ ). Therefore, using Theorem 6.4.2, we get reg(R/I) = reg(R/L) + r = (pd(R/L∨ ) − 1) + r = pd(R/I ∨ ) − 1.

2

Proposition 6.4.4 [434, Lemma 7] Let R1 = K[x] and R2 = K[y] be two polynomial rings over a ﬁeld K and let R = K[x, y]. If I1 and I2 are edge ideals of R1 and R2 , respectively, then reg R/(I1 R + I2 R) = reg(R1 /I1 ) + reg(R2 /I2 ). Proof. By abuse of notation, we will write Ii in place of Ii R for i = 1, 2 when it is clear from context that we are using the generators of Ii but extending to an ideal of the larger ring. Let x = {x1 , . . . , xn } and y = {y1 , . . . , ym } be two disjoint sets of variables. Notice that (I1 + I2 )∨ = I1∨ I2∨ = I1∨ ∩ I2∨ where Ii∨ is the Alexander dual of Ii . Hence, by Theorem 6.4.2 and using the Auslander–Buchsbaum formula, we need only show the equality depth(R/(I1∨ ∩ I2∨ )) = depth(R1 /I1∨ ) + depth(R2 /I2∨ ) + 1.

228

Chapter 6

The equality depth(R/(I1∨ + I2∨ )) = depth(R1 /I1∨ ) + depth(R2 /I2∨ ) holds thanks to Theorem 3.1.34. Thus the proof reduces to showing the equality depth(R/(I1∨ ∩ I2∨ )) = depth(R/(I1∨ + I2∨ )) + 1.

(6.3)

We may assume that depth(R/I1∨ ) ≥ depth(R/I2∨ ). There is an exact sequence of graded R-modules: ϕ

φ

0 −→ R/(I1∨ ∩ I2∨ ) −→ R/I1∨ ⊕ R/I2∨ −→ R/(I1∨ + I2∨ ) −→ 0,

(6.4)

where ϕ(r) = (r, −r) and φ(r 1 , r2 ) = r1 + r2 . From the inequality depth(R/I1∨ ⊕ R/I2∨ )

= max{depth(R/Ii∨ )}2i=1 = depth(R/I1∨ ) = depth(R1 /I1∨ ) + m > depth(R1 /I1∨ ) + depth(R2 /I2∨ ) = depth(R/(I1∨ + I2∨ ))

and applying the depth lemma to Eq. (6.4), we obtain Eq. (6.3).

2

Deﬁnition 6.4.5 Let S be a set of vertices of a clutter C. The induced subclutter on S, denoted by C[S], is the maximal subclutter of C with vertex set S. A clutter of the form C[S] for some S ⊂ V (C) is called an induced subclutter of C. Thus, the vertex set of C[S] is S and the edges of C[S] are exactly the edges of C contained in S. Notice that C[S] may have isolated vertices, i.e., vertices that do not belong to any edge of C[S]. If C is a discrete clutter, i.e., all the vertices of C are isolated, we set I(C) = 0 and α0 (C) = 0. Proposition 6.4.6 If D is an induced subclutter of C, then reg(R/I(D)) ≤ reg(R/I(C)). Proof. There is S ⊂ V (C) such that D = C[S]. Let p be the prime ideal of R generated by S. By duality (see Theorem 6.3.39), we have Ic (C) = (e) =⇒ Ic (C)p = (e)p = (e)p = Ic (D)p . e∈E(C)

e∈E(C)

e∈E(D)

Therefore, using Theorem 6.4.2 and Exercise 6.4.30, we get reg(R/I(C)) = pd(R/Ic (C)) − 1 ≥ pd(Rp /Ic (C)p ) − 1 = pd(Rp /Ic (D)p ) − 1 = pd(R /Ic (D)) − 1 = pd(R/Ic (D)) − 1 = reg(R/I(D)), where R = K[S]. Thus, reg(R/I(C)) ≥ reg(R/I(D)).

2

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229

Deﬁnition 6.4.7 An induced matching in a clutter C is a set of pairwise disjoint edges f1 , . . . , fr such that the only edges of C contained in ∪ri=1 fi are f1 , . . . , fr . The induced matching number , denoted by im(C), is the number of edges in the largest induced matching. Corollary 6.4.8 Let C be a clutter and let f1 , . . . , fr be an induced matching of C with di = |fi | for i = 1, . . . , r. Then ! r di − r ≤ reg(R/I(C)). i=1

Proof. Let D = C[∪ri=1 fi ]. Notice that I(D) = (xf1 , . . . , xfr ). Thus I(D) is a complete intersection and the regularity of R/I(D) is the degree of its h-polynomial. The Hilbert series of R/I(D) is given by ;r

+ t + · · · + tdi −1 ) . (1 − t)n−r r Thus, the degree of the h-polynomial equals ( i=1 di ) − r. Therefore, the inequality follows from Proposition 6.4.6. 2 F (t) =

i=1 (1

Corollary 6.4.9 If C is a clutter and R/Ic (C) is Cohen–Macaulay, then im(C) = 1. Proof. Let r be the induced matching number of C and let d be the cardinality of any edge of C. Using Theorem 6.4.2 and Corollary 6.4.8, we obtain d − 1 ≥ r(d − 1). Thus r = 1, as required. 2 Lemma 6.4.10 [128, Corollary 20.19] If 0 → N → M → L → 0 is a short exact sequence of graded ﬁnitely generated R-modules, then (a) reg(N ) ≤ max(reg(M ), reg(L) + 1). (b) reg(M ) ≤ max(reg(N ), reg(L)). (c) reg(L) ≤ max(reg(N ) − 1, reg(M )). The following result was shown by Kalai and Meshulam for square-free monomial ideals and by Herzog for arbitrary monomial ideals. Proposition 6.4.11 [269, 219] If I1 , I2 are monomial ideals of R. Then (a) reg R/(I1 + I2 ) ≤ reg(R/I1 ) + reg(R/I2 ), (b) reg R/(I1 ∩ I2 ) ≤ reg(R/I1 ) + reg(R/I2 ) + 1.

230

Chapter 6

Corollary 6.4.12 If C1 , . . . , Cs are clutters on the vertex set X, then reg(R/I(∪si=1 Ci )) ≤ reg(R/I(C1 )) + · · · + reg(R/I(Cs )). E(Ci ). By Proposition 6.4.11, Proof. The set of edges of C = ∪si=1 Ci is ∪si=1 s 2 it suﬃces to notice the equality I(∪si=1 Ci ) = i=1 I(Ci ). A clutter C is called co-CM if Ic (C) is Cohen–Macaulay. A co-CM clutter is uniform because Cohen–Macaulay clutters are unmixed. Corollary 6.4.13 If C1 , . . . , Cs are co-CM clutters with vertex set X, then reg(R/I(∪si=1 Ci )) ≤ (d1 − 1) + · · · + (ds − 1), where di is the number of elements in any edge of Ci . Proof. By Theorem 6.4.2, we get that reg R/I(Ci ) = di − 1 for all i. Thus the result follows from Corollary 6.4.12. 2 Deﬁnition 6.4.14 Let C be a clutter. We will denote the cardinality of a smallest maximal independent set of C by β0 (C). This number is called the small independence number of C. Lemma 6.4.15 Let C be a clutter and let Δ = ΔC be its independence complex. Then Δ[i] = Δi for i ≤ β0 (C) − 1. Proof. First we prove the inclusion Δ[i] ⊂ Δi . Let F be a face of Δ[i] . Then F is contained in a face of Δ of dimension i, and so F is in Δi . Conversely, let F be a face of Δi . Then dim(F ) ≤ i ≤ β0 (C) − 1 =⇒ |F | ≤ i + 1 ≤ β0 (C). Since β0 (C) is the cardinality of any smallest maximal independent set of C, we can extend F to an independent set of C with i + 1 vertices. Thus F 2 is in Δ[i] . Let us recall the following expression for the depth of K[ΔC ]; a simple proof can be found in [172]. Theorem 6.4.16 [388] Let C be a clutter and let Δ = ΔC be its independence complex. Then depth R/I(C) = 1 + max{i | K[Δi ] is Cohen–Macaulay}, where Δi = {F ∈ Δ | dim(F ) ≤ i} is the i-skeleton and −1 ≤ i ≤ dim(Δ). Deﬁnition 6.4.17 Let C be a clutter. The cardinality of a largest minimal vertex cover of C, denoted by bight(I(C)), is called the big height of I(C).

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Theorem 6.4.18 [326] Let C be a clutter with n vertices. Then (a) reg R/Ic (C) ≥ bight(I(C)) − 1, (b) pdR (R/I(C)) ≥ bight(I(C)), (c) depth R/I(C) ≤ n − bight(I(C)), with equality everywhere if R/I(C) is sequentially Cohen–Macaulay. Proof. (a): We set α0 (C) = bight(I(C)). Note that β0 (C) = n − α0 (C). By Theorem 6.4.2 and the Auslander–Buchsbaum formula, the proof reduces to showing: depth R/I(C) ≤ β0 (C), with equality if R/I(C) is sequentially Cohen–Macaulay. First we show depth R/I(C) ≤ β0 (C). Assume Δi is Cohen–Macaulay for some −1 ≤ i ≤ dim(Δ), where Δ is the independence complex of C. By Theorem 6.4.16, it suﬃces to prove that 1 + i ≤ β0 (C). We can pick a maximal independent set F of C with β0 (C) vertices. Since Δi is Cohen– Macaulay all maximal faces of Δ have dimension i. If 1 + i > β0 (C), then F is a maximal face of Δi of dimension β0 (C) − 1, a contradiction. Assume that R/I(C) is sequentially Cohen–Macaulay. By Lemma 6.4.15 Δ[i] = Δi for i ≤ β0 (C) − 1. Then by Theorem 6.3.30, the ring K[Δi ] is Cohen–Macaulay for i ≤ β0 (C) − 1. Therefore, applying Theorem 6.4.16, we get that the depth of R/I(C) is at least β0 (C). Consequently, in this case one has the equality depth R/I(C) = β0 (C). (b): It follows from the proof of (a). (c): It follows from (b) and the Auslander–Buchsbaum formula. 2 The inequality in part (a) of Theorem 6.4.18 also follows directly from the deﬁnition of regularity because reg(Ic (C)) is an upper bound for the largest degree of a minimal generator of Ic (C). Deﬁnition 6.4.19 Let I be a monomial ideal of R. The big height of I, denoted by bight(I), is max{ht(p)| p ∈ Ass(R/I)}. The following result underlines the advantage of viewing a square-free monomial ideal as the edge ideal of a clutter. Herzog pointed out that this result holds for any sequentially Cohen–Macaulay module, as is seen below. Corollary 6.4.20 [326] If I is a monomial ideal and R/I is sequentially Cohen–Macaulay, then pdR (R/I) = bight(I). Proof. Let I ⊂ R be the polarization of I ⊂ R. Since R/I and R /I have the same projective dimension and bight(I) = bight(I ), we may assume that I is a square-free monomial ideal. As any square-free monomial ideal is the edge ideal of a clutter. The formula follows from Theorem 6.4.18. 2 This formula can be applied to a wide variety of square-free monomial ideals as is seen below.

232

Chapter 6

Deﬁnition 6.4.21 A clutter C is called sequentially Cohen–Macaulay if R/I(C) is sequentially Cohen–Macaulay. Theorem 6.4.22 The following clutters are sequentially Cohen–Macaulay: (a) [432] graphs with no chordless cycles of length other than 3 or 5, (b) [228] clutters whose ideal of covers has linear quotients, (c) [211] clutters of paths of length t of directed rooted trees, (d) [155] totally balanced clutters, (e) [205] uniform admissible clutters whose covering number is 3. A clutter C is called shellable if ΔC is shellable. The clutters of parts (a)–(e) are in fact shellable, and the clutters of part (a) are in fact vertex decomposable; see [113, 228, 407, 409, 433, 432]. The families of clutters in (a)–(e) will be studied later in this book (see Chapter 7 and the index). Theorem 6.4.23 If M is a sequentially Cohen–Macaulay R-module, then pdR (M ) = max{ht(p)| p ∈ AssR (M )}. Proof. By Theorem 3.5.13 and [413, Proposition A.7.3] one has pdR (M ) = dim(R) − depth(M ) = sup{r | ExtrR (M, R) = 0},

(6.5)

for the second equality see also [65, Exercise 3.1.24, p. 96]. Since M is sequentially Cohen–Macaulay, there is a ﬁltration of graded R-modules 0 = M0 ⊂ M1 ⊂ · · · ⊂ Mr = M such that each Mi /Mi−1 is C–M and d1 < · · · < dr , where di is equal to dim(Mi /Mi−1 ) for i = 1, . . . , r. By [231, Proposition 2.5] one has AssR (Mi /Mi−1 ) = {p ∈ AssR (M ) | dim(R/p) = di }. In particular AssR (M ) = ∪i AssR (Mi /Mi−1 ). Hence n − d1 = max{ht(p)| p ∈ AssR (M )}, i where n = dim(R). By [231, Proposition 2.2], Extn−d (M, R) is C–M of R j dimension di for al i, ExtR (M, R) = (0) for j ∈ / {n − d1 , . . . , n − dr } and i i Extn−d (Extn−d (M, R), R) Mi /Mi−1 R R

for all i. Hence, by Eq. (6.5), we get pdR (M ) = n − d1 .

2

Deﬁnition 6.4.24 The number of maximal independent sets of C, denoted by arith-deg(I(C)), is called the arithmetic degree of I(C).

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The arithmetic rank of an ideal I, denoted by ara(I), is the least number of elements of R which generate the ideal I up to radical. By Krull’s principal ideal theorem ara(I) ≥ ht(I). If equality occurs, I is called a set-theoretic complete intersection. Theorem 6.4.25 [401] If ht(I(C)) ≥ 2, then reg(I(C)) ≤ arith-deg(I(C)). Theorem 6.4.26 [301, Proposition 3] pd(R/I(C)) ≤ ara(I(C)). There are some instances where the equality pd(R/I) = ara(I) holds; see [15, 16, 143, 279] and the references therein. Barile [15] conjectured that the equality holds for edge ideals of forests. This conjecture was recently shown by Kimura and Terai [278]. Since edge ideals of forests are sequentially Cohen–Macaulay (see Theorem 6.5.25), by Theorem 6.4.18, it follows that bight(I) = ara(I) if I is the edge ideal of a forest [278]. Conjecture 6.4.27 (Eisenbud–Goto [130]) If p ⊂ (x1 , . . . , xn )2 is a prime graded ideal, then reg(R/p) ≤ deg(R/p) − codim(R/p). The following gives a partial answer to the monomial version of the Eisenbud–Goto regularity conjecture. Theorem 6.4.28 [401] Let I be the edge ideal of a clutter C. If ΔC is connected in codimension 1, then reg(R/I) ≤ deg(R/I) − codim(R/I). The multiplicity or degree of the edge-ring R/I(C) is equal to the number of independent sets of C with β0 (C) vertices (see Exercise 6.7.10). Theorem 6.4.29 If C is a k-uniform clutter, then deg(R/I(C)) ≤ k α0 (C) . Proof. Let I be the edge ideal of C. We set r = α0 (C). One may assume that K is an inﬁnite ﬁeld by considering a ﬁeld extension K ⊂ L, with L an inﬁnite ﬁeld, and using the functor (·) ⊗K L. It is not hard to see that I is minimally generated by forms f1 , . . . , fq of degree k such that f1 , . . . , fr is a regular sequence (see Exercise 3.1.40). Consider the subideal I = (f1 , . . . , fr ). There is a graded epimorphism R/I → R/I → 0, where dim(R/I ) = dim(R/I). Therefore deg(R/I) ≤ deg(R/I ). On the other hand by Exercise 5.1.20 the Hilbert series of R/I is given by F (R/I , t) =

(1 + t + · · · + tk−1 )r . (1 − t)n−r

Making t = 1 in the numerator and using Remark 5.1.7 yields the required bound. 2

234

Chapter 6

Exercises 6.4.30 Let R = K[x1 , . . . , xn ] and I an ideal of R. If I ⊂ (x1 , . . . , xn−1 ), and R = R/(xn ) ∼ = K[x1 , . . . , xn−1 ], then reg(R/I) = reg(R /I) and pdR (R/I) = pdR (R /I). Similarly, if xn ∈ I and I = I/(xn ), then reg(R/I) = reg(R /I ) and pdR (R/I) = pdR (R /I ) + 1. 6.4.31 Let C be a clutter. If α0 (C) := bight(I(C)), then α0 (C) = max{|e| : e ∈ E(C ∨ )} and α0 (C ∨ ) = max{|e| : e ∈ E(C)}. 6.4.32 Let C be a clutter. If I(C) has linear quotients, then reg R/I(C) = max{|e| : e ∈ E(C)} − 1. 6.4.33 Let I ⊂ R be a monomial ideal and let I ⊂ R be its polarization. Prove that the following equalities hold: pd(R/I) = pd(R /I ), reg(R/I) = reg(R /I ) and bight(I) = bight(I ).

6.5

Unmixed and shellable clutters

In this section we study unmixed, Cohen–Macaulay and shellable clutters using some notions that come from combinatorial optimization. We are interested in determining what families of clutters have these properties. Let C be a clutter with vertex set X = {x1 , . . . , xn }, let I = I(C) be its edge ideal, and let ΔC be its independence complex. We shall always assume that C has no isolated vertices, i.e., each vertex occurs in at least one edge. Deﬁnition 6.5.1 If ΔC is pure (resp. Cohen–Macaulay, shellable), we say that C is unmixed (resp. Cohen–Macaulay, shellable). The following notions of contraction, deletion, and minor come from combinatorial optimization [373]. Deﬁnition 6.5.2 For xi ∈ X, the contraction C/xi and deletion C \ xi are the clutters constructed as follows: both have X \ {xi } as vertex set, E(C/xi ) is the set of minimal elements of {e \ {xi }| e ∈ E(C}, minimal w.r.t / e ∈ E(C)}. to inclusion, and E(C \ xi ) is the set {e| xi ∈ The edge ideals of a deletion and a contraction have a nice algebraic interpretation. For xi ∈ X, deﬁne the contraction and deletion of I as the ideals: 'i , . . . , xn ], (I : xi ) and I c = I ∩ K[x1 , . . . , x respectively. Notice that the clutter associated to the square-free monomial ideal (I : xi ) (resp. I c ) is the contraction C/xi (resp. deletion C \ xi ), i.e., the edge ideals of C/xi and C \ xi are (I : xi ) and I c , respectively.

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Deﬁnition 6.5.3 A minor of a clutter C is a clutter obtained from C by a sequence of deletions and contractions in any order. A c-minor of I is any ideal generated by the set of monomials obtained from G(I), the minimal set of generators of I, by making any sequence of variables equal to 1 in the monomials of G(I). If a c-minor I contains a variable xi and we remove this variable from I , we still consider the new ideal a c-minor of I. A c-minor of C is any clutter that corresponds to a c-minor of I. Proposition 6.5.4 Let C1 , . . . , Cs be the minimal vertex covers of a clutter C. Then the primary decomposition of I(C) is (C1 ) ∩ (C2 ) ∩ · · · ∩ (Cs ), and the facets of ΔC are X \ C1 , . . . , X \ Cs . Proof. This follows from Theorem 6.3.39.

2

Corollary 6.5.5 ΔC is pure if and only if all minimal vertex covers of C have the same cardinality. Deﬁnition 6.5.6 A perfect matching of K¨ onig type of C is a collection e1 , . . . , eg of pairwise disjoint edges whose union is X and such that g is the height of I(C). A set of pairwise disjoint edges is called independent or a matching and a set of independent edges of C whose union is X is called a perfect matching. A clutter C satisﬁes the K¨ onig property if the maximum number of independent edges of C equals the height of I(C). For uniform clutters, it is easy to check that if C has the K¨onig property and a perfect matching, then the perfect matching is of K¨ onig type. A vertex x of C is called isolated if x does not occur in any edge of C. A clutter with a perfect matching of K¨onig type has the K¨ onig property. Next we show the converse to be true for unmixed clutters. Lemma 6.5.7 If C is an unmixed clutter with the K¨ onig property and without isolated vertices, then C has a perfect matching of K¨ onig type. Proof. Let X be the vertex set of C. By hypothesis there are e1 , . . . , eg independent edges of C, where g is the height of I(C). If e1 ∪ · · · ∪ eg X, pick xr ∈ X \ (e1 ∪ · · · ∪ eg ). Since the vertex xr occurs in some edge of C, there is a minimal vertex cover C containing xr . Thus using that e1 , . . . , eg are mutually disjoint we conclude that C contains at least g + 1 vertices, a contradiction. 2 Proposition 6.5.8 Let C be an unmixed clutter with a perfect matching onig type and let C1 , . . . , Cr be any collection of minimal e1 , . . . , eg of K¨ vertex covers of C. If C is the clutter associated to I = ∩ri=1 (Ci ), then C has a perfect matching e1 , . . . , eg of K¨ onig type such that: (a) ei ⊂ ei for all i, and (b) every vertex of ei \ ei is isolated in C .

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Proof. We denote the minimal set of generators of the ideal I = I(C) by G(I). There are monomials xv1 , . . . , xvg in G(I) so that supp(xvi ) = ei for i = 1, . . . , g. Since xvi is in I and I ⊂ I , there is ei ⊂ ei such that ei is an edge of C . Let x be any vertex in ei \ ei . If x is not isolated in C , there would a minimal vertex cover Ck of C containing x. As Ck contains a vertex of ej for each 1 ≤ j ≤ g and since e1 , . . . , eg are pairwise disjoint, we get that Ck contains at least g + 1 vertices, a contradiction. Thus (a) and (b) are satisﬁed. Clearly g is the height of I by construction of I . Let X be the vertex set of C . To ﬁnish the proof we need only show that X = e1 ∪ · · · ∪ eg . Let x ∈ X , then x ∈ ei for some i and x belongs to at least one edge of C . By part (b) we get that x ∈ ei , as required. 2 Corollary 6.5.9 Let C be an unmixed clutter. If C has a perfect matching onig type, then C/xj has a perfect matching {ei }gi=1 of K¨ onig {ei }gi=1 of K¨ type such that: (a) ei ⊂ ei for all i, and (b) every vertex of ei \ ei is isolated in C/xj . Proof. Let C1 , . . . , Cs be the minimal vertex covers of C. Since I(C) is / equal to ∩si=1 (Ci ), one has (I(C) : xj ) = ∩xj ∈C / i (Ci ) for any vertex xj ∈ I(C). The clutter associated to (I(C) : xj ) is the contraction C/xj . Hence from Proposition 6.5.8, we get that C/xj has a perfect matching e1 , . . . , eg satisfying (a) and (b). 2 Lemma 6.5.10 Let C be an unmixed clutter with a perfect matching {ei }gi=1 of K¨ onig type and let I = I(C). If e1 = {x1 , . . . , xr } and C1 , . . . , Cs are the minimal vertex covers of C, then (Ci ) = (((· · · (((I : x2 ) : x3 ) : x4 ) · · · ) : xr−1 ) : xr ) , x1 ∈Ci

Proof. Let I be the ideal on the right-hand side of the equality. If xv1 , . . . , xvq generate I and we make xi = 1 for i = 2, . . . , r in xv1 , . . . , xvq , we obtain a generating set for I . Notice that I = (I : x2 · · · xr ) by the deﬁnition of the colon operation. ar+1 · · · xann in I . We may assume a1 = 0, “⊃”: Take a xa = xa1 1 xr+1 ar+1 a otherwise x is already in the left-hand side. Then x2 · · · xr xr+1 · · · xann is in I. Let Ci be any minimal vertex cover of C containing x1 . Observe that Ci cannot contain xj for 2 ≤ j ≤ r. Indeed if xj ∈ Ci for some 2 ≤ j ≤ r, then Ci would contain {x1 , xj } plus at least one vertex of each edge in the collection e2 , . . . , eg , a contradiction because Ci has exactly g vertices. ar+1 ar+1 Hence, using that x2 · · · xr xr+1 · · · xann is in I, we get that xr+1 · · · xann is a in (Ci ). Consequently x is in the left-hand side of the equality. “⊂”: Let xa be a minimal generator in the left-hand side of the equality. Then xa ∈ (Ci ) if x1 ∈ Ci . If x1 ∈ Ci , then x2 · · · xr ∈ (Ci ) since Ci covers e1 . Thus xa x2 · · · xr ∈ (Ci ) for all i, and so xa x2 · · · xr ∈ ∩si=1 (Ci ) = I. 2 Thus xa is in the right-hand side of the equality.

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Deﬁnition 6.5.11 We say xi is a free variable (resp. free vertex) of I (resp. C) if xi only appears in one of the monomials of G(I) (resp. in one of the edges of C), where G(I) denotes the minimal set of generators of I = I(C) consisting of monomials. Theorem 6.5.12 [325] Let C be a clutter with a perfect matching e1 , . . . , eg of K¨ onig type. If all c-minors of C have a free vertex and C is unmixed, then ΔC is pure shellable. Proof. The proof is by induction on the number of vertices. We may assume that C is a non-discrete clutter, i.e., it contains an edge with at least two vertices. Let z be a free vertex of C and let C1 , . . . , Cs be the minimal vertex covers of C. We may also assume that z ∈ em for some em = {z1 , . . . , zr }, with r ≥ 2. For simplicity of notation assume that z = z1 and m = g. Consider the clutters C1 and C2 associated with I1 = (Ci ) and I2 = (Ci ) (6.6) z1 ∈C / i

z1 ∈Ci

respectively. By Proposition 6.5.8, the clutter C2 has a perfect matching onig type such that: (a) ei ⊂ ei for all i, and (b) every e1 , . . . , eg of K¨ vertex x of ei \ ei is isolated in C2 , i.e., x does not occur in any edge of C2 . In particular all vertices of eg \ {z1 } are isolated vertices of C2 . Similar statements hold for C1 because of Proposition 6.5.8. By Lemma 6.5.10 and Corollary 6.5.9 we get I1 = (I : z1 ) and I2 = (((· · · (((I : z2 ) : z3 ) : z4 ) · · · ) : zr−1 ) : zr ) , that is, C1 = C/z1 and C2 = C/{z2 , . . . , zr }. Hence the ideals I1 and I2 are c-minors of I. The number of vertices of Ci is less than that of C for i = 1, 2. Thus ΔC1 and ΔC2 are shellable by the induction hypothesis. Consider the clutter Ci whose edges are the edges of Ci and whose vertex set is X. The minimal vertex covers of Ci are exactly the minimal vertex covers of Ci . Thus it follows that ΔCi is shellable for i = 1, 2. Let F1 , . . . , Fp be the facets of ΔC that contain z1 and let G1 , . . . , Gt be the facets of ΔC that do not contain z1 . Notice that the edge ideals of Ci and Ci coincide, the vertex set of Ci is equal to the vertex set of C, and I = I1 ∩ I2 . Hence from Eq. (6.6) we get that F1 , . . . , Fp are the facets of ΔC1 and G1 , . . . , Gt are the facets of ΔC2 . By the induction hypothesis we may assume F1 , . . . , Fp is a shelling of ΔC1 and G1 , . . . , Gt is a shelling of ΔC2 . We now prove that F1 , . . . , Fp , G1 , . . . , Gt is a shelling of ΔC . We need only show that given Gj and Fi there is v ∈ Gj \ Fi and F such that Gj \ F = {v}. We can write Gj = X \ Cj and Fi = X \ Ci ,

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where Cj (resp. Ci ) is a minimal vertex cover of C containing z1 (resp. not containing z1 ). Notice that z2 , . . . , zr are not in Cj because e1 , . . . , eg is a perfect matching and |Cj | = g. Thus z2 , . . . , zr are in Gj . Since z1 ∈ Fi and Fi cannot contain the edge eg , there is a zk so that zk ∈ / Fi and k = 1. Set v = zk and F = (Gj \ {zk }) ∪ {z1 }. Clearly F is an independent vertex set because z1 is a free vertex in eg and Gj is an independent vertex set. Thus F is a facet because C is unmixed. To complete the proof observe that Gj \ F = {zk }. 2 ; In what follows, we set xe = xi ∈e xi for any e ⊂ X. Next we give a characterization of the unmixed property of C. This characterization can be formulated combinatorially or algebraically. Theorem 6.5.13 [325] Let C be a clutter with a perfect matching e1 , . . . , eg of K¨ onig type. Then the following conditions are equivalent: (a) C is unmixed. (b) For any two edges e = e and for any two distinct vertices x ∈ e, y ∈ e contained in some ei , (e \ {x}) ∪ (e \ {y}) contains an edge. (c) For any two edges e = e and for any T ⊂ ei such that xT divides xe xe , supp(xe xe /xT ) contains an edge. (d) For any two edges e = e and for any ei , (xe xe : xei ) ⊂ I(C). (e) I(C) = (I(C)2 : xe1 ) + · · · + (I(C)2 : xeg ). Proof. (a) ⇒ (c): We may assume i = 1. Let T be a subset of e1 such that xT divides xe xe . If T ⊂ e, then e is an edge contained in S = supp(xe xe /xT ) and there is nothing to show. The proof is similar if T ⊂ e . So we can deﬁne T1 = e ∩ T and T2 = T \ T1 and we may assume neither T1 nor T2 is empty. Note that T1 ⊂ e and T2 ⊂ e . In fact, T2 ⊂ T ∩ e , but equality does not necessarily hold. Notice that S = (e \ T1 ) ∪ (e \ T2 ). If S does not contain an edge, its complement contains a minimal vertex cover C. We use c to denote complement. Then C ⊂ X \ S = S c = (e \ T1 )c ∩ (e \ T2 )c = (ec ∪ T1 ) ∩ (ec ∪ T2 ). Now C ∩ e = ∅, so there is an x ∈ C ∩ e. Then x ∈ ec ∪ T1 . This forces x ∈ T1 . Similarly there is a y ∈ C ∩ e , and so y ∈ ec ∩ T2 . Thus y ∈ T2 . By the deﬁnition of T2 , x = y. To derive a contradiction pick zk ∈ ek ∩ C for k ≥ 2 and notice that x, y, z2 , . . . , zg is a set of g + 1 distinct vertices in C, which is impossible because C is unmixed. (c) ⇒ (b): Let x ∈ e and y ∈ e be two distinct vertices contained in some ei . Let T = {x, y}. Then xT divides xe xe and S = supp(xe xe /xT ) ⊂ (e \ {x}) ∪ (e \ {y}).

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By (c), S contains an edge. Thus (e \ {x}) ∪ (e \ {y}) contains an edge. (b) ⇒ (a): Let C be a minimal vertex cover of C. Since the matching is perfect, there is a partition C = (C ∩ e1 ) ∪ · · · ∪ (C ∩ eg ). Hence it suﬃces to prove that |C ∩ei | = 1 for all i. We proceed by contradiction. For simplicity of notation assume i = 1 and |C ∩ e1 | ≥ 2. Pick x = y in C ∩ e1 . Since C is minimal, there are edges e, e such that e ∩ (C \ {x}) = ∅ and e ∩ (C \ {y}) = ∅.

(6.7)

/ e. Then by hypothesis the Clearly x ∈ e, y ∈ e , and e = e because y ∈ set S = (e \ {x}) ∪ (e \ {y}) contains an edge e . Take z ∈ e ∩ C, then z ∈ e \ {x} or z ∈ e \ {y}, which is impossible by Eq. (6.7). (c) ⇒ (d): Let xa ∈ (xe xe : xei ) be a monomial generator of the colon ideal. Then xa xei = mxe xe for some monomial m. Let T ⊂ ei be maximal such that xT divides xe xe . Then xei \T divides m, and xa = (m/xei \T )(xe xe /xT ). Since supp(xe xe /xT ) contains an edge, we have xe xe /xT ∈ I. Thus xa ∈ I(C) as desired. (d) ⇒ (c): Suppose T ⊂ ei is such that xT divides xe xe . Then (xe xe /xT )xei = xe xe xei \T , and so (xe xe /xT ) ∈ (xe xe : xei ) ⊂ I(C). Thus (xe xe /xT ) is a multiple of a monomial generator of I(C). Hence supp(xe xe /xT ) contains an edge. (e) ⇒ (d): If equality in (e) holds, then (I(C)2 : xei ) = I(C) for all i. Hence from (I(C)2 : xei ) ⊂ I(C) we get condition (d). (d) ⇒ (e): We set I = I(C). It suﬃces to show (I 2 : xei ) = I for all i. Since I is contained in (I 2 : xei ), we need only show the inclusion (I 2 : xei ) ⊂ I. Take xa ∈ (I 2 : xei ), then xa xei = mxe xe for some edges e, e of C and some monomial m. If e = e , then by hypothesis xa ∈ (xe xe : xei ) ⊂ I, i.e., xa ∈ I. If e = e , then xa xei = mx2e . Thus xe divides xa because xei is a square-free monomial, but this means that xa ∈ I, as required. 2 Deﬁnition 6.5.14 A matrix A with entries in {0, 1} is balanced if A has no square submatrix of odd order with exactly two 1’s in each row and column. Deﬁnition 6.5.15 Let A be the incidence matrix of a clutter C. A clutter C has a special cycle of length r if there is a square submatrix of A of order r ≥ 3 with exactly two 1’s in each row and column. A clutter with no special odd cycles is called balanced and a clutter with no special cycles is called totally balanced . This deﬁnition of special cycle is equivalent to the usual deﬁnition of special cycle in hypergraph theory [8, 226] (see Exercise 6.5.34). The next result classiﬁes all unmixed balanced clutters because balanced clutters have the K¨onig property (Corollary 14.3.11). In particular it gives a classiﬁcation of all unmixed bipartite graphs (cf. Theorem 7.4.19).

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Corollary 6.5.16 [325] A clutter C with the K¨ onig property is unmixed if onig type such that and only if there is a perfect matching e1 , . . . , eg of K¨ for any two edges e = e and for any two distinct vertices x ∈ e, y ∈ e contained in some ei , one has that (e \ {x}) ∪ (e \ {y}) contains an edge. Proof. ⇒) Assume that C is unmixed. By Theorem 6.5.13 it suﬃces to observe that any unmixed clutter with the K¨ onig property and without isolated vertices has a perfect matching of K¨ onig type; see Lemma 6.5.7. ⇐) This implication follows at once from Theorem 6.5.13. 2 Faridi [155] introduced the notion of a leaf for a simplicial complex Δ. Precisely, a facet F of Δ is a leaf if F is the only facet of Δ, or there exists a facet G = F such that F ∩ F ⊂ F ∩ G for all facets F = F . Δ is called a simplicial forest if every non-empty subcollection, i.e., a subcomplex whose facets are also facets of Δ, of Δ contains a leaf. A graph G is called chordal if every cycle Cr of G of length r ≥ 4 has a chord in G. A chord of Cr is an edge joining two non-adjacent vertices of Cr . A chordal graph is called strongly chordal if every cycle Cr of even length at least six has a chord that divides Cr into two odd length paths. A clique of a graph G is a set of vertices inducing a complete subgraph. The clique clutter of G, denoted by cl(G), is the clutter on V (G) whose edges are the maximal cliques of G (maximal with respect to inclusion). If all the minors of a clutter C have free vertices, we say that C has the free vertex property. Note that if C has the free vertex property, then so do all of its minors. Theorem 6.5.17 A clutter C is totally balanced if and only if any of the following equivalent conditions hold: (a) [226, Theorem 3.2] C is the clutter of the facets of a simplicial forest. (b) [389, Corollary 3.1] C has the free vertex property. (c) [152] C is the clique clutter of a strongly chordal graph. Theorem 6.5.18 [325] Let C be a clutter with a perfect matching e1 , . . . , eg of K¨ onig type. If C has no special cycles of length 3 or 4 and C is unmixed, then for any two edges f1 , f2 of C and for any ei , one has that f1 ∩ei ⊂ f2 ∩ei or f2 ∩ ei ⊂ f1 ∩ ei . Proof. For simplicity assume i = 1. We proceed by contradiction. Assume there are x1 ∈ f1 ∩ e1 \ f2 ∩ e1 and x2 ∈ f2 ∩ e1 \ f1 ∩ e1 . As C is unmixed, by Theorem 6.5.13(b) there is an edge e of C such that e ⊂ (f1 \ {x1 }) ∪ (f2 \ {x2 }) = (f1 ∪ f2 ) \ {x1 , x2 }. Since e ⊂ e1 , there is x3 ∈ e \ e1 . Then either x3 ∈ f1 or x3 ∈ f2 . Without loss of generality we may assume x3 ∈ f1 \ e1 . For use below we denote the incidence matrix of C by A.

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Case(I): x3 ∈ f2 . Then the matrix x1 x2 x3

f1 1 0 1

f2 0 1 1

e1 1 1 0

is a submatrix of A, a contradiction. Case(II): x3 ∈ / f2 . Notice that e ⊂ f1 , otherwise e = f1 which is impossible because x1 ∈ f1 \ e. Thus there is x4 ∈ e \ f1 and x4 ∈ (e ∩ f2 ) \ f1 . Subcase(II.a): x4 ∈ e1 . Then the matrix f1 1 1 0

x1 x3 x4

e 0 1 1

e1 1 0 1

is a submatrix of A, a contradiction. Subcase(II.b): x4 ∈ / e1 . Then the matrix x1 x2 x3 x4

f1 1 0 1 0

is a submatrix of A, a contradiction.

e f2 0 0 0 1 1 0 1 1

e1 1 1 0 0 2

Proposition 6.5.19 Let C be an unmixed clutter with no special cycles of length 3 or 4. If e1 , . . . , eg is a perfect matching of C of K¨ onig type, then ei has a free vertex for all i. Proof. Fix an integer i in [1, g]. We may assume that ei has at least one non-free vertex. Consider the set: F = {f ∈ E(C)| ei ∩ f = ∅; f = ei }. By Theorem 6.5.18, the edges of F can be listed as f1 , . . . , fr so that they satisfy the inclusions f1 ∩ ei ⊂ f2 ∩ ei ⊂ · · · ⊂ fr ∩ ei ei . Thus any vertex of ei \ (fr ∩ ei ) is a free vertex of ei . 2 Theorem 6.5.20 Let C be an unmixed clutter with a perfect matching onig type. If C has no special cycles of length 3 or 4, then ΔC e1 , . . . , eg of K¨ is pure shellable. Proof. All hypotheses are preserved under contractions, i.e., under cminors. This follows from Corollary 6.5.9 and the fact that the incidence matrix of a contraction of C is a submatrix of the incidence matrix of C. Thus by Proposition 6.5.19 any c-minor has a free vertex and the result follows from Theorem 6.5.12. 2

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Theorem 6.5.21 [325] Let C be a clutter with a perfect matching e1 , . . . , eg of K¨ onig type. If for any two edges f1 , f2 of C and for any edge ei of the perfect matching, one has that f1 ∩ ei ⊂ f2 ∩ ei or f2 ∩ ei ⊂ f1 ∩ ei , then ΔC is pure shellable. Proof. First we show that ΔC is pure or equivalently that C is unmixed. It suﬃces to verify condition (b) of Theorem 6.5.13. Let f1 = f2 be two edges and let x ∈ f1 , y ∈ f2 be two distinct vertices contained in some ei . For simplicity we assume i = 1. Set B = (f1 \ {x}) ∪ (f2 \ {y}). Then f2 ∩ e1 ⊂ f1 ∩ e1 or f1 ∩ e1 ⊂ f2 ∩ e1 . In the ﬁrst case we have that f2 ⊂ B. Indeed let z ∈ f2 . If z = y, then z ∈ f2 \ {y} ⊂ B, and if z = y, then z ∈ f2 ∩ e1 ⊂ f1 ∩ e1 and z = x, i.e., z ∈ f1 \ {x} ⊂ B. In the second case f1 ⊂ B. This proves that C is unmixed. Next we show that ΔC is shellable. Notice that (i) ei has a free vertex for all i, which follows from the proof of Proposition 6.5.19. Thus, as C is unmixed, by Theorem 6.5.12 we need only show that any c-minor has a free vertex. By (i) it suﬃces to show that our hypotheses are closed under contractions. Let x be a vertex of C and let C = C/x. By Corollary 6.5.9, we get that C/x has a perfect matching e1 , . . . , eg satisfying: (a) ei ⊂ ei for all i, and (b) every vertex of ei \ ei is isolated in C . Let e, e be two edges of C and let ei be an edge of the perfect matching of C . There are edges f, f of C such that one of the following is satisﬁed: e = f and e = f \ {x}, e = f \ {x} and e = f , e = f \ {x} and e = f \ {x}, e = f and e = f . We may assume f ∩ ei ⊂ f ∩ ei . To ﬁnish the proof we now show that e ∩ ei ⊂ e ∩ ei . Take z ∈ e ∩ ei . Then z ∈ f ∩ ei and consequently z ∈ f ∩ ei . Since x ∈ / ei , one has z = x. It follows that z ∈ e ∩ ei . 2 Let G be a graph and let V be its vertex set. For use below consider the graph G ∪ W (V ) obtained from G by adding new vertices {yi | xi ∈ V } and new edges {{xi , yi } | xi ∈ V }. Corollary 6.5.22 If G is a graph and H = G ∪ W (V ), then ΔH is pure shellable. Proof. It follows at once from Theorem 6.5.21. Indeed if V = {x1 , . . . , xn }, then {x1 , y1 }, . . . , {xn , yn } is a perfect matching of H satisfying the ordering condition in Theorem 6.5.21. 2 Lemma 6.5.23 Let C be a clutter with minimal vertex covers C1 , . . . , Cs . If ΔC is shellable, A ⊂ VC is a set of vertices, and (Ci ), I = Ci ∩A=∅

then ΔI is shellable with respect to the linear ordering of the facets of ΔI induced by the shelling of the simplicial complex ΔC .

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Proof. Let H1 , . . . , Hs be a shelling of ΔC . We may assume that Hi is equal to VC \ Ci for all i. Let Hi and Hj be two facets of ΔI with i < j, i.e., A∩Ci = ∅ and A∩Cj = ∅. By the shellability of ΔC , there is an x ∈ Hj \ Hi and an < j such that Hj \ H = {x}. It suﬃces to prove that C ∩ A = ∅. / Ci ∪ Cj and z ∈ Hi ∩ Hj . Since If C ∩ A = ∅, pick z ∈ C ∩ A. Then z ∈ z∈ / H (otherwise z ∈ / C , a contradiction), we get z ∈ Hj \ H , i.e., z = x, 2 a contradiction because x ∈ / Hi . Lemma 6.5.24 Let xn be a free variable of I(C) = (xv1 , . . . , xvq−1 , xvq ), and let xvq = xn xu . Then the following hold. (a) If C1 is the clutter associated to (xv1 , . . . , xvq−1 ), then C is a minimal vertex cover of C containing xn if and only if C ∩ supp(xu ) = ∅ and C = {xn } ∪ C for some minimal vertex cover C of C1 . (b) If C2 is the clutter associated to (xv1 , . . . , xvq−1 , xu ), then C is a minimal vertex cover of C not containing xn if and only if C is a minimal vertex cover of C2 . Proof. (a) Assume that C is a minimal vertex cover of C containing xn . If C ∩ supp(xu ) = ∅, then C \ {xn } is a vertex cover of C, a contradiction. Thus C ∩ supp(xu ) = ∅. Hence it suﬃces to notice that C = C \ {xn } is a minimal vertex cover of C1 . The converse also follows readily. (b) Assume that C is a minimal vertex cover of C not containing xn . Let xa be a minimal generator of I(C2 ), then either xu divides xa or xa = xvi for some i < q. Then clearly C ∩ supp(xa ) = ∅ because C ∩ A = ∅, where A = supp(xu ). Thus C is a vertex cover of C2 . To prove that C is minimal take C C. We must show that there is an edge of C2 not covered by C . As C is a minimal vertex cover of C, there is xvi such that supp(xvi )∩C = ∅. If xvi is a minimal generator of C2 there is nothing to prove, otherwise xu divides xvi and the edge A of C2 is not covered by C . The converse also follows readily. 2 If all c-minors have a free vertex and C is unmixed, then ΔC is pure shellable (see Theorem 6.5.12). The next result complements this fact. Theorem 6.5.25 [409] If the clutter C has the free vertex property, then the independence complex ΔC is shellable. Proof. We proceed by induction on the number of vertices of C. Let xn be a free variable of I = I(C) = (xv1 , . . . , xvq−1 , xvq ). We may assume that xn occurs in xvq . Hence we can write xvq = xn xu for some xu such that / supp(xu ). For use below we set A = supp(xu ). Consider the ideals xn ∈ J = (xv1 , . . . , xvq−1 ) and L = (J, xu ). Then J = I(C1 ) and L = I(C2 ), where C1 and C2 are the clutters deﬁned by the ideals J and L, respectively. Notice that J and L are minors of the ideal I obtained by setting xn = 0

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and xn = 1, respectively. The vertex set of Ci is VCi = X \ {xn } for i = 1, 2. Thus ΔC1 and ΔC2 are shellable by the induction hypothesis. Let F1 , . . . , Fr be the facets of ΔC that contain xn and let G1 , . . . , Gs be the facets of ΔC that do not contain xn . Set Ci = X \ Gi and Ci = Ci \ {xn } for i = 1, . . . , s. Then C1 , . . . , Cs is the set of minimal vertex covers of C that contain xn , and by Lemma 6.5.24(a) C1 , . . . , Cs is the set of minimal vertex covers of C1 that do not intersect A. One has the equality Gi = VC1 \ Ci for all i. Hence, by the shellability of ΔC1 and using Lemma 6.5.23, we may assume that G1 , . . . , Gs is a shelling for the simplicial complex generated by G1 , . . . , Gs . By Lemma 6.5.24(b) one has that C is a minimal vertex cover of C not containing xn if and only if C is a minimal vertex cover of C2 . Thus, F is a facet of ΔC that contains xn , i.e., F = F ∪ {xn } if and only if F is a facet of ΔC2 . By induction we may also assume that F1 = F1 \ {xn }, . . . , Fr = Fr \ {xn } is a shelling of ΔC2 . We now prove that F1 , . . . , Fr , G1 , . . . , Gs with Fi = Fi ∪ {xn } is a shelling of ΔC . We need only show that given Gj and Fi there is a ∈ Gj \ Fi and F such that Gj \ F = {a}. We can write Gj = X \ Cj and Fi = X \ Ci , where Cj (resp. Ci ) is a minimal vertex cover of C containing xn (resp. not containing xn ). Recall that A = supp(xu ) is an edge of C2 . Notice the following: (i) Cj = Cj ∪ {xn } for some minimal vertex cover Cj of C1 such that A ∩ Cj = ∅, and (ii) Ci is a minimal vertex cover of C2 . From (i) we get that A ⊂ Gj . Observe that A ⊂ Fi , otherwise A ∩ Ci = ∅, a contradiction because Ci must cover the edge A = supp(u). Hence there is a ∈ A \ Fi and a ∈ Gj \ Fi . Since Cj ∪ {a} is a vertex cover of C, there is a minimal vertex cover C of C contained in Cj ∪ {a}. Clearly a ∈ C because C has to cover xu and Cj ∩ A = ∅. Thus F = X \ C is a facet of ΔC containing xn . To ﬁnish the proof we now prove that Gj \ F = {a}. We know that a ∈ Gj . If a ∈ F , then a ∈ / C , a contradiction. Thus a ∈ Gj \ F . Conversely take / Cj ∪ {xn } and z ∈ C ⊂ Cj ∪ {a}. Hence z = a, as z ∈ Gj \ F . Then z ∈ required. 2 According to Theorem 6.5.17, a totally balanced clutter satisﬁes the free vertex property. Thus, we obtain: Corollary 6.5.26 If C is a totally balanced clutter, then ΔC is shellable. By Theorem 6.5.17, a clutter C is a simplicial forest if and only if C is a totally balanced clutter. Thus, we also obtain: Corollary 6.5.27 If C is the clutter of facets of a simplicial forest, then ΔC is shellable.

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Deﬁnition 6.5.28 [155] Let Δ be a simplicial complex. The facet ideal of Δ, denoted by I(Δ), is the edge ideal of the clutter of facets of Δ. Facet ideals were introduced and studied by Faridi in a series of papers [153, 154, 155, 156, 157]. Corollary 6.5.29 [155] Let I = I(Δ) be the facet ideal of a simplicial forest. Then R/I(Δ) is sequentially Cohen–Macaulay. Proof. If Δ = F1 , . . . , Fs , then I(Δ) is also the edge ideal of the clutter C whose edge set is EC = {F1 , . . . , Fs }. Now apply Corollary 6.5.27 and Theorem 6.3.27. 2

Exercises 6.5.30 Prove that a clutter with a perfect matching of K¨onig type has the K¨ onig property. 6.5.31 Let C be a uniform clutter. If C has the K¨onig property and a perfect matching, then the perfect matching is of K¨onig type. 6.5.32 Consider the clutter C whose edges are e1 = {x1 , x2 }, e2 = {x3 , x4 , x5 , x6 }, e3 = {x7 , x8 , x9 }, f4 = {x1 , x3 }, f5 = {x2 , x4 }, f6 = {x5 , x7 }, f7 = {x6 , x8 }. Prove that C has the K¨ onig property and that C has no perfect matching of K¨ onig type. 6.5.33 [185, Proposition 5.8] If C is a uniform totally balanced clutter, there is a partition X 1 , . . . , X d of X such that any edge of C intersects any X i in exactly one vertex. 6.5.34 Let H be a hypergraph and let A be its incidence matrix. Recall that a cycle (resp. special cycle) of H is a sequence x1 e1 x2 e2 · · · xr er x1 of r distinct vertices xi and r distinct edges ej (r ≥ 3) such that {xi , xi+1 } ⊂ ei (resp. {x1 , . . . , xr } ∩ ei = {xi , xi+1 }) for i = 1, . . . , r (xr+1 = x1 ). The value r is the length of the cycle. Prove that H has a special cycle of length r if and only if there is a square submatrix of A of order r ≥ 3 with exactly two 1’s in each row and column. 6.5.35 If e1 = {x1 , x2 , x4 }, e2 = {x2 , x3 , x5 }, e3 = {x1 , x3 , x6 }, the clutter with edges e1 , e2 , e3 has a special cycle x1 e1 x2 e2 x3 e3 x1 of order 3. 6.5.36 [8] If C is a totally balanced clutter with n vertices, then C has at most n2 + n edges.

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6.6

Admissible clutters

Let X 1 , . . . , X d and e1 , . . . , eg be two partitions of a ﬁnite set X such that |ei ∩ X j | ≤ 1 for all i, j. The variables of K[X] are linearly ordered by: x ≺ y iﬀ (x ∈ X i , y ∈ X j , i < j) or (x, y ∈ X i , x ∈ ek , y ∈ e , k < ). Let e ⊂ X such that |e| = k and |e ∩ X i | ≤ 1 for all i. There are unique integers 1 ≤ i1 < · · · < ik ≤ d and integers j1 , . . . , jk ∈ [1, g] such that ∅ = e ∩ X i1 = {x1 }, ∅ = e ∩ X i2 = {x2 }, . . . , ∅ = e ∩ X ik = {xk } and x1 ∈ ej1 , . . . , xk ∈ ejk . We say that e is an admissible set if one has i1 = 1, i2 = 2, . . . , ik = k and j1 ≤ · · · ≤ jk . We can represent an admissible set e = {x1 , . . . , xk } as e = x1j1 · · · xkjk , i.e., xi = xiji and xiji ∈ X i ∩ eji for all i. A monomial xa is admissible if supp(xa ) is admissible. A clutter C is called admissible if e1 , . . . , eg are edges of C, ei is admissible for all i, and all other edges are admissible sets not contained in any of the ei ’s. We can think of X 1 , . . . , X d as color classes that color the edges. Lemma 6.6.1 If C is an admissible clutter, then e1 , . . . , eg is a perfect matching of K¨ onig type. Proof. It suﬃces to prove that g = ht I(C). Clearly ht I(C) ≥ g because any minimal vertex cover of C must contain at least one vertex of each ei and the ei ’s form a partition of X. For each 1 ≤ i ≤ g there is yi = x1i so that ei ∩ X 1 = {yi }. Since the ei ’s form a partition we have the equality (e1 ∩ X 1 ) ∪ · · · ∪ (eg ∩ X 1 ) = X 1 . Thus |X 1 | = g. To complete the proof notice that X 1 is a vertex cover of C because all edges of C are admissible. This shows ht I(C) ≤ g. 2 Admissible clutters with two color classes X 1 , X 2 are special types of bipartite graphs (see Section 7.4). Example 6.6.2 Let C be the Cohen–Macaulay admissible balanced clutter with color classes X 1 , X 2 , X 3 and edges e1 , e2 , e3 , f1 , f2 , f3 . e1 e2 e3

= = =

X1 x1 x2 x3

X2 y1 y2 y3

X3 z2

f1 f2 f3

= = =

X1 x1 x1 x2

X2 y2 y3 y3

X3 z2

The edges e1 , e2 , e3 form a perfect matching of K¨ onig type of C.

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Example 6.6.3 Consider the clutter C whose edge ideal is generated by: a1 b1 c1 d1 g1 h1 k1 , a2 b2 c2 d2 g2 h2 k2 , a3 b3 c3 d3 g3 h3 k3 , a4 b4 c4 d4 g4 h4 k4 , a1 b1 c1 d1 g2 h3 k4 , a1 b2 c3 d4 g2 h3 k4 , where ai , bi , . . . are variables. This clutter is balanced because its incidence matrix is totally unimodular and I(C) is Cohen–Macaulay. However, we cannot order its vertices so that it becomes an admissible uniform clutter. Deﬁnition 6.6.4 If e1 , . . . , eg are admissible subsets of X, the clutter C on X whose set of edges is:

ei ⊂ e for i = 1, . . . , g, e is admissible, E(C) = e ⊂ X ∪ {e1 , . . . , eg } e ⊂ e for any admissible set e = e is called a complete admissible clutter (cf. Deﬁnition 14.5.11). The edges of this clutter are the maximal admissible sets with respect to inclusion. By Lemma 6.6.1, e1 , . . . , eg is a perfect matching of K¨onig type. In Section 14.5 we show an optimization property of complete admissible uniform clutters (see Theorems 14.3.6 and 14.5.12). Proposition 6.6.5 If C is a complete admissible clutter, then C is unmixed. Proof. To show that C is unmixed it suﬃces to verify condition (b) of Theorem 6.5.13. Let e = e be two edges of C and let x = y be two vertices such that {x, y} ⊂ ei for some ei , x ∈ e, and y ∈ e . Since e, e , ei are admissible we can write e = {x1 , . . . , xk }, e = {y1 , . . . , yk }, ei = {z1 , . . . , zk }, where xi ∈ X i , yi ∈ X i , zi ∈ X i . There are i1 , i2 such that x = xi1 , y = yi2 , x = zi1 , and y = zi2 . We may assume i1 < i2 . One has i1 < k, because if k = i1 , then e e ∪ {zi1 +1 , . . . , zi2 } and the right-hand side is admissible, a contradiction. Set f = {y1 , . . . , yi1 , xi1 +1 , . . . , xk }. Then f ⊂ e \ {x} ∪ e \ {y}. Thus to ﬁnish the proof we need only show that f is an edge of C. Since yi2 ∈ ei and xi1 ∈ ei , then yi1 ∈ e for some ≤ i and xi1 +1 ∈ et for some i ≤ t. Hence f is admissible. Next we show that f is maximal. Assume that f is not maximal. Then there exists an admissible subset f that properly contains f . Then there is z ∈ f ∩ X k+1 and since f ∪ {z} ⊂ f , we get that e ∪ {z} = {x1 , . . . , xk , z} is admissible, but e e ∪ {z}, a contradiction. Hence f is maximal. 2 Theorem 6.6.6 [325] If C is a complete admissible uniform clutter, then the simplicial complex generated by the edges of C is pure shellable.

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Proof. Order the variables of K[X] as in the beginning of Section 6.6. We can represent the edges of C as Fi = x1i1 x2i2 · · · xdid , where xiij ∈ X i ∩ eij . Then, we order the edges of C lexicographically, that is Fi = x1i1 x2i2 · · · xdid < Fj = x1j1 x2j2 · · · xdjd if the ﬁrst non-zero entry of (j1 , j2 , . . . , jd )−(i1 , i2 , . . . , id ) = j−i is positive. Under this order, we show that C is shellable. Suppose Fi and Fj are edges of C with Fi < Fj . Suppose the ﬁrst nonzero entry of j− i is jt − it . Then 1 ≤ it < jt . Let Fk = Fj \ {xtjt } ∪{xtit } and let v = xtjt . Since j1 = i1 ≤ · · · ≤ jt−1 = it−1 ≤ it < jt ≤ jt+1 ≤ · · · ≤ jd , 2 Fk is maximal admissible, v ∈ Fj \ Fi , Fk < Fj and Fj \Fk = {v}. The next example illustrates the construction of the lexicographical shelling used in the proof of Theorem 6.6.6. Example 6.6.7 Let C be the complete admissible uniform clutter with color classes X 1 = {x1 , x2 , x3 }, X 2 = {y1 , y2 , y3 }, X 3 = {z1 , z2 , z3 }. Then the shelling of the simplicial complex generated by the edges of C is: F1 = {x1 , y1 , z1 } < F2 = {x1 , y1 , z2 } F4 = {x1 , y2 , z2 } < F5 = {x1 , y2 , z3 } F7 = {x2 , y2 , z2 } < F8 = {x2 , y2 , z3 } F10 = {x3 , y3 , z3 }.

< < <

F3 = {x1 , y1 , z3 } < F6 = {x1 , y3 , z3 } < F9 = {x2 , y3 , z3 } <

Lemma 6.6.8 If C is a complete admissible uniform clutter, then the simplicial complex ΔC ∨ generated by {X \ F | F ∈ E(C)} is pure shellable. Proof. Let F1 , . . . Fr be the shelling of the edges of C deﬁned in Theorem 6.6.6. Let G1 = X \ F1 , . . . , Gr = X \ Fr be the facets of ΔC ∨ . We claim that G1 , . . . , Gr is the desired shelling. Suppose Gi < Gj . Then Fi < Fj . Using the notation deﬁned in Theorem 6.6.6, let v = xtjt and deﬁne u = xtit . 2 Then u ∈ Gj \ Gi and Gj \ Gk = {u} as required. Theorem 6.6.9 [325] If C is a complete admissible d-uniform clutter, then the face ring R/I(C) is a Cohen–Macaulay ring with a d-linear resolution and |E(C)| = d+g−1 g−1 . Proof. Consider the clutter C ∨ of minimal vertex covers of C. By Lemma 6.6.8 and Exercise 6.6.14 we have that ΔC ∨ is pure shellable. Now recall that the Stanley–Reisner ideal of ΔC ∨ is I(C ∨ ) and that I(C ∨ ) is the Alexander dual of I(C). Thus R/I(C ∨) is Cohen–Macaulay, and by Theorem 6.3.41 the ideal I(C) has a linear resolution. Since the Alexander dual of a complete admissible uniform clutter is also a complete admissible uniform clutter and since (C ∨ )∨ = C it follows that R/I(C) is Cohen–Macaulay. The formula for the number of edges of C follows from the explicit formula given in

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Theorem 3.5.17 for the Betti numbers of a Cohen–Macaulay ideal with a pure resolution; see also Exercise 5.3.17. 2 Let C be a complete admissible uniform clutter. For each edge of C, e = x1j1 x2j2 · · · xdjd , consider all pairs (xiji , xkjk ) with i < k and consider the union of all these pairs with e = x1j1 x2j2 · · · xdjd running through all edges of C. This deﬁnes a poset P = (X, ≺) on X whose comparability graph G is deﬁned by all the unordered pairs {xiji , xkjk }. Corollary 6.6.10 If G is the complement of the comparability graph G deﬁned above, then R/I(G ) is Cohen–Macaulay. Proof. Notice that ΔG = {Kr | Kr is a clique of G} = O(P ), where O(P ) is the order complex of P . Since the maximal faces of O(P ) are precisely the edges of C, by Theorem 6.6.6, we obtain that O(P ) is a pure shellable complex whose Stanley–Reisner ring is equal to R/I(G ). Hence R/I(G ) is Cohen–Macaulay by Theorem 6.3.23. 2 Let C be a clutter and let xv1 , . . . , xvq be the minimal set of generators of I(C). Consider the ideal I ∗ = (xw1 , . . . , xwq ), where vi + wi = (1, . . . , 1). Following the terminology of matroid theory we call I ∗ the dual of I. If I ∗ has linear quotients and all xwi have the same degree, then I ∗ has a linear resolution (see Proposition 6.3.46). Corollary 6.6.11 If C is a complete admissible uniform clutter, then I(C)∗ has linear quotients. Proof. Let xv1 , . . . , xvq be the minimal set of generators of I = I(C). We set Fi = supp(xvi ) for i = 1, . . . , q. By Theorem 6.6.6, we may assume that F1 , . . . , Fq is a shelling for the simplicial complex F1 , . . . , Fq generated by ∗ c c the Fi ’s. Thus, according to Exercise 6.3.66, the ideal ; I = (xF1 , . . . , xFq ) c has linear quotients, where Fk = X \ Fk and xFkc = xi ∈F c xi . 2 k

Exercises 6.6.12 Consider the following clutter with edges e1 , e2 , f1 , f2 and color classes X 1 , X 2 , X 3

e1 e2

= =

X1 x1

X2 y1 y2

X3 z2

X1 f1 f2

= =

x1

X2 y1 y2

X3 z2

Prove that this clutter is unmixed, is not Cohen–Macaulay, has a perfect onig type, and the height of I(C) is two. matching e1 , e2 of K¨

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6.6.13 Prove that the uniform admissible clutters with three color classes X 1 = {x1 , . . . , xg }, X 2 = {y1 , . . . , yg }, X 3 = {z1 , . . . , zg } are, up to permutation of variables, exactly the clutters with a perfect matching ei = {xi , yi , zi } for i = 1, . . . , g such that all edges of C have the form {xi , yj , zk }, with 1 ≤ i ≤ j ≤ k ≤ g. 6.6.14 [325] If C is a complete admissible uniform clutter, then the blocker C ∨ of C is also a complete admissible uniform clutter.

6.7

Hilbert series of face rings

For a Stanley–Reisner ring K[Δ] there are formulas for its Hilbert function and Hilbert series in terms of the combinatorial data of the complex Δ. If I(C) is the edge ideal of a clutter C, its Hilbert series can be obtained from the so-called edge induced polynomial of C [193, 347]. Hilbert series with the ﬁne and standard grading Let K be a ﬁeld. Note that the polynomial ring R = K[x1 , . . . , xn ] can be endowed with a ﬁne Zn -grading as follows. For a = (a1 , . . . , an ) ∈ Zn , set

Kxa , if ai ≥ 0 for i = 1, . . . , n, Ra = 0, if ai < 0 for some i. Let I ⊂ R be an ideal generated by monomials. Since I is Zn -graded, the quotient ring R/I inherits the Zn -grading given by (R/I)a = Ra /Ia for all a ∈ Zn . In particular Stanley–Reisner rings have a ﬁne grading. Let M be a Zn -graded R-module. Each homogeneous component Ma of M is an R0 -module. Deﬁne the Hilbert function H(M, a) = (Ma ), provided that the length (Ma ) of Ma is ﬁnite for all a, and call F (M, t) = H(M, a)ta a∈Zn

the Hilbert–Poincar´e series of M . Here t = (t1 , . . . , tn ), where the ti are indeterminates and ta = ta1 1 · · · tann for a = (a1 , . . . , an ) ∈ Zn . By induction on n it follows that the polynomial ring R = K[x1 , . . . , xn ] with the ﬁne grading has Hilbert–Poincar´e series: F (R, t) =

a∈Nn

ta =

n 7 i=1

1 . 1 − ti

Let Δ be a simplicial complex with vertices X1 , . . . , Xn , permitting an abuse of notation we also denote by Xi the residue class of xi in K[Δ].

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Thus K[Δ] = K[X1 , . . . , Xn ]. The support of a ∈ Zn , denoted by supp(a), is deﬁned as the set {Xi | ai > 0}. Let X a ∈ K[Δ] and let supp(a) = {Xi1 , . . . , Xim }. Since IΔ is generated by square free monomials we have / IΔ ⇔ supp(a) ∈ Δ. X a = 0 ⇔ Xi1 · · · Xim = 0 ⇔ xi1 · · · xim ∈ Hence the non-zero monomials X a form a K-basis of K[Δ]. Therefore F (K[Δ], t) = ta = ta F ∈Δ

a∈Nn supp(a)∈Δ

Let F ∈ Δ. If F = ∅, then

ta =

a ∈ Nn

7 xi ∈F

supp a=F

ta = 1, and if F = ∅, then

1 =⇒ 1 − ti

supp a⊂F

a∈Nn supp(a)=F

ta =

a ∈ Nn

7 xi ∈F

ti . 1 − ti

supp a=F

Altogether we obtain that the expression for F (K[Δ], t) simpliﬁes to F (K[Δ], t) =

7 F ∈Δ xi ∈F

ti , 1 − ti

(6.8)

where the product over an empty index set is equal to 1. Deﬁnition 6.7.1 Given a simplicial complex Δ of dimension d its f -vector is the (d + 1)-tuple: f (Δ) = (f0 , . . . , fd ), where fi is the number of i-faces of Δ. Note f−1 = 1. To compute the Hilbert series of K[Δ], as a standard N-graded algebra, note that for i ∈ Z we have ) K[Δ]i = K[Δ]a a∈Zn |a|=i

where |a| = a1 + · · · + an for a = (a1 , . . . , an ). Observe that the Hilbert series of K[Δ] with the ﬁne grading specializes to the Hilbert series of K[Δ] with the Z-grading, that is, if ti = t for all i, then F (K[Δ], t) = F (K[Δ], t). Thus, by Eq. (6.8), we obtain a formula for the Hilbert series of K[Δ]:

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Theorem 6.7.2 [395] If Δ is a simplicial complex and f (Δ) = (f0 , . . . , fd ) is its f -vector, then the Hilbert series of K[Δ] is given by F (K[Δ], z) =

d

fi z i+1 , d = dim(Δ). (1 − z)i+1 i=−1

The Hilbert function of K[Δ] can be read oﬀ from its Hilbert series: Proposition 6.7.3 If Δ is a simplicial complex of dimension d and (fi ) is its f -vector, then the Hilbert function of K[Δ] is: d j−1 H(K[Δ], j) = fi , i i=0

for j ≥ 1 and H(K[Δ], 0) = 1.

A result of Kruskal–Katona Kruskal and Katona [283] showed that (f0 , f1 , . . . , fd ) ∈ Zd+1 is the f -vector of some d-dimensional simplicial complex if and only if (i+1)

0 < fi+1 ≤ fi (i+1)

where fi

, 0 ≤ i ≤ d − 1,

is deﬁned according to Eq. (5.7), cf. [395, Theorem 2.1, p. 55].

Simplicial complexes and their h-vectors Next we derive formulas for the h-vector of K[Δ]. If Δ is Cohen–Macaulay, we present some numerical constraints for the h-vector which are central for applications of commutative algebra to combinatorics [391] (see Theorem 6.7.7). Deﬁnition 6.7.4 The h-vector of a simplicial complex Δ, denoted by h(Δ),, is deﬁned as the h-vector of the standard graded algebra K[Δ]. Theorem 6.7.5 [395, p. 58] Let Δ be a simplicial complex of dimension d with h-vector (hi ) and f -vector (fi ). Then hk = 0 for k > d + 1 and hk =

k i=0

k−i

(−1)

d+1−i fi−1 for 0 ≤ k ≤ d + 1 . k−i

(6.9)

Proof. The idea is to write the Hilbert series of K[Δ] in two ways. By Theorems 6.7.2 and 5.1.4 there is a polynomial h(t) = h0 + h1 t + · · · + hr tr with integral coeﬃcients so that h(1) = 0 and satisfying F (K[Δ], t) =

d i=−1

fi ti+1 h(t) = . (1 − t)i+1 (1 − t)d+1

Comparing the series yields the asserted equalities.

(6.10) 2

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Theorem 6.7.6 Let Δ be a simplicial complex of dimension d with h-vector (hi ) and f -vector (fi ). Then fk−1 =

k d+1−i k−i

i=0

fd =

d+1 i=0

hi , for 1 ≤ k ≤ d + 1,

hi , h1 = f0 − (d + 1) and hd+1 = (−1)d

(6.11)

d

i i=−1 (−1) fi .

Proof. By the substitution t = z/(1 + z) in the equality fi−1 ti h0 + h1 t + · · · + hd+1 td+1 = , d+1 (1 − t) (1 − t)i i=0 d+1

we obtain

d+1

fk−1 =

i=0

hi z i (1 + z)d+1−i =

d+1 i=0

fi−1 z i . Hence

d+1 k d+1−i d+1−i hi = hi , for 1 ≤ k ≤ d + 1. k−i k−i i=0 i=0

From Eqs. (6.9) and (6.11) we derive the formulas for fd , h1 and hd+1 . 2 Computation of the h-vector of face rings To compute the h-vector (of face rings) the following procedure of Stanley [393] can be used. Consider the 3-dimensional convex polytope P consisting of two solid tetrahedrons joined by a 2-face: ppsx1 ppp\ pp \ p p p p p p4p \ p p p p p p p p p p p p ppppsx p p p p p p p sx2 pp p p p p p p\ x3 p sp p p p p p pp pp \ p \ ppp \ pppp \ sx5 one has f (P) = (5, 9, 6). Write down the f -vector on a diagonal, and put a 1 to the left of f0 . 1

5 9 6

Complete this array constructing a table, by placing below a pair of consecutive entries their diﬀerence, and placing a 1 on the left edge:

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Chapter 6

1 1 1 h(P) = (1

5 4

9

3

5

2

6

2

1)

After completing the table the next row of diﬀerences will be the h-vector. Hence the boundary simplicial complex Δ = Δ(P) is 2-dimensional, K[Δ] has multiplicity 6, and IΔ = (x1 x5 , x2 x3 x4 ). Now we present a result that plays a role in the proof of the upper bound conjecture for spheres [391]. Theorem 6.7.7 Let Δ be a simplicial complex of dimension d with n vertices. If K[Δ] is Cohen–Macaulay and K is an inﬁnite ﬁeld, then the hvector of Δ satisﬁes i+n−d−2 0 ≤ hi (Δ) ≤ for 0 ≤ i ≤ d + 1. (6.12) i Proof. Set S = K[Δ] = R/I, where R = K[x1 , . . . , xn ] and I = IΔ . By Lemma 3.1.28 there exists a regular homogeneous system of parameters θ = θ1 , . . . , θd+1 for S such that each θi has degree one. Since A = S/(θ)S is Artinian, in this case Theorem 5.2.5 says that hi (Δ) = H(A, i). Note that S/(θ)S R/I, where R = R/(θ) is a polynomial ring in n − d − 1 variables and I is the image of I in R. Therefore we have i+n−d−2 hi (Δ) = H(A, i) = H(R/I, i) ≤ H(R, i) = . 2 i Example 6.7.8 Let Δ be a 2-dimensional complex with f (Δ) = (5, 6, 2). Thus K[Δ] is not Cohen–Macaulay because its h vector has a negative entry: 1 1 1 h(Δ) = (1

5 4

3 2

6 2

−1

2 0).

Exercises 6.7.9 Let S = K[Δ] be the Stanley–Reisner ring of a simplicial complex Δ over a ﬁeld K and ϕ(t) the Hilbert polynomial of S. If χ .(Δ) is the reduced Euler characteristic, prove: (a) ϕ(i) = dimK (Si ) for all i ≥ 1, and .(Δ) = 0. (b) ϕS (0) = 1 if and only if χ

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255

6.7.10 Let Δ be a simplicial complex of dimension d and (f0 , . . . , fd ) its f -vector. Prove that fd is equal to deg(K[Δ]), the degree of K[Δ]. 6.7.11 Let S = K[Δ] be the Stanley–Reisner ring of a simplicial complex Δ with vertices x1 , . . . , xn and A = S/(x21 , . . . , x2n ). Prove that Δ is pure if and only if all the non-zero elements of the socle of A have the same degree. 6.7.12 If Δ is a connected simplicial complex having all its vertices of degree two, then Δ is a cycle. 6.7.13 [23] Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K. If I is a monomial Gorenstein ideal and rad (I) = (x1 , . . . , xn ), prove that I is generated by xa1 1 , . . . , xann for some a1 , . . . , an in N+ . 6.7.14 Give an example of a monomial Gorenstein ideal which is not a complete intersection. 6.7.15 Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and let Δ be a simplicial complex with vertices x1 , . . . , xn . Associated to the Stanley–Reisner ideal IΔ deﬁne another ideal I (Δ) = (IΔ , x1 y1 , . . . , xn yn ) ⊂ K[x1 , . . . , xn , y1 , . . . , yn ], and denote by S(Δ) the Stanley–Reisner complex corresponding to this ideal I (Δ). Prove the following: (a) The simplicial complex S(Δ) is pure shellable. (b) The f -vector of Δ is related to the h-vector of S(Δ) by fi (Δ) = hi+1 (S(Δ)) for all i. 6.7.16 Let k and n be two positive integers. A k-partition of n is an element α = (α1 , . . . , αk ) in Nk+ so that n = α1 + · · · + αk . Prove that the number of k-partitions of n is equal to n−1 k−1 .

6.8

Simplicial spheres

Let Δ be a ﬁnite simplicial complex with vertices x1 , . . . , xn and let ei be the ith unit vector in Rn . Given a face F ∈ Δ, we set |F | = conv{ei | xi ∈ F }, where “conv” is the convex hull. Deﬁne the geometric realization of Δ as ( |Δ| = |F |. F ∈Δ

Then |Δ| is a topological space with the induced usual topology of Rn . Note . i (|Δ|; A) for all i, where . i (Δ, A) ∼ that there is a canonical isomorphism H =H . Hi are the corresponding reduced homology modules.

256

Chapter 6

Deﬁnition 6.8.1 A q-simplex is a polytope generated by a set of q + 1 aﬃnely independent points. A polytope is simplicial if everyone of its proper faces is a simplex. Let P be a polytope of dimension d+1 and denote the number of faces of dimension i of P by fi , thus f0 is the number of vertices of P. The f -vector of P is the vector f (P) = (f0 , . . . , fd ) and we set f−1 = 1. The boundary complex Δ(P) of P is by deﬁnition the “abstract simplicial complex” whose vertices are the vertices of P and whose faces are those sets of vertices that span a proper face of P, thus Δ(P) has dimension d. Notice that P and Δ(P) have the same f -vector and accordingly one deﬁnes the h-vector of P as the h-vector of Δ(P). The geometric realization |Δ(P)| is therefore homeomorphic to a sphere S d of dimension d. This suggests the following more general concept. Deﬁnition 6.8.2 A simplicial complex Δ of dimension d is a simplicial sphere if |Δ| S d , in this case Δ is said to be a triangulation of S d . Remark 6.8.3 There are simplicial d-spheres with d = 3 and n = 8 which are not the boundary complex of a simplicial (d + 1)-polytope [393]. Moreover it follows from a result of Steinitz (a graph is the 1-skeleton of a 3polytope in R3 if and only if it is a 3-connected planar graph) that such an example does not exist for d = 2; see [438, Theorem 4.1]. Theorem 6.8.4 [65, Chapter 5] Let Δ be a simplicial complex of dimension d. If |Δ| ∼ = S d , then

K for i = dim(lk(F )), . i (lk F ; K) ∼ H = 0 otherwise. Proposition 6.8.5 If Δ is a simplicial sphere, then Δ is Cohen–Macaulay and the following equality holds χ .(lk F ) = (−1)dim lk F for F ∈ Δ. Proof. If follows from Reisner’s criterion and Theorem 6.8.4.

2

Remark 6.8.6 If P is a simplicial d-polytope, then Δ(P) is a simplicial sphere of dimension d − 1. Hence, from Proposition 6.8.5, we obtain the Euler’s formula χ .(P) = χ .(Δ(P)) =

d−1

(−1)i fi = (−1)d−1 = −1 + χ(P),

i=−1

where f = (f0 , . . . , fd−1 ) is the f -vector of P. The Euler’s formula is valid for any d-polytope, not just the simplicial ones (see [57, Theorem 16.1]).

Stanley–Reisner Rings and Edge Ideals of Clutters

257

Corollary 6.8.7 If Δ is a simplicial sphere, then K[Δ] is Gorenstein. Proof. It follows from [395, Theorem 5.1] and Theorem 6.8.4.

2

Theorem 6.8.8 If Δ is a simplicial sphere of dimension d, then its hvector satisﬁes hi = hd+1−i for i = 0, . . . , d + 1. Proof. Since K[Δ] is a Gorenstein ring, its h-vector is symmetric according to Proposition 5.3.9 and Corollary 5.3.10. 2 Remark 6.8.9 If Δ is a simplicial sphere, then hi = hd+1−i and one can recover Euler’s formula (make i = 0 and use Theorem 6.7.6). Those relations are the Dehn–Sommerville equations. Corollary 6.8.10 [314] If h = (h0 , . . . , hd+1 ) is the h-vector of a simplicial polytope of dimension d + 1, then hi = hd+1−i for 0 ≤ i ≤ d + 1. Proof. It follows from Theorem 6.8.8.

2

The next two conjectures were raised around 1993 in connection with the problem of bounding the Betti numbers of graded Gorenstein ideals [318]. Conjecture 6.8.11 Let Δ be a simplicial sphere and I the Stanley–Reisner ideal of Δ. If I has initial degree p and height g, then for 1 ≤ i < g the ith Betti number βi of K[Δ] satisﬁes: p+g−1 p+i−2 βi ≤ p+i−1 i−1 p+g−1 p+g−i−2 g p+g−2 + − . i p−1 i p−1 Conjecture 6.8.12 Let Δ be a simplicial sphere and I its Stanley–Reisner ideal. If I has initial degree p and height g, then p+g−1 p+g−3 μ(I) ≤ − . g−1 g−1

Exercises 6.8.13 If Δ is a simplicial sphere on n vertices whose geometric realization |Δ| is isomorphic to the unit circle S 1 , then (a) Δ is a cycle of length. (b) μ(IΔ ) = n2 − n for n ≥ 4.

258

6.9

Chapter 6

The upper bound conjectures

In this section we present the main steps in the proof of the upper bound conjecture for simplicial spheres. An excellent reference for a detailed proof of this conjecture is [65]. Let P be a polytope of dimension d + 1 and let fi = fi (P) be the number of faces of dimension i of P. In 1893 Poincar´e proved the Euler characteristic formula d

(−1)i fi = 1 + (−1)d ,

(6.13)

i=0

see [57, Theorem 16.1]. In [393] there are some historical comments about this formula. It follows from Proposition 6.8.5 that this formula also holds for simplicial spheres. Are there any optimal bounds for fi (P)? As it will be seen there is a positive answer to this question which is valid in a more general setting. In order to formulate the upper bound theorem for simplicial spheres and the upper bound theorem for convex polytopes we need to introduce some results on cyclic polytopes, the reader is referred to [57] for a detailed discussion on this topic. Consider the monomial curve Γ ⊂ Rd+1 given parametrically by Γ = {(τ, τ 2 , . . . , τ d+1 )| τ ∈ R}. A cyclic polytope, denoted by C(n, d+1), is the convex hull of any n distinct points in Γ such that n > d + 1. The f -vector of C(n, d + 1) depends only on n and d and not on the points chosen, and dim C(n, d + 1) = d + 1. The cyclic polytope C(n, d + 1) is simplicial and has the remarkable property that its f -vector satisﬁes: ? @ n d+1 fi (C(n, d + 1)) = for 0 ≤ i < , (6.14) i+1 2 this means that C(n, d + 1) has the highest possible number of i-faces when i is within the speciﬁed rank. Hence for any polytope P of dimension d + 1 with n vertices we have n = fi (C(n, d + 1)), fi (P) ≤ i+1 for all 0 ≤ i < ((d + 1)/2). The upper bound conjecture for polytope (UBCP for short) states that fi (P) ≤ fi (C(n, d + 1)) for 0 ≤ i ≤ d.

Stanley–Reisner Rings and Edge Ideals of Clutters

259

This conjecture was posed by Motzkin [327] in 1957 and proved by McMullen [314] in 1970. It was motivated by the performance of the simplex algorithm in linear programming. Proposition 6.9.1 If h = (h0 , . . . , hd+1 ) is the h-vector of a cyclic polytope C(n, d + 1), then n−d+k−2 hk = , 0 ≤ k ≤ ((d + 1)/2). k Proof. Let 0 ≤ k ≤ ((d + 1)/2). One has k k d+1−i k−d−2 n hk = (−1)k−i (−1)2(k−i) fi−1 = k−i k−i i i=0 i=0 n−d+k−2 = . k The second equality follows from Exercise (6.9.5).

2

Since a cyclic polytope C(n, d + 1) is simplicial of dimension d + 1 its boundary complex Δ(C(n, d + 1)) is a d-sphere. In 1964 V. Klee pointed out that the upper bound conjecture for polytopes may as well be made for simplicial spheres. This conjecture was proved by Stanley [391] in 1975 using techniques from commutative and homological algebra. Next we prove the upper bound theorem for simplicial spheres. Theorem 6.9.2 [391] Let Δ be simplicial complex of dimension d with n vertices. If |Δ| ∼ = S d , then fi (Δ) ≤ fi (Δ(C(n, d + 1))) for i = 0, . . . , d. Proof. Consider the following four conditions: (a) hi (Δ) ≤ hi (Δ(C(n, d + 1))) for 0 ≤ i ≤ ((d + 1)/2). (b) hi (Δ(C(n, d + 1))) = hd+1−i (Δ(C(n, d + 1))) for 0 ≤ i ≤ d + 1. all i. (c) hi (Δ) ≤ i+n−d−2 i (d) hi (Δ) = hd+1−i (Δ) for 0 ≤ i ≤ d + 1. We claim that (a), (b) and (d) imply the theorem. To show it notice that if ((d + 1)/2) < i ≤ d + 1, then 0 ≤ d + 1 − i ≤ ((d + 1)/2) and hi (Δ) = hd+1−i (Δ) ≤ hd+1−i (Δ(C(n, d + 1))) = hi (Δ(C(n, d + 1))), hence hi (Δ) ≤ hi (Δ(C(n, d + 1))) for 0 ≤ i ≤ d + 1. Therefore d+1 d + 1 − i hi (Δ) fk−1 (Δ) = k−i i=0 d+1 d + 1 − i ≤ hi (Δ(C(n, d + 1))) k−i i=0 =

fk−1 (Δ(C(n, d + 1))) for 1 ≤ k ≤ d + 1,

260

Chapter 6

and the proof of the claim is completed. By Proposition 6.9.1 (c) ⇒ (a), and by Lemma 6.8.10 (b) is satisﬁed due to the fact that C(n, d + 1) is a simplicial polytope. Hence the theorem is reduced to prove (c) and (d). To complete the proof notice that (c) follows from Lemma 6.7.7, and (d) follows from Theorem 6.8.8. 2 As a particular case of the upper bound theorem for spheres we obtain the upper bound theorem for convex polytopes: Theorem 6.9.3 [314] Let P be a convex polytope of dimension d with n vertices, then fi (P) ≤ fi (C(n, d)) for 0 ≤ i ≤ d − 1. Proof. By pulling the vertices P can be transformed into a simplicial polytope with the same number of vertices as P and at least as many faces of higher dimension; see [315] and [438, Lemma 8.24]. Hence, we may assume that P is simplicial. Since P is simplicial, the boundary complex Δ(P) is a simplicial sphere, that is, |Δ(P)| ∼ = S d−1 . Using Theorem 6.9.2 we obtain fi (P) = fi (Δ(P)) ≤ fi (Δ(C(n, d))) = fi (C(n, d)), for i = 0, . . . , d.

2

Theorem 6.9.4 [57, Corollary 18.3] If P is a simplicial d-polytope with n vertices, then fj (P) ≤ ϕd−j−1 (d, n)

(j = 1, . . . , d − 1),

where d/2

ϕd−j−1 (d, n) =

i=0

i d−j−1

n−d+i−1 i

(d−1)/2

+

i=0

d−i d−j−1

n−d+i−1 . i

Exercises 6.9.5 If r ∈ R and k ∈ N, then r k −r + k − 1 = (−1) . k k 6.9.6 [57, p. 79] Prove that a d-polytope P is simplicial if and only if each facet of P is a simplex.

Chapter 7

Edge Ideals of Graphs In this chapter we give an introduction to graph theory and study algebraic properties and invariants of edge ideals using the combinatorial structure of graphs. We present classiﬁcations of the following families: unmixed bipartite graphs, Cohen–Macaulay bipartite graphs, Cohen–Macaulay trees, shellable bipartite graphs, sequentially Cohen–Macaulay bipartite graphs. The invariants examined here include regularity, depth, Krull dimension, and multiplicity. Edge ideals are shown to have the persistence property. For those readers interested in a comprehensive view of graph theory itself, we recommend the books of Bollob´ as [48], Diestel [111], and Harary [208], which we take as our main references for the subject.

7.1

Graph theory

A graph G is an ordered pair of disjoint ﬁnite sets (V, E) such that E is a subset of the set of unordered pairs of V . The set V is the set of vertices and the set E is called the set of edges. The number of vertices of a graph is its order. An edge e = {u, v}, with u, v ∈ V , is said to join the vertices u and v. When working with several graphs it is convenient to write V (G) and E(G) for the vertex set and edge set of G, respectively. Sometimes we will also write VG and EG for the vertex set and edge set of G, respectively. Let G be a graph on the vertex set V . If e = {u, v} is an edge of G one says that the vertices u and v are adjacent or neighboring vertices of G. In this case it is also usual to say that the edge e is incident with u and v. The degree of a vertex v in V denoted by deg(v), is the number of edges incident with v. A vertex with degree zero is called an isolated vertex . Note that by deﬁnition a graph does not contain loops, a pair {v, v} (“an edge joining a vertex to itself”); neither does it contain a pair {u, v} that occurs several times (“that is, several edges joining the same two vertices”).

262

Chapter 7

If we allow these type of relations as edges we will call G a multigraph. Most results on graphs carry over to multigraphs in a natural way. There are areas in graph theory (such as plane duality [111]) where multigraphs arise more naturally than graphs. Terminology introduced earlier for graphs can be used correspondingly for multigraphs. Given two graphs G and H, a mapping ϕ, from V (G) to V (H) is called a homomorphism if {ϕ(u), ϕ(v)} ∈ E(H) whenever {u, v} ∈ E(G). Two graphs G and H are isomorphic if there is a bijective map ψ from V (G) to V (H) such that {u, v} ∈ E(G) if and only if {ψ(u), ψ(v)} ∈ E(H). A map taking graphs as arguments is called a graph invariant if it assigns equal values to isomorphic graphs. The number of vertices and the number of edges are two simple examples of graph invariants. Let H and G be two graphs, then H is called a subgraph of G if V (H) ⊂ V (G) and E(H) ⊂ E(G). A subgraph H is called an induced subgraph if H contains all the edges {u, v} ∈ E(G) with u, v ∈ V (H). In this case H is said to be the subgraph induced by V (H). An induced subgraph is denoted by H = G[V (H)] or H = V (H). It is also denoted by H = GV (H) . A spanning subgraph is a subgraph H of G containing all the vertices of G. A walk of length n in G is an alternating sequence of vertices and edges, written as w = {v0 , z1 , v1 , . . . , vn−1 , zn , vn }, where zi = {vi−1 , vi } is the edge joining the vertices vi−1 and vi . A walk may also be written {v0 , . . . , vn } with the edges understood, or {z1 , z2 , . . . , zn } with the vertices understood. If v0 = vn , the walk w is called a closed walk . A path is a walk with all its vertices distinct. We say that G is connected if for every pair of vertices u and v there is a path from u to v. Notice that G has a vertex disjoint decomposition G = ∪ri=1 Gi where G1 , . . . , Gr are the maximal (with respect to inclusion) connected subgraphs of G, the Gi ’s are called the connected components of G. A component is called even (resp. odd) if its order is even (resp. odd). A cycle of length n is a closed path {v0 , . . . , vn } in which n ≥ 3. A cycle is even (resp. odd) if its length is even (resp. odd). We denote by Cn the graph consisting of a cycle with n vertices, C3 will be called a triangle, C4 a square and so on. If all the vertices of G are isolated, G is called a discrete graph. A forest is an acyclic graph and a tree is a connected forest. The complete graph Kn has every pair of its n vertices adjacent. For instance: sx3 @ @ @sx4

x2 s @ @ @s x1 K4

s x3 A A x2 As

xs4 @ @

@s x5 s x1

C5

x2 s

sx3 @ @ @sx4

xs5

s x1 Tree

A coloring of the vertices of a graph G is an assignment of colors to the

Edge Ideals of Graphs

263

vertices of G in such a way that adjacent vertices have distinct colors. The chromatic number of a graph G, denoted by χ(G), is the minimal number of colors in a coloring of G. There are algebraic methods to compute the chromatic number [165] (see Theorem 7.7.19). A clique of a graph G is a set of vertices that induces a complete subgraph. We will also call a complete subgraph of G a clique. A maximal clique of G is a clique which is maximal with respect to inclusion. The clique number of G, denoted by ω(G), is the size of the largest complete subgraph of G. The clique number and the chromatic number are related by the inequality ω(G) ≤ χ(G). A graph is called perfect if for every induced subgraph H, the chromatic number of H is equal to the size of the largest complete subgraph of H, i.e., ω(H) = χ(H) for every induced subgraph of G. This notion was introduced by Berge [25, Chapter 16]. An excellent reference for the theory of perfect graphs is the book of Golumbic [191]; see also [93, 373]. A graph G is bipartite if its vertex set V (G) can be partitioned into two disjoint subsets V1 and V2 such that every edge of G has one vertex in V1 and one vertex in V2 . The pair (V1 , V2 ) is called a bipartition of G. If G is connected such a bipartition is uniquely determined. A bipartite graph G with bipartition (V1 , V2 ) is called a complete bipartite graph if {u, v} ∈ E(G) for all u ∈ V1 and v ∈ V2 . If V1 and V2 have m and n vertices, respectively, we denote a complete bipartite graph by Km,n . A star is a complete bipartite graph of the form K1,n . For instance: s s s H V1 @ H @ @ HH @ @ HH @ s V2 H @s @s K3,3 Deﬁnition 7.1.1 The distance d(u, v) between two vertices u and v of a graph G is deﬁned to be the minimum of the lengths of all possible paths from u to v. If there is no path joining u and v, then d(u, v) = ∞. Proposition 7.1.2 A graph G is bipartite if and only if all the cycles of G are even. In particular every forest is a bipartite graph. Proof. ⇒) Let (V1 , V2 ) be a bipartition of G. If {v0 , . . . , vn } is a cycle of G, one may assume v0 ∈ V1 . Since v1 ∈ V2 , it follows at once that vi is in V1 if and only if i is even, thus n must be even. ⇐) It suﬃces to prove that each connected component of G is bipartite. Thus one may assume that G is connected. Pick a vertex v0 ∈ V . Set V1 = {v ∈ V (G)|d(v, v0 ) is even} and V2 = V (G) \ V1 .

264

Chapter 7

It follows that no two vertices of Vi are adjacent for i = 1, 2, otherwise G would contain an odd cycle. Therefore (V1 , V2 ) is a bipartition of G and the graph G is bipartite. 2 If e is an edge, we denote by G\{e} the spanning subgraph of G obtained by deleting e and keeping all the vertices of G. The removal of a vertex v from a graph G results in a subgraph G \ {v} of G consisting of all the vertices in G except v and all the edges not incident with v. For instance: sx3 @ @sx4 e@

x2 s G @ @ @s x1

sx3 @ @ x2 s @sx G \ {x1 } 4

sx3 @ @ x2 s @sx4 G \ {e} @ @ @s x1

Given a set of vertices A, by G \ A, we mean the graph formed from G by deleting all the vertices in A, and all edges incident to a vertex in A. Let G be a graph. A vertex v (resp. an edge e) of G is called a cutvertex or cutpoint (resp. a bridge) if the number of connected components of G\{v} (resp. G \ {e}) is larger than that of G. A maximal connected subgraph of G without cut vertices is called a block . A graph G is 2-connected if |V (G)| > 2 and G has no cut vertices. Thus a block of G is either a maximal 2-connected subgraph, a bridge, or an isolated vertex. By their maximality, diﬀerent blocks of G intersect in at most one vertex, which is then a cutvertex of G. Therefore every edge of G lies in a unique block, and G is the union of its blocks. Deﬁnition 7.1.3 A set of edges in a graph G is called independent or a matching if no two of them have a vertex in common. Deﬁnition 7.1.4 Let A be a set of vertices of a graph G. The neighbor set of A, denoted by NG (A) or simply by N (A) if G is understood, is the set of vertices of G that are adjacent with at least one vertex of A. In particular, if G is a graph and v ∈ V (G), the neighbor set of v, denoted by NG (v) or N (v), is the set of vertices of G adjacent to v. Notice that if Cr is a cycle of a graph G, then Cr ⊂ NG (Cr ). Proposition 7.1.5 Let G be a bipartite graph with bipartition (V1 , V2 ) and let m, n be the number of vertices in V1 , V2 , respectively. If |A| ≤ |NG (A)| for all A ⊂ V1 , then there are m independent edges in G. Proof. By induction on m. If m = 1, then there is at least one vertex in V2 connected to V1 , thus there is one independent edge. Assume m ≥ 2. Case (I): Assume that |A| < |NG (A)| for all A ⊂ V1 with |A| < m. Let x1 ∈ V1 and y1 ∈ V2 be two adjacent vertices. Consider the subgraph

Edge Ideals of Graphs

265

H = G \ {x1 , y1 } obtained from G by removing x1 and y1 . Note that / NG (A) and for every A ⊂ V1 \ {x1 } one has NH (A) = NG (A) if y1 ∈ NH (A) = NG (A)\{y1 } otherwise. Hence by induction there are independent edges z2 , . . . , zm in H, which together with z1 = {x1 , y1 } yield the required number of independent edges. Case (II): Assume that |A| = |NG (A)| for some A ⊂ V1 with |A| < m. Set r = |A|. Consider the subgraph H = G \ (V1 \ A). As NH (S) = NG (S) for all S ⊂ A, by induction there are independent edges z1 , . . . , zr in H. On the other hand consider F = G \ (A ∪ NG (A)). If S ⊂ V1 \ A, then |NG (A ∪ S)| ≥ |A ∪ S| = |S| + |A|, since NG (S ∪ A) = NG (A) ∪ NF (S) and r = |NG (A)| we get |NF (S)| ≥ |S|. Applying induction there are independent edges w1 , . . . , wm−r in F . Putting altogether it is seen that there are m independent edges in G, as required. This proof is due to Halmos and Vaughn. 2 Corollary 7.1.6 Let G be a bipartite graph with bipartition (V1 , V2 ). If there is an integer p ≥ 0 such that |A| − p ≤ |NG (A)| for all A ⊂ V1 , then there are m − p independent edges in G, where m = |V1 |. Proof. By Proposition 7.1.5 the assertion holds for p = 0. Consider the graph F obtained from G by adding a new vertex y to V2 and joining every vertex of V1 to y. Let A ⊂ V1 , since NF (A) = NG (A) ∪ {y} one concludes |NF (A)| ≥ |A|− (p− 1). Hence by induction there are m− p+ 1 independent edges in F , it follows that there are m − p independent edges in G. 2 Let G be a graph with vertex set V . A subset C ⊂ V is a minimal vertex cover of G if: (i) every edge of G is incident with at least one vertex in C, and (ii) there is no proper subset of C with the ﬁrst property. If C satisﬁes condition (i) only, then C is called a vertex cover of G. The following example illustrates the notion of a minimal vertex cover: x5 qP q x3 P BB PPx qd4 @ B @ Bqd G @qd x2 x6 A A qd x1 x7Aq

C = {x1 , x2 , x4 , x6 } is a minimal vertex cover of G.

It is convenient to regard the empty set as a minimal vertex cover for a graph with all its vertices isolated. A set of vertices of G is called independent or stable if no two of them are adjacent. Notice the following duality: a set of vertices in G is a maximal independent set (with respect to inclusion) if and only if its complement is a minimal vertex cover for G.

266

Chapter 7

Deﬁnition 7.1.7 Let G be a graph. The vertex covering number , denoted by α0 (G), is the number of vertices in any smallest vertex cover. The matching number , denoted by β1 (G), is the number of edges in any largest independent set of edges. The vertex independence number, denoted by β0 (G), is the number of vertices in any largest independent set of vertices. In [26] the integer α0 (G) (resp. β0 (G)) is denoted by τ (G) (resp. α(G)) and is called the transversal number (resp. stability number ) of G. Theorem 7.1.8 (K¨onig) If G is a bipartite graph, then β1 (G) = α0 (G). Proof. Let (V1 , V2 ) be a bipartition of G. Set r = β1 (G) and m = |V1 |. Since there are r independent edges in G one obtains α0 (G) ≥ r. There exists a subset A ⊂ V1 such that |NG (A)| ≤ |A|−(m−r), otherwise if |NG (A)| > |A| − (m − r) for all A ⊂ V1 , then by Corollary 7.1.6 one would have r + 1 independent edges which is impossible. Hence NG (A) ∪ (V1 \ A) is a vertex cover of G with at most r elements, thus we get α0 (G) ≤ r. 2 A pairing by an independent set of edges of all the vertices of a graph G is called a perfect matching or a 1-factor. Thus G has a perfect matching if and only if G has an even number of vertices and there is a set of independent edges containing all the vertices. Next we present a criterion for the existence of a perfect matching. Theorem 7.1.9 (Marriage theorem) Let G be a bipartite graph. Then the following are equivalent (a) G has a perfect matching. (b) |A| ≤ |NG (A)| for all A ⊂ VG independent set of vertices. Proof. Let (V1 , V2 ) be a bipartition of G. Set r = β1 (G). It is easy to see that (a) implies (b). To prove that (b) implies (a), we begin by noticing that |V1 | = |V2 | because Vi is an independent set for i = 1, 2. By K¨ onig theorem there are z1 , . . . , zr independent edges and a minimal vertex cover A of G with r elements. Hence zi ∩ A has exactly one vertex for any i and A is an independent set. Using that VG \ A is an independent set of vertices and from the equality NG (VG \ A) = A we conclude that |VG \ A| ≤ |NG (VG \ A)| = |A|. It follows that |V1 | = r, and thus z1 , . . . , zr yields a perfect matching. 2 What happens if the graph G is not bipartite? The answer is given by the following theorem of Tutte. A very readable and comprehensive reference about matchings in ﬁnite graphs is the book of Lov´ asz and Plummer [297]. Theorem 7.1.10 (Tutte; see [111, Theorem 2.2.1]) A graph G has a perfect matching if and only if c0 (G \ S) ≤ |S| for all S ⊂ VG , where c0 (G) denotes the number of odd components (connected components with an odd number of vertices) of G.

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A directed graph or digraph D consists of a ﬁnite set V (D) of vertices together with a prescribed collection E(D) of ordered pairs of distinct points called edges or arrows. An oriented graph is a digraph having no-cycles of length two. In other words an oriented graph is a graph together with an orientation of its edges. A tournament is a complete oriented graph. Any tournament has a spanning directed path [14, Theorem 1.4.5].

Exercises 7.1.11 Let G be a graph with vertex set VG = {x1 , . . . , xn } and edge set EG , prove Euler’s identity: 2|EG | = ni=1 deg(xi ). 7.1.12 If G is a connected bipartite graph, prove that G has a unique bipartition. 7.1.13 If u, v are two vertices at maximum distance in a connected graph G, then u, v are not cutpoints. 7.1.14 If G is a regular bipartite graph (all vertices have the same degree), then G has a perfect matching. 7.1.15 An edge cover in a graph G is a set of edges collectively incident with each vertex of G. The edge covering number of G denoted by α1 (G), is the number of edges in any smallest edge cover in G. Prove: (a) If G has no isolated vertices, then β1 (G) + α1 (G) = |VG |. (b) For any graph G, β1 (G) ≤ α0 (G). If G is bipartite, α1 (G) = β0 (G). 7.1.16 Let G be a bipartite graph with minimum degree r. Then G is the union of r edge disjoint edge covers. 7.1.17 Deﬁne a 2-coloration of a set A as a surjective mapping from A onto a set of two elements. A 2-coloration of a graph is a 2-coloration of its vertex set. Prove that a graph is connected if for any 2-coloration of the graph there exists a heterochromatic edge. 7.1.18 If G is a bipartite graph with bipartition (V1 , V2 ), prove that the mapping from V (G) to V (K2 ) that sends all the vertices from Vi to vertex i is a homomorphism from G to K2 . 7.1.19 Prove that the clique number and the chromatic number of a graph G satisfy ω(G) ≤ χ(G). 7.1.20 Let G be a connected perfect graph with vertex set V (G). Then there are cliques K1 , . . . , Ks of G such that V (G) is the disjoint union of K1 , . . . , Ks and α0 (G) = α0 (G[K1 ]) + · · · + α0 (G[Ks ]). 7.1.21 Prove that any tournament has a spanning directed path.

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7.2

Chapter 7

Edge ideals and B-graphs

In this section we introduce edge ideals and edge rings of graphs. We will compare several families of graphs and examine some of their numerical invariants. Some bounds for the Krull dimension are given. Let G be a graph with vertices x1 , . . . , xn and let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K. Here we are permitting an abuse of notation by using xi to denote a vertex of G and also a variable of R. Deﬁnition 7.2.1 The edge ideal I(G) associated to the graph G is the ideal of R generated by the set of all square-free monomials xi xj such that xi is adjacent to xj . If all the vertices of G are isolated we set I(G) = (0). The ring R/I(G) is called the edge ring of G. Edge ideals are algebraic representations of graphs. These ideals occur in computational chemistry [291], physics [51], and combinatorics [395]. The non-zero edge ideals are precisely those ideals generated by square-free monomials of degree two. A prime example of an edge ideal comes from the theory of posets: Deﬁnition 7.2.2 Given a poset P with vertex set X = {x1 . . . . , xn }, its order complex , denoted by Δ(P), is the simplicial complex on X whose faces are the chains (linearly ordered sets) in P, thus K[Δ(P)] = K[X]/(xi xj | xi ∼ xj ) is the Stanley–Reisner ring of Δ(P). Here xi ∼ xj means that xi is not comparable to xj . The next result establishes a one-to-one correspondence between the minimal vertex covers of G and the minimal primes of I(G). Proposition 7.2.3 A prime ideal p ⊂ R is a minimal prime of I(G) if and only if p is generated by a minimal vertex cover of G. Proof. It follows at once from Lemma 6.3.37 and Theorem 6.1.4.

2

Corollary 7.2.4 If G is a graph, then the vertex covering number α0 (G) is equal to the height of the edge ideal I(G). Proof. It follows at once from Proposition 7.2.3.

2

Corollary 7.2.5 Let G be a graph with n vertices and let I(G) be its edge ideal. Then n = α0 (G) + β0 (G) = ht(I(G)) + dim R/I(G). Proof. It follows from Corollary 7.2.4.

2

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269

Deﬁnition 7.2.6 Let I ⊂ R be a graded ideal. The quotient ring R/I is called Cohen–Macaulay (C–M for short) if depth(R/I) = dim(R/I). The ideal I is called Cohen–Macaulay if R/I is Cohen–Macaulay. Deﬁnition 7.2.7 A graph G is called Cohen–Macaulay over the ﬁeld K if R/I(G) is a Cohen–Macaulay ring. Deﬁnition 7.2.8 A graph G is called unmixed graph or well-covered if any two minimal vertex covers of G have the same cardinality. Proposition 7.2.9 If G is a Cohen–Macaulay graph, then G is unmixed. Proof. By Theorem 6.1.4 I(G) is the intersection of its minimal primes. Therefore I(G) is an unmixed ideal by Corollary 3.1.17, thus G is unmixed by the correspondence between minimal covers and minimal primes. 2 Deﬁnition 7.2.10 A vertex v is critical if α0 (G \ {v}) < α0 (G). A graph G is called vertex-critical if all its vertices are critical. Proposition 7.2.11 Let G be a graph. If α0 (G \ {v}) < α0 (G) for some vertex v, then α0 (G \ {v}) = α0 (G) − 1. Proof. Set r = α0 (G \ {v}) and note that r + 1 ≤ α0 (G). Pick a minimal vertex cover C of G \ {v} with r vertices. Since α0 (G \ {v}) < α0 (G), v is not an isolated vertex and NG (v) = C. Thus C ∪ {v} is a minimal vertex cover of G. Hence α0 (G) ≤ r + 1 and one has the asserted equality. 2 Proposition 7.2.12 Let G be a graph without isolated vertices. Then G is vertex critical if and only if any of the following equivalent conditions hold. (a) Any vertex is in some minimal vertex cover with α0 (G) vertices. (b) α0 (G) = α0 (G \ {v}) + 1 for any vertex v of G. (c) β0 (G) = β0 (G \ {v}) for any vertex v of G. Proof. It follows readily from Proposition 7.2.11.

2

Proposition 7.2.13 Let I be an ideal of a catenary local domain (R, m) and let x ∈ m. If ht (I) < ht (I, x), then ht (I, x) = ht (I) + 1. Proof. Let p be a minimal prime of I such that ht (I) = ht (p). Note that the image of (x) + p in R/p is a principal ideal, hence by Krull’s principal ideal theorem (see Theorem 2.3.16) there is a minimal prime ideal q of (x)+p such that ht (q/p) ≤ 1. As R is a catenary domain one has ht (I, x) − ht (I) ≤ ht (q) − ht (p) = ht (q/p) ≤ 1. Therefore ht (I, x) ≤ ht (I) + 1, and the required equality follows.

2

270

Chapter 7

Deﬁnition 7.2.14 A family F consisting of independent sets 0 of a graph G is said to be a maximal independent cover of G if V (G) = C∈F C and |C| = β0 (G) for all C ∈ F . A graph G without isolated vertices having a maximal independent cover is called a B-graph [26]. Proposition 7.2.15 [26] If G is a B-graph, then G is vertex critical. Proof. Let v be a vertex of G and let e = {v, w} be an edge of G. Since w is contained in an independent set A of G with β0 (G) vertices, then v is in 2 V (G) \ A. Hence α0 (G \ {v}) < α0 (G). Proposition 7.2.16 [26] If G is an unmixed graph, then G has a maximal independent cover. Proof. The result follows readily because any two maximal independent sets have the same cardinality. 2 Let us summarize some of the results obtained thus far. The following implications hold for any graph without isolated vertices: C–M =⇒ unmixed =⇒ B-graph =⇒ vertex-critical Proposition 7.2.17 If G is a vertex critical graph and G\{v} is not vertex critical for all v ∈ V (G), then G has a maximal independent cover. Proof. We use induction on the number of vertices of G. If G has two vertices or α0 (G) = 1, then for both cases the assumptions on G imply that G has a single edge and the result is clear. Thus one may assume G has at least three vertices and α0 (G) ≥ 2. Assume G has no maximal independent cover. Let C1 , . . . , Cr be the set of all minimal vertex covers of G with α0 (G) vertices. Pick a vertex v ∈ V (G) such that v is in Ci for all i. By hypothesis V (G) = ∪ri=1 Ci because G is vertex-critical (Proposition 7.2.12). Hence degG (x) ≥ 2 for all x ∈ V (G) adjacent to v, indeed if degG (x) = 1, then x ∈ Cj for some j and Cj \ {x} is a vertex cover of G which is impossible. Thus G \ {v} has no isolated vertices. Since α0 (G \ {v}) is equal to α0 (G) − 1, the set Ci \ {v} is a minimal vertex cover of G \ {v} with α0 (G \ {v}) vertices for all i. Thus, from the equality V (G) \ {v} = (C1 \ {v}) ∪ · · · ∪ (Cr \ {v}) we conclude that G \ {v} is vertex critical, a contradiction. 2 Remark 7.2.18 Let G be a graph and let A be the intersection of all minimal vertex covers of G with α0 (G) vertices. Note that G has a maximal independent cover if and only if A = ∅. Proposition 7.2.19 Let G be a graph and let A be the intersection of all the minimal vertex covers of G with α0 (G) vertices. If G is vertex critical, then the graph G \ A is vertex-critical, has a maximal independent cover, and β0 (G) = β0 (G \ A).

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Proof. Let A = {v1 , . . . , vs }. If A = ∅, then G has a maximal independent cover and there is nothing to prove. Assume A = ∅. By the arguments in the proof of Proposition 7.2.17 the graph G \ {v1 } is vertex-critical and β0 (G) = β0 (G\{v1 }). Since the intersection of all the minimal vertex covers of G \ {v1 } with α0 (G \ {v1 }) vertices is equal to A \ {v1 }, the result follows by induction. 2 Theorem 7.2.20 [186] If G is a B-graph, then β0 (G) ≤ α0 (G). Let G be a graph. In what follows ΔG is the independence complex of G and K[ΔG ] is the Stanley–Reisner ring of ΔG over a ﬁeld K. Theorem 7.2.21 (Erd¨ os–Gallai [144]) If G is a vertex-critical graph with A B n vertices, then β0 (G) = dim R/I(G) = dim K[ΔG ] ≤ n2 . Proof. By induction on n. The case n = 2 is clear. Assume n ≥ 3. One may assume that G has no maximal independent cover, otherwise the result follows directly from Theorem 7.2.20 and α0 (G) + β0 (G) = n. By Proposition 7.2.17 there is v ∈ V (G) such that G \ {v} is vertex-critical. Thus by the induction hypothesis β0 (G \ {v}) ≤ (n − 1)/2. Since β0 (G) is β0 (G \ {v}) (see Proposition 7.2.12), the assertion follows. 2 Theorem 7.2.22 Let G be a graph and let A be the intersection of all the minimal vertex covers of G with α0 (G) vertices. If G is vertex critical, then dim K[ΔG ] ≤ n − |A|/2. Proof. It follows from Proposition 7.2.19 and Theorem 7.2.21.

2

Example 7.2.23 Given n, k positive integers such that n ≥ 2(k − 1), let G be the following graph G on n vertices.

xn−1

xn x2k−1 rH r G

J H

H J

J HHJ J

r H

Jr x2k H J HH J

J H J

H Jr

pp pH Jr

xn−2 x2k+1

x1 r .

x2 r

x3 r

xk−2 xk−1 r r ppp

r r r xk xk+1 xk+2

r r x2k−3 x2k−2

Kn−2k+2 Using Proposition 7.3.2 it follows that G is Cohen–Macaulay. Notice that β0 (G) = k. In particular if n = 2k or n = 2k + 1 one has β0 (G) = ( n2 ). Lemma 7.2.24 If G is a bipartite graph with n vertices, then α0 (G) ≤ ( n2 ).

272

Chapter 7

Proof. Let (V1 , V2 ) be a bipartition of G. Since n = |V1 | + |V2 |, then 2 |Vi | ≤ ( n2 ) for some i. Thus α0 (G)) ≤ n2 . Example 7.2.25 Let G be the graph below and let I be its edge ideal. xr1 Q S Q S Q SSrxQ 5 Q x4 r Qr x6 x3 r Q Q S Q S Q S Q Sr Q x2 Using the following simple procedure for Macaulay2 R=ZZ/101[x1,x2,x3,x4,x5,x6] I=ideal(x1*x2, x1*x3, x1*x4, x1*x5, x1*x6, x2*x3, x2*x4, x2*x5, x2*x6, x3*x4, x4*x5, x5*x6) primaryDecomposition I we obtain that the minimal vertex covers of G are: {x6, x5, x4, x3, x1}, {x6, x5, x4, x3, x2}, {x5, x3, x2, x1}, {x5, x4, x2, x1}, {x6, x4, x2, x1}. The graph G is vertex critical, β0 (G) = 2, and α0 (G) = 4. Thus the set A = {x1 , x2 } is the intersection of the minimal vertex cover of G with four vertices. Hence the bound in Theorem 7.2.22 gives β0 (G) ≤ 2 and the Erd¨ os–Gallai bound gives β0 (G) ≤ 3.

Exercises 7.2.26 If I is the ideal (x1 x2 x3 − 1) ∩ (x1 , x2 , x3 ) of the polynomial ring K[x1 , x2 , x3 ], then I = (x1 x2 x23 − x3 , x1 x22 x3 − x2 , x21 x2 x3 − x1 ), ht(I, xi ) is strictly greater than ht(I) + 1 and V (I, xi ) = {0} for all i. 7.2.27 Let C be a clutter with vertex set X = {x1 , . . . , xn } and let I be the ideal I = I(C) + (x21 , . . . , x2n ) ⊂ R, where R = K[X] is a polynomial ring over a ﬁeld K. Prove the following: (a) the standard monomials of R/I are in one-to-one correspondence with the independent sets of C, (b) the Hilbert series of R/I is a polynomial a0 + a1 t + · · · + ad td of degree d = β0 (C), where ai is the number of independent sets of C of size i. This polynomial is called the independence polynomial of C. 7.2.28 Let G be a graph. Construct a graph H by appending an isolated vertex x to G. Prove that R/I(G)[x] = R[x]/I(H), α0 (G) = α0 (H) and β0 (H) = β0 (G) + 1.

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7.2.29 Let G be a graph. If e is an edge of G such that α0 (G\{e}) < α0 (G), then α0 (G) = α0 (G \ {e}) + 1. 7.2.30 A graph G is edge-critical if α0 (G \ {e}) < α0 (G) for all e ∈ E(G). If G is an edge-critical graph, then G has a maximal independent cover. 7.2.31 [145] If G is an edge-critical graph, then q ≤ (α0 (G) + α0 (G)2 )/2 where q is the number of edges of G. 7.2.32 If G is a graph, then there is a spanning subgraph H of G such that H is edge-critical and β0 (H) = β0 (G). 7.2.33 Prove that every edge-critical bipartite graph is unmixed. 7.2.34 Prove that a cycle with nine vertices is an edge-critical graph which is not unmixed. 7.2.35 Let G be a graph with q edges. If G has a maximal independent cover, then q ≤ α0 (G)2 . 7.2.36 Prove that the complete bipartite graph G = Kg,g is an unmixed graph with α0 (G)2 edges and has a maximal independent cover. 7.2.37 Let G be a graph with n vertices. If G is an unmixed graph without isolated vertices such that 2α0 (G) = n, then G has a perfect matching. 7.2.38 Prove that the following graph is neither edge-critical nor vertexcritical, is not unmixed, and {x1 , x4 , x6 , x7 }, {x2 , x3 , x5 , x6 } is a maximal independent cover. xr 1 G @ @ x4 @ x6 rx3 r r r @r x5 x2 H H @ isolated vertex HH @ HH @ H @r x7 7.2.39 (N. Terai) Let R = K[a, . . . , j] be a polynomial ring on 10 variables over a ﬁeld K and let I = (abc, abd, acd, bcd, abe, acf, adg, bch, bdi, cdj) be the edge ideal of the clutter which consists of: the boundary of a tetrahedron (abc, abd, acd, bcd) and a new triangle on each edge of the tetrahedron using a new vertex (e.g., ab → abe). Prove that this ideal is unmixed of height 3 (in fact Cohen–Macaulay).

274

Chapter 7

7.2.40 Let G be a bipartite graph with n vertices. If G has no isolated vertices and G has a maximal independent cover, then n is an even integer. 7.2.41 Prove that the graph G shown below is vertex-critical, α0 (G) = 3, β0 (G) = 2, and G does not have a maximal independent cover. xr1 TT

T

T

Tr x4

xr 2 TT T

T

xr 3

Tr x5

7.2.42 (Dilworth’s theorem) Using K¨ onig’s theorem, prove that in any ﬁnite poset the cardinality of any largest antichain equals the cardinality of any smallest chain partition. 7.2.43 Prove that in a ﬁnite poset the cardinality of any longest chain equals the cardinality of any smallest antichain partition.

7.3

Cohen–Macaulay and chordal graphs

In this section we study Cohen–Macaulay graphs, chordal graphs, and the Cohen–Macaulay property of the ideal of covers of a graph. A classiﬁcation of Cohen–Macaulay trees is presented. Constructions of Cohen–Macaulay graphs We begin by showing how large classes of Cohen–Macaulay graphs can be produced and give some obstructions for a graph to be Cohen–Macaulay. rw

First construction Let H be a graph with vertex set V (H) = {x1 , . . . , xn , z, w} and J its edge ideal. rz Assume that z is adjacent to w with deg(z) ≥ 2 and @ @ deg(w) = 1. We label the vertices of H such that r @r r x1 xk x , . . . , x , w are the vertices of H adjacent to z, as x2 1 k shown in the ﬁgure. The next two results describe how the Cohen–Macaulay property of H relates to that of the two subgraphs G = H \ {z, w} and F = G \ {x1 , . . . , xk }. One has the equalities J = (I, x1 z, . . . , xk z, zw) and (I, x1 , . . . , xk ) = (L, x1 . . . , xk ), where I and L are the edge ideals of G and F , respectively. Assume H is unmixed with height of J equal to g + 1. Since z is not isolated, there is a minimal prime p over I containing {x1 , . . . , xk } and such that ht(I) = ht(p) = g. It is not diﬃcult to prove that k < n and deg(xi ) ≥ 2 for i = 1, . . . , k.

Edge Ideals of Graphs

275

Proposition 7.3.1 [416] If H is a Cohen–Macaulay graph, then F and G are Cohen–Macaulay graphs. Proof. Set A = K[x1 , . . . , xn ] and R = A[z, w]. By Proposition 3.1.23 there exists a homogeneous system of parameters {f1 , . . . , fd } for A/I, where fi ∈ A+ for all i. Because of the hypothesis and the equalities z(z − w) + zw = z 2 and w(w − z) + zw = w2 , the set {f1 , . . . , fd , z − w} is a regular system of parameters for R/J. Hence f1 , . . . fd is a regular sequence on R/I, that is, G is Cohen–Macaulay. To show the corresponding property for F we consider the exact sequence z

ψ

0 −→ R/(I, x1 , . . . , xk , w)[−1] −→ R/J −→ R/(I, z) −→ 0 where the ﬁrst map is multiplication by z and ψ is induced by a projection. The exactness of this sequence follows from Theorem 6.1.4. Taking depths with respect to the maximal ideal of R, by the depth lemma one has n − g + 1 ≤ depth R/(I, x1 , . . . , xk , w), where g = ht(I). Since (I, x1 , . . . , xk , w) = (L, x1 , . . . , xk , w), F is Cohen–Macaulay.

2

Proposition 7.3.2 If F and G are Cohen–Macaulay graphs and x1 , . . . , xk are in some minimal vertex cover of G, then H is a Cohen–Macaulay graph. Proof. Consider the exact sequence of Proposition 7.3.1. Since the ends of this sequence have R-depth equal to dim(R/J), by the depth lemma H is a Cohen–Macaulay graph. 2 Corollary 7.3.3 If G is Cohen–Macaulay and {x1 , . . . , xk } is a minimal vertex cover of G, then H is Cohen–Macaulay. Proof. Note that in this case F is C–M because I(F ) = (0).

2

Second construction For the discussion of the second construction we change our notation. Let H be a graph on the vertex set V (H) = {x1 , . . . , xn , z}. r Let {x1 , . . . , xk } be the vertices of H adjacent to z, x1 as shown in the ﬁgure. Taking into account the ﬁrst construction one may assume deg(xi ) ≥ 2 for i = 1, . . . , k and deg(z) ≥ 2. Setting G = H \ {z} and F = G \ {x1 , . . . , xk }, notice that the ideals J, I, and L associated to H, G, and F , respectively are related by the equalities J = (I, x1 z, . . . xk z) and (I, x1 , . . . , xk ) = (L, x1 , . . . , xk ). rz @ r @ @r xk x2

276

Chapter 7

Proposition 7.3.4 [416] If H is a Cohen–Macaulay graph, then F is a Cohen–Macaulay graph. Proof. We set R = K[x1 , . . . , xn , z], A = K[x1 , . . . , xn ] and ht(J) = g + 1. The polynomial f = z − x1 − · · · − xk is regular on R/J because it is clearly not contained in any associated prime of J. Therefore there is a sequence {f, f1 , . . . , fm } regular on R/J so that {f1 , . . . , fm } ⊂ A+ , where m = n − g − 1. Observe that {f1 , . . . , fm } is in fact a regular sequence on A/I, which gives depth(A/I) ≥ n − g − 1. Next, we use the exact sequence z

0 −→ R/(I, x1 , . . . , xk )[−1] −→ R/J −→ R/(I, z) −→ 0 and ht(I, x1 , . . . , xk ) = g + 1 to conclude that F is Cohen–Macaulay.

2

Proposition 7.3.5 Assume that x1 , . . . , xk are not contained in any minimal vertex cover for G and ht(I, x1 , . . . , xk ) = ht(I) + 1. If F and G are Cohen–Macaulay, then H is Cohen–Macaulay. Proof. The assumption on {x1 , . . . , xk } forces ht(J) = ht(I) + 1. From the exact sequence z

0 −→ R/(I, x1 , . . . , xk )[−1] −→ R/J −→ R/(I, z) −→ 0 we obtain that H is Cohen–Macaulay.

2

Corollary 7.3.6 If G is Cohen–Macaulay and {x1 , . . . , xk−1 } is a minimal vertex cover for G, then H is Cohen–Macaulay. Connected components of Cohen–Macaulay graphs An important property of the family of Cohen–Macaulay graphs is their additivity with respect to connected components. Lemma 7.3.7 Let K[x] and K[y] be two polynomial rings over a ﬁeld K. If I1 and I2 are graded ideals of K[x] and K[y], respectively, then depth(K[x]/I1 ) + depth(K[y]/I2 ) = depth(K[x, y]/(I1 + I2 )). Proof. The formula is a direct consequence of Proposition 3.1.33 and Theorem 3.1.34. 2 Proposition 7.3.8 A graph G is Cohen–Macaulay if and only if all its connected components are Cohen–Macaulay. Proof. The result follows from Lemma 7.3.7.

2

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Whisker graphs in Cohen–Macaulay trees In this part we introduce whisker graphs, present a characterization of all Cohen–Macaulay trees, and construct a family of graphs containing all Cohen–Macaulay trees. Proposition 7.3.9 Let R = K[x1 , . . . , xn , y1 , . . . , yn ] be a polynomial ring over a ﬁeld K and M ⊂ {(i, j)| 1 ≤ i < j ≤ n}. Then, the ideal I = ({xi yi , yj y | i = 1, . . . , n and (j, ) ∈ M }), is Cohen–Macaulay. Proof. The polarization of (x2i , xj x | 1 ≤ i ≤ n, (j, ) ∈ M ) is equal to I. Hence, the result follows applying Proposition 6.1.6. 2 Deﬁnition 7.3.10 Let G0 be a graph on the vertex set Y = {y1 , . . . , yn } and take a new set of variables X = {x1 , . . . , xn }. The whisker graph or suspension of G0 , denoted by G0 ∪ W (Y ), is the graph obtained from G0 by attaching to each vertex yi a new vertex xi and the edge {xi , yi }. The edge {xi , yi } is called a whisker. Proposition 7.3.11 Let G0 be a graph and let G = G0 ∪ W (Y ) be its d whisker graph. Then the multiplicity of R/I(G) is i=−1 fi , where fi is the number of i-faces of the Stanley–Reisner complex of I(G0 ). Proof. Let Y = {y1 , . . . , yn } be the vertex set of G0 . By Proposition 7.3.9, the set of linear forms {yi − xi }ni=1 is a regular system of parameters for R/I(G). The reduction of R/I(G) modulo these forms is S = K[Y ]/(y12 , . . . , yn2 , I(G0 )), whose K-vector space basis is formed by the standard monomials, that is, by the faces of ΔI(G0 ) . To complete the proof notice that the degree of R/I(G) is the degree of S by Corollary 5.1.10. 2 Proposition 7.3.12 Let G0 and G1 be two copies of the same graph on disjoint sets of vertices {y1 , . . . , yn } and {x1 , . . . , xn }, respectively. If G∨ 1 is the clutter of minimal vertex covers of G1 and L = ({xi1 · · · xis | {xi1 , . . . , xis } ∈ E(G∨ 1 )}), then ((z) : I(G)) = (z, L), where z = {xi yi }ni=1 and G = G0 ∪ W (Y ). Proof. Take {xi1 , . . . , xir } ∈ E(G∨ 1 ) and yk y ∈ I(G0 ). Since is ∈ {k, } for some s we obtain xi1 · · · xir yk y ∈ (z). This shows (z, L) ⊂ ((z) : I(G)). Conversely, assume M is a monomial in ((z) : I(G)) \ (z). Let {yk , y } be any edge in G0 . Since M yk y = xt yt M1 , either xk divides M or x divides M ; in either case, M ∈ L. 2

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Corollary 7.3.13 The type of K[X, Y ]/I(G0 ∪ W (Y )) is |E(G∨ 0 )|. Example 7.3.14 Let I(G0 ) = (y1 y3 , y1 y4 , y2 y4 , y2 y5 , y3 y5 , y3 y6 , y4 y6 ) and let G be the whisker graph of G0 . The ideal ((z) : I(G)) is generated by {xi yi }6i=1 ∪ {x2 x3 x4 , x3 x4 x5 , x1 x2 x5 x6 , x1 x2 x3 x6 , x1 x4 x5 x6 }. The last ﬁve monomials correspond to the minimal primes of I(G0 ), the type of R/I(G) is 5 and its multiplicity is 17. Lemma 7.3.15 Let G be a tree. If v and w are two adjacent vertices of degree at least two, then there is a minimal vertex cover of G containing both v and w. Proof. Let x be the edge joining v and w. Since G \ {x} has exactly two components, say G1 and G2 , there are minimal vertex covers A and B for G1 and G2 , respectively, so that v ∈ A and w ∈ B. Therefore A ∪ B is the required cover for G. 2 Theorem 7.3.16 [416] Let G be a tree. Then, G is Cohen–Macaulay if and only if |V (G)| ≤ 2 or 2 < |V (G)| = 2r and G has a perfect matching {x1 , y1 }, . . . , {xr , yr } so that deg(xi ) = 1, deg(yi ) ≥ 2 for i = 1, . . . , r. Proof. ⇒) Let (V1 , V2 ) be a bipartition of G. Since V1 , V2 are minimal vertex covers of G, and G is unmixed, we have r = |V1 | = |V2 | = α0 (G). By Theorem 7.1.8 there are r independent edges. Therefore, we may assume V1 = {zi }ri=1 , V2 = {wi }ri=1 , and {zi , wi } ∈ E(G) for i = 1, . . . , r. To complete the proof it suﬃces to show that for each i either zi or wi has degree one. Assume deg(zi ) ≥ 2 and deg(wi ) ≥ 2 for some i. By Lemma 7.3.15 there is a minimal vertex cover for T , say A, containing zi and wi . Since {zj , wj } A = ∅ for j = i we conclude |A| ≥ r + 1, which is a contradiction. ⇐) The suﬃciency follows from Proposition 7.3.9. 2 The signiﬁcance of the notion of a whisker graph lies partly in the next restatement of Theorem 7.3.16. Theorem 7.3.17 If G is a graph, then G is a Cohen–Macaulay tree if and only if G = G0 ∪ W (Y ) for some tree G0 with vertex set Y . Corollary 7.3.18 If G is a tree, then G is Cohen–Macaulay if and only if G is unmixed. Proof. If G is Cohen–Macaulay, then G is unmixed by Proposition 7.2.9. If G is unmixed, using the proof above it follows that I(G) is a Cohen– Macaulay ideal of the form described in Proposition 7.3.9. 2

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Corollary 7.3.19 The only Cohen–Macaulay cycles are the triangle and the pentagon. Proof. Let Cn = {x0 , x1 , . . . , xn = x0 } be a Cohen–Macaulay cycle of length n. The cases n = 4, 6, and 7 can be treated separately, so we assume n ≥ 8. By Proposition 7.3.4 Cn \ {x0 , x1 , xn−1 } is a Cohen–Macaulay path of length n − 3 which is impossible by Theorem 7.3.16 2 The next result generalizes the Cohen–Macaulay criterion for trees given in Theorem 7.3.16 and is a generalization of Faridi’s characterization of unmixed simplicial trees [156]. Theorem 7.3.20 [325] Let C be a clutter with the K¨ onig property and with no special cycles of length 3 or 4. The following conditions are equivalent : (a) C is unmixed. (b) There is a perfect matching e1 , . . . , eg , g = ht I(C), such that ei has a free vertex for all i, and for any two edges f1 , f2 of C and for any edge ei , one has that f1 ∩ ei ⊂ f2 ∩ ei or f2 ∩ ei ⊂ f1 ∩ ei . (c) R/I(C) is Cohen–Macaulay. (d) ΔC is a pure shellable simplicial complex. Proof. Using Lemma 6.5.7, Theorems 6.5.18, and 6.5.21, and Proposition 6.5.19 it follows readily that conditions (a) and (b) are equivalent. Since (a) is equivalent to (b), from Theorem 6.5.20 we get that (b) implies (d). That (d) implies (c) and (c) implies (a), were shown in Theorem 6.3.23 and Corollary 6.3.7. 2 Simplicial complexes of edge ideals Given a graph G on the vertex set V one deﬁnes a simplicial complex Simp(G) on the same set of vertices: Simp(G) = {F | ∃ H a subgraph of G such that F = V (H) and H Kr }, Conversely a simplicial complex Δ deﬁnes a graph Skel(Δ), the 1-skeleton of Δ, on the same set of vertices: Skel(Δ) = {F ∈ Δ| dim(F ) ≤ 1}. In general one has Δ ⊂ Simp(Skel(Δ)). Proposition 7.3.21 Let Δ be a simplicial complex and IΔ its Stanley– Reisner ideal. Then Δ = Simp(Skel(Δ)) if and only if IΔ is generated by square-free monomials of degree two.

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Proof. Let V = {x1 , . . . , xn } be the vertex set of Δ. Assume IΔ is generated by square-free monomials of degree two. Take F ∈ Simp(Skel(Δ)), for simplicity set F = {x1 , . . . , xr }, that is, {xi , xj } ∈ Δ for all 1 ≤ i < j ≤ r. If F ∈ Δ, then x1 · · · xr is in IΔ , and by assumption xi xj is in IΔ for some 1 ≤ i < j ≤ r. Hence {xi , xj } ∈ Δ, which is a contradiction. This shows Δ = Simp(Skel(Δ)). The converse also follows readily. 2 Simplicial complexes associated to edge ideals have been studied in the literature. They are called ﬂag complexes. (See [395, Chapter III] and the references there for a series of amusing related problems). Deﬁnition 7.3.22 The complement G of a graph G has vertex set V (G) and two vertices are adjacent in G if and only if they are not adjacent in G. The complement of G is also denoted by G . Proposition 7.3.23 Let G be a graph and let ΔG be its independence complex. Then G ⊂ ΔG , with equality if and only if G has no triangles. Proof. If x = {v1 , v2 } is an edge of G, then {v1 , v2 } is a stable set of G and x ∈ ΔG . Hence G ⊂ ΔG . Assume G has no triangles. If A = {v1 , . . . , vn } is a stable set of G, then n = 1 or n = 2; otherwise if n > 2, then G would contain the triangle {v1 , v2 , v3 }. Therefore A is either a point or an edge of G, thus G = ΔG . The converse also follows readily. 2 Chordal graphs The use of graph theoretical methods in commutative algebra is implicit in Lyubeznik thesis [299] and explicit in the work of Fr¨ oberg [172]. In both cases the central notion is that of a chordal graph. Let G be a graph and let G be its complement. Recall that G is said to be chordal if every cycle Cn in G of length n ≥ 4 has a chord in G. A chord of Cn is an edge joining two non-adjacent vertices of Cn . Chordal graphs have been extensively studied [112, 228, 403]. Lemma 7.3.24 Let G be a connected graph and let d be the minimum degree among the vertices of G. If G is not a complete graph, then there is S ⊂ V (G) such that G \ S is disconnected and d = |S|. Proof. Let v be a vertex of G of degree d and S = NG (v) the neighbor set of v. Since G is not a complete graph and v has minimum degree there is a vertex w ∈ S ∪ {v}. Thus G \ S has at least two components. 2 Proposition 7.3.25 (Dirac [112]) If H is a chordal non-complete graph, then it can be constructed out of two smaller disjoint chordal graphs H1 and H2 by identifying two (possibly empty) complete subgraphs of the same size in H1 and H2 .

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Proof. One may assume that H is connected and has n vertices. Since H has at least one vertex of degree less than n − 1, by Lemma 7.3.24 there is a minimal set of vertices S such that H \ S is disconnected and |S| ≤ n − 2. Hence H \ S = F1 ∪ F2 , where F1 and F2 are disjoint subgraphs of H. Consider the induced subgraphs H1 = H[V (F1 ) ∪ S] and H2 = H[V (F2 ) ∪ S]. Notice that H1 ∪ H2 = H and H1 ∩ H2 = H[S]. Next we show that H[S] is a complete graph. Assume there are vertices x1 and x2 in S not connected by an edge of H. Since H \ (S \ {xi }) is connected for i = 1, 2, there are paths p1 and p2 in H1 and H2 , respectively, joining x1 with x2 , whose only vertices in S are x1 and x2 . If p1 and p2 are the shortest paths with this property, then p1 ∪ p2 is a cycle of length greater than or equal to four without a chord, a contradiction. 2 Theorem 7.3.26 [300] Let G be a graph and let G be its complement. Then Ic (G) is Cohen–Macaulay if and only if G is a chordal graph. Proposition 7.3.27 If Ic (G) is Cohen–Macaulay, then μ(Ic (G)) ≤ g + 1. where μ(Ic (G)) is the minimum number of generators of Ic (G) and g is the height of I(G). Proof. By the Hilbert–Burch Theorem (see Theorem 3.5.16) Ic (G) is generated by the m − 1 minors of an m × (m − 1) matrix A with homogeneous entries, where m = μ(Ic (G)). Take f in Ic (G) of degree g, since any m − 1 minor of A has degree at least m − 1, one obtains deg(f ) = g ≥ m − 1. 2

Exercises 7.3.28 Show that a cycle Cn of length n ≥ 3 is unmixed if and only if n = 3, 4, 5, 7. 7.3.29 Let G1 , G2 be two graphs with vertex set V . If their complements are triangle free and they have the same number of edges, then the Hilbert series of K[V ]/I(G1 ) and K[V ]/I(G2 ) are identical. 7.3.30 If G is a graph and ht(I(G)) = |V (G)| − 2, then (a) G is Cohen–Macaulay if and only if ΔG is connected, and (b) if I(G) is not Cohen–Macaulay, then I(G) is unmixed if and only if ΔG has no isolated vertices. 7.3.31 If G is a Cohen–Macaulay graph with q edges, then q ≤ (g 2 + g)/2, where g is the height of I(G).

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7.3.32 If G is a Cohen–Macaulay graph and u, v are two vertices of degree 1, prove that they cannot have an adjacent vertex in common. 7.3.33 Let G be a connected graph and let {v1 , . . . , vn } be a path with n ≥ 4. If deg(v1 ) = 1, deg(vi ) = 2 for i = 2, . . . n − 1 and deg(vn ) ≥ 3, prove that G cannot be unmixed. 7.3.34 Prove that H = Skel(Simp(H)), for any graph H. 7.3.35 Let T be a tree with 2r ≥ 4 vertices. Then T is Cohen–Macaulay if and only if ht I(T ) = r and T has exactly r vertices of degree 1. 7.3.36 Prove that the ideal I = IΔ of Example 6.3.65 is generated by x1 x3 , x1 x4 , x1 x7 , x1 x10 , x1 x11 , x2 x4 , x2 x5 , x2 x8 , x2 x10 , x2 x11 , x3 x5 , x3 x6 , x3 x8 , x3 x11 , x4 x6 , x4 x9 , x4 x11 , x5 x7 , x5 x9 , x5 x11 , x6 x8 , x6 x9 , x7 x9 , x7 x10 , x8 x10 . Let Ri = K[x1 , . . . , x 'i , . . . , x11 ] and Gi = G \ {xi }. Then I = I(G) for some graph G. Prove that the f -vector of ΔGi is (10, 25, 15) or (10, 24, 14), the h-vector of Ri /I(Gi ) is (1, 7, 8, −1) or (1, 7, 7, −1), and Gi is not a C–M graph for all i. 7.3.37 Let I = (xi yj | 1 ≤ i ≤ j ≤ n) ⊂ R = K[X, Y ]. Prove that R/I has type n and its a-invariant is −(n − 1).

7.4

Shellable and sequentially C–M graphs

In this section we classify all sequentially Cohen–Macaulay bipartite graphs (they are precisely the shellable bipartite graphs) and give a recursive procedure to verify if a bipartite graph is shellable. Then we present some structure theorems for unmixed and Cohen–Macaulay bipartite graphs. Deﬁnition 7.4.1 Let G be a graph whose independence complex is ΔG . We say G is a shellable graph if ΔG is a shellable simplicial complex. To prove that a graph G is shellable, it suﬃces to prove each connected component of G is shellable. Lemma 7.4.2 [409] Let G1 and G2 be two graphs with disjoint sets of vertices and let G = G1 ∪ G2 . Then G1 and G2 are shellable if and only if G is shellable.

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Proof. ⇒) Let F1 , . . . , Fr and H1 , . . . , Hs be the shellings of ΔG1 and ΔG2 , respectively. Then if we order the facets of ΔG as F1 ∪ H1 , . . . , F1 ∪ Hs ; F2 ∪ H1 , . . . , F2 ∪ Hs ; . . . ; Fr ∪ H1 , . . . , Fr ∪ Hs we get a shelling of ΔG . Indeed if F < F are two facets of ΔG we have two cases to consider. Case (i): F = Fi ∪ Hk and F = Fj ∪ Ht , where i < j. Because ΔG1 is shellable there is v ∈ Fj \ Fi and < j with Fj \ F = {v}. Hence v ∈ F \ F , F ∪ Ht < F , and F \ (F ∪ Ht ) = {v}. Case (ii): F = Fk ∪ Hi and F = Fk ∪ Hj , where i < j. This case follows from the shellability of ΔG2 . ⇐) Note that if F is a facet of ΔG , then F = F ∩ VG1 , respectively, F = F ∩ VG2 , is a facet of ΔG1 , respectively, ΔG2 . We now show that G1 is shellable and omit the similar proof for the shellability of G2 . Let F1 , . . . , Ft be a shelling of ΔG , and consider the subsequence Fi1 , . . . , Fis

with 1 = i1 < i2 < · · · < is

where F1 ∩ VG2 = Fij ∩ VG2 for ij ∈ {i1 , . . . , is }, but F1 ∩ VG2 = Fk ∩ VG2 for any k ∈ {1, . . . , t} \ {i1 , . . . , is }. We then claim that F1 = Fi1 \ VG2 , F2 = Fi2 \ VG2 , . . . , Fs = Fis \ VG2 is a shelling of ΔG1 . We ﬁrst show that this is a complete list of facets; indeed, each Fj = Fij ∩VG1 is a facet of ΔG1 , and furthermore, for any facet F ∈ ΔG1 , F ∪ (F1 ∩ VG2 ) is a facet of ΔG , and hence F ∪ (F1 ∩ VG2 ) = Fij for some ij ∈ {i1 , . . . , is }. Because the Fi ’s form a shelling, if 1 ≤ k < j ≤ s, there exists v in Fij \ Fik = (Fij \ VG2 ) \ (Fik \ VG2 ) = Fj \ Fk such that {v} = Fij \ F for some 1 ≤ < ij . It suﬃces to show that F is among Fi1 , . . . , Fis . Now because Fij ∩ VG2 ⊂ Fij and v ∈ Fij ∩ VG2 , we must have Fij ∩ VG2 ⊂ F . So, F ∩ VG2 ⊃ Fij ∩ VG2 . But F ∩ VG2 is a facet of ΔG2 , so we must have F ∩ VG2 = Fij ∩ VG2 . So F = Fir for some r < j, and hence, {v} = Fj \ Fr , as desired. 2 If F is a face of a simplicial complex Δ, the link of F is deﬁned to be lkΔ (F ) = {G | G ∪ F ∈ Δ, G ∩ F = ∅}. When F = {x}, then we shall abuse notation and write lkΔ (x) instead of lkΔ ({x}). Lemma 7.4.3 Let x be a vertex of G and let G = G\ ({x} ∪NG (x)). Then ΔG = lkΔG (x), / F and F ∪ {x} is a facet of ΔG . and F is a facet of ΔG if and only if x ∈ Proof. If F ∈ lkΔG (x), then x ∈ F , and F ∪ {x} ∈ ΔG implies that F ∪ {x} is an independent set of G. So (F ∪ {x}) ∩ NG (x) = ∅. But this means that

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F ⊂ VG because VG = VG \ ({x} ∪ NG (x)). Thus F ∈ ΔG since F is also an independent set of the smaller graph G . Conversely, if F ∈ ΔG , then F is an independent set of G that does not contain any of the vertices of {x} ∪ NG (x). But then F ∪ {x} is an independent set of G, i.e., F ∪ {x} ∈ ΔG . So F ∈ lkΔG (x). The last statement follows readily from the fact that F is a facet of / F and F ∪ {x} is a facet of ΔG . 2 lkΔG (x) if and only if x ∈ Theorem 7.4.4 [409] Let x be a vertex of a graph G and let G be the graph G = G \ ({x} ∪ NG (x)). If G is shellable, then G is shellable. Proof. Let F1 , . . . , Fs be a shelling of ΔG . Suppose the subsequence Fi1 , Fi2 , . . . , Fit with i1 < i2 < · · · < it is the list of all the facets with x ∈ Fij . Setting Hj = Fij \ {x} for each j = 1, . . . , t, Lemma 7.4.3 implies that the Hj ’s are the facets of ΔG . We claim that H1 , . . . , Ht is a shelling of ΔG . As the Fi ’s form a shelling, if 1 ≤ k < j ≤ t, there is v ∈ Fij \ Fik = (Fij \ {x}) \ (Fik \ {x}) = (Hj \ Hk ) such that {v} = Fij \ F for some 1 ≤ < ij . It suﬃces to show that F is among the list Fi1 , . . . , Fit . But because x ∈ Fij and x = v, we must have x ∈ F . Thus F = Fik for some k ≤ j. But then {v} = Fij \ F = Hj \ Hk . 2 So, the Hi ’s form a shelling of ΔG . Let G be a graph and let S ⊂ VG . The graph G ∪ WG (S) obtained from G by adding new vertices {yi | xi ∈ S} and new edges {{xi , yi } | xi ∈ S} is called the whisker graph of G over S. The edges {xi , yi } are called whiskers. Corollary 7.4.5 [409] Let G be a graph and let S ⊂ VG . If G ∪ WG (S) is shellable, then G \ S is shellable. Proof. We may assume that S = {x1 , . . . , xs }. Set G0 = G ∪ WG (S) and Gi = Gi−1 \ ({yi } ∪ NG (yi )) for i = 1, . . . , s. Notice that Gs = G \ S. Hence, by repeatedly applying Theorem 7.4.4, the graph G \ S is shellable. 2 Lemma 7.4.6 Let G be a bipartite graph with bipartition {x1 , . . . , xm }, {y1 , . . . , yn }. If G is shellable and G has no isolated vertices, then there is v ∈ VG with deg(v) = 1. Proof. Let F1 , . . . , Fs be a shelling of ΔG . We may assume that Fi = {y1 , . . . , yn }, Fj = {x1 , . . . , xm } and i < j. Then there is xk ∈ Fj \ Fi and F with ≤ j − 1 such that Fj \ F = {xk }. For simplicity assume that xk = x1 . Then {x2 , . . . , xm } ⊂ F and there is yt in F for some 1 ≤ t ≤ n. Since {yt , x2 , . . . , xm } is an independent set of G, we get that yt can only be adjacent to x1 . Thus deg(yt ) = 1 because G has no isolated vertices. 2

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Theorem 7.4.7 [409] Let G be a graph and let {x1 , x2 } be an edge of G with deg(x1 ) = 1. If Gi = G \ ({xi } ∪ NG (xi )) for i = 1, 2, then G is shellable if and only if G1 and G2 are shellable. Proof. If G is shellable, then G1 and G2 are shellable by Theorem 7.4.4. So it suﬃces to prove the reverse direction. Let F1 , . . . , Fr be a shelling of ΔG1 and let H1 , . . . , Hs be a shelling of ΔG2 . It suﬃces to prove that F1 ∪ {x1 }, . . . , Fr ∪ {x1 }, H1 ∪ {x2 }, . . . , Hs ∪ {x2 } is a shelling of ΔG . One ﬁrst shows that this is the complete list of facets of ΔG using Lemma 7.4.3. Indeed, take any facet F of ΔG . If x2 ∈ F , then x1 ∈ F because {x1 , x2 } is an edge of G, and by Lemma 7.4.3, F \{x2 } = Hi for some i. On the other hand, if x2 ∈ F , we must have x1 ∈ F , because if not, then {x1 }∪F is larger independent set of G because x1 is only adjacent to x2 . Again, by Lemma 7.4.3, we have F \ {x1 } = Fi for some i. Let F < F be two facets of ΔG . Consider the case in which F = Fi ∪ {x1 } and F = Hj ∪ {x2 }. Since Hj ∪ {x1 } is an independent set of G, it is contained in a facet of ΔG , i.e., Hj ∪ {x1 } ⊂ F ∪ {x1 } for some . Hence (Hj ∪ {x2 }) \ (F ∪ {x1 }) = {x2 }, x2 ∈ F \ F , and F ∪ {x1 } < F . The remaining two cases follow readily from the shellability of ΔG1 and ΔG2 . 2 The last two results yield a recursive procedure to verify if a bipartite graph is shellable. The next result is the ﬁrst combinatorial characterization of all shellable bipartite graphs. Corollary 7.4.8 [409] Let G be a bipartite graph. Then G is shellable if and only if there are adjacent vertices x and y with deg(x) = 1 such that the bipartite graphs G \ ({x} ∪ NG (x)) and G \ ({y} ∪ NG (y)) are shellable. Proof. By Lemma 7.4.2 it suﬃces to verify the statement when G is connected. By Lemma 7.4.6 there exists a vertex of x1 with deg(x1 ) = 1. Now apply the previous theorem. 2 A clutter is called vertex decomposable if its independence complex is vertex decomposable. The following major result of Woodroofe gives a suﬃcient condition for vertex decomposability of graphs and is an extension of the fact that chordal graphs are shellable [409]. Theorem 7.4.9 [432, Theorem 1.1] If G is a graph with no chordless cycles of length other than 3 or 5, then G is vertex decomposable (hence shellable and sequentially Cohen–Macaulay). Deﬁnition 7.4.10 A clutter C is called sequentially Cohen–Macaulay if R/I(C) is sequentially Cohen–Macaulay. We now give a sequentially Cohen–Macaulay analog of Theorem 7.4.4.

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Theorem 7.4.11 [409] Let x be a vertex of a graph G and let G be the graph G = G \ ({x} ∪ NG (x)). If G is sequentially Cohen–Macaulay, then G is sequentially Cohen–Macaulay. Proof. Let F1 , . . . , Fs be the facets of Δ = ΔG , and let F1 , . . . , Fr be the facets of Δ that contain x. Set Γ = ΔG ; by Lemma 7.4.3, the facets of Γ are F1 = F1 \ {x}, . . . , Fr = Fr \ {x}. Consider the pure simplicial complexes Δ[k]

=

{F ∈ Δ| dim(F ) = k}; −1 ≤ k ≤ dim(Δ),

[k]

=

{F ∈ Γ| dim(F ) = k}; −1 ≤ k ≤ dim(Γ),

Γ

where F denotes the subcomplex generated by the set of faces F . Recall that H is a face of F if and only if H is contained in some F in F . Take a facet Fi of Γ of dimension d = dim(Γ). Then Fi ∪ {x} ∈ Δ[d+1] and consequently {x} ∈ Δ[k+1] for k ≤ d. Because the facets of Γ are F1 = F1 \ {x}, . . . , Fr = Fr \ {x}, we have the equality Γ[k] = lkΔ[k+1] (x) for k ≤ d. By Theorem 6.3.30, the complex Δ is sequentially Cohen– Macaulay if and only if Δ[k] is Cohen–Macaulay for −1 ≤ k ≤ dim(Δ). Because Δ[k] is Cohen–Macaulay, by Proposition 6.3.15 lkΔ[k] (F ) is Cohen– Macaulay for any F ∈ Δ[k] . Thus, Γ[k] = lkΔ[k+1] (x) is Cohen–Macaulay for any −1 ≤ k ≤ dim(Γ) ≤ dim(Δ) − 1. Therefore Γ is sequentially Cohen– Macaulay by Theorem 6.3.30, as required. 2 The following result is due to Francisco and H`a. Corollary 7.4.12 [159, Theorem 4.1] Let G be a graph and let S ⊂ VG . If the whisker graph G ∪ WG (S) of G over S is sequentially Cohen–Macaulay, then G \ S is sequentially Cohen–Macaulay. Proof. We may assume that S = {x1 , . . . , xs }. Set G0 = G ∪ WG (S) and Gi = Gi−1 \ ({yi } ∪ NG (yi )) for i = 1, . . . , s where yi is the degree 1 vertex adjacent to xi . Notice that Gs = G \ S. Hence by Theorem 7.4.11 the graph G \ S is sequentially Cohen–Macaulay. 2 Theorem 7.4.13 [229] If I is the edge ideal of a graph, then I has linear quotients if and only if I has a linear resolution if and only if each power of I has a linear resolution Lemma 7.4.14 Let G be a bipartite graph with bipartition {x1 , . . . , xm }, {y1 , . . . , yn }. If G is sequentially Cohen–Macaulay and G is not a discrete graph, then there is v ∈ VG with deg(v) = 1.

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Proof. We may assume that m ≤ n and that G has no isolated vertices. Let J be the Alexander dual of I = I(G) and let L = J[n] be the monomial ideal generated by the square-free monomials of J of degree n. We may assume that L is generated by g1 , . . . , gq , where g1 = y1 y2 · · · yn and g2 = x1 · · · xm y1 · · · yn−m . Consider the linear map ϕ

Rq −→ R (ei → gi ). The kernel of this map is generated by syzygies of the form (gj / gcd(gi , gj ))ei − (gi / gcd(gi , gj ))ej . Since the vector α = x1 · · · xm e1 − yn−m+1 · · · yn e2 is in ker(ϕ) and since ker(ϕ) is generated by linear syzygies (see Theorem 6.3.44), there is a linear syzygy of L of the form xj e1 − zek , where z is a variable, k = 1. Hence xj (y1 · · · yn ) = z(gk ) and gk = xj y1 · · · yi−1 yi+1 · · · yn for some i. Because the support of gk is a vertex cover of G, we get that the complement of the support of gk , i.e., {yi , x1 , . . . , xj−1 , xj+1 , . . . , xm }, is an independent set of 2 G. Thus yi can only be adjacent to xj , i.e., deg(yi ) = 1. Proposition 7.4.15 If G is a C–M bipartite graph, then G \ {u} is C–M for some vertex u of G. Proof. By Lemma 7.4.14, there is an edge {u, v} of the graph G such that deg(v) = 1. Then by Proposition 7.3.1 we obtain that G \ {u} is Cohen–Macaulay. 2 Theorem 7.4.16 [409] Let G be a bipartite graph. Then G is shellable if and only if G is sequentially Cohen–Macaulay. Proof. Assume that G is sequentially Cohen–Macaulay. The proof is by induction on the number of vertices of G. By Lemma 7.4.14 there is a vertex x1 of G of degree 1. Let x2 be the vertex of G adjacent to x1 . For i = 1, 2 consider the subgraph Gi = G\ ({xi } ∪NG (xi )). By Theorem 7.4.11 G1 and G2 are sequentially Cohen–Macaulay. Hence ΔG1 and ΔG2 are shellable by the induction hypothesis. Therefore ΔG is shellable by Theorem 7.4.7. The converse follows at once from Theorem 6.3.27. 2 Van Tuyl [407] has shown that the independence complex ΔG must be vertex decomposable for any bipartite graph G whose edge ring R/I(G) is sequentially Cohen–Macaulay. Thus Theorem 7.4.16 remains valid if we replace shellable by vertex decomposable. As we saw in Corollary 7.4.8, one can verify recursively that a bipartite graph is shellable. The above theorem, therefore, implies the same for sequentially Cohen–Macaulay bipartite graphs.

288

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Lemma 7.4.17 The complete bipartite graph Km,n is Cohen–Macaulay if and only if m = n = 1. Proof. By Exercise 6.3.51 it is enough to observe that the independence complex of Km,n is disconnected for m + n ≥ 3. 2 Lemma 7.4.18 Let G be an unmixed bipartite graph and let I(G) be its edge ideal. If I(G) has height g, then there are disjoint sets of vertices V1 = {x1 , . . . , xg } and V2 = {y1 , . . . , yg } such that: (i) {xi , yi } is an edge of G for all i, and (ii) every edge of G joins V1 with V2 . Proof. The statement follows by using that g is equal to the maximum number of independent edges of G; see Theorem 7.1.8. 2 The following is a combinatorial characterization of all unmixed bipartite graphs. Unmixed graphs are also called well-covered [340]. Theorem 7.4.19 [421] Let G be a bipartite graph. Then G is unmixed if and only if there is a perfect matching e1 , . . . , eg such that for any two edges e = e and for any two distinct vertices x ∈ e, y ∈ e contained in some ei , one has that (e \ {x}) ∪ (e \ {y}) is an edge. Proof. It follows at once from Corollary 6.5.16 because, by Theorem 7.1.8, bipartite graphs satisfy the K¨ onig property. 2 This result can be reformulated as: Theorem 7.4.20 [421] Let G be a bipartite graph without isolated vertices. Then G is unmixed if and only if G has a bipartition V1 = {x1 , . . . , xg }, V2 = {y1 , . . . , yg } such that: (a) {xi , yi } ∈ E(G) for all i, and (b) if {xi , yj } and {xj , yk } are in E(G) and i, j, k are distinct, then {xi , yk } ∈ E(G). The following nice result of Herzog and Hibi classiﬁes the family of Cohen–Macaulay bipartite graphs. This family is contained in the class of uniform admissible clutters studied in [160, 205, 325]. Theorem 7.4.21 [222] Let G be a bipartite graph without isolated vertices. Then G is a Cohen–Macaulay graph if and only if there is a bipartition V1 = {x1 , . . . , xg }, V2 = {y1 , . . . , yg } of G such that: (i) {xi , yi } ∈ E(G) for all i, (ii) if {xi , yj } ∈ E(G), then i ≤ j, and (iii) if {xi , yj } and {xj , yk } are in E(G) and i < j < k, then {xi , yk } ∈ E(G). Proof. By Proposition 7.3.8 we may assume that G is connected. ⇐) The proof is by induction on g. The case g = 1 is clear. Assume g ≥ 2. Let S = {xr1 , . . . , xrs } be the set of all vertices of G adjacent to yg . Consider the subgraph G = G \ S obtained from G be removing the

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vertices in S. We claim that yr1 , . . . , yrs are isolated vertices of G . Indeed if yrj is not isolated, there is an edge {xi , yrj } in G with i < rj . Hence by condition (iii) we get that {xi , yg } is an edge of G and xi must be a vertex in S, a contradiction. Thus by induction the graphs G \ {yr1 , . . . , yrs } and G \ {xg , yg } are Cohen–Macaulay. Applying Proposition 7.3.2 we get that G is Cohen–Macaulay. ⇒) By Lemma 7.4.18 there is a bipartition (V1 , V2 ) satisfying (i). By Lemma 7.4.14 we may assume that deg(xg ) = 1 and deg(yg ) ≥ 2. Since G is Cohen–Macaulay we get that G \ {yg } is Cohen–Macaulay; see Proposition 7.3.1. Using induction on G it is seen that there is a bipartition of G satisfying (i) and (ii). To complete the proof notice that G must also satisfy (iii) according to Theorem 7.4.20. 2 The following result characterizes all bipartite graphs with a perfect matching that satisﬁes condition (ii) of Theorem 7.4.21. Theorem 7.4.22 [325, Theorem 4.3] Let G be a bipartite graph with a [g] perfect matching {ei }gi=1 such that ei = {xi , yi } for all i and let Γ = ΔG be the pure g-skeleton of ΔG . Then, Γ is pure shellable if and only if we can order e1 , . . . , eg such that {xi , yj } ∈ E(G) implies i ≤ j. Directed graphs in sequentially C–M graphs Let G be an unmixed bipartite graph without isolated vertices. Then, by Theorem 7.4.20, there is a bipartition V1 , V2 of G such that (1) if V1 = {x1 , . . . , xg } and V2 = {y1 , . . . , yg }, where g = α0 (G), (2) for i = 1, . . . , g, (after relabeling) {xi , yi } is an edge of G. Following Carr´a Ferro and Ferrarello [78], we deﬁne a directed graph D with vertex set V1 as follows: (xi , xj ) is a directed edge of D if i = j and {xi , yj } is an edge of G. We say that a cycle C of D is oriented if all the arrows of C are oriented in the same direction. Recall that D is called transitive if for any two (xi , xj ), (xj , xk ) in ED with i, j, k distinct, we have that (xi , xk ) ∈ ED . Example 7.4.23 If G = C4 is a four cycle with edge set {{x1 , y1 }, {x2 , y2 }, {x1 , y2 }, {x2 , y1 }}, then D has two vertices x1 , x2 and two arrows (x1 , x2 ), (x2 , x1 ) forming an oriented cycle of length two. Lemma 7.4.24 [208, Theorem 16.3(4), p. 200] Let D be the directed graph described above. D is acyclic, i.e., D has no oriented cycles, if and only if there is a linear ordering of the vertex set V1 such that all the edges of D are of the form (xi , xj ) with i < j.

290

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Theorem 7.4.25 Let G be a bipartite graph satisfying (1) and (2). If G is sequentially Cohen–Macaulay, then the directed graph D is acyclic. Proof. We proceed by induction on the number of vertices of G. Assume that D has an oriented cycle Cr with vertices {xi1 , . . . , xir }. This means that the graph G has a cycle C2r = {yi1 , xi1 , yi2 , xi2 , yi3 , . . . , yir−1 , xir−1 , yir , xir } of length 2r. By Lemma 7.4.14, the graph G has a vertex v of degree 1. Notice that v ∈ / {xi1 , . . . , xir , yi1 , . . . , yir }. Furthermore, if w is the vertex adjacent to v, we also have w ∈ / {xi1 , . . . , xir , yi1 , . . . , yir }. Hence by Theorem 7.4.11 the graph G = G \ ({v} ∪ NG (v)) is sequentially Cohen– Macaulay and DG has an oriented cycle, a contradiction to the induction hypotheses. Thus D has no oriented cycles, as required. 2 Cohen–Macaulay trees can also be described in terms of D: Theorem 7.4.26 Let G be a tree satisfying (1) and (2). Then G is a Cohen–Macaulay tree if and only if D is a tree such that every vertex xi of D is either a source (i.e., has only arrows leaving xi ) or a sink (i.e., has only arrows entering xi ). Proof. ⇒) Since a tree is bipartite, D is both acyclic and transitive. Suppose there is a vertex xi that is not a sink or source. i.e., there is an arrow entering xi and one leaving xi . Suppose the arrow entering xi originates at xj , and the arrow leaving xi goes to xk . Note that xj = xk because otherwise we would have a cycle in the acyclic graph D. Because D is transitive, the directed edge (xj , xk ) also belongs to D. But then the induced graph on the vertices {xj , yi , xi , yk } in G forms the cycle C4 , contradicting the fact that G is a tree. ⇐) The hypotheses on D imply D is acyclic and transitive, so apply Exercise 7.4.36(d). 2 Computing the type Assume that G is a Cohen–Macaulay bipartite graph. Note that {xi − yi }gi=1 is a regular system of parameters for R/I(G), where R = K[x, y] and R/I(G) modulo ({xi − yi }gi=1 ) reduces to: A = K[x]/(x21 , . . . , x2g , I(H)), where H is a graph on the vertex set V1 . As a K-vector space, A is generated by the image of 1 and the images of all the monomials xi1 · · · xir such that {xi1 , . . . , xir } is an independent set of H. If H ∨ denotes the clutter of minimal vertex covers of H, then the type of R/I(G) is equal to |E(H ∨ )|, because Soc(A) is generated by the images of all the monomials xi1 · · · xir such that {xi1 , . . . , xir } is a maximal independent set of H.

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Example 7.4.27 Let G be the Cohen–Macaulay bipartite graph whose edge ideal is I(G) = (x1 y1 , x1 y2 , x1 y3 , x1 y4 , x2 y2 , x2 y3 , x2 y4 , x3 y3 , x4 y4 ). The graph H has edges x1 x2 , x1 x3 , x1 x4 , x2 x3 , x2 x4 . Since this graph has only three minimal vertex covers, the type of R/I(G) is equal to three.

Exercises 7.4.28 Let Km,n be the complete bipartite graph. Prove that if m, n ≥ 2, then Km,n is not shellable. If m = 1 and n ≥ 1, prove that Km,n is shellable. 7.4.29 If G is a chordal graph, then the ideal I(G)∨ has linear quotients. 7.4.30 [149] Prove that a bipartite graph G is Cohen–Macaulay if and only if ΔG is pure shellable. 7.4.31 Prove that no even cycle can be sequentially Cohen–Macaulay. 7.4.32 [167, Proposition 4.1] Prove that C3 and C5 are the only sequentially Cohen–Macaulay cycles. 7.4.33 Let G be a Cohen–Macaulay bipartite graph. Show that the h-vector of R/I(G) has length at most g = ht I(G). Can we replace Cohen–Macaulay by unmixed? 7.4.34 (Ravindra [344]) A connected bipartite graph G is unmixed if and only if there is a perfect matching such that for every edge {x, y} in the perfect matching, the induced subgraph G[NG (x) ∪ NG (y)] is a complete bipartite graph. 7.4.35 A bipartite graph G without isolated vertices is unmixed if and only if G has a perfect matching {x1 , y1 }, . . . , {xg , yg } such that (I(G)2 : xi yi ) is equal to I(G) for i = 1, . . . , g. 7.4.36 Let G be a connected bipartite graph with bipartition V1 = {xi }gi=1 and V2 = {yi }gi=1 such that {xi , yi } ∈ E(G) for all i and g ≥ 2. Deﬁne a directed graph D with vertex set V1 as follows: (xi , xj ) is an edge of D if i = j and {xi , yj } is an edge of G. Prove that the following hold. (a) If G is a square, then D has two vertices x1 , x2 and two arrows (x1 , x2 ), (x2 , x1 ) forming a cycle of length two. (b) D is acyclic, i.e., D has no directed cycles, if and only if there is a linear ordering of the vertex set V1 such that all the edges of D are of the form (xi , xj ) with i < j.

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(c) D is called transitive if for any two (xi , xj ), (xj , xk ) in E(D) with i, j, k distinct, we have that (xi , xk ) ∈ E(D). The digraph D is transitive if and only if G is unmixed. (d) (Carr´ a Ferro–Ferrarello [78]) G is Cohen–Macaulay if and only if D is acyclic and transitive. (e) If G is a Cohen–Macaulay graph, then D has at least one source xi and at least one sink xj . (f) If every vertex of D is either a source or a sink, then G is Cohen– Macaulay. (g) If D is transitive and has a cycle, then D has a cycle of length 2. (h) D is transitive and acyclic if and only if D is transitive and has no cycles of length 2. (i) D has no cycles of length 2 if and only if for any edge {xi , yj } ∈ E(G) with i = j, one has that {xj , yi } is not an edge of G. (j) (Zaare-Nahandi [435]) G is Cohen–Macaulay if and only if the following conditions hold: (1) The induced subgraph G[N (xi ) ∪ N (yi )] is a complete bipartite graph for all i, and (2) if xi is adjacent to yj for i = j, then xj is not adjacent to yi . (k) [435] G is Cohen–Macaulay if and only if G is unmixed and G has a unique perfect matching. 7.4.37 Let G be a connected bipartite graph satisfying the conditions of Theorem 7.4.21. Prove that deg(y1 ) = 1 and deg(xg ) = 1. 7.4.38 Let G be a Cohen–Macaulay bipartite graph. If G is not a discrete graph, prove that G has at least two vertices of degree one. 7.4.39 Prove that the following is the full list of Cohen–Macaulay connected bipartite graphs with eight vertices. r r rr r r rr r r r r r r rr Q Q Q Q @ @ A A QA A Q @Q @ @Q @ A@QA A@Q @QA@ @Q @ AQ @ AQ A @Q A @QQ @Q @ @QQ @ A @A Q A @ Q @A Q @ @ Q @ Qr r Qr r Ar @Ar Q Ar @r Q Qr @ Qr r r @Ar Q r r @r Q r r r r Q @ A Q@ A A@QA@ A @Q AQ @ A @A Q @ Ar @Ar Q r @ Qr

r r r r Q @ A Q@ A A@Q @ A A @QQ @A A @ Q @A Ar @r Q r @ Q Ar

r r r r Q @ A Q@ A A A@QA@ A A @Q AQ @A A @A Q @A Ar @Ar Q r @ Q Ar

7.4.40 Let I = (xi yj | 1 ≤ i ≤ j ≤ n), draw a picture of the bipartite graph G such that I = I(G) and prove that I is a Cohen–Macaulay ideal.

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7.5

293

Regularity, depth, arithmetic degree

In this section we study the regularity and depth of edge ideals of graphs. We introduce a device to estimate the regularity of edge ideals and show some upper bounds for the arithmetic degree. Let G be a simple graph with vertex set X = {x1 , . . . , xn } and let R = K[X] be a polynomial ring over a ﬁeld K. As usual we denote the vertex set and edge set of G by V (G) and E(G), respectively. Lemma 7.5.1 [274, Lemma 2.2] Let G be a graph and let im(G) be its induced matching number. Then im(G) ≤ reg(R/I(G)). Proof. It follows at once from Corollary 6.4.8.

2

Theorem 7.5.2 [326] Let F be a family of graphs containing any discrete graph and let β : F → N be a function satisfying that β(G) = 0 for any discrete graph G, and such that given G ∈ F , with E(G) = ∅, there is x ∈ V (G) such that the following two conditions hold: (i) G \ {x} and G \ ({x} ∪ NG (x)) are in F . (ii) β(G \ ({x} ∪ NG (x))) < β(G) and β(G \ {x}) ≤ β(G). Then reg(R/I(G)) ≤ β(G) for any G ∈ F . Proof. The proof is by induction on the number of vertices. Let G be a graph in F . If G is a discrete graph, then I(G) = (0), reg(R) = 0 and β(G) = 0. Assume that G has at least one edge. There is a vertex x ∈ V (G) such that the induced subgraphs G1 = G \ {x} and G2 = G \ ({x} ∪ NG (x)) satisfy (i) and (ii). There is an exact sequence of graded R-modules x

0 −→ R/(I(G) : x)[−1] −→ R/I(G) −→ R/(x, I(G)) −→ 0. Notice that (I(G) : x) = (NG (x), I(G2 )) and (x, I(G)) = (x, I(G1 )). The graphs G1 and G2 have fewer vertices than G. It follows directly from the deﬁnition of regularity that reg(M [−1]) = 1 + reg(M ) for any graded Rmodule M . Therefore applying the induction hypothesis to G1 and G2 , and using conditions (i) and (ii) and Exercise 6.4.30, we get reg(R/(I(G) : x)[−1]) reg(R/(x, I(G)))

= reg(R /I(G2 )) + 1 ≤ β(G2 ) + 1 ≤ β(G), ≤ β(G1 ) ≤ β(G)

where R = K[V (G2 )]. Hence, by Lemma 6.4.10, we get that the regularity of R/I(G) is bounded by the maximum of the regularities of R/(x, I(G)) and R/(I(G) : x)[−1]. Thus reg(R/I(G)) ≤ β(G), as required. 2 Deﬁnition 7.5.3 A vertex x of a graph G is called simplicial if the subgraph G[NG (x)] induced by the neighbor set NG (x) is a complete subgraph.

294

Chapter 7

Lemma 7.5.4 [403] Let H be a chordal graph and K a complete subgraph of H. If K = H, then there is x ∈ V (K) such that the subgraph H[NH (x)] induced by the neighbor set NH (x) of x is a complete subgraph. Proof. By induction on V (H). If H is complete, any x ∈ V (K) would satisfy the requirement. Otherwise, using Proposition 7.3.25, H = H1 ∪ H2 , where H1 , H2 are chordal graphs smaller than H such that H1 ∩ H2 is a complete subgraph. It follows that K is a subgraph of Hi for some i, say i = 1. As H1 ∩ H2 is a proper complete subgraph of H2 , by induction x can be chosen in H2 \ (H1 ∩ H2 ). 2 Let G be a graph. We let β (G) be the cardinality of any smallest maximal matching of G. H` a and Van Tuyl proved that the regularity of R/I(G) is bounded from above by the matching number of G and Woodroofe improved this result showing that β (G) is an upper bound for the regularity. For edge ideals whose resolutions are k-steps linear and for ideals of covers of graphs some new upper bounds for the regularity are given by Dao, Huneke, and Schweig [103]. There are also upper bounds given by Banerjee [13] for the regularity of powers of edge ideals of graphs whose complement does not have any induced four cycle. Theorem 7.5.5 Let G be a graph and let R = K[V (G)]. Then (a) [206, Corollary 6.9] reg(R/I(G)) = im(G) for any chordal graph G. (b) ([206, Theorem 6.7], [434]) reg(R/I(G)) ≤ β (G). (c) [407, Theorem 3.3] reg(R/I(G)) = im(G) if G is bipartite and R/I(G) is sequentially Cohen–Macaulay. Proof. (a) Let F be the family of chordal graphs and let G be a chordal graph with E(G) = ∅. By Lemma 7.5.1 and Theorem 7.5.2 it suﬃces to prove that there is x ∈ V (G) such that im(G1 ) ≤ im(G) and im(G2 ) < im(G), where G1 and G2 are the subgraphs G \ {x} and G \ ({x} ∪ NG (x)), respectively. The inequality im(G1 ) ≤ im(G) is clear because any induced matching of G1 is an induced matching of G. We now show the other inequality. By Lemma 7.5.4, there is y ∈ V (G) such that G[NG (y) ∪ {y}] is a complete subgraph. Pick x ∈ NG (y) and set f0 = {x, y}. Consider an induced matching f1 , . . . , fr of G2 with r = im(G2 ). We claim that f0 , f1 , . . . , fr is an induced matching of G. Let e be an edge of G contained in ∪ri=0 fi . We may assume that e ∩ f0 = ∅ and e ∩ fi = ∅ for some i ≥ 1, otherwise e = f0 or e = fi for some i ≥ 1. Then e = {y, z} or e = {x, z} for some z ∈ fi . If e = {y, z}, then z ∈ NG (y) and x ∈ NG (y). Hence {z, x} ∈ E(G) and z ∈ NG (x), a contradiction because the vertex set of G2 is disjoint from NG (x) ∪ {x}. If e = {x, z}, then z ∈ NG (x), a contradiction. This completes the proof of the claim. Hence im(G2 ) < im(G).

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(b) Let F be the family of all graphs and let G be a graph with E(G) = ∅. By Theorem 7.5.2 it suﬃces to prove that there is x ∈ V (G) such that β (G1 ) ≤ β (G) and β (G2 ) < β (G), where G1 and G2 are the subgraphs G \ {x} and G \ ({x} ∪ NG (x)), respectively. We leave the proof of this as an exercise. (c) Let F be the family of all bipartite graphs G such that R/I(G) is sequentially Cohen–Macaulay, and let β : F → N be the function β(G) = im(G). Let G be a graph in F with E(G) = ∅. By Lemma 7.5.1 and Theorem 7.5.2 it suﬃces to observe that, according to Corollary 7.4.8 and Theorem 7.4.16, there are adjacent vertices x1 and x2 with deg(x2 ) = 1 such that the bipartite graph G\({xi }∪NG (xi )) is sequentially Cohen–Macaulay for i = 1, 2. Thus conditions (i) and (ii) of Theorem 7.5.2 are satisﬁed. 2 Corollary 7.5.6 [437] If G is a forest, then reg(R/I(G)) = im(G). Proof. Any forest is a bipartite graph. Thus, by Theorem 6.5.25, ΔG is shellable and R/I(G) is sequentially Cohen–Macaulay. Hence the result follows from Theorem 7.5.5. 2 A graph G is weakly chordal if every induced cycle in both G and its complement G has length at most 4. Theorem 7.5.7 [434] If G is weakly chordal, then reg(R/I(G)) = im(G). Proof. It is shown in [77] that a weakly chordal graph G can be covered by im(G) co-CM graphs. Thus, the result follows from Corollary 6.4.13. 2 If G is an unmixed graph, Kummini [285] showed that reg(R/I(G)) is equal to the induced matching number of G. This result was later extended to all very well-covered graphs [303] (a graph G is very well-covered if it is unmixed without isolated vertices and 2 ht(I(G)) = |V (G)|). If G is claw-free and G, the complement of G, has no induced 4-cycles, then reg(R/I(G)) ≤ 2 with equality if G is not chordal [331], in this case reg(R/I(G)) = im(G) + 1. Formulas for the regularity of ideals of mixed products are given in [260]. The regularity and depth of lex segment edge ideals are computed in [143]. The regularity and other algebraic properties of edge ideal of Ferrers graphs are studied in detail in [96]. Corollary 7.5.8 Let G be a bipartite without isolated vertices. If G B A graph has n vertices, then depth K[ΔG ] ≤ n2 . Proof. Let (V1 , V2 ) be a bipartition of G with |V1 | ≤ |V2 |. Note 2|V1 | ≤ n because |V1 | + |V2 | = n. Since V1 is a maximal independent set of vertices one has β0 (G) ≤ |V1 |. Therefore using Theorem 6.4.18 we conclude depth K[ΔG ] ≤ β0 (G) ≤ |V1 | ≤ n/2.

2

The bound above does not extend to non-bipartite graphs. The depth of powers of edge ideals of trees is studied in [324].

296

Chapter 7

Proposition 7.5.9 Let Kr be the complete graph with r vertices and let Gr,i be the graph obtained by attaching i lines at each vertex of Kr . Then depth(R/I(Gr,i )) = (r − 1)i + 1. Proof. Let R = K[x1 , . . . , xn ] be a polynomial ring and let x1 , . . . , xn be the vertices of Gr,i . Note that Gr,i has n = r(i + 1) vertices. We proceed by induction on r. If r = 1, then G1,i is a star and clearly K[ΔG1,i ] has depth 1. Assume r ≥ 2. Set G = Gr,i and I = I(G). Let x1 be any vertex of Kr and consider the exact sequence x

1 R/I −→ R/(x1 , I) −→ 0. 0 −→ R/(I : x1 ) −→

(∗)

It is not hard to see that (I : x1 ) is a face ideal generated by i + (r − 1) variables. Hence depth(R/(I : x1 )) = (r − 1)i + 1. On the other hand note the equality (I, x1 ) = (x1 ) + I(Gr−1,i ). Since x1 is adjacent in G to exactly i-vertices of degree 1 and those vertices do not occur in (I, x1 ), by induction hypothesis we derive depth(R/(x1 , I)) = i + [(r − 2)i + 1] = (r − 1)i + 1. Altogether the ends of the exact sequence (∗) have depth equal to (r−1)i+1. Thus, by the depth Lemma 2.3.9, we get depth(R/I) = (r − 1)i + 1. 2 Example 7.5.10 Let K = Q and let G = G4,3 be the following graph x8 xr10 xr7 r rx11 x x @ 5 9 r r r rx12 x6 @ @ @ @r r r x14 x4 r x1 x13 @ @rx r 15 r x3 r x2 x16 Then, by Proposition 7.5.9, depth R/I(G) = 10 and pd(R/I(G)) = 6. Theorem 7.5.11 [189, Theorem 2.7.14] Let G be a graph without isolated vertices and let β0 (C) be the cardinality of a smallest maximal independent set of G. If G has n vertices, then n + β0 (G)n − 2β0 (G) − β0 (G)2 , (n − β0 (G)) β0 (G) ≤ α0 (G)[1 + (β0 (G) − β0 (G))].

depth K[ΔG ] ≤ β0 (G) ≤

Lemma 7.5.12 Let G be a graph and let G∨ be its blocker. Then the arithmetic degree of G is given by arith-deg(I(G)) = |E(G∨ )|.

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Proof. It suﬃces to recall that a set of vertices of G is a maximal stable set if and only if its complement is a minimal vertex cover. 2 The basic problem of enumeration and counting of the stable sets in a graph—using an algebraic perspective—is addressed in [110]. The following estimate is due to E. Sperner (cf. [168, Theorem 3.1]). Theorem 7.5.13 [390] If G is a graph with n vertices, then n ∨ arith-deg(I(G)) = |E(G )| ≤ . (n/2) Theorem 7.5.14 [212] If G is a graph, then |E(G∨ )| ≤ 2α0 (G) . Proof. We may assume G has no isolated vertices. We set C = G∨ . Let C be a ﬁxed vertex cover with α0 (G) vertices. For 0 ≤ i ≤ α0 (G) consider: Ci = {C | C ∈ C and |C ∩ C | = i}. We claim that the following inequality holds: |Ci | ≤ α0 (G) for all i ≥ 0. i We are going to show that if C and C are two vertex covers in Ci such that C ∩ C = C ∩ C , then C = C . Using C ∩ C = C ∩ C we obtain that the neighbors sets satisfy: (∗) NG (C \ C ) = NG (C \ C ). Since C and C are vertex covers of G one has: NG (C \ C ) ∪ (C ∩ C ) ⊂ C and NG (C \ C ) ∪ (C ∩ C ) ⊂ C . Observe that NG (C \ C ) ∪ (C ∩ C ) and NG (C \ C ) ∪ (C ∩ C ) are vertex covers of G, because C is a vertex cover of G. From the minimality of C and C one concludes the equalities: NG (C \ C ) ∪ (C ∩ C ) = C and NG (C \ C ) ∪ (C ∩ C ) = C , which together with (∗) allow us to derive the equality C = C , as required. Therefore since there are only α0 (G) subsets in C with i vertices, one α0 (G)i obtains the inequality |Ci | ≤ , as claimed. Hence i ∨

|E(G )| = |C| =

α0 (G)

α0 (G)

i=0

|Ci | ≤

i=0

α0 (G) i

= 2α0 (G) .

2

Exercises 7.5.15 A graph whose complement is chordal is called co-chordal. A complex Δ is called a quasi-forest if Δ is the clique complex of a chordal graph. Prove that an edge ideal I(C) of a clutter C has regularity 2 if and only if ΔC is the independence complex of a co-chordal graph.

298

Chapter 7

7.5.16 Let G be the complement of a cycle of length six. Use Macaulay2, to show that reg(R/I(G)) = 2 and im(G) = 1. 7.5.17 Let Kr be a complete graph and let G be the graph obtained by attaching ij lines at each vertex vj of Kr . If ij ≤ ij+1 for all j, then depth(R/I(G)) = i1 + · · · + ir−1 + 1. 7.5.18 Let γn = sup {|E(G∨ )| | G is a graph with n vertices}. If {Fn } is the Fibonacci sequence, F0 = F1 = 1 and Fn = Fn−1 + Fn−2 , then γn ≤ Fn for n ≥ 1. If Pn is the path {x1 , . . . , xn }, n ≥ 3, then Fn = |E(Pn∨ )|. 7.5.19 Find e(G), the multiplicity of R/I(G), and arith-deg(I(G)), the arithmetic degree of I(G) if G is a disjoint union of stars. 7.5.20 Let G be the graph which is the disjoint union of k complete graphs with two vertices. Then, |E(G∨ )| = e(G) = 2α0 (G) and k = α0 (G).

7.6

Betti numbers of edge ideals

Let G be a graph and I(G) ⊂ R its edge ideal. An interesting problem is to express some of the ﬁrst initial Betti numbers and invariants of I(G) in terms of the graph theoretical data of G (see [138, 141, 276, 321, 362, 437] and the references therein). Kimura [277] has studied the projective dimension of edge ideals of unmixed bipartite graphs. Let R be a polynomial ring over a ﬁeld K with the usual grading and I a graded ideal of R. The minimal graded resolution of R/I by free R-modules can be expressed as: 0 −→

cg )

ϕg

Rbgi (−dgi ) −→ · · · −→

i=1

c1 )

ϕ1

Rb1i (−d1i ) −→ R −→ R/I −→ 0,

i=1

where all the maps are degree preserving and the dij are positive integers. One may assume dj1 < · · · < djcj for all j = 1, . . . , g. The integer g is equal to pdR (R/I), the projective dimension of R/I. To simplify notation set Fj =

cj )

Rbji (−dji ).

i=1

The rank of Fj is the jth Betti number of I. From the minimality of the resolution d11 < · · · < dg1 ; see Remark 3.5.9. We deﬁne the jth initial Betti number of I as γj = bj1 and the jth initial virtual Betti number of I as vj = dimK (Fj )d11 +j−1 .

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299

Note that some of the initial virtual Betti numbers may be zero. For a general monomial ideal I there is an explicit and elegant description of a minimal set of generators for the ﬁrst module of syzygies of I due to Eliahou [137]; see also [64]. Deﬁnition 7.6.1 The line graph of G, denoted by L(G), has vertex set E = E(G) with two vertices of L(G) adjacent whenever the corresponding edges of G have exactly one common vertex. Proposition 7.6.2 If G is a graph with vertices x1 , . . . , xn and edge set E(G), then the number of edges of the line graph L(G) is given by |E(L(G))| =

n deg(xi ) i=1

2

= −|E(G)| +

n deg2 xi i=1

2

.

(7.1)

Proof. To prove the ﬁrst equality note that each vertex xi of G of degree di contributes with d2i edges to the number of edgesof L(G). The second n 2 equality follows from the Euler’s identity 2|E(G)| = i=1 deg(xi ). Let us now present a formula, in graph theoretical terms, for the second initial Betti number of I(G). Proposition 7.6.3 [138] Let I ⊂ R be the edge ideal of a graph G, let V be the vertex set of G, and let L(G) be the line graph of G. If ψ

· · · −→ Rc (−4) ⊕ Rb (−3) −→ Rq (−2) −→ R −→ R/I −→ 0 is the minimal graded resolution of R/I. Then b = |E(L(G))| − Nt , where Nt is the number of triangles of G and c is the number of unordered pairs of edges {f, g} such that f ∩ g = ∅ and f and g cannot be joined by an edge. Proof. Let f1 , . . . , fq be the set of all xi xj such that {xi , xj } is an edge of G. We may assume ψ(ei ) = fi . Let Z1 be the set of elements in ker(ψ) of degree 3. We regard the fi ’s as the vertices of L(G). Every edge e = {fi , fj } in L(G) determines a syzygy syz(e) = xj ei −xi ej , where fi = xi z and fj = xj z for some z, xi , xj in V . By Theorem 3.3.19 the set of those syzygies generate Z1 . Given a triangle C3 = {z1 , z2 , z3 } in G we set φ(C3 ) = {z1 ej − z3 ei , z1 ej − z2 ek , z2 ek − z3 ei }, where fi = z1 z2 , fj = z2 z3 , and fk = z1 z3 . Notice that φ(C3 ) ∩ φ(C3 ) = ∅ if C3 = C3 . For every triangle C3 in G choose an element ρ(C3 ) ∈ φ(C3 ). It is not hard to show that the set B = {syz(e)| e ∈ L(G)} \ {ρ(C3 )| C3 is a triangle in G}

300

Chapter 7

is a minimal generating set for Z1 . Thus b = |E(L(G))| − Nt . The proof of the expression for c is left as an exercise. 2 The 2nd Betti number of a Stanley–Reisner ideal (resp. monomial ideal) is independent of the ground ﬁeld [244] (resp. [64, 137]); moreover for edge ideals the 3rd and 4th Betti numbers and the 5th and 6th Betti numbers are also independent of the ﬁeld [243] and [274], respectively. An application to Hilbert series Let Δ be a simplicial complex of dimension d and f = (f0 , . . . , fd ) its f -vector, where fi is the number of faces of dimension i in Δ and f−1 = 1. By Theorem 6.7.2 the Hilbert series of the Stanley–Reisner ring S = R/IΔ can be expressed as F (t) = f−1 +

fd td+1 f0 t + ···+ . (1 − t) (1 − t)d+1

(7.2)

We now use this formula to compute a particular instance. Corollary 7.6.4 Let G be a graph with q edges, V = {x1 , . . . , xn } its vertex set and F (t) the Hilbert series of S = R/I(G). If S has dimension 3, then F (t)(1 − t)3 is equal to 1 + gt +

2g + 3g 2 + g 3 − 6q − 6gq − 6N + 3v 3 g+1 − q t2 + t , 2 6

where g = n − 3, N is the number of triangles of G, and v =

n i=1

deg2 xi .

Proof. Let Δ = ΔG be the Stanley–Reisner complex of I = I(G). The − q. From ﬁrst entries of the f -vector of Δ are f−1 = 1, f0 = n, f1 = n(n−1) 2 − qn + b and the resolution above and Proposition 6.7.3, dim(R/I)3 = n+2 n−1 n+2 f2 = n−1 − qn + b − 2f1 − n. The desired formula follows by substitution of the fi ’s in Eq. (7.2) and using Eq. (7.1). 2 The number of triangles of a graph Let G be a graph with q edges and vertex set V = {x1 , . . . , xn }. As the number of edges of the line graph L(G) of G is given by the formula |E(L(G))| =

n deg(xi ) i=1

2

= −q +

n deg2 xi i=1

2

,

observe that Proposition 7.6.3 provides a method to compute the number of triangles of a graph by computing syzygies with Macaulay2. An alternative method using algebraic graph theory is recalled next.

Edge Ideals of Graphs

301

The adjacency matrix of G is the n × n matrix A = (aij ) with entries

1 if xi is adjacent xj , aij = 0 otherwise. It follows directly that A is symmetric and that its trace is equal to zero. If f (x) = xn + c1 xn−1 + c2 xn−2 + c3 xn−3 + · · · + cn , is the characteristic polynomial of A, then c1 = 0, −c2 is the number of edges of G, and −c3 is twice the number of triangles of G. For an interpretation of all the ci ’s see [40]. Example 7.6.5 Let G be the graph whose complement is the union of two disjoint copies of a path of length two. The edge ideal of G is equal to: I(G) = (x1 x4 , x1 x5 , x1 x6 , x2 x3 , x2 x4 , x2 x5 , x2 x6 , x3 x4 , x3 x5 , x3 x6 , x5 x6 ), and part of its minimal resolution is: · · · → R24 (−3) → R11 (−2) → R → R/I(G) → 0. Hence the number of triangles of G is equal to 6. The same number is obtained noticing that f (x) = x6 − 11x4 − 12x3 + 3x2 + 4x − 1 is the characteristic polynomial of the adjacency matrix of G. Linear resolutions of edge ideals First we introduce d-trees and state an interesting result of Fr¨oberg. Deﬁnition 7.6.6 A d-tree is deﬁned inductively: (i) a complete graph Kd+1 is a d-tree, (ii) let G be a d-tree and H a subgraph of G with H Kd , if v is a new vertex connected to all vertices in H, then G ∪ {v} is a d-tree. Theorem 7.6.7 [172] Let G be a simple graph, let ΔG be its independence complex, and let G be the complement of G. The following hold. (a) K[ΔG ] has a 2-linear resolution if and only if G is a chordal graph. (b) If G is Cohen–Macaulay over a ﬁeld K, then K[ΔG ] has a 2-linear resolution if and only if G is a d-tree, where d = dim ΔG . The problem of determining the simplicial complexes Δ whose Stanley– Reisner ideal IΔ has a pure resolution was solved by Bruns and Hibi [66, 67]. Corollary 7.6.8 Let G be a connected Cohen–Macaulay bipartite graph. If the number of edges of G is g(g + 1)/2, where g = ht I(G), then I(G) = (xi yj | 1 ≤ i ≤ j ≤ g). Proof. The ring R/I(G) has a linear resolution by Theorem 5.3.7. Hence, according to Theorem 7.6.7, the 1-skeleton of ΔG is a (g − 1)-tree, which implies that there is a vertex w of G of degree g. We now use Exercise 7.4.37 to conclude that deg(xi ) = 1 and deg(yj ) = 1, for some i, j. To ﬁnish the proof consider the graph G \ {w} and use induction. 2

302

Chapter 7

Exercises 7.6.9 Let G be a graph and let k = max{deg(v)| v ∈ V (G)}. If G is connected, then G is k-regular, i.e., its vertices have degree k, if and only if k is a characteristic value of the adjacency matrix of G. 7.6.10 Let G be a graph with vertex set x and FG (t) the Hilbert series of K[x]/I(G), where K is a ﬁeld. For any vertex x ∈ V (G) one has: t FG\(N (x)∪{x}) (t), with F∅ (t) = 1. FG (t) = FG\{x} (t) + 1−t 7.6.11 If Kr,s is a complete bipartite graph, then FKr,s (t) =

1 1 + − 1. (1 − t)r (1 − t)s

7.6.12 [426] Let Pn be a path graph with vertices x = x1 , . . . , xn . Prove that the Hilbert series and the Hilbert function of K[x]/I(Pn ) are given by n+1 2 FPn (t)

=

∞ n − j + 1 j + i − 1 j+i t j i i=0

j=0

min(m, n+1 2 )

HLn (m)

=

j=0

n−j +1 m−1 . j m−j

7.6.13 Let G = C7 be a heptagon. Prove the formula: FG (t) =

−t3 + 3t2 + 4t + 1 . (1 − t)3

What is the number of minimal primes of I(G)? 7.6.14 Let G = C7 be a heptagon and let ΔG be its independence complex. Prove that ΔG is a triangulation of a M¨obius band whose reduced Euler characteristic is χ .(ΔG ) = −1. 7.6.15 If G is a graph and I(G) ⊂ R is its edge ideal, then a(R/I(G)) ≤ 0. 7.6.16 If G is a graph and H is its whisker graph, then β0 (G) equals the induced matching number of H. 7.6.17 Let Δ be a simplicial complex of dimension d with n vertices and f = (f0 , . . . , fd ) its f -vector. If K is a ﬁeld and K[Δ] is a Cohen–Macaulay ring with a 2-linear resolution, then d+1 d + (n − d − 1) , 0 ≤ k ≤ d. fk = k+1 k 7.6.18 Let I be the ideal (abc, abd, ace, adf, aef, bcf, bde, bef, cde, cdf ) of the polynomial ring K[a, . . . , f ]. Prove that the third Betti number of I depends on the base ﬁeld, and that I is equal to its Alexander dual.

Edge Ideals of Graphs

7.7

303

Associated primes of powers of ideals

In this section we show that the sets of associated primes of the powers of the edge ideal of a graph form an ascending chain. Two excellent references for the general theory of asymptotic prime divisors in commutative Noetherian rings are [259] and [313]. Let G be a simple graph with ﬁnite vertex set X = {x1 , . . . , xn } and without isolated vertices, let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and let I = I(G) be the edge ideal of G. In [56], Brodmann showed that there exists a positive integer N1 such that Ass(R/I k ) = Ass(R/I N1 ) for all k ≥ N1 . A minimal such N1 is called the index of stability of I. Proposition 7.7.1 [81, Corollary 4.3] If G is a connected non-bipartite graph with n vertices, s leaves, and the smallest odd cycle of G has length 2k + 1, then N1 ≤ n − k − s. Deﬁnition 7.7.2 [165] An ideal I ⊂ R has the persistence property if Ass(R/I k ) ⊂ Ass(R/I k+1 ) for all k ≥ 1, i.e., the sets Ass(R/I k ) form an ascending chain. There are some interesting cases where associated primes are known to form ascending chains [36, 163, 165, 232, 326]. The ﬁrst listed is quite general, but has applications to square-free monomial ideals. Theorem 7.7.3 ([313, Proposition 3.9], [342, 343]) If R is a Noetherian ring and I is an ideal, then the sets Ass(R/I k ) form an ascending chain. In particular if I is normal, then I has the persistence property. Deﬁnition 7.7.4 Following Schrijver [373], the duplication of a vertex xi of a graph G means extending its vertex set X by a new vertex xi and replacing E(G) by E(G) ∪ {(e \ {xi }) ∪ {xi }| xi ∈ e ∈ E(G)}. The duplication of the vertex xi is denoted by Gxi . The deletion of xi is the subgraph G \ {xi }. A graph obtained from G by a sequence of deletions and duplications of vertices is called a parallelization of G. These two operations commute. If a = (ai ) is a vector in Nn , we denote by Ga the graph obtained from G by successively deleting any vertex xi with ai = 0 and duplicating ai − 1 times any vertex xi if ai ≥ 1. Lemma 7.7.5 Any parallelization of G has the form Ga for some a.

304

Chapter 7

Proof. It suﬃces to notice that, by Exercise 7.7.23, any parallelization of G can be obtained by duplications and deletions of vertices of G. 2 The subring K[G] = K[xi xj | {xi , xj } is an edge of G] ⊂ R is called the edge subring of G. This subring will be studied in detail in Chapter 10. Lemma 7.7.6 Let G be a graph and let K[G] be its edge subring. Then, Ga has a perfect matching if and only if xa ∈ K[G]. Proof. We may assume that a = (a1 , . . . , an ) and ai ≥ 1 for all i, because if a has zero entries we can use the induced subgraph on the vertex set {xi | ai > 0}. The vertex set of Ga is X a = {x11 , . . . , xa1 1 , . . . , x1i , . . . , xai i , . . . , x1n , . . . , xann } k

and the edges of Ga are exactly those pairs of the form {xki i , xj j } with i = j, ki ≤ ai , kj ≤ aj , for some edge {xi , xj } of G. We can regard xa as an ordered multiset xa = xa1 1 · · · xann = (x1 · · · x1 ) · · · (xn · · · xn ) a1

an

on the set X; that is, we can identify the monomial xa with the multiset Xa = {x1 , . . . , x1 , . . . , xn , . . . , xn } a1

an

on X in which each variable is uniquely identiﬁed with an integer between 1 and |a|. This integer is the position, from left to right, of xi in Xa . There is a bijective map 1 ↓ x1 ↓ x11

2 ↓ x1 ↓ x21

··· ··· ··· ··· ···

a1 ↓ x1 ↓ xa1 1

··· ··· ··· ··· ···

a1 + · · · + an−1 + 1 ↓ xn ↓ x1n

··· ··· ··· ··· ···

a1 + · · · + an ↓ xn ↓ xann .

Hence if Ga has a perfect matching, then the perfect matching induces a factorization of xa in which each factor corresponds to an edge of G, i.e., xa ∈ K[G]. Conversely, if xa ∈ K[G] we can factor xa as a product of monomials corresponding to edges of G and this factorization induces a perfect matching of Ga . 2 Given an edge f = {xi , xj } of G, we denote by Gf or G{xi ,xj } the graph obtained from G by successively duplicating the vertices xi and xj , i.e., Gf = G1+ei +ej , where ei is the ith unit vector and 1 = (1, . . . , 1). Recall that def(G), the deﬁciency of G, is given by def(G) = |V (G)| − 2β1 (G),

Edge Ideals of Graphs

305

where β1 (G) is the matching number of G. Hence def(G) is the number of vertices left uncovered by any maximum matching. Notation In what follows F = {f1 , . . . , fq } denotes the set of all monomials c xi xj such that {xi , xj } ∈ E(G). We use f c as an abbreviation for f1c1 · · · fq q , q where c = (ci ) ∈ N . Lemma 7.7.7 Let G be a graph and let a ∈ Nn and c ∈ Nq . Then (a) xa = xδ f c , where |δ| = def(Ga ) and |c| = β1 (Ga ). (b) xa belongs to I(G)k \ I(G)k+1 if and only if k = β1 (Ga ). (c) (Ga )f = (Ga ){xi ,xj } for any edge f = {xki i , xj j } of Ga . k

Proof. Parts (a) and (b) follow using the bijective map used in the proof of Lemma 7.7.6. To show (c) we use the notation used in the proof of Lemma 7.7.6. We now prove the inclusion E((Ga )f ) ⊂ E((Ga ){xi ,xj } ). Let k yi and yj be the duplications of xki i and xj j , respectively. We also denote the duplications of xi and xj by yi and yj , respectively. The common vertex set of (Ga )f and (Ga ){xi ,xj } is V (Ga ) ∪ {yi , yj }. Let e be an edge of (Ga )f . If e = {yi , yj } or e ∩ {yi , yj } = ∅, then clearly e is an edge of (Ga ){xi ,xj } . Thus, we may assume that e = {yi , xk }. Then {xki i , xk } ∈ E(Ga ), so {xi , x } ∈ E(G). Hence {xi , xk } is in E(Ga ), so e = {yi , xk } is an edge of (Ga ){xi ,xj } . This proves the inclusion “⊂”. The other inclusion follows using similar arguments (arguing backwards). 2 Theorem 7.7.8 (Berge [297, Theorem 3.1.14]) Let G be a graph. Then def(G) = max{c0 (G \ S) − |S| | S ⊂ V (G)}, where c0 (G) denotes the number of odd components (components with an odd number of vertices) of a graph G. The theorem of Berge is equivalent to a classical result of Tutte that describes perfect matchings [297] (see Theorem 7.1.10). Theorem 7.7.9 Let G be a graph. Then def(Gf ) = δ for all f ∈ E(G) if and only if def(G) = δ and β1 (Gf ) = β1 (G) + 1 for all f ∈ E(G). Proof. Assume that def(Gf ) = δ for all f ∈ E(G). In general, def(G) ≥ def(Gf ) for any f ∈ E(G). We proceed by contradiction. Assume that def(G) > δ. Then, by Berge’s theorem, there is an S ⊂ V (G) such that c0 (G \ S) − |S| > δ. We set r = c0 (G \ S) and s = |S|. Let H1 , . . . , Hr be the odd components of G \ S. Case (I): |V (Hk )| ≥ 2 for some 1 ≤ k ≤ r. Pick an edge f = {xi , xj } of Hk . Consider the parallelization Hk obtained from Hk by duplicating the

306

Chapter 7

vertices xi and xj , i.e., Hk = Hkf . The odd connected components of Gf \ S are H1 , H2 , . . . , Hk−1 , Hk , Hk+1 . . . , Hr . Thus c0 (Gf \ S) − |S| > δ = def(Gf ). This contradicts Berge’s theorem when applied to Gf . Case (II): |V (Hk )| = 1 for 1 ≤ k ≤ r. Notice that S = ∅ because G has no isolated vertices. Pick f = {xi , xj } an edge of G with {xi } = V (H1 ) and xj ∈ S. Let yi and yj be the duplications of xi and xj , respectively. The odd components of Gf \ (S ∪ {yj }) are H1 , . . . , Hr , {yi }. Thus c0 (Gf \ (S ∪ {yj })) − |S ∪ {yj }| = c0 (G \ S) − |S| > δ = def(Gf ). This contradicts Berge’s theorem when applied to Gf . Therefore def(G) = def(Gf ) for all f ∈ E(G). Hence β1 (Gf ) = β1 (G) + 1 for all f ∈ E(G). The 2 converse follows using the deﬁnition of def(G) and def(Gf ). Corollary 7.7.10 Let G be a graph. Then G has a perfect matching if and only if Gf has a perfect matching for every edge f of G. Proof. Assume that G has a perfect matching. Let f1 , . . . , fn/2 be a set of edges of G that form a perfect matching of V (G), where n is the number of vertices of G. If f = {xi , xj } is any edge of G and yi , yj are the duplications of the vertices xi and xj , respectively, then clearly f1 , . . . , fn/2 , {yi , yj } form a perfect matching of V (Gf ). Conversely, if Gf has a perfect matching for all f ∈ E(G), then def(Gf ) = 0 for all f ∈ E(G). Hence, by Theorem 7.7.9, we get that def(G) = 0, so G has a perfect matching. 2 If I = (0) is an ideal of a commutative Noetherian domain, Ratliﬀ showed that (I k+1 : I) = I k for all large k [342, Corollary 4.2] and that equality holds for all k when I is normal [342, Proposition 4.7]. The next lemma shows that equality holds for all k when I is an edge ideal. Lemma 7.7.11 Let I be the edge ideal of a graph G. Then (I k+1 : I) = I k for k ≥ 1. Proof. Let f1 , . . . , fq be the set of all xi xj such that {xi , xj } ∈ E(G). c Given c = (ci ) ∈ Nq , we set f c = f1c1 · · · fq q . The colon ideal (I k+1 : I) is a monomial ideal. Clearly I k ⊂ (I k+1 : I). To show the other inclusion take a monomial xa in (I k+1 : I). Then, fi xa ∈ I k+1 for all i. We may assume that fi xa ∈ / I k+2 , otherwise xa ∈ I k as required. Thus xa+ei +ej ∈ I k+1 \ I k+2 for any ei + ej such that {xi , xj } ∈ E(G). Hence, by Lemma 7.7.7(b), β1 (Ga+ei +ej ) = k + 1 for any {xi , xj } ∈ E(G), that is, (Ga ){xi ,xj } has a maximum matching of size k + 1 for any edge {xi , xj } of G. With the k notation of the proof of Lemma 7.7.6, for any edge {xki i , xj j } of Ga we have k ki ,xj j }

(Ga ){xi

= (Ga ){xi ,xj } ,

Edge Ideals of Graphs

307

see Lemma 7.7.7(c). Then, (Ga )f has a maximum matching of size k + 1 for any edge f of Ga . As a consequence def((Ga )f ) = (|a| + 2) − 2(k + 1) = |a| − 2k for any edge f of Ga . Hence, by Theorem 7.7.9, def(Ga ) = |a| − 2k. Using Lemma 7.7.7(a), we can write xa = xδ f c , where |δ| = def(Ga ) and |c| = β1 (Ga ). Taking degrees in xa = xδ f c gives |a| = |δ| + 2|c| = (|a| − 2k) + 2|c|, that is, |c| = k. Then xa ∈ I k and the proof is complete. 2 Proposition 7.7.12 Let I = I(G) be the edge ideal of a graph G and let m = (x1 , . . . , xn ). If m ∈ Ass(R/I k ), then m ∈ Ass(R/I k+1 ). Proof. As m is an associated prime of R/I k , there is xa ∈ / I k such that a k mx ⊂ I . By Lemma 7.7.11 there is an edge {xi , xj } of G such that / I k+1 . Then, x (xi xj xa ) ∈ I k+1 for = 1, . . . , n; that is, m is an xi xj xa ∈ associated prime of R/I k+1 . 2 Lemma 7.7.13 [204, Lemma 3.4] Let I be a square-free monomial ideal in S = K[x1 , . . . , xm , xm+1 , . . . , xr ] such that I = I1 S + I2 S, where I1 ⊂ S1 = K[x1 , . . . , xm ] and I2 ⊂ S2 = K[xm+1 , . . . , xr ]. Then p ∈ Ass(S/I k ) if and only if p = p1 S + p2 S, where p1 ∈ Ass(S1 /I1k1 ) and p2 ∈ Ass(S2 /I2k2 ) with (k1 − 1) + (k2 − 1) = k − 1. If pi is an ideal of R, the generators of pi will generate a prime ideal in any ring that contains those variables. We will abuse notation by denoting the ideal generated by the generators of pi in any other ring by pi as well. Theorem 7.7.14 [305] Let G be a graph and let I = I(G) be its edge ideal. Then Ass(R/I k ) ⊂ Ass(R/I k+1 ) for all k. That is, I has the persistence property. Proof. Take p in Ass(R/I k ). We set m = (x1 , . . . , xn ). We may assume that p = (x1 , . . . , xr ). By Proposition 7.7.12, we may assume that p m. Write Ip = (I2 , I1 )p , where I2 is the ideal of R generated by all xi xj , i = j, whose image, under the canonical map R → Rp , is a minimal generator of Ip , and I1 is the ideal of R generated by all xi whose image is a minimal generator of Ip , which correspond to the isolated vertices of the graph associated to Ip . The minimal generators of I2 and I1 lie in S = K[x1 , . . . , xr ], and the two sets of variables occurring in the minimal generating sets of I1 and I2 (respectively) are disjoint and their union is {x1 , . . . , xr }. If I2 = (0), then p is a minimal prime of I so it is an associated prime of R/I k+1 . Thus, we may assume I2 = (0). An important fact is that localization preserves associated primes; that is p ∈ Ass(R/I k ) if and only

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if pRp ∈ Ass(Rp /(Ip Rp )k ). Hence, p is in Ass(R/I k ) if and only if p is in Ass(R/(I1 , I2 )k ) if and only if p is in Ass(S/(I1 , I2 )k ). By Proposition 7.7.12 and Lemma 7.7.13, p is an associated prime of S/(I1 , I2 )k+1 . Hence, we can argue backwards to conclude that p is an associated prime of R/I k+1 . 2 If G is a graph with loops, then its edge ideal I(G) has the persistence property [353]. Corollary 7.7.15 If I is a square-free monomial ideal and (I k+1 : I) = I k for k ≥ 1. Then, I has the persistence property. Proof. As in Proposition 7.7.12, we ﬁrst show that m ∈ Ass(R/I k ) implies m ∈ Ass(R/I k+1 ). Assume m ∈ Ass(R/I k ). Then there is a monomial xa ∈ I k with xi xa ∈ I k+1 for all i. By the hypothesis, xa ∈ (I k+1 : I), so there is a square-free monomial generator e of I (which can be viewed as the edge of a clutter associated to I) with exa ∈ I k+1 . But xi exa = e(xi xa ) ∈ I k+1 for all i, so m ∈ Ass(R/I k+1 ). Recall that since I is ﬁnitely generated, (I k+1 : I)p = (Ipk+1 : Ip ). Thus (Ipk+1 : Ip ) = Ipk . The remainder of the argument now follows from localization, as in the proof of Theorem 7.7.14. 2 Lemma 7.7.16 If I is a monomial ideal, then Ass(I k−1 /I k ) = Ass(R/I k ). Proof. Suppose that p ∈ Ass(R/I k ). Then p = (I k : c) for some monomial c ∈ R. But since p is necessarily a monomial prime, generated by a subset of the variables, then if xc ∈ I k for a variable x ∈ p, then c ∈ I k−1 and so p ∈ Ass(I k−1 /I k ). The other inclusion is automatic. 2 Conjecture 7.7.17 (Persistence problem) If I is a square-free monomial ideal, then I has the persistence property. The next counterexample for this conjecture is due to Kaiser, Stehl´ık, ˇ and Skrekovski [268] (cf. Exercise 7.7.25). Example 7.7.18 [268] Let I ∨ be the Alexander dual of the edge ideal I=

(x1 x2 , x2 x3 , x3 x4 , x4 x5 , x5 x6 , x6 x7 , x7 x8 , x8 x9 , x9 x10 , x1 x10 , x2 x11 , x8 x11 , x3 x12 , x7 x12 , x1 x9 , x2 x8 , x3 x7 , x4 x6 , x1 x6 , x4 x9 , x5 x10 , x10 x11 , x11 x12 , x5 x12 ).

Using Macaulay2, one can readily verify that Ass(R/(I ∨ )3 ) is not a subset of Ass(R/(I ∨ )4 ) (see Exercise 7.7.27). Theorem 7.7.19 [165] If G is a simple graph, then the chromatic number of G is the minimal k such that (x1 · · · xn )k−1 ∈ Ic (G)k .

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309

Note that one can use Normaliz [68] inside Macaulay2 to compute the integral closure of a monomial ideal and also the normalization of the Rees algebra of a monomial ideal. Procedure 7.7.20 The following simple procedure for Macaulay2 decides whether Ass(R/I 3 ) is contained in Ass(R/I 4 ) and whether we have the equality Ass(R/I 3 ) = Ass(R/I 4 ). It also computes I 4 and decides whether Ass(R/I 4 ) is equal to Ass(R/I 4 ). R=QQ[x1,x2,x3,x4,x5,x6,x7,x8,x9]; load "normaliz.m2"; I=monomialIdeal(x1*x2,x2*x3,x1*x3,x3*x4,x4*x5,x5*x6, x6*x7,x7*x8,x8*x9,x5*x9); isSubset(ass(I^3),ass(I^4)) ass(I^3)==ass(I^4) (intCl4,normRees4)=intclMonIdeal I^4; intCl4’=substitute(intCl4,R); ass(monomialIdeal(intCl4’))==ass(I^4)

Exercises 7.7.21 Let I be the square-free monomial ideal generated by x1 x2 x5 , x1 x3 x4 , x1 x2 x6 , x1 x3 x6 , x1 x4 x5 , x2 x3 x4 , x2 x3 x5 , x2 x4 x6 , x3 x5 x6 , x4 x5 x6 . Use Normaliz [68] and Macaulay2 [199], to show that I is a non-normal ideal such that (I 2 : I) = I and (I 3 : I) = I 2 . Then, prove that I has the persistence property and that the index of stability of I is equal to 3. 7.7.22 Let I = I(G) be the edge ideal of a graph G. Prove that the sets Ass(I k−1 /I k ) form an ascending chain for k ≥ 1. 7.7.23 Let G be a graph and let xi be a vertex. If y ∈ V (Gxi ), then (Gxi )y = (Gxi )xj for some xj ∈ V (G). 7.7.24 If I = (I1 , . . . , Is ), where the Ii are square-free monomial ideals in disjoint sets of variables, then p ∈ Ass(R/I k ) if and only if p = (p1 , . . . , ps ), where pi ∈ Ass(R/Iiki ) with (k1 − 1) + · · · + (ks − 1) = k − 1. 7.7.25 [326] Let I = (x1 x22 x3 , x2 x23 x4 , x3 x24 x5 , x4 x25 x1 , x5 x21 x2 ) be the ideal obtained by taking the product of consecutive edges of a pentagon. Then m ∈ Ass(R/I k ) for k = 1, 4, m ∈ Ass(R/I k ) for k = 2, 3, m = (x1 , . . . , x5 ). 7.7.26 [323, Lemma 2.3] If I is a monomial ideal of R and a ∈ I/I 2 is a regular element of the associated graded ring grI (R). Then the sets Ass(I k−1 /I k ) form an ascending chain.

310

Chapter 7

7.7.27 Use the following procedure for Macaulay2 to verify that the ideal I ∨ of Example 7.7.18 does not have the persistence property. R=QQ[x1,x2,x3,x4,x5,x6,x7,x8,x9,x10,x11,x12] I=monomialIdeal(x1*x2,x2*x3,x3*x4,x4*x5,x5*x6,x6*x7,x7*x8, x8*x9,x9*x10,x1*x10,x2*x11,x8*x11,x3*x12,x7*x12,x1*x9,x2*x8, x3*x7,x4*x6,x1*x6,x4*x9,x5*x10,x10*x11,x11*x12,x5*x12) J=dual I isSubset(ass(J^3),ass(J^4)) 7.7.28 Consider the graph G of the ﬁgure below, where vertices are labeled with i instead of xi . The duplication of the vertices x3 and x4 of G is shown below. If f = {x3 , x4 }, prove that deﬁciencies of G and Gf are not equal. 1

2

s @ @ 3 s

s5 s4 @ @s6

s G

s 3s s4 s5 @ QQ A s Qs4A 3 @ @A As 6 s @ 2 1

Gf

7.7.29 Load the package EdgeIdeals [166] in Macaulay2 [199] to verify that the chromatic number χ(G) of the graph G of Example 7.7.18 is equal to 4. 7.7.30 Pick any planar graph and use EdgeIdeals [166] to verify the four color theorem (any planar graph has chromatic number at most 4).

Chapter 8

Toric Ideals and Aﬃne Varieties In this chapter we study algebraic and geometric aspects of three important types of polynomial ideals and their quotient rings: toric ideals → lattice ideals → binomial ideals. These ideals are interesting from a computational point of view and are related to diverse ﬁelds, such as, numerical semigroups [173, 341], semigroup rings [180], commutative algebra and combinatorics [61, 317, 395], algebraic geometry [176, 317], linear algebra and polyhedral geometry [354, 420], integer programming [400], graph theory and combinatorial optimization [405], algebraic coding theory [348, 367], and algebraic statistics [142].

8.1

Binomial ideals and their radicals

Ideals generated by binomials are an interesting family of polynomial ideals. In this section we study the radical of those ideals. Deﬁnition 8.1.1 Let S = K[t1 , . . . , tq ] be a polynomial ring over a ﬁeld K. A binomial of S is an element of the form f = tα − tβ , for some α, β in Nq . An ideal generated by binomials is called a binomial ideal . A polynomial with at most two terms, say λta − μtb , where λ, μ ∈ K and a, b ∈ Nq , is called a non-pure binomial. Accordingly an ideal generated by non-pure binomials is called a non-pure binomial ideal . The essence of the Eisenbud–Sturmfels paper [134] is that the associated primes, the primary components and the radical of a non-pure binomial ideal are non-pure binomial ideals if K is algebraically closed. Binomial

312

Chapter 8

primary decompositions can be recovered from the theory of mesoprimary decomposition of congruences [267]. Computing a primary decomposition of a binomial ideal over a ﬁeld of characteristic zero can be done using “Binomials”[266], a package for the computer algebra system Macaulay2 [199]. To study binomial ideals over arbitrary ﬁelds we introduce the notion of congruence that comes from semigroup theory, this allows us to link general semigroup rings with binomial ideals (see Proposition 8.1.6). An equivalence relation on a set S is a subset R of S × S which is reﬂexive: (a, a) ∈ R, symmetric: (a, b) ∈ R ⇒ (b, a) ∈ R, and transitive: (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R. As usual we write a ∼ b if (a, b) ∈ R. Deﬁnition 8.1.2 A congruence in a commutative semigroup with identity (S, +) is an equivalence relation ∼ on S compatible with +. Example 8.1.3 Let L ⊂ Zq be a subgroup. The relation, a ∼L b if a − b is in L, deﬁnes a congruence in the additive semigroup Nq . Given a congruence ∼ on a semigroup S, it is usual to denote the quotient semigroup of S by S/ ∼:= {a| a ∈ S}, where a is the equivalence class of a. Lemma 8.1.4 If ∼ is a congruence on a semigroup (S, +) and a ∼ b, then ra ∼ rb for r ∈ N. Proof. By induction on r. If (r − 1)a ∼ (r − 1)b, then (r − 1)a + b ∼ rb. Since a ∼ b, we get ra ∼ b + (r − 1)a. Thus ra ∼ rb. 2 Lemma 8.1.5 Let S be a semigroup with a congruence ∼ and let a, b ∈ S. If m and n are relatively prime positive integers such that ma ∼ mb and na ∼ nb, then ra ∼ rb for all r > mn − m − n. Proof. By Lemma 8.7.8 we can write r = η1 m + η2 n for some η1 , η2 in N. 2 As na = nb and ma = mb, it follows that ra = rb as required. The radical of a binomial ideal Let S = K[t1 , . . . , tq ] be a polynomial ring over a ﬁeld K and let I = (ta1 − tb1 , . . . , tas − tbs ) be a binomial ideal of S. We set 0 := {(a1 , b1 ), . . . , (as , bs )}. The intersection of all congruences in (Nq , +) that contain 0 is called the congruence generated by 0 and it will be denoted by in what follows. According to [180] we can construct as follows: 1 2

= 0 ∪ {(a, a)| a ∈ Nq } ∪ {(b, a)| (a, b) ∈ 0 }, = {(a + c, b + c)| (a, b) ∈ 1 , c ∈ Nq }, = {(a, b)| ∃a0 , . . . , ar ∈ Nq with a0 = a, ar = b, (ai , ai+1 ) ∈ 2 ∀i}.

Since is clearly a congruence contained in any congruence containing 0 , we have that is the congruence generated by 0 .

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313

Proposition 8.1.6 [180, Corollary 7.3] I = ({ta − tb | (a, b) ∈ }). Proof. It follows rapidly from the construction of .

2

Let ∼ be a congruence in Nq . We associate to ∼ an equivalence relation on the monomials in S, by ta ∼ tb if a ∼ b. Note that this relation is compatible with the product, i.e., ta ∼ tb implies tc ta ∼ tc tb for all c ∈ Nq . Deﬁnition 8.1.7 A non-zero polynomial f = α λα tα in S, with λα ∈ K, is called simple (with respect to ∼) if all its monomials, i.e., those tα with non-zero coeﬃcient λα , are equivalent under ∼. Given any polynomial f ∈ S \ {0}, we can group together its monomials by equivalence classes under ∼, thereby obtaining a decomposition f = h1 + · · · + hm with the property that each summand hi is simple, and that no monomial in hi is equivalent with a monomial in hj if j = i. Such a decomposition of f as a sum of maximal simple subpolynomials is unique up to order. We will refer to the hi ’s as the simple components of f (with respect to ∼). These concepts were introduced by Eliahou [135]. Lemma 8.1.8 Let ∼ be a congruence in Nq . If J is an ideal of S generated by a set G of simple polynomials and f ∈ J, then every simple component of f also belongs to J. Proof. Each generator g ∈ G is simple. Moreover, tc g remains simple for any c ∈ Nq , since the relation ∼ on monomials is compatible with the product. Let f be any element in the ideal J. Then, f is a ﬁnite linear combination of polynomials of the form tci gi , with gi ∈ G simple and ci ∈ Nq , which are simple. Hence, every simple component of f is also a linear 2 combination of some tci gi and therefore belongs to J. Deﬁnition 8.1.9 Let a, b ∈ Nq and let p be a positive prime number. If (pr a, pr b) ∈ for some integer r ≥ 0, we say that a and b are p-equivalent modulo and write a ∼p b. The radical of a binomial ideal is generated by binomials as is seen below. If K is algebraically closed, then the radical of a non-pure binomial ideal is a non-pure binomial ideal [134, Theorem 3.1] (see Exercise 8.1.15). Theorem 8.1.10 [180] If p = char(K) = 0, then rad(I) = ({ta − tb | a ∼p b}).

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Chapter 8 r

r

Proof. “⊃”: If a ∼p b, then tp a − tp b ∈ I by Proposition 8.1.6. Thus r (ta − tb )p belongs to I and ta − tb ∈ rad(I). m “⊂”: Take f ∈ rad(I), we can write f = i=1 λi tαi , with i ∈ K for λ r r r m all i. For r 0 we have f p ∈ I. From the equality f p = i=1 μi tp αi , r r p μi = λpi , and using Lemma 8.1.8 we may assume that mf is a simple component with respect to the congruence . Note that i=1 μi = 0, hence r

r

f p = μ2 (tp

α2

r

− tp

α1

r

) + · · · + μm (tp

αm

r

− tp

α1

),

with (pr αi , pr α1 ) ∈ for all i ≥ 2, as required.

2

Lemma 8.1.11 Let a, b ∈ Nq . If there is an integer s > 0 such that (ra, rb) is in for all r ≥ s, then f = ta − tb ∈ rad(I). Proof. Set r = 2s + 1. We claim f r ∈ I. From the binomial expansion: fr =

s i=0

(−1)i

r r (ta )r−i (tb )i + (−1)r−i (ta )i (tb )r−i . i r−i

(8.1)

Note s ≤ r − i for 0 ≤ i ≤ s. Hence, from ((r − i)a, (r − i)b) ∈ , we get (ra, (r − i)b + ia) ∈ and ((r − i)a + ib, rb) ∈ . Thus, as (ra, rb) ∈ , we get ((r − i)b + ia, (r − i)a + ib)) ∈ . Therefore any summand in Eq. (8.1) is in I, as required. 2 Following [180], two elements a, b in Nq that satisfy the conditions of Lemma 8.1.11 are called asymptotically equivalent modulo . Theorem 8.1.12 [180] If char(K) = 0, then rad(I) = ({ta − tb | ∃ 0 = s ∈ N such that (ra, rb) ∈ ; ∀ r ≥ s}). Proof. By Lemma 8.1.11 we need only show the inclusion “⊂”. Take 0 = f ∈ rad (I). We can write f = λ1 tα1 + · · · + λm tαm with 0 = λi ∈ K for all i and λ1 + · · · + λm = 0. If α1 is asymptotically equivalent modulo to α2 , from the equality f1 = f − λ2 (tα2 − tα1 ) = λ3 (tα3 − tα1 ) + · · · + λm (tαm − tα1 ) and Lemma 8.1.11 we get f1 ∈ rad(I). Thus we may either assume that αi is not asymptotically equivalent modulo to αj for i = j or that f = 0. To complete the proof we assume f = 0 and derive a contradiction. Let E be the set of prime numbers p > 0 such that αi is p-equivalent modulo to αj . By Lemma 8.1.5 the set E is ﬁnite. Consider the ring A0 = Z[λ1 , . . . , λm ] ⊂ K and the set {mi }i∈I of maximal ideals of A0 . Since

Toric Ideals and Aﬃne Varieties

315

A0 is a Hilbert ring, using [65, Corollary A.18], it follows that (0) = ∩i∈I mi and A0 /mi is a ﬁnite ﬁeld for i ∈ I. If I0 is the subset of I consisting of all i such that char(A ;0 /mi ) is not in E, then;(0) = ∩i∈I0 mi . Indeed take a ∈ ∩i∈I0 mi , then ( p∈E p)a = 0 because ( p∈E p)a ∈ mi for all i ∈ I, thus a = 0 because char(A0 ) = 0. ∈ / E. Pick Let i ∈ I0 , we set m = mi and p = char(A0 /m). Note p r r r suﬃciently large such that f p is in I. We can write f p = ni=1 μi tβi , with μi ∈ A0 \ {0} and tβ1 , . . . , tβn distinct. Since the n simple components r of f p , with respect to , are again in I using that i=1 μi = 0 (on every r component of f p ) we readily see that r

f p = η1 (tγ1 − tδ1 ) + · · · + ηs (tγs − tδs )

((γi , δi ) ∈ )

with ηi ∈ A0 for all i. If h ∈ A0 [t], we denote by h the image of h under the canonical map A0 [t] → (A0 /m)[t]. If f = 0, then from the equation f pr = η1 (tγ1 − tδ1 ) + · · · + ηs (tγs − tδs )

((γi , δi ) ∈ )

we get αi ∼p αj modulo for some i = j, a contradiction because p is not in E. Thus f ∈ m[t]. Therefore λ1 , . . . , λm are in the maximal ideal mi for every i ∈ I0 and consequently λi = 0 for all i, a contradiction. This proof was adapted from [180]. 2

Exercises 8.1.13 Let I = (ta1 − tb1 , . . . , tas − tbs ) ⊂ S and let L be the subgroup of Zq generated by a1 − b1 , . . . , as − bs . Prove the following: (a) The congruence generated by 0 = {(a1 , b1 ), . . . , (as , bs )} is contained in the congruence = {(a, b)| a, b ∈ Nq ; a − b ∈ L}. (b) = if ti is not a zero divisor of S/I for all i. (c) If char(K) = p > 0, then I is a radical ideal if and only if Nq / is a semigroup that is p-torsion-free. 8.1.14 Let I = (y 2 z−x2 t, z 4 −xt3 ) ⊂ Q[x, y, z, t]. Prove that I is a complete intersection whose primary decomposition is given by I = q1 ∩q2 ∩q3 , where q1 = (y 2 z − x2 t, z 4 − xt3 , xz 3 − y 2 t2 , x3 z 2 − y 4 t, y 6 − x5 z), q2 = (t3 , zt2 , z 2 t, z 3 , y 2 z − x2 t) and q3 = (x, z). Prove that q1 is prime. 8.1.15 (I. Swanson) If K = Z2 (t) is a ﬁeld of rational functions in one variable and R = K[x, y] is a polynomial ring, prove that the radical of the ideal (x2 + t, y 2 t + 1) is equal to (x2 + t, x + y + 1).

316

8.2

Chapter 8

Lattice ideals

In this section we study lattice ideals and their radicals. We present some classiﬁcations of lattice and toric ideals. In Chapter 9 we will study primary decompositions and the degree of lattice ideals. The class of lattice ideals has been studied in several places; see for instance [22, 134, 251, 337] and the references therein. Let S = K[t1 , . . . , tq ] be a polynomial ring over a ﬁeld K and let L be a lattice in Zq ; that is, L is an additive subgroup of Zq . The rank of L, denoted by rank(L), is the nonnegative integer r such that L Zr . A partial character is a group homomorphism from L to the multiplicative group K ∗ = K \ {0}. Deﬁnition 8.2.1 Let L be a lattice in Zq and let ρ : L → K ∗ be a partial character. The lattice ideal of L relative to ρ is the ideal: Iρ (L) = ({ta+ − ρ(a)ta− | a ∈ L}) ⊂ S. If ρ is the trivial character ρ(a) = 1 for a in L, we denote Iρ (L) simply by I(L). If no partial character is speciﬁed we shall always assume that ρ is the trivial character. Theorem 8.2.2 [134] Let L be a lattice in Zq of rank r and let ρ : L → K ∗ be a partial character. The following hold. (i) Iρ (L) contains no monomials. (ii) dim(S/Iρ (L)) = q − r, that is, ht(I(L)) = rank(L). (iii) ti is a non-zero divisor of S/Iρ (L) for all i. (iv) If Zq /L is torsion-free, then Iρ (L) is prime. Proof. Let α1 , . . . , αr be a Z-basis of L and let A be the r × q matrix with rows α1 , . . . , αr . Consider unimodular integral matrices P = (pij ) and Q = (qij ) of orders q and r, respectively, and make a change of basis αi =

r

qij αj and ei =

j=1

q

pij ej .

j=1

By Theorem 1.2.2, QAP −1 = diag{d1 , . . . , dr , 0, . . . , 0} for some P and Q. By the change of basis theorem the new basis is related by αi = di ei for i = 1. . . . , r, where ei = ei P −1 . Thus L = d1 e1 , . . . , dr er . Note that the isomorphism φ : Zq → Zq , α→αP −1 , induces a K-algebra isomorphism ±1 ϕ : S = K[t±1 1 , . . . , tq ] −→ S

ϕ

(tα −→ tφ(α) ).

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Consider I = (tα − ρ(α)| α ∈ L) ⊂ S . If α ∈ L, then we can rthe ideal write α = i=1 ai αi for some a1 , . . . , ar in Z and tα − ρ(α) maps under ϕ to ta1 1 d1 · · · tar r dr − ρ(α1 )a1 · · · ρ(αr )ar . Therefore, setting hi = tdi i − ρ(αi ) for i = 1, . . . , r, it follows that ϕ(I ) is equal to (h1 , . . . , hr ). This expression for ϕ(I ) is useful to study the structure of Iρ (L) as is seen below. (i): Let K be the algebraic closure of K and write ρ(αi ) = γidi for some γi in K. If Iρ (L) contains a monomial, then ϕ(I ) contains a Laurent monomial since Iρ (L) ⊂ I , a contradiction because tdi i − γidi = 0 if ti = γi . (ii): As h1 , . . . , hr form a regular sequence, we get ht(ϕ(I )) = r. Since S /I is isomorphic to S /ϕ(I ), by Exercise 8.2.31, it follows that the Krull dimension of S/Iρ (L) is equal to q − r. (iii): Let G = {g1 , . . . , gs } be a Gr¨obner basis for Iρ (L) with respect to a term order ≺. For simplicity we set i = 1. Assume that there is 0 = h ∈ S such that t1 h ∈ Iρ (L). By the division algorithm (see Theorem 3.3.6) we can write h = h1 g1 +· · ·+hs gs +f , where no term in f is divisible by in≺ (gi ) for all i; that is, the distinct terms ta1 , . . . , tar occurring in f are standard monomials. To show that t1 is not a zero divisor of S/Iρ (L), it suﬃces to prove that f = 0. Assume that f = 0. Let ∼ be the congruence given by a ∼ b if a − b ∈ L. Since Iρ (L) cannot contain monomials (see part (i)), we can decompose t1 f into simple components with respect to ∼ and apply Lemma 8.1.8 to derive that (e1 + ai ) − (e1 + aj ) ∈ L for some i = j. Hence tai − ρ(ai − aj )taj belongs to Iρ (L) and either tai or taj is not standard, a contradiction. (iv): If Zq /L is torsion-free, then di = 1 for i = 1, . . . , r. Thus ϕ(I ) is prime because hi = ti − ρ(αi ) for all i. Hence I is also prime. By part (iii) it follows that I ∩ S = Iρ (L). Thus Iρ (L) is prime as well. 2 Deﬁnition 8.2.3 The saturation of a lattice L ⊂ Zq , denoted by Ls , is the lattice consisting of all α ∈ Zq such that ηα ∈ L for some η ∈ Z \ {0}. From the proof of Theorem 8.2.2 we obtain the following expression for the saturation (cf. Lemma 8.2.5). Corollary 8.2.4 If L is a lattice in Zq , then Ls = Z{e1 , . . . , er } Proof. Since αi = di ei for i = 1, . . . , r, we need only show that Ls is contained in Z{e1 , . . . , er }. Take α ∈ Ls , then α = ri=1 (ai di /n)ei with n, ai ∈ Z; n = 0. Using that the set {e1 , . . . , eq } is a Z-basis for Zq we obtain that all the coeﬃcients ai di /n are in Z, thus α is in Z{e1 , . . . , er }. 2 The next result gives another method to compute the saturation. Lemma 8.2.5 If L is a lattice of rank r in Zq , then there is an integral matrix A of size (q − r) × q and rank(A) = q − r such that L ⊂ kerZ (A), with equality if and only if Zq /L is torsion-free. Furthermore kerZ (A) is the saturation of L.

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Proof. We may assume that L = Zα1 ⊕ · · · ⊕ Zαr , where α1 , . . . , αq is a basis of the vector space Qq . Consider the hyperplane Hi of Qq generated by α1 , . . . , α 'i , . . . , αq . Note that the subspace of Qq generated by α1 , . . . , αr is equal to Hr+1 ∩ · · · ∩ Hq . There is a normal vector wi ∈ Zq such that Hi = {α ∈ Qq | α, wi = 0}. It is not hard to see that the matrix A with rows wr+1 , . . . , wq is the matrix with the required conditions because by construction αi ∈ Hj for i = j and consequently wr+1 , . . . , wq are linearly independent. In particular we have the equality rank(L) = rank(kerZ (A)). Since kerZ (A)/L is a ﬁnite group it follows that Ls , the saturation of L, is equal to kerZ (A). 2 Lemma 8.2.6 Let L be a lattice in Zq . If f = ta −λtb ∈ S with 0 = λ ∈ K, then f ∈ Iρ (L) if and only if a − b ∈ L and λ = ρ(a − b). Proof. Assume that f ∈ Iρ (L). Consider the congruence in Nq given by a ∼L b if a − b ∈ L. By Theorem 8.2.2, Iρ (L) contains no monomials. Hence, by Lemma 8.1.8, f is simple, i.e., a − b ∈ L. As ta − ρ(a − b)tb is in Iρ (L), we get λ = ρ(a − b). The converse is clear. 2 Proposition 8.2.7 Let Iρ (L) ⊂ S be a lattice ideal and let ≺ be a term order. Then the elements in the reduced Gr¨ obner basis of Iρ (L) are of the form ta+ − ρ(a)ta− with a ∈ L. Proof. Let G be a ﬁnite set of generators of Iρ (L) consisting of polynomials of the form λ1 ta − λ2 tb with λi ∈ K ∗ for i = 1, 2 and let f, g ∈ G. As Iρ (L) contains no monomials (see Theorem 8.2.2), the S-polynomial S(f, g) and the remainder of S(f, g) with respect to G are both also of this form. Then it follows that the output of the Buchberger’s algorithm (see Theorem 3.3.12) is a Gr¨obner basis of Iρ (L) consisting of polynomials of the same form. Hence, by Lemma 8.2.6 and Theorem 8.2.2(iii), the elements in the reduced 2 Gr¨ obner basis of Iρ (L) are of the form ta+ − ρ(a)ta− with a ∈ L. Theorem 8.2.8 If I is a non-pure binomial ideal of S, then I is a lattice ideal of the form Iρ (L) for some lattice L and some partial character ρ if and only if I S and ti is a non-zero divisor of S/I for all i. Proof. ⇒) This part follows from items (i) and (iii) of Theorem 8.2.2. ⇐) Let L be the set of a ∈ Zq such that ta+ − λta− ∈ I for some λ ∈ K ∗ and let a, b be two points in L. Then tb+ − λ1 tb− ∈ I for some λ1 in K ∗ . We claim that L is a lattice. Since ta− − λ−1 ta+ ∈ I, we get −a ∈ L. Notice that ta+ +b+ − λλ1 ta− +b− is in I. Factoring the gcd of ta+ +b+ and ta− +b− , we get a + b ∈ L. Given a ∈ L there is a unique λ ∈ K ∗ such that ta+ − λta− ∈ I. Setting ρ(a) = λ, we get a partial character. Thus I is the 2 lattice ideal Iρ (L). This part was adapted from [134].

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Let L ⊂ Zq be a lattice. We associate to L an equivalence relation on the monomials in S = K[t1 , . . . , tq ], by tα ∼L tβ if α − β ∈ L. Lemma 8.2.9 Let Iρ (L) be a lattice ideal and let I be an ideal generated by a subset {tai − ρ(ai − bi )tbi }ri=1 of Iρ (L). If G = Z{ai − bi }ri=1 and f ∈ I, then every simple component of f with respect to ∼G also belongs to I. Proof. It follows from Lemma 8.1.8 because the relation, a ∼G b if a−b ∈ G deﬁnes a congruence in Nq and tai − ρ(ai − bi )tbi is simple for all i. 2 Lemma 8.2.10 Let a, αi , b, βi be in Nq for i = 1, . . . , r, let a − b be in the lattice L = Z{α1 −β1 , . . . , αr −βr } and let ρ : L → K ∗ be a partial character. Then there is tδ ∈ S such that tδ (ta − ρ(a − b)tb ) ∈ (tα1 − ρ(α1 − β1 )tβ1 , . . . , tαr − ρ(αr − βr )tβr ). Proof. We set f = ta −ρ(a−b)tb and gi = tαi −ρ(αi −βi )tβi for i = 1, . . . , r. There are integers n1 , . . . , nr such that n1 nr · · · tαr /tβr − ρ(a − b). (∗) (ta /tb ) − ρ(a − b) = tα1 /tβ1 We may assume that ni ≥ 0 for all i by replacing, if necessary, tαi /tβi by its inverse. Writing tαi /tβi = ((tαi /tβi ) − ρ(αi − βi )) + ρ(αi − βi ) and using the binomial theorem, from Eq. (∗), it follows that tδ f ∈ (g1 , . . . , gr ) for some monomial tδ . 2 Lemma 8.2.11 A lattice L ⊂ Zq is generated by a1 , . . . , am if and only if +

−

+

−

Iρ (L) = ((ta1 − ρ(a1 )ta1 , . . . , tam − ρ(am )tam ) : (t1 · · · tq )∞ ). +

−

+

−

Proof. ⇒) We set I = (ta1 − ρ(a1 )ta1 , . . . , tam − ρ(am )tam ). “⊂”: Take + − α ∈ L. By Lemma 8.2.10 there is tδ ∈ S such that tδ (tα − ρ(α)tα ) ∈ I. + − Thus tα − ρ(α)tα ∈ (I : (t1 · · · tq )∞ ). “⊃”: This inclusion follows readily from Theorem 8.2.8. ⇐) This part follows from Lemma 8.2.9 2 The next result can be used to compute the presentation ideal of a subring generated by rational functions. Proposition 8.2.12 Let F = {f1 /g1 , . . . , fq /gq } ⊂ K(x) be a ﬁnite set of rational functions, with fi , gi ∈ K[x] = K[x1 , . . . , xn ] and gi = 0 for all i, and let ϕ be the homomorphism of K-algebras ϕ : S = K[t1 , . . . , tq ] −→ K[F ], induced by ϕ(ti ) = fi /gi . Then ker(ϕ) = (g1 t1 − f1 , . . . , gq tq − fq , yg1 · · · gq − 1) ∩ S, where y is a new variable. If gi = 1 for all i, then ker(ϕ) = (t1 − f1 , . . . , tq − fq ) ∩ S.

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Proof. We set J = (g1 t1 − f1 , . . . , gq tq − fq , yg1 · · · gq − 1) and hi = fi /gi for i = 1, . . . , q. Let us show the ﬁrst equality. The second equality will follow by similar arguments. To show that ker(ϕ) ⊂ J ∩ S take f ∈ ker(ϕ). Making ti = (ti − hi ) + hi and using the binomial theorem we can write: f = f (t1 , . . . , tq ) =

γ

aγ (t1 − h1 )γ1 · · · (tq − hq )γq +

β

β

aβ hβ1 1 · · · hq q ,

with aγ ∈ K[F ] for all γ = (γ1 , . . . , γq ) and all γ’s in Nq \ {0}. As βq β1 = 0. Therefore multiplying f (h1 , . . . , hq ) = 0, we get β aβ h1 · · · hq the equality above by an appropriate positive power of g1 · · · gq we get (g1 · · · gq )s f = γ bγ (g1 t1 − f1 )γ1 · · · (gq tq − fq )γq , (8.2) where bγ is polynomial in K[x] for all γ. Making z = yg1 · · · gq − 1 we get g1 · · · gq = (z + 1)/y. Thus from Eq. (8.2), we obtain that (z + 1)s f ∈ J. Thus f ∈ J ∩ S. Conversely let f ∈ J ∩ S. Then f = f (t1 , . . . , tq ) = a1 (g1 t1 − f1 ) + · · · + aq (gq tq − fq ) + b(g1 · · · gq y − 1). Hence f (h1 , . . . , hq ) = b (g1 · · · gq y − 1). The polynomial f is independent of y. Making y = 1/g1 · · · gq , we get f (h1 , . . . , hq ) = 0, i.e., f ∈ ker(ϕ). 2 Let us illustrate how to compute presentation ideals using Gr¨obner bases and elimination of variables. Example 8.2.13 Let F = {xi xj | 1 ≤ i < j ≤ 4} and let ϕ : K[t] → K[F ] be the map of K-algebras induced by ϕ(tij ) = xi xj , where t is the set of variables {tij | 1 ≤ i < j ≤ 4}. By Proposition 8.2.12 one has PF = ker(ϕ) = L ∩ K[tij ], L = (tij − xi xj | 1 ≤ i < j ≤ 4). Using Macaulay2 [199] we get that the reduced Gr¨ obner basis of L with respect to the elimination ordering in the ﬁrst variables x1 , . . . , x4 is: t14 t23 − t12 t34 , x24 t12 − t14 t24 , x23 t12 − t13 t23 , x2 x4 − t24 , x1 t23 − x3 t12 , x3 t12 t34 − x4 t13 t23 .

t13 t24 − t12 t34 , x3 t24 − x4 t23 , x2 t34 − x4 t23 , x2 x3 − t23 , x1 x4 − t14 ,

x24 t23 − t24 t34 , x3 t14 − x4 t13 , x2 t14 − x4 t12 , x1 t34 − x4 t13 , x1 x3 − t13 ,

x24 t13 − t14 t34 , x3 x4 − t34 , x2 t13 − x3 t12 , x1 t24 − x4 t12 , x1 x2 − t12 ,

Therefore PF = (t14 t23 − t12 t34 , t13 t24 − t12 t34 ) by Theorem 3.3.20. The ring of Laurent polynomials in the variables x1 , . . . , xn , denoted by ±1 K[x±1 ] or K[x±1 1 , . . . , xn ], is generated as a K-vector space by the set of Laurent monomials, i.e., by the set of all xa with a ∈ Zn .

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Deﬁnition 8.2.14 Let A = {v1 , . . . , vq } be a set of points in Zn . The monomial subring generated or spanned by F = {xv1 , . . . , xvq } is: K[F ] := D, D∈F

where F is the family of all subrings D of R such that K ∪ F ⊂ D. a a The elements of K[F ] have the form ca (xv1 ) 1 · · · (xvq ) q with ca ∈ K and all but a ﬁnite number of ca ’s are zero. As a K-vector space K[F ] is generated by the set of monomials of the form xa , with a in the semigroup NA generated by A. That is K[F ] = K[xa | a ∈ NA]. This means that K[F ] coincides with K[NA], the semigroup ring of the semigroup NA (see [180]). Deﬁnition 8.2.15 An ideal P ⊂ S is called a toric ideal if there is a ﬁnite set F = {xv1 , . . . , xvq } in K[x±1 ] such that P is the kernel of the epimorphism of K-algebras: ϕ : S = K[t1 , . . . , tq ] −→ K[F ] −→ 0, induced by ϕ(ti ) = xvi . The ideal P is called the toric ideal of K[F ] and is denoted by PF . Proposition 8.2.16 [400, p. 32] If F = {xv1 , . . . , xvq } is a set of Laurent monomials in K[x±1 ], then the corresponding toric ideal is given by −

+

−

+

PF = (xv1 t1 − xv1 , . . . , xvq tq − xvq , yx1 · · · xn − 1) ∩ S. Proof. It follows adapting the proof of Proposition 8.2.12.

2

This result gives a method to compute toric ideals. Another method can be found in [109]. In [38] some algorithms are devised, that in many respects improve existing algorithms for the computation of toric ideals. Lemma 8.2.17 Let B = K[y1 , . . . , yn , t1 , . . . , tq ] be a polynomial ring over a ﬁeld K. If I is a binomial ideal of B, then the reduced Gr¨ obner basis of I with respect to any term order ≺ consists of binomials. Furthermore I ∩ K[t1 , . . . , tq ] is a binomial ideal. Proof. Let B be a ﬁnite set of generators of I consisting of binomials and let f, g ∈ B. Since the S-polynomial S(f, g) is again a binomial and the remainder of S(f, g) with respect to B is also a binomial, the output of the Buchberger’s algorithm (Theorem 3.3.12) is a Gr¨obner basis of I consisting of binomials. Hence the reduced Gr¨ obner basis of I consists of binomials. If ≺ is the lex order y1 " · · · " yn " t1 " · · · " tq and K[t] is the ring K[t1 , . . . , tq ], then by elimination theory (see Theorem 3.3.20) G ∩ K[t] is a Gr¨ obner basis of I ∩ K[t]. Hence I ∩ K[t] is a binomial ideal. 2

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Corollary 8.2.18 The reduced Gr¨ obner basis of any toric ideal P ⊂ S consists of binomials of the form ta+ − ta− with respect to any term order. Proof. By Proposition 8.2.16 and Lemma 8.2.17, P is a binomial ideal. Therefore, by Theorem 8.2.8, P is a lattice ideal because P is prime. Hence the result follows at once from Proposition 8.2.7. 2 Deﬁnition 8.2.19 Let F = {xv1 , . . . , xvq } be a set of Laurent monomials. The associated matrix of the monomial subring K[F ], denoted by A, is the matrix whose columns are the vectors v1 , . . . , vq . Corollary 8.2.20 If A is the associated matrix of a monomial subring K[F ], then the toric ideal PF is the lattice ideal of kerZ (A). Proof. By Corollary 8.2.18, PF is generated by binomials of the form ta+ − ta− . Since a binomial ta+ − ta− is in PF if and only if A(a+ ) = A(a− ), we get that PF is generated by the set of all ta+ − ta− such that Aa = 0, i.e., PF is the lattice ideal of kerZ (A). 2 Corollary 8.2.21 [216] Let K[F ] be a Laurent monomial subring and let A be its associated matrix. Then dim K[F ] = rank(A). Proof. The toric ideal P of K[F ] is the lattice ideal of L = kerZ (A) by Corollary 8.2.20. Since the rank of L is q − rank(A), using Theorem 8.2.2, we get that dim K[F ] = rank(A). 2 Toric ideals form a special class of lattice ideals. Theorem 8.2.22 Let L ⊂ Zq be a lattice. The following are equivalent : (a) I(L) is a toric ideal. (b) I(L) is a prime ideal. (c) Zq /L is torsion-free. (d) L = kerZ (A) for some integral matrix A. Proof. (a) ⇒ (b): If I(L) is a toric ideal, then S/I(L) K[F ] for some ﬁnite set of Laurent monomials of K[x±1 ]. Since K[F ] is an integral domain, the ideal I(L) is prime. (b) ⇒ (c): Assume nα ∈ L for some integer n > 0 and α ∈ Zq . First we assume that either p = char(K) = 0 or that p = 0 is relatively prime to n. + − Since tnα − tnα is in I(L) and n · 1K = 0, from the equation xn − y n = (x − y)(xn−1 + xn−2 y + · · · + xy n−2 + y n−1 ) +

−

+

−

with x = tα and y = tα we get tα − tα ∈ I(L). Thus α ∈ L. If p > 0 and gcd(p, n) = p, we write n = pr m, where p and m are relatively prime. + − Since tmα − tmα ∈ I(L), by the previous argument we obtain α ∈ L.

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(c) ⇒ (d): Let A be as in Lemma 8.2.5. Thus kerZ (A)/L is a ﬁnite group, which must be zero because it is torsion-free. (d) ⇒ (a): Let v1 , . . . , vq be the columns of A. If F = {xv1 , . . . , xvq }, by Corollary 8.2.20, PF is the lattice ideal of L = kerZ (A). 2 g = α − β. Given a non-pure binomial g = tα − λtβ , with λ ∈ K ∗ , we set ' If B ⊂ Zq , we denote by B or ZB the subgroup of Zq generated by B. Lemma 8.2.23 Let L ⊂ Zq be a lattice. If g1 , . . . , gr is a set of non-pure g1 , . . . , ' gr }. In particular if I is binomials that generate Iρ (L), then L = Z{' a lattice ideal, there are unique L and ρ such that I = Iρ (L). Proof. Consider the lattice G = Z{' g1 , . . . , g'r }. First we show the inclusion + − L ⊂ G. Take 0 = a ∈ L. Then f = ta − ρ(a)ta is in Iρ (L) = (g1 , . . . , gr ). By Lemma 8.2.9 any simple component of f , with respect to ∼G , is also in + − (g1 , . . . , gr ). Since ta and ta are not in Iρ (L), then f simple with respect to ∼G , i.e., a = a+ − a− ∈ G. Thus L ⊂ G. To show the other inclusion notice that, by Lemma 8.2.6, g'i ∈ L for all i, i.e., G ⊂ L. 2 Lemma 8.2.24 Let I be an ideal generated by a set of binomials g1 , . . . , gr of S and let h1 , . . . , hs be a set of binomials of S that generate rad(I). If char(K) = 0, then Z{' g1 , . . . , ' gr } = Z{' h1 , . . . , ' hs }. Proof. We begin by writing the binomials gi and hi as gi = tαi − tβi and g1 , . . . , g'r } and L2 = Z{' h1 , . . . , ' hs }. hi = tγi − tδi . We set L1 = Z{' “⊂” Since gi ∈ rad(I), then by Lemma 8.2.9, any simple component of gi with respect to L2 belongs to rad(I). Therefore αi ∼L2 βi otherwise rad(I) would contain tαi , which is impossible. This proves that L1 ⊂ L2 . m “⊃” Since hi = tγi − tδi is in rad(I), then hpi is in I for m 0 and for m m an arbitrary odd prime number p. We claim that tp γi ∼L1 tp δi . Consider the following equality which is obtained from the binomial theorem m

m hpi

=

p

(−1)k

k=0 m

m m p (tγi )p −k (tδi )k . k m

m

If tp γi and tp δi are not in the same simple component of hpi with respect to L1 , then there is C ⊂ {1, . . . , pm − 1}, C = ∅, such that the polynomial m m m p (−1)k f = tp γ i + (tγi )p −k (tδi )k k k∈C

m

is a simple component of hpi with respect to L1 . By Lemma 8.2.9, f ∈ I. Hence, since char(K) = 0, we get m p (−1)k f (1, . . . , 1) = 0 = 1 + , k k∈C

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m a contradiction to pk ≡ 0 mod(p) for 1 ≤ k ≤ pm − 1; see Exercise 8.3.43. m m Therefore tp γi ∼L1 tp δi . Thus pm (γi − δi ) ∈ L1 . If we pick another odd prime number q = p and t 0, repeating the previous argument, we get 2 q t (γi − δi ) ∈ L1 , and hence γi − δi ∈ L1 , as required. Theorem 8.2.25 Let I be a binomial ideal and let I = q1 ∩ · · · ∩ qm be an irredundant primary decomposition of I. If q1 , . . . , qm are binomial ideals and char(K) = 0, then rad(I) = I. Proof. There are binomials g1 , . . . , gr and h1 , . . . , hs that generate I and rad(I), respectively. Say gi = tαi − tβi and hi = tγi − tδi . Since rad(I) is equal to rad(q1 ) ∩ · · · ∩ rad(qm ), it suﬃces to prove the case m = 1. Thus, i let us assume that I is primary and show that hi = tγi − tδ belongs to I for all i = 1, . . . , s. By Lemma 8.2.24, we can write γi − δi = rj=1 ηj (αj − βj ) with ηj ∈ Z for all j, and by Lemma 8.2.10 there is a monomial tγ such that tγ hi ∈ I. If hi ∈ / I, then (tγ ) ∈ I for some ≥ 1 because I is primary, 2 but this is impossible. Thus hi ∈ I, as required. Corollary 8.2.26 Let I be a binomial ideal without embedded primes. If char(K) = 0 and rad(I) is prime, then rad(I) = I and I is a prime ideal. Proof. Set p = rad(I). Since I has no embedded primes, by Exercise 2.1.48, one has Ass(S/I) = {p}. Thus I is a p-primary ideal and I is a radical ideal by Theorem 8.2.25. Thus I = p and I is a prime ideal. 2 Theorem 8.2.27 Let I(L) be the lattice ideal of L and let p = char(K). (a) If p = 0, then rad(I(L)) = I(L). (b) If p = 0, then rad(I(L)) = (ta − tb | pr (a − b) ∈ L for some r ∈ N). Proof. (a) Assume tra − trb ∈ I(L) for r ≥ s0 > 0. By Theorem 8.1.12 and Exercise 8.1.13 it suﬃces to prove that a − b ∈ L. Note that the relation “α ∼ β if and only if α − β ∈ L” is a congruence in Nq , thus r(a − b) ∈ L for r ≥ s0 and consequently a − b ∈ L. (b) It follows from Theorem 8.1.10 and Lemma 8.1.8. 2 Deﬁnition 8.2.28 An ideal I ⊂ S is called a binomial set-theoretic complete intersection if there are binomials f1 , . . . , fr in I such that rad (I) is equal to rad (f1 , . . . , fr ), where r = ht (I). Theorem 8.2.29 [293] Let I(L) ⊂ S be a lattice ideal of height r and let I be an ideal of S generated by binomials g1 , . . . , gr . If rad(I(L)) = rad(I) and char(K) = 0, then I(L) = I.

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Proof. By Theorem 8.2.27 the ideal I(L) is radical. Let I = q1 ∩ · · · ∩ qs be a primary decomposition of I. Since I is an ideal of height r generated by r elements, by the unmixedness theorem I has no embedded primes; see Theorem 2.3.27. Hence rad(qi ) = pi is a minimal prime of both I and I(L). Take a binomial h in I(L). By Lemma 8.2.24 and Lemma 8.2.10, there is a monomial tα so that tα h ∈ I. Thus tα h ∈ qi for all i. It suﬃces to prove / qi for some i, then (tα ) ∈ qi and that h belongs to qi for all i. If h ∈ consequently pi must contain a variable, a contradiction to Theorem 8.2.8 because none of the variables of S is a zero divisor of S/I(L). 2

Exercises 8.2.30 If L is a lattice of Zq and Iρ (L) ⊂ S is a lattice ideal. Prove that Iρ (L) = ({ta − ρ(a)| a ∈ L}) ∩ S ± and ({ta − ρ(a)| a ∈ L}) = Iρ (L)S , where S = K[t± 1 , . . . , tq ].

8.2.31 Let T be the set of monomials of S. If L = Iρ (L) is a lattice ideal of S and p ∈ AssS (S/L), then T ∩ p = ∅. Prove that ht(L) is equal to ± ht(T −1 (L)) and that T −1 (L) = ({ta − ρ(a)| a ∈ L}) ⊂ K[t± 1 , . . . , tq ]. 8.2.32 Let I be a binomial ideal of S = K[t1 , . . . , tq ]. Then (a) I is a lattice ideal of S if and only if I = (I : (t1 · · · tq )∞ ). (b) If I = (ta1 − tb1 , . . . , tas − tbs ), then (I : (t1 · · · tq )∞ ) = I(L), where L is the lattice L = Z{ai − bi }si=1 in Zq generated by {ai − bi }si=1 . 8.2.33 Let S = K[t1 ] be a polynomial ring in one variable. Prove that the lattice ideals of S are the ideals of the form (tm 1 − 1), where m ≥ 0. +

−

8.2.34 Let I ⊂ K[t1 , . . . , tq ] be a binomial ideal generated by tci − tci , i = 1, . . . , m, and let L ⊂ Zq be the lattice generated by c1 , . . . , cs . Prove that I(L) = IK[t±1 ] ∩ S. 8.2.35 Let K ⊂ F be a ﬁeld extension, let S = K[t] and B = F [t] be polynomial rings and let I be an ideal of S. Then the following hold. (a) IB∩S = I. (b) If ∩ri=1 qi is a primary decomposition of IB, then ∩ri=1 (qi ∩S) is a primary decomposition of I such that rad(qi ∩ S) = rad(qi ) ∩ S. (c) If I is the lattice ideal of L in S, then IB is the lattice ideal of L in B. 8.2.36 Let L be a lattice in Zq and let ρ : L → K ∗ be a partial character. If Iρ (L) is prime and K is algebraically closed, then Zq /L is torsion-free. 8.2.37 Let L be the lattice 2Z and let ρ : L → Q∗ be the partial character given by ρ(2p) = (−1)p . If K = Q, prove that Iρ (L) = (t21 + 1).

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Chapter 8

Monomial subrings and toric ideals

Let R = K[x] = K[x1 , . . . , xn ] = ⊕i≥0 Ri be a polynomial ring over a ﬁeld K with the standard grading. As usual, x denotes the set of indeterminates x1 , . . . , xn of the ring R. In this section we focus on monomial subrings generated by monomials in the polynomial ring R. Let F = {f1 , . . . , fq } be a ﬁnite set of monomials in R with fi = xvi and fi = 1 for all i. The set F has a corresponding set of vectors in Nn : F = {xv1 , . . . , xvq } ←→ A = {v1 , . . . , vq }. The monomial subring K[F ] is a graded subring of R with the grading given by K[F ]i = K[F ] ∩ Ri . There is a graded epimorphism of K-algebras ϕ : S = K[t1 , . . . , tq ] −→ K[F ] −→ 0, induced by ϕ(ti ) = fi , where S = ⊕i≥0 Si is a positively graded polynomial ring with the grading induced by setting deg(ti ) = deg(xvi ) = |vi |. We also denote S by K[t]. The kernel of ϕ, denoted by PF , is the so called toric ideal of K[F ] with respect to f1 , . . . , fq . We also denote the toric ideal of K[F ] by IA . In this case we say that IA is the toric ideal of A. Let A be the n × q matrix with column vectors v1 , . . . , vq . This matrix is called the associated matrix of K[F ]. Closely related to the map ϕ is the homomorphism ψ : Zq −→ Zn determined by the matrix A in the standard bases of Zq and Zn . Indeed, we have ϕ(tα ) = xψ(α) for all α ∈ Nq . Recall that given a binomial g = tα − tβ , we set g' = α − β. Proposition 8.3.1 If g1 , . . . , gr is a set of binomials that generate the toric gr generate kerZ (A). ideal PF , then g'1 , . . . , ' Proof. By Corollary 8.2.20, PF is the lattice ideal of kerZ (A). Thus the result follows from Lemma 8.2.23. 2 Deﬁnition 8.3.2 A binomial tα − tβ ∈ PF is called primitive if there is no other binomial tγ − tδ ∈ PF such that tγ divides tα and tδ divides tβ . Lemma 8.3.3 If f is a binomial in the reduced Gr¨ obner basis of PF with respect to some term order ≺, then f is a primitive binomial. Proof. Let G be the reduced Gr¨ obner basis of PF . One may assume f = tα+ − tα− and α+ " α− . Assume g = tβ+ − tβ− = f is a binomial in PF such that tβ+ divides tα+ and tβ− divides tα− . Note β+ = α+ and β− = α− , otherwise f = g. If β+ " β− , then tβ+ ∈ in(G \ {f }) and consequently tα+ ∈ in(G \ {f }), a contradiction. If β− " β+ , then tβ− ∈ in(G \ {f }) and consequently tα− ∈ in(G \ {f }), a contradiction. 2 According to a result of [400], the set GPF of all primitive binomials in a toric ideal PF is ﬁnite and is called a Graver basis of PF .

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Deﬁnition 8.3.4 The universal Gr¨ obner basis of a toric ideal PF is the union of all reduced Gr¨ obner basis of PF . It is denoted by UPF . Deﬁnition 8.3.5 If α is a circuit of ker(A) we call the binomial tα+ − tα− a circuit of PF . The set of circuits of PF is denoted by CPF . Proposition 8.3.6 [400] CPF ⊂ UPF ⊂ GPF . Proof. The second inclusion follows from Lemma 8.3.3. To show the ﬁrst inclusion take a circuit f = tα+ − tα− of the toric ideal PF . We may assume −αr+1 αr 1 s that f = tα · · · t−α , α = (αi ) = α+ − α− . We reorder the s 1 · · · tr − tr+1 set of variables t as t \ {t1 , . . . , ts }, t1 , . . . , ts . Consider the lex order: ta " tb if the ﬁrst non-zero entry of a − b is positive. Let G be the reduced Gr¨ obner basis of PF with respect to " and let g = tβ+ −tβ− be a binomial in G so that in(g) = tβ+ divides in(f ). Clearly supp(tβ+ ) ⊂ supp(tα+ ). By the choice of term order it follows that supp(tβ− ) ⊂ supp(tα ). Thus supp(β) ⊂ supp(α). Since α is a circuit we get equality. Thus using Lemma 1.9.3 we obtain that β = λα for some λ ∈ Q, consequently α = β because the non-zero entries of α (resp. β) are relatively prime. 2 Corollary 8.3.7 If in every circuit ta+ − ta− of PF both monomials ta+ and ta− are square-free, then CPF = UPF = GPF . Proof. By Proposition 8.3.6 it suﬃces to show the inclusion GPF ⊂ CPF . Let f = tβ+ − tβ− ∈ GPF , i.e., f is primitive. By Lemma 1.9.5 there is a circuit α, in harmony with β, whose support is contained in the support of β. Since tα+ and tα− are square-free, we conclude that tα+ (resp. tα− ) 2 divides tβ+ (resp. tβ− ). Hence f = tα+ − tα− because f is primitive. Corollary 8.3.8 If A is t-unimodular, then CPF = UPF = GPF . +

Proof. By Corollary 1.9.11 in every circuit ta+ − ta− of PF both terms ta − and ta are square-free. Thus the equality follows from Corollary 8.3.7. 2 Deﬁnition 8.3.9 A universal Gr¨ obner basis of a polynomial ideal I is a ﬁnite set U ⊂ I which is a Gr¨obner basis of I with respect to all term orders. Proposition 8.3.10 GPF is a universal Gr¨ obner basis of PF . Proof. It follows from Proposition 8.3.6. Theorem 8.3.11 The following are equivalent: (a) In every circuit ta+ − ta− of PF both monomials are square-free. (b) Every initial monomial ideal of PF is square-free.

2

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Proof. That (a) implies (b) follows from Corollary 8.3.7. Conversely let a as+1 · · · tq q . f be a circuit of PF . We may assume that f = tar r · · · tas s − ts+1 Consider the lex order " with t1 " · · · " tq and consider the weight vector ω = e1 + · · · + er−1 ∈ Nq . We deﬁne the term order: a "ω b if and only if + − ω · a > ω · b or ω · a = ω · b and a " b. Then there is g = tb − tb , in the + + reduced Gr¨ obner basis of PF , such that tb is square-free, in(g) = tb , and + + + + tb divides ta . Hence supp(tb ) is contained in supp(ta ). Consequently, − by deﬁnition of "ω , we have that supp(tb ) is contained in {tr , . . . , tq }. Since a is a circuit we get that a and b have the same support. Thus, by Lemma 1.9.3, a = λb for some rational number λ. Using that a (resp. b) has + + relatively prime non-zero entries we obtain that a = ±b and tb = ta , i.e., + − ta is square-free. A similar argument shows that ta is also square-free. 2 Corollary 8.3.12 [17, Theorem 4] If char(K) = 0 and PF is a binomial set theoretic complete intersection, then PF is a complete intersection. Proof. Set r = dim S/PF . By hypothesis, there are g1 , . . . , gq−r binomials of S such that rad(g1 , . . . , gq−r ) = PF . Since the ideal (g1 , . . . , gq−r ) is C–M (see Proposition 2.3.24), the result follows from Corollary 8.2.26. 2 Lemma 8.3.13 [134] Let R be a ring and let x1 , . . . , xn ∈ R. If I is an ideal of R, then the radical of I satisﬁes rad(I) = rad(I : (x1 · · · xn )∞ ) ∩ rad(I, x1 ) ∩ · · · ∩ rad(I, xn ). Proof. The inclusion “⊂” is clear. To show the inclusion“⊃” take a prime p ⊃ I, it suﬃces to show that p contains one of the ideals on the right-hand side. If rad(I : (x1 · · · xn )∞ ) ⊂ p, there is f ∈ R \ p such that f (x1 · · · xn )s is in I for some s ≥ 1. Hence xi ∈ p for some i and p ⊃ rad(I, xi ). 2 Proposition 8.3.14 Let p be a prime ideal of a ring R and let x1 , . . . , xn be a sequence in R \ p. If I ⊂ p is an ideal, then rad(I) = p if and only if p = rad(I : (x1 · · · xn )∞ ) and rad(I, xi ) = rad(p, xi ) for all i. Proof. If J is any ideal of R and y1 , . . . , yn ∈ R, then by Lemma 8.3.13 rad(J) is equal to rad(J : (y1 · · · yn )∞ ) ∩ rad(J, y1 ) ∩ · · · ∩ rad(J, yn ). Hence the result follows by applying this equality to I and p, together with the fact that x1 · · · xn is regular on R/p. 2 Proposition 8.3.15 [139] Let I be an ideal generated by a set of binomials g1 , . . . , gr in the toric ideal PF . If char(K) = p = 0 (resp. char(K) = 0) and L is the lattice Z{' gi }ri=1 , then the following conditions are equivalent : (a1 ) PF = rad(I : (t1 · · · tq )∞ ). (a2 ) pu ker(ψ) ⊂ L for some u ∈ N (resp. ker(ψ) = L).

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Proof. (a1 ) ⇒ (a2 ): First we consider the case char(K) = p = 0. Setting u z = t1 · · · tq , by hypothesis, there exists u ≥ 0 such that f p ∈ (I : z ∞ ) for all f ∈ PF . Let γ ∈ ker(ψ). Write γ = α − β with α, β ∈ Nq and g = tα − tβ . Thus g ∈ PF and g' = α − β = γ. The lattice L deﬁnes a congruence in u Nq given by a ∼L b if a − b ∈ L. We know that tδ g p ∈ I for some δ ∈ Nq u u u and using that p = char(K), one has tδ g p = tαp +δ − tβp +δ . Since the u simple components of tδ g p with respect to L belong to I and hence to PF u (see Lemma 8.2.9), and since tδ tαp does not belong to I because it does u not belong to PF , it follows that tδ g p is simple, i.e., δ + pu γ ∼L pu β + δ. Hence pu α − pu β ∈ L. This shows the inclusion pu ker(ψ) ⊂ L. Next consider the case char(K) = 0. By Lemma 8.2.11 one has the equality (I : z) = I(L). Thus PF = rad(I(L)). As PF is the lattice ideal of ker(ψ) (Corollary 8.2.20), by Lemmas 8.2.23 and 8.2.24 we get L = ker(ψ). (a1 ) ⇐ (a2 ): It suﬃces to prove the inclusion PF ⊂ rad(I : z ∞ ), the other inclusion is clear. Assume char(K) = p > 0. Let f = tc − te be a binomial u u u in PF . By (a2 ) and Lemma 8.2.10, we get that tγ (tp c − tp e ) = tγ f p is in I for some monomial tγ . Thus f is in rad(I : z ∞ ). If char(K) = 0, the 2 same argument works if we replace pu by 1. Theorem 8.3.16 [139] Let I ⊂ PF be an ideal generated by a ﬁnite set of binomials g1 , . . . , gr . If char(K) = p = 0 (resp. char(K) = 0), then rad(I) = PF if and only if (a) pu ker(ψ) ⊂ Z{' gi }ri=1 for some u ∈ N (resp. ker(ψ) = Z{' gi }ri=1 ), and (b) rad(I, ti ) = rad(PF , ti ) for all i. Proof. It is a consequence of Propositions 8.3.14 and 8.3.15.

2

Next we give a result that holds over a ﬁeld of arbitrary characteristic. It is a consequence of the proofs of Theorem 8.2.25 and Corollary 8.2.26. Proposition 8.3.17 Let I ⊂ S be an ideal, and let {gi }ri=1 , {hi }si=1 be sets of binomials that generate I, rad(I), respectively. If Z{' gi }ri=1 is equal to Z{' hi }si=1 , I has no embedded primes, and rad(I) is prime, then I = rad(I). Corollary 8.3.18 A toric ideal PF of height r is a complete intersection if and only if there are binomials g1 , . . . , gr in PF such that (a) ker(ψ) = Z{' g1 , . . . , ' gr }, and (b) (g1 , . . . , gr , ti ) = (PF , ti ) for all i. Proof. ⇒) It follows readily from Proposition 8.3.1. ⇐) By Theorem 8.3.16 we get rad(g1 , . . . , gr ) = PF . Notice that the ideal I = (g1 , . . . , gr ) is a complete intersection because r = ht(PF ) and I is quasi-homogeneous. Hence applying Proposition 8.3.17 yields I = PF . 2 Given a subset I ⊂ S we denote its zero set in AqK by V (I), and given a subset X ⊂ AqK we denote its vanishing ideal in S by I(X).

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Corollary 8.3.19 Let I be an ideal generated by binomials g1 , . . . , gr in gi }ri=1 for PF . If char(K) = p = 0 (resp. char(K) = 0) and pm ker(ψ) ⊂ Z{' r some m ∈ N (resp. ker(ψ) = Z{' gi }i=1 ), then V (I) ⊂ V (PF ) ∪ V (t1 · · · tq ). Proof. By Proposition 8.3.15, there is a monomial tδ and an integer N such that tδ PFN ⊂ I. It follows that V (I) ⊂ V (PFN ) ∪ V (tδ ) ⊂ V (PF ) ∪ V (t1 · · · tq ).

2

Lemma 8.3.20 Let I be a binomial ideal of S such that V (I, ti ) = {0} for all i. If p is a prime ideal containing (I, tm ) for some 1 ≤ m ≤ q, then p = (t1 , . . . , tq ). Proof. Let h1 , . . . , hr be a set of binomials that generate I. For simplicity of notation assume that m = 1. We may assume that t1 , . . . , tk are in p and ti ∈ / p for i > k. If ti ∈ supp(hj ) for some 1 ≤ i ≤ k, say hj = taj − tbj and ti ∈ supp(taj ), then tbj ∈ p and there is 1 ≤ ≤ k such that t is in the support of tbj . Thus, hj ⊂ (t1 , . . . , tk ). Hence, for each 1 ≤ j ≤ r, either (i) supp(hj ) ∩ {t1 , . . . , tk } = ∅ or (ii) hj ∈ (t1 , . . . , tk ). Consider the point c = (ci ) ∈ AqK , with ci = 0 for i ≤ k and ci = 1 for i > k. If (i) occurs, then hj (c) = (taj − tbj )(c) = 1 − 1 = 0. If (ii) occurs, then hj (c) = (taj − tbj )(c) = 0 − 0 = 0. Clearly the polynomial t1 vanishes at c. Hence, c ∈ V (I, t1 ) = {0}. Therefore, k = q. Thus, p contains all the variables of S, i.e., p = (t1 , . . . , tq ). 2 Corollary 8.3.21 [139] Let I be an ideal generated by binomials g1 , . . . , gr d in PF . If F = {xd11 , . . . , x1q } and char(K) = p = 0 (resp. char(K) = 0), then rad(I) = PF if and only if (a) pm ker(ψ) ⊂ Z{' gi }ri=1 for some m ∈ N (resp. ker(ψ) = Z{' gi }ri=1 ) (b) V (g1 , . . . , gr , ti ) = {0}, for all i. Proof. ⇒) By Theorem 8.3.16 condition (a) holds and rad(I, ti ) is equal to rad(PF , ti ) for all i. On the other hand rad(PF , ti ) = (t1 , . . . , tq ), because the height of PF is q − 1 and ti is regular on S/PF . Hence V (I, ti ) = {0}. ⇐) By Proposition 8.3.15 one has PF = rad(I : z ∞ ), where z = t1 · · · tq . Hence there is monomial tδ and an integer N such that tδ PFN ⊂ I. To prove rad(I) = PF , it suﬃces to show that every prime ideal p containing I also contains PF . If p contains no variables, then the inclusions tδ PFN ⊂ I ⊂ p imply that PF ⊂ p, as desired. If p contains at least one variable, then p contains all the variables by Lemma 8.3.20 and consequently 2 contains PF , as desired.

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Proposition 8.3.22 Let I ⊂ S be a graded binomial ideal. The following hold. (a) If V (I, ti ) = {0} for all i, then ht(I) = q − 1. (b) If I is a lattice ideal and ht(I) = q − 1, then V (I, ti ) = {0} for all i. Proof. (a) As I is graded, all associated prime ideal of S/I are graded. Thus, all associated prime ideals of S/I are contained in m = (t1 , . . . , tq ). If ht(I) = q, then m would be the only associated prime of S/I, that is, m is the radical of I, a contradiction because I cannot contain a power of ti for any i. Thus, ht(I) ≤ q − 1. On the other hand, by Lemma 8.3.20, the ideal (I, tq ) has height q. Hence, q = ht(I, tq ) ≤ ht(I) + 1 (here we use the fact that I is graded). Altogether, we get ht(I) = q − 1. (b) Let L be the lattice that deﬁnes I and g1 , . . . , gr a set of homogeneous binomials that generate I. By Lemma 8.2.23, we get L = ' g1 , . . . , ' gr . Notice that q − 1 = ht(I) = rank(L). Given two integers 1 ≤ i, k ≤ q, the g1 , . . . , g'r . Hence, as L is homogeneous vector space Qq is generated by ek , ' with respect to ω = (ω1 , . . . , ωq ), there are positive integers ri and rk such that ri ei − rk ek ∈ L and ri ωi − rk ωk = 0. By Lemma 8.2.10, there is tδ such that tδ (tri i − trkk ) is in I. Hence, by Theorem 8.2.8, tri i − trkk is in I. Therefore, V (I, ti ) = {0} for all i. 2 Lemma 8.3.23 Let I ⊂ S be a graded binomial ideal. If V (I, ti ) = {0} for all i and I is a complete intersection, then I is a lattice ideal. Proof. By Proposition 8.3.22(a), the height of I is q −1. It suﬃces to prove that ti is a non-zero divisor of S/I for all i (see Theorem 8.2.8). If ti is a zero divisor of S/I for some i, there is an associated prime ideal p of S/I containing (I, ti ). Hence, using Lemma 8.3.20, we get that p = m, a contradiction because I is a complete intersection of height q −1 and all associated prime ideals of I have height equal to q − 1 (see Proposition 2.3.24). 2 Theorem 8.3.24 [293] If L ⊂ S is a graded lattice ideal and V (L, ti ) = {0} for all i, then L is a complete intersection if and only if there are homogeneous binomials h1 , . . . , hq−1 in L satisfying the following conditions: (i) L = ' h1 , . . . , ' hq−1 , where L is the lattice that deﬁnes L. (ii) V (h1 , . . . , hq−1 , ti ) = {0} for all i. +

−

(iii) hi = tai − tai for i = 1, . . . , q − 1. Proof. As L is graded, By Proposition 8.3.22, the height of L is q − 1. ⇒) By Lemma 3.1.29 and Exercise 3.3.26, L is generated by homogeneous binomials h1 , . . . , hq−1 . Then, by Lemma 8.2.23 and Theorem 8.2.8, (i) and (iii) hold. From the equality (L, ti ) = (h1 , . . . , hq−1 , ti ), it follows readily that (ii) holds. ⇐) We set I = (h1 , . . . , hq−1 ). By hypothesis I ⊂ L. Thus, we need only show the inclusion L ⊂ I. Let g1 , . . . , gm be a generating set of L consisting

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of binomials, then g'i ∈ L for all i. Using condition (i) and Lemma 8.2.10, for each i there is a monomial tγi such that tγi gi ∈ I. Hence, tγ L ⊂ I, where tγ is equal to tγ1 · · · tγm . By (ii) and Proposition 8.3.22, the height of I is q − 1. This means that I is a complete intersection. As tγ L ⊂ I, to show the inclusion L ⊂ I, it suﬃces to notice that by (ii), Lemma 8.3.23 2 and Theorem 8.2.8 ti is a non-zero divisor of S/I for all i. Remark 8.3.25 The result remains valid if we remove condition (iii), i.e., condition (iii) is redundant. In both implications of the theorem the set h1 , . . . , hq−1 is shown to generate L. Deﬁnition 8.3.26 The circuit ideal of the matrix A is given by I = ({tα+ − tα− | α is a circuit of ker(A)}). The next lemma is useful in induction arguments. Lemma 8.3.27 Let A be the matrix obtained from A by removing its ith column. Then the inclusion map j : Zq−1 → Zq given by j

(a1 , . . . , ai−1 , ai+1 , . . . , aq ) −→ (a1 , . . . , ai−1 , 0, ai+1 , . . . , aq ) induces a bijection between the circuits of ker(A ) and the circuits of ker(A) whose ith entry is equal to zero. Proof. Since the inclusion map j takes ker(A ) into ker(A), the result follows readily. 2 Proposition 8.3.28 [134] If I is the circuit ideal of A, then rad (I) = PF . Proof. We proceed by induction on q, the number of columns of A. If q = 1 or q = 2, then PF is a principal ideal and the result is clear. According to Theorems 8.3.16 and 1.9.10 it suﬃces to prove that rad (I, ti ) is equal to rad (PF , ti ) for all i. For simplicity of notation we assume i = 1. Let + − γ1 , . . . , γr be the circuits of ker(A). We set fi = tγi −tγi for i = 1, . . . , r and f0 = fr+1 = 0. If p is a prime ideal containing (I, t1 ), then after relabeling t1 , . . . , tq and f1 , . . . , fr there are t1 , . . . , ts ∈ p and 0 ≤ k ≤ r such that (i) (fk+1 , . . . , fr ) ⊂ (t1 , . . . , ts ) and (ii) ti ∈ / ∪kj=1 supp(fj ) for i = 1, . . . , s. + − tα in PF . Setting We claim that p contains (PF , t1 ). Take f = tα − r α = α+ − α− , by Theorem 1.9.6, we can write pα = i=1 ni γi for some p ∈ N+ and n1 , . . . , nr ∈ N, such that for each i either ni = 0 or ni > 0 and γi is in harmony with α and supp(γi ) ⊂ supp(α). Case (I): If ti ∈ supp(f ) for some 1 ≤ i ≤ s, then ti ∈ supp(fj ) for some j such that nj > 0. Thus rby (ii) one has j ≥ k + 1 and fj ∈ (t1 , . . . , ts ). From the equality pα = i=1 ni γi it follows that f ∈ (t1 , . . . , ts ) ⊂ p.

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Case (II): If ti ∈ / supp(f ) for 1 ≤ i ≤ s. Let ϕ (resp. ψ ) be the restriction of ϕ (resp. ψ) to K[ts+1 , . . . , tq ] (resp. Zes+1 + · · · + Zeq ). We set F = {xvs+1 , . . . , xvq }. Note that γ1 , . . . , γk are the circuits of ker(ψ ) and PF = K[ts+1 , . . . , tq ]∩PF is the toric ideal of K[F ]; see Exercise 8.3.33 and Lemma 8.3.27. Hence by induction one has rad (I ) = PF = K[ts+1 , . . . , tq ] ∩ PF , where I is a circuit ideal generated by f1 , . . . , fk . Since f ∈ PF , then there is m ≥ 1 such that f m ∈ I ⊂ p. Thus f ∈ p. Altogether p contains (PF , t1 ). As rad (PF , t1 ) is the intersection of the minimal primes of (PF , t1 ) one concludes rad (PF , t1 ) ⊂ rad (I, t1 ). Since the other inclusion is clear one has equality. 2

Exercises 8.3.29 If the matrix A is t-unimodular, then the reduced Gr¨obner basis of PF with respect to any term order consists of square-free binomials, i.e., of + − + − binomials ta − ta with ta and ta square-free. 8.3.30 Let B be a set of binomials of S = K[t1 , . . . , tq ] and let f be a binomial. Prove that dividing f by B produces a residue h = tγ − tδ which is independent of the ﬁeld K as long as we follow identical steps in the division process when changing ﬁelds. 8.3.31 Let F be a ﬁnite set of monomials in a polynomial ring Q[x1 , . . . , xn ] and let P be the toric ideal of Q[F ]. Consider the toric ideal PF of K[F ] over a ﬁeld K. If G is a Gr¨obner basis of P consisting of binomials with respect to a ﬁxed term order, prove that G is also a Gr¨ obner basis for PF . 8.3.32 Use the previous exercise to draw some conclusion if P is a toric ideal minimally generated by a Gr¨ obner basis over a ﬁeld K. 8.3.33 Let F = {f1 , . . . , fq } be a set of monomials of a polynomial ring R over a ﬁeld K. Consider the homomorphism ϕ : S = K[t1 , . . . , tq ] −→ K[F ] −→ 0, induced by ϕ(ti ) = fi . Prove that ker(ϕ)∩K[t1 , . . . , tr ] is a binomial ideal which is equal to ker(ϕ ), where ϕ is the restriction of ϕ to K[t1 , . . . , tr ]. 8.3.34 If I is an ideal of S = K[t1 , . . . , tq ] and rad(I) = PF , prove that ±1 rad(IS ) = PF S , where S = K[t±1 1 , . . . , tq ]. 8.3.35 Let A be an n × q integral matrix. If L is a subgroup of kerZ (A), prove that T(Zq /L) = T(kerZ (A)/L), where T denotes torsion.

334

Chapter 8

8.3.36 Let A be an n × q integral matrix with non-zero columns. If L is a subgroup of kerZ (A) and p ≥ 2 a prime number, then pu kerZ (A) ⊂ L for some u ∈ N if and only if kerZ (A)/L is a ﬁnite p-group or kerZ (A)/L = (0). 8.3.37 Let d1 , . . . , dq be a sequence of relatively prime integers and ψ the linear map from Zq to Z induced by ψ(ei ) = di . If L is a subgroup of ker(ψ), then T(Zq /L) ⊂ ker(ψ)/L, with equality if ker(ψ)/L is a ﬁnite group. 8.3.38 Let A be an n×q integer matrix. If L is a subgroup of kerZ (A), then L = kerZ (A) if and only if Zq /L is torsion-free of rank equal to rank(A). 8.3.39 Let P be the toric ideal of the subring K[x61 , x81 , x91 ] and let I be the ideal generated by g1 , g2 , where g1 = t31 − t23 , g2 = t92 − t83 . Then (a) P = (g1 , t32 − t1 t23 ) for any ﬁeld K, (b) Z3 /' g1 , g'2 Z × Z3 , (c) rad(I) = P if char(K) = 3, and rad(I) = P if char(K) = 2, (d) Γ = {(x6 , x8 , x9 )| x ∈ K} = {(0, 0, 0), (1, 1, 1)} = V (I), if K = Z2 . 8.3.40 Let d1 , . . . , dn be a sequence of relatively prime integers and let ψ be the linear map from Zn to Z induced by ψ(ei ) = di , then the set

di dj ei − ej 1 ≤ i < j ≤ n L= gcd(di , dj ) gcd(di , dj ) is a generating set for ker(ψ), where ei is the ith unit vector. 8.3.41 Let d1 , . . . , dn be a sequence of non-zero integers and let ψ be the linear map from Zn to Z induced by ψ(ei ) = di . Prove that the set

di dj L= ei − ej 1 ≤ i < j ≤ n d d is a generating set for ker(ψ), where d = gcd(d1 , . . . , dn ). 8.3.42 Let k ≥ 1 be an integer and p be a prime number. Consider mlet 1 the expansion of k in base p: k = i=0 ai pi , where am1 = 0, the ai ’s are integers and 0 ≤ ai ≤ p − 1 for all i. If pu1 is the largest power of p dividing k! (the factorial of k), prove the equalities u1 =

m1 i=1

ai

pi − 1 p−1

and u1 =

pm − 1 if k = pm . p−1

8.3.43 If p > 1 is a prime number m and m ≥ 1 is an integer. Prove that p divides the binomial coeﬃcient pk for 1 ≤ k ≤ pm − 1.

Toric Ideals and Aﬃne Varieties

8.4

335

Toric varieties

In this section we use linear algebra methods to characterize when a toric set is an aﬃne toric variety in terms of the existence of certain roots in the base ﬁeld and a vanishing condition. First we ﬁx some notation. Let K be a ﬁeld and let A be an n × q matrix with entries in N and with non-zero columns v1 , . . . , vq . The toric set parameterized by F = {xv1 , . . . , xvq } ⊂ K[x] is the set given by D C v Γ = (av111 · · · avnn1 , . . . , a11q · · · avnnq ) ∈ AqK a1 , . . . , an ∈ K , where vi = (v1i , . . . , vni ) for all i. Similarly one can deﬁne the toric set Γ parameterized by xv1 , . . . , xvq , with v1 , . . . , vq in Zn , as the set of all points v v (av111 · · · avnn1 , . . . , a11q · · · annq ) in AqK that are well-deﬁned for some ai ’s in K. As usual the toric ideal of K[F ] is denoted by PF . Recall that PF is a prime ideal of the polynomial ring S = K[t1 , . . . , tq ] over the ﬁeld K. There are unimodular integral square matrices U = (uij ) and Q = (qij ) of orders n and q, respectively, such that D = U AQ = diag(λ1 , . . . , λs , 0, . . . , 0), where s is the rank of A and λ1 , . . . , λs are the invariant factors of A; that is, λi divides λi+1 and λi > 0 for all i (see Theorem 1.2.2). For use below we set U −1 = (fij ) and Q−1 = (bij ). In what follows ei will denote the ith unit vector in Zq . Lemma 8.4.1 kerZ (A) = Zqs+1 ⊕ · · · ⊕ Zqq , where qi is the ith column of the matrix Q. Proof. Let x ∈ Zq . Make the change of variables y = Q−1 x. As D = U AQ, it follows that Ax = 0 if and only if Dy = 0. Set y = (y1 , . . . , yq ). = Dei . If x First note that qi ∈ kerZ (A) for i ≥ s + 1, because DQ−1 qi q is in kerZ (A), then λi yi = 0 for i = 1, . . . , s. Thus x = Qy = i=s+1 yi qi . To complete the proof observe that the columns of Q are a basis for Zq . 2 Proposition 8.4.2 kerZ (A) = Z{wj − ej }qj=1 , where wj = qi is the ith column of Q. Proof. Note that ej = wj =

s i=1

q

i=1 bij qi

bij qi = ej −

s

i=1 bij qi

for all j, because QQ−1 = I. Hence q

i=s+1

bij qi

(j = 1, 2, . . . , q).

and

336

Chapter 8

Then, using Lemma 8.4.1, we obtain wj − ej ∈ kerZ (A) for j = 1, . . . , q. By Lemma 8.4.1, it suﬃces to observe that from the equality above one has ⎛ ⎞ ! q q q q q qjk (ej − wj ) = qjk bij qi = qi ⎝ qjk bij ⎠ j=1

j=1

i=s+1

q

=

i=s+1

j=1

qi δik = qk ,

i=s+1

for all k ≥ s + 1, where δik = 1 if i = k and δik = 0 otherwise.

2

Deﬁnition 8.4.3 An aﬃne toric variety V is deﬁned as the zero set of a toric ideal, that is, V = V (PF ) for some toric ideal PF . Theorem 8.4.4 [354] Let Γ ⊂ K q be the toric set parameterized by F . Then Γ = V (PF ) if and only if the following two conditions hold : q

(a) If (ai ) ∈ V (PF ) and ai = 0 ∀i, then aq11i · · · aqqi has a λi -root in K for i = 1, . . . , s, where λ1 , . . . , λs are the invariant factors of A, (b) V (PF , ti ) ⊂ Γ for i = 1, . . . , q. Proof. ⇐) One invariably has Γ ⊂ V (PF ). To prove the other inclusion take a point a = (a1 , . . . , aq ) in V (PF ), by condition (b) one may assume ai = 0 for all i. Thus using (a) there are x1 , . . . , xs in K such that (xi )λi = aq11i · · · aqqqi = aqi

(i = 1, . . . , s).

(8.3)

For convenience of notation we extend the deﬁnition of xi by putting xi = 1 for i = s + 1, . . . , n and x = (x1 , . . . , xn ). Set xj = (x1 )u1j · · · (xn )unj

(j = 1, . . . , n).

(8.4)

We claim that xvk = xv11k · · · xvnnk = ak for k = 1, . . . , q. Setting U −1 = (fij ) and comparing columns in the equality U −1 D = AQ one has: λi fi =

q

qji vj

(i = 1, 2, . . . , s),

(8.5)

j=1

where fi = (f1i , . . . , fni ) and vj = (v1j , . . . , vnj ) denote the ith and jth columns of U −1 and A. Comparing columns in A = (U −1 D)Q−1 we get: vk =

s j=1

λj bjk fj

(k = 1, 2, . . . , q),

(8.6)

Toric Ideals and Aﬃne Varieties

337

where Q−1 = (bij ). Using U U −1 = I and Eq.( 8.4) we rapidly conclude: xfk = xk

(k = 1, . . . , n).

(8.7)

From Proposition 8.4.2 we derive Awj = Aej = vj for j = 1, . . . , q, where ! s s s bij qi = q1 bj , . . . , qn bj (j = 1, . . . , q). (8.8) wj = i=1

=1

=1 +

−

Hence Awj+ = A(ej + wj− ), that is, twj − tej +wj belongs to the toric ideal +

−

PF . Using that a ∈ V (PF ) yields awj = aej +wj , thus awj = aej = aj

(j = 1, . . . , q).

(8.9)

Therefore putting everything together: xvk

8.6

s

λj bjk fj

= (xf1 )λ1 b1k · · · (xfs )λs bsk = (x1 )λ1 b1k · · · (xs )λs bsk 8.7

=

x

8.3

(aq1 )b1k · · · (aqs )bsk = aq1 b1k +···+qs bsk = awk = ak

=

j=1

8.8

8.9

for k = 1, . . . , q. Thus a ∈ Γ, as required. ⇒) It is clear that (b) holds because V (PF , ti ) ⊂ V (PF ). To prove (a) take (ai ) ∈ V (PF ) with ai = 0 for all i, then by deﬁnition of Γ there are x1 , . . . , xn in K such that aj = xvj for all j. Therefore by Eq. (8.5) one has: (xfi )λi = xλi fi = xq1i v1 · · · xqqi vq = aq11i · · · aqqqi .

2

The proof of Theorem 8.4.4 works for arbitrary Laurent monomials. Another classiﬁcation, that uses polyhedral geometry, of the aﬃne toric varieties that are parameterized by Laurent monomials is given in [272]. Corollary 8.4.5 V (PF ) ⊂ Γ ∪ V (t1 · · · tq ) if K is algebraically closed. Proof. Let a = (ai ) ∈ V (PF ) such that ai = 0 for all i. Since K is algebraically closed condition (a) above holds. Therefore one may proceed as in the ﬁrst part of the proof of Theorem 8.4.4 to get a ∈ Γ. 2 Next we present another consequence that can be used to prove that monomial curves over arbitrary ﬁelds are aﬃne toric varieties. Corollary 8.4.6 If the columns of A generate Zn as Z-module, then the equality Γ = V (PF ) holds if and only if V (PF , ti ) ⊂ Γ for all i. Proof. Since Zv1 + · · · + Zvq = Zn , one has λi = 1 for all i, thus condition (a) holds. Therefore Γ is an aﬃne toric variety if and only if (b) holds. 2

338

Chapter 8

Corollary 8.4.7 If K is algebraically closed, then Γ = V (PF ) if and only if V (PF , ti ) ⊂ Γ for all i. Proof. If K is algebraically closed, then (a) is satisﬁed, thus Γ is a toric 2 variety if and only if V (PF , ti ) ⊂ Γ for all i. Proposition 8.4.8 [139] If A has only one row and v1 , . . . , vq are relatively prime positive integers, then Γ = V (PF ) over any ﬁeld K. Proof. As Z = Zv1 + · · · + Zvq by Corollary 8.4.6 it suﬃces to show v V (PF , ti ) ⊂ Γ. Let a ∈ V (PF , ti ) since all the binomials ti j − tvj i vanish on a one obtains a = 0 and a ∈ Γ. 2 A natural question is whether a toric set Γ can be a variety but not a toric variety (see Exercise 8.4.18): Example 8.4.9 Let K be the ﬁeld Z3 and let A be the matrix (2, 4). Then Γ = {(0, 0)} ∪ {(1, 1)} = V (t1 − t2 , t22 − t2 ). On the other hand PF = (t1 − t22 ) and (1, 2) ∈ V (PF ). Thus Γ = V (PF ). Aﬃne toric varieties are fully parameterized by monomials when K is algebraically closed or when the variety is normal. Theorem 8.4.10 [271, 272] Let F = {xv1 , . . . , xvq } be a set of Laurent monomials such that K is algebraically closed or K[F ] is normal. Then there exists a toric set ΓG parameterized by G = {x1 , . . . , xq } with i ∈ Zn such that V (PG ) = ΓG and PF = PG , where PG is the toric ideal of K[G]. Theorem 8.4.11 (Combinatorial Nullstellensatz [7]) Let S = K[t1 , . . . , tq ] be a polynomial ring over a ﬁeld K, let f ∈ S, and let c = (ci ) ∈ Nq . Suppose that the coeﬃcient of tc in f is non-zero and deg (f ) = c1 + · · ·+ cq . If A1 , . . . , Aq are subsets of K, with |Ai | > ci for all i, then there are a1 ∈ A1 , . . . , aq ∈ Aq such that f (a1 , . . . , aq ) = 0. Lemma 8.4.12 Let f be a polynomial in K[x1 , . . . , xn ]. If K is an inﬁnite ﬁeld and f (a) = 0 for all a ∈ K n , then f is the zero polynomial. Proof. It follows readily from Theorem 8.4.11.

2

Corollary 8.4.13 Let R = K[x] be a polynomial ring over a ﬁeld K and let X ⊂ K q be a set parameterized by ti = fi (x), where F = {f1 , . . . , fq } ⊂ R. If K is inﬁnite and I(X) is the vanishing ideal of X, then (a) I(X) is equal to PF , the presentation ideal of K[F ], and (b) the Zariski closure of X is equal to V (PF ).

Toric Ideals and Aﬃne Varieties

339

Proof. (a): As PF = (t1 −f1 , . . . , tq −fq )∩K[t1 , . . . , tq ], one has PF ⊂ I(X). Conversely take f ∈ I(X) and consider the revlex ordering such that ti > xj for all i, j. By the division algorithm f (t1 , . . . , tq ) = i gi (ti −fi )+r, where r is a polynomial in R. Hence for any a ∈ K n one has f (f1 (a), . . . , fq (a)) = i gi (a, x)(fi (a) − fi ) + r = 0. Thus r(a) = 0 for any a ∈ K n . Therefore r is the zero polynomial because K is an inﬁnite ﬁeld (see Lemma 8.4.12). (b): By part (a) and Proposition 3.2.3 we get X = V (PF ). 2 Aﬃne normal toric varieties Let A = {v1 , . . . , vq } ⊂ Zn be a point conﬁguration such that R+ A ∩ ZA = NA. Hence the monomial subring ±1 K[xv1 , . . . , xvq ] ⊂ K[x±1 1 , . . . , xn ]

is normal (Corollary 9.1.3). Pick a Z-basis B = {w1 , . . . , wr } of ZA, where r = rank(ZA), and consider the linear map T : ZA −→ Zr , wi → ei . For 1 ≤ i ≤ q we can write vi = u1i w1 + · · · + uri wr . Notice that T (vi ) = ui , where ui = (u1i , . . . , uri ). Setting H = {u1 , . . . , uq }, we have ZH = Zr . Lemma 8.4.14 Zr ∩ R+ H = NH. Proof. The equality follows readily by observing that T can be extended 2 to an isomorphism from QA onto Qr . Let P and P1 be the toric ideals of K[xv1 , . . . , xvq ] and K[xu1 , . . . , xuq ], respectively. Lemma 8.4.15 P = P1 and K[xv1 , . . . , xvq ] K[xu1 , . . . , xuq ]. Proof. Let K[t1 , . . . , tq ] be a polynomial ring over the ﬁeld K. Consider the following ring maps

ϕ

1 ll ll K[xv1 , . . . , xvq ]

ϕ

/ K[xu1 , . . . , xuq ] l6 ll

ϕ1

K[t1 , . . . , tq ]

induced by ϕ1 (ti ) = xui and ϕ(ti ) = xvi . Since the rings K[xv1 , . . . , xvq ] and K[xu1 , . . . , xuq ] have dimension r it suﬃces to prove that ker(ϕ) ⊂ ker(ϕ1 ) a c and to observe that P = ker(ϕ), P1 = ker(ϕ1 ). Let f = ta1 1 · · · tq q −tc11 · · · tqq be a binomial in ker(ϕ). Then (xv1 )a1 · · · (xvq )aq = (xv1 )c1 · · · (xvq )cq

340 and

Chapter 8 q

i=1 (ai

− ci )vi = 0. Applying T we get u1 a1

(x )

· · · (x )

uq aq

q

i=1 (ai

− ci )ui = 0. Thus

= (x ) · · · (x ) . u1 c1

uq cq

2

This proves that f ∈ ker(ϕ1 ), as required.

Lemma 8.4.16 The map φ : (K ∗ )r → (K ∗ )q ∩ V (P ), x → (xu1 , . . . , xuq ), is bijective, where K ∗ = K \ {0}. Proof. As ei ∈ ZH = Zr for 1 ≤ i ≤ r, there is αi ∈ NH such that ei + αi = βi and βi ∈ NH. Then the monomial xα = xα1 · · · xαr satisﬁes that xα ∈ NH and xi xα = xγi ∈ NH. Thus we can write xα = (xu1 )r1 · · · (xuq )rq ; xγi = (xu1 )si1 · · · (xuq )siq ; (ri ∈ N; sij ∈ N). r

s

Setting hi (t1 , . . . , tq ) = tr11 · · · tqq and h0 (t1 , . . . , tq ) = ts1i1 · · · tqiq , we have xi hi (xu1 , . . . , xuq ) = h0 (xu1 , . . . , xuq )

(8.10)

for 1 ≤ i ≤ r. If φ(x1 , . . . , xr ) = φ(y1 , . . . , yr ), then x = y for all i. Thus, from Eq. (8.10), xi = yi for all i. This proves that φ is injective. To prove that φ is onto take a = (ai ) ∈ V (P ) ∩ (K ∗ )q . Let Au be the matrix q with column vectors u1 , . . . , uq . There are qij in Z such that ej = i=1 qij ui for j = 1, . . . , r. Thus A ur(qj ) = ej for j = 1, . . . , r, where qj = (q1j , . . . , qqj ). Hence, setting wi = j=1 uji qj for i = 1, . . . , q, we get ui

ui

Au (wi ) = ui = Au (ei ) (i = 1, . . . , q) +

−

Hence Awi+ = A(ej +wi− ), that is, twi −tei +wi is in the toric ideal P . Using + − that a ∈ V (P ) yields awi = aei +wi , thus awi = aei = ai for i = 1, . . . , q. q q Setting xj = aqj = a11j · · · aqqj for j = 1, . . . , r, we get xui = xu1 1i · · · xur ri = (aq1 )u1i · · · (aqr )uri = awi = ai for i = 1, . . . , q. Thus a is in the image of φ, as required.

2

Theorem 8.4.17 If K is algebraically closed, then (a) V (P ) contains the torus (K ∗ )r as a dense open subset, and (b) there is an action of the torus (K ∗ )r on the toric variety V (P ): (K ∗ )r × V (P ) −→ V (P ) (x, (y1 , . . . , yq )) −→ (xu1 y1 , . . . , xuq yq ) Proof. The set Γ∗ := im(φ) = (K ∗ )r ∩ V (P ) is an open subset of V (P ) because the set V (t1 · · · tq ) is closed in the Zariski topology of K q , i.e., the set (K ∗ )q = K q \ V (t1 · · · tq ) is open in K q . By Hilbert Nullstellensatz (see Theorem 3.2.10) V (P ) is irreducible. Hence Γ∗ is dense in V (P ). Thus (a) follows from Lemma 8.4.16. Part (b) is left as an exercise. 2

Toric Ideals and Aﬃne Varieties

341

Exercises 8.4.18 Let Γ be a toric set parameterized by F . If K is an inﬁnite ﬁeld and Γ = V (J) for some ideal J, then J ⊂ PF and Γ = V (PF ). 8.4.19 (Rational quartic curve in P3 ) Let K be an algebraically closed ﬁeld and let Γ be the curve parameterized by F = {x41 , x31 x2 , x1 x32 , x42 }. Then rad (t2 t3 − t1 t4 , t33 − t2 t24 , t32 − t21 t3 ) = PF and Γ is an aﬃne toric variety. 8.4.20 Prove that Corollaries 8.4.5 and 8.4.7 are valid assuming condition (a) of Theorem 8.4.4, instead of assuming K algebraically closed. 8.4.21 Let S = K[t1 , t2 , t3 , t4 ] be a polynomial ring over a ﬁeld K. Then V (t1 t3 − t2 t4 ) = {(x1 x2 , x2 x3 , x3 x4 , x1 x4 )| x1 , . . . , x4 ∈ K}. 8.4.22 (Segre variety) Let m, n be two positive integers. Prove that Γ = {(ai bj ) ∈ K mn | ai ∈ K, bj ∈ K ∀ i, j} ⊂ Amn K is an aﬃne toric variety over any ﬁeld K. 8.4.23 (Veronese variety) Let d be a positive integer and let A be the set of k-partitions of d consisting of all a ∈ Nk such that |a| = d. If K is an algebraically closed ﬁeld and F = {xa | a ∈ A}, then the toric set Γ parameterized by F is an aﬃne toric variety. 8.4.24 Let A = {v1 , . . . , vq } ⊂ Nn and let K be a ﬁeld. Prove that K[xv1 , . . . , xvq ] is the coordinate ring of an aﬃne toric variety, in the sense of [176], if and only if R+ A ∩ Zn = NA and rank(A) = n. 8.4.25 Let A = {v1 , . . . , vq } ⊂ Zn and let K be a ﬁeld. Prove that if K[F ] = K[xv1 , . . . , xvq ] is normal and rank(A) = n, then K[F ] is isomorphic to the coordinate ring of an aﬃne toric variety, in the sense of [176]. 8.4.26 An aﬃne variety X ⊂ K q is called a set-theoretic complete intersection if I(X) is a set-theoretic complete intersection. If K is algebraically closed, then X is a set-theoretic complete intersection if and only if X can be deﬁned by q − dim(X) polynomials in q variables with coeﬃcients in K. 8.4.27 Let K = Z2 and let X ⊂ K 3 be a space curve given parametrically by t1 = x71 , t2 = x81 , t3 = x91 . If F = {x71 , x81 , x91 }, prove that PF = I(X) and PF = (t51 − t2 t33 , t22 − t1 t3 , t41 t2 − t43 ). 8.4.28 [272] Let F = {x21 x3 , x42 x23 , x1 x2 x3 } and let K = R be the ﬁeld of real numbers. Prove that V (PF ) is never fully parameterized by Laurent monomials in the sense of Theorem 8.4.10.

342

8.5

Chapter 8

Aﬃne Hilbert functions

In this section, we introduce the notion of degree via aﬃne Hilbert functions and show additivity of the degree. For lattice ideals the degree turns out to be independent of the base ﬁeld and the partial character. Let S = K[t1 , . . . , tq ] be a polynomial ring over a ﬁeld K and let I be an ideal of S. The vector space of polynomials in S (resp. I) of degree at most i is denoted by S≤i (resp. I≤i ). The functions HIa (i) = dimK (S≤i /I≤i ) and HI (i) = HIa (i) − HIa (i − 1) are called the aﬃne Hilbert function and the Hilbert function of S/I. Let u = ts+1 be a new variable and let I h ⊂ S[u] be the homogenization of I, where S[u] is given the standard grading. Lemma 8.5.1 If m = (t1 , . . . , tq ) is the irrelevant maximal ideal of S and I ⊂ S is a graded ideal of dimension zero, then Sm (Sm /Im ) = S (S/I) = S/I (S/I) = dimK (S/I). Proof. Let m be the image of m in the quotient ring S = S/I and let r be the least positive integer such that mr = (0). From the ascending chain (0) = mr ⊂ mr−1 ⊂ · · · ⊂ m2 ⊂ m ⊂ S and using that S/m = K, we get (S) =

r−1 i=0

S (mi /mi+1 ) =

r−1 i=0

S/m (mi /mi+1 ) =

r−1

dimK (mi /mi+1 ).

i=0

As the last summation is equal to dimK (S), we get that (S) is equal to dimK (S). The equality Sm (Sm /Im ) = S (S/I) follows readily by localizing at the maximal ideal m. 2 Remark 8.5.2 If S = ⊕∞ i=0 Si has the standard grading and I ⊂ S is a graded ideal, then i HIa (i) = k=0 dimK (Sk /Ik ) where Ik = I ∩ Sk . Thus, one has HI (i) = dimK (Si /Ii ) for all i. There are isomorphisms of K-vector spaces S/mi+1 S S/(mi+1 + I) S≤i /I≤i , where S = S/I and m = (t1 , . . . , tq ). By Lemma 8.5.1, one has S (S/(mi+1 + I)) = dimK (S/(mi+1 + I)). Thus, HIa is the Samuel or Hilbert–Samuel function of S with respect to m (see Section 4.4 and [413, Deﬁnition B.3.1]).

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Deﬁnition 8.5.3 The graded reverse lexicographical order (GRevLex for short) on the monomials of S is deﬁned as tb " ta if deg(tb ) > deg(ta ) or deg(tb ) = deg(ta ) and the last non-zero entry of b − a is negative. Lemma 8.5.4 Let I be an ideal of S. Then the following hold : (a) dim(S[u]/I h ) = dim(S/I) + 1. (b) HIa (i) = HI h (i) for i ≥ 0. Proof. (a): Let ≺ be the GRevLex order and let G be a Gr¨obner basis of I. Then {g h | g ∈ G} is a Gr¨obner basis of I h and in≺ (g) = in≺ (g h ) for g ∈ G (see Proposition 3.4.2). Since dim(S/I) = dim(S/in≺ (I)) and dim(S[u]/I) = dim(S[u]/in≺ (I h )), the equality follows. (b): Fix i ≥ 0. The mapping S[u]i → S≤i induced by mapping u → 1 is a K-linear surjection. Consider the induced composite K-linear surjection S[u]i → S≤i → S≤i /I≤i . An easy check shows that this has kernel Iih . Hence, we have an isomorphism of K-vector spaces S[u]i /Iih S≤i /I≤i ⇒ HIa (i) = HI h (i).

2

The degree of aﬃne algebras and lattice ideals Let d be the Krull dimension of S/I. By Lemma 8.5.4 and Proposition 5.1.6, there are unique d d−1 i i polynomials haI (t) = i=0 ai t ∈ Q[t] and hI (t) = i=0 ci t ∈ Q[t] of degrees d and d − 1, respectively, such that haI (i) = HIa (i) and hI (i) = HI (i) for i 0. By convention, the zero polynomial has degree −1. Notice that d! ad = (d − 1)! cd−1 for d ≥ 1. If d = 0, then HIa (i) = dimK (S/I) for i 0. Deﬁnition 8.5.5 The integer d! ad , denoted by deg(S/I) or e(S/I), is called the degree or multiplicity of S/I. Proposition 8.5.6 deg(S/I) = deg(S[u]/I h ). Proof. From Lemma 8.5.4(a), dim(S[u]/I h ) is equal to dim(S/I) + 1. Hence, the equality follows from Lemma 8.5.4(b). 2 Lemma 8.5.7 Let F be a ﬁeld extension of K, let B = F [t1 , . . . , tq ] be a polynomial ring, and let I be an ideal of S, then deg(S/I) = deg(B/IB). Proof. Let G be a Gr¨obner basis of I w.r.t the GRevLex order. Thanks to Buchberger’s criterion, G is a Gr¨obner basis for IB. By Proposition 3.4.2, I h and IB h are both generated by G h = {g h | g ∈ G}, and G h is a Gr¨obner basis for IB h . Hence to show the equality deg(S/I) = deg(B/IB) notice that, by Lemma 8.5.4, one has HIa (i) = HI h (i) = dimK (S[u]/I h )i = a (i) for i ≥ 0. dimF (B[u]/IB h )i = HIB h (i) = HIB

2

344

Chapter 8

For a lattice ideal Iρ (L) the degree depends only on the lattice L and is independent of the base ﬁeld K and the character ρ. In Section 9.4 we show eﬀective methods to compute the degree of any lattice ideal. Proposition 8.5.8 Let ρ : L → K ∗ be a partial character of a lattice L in Zq and let IQ (L) be the lattice ideal over the ﬁeld Q. Then deg(S/Iρ (L)) = deg(S/I(L)) = deg(Q[t1 , . . . , tq ]/IQ (L)). Proof. First we show the equality on the left. The ideal Iρ (L) contains no monomials (Theorem 8.2.2). Let G = {g1 , . . . , gm } be the reduced Gr¨ obner + − basis of I(L) w.r.t the GRevLex order ". We can write gi = tai − tai with + + − in≺ (gi ) = tai . We set hi = tai − ρ(ai )tai and H = {h1 , . . . , hm }. Next, we show that H is a Gr¨ obner basis of Iρ (L). Let f = 0 be any non-pure binomial of Iρ (L). We claim that f reduces to zero w.r.t H. By the Division m Algorithm, f = i=1 fi hi + g, where in≺ (f ) " in≺ (fi hi ) for all i, and g is a non-pure binomial in Iρ (L) such that none of the terms of g is divisible + + by any of the terms ta1 , . . . , tam . Then we can write g = μ(ta − λtb ) = μtδ (tu+ − λtu− ), with μ, λ ∈ K ∗ and a − b = u+ − u− . As tu+ − λtu− is in Iρ (L), by Lemma 8.2.6, u = u+ − u− is in L and λ = ρ(u). If g = 0, we obtain that + + tu+ − tu− , being in I(L), has one of its terms in (ta1 , . . . , tam ) = in≺ (I(L)), a contradiction. Thus, g must be zero, i.e., f reduces to zero w.r.t H. This proves the claim. In particular, Iρ (L) is generated by H. Note that the S-polynomial of hi and hj is a binomial; thus, by the claim, it reduces to zero with respect to H. Therefore by Buchberger’s criterion, H is a Gr¨obner basis of Iρ (L). Hence, S/I(L) and S/Iρ (L) have the same degree. Now we show the equality on the right. Let " be the GRevLex order on S and SQ = Q[t1 , . . . , tq ], and on the extensions S[u] and SQ [u]. Let GQ be the reduced Gr¨ obner basis of IQ (L). We set I = I(L) and IQ = IQ (L). Notice that Z ⊂ K if char(K) = 0 and Zp ⊂ K if p = char(K) > 0. In the second case, one has a map Z → K, 1 → 1K . Hence SZ = Z[t1 , . . . , tq ] embeds into S if char(K) = 0 and SZ maps into S if p = char(K) > 0. If G denotes the image of GQ under either of these two maps, by Buchberger’s criterion, it is seen that G is a Gr¨obner basis of I. Hence, by Proposition 3.4.2, GQh and G h are Gr¨ obner basis of IQh and I h , where G h is the set of f h with f ∈ G. Thus, the standard monomials of SQ [u]/IQh and S[u]/I h are the “same.” Therefore, these two rings have the same Hilbert function and the same degree. Thus, by Proposition 8.5.6, the result follows. 2 Proposition 8.5.9 (Additivity of the degree) If I is an ideal of S and I = q1 ∩ q2 ∩ · · · ∩ qm is an irredundant primary decomposition, then deg(S/qi ). deg(S/I) = ht(qi )=ht(I)

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Proof. We may assume that p1 , . . . , pr are the associated primes of I of height ht(I) and that rad(qi ) = pi for i = 1, . . . , r. The proof is by induction on m. We set J = ∩m i=2 qi . There is an exact sequence of K-vector spaces ϕ

φ

0 → S≤i /(q1 ∩ J)≤i → S≤i /(q1 )≤i ⊕ S/(J)≤i → S≤i /(q1 + J)≤i → 0, where ϕ(f ) = (f , −f ) and φ(f 1 , f 2 ) = f1 + f2 . Hence Hqa1 (i) + HJa (i) = HIa + Hqa1 +J (i).

(8.11)

As the decomposition of I is irredundant, by Exercise 8.5.10, one has dim(S/(q1 + J)) < dim(S/J). If r = 1, then dim(S/J) < dim(S/I). Hence, from Eq. (8.11), we get that deg(S/I) = deg(S/q1 ). If r > 1, then dim(S/J) = dim(S/I) = dim(S/q1 ). Hence, from Eq. (8.11), we get that deg(S/I) is deg(S/q1 ) + deg(S/J). Therefore, by induction, we get the required formula. 2

Exercises 8.5.10 Let I ⊂ S be an ideal and let I = q1 ∩q2 ∩· · ·∩qm be an irredundant primary decomposition. If J = ∩m i=2 qi , then dim(S/(q1 + J)) < dim(S/J). 8.5.11 Let I ⊂ S be an ideal. If q is a primary component of I of height ht(I), then qh is a primary component of I h of height ht(I h ).

8.6

Vanishing ideals over ﬁnite ﬁelds

Let K = Fq be a ﬁnite ﬁeld with q elements and X a subset of a projective space Ps−1 over K. Let S = K[t1 , . . . , ts ] be a polynomial ring over the ﬁnite ﬁeld K with the standard grading S = ⊕∞ d=0 Sd and let I(X) be the vanishing ideal of X which is the ideal of S generated by the homogeneous polynomials of S that vanish at all points of X. The following family arises in algebraic coding theory [348]. Proposition 8.6.1 [330] I(X) is a lattice ideal if and only if X = {[(xv1 , . . . , xvs )]| xi ∈ K ∗ = K \ {0} ∀ i} ⊂ Ps−1 ,

(8.12)

for some monomials xvi := xv1i1 · · · xvnin , i = 1, . . . , s. In what follows we assume that X is a subset of Ps−1 parameterized by monomials as in Eq. (8.12). The following formula allows us to compute the algebraic invariants of I(X) (degree, regularity, Hilbert function) using Macaulay2 [199].

346

Chapter 8

Theorem 8.6.2 [348] I(X) = ({ti − xvi z}si=1 ∪ {xq−1 − 1}ni=1 ) ∩ S. i Lemma 8.6.3 The degree of S/I(X) is |X|. Proof. For [P ] ∈ X, let I[P ] be its vanishing ideal. Since I(X) = ∩[P ]∈X I[P ] and since deg(S/I[P ] ) = 1 (see Exercise 8.6.10), the lemma follows from the additivity of the degree (see Proposition 8.5.9). 2 In this context it is usual to denote the Hilbert function of S/I(X) simply by HX . Proposition 8.6.4 There is an integer r ≥ 0 such that 1 = HX (0) < HX (1) < · · · < HX (r − 1) < HX (d) = |X| for d ≥ r. Proof. As S/I(X) is Cohen–Macaulay of dimension 1, by Theorem 2.2.4, its Hilbert polynomial has degree 0. Hence, by Lemma 8.6.3, HX (d) = |X| for d 0. Consequently the result follows readily from Theorem 5.6.4. 2 Recall from Chapter 5 that the integer r is the index of regularity of S/I(X) and that by Theorem 6.4.1 this number is the so-called Castelnuovo– Mumford regularity of S/I(X). For these reasons r is simply called the regularity and is denoted by reg(S/I(X)). Deﬁnition 8.6.5 Let X = {[P1 ], . . . , [Pm ]} and let f0 (t1 . . . , ts ) = td1 , where d ≥ 1. The linear map of K-vector spaces: f (P1 ) f (Pm ) ,..., f → evd : Sd → K |X| , f0 (P1 ) f0 (Pm ) is called an evaluation map. Notice that ker(evd ) = I(X)d . The image of evd , denoted by CX (d), is a linear code and CX (d) is called a parameterized code of order d. The algebraic invariants of S/I(X) are closely related to the so-called basic parameters of parameterized codes [120, 348]. Deﬁnition 8.6.6 The basic parameters of the linear code CX (d) are: (b1 ) dimK CX (d) = HX (d), the dimension, (b2 ) |X| = deg(S/I(X)), the length, (b3 ) δX (d) = min{ v : 0 = v ∈ CX (d)}, the minimum distance, where v is the number of non-zero entries of v. Lemma 8.6.7 δX (d) = 1 for d ≥ reg(S/I(X)).

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Proof. As CX (d) is a linear subspace of K |X| of dimension equal to |X| for d ≥ reg(S/I(X)), we get the equality CX (d) = K |X| . Thus δX (d) is equal to 1 for d ≥ reg(S/I(X)). 2 A good linear code should have large |X| together with dimK CX (d)/|X| and δX (d)/|X| as large as possible. From Lemma 8.6.7 we conclude that the potentially good codes CX (d) can occur only if 1 ≤ d < reg(S/I(X)). Problem 8.6.8 Find explicit formulas, in terms of the numerical data n, s, q, d, and the combinatorics of xv1 , . . . , xvs , for the basic parameters: (a) HX (d),

(b) deg(S/I(X)),

(c) δX (d),

(d) reg(S/I(X)).

Formulas for (a)-(d) are known when X is a projective torus [367] and when X is parameterized by xd11 , . . . , xds s , 1, where di ∈ N+ for al i [192, 292]. 36 20 ∗ Example 8.6.9 If X = {[(x90 1 , x2 , x3 , 1)]| xi ∈ F181 for i = 1, 2, 3}, then reg(S/I(X)) = 13. The basic parameters of the family {CX (d)}d≥1 are:

d |X| HX (d) δX (d)

1 2 3 4 5 6 7 8 9 10 11 12 13 90 90 90 90 90 90 90 90 90 90 90 90 90 4 9 16 25 35 45 55 65 74 81 86 89 90 45 36 27 18 9 8 7 6 5 4 3 2 1

Exercises 8.6.10 Let X be a ﬁnite subset of a projective space Ps−1 over a ﬁeld K. Let [P ] be a point in X, with P = (α1 , . . . , αs ) and αk = 0 for some k, and let I[P ] be the ideal generated by the homogeneous polynomials of S that vanish at [P ]. Prove that I[P ] is a prime ideal of height s − 1, I[P ] = ({αk ti − αi tk | k = i ∈ {1, . . . , s}), deg(S/I[P ] ) = 1 and that I(X) = [Q]∈X I[Q] .

8.7

Semigroup rings of numerical semigroups

Let S = (0) be a semigroup of (N, +), that is, S is a subset of N which is closed under addition and 0 ∈ S. Note that there exists a sequence d1 < · · · < dq of positive integers such that S = d1 N + · · · + dq N. If gcd(d1 , . . . , dq ) = 1, S is called a numerical semigroup [360]. Let K[x] be a polynomial ring in one variable over a ﬁeld K, the semigroup ring of S, denoted by K[S], is equal to the monomial subring K[xd1 , . . . , xdq ].

348

Chapter 8

Lemma 8.7.1 If S is a numerical semigroup, then c + N ⊂ S for some c. Proof. There are integers r1 , . . . , rq such that 1 = r1 d1 + · · · + rq dq . The integer c = (|r1 |d1 + · · · + |rq |dq )dq satisﬁes the required property. 2 Deﬁnition 8.7.2 Let S be a numerical semigroup. The Frobenius number of S, denoted by g(S), is the largest integer not in S. The semigroup S is called symmetric if g(S) − z ∈ S for all z ∈ / S. Proposition 8.7.3 A numerical semigroup S is symmetric if and only if (g(S) + 1)/2 = |N \ S|. Proof. We set c = g(S). Consider the map ϕ : N \ S → [0, c] ∩ N, z → c − z. Note that S is symmetric if and only if ϕ(N \ S) = S ∩ [0, c). Because of the decomposition [0, c] = (N \ S) ∪ (S ∩ [0, c)), one obtains that S is symmetric if and only if c + 1 = 2|N \ S|. 2 Proposition 8.7.4 [173] Let S be a numerical semigroup. If T (S) = {y ∈ N \ S | y + s ∈ S, ∀s ∈ S, s > 0}, then S is symmetric if and only if T (S) = {g(S)}. Proof. ⇒) We set c = g(S). Let y ∈ T (S) and assume y < c, then 0 < c − y ∈ S and y + (c − y) = c is in S, which is impossible, hence y = c. ⇐) Let z ∈ / S, z > 0, one must show c − z ∈ S. If c − z is not in S, choose the least positive integer z such that c−z is not in S. Since c−z = c, by deﬁnition of T (S) there is s ∈ S, s > 0 with (c − z) + s not in S, hence z − s > 0, which contradicts the choice of z. 2 The following nice result is due to Fr¨oberg. It gives an expression for the Cohen–Macaulay type for the semigroup ring of a numerical semigroup. Theorem 8.7.5 (Fr¨oberg [173]) Let S be a numerical semigroup. If T (S) = {y ∈ N \ S | y + s ∈ S, ∀s ∈ S, s > 0}, then |T (S)| is equal to the Cohen–Macaulay type of K[S]. Proof. We set D = K[S]. Let d1 < · · · < dq be a minimal generating set of S with d1 , . . . , dq relatively prime and s ∈ S, s > 0. Consider the mapping ϕ : T (S) → Soc(D/xs D), y → xy+s + xs D. The type of K[S] is, by deﬁnition, equal to dimK (Soc(D/xs D)) (see the last paragraph of Section 2.3). Thus it is enough to show that the image

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of T (S) is a K-vector space basis of Soc(D/xs D), because ϕ is injective. Note that Soc(D/xs D) has a K-basis consisting of monomials of the form xw + xs D, with w − s ∈ / S and xdi (xw + xs D) = xs D for all i, write di + w = s + si , where si is in S. We claim w − s ∈ T (S). It follows that (w − s) + s is in S for all s ∈ S, s > 0. Observe that w − s > 0; otherwise from w − s = s1 − d1 = s2 − d2 we derive s1 = 0 and d2 = s2 + d1 , hence d2 = μd1 , which is impossible because d1 and d2 are part of a minimal generating set of S. Hence w − s ∈ T (S) and ϕ(w − s) = w. To ﬁnish note that ϕ(T (S)) is linearly independent. 2 Theorem 8.7.6 [287] Let S be a numerical semigroup. Then the ring K[S] is Gorenstein if and only if S is symmetric. 2

Proof. It follows from Proposition 8.7.4 and Theorem 8.7.5.

Proposition 8.7.7 Let S be a numerical semigroup. If K[S] ⊂ K[x] has the induced grading, then the Frobenius number of S is equal to the degree, as a rational function, of the Hilbert series of K[S]. Proof. Let c = g(S) be the Frobenius number of S. Since K[S]i = Kxi if i ∈ S and K[S]i = (0) otherwise, one has F (K[S], z) =

i∈S

z i = f (z) +

∞ i=c+1

1 − zi (1 − z) i=0 c

z i = f (z) +

where f (z) is a polynomial with coeﬃcients in {0, 1} of degree at most c− 1. Hence deg(F (K[S], z)) = c. 2 Arithmetical symmetric semigroups Let S be a numerical semigroup of N. The problem of Frobenius consists in determining g(S), the Frobenius number of S. This problem can be explained as follows. Given a suﬃcient supply of coins of various denominations ﬁnd the largest amount that cannot be formed with these coins. The problem of computing g(S) has been examined by several authors [173, 359, 365, 374]. In [364, 365] the famous formula of Scarf and Shallcross [368], for the Frobenius number, is expressed in algebraic terms using homological algebra and lattice theory. Lemma 8.7.8 Let u, v ∈ N+ . If gcd(u, v) = 1 and S = uN + vN, then g(S) = uv − u − v. Proof. We set c = uv − u − v. It is easy to see that c is not in S. Hence, to show that c = g(S), it suﬃces to show that c + i ∈ S for i ∈ N+ . One can write 1 = μ1 u + μ2 v for some μ1 , μ2 in Z, hence c + i = (v − 1 + iμ1 )u + (−1 + iμ2 )v.

350

Chapter 8

By the division algorithm we can write c + i = b1 u + b2v, where 0 ≤ b2 < u. From the identity / S, c − b1 u − b2 v = (−b1 − 1)u + (u − 1 − b2 )v = −i ∈ one concludes b1 ≥ 0 and consequently c + i ∈ S.

2

Lemma 8.7.9 Let q ≥ 3 and v ≥ 1 be integers and let S be the semigroup S = d1 N + · · · + dq N, where di = d1 + (i − 1)v and d1 = r + k(q − 1) for some k, r ∈ N. If gcd(d1 , v) = 1 and 2 ≤ r ≤ q, then g(S) is equal to (k + v)d1 − v. Proof. Setting c = (k + v)d1 − v, we will show that c + i ∈ S for i ≥ 1. One may assume 1 ≤ i ≤ d1 . Notice c + i = (k + 1)d1 + λ + i, where λ = vd1 − d1 − v. Consider U = d1 N + vN. By Lemma 8.7.8, λ is equal to g(U ). Hence λ + i = ad1 + bv, for some a, b in N. One has vd1 + i = (a + 1)d1 + (b + 1)v. Hence b ≤ (q − 1)(k + a + 1) and we can write b = j(k + a + 1) + , for some 1 ≤ ≤ k + a + 1 and 0 ≤ j ≤ q − 2. Therefore the equality c + i = (k + a + 1)d1 + bv = (k + a + 1 − )dj+1 + dj+2 , q gives c + i in S. It remains to c∈ / S. If c ∈ S, then c = i=1 ai di , for show q some ai ’s in N. Setting s = i=1 ai , note that one can rewrite c as: c = (k + 1)d1 + λ = a1 d1 + · · · + aq dq = sd1 + pv, for some p ∈ N. Since λ ∈ / U , one derives s ≤ k and c ≤ sdq ≤ kdq , which is a contradiction since c = kdq + (r − 1)v. 2 Theorem 8.7.10 [148, 265] Let r, k, q, v ∈ N and let d1 = r + k(q − 1) and di = d1 + (i − 1)v be an arithmetical sequence. If 2 ≤ r ≤ q = 2 and gcd(d1 , v) = 1, then r = 2 if and only if the semigroup S = d1 N + · · · + dq N is symmetric. Proof. ⇒) Let d1 = 2 + (q − 1)k. Set c = (k + v)d1 − v = kdq + v and U = d1 N + vN. Assume z ∈ / S. First consider the case z ∈ U , write z = rd1 + sv, r, s ∈ N. By the Euclidean algorithm and Lemma 8.7.9 one can write z = adq + bv, where a, b are integers so that 0 ≤ a ≤ k

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and 1 ≤ b < d1 . Writing a = k − i, where 0 ≤ i ≤ k, we claim that 1 ≤ b ≤ (q − 1)i + 1. If b > (q − 1)i + 1, then z = x1 d1 + x2 v, with x1 = k + v − i and x2 = b − (q − 1)i − 2. Since 0 ≤ x2 ≤ (q − 1)x1 , it is not hard to show that z ∈ S, which is impossible. Hence 1 ≤ b ≤ (q − 1)i + 1. Set y1 = i and y2 = (q − 1)i − b + 1. Since 0 ≤ y2 ≤ (q − 1)y1 from the equality c − z = y1 d1 + y2 b, one gets c − z ∈ S. Now consider the case z ∈ / U . One can write z = ad1 + bv, with 1 ≤ b < d1 and a < 0. Set w1 = k + 1 and w2 = d1 − b − 1. Using c − z = w1 d1 − (1 + a)d1 + w2 v and 0 ≤ w2 ≤ (q − 1)w1 , gives c − z ∈ S. Therefore S is symmetric. ⇐) Assume that S is symmetric and keep q the same notations as above. As v ∈ / Sone has c − v ∈ S and c − v = i=1 bi di , where bi ∈ N for all i. q Set a = i=1 bi and notice that c − v = (k + 1)d1 + λ − v = ad1 + q1 v, for some q1 ∈ N. Therefore a ≤ k, otherwise one obtains λ ∈ U . From the 2 inequality c − v ≤ adq ≤ kdq , it follows that r = 2.

Exercises 8.7.11 Let S be a numerical semigroup of N. Prove that there is a unique minimal set of generators {d1 , . . . , dq } of S with gcd(d1 , . . . , dq ) = 1. 8.7.12 Let S = (0) be a semigroup of N. Prove that the following conditions are equivalent: (a) S is a numerical semigroup. (b) c + N ⊂ S for some c ∈ S. (c) |N \ S| < ∞. 8.7.13 Prove that the semigroup S = 4N + 5N + 6N is symmetric. 8.7.14 If S = 5N + 6N + 7N + 8N and K is a ﬁeld, prove that K[S] is a Gorenstein ring which is not a complete intersection. 8.7.15 Let S ⊂ N be a numerical semigroup with Frobenius number c and let K be a ﬁeld. If D = K[S] ⊂ K[x] has the induced grading, prove that the degree of the Hilbert series of D/xc+1 D is 2c + 1. 8.7.16 Let K be a ﬁeld and let D = K[t1 , t2 ]/(td12 − td21 , ta1 1 ta2 2 ), where ti has degree di ∈ N+ . If gcd(d1 , d2 ) = 1 and c + 1 = a1 d1 + a2 d2 , prove that the Hilbert series of D is (1 − z c+1 )(1 − z d1 d2 ) , (1 − z d1 )(1 − z d2 ) where c is the Frobenius number of the proper semigroup S = d1 N + d2 N. Note D D/xc+1 D, where D = K[xd1 , xd2 ].

352

8.8

Chapter 8

Toric ideals of monomial curves

In this section we study toric ideals of monomial curves. Using binary trees, we characterize when the toric ideal of an aﬃne monomial curve over an arbitrary ﬁeld is a complete intersection. If the base ﬁeld has positive characteristic, we show that 1-dimensional graded lattice ideals are set theoretic complete intersections. Let S = Nd1 + · · · + Ndq be a numerical semigroup of N generated by a sequence d = {d1 , . . . , dq } of relatively prime positive integers. As usual S = K[t1 , . . . , tq ] denotes a polynomial ring over a ﬁeld K. The kernel of the homomorphism of K-algebras ϕ : S = K[t1 , . . . , tq ] −→ K[S] = K[xd1 , . . . , xdq ], induced by ϕ(ti ) = xdi , is called the toric ideal of K[S] and will be denoted by P . The map ϕ is graded if we endow S and K[x] with the gradings induced by setting deg(ti ) = di and deg(x) = 1. The corresponding monomial curve in AqK parameterized by ti = xdi , i = 1, . . . , q, will be denoted by Γ. If q = 3, Γ is called a monomial space curve. In this context the ideal P is also called the toric ideal of Γ. Lemma 8.8.1 P is a binomial graded prime ideal of S of dimension 1 and Γ = V (P ). If K is an inﬁnite ﬁeld, then I(Γ) = P . Proof. As K[x] is an integral domain, we get that P is prime. Since K[x] is integral over K[S], by Proposition 2.4.13, we have ht(P ) = q − 1. By Corollary 8.2.18, the toric ideal P is generated by binomials. According to Proposition 8.4.8, if gcd (d) = 1, Γ is an aﬃne toric variety, that is Γ = V (P ). If K is inﬁnite, by Corollary 8.4.13, we get I(Γ) = P . 2 ; aij mi Deﬁnition 8.8.2 A binomial fi = ti − j=i tj ∈ P is called critical with respect to ti or ti -critical if mi is the least positive integer such that mi di ∈ dj N. j=i

The notion of a critical binomial was ﬁrst introduced by Eliahou [135], and later studied in [4]. This notion—and some of the results of this section—can be extended to binomial and lattice ideals [335]. Deﬁnition 8.8.3 A set {f1 , . . . , fq } is called a full set of critical binomials if fi is a ti -critical binomial for all i. The interest in studying ideals generated by critical binomials comes from a famous result of Herzog [216] showing that the toric ideal of any monomial space curve is generated by a full set of critical binomials (see Theorem 8.8.8). Herzog’s result is no longer true for toric ideals of monomial curves in higher dimensions.

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Example 8.8.4 [135] Let m be a positive integer. If d1 = m2 , d2 = 2m2 −1, d3 = 3m2 + m, d4 = 4m2 + m − 1, then μ(P ) ≥ m, where μ(P ) is the minimum number of generators of P . If S is symmetric and generated non-redundantly by four integers, then P is minimally generated by 3 or 5 elements [53]; thus even in this case P is not generated by critical binomials. A main open problem is to determine when a given full set of critical binomials generates P . If the toric ideal of a monomial curve in A4K is generated by a full set of critical binomials, produced by any critical-binomial-algorithm, then one can predict how these generators should look [4]. In what follows, for simplicity of notation, fi will stand for a ti -critical binomial for i = 1, . . . , q. Note that we do not assume ±f1 , . . . , ±fq distinct. Theorem 8.8.5 [216] Let B = {f1 , f2 , f3 } be a full set of critical binomials and let I = (B). If f1 = ta1 1 − ta2 2 ta3 3 , f2 = tb22 − tb11 tb33 , f3 = −tc11 tc22 + tc33 and bi > 0, ci > 0 for all i, then (a) ai > 0 for all i and a1 = b1 + c1 , b2 = a2 + c2 , c3 = a3 + b3 . (b) B is a Gr¨ obner basis for I w.r.t the revlex order t1 " t2 " t3 . (c) The ideal I = (f1 , f2 , f3 ) is equal to the toric ideal P of K[S]. Proof. (a) If a2 = 0, then a3 ≥ c3 . As f1 + ta3 3 −c3 f3 = ta1 1 − tc11 tc22 ta3 3 −c3 , one obtains a contradiction, hence a2 > 0, by a similar argument a3 > 0. We set α = (−a1 , a2 , a3 ), β = (b1 , −b2 , b3 ), γ = (c1 , c2 , −c3 ). Consider the vector (v1 , v2 , v3 ) = α + β + γ. We now show that vi ≥ 0 for all i. Note a1 > b1 , otherwise using tb11 −a1 tb33 f1 +f2 one derives a contradiction. Hence, using tb33 f1 + ta1 1 −b1 f2 , one has a3 + b3 ≥ c3 and v3 ≥ 0. Similarly, one shows that v1 , v2 ≥ 0. As v1 d1 + v2 d2 + v3 d3 = 0, we get vi = 0 for all i. (b) It follows from part (a) and Buchberger’s criterion (Theorem 3.3.17). (c) Let f be a binomial in P and assume we order the monomials with the revlex ordering t1 " t2 " t3 . As f1 , f2 , f3 are binomials, by the division algorithm (Theorem 3.3.6), we can write f = h1 f1 + h2 f2 + h3 f3 + r, where r = tr11 tr22 tr33 − ts11 ts22 ts33 ∈ P, r1 , s1 < a1 , r2 , s2 < b2 and such that none of the monomials occurring in r is divisible by the leading term of f3 . It suﬃces to show that r = 0. Assume r = 0. As tr33 or ts33 is a factor of r, one may assume that s3 = 0. If s1 ≤ r1 or s2 ≤ r2 we get a contradiction with the minimality of b2 or a1 , respectively; hence s1 > r1 and s2 > r2 . From the equality r = tr11 tr22 h, where h = tr33 − ts11 −r1 ts22 −r2 , we get h ∈ P and r3 ≥ c3 . Using the identity h − tr33 −c3 f3 = tr33 −c3 tc11 tc22 − ts11 −r1 ts22 −r2 and the minimality of b2 and a1 allows us to conclude s1 − r1 > c1 and 2 s2 − r2 > c2 , a contradiction because ts11 ts22 is not divisible by tc11 tc22 .

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Proposition 8.8.6 If f1 = ta1 1 − ta2 2 = −f2 , f3 = tc33 − tc11 tc22 is a full set of critical binomials, then (f1 , f3 ) is equal to the toric ideal P of K[S]. Proof. Let f be a binomial in P and assume we order the terms with the revlex order t3 " t1 " t2 . As f1 , f3 are binomials, by the division algorithm, we can write f = h1 f1 + h3 f3 + r, where r = tr11 tr22 tr33 − ts11 ts22 ts33 ∈ P , such that r1 , s1 < a1 and r3 , s3 < c3 . Assume r = 0. As tr22 or ts22 is a factor of r, one may assume that s2 = 0. Note s1 > r1 and s3 > r3 . From the equality r = tr11 tr33 h, where h = tr22 − ts11 −r1 ts33 −r3 , we get h ∈ P and r2 ≥ a2 . Therefore the identity h + tr22 −a2 f1 = ta1 1 tr22 −a2 − ts11 −r1 ts33 −r3 contradicts the choice of c3 because one has the inequality a1 > s1 − r1 . 2 Deﬁnition 8.8.7 The support of a binomial f = tα − tβ , denoted by supp(f ), is deﬁned as supp(tα ) ∪ supp(tβ ), where supp(tα ) = {ti | αi > 0}. The next result also holds for graded lattice ideals [335]. Theorem 8.8.8 [216] If {f1 , f2 , f3 } is a full set of critical binomials, then the ideal (f1 , f2 , f3 ) is equal to the toric ideal P of K[S]. Proof. Let f1 = ta1 1 − ta2 2 ta3 3 , f2 = tb22 − tb11 tb33 , f3 = −tc11 tc22 + tc33 . By Theorem 8.8.5(c) one may assume supp(fi ) = {t1 , t2 , t3 } for at least two fi . Hence one may assume a3 = 0 and the cases to consider are the following: (i) b1 = 0, (ii) b3 = 0, (iii) c1 = 0, and (iv) c2 = 0. (i) Note b3 ≥ c3 , if b3 = c3 , then f2 is a critical binomial w.r.t t3 and by Proposition 8.8.6 one concludes that P is generated by {f1 , f2 }. If b3 > c3 , from the identity f2 + tb33 −c3 f3 = tb22 − tb33 −c3 tc11 tc22 , we get c2 = 0. Thus f2 = f2 + tb33 −c3 f3 = tb22 − tb33 −c3 tc11 and f3 = f3 + tc11 −a1 f1 = tc33 − tc11 −a1 ta2 2 . Applying Theorem 8.8.5(a) to f1 , f2 , f3 yield c1 = a1 . Hence, f3 is t1 critical. Using Proposition 8.8.6, we get that P is generated by {f2 , f3 }. (ii) As a2 ≥ b2 and b1 ≥ a1 , from f1 − ta2 2 −b2 f2 = ta1 1 − ta2 2 −b2 tb11 we get f1 = −f2 and P = (f1 , f3 ) by Proposition 8.8.6. One may now assume b1 > 0 and b3 > 0. (iii) Using tc22 −b2 f2 + f3 we readily see that this case cannot occur. (iv) Using ta2 2 −b2 f2 + f1 we derive a contradiction. 2

Binary trees and complete intersections Deﬁnition 8.8.9 A binary tree is a connected directed rooted tree such that: (i) two edges leave the root and every other vertex has either degree

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1 or 3, (ii) if a vertex has degree 3, then one edge enters the vertex and the other two edges leave the vertex, and (iii) if a vertex has degree 1, then one edge enters the vertex. The vertices of degree 1 are called terminal . Proposition 8.8.10 If G is a binary tree with q terminal vertices, then the number of non-terminal vertices of G is q − 1. Proof. It follows by induction on q.

2

Deﬁnition 8.8.11 A binary tree G is said to be labeled by [q] := {1, . . . , q} if its terminal vertices are labeled by {1}, . . . , {q}. Extending this deﬁnition, we will also consider binary trees with q terminal vertices labeled by arbitrary ﬁnite subsets of N with q elements. If G is a binary tree labeled by [q] and v is a non-terminal vertex of G, consider v1 and v2 , the two vertices of G such that rv @ @ R @ @r v2 v1 r is a subgraph of G, and denote by G1 , resp. G2 , the subtree of G whose root is v1 , resp. v2 . We denote by 1 [v] and 2 [v] the two disjoint subsets of [n] formed by the union of the labels of the terminal vertices of G1 and G2 , respectively. Deﬁnition 8.8.12 Let B = {g1 , . . . , gq−1 } be a set of binomials of S with gi = tαi − tβi , supp(tαi ) ∩ supp(tβi ) = ∅, and αi = 0, βi = 0 for all i = 1, . . . , q − 1, and let G be a binary tree labeled by [q]. We say that G is compatible with B if, denoting by F the set of non-terminal vertices of G, there is a bijection f B −→ F such that supp(tαi ) ⊂ 1 [f (gi )] and supp(tβi ) ⊂ 2 [f (gi )] for all i ∈ [q − 1]. Example 8.8.13 The following binary tree G labeled by [5] is compatible with B = {g1 = t21 t42 − t4 t5 , g2 = t1 − t2 t3 , g3 = t44 − t25 , g4 = t2 − t73 } s g1 @ @ R @ @ sg g2 s 3 A @ @ UA A R @ @ @ g4 AAs {4} s @s {5} {1} s A AUA AAs {3} {2} s and if v is the root of G, then 1 [v] = {1, 2, 3} and 2 [v] = {4, 5}.

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Theorem 8.8.14 Let B = {g1 , . . . , gq−1 } be a set of binomials of S such that gi = tαi − tβi , supp(tαi ) ∩ supp(tβi ) = ∅, and αi = 0, βi = 0 for all i = 1, . . . , q − 1. Then the following two conditions are equivalent: (1) V (B, ti ) = {0} for all i = 1, . . . , q. (2) There is a binary tree G labeled by [q] which is compatible with B. Proof. (1) ⇒ (2): Set V1 := {1}, . . . , Vq := {q} and consider the partition F1 := {V1 , . . . , Vq } of [q]. Let us show that there exist Vq+1 , . . . , V2q−1 , subsets of [q], and F2 , . . . , Fq , partitions of [q], such that, reindexing g1 , . . . , gq−1 if necessary, the following assertions hold for all i ∈ [q − 1]: (a) Vq+i = Vj ∪ Vk for some Vj , Vk ∈ Fi , j = k. (b) supp(tαi ) ⊂ Vj and supp(tβi ) ⊂ Vk . (c) Fi+1 = (Fi \ {Vj , Vk }) ∪ {Vq+i }. Consider the digraph G with 2q−1 vertices, denoted by v1 , . . . , v2q−1 , where we connect vq+i with vj and vk as follows: r vq+i @ @ R @ @r vk vj r whenever Vq+i = Vj ∪ Vk in (a), it is not hard to see that G is a binary tree labeled by [q]. The root of G is v2q−1 , and the set of its non-terminal vertices is F := {vq+1 , . . . , v2q−1 }. Moreover, by construction, for i ∈ {1, . . . , q − 1}, one has that 1 [vq+i ] = Vj and 2 [vq+i ] = Vk for Vj and Vk in (a). Hence, by (b), G is compatible with B via the map f : B → F , gi → vq+i , and (2) will follow. Let us ﬁrst construct Vq+1 and F2 satisfying (a), (b), and (c). We ﬁrst claim that for all i ∈ [q], there exists an element gj ∈ B such that either supp(tαj ) ⊂ Vi or supp(tβj ) ⊂ Vi because otherwise, we have that the ith unit vector ei of AqK belongs to V (B, t1 , . . . , ti−1 , ti+1 , . . . , tq ) which is {0} by (1). Since |F1 | = q and |B| = q − 1, by the pigeonhole principle there exists an element in B, say g1 , and Vj , Vk ∈ F1 with j = k, such that supp(tα1 ) ⊂ Vj and supp(tβ1 ) ⊂ Vk . Setting Vq+1 := Vj ∪ Vk and F2 := (F1 \ {Vj , Vk }) ∪ {Vq+1 }, (a), (b), and (c) hold for i = 1. Assume now that for i ∈ [[2, q − 1]], we have constructed Vq+1 , . . . , Vq+i−1 and F2 , . . . Fi such that (a), (b), and (c) hold, and let us construct Vq+i and Fi+1 satisfying (a), (b), and (c). Observe ﬁrst that for all j ≤ i − 1, supp(gj ) is contained in some element of Fi . Set Bi := B \ {g1 , . . . , gi−1 }. We claim that for each Vk ∈ Fi , there exists gj ∈ Bi such that either supp(tαj ) ⊂ Vk or supp(tβj ) ⊂ Vk . In order to prove this, we show that if there exists an element in Fi , say Vs =

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{i1 , . . . , im }, that does not satisfy the claim, then α := ei1 +· · ·+eim belongs to V (B), which is a contradiction by (1). Take gj ∈ B. If gj ∈ Bi , then supp(tαj ) ⊂ Vs and supp(tβj ) ⊂ Vs by deﬁnition of Vs , and hence gj (α) = 0. If gj ∈ / Bi , i.e., if j ≤ i − 1, then supp(gj ) is contained in some element of Fi , say Vt . If t = s, i.e., if supp(gj ) ⊂ Vs , then gj (α) = 1 − 1 = 0. Otherwise, since Fi is a partition of [q] and Vs , Vt ∈ Fi , one has that Vs ∩ Vt = ∅, and hence supp(gj ) ∩ Vs = ∅. Thus, gj (α) = 0, and the claim is proved. We have proved that for each Vk ∈ Fi , there exists gj ∈ Bi such that either supp(tαj ) ⊂ Vk or supp(tβj ) ⊂ Vk . Since |Fi | = q − i + 1 and |Bi | = q − i, and using that Fi is a partition of [q], we get by the pigeonhole principle that there exists an element in Bi , say gi , and Vj , Vk ∈ Fi such that supp(tαi ) ⊂ Vj and supp(tβi ) ⊂ Vk . Setting Vq+i := Vj ∪ Vk and Fi+1 = (Fi \ {Vj , Vk }) ∪ {Vq+i }, the statements (a), (b), and (c) hold, and we are done. (2) ⇒ (1): The proof is by induction on q. The result is clear if q = 2. Denoting by v the root of G, we may assume without loss of generality, that 1 [v] = [r] and 2 [v] = [[r + 1, q]] for some r ∈ {1, . . . , q − 1}. Then, if G1 and G2 are the two connected components of the digraph G \ {v} obtained from G by removing the vertex v, one has that G1 and G2 are binary trees (for convenience we regard an isolated vertex as a binary tree), labeled by [r] and [[r + 1, q]], respectively. Reindexing the gi ’s if necessary, we may also assume that G1 is compatible with B1 := {g2 , . . . , gr }, G2 is compatible with B2 := {gr+1 , . . . , gq−1 }, and g1 = tα1 − tβ1 with supp(tα1 ) ⊂ [r] and supp(tβ1 ) ⊂ [[r + 1, q]]. Then, supp(gi ) ⊂ [r] if i = 2, . . . , r, and supp(gi ) ⊂ [[r +1, q]] if i = r +1, . . . , q −1. Moreover, applying the induction hypothesis, one has that {0} = V (B1 , ti ) ⊂ K r for i = 1, . . . , r, and {0} = V (B2 , ti ) ⊂ K q−r for i = r + 1, . . . , q. Fix i ∈ [q] and take a ∈ V (B, ti ). The result will be proved if we show that a = 0. By symmetry, we may assume that 1 ≤ i ≤ r. The vector a = (a1 , . . . , aq ) can be decomposed as a = b + c, where b = (a1 , . . . , ar , 0, . . . , 0). Then b ∈ V (B1 , ti ), and hence b = 0. On the other hand, g1 (a) = 0 implies that aj = 0 for some j ∈ {r + 1, . . . , q}. 2 Thus c ∈ V (B2 , tj ) which is {0}, and hence a = 0, as required. Complete intersections Let d = {d1 , . . . , dq } be a set of distinct positive integers and let P ⊂ S be the corresponding toric ideal deﬁned at the beginning of the section. The exact sequence ψ

0 −→ ker(ψ) −→ Zq −→ Z −→ 0;

ψ

ei −→ di

is related to P as follows. If g = ta − tb is a binomial, then g ∈ P if and only if g' = a − b ∈ ker(ψ). Deﬁnition 8.8.15 Let G be a binary tree labeled by [q], and consider a set of vectors in Zq , W = {w1 , . . . , wq−1 }. We say that G is compatible with W + − if G is compatible with the set of binomials {twi − twi ; i = 1, . . . , q − 1}.

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Theorem 8.8.16 Let G be a binary tree labeled by [q] and denote by F the set of its non-terminal vertices. The following two conditions are equivalent: (1) There exist vectors w1 , . . . , wq−1 ∈ Zq such that G is compatible with W = {w1 , . . . , wq−1 }, and ker(ψ) = ZW . (2) For all v ∈ F , 2 gcd(dj ; j ∈ 1 [v]) gcd(dj ; j ∈ 2 [v]) ∈ N{dj ; j ∈ i [v]}. gcd(dj ; j ∈ 1 [v] ∪ 2 [v]) i=1

Proof. Let v be the root of G and let G1 and G2 be the two connected components of the digraph G\{v} obtained from G by removing v. We may assume that 1 [v] = [r] and 2 [v] = [[r + 1, q]] for some r ∈ {1, . . . , q − 1}. Then G1 and G2 are binary trees (for convenience we regard an isolated vertex as a binary tree) labeled by [r] and [[r + 1, q]]. The result is clear if q = 2, and we will prove both implications by induction on q. (1) ⇒ (2): Reindexing the wi ’s if necessary, we may assume that wq−1 is the element of W associated to v through the map that makes G compatible with W , and that W1 = {w1 , . . . , wr−1 } and W2 = {wr , . . . , wq−2 } are the set of vectors in W such that Gi is compatible with Wi for i = 1, 2. There is a decomposition Zq = Zr ⊕ Zq−r , where Zr := Zr × {0}q−r and q−r := {0}r × Zq−r . Consider the linear maps Z

di if 1 ≤ i ≤ r, ψ 1 : Zq −→ Z, ψ 1 (ei ) = 0 if r < i ≤ q, and ψ 2 = ψ − ψ 1 . Let ψ1 (resp. ψ2 ) be the restriction of ψ 1 (resp. ψ 2 ) to Zr (resp. Zq−r ). It is not hard to see that ker(ψ1 ) = ZW1 and ker(ψ2 ) = ZW2 . Hence, setting d = gcd(d1 , . . . , dq ), d = gcd(d1 , . . . , dr ), d = gcd(dr+1 , . . . , dq ), d = {d1 , . . . , dr }, d = {dr+1 , . . . , dq }, d = {d1 , . . . , dq }, and using induction we need only show (d d )/d ∈ Nd ∩ Nd . For 1 ≤ j ≤ r and r + 1 ≤ k ≤ q we can write dk dj ej − ek = λ1jk w1 + · · · + λq−1 jk wq−1 , d d for some λ1kj , . . . , λq−1 kj in Z. We can write the last vector in the set W as + − wq−1 = (a1 , . . . , ar , −ar+1 , . . . , −aq ) = wq−1 − wq−1 .

Hence, we get −(dj /d)ek = λrjk wr + · · · + λq−1 jk wq−1 . This implies (dj /d)dk

=

λq−1 jk (ar+1 dr+1 + · · · + aq dq ).

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Set h = ar+1 dr+1 + · · · + aq dq . If we ﬁx k and vary j, we get gcd ((d1 /d)dk , . . . , (dr /d)dk ) = μk h

(μk ∈ Z) ⇒ dk d = μk hd.

Therefore varying k yields gcd (dr+1 d , . . . , dq d ) = gcd(μr+1 hd, . . . , μq hd) = hdμ (μ ∈ Z). As a consequence (d d )/d = (hdμ)/d ∈ Nd . A symmetric argument gives (d d )/d ∈ Nd , as required. (2) ⇒ (1): By induction hypothesis there are W1 = {w1 , . . . , wr−1 } and W2 = {wr , . . . , wq−2 } such that Gi is compatible with Wi and ker(ψi ) = ZWi for i = 1, 2. The + ) ⊂ [r], result will be proved if we give wq−1 ∈ Zq such that supp(wq−1 − supp(wq−1 ) ⊂ [[r + 1, q]], and ker(ψ) = ZW for W = W1 ∪ W2 ∪ {wq−1 }. By hypothesis there are where a1 , . . . , aq in N such that gcd(d1 , . . . , dr ) gcd(dr+1 , . . . , dq ) d d = d gcd(d1 , . . . , dq )

=

r

q

ai di =

i=1

ai di .

i=r+1

Setting wq−1 := (a1 , . . . , ar , −ar+1 , . . . , −aq ) and W := W1 ∪ W2 ∪ {wq−1 }, + − we get supp(wq−1 ) ⊂ [r] and supp(wq−1 ) ⊂ [[r + 1, q]]. Thus G is compatible with W . To complete the proof it remains to prove that ZW = ker(ψ). Clearly ZW ⊂ ker(ψ). To prove the reverse inclusion deﬁne σjk = (dj /d)ek − (dk /d)ej ;

j, k ∈ [q].

By Exercise 8.3.41 the set {σjk | j, k ∈ [q]} generates ker(ψ). Thus we need only show that σjk ∈ ZW for all j, k ∈ [q]. If j, k ∈ [r] or j, k ∈ [[r+1, q]], then σjk ∈ ker(ψ1 ) ⊂ ZW or σjk ∈ ker(ψ2 ) ⊂ ZW . Assume j ∈ [r] and k ∈ [[r + 1, q]]. From the equalities r r di dj dj d e S1 = ai e − e − ai e i , = j i j d d d d i=1 i=1 q q di dk dk d S2 = ai ek − ei = ek − ai e i d d d d i=r+1 i=r+1 we conclude dj dk S1 − S2 d d

=

dk dj ej − ek d d

−

dj dk wq−1 . d d

Since Si ∈ ker(ψi ) ⊂ ZW we obtain σjk ∈ ZW , as required. g = a − b. Notation Given a binomial g = ta − tb , we set '

2

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Theorem 8.8.17 Let K be an arbitrary ﬁeld and let B = {g1 , . . . , gq−1 } be a set of binomials in the toric ideal P . Then P = (B) if and only if gq−1 } and (a) ker(ψ) = Z{' g1 , . . . , ' (b) V (g1 , . . . , gq−1 , ti ) = {0} for i = 1, . . . , q. Proof. If P = (B), then (a) follows at once from Proposition 8.3.1, and (b) follows from Corollary 8.3.21(b). Conversely, if (a) and (b) hold then by Corollary 8.3.21, one has rad(B) = P . Let {h1 , . . . , hs } be a set of generators gr } by (a), of P consisting of binomials. Notice that ker (ψ) = Z{' g1 , . . . , ' ' ' and ker (ψ) = Z{h1 , . . . , hs } by Proposition 8.3.1. Thus using that (B) is a quasi homogeneous complete intersection and applying Proposition 8.3.17, (B) is a radical ideal, and hence P = (B). 2 Theorem 8.8.18 [33] The toric ideal P is a complete intersection if and only if there is a binary tree G labeled by [q] such that, for all non-terminal vertex v of G, one has that gcd(dj , j ∈ 1 [v]) gcd(dj , j ∈ 2 [v]) ∈ N{dj , j ∈ 1 [v]} ∩ N{dj , j ∈ 2 [v]}. gcd(dj , j ∈ 1 [v] ∪ 2 [v]) Proof. ⇒) There are binomials g1 , . . . , gq−1 such that P = (g1 , . . . , gq−1 ). We may assume that gi = tαi − tβi and supp(tαi ) ∩ supp(tβi ) = ∅ for all i. By Theorem 8.8.17(b) and Theorem 8.8.14 there exists a binary tree G labeled by [q] which is compatible with {g1 , . . . , gq−1 }. Then G is compatible with W = {' g1 , . . . , g'r } and ker(ψ) = Z{' g1 , . . . , ' gr } (see Theorem 8.8.17(a)). Thus applying Theorem 8.8.16 we obtain the required conditions. ⇐) By Theorem 8.8.16, there is W = {w1 , . . . , wq−1 } ⊂ Zq such that W + − is compatible with G and ker(ψ) = ZW . Setting gi := twi − twi , one has that G is compatible with {g1 , . . . , gq−1 }. Hence, by Theorem 8.8.14, we get V (g1 , . . . , gq−1 , ti ) = {0} for i = 1, . . . , q. Therefore, by Theorem 8.8.17, we deduce the equality P = (g1 , . . . , gq−1 ). 2 There is a characterization of complete intersection semigroups of N given in [107] (see also [158] for a generalization of this description to semigroups of arbitrary dimension). The complete intersection property of toric ideals of monomial curves has been studied in [30] from a computational point of view (showing an eﬃcient algorithm that checks this property). In the area of complete intersection toric ideals there are some other papers (see the introductions of [182, 184, 322] and the references therein). Remark 8.8.19 If P is a complete intersection and G is a binary tree labeled by [q] such that the conditions of Theorem 8.8.18 hold, then (i) The generators {g1 , . . . , gq−1 } of P and their degrees D1 , . . . , Dq−1 can be obtained as shown in the proofs of Theorems 8.8.16 and 8.8.18.

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361

(ii) The Frobenius number g(S) of a numerical semigroup S = Nd can be expressed entirely in terms of {d1 , . . . , dq }. This last assertion is a consequence of the following. Recall that the quasi-homogeneous Hilbert series of S/P is F (S/P, z) =

f (z) (1 −

z d1 ) · · · (1

− z dq )

for some polynomial f ∈ Z[z]; see Theorem 5.1.4. When d1 , . . . , dq are relatively prime, by Proposition 8.7.7 and its proof, one can write F (S/P, z) =

h(z) 1−z

for some polynomial h of degree g(S) + 1. If P is a complete intersection, one can write f (z) = (1 − z D1 ) · · · (1 − z Dq−1 ) where D1 , . . . , Dq−1 are the degrees of the minimal generators of P (see Exercise 5.1.20), and hence g(S) = D1 + · · · + Dq−1 − (d1 + · · · + dq ). Denoting by {v1 , . . . , vq−1 } the set of non-terminal vertices of G and using (i), we get : ! ! q−1 q gcd(dj , j ∈ 1 [vi ]) gcd(dj , j ∈ 2 [vi ]) − g(S) = di . gcd(dj , j ∈ 1 [vi ] ∪ 2 [vi ]) i=1 i=1 Example 8.8.20 Let K be a ﬁeld, and consider d1 = 16, d2 = 27, d3 = 45, and d4 = 56. The corresponding toric ideal P ⊂ K[t1 , t2 , t3 , t4 ] is a complete intersection because using the following binary tree labeled by [4] r HH j H H HHr r H HH j @ R @ @ H HHr{3} {1} r @r{4} {2} r the arithmetical conditions in Theorem 8.8.18 are satisﬁed: (16)(56) ∈ 16N ∩ 56N; gcd(16, 56) (27)(45) ∈ 27N ∩ 45N; 135 = gcd(27, 45) gcd(16, 56) gcd(27, 45) ∈ (16, 56)N ∩ (27, 45)N; 72 = gcd(16, 27, 45, 56)

112 =

(1)

2(56) =

(2)

3(45) =

7(16) 5(27)

(3)

1(16) + 1(56) = 1(27) + 1(45). Moreover, the equalities (1), (2) and (3) provide, by Remark 8.8.19(i), a set of minimal generators of P : g1 = t24 − t71 , g2 = t33 − t52 , g3 = t1 t4 − t2 t3 .

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Finally, by Remark 8.8.19(ii), the Frobenius number of the numerical semigroup S = N{16, 27, 45, 56} is g(S) = 112 + 135 + 72 − (16 + 27 + 45 + 56) = 175 . Monomial curves in positive characteristic Let K be a ﬁeld and let P ⊂ S be the toric ideal of a monomial curve. In characteristic zero it is an open problem whether P is a set theoretic complete intersection. Several authors have studied this problem; see [135, 302, 319, 402]. The solution of the case n = 3 is treated in [54]. The next result gives a nice family of binomial set theoretic complete intersections. We show this result using a theorem of Katsabekis, Morales, and Thoma [270, Theorem 4.4(2)]. Proposition 8.8.21 [293] If K is a ﬁeld of positive characteristic and L is a graded lattice ideal of S of dimension 1, then L is a binomial set theoretic complete intersection. Proof. Let L be the homogeneous lattice of Zq such that L = I(L). Notice that L is a lattice of rank q − 1 because ht(L) = rank(L). Thus, there is an isomorphism of groups ψ : Zq /Ls → Z, where Ls is the saturation of L consisting of all a ∈ Zq such that da ∈ L for some 0 = d ∈ Z. For each 1 ≤ i ≤ q, we set ai = ψ(ei + Ls ), where ei is the ith unit vector in Zq . Following [270], the multiset A = {a1 , . . . , aq } is called the conﬁguration of vectors associated to L. Recall that q − 1 = rank(L). Hence, as L is homogeneous with respect to d = (d1 , . . . , dq ), there are positive integers ri and rk such that ri ei − rk ek ∈ L and ri di − rk dk = 0. Thus, ri ai = rk ak and ai has the same sign as ak . This means that a1 , . . . , aq are all positive or all negative. It follows that A is a full conﬁguration in the sense of [270, Deﬁnition 4.3]. Thus, I(L) is a binomial set theoretic complete intersection by [270, Theorem 4.4(2)] and its proof. 2 Corollary 8.8.22 [319] Let P ⊂ S be the toric ideal of a monomial curve. If char(K) > 0, then P is a binomial set theoretic complete intersection. Proof. By Lemma 8.8.1 P is a 1-dimensional graded lattice ideal. Thus, the result follows at once from Proposition 8.8.21. 2 Remark 8.8.23 Working over an algebraically closed ﬁeld K of characteristic zero Eliahou [136] proved that the toric ideal P of a monomial curve in AnK is the radical of an ideal generated by n binomials.

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363

Exercises 8.8.24 Let P be the toric ideal of K[xd1 , xd2 , xd3 ], where K is a ﬁeld and d1 , d2 , d3 are relatively prime positive integers. Prove that P is a complete intersection if and only if P is Gorenstein. 8.8.25 Let f1 = ta1 1 − ta2 2 ta3 3 , f2 = tb22 − tb11 tb33 , f3 = −tc11 tc22 + tc33 be a full set of critical binomials. If bi > 0, ci > 0 for all i, prove the equalities d1 = a2 b3 + a3 b2 , d2 = a1 b3 + a3 b1 , d3 = a1 b2 − a2 b1 . 8.8.26 (a) If d1 = 7, d2 = 8, d3 = 9, d4 = 17. Prove that f1 = t51 − t23 t4 , f2 = t22 − t1 t3 , f3 = t43 − t41 t2 , f4 = t4 − t2 t3 is a set of critical binomials that generate a height 3 prime ideal of type 2. (b) If d1 = 204, d2 = 855, d3 = 1216, d4 = 1260. Prove that 4 3 2 8 15 3 6 19 4 5 13 3 f1 = t47 1 − t2 t3 t4 , f 2 = t2 − t1 t4 , f 3 = t3 − t1 t2 , f 4 = t4 − t1 t3

is a set of critical binomials that generate a height 3 prime ideal of type 3. (c) If d1 = 9, d2 = 12, d3 = 18, d4 = 19. Prove that f1 = −f3 = t21 − t3 , f2 = t32 − t23 , f4 = t34 − t1 t2 t23 is a complete intersection prime ideal. 8.8.27 Let f1 = ta1 1 − ta2 2 ta3 3 , f2 = tb22 − tb11 tb33 , f3 = −tc11 tc22 + tc33 be a full set of critical binomials and I = (f1 , f2 , f3 ). If bi > 0, ci > 0 for all i and S = K[t1 , t2 , t3 ] has the grading induced by deg(ti ) = di , then the minimal resolution of S/I is given by 0 → S(−d2 b2 − d3 a3 ) ⊕ S(−d3 c3 − d2 a2 ) → S(−d1 a1 ) ⊕ S(−d2 b2 ) ⊕ S(−d3 c3 ) → S → S/I → 0 and max{d2 a2 + d3 c3 − (d1 + d2 + d3 ), d2 b2 + d3 a3 − (d1 + d2 + d3 )} is the Frobenius number of S = N{d1 , d2 , d3 }. Hint Use Theorem 8.8.5(b) and Proposition 8.7.7. 8.8.28 Let K be a ﬁeld, let P be the toric ideal of K[td1 , . . . , tdq ] and let d

I = ({ti j − tdj i | 1 ≤ i < j ≤ q}). If gcd(d1 , . . . , dq ) = 1 and Γ is the monomial curve in AqK parameterized by ti = xdi , then rad(I) = P and Γ = V (I).

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Chapter 8

8.8.29 Let K be an algebraically closed ﬁeld and let X be the curve in AqK given parametrically by ti = fi (x), where fi (x) ∈ K[x] \ K for i = 1, . . . , q. Use the extension theorem [99, Chapter 3] to prove that V (P ) = X, where P is the presentation ideal of K[f1 (x), . . . , fq (x)]. 8.8.30 Prove that part (c) of Theorem 8.8.5 is valid for arbitrary distinct positive integers d1 , d2 , d3 . 8.8.31 [216] If P is the toric ideal of K[xd1 , xd2 , xd3 ], prove that P is a set theoretic complete intersection. 8.8.32 A rational polyhedral cone C ⊂ Rn is unimodular if there exists Γ = {γ1 , . . . , γn } a Z-basis of Zn such that C = R+ B for some B ⊂ Γ. A lattice polytope P ⊂ Rn with vertex set A is smooth [61, p. 371] if the cone R+ (A − v) is unimodular for any v ∈ A, where A − v = {a − v| a ∈ A}. Let P ⊂ Rn be a lattice polytope of dimension n and let K[P] := K[{xa t| a ∈ Zn ∩ P}] ⊂ K[x1 , . . . , xn , t] be its polytopal subring, where K[x1 , . . . , xn , t] is a polynomial ring over a ﬁeld K. An open problem posed by Bøgvad is whether for a smooth normal polytope P the toric ideal of K[P] is generated by quadrics (see [61, 400]). If n = 1, prove that P is smooth, K[P] is normal, and deg(K[P]) = vol(P). Hint Use Theorem 9.3.5, Lemma 9.3.7, and Theorem 9.3.25. 8.8.33 If P = conv(v1 , v1 + (s − 1)) ⊂ R, with v1 ∈ N+ and s ≥ 3 an integer, prove that the toric ideal P of K[P] is the ideal I2 (M ) generated by the 2 × 2 minors of the following 2 × (s − 1) generic Hankel matrix t1 t2 · · · ts−1 . M= t2 t3 · · · ts Hint The ideal I2 (M ) is a prime ideal of height s − 2 [128, p. 612]. 8.8.34 If P = conv((−1, −1), (1, 0), (0, 1)) ⊂ R2 , prove that the polytope P is not smooth, K[P] is normal, and the toric ideal P of K[P] is generated by t1 t2 t3 − t34 . Then prove that the point [(0, 0, 1, 0)] is a singular point of the projective toric variety V (P ) ⊂ P3 deﬁned by P . Hint Use the Jacobian criterion of [210, 5.8, p. 37].

Chapter 9

Monomial Subrings This chapter deals with point conﬁgurations and their lattice polytopes. We consider various subrings and ideals associated with them, e.g., Ehrhart rings, Rees algebras, homogeneous subrings, lattice and toric ideals, matrix and Laplacian ideals of graphs. Throughout we shall use the following symbology and terminology which is consistent with the notation introduced thus far: K; R A; F If A ⊂ Nn A P := conv(A) ⊂ Rn R[t] A(P) K[P] R[F t] K[F ] S PF ⊂ S; I(L) ⊂ S in≺ (PF ) K[F t] K[F t]

±1 ﬁeld; Laurent polynomial ring K[x±1 1 , . . . , xn ] n v A = {v1 , . . . , vq } ⊂ Z ; F = {x ∈ R | v ∈ A} set R = K[x] = K[x1 , . . . , xn ], I = (F ) ⊂ R n × q matrix with column vectors v1 , . . . , vq lattice polytope the polynomial ring K[x1 , . . . , xn , t] Ehrhart ring K[{xa ti | a ∈ Zn ∩ iP, i ∈ N}] polytopal subring K[{xa t | a ∈ P ∩ Zn }] Rees algebra R[xv1 t, . . . , xvq t] monomial subring K[xv1 , . . . , xvq ] polynomial ring K[t] = K[t1 , . . . , tq ] toric ideal of K[F ]; lattice ideal of L ⊂ Zq initial ideal of PF with respect to ≺ monomial subring K[xv1 t, . . . , xvq t] integral closure of K[F t].

Following [146, 189, 383], we shall be interested in comparing these monomial subrings and their algebraic properties and invariants. One has the following inclusions: K[F t] ⊂ K[P] ⊂ A(P) ⊂ R[F t]

and

K[F t] ⊂ A(P).

Using Hilbert functions, polyhedral geometry and Gr¨ obner bases, we will study normalizations of monomial subrings as well as initial ideals, special

366

Chapter 9

generating sets, primary decompositions and multiplicities of lattice and toric ideals. The reciprocity law of Ehrhart for integral polytopes and the Danilov–Stanley formula for canonical modules of monomial subrings will be introduced here. Applications of these results will be given.

9.1

Integral closure of monomial subrings

Here we present a description of the integral closure of a monomial subring using polyhedral geometry and integer programming techniques. Normality of monomial subrings If A is an integral domain with ﬁeld of fractions KA , recall that the normalization or integral closure of A is the subring A consisting of all the elements of KA which are integral over A. Normal monomial subrings arise in the theory of toric varieties [100]. The next result is essentially shown in the book of Fulton [176, pp. 29–30 ]. Theorem 9.1.1 If A ⊂ Zn is a ﬁnite set of points and S = ZA ∩ R+ A, then the following hold: (a) K[S] := K[{xa | a ∈ S}] is normal, (b) K[F ] = K[S], where F = {xa | a ∈ A}. Proof. (a): By Theorem 1.1.29, there are non-zero vectors a1 , . . . , ap in Zn such that R+ A = Ha+1 ∩ · · · ∩ Ha+p . We set Si = ZA ∩ Ha+i . Since K[S] is equal to K[S1 ] ∩ · · · ∩ K[Sp ] and since the intersection of normal domains is a normal domain, it suﬃces to show that K[Si ] is normal for all i. If ±1 ZA ⊂ Hai , then Si = ZA Zr . Hence, K[Si ] K[Zr ] = K[x±1 1 , . . . , xr ] which is normal by Corollary 4.3.9. Thus, we may assume that ZA ⊂ Hai . Setting r = rank(ZA) and L = ZA ∩ Ha , one has rank(L) = r − 1. This follows by noticing that QL = QA ∩ Ha , and using the equality n = dimQ (QA + Ha ) = dimQ (QA) + dimQ (Ha ) − dimQ (QA ∩ Ha ) together with the fact that the ranks of ZA and L are equal to dimQ (QA) and dimQ (QL), respectively. The quotient group H = ZA/L is torsion-free of rank 1. Thus, H is a free abelian group of rank 1 and we can write H = Zα for some is 0 = α ∈ ZA such that α, ai > 0. As a consequence, we get that Si = L ⊕ Nα Zr−1 ⊕ N and K[Si ] K[Zr−1 ⊕ N]. Therefore ±1 Si is normal because K[Zr−1 ⊕ N] is equal to K[x±1 1 , . . . , xr−1 , xr ] and this ring is normal again by Corollary 4.3.9. (b): Since K[F ] ⊂ K[S], taking integral closures and using part (a) gives the inclusion K[F ] ⊂ K[S]. To show the reverse inclusion, note the equality ZA ∩ R+ A = ZA ∩ Q+ A (see Corollary 1.1.27). A straightforward calculation shows that xα is in the ﬁeld of fractions of K[F ] if α ∈ ZA, and xα is an integral element over K[F ] if α ∈ Q+ A. Hence K[S] ⊂ K[F ]. 2

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367

Theorem 9.1.2 Let A be a ﬁnite set of points in Zn . If R+ A is a pointed cone, then the integral closure of K[F ] in R = K[x±1 ] is equal to K[{xa | a ∈ Zn ∩ R+ A}]. Proof. We set B = K[{xa | a ∈ Zn ∩ R+ A}]. Let K[F ] be the integral closure of K[F ] in R. To show K[F ] ⊂ B, take z ∈ K[F ]. One can write z = d1 xγ1 + · · · + dr xγr , where di ∈ K \ {0} for all i and γ1 , . . . , γr distinct non-zero points in Zn . Write γi = ni βi , with ni equal to the gcd of the entries of γi (observe that β1 , . . . , βr are not necessarily distinct). Consider the cone C spanned by A and {β1 , . . . , βr }. It suﬃces to show that βi ∈ R+ A for all i. Assume on the contrary that C is not equal to R+ A. By Theorem 1.1.54 one may assume that R+ β1 is a face of C not contained in R+ A. Let Ha be a hyperplane so that Ha ∩ C = R+ β1 and C ⊂ Ha− . There is 1 ≤ ≤ r such that n = sup {nj | γj = nj β1 }. 1≤j≤r

Since z is integral over K[F ] it satisﬁes a monic polynomial f of degree m with coeﬃcients in K[F ]. The monomials that occur in the expansion of α γi1 mi1 · · · (xγit )mit , where f (z) as a sum of monomials t are of the form x (x ) mij > 0 for all j, m ≥ j=1 mij and α ∈ R+ A. To derive a contradiction we claim that the term xmγ occurs only once in the expansion of f (z) as a sum of monomials. Assume the equality (xγ )m = xα (xγi1 )mi1 · · · (xγit )mit , t where mij > 0 for all j, m ≥ j=1 mij , and α ∈ R+ A. As γ = n β1 , one has mγ , a = 0 and from this equality one rapidly derives α, a = βi1 , a = · · · = βit , a = 0. / R+ A. Thus α = 0 and Hence α, βii , . . . , βit ∈ Ha ∩ C = R+ β1 . Note β1 ∈ βij = β1 for all j. Therefore t = 1 and γ = γi1 , as claimed. Altogether we get f (z) = 0, which is impossible. Now we show the inclusion B ⊂ K[F ]. By Corollary 1.1.27 one has the equality Zn ∩ R+ A = Zn ∩ Q+ A. A straightforward calculation shows that xa is an integral element over K[F ] if a ∈ Q+ A ∩ Zn . Hence B ⊂ K[F ]. 2 Corollary 9.1.3 (a) K[F ] is normal if and only if NA = ZA ∩ R+ A. (b) K[F ] = K[xγ1 , . . . , xγr ] for some γ1 , . . . , γr in ZA ∩ Q+ A. (c) K[F ] = xβ1 K[F ] + · · · + xβm K[F ] for some xβ1 , . . . , xβm in K[F ]. (d) If A is t-unimodular, then K[F ] is normal.

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Proof. (a): It suﬃces to observe the equality K[F ] = K[{xα |α ∈ NA}] and to use the description of K[F ] given in Theorem 9.1.1(b). (b): It follows from Corollary 1.1.27, Lemma 1.3.2, and Theorem 9.1.1. (c): By part (b) K[F ] is generated, as a K[F ]-algebra, by a ﬁnite set of Laurent monomials. As K[F ] is integral over K[F ], by Corollary 2.4.4, we get that K[F ] is ﬁnite over K[F ] and the result follows readily. (d): It follows from Theorem 1.6.7 and part (a). 2 8∞ Deﬁnition 9.1.4 A decomposition K[F ] = i=0 K[F ]i of the K-vector space K[F ] is an admissible grading if K[F ] is a positively graded K-algebra with respect to this decomposition and each component K[F ]i has a ﬁnite K-basis consisting of monomials. Theorem 9.1.5 (Danilov, Stanley [65, Theorem 6.3.5]) Let A ⊂ Zn be a ﬁnite set of points and let ri(R+ A) be the relative interior of R+ A. If K[F ] is normal and R+ A is a pointed cone, then the canonical module ωK[F ] of K[F ], with respect to an admissible grading, can be expressed as ωK[F ] = ({xa | a ∈ NA ∩ ri(R+ A)}).

(9.1)

The formula above represents the canonical module of K[F ] as an ideal of K[F ] generated by Laurent monomials. For a comprehensive treatment of the Danilov–Stanley theorem, see [61, 65, 102]. Theorem 9.1.6 (Hochster [248], [65, Theorem 6.3.5]) If A ⊂ Zn is a ﬁnite set of points and K[F ] is normal, then K[F ] is Cohen–Macaulay. Lemma 9.1.7 Let A ⊂ Zn be a ﬁnite set. If K[F ] is a normal positively graded K-algebra and ωK[F ] is the canonical module of K[F ], then a(K[F ]) = −min{ i | (ωK[F ])i = 0}.

(9.2)

Proof. Let PF be the toric ideal of K[F ]. Recall that K[F ] is isomorphic, as a K-algebra, to K[t1 , . . . , tq ]/PF , where K[t1 , . . . , tq ] has the grading induced by deg(ti ) = deg(xvi ). By Theorem 9.1.6, the algebra K[F ] is Cohen–Macaulay. Therefore by Proposition 5.2.3 the formula follows. 2 Theorem 9.1.8 If A ⊂ Nn and K[h] := K[h1 , . . . , hd ] → K[F ] is a homogeneous Noether normalization of K[F ], then K[F ] is a free K[h]-module d whose generators have degree at most i=1 deg(hi ). Proof. Let R = ⊕i≥0 Ri be the standard grading of R and endow K[F ] with the grading K[F ]i = K[F ] ∩ Ri . The composite K[h] → K[F ] → K[F ] is a homogeneous Noether normalization. Since K[F ] is Cohen–Macaulay (see Theorem 9.1.6), by Proposition 3.1.27, we get a direct sum decomposition K[F ] = K[h]xβ1 ⊕ · · · ⊕ K[h]xβm .

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369

Therefore the Hilbert series of K[F ] can be expressed as !E d m m 7 F (K[F ], t) = F (K[h]xβi , t) = tdeg βi (1 − tdeg hi ). i=1

i=1

i=1

Applying Theorem 9.1.5 one derives that a(K[F ]) is negative. Using that a(K[F ]) is equal to the degree of F (K[F ], t)m as a rational function (see Proposition 5.2.3) one concludes deg xβi ≤ i=1 deg hi . 2 Applications of the Danilov–Stanley theorem Next we introduce some techniques to compute the canonical module and the a-invariant of a wide class of monomial subrings. Let A be a ﬁnite set of points in Zn . The dual cone of R+ A is the polyhedral cone given by (R+ A)∗ = {x | x, y ≥ 0; ∀ y ∈ R+ A}. A set H ⊂ Zn \ {0} is called an integral basis of (R+ A)∗ if (R+ A)∗ = R+ H. Theorem 9.1.9 Let c1 , . . . , cr be an integral basis of (R+ A)∗ and let b = (bi ) be the vector given by bi = 0 if R+ A ⊂ Hci and bi = −1 if R+ A ⊂ Hci . Assume there is x0 ∈ Qn such that x0 , vi = 1 for all i. If NA = Zn ∩ R+ A and B is the matrix with column vectors −c1 , . . . , −cr , then (a) ωK[F ] = ({xa | a ∈ Zn ∩ {x|xB ≤ b}). (b) a(K[F ]) = −min {x0 , x| x ∈ Zn ∩ {x| xB ≤ b}}. Proof. K[F ] is a standard graded K-algebra with the grading induced by declaring that a monomial xa ∈ K[F ] has degree i if and only if a, x0 = i. We set H = {c1 , . . . , cr }. As R+ H = (R+ A)∗ = Hv+1 ∩ · · · ∩ Hv+q , by duality (see Corollary 1.1.30), we have the equality R+ A = Hc+1 ∩ · · · ∩ Hc+r .

(9.3)

Observe that R+ A ∩ Hci is a proper face if bi = −1 and it is an improper face otherwise. From Eq. (9.3) we get that each facet of R+ A has the form R+ A ∩ Hci for some i. The relative boundary of the cone R+ A is the union of its facets (see Theorem 1.1.44). Hence, using that H is an integral basis, we obtain the equality Zn ∩ (R+ A)o = Zn ∩ {x| xB ≤ b}.

(9.4)

Now, part (a) follows readily from Eq. (9.1) of Theorem 9.1.5 and Eq. (9.4). Part (b) follows from Eq. (9.2) of Lemma 9.1.7 and part (a). 2

370

Chapter 9

Deﬁnition 9.1.10 If xa is a monomial of K[x], we set log(xa ) = a. The vector a is called the exponent vector of xa . Given a set F of monomials, the log set of F , denoted log(F ), consists of all log(xa ) with xa ∈ F . Example 9.1.11 If F = {x1 , . . . , x4 , x1 x2 x5 , x2 x3 x5 , x3 x4 x5 , x1 x4 x5 } and A = log(F ), then A is a Hilbert basis and x0 , v = 1 for v ∈ A, where x0 = (1, 1, 1, 1, −1). An integral basis for (R+ A)∗ is given by {e1 , e2 , e3 , e4 , e5 , (0, 1, 0, 1, −1), (1, 0, 1, 0, −1)}. Then, using Theorem 9.1.9, it is easy to verify that ωK[F ] is generated by the set of all xa such that a = (ai ) satisﬁes the system of linear inequalities: ai ≥ 1 ∀ i; a1 + a3 − a5 ≥ 1; a2 + a4 − a5 ≥ 1.

(∗)

The only vertex of the polyhedron deﬁned by Eq. (∗) is v0 = (1, 1, 1, 1, 1). Thus, by Theorem 9.1.9, the a-invariant of K[F ] is −x0 , v0 = −3. Let R+ A = aﬀ(R+ A) ∩ Ha−1 ∩ · · · ∩ Ha−r be an irreducible representation of R+ A such that ai ∈ Zn and the non-zero entries of ai are relatively prime for all i. The existence of this representation follows from Theorem 1.1.29. Deﬁnition 9.1.12 If C is the integral matrix with rows a1 , . . . , ar , the polyhedron Q = aﬀ(R+ A) ∩ {x ∈ Rn | Cx ≤ −1} is called the shift polyhedron of R+ A relative to C. If Zn ∩ ri(R+ A) = ∅, then Q = ∅. The shift polyhedron will be used to give a technique to compute the canonical module of K[F ]. Example 9.1.13 A shift polyhedron of a cone is a shift of the cone as drawn below. 76

ppp p p pppp pppp p p pppp pp pp pp p ppp p p pp pp pp p p pp pp p 5 ppppppppp ppppp pp p pp p pp ppp ppp p pp pppp ppppp ppppp pppppp pppp pppp 4 p pppppp ppppp pp ppppp pp ppppppp p p pp pp pp pp p ppppp pppp pp p pppppp rpppppppp pppp 3 ppppp ppppp pppppp pp pp p p pp pp pp p ppppp p p ppppp pppppppp 2 pppppp pppp pp pp p pp pppp p 1 p p pp r 6

0

1

2

3

4

5

6

7

8

C=

1 2 2 3

R+ A = {x| Cx ≤ 0}

9 10

Here the shift polyhedron of the cone R+ A is Q = {x| Cx ≤ −1}.

Monomial Subrings

371

Proposition 9.1.14 If Q is the shift polyhedron of R+ A with respect to C and R+ A is a pointed cone, then Q is a pointed polyhedron, Zn ∩ Q = Zn ∩ ri(R+ A),

(9.5)

and conv(Z ∩ ri(R+ A)) is an integral polyhedron if Z ∩ ri(R+ A) = ∅. If Q is integral, then Q = conv(Zn ∩ ri(R+ A)). n

n

Proof. It is not hard to see that the lineality spaces of R+ A and Q are equal. As R+ A is pointed, so is Q by Exercise 1.1.80. To prove Eq. (9.5) it suﬃces to note that a point α is in ri(R+ A) if and only if α ∈ R+ A and α, ai < 0 for i = 1, . . . , r, see Theorem 1.1.44. The third assertion follows by taking convex hulls in Eq. (9.5) and observing that conv(Zn ∩ Q) is the integer hull of Q, which is a polyhedron by Proposition 1.1.65(a) and is integral by Proposition 1.1.65(b). The last assertion follows by taking convex hulls in Eq. (9.5) and using that Q is integral. 2 Theorem 9.1.15 Let K[F ] be a monomial subring such that (a) there is 0 = x0 ∈ Qn such that x0 , vi = 1 for all i, (b) NA = Zn ∩ R+ A, and (c) the shift polyhedron Q, relative to C, of the cone R+ A is integral. Then (i) K[F ] is a normal standard graded K-algebra whose canonical module and a-invariant are given by ωK[F ] = ({xa | a ∈ Zn ∩ Q}) and a(K[F ]) = −min {x0 , x| x ∈ Q}. (ii) xβ is a minimal generator of ωK[F ] for any vertex β of Q. Proof. (i): K[F ] is graded as follows: a monomial xa with a ∈ NA has degree i if and only if a, x0 = i. Using (a), it is seen that K[F ] is a standard graded K-algebra. That K[F ] is normal follows from (b). The canonical module is the ideal of K[F ] given by ωK[F ]

(9.1)

=

({xa | a ∈ NA ∩ ri(R+ A)})

(b)

({xa | a ∈ Zn ∩ ri(R+ A)})

(9.6)

(9.5)

({xa | a ∈ Zn ∩ Q}).

(9.7)

= =

To show the formula for the a-invariant observe that a monomial xa of K[F ] has degree x0 , a. Thus from Eq. (9.7) and Lemma 9.1.7 we obtain min {x0 , x| x ∈ Q} ≤ −a(K[F ]). As the minimum is attained at a vertex β of Q (see Proposition 1.1.41), to show equality, it suﬃces to observe that β is integral (see Corollary 1.1.64). (ii): By condition (c), β is integral. If F is a proper face of R+ A, then β ∈ F because β is in ri(R+ A) (see Theorem 1.1.44). Then, by Eq. (9.6), we get xβ ∈ ωK[F ] . There are c ∈ Qn and b ∈ Q such that

372

Chapter 9

(1) {β} = {x|x, c = b} ∩ Q and Q ⊂ {x|x, c ≤ b}. Then, by deﬁnition of Q, one has vi + β ∈ Q for all i. Thus vi + β, c = vi , c + b ≤ b ⇒ vi , c ≤ 0

(i = 1, . . . , q).

Assume there is α ∈ Q and η1 , . . . , ηq in N such that β = η1 v1 + · · · + ηq vq + α, then b = β, c = η1 v1 , c + · · · + ηq vq , c + α, c ≤ α, c ≤ b. Hence α, c = b and by (1) we get α = β. Thus xβ is a minimal generator of the canonical module ωK[F ] . 2

Exercises 9.1.16 If K[F ] = K[x3 , y 3 , x2 y], prove that K[F ] = K[F ][xy 2 ]. 9.1.17 Prove that the ring K[F ] is normal if and only if any element α ∈ ZA satisfying rα ∈ NA for some integer r ≥ 1 belongs to NA. 9.1.18 If S = R+ A ∩ Zn , then K[S] = K[{xα |α ∈ S}] is normal. 9.1.19 The shift polyhedron Q of R+ A, relative to C, can be written as Q = R+ A + P, i.e., R+ A is the characteristic cone of Q. 9.1.20 Let A ⊂ Nn be a ﬁnite set. If K[F ] is normal and F is the set of all xvi ∈ F such that x1 is not in the support of xvi , then K[F ] is normal. 9.1.21 If there is x0 ∈ Qn such that x0 , v = 1 for all v ∈ A, then R+ A is a pointed cone.

9.2

Homogeneous monomial subrings

The class of homogeneous monomial subrings will be studied here. Examples in this class include Rees algebras of ideals generated by monomials of the same degree and polytopal subrings. Deﬁnition 9.2.1 The subring K[F ] (resp. A) is said to be homogeneous if there is x0 ∈ Qn such that vi , x0 = 1 for i = 1, . . . , q. Proposition 9.2.2 K[F ] is homogeneous if and only if K[F ] is a standard graded algebra with the grading K(xv1 )a1 · · · (xvq )aq , where a = (a1 , . . . , aq ) ∈ Nq . K[F ]i = |a|=i

Monomial Subrings

373

Proof. ⇒) There is a vector x0 ∈ Qn such that vi , x0 = 1 for all i. One clearly has 8∞K[F ]i K[F ]j ⊂ K[F ]i+j for all i, j. Thus, it suﬃces to prove K[F ] = i=0 K[F ]i . First let us prove that K[F ]i ∩ K[F ]j = {0} for i = j. If this intersection is not zero, one has an equality (xv1 )a1 · · · (xvq )aq = (xv1 )b1 · · · (xvq )bq , with a = (a1 , . . . , aq ) ∈ Nq , b = (b1 , . . . , bq ) ∈ Nq , |a| = i, |b| = j. Hence i = |a| = a1 v1 + · · · + aq vq , x0 = b1 v1 + · · · + bq vq , x0 = |b| = j, a contradiction. To ﬁnish the proof assume fi1 + · · · + fir = 0, with fij in K[F ]ij for all j and i1 < · · · < ir . If fir = 0, using that the monomials of R form a K-basis, one has K[F ]ir ∩ K[F ]ij = {0} for some j < r, a contradiction. Thus fir = 0. By induction fij = 0 for all j, as required. ⇐) Set V = aﬀ Q (v1 , . . . , vq ) and r = dim(V ). To begin with we claim that 0 is not in V . Otherwise if 0 ∈ V , write 0 = λ1 v1 + · · · + λq vq with λi ∈ Q for all i and qi=1 λi = 1. There is 0 = p ∈ Z such that pλi ∈ Z for all i. One may assume, by reordering the vi , that pλi ≥ 0 for i ≤ s and pλi ≤ 0 for i > s. As K[F ] is graded, from the equality (xv1 )pλ1 · · · (xvs )pλs = (xvs+1 )−pλs+1 · · · (xvq )−pλq , q / V , and consequently one derives p i=1 λi = 0, a contradiction. Thus 0 ∈ r < n. It is well known that a linear variety V in Qn of dimension r can n−r be written as an intersection of hyperplanes V = i=1 H(yi , ci ), where n 0 = yi ∈ Q and ci ∈ Q for all i; see [427, Corollary 1.4.2]. Since 0 ∈ / V, 2 one has cj = 0 for some j. To ﬁnish the proof we set x0 = yj /cj . Corollary 9.2.3 PF is a graded ideal with respect to the standard grading of S if and only if K[F ] is homogeneous. Proof. As PF is generated by a ﬁnite set of binomials (Corollary 8.2.18), the result follows readily from Proposition 9.2.2. 2 Proposition 9.2.4 If ψ, ϕ are the maps of K-algebras deﬁned by ϕ

K[t1 , . . . , tq ] ψ

r K[F t]

ϕ

r

r

r

/ K[F ] r8

ti ψ

y xvi t

ϕ ϕ

y

y

/ xvi y<

then there is a unique epimorphism ϕ such ϕ = ϕψ. In addition ϕ is an isomorphism if and only if K[F ] and K[F t] have the same Krull dimension.

374

Chapter 9

Proof. To show the existence of ϕ it suﬃces to prove ker(ψ) ⊂ ker(ϕ). This inclusion follows at once because any binomial in ker(ψ) clearly belongs to ker(ϕ), and ker(ψ) is a binomial ideal. Note that ker(ψ) and ker(ϕ) are both prime ideals. The map ϕ is an isomorphism if and only if ker(ψ) = ker(ϕ); thus to ﬁnish the proof we need only observe that the last equality holds if and only if K[F ] and K[F t] have the same dimension. 2 Corollary 9.2.5 The map ϕ : K[F t] → K[F ] is an isomorphism if and only if K[F ] is homogeneous. Proof. ⇒) As ker(ϕ) = ker(ψ), by Corollary 9.2.3, K[F ] is homogeneous because ker(ψ) is homogeneous in the standard grading of S. ⇐) By Corollary 9.2.3, ker(ϕ) is homogeneous in the standard grading of S. Note that any homogeneous binomial in ker(ϕ) is also in ker(ψ), thus 2 one has ker(ϕ) = ker(ψ) and ϕ is an isomorphism. Remark 9.2.6 The subring K[F t] is always homogeneous because (vi , 1) lies in the hyperplane xn+1 = 1 for all i. If K[F ] is homogeneous, there is x0 ∈ Qn such that x0 , vi = 1 for all i. Then K[F ] = ⊕i∈N K[F ]i is a standard graded K-algebra with ith graded component deﬁned by K[F ]i = K{xa |xa ∈ K[F ]}. a,x0 =i

The isomorphism of K-algebras ϕ is degree preserving, that is, ϕ maps K[F t]i into K[F ]i for all i ∈ N. In particular e(K[F ]) = e(K[F t]). Lemma 9.2.7 There is n0 ∈ N+ such that (xα )n0 ∈ K[F ] for xα ∈ K[F ]. Proof. By Lemma 1.3.2 and Theorem 9.1.1, there are γ1 , . . . , γr in the semigroup ZA∩Q+ A such that K[F ] = K[xγ1 , . . . , xγr ]. For each i there is a positive integer ni such that xni γi ∈ K[F ]. The integer n0 = lcm(n1 , . . . , nr ) satisﬁes the required condition, because any xα ∈ K[F ] can be written as 2 xα = xa1 γ1 · · · xar γr , for some a1 , . . . , ar ∈ N. Lemma 9.2.8 There is xγ such that xγ K[F ] ⊂ K[F ]. Proof. By Lemma 1.3.2 and Theorem 9.1.1, there are γ1 , . . . , γr in the semigroup ZA ∩ Q+ A such that K[F ] = K[xγ1 , . . . , xγr ]. For each i, we can write xγi = xβi /xδi , where xβi and xδi are in K[F ]. By Lemma 9.2.7 there is 0 = n0 ∈ N such that xn0 α ∈ K[F ] for all xα ∈ K[F ]. We set xγ = xn0 δ1 · · · xn0 δr . Take xα in K[F ] and write xα = (xγ1 )a1 · · · (xγr )ar ,

Monomial Subrings

375

where ai ∈ N for all i. By the division algorithm, for each i there are qi , ci in N such that ai = qi n0 + ci and 0 ≤ ci < n0 . From the equality xγ xα =

r 7

r 7

(xn0 γi )qi

i=1

(xδi +γi )ci

i=1

r 7

(xδi )n0 −ci

i=1

is in K[F ] is in K[F ] is in K[F ] and using that δi + γi = βi for all i, we get xγ xα ∈ K[F ]. As xα was an arbitrary monomial in K[F ], we get xγ K[F ] ⊂ K[F ]. 2 The normalized Ehrhart ring of P = conv(A) is the graded algebra AP =

∞ ) (AP )i ⊂ R[t], i=0

where the ith component is given by (AP )i =

Kxα ti .

α∈ZA∩iP

To prove that AP is a graded algebra note that because of the convexity of the polytope P one has • (AP )i (AP )j ⊂ (AP )i+j and • (AP )i ∩ (AP )j = {0} for i = j. Proposition 9.2.9 K[F t] ⊂ AP , with equality if K[F ] is homogeneous. Proof. We set B = {(v1 , 1), . . . , (vq , 1)} ⊂ Zn+1 . First we show: ZB ∩ R+ B ⊂ {(α, i)| α ∈ ZA ∩ iP and i ∈ N}.

(∗)

Take z = (α, i) in ZB ∩ R+ B . There are ni ’s in Z and λi ’s in R+ such that z = n1 (v1 , 1) + · · · + nq (vq , 1) = λ1 (v1 , 1) + · · · + λq (vq , 1). Note α = 0 if i = 0, and α/i ∈ P if i ≥ 1. Hence z is in the right-hand side of Eq. (∗). Thus, by Theorem 9.1.1, we get K[F t] ⊂ AP . Assume that K[F ] is homogeneous. Let x0 be a vector in Qn such that vj , x0 = 1 for all j. By Theorem 9.1.1 it suﬃces to prove that equality holds in Eq. (∗). Take z = (α, i) in the right-hand side of Eq. (∗) and write α = n1 v1 + · · · + nq vq = i(λ1 v1 + · · · + λq vq ), where nj ∈ Z, λj ≥ 0, and j λj = 1. Hence α, x0 = j nj = i, and z = (α, i) = =

n1 (v1 , 1) + · · · + nq (vq , 1) iλ1 (v1 , 1) + · · · + iλq (vq , 1).

Thus z ∈ ZB ∩ R+ B , as required.

2

376

Chapter 9

Corollary 9.2.10 If K[F ] is homogeneous, then K[F ] is normal if and only if K[F t] = AP . Proof. It follows from Corollary 9.2.5 and Proposition 9.2.9.

2

The normalized Ehrhart function of P is deﬁned as E (i) = dimK (AP )i = |ZA ∩ iP|,

i ∈ N.

Proposition 9.2.11 If K[F ] is a homogeneous subring of dimension d, then E is a polynomial function of degree d − 1. Proof. As the normalized Ehrhart ring AP is equal to K[F t], thanks to Corollary 9.1.3 one has that AP is a ﬁnitely generated K-algebra. Hence AP is a ﬁnitely generated graded K[F t]-module. Using Theorem 2.2.4 and Proposition 2.4.13 one concludes that the 2 Hilbert function of AP is a polynomial function of degree d − 1. Example 9.2.12 Let F = {x41 , x31 x2 , x1 x32 , x42 } and let P be the convex hull of A = log(F ). As K[F ] is homogeneous, using Proposition 9.2.9 and Normaliz [68], we get that the normalized Ehrhart ring is AP = K[x41 t, x31 x2 t, x1 x32 t, x42 t, x21 x22 t]. The subrings K[F ] and AP have diﬀerent Hilbert functions, but they have the same Hilbert polynomial, which is equal to 4t + 1. Proposition 9.2.13 If K[F ] is homogeneous, then K[F ] is a positively graded K-algebra and its multiplicity is equal to the multiplicity of K[F ]. Proof. We set P = conv(A). As K[F ] is a homogeneous using 8subring, ∞ Proposition 9.2.9, we get K[F t] = AP . Hence K[F ] = i=0 K[F ]i is a graded K-algebra whose ith component is given by K[F ]i = Kxα . α∈ZA∩iP

Recall that K[F ] is graded as in Proposition 9.2.2. Let h and h be the Hilbert functions of K[F ] and K[F ], respectively. Since K[F ] and K[F ] have the same dimension d, one can write h(i) = a0 id−1 + terms of lower degree, h(i) = c0 id−1 + terms of lower degree, for i 0. One clearly has a0 ≤ c0 because K[F ]i ⊂ K[F ]i for all i. On the other hand by Lemma 9.2.8 there is xγ ∈ K[F ] of degree m such that xγ K[F ] ⊂ K[F ]. Thus for each i there is a one-to-one map xγ

K[F ]i −→ K[F ]i+m .

Monomial Subrings

377

Hence h(i) ≤ h(i + m) for all i, and consequently c0 ≤ a0 . Altogether a0 = c0 . Therefore e(K[F ]) = a0 (d − 1)! = c0 (d − 1)! = e(K[F ]).

2

Corollary 9.2.14 [72] If K[F ] is homogeneous and Cohen–Macaulay, then a(K[F ]) ≤ a(K[F ]). Proof. As K[F ] is normal, by Theorem 9.1.6, K[F ] is Cohen–Macaulay. Hence, the inequality follows from Propositions 5.2.8 and 9.2.13. 2 Deﬁnition 9.2.15 A binomial tα − tβ in K[t1 , . . . , tq ] is said to have a square-free term if at least one of its two terms tα , tβ is square-free. The following result gives a necessary condition for the normality of a homogeneous subring K[F ] in terms of its toric ideal. Theorem 9.2.16 [385] Let B be a ﬁnite set of binomials in the toric ideal PF of K[F ]. If K[F ] is a normal homogeneous subring and PF is minimally generated by B, then every element of B has a square-free term. Proof. Recall that the toric ideal of K[F ] is the kernel of the epimorphism ϕ : S = K[t1 , . . . , tq ] −→ K[F ]

ϕ

(ti −→ xvi )

of K algebras. Since K[F ] is homogeneous, PF is a graded ideal in the standard grading of S (see Corollary 9.2.3). We set B = {g1 , . . . , g } and fi = xvi . Let g be a binomial in B. We proceed by contradiction assuming that g has no square free-term. After permuting variables one can write a

r+1 g = ta1 1 · · · tar r − tr+1 · · · tas s ,

with ai ≥ 1 for all i and 2 ≤ a1 = max{ai }ri=1 ≤ as = max{ai }si=r+1 . ar+1 · · · fsas , we obtain f1 · · · fr /fs ∈ K[F ] From the equality f1a1 · · · frar = fr+1 because K[F ] is normal. Hence there exists a binomial h1 in PF of the form h 1 = t1 · · · tr − ts tα , with r = deg(h1 ) = deg(tα ) + 1 < deg(g). Note the equality g − ta1 1 −1 · · · tar r −1 h1 = ts h2 ,

(9.8)

where h2 is a binomial in PF with deg(h2 ) < deg(g) or h2 = 0. Writing hj = cij gi (j = 1, 2), gi =g

378

Chapter 9

and using Eq. (9.8) we conclude g = gi =g ci gi , ci ∈ S, a contradiction 2 because PF is minimally generated by g1 , . . . , g . A full converse to Proposition 9.2.16 is not true even if the monomials have the same positive total degree. Notice however that there is a partial converse due to Sturmfels (see Theorem 9.6.16). Example 9.2.17 Let F = {x1 x2 , x2 x3 , x3 x4 , x1 x4 , x21 , x22 , x23 , x24 } ⊂ K[x]. Using Macaulay2 [199], one obtains that PF is minimally generated by B = {t24 − t5 t8 , t23 − t7 t8 , t1 t3 − t2 t4 , t22 − t6 t7 , t21 − t5 t6 , t3 t4 t6 − t1 t2 t8 , t2 t3 t5 − t1 t4 t7 }. Using Normaliz [68], we obtain K[F ] = K[F ][x1 x3 , x2 x4 ]. The next result complements [400, Theorem 13.14]. Theorem 9.2.18 If K[F ] is normal and homogeneous, then the toric ideal PF is generated by homogeneous binomials of degree at most rank(A). Proof. Let B = {g1 , . . . , g } be a minimal generating set of PF consisting of binomials. Set r = maxi {deg(gi )}. Claim (I): If deg(gi ) = r for some i and r > rank(A), then gi is a linear combination of two binomials in PF of degree strictly less than r. By Theorem 9.2.16 we may assume that gi can be written as: a

r+1 · · · tamm with aj ≥ 1 for all j. gi = t1 · · · tr − tr+1

One has v1 + · · · + vr = ar+1 vr+1 + · · · + am vm . Applying Carath´eodory’s theorem (see Theorem 1.1.18) to both sides of this equality and permuting variables if necessary, we obtain an equation: b1 v1 + · · · + br1 vr1 = cr+1 vr+1 + · · · + cm1 vm1 with bj , ck ∈ N \ {0} for all j, k, r1 ≤ rank (A), m1 − r ≤ rank (A), and b c cr+1 · · · tmm11 belongs to PF . m1 ≤ m. Thus the binomial f = tb11 · · · trr11 − tr+1 Case (A): bi = 1 for all i. We can write gi − (tr1 +1 · · · tr )f = tr+1 h for some h. Hence gi is a linear combination of f and h, which in this case have degree less than r. Case (B): ci = 1 for all i. Notice that deg(f ) = m1 − r ≤ rank(A) < r. ar+1 · · · tamm = tγ (tr+1 · · · tm1 ). We can write gi − tγ f = t1 h Pick tγ so that tr+1 for some h. Thus gi is a linear combination of f and h, which in this case also have degree less than r. Case (C): maxi {ci } ≥ maxi {bi } ≥ 2. After permutation of variables we may assume cr+1 = maxi {ci } ≥ maxi {bi } = b1 ≥ 2. From the equality b

c

c

r+1 · · · fmm1 1 f1b1 · · · fr1r1 = fr+1

(fi = xvi ),

Monomial Subrings

379

we get f1 · · · fr1 /fr+1 ∈ K[F ] because K[F ] is normal. Hence there is h1 in PF of the form h1 = t1 · · · tr1 −tr+1 tα , with deg(h1 ) = deg(tα )+1 < deg(gi ). Note that r1 = deg(h1 ). We can write gi − tr1 +1 · · · tr h1 = tr+1 h2 , where h2 is a binomial in PF with deg(h2 ) < deg(g) or h2 = 0. Thus once again gi is a linear combination of binomials of degree less than r. The remaining cases follow by symmetry. This completes the proof of the claim. Applying Claim (I) to each binomial gi of degree r, it follows that PF is minimally generated by binomials of degree less than r. Thus, by induction, it follows that PF is generated by binomials of degree at most rank (A). 2

Exercises 9.2.19 If K[F ] is a homogeneous monomial subring over a ﬁeld K, use the Danilov–Stanley formula to show a(K[F ]) < 0. 9.2.20 Prove that the following conditions are equivalent: (a) K[F ] = K[xv1 , . . . , xvq ] is homogeneous. (b) 0 ∈ / aﬀ Q (v1 , . . . , vq ) ⊂ Qn . (c) dimQ Q{(v1 , 1), . . . , (vq , 1)} < dimQ Q{(v1 , 1), . . . , (vq , 1), (0, 1)}. (d) dimQ Q{(v1 , 1), . . . , (vq , 1)} = dimQ Q{v1 , . . . , vq }. 9.2.21 Is the integral closure of a homogeneous subring homogeneous? 9.2.22 Is the monomial subring K[x2 , xy, y 3 ] ⊂ K[x, y] homogeneous? 9.2.23 Let K[F ] be a normal homogeneous subring. If rank(A) = 2, then PF is generated by homogeneous binomials of degree at most 2. 9.2.24 Consider the monomial subring Q[F ] = Q[x31 x2 , x2 x33 , x21 x22 , x31 x3 ]. Prove that Q[F ] is equal to Q[F ][x21 x2 x3 , x1 x2 x23 ]. Prove that the toric ideal of Q[F ] is equal to (t51 t2 − t33 t34 ). 9.2.25 If K[F ] is a homogeneous and h (resp. E ) is the Hilbert function of K[F ] (resp. normalized Ehrhart function of P), then (a) h(i) ≤ E (i) for all i ≥ 0, and (b) h(i) = E (i) for all i ≥ 0 if and only if K[F ] is normal. 9.2.26 If K[F ] is a two-dimensional homogeneous monomial subring, prove that the Hilbert polynomials of K[F t] and K[F t] are equal. 9.2.27 Let K[F ] be a Cohen–Macaulay homogeneous monomial subring. If dim(K[F ]) = 2 and a(K[F ]) < 0, prove that K[F ] is normal.

380

9.3

Chapter 9

Ehrhart rings

In this section we introduce the Ehrhart ring and the polytopal subring of a lattice polytope. We study some of its properties. Let A = {v1 , . . . , vq } be a set of points in Zn and let P be the lattice polytope P = conv(A) ⊂ Rn . The set A is called a point conﬁguration. A point conﬁguration A is called normal if NA = ZA ∩ R+ A, and it is called homogeneous if A lies on an aﬃne hyperplane of Rn not containing the origin. In [106] by a “point conﬁguration” they mean a ﬁnite multiset of vectors in Rn . In [106] they also introduce “vector conﬁgurations” and “homogeneous point conﬁgurations.” To link the lattice polytope P and the point conﬁguration A to ring ±1 theory consider a Laurent polynomial ring R = K[x±1 1 , . . . , xn ] over a ﬁeld n K. If A ⊂ N , we set R = K[x1 , . . . , xn ]. There is an isomorphism between the group Zn and the multiplicative group of monomials of R given by a = (a1 , . . . , an ) ←→ xa = xa1 1 · · · xann . In this correspondence: A = {v1 , . . . , vq } ←→ F = {xv1 , . . . , xvq }. The monomial subring K[P] = K[{xa t| a ∈ P ∩ Zn }] ⊂ R[t] is called the polytopal subring of P. Notice the equality K[P] = K[{xa t| (a, 1) ∈ Q+ {(v1 , 1), . . . , (vq , 1)} ∩ Zn+1 }], which readily implies that K[F t] ⊂ K[P] is an integral extension. The polytope P is said to be normal if K[P] is normal [62] (cf. [100, 203]). Proposition 9.3.1 dim K[P] = dim(P) + 1. Proof. We set V = R(v2 − v1 ) + · · · + R(vq − v1 ). From the equality aﬀ(P) = aﬀ(v1 , . . . , vq ) = v1 + V, we get dim(P) = dimR (V ). Let B be the matrix with rows (v1 , 1), . . . , (vq , 1) and let B be the matrix with rows (v2 − v1 , 0), . . . , (vq − v1 , 0). Using Corollary 8.2.21, we obtain dim K[F t] = rank(B) = rank(B ) + 1 = dim(P) + 1. Since K[F t] ⊂ K[P] is an integral extension, by Proposition 2.4.13, we get that dim K[F t] is equal to dim K[P]. Lemma 9.3.2 Let A = {v1 , . . . , vd+1 } be an aﬃnely independent set in Zn . If P = conv(A) is a unimodular d-simplex, then P ∩ Zn = A and K[P] is a polynomial ring.

Monomial Subrings

381

Proof. By Corollary 1.2.22 we have P ∩ Zn = A. Hence K[P] is equal to K[xv1 t, . . . , xvd+1 t]. Thus K[P] has dimension d + 1 and by Lemma 3.1.5 K[P] is a polynomial ring. 2 Proposition 9.3.3 [62] Let {S, Si } be a family of ﬁnitely generated semigroups of Zn such that S = ∪i Si . If ZS = ZSi and K[Si ] is normal for all i, then K[S] is normal. Proof. By Corollary 9.1.3 it suﬃces to show that ZS ∩ R+ S is contained in S. Take a ∈ ZS ∩ R+ S. There is an integer s ≥ 1 such that sa ∈ S, thus sa ∈ Si for some i. As ZS = ZSi and K[Si ] is normal we get a ∈ Si ⊂ S, as required. 2 Deﬁnition 9.3.4 A collection {Δ } of unimodular lattice simplices in Rn is called a unimodular covering of P if P = ∪ Δ and dim(Δ ) = d for all , where d is the dimension of P. Theorem 9.3.5 [62] If P has a unimodular covering, then K[P] is normal. Proof. Let {Δ } be a unimodular covering of P. Consider the semigroups S and S of Zn+1 deﬁned as: S = N{(α, 1)| α ∈ Zn ∩ P}

and S = N{(α, 1)| α ∈ Zn ∩ Δ }.

By Lemma 9.3.2 K[S ] is a polynomial ring. In particular K[S ] is normal for all . Hence, by Proposition 9.3.3, we need only show that S = ∪ S and ZS = ZS . One clearly has the inclusion ∪ S ⊂ S, because P = ∪ Δ . To show the reverse inclusion take 0 = z ∈ S. Then z = (β, s) = n1 (β1 , 1) + · · · + nr (βr , 1),

where ni ∈ N and βi ∈ Zn ∩ P for all i. Note β/s ∈ P, where β = i ni βi and s = i ni . Hence β/s ∈ Δ for some . There is an aﬃnely independent d set Γ = {γ0 , . . . , γd } in Zn such that Δ = conv(Γ). Thus β/s = i=0 μi γi , for some μi ’s satisfying μi ≥ 0 and i μi = 1. Hence z = (β, s) = sμ0 (γ0 , 1) + · · · + sμd (γd , 1) ∈ R+ {(γ0 , 1), . . . , (γd , 1)}. As Δ is unimodular, by Corollary 1.1.27 and Proposition 1.2.21, one has z ∈ Z{(γ0 , 1), . . . , (γd , 1)} Hence using that K[S ] = K[xγ0 t, . . . , xγd t] is normal yields z ∈ S . To show the equality ZS = ZS note that ZS and ZS are free groups of rank d + 1. This follows from the assumption that P and Δ have the same dimension. Since Δ is unimodular, by Proposition 1.2.21, we get that Zn+1 /ZS is 2 torsion-free. Thus, ZS = ZS . Recall that the Ehrhart ring of P is: A(P) = K[xα ti |α ∈ Zn ∩ iP] ⊂ R[t].

382

Chapter 9

Theorem 9.3.6 The following assertions hold: (a) A(P) = K[{xα ti | (α, i) ∈ Zn+1 ∩ R+ B }], B = {(α, 1)| α ∈ Zn ∩ P}, (b) A(P) = K[{xα ti | (α, i) ∈ Zn+1 ∩ R+ B}], B = {(vi , 1)}ni=1 , (c) A(P) is a ﬁnitely generated K-algebra and a normal domain, (d) K[F t] ⊂ A(P) is a ﬁnitely generated integral extension, (e) K[F t] = A(P) ⇔ A(P) is contained in the ﬁeld of fractions of K[F t], (f) dim(K[F t]) = dim(K[F t]) = dim(A(P)) = dim(P) + 1, (g) K[F t] = A(P) if and only if B = {(vi , 1)}qi=1 is a Hilbert basis. Proof. (a): “⊂” Take a monomialxα ti in A(P), then α Zn ∩ iP and ∈ q q i ∈ N. Thus we can write (α, i) = j=1 iλj (vj , 1), with j=1 λj = 1 and λj ≥ 0 for all j. This proves that (α, i) is in the semigroup Zn+1 ∩ R+ B . “⊃” Take a monomial xα ti with (α, i) in Zn+1 ∩ R+ B .Any vector of the form (β, 1) with β ∈ Zn ∩ P can be written as (β, 1) = j μj (vj , 1) for some μj ’s in R+ . Hence we can write (α, i) = qj=1 ρj (vj , 1), with ρj ∈ R+ for all j. Consequently α ∈ Zn ∩ iP. Thus xα ti ∈ A(P). (b): This follows using part (a). (c): By Proposition 1.3.6 there is a ﬁnite Hilbert basis H ⊂ Zn such that R+ B ∩ Zn = NH and R+ B = R+ H. Thus, by Proposition 1.5.2, NH is normal. Using part (b) and Corollary 9.1.3, we get that A(P) is normal. (d): If xa ti is in A(P), then (xa ti )s ∈ K[F t] for some 0 = s ∈ N. Hence A(P) is integral over K[F t]. As a result, using (c) together with Corollary 2.4.4, we obtain that A(P) is a ﬁnitely generated K[F t]-module. (e): This follows directly from (c) and (d). (f): The ﬁrst two equalities follow from Proposition 2.4.13 because the extensions K[F t] ⊂ K[F t] ⊂ A(P) are integral. The third equality follows from Proposition 9.3.1. (g): This follows from part (a) because K[F t] = K[NB]. 2 Multiplicity of monomial subrings The subrings K[F t], K[F t] and A(P ) are graded as follows. We denote the multiplicity by e(·) or deg(·). Notice that K[F t] is homogeneous because (vi , 1) lies in the hyperplane xn+1 = 1 for all i. The rings K[F t] and K[F t] are graded K-algebras with ith graded components given by Kxa ti and K[F t]i = Kxa tb , (9.9) K[F t]i = (a,i)∈NB

(a,b)∈ZB∩iQ

respectively, where B = {(v1 , 1), . . . , (vq , 1)} and Q = conv(B).

Monomial Subrings

383

The Ehrhart ring A(P) is a normal ﬁnitely generated graded K-algebra with ith component given by Kxα ti . A(P)i = α∈Zn ∩iP

Since A(P) is a ﬁnitely generated module over the homogeneous ring K[F t], by Proposition 5.1.6, we obtain that its Hilbert function: χP (i) := |Zn ∩ iP| = cd id + · · · + c1 i + c0

(i

0)

is a polynomial function of degree d = dim(P) such that ci ∈ Q for all i and d!cd is an integer, which is the multiplicity of A(P). The function χP is the Ehrhart function of P. The Hilbert polynomial of A(P), denoted by EP (x), is called the Ehrhart polynomial of P. Recall that the relative volume of P is given by |Zn ∩ iP| , i→∞ id

vol(P) = lim

where d = dim(P); see Proposition 1.2.10. Hence vol(P) is equal to cd , the leading coeﬃcient of the Ehrhart function of P. For this reason d!cd is called the normalized volume of P. Altogether one has: Lemma 9.3.7 e(A(P)) = d!cd = d!vol(P). By the Hilbert–Serre theorem (Theorem 5.1.4) the Hilbert series of A(P) is a rational function that can be written uniquely as: F (A(P), x) :=

∞ i=0

|Zn ∩ iP|xi =

h0 + h1 x + · · · + hs xs , (1 − x)d+1

with h0 + h1 + · · · + hs = 0, hi ∈ Z for all i and d = dim(P). This series is called the Ehrhart series of P. Lemma 9.3.8 (a) a(A(P)) = s − (d + 1) < 0, (b) EP (i) = χP (i) for all integers i ≥ 0, (c) hi ≥ 0 for all i, (d) h0 + h1 + · · · + hs = d!vol(P). Proof. (a): This follows from Theorem 9.1.5. (b): This follows from Corollary 5.1.9. (c): By Theorem 9.3.6, A(P) is a normal ﬁnitely generated K-algebra. Hence, by Theorem 9.1.6, A(P) is Cohen–Macaulay. Therefore hi ≥ 0 for all i; see Exercise 5.2.9. (d): This follows from Remark 5.1.7. 2

384

Chapter 9

Proposition 9.3.9 The Ehrhart function of P can be expressed as:

EP (i) = e0

i+d d

− e1

i+d−1 i+1 d−1 ed−1 + · · · + (−1) + (−1)d ed , d−1 1

where ei = h(i) (1)/i! for all i and h(x) = h0 + h1 x + · · · + hs xs . Proof. This follows from Proposition 5.1.6 and its proof.

2

Deﬁnition 9.3.10 e0 , . . . , ed are called the Hilbert coeﬃcients. + Theorem 9.3.11 (Reciprocity law of Ehrhart [127]) If EP is the function + n EP (i) = |Z ∩ ri(iP)|, where ri(iP) is the relative interior of iP, then + EP (i) = (−1)d EP (−i)

∀ i ≥ 1.

Proof. By Exercise 5.2.10 and [394, Proposition 4.2.3], we have F (ωA(P ) , t) = (−1)d−1 F (A(P ), t−1 ) = (−1)d EP (−i)ti . i≥1

Therefore the required identity follows by comparing coeﬃcients.

2

Example 9.3.12 To illustrate the reciprocity law consider the polytope P: y

6

6 5 4 3 2 1 0

ppppppppppppppppppppppppppppppprpppppppppp pppppppppp ppp p p p p p p p ppp pp pp pJ p pppppppppppppp p p p p p p rpppppppppppppppppppppppppppppppppppppppppppppppppppppprpppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppprpppppppppppppppppppppppJ pp pp p pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp p p p p p pppppppppprpppppppppppp rppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppprpppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppprppppppppppppppppppppppppppppppppppppppppppppppppJ ppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppJ pp p ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppJ rpppppppppppppppppppppppppppppppppppprpppppppppppppppppppppppppppppppppppppprppppppppppppppppppppppppppppppppppprppppppppppppppppppppppppppppppppppppppp r p p p p p p p p p p p p p p p p p p p p p p p p p pppppppppppppp AAppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppprpppppppppppppppppppppppppppppppppppppprppppppp ppppp p p p p p p p p p p p p A ppppppppppppppppppppppppppppppppppppppp ppppppp ppppp p Apppp r 1 2 3 4 5

|Z2 ∩ P| = 16 |Z2 ∩ ∂P| = 6 EP (x) = 12x2 + 3x + 1 + EP (1) = (−1)2 EP (−1) = 10

6

x

Proposition 9.3.13 If EP (x) = cd xd + cd−1 xd−1 + · · · + c1 x + c0 is the Ehrhart polynomial and F1 , . . . , Fs are the facets of P, then cd = vol(P), s cd−1 = 12 i=1 vol(Fi ) and c0 = 1. Proof. By Lemma 9.3.7, cd = vol(P). Since EP (0) = χP (0) = 1, we get c0 = 1. Now we show the formula for cd−1 . Notice the equalities iP = ri(iP) ∪ ∂(iP);

∂(P) = F1 ∪ · · · ∪ Fs ,

Monomial Subrings

385

where ∂(P) is the relative boundary. Using the reciprocity law and the inclusion-exclusion principle [2, p. 38, Formula 2.12] we get |∂(iP) ∩ Zn | = EP (i) − E + (i) = EP (i) − (−1)d EP (−i) = 2cd−1 id−1 + terms of lower degree = EF1 (i) + · · · + EFs (i) + h(i), where h is a polynomial of degree at most d − 2. Comparing leading terms, the required formula follows. 2 In general some of the coeﬃcients c0 , . . . , cd of EP (x) may be negative; see [21, Example 3.22] and [241]. It is unknown whether the coeﬃcients are nonnegative if the vertices of P have {0, 1}-entries. Theorem 9.3.14 If P ⊂ R2 is a lattice polytope of dimension 2, then the Ehrhart polynomial is given by: EP (x) = area(P)x2 +

|Z2 ∩ ∂P| x + 1. 2

Proof. Let EP (x) = c2 x2 + c1 x + c0 be the Ehrhart polynomial of P. By Proposition 9.3.13, c2 = area(P) and c0 = 1. Writing P = ri(P) ∪ ∂(P), where ∂(P) is the boundary of P, we obtain by the reciprocity law: |∂(P) ∩ Z2 | =

Thus one has c1 =

|P ∩ Z2 | − |ri(P) ∩ Z2 |

=

+ EP (1) − EP (1) = EP (1) − EP (−1)

=

(c2 + c1 + c0 ) − (c2 − c1 + c0 ) = 2c1 .

|Z2 ∩∂(P)| , 2

2

as required.

Corollary 9.3.15 (Pick’s Formula) If P ⊂ R2 and dim(P) = 2, then |Z2 ∩ P| = area(P) +

|Z2 ∩ ∂P| + 1. 2

Proof. It follows making x = 1 in the formula of Theorem 9.3.14.

2

Lemma 9.3.16 If v1 , . . . , vq have {0, 1}-entries, then P ∩ Zn = A. Proof. Take α ∈ P ∩ Zn . Then we can write α = j λj vj with j λj = 1 and λj ≥ 0 for all j. We may assume λj > 0 for all j and v1 = 0. We set vj = (vj1 , . . . , vjn ). As the ith entry of v1 is 1 for some i, one has 1 ≤ λ1 v1i + λ2 v2i + · · · + λq vqi ≤ λ1 + · · · + λq = 1. Thus q = 1 and λ1 = 1, i.e., α = v1 and α ∈ A.

2

386

Chapter 9

Corollary 9.3.17 If v1 , . . . , vq have {0, 1}-entries, then EP (−1) = 0. + Proof. By Lemma 9.3.16 and Theorem 9.3.11, (−1)d EP (−1) = EP (1) = 0. 2 Thus EP (−1) = 0, as required.

Corollary 9.3.18 If xv1 , . . . , xvq are square-free, then K[F t] = K[P]. 2

Proof. It follows from Lemma 9.3.16.

Proposition 9.3.19 If A = {v1 , . . . , vq } ⊂ Zn and P = conv(A), then the multiplicities of A(P) and K[F t] are related by e(A(P)) = |T (Zn /(v2 − v1 , . . . , vq − v1 ))| e(K[F t]). Proof. We set B = {(v1 , 1)}qi=1 and Q = conv(B). Note that there is a bijective map Zn ∩ iP → Zn+1 ∩ iQ, α → (α, i), and d = dim(P) = dim(Q). Thus vol(P) = vol(Q). Since B lies in the aﬃne hyperplane xn+1 = 1, applying Theorem 1.2.14 gives vol(P) = |T (Zn+1 /((v2 − v1 , 0), . . . , (vq − v1 , 0)))| lim

i→∞

|ZB ∩ iQ| . (9.10) id

On the other hand, one has: |ZB ∩ iQ| . i→∞ id

e(K[F t]) = e(K[F t]) = d! lim

(9.11)

The ﬁrst equality follows from Proposition 9.2.13, while the second equality follows by deﬁnition of multiplicity noticing that K[F t] is graded as in Eq. (9.9) and using Theorem 9.3.6(f). Hence the asserted equality follows from Eq. (9.10), Eq. (9.11), and observing that the map T (Zn /(v2 − v1 , . . . , vq − v1 )) −→ T (Zn+1 /((v2 − v1 , 0), . . . , (vq − v1 , 0))) given by v −→ (v, 0) is bijective.

2

Lemma 9.3.20 Let P h ⊂ S[u] be the homogenization of P = PF . Then (a) P h is the toric ideal of K[F ] := K[z, xv2 −v1 z, . . . , xvq −v1 z, x−v1 z], where z is a new variable. (b) The toric ideal of K[xv1 z, . . . , xvq z, z] is the toric ideal P h of K[F ]. Proof. (a): The toric ideal of K[F ], denoted by P , is the kernel of the map ϕ : S[u] → K[F ] induced by ti → xvi −v1 z for i = 1, . . . , q + 1, where vq+1 = 0 and tq+1 = u. Let G be the reduced Gr¨ obner basis of P with respect to the GRevLex order ≺. First, we show the inclusion P h ⊂ P . Take an element f of G. By Proposition 3.4.2, it suﬃces to show that f h is

Monomial Subrings

387 +

−

+

in P . We can write f = ta − ta with in≺ (f ) = ta . Thus, |a+ | ≥ |a− | + − and f h = ta − ta u|a| , where a = a+ − a− . We set a = (a1 , . . . , aq ). Using 0 = a1 v1 + · · · + aq vq = a2 (v2 − v1 ) + · · · + aq (vq − v1 ) + |a|v1 , we get that f h ∈ P , as required. Let G be the reduced Gr¨ obner basis of P with respect to the GRevLex order. Next we show the inclusion P ⊂ P h . Take an element f of G . It suﬃces to show that f is in P h . As f is + − homogeneous in the standard grading of S[u], we can write f = tc −tc u|c| + with in≺ (f ) = tc and c = (c1 , . . . , cq ). Since f ∈ P , we get +

+

+

(z c1 )(xv2 −v1 z)c2 · · · (xvq −v1 z)cq −

−

−

= (z c1 )(xv2 −v1 z)c2 · · · (xvq −v1 z)cq (x−v1 z)|c|. Hence, c2 (v2 − v1 ) + · · · + cq (vq − v1 ) = −|c|v1 . Thus, c1 v1 + · · · + cq vq = 0, + − that is, the binomial f = tc − tc is in P . As f = f h , we get f ∈ P h . (b): The mapping that sends z to xv1 z induces an isomorphism of K2 algebras K[F ] → K[F ]. So this part is a consequence of (a). Recall that the multiplicity of an aﬃne algebra S/I is also denoted by deg(S/I) (see Deﬁnition 8.5.5). Thus, by Lemma 9.3.7, one can conveniently restate Proposition 9.3.19 as follows: Proposition 9.3.21 Let B = {β1 , . . . , βm } be a set of points of Zn and let Q = conv(B) be its convex hull. If r = dim(Q), then |T (Zn /β1 − βm , . . . , βm−1 − βm )| deg(K[xβ1 z, . . . , xβm z]) = r!vol(Q). Theorem 9.3.22 Let P be the toric ideal of K[F ] = K[xv1 , . . . , xvq ], let A be the matrix whose columns are v1 , . . . , vq and let r be the rank of A. Then |T (Zn /v1 , . . . , vq )| deg(S/P ) = r!vol(conv(v1 , . . . , vq , 0)). Proof. By Proposition 8.5.6, deg(S/P ) = deg(S[u]/P h ). On the other hand, by Lemma 9.3.20(b), P h is the toric ideal of the monomial subring: K[F ] = K[xv1 z, . . . , xvq z, z], where z is a new variable. Hence S[u]/P h K[F ]. Therefore, setting m = q + 1, βi = vi for i = 1, . . . , m − 1, βm = 0 and B = {β1 , . . . , βm }, the result follows readily from Proposition 9.3.19. 2 Corollary 9.3.23 If v1 , . . . , vq are positive integers and P is the toric ideal of K[xv1 , . . . , xvq ], then gcd(v1 , . . . , vq ) deg(S/P ) = max{v1 , . . . , vq }.

388

Chapter 9

Proof. We may assume that v1 ≤ · · · ≤ vq . The order of the group T (Z/Z{v1 , . . . , vq }) is equal to gcd(v1 , . . . , vq ). Then, by Theorem 9.3.22, we get that gcd(v1 , . . . , vq ) deg(S/P ) is vol([0, vq ]) = vq . 2 Example 9.3.24 Let P be the toric ideal of the monomial subring −1 −1 −1 −1 −1 −1 K[H] = K[x2 x−1 1 , x3 x2 , x4 x3 , x1 x4 , x5 x2 , x3 x5 , x4 x5 ].

We employ the notation of Theorem 9.3.22. The rank of A is 4 and the height of P is 3. Using Normaliz, we get 4!vol(conv(v1 , . . . , v7 , 0)) = 11. As the group Z5 /Z{v1 , . . . , v7 } is torsion- free, by Theorem 9.3.22, we have that deg(K[t1 , . . . , t7 ]/P ) = 11. Notation If B is an integral matrix, Δi (B) will denote the greatest common divisor of all the non-zero i × i minors of B. Theorem 9.3.25 [146] If A = {v1 , . . . , vq } ⊂ Zn and B is the matrix whose columns are the vectors in B = {(vi , 1)}qi=1 , then the following hold: (a) e(A(P)) = Δr (B)e(K[F t]), where r = rank(B). (b) Δr (B) = 1 if and only if K[F t] = A(P). (c) K[F t] = A(P) if and only if R+ B ∩ ZB = NB and T (Zn /ZB) = (0). Proof. (a): It follows at once by successively applying Proposition 9.3.19, Lemma 1.2.20 and Lemma 1.3.17. (b): ⇒) By Theorem 9.3.6, K[F t] ⊂ A(P) is an integral extension of rings and A(P) is normal. Thus K[F t] ⊂ A(P). For the other inclusion, by Theorem 9.3.6(e), it suﬃces to prove that A(P) is contained in the ﬁeld of fractions of K[F t]. Let xα ti ∈ A(P)i , that is, α ∈ Zn ∩ iP and i ∈ N. Hence the system of equations By = α has a rational solution, where α is the column vector (α, i) . Thus, the augmented matrix [B α ] has rank r. In general Δr ([B α ]) divides Δr (B), so in this case they are equal because Δr (B) = 1. By Theorem 1.6.3, the linear system By = α has an integral solution. Hence xα ti is in the ﬁeld of fractions of K[F t], as required. (b): ⇐) By hypothesis K[F t] = A(P). Hence, taking multiplicities and using Proposition 9.2.13 and part (a), we get Δr (B) = 1. (c): By Theorem 9.3.6(g), K[F t] = A(P) if and only if B = {(vi , 1)}qi=1 is a Hilbert basis. Hence this part follows readily from Corollary 1.3.21 because R+ B is a pointed cone (see Exercise 9.1.21). 2 Corollary 9.3.26 If Δr (B) = 1, then K[F t] is normal if and only if a(K[F t]) < 0 and the Hilbert polynomial of K[F t] is equal to EP (x).

Monomial Subrings

389

Proof. ⇒) It follows from Theorems 9.1.5 and 9.3.25. ⇐) Since the a-invariants of K[F t] and A(P) are both negative, then dimK K[F t]i is equal to dimK A(P)i for i ∈ N. Hence, noticing that K[F t]i ⊂ A(P)i , we get K[F t]i = A(P)i for i ∈ N. Therefore K[F t] = A(P) and consequently K[F t] is normal. 2 Deﬁnition 9.3.27 If K[F t] = A(P) and A is the incidence matrix of a clutter C, we say that C is an Ehrhart clutter. If K[F t] = A(P), we say that A is an Ehrhart point conﬁguration. There are many interesting families of Ehrhart point conﬁgurations that will be presented later; see Theorems 9.3.29, 9.3.31, and Section 14.8. Corollary 9.3.28 If the subgroup L = ({vi − vj | i, j = 1, . . . , q}) of Zn has rank n, then K[F t] = A(P) if and only if L = Zn . Proof. It follows from Theorem 9.3.25 and Lemma 1.2.20.

2

Theorem 9.3.29 If P has a unimodular covering with support in A, then K[F t] = A(P). Proof. Let Δ = conv(vi1 , . . . , vid+1 ) be any simplex in the unimodular covering of P, where d = dim(P). Note that Zn+1 /Z{(vij , 1)}d+1 j=1 is torsionfree by Proposition 1.2.21. Hence, by Lemma 1.3.17 and Theorem 9.3.25, we obtain the equality K[F t] = A(P). To ﬁnish the proof note that K[F t] is normal by Exercise 1.6.10. 2 Proposition 9.3.30 [146] If A is a unimodular matrix and the vectors v1 , . . . , vq lie on a hyperplane not containing the origin, then K[F t] = A(P). Proof. Let B be the matrix with column vectors (v1 , 1), . . . , (vq , 1). By Corollary 9.2.5, we have K[F ] K[F t]. Hence r = rank(A) = rank(B). Using Corollary 9.1.3(d), one obtains that K[F t] is normal. As Δr (A) = 1, applying Theorem 9.3.25 yields the asserted equality. 2 The following result is a useful generalization of [383, Theorem 7.1]. Theorem 9.3.31 [146] If the Rees algebra R[F t] is a normal domain and v1 , . . . , vq lie on a hyperplane b1 x1 + · · · + bn xn = 1 then K[F t] = A(P).

(bi > 0 ∀i),

390

Chapter 9

Proof. The inclusion K[F t] ⊂ A(P) is clear. To show the reverse inclusion take xα ti ∈ A(P). By Theorem 9.3.6(b), one can write (α, i) = λ1 (v1 , 1) + · · · + λq (vq , 1)

( λj ≥ 0 ∀j ).

(9.12)

Hence (α, i) ∈ Zn+1 ∩ R+ A = ZA ∩ R+ A, where A = {(e1 , 0), . . . , (en , 0), (v1 , 1), . . . , (vq , 1)} and ei is the ith unit vector in Rn . By the normality of R[F t] one has (α, i) ∈ NA . Thus one can write (α, i) = m1 (e1 , 0) + · · · + mn (en , 0) + n1 (v1 , 1) + · · · + nq (vq , 1),

(9.13)

where mi , nj are in N for all i, j. From Eqs. (9.12) and (9.13): α = =

m1 e1 + · · · + mn en + n1 v1 + · · · + nq vq λ1 v1 + · · · + λq vq ,

i

n1 + · · · + nq .

=

Taking the inner product of α with b = (b1 , . . . , bn ) yields α, b = m1 b1 + · · · + mn bn + i and α, b = λ1 v1 , b + · · · + λq vq , b = i. Hence mi = 0 for all i. Since xα ti = (xv1 t)n1 · · · (xvq t)nq , we get xα ti ∈ K[F t], as required. 2 Proposition 9.3.32 If xv1 , . . . , xvq are monomials of degree k ≥ 2 in the polynomial ring R = K[x1 , . . . , xn ] and R[F t] is normal, then e(R[F t]) = n!vol(P ), where P = conv(A ) and A = {(v1 , 1), . . . , (vq , 1), e1 , . . . , en }. Proof. The aﬃne hyperplane x1 + · · · + xn − (k − 1)xn+1 = 1 contains A . Then R[F t] is a standard graded K-algebra and dim(P ) = n. Thus, using Theorem 9.3.25, we get an isomorphism R[F t] A(P ), as graded K-algebras. By Lemma 9.3.7 the multiplicity of A(P ) is n!vol(P ). 2 Proposition 9.3.33 [385] If R[F t] is normal and there is x0 ∈ Nn+ such that x0 , vi = k, i = 1, . . . , q, for some k ≥ 2, then the following hold: (a) The torsion subgroup of Zn /ZA is cyclic and its order divides k. (b) If Zn /ZA is a ﬁnite group and x0 = (1, . . . , 1), then Zn /ZA Zk .

Monomial Subrings

391

Proof. We set B = {(v1 , 1)}qi=1 . There is an exact sequence of ﬁnite groups ϕ

ψ

0 −→ T (Zn+1 /ZB) −→ T (Zn /ZA) −→ Zk ,

(∗)

where ϕ and ψ are given by ϕ((α, b)) = α and ψ(α) = α, x0 , α ∈ Zn , b ∈ Z. Consider the matrix B whose columns are the vectors in the set B. By Theorems 9.3.31 and 9.3.25 we get Δr (B) = 1, where r is the rank of ZB. Since Δr (B) is the order of T (Zn+1 /ZB), part (a) follows using Eq. (∗). To prove (b), note that Zn /ZA is a torsion group and the ith unit vector ei maps into the element 1 under the map ψ. Hence ψ is onto, and ψ gives the required isomorphism. 2

Exercises 9.3.34 Let P be a lattice polytope in Zn . Then K[P] = A(P) if and only if kP ∩ Zn = P ∩ Zn + · · · + P ∩ Zn (k-times) for any k ∈ N+ . If this holds P is said to satisfy the integer decomposition property (cf. Deﬁnition 14.6.3). 9.3.35 Let F = {xv1 , . . . , xvq } ⊂ K[x] be a set of square-free monomials of degree k. If P = conv(v1 , . . . , vq ), then K[P] K[F ] as graded K-algebras. 9.3.36 Let P = conv((0, 0), (0, 1), (1, 0), (1, 1)) be the unit square. In the ﬁgure below we show the integral points of 4P. Verify the following: |Z2 ∩ 4P| = 25

6 4 s

s

s

s

s

EP (x) = (x + 1)2

3 s

s

s

s

s

vol(P) = 1

2 s

s

s

s

s

A(P) = K[F t]

1 s

s

s

s

s

F (A(P), x) = (1 + x)/(1 − x)3

s

s 1

s 2

s 3

s 4

0

-

9.3.37 Let A be an integral matrix of size n × n with det(A) = 0 and let A = {v1 , . . . , vn } be the set of columns of A. Consider the nth unit cube Q = [0, 1]n and the parallelotope P = {λ1 v1 + · · · + λn vn | 0 ≤ λi ≤ 1}. If A is a Hilbert basis, prove that the Ehrhart polynomials of Q and P are equal to (x + 1)n . Prove that the normalized volume of P is equal to n!.

392

Chapter 9

9.3.38 If P = conv((0, 0, 0), (12, 1, 0), (0, 1, 1), (1, 0, 1)), use Normaliz [68] to show that the Ehrhart polynomial of P is EP (x) = 13/6x3 +x2 −1/6x+1. 9.3.39 Let R = K[x1 , . . . , xn ] be a polynomial ring. If m = (x1 , . . . , xn ), prove that e(K[md ]), the multiplicity of K[md ], is equal to dn−1 . 9.3.40 Let P be the convex hull of A = {(3, 1), (1, 3)} in R2 , let K be the real numbers ﬁeld and let F = {x31 x2 , x1 x32 } ⊂ K[x1 , x2 ]. Prove that A is a weakly unimodular covering of P and K[F t] = A(P). 9.3.41 If K[F ] is homogeneous and the lattice polytope P has a weakly unimodular covering with support in A, then K[F ] is normal. 9.3.42 If F = {xv1 , . . . , xvq } is a set of monomials in R = K[x1 , . . . , xn ] and P is the toric ideal of R[F t], then deg(K[x1 , . . . , xn , t1 , . . . , tq ]/P ) = (n + 1)!vol(P ), where P = conv(A ∪ {0}) and A = {(v1 , 1), . . . , (vq , 1), e1 , . . . , en }. 9.3.43 If {(v1 , 1), . . . , (vq , 1)} ⊂ Nn+1 is a Hilbert basis with |vi | = d for all i and rank(v1 , . . . , vq ) = n, prove that Zn /Z{v1 , . . . , vq } Zd .

9.4

The degree of lattice and toric ideals

In this section we give formulas to compute the degree of a lattice ideal in terms of the torsion of certain factor groups of Zq and in terms of relative volumes of polytopes. Then we study primary decompositions of lattice ideals over an arbitrary ﬁeld. In this section we will use the Eisenbud– Sturmfels theory of binomial ideals over algebraically closed ﬁelds [134]. q+1 Let L ⊂ Zq be a lattice and let ei be the q ith unit vector in Z . For q a = (ai ) ∈ Z deﬁne the value of a as |a| = i=1 ai , and the homogenization of a with respect to eq+1 as ah = (a, 0) − |a|eq+1 if |a| ≥ 0 and ah = (−a)h if |a| < 0. The homogenization of L, denoted by Lh , is the lattice of Zq+1 generated by all ah such that a ∈ L. Lemma 9.4.1 Let I(L) ⊂ S be a lattice ideal. Then the following hold. (a) I(L)h ⊂ S[u] is a lattice ideal. (b) If L = Z{b1 , . . . , br }, then Lh = Z{bh1 , . . . , bhr }. (c) I(Lh ) = I(L)h . Proof. It is left as an exercise.

2

Monomial Subrings

393

Theorem 9.4.2 [294, 335] If L ⊂ Zq is a lattice of rank r and Iρ (L) is its lattice ideal with respect to a partial character ρ, then (a) If r < q, there is an integer matrix A of size (q − r) × q and rank q − r such that we have the containment of rank r lattices L ⊂ kerZ (A), with equality if and only if Zq /L is torsion-free. (b) If r < q and v1 , . . . , vq are the columns of A, then deg(S/Iρ (L))

=

|T (Zq /L)|(q − r)!vol(conv(0, v1 , . . . , vq )) . |T (Zq−r /v1 , . . . , vq )|

(c) If r = q, then deg(S/Iρ (L)) = |Zq /L|. Proof. (a): This follows at once from Lemma 8.2.5. (b): By Proposition 8.5.8, we may assume that K is algebraically closed of characteristic zero and that ρ(a) = 1 for a ∈ L, i.e., Iρ (L) = I(L). Let P be the toric ideal of K[xv1 , . . . , xvq ]. By Theorem 9.3.22, we need only show the equality deg(S/I(L)) = |T (Zq /L)| deg(S/P ). Let Ls be the saturation of L. By part (a), Ls is equal to kerZ (A). We set c = |T (Zq /L)|. Notice that T (Zq /L) = Ls /L. According to [134, Corollaries 2.2 and 2.5], there are distinct partial characters ρ1 , . . . , ρc of Ls , extending the trivial character ρ(a) = 1 for a in L, such that the minimal primary decomposition of I(L) is given by I(L) = Iρ1 (Ls ) ∩ · · · ∩ Iρc (Ls ), By part (a), P is a minimal prime of I(L). Thus we may assume ρ1 (a) = 1 for a in Ls , i.e., P is equal to the lattice ideal Iρ1 (Ls ). By additivity of the degree (see Proposition 8.5.9), we get deg(S/I(L)) = deg(S/Iρ1 (Ls )) + · · · + deg(S/Iρc (Ls )). Therefore, it suﬃces to recall that the degree is independent of the character, i.e., deg(S/P ) = deg(S/Iρk (Ls )) for k = 1, . . . , c (see Proposition 8.5.8). + − + − (c): Let tc1 − tc1 , . . . , tcm − tcm be a set of generators of I(L). By Lemma 8.2.23, L is generated by c1 , . . . , cm . We may assume that |ci | ≥ 0 for all i. Then, by Lemma 9.4.1(b), Lh is generated by ch1 , . . . , chm . Let C and C h be the matrices with rows c1 , . . . , cm and ch1 , . . . , chm , respectively. Notice that C h is obtained from C by adding the column vector given by b = (−|c1 |, . . . , −|cm |) . Since b is a linear combination of the columns of C, using the fundamental theorem of ﬁnitely generated abelian group (Theorem 1.3.16), we get that the groups Zq /L and Zq+1 /Lh have the

394

Chapter 9

same torsion. Thus, |Zq /L| is equal to |T (Zq+1 /Lh )|. By Proposition 8.5.6, deg(S/I(L)) = deg(S[u]/I(L)h ), and by Lemma 9.4.1(c), I(L)h = I(Lh ). Altogether it suﬃces to show that the degree of S[u]/I(Lh ) is equal to |T (Zq+1 /Lh )|. Since S[u]/I(Lh ) has dimension 1, we get q = rank(Lh ). By ) such that Lh ⊂ kerZ (A ). part (a), there is a row vector A = (v1 , . . . , vq+1 h Since the Q-vector spaces generated by L and kerZ (A ) are equal and have dimension q and since α, 1 = 0 for all α in Lh , it follows readily, using orthogonal complements, that v1 = vi for all i. Thus, by part (b), the required equality follows. 2 Corollary 9.4.3 If I(L) is a lattice ideal of dimension 1 which is homogeneous with respect to a positive vector d = (d1 , . . . , dq ), then gcd(d1 , . . . , dq ) deg(S/I(L)) = max{d1 , . . . , dq }|T (Zq /L)|. Proof. Let A be the 1 × q matrix (d1 , . . . , dq ). By hypothesis L ⊂ kerZ (A). The order of T (Z/Z{d1 , . . . , dq }) is equal to gcd(d1 , . . . , dq ). Therefore, by Theorem 9.4.2, we get deg(S/I(L)) =

|T (Zq /L)| max{d1 , . . . , dq } T (Zq /L)vol(conv(0, d1 , . . . , dq )) = , T (Z/Z{d1 , . . . , dq }) gcd(d1 , . . . , dq )

2

as required.

Corollary 9.4.4 Let L ⊂ Zq be a lattice of rank q − 1 and let α1 , . . . , αq−1 be a Z-basis of L. If αi = (αi,1 , . . . , αi,q ), for i = 1, . . . , q − 1, and ⎛

α1,1 ⎜ ni = (−1)i det ⎝ ... αq−1,1

... ...

α1,i−1 .. . αq−1,i−1

α1,i+1 .. . αq−1,i+1

... .. . ...

α1,q

⎞ ⎟ ⎠ for 1 ≤ i ≤ q,

αq−1,q

then deg(S/I(L)) = max{n1 , . . . , nq , 0} − min{n1 , . . . , nq , 0}. Proof. Let B be the (q − 1) × q matrix with rows α1 , . . . , αq−1 and let A be the 1 × q matrix (n1 , . . . , nq ). We set r = q − 1. The order of T (Zq /L) is equal to gcd(n1 , . . . , nq ), the gcd of the r × r minors of B. The order of T (Z/Z{n1 , . . . , nq }) is also equal to gcd(n1 , . . . , nq ). Since αi ∈ ker(A) for all i, we obtain that L ⊂ ker(A). Hence, by Theorem 9.4.2, we get that deg(S/I(L)) is equal to vol(conv(0, n1 , . . . , nq )) which is equal to max{n1 , . . . , nq , 0} − min{n1 , . . . , nq , 0}. 2 Normaliz [68] computes normalized volumes of lattice polytopes using polyhedral geometry. Thus we can compute the degree of any lattice ideal using Theorem 9.4.2.

Monomial Subrings

395

Example 9.4.5 Let K be the ﬁeld of rational numbers and let L be the lattice generated by a1 = (2, 1, 1, 1, −1, −1, −1, −2), a3 = (2, −1, 1, −2, 1, −1, 1, −1),

a2 = (1, 1, −1, −1, 1, 1, −1, −1), a4 = (5, −5, 0, 0, 0, 0, 0, 0).

We use the notation of Theorem 9.4.2. The lattice L has rank 4 and a1 , . . . , a4 , e1 , e3 , e4 , e5 form a Q-basis of Q8 . In this case we obtain the matrix ⎞ ⎛ 4 4 0 0 0 −1 1 6 ⎜ 0 0 1 0 0 1 0 0 ⎟ ⎟ A=⎜ ⎝ 0 0 0 4 0 7 9 −6 ⎠ 0 0 0 0 2 −3 −3 2 whose columns are denoted by v1 , . . . , v8 . Let P be the toric ideal of K[xv1 , . . . , xv8 ]. Therefore, by Theorem 9.4.2, we get deg(S/I(L)) =

(5)(200) 5(4)!vol(conv(0, v1 , . . . , v8 )) = = 125. |T (Z4 /v1 , . . . , v8 )| 8

The normalized volume of the polytope conv(0, v1 , . . . , v8 ) was computed using Normaliz [68]. Theorem 9.4.6 [335] Let I(L) be a lattice ideal of S over an arbitrary ﬁeld K of characteristic p, let c be the number of associated primes of I(L), and for p > 0, let G be the unique largest subgroup of T (Zq /L) whose order is relatively prime to p. Then (a) All associated primes of I(L) have height equal to rank(L). (b) |T (Zq /L)| ≥ c if p = 0 and |G| ≥ c if p > 0, with equality if K is algebraically closed. (c) deg(S/I(L)) ≥ |T (Zq /L)| if p = 0 and deg(S/I(L)) ≥ |G| if p > 0. Proof. Let K be the algebraic closure of K and let S = K[t1 , . . . , tq ] be the corresponding polynomial ring with coeﬃcients in K. Thus, we have an integral extension S ⊂ S of normal domains. We set I = I(L) and I = IS, where the latter is the extension of I to S. The ideal I is the lattice ideal of L in S (see Exercise 8.2.35(c)). Hence, as K is algebraically closed, by [134, Corollaries 2.2 and 2.5] I has a unique irredundant primary decomposition I = q1 ∩ · · · ∩ qc1 ,

(†)

where c1 = |T (Zq /L)| if p = 0 and c1 = |G| if p > 0. Notice that c1 = |T (Zq /L)| if p is relatively prime to |T (Zq /L)|. Furthermore, also by [134, Corollaries 2.2 and 2.5], one has that if pi = rad(qi ) for i = 1, . . . , c1 ,

396

Chapter 9

then p1 , . . . , pc1 are the associated primes of I and ht(pi ) = rank(L) for i = 1, . . . , c1 . Hence, by Exercise 8.2.35(b), one has a primary decomposition I = (q1 ∩ S) ∩ · · · ∩ (qc1 ∩ S)

(‡)

such that rad(qi ∩ S) = rad(qi ) ∩ S = pi ∩ S. We set pi = pi ∩ S and qi = qi ∩ S for i = 1, . . . , c1 . (a): Since S is a normal domain and S ⊂ S is an integral extension, we get ht(pi ) = ht(pi ) = rank(L) for all i (see Proposition 2.4.14). (b): By Eq. (‡), the associated primes of I are contained in {p1 , . . . , pc1 }. Thus, c1 ≥ c, which proves the ﬁrst part. Now, assume that K = K. By (a), we may assume that p1 , . . . , pc are the minimal primes of I. Consequently I has a unique minimal primary decomposition I = Q1 ∩ · · · ∩ Qc such that Qi is pi -primary and ht(pi ) = rank(L) for i = 1, . . . , c. As I = I, from Eq. (†), we get that c1 = c. (c): Using Lemma 8.5.7, Eq. (†) that was stated at the beginning of the proof, and the additivity of the degree (Proposition 8.5.9), we get c1 deg(S/qi ) ≥ c1 . 2 deg(S/I) = deg(S/I) = i=1

Exercises 9.4.7 Let L = (2, −1, −1), (−3, 1, −1). Prove that I(L) is a non-graded lattice ideal of height 2 given by I(L) = ((t21 − t2 t3 , t2 − t31 t3 ) : (t1 t2 t3 )∞ ) = (t21 − t2 t3 , t1 t23 − 1). Then, apply Corollary 9.4.4 with v1 = −2, v2 = −5 and v3 = 1, to show that K[t1 , t2 , t3 ]/I(L) has degree 6.

9.5

Laplacian matrices and ideals

In this section we use algebraic graph theory to study pure binomial matrices and their matrix ideals. We are interested in relating the combinatorics of a graph with the algebraic invariants and properties of the Laplacian ideal associated to the Laplacian matrix of the graph. If d = (d1 , . . . , dq ) ∈ Nq+ , then S = ⊕∞ k=0 Sk has a grading induced by setting deg(ti ) = di for i = 1, . . . , q. A graded ideal of S = K[t1 , . . . , tq ] is an ideal which is graded with respect to some vector d in Nq+ . Deﬁnition 9.5.1 A matrix ideal is an ideal of the form +

−

I(L) = (tai − tai | i = 1, . . . , m) ⊂ S, where the ai ’s are the columns of an integer matrix L of size q × m.

Monomial Subrings

397

Note that I(L) is graded if and only if dL = 0 for some d ∈ Nq+ . Example 9.5.2 If L is the matrix ⎛ ⎞ 2 −1 −1 1 −1 ⎠ , L=⎝ 0 0 −1 1 then I(L) = (t21 − 1, t2 − t1 t3 , t3 − t1 t2 ). Proposition 9.5.3 [335] Let I = I(L) be a graded matrix ideal and let L be the lattice spanned by the columns of L. Suppose that V (I, ti ) = {0} for all i. Then we have the following. (a) I = I(L) if I is unmixed, or else I = I(L) ∩ q, if I is not unmixed, where q is an m-primary component of I and m = (t1 , . . . , tq ). (b) (gcd{di }qi=1 ) deg(S/I) = maxi {di }|T (Zq /L)|. Proof. By Lemma 8.2.11, one has (I : (t1 · · · tq )∞ ) = I(L). By considering a primary decomposition of I, part (a) follows from Lemma 8.3.20 and Proposition 8.3.22. Part (b) follows from Corollary 9.4.3 and part (a). 2 Deﬁnition 9.5.4 Let ai,j ∈ N, i, j = 1, . . . , q, and let L be a q × q matrix of the following special form: ⎞ ⎛ a1,1 −a1,2 · · · −a1,q ⎜ −a2,1 a2,2 · · · −a2,q ⎟ ⎟ ⎜ L=⎜ .. .. .. ⎟ . ⎝ . . ··· . ⎠ −aq,1

−aq,2

···

aq,q

L is called a pure binomial matrix (PB matrix) if aj,j > 0 for all j, and for each row and each column of L at least one oﬀ-diagonal entry is non-zero. Deﬁnition 9.5.5 Let B = (bi,j ) be a q × q real matrix. The underlying digraph GB of B has vertex set {t1 , . . . , tq }, with an arc from ti to tj iﬀ bi,j = 0. If bi,i = 0, we put a loop at ti . A digraph is strongly connected if for any two vertices ti and tj there is a directed path form ti to tj and a direct path from tj to ti . Theorem 9.5.6 (Duality Theorem [335]) Let L be a PB matrix of size q × q such that Lc = 0 for some c ∈ Nq+ . If GL is strongly connected, then rank(L) = q − 1 and there is d in Nq+ such that dL = 0. Proof. Let L be a PB matrix as in Deﬁnition 9.5.4. First we treat the case c = 1 = (1, . . . , 1). Let L be a PB matrix as in Deﬁnition 9.5.4. We can write L = D − A, where D = diag(a1,1 , . . . , aq,q ) and A is the matrix whose

398

Chapter 9

i, j entry is ai,j if i = j and whose diagonal entries are equal to zero. We set δi = ai,i for i = 1, . . . , q. By hypothesis L1 = 0, hence rank(L) ≤ q − 1. There exists a non-zero vector d ∈ Zq such that dL = 0. Therefore dD = dA = dD(D−1 A). Since L1 = 0, we get that D1 = A1 or equivalently (D−1 A)1 = 1 . Thus, as the entries of D−1 A are nonnegative, the matrix B := D−1 A is stochastic. It is well-known that the spectral radius ρ(B) of a stochastic matrix B is equal to 1 [28, Theorem 5.3], where ρ(B) is the maximum of the moduli of the eigenvalues of B. As the diagonal entries of B are zero and δi > 0 for all i, the underlying digraph GB of B is equal to the digraph obtained from GL by removing all loops of G. Since GL is strongly connected so is GB , and by the Perron–Frobenius Theorem for nonnegative matrices [190, Theorem 8.8.1, p. 178], ρ(B) = 1 and 1 is a simple eigenvalue of B (i.e., the eigenspace of B relative to ρ(B) = 1 is 1-dimensional), and if z is an eigenvector for ρ(B) = 1, then no entries of z are zero and all have the same sign. Applying this to z = dD, we get that di = 0 for all i and all entries of d have the same sign. Hence, ker(L ) = (d ) for any non-zero vector d such that dL = 0 and L has rank q − 1. . = Ldiag(c1 , . . . , cq ), The general case follows by considering the matrix L . = 0 because Lc = 0. 2 where c = (c1 , . . . , cq ). Notice that L1 Lemma 9.5.7 Let L be an integer matrix of size q × q with rows 1 , . . . , q and let adj(L) = (Li,j ) be the adjoint matrix of L. Suppose Lc = 0 for some c in Nq+ . The following hold. (a) If rank(1 , . . . , 'i , . . . , q ) = q − 1 and Li,i ≥ 0 for all i, then dL = 0 for some d ∈ Nq+ . (b) If Li,i > 0 for all i, then dL = 0 for some d ∈ Nq+ . (c) If L is a PB matrix and rank(1 , . . . , 'i , . . . , q ) = q − 1 for all i, then dL = 0 for some d ∈ Nq+ . Proof. (a): Let Li be the ith column of adj(L). Since 1 , . . . , 'i , . . . , q are linearly independent, we get Li = 0. The vector c generates kerQ (L) because L has rank q − 1 and Lc = 0. Then, because of the equality Ladj(L) = 0, we can write Li = μi c for some μi ∈ Q. Notice that μi > 0 because Li,i ≥ 0 and c ∈ Nq+ . Hence, all entries of adj(L) are positive integers. If d is any row of adj(L), we get dL = 0 because adj(L)L = 0. (b): For any i, the vectors are 1 , . . . , 'i , . . . , q are linearly independent because Li,i > 0. Thus this part follows from (a). (c): Let L be as in Deﬁnition 9.5.4. (c1 ): First we treat the case c = 1. By part (a) it suﬃces to show that Li,i ≥ 0 for all i. Let Hi,i be the

Monomial Subrings

399

submatrix of L obtained by eliminating the ith row and ith column. By the Gershgorin Circle Theorem, every (possibly complex) eigenvalue λ of Hi,i lies within at least one of the discs {z ∈ C| z − aj,j ≤ rj }, j = i, where rj = u=i,j | − aj,u | ≤ aj,j since L1 = 0 and ai,j ≥ 0 for all i, j. If λ ∈ R, / R, then since Hi,i we get |λ − aj,j | ≤ aj,j , and consequently λ ≥ 0. If λ ∈ is a real matrix, its conjugate λ must also be an eigenvalue of Hi,i . Since det(Hi,i ) is the product of the q − 1 (possibly repeated) eigenvalues of Hi,i , we get Li,i ≥ 0. (c2 ): Now, we treat the general case. Let B be the q × q . = LB. diagonal matrix diag(c1 , . . . , cq ), where c = (c1 , . . . , cq ), and let L . is a PB matrix because L . = 0 because Lc = 0, and L Notice that L1 . i,j ) be the adjoint matrix of L. . Since ci > 0 for all is a PB matrix. Let (L i, by the multilinearity of the determinant, it follows that Li,j = 0 if and . is linearly independent. . i,j = 0. Hence, any set of q − 1 rows of L only if L . Therefore, applying case (c1 ) to L, we obtain that there is d ∈ Nq+ such . = 0. Then, dL = 0. that dL 2 Theorem 9.5.8 [335] Let L be a PB matrix of size q × q such that Lc = 0 for some c ∈ Nq+ . The following conditions are equivalent: (a) GL is strongly connected. (b) V (I(L), ti ) = {0} ∀ i, where V (I(L), ti ) is the zero set of (I(L), ti ). (c) Li,j > 0 ∀ i, j; adj(L) = (Li,j ) is the adjoint of L. Proof. (a) ⇒ (b): Let L be as in Deﬁnition 9.5.4. For 1 ≤ k ≤ q, let fk be the binomial deﬁned by the kth column of L. Set I = I(L) and ﬁx i such that 1 ≤ i ≤ q. Clearly {0} is contained in V (I, ti ) because I is graded by Theorem 9.5.6. To show the reverse containment, let α = (α1 , . . . , αq ) be a point in V (I, ti ). Consider the set E(ti ) of all arrows (ti , tk ) leaving ti . If (ti , tk ) ∈ E(ti ), i.e., ai,k = 0, we claim that αk = 0. We can write a

fk = tkk,k −

7

a

tj j,k .

j=k

; a a Using that fk (α) = 0, we get αkk,k = j=k αj j,k . Since ai,k = 0 and using that αi = 0, we obtain that αk = 0, as claimed. Let be an integer in [q] := {1, . . . , q}. Since the digraph GL is strongly connected, there is a directed path {v1 , . . . , vr } joining ti and t , i.e., v1 = ti , vr = t and (vj , vj+1 ) ∈ E(GL ) for all j. There is a permutation π of [q] such that π(1) = i, π(r) = and vj = tπ(j) for j = 1, . . . , r. Applying the claim successively for j = 2, . . . , r, we obtain απ(2) = 0, απ(3) = 0, . . . , απ(r) = 0. Thus, α = 0. This proves that α = 0. (b) ⇒ (a): We proceed by contradiction. Assume that GL is not strongly connected. Without loss of generality we may assume that there is no

400

Chapter 9

directed path from t1 to tq . Let W be the set of all vertices ti such that there is a directed path from t1 to ti , the vertex t1 being included in W . The set W is non-empty because GL has no sources or sinks by deﬁnition of a PB matrix, and the vertex tq is not in W . Consider the vector α ∈ K q deﬁned as αi = 1 if ti ∈ / W and αi = 0 if ti ∈ W . To derive a contradiction it suﬃces to show that all binomials;of I(L) vanish at α. Let L be as in a a Deﬁnition 9.5.4 and let fk = tkk,k − j=k tj j,k be the binomial deﬁned by the k-columns of L. If tk ∈ W , there is a directed path P from t1 to tk . Then tj is part of the path P for some j such that aj,k > 0 because the last arrow of the path P has the form (tj , tk ). Thus, since tj ∈ W , fk (α) = 0. If tk ∈ / W , then tj is not in W for any j such that aj,k > 0, because if aj,k > 0, the pair (tj , tk ) is an arc of GL . Thus, fk (α) = 0. (a) ⇒ (c): By the proof of Lemma 9.5.7(c), one has that Li,i ≥ 0 for all i. By Theorem 9.5.6 and using the proof of Lemma 9.5.7(a), we get that Li,j > 0 for all i, j. (c) ⇒ (a): We proceed by contradiction. Assume that GL is not strongly connected. We may assume that there is no directed path from t1 to tq . / W } be its complement. We Let W be as above and let W c = {ti | ti ∈ can write W c = {t1 , . . . , tr }. Consider the r × r submatrix B obtained from L by ﬁxing rows 1 , . . . , r and columns 1 , . . . , r . Notice that by the arguments above W c = ∪tk ∈W c (supp(fk )). Hence any column of B extends to a column of L by adding 0’s only, and consequently since dL = 0 for some d ∈ Ns , the rows of B are linearly dependent. Hence det(B) = 0. By permuting rows and columns, L can be brought to the form C 0 L = C C where C and C are square matrices of orders r and q − r, respectively, and det(C) = 0. Hence the adjoint of L has a zero entry, and so does the adjoint of L, a contradiction. 2 Laplacian matrices Let G be a connected graph, where V = {t1 , . . . , tq } is the set of vertices, E is the set of edges, and let w be a weight function w : E → N+ ,

e → we .

Let E(ti ) be the set of edges incident to ti . The Laplacian matrix L(G) of G is the q × q matrix whose (i, j)-entry L(G)i,j is given by ⎧ if i = j, ⎨ e∈E(ti ) we L(G)i,j := −we if i = j and e = {ti , tj } ∈ E, ⎩ 0 otherwise. Notice that L(G) is symmetric and 1L(G) = 0.

Monomial Subrings

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The notion of a Laplacian matrix can be extended to weighted digraphs; see [95] and the references there. Let G = (V, E, w) be a weighted digraph without loops and with vertices t1 , . . . , tq , let w(ti , tj ) be the weight of the directed arc from ti to tj and let A(G) be the adjacency matrix of G given by A(G)i,j = w(ti , tj ). The Laplacian matrix of G is given by L(G) = D+ (G) − A(G), where D+ (G) is the diagonal matrix with the outdegrees of the vertices of G in the diagonal entries. Note that L(G)1 = 0 and that the Laplacian matrix of a digraph may not be symmetric. If ti is a sink, i.e., there is no arc of the form (ti , tj ), then the ith row of L(G) is zero. Thus, the rank of L(G) may be much less than q − 1. Deﬁnition 9.5.9 The matrix ideal IG of L(G) is called the Laplacian ideal of G. If L is the lattice generated by the columns of L(G), the group K(G) := T (Zq /L) is called the critical group or the sandpile group of G. The structure, as a ﬁnite abelian group, of K(G) is only known for a few families of graphs [5, 295]. Theorem 9.5.10 (the Matrix-Tree Theorem [396, Theorem 9.8]) If G is regarded as a multigraph (where each edge e occurs we times), then the number of spanning trees of G is the (i, j)-entry of the adjoint matrix of L(G) for any (i, j). Remark 9.5.11 The order of T (Zq /L) is the gcd of all (q − 1)-minors of L(G). Thus the order of K(G) is the number of spanning trees of G. Proposition 9.5.12 [335] Let G be a connected graph, IG ⊂ S its Laplacian ideal, and L the lattice spanned by the columns of L(G). Then: (a) V (IG , ti ) = {0} for all i and rank(L(G)) = q − 1. (b) deg(S/IG ) = deg(S/I(L)) = |K(G)|. (c) If G = Kq is a complete graph, then deg(S/IG ) = q q−2 . (d) Hull(IG ) = I(L), where Hull(IG ) is the intersection of the isolated primary components of IG . Proof. The underlying digraph GL(G) of the Laplacian matrix L(G) is strongly connected because L(G) is symmetric and G is connected. Hence, by Theorems 9.5.6 and 9.5.8, (a) holds. Parts (b) and (d) follow from (a) and Proposition 9.5.3. Part (c) follows from (b) and the fact that a complete 2 graph with q vertices has q q−2 spanning trees [396, p. 143].

402

Chapter 9

Exercises 9.5.13 Let L be the following PB matrix ⎛ 4 −1 −1 ⎜ −1 4 −1 ⎜ 0 −1 2 L=⎜ ⎜ ⎝ −1 −1 −1 −1 0 0

and adj(L) its adjoint: ⎞ −1 −1 −1 −1 ⎟ ⎟ −1 0 ⎟ ⎟. 4 −1 ⎠ 0 1

Prove that V (I(L), ti ) = {0} for all i and that GL is strongly connected. 9.5.14 Let G be a connected graph and let IG ⊂ S be its Laplacian ideal. Prove that the following hold. (a) If degG (ti ) ≥ 3 for all i, then IG is not a lattice ideal. (b) If degG (ti ) ≥ 2 for all i, then IG is not a complete intersection. 9.5.15 Let G be the following weighted digraph and L its Laplacian matrix. Prove that the underlying digraph GL is not strongly connected, the matrix ideal I(L ) is graded but I(L) is not. 3 t1

t4

1 4 t2

1 1

1 t3

⎛

5 −4 0 ⎜ 0 1 −1 L = L(G) = ⎜ ⎝ 0 −1 1 −3 0 −1

⎞ −1 0 ⎟ ⎟ 0 ⎠ 4

9.5.16 Let G be the complete graph on 3 vertices. Prove that ⎞ ⎛ 2 −1 −1 2 −1 ⎠ , L(G) = ⎝ −1 −1 −1 2 IG = (t21 − t2 t3 , t22 − t1 t3 , t23 − t1 t2 ), V (IG , ti ) = {0} for all i, IG is not a complete intersection, is a lattice ideal, is not a prime ideal, has degree 3 and dimension 1. 9.5.17 Use the package “Binomials” [266] and the following procedure for Macaulay2 [199] to verify that over C the ideal (t21 − t2 t3 , t22 − t3 t1 , t23 − t24 ) is a complete intersection vanishing ideal and is not a lattice ideal. load "Binomial.m2" R=QQ[t1,t2,t3,t4] I=ideal(t1^2-t2*t3,t2^2-t3*t1,t3^2-t4^2) binomialAssociatedPrimes I saturate(I,t1*t2*t3*t4)==I

Monomial Subrings

9.6

403

Gr¨ obner bases and normal subrings

±1 Let R = K[x±1 1 , . . . , xn ] be a Laurent polynomial ring over a ﬁeld K and consider a ﬁnite set F = {xvi }qi=1 of distinct monomials with vi = 0 for all i. There is a homomorphism of K-algebras: ϕ

S = K[t1 , . . . , tq ] −→ K[F ]

ϕ

(ti −→ xvi ),

where S is a polynomial ring. The kernel of ϕ, denoted by P := PF , is the toric ideal of K[F ]. In what follows A will denote the n × q matrix with column vectors v1 , . . . , vq and A will denote the set {v1 , . . . , vq }. The main reference for this section is the book of Sturmfels on Gr¨ obner bases and convex polytopes [400]. For the rest of this section we assume that ≺ is a ﬁxed term order for the set of monomials of K[t1 , . . . , tq ]. We denote the initial ideal of P by in≺ (P ) or simply by in(P ). The Stanley–Reisner complex of the square-free monomial ideal rad (in≺ (P )) will be denoted by Δ. The following result was pointed out by Sturmfels. It can be shown using essentially “line shellings” of polytopes (as explained in [438, pp. 240–242]). Theorem 9.6.1 [106, Theorem 9.5.10] Δ is pure shellable. Corollary 9.6.2 (Sturmfels) rad(in≺ (P )) is a Cohen–Macaulay ideal. Proof. The simplicial complex Δ is pure shellable by Theorem 9.6.1. Hence Δ is Cohen–Macaulay by Theorem 6.3.23. 2 Lemma 9.6.3 If σ = {t1 , . . . , tr } is a maximal face of Δ, then v1 , . . . , vr are linearly independent. Proof. If v1 , . . . , vr are linearly dependent, then after permutation of the variables we can write λ1 v1 + · · · + λj vj = λj+1 vj+1 + · · · + λr vr , λi ∈ N λ λj+1 · · · tλr r , then in(f ) ∈ in(P ), a and λr = 0. Thus if f = tλ1 1 · · · tj j − tj+1 contradiction because rad(in(P )) ⊂ (tr+1 , . . . , tq ). 2 Deﬁnition 9.6.4 [400, p. 4] Let I be an ideal of S and let ω = (ω1 , . . . , ωq ) be a vector in Rq . If f = λ1 ta1 + · · · + λs tas , deﬁne inω (f ), the initial form of f relative to ω, as the sum of all terms λi tai such that ω, ai is maximal. The ideal generated by {inω (f )| f ∈ I} is denoted by inω (I). If ω ∈ Nq , then inω (f ) is the leading coeﬃcient in the x-variable of the ωq 1 univariate polynomial h(x) = f (xtω 1 , . . . , xtq ) ∈ S[x]. If ω ≥ 0, the following rule deﬁnes a term order ≺ω for S: ta ≺ω tb if and only if ω, a < ω, b, or ω, a = ω, b and ta ≺ tb .

404

Chapter 9

Proposition 9.6.5 [400, Proposition 1.8] Let I be an ideal of S. If ω ≥ 0, then in(inω (I)) = in≺ω (I). In addition if inω (I) is a monomial ideal, then inω (I) = in≺ω (I). Proposition 9.6.6 Let I be an ideal of S and let G be the reduced Gr¨ obner basis of I. If ω ∈ Rq+ , then in(I) = inω (I) if and only if in(g) = inω (g) for all g ∈ G. Proof. ⇒) Let g ∈ G. There are tas , . . . , tas distinct monomials in S such that g = λ1 ta1 + · · · + λs tas and inω (g) = λ1 ta1 + · · · + λr tar , where r ≤ s and 0 = λi ∈ k for all i. Since G is reduced there is j satisfying in(g) = taj ∈ in(I) and tai ∈ / in(I) for i = j. Using that in(I) is a monomial ideal we get tai ∈ in(I) for i = 1, . . . , r. Thus r = 1 and j = 1, that is, in(g) = inω (g). ⇐) First let us prove in(I) = in≺ω (I). Clearly in(I) ⊂ inω (I). Hence taking initial ideals with respect to ≺ in both sides and using the previous proposition we get in(I) ⊂ in≺ω (I). To show the reverse inclusion take ta ∈ in≺ω (I). Assume ta ∈ / I, otherwise ta ∈ in(I). Since the standard monomials with respect to ≺ω are a K-basis for S/I (see Proposition 3.3.13), / in≺ω (I). we have ta +I = (λ1 ta1 +I)+· · ·+(λs tas +I), λi ∈ K ∗ , where tai ∈ Therefore ta ∈ in(I), as required. Let inω (f ) with f ∈ I. There are tas , . . . , tas distinct monomials in S and non-zero scalars λ1 , . . . , λs such that f = λ1 ta1 +· · ·+λr tar +· · ·+λs tas , where inω (f ) = λ1 ta1 + · · · + λr tar and ta1 " · · · " tar . Note in≺ω (f ) = ta1 , hence ta1 ∈ in(I). There is g ∈ G such that λ1 ta1 = λ1 tδ in(g) = λ1 tδ inω (g) = λ1 inω (tδ g). Set h = f − λ1 tδ g. As inω (h) = λ2 ta2 + · · · + λr tar , we can repeat the same 2 argument to get λi tai ∈ in(I) for 2 ≤ i ≤ r, that is, inω (f ) ∈ in(I). The next result shows that in(P ) can be represented by a weight vector. Proposition 9.6.7 [400, Proposition 1.11] in(P ) = inω (P ) for some nonnegative integer weight vector ω ∈ Nq . A fundamental result linking Gr¨obner basis theory with “regular triangulations” of point conﬁgurations is the following result of Sturmfels. Theorem 9.6.8 [399, 400] If in(P ) = inω (P ), then Δ = {σ| ∃ c ∈ Rn such that vi , c = ωi if ti ∈ σ & vi , c < ωi if ti ∈ / σ}. In what follows ω denotes a nonnegative integer vector that represents the initial ideal of the toric ideal P , that is, in(P ) = inω (P ).

Monomial Subrings

405

Lemma 9.6.9 If σ = {t1 , . . . , tr } is a face of Δ, then in(P ) = inω (P ) = in≺ω (P ) / σ. for some ω = (ωi ) ∈ Rq such that ωi = 0 if ti ∈ σ and ωi > 0 if ti ∈ Proof. According to Theorem 9.6.8 there is c ∈ Rq such that vi , c = ωi if ti ∈ σ and vi , c < ωi if ti ∈ / σ. Let g = ta − tb be an element in the reduced Gr¨ obner basis of P with ta " tb . Note inω (g) = ta , otherwise if b inω (g) = t or inω (g) = g, the using in(P ) = inω (P ) we get that tb is not a standard monomial, a contradiction. Thus ω, a > ω, b. Consider the vector ω = (ωi ) − (vi , c). By Proposition 9.6.5 and Proposition 9.6.6 we need only show the inequality ω , a > ω , b. If a = (ai ) and b = (bi ), from a1 v1 + · · · + aq vq = b1 v1 + · · · + bq vq , we get a1 c, v1 + · · · + aq c, vq = b1 c, v1 + · · · + bq c, vq . Hence ω , a − ω , b = ω, a − ω, b > 0.

2

Lemma 9.6.10 If in(P ) is square-free and 0 = v ∈ NA, then there is σ a face of Δ such that v ∈ NA , where A = {vi | ti ∈ σ }. Proof. Let {tβ1 − tγ1 , . . . , tβp − tγp } be the reduced Gr¨ obner basis of the toric ideal P , where tβi " tγi for all i. There is λ = (λi ) in Nq such that v is equal to λ1 v1 + · · · + λq vq . Consider the monomial h0 = tλ and its support σ0 = supp(h). If σ0 ∈ Δ, set σ = σ0 . Assume σ0 ∈ / Δ, then h0 ∈ in(P ) and we may assume h0 = tδ1 tβ1 . Set h1 = tδ1 tγ1 and σ1 = supp(h1 ). Note f λ := f1λ1 · · · fqλq = f δ1 f β1 = f δ1 f γ1 . Thus one may assume σ1 ∈ / Δ, otherwise we set σ = σ1 . Since h0 " h1 and using that ≺ is Noetherian, the existence of σ follows by induction. 2 Lemma 9.6.11 Let A = {v1 , . . . , vr } and let π be the projection map π : Zq −→ Zq−r

(π(ai ) = (ar+1 , . . . , aq )).

If L = π(ker(A) ∩ Zq ), then Zq−r /L ZA/ZA . ψ

Proof. It suﬃces to note that the linear map Zq−r −→ ZA/ZA given by ψ(ar+1 , . . . , aq ) = ar+1 vr+1 + · · · + aq vq + ZA is an epimorphism whose kernel is equal to L. 2 For use below consider the epimorphism of K-algebras: φ

S = K[t1 , . . . , tq ] −→ S = K[tr+1 , . . . , tq ] induced by φ(ti ) = 1 for i ≤ r and φ(ti ) = ti otherwise. Note that φ is related to the map π by φ(ti ) = tπ(ei ) for all i.

406

Chapter 9

Lemma 9.6.12 Let I be a monomial ideal of S and let p = (tr+1 , . . . , tq ) be a minimal prime of I, then (a) Ass(S /I ) = {p }, where I = φ(I) and p = p ∩ S , (b) dim(S /I ) = 0, and (c) I = p , if I is generated by square-free monomials. Proof. (a): Let q be an associated prime of S /I and set q = q S. Note that I ⊂ I S and q is generated by variables in V = {tr+1 , . . . , tq }. Since I ⊂ q, there is a minimal prime qi of I contained in q, thus qi is generated by variables in V . Hence by the minimality of p we get q = p and q = p . (b): From (a) the radical of I is equal to the irrelevant maximal ideal of S , thus S /I is Artinian. Part (c) follows by noticing that I is also a square-free monomial ideal and applying (a). 2 Deﬁnition 9.6.13 Let I be a monomial ideal of S and let p be a minimal prime of I. The multiplicity of p in I is: mI (p) = Sp (Sp /Ip ). Proposition 9.6.14 Let I be a monomial ideal of S and p = (tr+1 , . . . , tq ) a minimal prime of I. If I = φ(I), then mI (p) = dimK (S /I ). Proof. Since Sp /Ip (S )p /(I )p , the result follows from Lemma 8.5.1 2 Proposition 9.6.15 If L is a lattice in Zq−r of rank q − r, then dimK (S /I(L)) = |Zq−r /L|. Proof. It follows at once from Theorem 9.4.2 (c) because dimK (S /I(L)) is equal to deg S /I(L). 2 For use below recall that the map σ → (σ c ) induces a bijection between the maximal faces of Δ and the minimal primes of rad(in(P )), where σ c is / σ} (see Proposition 6.3.4). equal to {ti | ti ∈ Theorem 9.6.16 [400] If the initial ideal in(P ) is generated by square-free monomials, then K[F ] is normal. Proof. Let α ∈ R+ A ∩ ZA. By Corollary 1.1.27, pα ∈ NA for some 0 = p ∈ N. By Lemma 9.6.10 there is σ , a maximal face of Δ, such that pα is in NA , where A = {vi | ti ∈ σ }. We may assume that σ = {t1 , . . . , tr }. Consider the prime ideal p of S generated by {tr+1 , . . . , tq }. As p is a minimal prime of the initial ideal in(P ), using Lemma 9.6.12(c), we get φ(in(P )) = p = (tr+1 , . . . , tq ) ⊂ S .

Monomial Subrings

407

By Lemma 9.6.9 we may assume that in(P ) = inω (P ) = in≺ω (P ) for some / σ . Therefore ω = (ωi ) ∈ Rq such that ωi = 0 if ti ∈ σ and ωi > 0 if ti ∈ p = φ(in(P )) = inω (φ(P )) = in≺ω (φ(P )).

(9.14)

Indeed the ﬁrst equality follows from Proposition 9.6.6 and noticing that inω (φ(P )) ⊂ p . The second equality follows at once from Proposition 9.6.5 because inω (φ(P )) is a monomial ideal by the ﬁrst equality. If ta − tb is a binomial in the reduced Gr¨ obner basis of φ(P ) with respect to ≺ω , then ta b or t is a standard monomial. Thus by Eq. (9.14) ta = 1 or tb = 1. Hence φ(P ) = (tr+1 − 1, . . . , tq − 1) ⊂ S . Let L be the lattice π(ker(A) ∩ Zq ). As a result S /φ(P ) is Artinian, L has rank q −r and dimK (S /φ(P )) = 1. Since φ(P ) = I(L), using Lemma 9.6.11 and Proposition 9.6.15 we conclude ZA = ZA . By Lemma 9.6.3 the set A is linearly independent, thus α ∈ NA, as required. This proof was adapted from [400]. 2 Proposition 9.6.17 Let I be an ideal of S. The following hold. (a) [336] If in≺ (I) is square-free, then rad(I) = I. (b) [142, Corollary 6.9] If I is graded and in≺ (I) is Cohen–Macaulay (resp. Gorenstein), then I is Cohen–Macaulay (resp. Gorenstein). Proposition 9.6.18 Let I = I(L) ⊂ S be a standard graded lattice ideal. If the initial ideal in≺ (I) is square-free, then I is a prime ideal and S/I is normal and Cohen–Macaulay. Proof. By Theorem 9.4.6 and Proposition 9.6.17 all associated prime ideals of I have height r = rank(L) and I is a radical ideal. Then I has an irredundant primary decomposition I = p1 ∩ · · · ∩ pm , where pi is a prime ideal of height r for all i. Let Ls be the saturation of L consisting of all a ∈ Zq such that pa ∈ L for some 0 = p ∈ N and let I(Ls ) be its lattice ideal. Since rank(L) is equal to rank(Ls ), by Theorem 8.2.2, we get that r is also the height of I(Ls ). As Zq /Ls is torsion-free, by Theorem 8.2.22, I(Ls ) is a prime toric ideal. Then we may assume that p1 = I(Ls ). We claim that in≺ (I) = in≺ (I(Ls )). Clearly in≺ (I) ⊂ in≺ (I(Ls )) because I ⊂ I(Ls ). To show the reverse inclusion take any element f in the reduced Gr¨ obner basis of I(Ls ). It suﬃces to show that in≺ (f ) ∈ in≺ (I). By Lemma 8.3.3, + − we can write f = ta − ta for some a = a+ − a− in Ls . We may assume + that in≺ (f ) = ta . There is p ∈ N+ such that pa ∈ L. The binomial + − + + g = tpa − tpa is in I = I(L) and in≺ (g) = tpa . Thus tpa ∈ in≺ (I) + and since this ideal is square-free we get that ta ∈ in≺ (I). This proves the claim. Hence deg(S/I) is deg(S/I(Ls )) because S/I and S/I(Ls ) have the

408

Chapter 9

same Hilbert function. Therefore, by additivity of the degree, we get that m = 1. Consequently, by Theorems 9.6.16 and 9.1.6, S/I is normal and Cohen–Macaulay. 2 Interesting families of graded ideals with square-free initial ideals are given in [196]. For these families the corresponding simplicial complexes associated to the initial ideals are vertex decomposable and hence shellable. Regular triangulations of cones Consider the primary decomposition of the radical of in(P ) as a ﬁnite intersection of face ideals: rad (in(P )) = p1 ∩ p2 ∩ · · · ∩ pr , see Theorem 6.1.4. Recall that, by Proposition 6.3.4, the facets of Δ are given by σi = {tj | tj ∈ / pi }, or equivalently by Ai = {vj | tj ∈ / pi } if one identiﬁes ti with vi . According to [400, Theorem 8.3], the family of cones (resp. family of simplices if A is homogeneous) {R+ A1 , . . . , R+ Ar }

(resp. {conv(A1 ), . . . , conv(Ar )})

is a regular triangulation of the cone R+ A (resp. the polytope conv(A)) in the sense of [400, pp. 63-64] (resp. in the sense of [438, pp. 129-130]). This means that R+ A1 , . . . , R+ Ar (resp. conv(A1 ), . . . , conv(Ar )) are obtained by projection onto the ﬁrst n coordinates of the lower facets of Q = R+ {(v1 , ω1 ), . . . , (vq , ωq )} (resp. conv((v1 , ω1 ), . . . , (vq , ωq ))), a facet of the cone Q is lower if it has a normal vector with negative last entry, i.e., a lower facet of Q has the form F = {x ∈ Q | α, x = 0},

α, x ≥ 0 valid for Q , αn+1 < 0,

see Example 9.6.19. Furthermore all regular triangulations of R+ A (resp. conv(A) if A is homogeneous) arise in this way, i.e., the regular triangulations of R+ A (resp. conv(A) if A is homogeneous) are in one to one correspondence with the radicals of the monomial initial ideals of the toric ideal P ; see [400, Theorem 8.3] and [106, Theorem 9.4.5]. The simplicial complex Δ and also its set of facets {A1 , . . . , Ar } is called a regular triangulation of the cone R+ A (resp. polytope conv(A) if A is homogeneous). For a thorough study of triangulations we refer to [106] and [438]. The regular triangulation {R+ A1 , . . . , R+ Ar } is called weakly unimodular (resp. unimodular) if ZAi = ZA for all i (resp. the simplex conv(Ai ) is unimodular for all i). If A is homogeneous, this regular triangulation is weakly unimodular if and only if the monomial ideal in(P ) is square-free; see [400, Corollary 8.9].

Monomial Subrings

409

Example 9.6.19 The facets of the cone generated by (1, 1, 0, 0, 1), (0, 1, 1, 0, 1), (0, 0, 1, 1, 1), (1, 0, 0, 1, 1), (1, 0, 0, 0, 0), (0, 1, 0, 0, 0), (0, 0, 1, 0, 0), (0, 0, 0, 1, 0). are given by the system of linear inequalities xi ≥ 0, i = 1, . . . , 5 x1 + x3 − x5 ≥ 0, x2 + x4 − x5 ≥ 0, Thus the lower facets are deﬁned by the last two inequalities. Example 9.6.20 Let R = Q[x1 , x2 , x3 , y1 , y2 , y3 ] be a polynomial ring with the default monomial order ≺ in Macaulay2 [199], let F = {f1 , . . . , f10 } be the set of monomials f1 = y1 y2 y3 , f2 = x1 x2 y1 , f3 = x2 x3 y1 , f4 = x1 x3 y1 , f5 = x1 x2 y2 f6 = x2 x3 y2 , f7 = x1 x3 y2 , f8 = x1 x2 y3 , f9 = x2 x3 y3 , f10 = x1 x3 y3 , and let A = {v1 , . . . , v10 } be the set of exponent vectors of f1 , . . . , f10 . The initial ideal of P is in(P ) = (t4 t5 , t3 t5 , t4 t6 , t4 t8 , t3 t8 , t4 t9 , t7 t8 , t6 t8 , t7 t9 ). The primary decomposition of the initial ideal in(P ) gives the following regular triangulation of conv(A): A1 = {v1 , v2 , v5 , v8 , v9 , v10 }, A2 = {v1 , v2 , v5 , v6 , v9 , v10 }, A3 = {v1 , v2 , v3 , v6 , v9 , v10 }, A4 = {v1 , v2 , v5 , v6 , v7 , v10 }, A5 = {v1 , v2 , v3 , v6 , v7 , v10 }, A6 = {v1 , v2 , v3 , v4 , v7 , v10 }. It is easy to verify that ZA = ZAi for all i, i.e., the regular triangulation is weakly unimodular. However notice that this triangulation is not unimodular in the sense that the simplex Δi = conv(Ai ) has normalized volume equal to 2 for all i, i.e., vol(Δi ) = 2/5! and dim(Δi ) = 5 for all i. The Ehrhart function of P = conv(A) is given by |Z6 ∩ iP| =

12 5 1 5 1 4 11 3 7 2 46 i + i + i + i + i+1= i + ··· 10 2 6 2 15 5!

Thus the normalized volume of P is equal to 12. The notion of weakly unimodular regular triangulation is related to TDI systems and to Hilbert bases [252]. Proposition 9.6.21 [252] Let A = {v1 , . . . , vq } and let A be the matrix with column vectors v1 , . . . , vq . If ZA = Zn , then the system xA ≤ ω is TDI if and only if Δ is a weakly unimodular regular triangulation of R+ A. Proof. It follows from Theorems 1.3.22 and 9.6.8, and Proposition 1.3.14. We leave the details as an exercise. 2

410

Chapter 9

Conjecture 9.6.22 [119] If A is the incidence matrix of a uniform clutter C that satisﬁes the max-ﬂow min-cut property, then the rational polyhedral cone R+ {v1 , . . . , vq } has a weakly unimodular regular triangulation. This conjecture holds for some interesting classes of uniform clutters coming from combinatorial optimization (see Theorem 14.4.17). Proposition 9.6.23 [400] If A is a t-unimodular matrix, then the regular triangulation {R+ A1 , . . . , R+ Ar } of R+ A is weakly unimodular. Proof. From Proposition 1.6.6, we get ZAi = ZA for all i, that is, the triangulation is weakly unimodular. 2

Exercises 9.6.24 Let A = {v1 , . . . , vq } ⊂ Nn be a homogeneous conﬁguration. Then F = conv(v1 , . . . , vs ) is a face of conv(A) if and only if F = R+ {v1 , . . . , vs } is a face of R+ A. 9.6.25 Let xA ≤ c be a rational system. Assume that Q = {x| xA ≤ c} is pointed and rank(A) < rank ((v1 , c1 ), . . . (vq , cq )), where the vi ’s are the columns of A. Prove that the map α → (α, −1) gives a bijection between the vertices of Q and the lower facets of the cone spanned by the (vi , ci )’s.

9.7

Toric ideals generated by circuits

In this section we give suﬃcient conditions for the normality of K[F ] and classify when the toric ideal of a homogeneous normal monomial subring is generated by circuits. Theorem 9.7.1 [49] If each circuit of ker(A) has a positive or negative part with entries consisting of 0’s and 1’s, then K[F ] is normal. Proof. The proof is by induction on q, the number of generators of K[F ]. The case q = 1 is clear. Assume q ≥ 2 and that the result holds for monomial subrings with less than q generators. To simplify the notation we set fi = xvi for i = 1, . . . , q. Using Theorem 9.1.1 one has: K[F ] = K[{xa | a ∈ ZA ∩ R+ A}]. Let z = xa ∈ K[F ] be a minimal generator of the integral closure of K[F ]. λ One can write z = xa = f1λ1 · · · fq q for some λ1 , . . . , λq ∈ Z. There is a positive integer m such that z m = f1mλ1 · · · fqmλq = f1b1 · · · fqbq ,

(9.15)

Monomial Subrings

411

for some b1 , . . . , bq ∈ N. First we assume that bi = λi = 0 for some i. Consider the matrix A obtained from A by removing its ith column, then Lemma 8.3.27 and the induction hypothesis yield z ∈ K[F \ {fi }]. Hence, we may assume that for each i either λi = 0 or bi > 0. If ker(A) = (0), the columns of A are linearly independent and K[F ] is normal because it is a polynomial ring. Assume ker(A) = (0) and take a circuit 0 = α ∈ ker(A), one can write α = (ei1 + · · · + eik ) − (1 ej1 + · · · + t ejt ),

α+

α−

where ei is the ith unit vector, i ∈ N for all i, and i1 , . . . , ik , j1 , . . . , jt distinct. Thus one has the equality k

vi =

=1

t

vj .

(9.16)

=1

If bir = 0 for some 1 ≤ r ≤ k, then using Eq. (9.16) one can rewrite mμ b Eq. (9.15) as z m = f1mμ1 · · · fq q = f1b1 · · · fq q , where μi ∈ Z for all i and μir = bir = 0, which by induction yields that z ∈ K[F \{fir }]. It remains to consider the case bir > 0 for all 1 ≤ r ≤ k. One may assume bi1 ≤ · · · ≤ bik . Using Eqs. (9.15) and (9.16) we obtain (9.15)

ma =

q i=1

bi vi = bi1

k =1

vi +

(9.16)

ci vi = bi1

i=i1

t

vj +

=1

ci vi , (9.17)

i=i1

where the ci ’s are nonnegative integers. Therefore, using Eq. (9.16) and mδ d (9.17), we can rewrite Eq. (9.15) as z m = f1mδ1 · · · fq q = f1d1 · · · fq q , where δi ∈ Z, di ∈ N for all i, and di1 = δi1 = 0. Hence by the induction hypothesis 2 one obtains z ∈ K[F \ {fi1 }]. A non-zero binomial ta − tb is said to have a square-free term if ta is square-free or tb is square-free. If ta and tb are both not square-free monomials we say that the binomial ta − tb has nonsquare-free terms. a

b

Deﬁnition 9.7.2 A binomial g = ta1 1 · · · tq q − tb11 · · · tqq is called balanced if the following holds: max{a1 , . . . , aq } = max{b1 , . . . , bq }. If g is not balanced it is called unbalanced. Let g be an unbalanced binomial of the form: b

r+1 · · · tbss , g = tb11 · · · tbrr − tr+1

bi ≥ 1 ∀i,

where 1 ≤ m1 = max{b1 , . . . , br } < max{br+1 , . . . , bs } = m2 . A connector cj+1 m · · · tcim , ci ≥ 1 ∀i, with a square-free of g is a binomial: ti1 · · · tij − tij+1 term ti1 · · · tij such that {i1 , . . . , ij } ⊂ {1, . . . , r} and the intersection of {ij+1 , . . . , im } with {r + 1, . . . , s} is non-empty.

412

Chapter 9

Theorem 9.7.3 [308] If K[F ] is homogeneous and normal and PF is its toric ideal, then the following are equivalent: (a) PF is generated by a ﬁnite set of circuits. (b) PF is generated by a ﬁnite set of circuits with a square-free term. (c) Every unbalanced circuit of PF has a connector which is a linear combination (with coeﬃcients in K[t]) of circuits of PF with a square-free term. Proof. As K[F ] is homogeneous, we get that any binomial ta − tb in PF is homogeneous with respect to the standard grading of K[t] = K[t1 , . . . , tq ] induced by setting deg(ti ) = 1 for all i, a fact that will be used repeatedly below without any further notice. (a) ⇒ (b): The toric ideal PF is minimally generated by a ﬁnite set B of circuits. Thus, by Theorem 9.2.16, each binomial of B is a circuit with a square-free term. br+1 · · · tbss , bi ≥ 1 ∀i, be an unbalanced (b) ⇒ (c): Let g = tb11 · · · tbrr − tr+1 circuit of A, where 1 ≤ m1 = max{b1 , . . . , br } < max{br+1 , . . . , bs } = m2 . We may assume m2 = br+1 . Then (xv1 · · · xvr /xvr+1 )m1 ∈ K[F ].

(9.18)

The element xv1 · · · xvr /xvr+1 is in the ﬁeld of fractions of K[F ] and by Eq. (9.18) it is integral over K[F ]. Hence, by the normality of K[F ], the element xv1 · · · xvr /xvr+1 is in K[F ]. Since K[F ] is generated as a K-vector space by Laurent monomials of the form xa , with a ∈ NA, it is not hard to see that there is a monomial tγ such that t1 · · · tr − tr+1 tγ ∈ PF . This is a connector of g and by hypothesis it is a linear combination of circuits of PF with a square-free term. (c) ⇒ (a): By Theorem 9.2.16, the toric ideal PF is minimally generated by a ﬁnite set B = {f1 , . . . , fm } consisting of binomials with a square-free term. We will show, by induction on the degree, that each one of the fi ’s is a linear combination of circuits. The degree is taken with respect to the standard grading of K[t]. ap+1 · · · ta , Let f be a binomial in B. We may assume that f = t1 · · · tp −tp+1 ai ≥ 1 ∀i, ≤ q. Assume that f is not a circuit. Then by Lemma 1.9.5 there is a circuit in PF (permuting variables if necessary) of the form b

p+1 · · · tbss , g = tb11 · · · tbrr − tp+1

bi ≥ 1 ∀i,

with r < p or s < . Set m1 = max{b1 , . . . , br } and m2 = max{bp+1 , . . . , bs }. We claim that there exist binomials h and h1 (we allow h = h1 or h = 0) in PF of degree less than deg(f ) = p and a binomial h2 which is a linear combination of circuits of PF such that f is in the ideal of K[t] generated by g, h, h1 , h2 . To prove this we consider the following two cases.

Monomial Subrings

413

Case (A): r = p and s < . Then b

p+1 · · · tbss . g = tb11 · · · tbpp − tp+1

(9.19)

Subcase (A1 ): bi = 1 for i = 1, . . . , p. Then we can write b

a

p+1 p+1 f − g = tp+1 · · · tbss − tp+1 · · · ta = tp+1 h,

for some binomial 0 = h ∈ PF (PF is prime) with deg(h) < deg(f ) = p. b Subcase (A2 ): bi = 1 for i = p+1, . . . , s. Then g = tb11 · · · tpp −tp+1 · · · ts . By subcase (A1 ), we may assume that bi ≥ 2 for some 1p ≤ i ≤ p. Then, on the one hand, by the homogeneity of g, p + 1 ≤ i=1 bi = s − p, so 2p+1 ≤ s. On the other hand, by the homogeneity of f , p ≥ −p ≥ s−p+1, so 2p − 1 ≥ s. This is a contradiction. So this case cannot occur. Subcase (A3 ): bi ≥ 2 for some 1 ≤ i ≤ p, bp+j ≥ 2 for some 1 ≤ j ≤ s−p, ap+1 b bp+1 · · · ta , g = tb11 · · · tpp − tp+1 · · · tbss . and m1 ≥ m2 . Then f = t1 · · · tp − tp+1 For simplicity of notation we may assume that m1 = b1 . Using that g ∈ PF and m1 ≥ m2 ≥ 2, we get (xvp+1 · · · xvs /xv1 )m2 ∈ K[F ]. Hence, by the normality of K[F ], there is tγ such that h1 = tp+1 · · · ts − t1 tγ is in PF . The binomial h1 is non-zero and has degree less than deg(f ) because s − p < − p ≤ p; the second inequality follows from the homogeneity of f . Let ap+1 · · · ta /tp+1 · · · ts . We have tδ = tp+1 f + h1 tδ = f + (tp+1 · · · ts − t1 tγ )tδ = t1 · · · tp − t1 tγ tδ = t1 h,

(9.20)

where 0 = h ∈ PF and deg(h) < p. Subcase (A4 ): bi ≥ 2 for some 1 ≤ i ≤ p, bp+j ≥ 2 for some 1 ≤ j ≤ s−p, and m1 < m2 . Since g is an unbalanced circuit, by hypothesis g has a connector h2 = ti1 · · · tik − tik+1 tγ , i1 < · · · < ik , with ik+1 ∈ {p + 1, . . . , s}, {i1 , . . . , ik } ⊂ {1, . . . , p} and such that h2 is a linear combination of circuits of PF . Set tδ = t1 · · · tp /ti1 · · · tik . If f = h2 tδ , then tδ = 1 and f = h2 . If f = h2 tδ , then we can write a

p+1 · · · ta = tik+1 h, f − h2 tδ = tik+1 tγ tδ − tp+1

(9.21)

with 0 = h ∈ PF and deg(h) < p. Case (B): r < p and s ≤ . In this case a

p+1 f = t1 · · · tp − tp+1 · · · ta ,

b

p+1 g = tb11 · · · tbrr − tp+1 · · · tbss .

Subcase (B1 ): bi = 1 for i = 1, . . . , r. Then b

a

p+1 p+1 · · · tbss tr+1 · · · tp − tp+1 · · · ta = tp+1 h, f − gtr+1 · · · tp = tp+1

(9.22)

414

Chapter 9

where 0 = h ∈ PF and deg(h) < p. ap+1 · · · ta /tp+1 · · · ts . Subcase (B2 ): bi = 1 for i = p+1, . . . , s. Let tγ = tp+1 Then we have f − gtγ = t1 · · · tp − tb11 · · · tbrr tγ = t1 h (9.23) where 0 = h ∈ PF and deg(h) < p. Subcase (B3 ): bi ≥ 2 for some 1 ≤ i ≤ r, bj ≥ 2 for some p + 1 ≤ j ≤ s, and m1 ≤ m2 . We may assume m2 = bp+1 . Using that g ∈ PF and that m2 ≥ m1 ≥ 2, we get (xv1 · · · xvr /xvp+1 )m1 ∈ K[F ]. Hence, by the normality of K[F ], there is tγ such that h1 = t1 · · · tr − tp+1 tγ is in PF . The binomial h1 is non-zero and has degree less than deg(f ) because r < p. Then we have f − h1 tr+1 · · · tp

=

f − (t1 · · · tr − tp+1 tγ )tr+1 · · · tp

=

tp+1 t tr+1 · · · tp − γ

ap+1 tp+1

· · · ta

(9.24)

= tp+1 h,

where 0 = h ∈ PF and deg(h) < p. Subcase (B4 ): bi ≥ 2 for some 1 ≤ i ≤ r, bj ≥ 2 for some p + 1 ≤ j ≤ s, and m1 > m2 . Since g is an unbalanced circuit, by hypothesis g has a connector h2 = tik+1 · · · tik+t − tid tγ , ik+1 < · · · < ik+t , with {ik+1 , · · · , ik+t } ⊂ {p + 1, . . . , s}, id ∈ {1, . . . , r}, and such that h2 is ap+1 · · · ta /tik+1 · · · tik+t . a linear combination of circuits of PF . Set tδ = tp+1 If f = −h2 tδ , then tδ = 1 and f = −h2 . If f = −h2 tδ , then we can write f + h2 tδ = t1 · · · tp − tid tγ tδ = tid h,

(9.25)

with 0 = h ∈ PF and deg(h) < p. This completes the proof of the claim. We are now ready to show that each fi in B is a linear combination (with coeﬃcients in K[t]) of circuits. We proceed by induction on deg(fi ). Let p = min{deg(fi )| 1 ≤ i ≤ m} be the initial degree of PF . If fi is a binomial in B of degree p, then either fi is a circuit or fi is not a circuit and by the claim fi is a linear combination of circuits (notice that in this case h = h1 = 0 because there are no non-zero binomials in PF of degree less than p). Let d be an integer greater than p and let fk be a binomial of B of degree d (if any). Assume that each fi of degree less than d is a linear combination of circuits. If fk is a circuit there is nothing to prove. If fk is not a circuit, then by the claim (or more precisely by Eqs. (9.19)–(9.25)) we can write (9.26) fk = λg + μh + μ1 h1 + μ2 h2 , where λ, μ, μ1 , μ2 are monomials, h, h1 are binomials in PF of degree less than d = deg(fk ), h2 is a linear combination of circuits, and g is a circuit. Since PF is a graded ideal with respect to the standard grading of K[t1 , . . . , tq ], we get that h and h1 are linear combinations of binomials in

Monomial Subrings

415

B of degree less than d. Therefore by Eq. (9.26) and the induction hypothesis, we conclude that fk is a linear combination of circuits. Therefore the ideal PF is generated by a ﬁnite set of circuits. 2 The following result will be used below to show a class of toric ideals generated by circuits. See Theorem 10.3.14 for an application of this result. Corollary 9.7.4 Let K[F ] be a homogeneous and normal subring. If each circuit of PF with nonsquare-free terms is balanced, then PF is generated by a ﬁnite set of circuits with a square-free term. Proof. The circuits of PF satisfy condition (c) of Theorem 9.7.3. Indeed, let f be an unbalanced circuit of PF . Then f has a square-free term by hypothesis. Thus f is a circuit with a square-free term and it is a connector of f . Hence the result follows from Theorem 9.7.3. 2 The best known examples of toric ideals generated by circuits come from conﬁgurations whose matrix A is t-unimodular (see Corollary 8.3.8). Other interesting examples of toric ideals generated by circuits are the phylogenetic ideals studied in [82]. As noted in [328], these phylogenetic ideals actually represent the family of cut ideals of cycles. Corollary 9.7.5 If K[F ] is a homogeneous subring such that each circuit of PF has a square-free term, then K[F ] is normal and PF is generated by a ﬁnite set of circuits with a square-free term. Proof. The normality of K[F ] follows from Theorem 9.7.1. Since the circuits of PF satisfy condition (c) of Theorem 9.7.3, we get that PF is generated by a ﬁnite set of circuits with a square-free term. 2

Exercises 9.7.6 Let F = {x1 x2 , x2 x3 , x3 x4 , x1 x4 , x1 x3 , x21 , x22 , x23 , x24 }. Use Macaulay2 and the procedure below to show that PF is minimally generated by t3 t4 − t5 t9 , t4 t5 − t3 t6 , t3 t5 − t4 t8 , t2 t5 − t1 t8 , t24 − t6 t9 , t25 − t6 t8 , t1 t5 − t2 t6 , t1 t2 − t5 t7 , t1 t3 − t2 t4 , t22 − t7 t8 , t23 − t8 t9 , t21 − t6 t7 . Prove that Q[F ] is not normal. Notice that PF is minimally generated by “circuits” with square-free part. KK=ZZ/31991 R=KK[x_1..x_4,t_1..t_9,MonomialOrder=>Eliminate 4] I=ideal(t_1-x_1*x_2,t_2-x_2*x_3,t_3-x_3*x_4,t_4-x_1*x_4, t_5-x_1*x_3,t_6-x_1^2,t_7-x_2^2,t_8-x_3^2,t_9-x_4^2) J= ideal selectInSubring(1,gens gb I) mingens J

416

Chapter 9

9.7.7 We set F = {xv1 , . . . , xv7 }, where the vi ’s are given by: v1 = (0, 1, 1, 0, 1, 0), v2 = (1, 0, 0, 1, 1, 0), v3 = (1, 0, 1, 0, 0, 1), v4 = (1, 0, 0, 1, 0, 1), v5 = (0, 1, 1, 0, 0, 1), v6 = (0, 1, 0, 1, 1, 0), v7 = (0, 1, 0, 1, 0, 1). Let A be the matrix with column vectors v1 , . . . , v7 . Prove that the circuits of ker(A) are given by the rows of the matrix: ⎡ ⎤ −1 1 0 −1 1 0 0 ⎢ −1 0 1 −1 0 1 0 ⎥ ⎢ ⎥ ⎢ −1 ⎥ 1 1 −2 0 0 1 ⎢ ⎥ ⎢ 0 −1 ⎥ 1 0 −1 1 0 ⎢ ⎥ ⎢ 1 −1 ⎥ 1 0 −2 0 1 ⎢ ⎥ ⎢ 1 ⎥ 1 −1 0 0 −2 1 ⎢ ⎥ ⎢ 0 ⎥ 1 0 −1 0 −1 1 ⎢ ⎥ ⎣ 1 0 0 0 −1 −1 1 ⎦ 0 0 −1 1 1 0 −1 Then show that K[F ] is a normal monomial subring using the criterion of Theorem 9.7.1. 9.7.8 In the proof of Theorem 9.7.3 (from (c) to (a)), show that the subcases (A3 ), (B1 ), and (B3 ) cannot occur.

9.8

Divisor class groups of semigroup rings

In this section we present an algorithm to compute the divisor class group of a Krull semigroup ring of the form K[NA], where K is a ﬁeld and NA is a subsemigroup of Nn generated by a ﬁnite set of vectors. Krull semigroup rings In this part the semigroups are commutative, satisfy the cancellation law, have a unit, and the total ring of quotients is torsion-free. The semigroups will be written additively. Let S be a semigroup and let ∼ be the equivalence relation on S × S given by: (a, b) ∼ (c, d) if and only if a + d = c + b. We denote each equivalence class [(a, b)] by a − b or simply by a if b = 0. Let S be the set of equivalence classes with the operation (a − b) + (c − d) = (a + c) − (b + d). It follows readily that S is a group which is called the total quotient group of S or the group of diﬀerences of S.

Monomial Subrings

417

Deﬁnition 9.8.1 S is called normal if s ∈ S and ns ∈ S for some n ≥ 1 implies s ∈ S. S is called completely normal if s ∈ S, a ∈ S and a+ ns ∈ S for all n ≥ 1 implies s ∈ S. If S = NA is a semigroup of Nn generated by a ﬁnite set A, then the group of diﬀerences of S is S = ZA and S is normal if and only if NA = ZA ∩ R+ A. Remark 9.8.2 If S is completely normal, then S is normal. The converse holds if S satisﬁes the ascending chain condition on ideals, thus in particular when S is ﬁnitely generated (see [83, 161]). Deﬁnition 9.8.3 A discrete valuation of an abelian group G is a homomorphism of groups ν : G → Z. The set {x ∈ G | ν(x) ≥ 0} is the valuation semigroup of ν. The residue group of ν is ker(ν). We say that ν is normalized if ν(G) = Z. Proposition 9.8.4 If ν = 0 is a valuation of a group G, then the valuation semigroup of ν is isomorphic to ker(ν) × N. Proof. There is 0 = n0 ∈ N such that ν(G) = n0 Z and we can write n0 = ν(x0 ) for some x0 ∈ G. The following mapping gives the required isomorphism ϕ : ker(ν) × N → {x ∈ G | ν(x) ≥ 0},

(x, n) → x + nx0 .

2

Deﬁnition 9.8.5 S is a Krull semigroup if there is a family (νi )i∈I of discrete valuations of S such that S is the intersection of the valuation semigroups of the νi ’s and for each s ∈ S the set {i ∈ I | νi (s) > 0} is ﬁnite. 8 Notation Let F = i∈I Zei be the free abelian group with basis {ei | i ∈ I}. For each i ∈ I let πi the projection πi : F → Z given by πi (ak ek )k∈I = ai . The positive part of F is: F+ = {x ∈ F | πi (x) ≥ 0 ∀ i ∈ I}. Notice that 8 F+ = Ne i and if S is a subsemigroup of F, then ZS (the subgroup i∈I generated by S) is equal to S (the group of diﬀerences of S). Proposition 9.8.6 [83] S is a Krull semigroup if and only if S ∼ =8G × S1 , where G is a group and S1 is a subsemigroup of a free group F = i∈I Zei such that S1 = S1 ∩ F+ . Corollary 9.8.7 If S is a semigroup of F such that S = S ∩ F+ , then S is a Krull semigroup. Proof. Notice that S {0} × S and apply Proposition 9.8.6.

2

Let S be a semigroup and let A be a ring. The semigroup ring of S with coeﬃcients in A is denoted by A[S]. The elements of A[S] can be written in

418

Chapter 9

the form s∈S rs X s , rs ∈ A, where rs = 0 except for a ﬁnite number, with addition and multiplication deﬁned as for polynomials. See [180, p. 64] for a formal construction of a semigroup ring. Theorem 9.8.8 [83] A[S] is a Krull ring if and only if A is a Krull ring and S is a Krull semigroup such that S satisﬁes the ascending chain condition on cyclic subgroups. Corollary 9.8.9 Let A ⊂ Nn be a ﬁnite set. Then K[NA] is a Krull ring if and only if NA is a Krull semigroup. The divisor class group of a Krull semigroup Deﬁnition 9.8.10 A subsemigroup S of F is called reduced in F if the projection πi | S : S → Z is onto for all i ∈ I. 8 Theorem 9.8.11 [83] Let F = i∈I Zei be a free abelian group and let S be a subsemigroup of F such that S = S ∩ F+ . If any of the following two equivalent conditions hold, then Cl(K[S]) ∼ = F/S. (a) S is a reduced subsemigroup of F, and for all i, j ∈ I with i = j, there exists s ∈ S such that πj (s) > 0 = πi (s). (b) Ti = {xj + S | i = j ∈ I} generates F/S as a subsemigroup ∀ i ∈ I. Next we state a particular case of Theorem 9.8.11 that can be used to compute the divisor class group of an aﬃne semigroup. Theorem 9.8.12 [83] Let F = Ze1 ⊕ · · · ⊕ Zer be a free abelian group of rank r and let S = Nw1 + · · · + Nwq be a subsemigroup of F such that the following two conditions hold: (a) S = ZS ∩ F+ , where ZS is the subgroup of F generated by S. (b) πi (ZS) = Z for i = 1, . . . , r and for each pair i = j in {1, . . . , r} there exists w ∈ S such that πi (w) > 0 = πj (w). Then Cl(K[S]) ∼ = F/ZS for any ﬁeld K. Divisor class groups of aﬃne semigroups Let A = {v1 , . . . , vq } be a ﬁnite set of distinct vectors in Nn \ {0} and let NA be the aﬃne semigroup of Nn generated by A. If R = K[x1 , . . . , xn ] is a polynomial ring over a ﬁeld K and F = {xv1 , . . . , xvq }, then the elements in the monomial subring K[F ] are expressions of the form:

a1

ca (xv1 )

· · · (xvq )

aq

,

Monomial Subrings

419

where a = (a1 , . . . , aq ) ∈ Nq , ca ∈ K and ca = 0 except for a ﬁnite number. Therefore K[F ] is generated, as a K-vector space, by the set of all xa such that a ∈ NA, i.e., K[F ] is the semigroup ring of NA: K[F ] = K[NA] = K[{xa | a ∈ NA}]. Theorem 9.8.13 The following conditions are equivalent: (a) K[F ] is normal. (b) K[NA] is a Krull ring. (c) NA = R+ A ∩ ZA. Proof. Notice that K[F ] is a Noetherian domain because K[F ] is a ﬁnitely generated K-algebra. Thus, by the Mori–Nagata theorem (Theorem 2.6.2), part (a) is equivalent to part (b). From Corollary 9.1.3, we obtain that part (a) is equivalent to part (c). 2 Corollary 9.8.14 NA is a Krull semigroup if and only if NA is equal to R+ A ∩ ZA. 2

Proof. It follows from Theorem 9.8.13 and Corollary 9.8.9. Theorem 9.8.15 ZA ∩ R+ A is a ﬁnitely generated Krull semigroup. Proof. By the ﬁnite basis theorem there are a1 , . . . , ak in Qn such that R+ A = Ha+1 ∩ · · · ∩ Ha+k .

(†)

For 1 ≤ i ≤ k consider the valuation νi : ZA → Z given by νi (x) = ai , x. + We may assume that ai ∈ Zn for all i because Ha+ = Hra for r > 0. Hence ZA ∩ R+ A =

k

{x ∈ ZA | νi (x) ≥ 0}.

i=1

Hence ZA ∩ R+ A is a Krull semigroup because it is the intersection of the valuation semigroups of ν1 , . . . , νk . By Gordan’s lemma the semigroup is ﬁnitely generated. 2 Proposition 9.8.16 There exists an irreducible representation: R+ A = RA ∩ H+1 ∩ · · · ∩ H+r ,

(9.27)

with i ∈ Qn for all i and such that the following conditions are satisﬁed: (a) i , vj ∈ N for all i, j. (b) Z = i , v1 Z + · · · + i , vq Z for all i.

420

Chapter 9

Proof. The decomposition of R+ A given in Eq. (†) can be brought to the form of Eq. (9.27), where i ∈ Qn for all i and RA is the vector space generated by A. By multiplying each i by an appropriate positive integer, we can choose the i ’s such that condition (a) holds. Let 1 ≤ i ≤ r be a ﬁxed integer. We claim that there exists 1 ≤ j ≤ q such that i , vj > 0. Otherwise i , vj = 0 for all j and one has R+ A ⊂ Hi . By the irreducibility of the representation (see Theorem 1.1.44), the set R + A ∩ H i = R + A is a facet (proper face of maximum dimension), a contradiction. Let m be the greatest common divisor of the non-zero elements of the set {i , v1 , . . . , i , vq }. We can write i , vk = λk m. Setting i = i /m, conditions (a) and (b) are satisﬁed if we substitute i by i , this follows from the fact that the non-zero elements of the set λ1 , . . . λq are relatively prime. 2 For the rest of this section we assume that R+ A has an irreducible representation satisfying the conditions of Proposition 9.8.16. Consider the homomorphism of Z-modules ϕ : Zn → Zr given by: ϕ(v) = 1 , ve1 + · · · + r , ver . In what follows wi will denote the vector of Nr deﬁned as wi := ϕ(vi ) = 1 , vi e1 + · · · + r , vi er

(i = 1, . . . , q),

S will denote the semigroup S := Nw1 + · · · + Nwq , and F will denote the free abelian group F := Ze1 ⊕ · · · ⊕ Zer . Proposition 9.8.17 NA S. Proof. Clearly ϕ induces a homomorphism from NA onto S that we also denote by ϕ. Thus it suﬃces to prove that ϕ is an injective map. Let α = a1 v1 +· · ·+aq vq be a vector in ker(ϕ). It follows readily that α, i = 0 for i = 1, . . . , r. Therefore α belongs to the set R+ A ∩ H1 ∩ · · · ∩ Hr = {0} and consequently α = 0. The previous equality follows using that the vector 0 is a vertex of the pointed cone R+ A, hence {0} is an intersection of facets. From the irreducibility of the representation of the cone one obtains that its facets have the form R+ A ∩ Hi for i = 1, . . . , r (see Theorem 1.1.44). 2 Corollary 9.8.18 K[NA] K[S].

Monomial Subrings

421

Proof. By Proposition 9.8.17 we have NA S. Hence applying [180, Theorem 7.2, p. 69] we obtain the required isomorphism. 2 Theorem 9.8.19 If K[NA] is a Krull ring , then Cl(K[NA]) F/S = (Ze1 ⊕ · · · ⊕ Zer ) /(Zw1 + · · · + Zwq ) . In particular rank(F/S) = r − rank(S). Proof. Taking into account Theorem 9.8.12 and Corollary 9.8.18 it suﬃces to prove that the following two conditions are satisﬁed: (a) S = ZS ∩ F+ . (b) πi (ZS) = Z for i = 1, . . . , r and for every i = j in {1, . . . , r} there exists w ∈ S such that πi (w) > 0 = πj (w). First we prove (a). Since j , vi ∈ N for all i, j, we get that wi ∈ F+ and clearly wi ∈ ZS for all i. Therefore S ⊂ ZS ∩ F+ . Conversely let α ∈ ZS ∩ F+ . We can write α = a1 w1 + · · · + aq wq = η1 e1 + · · · + ηr er

(ai ∈ Z; ηj ∈ N ∀ i, j).

Substituting the wi ’s and matching the coeﬃcients of the ei ’s yields: = a1 i , v1 + · · · + aq i , vq , i = 1, . . . , r.

ηi

Consider the vector β = a1 v1 + · · · + aq vq ∈ ZA. Clearly ϕ(β) = α because ϕ(vi ) = wi for = 1, . . . , q. Using these equations we obtain β, i =

a1 i , v1 + · · · + aq i , vq = ηi ≥ 0, i = 1, . . . , r.

Therefore β ∈ RA ∩ H+1 ∩ · · · ∩ H+r = R+ A and as a consequence we get β ∈ ZA ∩ R+ A. By the normality of K[NA] we conclude that β ∈ NA. This q c means that we can rewrite β as β = i=1 i vi with ci ∈ N for all i. Hence q ϕ(β) = i=1 ci wi = α. Thus α ∈ S, as desired. Next we prove (b). Since πi (wj ) = i , vj for all i, j, one has πi (ZS)

= πi (Zw1 + · · · + Zwq ) = Zπi (w1 ) + · · · + Zπi (wq ) = Zi , v1 + · · · + Zi , vq = Z,

where the last equality is satisﬁed thanks to the choice of 1 , . . . , r (see Proposition 9.8.16). To prove the second part of condition (b) it suﬃces to prove that for i = j there exists wk = ϕ(vk ) ∈ S such that i , vk = πi (wk ) > 0 = πj (wk ) = j , vk . By the irreducibility of the representation of R+ A, the sets Fi = Hi ∩ R+ A and Fj = Hj ∩ R+ A are distinct facets. By Proposition 1.1.23, Fi is a cone generated by a subset of A and since Fj ⊂ Fi there exists vk ∈ Fj \ Fi . 2 Therefore j , vk = 0 and i , vk > 0, as desired.

422

Chapter 9

Algorithm To compute the divisor class group of the semigroup ring K[NA] proceed as follows: (1) First, using Normaliz [68], compute the integral closure of K[NA] in its ﬁeld of fractions to determine whether K[NA] is a Krull ring. (2) Second, using PORTA [84], compute an irreducible representation of R+ A and adjust this representation (if necessary) to satisfy conditions (a) and (b) of Proposition 9.8.16. (3) Third, using (2) determine the vectors w1 , . . . , wq . Then, using Maple [80], diagonalize (over the integers) the matrix B whose rows are w1 , . . . , wq , to obtain a “diagonal matrix” D = diag{d1 , . . . , ds }, where d1 , . . . , ds are the invariant factors of Zr /Z{w1 , . . . , wq }. (4) Fourth, use Theorem 9.8.19 together with the fundamental theorem for ﬁnitely generated abelian groups (see Theorem 1.3.16), to obtain: Cl(K[NA]) Zr /Z{w1 , . . . , wq } Zt ⊕ Z/Zd1 ⊕ · · · ⊕ Z/Zds , where t = r − rank(B) = r − s. Example 9.8.20 Let A = {(a1 , . . . , ad ) ∈ Nn | a1 + · · · + an = d} be the set of ordered partitions of an integer d ≥ 2. Assume A = {v1 , . . . , vq }. Notice that RA = Rn and the irreducible representation of R+ A is: R+ A = Rn+ = He+1 ∩ · · · ∩ He+n . In this case r = n and i = ei for i = 1, . . . , n. It is easy to see that ϕ(vi ) = wi = vi for all i. Thus S = ZA. Let B be the matrix whose rows are w1 , . . . , wq . Using elementary operations we obtain that the Smith normal form of B is diag{1, . . . , 1, d}, where the number of 1’s is n − 1. By Theorem 9.8.19 we get Cl(K[NA]) Zn /ZA Zd .

Exercises 9.8.21 A semigroup (0) = S of Nn is called full if Nn ∩ ZS = S. If S ⊂ Nn is a full semigroup, prove that S is a ﬁnitely generated semigroup. 9.8.22 [248] Let A ⊂ Nn be a ﬁnite set. If NA is a normal semigroup of Nn , then there is a full semigroup S ⊂ Nr such that NA S (as semigroups). In particular K[NA] K[S] (as rings). 9.8.23 If S ⊂ Nn is a full semigroup, prove that R+ S ∩ ZS = NS, that is, S is a normal semigroup. 9.8.24 Let F be the set of square-free monomials of degree d of a polynomial ring over a ﬁeld K and let K[F ] be the monomial subring spanned by F . Find a formula for Cl(K[F ]).

Chapter 10

Monomial Subrings of Graphs In this chapter we study monomial subrings associated to graphs and their toric ideals. We relate the even closed walks and circuits of the vector matroid of a graph with Gr¨obner bases. A description of the integral closure of the edge subring of a multigraph will be presented along with a description of the circuits of its toric ideal. The Smith normal form and the invariant factors of the incidence matrix of a graph are fully determined. As an application we compute the multiplicity of edge subrings in terms of relative volumes. The family of ring graphs is studied here. These graphs are characterized in algebraic and combinatorial terms. Several interesting connections between monomial subrings, polyhedral geometry and graph theory will occur in this chapter. We study in detail the irreducible representation of an edge cone of a graph and show some applications to graph theory, e.g., we show the marriage theorem.

10.1

Edge subrings and ring graphs

Let R = K[x18 , . . . , xn ] be a polynomial ring over a ﬁeld K with the standard ∞ grading R = i=0 Ri induced by deg(xi ) = 1 for all i and let G be a graph on the vertex set X = {x1 , . . . , xn }. Deﬁnition 10.1.1 The edge subring of the graph G, denoted by K[G], is the K-subalgebra of R given by: K[G] = K[{xi xj | xi is adjacent to xj }] ⊂ R.

424

Chapter 10

To obtain a presentation of the edge subring of G note that K[G] is a standard K-algebra with the normalized grading K[G]i = K[G] ∩ R2i and consider the set F = {f1 , . . . , fq } of all monomials xi xj such that xi is adjacent to xj . Since the elements in K[G] are polynomial expressions in F with coeﬃcients in K, there is a graded epimorphism of K-algebras ϕ : S = K[t1 , . . . , tq ] −→ K[G],

ti −→ fi ,

where S is a polynomial ring graded by deg(ti ) = 1 for all i. The kernel of ϕ, denoted by P (G), is a graded ideal of S called the toric ideal of K[G] with respect to f1 , . . . , fq . Deﬁnition 10.1.2 Let w = {x0 , x1 , . . . , xr = x0 } be an even closed walk of G such that fi = xi−1 xi . The binomial tw = t1 · · · tr−1 − t2 · · · tr is called the binomial associated to w. Remark 10.1.3 tw ∈ P (G) because f1 f3 · · · fr−1 = f2 f4 · · · fr . Deﬁnition 10.1.4 A closed walk of even length will be called a monomial walk . A monomial xi xj of R is said to be an edge generator if {xi , xj } is an edge of G. Notation In what follows Is denotes the set of all non-decreasing sequences α = (i1 , · · · , is ) of length s. If a1 , . . . , aq is a sequence and α = (i1 , · · · , is ) is in Is , then we set aα = ai1 · · · ais . Proposition 10.1.5 [417] If G is a graph and P = P (G) is the toric ideal of the edge subring K[G], then (a) P = ({tw | w is an even closed walk}), and (b) P = ({tw | w is an even cycle}) if G is bipartite. Proof. (a): We set B = {tw | w is an even closed walk}. As P is a graded ideal in the standard grading of S = K[t1 , . . . , tq ], we can write P =S·

∞ (

! Ps

.

s=2

One clearly has (B) ⊂ P . To prove the other inclusion we use induction on s. First recall that P is a binomial ideal by Corollary 8.2.18. If s = 2, it is enough to note that the binomials in P2 come from squares. Assume

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425

Ps−1 ⊂ (B). To show Ps ⊂ (B), take tα − tβ in Ps , where α = (i1 , . . . , is ) and β = (j1 , . . . , js ). Deﬁne G1 as the subgraph of G having vertex set V1 = {xi ∈ V | xi divides fα } and edge set E1 = {{x, y} ∈ E(G)| f = xy for some ∈ {i1 , . . . , is , j1 , . . . , js }}. Note that if fi1 · · · fim = fj1 · · · fjm for some m < s and for some ordering of the generators then tα − tβ ∈ (B); to prove it, notice that tα − tβ can be written as tα − tβ = tim+1 · · · tis (ti1 · · · tim − tj1 · · · tjm ) +tj1 · · · tjm (tim+1 · · · tis − tjm+1 · · · tjs ) and use induction hypothesis. Now we can assume fi1 · · · fim = fj1 · · · fjm for m < s and for any re-ordering of the fi ’s. Observe that this forces G1 to be connected. Take x0 ∈ V1 . Since fα = fβ , it is easy to check that after re-ordering we can write fik = x2k−2 x2k−1 and fjk = x2k−1 x2k , where 1 ≤ k ≤ s and x0 = x2s . Therefore the monomial walk w = {x0 , . . . , x2n } satisﬁes tw = tα − tβ and the induction is complete. (b): Assume that G is bipartite. This part can be proven similarly to part (a) if we notice that in this case the condition fi1 · · · fim = fj1 · · · fjm for m < s and for any re-ordering of the fi ’s implies that the graph G1 is an even cycle. 2 Corollary 10.1.6 If G is a graph and f = tα − tβ is a primitive binomial in P (G), then f = tw for some even closed walk w of G. Proof. It follows from the proof of Proposition 10.1.5.

2

Corollary 10.1.7 If G is a bipartite graph and f = tα − tβ is a primitive binomial in P (G), then f = tw for some even cycle w of G. Proof. It follows from the proof of Proposition 10.1.5.

2

Proposition 10.1.8 If G is a graph and f = tα −tβ is a primitive binomial in P (G), then the entries of α satisfy αi ≤ 2 for all i. Proof. By Corollary 10.1.6 one can write f = tw = t1 t3 · · · t−1 − t2 t4 . . . t , where is even, and w is an even closed walk in G of length : w = {x0 , x1 , . . . , x−1 , x = x0 }

426

Chapter 10

such that f1 f3 · · · f−1 = f2 f4 · · · f and fi = xi−1 xi for all i. Note that there may be some repetitions of the variables ti and accordingly some repetitions in the monomials fi . As usual we are identifying the vertices of G with variables. It suﬃces to verify that x0 occurs at most three times (including the end vertices) in the closed walk w. If x0 occurs more than three times in w, one can write w = {x0 , x1 , . . . , xt1 = x0 , xt1 +1 , . . . , xt1 +t2 = x0 , xt1 +t2 +1 , . . . , x = x0 }, note that t1 must be odd; otherwise using w1 = {x0 , x1 , . . . , xt1 = x0 } one has that tw1 = tγ − tδ , where tγ divides t1 t3 · · · t−1 and tδ divides t2 t4 . . . t , a contradiction because f is primitive. By a similar argument t2 must be odd. Thus t1 + t2 is even, to get a contradiction note that the closed walk w = {xt1 +t2 = x0 , xt1 +t2 +1 , . . . , x = x0 } is of even length, and apply the previous argument.

2

In [352] there are characterizations of when a binomial of a toric ideal of K[G] is primitive, minimal, indispensable, or fundamental in terms of the even closed walks. Lemma 10.1.9 Let G be a graph and let P = P (G) be the toric ideal of K[G]. If f is a polynomial in any reduced Gr¨ obner bases of P , then (a) f is a primitive binomial and f = tw for some even closed walk w of the graph G. (b) If G is bipartite, then f is primitive and f = tw for some even cycle w of the graph G. Proof. Let B be a reduced Gr¨ obner basis of P and take f ∈ B. Using Corollary 8.2.18 and the fact that normalized reduced Gr¨ obner bases are uniquely determined, we obtain f = tα − tβ . By Lemma 8.3.3 f is primitive, thus by Corollaries 10.1.6 and 10.1.7 f has the required form. 2 Proposition 10.1.10 If G is a graph, then the set {tw | tw is primitive and w is an even closed walk} is a universal Gr¨ obner basis of P (G). Proof. It follows from Lemma 10.1.9. Proposition 10.1.11 If G is a bipartite graph, then the set {tw | w is an even cycle} is a universal Gr¨ obner basis of P (G).

2

Monomial Subrings of Graphs

427 2

Proof. It follows from Lemma 10.1.9.

Deﬁnition 10.1.12 A chord of a cycle c of a graph G is any edge of G joining two non-adjacent vertices of c. A cycle without chords is called primitive. Proposition 10.1.13 If a is a cycle and ta ∈ (tc | c is a cycle, c = a), then a has a chord. Proof. If ta = tα − tβ ∈ (tc | c is a cycle, c = a), then each term of ta must be divisible by a term of tc for some c = a. This means that tα = tδ tα1 and tβ = tγ tα2 , where tαi are terms of some binomial of the form tci with ci = a and ci a cycle. Hence (after permuting variables) x1 · · · xr = xθ (x1 · · · xs ), where x1 , . . . , xr and x1 , . . . , xs are the vertices of the cycles a and c1 . It is seen that a must have a chord. 2 Proposition 10.1.14 If G is a bipartite graph, then P (G) is a prime ideal minimally generated by the set {tc | c is a primitive cycle of G}. Proof. It follows from Proposition 10.1.13.

2

Rees algebras of edge ideals We will look closely at the toric ideal of the Rees algebra of an edge ideal, and then show a few applications. Let I = I(G) be the edge ideal of a graph G, let F = {f1 , . . . , fq } be the set of edge generators of I, and let R[It] = R[f1 t, . . . , fq t] ⊂ R[t] be the Rees algebra of I. There is an epimorphism φ : R[t1 , . . . , tq ] → R[It], induced by ti → fi t. We set J = ker(φ) and B = R[t1 , · · · , tq ], notice that J = ⊕∞ i=1 Ji is a graded ideal (in the ti -variables) of B = ⊕∞ i=0 Bi . The ideal J is the presentation ideal or toric ideal of R[It] with respect to F . Recall that the ideal I is said to be of linear type if J = J1 B. Theorem 10.1.15 [417] Let I = I(G) be the edge ideal of a graph G and let J be the toric ideal of R[It]. Then Js = B1 Js−1 + RPs for s ≥ 2 and ! ∞ ( Ps , J = BJ1 + B · s=2

where Ps = {tα − tβ | fα = fβ , for some α, β ∈ Is }. Corollary 10.1.16 [417] Let G be a connected graph and let I be its edge ideal. Then I is an ideal of linear type if and only if G is a tree or G has a unique cycle of odd length.

428

Chapter 10

Proof. ⇒) If I is an ideal of linear type, then by Proposition 4.2.1, G is a tree or G has a unique cycle of odd length. ⇐) Assume that G is either a tree or G has a unique cycle of odd length. We claim that RPs = (0) for s ≥ 2, we proceed by induction on n. If n ≤ 3 this is easily veriﬁed, assume the claim true for graphs with less than n vertices. Using induction on s we now show RPs = (0) for s ≥ 2, notice that RP2 = (0) because G has no squares. Assume RPs−1 = (0) and RPs = (0), take a non-trivial relation fi1 · · · fis = fj1 · · · fjs . Since a tree has a vertex of degree one, by induction hypothesis it follows that G must be a cycle. Therefore, using induction again, we may write {fi1 , . . . , fis } = {g1 , . . . , gs }, where gcd(gi , gj ) = 1 for i = j and g1 · · · gk = x1 · · · xn . Notice that this equality cannot occur if n is odd, hence RPs = (0) and the proof of the claim is complete. Hence, by Proposition 10.1.15, I is of linear type. 2 Arbitrary square-free monomial ideals of linear type have been studied in [6, 162, 376] using simplicial complexes. These papers show a deep interplay between graph theory and the deﬁning equations of Rees algebras. An interesting family of square-free monomial ideals of linear type is given in [376]. This family includes edge ideals of totally balanced clutters and facet ideals of disjoint simplicial cycles of odd lengths. This family is closed under the operation of adding M -elements. Let I ⊂ R be a square-free monomial ideal minimally generated by monomials f1 , . . . , fq . The line graph of I, denoted by L(I), is the graph whose vertex set is f1 , . . . , fq and where {fi , fj } is an edge of L(I) if and only if fi and fj have at least one variable in common. Recently Fouli and Lin [162] have shown that if the line graph L(I) is a disjoint union of trees and graphs with a unique cycle of odd length, then I is of linear type. A generalization of this fact is given in [6]. A dimension formula Let us describe the cycle space of a graph G over the ﬁeld F = Z2 . We denote the edge set and vertex set of G by E(G) = {f1 , . . . , fq } and X = {x1 , . . . , xn }, respectively. Let C0 and C1 denote the vector spaces over F of 0-chains and 1-chains, respectively. Recall that a 0-chain of G is a formal linear combination ai xi of points and a 1-chain is a formal linear combination bi fi of edges, where ai ∈ F and bi ∈ F. The boundary operator ∂ is the linear transformation deﬁned by ∂(fk ) = ∂({xi , xj }) = xi + xj .

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A cycle vector is a 1-chain of the form t1 + · · · + tr where t1 , . . . , tr are the edges of a cycle of G. The cycle space Z(G) of G over F is equal to ker(∂). The vectors in Z(G) can be regarded as a set of edge-disjoint cycles. A cycle basis for G is a basis for Z(G) which consist entirely of cycle vectors, such a basis can be constructed as follows: Remark 10.1.17 If G is connected, then G has a spanning tree T . The subgraph of G consisting of T and any edge of G not in T has exactly one cycle, the collection of all the cycle vectors obtained in this way form a cycle basis for G. See [208] for details (cf. Exercise 6.2.8). In particular dimF Z(G) = q − n + 1. Deﬁnition 10.1.18 If G is a graph, the number dimF Z(G) is called the cycle rank of G and is denoted by rank(G). Lemma 10.1.19 Let G be a connected graph and Ze (G) the subspace of Z(G) of all cycle vectors of G with an even number of terms. Then

dimF Ze (G) =

q−n+1 q−n

if G is bipartite, and otherwise.

Proof. Let c1 , . . . , cl , cl+1 , . . . , cm be a cycle basis for G, where c1 , . . . , cl are even cycles and cl+1 , . . . , cm odd cycles. It is clear that a basis for Ze (G) 2 is given by {c1 , . . . , cl , cl+1 + cm , . . . , cm−1 + cm }. Proposition 10.1.20 [417] If G is a connected graph, then ht(P (G)) = dimF Ze (G). Proof. Let G be a connected graph with q edges and n vertices. Assume that I(G) is minimally generated by the monomials f1 , . . . , fq . There is a spanning tree T of G so that (after re-ordering) I(T ) = (f1 , . . . , fn−1 ). Since I(T ) is an ideal of linear type dim K[G] = tr.degK K[G] ≥ n − 1. If G is bipartite notice that fk ∈ k(f1 , . . . , fn−1 ) for k ≥ n; to prove it we write fk = xz and observe that the graph T ∪ {x, z} has a unique cycle of even length. This shows the equality dim K[G] = n − 1. If G is not bipartite, then for some fk = xz, k ≥ n, the subgraph T ∪ {x, z} has a unique cycle of odd length. By Corollary 10.1.16 the ideal (f1 , . . . , fn−1 , fk ) is of linear type, hence dim K[G] ≥ n; to show equality 2 recall that tr.degK K[G] ≤ n. To ﬁnish the proof use Lemma 10.1.19.

430

Chapter 10

Corollary 10.1.21 If G is a connected graph with n vertices and K[G] is its edge subring, then

n if G is not bipartite, and dim(K[G]) = n − 1 otherwise. Proof. It follows from the proof of Proposition 10.1.20.

2

Deﬁnition 10.1.22 The number of primitive cycles of a graph G, denoted by frank(G), is called the free rank of G. Proposition 10.1.23 If G is a graph, then Z(G) is generated by cycle vectors of primitive cycles. In particular rank(G) ≤ frank(G). Proof. Let c1 , . . . , cr be a cycle basis for the cycle space of G and let c1 , . . . , cr be the corresponding cycles of G. It suﬃces to notice that if some cj has a chord, we can write cj = cj + cj , where cj and cj are cycle vectors 2 of cycles of length smaller than that of cj . Corollary 10.1.24 Let G be a graph. The following are equivalent : (a) rank(G) = frank(G). (b) The set of cycle vectors of primitive cycles is a basis for Z(G). (c) The set of cycle vectors of primitive cycles is linearly independent. Proof. (a) ⇒ (b): By Proposition 10.1.23 there is a basis B of Z(G) consisting of cycle vectors of primitive cycles. By hypothesis rank(G) is frank(G). Thus B is the set of all cycle vectors of primitive cycles and B is a basis. That (b) implies (c) and (c) implies (a) are also easy to prove. 2 The family of graphs satisfying the equality rank(G) = frank(G) can be constructed as is seen later in this section. Ring graphs In this part we characterize ring graphs in algebraic and combinatorial terms. The reader is referred to Section 7.1 for the notions of cutvertex, bridge, and block. Lemma 10.1.25 Let G be a graph and let G1 , . . . , Gr be its blocks. Then rank(G) = frank(G) if and only if rank(Gi ) = frank(Gi ) for all i. Proof. ⇒) Let Gi be any block of G. We may assume |V (Gi )| > 2, otherwise rank(Gi ) = frank(Gi ) = 0. If c is a primitive cycle of Gi , then by the maximality condition of a block one has that c is also a primitive cycle of G. Thus by Corollary 10.1.24 the set of cycle vectors of primitive cycles of Gi is linearly independent and rank(Gi ) = frank(Gi ).

Monomial Subrings of Graphs

431

⇐) Let Bi and B be the set of cycle vectors of primitive cycles of Gi and G, respectively. As ∪ri=1 Bi is linearly independent, by Corollary 10.1.24 it suﬃces to prove that ∪ri=1 Bi = B. In the ﬁrst part of the proof we have already observed that ∪ri=1 Bi ⊂ B. To prove the equality take any cycle vector c of a primitive cycle c of G. Since c is a 2-connected subgraph, it must be contained in some block of G, i.e., in some Gi . Thus c is a primitive 2 cycle of Gi , so c is in Bi . Deﬁnition 10.1.26 Given a graph H, we call a path P an H-path if P is non-trivial and meets H exactly in its ends. Theorem 10.1.27 ([111, Proposition 3.1.2], [208, Theorem 5.10]) Let G be a graph. Then the following three conditions are equivalent: (a) G is 2-connected. (b) G can be constructed from a cycle by successively adding H-paths to graphs H already constructed. (c) Every pair of vertices of G is joined by at least 2 vertex-disjoint paths. This result suggests the following more restrictive notion: Deﬁnition 10.1.28 A graph G is called a ring graph if each block of G which is not a bridge can be constructed from a cycle by successively adding H-paths of length at least 2 that meet graphs H already constructed in two adjacent vertices. Families of ring graphs include forests and cycles. Remark 10.1.29 Let G be a 2-connected ring graph and let c be a ﬁxed primitive cycle of G, then G can be constructed from c by successively adding H-paths of length at least 2 that meet graphs H already constructed in two adjacent vertices. Deﬁnition 10.1.30 A graph H is called a subdivision of a graph G if H arises from G by replacing edges by paths. That is H is obtained by iteratively choosing an edge (u, v), introducing a new vertex w, deleting edge (u, v), and adding edges (u, w) and (w, v). Example 10.1.31 A complete bipartite graph K2,3 and a subdivision s @ @ @s s K2,3 @ H H @ HH s @ @ @s

s @ @s @ @s s H H @ HHss @ s @ @s

432

Chapter 10

Lemma 10.1.32 [2, Lemma 7.78, p. 387] Let G be a graph with vertex set V. If G is 2-connected and deg(v) ≥ 3 for all v ∈ V, then G contains a subdivision of K4 as a subgraph. Lemma 10.1.33 Let G be a graph. If rank(G) = frank(G) and x, y are two non-adjacent vertices of G, then there are at most two vertex disjoint paths joining x and y. Proof. Assume that there are three vertex disjoint paths joining x and y: P1 = {x, x1 , . . . , xr , y},

P2 = {x, z1 , . . . , zt , y},

P3 = {x, y1 , . . . , ys , y}, where r, s, t are greater than or equal to 1. We may assume that the sum of the lengths of the Pi ’s is minimal. Consider the cycles c1 c2

= {x, x1 , . . . , xr , y, zt , . . . , z1 , x}, = {x, z1 , . . . , zt , y, ys , . . . , y1 , x},

c3

= {x, x1 , . . . , xr , y, ys , . . . , y1 , x}.

Thus we are in the following situation: c3 s c1 z x s 1s s

x1 s

y1 s

s

s

c2 s

s

s xr

zs t sy s

sys

Observe that, by the choice of the Pi ’s, a chord of the cycle c1 (resp. c2 , c3 ) must join xi and zj (resp. zi and yj , xi and yj ) for some i, j. Therefore we can write n1 n2 n3 ai , c2 = bi , c3 = di c1 = i=1

i=1

i=1

where a1 , . . . , an1 , b1 , . . . , bn2 , d1 , . . . , dn3 are distinct cycle vectors of primitive cycles of G and ci is the cycle vector corresponding to ci . Thus from the equality c3 = c1 + c2 . we get that the set of cycle vectors of primitive cycles is linearly dependent, a contradiction to Corollary 10.1.24. 2 Deﬁnition 10.1.34 A graph G has the primitive cycle property (PCP) if any two primitive cycles intersect in at most one edge. Lemma 10.1.35 Let G be a graph. If rank(G) = frank(G), then G has the primitive cycle property.

Monomial Subrings of Graphs

433

Proof. Let c1 , c2 be two distinct primitive cycles. Assume that c1 and c2 intersect in at least two edges. Thus c1 and c2 must intersect in two non-adjacent vertices. Hence there are two non-adjacent vertices x, y in c1 and a path P = {x, x1 , . . . , xr , y} of length at least two that intersect c1 in exactly the vertices x, y: x1 s

s

x s s

P

s

s

s xr

s

s

s

sy

s

c1 s

s

s

This contradicts Lemma 10.1.33. Hence c1 and c2 have at most one edge in common. 2 Lemma 10.1.36 Let G be a graph. If G satisﬁes PCP and G does not contain a subdivision of K4 as a subgraph, then for any two non-adjacent vertices x, y there are at most two vertex disjoint paths joining x and y. Proof. Assume that there are three vertex disjoint paths joining x and y: P1 = {x, x1 , . . . , xr , y}, P2 = {x, z1 , . . . , zt , y}, P3 = {x, y1 , . . . , ys , y}, where r, s, t are greater than or equal to 1. We may assume that the sum of the lengths of the Pi ’s is minimal. Consider the cycles c1 = {x, x1 , . . . , xr , y, zt , . . . , z1 , x}, c2 = {x, z1 , . . . , zt , y, ys , . . . , y1 , x}, c3 = {x, x1 , . . . , xr , y, ys , . . . , y1 , x}. Thus we are in the following situation: c3 s c1 z x s 1s s

x1 s

y1 s

s

s

c2 s

s

s xr

zs t sy s

sys

Observe that, by the choice of the Pi ’s, a chord of the cycle c1 (resp. c2 , c3 ) must join xi and zj (resp. zi and yj , xi and yj ) for some i, j. Using that G does not contain a subdivision of K4 as a subgraph, it is seen that the cycles c1 and c3 are primitive. Thus, since c1 and c3 have at least two edges in common, we obtain that G does not satisfy PCP, a contradiction. 2

434

Chapter 10

Lemma 10.1.37 Let G be a graph. If rank(G) = frank(G), then G does not contain a subdivision of K4 as a subgraph. Proof. Assume there is a subgraph H ⊂ G which is a subdivision of K4 . If K4 is a subgraph of G, then G has four distinct triangles whose cycle vectors are linearly dependent, a contradiction to Corollary 10.1.24. If K4 is not a subgraph of G, then H is a strict subdivision of K4 , i.e., H has more than four vertices. It follows that there are two vertices x, y in V (H) which are non-adjacent in G. Notice that x, y can be chosen in K4 before subdivision. Therefore there are at least three non-adjacent paths joining x and y, a contradiction to Lemma 10.1.33. 2 Theorem 10.1.38 [184] Let G be a graph. The following are equivalent: (a) G is a ring graph. (b) rank(G) = frank(G). (c) G satisﬁes PCP and does not contain a subdivision of K4 as a subgraph. Proof. (a) ⇒ (b): By induction on the number of vertices it is not hard to see that any ring graph G satisﬁes the equality rank(G) = frank(G). (b) ⇒ (c): It follows at once from Lemmas 10.1.35 and 10.1.37. (c) ⇒ (a): Let G1 , . . . , Gr be the blocks of G. The proof is by induction on the number of vertices of G. If each Gi is either a bridge or an isolated vertex, then G is a forest and consequently a ring graph. Hence by Lemma 10.1.25 we may assume that G is 2-connected and that G is not a cycle. We claim that G has at least one vertex of degree 2. If deg(v) ≥ 3 for all v ∈ V (G), then by Lemma 10.1.32 there is a subgraph H ⊂ G which is a subdivision of K4 , which is impossible. Let v0 ∈ V (G) be a vertex of degree 2 as claimed. By the primitive cycle property there is a unique primitive cycle c = {v0 , v1 , . . . , vs = v0 } of G containing v0 . The graph H = G \ {v0 } satisﬁes PCP and does not have a subdivision of K4 as a subgraph. Consequently H is a ring graph. Thus we may assume that c is not a triangle, otherwise G is a ring graph because it can be obtained by adding the H-path {v2 , v0 , v1 } to H. Next we claim that if 1 ≤ i < j < k ≤ s − 1, then vi and vk cannot be in the same connected component of H \ {vj }. Otherwise there is a path of H \ {vj } that joins vi with vk . It follows that there is a path P of H \ {vj } with at least three vertices that joins a vertex of {vj+1 , . . . , vs−1 } with a vertex of {v1 , . . . , vj−1 } and such that P intersects c exactly in its ends, but this contradicts Lemma 10.1.36. This proves the claim. In particular vi is a cutvertex of H for i = 2, . . . , s − 2 and vi−1 , vi+1 are in diﬀerent connected components of H \ {vi }. For each 1 ≤ i ≤ s − 2 there is a block Ki of H such that {vi , vi+1 } is an edge of Ki . Notice that if 1 ≤ i < j < k ≤ s − 1, then vi , vj , vk cannot lie in some K . Indeed if the three vertices lie in some

Monomial Subrings of Graphs

435

K , then there is a path P in K \ {vj } that joins vi and vk . Since P is also a path in H \ {vj }, we get that vi and vk are in the same connected component of H \ {vj }, but this contradicts the last claim. In particular V (K ) intersects the cycle c in exactly the vertices v , v+1 for 1 ≤ ≤ s − 2. Observe that at least one of the edges of c not containing v0 is not a bridge of H. To show this pick x ∈ / c such that {x, vk } is an edge of H. We may assume that vk+1 = v0 (or vk−1 = v0 ). Since G = G \ {vk } is connected, there is a path P of G joining x and vk+1 (or vk−1 ). This readily yields a cycle of H containing an edge of c which is not a bridge of H. Hence at least one of the blocks K1 , . . . , Ks−2 , say Ki , contains vertices outside c. Next we show that two distinct blocks B1 , B2 of H cannot intersect outside c. We proceed by contradiction assuming that V (B1 )∩V (B2 ) = {z} for some z not in c. Let H1 , . . . , Ht be the connected components of H \{z}. Notice that t ≥ 2 because {z} is the intersection of two diﬀerent blocks of H. We may assume that {v1 , . . . , vs−1 } are contained in H1 . Consider the subgraph H1 of G \ {z} obtained from H1 by adding the vertex v0 and the edges {v0 , v1 }, {v0 , vs−1 }. It follows that the connected components of G \ {z} are H1 , H2 , . . . , Ht , which is impossible because G is 2-connected. Let Ki be a block of H that contains vertices outside of c for some 1 ≤ i ≤ s − 2. By induction hypothesis Ki is a ring graph. Thus by Remark 10.1.29 we can construct Ki starting with a primitive cycle c1 that contains the edge {vi , vi+1 }, and then adding appropriate paths. Suppose that P1 , . . . , Pm is the sequence of paths added to c1 to obtain Ki . If we remove the path Pm from G and use the fact that distinct blocks of H cannot intersect outside c, then again by induction hypothesis we obtain a ring graph. It follows that G is a ring graph as well. 2 A graph G that can be embedded in the (Euclidean) plane R2 is called a planar graph. The point set of a plane graph is compact. Therefore exactly one face of G (a region of R2 \ G) is unbounded. It is called the unbounded face of G (outer face, exterior face). A polygonal arc is a subset of R2 which is the union of a ﬁnite number of straight line segments {p + λ(q − p) | 0 ≤ λ ≤ 1} (p, q ∈ R2 ) and is homeomorphic to the closed unit interval [0, 1]. A ring graph is planar by construction. Thus an immediate consequence of Theorem 10.1.38 is: Corollary 10.1.39 Let G be a graph. If rank(G) = frank(G), then G is planar. Lemma 10.1.40 [320] If G is a planar graph, then G has a representation in the plane such that all edges are simple polygonal arcs.

436

Chapter 10

Proposition 10.1.41 [111, Proposition 4.2.5] If G is a 2-connected planar graph, then every face of G is bounded by a cycle. Theorem 10.1.42 (Euler’s formula; see [111, Theorem 4.2.7]) Let G be a connected planar graph with n vertices, q edges, and r faces. Then n−q+r =2 Proposition 10.1.43 (Kuratowski; see [48, p. 14, Theorem 17]) A graph is planar if and only if it does not contain a subdivision of K5 or a subdivision of K3,3 as a subgraph. Deﬁnition 10.1.44 A planar graph is outerplanar if it can be embedded in the plane so that all its vertices lie on some face; it is usual to choose this face to be the exterior. Deﬁnition 10.1.45 Two graphs H1 and H2 are called homeomorphic if there exists a graph G such that both H1 and H2 are subdivisions of G. Theorem 10.1.46 [208, Theorem 11.10] A graph is outerplanar if and only if it has no subgraph homeomorphic to K4 or K2,3 except K4 \ {e}, where e is an edge. Proposition 10.1.47 If G is outerplanar, then rank(G) = frank(G). Proof. By Theorem 10.1.38(c) it suﬃces to prove that G satisﬁes PCP and G does not contain a subdivision of K4 as a subgraph. If G contains a subdivision H of K4 as a subgraph, then G contains a subgraph, namely H, homeomorphic to K4 , but this is impossible by Theorem 10.1.46. To ﬁnish the proof we now show that G has the PCP property. Let c1 = {x1 , x2 , . . . , xm = x1 } and c2 = {y1 , y2 , . . . , yn = y1 } be two distinct primitive cycles having at least one common edge. We may assume / c1 because that xi = yi for i = 1, 2 and x3 = y3 . Notice that y3 ∈ otherwise {y2 , y3 } = {x2 , y3 } is a chord of c1 . We need only show that {x1 , x2 } = c1 ∩ c2 , because this implies that c1 and c2 cannot have more than one edge in common. Assume that {x1 , x2 } c1 ∩ c2 . Let r be the minimum integer such that yr belong to (c1 ∩ c2 ) \ {x1 , x2 }. Notice that yr = x3 because otherwise {x2 , x3 } is a chord of c2 . Hence c1 together with the path {x2 = y2 , y3 , . . . , yr } give a subgraph H of G which is a subdivision of K2,3 , a contradiction to Theorem 10.1.46. 2 Proposition 10.1.48 Let G be a bipartite graph and let G1 , . . . , Gr be their blocks, then K[G] K[G1 ] ⊗K · · · ⊗K K[Gr ] and P (G) = (P (G1 ), . . . , P (Gr )).

Monomial Subrings of Graphs

437 2

Proof. Since G is bipartite it follows from Exercise 10.1.71.

Deﬁnition 10.1.49 A graph G is called a complete intersection (over K) if the toric ideal P (G) of K[G] is a complete intersection. The complete intersection property of the edge subring of a bipartite graph was ﬁrst studied in [114], and later in [379]. Proposition 10.1.50 [379] If G is a bipartite graph, then G is a complete intersection if and only if rank(G) = frank(G). Proof. Let G1 , . . . , Gr be the blocks of G. By Proposition 10.1.48 G is a c.i. if and only if Gi is a c.i. for all i. On the other hand by Exercise 10.1.72 Gi is a c.i. if and only if rank(Gi ) = frank(Gi ). To ﬁnish the proof apply Lemma 10.1.25. 2 Corollary 10.1.51 If G is a bipartite graph and G is a complete intersection, then G is a planar graph. Proof. It follows from Proposition 10.1.50 and Corollary 10.1.39.

2

Corollary 10.1.52 If G is a bipartite graph, then G is a complete intersection if and only if G is a ring graph. Proof. By Proposition 10.1.50 G is a complete intersection if and only if rank(G) = frank(G) and the result follows from Theorem 10.1.38. 2 Theorem 10.1.53 [273] The edge subring K[G] of a bipartite graph G is a complete intersection if and only if G is planar and any two primitive cycles of G have at most one common edge. Proof. ⇒) It follows from Lemma 10.1.35 and Corollary 10.1.51. ⇐) See [273, Theorem 3.5].

2

Let us summarize some of the results obtained thus far. The following implications hold for any graph G: outerplanar ⇒

ring graph , rank = frank

⇔ PCP + contains no subdivision of K4 as a subgraph

⇒ planar

If G is bipartite, then ring graph ⇐⇒ complete intersection ⇐⇒ PCP + planar. If G is not bipartite, the complete intersection property of P (G) is hard to characterize in combinatorial terms. In [29] a polynomial time algorithm is presented that checks whether a given graph is a complete intersection.

438

Chapter 10

Deﬁnition 10.1.54 Let g1 = tα1 − tβ1 , . . . , gr = tαr − tβr be homogeneous binomials of degree at least 2 in the polynomial ring S = K[t1 , . . . , tq ]. We say that B = {g1 , . . . , gr } is a foliation if the following conditions hold: (a) tαi and tβi are square-free for all i, (b) supp(tαi ) ∩ supp(tβi ) = ∅ for all i, and (c) |(∪ji=1 Ci ) ∩ Cj+1 | = 1 for 2 ≤ j < r, where Ci = supp(gi ) for all i. Proposition 10.1.55 If B = {g1 , . . . , gr } is a foliation, then the binomial ideal I = (B) is a complete intersection. Proof. By the constructive nature of B we can order the variables t1 , . . . , tq so that the leading terms of g1 , . . . , gr , with respect to the lex order ≺, are relatively prime. Since dim(S/I) = dim(S/(in≺ (g1 ), . . . , in≺ (gr ))), we obtain that the height of I is equal to r, as required. Corollary 10.1.56 If G is a 2-connected bipartite graph with at least four vertices, then the toric ideal P (G) is a complete intersection if and only if it is generated by a foliation.

Exercises 10.1.57 [376] A clutter C is called of linear type if its edge ideal I(C) is of linear type. If all connected components of a clutter are of linear type, then the clutter is of linear type. 10.1.58 [376] If C is a clutter of linear type, then so are all minors of C. 10.1.59 Let R = Q[x1 , . . . , x7 ] and let I be the ideal of R generated by f1 = x1 x2 x3 , f2 = x2 x4 x5 , f3 = x5 x6 x7 , f4 = x3 x6 x7 . Use Macaulay2 to verify that the toric ideal of R[It] is minimally generated by the binomials x 3 t3 − x 5 t4 , x 6 x 7 t1 − x 1 x 2 t4 , x 6 x 7 t2 − x 2 x 4 t3 , x 4 x 5 t1 − x 1 x 3 t2 , x 4 t1 t3 − x 1 t2 t4 .

Prove that Theorem 10.1.15 does not extend to uniform square-free monomial ideals. 10.1.60 Let G be a graph. Then G is planar (resp. outerplanar) if and only if each block of G is planar (resp. outerplanar). 10.1.61 Let G be a ring graph. Prove that G is outerplanar if and only if G does not contain a subdivision of K2,3 as a subgraph.

Monomial Subrings of Graphs

439

10.1.62 Let G be a ring graph with n ≥ 3 vertices and q edges. If G is 2-connected and has no triangles, then q ≤ 2(n − 2). 10.1.63 Let G be a graph with q edges and P (G) the toric ideal of the edge subring K[G]. If f is a primitive binomial in P (G) prove:

q if q is even, and deg(f ) ≤ q − 1 otherwise. 10.1.64 Let G be a bipartite graph with bipartition (X, Y ). Prove that the largest degree of a binomial which is part of a minimal set of homogeneous generators of P (G) is less than or equal to min(|X|, |Y |). 10.1.65 If G is a connected graph, then G has a spanning tree T , that is, there exists a tree T which is a spanning subgraph of G. 10.1.66 [208] Prove that a graph G is bipartite if and only if every cycle in some cycle basis of G is even. 10.1.67 If G is a connected non-bipartite graph with n vertices, prove that there is a connected subgraph H of G with n vertices and n edges and with a unique cycle of odd length. 10.1.68 If G is a connected graph, prove that e is a bridge if and only if e is not on any cycle of G. 10.1.69 Let G be a connected graph and let f1 , . . . , fq be the edge generators of K[G]. If fr correspond to a bridge of G and f a = f1a1 · · · fqaq = f1b1 · · · fqbq = f b for some a, b ∈ Nq such that supp(a) ∩ supp(b) = ∅ and ar > 0, prove that ar is an even integer. 10.1.70 The graph below is planar but not outerplanar and it satisﬁes rank(G) = frank(G) = 3. The regions of this embedding are not bounded by primitive cycles. Thus, the converse of Proposition 10.1.47 fails. s s

s

s

s

s s @ @ @s

10.1.71 Let G be a graph that can be written as G = G1 ∪ G2 , where G1 and G2 are subgraphs with at most one vertex in common. If G1 is bipartite, then P (G) = (P (G1 ), P (G2 )) and K[G] ∼ = K[G1 ] ⊗K K[G2 ].

440

Chapter 10

10.1.72 [379] Let G be a connected bipartite graph with n vertices and q edges. Prove that frank(G) = q − n + 1 if and only if the toric ideal P (G) of K[G] is a complete intersection. 10.1.73 Let G be a bipartite graph. If G is a subdivision of K5 or a subdivision K3,3 prove that the binomials of P (G) that correspond to primitive cycles do not form a regular sequence. 10.1.74 Let G be a bipartite graph. Use Kuratowski’s theorem (see Theorem 10.1.43) to prove that if the toric ideal of K[G] is a complete intersection, then G is a planar graph. 10.1.75 If G is a ring graph and H is an induced subgraph of G, then H is a ring graph. 10.1.76 Prove that the complete graph K4 has the primitive cycle property and rank(G) = frank(G). Thus, the converse of Lemma 10.1.35 fails.

10.2

Incidence matrices and circuits

Let G be a simple graph with vertex set V = {x1 , . . . , xn } and with edge set E = {f1 , . . . , fq }, where every edge fi is an unordered pair of distinct vertices fi = {xij , xik }. The incidence matrix AG = (aij ) associated to G is the n × q matrix deﬁned by

1 if xi ∈ fj , and aij = / fj . 0 if xi ∈ Notice that each column of AG has exactly two 1’s and the rest of its entries equal to zero. If fi = {xij , xik } deﬁne vi = eij + eik , where ei is the ith canonical vector in Rn . Thus the columns of AG are precisely the vectors v1 , . . . , vq regarded as column vectors. In what follows we denote the set {v1 , . . . , vq } by A. Example 10.2.1 Consider a triangle dence matrix of G is: ⎛ 1 AG = ⎝ 1 0

G with vertices x1 , x2 , x3 . The inci⎞ 0 1 1 0 ⎠, 1 1

with the vectors v1 = e1 + e2 , v2 = e2 + e3 , and v3 = e1 + e3 corresponding to the edges f1 = {x1 , x2 }, f2 = {x2 , x3 }, and f3 = {x1 , x3 }. Proposition 10.2.2 If A is the incidence matrix of a bipartite graph G, then A is totally unimodular.

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Proof. By induction on the number of columns of A. Let (V1 , V2 ) be the bipartition of G and let B be a square submatrix of A. If a column of B has at most one entry equal to 1, then by induction det(B) = 0, ±1. Thus we may assume that all the columns of B have two entries equal to 1. Hence

Fi =

xi ∈V1

Fi = (1, . . . , 1),

xi ∈V2

where x1 , . . . , xn are the vertices of G and F1 , . . . , Fr are the rows of B. Therefore the rows of B are linearly dependent and det(B) = 0. 2 Theorem 10.2.3 [187] Let G be a simple bipartite graph with n vertices and q edges and let A = (aij ) be its incidence matrix. If e1 , . . . , en are the ﬁrst n unit vectors in Rn+1 and C is the matrix ⎛

a11 ⎜ .. ⎜ C=⎜ . ⎝ an1 1

... .. .

a1q .. .

e1

···

. . . anq ... 1

en

⎞ ⎟ ⎟ ⎟ ⎠

obtained from A by adjoining a row of 1’s and the column vectors e1 , . . . , en , then C is totally unimodular. Proof. Suppose that {1, . . . , m} and {m + 1, . . . , n} is the bipartition of the graph G. Let C be the matrix obtained by deleting the last n − m columns from C. It suﬃces to show that C is totally unimodular. First one successively subtracts the rows 1, 2, . . . , m from the row n+1. Then one reverses the sign in the rows m + 1, . . . , n. These elementary row operations produce a new matrix C . The matrix C is the incidence matrix of a directed graph, namely, consider G as a directed graph, and add one more vertex n + 1, and add the edges (i, n + 1) for i = 1, . . . , m. The matrix C , being the incidence matrix of a directed graph, is totally unimodular; see Exercise 1.8.10. As the last m column vectors of C are e1 − en+1 , . . . , em − en+1 , one can successively pivot on the ﬁrst non-zero entry of ei − en+1 for i = 1, . . . , m and reverse the sign in the rows m + 1, . . . , n to obtain back the matrix C . Here a pivot on the entry cst means transforming column t of C into the sth unit vector by elementary row operations. Since pivoting preserves total unimodularity [338, Lemma 2.2.20] one derives that C is totally unimodular, and hence so is C. In the next speciﬁc example we 2 display C and C .

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Chapter 10

Example 10.2.4 To illustrate the constructions in the proof above consider the complete bipartite graph G = K2,3 . In this case the two Z-row equivalent matrices C and C are ⎡ 1 ⎢0 ⎢ ⎢1 ⎢ ⎢0 ⎢ ⎣0 1

1 0 0 1 0 1

1 0 0 0 1 1

0 1 1 0 0 1

0 1 0 1 0 1

0 1 0 0 1 1

1 0 0 0 0 0

⎤ 0 1⎥ ⎥ 0⎥ ⎥, 0⎥ ⎥ 0⎦ 0

⎡

1 ⎢ 0 ⎢ ⎢−1 ⎢ ⎢ 0 ⎢ ⎣ 0 0

1 0 0 −1 0 0

1 0 0 0 −1 0

0 1 −1 0 0 0

0 1 0 −1 0 0

0 1 0 0 −1 0

1 0 0 0 0 −1

⎤ 0 1⎥ ⎥ 0⎥ ⎥, 0⎥ ⎥ 0⎦ −1

respectively. Note that pivoting in the last two columns of C amounts to adding row 1 to row 6, and then adding row 2 to row 6. Remark 10.2.5 Let G be a graph, let AG be the incidence matrix of G, and let B be a square submatrix of AG . In [201] it is shown that either det(B) = 0 or det(B) = ±2k , for some integer k such that 0 ≤ k ≤ τ0 , where τ0 is the maximum number of vertex disjoint odd cycles in G. Moreover for any such value of k there exists a minor equal to ±2k . The number of bipartite (resp. non-bipartite) connected components of a graph G will be denoted by c0 (resp. c1 ). Thus c = c0 + c1 is the total number of components of G. Lemma 10.2.6 [201] If G is a graph with n vertices and AG is its incidence matrix, then rank(AG ) = n − c0 . Proof. Let G1 , . . . , Gc be the connected components of G, and ni the number of vertices of Gi . After permuting the vertices we may assume that AG is a “diagonal” matrix AG = diag(AG1 , . . . , AGc ), where AGi is the incidence matrix of Gi . By Proposition 10.1.20 the rank of AGi is equal to ni − 1 if Gi is bipartite, and is equal to ni otherwise. 2 Hence the rank of AG is equal to n − c0 . The incidence matrix of G is denoted by AG or simply by A. Regarding the vi ’s as column vectors, we deﬁne the matrix B as: v1 · · · vq . B= 1 ··· 1 Observe that rank(B) = rank(AG ) because the last row of B is a linear combination of the ﬁrst n rows. Recall that Δr (B) denotes the gcd of all non-zero r-minors of B. For simplicity we keep the notation introduced above throughout the rest of this section. We shall assume, if need be, that the graph G has no isolated vertices.

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Lemma 10.2.7 If G is a unicyclic graph with a unique odd cycle and r is the rank of B, then Δr (B) = 1. Proof. If G is an odd cycle of length n, then the matrix B obtained from B by deleting any of its ﬁrst n rows has determinant ±1. The matrix B is in fact totally unimodular by Theorem 10.2.3. We proceed by induction. If G is not an odd cycle, then G has a vertex xj of degree 1. Thus the jth row of A has exactly one entry equal to 1, say ajk = 1. Consider the matrix C obtained from B by deleting the jth row and the kth column. Thus using induction Δr−1 (C) = 1, which gives Δr (B) = 1. 2 Lemma 10.2.8 Let G be a graph with connected components G0 , . . . , Gm . If G0 is non-bipartite, then there exists a subgraph H of G with connected components H0 , . . . , Hm such that H0 is a spanning unicyclic subgraph of G0 with a unique odd cycle and Hi is a spanning tree of Gi for all i ≥ 2. Proof. The existence of H2 , . . . , Hm is clear because any connected graph has a spanning tree. Take a spanning tree T0 of G0 . For each line e of G0 not in T0 , the graph T0 ∪{e} has exactly one cycle. According to Remark 10.1.17 the set Z(T ) of such cycles form a basis for the cycle space of G0 . Since a graph is bipartite if and only if every cycle in some cycle basis is even (see Exercise 10.1.66), there is an edge e of G0 such that T0 ∪ {e} is a unicyclic connected graph with a unique odd cycle. 2 Proposition 10.2.9 If G is a graph with exactly one non-bipartite connected component and r = rank(B), then Δr (B) = 1. Proof. Let G0 , . . . , Gm be the components of G, with G0 non-bipartite, and let H be as in Lemma 10.2.8. If A is the incidence matrix of H, then using that G and H have the same number of non-bipartite components one 2 has rank(A) = rank(A ). Therefore Δr (B) = 1 by Lemma 10.2.7. Proposition 10.2.10 Let G be a non-bipartite graph and let A be its incidence matrix. If A has rank r, then Δr (A) = 2Δr (B). Proof. We set A = {v1 , . . . , vq } and A = {(v1 , 1), . . . , (vq , 1)}. It suﬃces to prove that there is an exact sequence of groups ϕ

ψ

0 −→ T (Zn+1 /ZA ) −→ T (Zn /ZA) −→ Z2 −→ 0. For β = (a1 , . . . , an ) ∈ Zn and b ∈ Z, deﬁne ϕ(β, b) = β and ψ(β) = a1 + · · · + an . It is not hard to verify that ϕ is injective and that im(ϕ) = ker(ψ).

444

Chapter 10

To show that ψ is onto consider a non-bipartite component G1 of G. If are the columns of the x1 , . . . , xs are the vertices of G1 and v1 , v2 , . . . , vm s incidence matrix of G1 , then Z /(v1 , v2 , . . . , vm ) is a ﬁnite group. It follows that for any 1 ≤ j ≤ s, the element ej is in the torsion subgroup of Zn /ZA, 2 hence ψ(ej ) = 1, as required. Corollary 10.2.11 [201] If G is a graph and c0 (resp. c1 ) is the number of bipartite (resp. non-bipartite) components, then Zn /ZA Zn−r × Zc21 = Zc0 × Zc21 , where r = n − c0 is the rank of AG . Proof. Let G1 , . . . , Gc be the components of G and let Ai be the set of vectors in A corresponding to the edges of Gi . Since the vectors in Ai can be regarded as vector in Zηi , where ηi is the number of vertices in Gi , there is a canonical isomorphism Zn /ZA Zη1 /ZA1 ⊕ · · · ⊕ Zηc /ZAc . To ﬁnish the proof apply Propositions 10.2.9 and 10.2.10 to derive

Z2 if Gi is non-bipartite, ηi Z /ZAi Z if Gi is bipartite. In the second isomorphism we are using the fact that incidence matrices of bipartite graphs are totally unimodular; see Proposition 10.2.2. 2 Corollary 10.2.12 If c1 is the number of non-bipartite components of G and r is the rank of B, then

c −1 21 if c1 ≥ 1, Δr (B) = 1 if c1 = 0. Proof. It follows from Proposition 10.2.10 and Corollary 10.2.11.

2

Deﬁnition 10.2.13 A square matrix U , with entries in a commutative ring R, is called unimodular if det(U ) is a unit of R. Theorem 10.2.14 [201] If G is a graph with n vertices and AG is the incidence matrix of G, then there are unimodular integral matrices U , U such that D 0 S = U AG U = , 0 0 where D = diag(1, 1, . . . , 1, 2, 2, . . . , 2), n − c is the number of 1’s and c1 is the number of 2’s. Proof. It follows using Corollary 10.2.11 and the fundamental structure theorem for ﬁnitely generated abelian groups. See Theorem 1.3.16. 2 The matrix S is called the Smith normal form of AG and the numbers in the diagonal matrix D are the invariant factors of AG .

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445

Circuits of the kernel of an incidence matrix We give a geometric description of the elementary integral vectors or circuits of the kernel of the incidence matrix of a graph (see Section 10.3 for the multigraph case). Notation For α = (α1 , . . . , αq ) ∈ Nq and f1 , . . . , fq in a commutative ring α with identity we set f α = f1α1 · · · fq q . The support of f α is deﬁned as the set supp(f α ) = {fi | αi = 0}. Let G be a graph on the vertex set V with edge set E and incidence matrix A. We set N = ker(A), the kernel of A in Qq . Note that a vector α ∈ N ∩ Zq determine the subgraph Gα of G having vertex set Vα = {x ∈ V | x divide f α+ } and edge set Eα = {{x, y} ∈ E | xy ∈ supp(f α+ ) ∪ supp(f α− )}, where f1 , . . . , fq are the square-free monomials corresponding to the edges of G, that is, the edge generators of G. The next result can be generalized to multigraphs; see Section 10.3. Proposition 10.2.15 [417] Let G be a graph with incidence matrix A and let N be the kernel of A in Qq . Then a vector 0 = α ∈ Zq ∩ N is an elementary vector of N if and only if (i) Gα is an even cycle, or (ii) Gα is a connected graph consisting of two odd cycles with at most one common vertex joined by a path. Proof. Assume α ∈ Zq ∩ N is an elementary vector of N so that Gα is not an even cycle. From the equation f α+ = f α− we obtain a walk in Gα w = {x0 , x1 , . . . , xn = x0 , xn+1 , . . . , x }, so that x1 , . . . , x−1 are distinct vertices, n is odd and x is in {x1 , . . . , x−2 }. The walk w can be chosen so that xi xi+1 ∈ supp(f α+ ) if i is even and xi xi+1 ∈ supp(f α− ) otherwise. We claim that x is not in {x1 , . . . , xn−1 }. Assume x = xm for some 1 ≤ m ≤ n − 1. If m and − n are odd then the monomial walk w1 = {x0 , x1 , . . . , xm , x−1 , x−2 , . . . , xn = x0 } yields a relation f δ = f γ with supp(δ − γ) properly contained in supp(α), which is a contradiction. If m is odd and − m is even then we can consider the walk w1 = {xn , xn+1 , . . . , x = xm , xm+1 , . . . , xn } and derive a contradiction; the remaining cases are treated similarly. We may now assume x = xm for n ≤ m ≤ − 2. The cycle C1 = {xm , . . . , x = xm }

446

Chapter 10

must be odd, otherwise C1 would give a relation f δ = f γ with supp(δ − γ) a proper subset of supp(α). The walk w gives a relation f δ = f γ with supp(f δ ) ⊂ supp(f α+ ) and supp(f γ ) ⊂ supp(f α− ). Hence supp(δ − γ) is a subset of supp(α), and since the support of α is minimal one has the equality supp(δ − γ) = supp(α). Therefore Vα = {x1 , . . . , x } and Gα is as required. The converse follows from Corollary 10.1.16.

2

Deﬁnition 10.2.16 A circuit of a graph G is either an even cycle or a subgraph consisting of two odd cycles with at most one common vertex joined by a path. A family of connected graphs is called matroidal if given any graph G, the subgraphs of G isomorphic to a member of the family are the circuits of a matroid on the edge set of G. Thus Proposition 10.2.15 implies that the family of circuits is matroidal as was shown in [387]. Proposition 10.2.17 If A is the incidence matrix of a connected simple graph G, then the circuits of the vector matroid M [A] consists of even cycles, two edge disjoint odd cycles meeting at exactly one vertex, and two vertex disjoint odd cycles joined by an arbitrary path. Proof. It follows from Proposition 10.2.15 (see Section 1.9).

2

Circuits of a graph and Gr¨ obner bases Let G be a graph. We study the toric ideal P (G) of the edge subring K[G] from the view point of Gr¨obner bases. For general results on Gr¨ obner bases of toric ideals a standard reference is [400, Chapter 4]. Proposition 10.2.18 If G is a graph with incidence matrix A and P is the toric ideal of K[G], then P = ({tα+ − tα− | α ∈ Zq and Aα = 0}). Proof. The result is a consequence of Corollary 8.2.20.

2

Corollary 10.2.19 Let G be a graph with incidence matrix A. If λ = (λi ) is a circuit of N = ker(A), then |λi | ≤ 2 for all i. Proof. Let λ be a circuit of N . Since Gλ is an even cycle or two edge disjoint cycles joined by a path we obtain a circuit α with supp(α) = supp(λ) and so that |αi | ≤ 2 for all i. Because any two elementary vectors with the same support are dependent we obtain λ = α. 2

Monomial Subrings of Graphs

447

Theorem 10.2.20 Let G be a connected graph with q edges and let P be the toric ideal of K[G]. Then the total degree of a polynomial in any reduced Gr¨ obner basis of P is less than or equal to 2q dim Ze (G). Proof. We set N = {α ∈ Qq | Aα = 0}, where A is the incidence matrix of G. Assume F is the reduced Gr¨ obner basis of P and take f = f (t) ∈ F . Using Corollary 8.2.18 and Corollary 10.2.18, together with the fact that normalized reduced Gr¨obner bases are uniquely determined, we obtain f (t) = tα − tβ . Notice that supp(α) ∩ supp(β) = ∅, for otherwise take i in the intersection and write f (t) = ti (ta − tb ). Since ta − tb reduces with respect to F it follows that f (t) reduces with respect to F \ {f }, which is impossible. We can now write f (t) = tα+ − tα− for some α ∈ N ∩ Zq . By Theorem 1.9.6 we can write α = a1 λ1 + · · · + ar λr for some ai ∈ Q+ , and for some circuit λi each in harmony with α such that supp(λi ) ⊂ supp(α) for all i and r ≤ dim N . Hence α+ = a1 (λ1 )+ + · · · + ar (λr )+ and α− = a1 (λ1 )− + · · · + ar (λr )− . If ai > 1 for some i then αi + and αi − are componentwise larger than λi+ and λi− , respectively; note that this is not possible by the previous argument. Hence 0 < ai ≤ 1. Using Corollary 10.2.19 it now follows that the entries of α satisfy |αi | ≤ 2 dim Ze (G) for all i, which gives the asserted inequality. 2

Exercises 10.2.21 Let G be a cycle of length n and AG its incidence matrix. Prove

±2 if n is odd det(AG ) = 0 if n is even. 10.2.22 Let U be a square matrix with entries in a commutative ring R. Prove that U is unimodular if and only if U is invertible. 10.2.23 Let G be a graph with incidence matrix A and N = ker(A) ⊂ Qq . Give an example to show that the set of circuits of N does not determine the presentation ideal of K[G]. 10.2.24 Let G be a graph with incidence matrix A and N = ker(A) the kernel of A in Qq . Let λ1 , . . . , λs be the circuits of N . Deﬁne the elementary zonotope of N as EN = [0, λ1 ] + · · · + [0, λs ]. Prove that EN is a polytope.

448

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10.2.25 If N is the kernel of the incidence matrix of a graph G and P is the toric ideal of K[G], prove that {tα+ − tα− | α ∈ EN ∩ Zq } is a universal Gr¨ obner basis for P . 10.2.26 (I. Gitler) Prove that the complete graph K5 is not a circuit and consists of two edge disjoint odd cycles of length 5. 10.2.27 (K. Truemper) Consider the matrices ⎛ ⎛

1 ⎜ 0 ⎜ V =⎝ 0 1

0 1 0 1

0 0 1 1

⎞

1 1 ⎟ ⎟, 1 ⎠ 1

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ A=⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

0 0 0 0 1 1 1 0 0

0 0 1 1 0 0 0 1 0

1 1 0 0 0 0 0 0 1

0 0 0 0 0 0 1 1 1

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟, ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

B=

A 1111

.

Note det(V ) = −2. Prove that A is totally unimodular but B is not. Thus Theorem 10.2.3 does not extend to k-hypergraphs with k > 2.

10.3

The integral closure of an edge subring

Let G be a multigraph. In this section we unfold a construction for the integral closure of K[G] and study the circuits of the toric ideal of K[G]. Let G a multigraph with vertex set X = {x1 , . . . , xn }, i.e., G is obtained from a simple graph by allowing multiple edges and loops. Thus the edges of G have the form {xi , xj }. If e = {xi , xj } is an edge of G, its characteristic vector is given by ve = ei + ej , where ei is the ith unit vector in Rn . Notice that if e is a loop, i.e., if i = j, then ve = 2ei . The incidence matrix of G, denoted by A, is the matrix whose column vectors are the characteristic vectors of the edges and loops of G. Let v1 , . . . , vq be the characteristic vectors of the edges and loops of G and let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K. In what follows A denotes the set {v1 , . . . , vq } and F denotes the set {f1 , . . . , fq } of all monomials xi xj in R such that {xi , xj } is an edge of G. The edge subring of G is given by K[G] = K[f1 , . . . , fq ] ⊂ R. We may assume that fi = xvi for i = 1, . . . , q. One has the following description of the integral closure of K[G]: K[G] = K[{xa | a ∈ R+ A ∩ ZA}], where R+ A is the cone in Rn generated by A and ZA is the subgroup of Zn generated by A. See Theorem 9.1.1.

Monomial Subrings of Graphs

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Proposition 10.3.1 If G is a bipartite graph, then K[G] is normal and Cohen–Macaulay. Proof. By Proposition 10.2.2 the incidence matrix of a bipartite graph is totally unimodular. Then, by Corollary 1.6.8, we have NA = Zn ∩ R+ A. Thus K[G] = K[G], that is K[G] is normal. Hence, by Theorem 9.1.6, K[G] is Cohen–Macaulay. 2 Deﬁnition 10.3.2 A bow-tie of G is an induced submultigraph w of G consisting of two odd cycles with at most one common vertex Z1 = {z0 , z1 , . . . , zr = z0 } and Z2 = {zs , zs+1 , . . . , zt = zs } joined by a path {zr , . . . , zs }. In this case we set Mw = z1 · · · zr zs+1 · · · zt . We regard a loop of G as an odd cycle. Remark 10.3.3 A bow-tie w, as above, always deﬁnes an even closed walk: w = {z0 , z1 , . . . , zr , zr+1 , . . . , zs , zs+1 , . . . , zt = zs , zs−1 , . . . , zr = z0 }. Lemma 10.3.4 If w is a bow-tie of a multigraph G, then Mw is in the integral closure of K[G]. Proof. Let w be a bow-tie. With the notation above. If fi = zi−1 zi , then z12 · · · zr2 = f1 · · · fr , which together with the identities 7 7 fi Mw = fi−1 i even i odd r
2 zs2 · · · zt−1 = fs+1 · · · ft ,

and

Mw2 = f1 · · · fr fs+1 · · · ft

gives Mw ∈ K[G].

2

Remark 10.3.5 Putting the two equations above together one obtains 7 7 7 7 fi2 (10.1) fi2 fi = fi . i even i odd i odd i even r
(10.2)

If s = r + 1, that is Z1 and Z2 are joined by an edge {zr , zs }, then 7 7 fi = fs2 fi . i even i odd i = s

(10.3)

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Chapter 10

Example 10.3.6 A bow-tie formed with a triangle and a pentagon joined by a path of length 2: s z8 s z1 s J z2 J sz6 s Js Z

z0 = z5 z7 = z10 Z s z3 X

XX Z s z4 X

Zs z9 Example 10.3.7 A bow-tie w formed with two loops (odd cycles of length 1) joined by a path of length 2: x s s s2 x1 x3 We can write w as an even closed walk w = {x1 , x1 , x2 , x3 , x3 , x2 , x1 }. In this example Mw = x1 x3 . The cone generated by A is given by R+ A = R+ v1 + · · · + R+ vq . This cone will be studied in detail in Section 10.7 when G is a graph. Deﬁnition 10.3.8 A multigraph having just one cycle is called unicyclic. Theorem 10.3.9 [384] If G is a connected multigraph, then K[G] = K[{f1 , . . . , fq } ∪ {Mw | w is a bow-tie}]. Proof. We set B = {f1 , . . . , fq } ∪ {Mw | w is a bow-tie}. According to Theorem 9.1.1 and Corollary 1.1.27 one has: K[G] = K[{xa | a ∈ ZA ∩ R+ A}] = K[{xa | a ∈ ZA ∩ Q+ A}]. Hence, by Lemma 10.3.4, it suﬃces to show that K[G] ⊂ K[B]. Let xa be a minimal (irreducible) generator of K[G]. Then a ∈ Q+ A and a ∈ ZA. By Carath´eodory’s Theorem (see Theorem 1.1.18), and the irreducibility of xa , we can write (10.4) a = λ1 v1 + · · · + λr vr , where v1 , . . . , vr are linearly independent vectors in A and λi ∈ Q+ ∩ (0, 1) for i = 1, . . . , r. Permitting an abuse of notation, we will also denote the edges of G by v1 , . . . , vq , and refer to vi as an edge of G. Let Ga be the multigraph whose edges are v1 , . . . , vr and let (Ga )1 , . . . , (Ga )s be the components of Ga . For each i consider the graph Hi obtained from (Ga )i by removing loops and by replacing multiple edges by single edges.

Monomial Subrings of Graphs

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Each Hi is either a tree or a unicyclic graph with a unique odd cycle because v1 , . . . , vr are linearly independent. If Hi is a unicyclic graph with a unique odd cycle, then (Ga )i has no loops or multiple edges because v1 , . . . , vr are linearly independent and the rank of the incidence matrix of (Ga )i is equal to the number of vertices of (Ga )i . Thus in this case (Ga )i = Hi . If Hi is a tree, then (Ga )i is a tree with at most one loop. This follows using that the rank of the incidence matrix of (Ga )i is equal to |V ((Ga )i )| − 1 if (Ga )i is a tree. Since 0 < λi < 1, by Eq. (10.4), the multigraph Ga has no vertices of degree 1. Therefore the connected components of Ga are odd cycles. Hence, as 0 < λi < 1 for all i and the components of Ga are odd cycles, we get that the entries of a are in {0, 1}. The multigraph Ga has at least two components, i.e., two odd cycles Z1 and Z2 , because a ∈ ZA. Thus, by the irreducibility of xa , the components of Ga are Z1 and Z2 . As the graph is connected, there is a bow-tie w whose odd cycles are Z1 and Z2 and such that xa = Mw . So xa ∈ B as required. This proof was adapted from [263]. 2 There are two special types of bow-ties that can be omitted in the description of the integral closure of K[G]. Proposition 10.3.10 Let w be a bow-tie of G with two odd cycles Z1 , Z2 such that either Z1 and Z2 meet at exactly one vertex or Z1 and Z2 are disjoint and are connected by an edge, then Mw is in K[G]. Proof. It is left as an exercise.

2

Deﬁnition 10.3.11 A multigraph G has the odd cycle condition if every two vertex disjoint odd cycles can be joined by at least one edge. The odd cycle condition comes from graph theory [175]. This notion was used to study the normality of edge subrings [242, 383, 384] and the circuits of a graph [358, 417]. Corollary 10.3.12 Let G be a connected multigraph. Then K[G] is normal if and only if G satisﬁes the odd cycle condition. Proof. It follows from Theorem 10.3.9 and Proposition 10.3.10.

2

Normality and the circuits of a multigraph Let S = K[t1 , . . . , tq ] be a polynomial ring over a ﬁeld K. Recall that the toric ideal of K[G], denoted by P (G), is the kernel of the epimorphism of K-algebras: S = K[t1 , . . . , tq ] −→ K[G] induced by ti → xvi . Lemma 10.3.13 If f = ta −tb is a circuit of P (G), then f has a square-free term or f has nonsquare-free terms and maxi {ai } = maxi {bi } = 2.

452

Chapter 10

Proof. If G is a simple graph, the result was shown in Corollary 10.2.19. In the case of a multigraph it follows using an identical argument as the one given in the proof of this corollary. 2 Theorem 10.3.14 Let G be a multigraph. If K[G] is normal, then P (G) is generated by circuits with a square-free term. Proof. It follows at once applying Lemma 10.3.13 and Corollary 9.7.4. 2 The converse of this result is not true (see Exercise 10.3.25). Deﬁnition 10.3.15 A sub-multigraph H of G is called a circuit of G if H has one of the following forms: (a) H is an even cycle. (b) H consists of two odd cycles intersecting in exactly one vertex; a loop is regarded as an odd cycle of length 1. (c) H consists of two vertex disjoint odd cycles joined by a path. The circuits of G are in one-to-one correspondence with the circuits of P (G) as we now explain. (See Section 10.2 for a detailed discussion when G is a graph.) Any circuit H of G can be regarded as an even closed walk w = {w0 , w1 , . . . , wr , w0 }, where r is even, w0 , w1 , . . . , wr are the vertices of H (we allow repetitions) and {wi , wi+1 } is an edge (we allow loops) of G for all i. Then the binomial tw = t1 t3 · · · tr−1 − t2 t4 · · · tr is in P (G), where fi = wi−1 wi and ti maps to fi for all i. Remark 10.3.16 The circuits of P (G) with a square-free term correspond to the following types of circuits of G: (a) Even cycles, (b) two odd cycles intersecting in exactly one vertex, (c) two vertex disjoint odd cycles joined by an edge. Thus if K[G] is normal, by Theorem 10.3.14 we obtain a precise graph theoretical description of a generating set of circuits for P (G). Toric ideals of edge subrings of oriented graphs were studied in [184]. In this case the toric ideal is also generated by circuits and the circuits correspond to the cycles of the graph. In [182] they completely characterize the graphs that are complete intersections for all orientation.

Exercises 10.3.17 Prove that K[G] is normal for every connected graph G with six vertices, where K is a ﬁeld.

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10.3.18 Let G be a graph and let G1 , . . . , Gr be its connected components. Then K[G] is normal if and only if K[Gi ] is normal for each i. 10.3.19 Let G be a graph and let G1 , . . . , Gr be the connected components of G. If K is a ﬁeld, prove that K[G] is normal if and only if Gi satisﬁes the odd cycle condition for all i. 10.3.20 Let G1 and G2 be two ﬁnite sets of monomials in disjoint sets of indeterminates. Then K[G1 ∪ G2 ] K[G1 ] ⊗K K[G2 ] and K[G1 ∪ G2 ] K[G1 ] ⊗K K[G2 ] K[G1 ] K[G2 ], is the subring generated by the generators of K[G1 ] and K[G2 ]. 10.3.21 Prove that the smallest connected graph G such that K[G] is nonnormal is the graph consisting of two disjoint triangles joined by a 2-path (see Example 10.5.1). 10.3.22 Le G be a graph and let w be one of the following bow-ties of G with cycles Z1 and Z2 •??

?? Z • ?? 1 ?? ? ?? •? ??? ?? ? Z2 • •

•??

?? ?? ?? ? Z1 •

•

• Z ? •? 2 ?? ?? ?? ? •

prove that Mw is in K[G]. 10.3.23 If G is a graph and K[G] is normal, then P (G) is generated by the binomials that correspond to the circuits of the incidence matrix of G. 10.3.24 Let F = {x1 x2 , x2 x3 , x3 x4 , x1 x4 , x21 , x22 , x23 , x24 } ⊂ K[x]. Find all the bow-ties of the multigraph deﬁned by F . Use Theorem 10.3.9 to show that K[F ] = K[F ][x1 x3 , x2 x4 ]. 10.3.25 (A. Thoma) Let G be the graph whose edge subring is K[G] = K[x1 x2 , x2 x3 , x1 x3 , x1 x4 , x5 x4 , x5 x6 , x6 x4 , x5 x7 , x6 x7 ]. Prove that the toric ideal is a complete intersection generated by circuits but that K[G] is not normal.

454

Chapter 10

10.4

Ehrhart rings of edge polytopes

Let G be a graph on the vertex set V = {x1 , . . . , xn }. The edge polytope of G is the lattice polytope P = conv(A) ⊂ Rn , where A = {v1 , . . . , vq } is the set of vectors in Nn of the form ei + ej such that xi is adjacent to xj . Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K. Recall that the Ehrhart ring of P is A(P) = K[{xα ti |α ∈ Zn ∩ iP}] ⊂ R[t], the edge subring of G is the K-subalgebra K[G] = K[xv1 , . . . , xvq ] ⊂ R, the polytopal subring is given by K[P] = K[xα t| α ∈ Zn ∩ P] ⊂ R[t], and the incidence matrix of G, denoted by A, is the matrix with column vector v1 , . . . , vq . In what follows B denotes the matrix v1 · · · vq B= . 1 ··· 1 As was earlier observed rank(B) = rank(A). We shall keep these notations throughout the rest of this section. If need be, we assume that the graph G has no isolated vertices. Proposition 10.4.1 dim(P) = n − c0 − 1, where c0 denotes the number of bipartite components of G. Proof. By Lemma 10.2.6 and Proposition 9.3.1 one has the equalities dim K[G] = rank(A) = n − c0 and dim K[P] = dim(P) + 1. Thanks to Lemma 9.3.16 one has that K[P] is equal to K[xv1 t, . . . , xvq t]. Thus K[G] K[P] ⊂ A(P). Consequently P has dimension n − c0 − 1. 2 Theorem 10.4.2 K[P] = A(P) if and only if the graph G has at most one non-bipartite connected component. Proof. ⇒) If G has two non-bipartite components G1 , G2 , take Z1 = {x0 , x1 , . . . , xr = x0 },

Z2 = {xr+1 , xr+2 . . . , xr+s = xr+1 }

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two odd cycles of lengths r, s − 1 in G1 , G2 , respectively. Setting β = e1 + · · · + er + er+1 + · · · + er+s−1 , note the equality 9 ! r−1 er + e1 ei + ei+1 m + + β= 2 m m i=1

er+s−1 + er+1 ei + ei+1 + m m i=r+1 r+s−2

!:

where m = r + s − 1. Hence β ∈ Zn ∩ pP, where p = m/2, that is, xβ tp belongs to A(P) = K[P]. Hence (β, p) is in the ﬁeld of fractions of the polytopal subring K[P]. There are integers λ1 , . . . , λq such that e1 + e2 + · · · + er+s−1 + pen+1 = λ1 (v1 , 1) + · · · + λq (vq , 1). Since Z1 and Z2 are in diﬀerent connected components we obtain an equality e1 + · · · + er = j λj vj , where the sum is taken over all j such that the vector vj corresponds to an edge of G1 . Setting 1 = (1, . . . , 1) and taking inner products in the last equality one has r = e1 + · · · + er , 1 = 2 j λj , a contradiction because r is odd. ⇐) By Lemma 9.3.16 K[P] = K[xv1 t, . . . , xvq t]. If G is bipartite then, by Proposition 9.3.30, K[P] = A(P). If G is not bipartite, the equality K[P] = A(P) follows from Theorem 9.3.25 and Proposition 10.2.9. 2 Corollary 10.4.3 [242, p. 423] If G is connected, then K[P] = A(P). Theorem 10.4.4 The multiplicities of A(P) and K[G] are related by

c −1 2 1 e(K[G]) if G is non-bipartite, e(A(P)) = vol(P)d! = e(K[G]) if G is bipartite, where d = dim(P) and c1 is the number of non-bipartite components of G. Proof. By Theorem 9.3.25 one has e(A(P)) = Δr (B)e(K[P]), where r is the rank of A. From Corollary 10.2.11 one derives Δr (A) = |T (Zn /ZA)| = 2c1 . Thus the result follows readily from Proposition 10.2.10 and using the fact that A is totally unimodular if c1 = 0. 2 Lemma 10.4.5 Let G be a connected non-bipartite graph with n vertices and let B = {v1 , . . . , vn } be a set of linearly independent columns of A. If K[G] is normal and 0 = β ∈ R+ B ⊂ Rn , then β can be written as β = μ1 v1 + · · · + μn vn

(μi ≥ 0),

such that v1 , . . . , vn are linearly independent columns of A deﬁning a spanning subgraph of G which is unicyclic connected and non-bipartite.

456

Chapter 10

Proof. For simplicity we identify the edges of G with the columns of A. There are nonnegative real numbers λ1 , . . . , λn such that β = λ1 v1 + · · · + λn vn

(λi ≥ 0).

(10.5)

Consider the subgraph H of G deﬁned by the edges v1 , . . . , vn and the subgraph H of G deﬁned by the set of edges {vi |λi > 0}. Using that v1 , . . . , vn are linearly independent one rapidly derives that H is a spanning subgraph of G whose connected components are unicyclic and non-bipartite (cf. [400, Lemma 9.5]). Let H1 , . . . , Hs be the components of H and Zi the unique odd cycle of Hi , for i = 1, . . . , s. First assume that H has at most one odd cycle, say Z1 . For each cycle Zi with i ≥ 2 there is vji in Zi with λji = 0. Construct a graph K by removing vj1 , . . . , vjs from H and then adding an edge connecting Z1 with Zi for each i ≥ 2, such construction is possible thanks to Corollary 10.3.12. This new graph K is a spanning connected subgraph of G, is non-bipartite and has a unique cycle Z1 . Thus in this case Eq. (10.5) is itself the required expression for β. Next we assume that H has at least two odd cycles Z1 , Z2 . Let v be an edge joining Z1 with Z2 . In this case, using Remark 10.3.5, it is seen that we can rewrite β = μ1 v1 + · · · + μn vn

(μi ≥ 0),

where v1 , . . . , vn are linearly independent edges such that the graph H determined by the set {vi | μi > 0} has one cycle less than H . Thus, by induction, the result follows. 2 For connected graphs the next result was shown in [242]. Theorem 10.4.6 If K[G] is normal, then P has a unimodular covering with support in the vertices of P if and only if G has at most one nonbipartite component. Proof. ⇒) By Theorem 9.3.29 K[F t] = A(P). Hence using Theorem 10.4.4 we obtain that the number of non-bipartite components of G is at most 1. ⇐) Let G1 , . . . , Gc be the connected components of the graph G and let ηi be the number of vertices of Gi . Assume that G1 is non-bipartite and G2 , . . . , Gc are bipartite. We denote the set of columns of the incidence matrix of G by A = {v1 , . . . , vq }. To construct a unimodular covering of P consider any subgraph H of G with connected components H1 , . . . , Hc such that Hi is a spanning tree of Gi for i > 1 and H1 is a unicyclic connected spanning subgraph of G1 with a unique odd cycle. The edge polytope ΔH of H is a simplex of dimension d = n−c. From Proposition 10.2.9 and 1.2.21 ΔH is a simplex of normalized volume equal to 1. We claim that the family of all simplices ΔH is a unimodular covering of P. Let β ∈ P. By Carath´eodory’s Theorem (see Theorem 1.1.18) one

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457

m can write (after permutation of edges) β = i=1 λi vi with λi ≥ 0, where m = n−(c−1) and v1 , . . . , vm are linearly independent. Note that v1 , . . . , vm deﬁne a subgraph H of G. For each i there is a subgraph Hi of Gi such that H = H1 ∪· · ·∪Hc . Since v1 , . . . , vm are linearly independent, Hi is bipartite for i > 1 and hence the number of edges of Hi is at most ηi − 1 for all i > 1. On the other hand H1 has at most η1 edges. Altogether the number of edges of H is at most η1 + (η2 − 1) + · · · + (ηc − 1), but since the number of edges of H is exactly this number one concludes that the number of edges of Hi (resp. H1 ) is ηi − 1 (resp. η1 ) for i > 1 (resp. i = 1). Hence H2 , . . . , Hc are trees. Since G1 is connected and non-bipartite applying Lemma 10.4.5 we can rewrite β such that H1 is a spanning subgraph of G1 which is connected unicyclic and non-bipartite. Therefore β ∈ ΔH , as required. If G1 is bipartite the proof follows using similar arguments as above. 2 Indeed is suﬃces to make G1 = ∅ and proceed as above.

Exercise 10.4.7 Let G be an arbitrary graph and let P be its edge polytope. Prove that K[P] = A(P) if and only if G has at most one non-bipartite connected component and this component satisﬁes the odd cycle condition.

10.5

Integral closure of Rees algebras

Let G be a graph with vertex set X = {x1 , . . . , xn }, let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and let I(G) be the edge ideal of G. As usual we denote the Rees algebra of I(G) by R[I(G)t]. Example 10.5.1 (Hochster’s conﬁguration) Let G be the graph x2 sH H HHx s1 x3 s

x4 s

sx6 x5 s HH HHsx 7

Note that f = (x1 x2 x3 )(x5 x6 x7 ) satisﬁes f ∈ I(G)3 \ I(G)3 . Deﬁnition 10.5.2 A Hochster conﬁguration of order k of G consists of two odd cycles C2r+1 and C2s+1 satisfying the following conditions: (i) C2r+1 ∩ NG (C2s+1 ) = ∅ and k = r + s + 1. (ii) No chord of either C2r+1 or C2s+1 is an edge of G.

458

Chapter 10

Proposition 10.5.3 If G has a Hochster conﬁguration of order k, then I(G)k = I(G)k . Proof. We set I = I(G). The monomial obtained by multiplying all the variables in the two cycles of the conﬁguration is an element of I k \ I k . Indeed, let δ = z1 · · · z2r+1 and γ = w1 · · · w2s+1 be the monomials corresponding to the two cycles in the conﬁguration. By deﬁnition, we have r + s + 1 = k. Therefore δγ ∈ I k−1 \ I k . On the other hand δ 2 ∈ I 2r+1 and γ 2 ∈ I 2s+1 , thus (δγ)2 ∈ I 2(r+s+1) = I 2k . This shows 2 that δγ ∈ I k . Deﬁnition 10.5.4 The cone C(G), over the graph G, is obtained by adding a new vertex t to G and joining every vertex of G to t. Proposition 10.5.5 R[I(G)t] is isomorphic to K[C(G)]. Proof. Let R[I(G)t] = K[{x1 , . . . , xn , tfi | 1 ≤ i ≤ q}] be the Rees algebra of I(G) = (f1 , . . . , fq ), where f1 , . . . , fq are the monomials corresponding to the edges of G. Let K[C(G)] = K[{txi , fj | 1 ≤ i ≤ n, 1 ≤ j ≤ q}] be the monomial subring of the cone. R[I(G)t] and K[C(G)] are integral domains of the same dimension by Corollary 8.2.21 and Proposition 10.1.20. It follows that there is an isomorphism ϕ : R[I(G)t] −→ K[C(G)], induced by ϕ(xi ) = txi and ϕ(tfi ) = fi . ϕ

That is R[I(G)t] K[C(G)].

2

Corollary 10.5.6 Let G be a connected graph and I = I(G) its edge ideal. Then K[G] is normal if and only if the Rees algebra R[It] is normal. Proof. ⇒) Let C(G) be the cone over G. By Proposition 10.5.5 R[I(G)t] is isomorphic to K[C(G)]. Let Mw be a bow-tie of C(G) with edge disjoint cycles Z1 and Z2 joined by a path P . It is enough to verify that Mw is in K[C(G)]. If t ∈ / Z1 ∪ Z2 ∪ P , then w is a bow-tie of G and Mw ∈ K[G]. Assume t ∈ Z1 ∪ Z2 , say t ∈ Z1 . If Z1 ∩ Z2 = ∅, then Mw ∈ K[C(G)]. On the other hand if Z1 ∩ Z2 = ∅, then Mw ∈ K[C(G)] because in this case Z1 and Z2 are joined by the edge {t, z}, where z is any vertex in Z2 . It remains to consider the case t ∈ / Z1 ∪ Z2 and t ∈ P , since G is connected there is a path in G joining Z1 with Z2 . Therefore Mw = Mw1 for some bow-tie w1 of G and Mw ∈ K[G]. ⇐) Conversely if R[I(G)t] is normal, then by Proposition 4.3.42 we obtain that K[G] is normal. 2

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Example 10.5.7 Let F = {x1 x2 , x3 x4 x5 , x1 x3 , x2 x4 , x2 x5 , x1 x5 } and let I = (F ). Using Normaliz [68] we get that R[It] is normal but K[F ] is not. Proposition 10.5.8 Let G be a graph. Then R[I(G)t] is normal if and only if either (i) G is bipartite, or (ii) exactly one of the components, say G1 , is non-bipartite and K[G1 ] is a normal domain. Proof. It follows adapting the proof of Corollary 10.5.6.

2

Corollary 10.5.9 [383] The edge ideal of a graph G is normal if and only if G admits no Hochster conﬁgurations. Remark 10.5.10 (a) Let G be a graph consisting of two disjoint triangles. Note that G itself is a Hochster conﬁguration but G is not contained in a bow-tie, since the two triangles cannot be joined by a path. (b) If two disjoint odd cycles Z1 , Z2 of a graph G are in the same connected component, then the two cycles form a Hochster conﬁguration if and only if Z1 , Z2 cannot be connected by an edge of G and Z1 , Z2 have no chords in G Integral closure of Rees algebras Consider the endomorphism ϕ of the ﬁeld K(x1 , . . . , xn , t) deﬁned by xi → xi t, t → t−2 . If I = I(G) is the edge ideal of G, then by Proposition 10.5.5 it induces an isomorphism ϕ

R[It] −→ K[C(G)],

xi → txi ,

xi xj t → xi xj .

We thank Wolmer Vasconcelos for showing us how to get the description of the integral closure of R[It] given below. First we describe the integral closure of K[C(G)]. If Z1 = {x1 , x2 , . . . , xr = x1 } and Z2 = {z1 , z2 , . . . , zs = z1 } are two edge disjoint odd cycles in C(G). Then one has Mw = x1 · · · xr z1 · · · zs for some bow-tie w of C(G), this follows readily because C(G) is a connected graph. In particular Mw is in the integral closure of K[C(G)]. Observe that if t occurs in Z1 or Z2 , then Mw ∈ K[C(G)] because in this case either Z1 and Z2 meet at a point, or Z1 and Z2 are joined by an edge in C(G); see Proposition 10.3.10. Thus in order to compute the integral closure of K[C(G)] the only bow-ties that matter are those deﬁning the set: / supp(Mw )}. B = {Mw | w is a bow-tie in C(G) such that t ∈ Therefore by Theorem 10.3.9 we have proved: Proposition 10.5.11 K[C(G)] = K[C(G)][B].

460

Chapter 10

To obtain a description of the integral closure of the Rees algebra of I note that using ϕ together with the equality

(tx1 )Mw = (tz1 )

r+1 2 7

(x2i−1 x2i )

i=1

s−1 2 7

(z2i z2i+1 ),

i=1

where xr+1 = x1 , we see that Mw is mapped back in the Rees algebra in the element r+s r+s Mw t 2 = x1 x2 · · · xr z1 z2 · · · zs t 2 . If B is the set of all monomials of this form, then one obtains: Proposition 10.5.12 R[It] = R[It][B ].

Exercises 10.5.13 Let G be a graph. Prove that I(G)2 is integrally closed. 10.5.14 If G is a graph consisting of two disjoint triangles, then K[G] is normal and the Rees algebra of I(G) is not normal. 10.5.15 Let G be a graph. If H is the whisker graph of G or the cone of G and I(G) is normal, then I(H) is normal. 10.5.16 Let G be a graph and let m = (r + s)/2 be the least positive integer such that G has two vertex disjoint odd cycles of lengths r and s. If I = I(G) is the edge ideal of G, prove that I i is integrally closed for i < m. 10.5.17 If G is any non-discrete graph with n vertices and C(G) its cone, prove that the edge subring K[C(G)] has dimension n + 1. 10.5.18 Let G be a simple graph with vertex set X = {x1 , . . . , xn } and let I ⊂ R = K[X] be its edge ideal. If ϕ, ψ are the maps of K-algebras deﬁned by the diagram R[t1 , . . . , tq ] ψ

K[C(G)]

ϕ

/ R[It]

ϕ ti /o o/ /o / T fi O O ψ O fi

ϕ xi /o o/ /o / xi O O ψ O xti

where C(G) is the cone of G and f1 , . . . , fq are the edge generators, then ker(ϕ) = ker(ψ). 10.5.19 Let R = K[x1 , x2 , x3 , y1 , y2 , y3 , y4 ] and I = I1 J3 + I3 J1 , where Ir (resp. Jr ) denote the ideal of R generated by the square-free monomials of degree r in the xi variables (resp. yi variables). If z = x1 x2 x3 y12 y2 y3 y4 , then z 2 ∈ I 4 and z ∈ I 2 \ I 2 .

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10.5.20 Let G be a graph. Explain the diﬀerences between a circuit of G and a Hochster conﬁguration of G. 10.5.21 Let G be a bipartite graph and I(G) its edge ideal. Prove that the toric ideal of the Rees algebra of I(G) has a universal Gr¨ obner basis consisting of square-free binomials.

10.6

Edge subrings of complete graphs

In this section we treat two special types of edge subrings, attached to complete graphs, and compute their Hilbert series and a-invariant. Edge subrings of bipartite graphs One result that simpliﬁes the study of an edge subring associated to a complete bipartite graph is the fact that its toric ideal is a determinantal prime ideal. Let us begin with a classical formula for determinantal ideals. Theorem 10.6.1 [73, Theorem 2.5] Let T = (tij ) be an m × n matrix of indeterminates over a ﬁeld K and Ir (T ) the ideal generated by the r-minors of T . If m ≤ n and 1 ≤ r ≤ m + 1, then height(Ir (T )) = (m − r + 1)(n − r + 1). Proposition 10.6.2 Let K be a ﬁeld and let Km,n be the complete bipartite graph. Then, the toric ideal P of K[Km,n ] is generated by the 2 × 2 minors of a generic m × n matrix. Proof. Let V1 = {x1 , . . . , xm } and V2 = {y1 , . . . , yn } be the bipartition of Km,n and let S = K[{tij | 1 ≤ i ≤ m, 1 ≤ j ≤ n}] be a polynomial ring over a ﬁeld K in the indeterminates tij and consider the graded homomorphism of K-algebras ϕ : S −→ K[Km,n ] = K[xi yj ’s],

ϕ(tij ) = xi yj .

We claim that the kernel of ϕ is generated by the 2 × 2 minors of a generic m × n matrix. We set T = (tij ). It is clear that the ideal I2 (T ), generated by the 2×2 minors of T , is contained in P = ker(ϕ). On the other hand, since the graph is bipartite, one has that the dimension of K[Km,n ] is equal to m + n − 1; see Corollary 10.1.21. Therefore height(P ) = (mn) − (m + n − 1) = (m − 1)(n − 1) = height(I2 (T )), the last equality uses Theorem 10.6.1. Since they are both prime ideals, we 2 have I2 (T ) = P . By Proposition 10.3.1 the ring K[Km,n ] is normal and Cohen–Macaulay. The Gorensteiness of K[Km,n ] has been dealt with in great detail in [71].

462

Chapter 10

Proposition 10.6.3 [418] Let S/P be the presentation of the edge subring K[Km,n ]. If n ≥ m, then the Hilbert series of S/P is given by m−1

F (S/P, z) =

i=0

m−1 i

n−1 i z i

(1 − z)n+m−1

.

Corollary 10.6.4 If n ≥ m, then the a-invariant and the multiplicity of K[Km,n ] are given by: m+n−2 . a(K[Km,n ]) = −n and e(K[Km,n ]) = m−1 Proof. It follows using Remark 5.1.7.

2

Proposition 10.6.5 Let G be a spanning connected subgraph of Km,n . If n ≥ m, then the a-invariant of K[G] satisﬁes a(K[G]) ≤ a(K[Km,n ]) = −n. Proof. From Corollary 10.6.4 one has a(K[Km,n ]) = −n. Hence one may proceed as in the proof of Proposition 12.4.13 to rapidly derive the asserted inequality. 2 An application of Dedekind–Mertens formula ring and f = f (t) ∈ R[t] is a polynomial, say

If R is a commutative

f = a 0 + · · · + a m tm , the content of f is the R-ideal (a0 , . . . , am ). It is denoted by c(f ). Given another polynomial g, the Gaussian ideal of f and g is the R-ideal c(f g). This ideal bears a close relationship to the ideal c(f )c(g), one aspect of which is expressed in the classical lemma of Gauss: If R is a principal ideal domain, then c(f g) = c(f )c(g). In general, these two ideals are diﬀerent but one aspect of their relationship is given by a formula due to Dedekind–Mertens (see [126] and [334]): c(f g)c(g)m = c(f )c(g)m+1 .

(10.6)

We consider the ideal c(f g) in the case when f , g are generic polynomials. It turns out that some aspects of the theory of Cohen–Macaulay rings show up very naturally when we closely examine c(f g).

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One path to our analysis, and its applications to Noether normalizations, starts by multiplying both sides of (10.6) by c(f )m ; we get c(f g)[c(f )c(g)]m = c(f )c(g)[c(f )c(g)]m .

(10.7)

It will be shown that this last formula is sharp in terms of the exponent m = deg(f ). To make this connection, we recall the notion of a reduction of an ideal. Let R be a ring and I an ideal. A reduction of I is an ideal J ⊂ I such that, for some nonnegative integer r, the equality I r+1 = JI r holds. The smallest such integer is the reduction number rJ (I) of I relative to J, and the reduction number r(I) of I is the smallest reduction number among all reductions J of I. Thus (10.7) says that J = c(f g) is a reduction for I = c(f )c(g), and that the reduction number is at most min{deg(f ), deg(g)}. The following result solves the question of Noether normalizations for monomial subrings of complete bipartite graphs. Theorem 10.6.6 [126] Let R = K[x0 , . . . , xm , y0 , . . . , yn ] be a polynomial ring over a ﬁeld K and let hq = i+j=q xi yj . Then A = K[h0 , h1 , . . . , hm+n ] → S = K[{xi yj | 0 ≤ i ≤ m, 0 ≤ j ≤ n}] is a Noether normalization of S. Proof. Let R[t] be a polynomial ring in a new variable t and f, g ∈ R[t] the generic polynomials f = f (t) =

m

xi ti

and g = g(t) =

i=0

n

y j tj .

j=0

Set I = c(f )c(g) and J = c(f g). Note that J is a reduction of I by Eq. (10.7) and more precisely JI m = I m+1 . Therefore R[Jt] → R[It] is an integral extension; indeed R[It] is generated as an R[Jt]-module by a ﬁnite set of elements of t-degree at most m. On the other hand, as I and J are generated by homogeneous polynomials of the same degree, one has an integral embedding: R[Jt] ⊗R R/m = K[h0 , . . . , hm+n ] → R[It] ⊗ R/m = K[xi yj ’s], where m = (x0 . . . , xm , y0 , . . . , yn ). To complete the proof note that the 2 dimension of K[xi yj ’s] is equal to n + m + 1, by Corollary 10.1.21.

464

Chapter 10

Theorem 10.6.7 [97]Let R = K[x0 , . . . , xm , y0 , . . . , yn ] be a polynomial ring over a ﬁeld K and let f, g ∈ R[t] be the generic polynomials f=

m

xi ti

and

i=0

g=

n

y j tj

j=0

with n ≥ m. If I = c(f )c(g) and J = c(f g), then rJ (I) = m. Proof. By Theorem 10.6.6 there is a Noether normalization:

A = K[h0 , h1 , . . . , hm+n ] → S = K[xi yj ’s],

where hq = i+j=q xi yj . Set I = (f1 , . . . , fs ). We note that the ideal I = (xi yj ’s) is the edge ideal associated to a complete bipartite graph G which is the join of two discrete graphs, one with m + 1 vertices and another with n + 1 vertices. Since K[G] is Cohen–Macaulay (see Corollary 10.3.12), using Proposition 3.1.27 we get K[G] = Af β1 ⊕ · · · ⊕ Af βk , βi ∈ Ns . By Corollary 10.6.4 the a-invariant of K[G] is −n − 1, we can compute the Hilbert series of K[G] using the decomposition above to get that max {deg(f βi )} = m,

1≤i≤k

where the degree is taken with respect to the normalized grading. On the other hand assume r = rJ (I) and JI r = I r+1 , since we already know the inequality r ≤ m. It suﬃces to prove r ≥ m. Note Af α , K[G] = α∈I

s where I = { α | α = (v1 , . . . , vs ) ∈ Ns and |α| = i=1 vi ≤ r} . If r < m, this rapidly leads to a contradiction because, by Exercise 10.6.14, there is J ⊂ I such that K[G] can be written as ) K[G] = Af α . 2 α∈J

Edge subrings of complete graphs Let R = K[x1 , . . . , xn ] be a ring of polynomials over a ﬁeld K and let Kn be the complete graph on the vertex set X = {x1 , . . . , xn }. For i < j, we set fij = xi xj . Let K[Kn ] = K[{fij | 1 ≤ i < j ≤ n}] be the K-subring of R spanned by the fij . Consider the homomorphism ψ : An = K[{tij | 1 ≤ i < j ≤ n}] → K[Kn ], induced by tij → fij . The ideal Pn = ker(ψ) is the presentation ideal or toric ideal of K[Kn ].

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Proposition 10.6.8 [418] If the terms in An are ordered lexicographically by t12 ≺ · · · ≺ t1n < t23 ≺ · · · ≺ t2n ≺ · · · ≺ t(n−1)n , then the set B = {tij tk − ti tjk , tik tj − ti tjk | 1 ≤ i < j < k < ≤ n}. is a minimal generating set for Pn and B is a reduced Gr¨ obner basis for Pn . Example 10.6.9 If K[{tij | 1 ≤ i < j ≤ 5}] has the lex ordering t45 ≺ t34 ≺ t35 ≺ t23 ≺ t24 ≺ t25 ≺ t12 ≺ t13 ≺ t14 ≺ t15 . Then the reduced Gr¨ obner basis for the toric ideal P of K[K5 ] is equal to: t35 t24 − t45 t23 , t25 t13 − t35 t12 , t34 t15 − t45 t13 , t24 t13 − t34 t12

t34 t25 − t45 t23 , t45 t23 t13 − t34 t35 t12 , t35 t14 − t45 t13 , t23 t14 − t34 t12 , t23 t15 − t35 t12 , t24 t15 − t45 t12 . t25 t14 − t45 t12 .

Theorem 10.6.10 [418] Let Kn be the complete graph on n vertices and An /Pn be the presentation of K[Kn ]. If n ≥ 3, then the Hilbert series Fn (z) of An /Pn satisﬁes: 2 n n(n − 3) z+ (1 − z) Fn (z) = 1 + zm 2m 2 m=2 n

n

Corollary 10.6.11 The multiplicity e(K[Kn ]), of the ring K[Kn ], is equal to 2n−1 − n. Theorem 10.6.12 Let G be a connected graph with n vertices and q edges. Then the following are sharp upper bounds for the multiplicity of K[G]: e(K[G]) ≤

⎧ n−1 − n, ⎨2

if G is non-bipartite,

⎩ m+r−2 m−1 ,

if G is a spanning subgraph of Km,r .

Proof. It follows from Corollary 10.6.4 and Corollary 10.6.11.

2

It is easy to produce examples where those bounds are very lose. The extreme cases being the unicyclic connected non-bipartite graphs and the trees, because in both cases the multiplicity of K[G] is equal to 1. In our case adding edges to G increases the volume of the edge polytope (see Lemma 9.3.16); hence the multiplicity is an increasing function of the edges. For additional information on the multiplicity of edge subrings of bipartite planar graphs, see [189] and the references therein.

466

Chapter 10

Exercises 10.6.13 Let R[t] be a polynomial ring over a ring R. If f = f (t) and g = g(t) are two polynomials in R[t], prove the Dedekind–Mertens formula c(f g)c(g)m = c(f )c(g)m+1

(m = deg(f )).

10.6.14 Let M be a graded free module over a polynomial ring A such that M = ⊕ri=1 Afi , where fi is homogeneous for all i. If M is generated by homogeneous elements g1 , . . . , gs , then M = ⊕rj=1 Agij , for some i1 , . . . , ir . 10.6.15 Let A = K[x1 , . . . , xm ] and R = K[y1 , . . . , yn ] be two polynomial rings over a ﬁeld K, prove that the Segre product (A0 ⊗K R0 ) ⊕ (A1 ⊗K R1 ) ⊕ · · · ⊂ A ⊗K R is isomorphic to S = K[xi yj ’s] ⊂ K[xi ’s, yj ’s]. 10.6.16 Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and m = (x1 , . . . , xn ). Prove that the toric ideal of the Rees algebra R[mt] is the ideal generated by the 2-minors of the following matrix of indeterminates: x1 x2 · · · xn . X= t1 t2 · · · tn 10.6.17 Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and I = (x1 , . . . , xn )k . Set s = −n/k + n, and write n = qk + r, for some q, r ∈ N such that 0 ≤ k < r. By a “counting degrees” argument prove: · · · xk−1 ∈ Is xk−1 n 1 xk−1 1

k−1 · · · xk−1 xr−1 q q+1 xq+2

· · · xk−1 n

∈I

s

if r = 0, if r > 0.

10.6.18 Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and m = (x1 , . . . , xn ). If I = mk and J = (xk1 , . . . , xkn ), then J is a reduction of I and the reduction number of I relative to J is given by FnG + n, where x is the ceiling of x. rJ (I) = − k 10.6.19 Let S be a standard algebra over an inﬁnite ﬁeld K and A = K[h1 , . . . , hd ] → S a Noether normalization of S with hi ∈ S1 for all i. If b1 , . . . , bt is a minimal set of homogeneous generators of S as an A-module, prove (h1 , . . . , hd )Sr = Sr+1 , where r = maxi {deg(bi )}, and that r ≥ 0 is the minimum integer where equality occurs. The reduction number of S, denote by r(S), is the minimum r taken over all Noether normalizations. See [413, Chapter 9] for a study of several degrees of complexity associated to a graded module.

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10.6.20 Let S be a standard algebra over an inﬁnite ﬁeld K and a(S) its a-invariant. If S is Cohen–Macaulay, then r(S) = a(S) + dim(S). 10.6.21 Let φ : B → S be a graded epimorphism of standard K-algebras, where K is an inﬁnite ﬁeld. If dim(B) = dim(S) and ı

A = K[h1 , . . . , hd ] → B is a Noether normalization of B with hi ∈ S1 for all i, prove that S is integral over A = K[φ(h1 ), . . . , φ(hd )] and that the composite ı

φ

A = K[h1 , . . . , hd ] → B −→ S is a Noether normalization of S. 10.6.22 Let φ : B → S be a graded epimorphism of standard K-algebras, where K is an inﬁnite ﬁeld. If dim(B) = dim(S), prove r(B) ≥ r(S).

10.7

Edge cones of graphs

In this section we give explicit combinatorial descriptions of the edge cone of a graph. In Chapter 11 these descriptions will be used to compute the ainvariant and the canonical module of an edge subring. The main tools used to show these descriptions are linear algebra (Farkas’s Lemma, incidence matrices of graphs, Carath´eodory’s Theorem), graph theory and polyhedral geometry (ﬁnite basis theorem, facet structure of polyhedra). First we ﬁx the notation that will be used throughout this section and adapt the terminology and notation of Chapter 1 for the case of edge cones. Let G be a simple graph with vertex set V = {x1 , . . . , xn } and edge set E = {f1 , . . . , fq }. Every edge fi is an unordered pair of distinct vertices fi = {xij , xik }. If fi = {xij , xik } deﬁne vi = eij + eik , where ei is the ith unit vector in Rn . The incidence matrix of G, denoted by A, is the matrix of size n × q whose columns are precisely the vectors v1 , . . . , vq . We set AG (or simply A if G is understood) equal to the set {v1 , . . . , vq } of row vectors of the transpose of the incidence matrix of G. Since vi represents an edge of G, sometimes vi is called an edge or an edge vector . The edge cone of G, denoted by R+ A, is the cone generated by A. Notice that R+ A = (0) if G is not a discrete graph. By Corollary 10.2.11 one has n − c0 (G) = rank (A) = dim R+ A, where c0 (G) is the number of bipartite components of G. Lemma 10.7.1 If xi is not an isolated vertex of the graph G, then the set F = Hei ∩ R+ A is a proper face of the edge cone.

468

Chapter 10

Proof. Note F = ∅ because 0 ∈ F , and R+ A ⊂ He+i . Since xi is not an 2 isolated vertex R+ A ⊂ Hei . Let A be an independent set of vertices of G. The supporting hyperplane of the edge cone deﬁned by xi = xi xi ∈A

xi ∈N (A)

will be denoted by HA , where N (A) is the neighbor set of A. The sum over an empty set is deﬁned to be zero. Lemma 10.7.2 If A is an independent set of vertices of the graph G and F = R+ A∩HA , then either F is a proper face of the edge cone or F = R+ A. Proof. It suﬃces to prove R+ A ⊂ HA− . Take an edge {xj , x } of G. If {xj , x } ∩ A = ∅, then ej + e is in HA , else ej + e is in HA− . 2 Deﬁnition 10.7.3 The support of a vector β = (βi ) ∈ Rn is deﬁned as supp(β) = {βi | βi = 0}. Lemma 10.7.4 [419] Let G1 , . . . , Gr be the components of G. If G1 is a tree with at least two vertices and G2 , . . . , Gr are unicyclic non-bipartite n graphs, then ker(A G ) = (β) for some β ∈ R with supp(β) = {1, −1} such that V (G1 ) = {vi ∈ V | βi = ±1}. Proof. If G is a tree, using induction on the number of vertices it is not hard to show that ker(A ) = (β), where supp(β) = {1, −1}. On the other hand if G = Gi for some i ≥ 2, then A is non-singular and ker(A ) = (0). The general case follows readily by writing A as a “diagonal” matrix A = diag(A G1 , . . . , AGm ),

where AGi is the incidence matrix of the graph Gi . Observe the equality V (G1 ) = {xi ∈ V | βi = ±1}. 2 To determine whether HA deﬁnes a facet of R+ A consider the subgraph L = L1 ∪ L2 , where L1 is the subgraph of G with vertex set and edge set V (L1 ) = A ∪ N (A) and E(L1 ) = {z ∈ E| z ∩ A = ∅} respectively, and L2 = G[S] is the subgraph of G induced by S = V \ V (L1 ). The vectors in A ∩ HA correspond precisely to the edges of L. Remark 10.7.5 Let F be a facet of R+ A deﬁned by the hyperplane Ha . If the rank of A is n, then it is not hard to see that Ha is generated by a linearly independent subset of A. See Proposition 1.1.23.

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Theorem 10.7.6 [419] If rank(A) = n, then F is a facet of R+ A if and only if either (a) F = Hei ∩ R+ A for some i, where all the connected components of G \ {xi } are non-bipartite graphs, or (b) F = HA ∩ R+ A for some independent set A ⊂ V such that L1 is a connected bipartite graph, and the connected components of L2 are non-bipartite graphs. Proof. ⇒) Assume that F is a facet of R+ A. By Remark 10.7.5 we may assume that there is a ∈ Rn such that (i) F = R+ A ∩ Ha , a, vi ≤ 0 ∀ i ∈ {1, . . . , q}, and (ii) v1 , . . . , vn−1 are linearly independent vectors in Ha . Consider the subgraph D of G whose edges correspond to v1 , . . . , vn−1 and its vertex set is the union of the vertices in the edges of D. Let Γ be the transpose of the incidence matrix of D, and D1 , . . . , Dr the connected components of D. Without loss of generality one may assume ⎤ ⎡ Γ1 0 . . . 0 ⎢ 0 Γ2 . . . 0 ⎥ ⎥ ⎢ Γ=⎢ . . ⎥ , rank(Γ) = n − 1, .. . . ⎣ .. . .. ⎦ . 0

0

. . . Γr

where Γi is the transpose of the incidence matrix of Di . We set ni and qi equal to the number of vertices and edges of Di , respectively. Let Θ be the matrix with rows v1 , . . . , vn−1 . We consider two cases. Case (I): Assume Θ has a zero column. Hence it has exactly one zero column rand D has r vertex set V (D) = V \ {xi }, for some i. This implies that i=1 ni = i=1 qi = n − 1. Since ni ≤ qi for all i, one obtains ni = qi for all i. Note that c0 (D) = 0, by Lemma 10.2.6. Hence D1 , . . . , Dr are unicyclic non-bipartite graphs. As any component of G \ {xi } contains a component of D, we are in case (a). Case (II): Assume that all the columns of Θ are non-zero, that is, D has vertex set equal to V . Note that in this case n=

r i=1

ni = 1 +

r

qi .

(10.8)

i=1

Since rank (Γ) = n − c0 (D) = n − 1, we get that D has exactly one bipartite component, say D1 . Hence n1 − 1 ≤ q1 and ni ≤ qi for all i ≥ 2, which together with Eq. (10.8) yields q1 = n1 − 1 and ni = qi for all i ≥ 2. Altogether D1 is a tree and D2 , . . . , Dr are unicyclic non-bipartite graphs.

470

Chapter 10

By Lemma 10.7.4, ker(Γ) = (β) and supp(β) ⊂ {1, −1} for some β ∈ Rn . We may harmlessly assume a = β, because Ha = Hβ and a ∈ (β). Set A = {xi ∈ V | ai = 1} and B = {xi ∈ V | ai = −1}. Note that A = ∅, because D has no isolated vertices and its vertex set is V . We will show that A is an independent set in G and B = N (A), where the neighbor set of A is taken w.r.t G. On the contrary if {xi , xj } is an edge of G for some xi , xj in A, then vk = ei + ej and by (i) we get a, vk ≤ 0, which is impossible because a, vk = 2. This proves that A is an independent set. Next we show N (A) = B. If xi ∈ N (A), then vk = ei + ej for some xj in A, using (i) we obtain a, vk = ai + 1 ≤ 0 and ai = −1, hence xi ∈ B. Conversely if xi ∈ B, since D has no isolated vertices, there is 1 ≤ k ≤ n − 1 so that vk = ei + ej , for some j, by (ii) we obtain a, vk = −1 + aj = 0, which shows that xj ∈ A and xi ∈ N (A). From the proof of Lemma 10.7.4 we obtain V (D1 ) = A∪N (A). Therefore v1 , . . . , vn−1 are in HA and Ha = HA . Observe that L1 (see notation before Remark 10.7.5) cannot contain odd cycles because A is independent and every edge of L1 contains a vertex in A. Hence L1 is bipartite, that L1 is connected follows from the fact that D1 is a spanning tree of L1 . Note that every connected component of L2 is non-bipartite because it contains some Di , with i ≥ 2. ⇐) Assume that F is as in (a). Since the vectors in A ∩ Hei correspond precisely to the edges of G \ {xi } one obtains that F is a facet of the edge cone. If F is as in (b), a similar argument shows that F is also a facet. 2 Theorem 10.7.7 If G is a connected graph and F is a facet of the edge cone of G, then either (a) F = R+ A ∩ {x ∈ Rn | xi = 0} for some 1 ≤ i ≤ n, or (b) F = R+ A ∩ HA for some independent set A of G. Proof. By Theorem 10.7.6 one may assume that G is bipartite and n ≥ 3. Notice that F = R+ B, where B = {vi ∈ A | vi , a = 0}. Since dim(F ) is equal to n − 2, one may assume that v1 , . . . , vn−2 are linearly independent vectors in F . There is 0 = a ∈ Rn such that (i) F = R+ A ∩ Ha , a, vi = 0 ∀ i ∈ {1, 2, . . . , n − 2}, and (ii) a, vi ≤ 0 for i ∈ {1, 2, . . . , q}. Let D be the subgraph of G whose edges correspond to v1 , . . . , vn−2 and its vertex set is the union of the vertices in the edges of D. Setting k = |V (D)|, by Corollary 10.2.11 one has n − 2 = rank (AD ) = k − c0 (D), where AD is the incidence matrix of D and c0 (D) is the number of bipartite

Monomial Subrings of Graphs

471

components of D. Thus 0 ≤ n − k = 2 − c0 (D). This shows that either c0 (D) = 1 and k = n − 1 or c0 (D) = 2 and k = n. Case (I): Assume c0 (D) = 1 and k = n − 1. Set V (D) = {x1 , . . . , xn−1 }. As D is a tree with n − 2 edges and vi , a = 0 for i = 1, . . . , n − 2, applying Lemma 10.7.4, one may assume a = (a1 , . . . , an−1 , an ), where ai = ±1 for 1 ≤ i ≤ n − 1. Subcase(I.a): Assume that an = 0. Set A = {xi ∈ V | ai = 1} and B = {xi ∈ V | ai = −1}. Note that ∅ = A ⊂ V (D), because D is a tree with at least two vertices. We will show that A is an independent set of G and B = NG (A). If A is not an independent set of G, there is an edge {xi , xj } of G for some xi , xj in A. Thus vk = ei + ej and by (ii) we get a, vk ≤ 0, which is impossible because a, vk = 2. Next we will show NG (A) = B. If xi ∈ N (A), then vk = ei + ej for some xj in A, using (ii) we obtain a, vk = ai + 1 ≤ 0 and ai = −1, hence xi ∈ B. Conversely if xi ∈ B, since D has no isolated vertices, there is 1 ≤ k ≤ n − 2 so that vk = ei + ej , for some j, by (i) we obtain a, vk = −1 + aj = 0, which shows that xj ∈ A and xi ∈ N (A). Therefore Ha = HA and F is as in (b). Subcase(I.b): Assume an > 0. In this case consider a = −en . Then: (c) vi , a = 0 for i = 1, . . . , n − 2. (d) If vj ∈ R(v1 , . . . , vn−2 ) ⇒ vj , a = vj , a = 0. (e) If vj ∈ / R(v1 , . . . , vn−2 ), then vj , a = −1 and vj , a < 0. / R(v1 , . . . , vn−2 ), then vj = ek + en , because To prove (e) assume vj ∈ otherwise the “edge” vj added to the tree D form a graph with a unique even cycle, a contradiction. Thus vj , a = −1. On the other hand vj , a < 0, because if vj , a = 0, then Ha contains the linearly independent vectors v1 , . . . , vn−2 , vj , a contradiction to dim(F ) = n − 2. Thus given 1 ≤ j ≤ q, then vj , a = 0 if and only if vj , a = 0. Note R + A ∩ Ha = R + B ,

(10.9)

where B = {vi ∈ A | vi , a = 0}. Hence using F = R+ B and Eq. (10.9) we get F = R+ A ∩ Ha , as required. The case an < 0 can be treated similarly. Case (II): Assume c0 (D) = 2 and k = n. Let D1 and D2 be the two components of D and set V1 = V (D1 ) and V2 = V (D2 ). Using Lemma 10.7.4 we can relabel the vertices of D to write a = rb+sc, where 0 = r ≥ s ≥ 0 are rational numbers and b = (b1 , . . . , bm , 0, . . . , 0), c = (0, . . . , 0, cm+1 , . . . , cn ), where V1 = {x1 , . . . , xm }, bi = ±1 for i ≤ m, and ci = ±1 for i > m. Set a = b. Note the following: (f) vi , a = 0 for i = 1, . . . , n − 2. (g) If vj ∈ R(v1 , . . . , vn−2 ) ⇒ vj , a = vj , a = 0. / R(v1 , . . . , vn−2 ), then vj , a < 0 and vj , a = −1 . (h) If vj ∈

472

Chapter 10

Let us prove assertion (h). Assume vj ∈ / R(v1 , . . . , vn−2 ). Since dim(F ) is equal to = n − 2, by (ii) we have vj , a < 0. Observe that if an “edge” vk has vertices in V1 (resp. V2 ), then vk , a = 0. Indeed if we add the edge vk to the tree D1 (resp. D2 ) we get a graph with a unique even cycle and this implies that v1 , . . . , vn−2 , vk are linearly dependent, that is, vk , a = 0. Thus vj = ei + e for some xi ∈ V1 and x ∈ V2 . From the inequality vj , a = rvj , b + svj , c = rbi + sc < 0 we obtain bi = −1 = vj , a , as required. In particular note that Ha is a supporting hyperplane of R+ A because R+ A ⊂ Ha− . We may now proceed as in case (I) to obtain the equality F = R+ A ∩ Ha = R+ A ∩ Ha . Using that D1 has no isolated vertices, one may proceed as in case (I) to prove that Ha = HA for the independent set of vertices given by A = {xi | bi = 1}. 2 Theorem 10.7.8 If G is a connected graph, then ! ! n − + R+ A = HA Hei , A∈F

i=1

where F is the family of all the independent sets of vertices of G and He+i is the closed halfspace {x ∈ Rn | xi ≥ 0}. Proof. Set C = R+ A. From Theorem 10.7.7 and Corollary 1.1.32 we get ! ! n HA− He+i . C = aﬀ(C) ∩ A∈F

i=1

We may assume that G is bipartite, otherwise aﬀ(C) = Rn and there is nothing to prove. If (V1 , V 2 ) is the bipartition of G, then aﬀ(R+ A) is the hyperplane x = xi ∈V1 i xi ∈V2 xi , because dim(R+ A) = n − 1. Since − − 2 HV1 ∩ HV2 = HV1 = aﬀ(C), we obtain the required equality. Studying the bipartite case For connected bipartite graphs we will present more precise results about the irreducible representations of edge cones. Let G be a connected bipartite graph with bipartition (V1 , V2 ). To avoid repetitions we continue using the notation introduced at the beginning of the section. Proposition 10.7.9 If A is an independent set of G such that A = Vi for i = 1, 2, then F = R+ A ∩ HA is a proper face of the edge cone. Proof. Assume N (A) = V2 . Take any xi ∈ V1 \ A and any xj ∈ V2 adjacent to xi , then ei + ej ∈ / HA . This means that R+ A HA . Thus we may assume N (A) = Vi for i = 1, 2.

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Case (I): N (A) ∩ Vi = ∅ for i = 1, 2. If the vertices in N (A) ∩ Vi for i = 1, 2 are only adjacent to vertices in A, then pick vertices xi ∈ N (A) ∩ Vi and note that there is no path between x1 and x2 , a contradiction. Thus there must be a vector in the edge cone which is not in HA . Case (II): A V1 . If the vertices in N (A) are only adjacent to vertices in A. Then a vertex in A cannot be joined by a path to a vertex in V2 \ N (A), 2 a contradiction. As before we obtain R+ A ⊂ HA . Proposition 10.7.10 Let F be the family of all independent sets A of G such that HA ∩ R+ AG is a facet. If A is in F and Vi ∩ A = ∅ for i = 1, 2, then HA− is redundant in the following expression of the edge cone R+ A = aﬀ(R+ A) ∩

A∈F

! HA−

n

! He+i

.

i=1

Proof. Set A = {v1 , . . . , vq }. One can write A = A1 ∪ A2 with Ai ⊂ Vi for i = 1, 2. There are v1 , . . . , vn−2 linearly independent vectors in HA ∩ R+ A, where n is the number of vertices of G. Consider the subgraph D of G whose edges correspond to v1 , . . . , vn−2 and its vertex set is the union of the vertices in those edges. Note that D cannot be connected. Indeed there is no edge of D connecting a vertex in NG (A1 ) with a vertex in NG (A2 ) because all the vectors v1 , . . . , vn−2 satisfy the equation xi ∈A

xi =

xi .

xi ∈NG (A)

Hence D is a spanning subgraph of G with two components D1 , D2 , which are trees, such that V (Di ) = Ai ∪ NG (Ai ), i = 1, 2. Therefore HAi is a proper support hyperplane deﬁning a facet Fi = HAi ∩ R+ A, that is A1 , A2 are in F . Since HA−1 ∩ HA−2 is contained in HA− the proof is complete. 2 Proposition 10.7.11 If A2 V2 , F = HA2 ∩ R+ A is a facet of the edge cone of G and A = A ∪ {0}, then HA−2 ∩ aﬀ(A ) =

HA−1 ∩ aﬀ(A ), where A1 = V1 \ N (A2 ) = ∅, or He+i ∩ aﬀ(A ), for some xi with G \ {xi } connected.

Proof. Let us assume that G has n vertices x1 , . . . , xn and that V1 is the set of the ﬁrst m vertices of G. There are v1 , . . . , vn−2 linearly independent vectors in the hyperplane HA2 . Consider the subgraph D of G whose edges correspond to v1 , . . . , vn−2 and its vertex set is the union of the vertices in those edges. As G is connected, either D is a tree with n − 1 vertices or D is a spanning subgraph of G with two connected components.

474

Chapter 10

Assume that D is a tree (the other case is left as an exercise), write V (D) = V \ {xi } for some i. Note vj , vA2 = −vj , ei , j = 1, . . . , q, where vA2 = ei − ei . xi ∈A2

xi ∈N (A2 )

Indeed if the “edge” vj has vertices in V (D), then both sides of the equality are zero, otherwise write vj = ei + e . Observe xi ∈ / A2 and x ∈ N (A2 ) because as HA2 is a facet it cannot contain vj , thus both sides of the equality are equal to −1. As a consequence since aﬀ(A ) = R(v1 , . . . , vn−2 , vj ) for some vj = ei + e we rapidly obtain α, vA2 = −α, ei , ∀ α ∈ aﬀ(A ). Therefore HA−2 ∩ aﬀ(A ) = He+i ∩ aﬀ(A ), as required. 2 From the proof of Proposition 10.7.11 we get: Lemma 10.7.12 Let F = HA ∩ R+ A be a facet of R+ A with A V1 . (a) If N (A) = V2 , then A = V1 \{xi } for some xi ∈ V1 and F = Hei ∩R+ A. (b) If N (A) V2 , then F = HV2 \N (A) ∩ R+ A and N (V2 \ N (A)) = V1 \ A. Proposition 10.7.13 Let A V1 . Then F = HA ∩ R+ A is a facet of R+ A if and only if (a) G[A ∪ N (A)] is connected with vertex set V \ {v} for some v ∈ V1 , or (b) G[A ∪ N (A)] and G[(V2 \ N (A)) ∪ (V1 \ A)] are connected and their union is a spanning subgraph of G. Moreover any facet has the form F = HA ∩R+ A for some A Vi , i ∈ {1, 2}. Proof. The ﬁrst statement follows readily from Lemma 10.7.12 and using part of the proof of Theorem 10.7.8. The last statement follows combining Theorem 10.7.8 with Proposition 10.7.10. 2 Remark 10.7.14 In Proposition 10.7.13 the case (a) is included in case (b). To see this fact, take N (A) = V2 and note that G[(V2 \ N (A)) ∪ (V1 \ A)] must consist of a point. The condition in case (a) is equivalent to require G\ {v} connected and in this case F = Hei ∩R+ A, where v = xi correspond to the unit vector ei . Lemma 10.7.15 Let F be a facet of R+ A. If F = HA ∩R+ A = HB ∩R+ A with A V1 and B V1 , then A = B. Proof. It follows from Lemma 10.7.12 and Proposition 10.7.13.

2

Putting together the previous results we get the following canonical way of representing the edge cone. The uniqueness follows from Lemma 10.7.12 and Lemma 10.7.15.

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475

Theorem 10.7.16 There is a unique irreducible representation R+ A = aﬀ(R+ A) ∩ (∩ri=1 HA−i ) ∩ (∩i∈I He+i ) such that Ai V1 for all i and I = {i|xi ∈ V2 is not cut vertex }. Corollary 10.7.17 There is an irreducible representation − R+ A = aﬀ(R+ A) ∩ (∩ri=1 HA−i ) ∩ (∩si=1 HB ) i

such that Ai V1 for all i and Bi V2 for all i. Lemma 10.7.18 Zn ∩ R+ A = NA.In particular if (β1 , . . . , βn ) is an integral vector in the edge cone, then ni=1 βi is an even integer. Proof. Since G is bipartite, by Proposition 10.2.2 its incidence matrix is totally unimodular, the lemma follows from Corollary 1.6.8. 2 As an application we recover the marriage theorem (Theorem 7.1.9). For a generalized version of the marriage theorem; see Proposition 13.5.3. Recall that a pairing oﬀ of all the vertices of a graph G is called a perfect matching. Theorem 10.7.19 (Marriage Theorem) If G is a connected bipartite graph, then G has a perfect matching if and only if |A| ≤ |N (A)| for every independent set of vertices A of G. Proof. G has a perfect matching if and only if the vector β = (1, . . . , 1) is in NA. By Lemma 10.7.18 β is in NA if and only if β ∈ R+ A. Thus the result follows from Theorem 10.7.8. 2

Exercises 10.7.20 Let G be a graph with n vertices and let A be its incidence matrix. Prove that if rank(A) = n and A is an independent set of G, then the sets F = R+ A ∩ HA and F = Hei ∩ R+ A are proper faces of the edge cone. 10.7.21 Let G be a connected bipartite graph and let Q be the polyhedron deﬁned as the set of x = (xi ) ∈ Rn+1 satisfying xi ≥ 1 fori = 1, . . . , n + 1, and −xn+1 + vi ∈C xi ≥ 1 for any minimal vertex cover C of G. If G is unmixed and v0 ≥ 3, then a = (1, . . . , 1) and b = (1, . . . , 1, v0 − 1) are the vertices of Q.

476

Chapter 10

10.7.22 [72] Let G = Kn be the complete graph on n vertices and let A be the set of column vectors of its incidence matrix. Then a vector x ∈ Rn is in R+ A if and only if x = (x1 , . . . , xn ) satisﬁes xi −

−xi j=i xj

≤ ≤

0, i = 1, . . . , n 0, i = 1, . . . , n.

In addition if n ≥ 4, these equations deﬁne all the facets of R+ A. 10.7.23 If G is a triangle with vertices {x1 , x2 , x3 }, then the edge cone of G has three facets deﬁned by x1 ≤ x2 + x3 ,

x2 ≤ x1 + x3 ,

x3 ≤ x1 + x2 .

Note that x1 ≥ 0, x2 ≥ 0 and x3 ≥ 0 deﬁne proper faces of dimension 1. 10.7.24 If G is an arbitrary graph, then ! − R+ A = HA A∈F

n

! He+i

,

i=1

where F is the set of all the independent sets of vertices of G. 10.7.25 If G1 , . . . , Gr are the components of a graph G, then ! ! n − + HA Hei , R+ A = A

i=1

where the ﬁrst intersection is taken over all the independent sets of vertices of the components Gi of G. 10.7.26 Let G = Km,n be the complete bipartite graph with m ≤ n. If V1 = {x1 , . . . , xm } and V2 = V \ V1 is the bipartition of G, then a vector z ∈ Rm+n is in R+ A if and only if z = (x1 , . . . , xm , y1 , . . . , yn ) satisﬁes x1 + · · · + xm −xi −yi

= y1 + · · · + yn , ≤ 0, i = 1, . . . , m, ≤ 0, i = 1, . . . , n.

In addition if m ≥ 2, the inequalities deﬁne all the facets of R+ A. 10.7.27 If G = K1,3 is the star with vertices {v, x1 , x2 , x3 } and center x, then the edge cone of G has three facets deﬁned by xi ≥ 0,

(i = 1, 2, 3)

and x = 0 deﬁnes a proper face of dimension 1.

Monomial Subrings of Graphs

10.8

477

Monomial birational extensions

Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K. In this section we consider a ﬁnite set of distinct monomials F = {xv1 , . . . , xvq } ⊂ R of degree d ≥ 2 and consider the integer matrices: v1 · · · vq , A = (v1 , . . . , vq ) and B = 1 ··· 1 where the vi ’s are regarded as column vectors, A is sometimes called the log-matrix of F . In the sequel let xd denote the set of all monomials of degree d in R. Then K[xd ] is the dth Veronese subring R(d) of R. If L is an integer matrix with r rows, we denote by ZL (resp. QL) the subgroup of Zr (resp. subspace of Qr ) generated by the columns of L. Deﬁnition 10.8.1 An extension D1 ⊂ D2 of integral domains is said to be birational if D1 and D2 have the same ﬁeld of fractions. Proposition 10.8.2 Let F1 = {xβ1 , . . . , xβr } be a set monomials of R such that F ⊂ F1 . Then the ring extension K[F ] ⊂ K[F1 ] is birational if and only if ZA = ZL, where L is the matrix with column vectors β1 , . . . , βr . 2

Proof. It is left as an exercise.

Following [385] an aim here is to study the birationality of the extension K[F ] ⊂ R(d) in terms of the matrices: (a) the linear syzygy matrix Syz (F ) whose columns are the set of vectors of the form xi ej − xk e such that xi xvj = xk xv , (b) the numerical linear syzygy matrix L obtained from Syz (F ) by making the substitution xi = 1 for all i, (c) the matrix M whose columns are the set of vectors ei − ek ∈ Rn such that ei − ek = vj − v , i.e., M = AL, (d) the Jacobian matrix Θ(F ) = (∂fi /∂xj ). Notice that the matrices in the ﬁrst row of the diagram: Θ(F ) ↓ A

Syz (F ) ↓ L

M := Θ(F ) Syz (F ) ↓ M := AL

specialize to the matrices in the second row by making xi = 1 for all i. The matrices Syz (F ) and L have size q × r, while the matrices M and M have size n × r. Proposition 10.8.3 [385] If rank(M ) = n − 1, then K[F ] ⊂ K[xd ] is a birational extension.

478

Chapter 10

Proof. Let a = (ai ) ∈ Nn such that |a| = i ai = d. It suﬃces to prove that a ∈ ZA. Let w1 , . . . , wr be the column vectors of the matrix M . Each wm has the form wm = ei − ek = vj − v for some i = k and j = . Hence rank(A) = n because v1 ∈ / QM . Therefore we can write λa = λ1 w1 + · · · + λr wr + μv1

(λ, μ, λi ∈ Z).

Taking inner product with 1 = (1, . . . , 1) yields λd = λ|a| = λ1 |w1 | + · · · + λr |wr | + μ|v1 | = μd ⇒ λ = μ (10.10) ⇒ λ(a − v1 ) = λ1 w1 + · · · + λr wr . Consider the digraph D with vertex set X = {x1 , . . . , xn } such that (xi , xk ) is a directed edge iﬀ wm = ei − ek for some m. The incidence matrix of D is M , thus M is totally unimodular by Exercise 1.8.10 and consequently Zn /ZM is torsion-free. Hence from Eq. (10.10) we get a − v1 ∈ ZM and a ∈ ZM + v1 ⊂ ZA, as required. 2 Remark 10.8.4 Let D be the digraph in the proof of Proposition 10.8.3. Then according to [190, Theorem 8.3.1] we have rank(M ) = n − c, where c is the number of connected components of D. In particular M has rank n − 1 if and only if D is connected. A generalization of the next result is given in Theorem 10.8.12. Corollary 10.8.5 Let G be a connected non-bipartite graph with n vertices and let K[G] be its edge subring. Then K[G] ⊂ K[x2 ] is birational. Proof. There is a connected subgraph H of G with n vertices and n edges and with a unique cycle of odd length. This follows using that any connected graph has a spanning tree together with the fact that a graph is bipartite if and only if every cycle in some cycle basis is even; see Exercise 10.1.67. By induction on n it is not hard to see that the numerical matrix M associated to H has rank n−1. Thus K[H] ⊂ K[x2 ] is birational by Proposition 10.8.3, 2 which implies the birationality of K[G] ⊂ K[x2 ]. Proposition 10.8.6 [385] K[F ] ⊂ K[xd ] is a birational extension if and only if n = rank(A) and Δn (A) = d. Proof. Assume the given extension is birational. Then Zlog(xd ) = ZA by Proposition 10.8.2. By Proposition 9.3.33(b) one has Zn /Zlog(xd ) Zd . Therefore, Δn (A) = d. Conversely, if Δn (A) = d, the surjection Zn /ZA −→ Zd ,

a−→|a|,

is an isomorphism. Since log(xd ) maps to zero under this map, one obtains 2 ZA = Zlog(xd ), as required.

Monomial Subrings of Graphs

479

Proposition 10.8.7 [385] The following conditions are equivalent (a) K[F ] ⊂ K[xd ] is birational. (b) Zn /Z({vi − vj |1 ≤ i < j ≤ q}) is torsion-free of rank 1. (c) n = rank(A) and Z({vi − vj |1 ≤ i < j ≤ q}) = Z({ei − ej |1 ≤ i < j ≤ n}). Proof. If n = rank(A), then there is an exact sequence of ﬁnite groups ϕ

ψ

0 −→ T (Zn+1 /ZB) −→ T (Zn /ZA) −→ Zd −→ 0,

(∗)

where the maps ϕ and ψ are given by (α ∈ Zn , b ∈ Z),

ϕ((α, b)) = α ψ(α) = α, x0 .

Pick a basis {v1 , . . . , vn } of the column space of A and notice that {v1 − vn , v2 − vn , . . . , vn−1 − vn , vn } is also a basis because |vi | = d for all i. Therefore one has the equality Q({vi − vj |1 ≤ i < j ≤ q}) = Q({ei − ej |1 ≤ i < j ≤ n}). The equivalences now follow using Lemma 1.2.20, Proposition 9.3.33, the exact sequence (∗), together with the fact that the group Zn /Z({ei − ej |1 ≤ i < j ≤ n}) is torsion-free of rank 1, see the proof of Proposition 10.8.3.

2

Proposition 10.8.8 [385] If rank(L) = q − 1 and rank(A) = n, then rank(M ) = n − 1. In particular K[F ] ⊂ K[xd ] is birational. Proof. There are linear maps L

A

Qr −→ Qq −→ Qn . Hence we have a linear map A

1 im(AL) = im(M ) −→ 0, im(L) −→

where A1 is the restriction of A to im(L). By hypothesis dim(im(A)) = n and dim(im(L)) = q − 1. Hence q−1 =

dim(im(L)) = dim(ker(A1 )) + dim(im(M )),

q−n

dim(ker(A)) ≥ dim(ker(A1 )).

=

Therefore dim(im(M )) ≥ n − 1. Since im(M ) is generated by vectors of the form ei − ej it follows that im(M ) has rank n simply because e1 is not in 2 the linear space generated by {ei − ej |1 ≤ i < j ≤ n}.

480

Chapter 10

Lemma 10.8.9 If m is a non-zero minor of Syz (F ) of order s for some s ≥ 1, then (ai ∈ N). m = ±xa1 1 · · · xann Proof. By induction on s. The case s = 1 is clear because the entries of Syz (F ) are monomials. Assume s ≥ 2. Let L1 be a submatrix of Syz (F ) of size s × s. We may assume that L1 is obtained using the ﬁrst s rows and columns of Syz (F ). Let g1 , . . . , gq be the rows obtained from Syz (F ) by ﬁxing the ﬁrst s columns. Thus L1 is the matrix with rows g1 , . . . , gs . Case (I): If the vector gi has a non-zero entry for some s + 1 ≤ i ≤ q, say the jth entry of gi is non-zero, then expanding the determinant of L1 along the jth column and using the induction hypothesis we get that det(L1 ) = ±xa1 1 · · · xann

(ai ∈ N).

Case (II): Now we assume that gi is the zero vector for s + 1 ≤ i ≤ q, but this implies that the linear syzygy matrix of F = {xv1 , . . . , xvs } has rank at least s, a contradiction because from the minimal free resolution of I = (xv1 , . . . , xvs ) one has rank(Syz (F )) ≤ s − 1; see Exercise 10.8.17. 2 Theorem 10.8.10 [385] If rank(A) = n and rank(Syz (F )) = q − 1, then K[F ] ⊂ K[xd ] is birational. Proof. By Lemma 10.8.9 the matrix L obtained from Syz (F ) by making xi = 1 for all i has also rank q − 1. Therefore by Propositions 10.8.8 and 10.8.3 we get that K[F ] ⊂ K[xd ] is birational. 2 Corollary 10.8.11 [385] If I = (F ), rank(A) = n, and I has a linear presentation, then K[F ] ⊂ K[xd ] is birational. Proof. It follows at once from Theorem 10.8.10 because in this case the 2 rank of Syz (F ) is q − 1. Monomials of degree two The birational theory of monomials of degree two can be fully established. Theorem 10.8.12 gives a complete answer for the birationality of subrings of multigraphs. In what follows we assume that deg(xvi ) = 2 for all i. Consider the multigraph G on the vertex set X = {x1 , . . . , xn } whose edges (resp. loops) are the pairs {xi , xj } (resp. {x , x }) such that xi xj ∈ F (resp. x2 ∈ F ) for some i = j. Notice that, in our situation, the log-matrix A of F is the incidence matrix of G and the monomial subring K[F ] is the edge subring K[G] of the multigraph G.

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481

Theorem 10.8.12 K[G] ⊂ K[x2 ] is a birational extension if and only if the following two conditions hold : (a) G is connected. (b) G is bipartite and has at least one loop or G is non-bipartite. Proof. ⇐) Since G is connected, there is a spanning tree T of G containing all the vertices of G; see Exercise 10.1.65. Case (I): G is a bipartite and xn is a loop of G. We may then regard T as a tree with a loop at xn . Notice that T has exactly n − 1 simple edges plus a loop. The incidence matrix AT of T has order n, is non-singular, and we may assume that the last column of AT has the form (0, 0, . . . , 0, 2) . Consider the matrix AT obtained from AT by removing the last column. The matrix AT is totally unimodular because it is the incidence matrix of a bipartite graph; see Proposition 10.2.2. Therefore det(AT ) = ±2 and rank(A) = n. From Lemma 10.8.6 we obtain that K[T ] ⊂ K[x2 ] is birational. Hence K[G] ⊂ K[x2 ] is birational as well. Case (II): G is non-bipartite. This case follows from Corollary 10.8.5. For use below notice that in Cases (I) and (II) we have that rank(A) = n and Zn /ZA Z2 . ⇒) Let G1 , . . . , Gr be the connected components of G and let Ai be the incidence matrix of Gi . Thus we have A = diag(A1 , A2 , . . . , Ar ). Notice that for each i the multigraph Gi is either non-bipartite or bipartite with at least one loop, for otherwise rank(Ai ) < ni for some i, where ni is the number of vertices of Gi , a contradiction because n = n1 + · · · + nr and n = rank(A) = j rank(Aj ). Therefore by the observation at the end of the proof of Case (II) we have rank(Ai ) = ni and Zni /ZAi Z2 . This means that the Smith normal form of Ai is diag(1, . . . , 1, 2, 0) and consequently the Smith normal form of A has exactly r entries equal to 2 and the remaining entries equal to 1 and 0. Hence Zn /ZA Zr2 and |Zn /ZA| = |Δn (A)| = 2r . By the birationality of the extension, using Lemma 10.8.6, we get r = 1, i.e., G is connected, as required. 2 Corollary 10.8.13 If G is a connected graph, then K[G] ⊂ K[x2 ] is a birational extension if and only if G is non-bipartite. Proof. If G is bipartite, dim K[G] = rank(A) = n − 1, hence K[G] ⊂ K[x2 ] cannot be birational. The converse follows from Theorem 10.8.12. 2

Exercises 10.8.14 Let A be an integral matrix of size n × q with equal column sum d. Prove the equalities A A A = In = dIn , In (A) = In A1 + · · · + An d1 1

482

Chapter 10

where A1 , . . . , An are the rows of A and In (A) is the ideal of Z generated by the n-minors of A. Hint d1 = (d, . . . , d) = A1 + · · · + An . 10.8.15 Let F = {xv1 , . . . , xvq } and G = {xβ1 , . . . , xβr } be two sets of monomials of R such that F ⊂ G. Prove that K[F ] ⊂ K[G] is a birational extension if and only if Zlog(F) = Zlog(G). 10.8.16 Let F be a set of monomials of R = K[x1 , . . . , xn ] of the same degree d such that Zn /Zlog(F ) is a ﬁnite group. Prove that K[F t] = A(P ) if and only if Zn /Zlog(F ) Zd . 10.8.17 Let I = (xv1 , . . . , xvq ) be a monomial ideal of a polynomial ring R over a ﬁeld K. Prove that if ϕ

· · · −→ Rb2 −→ Rq −→ I −→ 0 is the minimal free resolution of I, then rank(ϕ) = q − 1 and the columns of the matrix ϕ have the form xai ej − xak e (cf. Theorem 3.3.19 and [137]).

Chapter 11

Edge Subrings and Combinatorial Optimization Let G be a connected bipartite graph. We present an approach to compute the canonical module of the edge subring K[G] using linear programming techniques and study the Gorenstein property and the type of K[G]. The a-invariant of K[G] will be interpreted in combinatorial optimization terms as the maximum number of edge disjoint directed cuts.

11.1

The canonical module of an edge subring

Let G be a connected bipartite graph on the vertex set V = {x1 , . . . , xp } and let R = K[x1 , . . . , xp ] = ⊕∞ i=0 Ri be a polynomial ring over a ﬁeld K with the standard grading induced by deg(xi ) = 1. The edge subring of G is the K-subalgebra K[G] = K[{xi xj | xi is adjacent to xj }] ⊂ R, we grade K[G] with the normalized grading K[G]i = K[G] ∩ R2i . Recall that, by Proposition 10.3.1, K[G] is normal and Cohen–Macaulay. The set of vectors vk = ei + ej ∈ Rp such that xi is adjacent to xj will be denoted by A = {v1 , . . . , vq }. Note that A is the set of column vectors of the incidence matrix of G. As G is bipartite, by Lemma 10.7.18, we have NA = Zp ∩ R+ A.

(11.1)

Thus, by a formula of Danilov–Stanley (see Theorem 9.1.5), the canonical

484

Chapter 11

module ωK[G] is the ideal given by ωK[G]

= (11.1)

=

({xa | a ∈ NA ∩ ri(R+ A)}) ({xa | a ∈ Zp ∩ ri(R+ A)}) ⊂ K[G],

(11.2)

where ri(R+ A) is the interior of R+ A relative to the aﬃne hull of R+ A. Recall, from Section 10.7, that R+ A is called the edge cone of G.

11.2

Integrality of the shift polyhedron

In what follows G will denote a connected bipartite graph with p = m + n vertices and bipartition (V1 , V2 ). We will assume that the vertices in V1 are x1 , . . . , xm and the vertices in V2 are xm+1 , . . . , xn+m , where 2 ≤ m ≤ n. Consider the family F = {A ∪ A | ∅ = A V1 ; N (A) ⊂ A ⊂ V2 } ∪ {A | ∅ = A ⊂ V2 }, where N (A) is the neighbor set of A. For each Y = A ∪ A ∈ F we associate the following vector ei − ei ∈ Rm+n , βY = xi ∈A

xi ∈A

note that if A = ∅ the vector βY is a {0, −1}-vector. Let C be the matrix whose rows are the vectors in {βY }Y ∈F . According to Corollary 10.7.17 the edge cone of G can be written as: R+ A = aﬀ(R+ A) ∩ {x| C x ≤ 0}. The shift polyhedron of the edge cone of G, with respect to C , is the rational polyhedron: Q = aﬀ(R+ A) ∩ {x| C x ≤ −1}, see Deﬁnition 9.1.12. From the ﬁnite basis theorem (see Theorem 1.1.33) the shift polyhedron can be written as the sum of a unique cone and a polytope. In our case: Q = R+ A + conv(β1 , . . . , βr ), where β1 , . . . , βr are the vertices of Q. Recall that Q is an integral polyhedron if Q = conv(Zp ∩ Q). As Q is a pointed polyhedron, it is integral if and only if β1 , . . . , βr are integral vectors; see Corollary 1.1.64. A reason for introducing the shift polyhedron is that its integral points deﬁne the canonical module of K[G].

Edge Subrings and Combinatorial Optimization

485

Proposition 11.2.1 Zp ∩ Q = Zp ∩ ri(R+ A). Proof. It follows from Eq. (11.2) and Corollary 1.1.45.

2

Thus, the shift polyhedron is a bridge which allows us to use combinatorial optimization techniques to study the edge subring K[G]. Theorem 11.2.2 The shift polyhedron Q = aﬀ(R+ A) ∩ {x| C x ≤ −1} is integral. Proof. If G is regarded as the digraph with all its arrows leaving the vertex set V2 , then it is seen that one has the equality C A = −B , where A is the incidence matrix of G and B is the one-way cut-incidence matrix of the family F ; see Deﬁnition 1.8.5. Let b be any vector in Rp , p = m + n, such that the following maximum is ﬁnite max{x, b| x ∈ Q}.

(11.3)

According to Theorem 1.1.63 it suﬃces to prove that the maximum in Eq. (11.3) is attained by an integral vector. As Q ⊂ R+ A, any vector x ∈ Q can be written as x = A. x for some x . ≥ 0, x . ∈ Rq , where q is the number of edges of G. Hence {b, x| x ∈ Q} = {bA, x .| x . ≥ 0; B x . ≥ 1}. Therefore max{x, b| x ∈ Q} = max{bA, x .| x . ≥ 0; B x . ≥ 1}.

(11.4)

By Corollary 1.8.8 and Lemma 1.8.9 the polyhedron Q = {. x ∈ Rq | x . ≥ 0; B x . ≥ 1} is integral. Hence, by Theorem 1.1.63, the maximum in the right-hand side of Eq. (11.4) is attained by an integral vector x .0 ∈ Q . Consequently the 2 maximum in Eq. (11.3) is attained by the integral vector A. x0 ∈ Q. Example 11.2.3 Consider the following bipartite graph G and make G a digraph with “edges” v1 , . . . , v6 as shown below. x1 x2 x3 s v s s Y H 2 6HH 6 6 v4 HH v1 v6 HH v3 s v5 Hs V2 s x4 x5 x6 V1

486

Chapter 11

The family F consists of the subsets that occur in the directed cuts: δ + ({x1 , x4 , x6 }) = {v3 , v6 }, δ + ({x5 }) = {v4 , v5 }, + δ + ({x6 }) = {v2 , v6 }, δ ({x2 , x4 , x5 }) = {v1 , v5 }, + + δ ({x3 , x5 , x6 }) = {v2 , v4 }, δ ({x4 , x5 }) = {v1 , v3 , v4 , v5 }, δ + ({x1 , x2 , x4 , x5 , x6 }) = {v5 , v6 }, δ + ({x4 , x6 }) = {v1 , v2 , v3 , v6 }, δ + ({x1 , x3 , x4 , x5 , x6 }) = {v3 , v4 }, δ + ({x5 , x6 }) = {v2 , v4 , v5 , v6 }, δ + ({x2 , x3 , x4 , x5 , x6 }) = {v1 , v2 }, δ + ({x4 , x5 , x6 }) = {v1 , . . . , v6 }. δ + ({x4 }) = {v1 , v3 }, As pointed out in the proof of Theorem 11.2.2 the one-way cut-incidence matrix B of F is related to the incidence matrix A of G and to the matrix C by the equality C A = −B . In this example one has: ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ C =⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

1 0 0 1 1 0 0 0 0 0 0 0 0

0 1 0 1 0 1 0 0 0 0 0 0 0

0 0 1 0 1 1 0 0 0 0 0 0 0

−1 −1 0 −1 −1 −1 −1 0 0 −1 −1 0 −1

0 −1 −1 −1 −1 −1 0 −1 0 −1 0 −1 −1

−1 0 −1 −1 −1 −1 0 0 −1 0 −1 −1 −1

⎤

⎡

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ B =⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

and

0 1 0 0 0 1 1 0 0 1 1 0 1

0 0 1 0 0 1 0 0 1 0 1 1 1

1 0 0 0 1 0 1 0 0 1 1 0 1

0 0 1 0 1 0 0 1 0 1 0 1 1

0 1 0 1 0 0 0 1 0 1 0 1 1

1 0 0 1 0 0 0 0 1 0 1 1 1

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

The polyhedron Q = {. x|B x . ≥ 1; x . ≥ 0} is integral and has two vertices (0, 1, 1, 0, 1, 0) and (1, 0, 0, 1, 0, 1) that map under A onto the vector 1 of Q. Corollary 11.2.4 If G is a connected bipartite graph and Q is the shift polyhedron of the edge cone of G, with respect to C , then Q = conv(Zp ∩ ri(R+ A)). Proof. The polyhedron Q is integral by Theorem 11.2.2. Hence the result follows from Proposition 9.1.14. 2

Exercises 11.2.5 Let T : Rq → Rn be a linear map and let Q ⊂ Rq be an integral polyhedron. If T (Zq ) ⊂ Zn , then T (Q ) is an integral polyhedron. 11.2.6 Prove that, in the proof Theorem 11.2.2, the linear transformation x . → A. x maps Q onto Q. Prove that dim(Q ) = q and dim(Q) = m + n − 1. Give a shorter proof of Theorem 11.2.2 using Exercise 11.2.5.

Edge Subrings and Combinatorial Optimization

11.3

487

Generators for the canonical module

Let S be a standard graded K-algebra over a ﬁeld K. The a-invariant of S, denoted by a(S), is the degree, as a rational function, of the Hilbert series of S. If S is Cohen–Macaulay and ωS is the canonical module of S, then a(S) = −min{ i | (ωS )i = 0}, see Proposition 5.2.3. In our situation, since G is bipartite, S = K[G] is a normal Cohen–Macaulay standard K-algebra. Thus this formula applies. Theorem 11.3.1 [405] If Q is the shift polyhedron of the edge cone of a connected bipartite graph G, then the a-invariant of K[G] is given by

|x| x∈Q . a(K[G]) = −min 2 Proof. Let C be the matrix deﬁning the shift polyhedron. Note that any row of C deﬁnes a proper face of the edge cone, except the row with the ﬁrst m entries equal to 0 and the last n entries equal to −1. Thus from Corollary 1.1.45 and Theorem 1.1.44 we have Zp ∩Q = Zp ∩ri(R+ A). Hence the inequality “≤” follows from the Danilov–Stanley formula of Eq. (11.2). By Proposition 1.1.41 the minimum above is attained at a vertex β of Q. Hence it suﬃces to observe that β has integral entries by Theorem 11.2.2 and Corollary 1.1.64. 2 Theorem 11.3.2 [405, Proposition 4.2] Let G be a connected bipartite graph. If G is regarded as the digraph with all its arrows leaving the vertex set V2 , then the following three numbers are equal: (a) −a(K[G]), minus the a-invariant of K[G]. (b) The minimum cardinality of an edge set that contains at least one edge of each directed cut. (c) The maximum number of edge disjoint directed cuts. Proof. Let A be the incidence matrix of G and let C be the matrix deﬁning the shift polyhedron. As C A = −B , then by Theorem 1.8.7 and LP duality (Theorem 1.1.56) one has that the optimum values in the equality . ≥ 1} = max{y, 1| y ≥ 0; yB ≤ 1} min{1, x .| x . ≥ 0; B x are attained by integral vectors. By looking at B as a one-way cut-incidence matrix it follows that the two numbers in (b) and (c) are equal. In general, for digraphs, it is known that the numbers in (b) and (c) are equal. See for instance [281, Theorem 19.10]. Noticing that .q vq /2 = (2. x1 + · · · + 2. xq )/2 = 1, x ., 1, A. x/2 = 1, x .1 v1 + · · · + x

488

Chapter 11

where v1 , . . . , vq are the column vectors of A, from the equalities min{1, x/2| x ∈ aﬀ(R+ A); C x ≤ −1} = . ≥ 1} = min{1, x .| x . ≥ 0; B x . ≥ 1}, min{1, A. x/2| x . ≥ 0; B x and Theorem 11.3.1 we get that the numbers in (a) and (b) are equal.

2

Proposition 11.3.3 If G is a connected bipartite graph and β is a vertex of the shift polyhedron Q, then xβ is a minimal generator of ωK[G] . Proof. By Eq. (11.2), Corollary 1.1.45, and Theorem 11.2.2 we get that xβ is in ωK[G] . There are c ∈ Qp and b ∈ Q such that (i) {β} = {x|x, c = b} ∩ Q and (ii) Q ⊂ {x|x, c ≤ b}. If v1 , . . . , vq are the columns of the incidence matrix of G, then by deﬁnition of Q one has vi + β ∈ Q for all i. Thus vi + β, c = vi , c + b ≤ b ⇒ vi , c ≤ 0

(i = 1, . . . , q). q Assume there are α ∈ Q, η1 , . . . , ηq ∈ N such that β = ( i=1 ηi vi ) + α, then b = β, c = η1 v1 , c + · · · + ηq vq , c + α, c ≤ α, c ≤ b. Hence α, c = b and by (i) we get α = β. Thus xβ is a minimal generator 2 of the canonical module ωK[G] . Deﬁnition 11.3.4 Let G be a bipartite graph with incidence matrix A and let C be the matrix deﬁning the shift polyhedron. If C A = −B , the polyhedron x ∈ Rq | x . ≥ 0; B x . ≥ 1} Q = {. is called the blocking polyhedron. Theorem 11.3.5 If G is a connected bipartite graph, then Q is integral. Proof. It follows from Corollary 1.8.8 and Lemma 1.8.9.

2

Proposition 11.3.6 Let G be a connected bipartite graph and let Q be the shift polyhedron with respect to C . If xβ is a minimal generator of ωK[G] and A is the incidence matrix of G, then there is a vertex α . of the blocking α = β. polyhedron Q such that A. Proof. Let v1 , . . . , vq be the column vectors of the matrix A and let C be the matrix deﬁning Q. As A is totally unimodular (see Proposition 10.2.2), using Carath´eodory’s theorem (see Theorem 1.1.18) and Heger’stheorem r (see Theorem 1.6.4), after permuting the vi ’s, we can write β = i=1 ηi vi

Edge Subrings and Combinatorial Optimization

489

with ηi ∈ N \ {0} and r ≤ p − 1 ≤ q, where p = m + n. We claim that ηi = 1 for all i. Assume ηi > 1. Take any row v of C . Observe that v, vk is equal to 0 or −1 for any vk . The vector β = η1 v1 + · · · + ηi−1 vi−1 + (ηi − 1)vi + ηi+1 vi+1 + · · · + ηr vr satisﬁes β , v ≤ −1 because β, v ≤ −1. Thus β ∈ Q, a contradiction because β = β − vi and xβ is minimal. Thus ηi = 1. Notice that the vector α = β, and from C A = −B we get α . ∈ Q . α . = e1 + · · · + er satisﬁes A. Consider the linear program .q min x .r+1 + · · · + x subject to B x . ≥ 1 and x .≥0 and notice that 0 is the optimum value of this linear program because α . is in Q . By Theorem 11.3.5 there is an integral vertex γ . of Q where the minimum is attained. Hence γ .i = 0 for i > r. By Exercise 11.3.8 the vector γ has {0, 1}-entries. If . γk = 0 for some 1 ≤ k ≤ r, then A. γ = i=k i vi ∈ Q, where i ∈ {0, 1} for all i = k, a contradiction to the minimality of xβ . Thus α .=. γ , as required. Observe that the last part of the argument works even if q = r, which is the case of a tree. 2 Theorem 11.3.7 If G is a connected bipartite graph with incidence matrix A and Q is the blocking polyhedron, then the canonical module ωK[G] of K[G] is generated by the set α |α . is a vertex of Q }. {xA

Proof. It follows by recalling that the blocking polyhedron Q is integral and using Proposition 11.3.6. 2

Exercise 11.3.8 If B is a {0, 1}-matrix, then any integral vertex of the rational polyhedron Q = {x | x ≥ 0; Bx ≥ 1} is a {0, 1}-vector.

11.4

Computing the a-invariant

Let G be a bipartite graph. In order to use the results of Section 11.3 in an eﬃcient way, we introduce other representations of the shift polyhedron. For each independent set of vertices A of G consider the vector αA = ei − ei . xi ∈A

xi ∈N (A)

490

Chapter 11

By Corollary 10.7.17, there exists an irreducible representation of the edge cone of G, as an intersection of closed half-spaces, of the form: − − ∩ · · · ∩ H−e , R+ A = aﬀ(R+ A) ∩ Hα−A ∩ · · · ∩ Hα−Ar ∩ H−e i1 is 1

(11.5)

where for each i either Ai V1 or Ai V2 and none of the half-spaces can be omitted from the intersection. Let us denote by C the matrix whose rows are the vectors αA1 , . . . , αAr , −ei1 , . . . , −eis and by C the matrix deﬁning the shift polyhedron, as deﬁned in Section 11.2. The next result says that the shift polyhedrons Q and Q1 with respect to C and C, respectively, are equal. Theorem 11.4.1 If G is a connected bipartite simple graph, then the shift polyhedron Q of the edge cone of G, with respect to C , is given by Q = aﬀ(R+ A) ∩ {x| Cx ≤ −1}. Proof. It follows from Exercises 11.4.7, 11.4.8, and Lemma 10.7.15.

2

Remark 11.4.2 If CA = −B and C A = −B , where A is the incidence matrix of G, then B and B deﬁne the same blocking polyhedron. There are linear programming techniques to convert the description of a rational polyhedron given by a “ﬁnite basis” into an irreducible representation as intersection of closed half-spaces and vice versa. These techniques have been converted into very eﬃcient routines in several programming environments; see for instance PORTA [84]. Thus, one can eﬀectively compute a generating set for the ideal ωK[G] (see Example 11.4.4 for an illustration). Remark 11.4.3 To compute the vertices of a shift polyhedron of an edge cone using PORTA we need a “valid” point. Note that if A = {v1 , . . . , vq } is theset of column vectors of the incidence matrix of G, then the point q α = i=1 vi = (deg(x1 ), . . . , deg(xn+m )) is valid , that is, α ∈ Q: Example 11.4.4 Consider the following bipartite simple graph G: x11 t A x4 t A A t A tXXX tx x tH H H x3 A XX2X10 x7 X H X XXXHH A XXX HX AAt t H x8 x6 x1 t

x9 t

x5 t

with bipartition V1 = {x1 , . . . , x6 } and V2 = {x7 , . . . , x11 }. The incidence matrix of G, denoted by A, is the transpose of the following matrix. We will display the data as input ﬁles for PORTA.

Edge Subrings and Combinatorial Optimization CONE_SECTION (columns ( 1) 1 0 0 0 0 0 1 0 ( 2) 1 0 0 0 0 0 0 1 ( 3) 0 1 0 0 0 0 1 0 ( 4) 0 1 0 0 0 0 0 0 ( 5) 0 0 1 0 0 0 0 1 ( 6) 0 0 1 0 0 0 0 0 ( 7) 0 0 1 0 0 0 0 0 ( 8) 0 0 0 1 0 0 0 0 ( 9) 0 0 0 0 1 0 0 0 ( 10) 0 0 0 0 0 1 0 0 ( 11) 0 0 0 0 0 1 0 0 ( 12) 0 0 0 0 1 0 0 0 ( 13) 0 0 0 1 0 0 0 0 ( 14) 1 0 0 0 0 0 0 0 ( 15) 0 1 0 0 0 0 0 0 ( 16) 0 0 0 0 0 1 1 0 ( 17) 0 0 0 0 1 0 0 0

of A) 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 1 0 0 1 0 0 0 1 1 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 1

491

corresponding monomial x1x7 x1x8 x2x7 x2x9 x3x8 x3x10 x3x11 x4x10 x5x10 x6x10 x6x11 x5x9 x4x11 x1x9 x2x10 x6x7 x5x11

Applying PORTA to this input ﬁle we obtain an irreducible representation of R+ A, which immediately yields the following representation of Q. VALID POINT: 3 3 3 2 3 3 3 2 3 5 4 (see previous Remark) INEQUALITIES_SECTION -xi<= -1 (for i=7,...,11) ( 1) x1+x2+x3+x4+x5+x6-x7-x8-x9-x10-x11 == 0 ( 1) x1+x3+x4+x5+ x6-x7-x8-x9-x10-x11 <= -1 ( 2) x1+x2+x4+x5+ x6-x7-x8-x9-x10-x11 <= -1 ( 3) x1+x2+x3+x5+ x6-x7-x8-x9-x10-x11 <= -1 ( 4) x1+x2+x3+x4+ x6-x7-x8-x9-x10-x11 <= -1 ( 5) x1+x2+x3+x4+x5 -x7-x8-x9-x10-x11 <= -1 ( 6) x4 -x10-x11 <= -1 ( 7) x2 -x7 -x9 -x10 <= -1 ( 8) x4+ x6-x7 -x10-x11 <= -1 ( 9) x4+x5 -x9-x10-x11 <= -1 (10) x3+x4 -x8 -x10-x11 <= -1 (11) x1+x2 -x7-x8-x9-x10 <= -1 (12) x3+x4 +x6-x7-x8 -x10-x11 <= -1 (13) x3+x4+x5 -x8-x9-x10-x11 <= -1 (14) x2 +x4+x5+x6-x7 -x9-x10-x11 <= -1 (15) x1 -x7-x8-x9 <= -1 (16) x2+x3+x4+x5+x6-x7-x8-x9-x10-x11 <= -1

Using PORTA we get that the vertices of the shift polyhedron are: (1) (2) (3) (4)

1 2 2 1

2 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

1 1 3 1

3 3 1 2

(5) 1 1 1 1 1 1 1 1 1 2 1 (6) 1 1 1 1 1 1 1 1 2 1 1 (7) 1 1 1 1 1 1 2 1 1 1 1

492

Chapter 11

From the equality CA = −B (Remark 11.4.2), we get that the corresponding blocking polyhedron is deﬁned by INEQUALITIES_SECTION xi>=0 (for i=1,...,17) ( 1) x15+x3+x4>=1 ( 2) x5+x6+x7>=1 ( 3) x13+x8>=1 ( 4) x12+x17+x9>=1 ( 5) x10+x11+x16>=1 ( 6) x1+x16+x3>=1 ( 7) x2+x5>=1 ( 8) x12+x14+x4>=1 ( 9) x10+x15+x6+x8+x9>=1 (10) x11+x13+x17+x7>=1

(11) (12) (13) (14) (15) (16) (17) (18) (19) (20) (21)

x10+x11+x15+x17+x6+x7+x9>=1 x1+x10+x12+x14+x16+x6+x8+x9>=1 x1+x15+x17+x3+x6+x7+x9>=1 x10+x11+x14+x15+x4+x6+x7>=1 x10+x11+x15+x17+x2+x9>=1 x10+x12+x16+x5+x6+x8+x9>=1 x1+x15+x17+x2+x3+x9>=1 x10+x11+x14+x15+x2+x4>=1 x1+x14+x6+x7>=1 x12+x16+x3+x4+x5>=1 x1+x14+x2>=1

Using PORTA we get that the blocking polyhedron Q has 173 vertices. The distinct images of those vertices under the incidence matrix A are: 1 2 2 1 1 1 1 2

2 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 2 1

1 1 1 1 1 1 1 1

1 1 1 1 1 2 1 1

1 1 3 1 2 1 1 2

3 3 1 2 1 1 1 2

1 1 1 1 1 2 1 1 2 2 1 1 1 2 1 1 1 1 1 2 2 1 .....................

vertex of the shift polyhedron Q vertex of Q vertex of Q vertex of Q vertex of Q vertex of Q vertex of Q not a vertex of Q but correspond to a minimal generator of the canonical module this and the following vectors correspond to redundant monomials in the canonical module

By Theorem 11.3.7, ωK[G] is minimally generated by the eight monomials corresponding to the ﬁrst eight vectors above. Thus type(K[G]) = 8 and the rank of the last module in the graded free resolution of K[G] is 8. Remark 11.4.5 From Theorem 11.3.1 and Theorem 11.4.1 we obtain an eﬀective method to compute the a-invariant of K[G] that only requires a description of the shift polyhedron by linear inequalities and to solve a linear program. See Example 11.4.6. Example 11.4.6 Consider the following bipartite graph G: sx4 bipartition: x7 s @ @sx8 V1 = {x1 , x2 , x3 , x4 } @ @sx3 V2 = {x5 , x6 , x7 , x8 } sx6 x1 s @ @sx2 @ @sx5

Edge Subrings and Combinatorial Optimization

493

In order to estimate the a-invariant of G we set up the next linear program using the following input ﬁle for Mathematica [431] vars:={x1,x2,x3,x4,x5,x6,x7,x8} f:=(x1+x2+x3+x4+x5+x6+x7+x8)/2 ieq:={x1+x2+x3+x4-x5-x6-x7-x8 == 0, -x1<= -1, -x2<= -1, -x4<= -1, -x5<= -1, -x7<= -1,-x8 <= -1,x4-x7-x8 <= -1, -x3-x4+x7+x8 <= -1,x3+x4-x6-x7-x8 <= -1} ConstrainedMin[f,ieq,vars] where the set of inequalities was found using PORTA and it comes from an irreducible representation of the edge cone as in Eq. (11.5). The optimal value of this linear program is equal to 5 and is attained at the vertex (1, 1, 2, 1, 1, 2, 1, 1). Hence by Theorems 11.3.1 and 11.4.1 we get that the a-invariant of K[G] is equal to −5.

Exercises 11.4.7 [189, Lemma 7.3.2] Let G = G(m, n) be a connected bipartite graph. If Q is the shift polyhedron of G, with respect to C , and x ∈ Q, then xi ≥ 1 for all i = 1, . . . , m + n. 11.4.8 [189, Proposition 7.3.3] Let G be a connected bipartite graph. If Q is the shift polyhedron of G, with respect to C , and x ∈ Q, then for A V1 (resp. A V2 ) such that N (A) ⊂ A ⊂ V2 (resp. N (A) ⊂ A ⊂ V1 ) one has xi − xi ≤ −1. xi ∈A

11.5

xi ∈A

Algebraic invariants of edge subrings

In this section we give upper bounds for the a-invariant of the edge subring K[G] valid for any bipartite connected graph G. We examine the Gorenstein property and the type of K[G]. As a tool we use a technique, introduced in Section 10.1, based on decomposing a graph into blocks. Proposition 11.5.1 If G is a connected bipartite graph with a bipartition (V1 , V2 ) and |V1 | = m ≤ |V2 | = n, then a(K[G]) ≤ −n. Proof. Let V1 = {x1 , . . . , xm }. Take a monomial xβ in the canonical module of K[G], that is, β = (βi ) is an integral vector in ri(R+ A). One may assume m ≥ 2 because if m = 1 the graph and the result is mG is a star n+m clear. Thus βi ≥ 1 by Corollary 1.1.45, and i=1 βi = i=m+1 βi because the edge cone lies in the hyperplane m i=1

xi =

m+n i=m+1

xi .

494

Chapter 11

Hence deg(xβ ), the normalized degree of xβ , is deg(xβ ) ≥ n, which proves the required inequality.

m+n i=m+1

βi .

Therefore 2

Corollary 11.5.2 [63] Let Km,n be the complete bipartite graph. If n ≥ m, then a(K[Km,n ]) = −n. Proof. By Exercise 10.7.26 and Proposition 11.5.1 it suﬃces to observe that the monomial xβ with β = (n − m + 1, 1, . . . , 1, 1, . . . , 1 )

m entries n entries is in the canonical module of K[G] and has degree equal to n.

2

There are 2-connected graphs where the a-invariant does not reach the upper bound of Proposition 11.5.1 (see Exercise 11.5.14). A formula for the type of a determinantal ring is due to Brennan; see [73, p. 115]. We show a special case of this formula. Proposition 11.5.3 Let G = Km,n be the complete bipartite graph. If n ≥ m, then K[G] is a level algebra and its type is given by: n−1 type(K[G]) = . m−1 Proof. From Exercise 10.7.26 it is seen that the canonical module of K[G] is minimally generated by the set of monomials of the form xa1 1 · · · xamm y1 · · · yn , such that a1 + · · · + am = n and ai ≥ 0 for all i. Thus the number of such partitions is the type. As the canonical module is generated in a single degree the algebra K[G] is level. 2 Proposition 11.5.4 If G is a bipartite connected graph with bipartition (V1 , V2 ) and |V1 | = m ≤ |V2 | = n, then a(K[G]) = −m if and only if m = n and 1 = (1, . . . , 1) is in ri(R+ A). Proof. ⇒) By Corollary 1.1.45, the canonical module ωK[G] of the ring K[G] is generated by monomials xβ such that all the entries of β are positive integers. As n = m we derive that x1 = x1 x2 · · · xm+n is in the canonical module and it is the only monomial of normalized degree n in ωK[G] . ⇐) It follows from Proposition 11.5.1. 2 Deﬁnition 11.5.5 Let G be a graph. A cycle containing all the vertices of G is said to be a Hamilton cycle, and a graph containing a Hamilton cycle is said to be Hamiltonian. Corollary 11.5.6 If G is a Hamiltonian bipartite graph with 2n vertices, then a(K[G]) = −n.

Edge Subrings and Combinatorial Optimization

495

Proof. Let H be a Hamiltonian cycle of G and let 2n be the number of vertices of G. Note that a(K[H]) = −n (because K[H] is the ring of a hypersurface of degree n) and dim K[H] = dim K[G]. Hence one has a(K[H]) ≤ a(K[G]) and consequently a(K[G]) = −n. 2 Corollary 11.5.7 If G = Kn,n is a complete bipartite graph, then the ainvariant of K[G \ {e}] is equal to −n. Proof. Note that for n ≥ 3 the graph G \ {e} is Hamiltonian.

2

Decomposing a bipartite graph into blocks Let G be a connected bipartite graph. Next we describe in a condensed form a technique that can be used to simplify the computation of the a-invariant and the type of K[G]. It is based in the graph theoretical notion of block (see Section 7.1) and the following result. Proposition 11.5.8 If B and C are two standard K-algebras, then a(B ⊗K C) = a(B) + a(C). If in addition B and C are Cohen–Macaulay, then type(B ⊗K C) = type(B) · type(C). Proof. The ﬁrst equality follows from the proof of Proposition 5.1.11. For the second equality, see [171]. 2 Proposition 11.5.9 If G1 , . . . , Gr are the blocks of a connected bipartite graph, then a(K[G]) =

r

a(K[Gi ]) and type(K[G]) =

i=1

r 7

type(K[Gi ]).

i=1

Proof. Since G is connected and bipartite one has K[G] K[G1 ] ⊗K K[G2 ] ⊗K · · · ⊗K K[Gr ]. Hence the result follows from Proposition 11.5.8.

2

The next theorem is a combinatorial obstruction for an edge subring to be Gorenstein. Because of Proposition 11.5.9 it suﬃces to consider the class of Gorenstein subrings of 2-connected bipartite graphs. Theorem 11.5.10 Let G be a Gorenstein connected bipartite graph with bipartition (V1 , V2 ), then G has a perfect matching and furthermore |A| < |N (A)| for all A V1 .

496

Chapter 11

Proof. Let xβ be the generator of ωK[G] . For each 1 ≤ i ≤ m + n choose a spanning tree of G \ {xi } and enlarge this to a spanning tree Ti of G. Note that the monomial xγi with γi = (degTi (x1 ), . . . , degTi (xn+m )) and degTi (xi ) = 1 is the generator of the canonical module of K[Ti ]. Hence since aﬀ(R+ Ai ) = aﬀ(R+ A), where R+ Ai is the edge cone of Ti , we get xγi ∈ ωK[G] . Therefore βi = 1, and consequently β = 1. Noting that β is in the aﬃne space generated by the edge cone of G it follows m = n. Thus 2 from Theorem 10.7.16 we obtain |A| < |N (A)| for all A V1 . The converse of Theorem 11.5.10 does not hold. Example 11.5.11 Consider the bipartite graph G: xr 5 xr2 T x6r T Trx8 x1 r JJr T x3 T Tr r x7 x4 Using the following procedure in Macaulay2 [199], we rapidly obtain that K[G] has type equal to three. On the other hand 1 = (1, . . . , 1) is in the relative interior of the edge cone. In this example the shift polyhedron has vertices 1, (2, 1, 1, 1, 1, 1, 1, 2) and (1, 2, 1, 1, 1, 1, 2, 1). KK=ZZ/31991 R=KK[x_1..x_8,t_1..t_11,MonomialOrder=>Eliminate 8] I=ideal(t_1-x_1*x_5, t_2-x_1*x_6, t_3-x_1*x_7, t_4-x_2*x_5, t_5-x_2*x_6, t_6-x_2*x_8, t_7-x_3*x_6, t_8-x_3*x_7, t_9-x_3*x_8, t_10-x_4*x_7, t_11-x_4*x_8) J= ideal selectInSubring(1,gens gb I) M=coker gens J res M Corollary 11.5.12 Let G be a bipartite connected graph with bipartition (V1 , V2 ) and |V1 | = m ≤ |V2 | = n. If G has no cut vertices and K[G] is Gorenstein, then G has a perfect matching and a(K[G]) = −n. Proof. It follows from the proof of Theorem 11.5.10.

2

If one of the blocks of a graph G is a bridge e one can say a bit more about the relationship between G and G \ {e}. Proposition 11.5.13 Let G be a graph with a bridge e. If G\{e} = G1 ∪G2 with G1 and G2 disjoint graphs such that K[Gi ] is Gorenstein for i = 1, 2, then K[G] is Gorenstein.

Edge Subrings and Combinatorial Optimization

497

Proof. There is an isomorphism K[G] (K[G1 ] ⊗K K[G2 ])[t], where t is a variable. Since the tensor product of Gorenstein rings is Gorenstein and a polynomial ring over a Gorenstein ring is Gorenstein, the result follows. 2

Exercises 11.5.14 Consider the following bipartite graph G: x6 s @ @ @sx2

x1 s @ @ @s x5

x3 s @ @ @s x8

x7 s @ @ @s x4

Using Theorems 11.3.1 and 11.4.1, show that a(K[G]) = −5 and that the shift polyhedron has four vertices. 11.5.15 Prove that the following are the equations of the edge cone of the graph G shown on the left. rx1 @ @ x3 rH @ rx2 x 4 @HH r @ @r x5

−x1 −x4 −x5 x3 x2 x1 + x4 + x5

≤ ≤ ≤ ≤ ≤ ≤

0, 0, 0, x1 + x2 + x4 + x5 , x1 + x3 + x4 + x5 , x2 + x3 .

Show that the a-invariant of K[G] is equal to −4. 11.5.16 Let G be the graph consisting of two squares joined by an edge. Prove that K[G] is a Gorenstein ring, its a-invariant is −5, it has a perfect matching, and α = (1, . . . , 1) is not in the relative interior of the edge cone of G. Prove that the monomial corresponding to (1, 1, 2, 1, 1, 2, 1, 1) generates the canonical module of K[G]. 11.5.17 Let G be a graph with a bridge e. If G \ {e} = G1 ∪ G2 with G1 and G2 disjoint graphs, then a(G) = a(G1 ) + a(G2 ) − 1. 11.5.18 Let G be a graph and let A be the set of column vectors of its incidence matrix. If K[G] is normal, then the canonical module of K[G] is: ωK[G] = ({xa | a ∈ NA ∩ ri(R+ A)}). Use this formula to prove the following:

498

Chapter 11

(a) The join G1 ∗ G2 of two disjoint graphs G1 and G2 consists of G1 ∪ G2 and all the lines joining G1 with G2 . If G1 , G2 are connected graphs on n vertices and K[G1 ], K[G2 ] are normal, then a(K[G1 ∗ G2 ]) = −n. (b) Let G be a connected non-bipartite graph on n vertices. If K[G] is normal, then H I n+1 a(K[G]) − 1 ≤ a(K[C(G)]) ≤ − . 2 (c) Let G1 , G2 be two connected non-bipartite graphs. If K[G1 ] and K[G2 ] are normal, then a(K[G1 ]) + a(K[G2 ]) ≤ a(K[G1 ∗ G2 ]).

Chapter 12

Normality of Rees Algebras of Monomial Ideals Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and let I be an ideal of R generated by a ﬁnite set F = {xv1 , . . . , xvq } of monomials. In this chapter we study the integral closure of I, and the normality and invariants of R[It], the Rees algebra of I. The normalization of a Rees algebra is examined using the Danilov–Stanley formula, Carath´eodory’s theorem, and Hilbert bases of Rees cones. Interesting classes of normal ideals, such as ideals of Veronese type and polymatroidal ideals, are introduced in the chapter. The divisor class group of a normal Rees algebra is computed using polyhedral geometry.

12.1

Integral closure of monomial ideals

Proposition 12.1.1 [275] Let I be a monomial ideal of a polynomial ring R over a ﬁeld K. Then I, the integral closure of I, is a monomial ideal. Proposition 12.1.2 Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and let I ⊂ R be a monomial ideal. Then (a) I = (xα | xmα ∈ I m for some m ≥ 1), and (b) I = (xα | α ∈ conv(log(I)) ∩ Zn ). Proof. (a): If xmα ∈ I m , then xα satisﬁes a polynomial of the form z m + am , with am ∈ I m , hence xα is in I. On the other hand if z = xα ∈ I, then there is an equation z r + a1 z r−1 + · · · + ar−1 z + ar = 0 with ai ∈ I i

500

Chapter 12

for all i. Since I is a monomial ideal one obtains z m ∈ I m for some m ≥ 1. Observing that I is a monomial ideal the asserted equality follows. q (b): “⊃”: Let Zn . One can write α = i=1 μi βi , qα ∈ conv(log I)β∩ where μi ∈ Q+ , i=1 μi = 1 and x i ∈ I. Pick m ∈ N+ such that mμi ∈ N for all i. Then xmα ∈ I m and xα ∈ I. “⊂”: Since I is generated by monomials for any xα ∈ I there is m ∈ N+ such that xmα ∈ I m . It follows readily that α ∈ conv(log I). 2 Example 12.1.3 If R = K[x1 , x2 , x3 ] and I = (x31 , x2 x23 ), then using the program Normaliz [68] we get I = (x31 , x2 x23 , x21 x2 x3 ). A geometric description of the integral closure Let α ∈ Qn+ , where Q+ is the set of nonnegative rational numbers. We deﬁne the upper right corner or ceiling of α as the vector α whose entries are given by

αi if αi ∈ N, αi = (αi ) + 1 if αi ∈ / N, where (αi ) stands for the integer part of αi . Accordingly we can deﬁne the ceiling of any vector in Rn or the ceiling of any real number. Let conv(v1 , . . . , vq ) be the convex hull (over the rationals), that is, q q λi vi λi = 1, λi ∈ Q+ conv(v1 , . . . , vq ) = i=1

i=1

is the set of all convex combinations of v1 , . . . , vq . Proposition 12.1.4 Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K. If I ⊂ R is an ideal generated by monomials xv1 , . . . , xvq , then *3 4+ I= xα α ∈ conv(v1 , . . . , vq ) and I is generated by all xa with a ∈ (conv(v1 , . . . , vq ) + [0, 1)n ) ∩ Nn . Proof. “⊃”: Let α = qi=i λi vi be a convex combination of v1 , . . . , vq with λi ∈ Q+ for all i. Since α ≥ α, there is β ∈ Qn+ such that α = β + α. Hence there is 0 = p ∈ N so that pβ ∈ Nn and pλi ∈ N for all i. Therefore pλ1

xpα = xpβ xpα = xpβ (xv1 )

pλq

· · · (xvq )

∈ I p ⇒ xα ∈ I.

“⊂”: Let xγ ∈ I, that is, xpγ ∈ I p for some 0 = p ∈ N. There are nonnegative integers s1 , . . . , sq such that s

s

xpγ = xδ (xv1 ) 1 · · · (xvq ) q and s1 + · · · + sq = p. Hence γ = (δ/p) + qi=1 (si /p) vi . We set α = qi=1 (si /p)vi . By dividing the entries of δ by p, we can write γ = θ + β + α, where 0 ≤ βi < 1 for all i and θ ∈ Nn . Notice β + α ∈ Nn . It is seen that α = β + α. Thus 2 xγ = xθ xα with α ∈ conv(v1 , . . . , vq ) as required.

Normality of Rees Algebras of Monomial Ideals

501

Example 12.1.5 If I = (x31 , x42 ) ⊂ K[x1 , x2 ], by Proposition 12.1.4, I is generated by the monomials of the points marked in the ﬁgure: 6 4s s S S 3 s Ss S S s 2 s S S 1 s S S S Ss 0 1 2 3

-

Thus I = (x31 , x42 , x1 x32 , x21 x22 ). Corollary 12.1.6 Let I ⊂ R = K[x1 , . . . , xn ] be an ideal generated by monomials of degree at most k. The following hold. (a) I is generated by monomials of degree at most k + n − 1. (b) If ht(I) = n, then I is generated by monomials of degree at most k. Proof. (a) Let G(I) = {xv1 , . . . , xvq } be the minimal set of generators of I. Take α = (αi ) ∈ conv(v1 , . . . , vq ) and set β = (βi ) = α. As βi is equal to αi + δi , for some δi in [0, 1), one has deg(xβ ) < k + n, as required. (b) One has (xk1 , . . . , xkn ) = (x1 , . . . , xn )k ⊂ I because xki ∈ I for all i. Therefore I does not require minimal generators of degree k + 1. 2 Complete and m-full ideals in dimension two Let R = K[x1 , x2 ] be a polynomial ring over an inﬁnite ﬁeld K, let m = (x1 , x2 ) and let I be an m-primary ideal of R minimally generated by q monomials, q = μ(I), that are listed lexicographically, b

b

a

I = (xa1 1 , xa1 2 x2q−1 , . . . , xa1 i x2q−i+1 , . . . , x1 q−1 xb22 , xb21 ) with a1 > a2 > · · · > aq−1 > aq := 0, b1 > b2 > · · · > bq−1 > bq := 0. We assume that I = mq−1 . The ideal I is said to be m-full if (mI : f ) = I for some f ∈ m \ m2 or equivalently I is m-full if (I : m) = (I : f ) [259, Proposition 14.1.6]. A useful characterization of m-full ideals is the following result. Theorem 12.1.7 (Rees–Watanabe [425, Theorem 4]) If I is an m-primary ideal of a regular local ring (R, m) of dimension two, then I is m-full if and only if for all ideals I ⊂ J, μ(J) ≤ μ(I).

502

Chapter 12

The order of I is ord(I) := max{k| I ⊂ mk }. In regular local rings of dimension two one has μ(I) ≤ ord(I) + 1 [259, Lemma 14.1.3]. Lemma 12.1.8 If μ(I) = ord(I) + 1, then I is m-full. Proof. Let J be an ideal of R. Note that I ⊂ J implies ord(J) ≤ ord(I). Since μ(J) ≤ ord(J) + 1 ≤ μ(I), I satisﬁes Theorem 12.1.7. 2 The following result classiﬁes all m-full monomial ideals of K[x1 , x2 ]. Theorem 12.1.9 [181] I is m-full if and only if there is 1 ≤ k ≤ q such that the following conditions hold (i) bq−i − bq−i+1 = 1 for 1 ≤ i ≤ k − 1, (ii) k = q or k < q and bq−k − bq−k+1 ≥ 2, (iii) ai − ai+1 = 1 for k ≤ i ≤ q − 1. Proof. ⇒) If I is m-full, ord(I) ≤ q − 1, as otherwise I ⊂ (x1 , x2 )q , which has q + 1 minimal generators, which would violate Theorem 12.1.7. Thus ord(I) ≤ q − 1. Let xk1 x2q−1−k be a monomial of I of degree q − 1. This means that there are at most q − 1 − k elements prior to xk1 x2q−1−k and at most k elements after. This gives a

b

, xk−1 x2q−k , . . . , x1 xb22 , xb21 ). I = (xa1 1 , xa1 2 x2 , . . . , x1 k−1 x2q−2−k , xk1 xq−1−k 2 1 By choosing k as small as possible we achieve all three conditions. ⇐) By Lemma 12.1.8 we need only show μ(I) ≥ ord(I) + 1. Since q = μ(I) its suﬃces to show that I has a monomial of degree q − 1. Notice that bq−i = i for 1 ≤ i ≤ k − 1 and ai = q − i for k ≤ i ≤ q − 1. In particular xk−1 belongs to I ak = q − k and bq−k+1 = k − 1. Thus the monomial xq−k 1 2 and has degree q − 1. 2 If I is complete, then I is m-full [425, Theorem 5]. In polynomial rings in two variables any complete ideal is normal; more generally one has the following result of Zariski. Theorem 12.1.10 [436, Appendix 5] If I1 , . . . , Ir are complete ideals in a regular local ring R of dimension two, then I1 · · · Ir is complete. In [101] Crispin Qui˜ nonez studied the normality of m-primary monomial ideals in K[x1 , x2 ] and established a criterion in terms of certain partial blocks and associated sequences of rational numbers. 21 Example 12.1.11 The ideal I = (x31 , x21 x82 , x1 x15 2 , x2 ) is m-full and is not 3 2 7 21 normal. The integral closure of I is I = (x1 , x1 x2 , x1 x14 2 , x2 ).

Normality of Rees Algebras of Monomial Ideals

503

Linear programming test We give a test to check whether or not a given monomial lies in the integral closure of a monomial ideal. Proposition 12.1.12 (Membership test) Let v1 , . . . , vq be a set of vectors in Nn and let A be the n×q matrix whose columns are the vectors v1 , . . . , vq . If I is the ideal generated by xv1 , . . . , xvq , then a monomial xb lies in the integral closure of I if and only if the linear program: Maximize x1 + · · · + xq Subject to Ax ≤ b and x ≥ 0 has an optimal value greater than or equal to 1, which is attained at a vertex of the rational polytope P = {x ∈ Rq | Ax ≤ b and x ≥ 0}. Proof. ⇒) Let xb ∈ I, that is, xpb ∈ I p for some positive integer p. There r r are nonnegative integers r1 , . . . , rq such that xpb = xδ (xv1 ) 1 · · · (xvq ) q and r1 + · · · + rq = p. Hence the column vector c with entries ci = ri /p satisﬁes Ac ≤ b and c1 + · · · + cq = 1. Thus the linear program has an optimal value greater than or equal to 1. ⇐) Since the vertices of P have rational entries (see Proposition 1.1.46) and the maximum of x1 + · · · + xq is attained at a vertex of the polytope P (see Proposition 1.1.41), there are c1 , . . . , cq in Q+ such that c1 +· · ·+cq ≥ 1 and c1 v1 +· · ·+c qqvq ≤ b. Hence there are 1 , . . . , q in Q such that 0 ≤ i ≤ ci for all i and i=1 i = 1. Therefore there is a vector δ ∈ Qn+ such that q b = δ + i=1 i vi . Thus there is an integer p > 0 such that pb = pδ + p1 v1 + · · · + pq vq ⇒ xb ∈ I. ∈Nn

∈N

2

∈N

Remark 12.1.13 By Theorem 1.1.56 one can also use the dual problem Minimize b1 y1 + · · · + bn yn Subject to yA ≥ 1 and y ≥ 0 to check whether xb is in I. In this case one has a ﬁxed polyhedron Q = {y ∈ Rn | yA ≥ 1 and y ≥ 0} that can be used to test membership of any monomial xb , while in the primal problem the polytope P depends on b. Example 12.1.14 Consider the ideal I = (x31 , x42 , x53 , x64 , x1 x2 x3 x4 ) and the exponent vector b = (0, 1, 1, 4). To check whether xb is in I we use the following procedure in Mathematica [431]:

504

Chapter 12

ieq:={3x1+x5<=0,4x2+x5<=1,5x3+x5<=1,6x4+x5<=4} vars:={x1,x2,x3,x4,x5} f:=x1+x2+x3+x4+x5 ConstrainedMax[f,ieq,vars] The answer that Mathematica gives is: {67/60, {x1 -> 0, x2 -> 1/4, x3 -> 1/5, x4 -> 2/3, x5 -> 0}} where the ﬁrst entry is the optimal value and the other entries correspond to a vertex of the polytope P. By Proposition 12.1.12, we get xb ∈ I. Using PORTA [84] one obtains that the vertices of the polyhedral set Q described in Remark 12.1.13 are the rows of the matrix: ⎡ ⎤ 1/3 1/4 1/5 13/60 ⎢ 1/3 1/4 1/4 1/6 ⎥ ⎥. M =⎢ ⎣ 1/3 3/10 1/5 1/6 ⎦ 23/60 1/4 1/5 1/6 This is a “membership test matrix” in the sense that a monomial xb lies in I if and only if M b ≥ 1. If xb = x2 x3 x44 , we get: M b = (79/60, 7/6, 7/6, 67/60) ≥ (1, 1, 1, 1) ⇒ xb ∈ I.

Exercises 12.1.15 Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and F a ﬁnite set of monomials of the same degree. If I = (F ) is complete, then conv(log(F )) ∩ Zn = log(F ). 12.1.16 Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K. If m = (x1 , . . . , xn ), then md is the integral closure of (xd1 , . . . , xdn ). 12.1.17 Let I = (xv1 , . . . , xvq ) be a monomial ideal of a polynomial ring over a ﬁeld K. Prove that I i = (xiv1 , . . . , xivq ) for i ≥ 1. 12.1.18 If I ⊂ K[x1 , x2 ] is an ideal generated by monomials of degree k, then so is I. 12.1.19 Prove that I ∩ J = I ∩ J, where I = (x51 , x32 ) and J = (x52 , x23 ). 12.1.20 If I = (x21 , x32 ) ⊂ K[x1 , x2 ] and P = conv((2, 0), (0, 3)), then P ∩Z2 is equal to {(2, 0), (0, 3)} and I = I + (x1 x22 ). 12.1.21 If I = (x21 , x1 x22 , x32 ) ⊂ K[x1 , x2 ], then I is normal.

Normality of Rees Algebras of Monomial Ideals

12.2

505

Normality criteria

In this section we present a simple normality criterion which is suitable to study the normality of monomial ideals. Proposition 12.2.1 Let (R, m) be a Noetherian local ring and let I be a proper ideal of R. Then I is normal if and only if (a) IRp is normal for any prime ideal I ⊂ p = m, and (b) I r ∩ (I r : m) = I r for all r ≥ 1. Proof. ⇒) That (a) is satisﬁed follows from the fact that integral closures and powers of ideals commute with localizations. Part (b) is clearly satisﬁed. ⇐) Assume I r = I r for some r ≥ 1. Take p ∈ AssR (M ), where M = r I /I r . Since p is in the support of M we get Mp = (0) and consequently, by (a), we must have p = m. Hence there is an embedding R/m → M

(1 → x0 ),

where x0 ∈ I r \ I r and m = ann(x0 ). From (b) one concludes that x0 ∈ I r , a contradiction. Thus I r = I r for all r ≥ 1. 2 Proposition 12.2.2 Let (R, m) be a Noetherian local ring and let I be a proper ideal of R. Then I is integrally closed if and only if IRp is integrally closed for any prime ideal I ⊂ p = m and I ∩ (I : m) = I. Proof. It follows from the proof of Proposition 12.2.1.

2

The next result proves that a normal monomial ideal of K[x1 , . . . , xn ] stays normal if we make any variable, say xn , equal to 1 or 0. Proposition 12.2.3 Let I = (xv1 , . . . , xvq ) be a normal ideal. If each xvi is written as xani gi , where gi is a monomial in R = K[x1 , . . . , xn−1 ], then (a) J = (g1 , . . . , gq ) ⊂ R is a normal ideal, and (b) L = ({gi | ai = 0}) ⊂ R is a normal ideal. Proof. (a) We will show that J r = J r for all r ≥ 1. Take a monomial xα of R that belong to J r . Then (xα )p ∈ J rp for some p ≥ 1 and we can write (xα )p = g1b1 · · · gqbq xβ , q β both sides of the equation where i=1 bi = rp and xq ∈ R . Multiplying sp s α r above by xn with s = i=1 ai bi , we get (xsn xα )p ∈ I rp . Hence qxn x ∈ I aq s α a1 c1 cq γ and we can write xn x = (xn g1 ) · · · (xn gq ) x , where i=1 ci = r. Evaluating the last equality at xn = 1 gives that xα ∈ J r , as required. Part (b) follows using similar arguments. 2

506

Chapter 12

Theorem 12.2.4 (Normality criterion) Let I = (xv1 , . . . , xvq ) be a monomial ideal of R = K[x1 , . . . , xn ] and let Ji be the ideal of R generated by all monomials obtained from {xv1 , . . . , xvq } by making xi = 1. Then I is normal if and only if (a) Ji is normal for all i (resp. Ip is normal for any prime ideal p = m), (b) I r ∩ (I r : m) = I r for all r ≥ 1, where m = (x1 , . . . , xn ). Proof. ⇒) Condition (a) follows from Proposition 12.2.3 (resp. that Ip is normal follows using that integral closures and powers of ideals commute with localizations; see Proposition 4.3.4). Part (b) is clearly satisﬁed. ⇐) To prove that I r = I r set M = I r /I r . We proceed by contradiction assuming I r I r . Take p ∈ AssR (M ). There are embeddings ϕ

R/p → M → R/I r . Thus p ∈ AssR (R/I r ). Consequently p is generated by a subset of {x1 , . . . , xn } and p ⊂ m. Note p = m, otherwise if xi ∈ / p for some i, then by (a) IRp = Ji Rp is normal and Mp = (0), a contradiction because p is in the support of M . Thus p = m. If x0 = ϕ(1), we get x0 ∈ I r \ I r and x0 ∈ (I r : m), which contradicts (b). 2 Proposition 12.2.5 Let R = ⊕∞ i=0 Ri be an N-graded ring and let I = R≥α be the ideal of R generated by the elements of degree at least α. If R is a domain, then I is integrally closed. Proof. For f = fs + fs+1 + · · · + fr ∈ R, with fi ∈ Ri for all i and fs = 0, we set deg(f ) = s. As R is a domain one has deg(gh) = deg(g) + deg(h) for g, h ∈ R and gh = 0. Take 0 = f ∈ I with s = deg(f ). There is an equation f n + a1 f n−1 + · · · + an−1 f + an = 0

(ai ∈ I i ).

Assume s < α. If ai f n−i = 0, then deg(ai f n−i ) ≥ iα + (n − i)s > ns. Thus one can rewrite the equation above as fsn + (terms of degree greater than ns) = 0, a contradiction because fs = 0 and R is graded. Hence deg(f ) ≥ α.

2

Theorem 12.2.6 [153] Let K be a ﬁeld and let R = K[x1 , . . . , xn ] be a positively graded K-algebra. If R is a domain and c is the least common multiple of the weights of the xi , then I = R≥nc is a normal ideal. If R is normal, then the Rees ring R[It] is normal. Proof. In [153] it is proved that I p is equal to R≥pnc for all p ≥ 1, where the latter is integrally closed. 2

Normality of Rees Algebras of Monomial Ideals

507

Ideals of mixed products Let R = K[x1 , . . . , xm , y1 , . . . , yn ] be a ring of polynomials over a ﬁeld K. Given k, r, s, t in N such that k + r = s + t, consider the square-free monomial ideal of mixed products given by L = Ik Jr + Is Jt , where Ik (resp. Jr ) is the ideal of R generated by the square-free monomials of degree k in the variables x1 , . . . , xm (resp. y1 , . . . , yn ). The Betti numbers and some of the invariants of L have been studied in [246, 260, 355]. The invariants of the symmetric algebra of L are studied in [288]. Next we present a complete classiﬁcation of the normal ideals of mixed products that can be shown using the normality criterion of Theorem 12.2.4. Theorem 12.2.7 [351] If L = R, then L is normal if and only if it can be written (up to permutation of k, s and r, t) in one of the following forms: (a) L = Ik Jr + Ik+1 Jr−1 , k ≥ 0 and r ≥ 1. (b) L = Ik Jr , k ≥ 1 or r ≥ 1. (c) L = Ik Jr + Is Jt , 0 = k < s = m, or 0 = t < r = n, or k = t = 0, s = 1. Deﬁnition 12.2.8 Let G be a simple graph (resp. digraph) with vertices x1 , . . . , xn and let t ≥ 2 be an integer. The path ideal of G, denoted by It (G), is the ideal of K[x1 , . . . xn ] generated by all square-free monomials xi1 · · · xit such that the xij is adjacent to xij+1 (resp. ej = (xij , xij+1 ) is a directed edge from xij to xij+1 ) for all 1 ≤ j ≤ t − 1. Path ideals of digraphs were introduced by Conca and De Negri [89] and later studied in [50, 211]. Kubitzke and Olteanu [284] have studied the algebraic properties and invariants of path ideals of certain types of posets. Corollary 12.2.9 If G is a complete bipartite graph, then the path ideal It (G) is normal for all t ≥ 2. Proof. Let x1 , . . . , xm , y1 , . . . , yn be the vertex set of G, one may assume that the edges of G are precisely the pairs of the form {xi , yj }. Therefore It (G) is Id Jd+1 + Id+1 Jd if t = 2d + 1 and It (G) is Id Jd if t = 2d. Hence the result follows from Theorem 12.2.7. 2

Exercises 12.2.10 Let R = K[x1 , . . . , xm , y1 , . . . , yn ] and L = Is Jt , where Is (resp. Jt ) denote the ideal of R generated by the square-free monomials of degree s (resp. t) in the xi (resp. yi ) variables. Prove that L is normal.

508

Chapter 12

12.2.11 Let R = K[x1 , . . . xn ] and R[x] be polynomial rings over a ﬁeld K and let Lt = It + xIt−2 , where It is the ideal of R generated by the square-free monomials of degree t ≥ 2. Then Lt is a normal ideal of R[x]. 12.2.12 Let I and J be two ideals of the polynomial rings K[x1 , . . . , xm ] and K[y1 , . . . , yn ], respectively. The join of I and J is: I ∗ J = I + J + (xi yj | 1 ≤ i ≤ m and 1 ≤ j ≤ n). If I and J are normal ideals generated by square-free monomials of the same degree t ≥ 2, then their join I ∗ J is normal. 12.2.13 Let X = {x1 , . . . , xm } and {y1 , . . . , yn } be two sets of variables over a ﬁeld K. Let I be a normal ideal of K[X] generated by square-free monomials of degree t and let L = I + (X)(Y ). Then L is a normal ideal. 12.2.14 Let R = K[x1 , . . . , x8 ] be a polynomial ring over a ﬁeld K of characteristic zero and let I be the ideal of R generated by x1 x2 x3 x4 , x5 x6 x7 x8 , x1 x2 x4 x5 , x3 x4 x6 x8 , x1 x3 x7 , x2 x7 x8 . Prove that I is normal, whereas the ideal J = I + (x1 y1 , . . . , x8 y1 ) is not. 12.2.15 Let R = K[x1 , . . . , x6 ] be a polynomial ring over a ﬁeld K. Prove that I = (x3 x5 x6 , x3 x4 x6 , x2 x5 x6 , x2 x4 x5 , x1 x3 x5 , x1 x2 x6 ) is not normal.

12.3

Rees cones and polymatroidal ideals

Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K. In this part we prove that Rees algebras of polymatroidal ideals are normal domains. In particular ideals of Veronese type are normal. Quasi-ideal Rees cones Let I be a monomial ideal of R minimally generated by the set F = {xv1 , . . . , xvq }. The Rees cone of I is the rational polyhedral cone in Rn+1 , denoted by R+ A or R+ (I), generated by A := {e1 , . . . , en , (v1 , 1), . . . , (vq , 1)} ⊂ Rn+1 , where ei is the ith unit vector. Consider the index set J = {1 ≤ i ≤ n| ei , vj = 0 for some j} ∪ {n + 1}. Notice that dim R+ A = n + 1. By Theorem 1.4.2, the Rees cone has a unique irreducible representation ! ! r + + Hei H i (12.1) R+ A = i∈J

i=1

Normality of Rees Algebras of Monomial Ideals

509

such that, for each i, 0 = i ∈ Zn+1 , the non-zero entries of i are relatively prime, the ﬁrst n entries of i are in N and the last entry of i is negative. Let P be the convex hull of A = {v1 , . . . , vq }. Recall that the Ehrhart ring of P and the Rees algebra of I are the monomial subrings given by A(P) = K[xa tb | a ∈ bP ∩ Zn ] and R[It] = K[x1 , . . . , xn , xv1 t, . . . , xvq t] respectively, where t is a new variable. Using Theorem 9.1.1, it follows readily that the normalization of R[It] is R[It] = K[xa tb | (a, b) ∈ R+ A ∩ Zn+1 ]. Notation For use below we set [n] = {1, . . . , n}. Theorem 12.3.1 If k has the form k = −dk en+1 + i∈Ak ei for some Ak ⊂ [n] for k = 1, . . . , r, then A(P)[x1 , . . . , xn ] = R[It]. Proof. “⊂”: This inclusion is clear because A(P) ⊂ R[It]. “⊃”: Let xa tb = xa1 1 · · · xann tb ∈ R[It] be a minimal generator, i.e., (a, b) cannot be written as a sum of two non-zero integral vectors in the Rees cone R+ A . We may assume ai ≥ 1 for 1 ≤ i ≤ m, ai = 0 for i > m, and b ≥ 1. Case (I): (a, b), i > 0 for all i. The vector γ = (a, b) − e1 satisﬁes γ, i ≥ 0 for all i, that is γ ∈ R+ A . Thus since (a, b) = e1 + γ we derive a contradiction. Case (II): (a, b), i = 0 for some i. We may assume {i | (a, b), i = 0} = {1 , . . . , p }. Subcase (II.a): ei ∈ H1 ∩ · · · ∩ Hp for some 1 ≤ i ≤ m. It is not hard to verify that the vector γ = (a, b) − ei satisﬁes γ, k ≥ 0 for all k. Thus γ ∈ R+ A , a contradiction because (a, b) = ei + γ. / H1 ∩ · · · ∩ Hp for all 1 ≤ i ≤ m. Since the vector Subcase (II.b): ei ∈ (a, b) belongs to the Rees cone it follows that we can write (a, b) = λ1 (v1 , 1) + · · · + λq (vq , 1)

(λi ≥ 0).

Therefore a ∈ bP, i.e., xa tb ∈ A(P).

2

If 1 , . . . , r satisfy the condition of Theorem 12.3.1 (resp. dk = 1 for all k), we say that the Rees cone of I is quasi-ideal (resp. ideal ). Corollary 12.3.2 If the Rees cone of I is quasi-ideal and K[F t] = A(P), then the Rees algebra R[It] is normal. Proof. By Theorem 12.3.1 we get R[It] = K[F t][x1 , . . . , xn ] = A(P)[x1 , . . . , xn ] = R[It].

2

510

Chapter 12

Polymatroidal sets of monomials Let A = {v1 , . . . , vq } ⊂ Nn be the set of bases of a discrete polymatroid of rank d, i.e., |vi | = d for all i and given any two a = (ai ), c = (ci ) in A, if ai > ci for some index i, then there is an index j with aj < cj such that a− ei + ej is in A. We refer to [221, 227] for the theory of discrete polymatroids. The set F = {xv1 , . . . , xvq } (resp. the ideal I = (F ) ⊂ R) is called a polymatroidal set of monomials (resp. a polymatroidal ideal ). Lemma 12.3.3 F = {xvi /x1 : x1 occurs in xvi } is also polymatroidal. Proof. It is left as an exercise.

2

Lemma 12.3.4 Let A = {e1 , . . . , en , (v1 , 1), . . . , (vq , 1)}. If E is a facet of R+ A , then E = R+ A ∩ Hb , where b = ei for some 1 ≤ i ≤ n + 1 or b = (bi ) ∈ Zn+1 , bi ∈ {0, 1} for i = 1, . . . , n and bn+1 ≥ −d. Proof. The proof is by induction on d. The case d = 1 is clear because the matrix whose columns are the vectors in A is totally unimodular. Assume that d ≥ 2. For 1 ≤ k ≤ q, we set fvi = (vi , 1), Bi = supp(vi ) and vk = (vk1 , . . . , vkn ). Let E be a facet of R+ A . There is a unique 0 = b = (bi ) in Zn+1 whose non-zero entries are relatively prime such that (a) E = R+ A ∩ Hb = R+ A , R+ A ⊂ Hb+ , and (b) there is a linearly independent set A ⊂ Hb ∩ A with |A| = n. If A = {e1 , . . . , en } (resp. A ⊂ {fv1 , . . . , fvq }), then b = en+1 (resp. b = (1, . . . , 1, −d)). Notice that bi ≥ 0 for i = 1, . . . , n. Thus we may assume that A is the set {e1 , . . . , es , fv1 , . . . , fvt }, where 1 ≤ s ≤ n − 1, s + t = n, and {1, . . . , s} is the set of all i ∈ [n] such that ei ∈ Hb . We may assume that 1 ∈ Bk for some 1 ≤ k ≤ q. Indeed if 1 is not in ∪qi=1 Bi , then the facets of R+ A diﬀerent from He1 ∩ R+ A are in one-to-one correspondence with the facets of R+ {e2 , . . . , en , fv1 , . . . , fvq }. Assume that 1 ∈ / B1 . Since vk1 > v11 for some k, by the symmetric exchange property of A [221, Theorem 4.1, p. 241] there is j with vkj < v1j such that v1 + e1 − ej = vi for some vi in A. Hence fv1 + e1 = fvi + ej ⇒ fvi , b = −ej , b ⇒ fvi , b = ej , b = 0, i.e., fvi and ej belong to Hb . Thus from the outset we may assume that / B2 . Since v11 > v21 , by the 1 ∈ B1 . Next assume that t ≥ 2 and 1 ∈ symmetric exchange property of A there is j with v1j < v2j such that v2 + e1 − ej = vi for some vi in A. Hence fv2 + e1 = fvi + ej ⇒ fvi , b = ej , b = 0,

Normality of Rees Algebras of Monomial Ideals

511

i.e., fvi and ej belong to Hb . Notice that i = 1. Indeed if i = 1, then from the equality above we get that {fv2 , e1 , fv1 , ej } is linearly dependent, a contradiction. Applying the arguments above repeatedly shows that we may assume that 1 belongs to Bi for i = 1, . . . , t. We may also assume that v1 , . . . , vr is the set of all vectors in A such that 1 ∈ Bi , where t ≤ r. For 1 ≤ i ≤ r, we set vi = vi − e1 . By Lemma 12.3.3 the set {v1 , . . . , vr } is again the set of basis of a discrete polymatroid of rank d − 1. Consider the Rees cone R+ A generated by A = {e1 , e2 , . . . , en , fv1 , . . . , fvr }. Notice that R+ A ⊂ Hb+ and that {e1 , e2 , . . . , es , fv1 , . . . , fvt } is a linearly independent set. Thus R+ A ∩ Hb is a facet of the cone R+ A and the result follows by induction. 2 Theorem 12.3.5 [221] If P = conv(v1 , . . . , vq ), then K[F t] = A(P). Theorem 12.3.6 [422] If I is a polymatroidal ideal, then I is normal. Proof. By Lemma 12.3.4 and Theorem 12.3.5 R+ A is quasi-ideal and K[F t] = A(P). Thus, by Corollary 12.3.2, we get that R[It] is normal. 2 Corollary 12.3.7 If I is a polymatroidal ideal, then I has the persistence property. Proof. By Theorems 12.3.6 and 7.7.3 I has the persistence property.

2

Normality of an ideal of Veronese type Let R = K[x1 , . . . , xn ] be a ring of polynomials over an arbitrary ﬁeld K. Given a sequence of integers n (s1 , s2 , . . . , sn ; d) such that 1 ≤ sj ≤ d ≤ i=1 si for all j, we deﬁne A as the set of partitions A = {(a1 , . . . , an ) ∈ Zn | a1 + · · · + an = d; 0 ≤ ai ≤ si ∀ i}, and F as the set of monomials F = {xa | a ∈ A} = {f1 , . . . , fq }. Deﬁnition 12.3.8 The ideal I = (F ) ⊂ R is said to be of Veronese type of degree d with deﬁning sequence (s1 , . . . , sn ; d). If si = 1 for all i we call I the dth square-free Veronese ideal , and if si = d for all i we call I the dth Veronese ideal . Similar terminology may be applied to subrings. The monomial subring K[F ] ⊂ R and its toric ideal have been studied by Sturmfels in [400]. In loc. cit. it is shown that the toric ideal of K[F ] has a quadratic Gr¨obner basis whose initial ideal is square-free. Proposition 12.3.9 [147] If I ⊂ R is an ideal of Veronese type, then I is normal.

512

Chapter 12

Proof. The ideal I is a polymatroidal ideal. Then, the Rees algebra of I is normal by Theorem 12.3.6, i.e., I is a normal ideal. 2 Proposition 12.3.10 [418] If I is the dth square-free Veronese ideal of R, then I is normal. Proof. The result follows at once from Proposition 12.3.9.

2

The basis monomial ring of a matroid Let M be a matroid on a ﬁnite set X = {x1 , . . . , xn } and let B be the collection of bases of M . The clutter with vertex set X and edge set B is called the clutter of bases of M . Deﬁnition 12.3.11 The set of all xi1 · · · xir ∈ R such that {xi1 , . . . , xir } is in B is denoted by FM . The basis monomial ring of M is the ring K[FM ]. The square-free monomial ideal (FM ) is the basis monomial ideal of M . It is well known [338] that all bases of a matroid M have the same number of elements; this common number is called the rank of the matroid. Thus K[FM ] is a standard graded K-algebra and all monomials of FM have the same degree; see Section 9.2. Corollary 12.3.12 If I = (FM ) is the basis monomial ideal of a matroid, then R[It] is normal. Proof. It follows from Theorem 12.3.6.

2

Corollary 12.3.13 [429] K[FM ] is normal. Proof. As K[FM t] K[FM ], by Theorems 12.3.5 and 9.3.6, K[FM ] is normal. 2 Deﬁnition 12.3.14 The standard graded K-algebra S = K[FM ] is said to be Koszul if the residue ﬁeld K has a linear S-free resolution. A suﬃcient condition for K[FM ] to be Koszul is that its toric ideal has a quadratic Gr¨ obner basis; see [46, 221]. In [430, Conjecture 12] White conjectured that the toric ideal of K[FM ] is generated by quadrics. More recently Blum [46] asked whether K[FM ] is Koszul. An open problem posed by Bøgvad is whether the toric ideal of a smooth projectively normal toric variety is generated by quadrics (see [61, 400]).

Exercises 12.3.15 Let G = {xu1 , . . . , xus } and let d = maxi {|ui |}. Suppose that F = {xu1 , . . . , xur } is the set of all xui of degree d. If Q = conv(u1 , . . . , us ) and K[Gt] = A(Q), then K[F t] = A(P), where P = conv(u1 , . . . , ur ). 12.3.16 If B = {{x1 , x2 }, {x2 , x3 }, {x3 , x4 }, {x1 , x4 }}, prove that B satisﬁes the bases exchange property of matroids.

Normality of Rees Algebras of Monomial Ideals

12.4

513

Veronese subrings and the a-invariant

In this section we present an explicit generating set for the canonical module of a square-free Veronese subring and compute its a-invariant. A sharp upper bound for the a-invariant of the normalization of a subring generated by square-free monomials of the same degree will be presented, this bound is attained at a square-free Veronese subring. Let R = K[x1 , . . . , xn ] = ⊕∞ i=0 Ri be a polynomial ring over a ﬁeld K with the standard grading and let k ∈ N+ . An element in Rik is said to have normalized degree i. The kth Veronese subring of R is given by: R(k) :=

∞ )

Rik ⊂ R,

i=0

and the kth square-free Veronese subring of R is given by: K[Vk ] := K[{xi1 · · · xik | 1 ≤ i1 < · · · < ik ≤ n}]. The kth Veronese subring of R is graded by (R(k) )i = Rki . Thus R(k) is a K-algebra generated by all the monomials of R of normalized degree 1 and K[Vk ] is a graded subring of R(k) with the normalized grading: K[Vk ] =

∞ )

(K[Vk ])i ,

i=0

where (K[Vk ])i = K[Vk ] ∩ (R(k) )i . In what follows we shall assume that K[Vk ] has the normalized grading. Let F a ﬁnite set of square-free monomials of the same degree k. In this case there are embeddings K[F ] ⊂ K[Vk ] ⊂ R(k) . Let us ﬁx some of the notation that will be used throughout the rest of the section. Let n ≥ 2k ≥ 4 be two integers (this is not an essential restriction; see Remark 12.4.11). We set V = {ei1 + · · · + eik | 1 ≤ i1 < · · · < ik ≤ n}, where e1 , . . . , en are the canonical vectors in Rn . The K-subring of R spanned by the set {xa | a ∈ V} is equal to K[Vk ]. Remark 12.4.1 dim K[Vk ] = n. This follows from Corollary 8.2.21. The vector space generated by V is equal to Rn and dim R+ V = n.

514

Chapter 12

Lemma 12.4.2 [72] Set N1 = {−e1 , . . . , −en }, N = N1 ∪ N2 and N2 = {−e1 − · · · − ei−1 + (k − 1)ei − ei+1 − · · · − en | 1 ≤ i ≤ n}. If H is a supporting hyperplane of R+ V such that H contains a set of n − 1 linearly independent vectors in V, then H = Ha for some a ∈ N . Remark 12.4.3 The converse of Lemma 12.4.2 is also true because we are assuming n ≥ 2k ≥ 4, and this implies n ≥ k + 2. Note that if n = 3 and k = 2, then the cone R+ V has only three facets. Proposition 12.4.4 [72] A point x = (x1 , . . . , xn ) ∈ Rn is in R+ V if and only if x is a feasible solution of the system of linear inequalities (k − 1)xi −

−xi j=i xj

≤ 0, i = 1, . . . , n ≤ 0, i = 1, . . . , n.

Proof. Let R+ V = Hb−1 ∩ · · · ∩ Hb−m be the irreducible representation of R+ V. By Theorem 1.1.44 the set Hbi ∩ R+ V is a facet of R+ V. Note that Hbi is generated by a set of linearly independent vectors in V; see Proposition 1.1.23. Hence, by Lemma 12.4.2, we get Hbi = Ha for some a ∈ N , where N is the set deﬁned in Lemma 12.4.2. 2 Lemma 12.4.5 Let a be a vector in C ∩ ri(R+ V) and set I = {i | ai ≥ 2}. If |I| ≥ k and i1 , . . . , ik are distinct integers in I, then a = a−ei1 −· · ·−eik also belongs to NV ∩ ri(R+ V). Proof. Without loss of generality one may assume a1 ≥ a2 ≥ · · · ≥ an , ak ≥ 2 and a = a − e1 − · · · − ek . As R+ V has dimension n, it is seen that a ∈ ri(R+ V). By Proposition 12.3.10 the subring K[Vk ] is a normal domain, 2 and therefore NV = ZV ∩ R+ V. Since a ∈ ZV, we conclude a ∈ NV. Theorem 12.4.6 [72] Let ωK[Vk ] be the canonical module of K[Vk ] and let B be the set of monomials xa1 1 · · · xann satisfying the following conditions: (a) ai ≥ 1 and (k − 1)ai ≤ −1 + j=i aj , for all i. n (b) i=1 ai ≡ 0 mod (k). (c) |{ i | ai ≥ 2}| ≤ k − 1. If n ≥ 2k ≥ 4, then B is a generating set for ωK[Vk ] . Proof. By Theorem 9.1.5 we get ωK[Vk ] = ({xa | a ∈ NV ∩ri(R+ V)}). Using the arguments of the proof of Lemma 12.4.5 and by a repeated use of this lemma it is enough to prove that B ⊂ ωK[Vk ] . Let M ∈ B; we may assume

Normality of Rees Algebras of Monomial Ideals

515

a

k−1 M = xa1 1 · · · xk−1 xk · · · xn , where a1 ≥ · · · ≥ ak−1 ≥ 1. The monomial ak−1 a1 N = x1 · · · xk−1 can be factored as

N=

k−1 7 i=1

Ni , where Ni =

(x1 · · · xi )ai −ai+1 , (x1 · · · xk−1 )ak−1 ,

if 1 ≤ i ≤ k − 2 if i = k − 1.

;k−1 ;n On the other hand, by (a) and (b), we can write i=k xi = N i=1 Ni , where deg(Ni ) = (k − i)(ai − ai+1 ) if 1 ≤ i ≤ k − 2, deg(Ni ) = ak−1 if ; i = k − 1, and deg(N ) ≡ 0 mod (k). Hence M = N k−1 i=1 (Ni Ni ) is in K[Vk ], which readily implies M ∈ ωK[Vk ] . 2 Let S = K[{ti1 ···ik | 1 ≤ i1 < · · · < ik ≤ n}] be a polynomial ring over a ﬁeld K with one variable ti1 ···ik for each monomial xi1 · · · xik . Here S has the standard grading. There is a graded homomorphism of K-algebras ϕ

ϕ : S −→ K[Vk ], induced by ti1 ···ik −→ xi1 · · · xik , the ideal P = ker(ϕ) is the toric ideal of K[Vk ]. Thus the Cohen–Macaulay type of the ring K[Vk ], denoted by type(K[Vk ]), is the last Betti number in the minimal free resolution of S/P as an S-module; see Proposition 5.3.4. Remark 12.4.7 To compute the type of K[Vk ] recall that this number is the minimal number of generators of the canonical module ωK[Vk ] of K[Vk ]; ak−1 see Corollary 5.3.6. Notice that a monomial xa1 1 · · · xk−1 xk · · · xn is in B if k−1 and only if for all 1 ≤ i ≤ k − 1 one has j=1 aj = mk − n + k − 1 and 1 ≤ ai ≤ m − 1, for some m ≥ 2. Therefore nk ≤ m ≤ n − 2k + 2. Hence, by Theorem 12.4.6, the computation of the type of K[Vk ] reduces to counting partitions of positive integers. Corollary 12.4.8 If k = 2, n ≥ 2k and n is odd, then ωK[Vk ] = ({x1 · · · xj−1 x2i j xj+1 · · · xn | 1 ≤ j ≤ n, 1 ≤ i ≤ (n − 3)/2}). Corollary 12.4.9 If k = 2 and n ≥ 2k, then type(K[Vk ]) is n(n − 3)/2 if n is odd, and type(K[Vk ]) is (n2 − 4n + 2)/2 if n is even. J K Corollary 12.4.10 [72] If n ≥ 2k ≥ 4, then the a((K[Vk ]) = − nk . J K Proof. Set D = K[Vk ] and m = nk . To compute the a-invariant of D we use the formula of Lemma 9.1.7. It follows from Remark 12.4.7 that the degree of the generators in least degree of ωD is at least m. To complete the proof we exhibit some generators of ωD living in degree m. Write n = qk+r, 0 ≤ r < k; note q ≥ 2. If r ≥ 1, observe that the monomials x21 · · · x2k−r xk−r+1 · · · xk−1 xk · · · xn and x1 · · · xn−k+r x2n−k+r+1 · · · x2n−1 x2n

516

Chapter 12

belong to (ωD )m . In particular, D cannot be a Gorenstein ring in this case. 2 If r = 0, then the monomial M = x1 · · · xn satisﬁes M ∈ (ωD )m . Remark 12.4.11 (Duality) Let 1 ≤ k ≤ n − 1 be an integer and let Dn,k = K[{xi1 · · · xik | 1 ≤ i1 < · · · < ik ≤ n}], be the K-subring of R spanned by the xi1 · · · xik ’s. Observe that there is a graded isomorphism of K-algebras of degree zero: ρ : Dn,k −→ Dn,n−k , induced by ρ(xi1 · · · xik ) = xj1 · · · xjn−k , where {j1 , . . . , jn−k } = {1, . . . , n} \ {i1 , . . . , ik }. Thus if n ≤ 2k, then H a(Dn,k ) = a(Dn,n−k ) = −

I n . n−k

Because of this duality one may always assume that n ≥ 2k. Corollary 12.4.12 [108] The subring Dn,k is a Gorenstein ring if and only if k ∈ {1, n − 1} or n = 2k. Proof. By duality one may assume n ≥ 2k ≥ 4. Set D = Dn,k . If D is Gorenstein, then by the proof of Corollary 12.4.10 we may assume n = qk. If q ≥ 3, then x1 · · · xn and x31 x22 · · · x2k−1 xk · · · xn belong to (ωD )q and (ωD )q+1 respectively, which is impossible. Therefore q = 2, as required. Conversely assume n = 2k. Let B be as in Theorem 12.4.6. Take a monomial xa in B; it suﬃces to verify that xa is equal to x1 · · · xn . One ak−1 may assume that xa = xa1 1 · · · xk−1 xk · · · xn , where ai ≥ ai+1 ≥ 1. By k−1 hypothesis j=1 aj = k(m − 1) − 1 for some m ≥ 2. On the other hand one has kai ≤ k + k−1 j=1 aj for all 1 ≤ i ≤ k − 1. Hence ai ≤ m − 1 for all i. Therefore using the previous equality again we rapidly derive k(m − 1) − 1 ≤ (k − 1)(m − 1), which yields m ≤ 2. As a consequence ai = 1 for all i, as required. 2 Theorem 12.4.13 [72] Let F be a ﬁnite set of square-free monomials of degree k in R such that dim(K[F ]) = n. The following hold. (i) a(K[F ]) ≤ a(K[Vk ]). G F J K n if n ≤ 2k, n = k. (ii) a(K[F ]) ≤ − nk if n ≥ 2k and a(K[F ]) ≤ − n−k (iii) If n ≥ 2k ≥ 4, then K[F ] is generated as a JK-algebra by elements of K normalized degree less than or equal to n − nk .

Normality of Rees Algebras of Monomial Ideals

517

Proof. (i): We set V = {ei1 + · · · + eik | 1 ≤ i1 < · · · < ik ≤ n} and D = K[Vk ]. Let C, CF be the semigroups of Nn generated by V, log(F ), respectively. By Proposition 12.3.10 D is normal. Hence, K[F ] ⊂ D. Let M1 = {xa | a ∈ C F ∩ ri(R+ C F )} and M2 = {xa | a ∈ C ∩ ri(R+ C)}, where C F = ZCF ∩ R+ CF . Notice R+ C F = R+ CF and aﬀ(R+ CF ) = Rn . Therefore the relative interior of R+ C F equals its interior in Rn . For similar reasons we have ri(R+ C) = (R+ C)o . Hence ri(R+ C F ) ⊂ ri(R+ C). Altogether we obtain M1 ⊂ M2 . Let xb be an element of minimal degree in M1 so that deg(xb ) = −a(K[F ]). Set r = −a(K[F ]). Since xb is in M2 and xb ∈ K[F ]r ⊂ Dr , we conclude −a(D) = min{deg(xa )|a ∈ M2 } ≤ r = −a(K[F ]). (ii): It follows from Corollary 12.4.10, part (i), and the duality given in Remark 12.4.11. (iii): It follows from the proof of part (i) and using part (ii). 2 Example 12.4.14 Let K[F ] be the subring of R = K[x1 , . . . , x8 ] spanned by the monomials of R deﬁning the edges of the graph shown below. x1

x s s 2 @ @ @s x3 s x4 s x5 @ @ @sx6

x8 s @ @ @s x7

f1 = x1 x2 , f2 = x2 x3 , f3 = x1 x3 , f4 = x3 x4 , f5 = x4 x5 , f6 = x5 x6 , f7 = x6 x7 , f8 = x5 x7 , f9 = x7 x8 , f10 = x5 x8 , f11 = x6 x8 ,

f12 f13 f14 f15

= x1 x2 x3 x5 x6 x8 , = x1 x2 x3 x6 x7 x8 , = x1 x2 x3 x5 x6 x7 , = x1 x2 x3 x5 x7 x8 .

F := {f1 , . . . , f11 }, K[F ] = K[F ][f12 , f13 , f14 , f15 ].

The generators of K[F ] can be computed using the program Normaliz [68]. See also [413, Section 7.3]. A Noether normalization for K[F ] is given by A0 = K[h1 , . . . , h8 ] → K[F ] → K[F ], f3 , h7 = f5 − f7 − f9 − f11 , h2 = f6 , where h1 = f1 , h3 = f8 − f11 , h5 = f2 − h4 = f9 − f10 , h6 = f3 − f5 , h8 = f1 − 11 i=2 fi . As K[F ] is Cohen–Macaulay, by Proposition 3.1.27, we get K[F ] =

A0 1 ⊕ A0 f7 ⊕ A0 f10 ⊕ a0 f11 ⊕ 2 A0 f72 ⊕ A0 f10 ⊕ A0 f10 f11 ⊕ A0 f7 f11 ⊕ 3 3 2 ⊕ A0 f10 f11 ⊕ A0 f7 ⊕ A0 f72 f11 ⊕ A0 f10 3 A0 f12 ⊕ A0 f13 ⊕ A0 f14 ⊕ A0 f15 ⊕ A0 f10 f11 .

518

Chapter 12

Therefore the Hilbert series of K[F ] is equal to H(K[F ], z) =

1 + 3z + 4z 2 + 8z 3 + z 4 and a(K[F ]) = −4. (1 − z)8

Exercises 12.4.15 Let P be the toric ideal of the Veronese subring R(k) or the toric ideal of the kth square-free Veronese subring K[Vk ]. If k ≥ 2, then P is generated by homogeneous binomials of degree two. 12.4.16 [197, 311] Prove the following: (a) R(k) is Gorenstein if and only if k divides n. (b) a(K[Vk ]) ≤ a(R(k) ). (c) If nJ=K 5 and k = 3, then a(K[Vk ]) = −3 and a(R(k) ) = −2. (d) a(R(k) ) = − nk . 12.4.17 (a) If n = 2k + 1 ≥ 5, then the canonical module of K[Vk ] is generated by the set of all xa , a = (a1 , . . . , an ), such that |{i|ai = 2}| = k−1 and |{i|ai = 1}| = n − k + 1. (b) If h1 , . . . , hn is a system of parameters for K[Vk ] with each hi a form of degree 1, then K[Vk ]/(h1 , . . . , hn ) is a level algebra, i.e., all the non-zero elements of its socle have the same degree. n 12.4.18 If n = 2k + 1 ≥ 5, then type(K[Vk ]) = k−1 . 12.4.19 Let F be the set of all monomials xi xj such that xi and xj are connected by an edge of the following graph: x2 s HH HHx s1 x3 s

x4 s

sx 6 x5 s HH HHsx 7

The monomial subring A = K[F ] is a graded subring of A, where both rings are endowed with the normalized grading. Prove that the Hilbert series and Hilbert polynomial of A and A are given by: F (A, t) =

1 + t + t2 + t3 + t4 , (1 − t)7

ϕ(A, t) =

1 6 1 55 4 65 3 47 2 31 t + t5 + t + t + t + t + 1, 144 16 144 48 18 12

ϕ(A, t) =

1 6 17 5 55 4 21 3 47 2 157 t + t + t + t + t + t + 1. 144 240 144 16 18 60

F (A, t) =

1 + t + t2 + 2t3 (1 − t)7

In particular e(A) = e(A) = 5, a(A) = −3 and a(A) = −4.

Normality of Rees Algebras of Monomial Ideals

12.5

519

Normalizations of Rees algebras

Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K, and let I be a monomial ideal of R generated by xv1 , . . . , xvq . In this section we study the normality of R[It] and the behavior of the ﬁltration F = {I i }i≥0 . Proposition 12.5.1 If I = (xv1 , . . . , xvq ) is a monomial ideal of the ring of polynomials R = K[x1 , . . . , xn ], then R[It] is generated as a K-algebra by monomials of degree in t at most n − 1. Proof. If H is the minimal integral Hilbert basis of the Rees cone of I, 2 then R[It] = K[NH] and the result follows from Proposition 1.5.4. Lemma 12.5.2 Let I be a monomial ideal of R = K[x1 , . . . , xn ]. If R[It] is generated as a K-algebra by monomials of degree in t at most s and I i is integrally closed for i ≤ s, then R[It] is normal. Proof. Let xα tb be a generator of R[It] with b ≤ s. Note that xα ∈ I b = I b . Hence xα tb ∈ R[It], as required. 2 Corollary 12.5.3 [345] If I is a monomial ideal of a polynomial ring in n variables and I i is integrally closed for i ≤ n − 1, then I is normal. Proof. It follows from Proposition 12.5.1 and Lemma 12.5.2.

2

Theorem 12.5.4 Let I = (xv1 , . . . , xvq ) ⊂ K[x1 , . . . , xn ] be a monomial ideal and let r0 = rank(v1 , . . . , vq ). The following hold. (i) I b = II b−1 for b ≥ n. (ii) If A = {v1 , . . . , vq } is homogeneous, then I b = II b−1 for b ≥ r0 . Proof. We set A = {(v1 , 1), . . . , (vq , 1), e1 , . . . , en }. Assume b ≥ n (resp. b ≥ r0 and A homogeneous). Notice that we invariably have II b−1 ⊂ I b . To show the reverse inclusion take xα ∈ I b . Thus xα tb ∈ R[It] and (α, b) is in the Rees cone R+ A . By the proof of Proposition 1.5.4 and by Carath´eodory’s theorem (see Theorem 1.1.18), we can write (α, b) = λ1 (vi1 , 1) + · · · + λr (vir , 1) + μ1 ej1 + · · · + μs ejs

(λi , μk ∈ Q+ ),

where {(vi1 , 1), . . . , (vir , 1), ej1 , . . . , ejs } is a set of vectors (resp. linearly independent set of vectors) contained in A and r ≤ n (resp. r ≤ r0 ). Since b = λ1 + · · · + λr , we obtain that λi ≥ 1 for some i. It follows readily that 2 xα ∈ II b−1 , as required.

520

Chapter 12

Normalizations and Hilbert functions Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K with n ≥ 2 and let I be a zero dimensional monomial ideal of R. The integral closure of I i , denoted by I i , is again a monomial ideal for all i ≥ 1 (see Proposition 12.1.2). The length of R/I i will be denoted simply by (R/I i ). By Lemma 8.5.1 the length of R/I i is dimK (R/I i ). We are interested in computing the Hilbert function HF (i) := (R/I i ) = dimK (R/I i );

i ∈ N \ {0}; HF (0) = 0,

of the ﬁltration F = {I i }∞ i=0 . The Hilbert function of F is a polynomial function of degree n: HF (i) = cn in + cn−1 in−1 + · · · + c1 i + c0

(i

0),

where c0 , . . . , cn ∈ Q and cn = 0 (see the proof of Proposition 12.5.16). The polynomial ϕF (x) := cn xn + · · · + c0 is the Hilbert polynomial of F . By the results of [104], one has the equality n!cn = e(I), where e(I) is the multiplicity of the ideal I in the sense of Section 4.4. There is a unique set of monomials F = {xu1 , . . . , xur } that minimally generate I. For 1 ≤ i ≤ n, we may assume that the ith entry of ui is ai and that all other entries of ui are zero, where a1 , . . . , an are positive integers. We set α0 = (1/a1 , . . . , 1/an ). Let {un+1 , . . . , us } be the set of ui such that ui , α0 < 1, and let {us+1 , . . . , ur } be the set of ui such that i > n and ui , α0 ≥ 1. Consider the convex rational polyhedron Q

= =

Qn+ + conv(u1 , . . . , ur ) Qn+ + conv(u1 , . . . , un , un+1 , . . . , us ).

The second equality follows from the ﬁnite basis theorem and the equality Qn+ + conv(u1 , . . . , un ) = {x | x ≥ 0; x, α0 ≥ 1}. Lemma 12.5.5 I i = ({xa | a ∈ iQ ∩ Zn }) for 0 = i ∈ N. Proof. Let xv1 , . . . , xvq be a set of generators of I. Let xα ∈ I i , i.e., xpα ∈ I ip for some 0 = p ∈ N. Hence α/i ∈ conv(v1 , . . . , vq ) + Qn+ ⊂ conv(u1 , . . . , ur ) + Qn+ = Q and α ∈ iQ ∩ Zn . Conversely let α ∈ iQ ∩ Zn . Hence xpα ∈ (I)ip for some 0 = p ∈ N, this yields xα ∈ (I)i . Since (I)i ⊂ I i and the latter is complete 2 we get xα ∈ I i . Corollary 12.5.6 (R/I i ) = |Nn \ iQ| for i ≥ 1.

Normality of Rees Algebras of Monomial Ideals

521

Proof. It follows from Lemmas 8.5.1 and 12.5.5.

2

In what follows we set P := conv(u1 , . . . , us ), S := conv(0, u1 , . . . , un ) and d = dim(P). The Ehrhart function of P is given by χP (i) := |Zn ∩ iP|, i ∈ N. This is a polynomial function of degree d. The Ehrhart polynomial of P is the unique polynomial EP (x) of degree d such that EP (i) = χP (i) for i 0. Ehrhart functions are studied in Section 9.3. Proposition 12.5.7 HF (i) = ES (i) − EP (i) for i ∈ N. Proof. Since EP (0) = ES (0) = 1, we get HF (0) = 0. Assume i ≥ 1. Notice that from the decomposition Q = (Qn+ \ S) ∪ P we get iQ n N \ iQ

= (Qn+ \ iS) ∪ iP, = [Nn ∩ (iS)] \ [Nn ∩ (iP)].

Hence, by Corollary 12.5.6, (R/I i ) = |Nn \ iQ| = ES (i) − EP (i).

2

The Hilbert function of F and its Hilbert polynomial are equal for nonnegative integer values: Corollary 12.5.8 HF (i) = cn in + · · · + c1 i + c0 for i ∈ N and c0 = 0. Proof. By Lemma 9.3.8, EP (i) = χP (i) and ES (i) = χS (i) for i ∈ N. Thus the result follows from Proposition 12.5.7. 2 Example 12.5.9 If I is the monomial ideal (x21 , x32 ), then I = I + (x1 x22 ). Notice that P = conv((2, 0), (0, 3)) because (1, 2) lies above the hyperplane x1 /2 + x2 /3 = 1 and S = conv(0, (2, 0), (0, 3)). Using Normaliz [68] we get ES (i) = 3i2 + 3i + 1 and EP (i) = i + 1. Thus HF (i) = 3i2 + 2i. Example 12.5.10 If I = (x41 , x52 , x63 , x1 x2 x23 ), using Normaliz [68] we get: I = I + (x1 x42 , x21 x32 , x31 x22 , x21 x22 x3 , x42 x23 , x1 x53 , x32 x33 , x21 x33 , x31 x2 x3 , x22 x43 , x1 x32 x3 , x31 x23 , x2 x53 ). Notice that P = conv((4, 0, 0), (0, 5, 0), (0, 0, 6), (1, 1, 2)) because (1, 1, 2) is the only ui lying strictly below the hyperplane x1 /4 + x2 /5 + x3 /6 = 1 and S = conv(0, (4, 0, 0), (0, 5, 0), (0, 0, 6)). Using Normaliz [68] we get HF (i) = = =

ES (i) − EP (i) (1 + 6i + 19i2 + 20i3 ) − (1 + (1/6)i + (3/2)i2 + (13/3)i3 ) (35/6)i + (35/2)i2 + (47/3)i3 .

522

Chapter 12

Remark 12.5.11 For i ≥ n, I i = II i−1 , see Theorem 12.5.4. Thus we can use polynomial interpolation and Corollary 12.5.8 to determine c1 , . . . , cn . Example 12.5.12 Consider the ideal of Example 12.5.10. For i = 0, 1, 2, 3 the values of HF (i) = (R/I i ) are 0, 39, 207, 598. To compute the Hilbert polynomial ϕF (x) of F , using polynomial interpolation, one can use the following command in Maple [80]: interp([0,1,2,3],[0,39,207,598],x); to get that ϕF (x) = (47/3)x3 + (35/2)x2 + (35/6)x. 8 5 Example 12.5.13 Consider the ideal I = (x10 1 , x2 , x3 ) ⊂ R. The values of HF (i) at i = 0, 1, 2, 3 are 0, 112, 704, 2176. Using interpolation we obtain:

HF (i) = (R/I i ) =

200 3 16 i + 40x2 + i. 3 3

Proposition 12.5.14 [424] Let Ii be the ideal obtained from I by making xi = 0 in the generators of I and let e(Ii ) be its multiplicity. Then 2cn−1 ≥

n−1 i=1

e(Ii ) . (n − 1)!

Let e0 , e1 , . . . , en be the Hilbert coeﬃcients of HF . Recall that we have:

HF (i) = e0

i+n−1 n

i+n−2 i n−1 en−1 − e1 + · · · + (−1) + (−1)n en , n−1 1

where e0 = e(I) is the multiplicity of I and cn = e0 /n!. Notice that en = 0 because HF (0) = 0. Corollary 12.5.15 e0 (n − 1) − 2e1 ≥ e(I1 ) + · · · + e(In−1 ) ≥ n − 1. Proof. From the equality n e0 (n − 1) 1 1 cn−1 = − e1 − ne1 = e0 2 n! (n − 1)! 2 and using Proposition 12.5.14 we obtain the desired inequality.

2

Proposition 12.5.16 ei ≥ 0 for all i. Proof. Since the Rees algebra of F is R[It] and this algebra is normal and Cohen–Macaulay, we get that the ring grF (R) = ⊕i≥0 I i /I i+1 is Cohen– Macaulay (see [198, Corollary 2.1, p. 74]). Recall that grF (R) is a ﬁnitely generated (R/I)-algebra, i.e., grF (R) = (R/I)[z 1 , . . . , z m ], where z 1 , . . . , z m

Normality of Rees Algebras of Monomial Ideals

523

are homogeneous elements of positive degree (see [65, Proposition 4.5.4]). The canonical map grI (R) → grF (R) is integral. Thus grF (R) is a positively graded Noetherian module over the standard graded algebra grI (R). Hence, by the Hilbert–Serre theorem, we can write the Hilbert series of grF (R) as h(x)/(1 − x)n , where h(x) is a polynomial with nonnegative integer coeﬃcients because grF (R) is Cohen–Macaulay. To ﬁnish the proof 2 notice that ei = h(i) (1)/i! (see [65, Proposition 4.1.9]). Example 12.5.17 Let m and let I = mk be the kth ki + n − 1 = HF (i) = n

= (x1 , . . . , xn ) be the irrelevant maximal ideal Veronese ideal of R. Then kn n k n−1 n−1 i + i + terms of lower degree n! (n − 2)!2

because I is normal. Here e0 = e(I) = k n , e1 = (n − 1)(k n − k n−1 )/2 and cn−1 =

n−1 1 e(Ii ) . 2 i=1 (n − 1)!

Exercises 12.5.18 Let I = (xv1 , . . . , xvq ) ⊂ R be a monomial ideal and let r0 be the rank of A = (v1 , . . . , vq ). If A = {v1 , . . . , vq } is homogeneous and I i is integrally closed for i = 1, . . . , r0 − 1, then R[It] is a normal domain. 12.5.19 If R = Q[x1 , x2 ] and I = (x21 , x32 , x1 x2 ), ﬁnd a presentation of grI (R) and ﬁnd the multiplicity of I using the CoCoA [88] procedure: Use R::=Q[t[1..3],x[1..2]]; J:=Toric([[2,0,1,1,0],[0,3,1,0,1],[1,1,1,0,0]]); I:=Ideal(x[1]^2,x[2]^3,x[1]*x[2]); L:=I+J; Multiplicity(R/LT(L)); 12.5.20 If I = (x21 , x32 , x1 x2 ) ⊂ Q[x1 , x2 ], use Exercise 12.5.19 to prove:

5i + 4 if 0 ≤ i = 2, i i+1 (I /I ) = 27 if i = 2. Prove that the Samuel function of I is given by χIR (i) := (R/I i+1 ) =

5 2 13 i + i+4 2 2

(i ≥ 0).

12.5.21 If R = Q[x1 , x2 ] and I = (x1 x2 , x21 + x22 ), prove that e(I) = 4. 12.5.22 If I = (x2 , y 3 , z 5 , w6 , xy, z 2 w, xz 2 , y 2 z 2 ) ⊂ R = K[x, . . . , w], use the methods of this section to prove: (R/I i ) = (7/2)i4 + (65/6)i3 + (21/2)i2 + (19/6)i;

∀ i ∈ N.

524

Chapter 12

12.6

Rees algebras of Veronese ideals

In this section we compute the a-invariant of the Rees algebra of a Veronese ideal. This will be used to show some degree bounds for the normalization of a uniform ideal; see Theorem 12.6.7. The Rees algebra of a square-free Veronese ideal Let K be a ﬁeld and let R = K[x1 , . . . , xn ] be a polynomial ring with coeﬃcients in K. Given two positive integers k, n with k ≤ n, we deﬁne Vk = {xi1 · · · xik | 1 ≤ i1 < · · · < ik ≤ n}. The ideal L := (Vk ) is called the kth square-free Veronese ideal of R. Recall that the Rees algebra of L is given by R[Lt] = K[x1 , . . . , xn , f1 t, . . . , f(n) t] ⊂ R[t], k

where t is a new variable and Vk = {f1 , . . . , f(n) }. We can make R[Lt] k

a standard K-algebra with the grading induced by setting δ(xi ) = 1 and δ(t) = 1 − k. In this grading δ(fi t) = 1 for all i. The Rees algebra of L satisﬁes the hypothesis of Theorem 9.1.5 because: (i) R[Lt] is a normal domain by Proposition 12.3.10, and (ii) R[Lt] is a standard K-algebra. The next expression for the a-invariant follows from Theorem 9.1.5. Theorem 12.6.1 [3] If 1 ≤ k < n and n = pk + r, where 0 ≤ r < k, then ⎧ ⎪ ⎪ −(p + 2) if r ≥ 2, ⎨ −(p + 1) if r = 1, a(R[Lt]) = −(p + 1) if r = 0 and k > 1, ⎪ ⎪ ⎩ −p if r = 0 and k = 1. There is an alternative expression for a Rees algebra of a monomial ideal generated by monomials of the same degree k. Proposition 12.6.2 Let F = {xv1 , . . . , xvq } be a set of monomials in R of ﬁxed degree k ≥ 1. If x = {x1 , . . . , xn }, then R[F t] K[F, xt]. Proof. Consider the epimorphisms of K-algebras: φ : R[t1 , . . . , tq ] → R[tF ], induced by φ(ti ) = xvi t and φ(xi ) = xi , ψ : R[t1 , . . . , tq ] → K[F, xt], induced by ψ(ti ) = xvi and ψ(xi ) = xi t. It suﬃces to notice that ker(φ) = ker(ψ). This equality follows from the fact that ker(φ) and ker(ψ) are binomial ideals (see Corollary 8.2.18). 2 Corollary 12.6.3 [72] The a-invariant Jof the K edge subring of a complete . graph Kn+1 is given by a(K[Kn+1 ]) = − n+1 2 Proof. It follows from Theorem 12.6.1 and Proposition 12.6.2.

2

Normality of Rees Algebras of Monomial Ideals

525

The Rees cone of a Veronese ideal In this part, we refer to Chapter 1 for standard terminology and notation on polyhedral cones. Setting A = A =

{e1 , . . . , en , ke1 + en+1 , . . . , ken + en+1 }, {e1 , . . . , en , a1 e1 + · · · + an en + en+1 | ai ∈ N; |a| = k},

where ei is the ith unit vector in Rn+1 and |a| := a1 + · · · + an , one has the equality R+ A = R+ A . This cone is the so-called Rees cone. Lemma 12.6.4 The irreducible representation of R+ A is: R+ A = He+1 ∩ · · · ∩ He+n ∩ He+n+1 ∩ Ha+ , a = (1, . . . , 1, −k). Proof. Set N = {e1 , . . . , en+1 , a}. It suﬃces to prove that F is a facet of R+ A if and only if F = Hb ∩ R+ A for some b ∈ N . Let 1 ≤ i ≤ n + 1. Consider the following sets {e1 , e2 , . . . , ' ei , . . . , en , en+1 } and {ke1 + en+1 , . . . , ken + en+1 }, where e'i means to omit ei from the list. Since those sets are linearly independent it follows that F = Hb ∩ R+ A is a facet for b ∈ N . Conversely let F be a facet of R+ A. There are linearly independent vectors α1 , . . . , αn ∈ A and 0 = b = (b1 , . . . , bn+1 ) ∈ Rn+1 such that (i) F = R+ A ∩ Hb , (ii) Rα1 + · · · + Rαn = Hb , and (iii) R+ A ⊂ Hb+ . Since e1 , . . . , en are in A, by (iii), one has ei , b = bi ≥ 0 for i = 1, . . . , n. Set B = {α1 , . . . , αn } and consider the matrix MB whose rows are the vectors in B. Let us prove that there exists c ∈ N such that Hb = Hc and R+ A ⊂ Hc+ . Consider the following cases. Case (1): If the ith column of MB is zero for some 1 ≤ i ≤ n + 1, it suﬃces to take c = ei . Case (2): If B = {ke1 + en+1 , . . . , ken + en+1 }, then take c = a. Case (3): If B = {ei1 , . . . , eis , kej1 + en+1 , . . . , kejt + en+1 }, where s, t > 0, s + t = n, 1 ≤ i1 < · · · < is ≤ n, 1 ≤ j1 < · · · < jt ≤ n, and MB has all its columns diﬀerent from zero. Since bi1 = 0, using kei1 + en+1 , b = bn+1 ≥ 0, kej1 + en+1 , b = kbj1 + bn+1 = 0, and bj1 ≥ 0, we obtain bn+1 = 0. Then en+1 ∈ Hb . It follows that Hb is generated by the set {e1 , e2 , . . . , e'i , . . . , en , en+1 } for some 1 ≤ i ≤ n and we 2 take c = ei .

526

Chapter 12

Theorem 12.6.5 Let R[Jt] be the Rees algebra of the kth Veronese ideal J. (a) If 2 ≤ k < n and n = pk + s, where 0 ≤ s < k, then

−(p + 2) if s ≥ 2, a(R[Jt]) = −(p + 1) if s = 0 or s = 1. (b) If k ≥ n, then a(R[Jt]) = −2. Proof. Let M = xa xb tc be a monomial in the canonical module ωR[Jt] of R[Jt], where xb tc = (f1 t) · · · (fc t) and fi is a monomial in R of degree k for all i. Note δ(M ) = |a| + c. Recall that log(M ) = (a + b, c) is in the interior of R+ A by Theorem 9.1.5. Therefore, using Lemma 12.6.4, one has c ≥ 1, ai + bi ≥ 1 for all i, and |a| + |b| ≥ kc + 1. As |b| = kc, altogether we get: |a| + |b| ≥ n and |a| ≥ 1.

(12.2)

In particular δ(M ) ≥ 2 and a(R[Jt]) ≤ −2. To prove (b) note that by x2 · · · xn t is in ωR[Jt] and δ(m) is Lemma 12.6.4 the monomial m = xk−n+2 1 equal to 2. Hence a(R[Jt]) = −2. To prove (a) there are the following three cases to consider. Case s ≥ 2: First we show δ(M ) ≥ p + 2. If c > p, then δ(M ) ≥ p + 2 follows from Eq.(12.2). On the other hand assume c ≤ p. Observe: k(p − c) + s ≥ (p − c) + 2.

(∗)

From Eq.(12.2) one has |a| + |b| = |a| + kc ≥ n = kp + s. Consequently δ(M ) = |a| + c ≥ k(p − c) + s + c.

(∗∗)

Hence from (∗) and (∗∗) we get δ(M ) ≥ p + 2, as required. Therefore one has the inequality a(R[Jt]) ≤ −(p + 2), to show equality we claim that m = x21 x22 · · · x2k−s+1 xk−s+2 · · · xn tp+1 is in ωR[Jt] and δ(m) = p + 2. An easy calculation shows that m is in R[Jt] and δ(m) = p + 2. Finally let us see that m is in ωR[Jt] via Lemma 12.6.4. That the entries of log(m) satisfy yi > 0 for all i is clear. The inequality y1 + y2 + · · · + yn > kyn+1 ,

(∗ ∗ ∗)

after making yi equal to the ith entry of log(m), transforms into 2(k − s + 1) + (n − (k − s + 1)) > k(p + 1) but the left-hand side is k(p + 1) + 1, hence log(m) satisﬁes (∗ ∗ ∗). Case s = 1: If c ≥ p, then clearly |a| + c ≥ p + 1, so let us suppose that c < p, then n + c(1 − k) ≥ n + p(1 − k) and using Eq. (12.2) we get |a| + c = (|a| + |b| − n) + (n + c(1 − k)) ≥ n + c(1 − k) ≥ p + 1

Normality of Rees Algebras of Monomial Ideals

527

so a(R[Jt]) ≤ −(p + 1). Now observe that the monomial m = x1 x2 · · · xn tp has δ(m) = p + 1 and m ∈ R[Jt]. Finally, an argument similar to the previous case shows that m is in ωR[Jt] . Thus a(R[Jt]) = −(p + 1). Case s = 0: First note that p > 1 because n > k. If c ≥ p, then clearly |a| + c ≥ p + 1. So let us suppose c < p. Then, using Eq. (12.2), we get |a| ≥ k(p − c) > (p − c). Then adding c we derive |a| + c ≥ p + 1. Hence a(R[Jt]) ≤ −(p + 1). Note that the monomial m = x21 x2 · · · xn tp has δ(m) = p + 1 and belongs to ωR[Jt] . Thus, one has the required equality. 2 Corollary 12.6.6 Let I be an ideal of Veronese type of degree k. If 2 ≤ k < n and n = pk + s, where 0 ≤ s < k, then

−(p + 2) if s ≥ 2, a(R[It]) = −(p + 1) if s = 0 or s = 1. Proof. Let L and J be the kth square-free Veronese ideal and the kth Veronese ideal, respectively. From Theorems 12.6.1 and 12.6.5 we obtain that a(R[Lt]) and a(R[Jt]) can be computed using the formula above. Since R[It] is normal, using the proof of Theorem 12.4.13, we get a(R[Lt]) ≤ a(R[It]) ≤ a(R[Jt]). Therefore a(R[It]) is given by the formula above.

2

Normalization of a uniform ideal Let R = K[x1 , . . . , xn ] be a ring of polynomials over a ﬁeld K and let I be an ideal of R generated by monomials xv1 , . . . , xvq of degree k ≥ 2. Ideals of this type are called uniform. In this case the Rees algebra of I is a standard graded K-algebra with the grading δ induced by δ(xi ) = 1 and δ(t) = 1 − k. Notice that δ(xvi t) = 1 for all i. Notice that we can embed R[It] in R[Jt], where J is the kth Veronese ideal of R. The next result complements [72, Theorem 3.4] for the class of uniform ideals. Theorem 12.6.7 Let R = R[It] be the Rees algebra of I. If 2 ≤ k < n, then the normalization R of I is generated as an R-module by monomials g ∈ R[t] of degree in t at most n − (n/k). Proof. We set A = {e1 , . . . , en , (v1 , 1), . . . , (vq , 1)}. As ZA = Zn+1 , according to Theorem 9.1.1, the normalization of I can be expressed as: R = K[{xa tb | (a, b) ∈ Zn+1 ∩ R+ A }].

528

Chapter 12

n+1 If m = xa tb with ∩ R+ A , then δ(m) ≥ b. To show this n 0 = (a, b)∈q Z write (a, b) = i=1 λi ei + i=1 μi (vi , 1) for some λi ≥ 0, μi ≥ 0. Hence

|a| = b =

λ1 + · · · + λn + (μ1 + · · · + μq )k μ1 + · · · + μ q ,

and consequently δ(m) = |a| + (1 − k)b = (λ1 + · · · + λn ) + b ≥ b. Therefore R is generated as a K-algebra by monomials of positive degree. There is a Noether normalization of R ϕ

ψ

A = K[z1 , . . . , zn+1 ] → R → R, where z1 , . . . , zn+1 ∈ R1 . Notice that the composite A → R is a Noether normalization of R. By Theorem 9.1.6 R is Cohen–Macaulay. Hence, by Proposition 3.1.27, R is a free module over A and one can write R = Am1 ⊕ · · · ⊕ Amp ,

(12.3)

where mi = xβi tbi . Using that the length is additive one has the following expression for the Hilbert series F (R, z) =

p i=0

z δ(mi ) h0 + h1 z + · · · + hs z s = , (1 − z)n+1 (1 − z)n+1

where hi = |{j | δ(mj ) = i}|. Note that a(R), the a-invariant of R, is equal to s − (n + 1). On the other hand one has a(R) = − min{i | (ωR )i = 0}, where ωR is the canonical module of the normalization R. Using the proof of Theorem 12.4.13 together with Theorem 12.6.5 yields LnM −1 a(R) = s − (n + 1) ≤ a(R[Jt]) ≤ − k and s ≤ n−(n/k). Altogether if mi = xβi tbi one has bi ≤ δ(mi ) ≤ n−(n/k), 2 that is, the degree in t of mi is at most n − (n/k), as required. Proposition 12.6.8 I i = II i−1 for i ≥ n + 2 + a(R[Jt]). Furthermore equality holds for i ≥ n − (n/k) + 1. Proof. It follows for the proof of Theorem 12.6.7 and using Eq.(12.3). 2 Deﬁnition 12.6.9 The index of normalization of I, denoted by s(I), is the smallest integer s such that I i = II i−1 for i > s. Corollary 12.6.10 s(I) ≤ n − (n/k).

Normality of Rees Algebras of Monomial Ideals

529

Exercises 12.6.11 Let B = K[x1 , . . . , xn , t1 , . . . , tn ] be a polynomial ring, let A be the following matrix of indeterminates x1 x2 · · · xn A= , t1 t2 · · · tn and let I2 (A) be the ideal generated by the 2-minors of A. If R is the polynomial ring K[x1 , . . . , xn ] and m = (x1 , . . . , xn ), then the toric ideal of R[mt] is I2 (A) and a(R[mt]) = a(B/I2 (A)) = −n. 12.6.12 [72] Let K[F ] ⊂ R be a subring generated by a set F of monomials of degree k. Prove that K[F ] is generated as a K[F ]-module by elements g ∈ R of normalized degree at most dim(K[F ]) − 1.

12.7

Divisor class group of a Rees algebra

In this section we determine the divisor class group of the Rees algebra of a normal monomial ideal. Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and let I be an ideal of R of height g ≥ 2 minimally generated by monomials xv1 , . . . , xvq of degree at least two. For technical reasons we shall assume that each variable xi occurs in at least one monomial xvj . The Rees cone of the point conﬁguration A = {v1 , . . . , vq } is the rational polyhedral cone in Rn+1 , denoted by R+ A , consisting of the nonnegative linear combinations of the set A := {e1 , . . . , en , (v1 , 1), . . . , (vq , 1)} ⊂ Rn+1 , where ei is the ith unit vector. Notice that dim(R+ A ) = n + 1. Thus according to Corollary 1.1.50 there is a unique irreducible representation R+ A = He+1 ∩ · · · ∩ He+n+1 ∩ H+1 ∩ · · · ∩ H+r

(12.4)

such that 0 = i ∈ Zn+1 and the non-zero entries of i are relatively prime for all i. Lemma 12.7.1 If j = (a1 , . . . , an , −an+1 ), then ai ≥ 0 for i = 1, . . . , n and an+1 > 0. Proof. Clearly ai ≥ 0 because ei ∈ R+ A for i = 1, . . . , n. Notice that ai > 0 for some 1 ≤ i ≤ n. Assume an+1 ≤ 0. Then He+1 ∩ · · · ∩ He+n+1 ⊂ He+1 ∩ · · · ∩ He+n+1 ∩ H+j .

530

Chapter 12

To show the inclusion take x = (xi ) is in the left-hand side, then xi ≥ 0 for all i and a1 x1 + · · · + an xn ≥ 0 ≥ an+1 xn+1 . Thus x ∈ H+j . Hence H+j can be deleted from Eq. (12.4), a contradiction. Thus an+1 > 0. 2 Lemma 12.7.2 If 1 ≤ i ≤ r, then the following equality holds Z = e1 , i Z + · · · + en , i Z + (v1 , 1), i Z + · · · + (vq , 1), i Z. (12.5) Proof. Fix an integer 1 ≤ i ≤ r. Since Hi ∩ R+ A is a facet of the Rees cone there is (vj , 1) such that (vj , 1), i = 0. Setting vj = (c1 , . . . , cn ) and i = (a1 , . . . , an , −an+1 ), we get c1 a1 + · · · + cn an − an+1 = 0. Notice that the right-hand side of Eq. (12.5) is equal to a1 Z + · · · + an Z + (v1 , 1), i Z + · · · + (vq , 1), i Z. Thus we need only show the equality Z = a1 Z+· · ·+an Z. By Lemma 12.7.1 n an+1 > 0. Thus, from the equality i=1 ci ai = an+1 , and using that the non-zero entries of j are relatively prime, we get that the non-zero elements 2 in a1 , . . . , an are relatively prime, as desired. Next we prove a result about the torsion freeness of the divisor class group of R[It]. It shows that the rank of this group can be read oﬀ from the irreducible representation of the Rees cone given in Eq (12.4). Proposition 12.7.3 [380, Theorem 1.1] If R[It] is normal, then its divisor class group Cl(R[It]) is a free abelian group of rank r. Proof. Set p = n + 1 + r. Consider the map ϕ : Zn+1 → Zp given by ϕ(v) = v, e1 e1 + · · · + v, en+1 en+1 + v, 1 en+2 + · · · + v, r ep and consider the matrix B whose rows are the vectors in the set B = ϕ(A ) = {ϕ(e1 ), . . . , ϕ(en ), ϕ(v1 , 1), . . . , ϕ(vq , 1)}. The matrix B is given by: ⎡ e1 0 ⎢ e2 0 ⎢ ⎢ ··· ··· ⎢ B=⎢ ⎢ en 0 ⎢ v1 1 ⎢ ⎣ ··· ··· vq 1

11 12 ··· 1n (v1 , 1), 1 ··· (vq , 1), 1

··· ··· ··· ··· ··· ··· ···

r1 r2 ··· rn (v1 , 1), r ··· (vq , 1), r

⎤ ⎥ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎥ ⎦

where i = (i1 , . . . , i(n+1) ). It is not hard to see that the Smith normal form of B is equal to diag(1, . . . , 1, 0, . . . , 0), where the number of 1’s is 2 n + 1. Hence, using Theorem 9.8.19, we get Cl(R[It]) Zp /ZB Zr .

Normality of Rees Algebras of Monomial Ideals

531

Exercise 12.7.4 Let I = (x1 x2 , x2 x3 , x3 x4 , x1 x4 ) be the edge ideal of a cycle of length 4. Prove that Cl(R[It]) is a free abelian group of rank 2.

12.8

Stochastic matrices and Cremona maps

In this section we discuss a conjecture on the normality of ideals arising from doubly stochastic matrices and study Cremona monomial maps. Let A = (aij ) be a non-singular matrix of order n with entries in N. The matrix A is called doubly stochastic of degree d if n k=1

aki =

n

ajk = d ≥ 2 for all 1 ≤ i, j ≤ n.

k=1

If nk=1 aki = d ≥ 2 for all i, we say that A is a d-stochastic matrix by columns. The set A = {v1 , . . . , vn } will denote the set of columns of A and I will denote the monomial ideal of the polynomial ring R = K[x1 , . . . , xn ] generated by xv1 , . . . , xvn . Proposition 12.8.1 If A = (aij ) is a d-stochastic matrix by columns and R[It] is normal, then det(A) = ±d. Proof. By Proposition 9.3.33 we have H = Zn /(v1 , . . . , vn ) Zd . By Lemma 1.3.17 the order of H is | det(A)|. Thus d = | det(A)|. 2 Conjecture 12.8.2 The Rees algebra R[It] is normal for every doubly stochastic matrix A of degree d with entries in {0, 1} and det(A) = ±d. To prove this conjecture one could try to use the Birkhoﬀ von Neumann theorem: “A matrix is doubly stochastic if and only if it is a convex combination of permutation matrices multiplied by d” [372, Theorem 8.6]. Proposition 12.8.3 If A = (aij ) is a d-stochastic matrix by columns, then det(A) is a multiple of d. Proof. Let F the matrix with rows 1 , . . . , Fn be the rows of A and let B be n n F1 , . . . , Fn−1 , i=1 Fi . Since det(B) = det(A) and i=1 Fi = (d, . . . , d), the assertion follows. 2 Proposition 12.8.4 Let A = (aij ) be a 2-stochastic matrix by columns with entries in {0, 1}. If det(A) = ±2, then R[It] is normal.

532

Chapter 12

Proof. Since d = 2, A is the incidence matrix of a simple graph G = (V, E) with vertex set V = {x1 , . . . , xn } and edge set E, where {xi , xj } ∈ E if and only if ei + ej ∈ A. Notice that |V | = |E| because A is a square matrix. We claim that G is a connected. Let G1 , . . . , Gk be the connected components of G. By re-ordering the vertices of G we can write A = diag(M1 , . . . , Mk ), where Mi is the incidence matrix of Gi . Let ni be the number of vertices of Gi . By Lemma 10.2.6, rank(Mi ) = ni if Gi is non-bipartite and rank(Mi ) = ni − 1 if Gi is bipartite. As G is a disjointunion of G1 , . . . , Gk , we get n = n1 +· · ·+nk . Thus, using that rank(A) is ki=1 rank(Mi ), it follows that Gi is a connected non-bipartite graph with at least ni edges for any i. Since the number of edges of G is n we obtain that Mi is a square matrix of order ni for all i. Then det(A) = det(M1 ) · · · det(Mk ). By Proposition 12.8.3 we get det(Mi ) = 2mi for 1 ≤ i ≤ k, where mi ∈ Z for all i. Hence k = 1. Let T = (VT , ET ) be a spanning tree of G (see Exercise 10.1.65). Then |ET | + 1 = |VT | = |V | = |E| and G is obtained from T by adding one edge. This means that G has a unique odd cycle and some branches. Using Corollary 10.3.12 we get K[G] = K[{xa | a ∈ A}] is normal and according to Corollary 10.5.6 we have that the Rees algebra R[It] is normal. 2 Example 12.8.5 Consider the following doubly stochastic matrix ⎤ ⎡ 0 0 1 1 1 1 ⎢ 0 1 0 1 1 1 ⎥ ⎥ ⎢ ⎢ 1 1 1 0 1 0 ⎥ ⎥ At = ⎢ ⎢ 1 1 1 1 0 0 ⎥ ⎥ ⎢ ⎣ 1 0 0 1 1 1 ⎦ 1 1 1 0 0 1 whose rows are v1 , . . . , v6 . Notice that det(A) = ±8 = ±4. The ideal I is minimally non-normal , that is I is not normal and I is normal for every proper minor I of I (see Deﬁnition 14.2.23 for the notion of minor). Lemma 12.8.6 If A is a d-stochastic matrix by columns and det(A) = ±d, then A−1 (ei − ej ) ∈ Zn for all i, j. n Proof. Fixing indices i, j, we can write A−1 (ei − ej ) = k=1 λk ek for some λ1 , . . . , λn in Q. Notice that A−1 (ei ) is the ith column of A−1 . Since 1A = d1, we get 1/d = 1A−1 . Therefore |A−1 (ei )| = |A−1 (ej )| = 1/d and k λk = 0. Then we can write −1

A

(ei − ej ) =

n k=2

λk (ei − e1 ) =⇒ ei − ej =

n k=2

λk (vk − v1 ).

Normality of Rees Algebras of Monomial Ideals

533

Thus there is 0 = s ∈ N such that s(ei − ej ) belong to Z{v1 − vk }nk=2 . By n n Proposition n10.8.7, the group Z /Z{v1 − vk }k=2 is free. Then wencan write ei − ej = k=2 ηk (vk − v1 ), for some ηi ’s in Z. Since Z{v1 − vk }k=2 is also free (of rank n − 1), the vectors v2 − v1 , . . . , vn − v1 are linearly independent. 2 Thus λk = ηk ∈ Z for all k ≥ 2, hence ultimately A−1 (ei − ej ) ∈ Zn . Theorem 12.8.7 [386] If A is a d-stochastic matrix by columns such that det(A) = ±d, then there are unique vectors β1 , . . . , βn , γ ∈ Nn such that the following two conditions hold: (a) Aβi = γ + ei for all i, where βi , γ and ei are column vectors ; (b) The matrix B with columns β1 , . . . , βn has at least one zero entry in every row. Moreover, det(B) = ±(|γ| + 1)/d = ±|βi | for all i. Proof. First we show the uniqueness. Assume that β1 , . . . , βn , γ is a set of vectors in Nn such that: (a ) Aβi = γ + ei for all i, and (b ) The matrix B whose column vectors are β1 , . . . , βn has at least one zero entry in every row. Let Δ = (Δi ) and Δ = (Δi ) be nonnegative vectors such that A−1 (γ − γ ) = Δ − Δ. Then from (a) and (a ) we get βi −βi = A−1 (γ −γ ) = Δ −Δ, ∀ i =⇒ βik −βik = Δk −Δk , ∀ i, k, (12.6) where βi = (βi1 , . . . , βin ) and βi = (βi1 , . . . , βin ). It suﬃces to show that Δ = Δ . If Δk > Δk for some k; then, by Eq. (12.6), we obtain βik > 0 for i = 1, . . . , n, which contradicts (b). Similarly if Δk < Δk for some k; then, by Eq. (12.6), we obtain βik > 0 for i = 1, . . . , n, which contradicts (b ). Thus Δk = Δk for all k, i.e., Δ = Δ . Next we prove the existence of β1 , . . . , βn and γ. By Lemma 12.8.6, for i ≥ 2 we can write − 0 = αi = A−1 (e1 − ei ) = α+ i − αi − + − n where α+ i and αi are in N . Notice that αi = 0 and αi = 0. Indeed the − −1 sum of the entries of A (ei ) is equal to 1/d. Thus |αi | = |α+ i | − |αi | = 0, and consequently the positive and negative part of αi are both non-zero for + + + i ≥ 2. The vector α+ i can be written as αi = (αi1 , . . . , αin ) for i ≥ 2. For 1 ≤ k ≤ n consider the integers given by mk = max2≤i≤n {α+ ik } and set n β1 = (m1 , . . . , mn ). Since β1 ≥ α+ i , for each i ≥ 2 there is θi ∈ N such + that β1 = θi + αi . Therefore − − αi = A−1 (e1 − ei ) = α+ i − αi = β1 − (θi + αi ).

We set βi = θi + α− i for i ≥ 2. Since we have Aβ1 − e1 = Aβi − ei for i ≥ 2, it follows readily that Aβ1 − e1 ≥ 0 (make i = 2 in the last equality and

534

Chapter 12

compare entries). Thus, setting γ := Aβ1 − e1 , it follows that β1 , . . . , βn and γ satisfy (a). If each row of B has some zero entry the proof of the existence is complete. If every entry of a row of B is positive we subtract the vector 1 = (1, . . . , 1) from that row and change γ accordingly so that (a) is still satisﬁed. Applying this argument repeatedly we get a sequence β1 , . . . , βn , γ satisfying (a) and (b). We now prove the last part of the assertion. If βij denotes the j-entry of βi , then Aβi = γ + ei is equivalent to βi1 v1 + · · · + βin vn = γ + ei . Thus |βi |d = |γ| + 1. Condition (a) is equivalent to AB = Γ + I, where Γ is the matrix all of whose columns are equal to γ. Since det(B) = ± det(Γ + I)/d it suﬃces to show that det(Γ + I) = |γ| + 1. By Exercise 12.8.16, if Γ has rank at most one and D is the identity, we get det(Γ + I) = trace(Γ) + 1. 2 Example 12.8.8 Consider the following matrix A and its inverse: ⎛ ⎛ ⎞ ⎞ d d−1 0 1 1 − d (d − 1)2 1 1 d − 1⎠ ; A−1 = ⎝0 d d(1 − d)⎠ . A = ⎝0 d 0 0 1 0 0 d To compute the βi ’s and γ we follow the proof of Theorem 12.8.7. Then β1 = (2, d, 0), β2 = (1, d + 1, 0), β3 = (d, 1, 1), γ = (d2 + d − 1, d, 0), and ⎛ ⎞ 2 1 d B = ⎝d d + 1 1⎠ . 0 0 1 By subtracting the vector (1, 1, 1) from rows 1 and 2, we get ⎛ ⎞ 1 0 d−1 0 ⎠. B = ⎝d − 1 d 0 0 1 The column vectors β1 = (1, d − 1, 0), β2 = (0, d, 0), β3 = (d − 1, 0, 1), γ = (d2 − d, d − 1, 0) satisfy (a) and (b). Cremona maps Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K. In what follows we assume that A is a d-stochastic matrix by columns and that the corresponding set of monomials F = {xv1 , . . . , xvn } ⊂ R have no non-trivial common factor. We also assume throughout that every xi divides at least one member of F , a harmless condition. Deﬁnition 12.8.9 F deﬁnes a rational (monomial) map Pn−1 Pn−1 denoted again by F and written as a tuple F = (xv1 , . . . , xvn ). F is called a Cremona map if it admits an inverse rational map with source Pn−1 .

Normality of Rees Algebras of Monomial Ideals

535

A rational monomial map F is deﬁned everywhere if and only if the deﬁning monomials are pure powers of the variables, in which case it is a Cremona map if and only if F = (xσ(1) , . . . , xσ(n) ) for some permutation σ. Proposition 12.8.10 [385, Proposition 2.1] F is a Cremona map if and only if det(A) = ±d. Thus Conjecture 12.8.2 has the following reformulation: “the rational map F : Pn−1 Pn−1 is a Cremona map if and only if R[It] is normal.” For d = 2 this conjecture was proved in [385]; see Proposition 12.8.4. Cremona maps deﬁned by monomials of degree d = 2 are thoroughly analyzed and classiﬁed via integer arithmetic and graph combinatorics in [98]. Theorem 12.8.11 If F : Pn−1 Pn−1 is a Cremona map then its inverse is also deﬁned by monomials of ﬁxed degree. Proof. We set fi = xvi for i = 1, . . . , n. By Proposition 12.8.10, A has determinant ±d. Therefore Theorem 12.8.7 implies the existence of an n×n matrix B such that AB = Γ+I, where Γ is a matrix with repeated column γ throughout. Let g1 , . . . , gn denote the monomials deﬁned by the columns of B and call G the corresponding rational monomial map. The above matrix equality translates into the equality (f1 (g1 , . . . , gn ), . . . , fn (g1 , . . . , gn )) = (xγ · x1 , . . . , xγ · xn ). Thus the left-hand side is proportional to the vector (x1 , . . . , xn ) which means that the composite map F ◦ G is the identity map wherever the two are deﬁned. On the other hand Theorem 12.8.7 also says that G is also a Cremona map. Therefore G has to be the inverse of F , as required. Notice that the proof of Theorem 12.8.7 provides an algorithm to compute B and γ. The input for this algorithm is the matrix A. 2

Exercises 12.8.12 If B is a real non-singular matrix such that the sum of the elements in each row is d = 0, then the sum of the elements in each row of B −1 is equal to 1/d 12.8.13 Let A = (aij ) be a d-stochastic matrix by columns of order n. If B is the matrix (bij ) = (1 − aij ), then (n − d) det(A) = (−1)n−1 d det(B). 12.8.14 Let A = (aij ) be a doubly stochastic matrix of order n with n i=1

aik =

n

aj = d

j=1

If | det(A)| = d ≥ 1, then gcd{n, d} = 1.

(∀ k, ).

536

Chapter 12

12.8.15 Let F be a Cremona map of the form F = (xa1 1 xa2 2 , xb22 xb33 , xc11 xc33 ). Prove that, up to permutation of the variables and the monomials, F is one of the following two kinds: F = (x1 x2 , x2 x3 , x1 x3 ) or F = (xd1 , x2 xd−1 , xd−1 x3 ). 3 1 12.8.16 Let Γ = (γi,j ) be an n × n matrix over a commutative ring, and let D = diag(d1 , . . . , dn ) be a diagonal matrix. Then det(Γ + D)

= det(Γ) + di Δ[n]\{i} + i

+ ···+

di1 di2 Δ[n]\{i1 ,i2 }

1≤i1
di1 · · · din−1 Δ[n]\{i1 ,...,in−1 }

1≤i1 <···
+ det D, where Δ[n]\{i1 ,...,ik } denotes the principal (n − k) × (n − k)-minor of Γ with rows and columns [n] \ {i1 , . . . , ik }. Here [n] = {1, . . . , n}.

Chapter 13

Combinatorics of Symbolic Rees Algebras of Edge Ideals of Clutters In this chapter we give a description—using notions from combinatorial optimization and polyhedral geometry— of the minimal generators of the symbolic Rees algebra of the edge ideal of a clutter and show a complete graph theoretical description of the minimal generators of the symbolic Rees algebra of the ideal of covers of a graph. For a connected non-bipartite graph G whose edge subring is normal, we give conditions for G to have a perfect matching. The notion of a indecomposable graph is related to the strong perfect graph theorem. We give a description—in terms of cliques—of the symbolic Rees algebra and the Simis cone of the edge ideal of a perfect graph.

13.1

Vertex covers of clutters

In this section we characterize the notion of a minimal vertex cover of a clutter in algebraic and combinatorial terms and relate the matching number and the vertex covering number of a clutter to the algebraic invariants of the corresponding edge ideal. Let C be a clutter with vertex set X = {x1 , . . . , xn } and let R = K[X] be a polynomial ring over a ﬁeld K. We denote the incidence matrix of C by A and denote the column vectors of A by v1 , . . . , vq . The edge ideal of C, denoted by I(C), is the ideal of R generated by xv1 , . . . , xvq . Deﬁnition 13.1.1 A subset C ⊂ X is a minimal vertex cover of a clutter C if: (c1 ) every edge of C contains at least one vertex of C, and (c2 ) there is

538

Chapter 13

no proper subset of C with the ﬁrst property. If C only satisﬁes condition (c1 ), then C is called a vertex cover of C. Proposition 13.1.2 Let Q(A) = {x| x ≥ 0; xA ≥ 1} be the set covering polyhedron of C. The following are equivalent : (a) p = (x1 , . . . , xr ) is a minimal prime of I = I(C). (b) C = {x1 , . . . , xr } is a minimal vertex cover of C. (c) α = e1 + · · · + er is a vertex of Q(A). Proof. (a) ⇔ (b): This follows at once from Lemma 6.3.37. (b) ⇒ (c): Fix 1 ≤ i ≤ r. To make notation simpler ﬁx i = 1. We may assume that there is s1 such that xvj = x1 mj for j = 1, . . . , s1 and x1 ∈ / supp(xvj ) for j > s1 . Notice that supp(mk1 ) ∩ (C \ {x1 }) = ∅ for some 1 ≤ k1 ≤ s1 , otherwise C \ {x1 } is a vertex cover of C strictly contained in C, a contradiction. Hence for each 1 ≤ i ≤ r there is vki in {v1 , . . . , vq } such that xvki = xi mki and supp(mki ) ⊂ {xr+1 , . . . , xn }. The vector α is clearly in Q(A), and since {ei }ni=r+1 ∪ {vk1 , . . . , vkr } is linearly independent, and α, ei = 0

(i = r + 1, . . . , n),

α, vki = 1 (i = 1, . . . , r),

we get that the vector α is a basic feasible solution of the linear system x ≥ 0; xA ≥ 1. Therefore α is a vertex of Q(A) by Corollary 1.1.49. (c) ⇒ (b): If C C is a vertex cover of C, then α = xi ∈C ei satisﬁes α A ≥ 1 and α ≥ 0. Using that α is a basic feasible solution of the linear system x ≥ 0; xA ≥ 1, it is not hard to verify that α is also a vertex of Q(A). If V is the vertex set of Q(A), by Theorem 1.1.42, we can write Q(A) = Rn+ + conv(V ). Since α = β + α , for some β ∈ Rn+ , we obtain Q(A) = Rn+ + conv(V \ {α}), a contradiction (see Propositions 1.1.36 and 1.1.39). Thus C is a minimal vertex cover, as required. 2 Corollary 13.1.3 A vector α ∈ Rn is an integral vertex of Q(A) if and only if α = ei1 + · · · + eis for some minimal vertex cover {xi1 , . . . , xis } of C. Proof. By Proposition 13.1.2 it suﬃces to observe (see Exercise 11.3.8) that any integral vertex of Q(A) has entries in {0, 1}. 2 Let C be a clutter. A set of edges of C is called independent or a matching if no two of them have a common vertex. We denote the smallest number of vertices in any minimal vertex cover of C by α0 (C) and the maximum number of independent edges of C by β1 (C). We call α0 (C) the vertex covering number and β1 (C) the matching number of C. These numbers are related to min-max problems (see Exercise 13.1.7).

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Deﬁnition 13.1.4 If α0 (C) = β1 (C) we say that the clutter C (resp. the edge ideal I(C)) has the K¨ onig property. Deﬁnition 13.1.5 The monomial grade of I(C), denoted by mgrade(I(C)), is the maximum integer r such that there exists a regular sequence of monomials xα1 , . . . , xαr in I(C). The combinatorial invariants α0 (C) and β1 (C) can be interpreted in terms of algebraic invariants of I(C). Proposition 13.1.6 ht I(C) = α0 (C) and β1 (C) = mgrade(I(C)). Proof. The ﬁrst equality follows at once from Proposition 6.5.4. The second equality follows readily from Exercise 13.1.10. 2

Exercises 13.1.7 Let A be the incidence matrix of a clutter C. Prove that the vertex covering number and the matching number of C satisfy: α0 (C) ≥ min{1, x| x ≥ 0; xA ≥ 1} = max{y, 1| y ≥ 0; Ay ≤ 1} ≥ β1 (C). Prove that α0 (C) = β1 (C) if and only if both sides of the equality have integral optimum solutions. 13.1.8 If C is a clutter, prove that the following conditions are equivalent: (a) α0 (C) = β1 (C). (b) x1 · · · xn tg belongs to R[It], where g = ht(I) and I = I(C). 13.1.9 Let f = {f1 , . . . , fr } be a sequence of elements of a ring R and let f0 = 0. Prove that f is a regular sequence if and only if ((f1 , . . . , fi−1 ) : fi ) = (f1 , . . . , fi−1 ) for all i ≥ 1. 13.1.10 A sequence xα1 , . . . , xαr of monomials of R is a regular sequence if and only if xαi and xαj have no common variables for all i = j. 13.1.11 Let C be a clutter and let C be a minimal vertex cover of C. Prove that |C ∩ e| = 1 for some edge e of C.

540

Chapter 13

13.2

Symbolic Rees algebras of edge ideals

Let C be a clutter with vertex set X = {x1 , . . . , xn } and edge set E(C), let R = K[X] be a polynomial ring over a ﬁeld K, and let I = I(C) be the edge ideal of C. The blowup algebra studied here is the symbolic Rees algebra Rs (I) = R ⊕ I (1) t ⊕ · · · ⊕ I (i) ti ⊕ · · · ⊂ R[t] of I, where t is a new variable and I (i) is the ith symbolic power of I. The main theorem of this section is a description—in combinatorial optimization terms—of the minimal set of generators of Rs (I) as a K-algebra. As usual, we denote by C ∨ the clutter whose edges are the minimal vertex covers of C. If C is a subset of X, its characteristic vector is the vector v = xi ∈C ei , where ei is the ith unit vector in Rn . Let C1 , . . . , Cs be the minimal vertex covers of C and let uk be the characteristic vector of Ck for 1 ≤ k ≤ s. In our situation, according to Propositions 4.3.24 and 6.5.4, the bth symbolic power of I has a simple expression: I (b) = pb1 ∩ · · · ∩ pbs = ({xa | a, uk ≥ b for k = 1, . . . , s}),

(13.1)

where pk is the prime ideal of R generated by Ck . In particular, if b = 1, we obtain the primary decomposition of I because I (1) = I. Deﬁnition 13.2.1 The Simis cone of I = I(C) is the polyhedral cone: + + Cn(I) = He+1 ∩ · · · ∩ He+n+1 ∩ H(u ∩ · · · ∩ H(u . 1 ,−1) s ,−1)

The term Simis cone was coined in [147] to recognize the pioneering work of Aron Simis on symbolic powers of monomial ideals [377]. The Simis cone is a ﬁnitely generated rational cone (Theorem 1.1.29). Hence, by Theorem 1.3.9, there is a unique minimal ﬁnite set of integral vectors H ⊂ Zn+1 such that Zn+1 ∩ R+ H = NH and Cn(I) = R+ H (minimal relative to taking subsets). The set H is called the minimal Hilbert basis of Cn(I). See Section 1.3 for a detailed study of Hilbert bases. Theorem 13.2.2 H is the set of all integral vectors 0 = α ∈ Cn(I) such that α is not the sum of two other non-zero integral vectors in Cn(I). Proof. It follows at once from Theorem 1.3.9.

2

Theorem 13.2.3 [147] Let H ⊂ Nn+1 be a Hilbert basis of Cn(I). If K[NH] is the semigroup ring of NH, then Rs (I) = K[NH]. Proof. Recall that K[NH] = K[{xa tb | (a, b) ∈ NH}]. Take xa tb ∈ Rs (I), that is, xa ∈ pbi for all i. Hence (a, b), (ui , −1) ≥ 0 for all i or equivalently (a, b) ∈ Cn(I). Thus (a, b) ∈ NH and xa tb ∈ K[NH]. Conversely take xa tb ∈ K[NH], then (a, b) is in Cn(I) and (a, b), (ui , −1) ≥ 0 for all i. 2 Hence xa ∈ pbi for all i and xa ∈ I (b) , as required.

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Corollary 13.2.4 [301] Rs (I) is a ﬁnitely generated K-algebra. Proof. Let Cn(I) be the Simis cone of I and let H ⊂ Nn+1 be a Hilbert basis of Cn(I). Applying Theorem 13.2.3, we get Rs (I) = K[NH]. Thus, 2 Rs (I) is a ﬁnitely generated K-algebra. Deﬁnition 13.2.5 Let a = (ai ) = 0 be a vector in Nn and let b ∈ N. If a, b satisfy a, uk ≥ b for k = 1, . . . , s, we say that a is a b-cover of C ∨ . Lemma 13.2.6 xa tb is in Rs (I) if and only if a is a b-cover of C ∨ . Proof. It follows from the description of I (b) given in Eq. (13.1).

2

The notion of a b-cover occurs in combinatorial optimization (see for instance [373, Chapter 77, p. 1378] and the references therein) and algebraic combinatorics [147, 225]. Deﬁnition 13.2.7 A b-cover a of C ∨ is called reducible if there exists an i-cover c and a j-cover d of C ∨ such that a = c + d and b = i + j. If a is not reducible, we call a irreducible. The irreducible 0 and 1 covers of C ∨ are the unit vectors e1 , . . . , en and the characteristic vectors v1 , . . . , vq of the edges of C, respectively. Lemma 13.2.8 A monomial xa tb is a minimal generator of Rs (I), as a K-algebra, if and only if a is an irreducible b-cover of C ∨ . Proof. It follows from the discussion above, by decomposing any b-cover into irreducible ones. 2 Let S be a set of vertices of a clutter C. The induced subclutter on S, denoted by C[S], is the maximal subclutter of C with vertex set S. Thus the vertex set of C[S] is S and the edges of C[S] are exactly the edges of C contained in S. Notice that C[S] may have isolated vertices, i.e., vertices that do not belong to any edge of C[S]. If C is a discrete clutter, i.e., all the vertices of C are isolated, we set I(C) = 0 and α0 (C) = 0. Let C be a clutter and let X1 , X2 be a partition of its vertex set V (C) into non-empty sets. Clearly, one has the inequality α0 (C) ≥ α0 (C[X1 ]) + α0 (C[X2 ]).

(13.2)

If C is a graph and equality occurs, Erd¨os and Gallai [144] call C a decomposable graph. This motivates the following similar notion for clutters. Deﬁnition 13.2.9 A clutter C is called decomposable if there are nonempty vertex sets X1 , X2 such that X = V (C) is the disjoint union of X1 and X2 , and α0 (C) = α0 (C[X1 ]) + α0 (C[X2 ]). If C is not decomposable, it is called indecomposable.

542

Chapter 13

Examples of indecomposable graphs include complete graphs, odd cycles and complements of odd cycles of length at least ﬁve (see Lemma 13.3.1). Deﬁnition 13.2.10 (Schrijver [373]) The duplication of a vertex xi of a clutter C means extending its vertex set X by a new vertex xi and replacing the edge set E(C) by E(C) ∪ {(e \ {xi }) ∪ {xi }| xi ∈ e ∈ E(C)}. The deletion of xi , denoted by C \{xi }, is the clutter formed from C by deleting the vertex xi and all edges containing xi . A clutter obtained from C by a sequence of deletions and duplications of vertices is called a parallelization. It is not diﬃcult to verify that these two operations commute. If a = (ai ) is a vector in Nn , we denote by C a the clutter obtained from C by successively deleting any vertex xi with ai = 0 and duplicating ai − 1 times any vertex xi if ai ≥ 1 (for graphs cf. [191, p. 53]). Example 13.2.11 Let G be the graph whose only edge is {x1 , x2 } and let a = (3, 3). We set x1i = xi for i = 1, 2. The parallelization Ga is a complete bipartite graph with bipartition V1 = {x11 , x21 , x31 } and V2 = {x12 , x22 , x32 }. Note that xki is a vertex, i.e., k is an index not an exponent. x11 s

x1 s

x2

s

G x12

Graph

x21 s

x31 s G(3,1) s

Duplications of x1

x21 s x31 s x11 s H H @ @ @ HH @ @ HH @ G(3,3) s @ H s @s 1 2 3 x2 x2 x2 Duplications of x1 and x2

Lemma 13.2.12 Let C be a clutter and let A be its incidence matrix. If a = (ai ) is a vector in Nn , then β1 (C a ) ≤ max{y, 1| y ∈ Nq ; Ay ≤ a}. Proof. We may assume that a = (a1 , . . . , am , 0, . . . , 0), where ai ≥ 1 for i = 1, . . . , m. Recall that for each 1 ≤ i ≤ m the vertex xi is duplicated ai − 1 times and for each i > m the vertex xi is deleted. We denote the duplications of xi by x2i , . . . , xai i and set x1i = xi . Thus, we can write V (C a ) = {x11 , . . . , xa1 1 , . . . , x1i , . . . , xai i , . . . , x1m , . . . , xamm }. There are f1 , . . . , fβ1 independent edges of C a , where β1 = β1 (C a ). Each fi has the form j

j

j

fk = {xkk11 , xkk22 , . . . , xkkrr }

(1 ≤ k1 < · · · < kr ≤ m; 1 ≤ jki ≤ aki ).

Combinatorics of Symbolic Rees Algebras of Edge Ideals

543

We set gk = {x1k1 , x1k2 , . . . , x1kr } = {xk1 , xk2 , . . . , xkr }. By deﬁnition of C a we get that gk ∈ E(C) for all k. We may re-order the fi ’s so that g1 = g2 = · · · = gs1 , gs1 +1 = · · · = gs2 , . . . , gsr−1 +1 = · · · = gsr

s2 −s1

s1

sr −sr−1

and gs1 , . . . , gsr distinct, where sr = β1 . Let vi be the characteristic vector of gsi . Set y = s1 e1 + (s2 − s1 )e2 + · · · + (sr − sr−1 )er . We may assume that the incidence matrix A of C has column vector v1 , . . . , vq . Then y satisﬁes y, 1 = β1 . For each ki the number of variables of the form xki that occur in f1 , . . . , fβ1 is at most aki because the fi are pairwise disjoint. Hence for each ki the number of times that the variable x1ki occurs in g1 , . . . , gβ1 is at most aki . Then Ay = s1 v1 + (s2 − s1 )v2 + · · · + (sr − sr−1 )vr ≤ a.

2

Deﬁnition 13.2.13 The clutter of minimal vertex covers of C, denoted by b(C) or C ∨ , is called the blocker of C or the Alexander dual of C. Lemma 13.2.14 ([116, Lemma 2.15], [373, p. 1385, Eq. (78.6)]) Let C be a clutter and let C ∨ be the blocker of C. If a = (ai ) ∈ Nn , then ∨ min ai C ∈ C = α0 (C a ). xi ∈C

Proof. We may assume that a = (a1 , . . . , am , am+1 , . . . , am1 , 0, . . . , 0), where ai ≥ 2 for i = 1, . . . , m, ai = 1 for i = m + 1, . . . , m1 , and ai = 0 for i > m1 . Thus for i = 1, . . . , m the vertex xi is duplicated ai − 1 times. We denote the duplications of xi by x2i , . . . , xai i and set x1i = xi . We prove ﬁrst the inequality “≤”. Let C a be a minimal vertex cover of a C with α0 elements, where α0 is equal to α0 (C a ). We may assume that C a ∩ {x1 , . . . , xm1 } = {x1 , . . . , xs }. Note that x1i , . . . , xai i are in C a for i = 1, . . . , s. Indeed since C a is a minimal vertex cover of C a , there exists an edge e of C a such that e ∩ C a = {x1i }. Then (e \ {x1i }) ∪ {xji } is an edge of C a for j = 1, . . . , ai . Consequently xji ∈ C a for j = 1, . . . , ai . Hence a1 + · · · + as ≤ |C a | = α0 .

(13.3)

On the other hand the set C = {x1 , . . . , xs } ∪ {xm1 +1 , . . . , xn } is a vertex cover of C. Let D be a minimal vertex cover of C contained in C . Let eD denote the characteristic vector of D. Then, since ai = 0 for i > m1 , using Eq. (13.3) we get a, eD = ai = ai ≤ ai ≤ α0 . xi ∈D

xi ∈D∩{x1 ,...,xs }

xi ∈{x1 ,...,xs }

544

Chapter 13

This completes the proof of the inequality “≤”. Next we show the inequality “≥”. Let C be a minimal vertex cover of C. Note that the set C = ∪xi ∈C {x1i , . . . , xai i } is a vertex cover of C a . Indeed any edge ea of the clutter C a is of the form ea = {xji11 , . . . , xjirr } for some edge e = {xi1 , . . . , xir } of C and since e is a covered by C, we have that e is covered by C . Therefore one has that a α0 (C ) ≤ |C | = xi ∈C ai . As C was an arbitrary vertex cover of C we get the asserted inequality. 2 Theorem 13.2.15 [307] Let C be a clutter with vertex set X = {x1 , . . . , xn } and let 0 = a ∈ Nn , b ∈ N. Then xa tb is a minimal generator of Rs (I(C)), as a K-algebra, if and only if C a is indecomposable and b = α0 (C a ). Proof. We may assume that a = (a1 , . . . , am , 0, . . . , 0), where ai ≥ 1 for i = 1, . . . , m. For each 1 ≤ i ≤ m the vertex xi is duplicated ai − 1 times, and the vertex xi is deleted for each i > m. We denote the duplications of xi by x2i , . . . , xai i and set x1i = xi for 1 ≤ i ≤ m. The vertex set of C a is X a = {x11 , . . . , xa1 1 , . . . , x1i , . . . , xai i , . . . , x1m , . . . , xamm } = X a1 ∪ · · · ∪ X am , where X ai = {x1i , . . . , xai i } for 1 ≤ i ≤ m and X ai ∩ X aj = ∅ for i = j. ⇒) Assume that xa tb is a minimal generator of Rs (I(C)). Then, by Lemma 13.2.8, a is an irreducible b-cover of C ∨ . First we prove that b is equal to α0 (C a ). There is k such that ak = 0. We may assume that a − ek = 0. By Lemma 13.2.14 we need only show the equality ∨ b = min ai C ∈ C . xi ∈C

As a is a b-cover of C ∨ , the minimum is greater than or equal to b. If the minimum is greater than b, then we can write a = (a− ek )+ ek , where a− ek is a b-cover and ek is a 0-cover of C ∨ , a contradiction. Next we show that C a is indecomposable. We proceed by contradiction. Assume that C a is decomposable. Then there is a partition X1 , X2 of X a such that α0 (C a ) = α0 (C a [X1 ]) + α0 (C a [X2 ]). For 1 ≤ i ≤ n, we set i = |X ai ∩ X1 | and pi = |X ai ∩ X2 | if 1 ≤ i ≤ m, and i = pi = 0 if i > m. Consider the vectors = (i ) and p = (pi ). Notice that a has a decomposition a = + p because one has a partition X ai = (X ai ∩ X1 ) ∪ (X ai ∩ X2 ) for 1 ≤ i ≤ m. To derive a contradiction we now claim that (resp. p) is an α0 (C a [X1 ])-cover (resp. α0 (C a [X2 ])-cover) of C ∨ . Take an arbitrary C in C ∨ . The set ( ( {x1i , . . . , xai i } = X ai Ca = xi ∈C

xi ∈C

Combinatorics of Symbolic Rees Algebras of Edge Ideals

545

is a vertex cover of C a . Indeed, if fk is any edge of C a , then fk has the form j

j

fk = {xkk11 , . . . , xkkrr }

(1 ≤ k1 < · · · < kr ≤ m; 1 ≤ jki ≤ aki )

(13.4)

for some edge {xk1 , . . . , xkr } of C. Since {xk1 , . . . , xkr } ∩ C = ∅, we get fk ∩Ca = ∅. Thus Ca is a vertex cover of C a . Therefore Ca ∩X1 and Ca ∩X2 are vertex covers of C a [X1 ] and C a [X2 ], respectively, because E(C a [Xi ]) is contained in E(C a ) for i = 1, 2. Hence using the partitions ( ( (X ai ∩ X1 ) and Ca ∩ X2 = (X ai ∩ X2 ) Ca ∩ X1 = xi ∈C

xi ∈C

we obtain α0 (C a [X1 ]) ≤ |Ca ∩ X1 | =

i

and α0 (C a [X2 ]) ≤ |Ca ∩ X2 | =

xi ∈C

pi .

xi ∈C

This completes the proof of the claim. Consequently a is a reducible b-cover of C ∨ , where b = α0 (C a ), a contradiction to the irreducibility of a. ⇐) Assume that C a is an indecomposable clutter and b = α0 (C a ). We set a = (a1 , . . . , an ). To show that xa tb is a minimal generator of Rs (I(C)) we need only show that a is an irreducible b-cover of C ∨ . To begin with, notice that a is a b-cover of C ∨ by Lemma 13.2.14. We proceed by contradiction assuming that there is a decomposition a = + p, where = (i ) is a ccover of C ∨ , p = (pi ) is a d-cover of C ∨ , and b = c + d. Each X ai can be decomposed as X ai = X i ∪ X pi , where X i ∩ X pi = ∅, i = |X i |, and pi = |X pi |. We set X = X 1 ∪ · · · ∪ X m

and X p = X p1 ∪ · · · ∪ X pm .

Then one has a decomposition X a = X ∪ X p of the vertex set of C a . We now show that α0 (C a [X ]) ≥ c and α0 (C a [X p ]) ≥ d. By symmetry, it suﬃces to prove the ﬁrst inequality. Take an arbitrary minimal vertex cover C of C a [X ]. Then C ∪ X p is a vertex cover of C a because if f is an edge of C a contained in X , then f is covered by C , otherwise f is covered by X p . Hence there is a minimal vertex cover Ca of C a such that Ca ⊂ C ∪ X p . We set Vm = {x1 , . . . , xm }. Since C[Vm ] is a subclutter of C a , there is a minimal vertex cover C1 of C[Vm ] contained in Ca . Then the set C1 ∪ {xi | i > m} is a vertex cover of C. Therefore there is a minimal vertex cover C of C such that C ∩ Vm ⊂ Ca . Altogether one has: C ∩ Vm ⊂ Ca ⊂ C ∪ X p ⇒ C ∩ Vm ⊂ Ca ∩ Vm ⊂ (C ∪ X p ) ∩ Vm . (13.5) We may assume that Ca ∩ Vm = {x1 , . . . , xs }. Next we claim that X ai ⊂ Ca for 1 ≤ i ≤ s. Take an integer i between 1 and s. Since Ca is a minimal vertex cover of C a , there exists an edge e of C a such that e ∩ Ca = {x1i }.

546

Chapter 13

Then (e \ {x1i }) ∪ {xji } is an edge of C a for j = 1, . . . , ai , this follows using that the edges of C a are of the form described in Eq. (13.4). Consequently xji ∈ Ca for j = 1, . . . , ai . This completes the proof of the claim. Thus one has X i ⊂ X ai ⊂ Ca for 1 ≤ i ≤ s. Hence, by Eq. (13.5), and noticing that X i ∩ X p = ∅, we get X i ⊂ C for 1 ≤ i ≤ s. So, using that i = 0 for i > m, we get α0 (C a [X ]) ≥ |C | ≥

s i=1

i ≥

i =

xi ∈C∩Vm

i ≥ c.

xi ∈C

Therefore α0 (C a [X ]) ≥ c. Similarly α0 (C a [X p ]) ≥ d. Thus α0 (C a [X ]) + α0 (C a [X p ]) ≥ c + d = b = α0 (C a ), and consequently, by Eq. (13.2), we have equality. Thus we have shown that C a is a decomposable clutter, a contradiction. 2 Lemma 13.2.16 If C is an indecomposable clutter with the K¨ onig property, then either C has no edges and has exactly one isolated vertex or C has only one edge and no isolated vertices. Proof. Let f1 , . . . , fg be a set of independent edges and let X = ∪gi=1 fi , where g = α0 (C). Note that g = 0 if C has no edges. Then V (C) has a partition V (C) = (∪gi=1 fi ) ∪ ∪xi ∈V (C)\X {xi } . As C is indecomposable, we get that either g = 0 and V (C) = {xi } for some vertex xi or g = 1 and V (C) = fi for some i. Thus in the second case, as C is a clutter, we get that C has exactly one edge and no isolated vertices. 2 Corollary 13.2.17 Let C be a clutter and let I = I(C) be its edge ideal. Then all indecomposable parallelizations of C satisfy the K¨ onig property if and only if I i = I (i) for i ≥ 1. Proof. ⇒) It suﬃces to prove that R[It] = Rs (I). Clearly R[It] ⊂ Rs (I). To prove the reverse inclusion take a minimal generator xa tb of Rs (I). If b = 0, then a = ei for some i and xa tb = xi is in R[It]. Assume b ≥ 1. By Theorem 13.2.15 C a is an indecomposable clutter such that b = α0 (C a ). As C a satisﬁes the K¨onig property, using Lemma 13.2.16, it is not hard to see that b = 1 and that;E(C a ) = {e} consists of a single edge e of C, i.e., xa tb = xe t, where xe = xi ∈e xi . Thus xa tb ∈ R[It]. ⇐) Since R[It] = Rs (I), by Theorem 13.2.15, we obtain that the only indecomposable parallelizations are either induced subclutters of C with exactly one edge and no isolated vertices or subclutters consisting of exactly one isolated vertex. Thus in both cases they satisfy the K¨ onig property. 2

Combinatorics of Symbolic Rees Algebras of Edge Ideals

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Corollary 13.2.18 Let C be a clutter with vertex set X = {x1 , . . . , xn } and let S ⊂ X. Then the ; induced clutter H = C[S] is indecomposable if and only if the monomial ( xi ∈S xi )tα0 (H) is a minimal generator of Rs (I(C)). Proof. We set a = xi ∈S ei . Since C a = C[S], the result follows from Theorem 13.2.15. 2 Corollary 13.2.19 Let H be the Hilbert basis of Cn(I(C)). Then H

=

{(a, b)| xa tb is a minimal generator of Rs (I(C))}

= =

{(a, α0 (C))| C a is an indecomposable parallelization of C} {(a, b)| a is an irreducible b-cover of C ∨ }.

Proof. The ﬁrst and third equalities follow readily from Lemma 13.2.8 and Theorem 13.2.2. The second equality follows from Theorem 13.2.15. 2 This result allows us to compute all indecomposable parallelizations of C and all indecomposable induced subclutters of C using Hilbert bases (see also Exercise 13.2.31). Computing the Hilbert basis of a Simis cone Let xv1 , . . . , xvq be the minimal set of generators of the edge ideal I(C) of a clutter C and let R+ A be the Rees cone of A = {v1 , . . . , vq }. Lemma 13.2.20 If i = (ui , −1) for i = 1, . . . , s, then the halfspace H+i occurs in the irreducible representation of R+ A . Proof. By Corollary 13.1.3 the integral vertices of Q(A) are u1 , . . . , us , where A is the incidence matrix of the clutter C. Hence the result follows from Theorem 1.4.2. 2 Procedure 13.2.21 To determine the Hilbert basis of Cn(I) and the generators of Rs (I(C)) we proceed as follows. Using Normaliz , one computes the irreducible representation of R+ A and apply Lemma 13.2.20 to obtain the set of “inequalities” Γ = {e1 , . . . , en+1 , 1 , . . . , s } that deﬁne Cn(I). Then one can use Γ as input for Normaliz [68] to obtain the Hilbert basis. Then one uses Corollary 13.2.19 to obtain the generators of Rs (I(C)). This procedure is based on the fact that the program Normaliz [68] is able to compute the minimal Hilbert basis of a pointed cone given by a linear system of inequalities (cf. [189, Procedure 4.6.10]). Example 13.2.22 Using Procedure 13.2.21 and Normaliz [68], we compute the minimal generators of the symbolic Rees algebra of the edge ideal I = I(C) = (x1 x2 , x2 x3 , x1 x3 ). The input ﬁle for Normaliz is:

548

Chapter 13

3 3 1 1 0 0 1 1 1 0 1 3

Using Normaliz, with mode 3, and Lemma 13.2.20, we obtain that Cn(I) is deﬁned by the “inequalities” given by the rows of the matrix: 7 4 0 0 1 1 0 0 1 4

0 0 0 0 1 1 1

1 0 0 1 1 -1 0 0 1 -1 0 0 0 -1

Using this input ﬁle for Normaliz, with mode 4, to compute the Hilbert basis of Cn(I), we get: 7 Hilbert basis elements: 0 1 1 1 1 0 1 1 1 1 1 2 0 0 1 0 0 1 0 0 1 0 0 0 1 1 0 1

Thus, by Corollary 13.2.19, the symbolic Rees algebra of the edge ideal I is minimally generated as a K-algebra by x1 , x2 , x3 , It, x1 x2 x3 t2 . Algebras of vertex covers In this part we will further examine minimal sets of generators of symbolic Rees algebras of edge ideals using polyhedral geometry and vertex covers. Let I = I(C) be the edge ideal of a clutter C and let v1 , . . . , vq be the characteristic vectors of the edges of C. Recall that Rees cone of I, denoted by R+ (I) or R+ A , is the rational cone generated by the set A = {e1 , . . . , en , (v1 , 1), . . . , (vq , 1)} ⊂ Rn+1 , where ei is the ith unit vector. By Proposition 1.1.51 and Theorem 1.4.2, the Rees cone has a unique irreducible representation R+ (I) = He+1 ∩ He+2 ∩ · · · ∩ He+n+1 ∩ H+1 ∩ H+2 ∩ · · · ∩ H+r

(13.6)

Combinatorics of Symbolic Rees Algebras of Edge Ideals

549

such that, for each k, k ∈ Zn+1 , the non-zero entries of k are relatively prime, the ﬁrst n entries of k are in N, the last entry of k is negative, and none of the closed halfspaces He+1 , . . . , He+n+1 , H+1 , . . . , H+r can be omitted from the intersection. The facets (i.e., the proper faces of maximum dimension or equivalently the faces of dimension n) of the Rees cone are exactly: He1 ∩ R+ (I), . . . , Hen+1 ∩ R+ (I), H1 ∩ R+ (I), . . . , Hr ∩ R+ (I). covers of C and let u1 , . . . , us Let C1 , . . . , Cs be the minimal vertex be their characteristic vectors, i.e., uk = xi ∈Ck ei for k = 1, . . . , s. By Lemma 13.2.20 we may always assume that k = (uk , −1) for 1 ≤ k ≤ s, and that every facet Hk ∩ R+ (I), with k > s, satisﬁes k , en+1 < −1. Thus one has: Remark 13.2.23 The primary decomposition of I(C) as well as all the minimal vertex covers of C can be read oﬀ from the irreducible representation of the Rees cone R+ (I(C)). The ideal of covers of C, denoted by Ic (C), is the ideal of R generated by xu1 , . . . , xus . We also denote Ic (C) by I(C)∨ and call I(C)∨ the Alexander dual of I(C). Notice that Ic (C) = I(C ∨ ) = I(C)∨ . The symbolic Rees algebra of Ic (C) can also be interpreted in terms of “b-vertex covers” because Rs (Ic (C)) is the symbolic Rees algebra of the edge ideal of C ∨ and (C ∨ )∨ = C. Let a = (a1 , . . . , an ) = 0 be a vector in Nn and let b ∈ N. Recall that a is a b-vertex cover of C if vi , a ≥ b for i = 1, . . . , q. Deﬁnition 13.2.24 The algebra of vertex covers of I, denoted by Rc (I), is the K-subalgebra of R[t] generated by all monomials xa tb such that a is a b-cover of C. This algebra turns out to be equal to Rs (Ic (C)). The irreducible 0 and 1 covers of C are the unit vector e1 , . . . , en and the vectors u1 , . . . , us . The minimal generators of Rc (I), as a K-algebra, correspond to the irreducible covers of C (see Lemma 13.2.8). Notice the following dual descriptions: I (b)

= ({xa | a, ui ≥ b for i = 1, . . . , s}),

Ic (C)(b)

= ({xa | a, vi ≥ b for i = 1, . . . , q}).

Thus, we have the following duality between symbolic Rees algebras. Theorem 13.2.25 Rc (I) = Rs (Ic (C)) and Rc (Ic (C)) = Rs (I).

550

Chapter 13

Theorem 13.2.26 Rc (I) is a ﬁnitely generated K-algebra. Proof. It follows from Corollary 13.2.4 because Rc (I) is the symbolic Rees algebra of the edge ideal Ic (C). 2 Lemma 13.2.27 If k = (ak , −dk ) is any of the vectors of Eq. (13.6), where ak ∈ Nn , dk ∈ N+ , then ak is an irreducible dk -cover of C and xak tdk is a minimal generator of Rs (Ic (C)). Proof. We proceed by contradiction and assume that there is a dk -cover ak and a dk -cover ak such that ak = ak + ak and dk = dk + dk . Set F = H(ak ,−dk ) ∩ R+ (I) and F = H(ak ,−dk ) ∩ R+ (I). Clearly F , F are proper faces of R+ (I) and F = R+ (I) ∩ Hk = F ∩ F . Applying Theorem 1.1.44(d) to F and F it is seen that F ⊂ F or F ⊂ F , i.e., F = F or F = F . We may assume F = F . Hence H(ak ,−dk ) = Hk . Taking orthogonal complements we get that (ak , −dk ) = λ(ak , −dk ) for some λ ∈ Q+ , because the orthogonal complement of Hk is generated by k . Since the non-zero entries of k are relatively prime, we may assume that λ ∈ N. Thus dk = λdk ≥ dk ≥ dk and λ must be 1. Hence ak = ak and ak must be zero, a contradiction. 2 Remark 13.2.28 Let Fn+1 be the facet of the Rees cone R+ (I) determined by the hyperplane Hen+1 . Thus, by Lemma 13.2.27, we have a map ψ: ψ

{Facets of R+ (I(C))} \ {Fn+1 } −→ Rs (Ic (C)) ψ

Hk ∩ R+ (I) −→ xak tdk , where k = (ak , −dk ) ψ

Hei ∩ R+ (I) −→ xi whose image is a good approximation for the minimal set of generators of Rs (Ic (C)) as a K-algebra. Likewise, from the facets of R+ (Ic (C)), we obtain an approximation for the minimal set of generators of Rs (I(C)). The next example shows a connected graph G for which the image of the map ψ does not generate Rs (Ic (G)). Example 13.2.29 Consider the following graph G:

s x1

sx2 @ @ @s x3

s x4 @ @ @s x5

sx6 @ @ @s x7

s x8 @ @ @s x9

Using Normaliz [68] it is seen that the vector a = (1, 1, 2, 0, 2, 1, 1, 1, 1) is an irreducible 2-cover of G such that the supporting hyperplane H(a,−2) does not deﬁne a facet of the Rees cone of I(G).

Combinatorics of Symbolic Rees Algebras of Edge Ideals

551

For balanced clutters, the image of the map ψ generates Rs (Ic (C)). This follows from the next result (see Corollary 14.3.11). In particular the image of the map ψ generates Rs (Ic (C)) when C is a bipartite graph. Proposition 13.2.30 [183, 185] If C is a balanced clutter, then Rs (Ic (C)) = R[Ic (C)t]. This result was ﬁrst shown for bipartite graphs in [183, Corollary 2.6] and later generalized to balanced clutters in [185].

Exercises 13.2.31 Let C be a clutter with vertices x1 , . . . , xn and α = (a1 , . . . , an , b) a vector in {0, 1}n × N. Then α is in the minimal Hilbert basis of Cn(I(C)) if and only if the induced subclutter H = C[{xi | ai = 1}] is indecomposable with b = α0 (H). 13.2.32 Let C be a clutter. Prove that (1, . . . , 1, α0 (C)) is in the minimal Hilbert basis of Cn(I(C)) if and only if C is indecomposable.

13.3

Blowup algebras in perfect graphs

Harary and Plummer [209] studied indecomposable graphs. To the best of our knowledge there is no structure theorem for indecomposable graphs. In this section we relate these graphs to the theory of perfect graphs and to symbolic Rees algebras. Then we present some general properties of indecomposable clutters. Lemma 13.3.1 Let Cn = {x1 , . . . , xn } be a cycle of length n. (a) If n is odd and n ≥ 5, then the complement C n of Cn is indecomposable, (b) if n is odd, then Cn is indecomposable, and (c) any complete graph is indecomposable. Proof. (a) Assume that G = C n is decomposable. Then there are disjoint sets X1 , X2 such that V (G) = X1 ∪X2 and α0 (G) = α0 (G[X1 ])+α0 (G[X2 ]). Since β0 (G) = 2, it is seen that G[Xi ] is a complete graph for i = 1, 2. We may assume that x1 ∈ X1 . Then x2 must be in X2 , otherwise {x1 , x2 } is an edge of G[X1 ], a contradiction. By induction it follows that x1 , x3 , x5 , . . . , xn are in X1 . Hence, {x1 , xn } is an edge of G[X1 ], a contradiction. (b) This is left as an exercise. (c) It follows readily from the fact that the covering number of a complete graph in r vertices is r − 1. 2 For graphs, we can use this lemma together with Corollary 13.2.19 and Exercise 13.2.31 to locate all induced odd cycles (odd holes) and all induced complements of odd cycles (odd antiholes) of length at least ﬁve. Notice that,

552

Chapter 13

by Theorem 13.2.15, odd holes and odd antiholes correspond to minimal generators of the symbolic Rees algebra of the edge ideal of the graph. A graph G is called a Berge graph if and only if G has no odd holes or odd antiholes of length at least ﬁve. Theorem 13.3.2 (Strong perfect graph theorem [85]) A graph G is perfect if and only if G is a Berge graph. In commutative algebra odd holes occurred for the ﬁrst time in [377], and later in the description of I(G){2} , the join of an edge ideal of a graph G with itself [381], and in the description of the associated primes of powers of ideals of vertex covers of graphs [164]. The expository paper [163] surveys algebraic techniques for detecting odd cycles and odd holes in a graph and for computing the chromatic number of a hypergraph (see Theorem 7.7.19). Theorem 13.3.3 [164] If G is a graph and p is an associated prime of Ic (G)2 , then p = (xi , xj ) for some edge {xi , xj } of G or p = (xi | ∈ S), where S is a set of vertices of G that induces an odd hole. Example 13.3.4 Let G be the graph below. Computing the generators of Rs (I(G)) via Procedure 13.2.21 and Corollary 13.2.19, gives that the indecomposable parallelizations of G are: seven vertices, nine edges, one induced triangle, three induced pentagons, and the duplication shown below. x1 t @ @ @ tx6 @tx2 x5 t tx7 H H HHtx x4 t 3 Decomposable graph G

x1 t @ @ 1 tx X X @ X@ @ Xtx2 @txX x5 t 6 tx7 H H HHtx x4 t 3 Indecomposable graph G(2,1,1,1,1,1,1)

The next result shows that indecomposable graphs occur naturally in the theory of perfect graphs. Proposition 13.3.5 [117, Proposition 2.13] A graph G is perfect if and only if the indecomposable parallelizations of G are exactly the complete subgraphs of G The following was one of the ﬁrst deep results in the study of symbolic powers of edge ideals from the viewpoint of graph theory. Corollary 13.3.6 [383, Theorem 5.9] Let G be a graph and let I be its edge ideal. Then G is bipartite if and only if I i = I (i) for i ≥ 1.

Combinatorics of Symbolic Rees Algebras of Edge Ideals

553

Proof. ⇒) If G is a bipartite graph, then any parallelization of G is again a bipartite graph. This means that any parallelization of G satisﬁes the K¨onig property because bipartite graphs satisfy this property; see Theorem 7.1.8. Thus I i = I (i) for all i by Corollary 13.2.17. ⇐) Assume that I i = I (i) for i ≥ 1. Thanks to Corollary 13.2.17 all indecomposable induced subgraphs of G have the K¨ onig property. If G is not bipartite, then G has an induced odd cycle, a contradiction because induced odd cycles are indecomposable by Lemma 13.3.1 and do not satisfy the K¨onig property. 2 Basic properties of indecomposable clutters If e is a edge of a clutter C, we denote by C \ {e} the spanning subclutter of C obtained by deleting e and keeping all the vertices of C. Deﬁnition 13.3.7 A clutter C is called vertex critical (resp. edge critical ) if α0 (C \ {xi }) < α0 (C) (resp. α0 (C \ {e}) < α0 (C)) for all xi ∈ V (C) (resp. for all e ∈ E(C)). Lemma 13.3.8 Let xi be a vertex and let e be an edge of a clutter C. (a) If α0 (C \ {xi }) < α0 (C), then α0 (C \ {xi }) = α0 (C) − 1. (b) If α0 (C \ {e}) < α0 (C), then α0 (C \ {e}) = α0 (C) − 1. Proof. Part (a) is left as an exercise (cf. Proposition 7.2.11). Part (b) follows from Proposition 7.2.13. 2 Deﬁnition 13.3.9 A clutter C is called connected if there is no U ⊂ V (C) such that ∅ U V (C) and such that e ⊂ U or e ⊂ V (C) \ U for each edge e of C. Proposition 13.3.10 If a clutter C is indecomposable, then it is connected and vertex critical. Proof. Assume that C is disconnected. Then there is a partition X1 , X2 of V (C) such that (13.7) E(C) ⊂ E(C[X1 ]) ∪ E(C[X2 ]). For i = 1, 2, let Ci be a minimal vertex cover of C[Xi ] with α0 (C[Xi ]) vertices. Then, by Eq. (13.7), C1 ∪ C2 is a minimal vertex cover of C. Hence α0 (C[X1 ]) + α0 (C[X2 ]) is greater than or equal to α0 (C). So α0 (C) is equal to α0 (C[X1 ]) + α0 (C[X2 ]), a contradiction to the indecomposability of C. Thus C is connected. We now show that α0 (C \ {xi }) < α0 (C) for all i. If α0 (C \ {xi }) = α0 (C), then V (C) = X1 ∪ X2 , where X1 = V (C) \ {xi } and X2 = {xi }. Note that C[X1 ] = C \ {xi }. As α0 (C[X1 ]) = α0 (C) and α0 (C[X2 ]) = 0, we contradict 2 the indecomposability of C. Thus α0 (C \ {xi }) < α0 (C).

554

Chapter 13

Proposition 13.3.11 If C is a connected edge critical clutter, then C is indecomposable. Proof. Assume that C is decomposable. Then there is a partition X1 , X2 of V (C) into non-empty vertex sets such that α0 (C) = α0 (C[X1 ]) + α0 (C[X2 ]). Since C is connected, there is an edge e ∈ E(C) intersecting both X1 and X2 . Pick a minimal vertex cover C of C \ {e} with less than α0 (C) vertices. As E(C[Xi ]) is a subset of E(C \ {e}) = E(C) \ {e} for i = 1, 2, we get that C covers all edges of C[Xi ] for i = 1, 2. Hence C must have at least α0 (C) vertices, a contradiction. 2

Exercises 13.3.12 Let D be a clutter obtained from C by adding a new vertex v and some new edges containing v and some vertices of V (C). If a = 1 ∈ Nn is an irreducible α0 (C)-cover of C ∨ such that α0 (D) = α0 (C) + 1, then a = (a, 1) is an irreducible α0 (D)-cover of D∨ . 13.3.13 [12] If G is a complete graph and H is the graph obtained by taking a cone over a pentagon, then Rs (I(G)) Rs (I(H))

= =

K[{xa tb | xa is square-free ; deg(xa ) = b + 1}], R[I(H)t][x1 · · · x5 t3 , x1 · · · x6 t4 , x1 · · · x5 x26 t5 ].

13.3.14 [209] Prove that any odd cycle is indecomposable. 13.3.15 Using Procedure 13.2.21 and Corollary 13.2.19, show that the graph G below has exactly 103 indecomposable parallelizations, 92 of which correspond to subgraphs. The only indecomposable parallelization Ga , with ai > 0 for all i, is that obtained by duplication of the ﬁve outer vertices, i.e., a = (2, 2, 2, 2, 2, 1, 1, 1, 1, 1) and α0 (Ga ) = 11. s x1 L@ L@ !a sL @ ! x6 D a L a@ ! s ! s D Lsx7aa x5 ! @sx2 x10 @ D D Z L Z L Z x9@ D D x8 s D L Zs @D L H H @ D H @ HDs x3 x4 Ls Decomposable graph G

Combinatorics of Symbolic Rees Algebras of Edge Ideals

13.4

555

Algebras of vertex covers of graphs

Let G be a graph and let Ic (G) be its ideal of covers. In this section we give a graph theoretical description of the irreducible b-covers of G, i.e., we describe the minimal generators of the symbolic Rees algebra of Ic (G). Lemma 13.4.1 If a = (a1 , . . . , an ) ∈ Nn is an irreducible b-cover of G, then 0 ≤ b ≤ 2 and 0 ≤ ai ≤ 2 for i = 1, . . . , n. Proof. Recall that a is a b-cover of G if and only if ai + aj ≥ b for any edge {xi , xj } of G. If b = 0 or b = 1, then by the irreducibility of a it is seen that either a = ei for some i or a = ei1 + · · · + eir for some minimal vertex cover {xi1 , . . . , xir } of G. Thus we may assume that b ≥ 2. Case (I): ai ≥ 1 for all i. Clearly 1 = (1, . . . , 1) is a 2-cover. If a − 1 = 0, then a − 1 is a b − 2 cover and a = 1 + (a − 1), a contradiction. Hence a = 1. Pick any edge {xi , xj } of G. Since a is a b-cover, we get 2 = ai + aj ≥ b and b must be equal to 2. Case (II): ai = 0 for some i. We may assume ai = 0 for 1 ≤ i ≤ r and ai ≥ 1 for i > r. Notice that the set S = {x1 , . . . , xr } is independent because if {xi , xj } is an edge and 1 ≤ i < j ≤ r, then 0 = ai + aj ≥ b, a contradiction. Consider the neighbor set NG (S) of S. We may assume that NG (S) = {xr+1 , . . . , xs }. Observe that ai ≥ b ≥ 2 for i = r + 1, . . . , s, because a is a b-cover. Write a = (0, . . . , 0, ar+1 − 2, . . . , as − 2, as+1 − 1, . . . , an − 1)+ (0, . . . , 0, 2, . . . , 2, 1, . . . , 1) = c + d. r

s−r

n−s

Clearly d is a 2-cover. If c = 0, using that ai ≥ b ≥ 2 for r + 1 ≤ i ≤ s and ai ≥ 1 for i > s it is not hard to see that c is a (b − 2)-cover. This gives a contradiction, because a = c + d. Hence c = 0. Therefore ai = 2 for 2 r < i ≤ s, ai = 1 for i > s, and b = 2. Corollary 13.4.2 Rs (Ic (G)) is generated, as a K-algebra, by monomials of degree in t at most two and total degree at most 2n. Proof. Let xa tb be a minimal generator of Rs (Ic (G)). Then a = (ai ) is an irreducible b-cover of G. By Lemma 13.4.1, we get 0 ≤ b ≤ 2 and 0 ≤ ai ≤ 2 for all i. If b = 0 or b = 1, we get that thedegree of xa tb is at most n because when b = 0 or 1 one has a = ei or a = xi ∈Ck ei for some minimal vertex cover Ck of G, respectively. If b = 2, by the proof of Lemma 13.4.1, either a = 1 or ai = 0 for some i. Thus deg(xa ) ≤ 2(n − 1). 2 For use below consider the vectors 1 , . . . , r that occur in the irreducible representation of R+ (I(G)) given in Eq. (13.6).

556

Chapter 13

Corollary 13.4.3 If i = (i1 , . . . , in , −i(n+1) ), then 0 ≤ ij ≤ 2 for j = 1, . . . , n and 1 ≤ i(n+1) ≤ 2. Proof. It suﬃces to observe that (i1 , . . . , in ) is an irreducible i(n+1) -cover of G and to apply Lemma 13.4.1. 2 Lemma 13.4.4 a = (1, . . . , 1) is an irreducible 2-cover of G if and only if G is non-bipartite. Proof. ⇒) We proceed by contradiction. Assume that G is bipartite. Then G has a bipartition (V1 , V2 ). Set a = xi ∈V1 ei and a = xi ∈V2 ei . Since V1 and V2 are minimal vertex covers of G, we can decompose a as a = a + a , where a and a are 1-covers, which is impossible. ⇐) Notice that a cannot be the sum of a 0-cover and a 2-cover. Indeed if a = a + a , where a is a 0-cover and a is a 1-cover, then a has an entry ai equal to zero. Pick an edge {xi , xj } incident with xi , then a , ei + ej ≤ 1, a contradiction. Thus we may assume that a = c + d, where c, d are 1-covers. Let Cr be an odd cycle of G of length r. Notice that any vertex cover of Cr must contain a pair of adjacent vertices. Hence the vertex covers of G corresponding to c and d must contain a pair of adjacent vertices, a contradiction because c and d are complementary vectors and the complement of a vertex cover is an independent set. 2 Theorem 13.4.5 Let 0 = a = (ai ) ∈ Nn and let G∨ be the clutter of minimal vertex covers of G. Then the following hold. (i) If G is bipartite, then a is an irreducible b-cover of G ifand only if b = 0 and a = ei for some 1 ≤ i ≤ n or b = 1 and a = xi ∈C ei for some C ∈ E(G∨ ). (ii) If G is non-bipartite, then a is an irreducible b-cover if and only if a has one of the following forms: (a) (0-covers) b = 0 and a = ei for some 1 ≤ i ≤ n, (b) (1-covers) b = 1 and a = xi ∈C ei for some C ∈ E(G∨ ), (c) (2-covers) b = 2 and a = (1, . . . , 1), (d) (2-covers) b = 2 and up to permutation of vertices a = (0, . . . , 0, 2, . . . , 2, 1, . . . , 1) |A|

|NG (A)|

for some independent set of vertices A = ∅ of G such that (d1 ) NG (A) is not a vertex cover of G and V = A ∪ NG (A), (d2 ) the induced subgraph G[V \ (A ∪ NG (A))] has no isolated vertices and is not bipartite.

Combinatorics of Symbolic Rees Algebras of Edge Ideals

557

Proof. (i) ⇒) Since G is bipartite, by Proposition 13.2.30, we have the equality Rs (Ic (G)) = R[Ic (G)t]. Thus the minimal set of generator of Rs (Ic (G)) as a K-algebra is {x1 , . . . , xn , xu1 t, . . . , xur t}, where u1 , . . . , ur are the characteristic vectors of the minimal vertex covers of G. As a is an irreducible b-cover of G, xa tb is a minimal generator of Rs (Ic (C)). Therefore either a = ei for some i and b = 0 or a = ui for some i and b = 1. The converse follows readily and is valid for any graph or clutter. (ii) ⇒) By Lemma 13.4.1, 0 ≤ b ≤ 2 and 0 ≤ ai ≤ 2 for all i. If b = 0 or b = 1, then clearly a has the form indicated in (a) or (b), respectively. Assume b = 2. If ai ≥ 1 for all i, then ai = 1 for all i, otherwise if ai = 2 for some i, then a − ei is a 2-cover and a = ei + (a − ei ), a contradiction. Hence a = 1. Thus we may assume that a has the form a = (0, . . . , 0, 2, . . . , 2, 1, . . . , 1). We set A = {xi | ai = 0} = ∅, B = {xi | ai = 2}, and C = V \(A∪B). Observe that A is an independent set because a is a 2-cover and B = NG (A) because a is irreducible. Hence it is seen that conditions (d1 ) and (d2 ) are satisﬁed. By Lemma 13.4.4, the proof of the converse is straightforward. 2 If I ⊂ R is an ideal, the ring Fs (I) = Rs (I)/mRs (I) is called the symbolic special ﬁber of I (cf. Deﬁnition 14.2.10). If G is a graph, the algebraic properties of the ring Fs (I) are studied in [91], when G is bipartite the symbolic special ﬁber of Ic (G) turns out to be Koszul [91, 356] and in this case, by Proposition 13.2.30, one has Fs (Ic (G)) = R[Ic (G)t]/mR[Ic (G)t].

Exercises 13.4.6 Let G be a graph with vertex set V (G) = {x1 , . . . , xn } and let Hα+ be any of the closed halfspaces that occur in the irreducible representation of R+ (Ic (G)) with α = (a1 , . . . , an , −b), ai ∈ N for all i, 0 = b ∈ N, and the non-zero entries of α are relatively prime. (a) Let H be the cone over G. If ai ≥ 1 for all i and n n β = (a1 , . . . , an , ( i=1 ai ) − b, − i=1 ai ) = (β1 , . . . , βn+1 , −βn+2 ), β

n+1 βn+2 t is a then Hβ ∩ R+ (Ic (H)) is a facet of R+ (Ic (H)) and xβ1 1 · · · xn+1 minimal generator of Rs (I(H)). (b) Let G0 = G and let Gr be the cone over Gr−1 for r ≥ 1. If α is equal to (1, . . . , 1, −g) and b = n + (r − 1)(n − g), then

(1, . . . , 1, n − g, . . . , n − g)

n

r

is an irreducible b-cover of Gr and Rs (I(Gr )) has a generator of degree in t equal to b.

558

Chapter 13

(c) Let G = Cs be an odd cycle of length s = 2k + 1. Then, α0 (Cs ) is equal to (s + 1)/2 = k + 1 and x1 · · · xs xks+1 · · · xks+r trk+k+1 is a minimal generator of Rs (I(Gr )). This proves that the degree in t of the minimal generators of Rs (I(Gr )) is much larger than the number of vertices of the graph Gr [225].

13.5

Edge subrings in perfect matchings

In this section we present membership criteria and a generalized version of the marriage theorem. For a connected non-bipartite graph whose edge subring is normal, we give conditions for the existence of a perfect matching. Let G be a connected graph and let A = {v1 , . . . , vq } be the set of all vectors ei + ej such that {xi , xj } is an edge of G. The edge cone of G, denoted by R+ A, is deﬁned as the cone generated by A. Below we recover an explicit combinatorial description of the edge cone (see Section 10.7). Let A be an independent set of vertices of G. The supporting hyperplane of the edge cone of G deﬁned by xi − xi = 0 xi ∈A

xi ∈NG (A)

is denoted by HA . If aA = xi ∈A ei − xi ∈NG (A) ei , then HA = HaA . Edge cones and their representations by closed halfspaces are a useful tool to study a-invariants of edge subrings [419, 405]; see Chapter 11. The following result is a prototype of these representations (cf. Theorem 10.7.8 and Exercise 10.7.24). As an application we give a direct proof of the next result using Rees cones and Theorem 13.4.5. Corollary 13.5.1 [406, Corollary 2.8] A vector a = (a1 , . . . , an ) ∈ Rn is in R+ A if and only if a satisﬁes the following system of linear inequalities

xi ∈NG (A) ai −

ai

xi ∈A ai

≥ 0, i = 1, . . . , n; ≥ 0, for all independent sets A ⊂ V (G).

Proof. We set B = {(v1 , 1), . . . , (vq , 1)} and I = I(G). Notice the equality R+ (I) ∩ RB = R+ B,

(13.8)

where RB is the R-vector space spanned by B. Consider the irreducible representation of R+ (I) given in Eq. (13.6) and write i = (ai , −di ), where 0 = ai ∈ Nn , 0 = di ∈ N. Next we show the equality: + + R+ A = RA ∩ Rn+ ∩ H(2a ∩ · · · ∩ H(2a , 1 /d1 −1) r /dr −1)

(13.9)

Combinatorics of Symbolic Rees Algebras of Edge Ideals

559

where 1 = (1, . . . , 1). Take α ∈ R+ A. Clearly α ∈ RA ∩ Rn+ . We can write α = λ1 v1 + · · · + λq vq ⇒ |α| = 2(λ1 + · · · + λq ) = 2b. Thus (α, b) = λ1 (v1 , 1) + · · · + λq (vq , 1), i.e., (α, b) ∈ R+ B. Hence, from Eq. (13.8), we get (α, b) ∈ R+ (I) and (α, b), (ai , −di ) ≥ 0 ⇒ α, ai ≥ bdi = (|α|/2)di = |α|(di /2). Writing α = (α1 , . . . , αn ) and ai = (ai1 , . . . , ain ), the last inequality gives: α1 ai1 + · · · + αn ain ≥ (α1 + · · · + αn )(di /2) ⇒ α, ai − (di /2)1 ≥ 0. + for all i. This proves the Then α, 2ai /di − 1 ≥ 0 and α ∈ H(2a i /di −1) inclusion “⊂” of Eq. (13.9). The other inclusion follows similarly. Now, by Lemma 13.2.27, ai is an irreducible di -cover of G. Therefore, using Theorem 13.4.5, we readily get the equality ! ! n − + HA Hei , R+ A = A∈F

i=1

where F is the collection of all the independent sets of vertices of G. From this equality the assertion follows at once. 2 Proposition 13.5.2 [419] Let G be a connected non-bipartite graph with n vertices. If K[G] is a normal domain, then a monomial xβ1 1 · · · xβnn belongs to K[G] if and only if the following two conditions hold (i) β = (β1 , . . . , βn ) is in the edge cone of G, and (ii) ni=1 βi is an even integer. Proof. ⇐) Let A = {v1 , . . . , vq } be the set of column vector of the incidence matrix of G. Assume that β ∈ R+ A and deg(xβ ) even. We proceed by induction on deg(xβ ). Using Corollary 10.2.11 one has the isomorphism Zn /(v1 , . . . , vq ) Z2 . Hence 2β ∈ R+ A ∩ ZA = NA and one can write 2β = 2

q i=1

si vi +

q i=1

i vi ,

si ∈ N and i ∈ {0, 1},

by induction one may assume qi=1 si vi = 0. Therefore, from the equality above, one concludes that the subgraph whose edges are deﬁned by the set {vi | i = 1} is an edge disjoint union of cycles C1 , . . . , Cr . By induction one may further assume that all the Ci ’s are odd cycles. Note r ≥ 2, because deg(xb ) is even. As G is connected, using Corollary 10.3.12, it follows that xβ is in K[G]. ⇒) It follows readily by Corollary 9.1.3 because K[G] is normal. 2

560

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Proposition 13.5.3 Let G be a connected graph with n vertices. If n is even and K[G] is normal of dimension n, then x1 · · · xn is in K[G] if and only if |A| ≤ |N (A)| for every independent set of vertices A of G. Proof. ⇒) Let A = {v1 , . . . , vq } be the set of column vector of the incidence matrix of G. Since K[G] is normal: R+ A ∩ ZA = NA. Hence the vector a = (a1 , . . . , an ) = 1 is in R+ A. Using Corollary 13.5.1 we get that a satisﬁes the inequalities: |A| = ai ≤ ai = |N (A)| xi ∈A

xi ∈N (A)

for every independent set of vertices A of G, as required. ⇐) First we use Corollary 13.5.1 to conclude that a is in the edge cone, 2 then apply Proposition 13.5.2 to get xa ∈ K[G]. Corollary 13.5.4 (Marriage theorem) Let G be a graph with an even number of vertices. If G is connected and satisﬁes the odd cycle condition, then the following are equivalent (a) G has a perfect matching. (b) |A| ≤ |N (A)| for all A independent set of vertices of G. Proof. The proof follows using Theorem 7.1.9 and Proposition 13.5.3. 2

Exercises 13.5.5 Let G be a connected non-bipartite graph with n vertices and let v1 , . . . , vq be the columns of its incidence matrix. If n is even, prove that the vector (1, . . . , 1) is in L = Zv1 + · · · + Zvq . 13.5.6 Let G be a connected non-bipartite graph with n vertices. If n is even, prove that the monomial x1 · · · xn is in the ﬁeld of fractions of K[G]. 13.5.7 Let G be a connected non-bipartite graph with n vertices. If K[G] is a normal domain, then x1 · · · xn is in K[G] if and only if (1, . . . , 1) is in the edge cone of G and n is even. 13.5.8 Prove that the following graph does not have a perfect matching by exhibiting an independent set A such that |A| ≤ |N (A)|. r @ @ rH @ r H @ Hr @ @r

r rH H Hr

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561

13.5.9 Consider the graph G = G1 ∪G2 with six vertices x1 · · · x6 consisting of two disjoint triangles: •??

?? ?? ?? ? G1 •

•

• •? ?? G2 ?? ?? ?? •

Prove that the vector 1 = (1, . . . , 1) belongs to the edge cone of G and that x1 = x1 · · · x6 is not in K[G].

13.6

Rees cones and perfect graphs

Let G be a graph and let Ic (G) be its ideal of covers. In this section we characterize when G is perfect in terms of the irreducible representation of the Rees cone of Ic (G) and show an application to Ehrhart rings. Let G be a graph with vertex set X = {x1 , . . . , xn }. In what follows we shall always assume that G has no isolated vertices. We denote a complete subgraph of G with r vertices by Kr . The empty set is regarded as an independent set whose characteristic vector is the zero vector. Theorem 13.6.1 [86, 296] The following statements are equivalent : (a) G is a perfect graph. (b) The complement of G is perfect. (c) The independence polytope of G, i.e., the convex hull of the incidence vectors of the independent sets of G, is given by: C D (ai ) ∈ Rn+ | xi ∈Kr ai ≤ 1; ∀ Kr ⊂ G . The equivalence between (a) and (b) is due to Lov´asz [296] and this is called the weak perfect graph theorem. That (b) and (c) are equivalent is due to Fulkerson and Chv´ atal [86]. Lemma 13.6.2 Let G be a graph. Then the set F = {(ai ) ∈ Rn+1 | xi ∈V (Kr ) ai = (r − 1)an+1 } ∩ R+ (Ic (G)) is a facet of R+ (Ic (G)), where Kr is a complete subgraph of G. Proof. If Kr = ∅, then r = 0 and F = Hen+1 ∩ R+ (Ic (G)), which is a facet because e1 , . . . , en ∈ F . If r = 1, then F = Hei ∩ R+ (Ic (G)) for some / {i, n + 1} and there is 1 ≤ i ≤ n, which is a facet because ej ∈ F for j ∈

562

Chapter 13

at least one minimal vertex cover of G not containing xi . We may assume that V (Kr ) = {x1 , . . . , xr } and r ≥ 2. For each 1 ≤ i ≤ r there is a minimal vertex cover Ci of G not containing xi . Notice that Ci contains V (Kr )\{xi }. Let ui be the characteristic vector of Ci . Since rank(u1 , . . . , ur ) is r, the set {(u1 , 1), . . . , (ur , 1), er+1 , . . . , en } is linearly independent and contained in F , i.e., dim(F ) = n. Hence F is a facet of R+ (Ic (G)) because the hyperplane that deﬁnes F is a supporting hyperplane. 2 We regard K0 as the empty set with zero elements. A sum over an empty set is deﬁned to be 0. Proposition 13.6.3 Let J = Ic (G) be the ideal of covers of G. Then G is perfect if and only if the following equality holds C D R+ (J) = (ai ) ∈ Rn+1 | xi ∈Kr ai ≥ (r − 1)an+1 ; ∀ Kr ⊂ G . (13.10) Moreover this is the irreducible representation of R+ (J) if G is perfect. Proof. ⇒) The inclusion “⊂” is clear because any minimal vertex cover of G contains at least r − 1 vertices of any Kr . To show the reverse inclusion take a vector a = (ai ) satisfying b = an+1 = 0 and xi ∈Kr ai ≥ (r − 1)b; ∀ Kr ⊂ G ⇒ xi ∈Kr (ai /b) ≥ r − 1; ∀ Kr ⊂ G. This implication follows because by making r = 0 we get b > 0. We may assume that ai ≤ b for all i. Indeed if ai > b for some i, say i = 1, then we can write a = e1 + (a − e1 ). From the inequality ai = a1 + ai ≥ a1 + (r − 2)b ≥ 1 + (r − 1)b xi ∈Kr x1 ∈Kr

xi ∈Kr−1

it is seen that a − e1 belongs to the right-hand side of Eq. (13.10). Thus, if necessary, we may apply this observation again to a − e1 and so on till we get that ai ≤ b for all i. Hence, by Theorem 13.6.1(c), the vector γ = 1 − (a1 /b, . . . , an /b) belongs to the independence polytope of G. Thus we can write γ = λ1 w1 + · · · + λs ws ; (λi ≥ 0; i λi = 1), where w1 , . . . , ws are characteristic vectors of independent sets of G. Hence γ = λ1 (1 − u1 ) + · · · + λs (1 − us ), where u1 , . . . , us are characteristic vectors of vertex covers of G. For each i we can write ui = ui + i , where ui is the characteristic vector of a minimal vertex cover of G and i ∈ {0, 1}n. Therefore 1 − γ = λ1 u1 + · · · + λs us =⇒ a = bλ1 (u1 , 1) + · · · + bλs (us , 1) + bλ1 1 + · · · + bλs s .

Combinatorics of Symbolic Rees Algebras of Edge Ideals

563

Thus a ∈ R+ (J). If b = 0, clearly a ∈ R+ (J). Hence we get equality in Eq. (13.10), as required. The converse follows using similar arguments. To ﬁnish the proof notice that, by Lemma 13.6.2, the decomposition of Eq. (13.10) is irreducible. 2 Remark 13.6.4 Normaliz [68] determines the irreducible representation of a Rees cone. Thus Proposition 13.6.3 can be used to check whether a given graph is perfect. Deﬁnition 13.6.5 The clique clutter of a graph G, denoted by cl(G), is the clutter on V (G) whose edges are the maximal cliques of G (maximal with respect to inclusion). Deﬁnition 13.6.6 The incidence matrix A of a clutter is called perfect if the polytope deﬁned by the system x ≥ 0; xA ≤ 1 is integral. Theorem 13.6.7 ([24], [373, Corollary 83.1a(vii)]) Let C be a clutter and let A be its incidence matrix. Then C is balanced if and only if every submatrix of A is perfect. The vertex-clique matrix of a graph G is the incidence matrix of cl(G), the clique clutter of G. Theorem 13.6.8 ([296], [86]) Let A be the incidence matrix of a clutter. Then the following are equivalent: (a) The system x ≥ 0; xA ≤ 1 is TDI. (b) A is perfect. (c) A is the vertex-clique matrix of a perfect graph. Let v1 , . . . , vq be a set of points in Nn and let P = conv(v1 , . . . , vq ). The Ehrhart ring of the lattice polytope P is the K-subring of R[t] given by A(P) = K[{xa tb | a ∈ bP ∩ Zn }]. Corollary 13.6.9 Let A be a perfect matrix with column vectors v1 , . . . , vq . If there is x0 ∈ Rn such that all the entries of x0 are positive and vi , x0 = 1 for all i, then A(P) = K[xv1 t, . . . , xvq t]. a b Proof. The inclusion “⊃” is clear. To qshow the other inclusion take x t in A(P). Then we can write (a, b) = i=1 λi (vi , 1), where λi ≥ 0 for all i. Hence a, x0 = b. By Theorem 13.6.8 the system x ≥ 0; xA ≤ 1 is TDI. Therefore, applying Proposition 1.3.28, we have:

(a, b) = η1 (v1 , 1) + · · · + ηq (vq , 1) − δ1 e1 − · · · − δn en

(ηi ∈ N; δi ∈ N).

Then b = a, x0 = b − δ1 x0 , e1 − · · · − δn x0 , en . Using that x0 , ei > 0 for all i, we conclude that δi = 0 for all i, i.e., xa tb ∈ K[xv1 t, . . . , xvq t]. 2

564

Chapter 13

Exercises 13.6.10 Let A be an n × q integer matrix with column vectors v1 , . . . , vq . If the polyhedron {x| x ≥ 0; xA ≤ 1} is integral and R+ B ∩ Zn+1 = NB, where B = {(vi , 1)}qi=1 ∪ {−ei }ni=1 , then the system x ≥ 0; xA ≤ 1 is TDI. 13.6.11 Let G be a graph and let G be its complement. Then: (a) Ic (G) = ({xa | X \ supp(xa ) is a maximal clique of G}). (b) If G is perfect, then R+ (Ic (G)) is equal to C D (ai ) ∈ Rn+1 | xi ∈S ai ≥ (|S| − 1)an+1 ; ∀ S independent set of G . 13.6.12 Let G be a perfect graph with vertex set X and let β0 be its vertex independence number. Then there is a partition X1 , . . . , Xβ0 of X such that Xi is a clique of G for all i. If G is unmixed, then cl(G)∨ has a perfect matching, where G is the complement of G.

13.7

Perfect graphs and algebras of covers

Let G be a graph with vertex set X = {x1 , . . . , xn } and let I = I(G) be its edge ideal. The main purpose of this section is to study the symbolic Rees algebra of I and the Simis cone of I when G is a perfect graph, i.e., we study the algebra of vertex covers of G∨ . We show that the cliques of a perfect graph G completely determine both the Hilbert basis of the Simis cone and the symbolic Rees algebra of I(G). The Simis cone of I is denoted by Cn(I) (see Deﬁnition 13.2.1). Let C1 , . . . , Cs be the minimal vertex covers of G. For 1 ≤ k ≤ s let uk be the characteristic vector of Ck , i.e., uk = xi ∈Ck ei . If H is the minimal integral Hilbert basis of Cn(I), then Rs (I(G)) is equal to K[NH], the semigroup ring of NH (see Corollary 13.2.19). Next we describe H when G is perfect. Theorem 13.7.1 Let ω1 , . . . , ωp be the characteristic vectors of the nonempty cliques of a perfect graph G. If H = {(ω1 , |ω1 | − 1), . . . , (ωp , |ωp | − 1)}. Then NH = Cn(I) ∩ Zn+1 , that is, H is the integral Hilbert basis of Cn(I). Proof. The inclusion NH ⊂ Cn(I)∩Zn+1 is clear because any clique of size r intersects any minimal vertex cover in at least r − 1 vertices. Let us show the reverse inclusion. Let (a, b) be a minimal generator of Cn(I) ∩ Zn+1 , where 0 = a = (ai ) ∈ Nn and b ∈ N. Then (13.11) xi ∈Ck ai = a, uk ≥ b,

Combinatorics of Symbolic Rees Algebras of Edge Ideals

565

for all k. If b = 0 or b = 1, then (a, b) = ei for some i ≤ n or (a, b) = (ei + ej , 1) for some edge {xi , xj }, respectively. In both cases (a, b) ∈ H. Thus we may assume that b ≥ 2 and aj ≥ 1 for some j. Using Eq. (13.11) we obtain ai + ai = |a| ≥ b + ai = b + 1 − uk , a, (13.12) xi ∈Ck

xi ∈X\Ck

xi ∈X\Ck

for all k. Set c = |a| − b. Notice that c ≥ 1 because a = 0. Indeed if c = 0, from Eq. (13.12) we get xi ∈X\Ck ai = 0 for all k, i.e., a = 0, a contradiction. Consider the vertex-clique matrix of G : A = (1 − u1 · · · 1 − us ) , where 1 − u1 , . . . , 1 − us are regarded as column vectors. From Eq. (13.12) we get (a/c)A ≤ 1. Hence by Theorem 13.6.1(c) we obtain that a/c pbelongs to conv(ω0 , ω1 , . . . , ωp ), where ω = 0, i.e., we can write a/c = 0 i=0 λi ωi , where λi ≥ 0 for all i and i λi = 1. Thus we can write (a, c) = cλ0 (ω0 , 1) + · · · + cλp (ωp , 1). Using Theorem 13.6.8(a) it follows that the subring K[{xωi t|0 ≤ i ≤ p}] is normal. Hence there are η0 , . . . , ηp in N such that (a, c) = η0 (ω0 , 1) + · · · + ηp (ωp , 1). Thus |a| = η0 |ω0 | + · · · + ηp |ωp | and c = η0 + · · · + ηp = |a| − b, consequently: (a, b) = η0 (ω0 , |ω0 | − 1) + η1 (ω1 , |ω1 | − 1) + · · · + ηp (ωp , |ωp | − 1). Notice that there is u such that a, u = b; otherwise since aj ≥ 1, by Eq. (13.11) the vector (a, b) − ej would be in Cn(I) ∩ Zn+1 , contradicting the minimality of (a, b). Therefore from the equality p 0 = (a, b), (u , −1) = η0 + i=1 ηi (ωi , |ωi | − 1), (u , −1) we conclude that η0 = 0, i.e., (a, b) ∈ NH, as required.

2

Corollary 13.7.2 If G is a perfect graph, then Rs (I(G)) = K[xa tr | xa is square-free ; supp(xa ) = Kr+1 ; 0 ≤ r < n]. Proof. By Theorem 13.2.3, we have the equality Rs (I(G)) = K[NH], thus the formula follows from Theorem 13.7.1. 2 Lemma 13.7.3 Let G be a graph and let 0 = a = (ai ) ∈ Nn . If ai ∈ {0, 1} for all i and G[{xi | ai > 0}] = Kb+1 , then a is an irreducible b-cover of G∨ .

566

Chapter 13

+ Proof. By Lemma 13.6.2, the closed halfspace H(a,−b) must occur in the irreducible representation of R+ (Ic (G)). Hence a is an irreducible b-cover of G∨ by Lemma 13.2.27. 2

Corollary 13.7.4 [423] If G is a graph, then K[xa tr | xa square-free ; supp(xa ) = Kr+1 ; 0 ≤ r < n] ⊂ Rs (I(G)) with equality if and only if G is a perfect graph. Proof. The inclusion follows from Lemma 13.7.3. If G is a perfect graph, then by Corollary 13.7.2 the equality holds. Conversely if the equality holds, then by Lemmas 13.2.27 and 13.6.2 we have C D R+ (Ic (G)) = (ai ) ∈ Rn+1 | xi ∈Kr ai ≥ (r − 1)an+1 ; ∀ Kr ⊂ G . Hence an application of Proposition 13.6.3 gives that G is perfect.

2

Exercises 13.7.5 Let G be a graph with clique number ω(G) and chromatic number χ(G). Notice that the chromatic number is given by χ(G) = min{k| ∃ X1 , . . . , Xk independent sets of G whose union is X}. Prove that ω(G) ≤ χ(G). 13.7.6 If G is a perfect graph and C = G∨ , show that the image of ψ given in Remark 13.2.28 generates Rs (Ic (G∨ )). Hint Use that the irreducible b-covers of G∨ correspond to cliques of the graph G (see Corollary 13.7.4). Notice that Ic (G∨ ) is equal to I(G).

Chapter 14

Combinatorial Optimization and Blowup Algebras In this chapter we relate commutative algebra and blowup algebras with combinatorial optimization to gain insight on these research areas. A main goal here is to connect algebraic properties of blowup algebras associated to edge ideals with combinatorial and optimization properties of clutters and polyhedra. A conjecture of Conforti and Cornu´ejols about packing problems is examined from an algebraic point of view. We study max-ﬂow min-cut problems of clutters, packing problems, and integer rounding properties of systems of linear inequalities—and their underlying polyhedra—to analyze algebraic properties of blowup algebras and edge ideals (e.g., normality, normally torsion freeness, Cohen–Macaulayness, unmixedness). Systems with integer rounding properties and clutters with the max-ﬂow min-cut property come from linear optimization problems [372, 373]. The study of algebraic and combinatorial properties of edge ideal of clutters and hypergraphs is of current interest; see [119, 155, 206, 285, 326] and the references therein. A comprehensive reference for combinatorial optimization and hypergraph theory is the 3-volume book of Schrijver [373]. For a thorough study of clutters—that includes 18 conjectures in the area— from the point of view of combinatorial optimization, see [93]. In this chapter we make use of polyhedral geometry and combinatorial optimization to study blowup algebras and vice versa. We refer the reader to Chapter 1 for the undeﬁned terminology and notation regarding these areas. As a handy reference in Section 14.1 we introduce and ﬁx some of the notation and deﬁnitions that will be used throughout this chapter.

568

14.1

Chapter 14

Blowup algebras of edge ideals

In this section we introduce some of the notation and deﬁnitions that will be used throughout this chapter. In particular the irreducible representation of a Rees cone is introduced here. Let C be a clutter with vertex set X = {x1 , . . . , xn } and let R = K[X] be a polynomial ring over a ﬁeld K. The set of vertices and edges of C are denoted by V (C) and E(C), respectively. Let f1 , . . . , fq be the edges of C and let vk = xi ∈fk ei be the characteristic vector of fk for 1 ≤ k ≤ q. Recall that the edge ideal of C, denoted by I = I(C), is the ideal of R ; generated by all monomials xe = xi ∈e xi such that e is an edge of C. Thus, I is minimally generated by F = {xv1 , . . . , xvq }. In what follows we shall always assume that the height of I is at least 2. The blowup algebras and monomial algebras studied in this chapter are the following: (a) Rees algebra R[It] := R ⊕ It ⊕ · · · ⊕ I i ti ⊕ · · · ⊂ R[t], where t is a new variable, (b) extended Rees algebra R[It, t−1 ] := R[It][t−1 ] ⊂ R[t, t−1 ], (c) symbolic Rees algebra Rs (I) := R + I (1) t + I (2) t2 + · · · + I (i) ti + · · · ⊂ R[t], where I (i) is the ith symbolic power of I, (d) associated graded ring grI (R) := R/I ⊕ I/I 2 ⊕ · · · ⊕ I i /I i+1 ⊕ · · · R[It] ⊗R (R/I), with multiplication (a + I i+1 )(b + I j+1 ) = ab + I i+j+1

(a ∈ I i , b ∈ I j ),

(e) monomial subring K[F ] = K[xv1 , . . . , xvq ] ⊂ R spanned by F = {xv1 , . . . , xvq }, (f) homogeneous monomial subring K[F t] = K[xv1 t, . . . , xvq t] ⊂ R[t] spanned by F t, (g) homogeneous monomial subring K[F t ∪ {t}] = K[xv1 t, . . . , xvq t, t] ⊂ R[t] spanned by F t ∪ {t}, (h) homogeneous monomial subring S = K[xw1 t, . . . , xwr t] ⊂ R[t],

Combinatorial Optimization and Blowup Algebras

569

where {w1 , . . . , wr } is the set of all α ∈ Nn such that 0 ≤ α ≤ vi and we allow xvi to be an arbitrary monomial, (i) Ehrhart ring A(P) = K[{xa ti | a ∈ Zn ∩ iP; i ∈ N}] ⊂ R[t] of the lattice polytope P = conv(v1 , . . . , vq ), and (j) Stanley–Reisner ring K[ΔC ] = R/I(C). of the independence complex ΔC . This ring is also called the edge ring of C. The incidence matrix of C will be denoted by A; it is the n × q matrix with column vectors v1 , . . . , vq . We shall always assume that the rows and columns of A are diﬀerent from zero. Consider the set A = {e1 , . . . , en , (v1 , 1), . . . , (vq , 1)} ⊂ Nn+1 , where ei is the ith unit vector. The Rees cone of I, will be denoted by R+ (I), it is the rational polyhedral cone generated by A . The closed halfspace He+i occurs in the irreducible representation of R+ (I) for i = 1, . . . , n + 1 (see Exercise 14.2.34). Thus, by Theorem 1.4.2, there is a unique irreducible representation R+ (I) = He+1 ∩ · · · ∩ He+n+1 ∩ Hα+1 ∩ · · · ∩ Hα+r such that 0 = αi ∈ Qn+1 , αi , en+1 = −1 for all i, and none of the closed halfspaces can be omitted from the intersection. Let p1 , . . . , ps be the minimal primes of the edge ideal I = I(C) and let Ck = {xi | xi ∈ pk }

(k = 1, . . . , s)

be the corresponding minimal vertex covers of C (see Proposition 13.1.2). The set of column vectors of A will be denoted by A = {v1 , . . . , vq } and Q(A) will denote the set covering polyhedron Q(A) := {x| x ≥ 0; xA ≥ 1}. Notation Let dk be the unique positive integer such that dk αk has relatively prime integral entries. We set k = dk αk for k = 1, . . . , r. Theorem 14.1.1 The irreducible representation of the Rees cone is: (14.1) R+ (I) = He+1 ∩ · · · ∩ He+n+1 ∩ H+1 ∩ · · · ∩ H+r where k = −en+1 + xi ∈Ck ei for 1 ≤ k ≤ s. Moreover all the vertices of Q(A) are integral if and only if r = s.

570

Chapter 14

Proof. By Proposition 13.1.2 uk = xi ∈Ck ei is a vertex of Q(A). Since k + en+1 = (uk , 0) for k = 1, . . . , s, by Theorem 1.4.3, the result follows. 2 Notation In the sequel we shall always assume that u1 , . . . , us are the vectors given by uk = xi ∈Ck ei , 1 , . . . , s are the vectors deﬁned as k = −en+1 +

ei ,

xi ∈Ck

and that 1 , . . . , r are the vectors of Eq. (14.1). Notice that dk = 1 for k = 1, . . . , s and dk = −k , en+1 for 1 ≤ k ≤ r. The Simis cone of I is the polyhedral cone: Cn(I) = He+1 ∩ · · · ∩ He+n+1 ∩ H+1 ∩ · · · ∩ H+s , see Deﬁnition 13.2.1. The Simis and Rees cones are related to the normality and torsion freeness of I (Theorem 14.2.2 and Proposition 14.2.3).

14.2

Rees algebras and polyhedral geometry

In this section we describe when the integral closure of a Rees algebra of an edge ideal is equal to the symbolic Rees algebra in combinatorial and algebraic terms. If the Rees algebra is normal, we describe the primary decomposition of the zero ideal of the associated graded ring. We show that certain properties of clutters and edge ideals are closed under taking minors and Alexander duals. Lemma 14.2.1 If 1 ≤ j ≤ r and j , en+1 = −1, then 1 ≤ j ≤ s. In particular r = s if and only if j , en+1 = −1 for i = 1, . . . , r Proof. Let j = (a1 , . . . , an , −1). Since R+ (I) ⊂ H+j we get ai ≥ 0 for i = 1, . . . , n. Consider the ideal p of R generated by the set of xi such that ai > 0. Note that I ⊂ p because j , (vi , 1) ≥ 0 for all 1 ≤ i ≤ n. For simplicity of notation assume that j = (a1 , . . . , am , 0, . . . , 0, −1), where ai > 0 for all i. Take an arbitrary vector v ∈ A such that v ∈ Hj , then v satisﬁes the equation a1 y1 + · · · + am ym = yn+1 . Observe that v also satisﬁes the equation y1 + · · · + ym = yn+1 . Using that Hj contains n linearly independent vectors from A we conclude that Hj = Hb , where b = e1 + · · · + em − en+1 . Hence j = b and consequently ai = 1 for all i. It remains to show that p is a minimal prime of I, and this follows from the irreducibility of Eq. (14.1). 2

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Theorem 14.2.2 R[It] = Rs (I) ⇐⇒ i , en+1 = −1 for i = 1, . . . , r. Proof. The equality R[It] = Rs (I) holds if and only if Zn+1 ∩ R+ (I) is equal to Zn+1 ∩ Cn(I) if and only if the Rees and Simis cones are equal. By the irreducibility of the representations of both cones it follows that R[It] = Rs (I) if and only if r = s. Now, to ﬁnish the proof recall that by Lemma 14.2.1, r = s if and only if i , en+1 = −1 for i = 1, . . . , r. 2 Proposition 14.2.3 If I is a monomial ideal, then R[It] is normal if and only if any of the following two equivalent conditions hold: (a) A is a Hilbert basis, i.e., NA = Zn+1 ∩ R+ (I). (b) I is normal, i.e., I i = I i for all i ≥ 1. Proof. The Rees algebra of I can be written as R[It] = = =

K[{x1 , . . . , xn , xv1 t, . . . , xvq t}] K[{xa tb | (a, b) ∈ NA }] R ⊕ It ⊕ I 2 t2 ⊕ · · · ⊕ I i ti ⊕ · · · .

By Theorem 9.1.1 and Exercise 4.3.49, the integral closure of R[It] in its ﬁeld of fractions can be expressed as R[It] = K[{xa tb | (a, b) ∈ Zn+1 ∩ R+ (I)}] = R ⊕ It ⊕ I 2 t2 ⊕ · · · ⊕ I i ti ⊕ · · · , where I i is the integral closure of I i . Thus, R[It] is normal if and only if 2 A is a Hilbert basis if and only if I i = I i for all i ≥ 1. Let p be a prime ideal of R. In what follows we denote the localization of R[It] at the multiplicative set R \ p by R[It]p Lemma 14.2.4 If p ∈ Spec(R), then pR[It]p ∩ R = p. Proof. It is left as an exercise.

2

Lemma 14.2.5 If B = R[It] and p is a minimal prime of I, then pBp ∩ B is a minimal prime of IB. Proof. From the equality pBp = p(Rp [pRp t]) = pRp + p2 Rp t + · · · we get that pBp is a prime ideal. Hence its contraction pBp ∩B is also a prime ideal. To prove the minimality take P ∈ Spec(B) such that IB ⊂ P ⊂ pBp ∩ B. Contracting to R and using Lemma 14.2.4 we get P ∩ R = p. From this equality it follows that P Bp ∩ B = P . Therefore pBp ∩ B ⊂ P Bp ∩ B = P =⇒ P = pBp ∩ B.

2

572

Chapter 14 (dk )

Notation In what follows Jk (dk )

Jk

will denote the ideal of R[It] given by

= ({xa tb | (a, b), k ≥ dk }) ∩ R[It]

(k = 1, . . . , r)

and Jk will denote the ideal of R[It] given by Jk = ({xa tb | (a, b), k > 0}) ∩ R[It]

(k = 1, . . . , r), (d )

where dk = −k , en+1 . In general the ideal Jk k might not be equal to (1) the dk -th symbolic power of Jk ; see [415]. If dk = 1 we have Jk = Jk . By Lemma 14.2.1 we have that dk = 1 if and only if 1 ≤ k ≤ s. Proposition 14.2.6 J1 , . . . , Jr are distinct height one prime ideals of R[It] that contain IR[It] and Jk is equal to pk R[It]pk ∩ R[It] for k = 1, . . . , s. If Q(A) is integral, then rad(IR[It]) = J1 ∩ · · · ∩ Js . Proof. IR[It] is clearly contained in Jk for all k by deﬁnition of Jk . To show that Jk is a prime ideal of height one it suﬃces to notice that the right-hand side of the isomorphism: R[It]/Jk K[{xa tb ∈ R[It]| (a, b), k = 0}] is an n-dimensional integral domain, because Fk = R+ (I) ∩ Hk is a facet of the Rees cone for all k. The prime ideals J1 , . . . , Jr are distinct because the dimension of R+ (I) is n + 1 and F1 , . . . , Fr are distinct facets of the Rees cone. Set Pk = pk R[It]pk ∩ R[It] for 1 ≤ k ≤ s. By Lemma 14.2.5 this ideal is a minimal prime of IR[It] and admits the following description = pk Rpk [pk Rpk t] ∩ R[It] ∩ I i )ti + · · · = pk + (p2k ∩ I)t + (p3k ∩ I 2 )t2 + · · · + (pi+1 k Notice that xa ∈ pb+1 if and only if a, xi ∈Ck ei ≥ b + 1. Hence Jk = Pk . k Assume that Q(A) is integral, i.e., r = s. Take xα tb ∈ Jk for all k. Using Eq. (14.1) it is not hard to see that (α, b + 1) ∈ R+ (I); that is, xα tb+1 is in R[It] and xα tb+1 ∈ I b+1 tb+1 . It follows that xα tb is a monomial in the radical of IR[It]. This proves the asserted equality. 2 Pk

The following formulas, pointed out to us by Vasconcelos, show the diﬀerence between the symbolic Rees algebra and the normalization of I.

Rs (I) =

s k=1

R[It]pk ∩ R[t];

R[It] =

r

R[It]pk ∩ R[t],

k=1

where pk = Jk ∩ R for k = 1, . . . , r. These representations are related to the so-called Rees valuations of I; see [414, Chapter 8].

Combinatorial Optimization and Blowup Algebras

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Theorem 14.2.7 [185, 188] The following are equivalent: (a) Q(A) is integral. (b) R+ (I) = He+1 ∩ · · · ∩ He+n+1 ∩ H+1 ∩ · · · ∩ H+s , i.e., r = s. (c) Rs (I) = R[It]. (d) The minimal primes of IR[It] are of the form pR[It]p ∩ R[It], where p is a minimal prime of I. Proof. (a) ⇔ (b): It follows at once from Theorem 14.1.1. (b) ⇔ (c): It follows from Theorem 14.2.2 and Lemma 14.2.1. (a) ⇒ (d): It follows from Proposition 14.2.6. (d) ⇒ (b): From Proposition 14.2.6, J1 , . . . , Jr are minimal primes of IR[It] and Jk is equal to pk R[It]pk ∩R[It] for k = 1, . . . , s. Thus r = s. 2 Corollary 14.2.8 If Q(A) has only integral vertices and C has n vertices, then α0 (C a ) ≤ n − 1 for all indecomposable parallelizations C a of C. Proof. By Theorem 14.2.7 we have the equality R[It] = Rs (I). Take any indecomposable parallelization C a of C and consider the monomial m = xa tb , where b = α0 (C a ). By Theorem 13.2.15 m is a minimal generator of Rs (I). Now, according to Proposition 12.5.1, a minimal generator of R[It] has degree in t at most n − 1, i.e., b ≤ n − 1. 2 Example 14.2.9 Let I = I(C) = (x1 x2 x5 , x1 x3 x4 , x2 x3 x6 , x4 x5 x6 ) be the edge ideal associated to the clutter: x1 s

p p p p p pp pp p p p p p p p p pp p p p p p p p p p pp p p p p p s x2 p p pp p p pp pp p pp p p pp p p p p p pp p pp p p p p p p p p p p p p pp p pp p pp p p p pp pp p pp p p s x5 ppp pp pp p ppp p pp pp p pp p p p p p pp p p x p pp ppp pp p ppp p pp p p pp pp pp 4 p p p p pp pp ppp pp p p p pp pp pp p sp p pp pp p p p p p p pp p s p pp pp pp pp x6 p p pp p pp p p p pp p p pp p p pp pp p p p p pp p pp p p p C p pp ppp p p p pp p s

x3

This clutter, usually denoted by Q6 in the literature, plays an important role in combinatorial optimization [93]. Using Normaliz [68], Theorems 1.4.9, 14.2.7, and Remark 13.2.23, we obtain that Q(A) is integral, R[It] Rs (I) = R[It] = R[It][x1 · · · x6 t2 ],

574

Chapter 14

R[It] is not normal, the minimal primes of I are p2 = (x2 , x4 ), p3 = (x3 , x5 ), p1 = (x1 , x6 ), p4 = (x1 , x2 , x5 ), p5 = (x1 , x3 , x4 ), p6 = (x2 , x3 , x6 ), p7 = (x4 , x5 , x6 ), and C ∨ , the blocker of C, has the max-ﬂow min-cut property. The dual hypergraph H of a hypergraph H = (V, E) is the hypergraph with vertex set E and edges all sets {e ∈ E| v ∈ e} for v ∈ V . So the incidence matrix of H is the transpose of the incidence matrix of H. The clutter Q6 is the dual hypergraph of K6 , the complete graph on 6 vertices. Recall that the analytic spread of I, denoted by (I), is given by (I) = dim R[It]/mR[It]. This number satisﬁes ht(I) ≤ (I) ≤ dim R [412, Corollary 5.1.4]. The analytic spread of a monomial ideal I can be computed in terms of the Newton polyhedron of I [42]. Deﬁnition 14.2.10 The ring F (I) = R[It]/mR[It] is called the special ﬁber of I. Corollary 14.2.11 If Q(A) is integral, then (I) < n. Proof. By Theorem 14.2.7 we have r = s. If (I) = n, then the height of mR[It] is equal to 1. Hence there is a height one prime ideal P of R[It] such that IR[It] ⊂ mR[It] ⊂ P . By Proposition 14.2.6 the ideal P has the form pk R[It]pk ∩ R[It], this readily yields a contradiction. 2 This corollary also follows directly from [312, Theorem 3]. Proposition 14.2.12 If C is a uniform clutter, then (I) = rank(A). Proof. Since deg(xvi ) = d for all i. There are isomorphisms R[It]/mR[It] K[xv1 t, . . . , xvq t] K[xv1 , . . . , xvq ]. Thus, by Corollary 8.2.21, we get that (I) is the rank of A.

2

Theorem 14.2.13 inf i {depth(R/I i )} ≤ dim(R) − (I) with equality if the associated graded ring grI (R) is Cohen–Macaulay. The inequality is due to Burch [76]. If grI (R) is C–M, the equality comes from [131]. By a result of Brodmann [55], depth R/I k is constant for k 0. Brodmann improved the Burch inequality by showing that the constant value is bounded by dim(R) − (I). For a study of the initial and limit behavior of the numerical function f (k) = depth R/I k , see [223].

Combinatorial Optimization and Blowup Algebras

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Lemma 14.2.14 If xv1 = x1 xv1 , . . . , xvp = x1 xvp and x1 ∈ / supp(xvi ) for i > p, where x1 is a variable in Ck for some 1 ≤ k ≤ s, then there is xvj such that supp(xvj ) ∩ Ck = ∅.

Proof. If supp(xvj ) ∩ Ck = ∅ for all j, then Ck \ {x1 } is a vertex cover of 2 C, a contradiction because Ck is a minimal vertex cover. Proposition 14.2.15 If 1 ≤ i ≤ s, then Ji is the Ji -primary component of IR[It]. Proof. We set P = Ji . It suﬃces to show the equality R[It] ∩ IR[It]P = P . In general the left-hand side is contained in the right-hand side. To show the reverse inclusion we may assume that P can be written as P = (x1 , . . . , xm , xv1 t, . . . , xvp t)R[It], where p = (x1 , . . . , xm ) is a minimal prime of I. Set C = {x1 , . . . , xm }. Case (I): Consider xk with 1 ≤ k ≤ m. By Lemma 14.2.14 there is j such that xvj = xk xα and supp(xα ) ∩ C = ∅. Thus since xα is not in P (because of the second condition) we obtain xk ∈ R[It] ∩ IR[It]P . Case (II): Now, consider xvk t with 1 ≤ k ≤ p. Since (vk , 1), e1 + · · · + em − en+1 ≥ 1, the monomial xvk contains at least two variables in C. Thus we may assume that x1 , x2 are in the support of xvk . Again by Lemma 14.2.14 there are j, such that xvj = x1 xα , xv = x2 xγ , and the support of xα and xγ disjoint from C. Hence the monomial xvk xα+γ t belongs to I 2 t and xα+γ is not in 2 P . Writing xvk t = (xvk xα+γ t)/xα+γ , we get xvk t ∈ R[It] ∩ IR[It]P . (dk )

Lemma 14.2.16 rad(Jk

) = Jk for 1 ≤ k ≤ r. (d )

Proof. By construction one has rad(Jk k ) ⊂ Jk . The reverse inclusion (d ) follows by noticing that if xa tb ∈ Jk , then (xa tb )dk ∈ Jk k . 2 Theorem 14.2.17 [258, Theorem 1.11] If the height of I ≥ 2, then the following are equivalent: (i)

grI (R) is torsion-free over R/I.

(ii) grI (R) is reduced. (iii) R[It] is normal and Cl(R[It]), the divisor class group of R[It], is a free abelian group whose rank is the number of minimal primes of I. Proposition 14.2.18 ([70], [188]) If R[It] is normal, then (d1 )

IR[It] = J1

(d2 )

∩ J2

∩ · · · ∩ Jr(dr ) .

576

Chapter 14

Proof. “⊂”: Assume xα tb ∈ IR[It]. Since xα ∈ I b+1 we readily obtain (α, b + 1) ∈ NA . In particular we get (α, b + 1) ∈ R+ (I). Therefore 0 ≤ (α, b + 1), k = (α, b), k − dk and consequently xα tb ∈ J (dk ) for 1 ≤ k ≤ r. “⊃”: Assume xα tb ∈ J (dk ) for all k. Since the element (α, b + 1) is in R+ (I) ∩ Zn+1 and using that R[It] is normal yields (α, b + 1) ∈ NA . It 2 follows that xα tb ∈ I b+1 tb ⊂ IR[It]. Since the program Normaliz [68] computes the irreducible representation of the Rees cone and the integral closure of R[It], the following result is an eﬀective criterion for the reducedness of the associated graded ring. Theorem 14.2.19 ([258], [147]) The following are equivalent : (a) R[It] = Rs (I). (b) grI (R) is reduced. (c) R[It] is normal and i , en+1 = −1 for i = 1, . . . , r. Proof. (a) ⇒ (b) By Corollary 4.3.26 the Rees algebra R[It] is normal. Thus using Theorem 14.2.7 we obtain that r = s and that the minimal primes of IR[It] are J1 , . . . , Js . Hence from Proposition 14.2.18 we have (d1 )

IR[It] = J1

(d2 )

∩ J2

∩ · · · ∩ Js(ds ) .

Therefore IR[It] is a radical ideal because di = 1 for i = 1, . . . , s. Since grI (R) R[It]/IR[It] we get that grI (R) is reduced. (b) ⇒ (c) By Theorem 14.2.17, the ring R[It] is normal and Cl(R[It]), the divisor class group of R[It], is a free abelian group whose rank is the number of minimal primes of I, i.e., the rank of Cl(R[It]) is equal to s. On the other hand by Proposition 12.7.3, the rank of Cl(R[It]) is equal to r. Thus s = r and i , en+1 = −1 for i = 1, . . . , r. (c) ⇒ (a) By Lemma 14.2.1, r = s. Hence, by Theorem 14.2.7, we get R[It] = Rs (I). Thus R[It] = Rs (I). 2 Example 14.2.20 Let I = (x1 x5 , x2 x4 , x3 x4 x5 , x1 x2 x3 ). Using Normaliz [68] with the input ﬁle: 4 5 1 0 0 1 3

0 1 0 1

0 0 1 1

0 1 1 0

1 0 1 0

Combinatorial Optimization and Blowup Algebras

577

we get the output ﬁle: 9 generators of integral closure of Rees algebra: 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 1 1 0 1 0 1 0 1 0 0 1 1 1 1 1 1 1 0 0 1 10 support 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0 1 1 1

hyperplanes: 1 1 1 -1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 -1 0 1 0 0 0 0 1 0 0 0 1 -1 1 0 0 -1

The ﬁrst block of the output ﬁle shows the list of generators of R[It]. Thus R[It] is normal. The second block gives the irreducible representation of the Rees cone of I. This means that all the i ’s that occur in the irreducible representation of the Rees cone R+ (I) (see Eq. (14.1)) have its last entry equal to −1. Thus, by Theorem 14.2.19, the ring grI (R) is reduced. Proposition 14.2.21 If R[It] = Rs (I), then R[Ic (C)t] = Rs (Ic (C)). Proof. We set J = Ic (C). Using Theorem 14.2.7 one has R+ (I) = He+1 ∩ · · · ∩ He+n+1 ∩ H+1 ∩ · · · ∩ H+s , where i = (ui , −1) for all i. Then, by Corollary 1.1.30, we get: ! ! ! ! q n n+1 s + + Hei H(vi ,1) = R+ e i + R+ (ui , −1) . (14.2) i=1

i=1

i=1

i=1

By Theorem 14.2.7 we need only show the equality ! ! q n+1 + R+ (J) = He+i H(v . i ,−1) i=1

i=1

578

Chapter 14

Clearly the left-hand side is contained in the right-hand side. To show the reverse inclusion take (α, b) in the right-hand side. Then 0 ≤ (α, b), (vi , −1) = (α, −b), (vi , 1) for all i. Hence using Eq. (14.2) we can write (α, −b) =

s

λi (ui , −1) +

i=1

n+1

μi e i

(λi ; μj ≥ 0).

(14.3)

i=1

Since λ1 + · · · + λs − b = μn+1 ≥ 0, there are λ1 , . . . , λs such that 0 ≤ λi ≤ λi ; b = λ1 + · · · + λs ; μn+1 = (λ1 − λ1 ) + · · · + (λs − λs ). Therefore using Eq. (14.3) we conclude (α, b) =

s i=1

λi (ui , 1) +

n i=1

μi e i +

s

(λi − λi )ui ∈ R+ (J).

2

i=1

Corollary 14.2.22 [93, Theorem 1.17] If Q(A) is integral and A∨ is the incidence matrix of C ∨ , then Q(A∨ ) is integral. Proof. It follows at once from Theorem 14.2.7 and Proposition 14.2.21. 2 The following notion of minor of an edge ideal was introduced in [188]. It is inspired by the notion of a minor in combinatorial optimization [93] and is consistent with the terminology of Section 6.5 (Deﬁnition 6.5.3). Deﬁnition 14.2.23 A minor of I is an ideal (0) I R obtained from I by making any sequence of variables equal to 1 or 0 in F = {xv1 , . . . , xvq }. The ideal I is considered itself a minor. A minor of C is a clutter C that corresponds to a minor (0) I R of I. The clutter C is itself a minor. If C is a minor of C and E(C ) = E(C), we call C a proper minor. If I ⊂ R is a minor of I, notice that C is obtained from I by considering G(I ), the unique set of square-free monomials that minimally generate I . Indeed if G(I ) = {xα1 , . . . , xαm }, then I deﬁnes a clutter C (resp. matrix A ) whose vertices are the variables of R and whose edges (resp. columns) are the supports of the monomials xαi (resp. the column vectors αi ). We can also take as vertex set of C , the set of variables that occur in xα1 , . . . , xαm . This choice of vertex set is useful in induction arguments, where one needs to reduce the number of variables. Proposition 14.2.24 If I i = I (i) for some i ≥ 2 and J = I is a minor of I, then J i = J (i) .

Combinatorial Optimization and Blowup Algebras

579

Proof. Assume that J is the minor obtained from I by making x1 = 0. Take xα ∈ J (i) . Then xα ∈ I (i) = I i because J ⊂ I. Thus xα ∈ I i . Since x1 ∈ / supp(xα ) it follows that xα ∈ J i . This proves J (i) ⊂ J i . The other inclusion is clear because J (i) is integrally closed and J i ⊂ J (i) . Assume that J is the minor obtained from I by making x1 = 1. Take xα ∈ J (i) . Notice that xi1 xα ∈ I (i) = I i . Indeed if x1 ∈ pk , then xi1 ∈ pik , and / pk , then J ⊂ pk and xα ∈ pik . Thus xi1 xα ∈ I (i) . Hence (xi1 xα )p ∈ I ip if x1 ∈ for some p ∈ N+ . Since x1 ∈ / supp(xα ) it follows that xα ∈ J i . 2 Corollary 14.2.25 (a) If Rs (I) = R[It], then Rs (I ) = R[I t] for any minor I of I. (b) The integrality of Q(A) is preserved under taking minors and parallelizations. Proof. (a): It follows from Proposition 14.2.24. (b): This part follows from part (a), Theorem 14.2.7, and [373, p. 1390]. 2 Proposition 14.2.26 Let C ∨ be the blocker of C. If R[It] is equal to Rs (I) and |C ∩ B| ≤ 2 for C ∈ C and B ∈ C ∨ , then R[It] is normal. Proof. Let xa tb = xa1 1 · · · xann tb ∈ R[It] be a minimal generator, that is (a, b) cannot be written as a sum of two non-zero integral vectors in the Rees cone R+ (I). We may assume ai ≥ 1 for 1 ≤ i ≤ m, ai = 0 for i > m, and b ≥ 1. Case (I): (a, b), i > 0 for all i. The vector γ = (a, b) − e1 satisﬁes γ, i ≥ 0 for all i, that is γ ∈ R+ (I). Thus since (a, b) = e1 + γ we derive a contradiction. Case (II): (a, b), i = 0 for some i. We may assume {i | (a, b), i = 0} = {1 , . . . , p }. Subcase (II.a): ei ∈ H1 ∩ · · · ∩ Hp for some 1 ≤ i ≤ m. It is not hard to verify that the vector γ = (a, b) − ei satisﬁes γ, k ≥ 0 for all 1 ≤ k ≤ s. Thus γ ∈ R+ (I), a contradiction because (a, b) = ei + γ. Subcase (II.b): ei ∈ / H1 ∩ · · · ∩ Hp for all 1 ≤ i ≤ m. Since the vector (a, b) belongs to the Rees cone it follows that we can write (a, b) = λ1 (v1 , 1) + · · · + λq (vq , 1)

(λi ≥ 0).

(∗)

By the choice of xa tb we may assume 0 < λ1 < 1. Set γ = (a, b) − (v1 , 1) and notice that by Eq. (∗) this vector has nonnegative entries. We claim that γ is in the Rees cone. Since by hypothesis one has that j = uj (see Theorem 14.2.7) and 0 ≤ (v1 , 1), j ≤ 1 for all j we readily obtain

(a, b), k − (v1 , 1), k = 0 if 1 ≤ k ≤ p, γ, k = (a, b), k − (v1 , 1), k ≥ 0 otherwise. Thus γ ∈ R+ (I) and (a, b) = (v1 , 1)+γ. As a result γ = 0 and (a, b) ∈ R[It], as desired. 2

580

Chapter 14

Normality criteria and Rees cones For each proper face G of the rational polyhedral cone R+ A we set AG = {vi | vi ∈ G} and FG = {xvi | vi ∈ G}. If G is a proper face of R+ A, there is a supporting hyperplane Ha in Rn such that G = R+ A ∩ Ha = ∅, R+ A ⊂ Ha , and R+ A ⊂ Ha+ . Notice that AG = {vi ∈ A | vi , a = 0} = ∅ and FG = {xvi | vi , a = 0} = ∅. Lemma 14.2.27 If NA is a normal semigroup (resp. A is a Hilbert basis), then NAG is a normal semigroup (resp. AG is a Hilbert basis) for each proper face G of R+ A. Proof. Assume that NA is normal. Let α ∈ R+ AG ∩ ZAG . We may assume without loss of generality that vi , a = 0 for 1 ≤ i ≤ p and vi , a > 0 for p + 1 ≤ i ≤ q. Using that α ∈ R+ A ∩ ZA = NA we can write: α = λ1 v1 + · · · + λq vq = a1 v1 + · · · + ap vp , where λi ∈ N for 1 ≤ i ≤ q and ai ∈ R+ for 1 ≤ i ≤ p. Taking inner products we get the equality: 0 = a1 v1 , a + · · · + ap vp , a = λp+1 vp+1 , a + · · · + λq vq , a. Since the λi ’s are nonnegative and vi , a > 0 for p + 1 ≤ i ≤ q, one has λp+1 = · · · = λq = 0. Therefore α ∈ NAG . This shows that NAG is normal. The other assertion is shown similarly. 2 Lemma 14.2.28 If K[F ] is normal, then K[FG ] is normal for any proper face G of R+ A. Proof. Notice K[F ] = K[NA] and K[FG ] = K[NAG ]. By Corollary 9.1.3 K[F ] is normal if and only if NA is normal. Thanks to Lemma 14.2.27 the 2 semigroup is NAG is normal. Therefore K[FG ] is normal. The converse of Lemma 14.2.27 fails as the next example shows. Example 14.2.29 If A = {(1, 1, 0), (1, 0, 1), (0, 1, 1)} and 1 = (1, 1, 1), then 1 is in R+ A ∩ Z3 and 1 ∈ NA. Thus A is not a Hilbert basis. By Proposition 1.1.23 any proper face G of R+ A is a cone generated by a proper subset of A. It follows that AG is a Hilbert basis. Proposition 14.2.30 If K[NAG ] is normal for every proper face G of the Rees cone R+ (I) and there is a ﬁxed integer 1 ≤ i ≤ n such that the ith entry of j is either 0 or 1 for all 1 ≤ j ≤ r, then R[It] is normal.

Combinatorial Optimization and Blowup Algebras

581

Proof. Recall that R[It] = K[NA ]. Let (α, b) be an element in the minimal integral Hilbert basis of the aﬃne semigroup NA , where α ∈ Nn and b ∈ N. It suﬃces to prove that (α, b) belongs to NA . Case (I): Assume (α, b) ∈ rb(R+ (I)). By Theorem 1.1.44(b) the relative boundary of R+ (I) is given by: rb(R+ (I)) =

n+1 (

! R+ (I) ∩ Hei

i=1

∪

r (

! R+ (I) ∩ Hi

.

i=1

Hence (α, b) is in some facet of the Rees cone and by hypothesis it follows readily that (α, b) is in NA . Case (II): If (α, b) ∈ ri(R+ (I)), then (α, b), j > 0 for 1 ≤ j ≤ r and (α, b), ek > 0 for 1 ≤ k ≤ n + 1. For simplicity of notation we may assume i = 1. Write (α, b) = (α1 , . . . , αn , b). Notice (α, b) = β + e1 , where β = (α1 − 1, α2 , . . . , αn , b). Therefore β, j = (α, b), j − e1 , j ≥ 0 for all j, that is, β ∈ NA . Since e1 ∈ NA , we obtain a contradiction.

2

Theorem 14.2.31 If Q(A) is integral, then R[It] is normal if and only if K[NAG ] is normal for any facet G of R+ (I). Proof. By Theorem 14.1.1 r = s and the vectors 1 , . . . , r have their ﬁrst n entries in {0, 1} because i = ui for i = 1, . . . , s. Hence the result follows from Lemma 14.2.28 and Proposition 14.2.30. 2

Exercises 14.2.32 Prove that grI (R) is reduced if and only if Cn(I) ∩ Zn+1 = NA . 14.2.33 Consider a sequence 1 ≤ i1 < · · · < is ≤ n of integers. Prove that the following two conditions are equivalent: (a) ei1 + · · · + eis is a vertex of Q(A). (b) F = R+ (I)∩H(a,−1) is a facet of the Rees cone for some vector a = (ai ) in Zn such that {i| ai = 0} = {i1 , . . . , is }. 14.2.34 If height of I ≥ 2, prove the following assertions: (a) dim(R+ (I)) = n + 1. (b) Hei ∩ R+ (I) is a facet of R+ (I) for i = 1, . . . , n + 1. (c) He+1 , . . . , He+n+1 occur in the irreducible representation of R+ (I).

582

Chapter 14

14.2.35 Let A be a matrix with entries in Nn and let A be the matrix obtained from A by adjoining a row of 1’s. Prove that the vertices of Q(A) and Q(A ) are related by vert(Q(A )) = {en+1 } ∪ {(β, 0)| β ∈ vert(Q(A))}. In particular Q(A) is integral if and only if Q(A ) is integral. 14.2.36 Let C be a uniform clutter with n vertices and let A be its incidence matrix. If A has rank n and I = I(C), then grI (R) is not reduced. 14.2.37 Let I = (xv1 , . . . , xv4 ), where v1 , . . . , v4 denote the column vectors of the matrix: ⎛ ⎞ 2 1 2 1 ⎜ 9 6 8 4 ⎟ ⎟ A=⎜ ⎝ 0 0 1 1 ⎠. 1 6 1 2 Prove that R[It] is not normal and det(A) = ±1. 14.2.38 Let L1 , L2 be monomial ideals with disjoint sets of variables. If L1 , L2 are generated by monomials of degrees d1 and d2 , respectively, then (L1 + L2 ) = (L1 ) + (L2 ), where (Li ) is the analytic spread of Li . 14.2.39 Let I = I(Q6 ) = (x1 x2 x5 , x1 x3 x4 , x2 x3 x6 , x4 x5 x6 ) be the edge ideal of the clutter Q6 . Use Macaulay2 to show the equality I 2 = (I 2 , x1 x2 · · · x6 ). Prove that I 3 /I I 2 has four generators and prove that I (i) = II (i−1) for i ≥ (I) = 4, where (I) is the analytic spread of I. 14.2.40 Let I = (x6 x7 x8 , x2 x4 x7 , x1 x2 x6 , x1 x3 x8 , x1 x3 x5 ) be the edge ideal of a clutter C. Prove that R[It] R[It] = Rs (I), ht(I) = 3 and C satisﬁes the K¨onig property. 14.2.41 Let I = (xv1 , xv2 , xv3 , xv4 , xv5 ), where the vi ’s are: v1 = (1, 1, 1, 1, 0, 0, 0, 0, 0, 0), v2 = (1, 0, 0, 0, 1, 1, 1, 0, 0, 0), v3 = (0, 1, 0, 0, 1, 0, 0, 1, 1, 0), v4 = (0, 0, 1, 0, 0, 1, 0, 1, 0, 1), v5 = (0, 0, 0, 1, 0, 0, 1, 0, 1, 1). Prove that the Rees algebra R[It] is not normal, whereas for every proper face G of the Rees cone K[NAG ] is normal. See Proposition 14.2.30.

Combinatorial Optimization and Blowup Algebras

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14.2.42 Consider the matrix A whose transpose is: ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ A = ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

1 1 1 0 0 0 0 0 0

0 0 0 1 1 1 0 0 0

0 0 0 0 0 0 1 1 1

1 0 0 1 0 0 0 0 1

0 1 0 0 1 0 0 1 0

0 0 1 0 0 1 1 0 0

1 0 0 0 0 1 1 0 0

0 1 0 0 1 0 0 1 0

0 0 1 1 0 0 0 0 1

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

This is a doubly stochastic singular matrix of rank 6. Prove that Q(A) is integral, R[It] is not normal, A(P ) = K[F t] and K[F t] = K[F t].

14.3

Packing problems and blowup algebras

In this section, we present several characterizations of the max-ﬂow mincut property of a clutter in algebraic and combinatorial terms and prove that certain properties are closed under taking minors and parallelizations. We study packing problems and analyze the Conforti–Cornu´ejols conjecture about the packing property [90]. The following basic result about the nilpotent elements of the associated graded ring holds for arbitrary monomial ideals. Lemma 14.3.1 If I is a monomial ideal of R, then the nilradical of the associated graded ring of I is given by nil(grI (R)) = =

({g ∈ I i /I i+1 | g s ∈ I si+1 ; i ≥ 0; s ≥ 1}) ({xα ∈ I i /I i+1 | xsα ∈ I si+1 ; i ≥ 0; s ≥ 1}).

Proof. The ﬁrst equality holds because the radical of any graded algebra is generated by homogeneous elements. To prove the second equality take an equivalence class 0 = g ∈ I i /I i+1 such that g s ∈ I is+1 . There are non-zero scalars λ1 , . . . , λr in the ﬁeld K and monomials m1 , . . . , mr in I i such that g = λ1 m1 + · · · + λr mr and m1 ≺ · · · ≺ mr , where ≺ is the lexicographical ordering. We may / I i+1 for all j. Since msr is the leading term of g s and assume that mj ∈ is+1 since I is a monomial ideal, we get msr ∈ I is+1 . Thus mr is in the nilradical of grI (R). By induction it follows that mj is in nil(grI (R)) for all j, as required. 2

584

Chapter 14

Proposition 14.3.2 If grI (R) is reduced (resp. R[It] is normal) and I is a minor of I, then grI (R) is reduced (resp. R[I t] is normal). Proof. Assume that grI (R) is reduced. Notice that we need only show that grI (R) is reduced when I is a minor obtained from I by making x1 = 0 or x1 = 1. Using Lemma 14.3.1 both cases are quite easy to prove. We leave the details as an exercise. If R[It] is normal, then R[I t] is normal by Proposition 12.2.3. 2 Theorem 14.3.3 Let C be a clutter and let C be a parallelization of C. If I(C) is normal, then I(C ) is normal. Proof. By Proposition 14.3.2 normality is closed under taking minors. Thus we need only show that the normality of I = I(C) is preserved when we duplicate a vertex of C. Let V (C) = {x2 , . . . , xn } be the vertex set of C and let C be the duplication of the vertex x2 . We denote the duplication of x2 by x1 . We may assume that I = I(C) = (x2 xw1 , . . . , x2 xwr , xwr+1 , . . . , xwq ), where xwi ∈ K[x3 , . . . , xn ] for all i. We must show that the ideal I(C ) = I + (x1 xw1 , . . . , x1 xwr ) is normal. Consider the sets B0 = {(0, 1, w1 , 1), . . . , (0, 1, wr , 1)}, B

=

{e2 , . . . , en } ∪ B0 ∪ {(0, 0, wr+1 , 1), . . . , (0, 0, wq , 1)},

B

=

B ∪ {e1 , (1, 0, w1 , 1), . . . , (1, 0, wr , 1)}.

As I is normal, by Proposition 14.2.3, we have Zn+1 ∩ R+ B = NB and we need only show the equality Zn+1 ∩ R+ (I) = NB . It suﬃces to show the inclusion “⊂”. Take an integral vector w = (a, b, c, d) in R+ (I), where a, b, d ∈ Z and c ∈ Zn−2 . Then w = (a, b, c, d) =

r i=1

+

q

αi (0, 1, wi , 1) +

r

i=r+1 n

βi (1, 0, wi , 1) +

i=1

αi (0, 0, wi , 1) γi e i

i=1

for some αi , βi , γi in R+ . Comparing entries it follows that (0, a + b, c, d) =

r

(αi + βi )(0, 1, wi , 1) +

i=1

+(γ1 + γ2 )e2 +

q i=r+1

n i=3

γi e i ,

αi (0, 0, wi , 1)

Combinatorial Optimization and Blowup Algebras

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that is, the vector (0, a + b, c, d) is in Zn+1 ∩ R+ B = NB. Thus there are λi , μi in N such that. (0, a + b, c, d) =

r

μi (0, 1, wi , 1) +

i=1

q

μi (0, 0, wi , 1) +

n

i=r+1

λi ei .

i=2

Comparing entries we obtain the equalities a + b = μ1 + · · · + μr + λ2 , = μ1 w1 + · · · + μq wq + λ3 e3 + · · · + λn en , = μ1 + · · · + μ q . r Case (I): b ≤ i=1 μi . If b < μ1 , we set b = μ1 , μ1 < μ1 , and deﬁne μ1 = μ1 − μ1 . Otherwise pick s ≥ 2 such that c d

μ1 + · · · + μs−1 ≤ b ≤ μ1 + · · · + μs . Then b = μ1 + · · · + μs−1 + μs , where μs ≤ μs . Set μs = μs − μs . Hence a+b a w

= μ1 + · · · + μr + λ2 = a + μ1 + · · · + μs−1 + μs , = μs + · · · + μr + λ2 − μs = μs+1 + · · · + μr + μs + λ2 , q s−1 = μi (0, 1, wi , 1) + μs (0, 1, ws , 1) + μi (0, 0, wi , 1) i=1

i=r+1

+μs (1, 0, ws , 1) +

r

μi (1, 0, wi , 1) + λ2 e1 +

i=s+1

n

λi ei ,

i=3

that is, w = (a, b, c, ∈ NB . d) r r Case (II): b > i=1 μi . Then b = i=1 μi + λ2 . Since a + b = μ1 + · · · + μr + λ2 = a + μ1 + · · · + μr + λ2 we get a = λ2 − λ2 . In particular λ2 ≥ λ2 . Then w=

r i=1

μi (0, 1, wi , 1) +

q

μi (0, 0, wi , 1) + ae1 + λ2 e2 +

n

i=r+1

that is, w = (a, b, c, d) ∈ NB .

λi ei ,

i=3

2

Deﬁnition 14.3.4 The clutter C satisﬁes the max-ﬂow min-cut (MFMC) property if both sides of the LP-duality equation min{α, x| x ≥ 0; xA ≥ 1} = max{y, 1| y ≥ 0; Ay ≤ α}

(14.4)

have integral optimum solutions x and y for each nonnegative integral vector α. The system x ≥ 0; xA ≥ 1 is called totally dual integral (TDI) if the maximum in Eq. (14.4) has an integral optimum solution y for each integral vector α with ﬁnite maximum.

586

Chapter 14

Deﬁnition 14.3.5 A clutter C is called Mengerian if β1 (C a ) = α0 (C a ) for all a ∈ Nn , i.e., C is Mengerian if all its parallelizations have the K¨onig property. Theorem 14.3.6 [147, 188, 258, 373] Let C be a clutter and let A be its incidence matrix. The following are equivalent: (i)

grI (R) is reduced, where I = I(C) is the edge ideal of C.

(ii) R[It] is normal and Q(A) is an integral polyhedron. (iii) x ≥ 0; xA ≥ 1 is a TDI system. (iv) C has the max-ﬂow min-cut property. (v) I i = I (i) for i ≥ 1. (vi) I is normally torsion-free, i.e., Ass(R/I i ) ⊂ Ass(R/I) for i ≥ 1. (vii) C is Mengerian, i.e., β1 (C a ) = α0 (C a ) for all a ∈ Nn . Proof. By Theorem 14.2.19, grI (R) is reduced if and only if R[It] is normal and the vectors 1 , . . . , r occurring in Eq. (14.1) are integral. Thus (i) and (ii) are equivalent by Theorem 1.4.3. (ii)⇒(iii): Let α be a vector such that Eq. (14.4) has a ﬁnite maximum equal to b. We may assume b > 0, otherwise the maximum in Eq. (14.4) is attained at y = 0. Note α ≥ 0. As Q(A) is integral, b ∈ N and there is 0 = s ∈ N such that xsα ∈ I sb . Hence xα ∈ I b = I b and this implies that the maximum in Eq. (14.4) has an integral optimum solution y. (iii)⇒ (i): Let xα ∈ I i /I i+1 be a nilpotent element and let m = xα ; that is, m is a monomial such that: (i) m ∈ I i /I i+1 and (ii) ms ∈ I is+1 , for some 0 = s ∈ N. By Lemma 14.3.1 it suﬃces to prove that m ∈ I i+1 . From (ii) there are a1 , . . . , aq in N and δ ∈ Nn such that q xsα = (xv1 )a1 · · · (xvq )aq xδ and j=1 aj = is + 1. The rational vector y0 = (a1 /s, . . . , aq /s) satisﬁes Ay0 ≤ α, y0 ≥ 0 and the sum of its entries is equal to z0 . Hence the linear program q Maximize f (y) = i=1 yi Subject to y ≥ 0; Ay ≤ α has an optimal value greater than or equal to z0 = i + 1/s. Note that the polyhedron Q = {y| Ay ≤ α; y ≥ 0} is bounded. Indeed any y = (y1 , . . . , yq ) in Q must satisfy yj ≤ max{α1 , . . . , αq } for all j, where αj is the jth entry of α. Since the system x ≥ 0; xA ≤ 1 is TDI, the optimum value of the linear program above is attained by an integral vector b = (b1 , . . . , bq ). Thus b1 + · · · + bq ≥ i + 1. As b ∈ Q, we can write xα = (xv1 )b1 · · · (xvq )bq xγ ,

Combinatorial Optimization and Blowup Algebras

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for some γ ∈ Nn . This proves m ∈ I i+1 , as required. (iii)⇔ (iv): This is Proposition 1.4.8. (i)⇔ (v): This was shown in Theorem 14.2.19. (v) ⇔ (vi): This was shown in Proposition 4.3.29. (iv) ⇒ (vii): Let A be the incidence matrix of C = C a . We have shown that (ii) is equivalent to (iv). Thus I(C) is normal and Q(A) is integral. By Theorem 14.3.3 the ideal I(C ) is normal, and since the integrality of Q(A) is closed under minors and duplications (see Corollary 14.2.25), we get that Q(A ) is also integral. Thus, using once again that (ii) is equivalent to (iv), we get that C satisﬁes the max-ﬂow min-cut property. Therefore the LP-duality equation min{1, x| x ≥ 0; xA ≥ 1} = max{y, 1| y ≥ 0; A y ≤ 1} has optimum integral solutions x, y. Observe that the left-hand side of this equality is α0 (C ) and the right-hand side is β1 (C ). (vii) ⇒ (iv): Conversely if C a has the K¨onig property for all a ∈ Nn , then by Lemmas 13.2.12 and 13.2.14 both sides of the LP-duality equation min{a, x| x ≥ 0; xA ≥ 1} = max{y, 1| y ≥ 0; Ay ≤ a} have integral optimum solutions x and y for each nonnegative integral vector a, i.e., C has the max-ﬂow min-cut property. 2 Corollary 14.3.7 Let C be a parallelization or a minor of a clutter C. If C has the max-ﬂow min-cut property, then so does C . Proof. The max-ﬂow min-cut property of a clutter is closed under taking minors; this follows from Proposition 14.3.2 and Theorem 14.3.6. That the max-ﬂow min-cut property is closed under parallelizations follows from Theorem 14.3.6. 2 Corollary 14.3.8 If a clutter C has the max-ﬂow min-cut property, then all parallelizations and minors of C have the K¨ onig property. Proof. Assume that the clutter C has the max-ﬂow min-cut property. By Corollary 14.3.7 it suﬃces to prove that C has the K¨onig property. Let A be the incidence matrix of C. By hypothesis the LP-duality equation min{1, x| x ≥ 0; xA ≥ 1} = max{y, 1| y ≥ 0; Ay ≤ 1} has optimum integral solutions x, y. To complete the proof notice that the left-hand side of this equality is α0 (C) and the right-hand side is β1 (C). 2 Deﬁnition 14.3.9 A clutter C is called dyadic if |C ∩ B| ≤ 2 for C ∈ C and B ∈ C ∨ .

588

Chapter 14

Corollary 14.3.10 [94, Theorem 1.3] If Q(A) is integral and C is dyadic, then x ≥ 0; xA ≥ 1 is a TDI system. Proof. By Proposition 14.2.26 the Rees algebra R[It] is normal. Thus the proof follows from Theorem 14.3.6. 2 Corollary 14.3.11 [185] If C is a balanced clutter, then (i) R[I(C)t] = Rs (I(C)) and R[Ic (C)t] = Rs (Ic (C)). (ii) grI(C) (R) is reduced and all minors of C satisfy the K¨ onig property. Proof. By [373, Corollary 83.1a(iv), p. 1441], we get that both C and C ∨ are Mengerian. Thus the result follows at once from Theorem 14.3.6. 2 Corollary 14.3.12 [156, Theorem 5.3] If C is the clutter of facets of a simplicial forest, then C has the K¨ onig property. Proof. We set I = I(C). By Theorem 6.5.17, C is totally balanced. Hence, by Corollary 14.3.11, the associated graded ring grI (R) is reduced and all minors of C satisfy the K¨onig property. 2 Remark 14.3.13 Let C be the clutter of facets of an unmixed simplicial forest. By Theorem 6.5.17, C has no special cycles. As C is balanced, it has the K¨ onig property (Corollary 14.3.11), and by Lemma 6.5.7 it has a perfect matching of K¨onig type. Thus, Theorem 6.5.18 holds for C [156]. Corollary 14.3.14 If A is totally unimodular, then grI (R) is reduced. Proof. If A is totally unimodular, then A is balanced. Thus we can apply Corollary 14.3.11. 2 Corollary 14.3.15 [383, Theorem 5.9] If C is a bipartite graph and I is its edge-ideal, then I is normally torsion-free and R[It] is normal. Proof. By Proposition 10.2.2, the incidence matrix of a bipartite graph is totally unimodular. Thus the result follows from Corollary 14.3.14 and Theorem 14.3.6. 2 Deﬁnition 14.3.16 A clutter C is said to satisfy the packing property (PP for short) if α0 (C ) = β1 (C ) for any minor C of C; that is, all minors of C satisfy the K¨onig property. Theorem 14.3.17 (A. Lehman [290], [93, Theorem 1.8]) If a clutter C has the packing property, then Q(A) is integral. The converse of this result is not true. A famous example is the clutter Q6 of Example 14.2.9. It does not pack and Q(A) is integral.

Combinatorial Optimization and Blowup Algebras

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Corollary 14.3.18 [93] If a clutter C has the max-ﬂow min-cut property, then C has the packing property. Proof. It follows at once from Corollary 14.3.8.

2

Conforti and Cornu´ejols [90] conjectured that the converse is also true; see also [93, Conjecture 1.6]. Conjecture 14.3.19 (Packing problem, Conforti–Cornu´ejols) If a clutter C has the packing property, then C has the max-ﬂow min-cut property. The following is an equivalent version of this conjecture [93]. Conjecture 14.3.20 (Duplication Conjecture, [93, Conjecture 4.24]) If C has the packing property and C is the clutter obtained from C by duplicating a vertex xi , then C has the K¨onig property. The Conforti–Cornu´ejols conjecture can be studied via blowup algebras. By Theorem 14.3.6, the following is an algebraic version of this conjecture. Conjecture 14.3.21 If α0 (C ) = β1 (C ) for all minors C of C, then the ring grI (R) is reduced. Using Theorems 14.3.17 and 14.3.6 this conjecture reduces to: Conjecture 14.3.22 If α0 (C ) = β1 (C ) for all minors C of C, then R[It] is normal. Proposition 14.3.23 Let Ji be the ideal obtained from I by making xi = 1 in F . If Q(A) is integral, then I is normal if and only if Ji is normal for i = 1, . . . , n and depth(R/I k ) ≥ 1 for all k ≥ 1. Proof. ⇒) The normality of an edge ideal is closed under taking minors (Proposition 12.2.3), hence Ji is normal for all i. Since R[It] is normal, we get that R[It] is C–M thanks to Theorem 9.1.6. Then by Theorem 4.3.19 the ring grI (R) is C–M. Hence using Theorem 14.2.13 and Corollary 14.2.11 we get that depth(R/I i ) ≥ 1 for all i. ⇐) It follows readily adapting the arguments given in the proof of the normality criterion presented in Theorem 12.2.4. 2 By Proposition 14.3.23, we get that Conjecture 14.3.21 also reduces to: Conjecture 14.3.24 If α0 (C ) = β1 (C ) for any minor C of C, then depth(R/I i ) ≥ 1 for all i ≥ 1. Deﬁnition 14.3.25 A clutter C is called minimally non-packing if it does not pack but all its proper minors do.

590

Chapter 14

The next conjecture implies Conjecture 14.3.19 [94]. Conjecture 14.3.26 [94] If C is a minimally non-packing clutter and the polyhedron Q(A) is integral, then α0 (C) = 2. The next result essentially says that, for uniform clutters, it suﬃces to prove Conjecture 14.3.19 for Cohen–Macaulay clutters. Theorem 14.3.27 [116] Let C be a d-uniform clutter. If C satisﬁes PP (resp. max-ﬂow min-cut), then there is a d-uniform Cohen–Macaulay clutter C satisfying PP (resp. max-ﬂow min-cut) such that C is a minor of C . Theorem 14.3.28 If R[It] = Rs (I) and K[F t] = A(P), then R[It] is normal. Proof. By Theorem 14.2.7 the Rees cone of I is quasi-ideal. Thus by Corollary 12.3.2 the Rees algebra of I is normal. 2 2 Proposition 14.3.29 If C is a d-uniform clutter such that I b = I (b) for all b and d ≥ 2, then I b is generated by monomials of degree bd for b ≥ 1. Proof. The monomial ideal I b has a unique minimal set of generators consisting of monomials. Take xa in this minimal set. Notice that (a, b) is in R+ (I). Thus we may proceed as in the proof of Theorem 12.3.1 to obtain that (a, b) is in the cone generated by {(v1 , 1), . . . , (vq , 1)}. This yields deg(xa ) = bd. 2 Proposition 14.3.30 [185] If C is a d-uniform clutter and d ≥ 2, then I i = I (i) for all i ≥ 1 if and only if Q(A) is integral and K[F t] = A(P). Proof. ⇒) By Theorem 14.2.7 the polyhedron Q(A) is integral. Since I (i) is integrally closed (see Corollary 4.3.26), we get that R[It] is normal. Therefore applying Theorem 9.3.31, we obtain K[F t] = A(P). Here the hypothesis on the degrees of xvi is essential. ⇐) By Theorem 14.2.7 I i = I (i) for all i, thus applying Theorem 14.3.28 gives that R[It] is normal and we get the required equality. In this part the 2 hypothesis on the degrees of xvi is not needed. For uniform clutters, using Proposition 14.3.30 and Theorem 14.3.17, we obtain another algebraic version of the Conforti–Cornu´ejols conjecture: Conjecture 14.3.31 If C is a uniform clutter with the packing property, then one has the equality K[F t] = A(P). Corollary 14.3.32 Let C be a uniform clutter and let A be its incidence matrix. If the polyhedra {x| x ≥ 0; xA ≤ 1} and {x| x ≥ 0; xA ≥ 1} are integral, then C has the max-ﬂow min-cut property.

Combinatorial Optimization and Blowup Algebras

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Proof. By Proposition 14.3.30, the clutter C has the max-ﬂow min-cut property if and only if Q(A) is integral and K[F t] = A(P). Thus, the result follows from Corollary 13.6.9. 2 Notation For an integral matrix B = (0), the greatest common divisor of all the non-zero r × r subdeterminants of B will be denoted by Δr (B). Corollary 14.3.33 Let B be the matrix obtained from A by adjoining a row of 1’s. If xv1 , . . . , xvq have degree d ≥ 2 and C has the max-ﬂow min-cut property, then Δr (B) = 1 with r = rank(B). Proof. By Theorem 14.3.6 and Proposition 14.3.30, we obtain the equality A(P) = K[F t]. Hence a direct application of Theorem 9.3.25 gives that 2 Δr (B) is equal to 1. Deﬁnition 14.3.34 Let M be a matroid on X and let C be the clutter of bases of M . The edge ideal I(C) is called the basis monomial ideal of M . Proposition 14.3.35 Let C be the clutter of bases of a matroid M on X. If C satisﬁes the packing property, then grI(C) (R) is reduced. Proof. Since Q(A) is integral (Theorem 14.3.17) and R[I(C)t] is normal (Corollary 12.3.12), we get that grI(C) (R) is reduced by Theorem 14.3.6. 2 Deﬁnition 14.3.36 Let X1 , . . . , Xd be a partition of X. The matroid M whose collection of bases is {{y1 , . . . , yd }| yi ∈ Xi for i = 1, . . . , d} is called the transversal matroid deﬁned by X1 , . . . , Xd . Proposition 14.3.37 Let X1 , . . . , Xd be a partition of X and let M be the transversal matroid on X deﬁned by X1 , . . . , Xd . If I is the basis monomial ideal of M, then grI (R) is reduced. Proof. Let C be the clutter of bases of M and let A be the incidence matrix of C. Then I = I(C). Notice that the edges of C ∨ are X1 , . . . , Xd , therefore the incidence matrix of C ∨ is totally unimodular. Hence C ∨ is a balanced clutter and (C ∨ )∨ = C. Thus, by Corollary 14.3.11, grI (R) is reduced. 2 Another property which is closed under taking minors is being the basis monomial ideal of a matroid. Proposition 14.3.38 Let I be the basis monomial ideal of a matroid M on X of rank d. If I is a minor of I, then I is again the basis monomial ideal of a matroid M .

592

Chapter 14

Proof. Let I be the minor obtained from I by making xn = 1 and let G(I ) be the unique minimal generating set of I consisting of monomials. First we prove that I is generated by monomials of degree d − 1. Take f ∈ G(I ). We proceed by contradiction. Assume that f = x1 x2 · · · xd with / supp(f ). Pick a monomial in I of the form g = xi1 xi2 · · · xid−1 xn . xn ∈ If supp(g) \ {xn } ⊂ {x1 , . . . , xd } we obtain a contradiction because of the / supp(f ). By the exchange minimality of G(I ). Thus we may assume xi1 ∈ property (Theorem 1.9.14) we may assume that the monomial g1 = x1 xi2 xi3 · · · xid−1 xn belongs to I. If supp(g1 ) \ {xn } ⊂ {x1 , . . . , xd } we obtain a contradiction. Thus we may assume xi2 ∈ / supp(f ). Again by the exchange property we may assume that g2 = x1 x2 xi3 xi4 · · · xid−1 xn is in I. Repeating this procedure we get x1 x2 · · · xd−1 xn ∈ I, a contradiction. Thus I is generated by monomials of degree d − 1, as claimed. We set G(I ) = {xα1 , . . . , xαr }, Bi = supp(xαi ), and Bi = Bi ∪ {xn }. The collection of bases {B1 , . . . , Br } satisﬁes the exchange property, hence so does the collection B = {B1 , . . . , Br }. Thus, by Theorem 1.9.14, there is a matroid M whose collection of bases is B . Hence I is the basis monomial ideal of M , as required. The case xn = 0 is also not hard to show. 2 Proposition 14.3.39 If C is a graph, then the following are equivalent: (a) grI (R) is reduced. (b) C is bipartite. (c) Q(A) is integral. (d) C has the packing property. Proof. (a) ⇒ (d): By Theorem 14.3.6, C has the max-ﬂow min-cut property. Then, by Corollary 14.3.18, C has the packing property. (d) ⇒ (b): It suﬃces to show that C has no induced odd cycles (Proposition 7.1.2). If Cr = {x1 , . . . , xr } is an induced cycle, then deleting all vertices outside Cr we obtain that Cr is a minor of C. Thus Cr has the K¨onig property. It follows that r must be even. (b) ⇒ (c): By Proposition 10.2.2, the matrix A is totally unimodular. Then by a result of Hoﬀman and Kruskal (see Theorem 1.5.6), Q(A) is integral. (c) ⇒ (a): Since any graph is a dyadic clutter, using Corollary 14.3.10, we get that x ≥ 0; xA ≥ 1 is a TDI system. 2 Thus by Theorem 14.3.6, the ring grI (R) is reduced. Deﬁnition 14.3.40 A clutter is binary if its edges and its minimal vertex covers intersect in an odd number of vertices. Theorem 14.3.41 [375] A binary clutter C has the max-ﬂow min-cut property if and only if Q6 is not a minor of C.

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Corollary 14.3.42 If C is a binary clutter with the packing property, then C has the max-ﬂow min-cut property. Proof. Any clutter with the packing property cannot have Q6 as a minor because Q6 does not satisfy the K¨onig property. Thus C has the max-ﬂow min-cut property by Theorem 14.3.41. 2 Lemma 14.3.43 If I is an unmixed ideal and C satisﬁes the K¨ onig property, then x1 = x1 x2 · · · xn belongs to the subring K[xv1 , . . . , xvq ]. Proof. We may assume x1 = xv1 · · · xvg xδ for some xδ , where g is the height of I. If δ = 0 pick xn ∈ supp(xδ ). Since the variable xn occurs in some monomial minimal generator of I, there is a minimal prime p containing xn . Thus using that xv1 , . . . , xvg have disjoint support we conclude that p contains at least g + 1 variables, a contradiction. 2 Proposition 14.3.44 Let Ii = I ∩ K[X \ {xi }]. If I is an unmixed ideal such that the following conditions hold (a1 ) Q(A) is integral, (a2 ) Ii is normal for i = 1, . . . , n, and onig property, (a3 ) C has the K¨ then R[It] is normal. Proof. Take xa tb = xa1 1 · · · xann tb ∈ R[It] a minimal generator; that is, xa tb cannot be written as a product of two non-constant monomials of R[It]. By condition (a2 ) we may assume ai ≥ 1 for all i. Notice that x1 · · · xn tg is always in R[It] if Q(A) is integral, where g is the height of I. We claim that b ≤ g. If b > g, consider the decomposition xa tb = (x1 · · · xn tg )(xa1 1 −1 · · · xann −1 tb−g ). To derive a contradiction consider the irreducible representation of the Rees cone R+ (I) (see Eq. (14.1)). Observe that using condition (a1 ) gives ai ≥ b (k = 1, . . . , s) i∈Ck

because (a, b) ∈ R+ (I). Now since I is unmixed we get (ai − 1) ≥ b − g (k = 1, . . . , s), i∈Ck

and consequently xa1 1 −1 · · · xann −1 tb−g ∈ R[It], a contradiction to the choice of xa tb . Thus b ≤ g. Using condition (a3 ) we get x1 · · · xn tg ∈ I g tg ⊂ R[It], 2 which readily implies xa tb ∈ R[It].

594

Chapter 14

Corollary 14.3.45 Let C be an unmixed clutter with the packing property and let I = I(C) be its edge ideal. If the ideal Ii = I ∩ K[X \ {xi }] is normal for all i, then C satisﬁes the max-ﬂow min-cut property. Proof. By Theorem 14.3.17, the set covering polyhedron Q(A) is integral. Then R[It] is normal by Proposition 14.3.44. Using Theorem 14.3.6, we get that C satisﬁes the max-ﬂow min-cut property. 2 Corollary 14.3.46 [204] If a minimal counterexample C to the Conforti– Cornu´ejols conjecture exists, then C cannot be unmixed. Proof. Assume that C is unmixed. Let I = I(C) be the edge ideal of C, let Ii = I ∩ K[X \ {xi }], and let Ci be the clutter that corresponds to Ii . Then Ci satisﬁes the max-ﬂow min-cut property, and by Theorem 14.3.6 we get that Ii is normal. Hence by Corollary 14.3.45, the clutter C has the max-ﬂow min-cut property, a contradiction. 2 Proposition 14.3.47 Let Y ⊂ X and let IY = I ∩ K[Y ]. If IY has the K¨ onig property for all Y and R[It] is generated as a K-algebra by monomials of the form xa tb , with xa square-free, then R[It] is normal. Proof. Take xa tb a generator of R[It], with xa square-free. By induction we may assume xa tb = x1 · · · xn tb . Hence, since (1, . . . , 1, b) is in the Rees cone, we get |Ck | ≥ b for k = 1, . . . , s. In particular g = ht(I) ≥ b. As I has the K¨onig property, we get that the monomial x1 · · · xn is in I g and consequently xa tb ∈ R[It]. 2 Proposition 14.3.48 Let Ii = I ∩ K[X \ {xi }]. If Ii is a normal ideal for i = 1, . . . , n and H1 ∩ H2 ∩ · · · ∩ Hr ∩ Rn+1 = (0), +

(14.5)

then R[It] is normal. Proof. Let xa tb = xa1 1 · · · xann tb ∈ R[It] be a minimal generator. It suﬃces to prove that 0 ≤ b ≤ 1 because this readily implies that xa is either a variable or a monomial in F . Assume b ≥ 2. Since Ii is normal we may assume that ai ≥ 1 for all i. As each variable occurs in at least one monomial of F , using Eq. (14.5) it follows that there is (vk , 1) such that (vk , 1), i = 0 for i = 1, . . . , r. Therefore (a − vk , b − 1), i ≥ 0

(i = 1, . . . , r).

Thus (a, b) − (vk , 1) ∈ R+ (I), a contradiction to the choice of xa tb .

2

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595

Rees algebras and their a-invariants In this part we compute some a-invariants in terms of combinatorial invariants of graphs and clutters, and exhibit some families of Gorenstein Rees algebras. Theorem 14.3.49 Let C be a d-uniform clutter and let a(R[It]) be the ainvariant of R[It] with respect to the grading of R[It] induced by deg(xi ) = 1 and deg(t) = −(d − 1). If grI (R) is reduced, then a(R[It]) ≥ − [n − (d − 1)(α0 (C) − 1)] , with equality if I is unmixed. Proof. By assigning deg(xi ) = 1 and deg(t) = −(d− 1), the Rees ring R[It] becomes a standard graded K-algebra. By Theorem 14.3.6 the Rees algebra R[It] is a normal domain. Then according to a formula of Danilov–Stanley (see Theorem 9.1.5) its canonical module is the ideal of R[It] given by ωR[It] = ({xa1 1 · · · xann tan+1 | a = (ai ) ∈ (R+ (I))o ∩ Zn+1 }). Set α0 = α0 (C). By Theorem 14.3.6, the Rees algebra of I is normal and Q(A) is integral. The ring R[It] is Cohen–Macaulay by Theorem 9.1.6. Then by Proposition 5.2.3, the a-invariant can be expressed as a(R[It]) = −min{ i | (ωR[It] )i = 0}. Using Eq. (14.1) it is seen that (1, . . . , 1, α0 − 1) ∈ (R+ (I))o . Thus the inequality follows by computing the degree of x1 · · · xn tα0 −1 . Assume that I is unmixed. Take any monomial xa tb = xa1 1 · · · xann tb in the ideal ωR[It] , that is, (a, b) ∈ (R+ (I))o . By Theorem 14.2.7 the vector (a, b) has positive entries and satisﬁes (14.6) −b + xi ∈Ck ai ≥ 1 (k = 1, . . . , s). If α0 ≥ b + 1 we readily obtain the inequality deg(xa tb ) = a1 + · · · + an − b(d − 1) ≥ n − (d − 1)(α0 − 1).

(14.7)

Now assume α0 ≤ b. Using the normality of R[It] and Eqs. (14.1) and (14.6) it follows that the monomial m = xa1 1 −1 · · · xann −1 tb−α0 +1 belongs to R[It]. Since xa tb = mx1 · · · xn tα0 −1 , the inequality (14.7) also holds in this case. Altogether we conclude the desired equality. 2 Corollary 14.3.50 [183] Let C be a d-uniform clutter. If I is unmixed with α0 (C) = 2 and grI (R) is reduced, then R[It] is a Gorenstein ring and a(R[It]) = −(n − d + 1).

596

Chapter 14

Proof. From the proof of Theorem 14.3.49 it follows that the monomial 2 x1 · · · xn t generates the canonical module. Remark 14.3.51 If α0 (C) ≥ 3, then R[It] is not a Gorenstein ring because x1 · · · xn tα0 −1 and x1 · · · xn t are distinct minimal generators of ωR[It] . This holds in a more general setting (see Proposition 4.3.39). Corollary 14.3.52 [183] Let J = Ic (C) be the ideal of covers of C. If C is a bipartite graph and I = I(C) is unmixed, then R[Jt] is Gorenstein and a(R[Jt]) = −(n − α0 (C) + 1). Proof. Notice that R[Jt] has the grading induced by assigning deg(xi ) = 1 and deg(t) = 1 − α0 (C). Thus the formula follows from Corollary 14.3.50 once we recall that grJ (R) is reduced according to Corollary 14.3.11. 2 Corollary 14.3.53 Let C be a bipartite graph. Then the Rees cone R+ (I) is the intersection of the closed halfspaces given by the linear inequalities −xn+1 +

vi ∈C

xi ≥ 0 xi ≥ 0

i = 1, . . . , n + 1, C is a minimal vertex cover of C,

and none of those halfspaces can be omitted from the intersection. Proof. Let A be the incidence matrix of C. By Proposition 10.2.2, the matrix A is totally unimodular. Then the vertices of Q(A) are integral by Theorem 1.5.6. Hence the result follows from Theorem 14.1.1. 2 Corollary 14.3.54 Let C be a bipartite graph. Then F is a facet of the Rees cone R+ (I) if and only if (a) F = R+ (I) ∩ Hei for some 1 ≤ i ≤ n + 1, or (b) F = R+ (I) ∩ {x ∈ Rn+1 | − xn+1 + vi ∈C xi = 0} for some minimal vertex cover C of C. Proof. The result follows from Theorem 1.1.44 and Corollary 14.3.53.

2

Theorem 14.3.55 [187] Let C be a bipartite graph. If a(R[It]) is the ainvariant of R[It] with respect to the grading of R[It] induced by deg(xi ) = 1 and deg(t) = −1, then a(R[It]) = −(β0 (C) + 1). Proof. By Corollary 14.3.15, the Rees algebra R[It] is a normal domain. Clearly R[It] is a standard graded K-algebra with the grading induced by

Combinatorial Optimization and Blowup Algebras

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deg(xi ) = 1 and deg(t) = −1. Then according to a formula of Danilov– Stanley (see Theorem 9.1.5) the canonical module of R[It] is the ideal of R[It] given by ωR[It] = ({xa1 1 · · · xann tan+1 | a = (ai ) ∈ (R+ (I))o ∩ Zn+1 }), where (R+ (I))o is the topological interior of the Rees cone. The ring R[It] is Cohen–Macaulay by Theorem 9.1.6. Therefore the a-invariant can be expressed as a(R[It]) = −min{ i | (ωR[It] )i = 0}, see Theorem 5.2.3. Let a = (ai ) be an arbitrary vector in (R+ (I))o ∩ Zn+1 . By Corollary 14.3.53 a satisﬁes ai ≥ 1 for 1 ≤ i ≤ n + 1 and −an+1 + vi ∈C ai ≥ 1 for any minimal vertex cover C of C. Let C be a vertex cover of C with α0 (C) elements and let A = V \ C. Note β0 (C) = |A|. Hence if m denotes the monomial xa1 1 · · · xann tan+1 , then deg(m)

= =

a1 + · · · + an − an+1 ai + ai − an+1 ≥ β0 (C) + 1. vi ∈A

vi ∈C

This proves the inequality a(R[It]) ≤ −(β0 (C) + 1). On the other hand using Corollary 14.3.53 and the assumption α0 (C) ≥ 2 we get that the monomial m1 = x1 · · · xn tα0 −1 is in ωR[It] and has degree β0 (C) + 1. Thus a(R[It]) ≥ −(β0 (C) + 1). 2 Corollary 14.3.56 If C is a bipartite graph, then type(R[It]) ≥ α0 (C) − 1. Proof. By Corollary 14.3.53 it is seen that the monomials x1 · · · xn ti , with i = 1, . . . , α0 (C) − 1, are distinct minimal generators of ωR[It] . 2

Exercises 14.3.57 Prove that Q6 is a binary non-dyadic clutter. 14.3.58 If a clutter C has the packing property and I = I(C), then I 2 = I 2 . 14.3.59 Let v1 and v2 be two vectors in Rn with entries in {0, 1} and let I = (xv1 , xv2 ) ⊂ R. Prove that the Rees algebra R[It] is normal. 14.3.60 Prove that the ideal I = (x21 , x22 ) ⊂ K[x1 , x2 ] is not normal. 14.3.61 Let I = (x1 x4 x7 , x3 x5 x8 , x2 x6 x9 , x1 x4 x8 , x3 x4 x7 , x1 x5 x7 ) be the edge ideal of a clutter C and let A be its incidence matrix. Prove that (a) grI (R) is reduced, and (b) A is not totally unimodular.

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14.3.62 Let I and I be monomial ideals of R and R = K[x2 , . . . , xn ], respectively. If I is square-free and I = x1 I , prove that grI (R) is reduced if and only if grI (R ) is reduced. 14.3.63 If C is a minimally non-packing clutter, then C is either ideal or minimally non-ideal. Recall that C is called ideal if Q(A) is integral, where A is the incidence matrix of C, and that C is called minimally non-ideal if C is not ideal but all its proper minors are ideal [94]. 14.3.64 If G is an odd cycle, then G does not satisfy the K¨onig property. 14.3.65 Let C be a clutter with vertex set X = {x1 , . . . , xn }. Consider the clutter C with vertex set X ∪ {y1 , . . . , yn } and whose edges are the edges of C together with {xi , yi } for i = 1, . . . , n. Prove that C satisﬁes the max-ﬂow min-cut property if and only if C does. 14.3.66 Let C be a clutter with vertex set X = {x1 , . . . , xn } such that all its edges have d ≥ 2 vertices. Construct a clutter C satisfying the following properties: (a) all the edges of C have d vertices. (b) C is unmixed. (c) C satisﬁes the max-ﬂow min-cut property if and only if C does. 14.3.67 Prove that condition (a3 ) of Proposition 14.3.44 is not needed when I is generated by monomials of the same degree (cf. Corollary 14.4.7). 14.3.68 Let R = K[x1 , . . . , xn ] and R[z1 , . . . , z ] be polynomial rings over a ﬁeld K. If I is an ideal of R generated by square-free monomials and I is normal, then J = (I, x1 z1 · · · z ) is a normal ideal of R[z1 , . . . , z ]. 14.3.69 If grI (R) is reduced and C is d-uniform, then any square submatrix B of A of order m such that the sum of the entries in any row or column of B is equal to d satisﬁes m ≡ 0 mod (d) and det(B) = 0. 14.3.70 Let xa1 , . . . , xar be a set of generators of I and let B be the matrix whose columns are a1 , . . . , ar . Then both sides of the equation min{1, x| x ≥ 0; xA ≥ 1} = max{y, 1| y ≥ 0; Ay ≤ 1} have integral optimum solutions if and only if both sides of the equation min{1, x| x ≥ 0; xB ≥ 1} = max{y, 1| y ≥ 0; By ≤ 1} have integral optimum solutions. If any of the two systems have integral optimum solutions x, y their optimal values are equal.

Combinatorial Optimization and Blowup Algebras

14.4

599

Uniform ideal clutters

A clutter is called ideal if its set covering polyhedron is integral. In this section we study the combinatorial structure of uniform ideal clutters. The max-ﬂow min-cut property of clutters is studied here from a linear algebra perspective. If a uniform clutter has the max-ﬂow min-cut property, we prove that its incidence matrix is equivalent over Z to an “identity matrix” and that its column vectors form a Hilbert basis. The structure of normally torsion-free Cohen–Macaulay edge ideals of height two is also studied here. Lemma 14.4.1 [185] If C is a d-uniform clutter and Q(A) is integral, then there exists a minimal vertex cover of C intersecting every edge of C in exactly one vertex. Proof. Let B be the incidence matrix of C ∨ . Using Corollary 14.2.22, we get that Q(B) is an integral polyhedron. We proceed by contradiction. Assume that for each column αk of B there exists vik in {v1 , . . . , vq } such that vik B ≥ 1 + ek . Consider the vector α = vi1 + · · · + vis . From the inequality αB ≥ (s + 1, . . . , s + 1) we obtain that α/(s+1) ∈ Q(B). Notice that Q(B) = Rn+ +conv(v1 , . . . , vq ); see Theorem 1.1.42. Thus, we can write λi = 1). α/(s + 1) = λ1 v1 + · · · + λq vq + μ1 e1 + · · · + μn en (λi , μj ≥ 0; Therefore, taking inner product with 1, we get |α| = sd ≥ (s + 1)d, a contradiction. 2 Deﬁnition 14.4.2 A graph G is called strongly perfect if every induced subgraph H of G has a maximal independent set of vertices A such that |A ∩ K| = 1 for any maximal clique K of H. Bipartite and chordal graphs are strongly perfect. If A is the vertexclique matrix of G, then G being strongly perfect implies that the clique polytope of G, {x| x ≥ 0; xA ≤ 1}, has a vertex that intersects every maximal clique. In this sense, uniform ideal clutters can be thought of as being analogous to strongly perfect graphs. Proposition 14.4.3 Let C be a d-uniform clutter. If R[It] = Rs (I) and I = (0). is unmixed, then H1 ∩ · · · ∩ Hr ∩ Rn+1 + Proof. By Proposition 14.2.21 one has R[Ic (C)t] = Rs (Ic (C)). Thus, by Lemma 14.4.1, there is vk such that |supp(vk ) ∩ Ci | = 1 for i = 1, . . . , r. 2 This means that (vk , 1) is in the intersection of H1 , . . . , Hr .

600

Chapter 14

Proposition 14.4.4 If C is a d-uniform clutter whose covering polyhedron Q(A) is integral, then there are X1 , . . . , Xd mutually disjoint minimal vertex covers of C such that X = ∪di=1 Xi . Proof. By induction on d. By Lemma 14.4.1 there is a minimal vertex cover X1 of C such that |supp(xvi ) ∩ X1 | = 1 for all i. Consider the ideal I obtained from I by making xi = 1 for xi ∈ X1 . Let C be the clutter corresponding to I and let A be the incidence matrix of C . The integrality of Q(A) is preserved under taking minors (Corollary 14.2.25), so Q(A ) is integral. Then C is a (d − 1)-uniform clutter with Q(A ) integral and V (C ) = X \ X1 . Therefore by induction hypothesis there are X2 , . . . , Xd pairwise disjoint minimal vertex covers of C such that X\X1 = X2 ∪· · ·∪Xd . To complete the proof we now show that X2 , . . . , Xd are minimal vertex covers of C. If e is an edge of C and 2 ≤ k ≤ d, then e ∩ X1 = {xi } for some i. Since e \ {xi } is an edge of C , we get (e \ {xi }) ∩ Xk = ∅. Hence Xk is a vertex cover of C. Furthermore if x ∈ Xk , then by the minimality of Xk there is an edge e of C disjoint from Xk \ {x}. Since e = e ∪ {y} is an edge of C for some y ∈ X1 , we obtain that e is an edge of C disjoint from Xk \ {x}. Therefore Xk is a minimal vertex cover of C, as required. 2 Corollary 14.4.5 If C is a d-uniform clutter and Q(A) is integral, then there is a partition X1 , . . . , Xd of X such that C is a subclutter of the clutter of bases of the transversal matroid M deﬁned by X1 , . . . , Xd . Proof. It suﬃces to notice that, by Proposition 14.4.4, any edge of C intersects Xi in exactly one vertex for any i. 2 Theorem 14.4.6 [205] Let C be a d-uniform clutter. Assume that I(C) is normally torsion-free of height two. Then C is Cohen–Macaulay if and only if (i) C is unmixed, and (ii) there is a partition X 1 = {x11 , x12 }, . . . , X d = {xd1 , xd2 } of X and a perfect matching e1 = {x11 , . . . , xd1 }, e2 = {x12 , . . . , xd2 } of C so that all edges of C have the form {x1i1 , . . . , xdid } for some non-decreasing sequence 1 ≤ i1 ≤ · · · ≤ id ≤ 2. Proof. ⇒) By Corollary 3.1.17 Cohen–Macaulay rings are unmixed, so (i) holds true. We shall prove (ii) by induction on d. We claim that C has a free vertex. By Theorem 14.3.6 Q(A) is integral. Hence, by Proposition 14.4.4, there are minimal vertex covers Z1 , . . . , Zd of C such that Z1 , . . . , Zd partition X and |Zi ∩ e| = 1 for all e ∈ E(C) and i = 1, . . . , d. Since C is unmixed, one has |Zi | = 2 for all i. By Theorem 14.3.6, C has the K¨onig property. Then, by Lemma 6.5.7, C has a perfect matching of K¨onig type, i.e., there are edges e1 , e2 of C such that

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e1 ∩ e2 = ∅ and e1 ∪ e2 = X. We may assume that e1 = {x1 , . . . , xd }, e2 = {y1 , . . . , yd }, and Zi = {xi , yi } for all i. Any minimal vertex cover of C has the form C = {x, y} for some x ∈ e1 and y ∈ e2 . Let G = C ∨ be the blocker of C, which, in this case, is a graph. The graph G is bipartite with bipartition e1 , e2 . As R/I(C) is Cohen–Macaulay, I(G) = I(C ∨ ) has a linear resolution (Theorem 6.3.41). Then, by Theorem 7.6.7, the complement graph G of G is chordal. By Lemma 7.5.4, G has a simplicial vertex z. We may assume that z = xk for some k. The induced subgraphs G [e1 ] and G [e2 ] of G are complete graphs on d vertices. Next we prove that / NG (e2 ) for any k. If xk ∈ NG (e2 ) for some k, then {xk , y } is an xk ∈ edge of G for some . Then y would have to be adjacent in G to any xi in e1 , in particular {x , y } ∈ E(G ), a contradiction. Thus {xk , yi } ∈ E(G) for all i. Note that yk is a free vertex of C. Indeed let e be any edge of C containing yk , then xk is not in e because |e ∩ Zk | = 1. Hence since {xk , yi } is a vertex cover of C for any i we get that yi ∈ e for any i, i.e., e = e2 . Let I be the edge ideal obtained from I(C) by making xk = 1 and yk = 1 and let C be the clutter on V = X \{xk , yk } such that I = I(C ). The ideal I is Cohen–Macaulay of height two, normally torsion-free, and is generated by monomials of degree d − 1. Therefore, by the induction hypothesis, there is a partition X 2 = {x21 , x22 }, . . . , X d = {xd1 , xd2 } of V such that all edges of C have the form {x2i2 , . . . , xdid } for some 1 ≤ i1 ≤ · · · ≤ id ≤ 2. To complete the proof we set x11 = xk , x12 = yk , and X 1 = {x11 , x12 }. ⇐) It is left as an exercise. Here, the assumption that I(C) is normally torsion-free is not needed. 2 Corollary 14.4.7 Let C be a d-uniform clutter. If R[It] = Rs (I) and I is unmixed, then C and C ∨ have the K¨ onig property. Proof. According to Proposition 14.2.21 one has R[Ic (C)] = Rs (Ic (C)). Thus, by duality (Theorem 6.3.39), we need only show that C has the K¨onig property, and this follows readily from Proposition 14.4.4. 2 Theorem 14.4.8 Let C be a d-uniform clutter and let Ii = I ∩ K[X \ {xi }]. If I is unmixed and Q(A) is integral, then grI (R) is reduced if and only if Ii is normal for i = 1, . . . , n. Proof. It follows from Corollary 14.4.7 and Proposition 14.3.44.

2

Proposition 14.4.9 If C is a d-uniform clutter with a perfect matching and Q(A) is integral, then C has the K¨ onig property and is vertex critical. Proof. By Theorem 1.1.42, we can write Q(A) = Rn+ + conv(u1 , . . . , us ), where the ui ’s are the characteristic vectors of the minimal vertex covers of C. As 1/d ∈ Q(A), using this equality, we get 1/d = δ + λ1 u1 + · · · + λs us ; (δ ∈ Rn+ ; λi ≥ 0; i λi = 1).

602

Chapter 14

Therefore n ≥ gd, where g = α0 (C). By hypothesis there are mutually disjoint edges f1 , . . . , fr such that X = f1 ∪ · · · ∪ fr and n = rd. Thus n = rd ≥ gd and r ≥ g. On the other hand g = α0 (C) ≥ β1 (C) ≥ r. Thus r = g and n = gd. Hence C has the K¨onig property. We now prove that C is vertex critical. By Proposition 14.4.4 there are X1 , . . . , Xd mutually disjoint minimal vertex covers of C such that X = ∪di=1 Xi . Hence n = gd = |X1 | + · · · + |Xd |. As |Xi | ≥ g for all i, we get |Xi | = g for all i. It follows easily that C is vertex critical. Indeed notice that each vertex xi belongs to a minimal vertex cover of C with g vertices. Hence α0 (G \ xi ) < α0 (G). 2 Proposition 14.4.10 Let C be a d-uniform clutter with a perfect matching f1 , . . . , fr . If Q(A) is integral, then r = α0 (C) and there are X1 , . . . , Xd disjoint minimal vertex covers of C of size α0 (C) such that X = ∪di=1 Xi . Proof. It follows from the proof of Proposition 14.4.9.

2

Theorem 14.4.11 [119] If C is a d-uniform clutter with the max-ﬂow mincut property, then (a) Δr (A) = 1, where r = rank(A). (b) NA = R+ A ∩ Zn , where A = {v1 , . . . , vq }. (c) A diagonalizes over Z to an identity matrix. Proof. (a): Let B be the matrix with column vectors (v1 , 1), . . . , (vq , 1). Since the clutter is uniform, the last row vector of B, i.e., the vector 1, is a Q-linear combination of the ﬁrst n rows of B. Thus A and B have the same rank. By Proposition 14.3.30 we obtain K[F t] = A(P). In particular K[F t] = A(P) because A(P) is normal. Hence, by Theorem 9.3.25, we have Δr (B) = 1. Recall that Δr (A) = 1 if and only if A is equivalent over Z to an “identity matrix.” We can regard A as a matrix of size (n + 1) × q by adding a row of zeros. Thus it suﬃces to prove that B is equivalent to A over Z. Thanks to Proposition 14.4.4, there are X1 , . . . , Xd mutually disjoint minimal vertex covers of C such that X = ∪di=1 Xi and |supp(xvi ) ∩ Xk | = 1

∀ i, k.

By permuting the variables we may assume that X1 is equal to {x1 , . . . , xr }. Hence the last row of B, which is the vector 1, is the sum of the ﬁrst |X1 | rows of B, i.e., B is equivalent to A over Z. (b): It suﬃces to prove the inclusion “⊃”. Let a be an integral vector in R+ A. Then a = λ1 v1 + · · · + λq vq , λi ≥ 0 for all i. Thus, setting b = i λi , one has |a| = bd. We claim that (a, b) belongs to R+ (I). Let u1 , . . . , us

Combinatorial Optimization and Blowup Algebras

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be the characteristic vectors of the minimal vertex covers of C. Since Q(A) is integral, by Theorem 14.1.1, we can write R+ (I) = He+1 ∩ He+2 ∩ · · · ∩ He+n+1 ∩ H+1 ∩ H+2 ∩ · · · ∩ H+s ,

(14.8)

where i = (ui , −1) for 1 ≤ i ≤ s. Notice that (a, b) ∈ R+ (I), thus using Eq. (14.8) we get that a, ui ≥ b for all i. Hence a, ui ≥ b for all i because a, ui is an integer for all i. Using Eq. (14.8) again we get that (a, b) ∈ R+ (I), as claimed. By Theorem 14.3.6 the Rees algebra R[It] is normal. Then, applying Proposition 14.2.3, we obtain that (a, b) is in NA . There are nonnegative integers η1 , . . . , ηq and ρ1 , . . . , ρn such that (a, b) = η1 (v1 , 1) + · · · + ηq (vq , 1) + ρ1 e1 + · · · + ρn en . Hence it is seen that |a| = bd + i ρi = bd and consequently ρi = 0 for all i and b = b. It follows at once that a ∈ NA as required. (c): By part (a) one has Δr (A) = 1. Thus the invariant factors of A are all equal to 1 (see Theorem 1.3.15), i.e., the Smith normal canonical form of A is an identity matrix. 2 This result and the Conforti–Cornu´ejols conjecture suggest the following weaker conjecture. Conjecture 14.4.12 [185] If C is a uniform clutter with the packing property, then either of the following equivalent conditions hold: (a) Zn /ZA is a free group. (b) A diagonalizes over Z to an identity matrix. This conjecture will be proved later for d-uniform clutters with a perfect matching and α0 (C) = 2 (see Theorem 14.4.15). Corollary 14.4.13 Let C be a uniform clutter. Then C satisﬁes the maxﬂow min-cut property if and only if Q(A) is integral and NA = R+ A ∩ Zn . Proof. ⇒) It follows from Theorems 14.4.11 and 14.3.6. ⇐) By Proposition 14.3.30 we need only show K[F t] = A(P). The a b t ∈ A(P). Then inclusion “⊂” is clear. To show the inclusion “⊃” take x q n n a ∈ bP ∩ Z . From NA = R A ∩ Z it is seen that a = + i=1 ηi vi for some ηi ∈ N such that i ηi = b. Thus xa tb ∈ K[F t], as required. 2 Corollary 14.4.14 Let C be a d-uniform clutter. If C has the max-ﬂow min-cut property, then C has a perfect matching if and only if n = dα0 (C). Proof. ⇒) As Q(A) is integral, from the proof of Proposition 14.4.9 we obtain the equality n = dα0 (C).

604

Chapter 14

⇐) We set g = α0 (C). Let B be the incidence matrix of C ∨ . As Q(A) is integral, by Corollary 14.2.22, we get that Q(B) is an integral polyhedron. Consequently, by Theorem 1.1.42, we can write Q(B) = Rn+ + conv(v1 , . . . , vq ), Therefore, since the rational vector 1/g is in Q(B), we can write 1/g = δ + μ1 v1 + · · · + μq vq (δ ∈ Rn+ ; μi ≥ 0; μ1 + · · · + μq = 1). q Hence n/g = |δ| + ( i=1 μi )d = |δ| + d. Since n = dg, we obtain that δ = 0. Thus the vector 1 is in R+ A ∩ Zn . By Theorem 14.4.11(b) this intersection is equal to NA. Then we can write 1 = η1 v1 + · · · + ηq vq , for some η1 , . . . , ηq in N. Hence it is readily seen that C has a perfect matching. 2 The next result gives some support to Conjecture 14.3.19. Theorem 14.4.15 [119] Let C be a d-uniform clutter with a perfect matching such that C has the packing property and α0 (C) = 2. Then: (a) A diagonalizes over Z to an “identity matrix.” (b) If K[F t] is normal, then C has the max-ﬂow min-cut property. (c) If A is linearly independent, then C has the max-ﬂow min-cut property. Proof. (a): By Lehman theorem Q(A) is integral; see Theorem 14.3.17. Thus by Proposition 14.4.10 there is a perfect matching f1 , f2 of X with X = f1 ∪ f2 and there is a partition X1 , . . . , Xd of X such that Xi is a minimal vertex cover of C for all i, |Xi | = 2 for all i, and |supp(xvi ) ∩ Xk | = 1

∀ i, k.

(14.9)

Thus we may assume that Xi = {x2i−1 , x2i } for i = 1, . . . , d. Notice that n = 2d because X = f1 ∪ f2 . By induction on r = rank(A). Since 1 is the sum of the ﬁrst two rows of A it suﬃces to prove that 1 is the only invariant factor of A or equivalently that the Smith normal form of A is the “identity.” Let w1 , . . . , w2d be the columns of A and let Vi be the linear space generated by w1 , . . . , w2i . For k odd, one has wk + wk+1 = 1. Hence if k is odd and we remove columns wk and wk+1 from A we obtain a submatrix whose rank is greater than or equal to r − 1. Thus after permuting columns we may assume

i + 1 if 1 ≤ i ≤ r − 1, dim(Vi ) = (14.10) r if r ≤ i. Let J be the monomial ideal deﬁned by the rows of [w1 , . . . , w2(r−1) ]. Then a minimal set of generators of J consists of monomials of degree r − 1 and will

Combinatorial Optimization and Blowup Algebras

605

still satisfy condition in Eq. (14.9) with d replaced by r − 1. Furthermore J satisﬁes the packing property because it is a minor of I. If [w1 , . . . , w2(r−1) ] diagonalizes (over the integers) to the identity matrix so does A . Therefore we may assume d = r − 1 and I = J. Let B be the matrix [w1 , . . . , w2(d−1) ] and let I be the monomial ideal deﬁned by the rows of B; that is, I is obtained from I making x2d−1 = 1 and x2d = 1. Notice that by induction B diagonalizes to an identity matrix [Ir−1 , 0] of order r − 1. Since I has the K¨onig property we may permute rows and columns and assume that the matrix A is written as:

10 01

10 01

10 01

.. .

.. .

.. .

.. .

··· ··· ···

10 01

.. . ··· ···

.. .

B

···

10 01 10 .. . 10 01 .. . 01

← ← .. . ←

where either a pair 1 0 or 0 1 must occur in the places marked with a circle and such that the number of 1 s in the last column is greater than or equal to the number of 1 s in any other column. Consider the matrix C obtained from A by removing the rows whose penultimate entry is equal to 1 (these are marked above with an arrow) and removing the last column. Let K be the monomial ideal deﬁned by the rows of C; that is, K is obtained from I by making x2d−1 = 0 and x2d = 1. By the choice of the last column and because of Eq. (14.10) it is seen that K has height two. Since K is a minor of I it has the K¨onig property. As a consequence using row operations A can be brought to the form:

10 10

10 10

10 10

.. .

.. .

.. .

.. .

··· ··· ···

10 10

.. . ··· ···

.. .

B1

···

10 01 10 .. . 10 01 .. . 01

606

Chapter 14

where B1 has rank r − 1 and diagonalizes to an identity. Therefore it is readily seen that this matrix reduces to Ir−1 0 0 .. . 0

0 0 0 .. . 0

0 0 0

··· ··· ···

0

···

0 0 0 .. . 0

0 1 a1 .. . ar

0 −1 b1 .. . br

To ﬁnish the proof observe that by rank considerations (see Eq. (14.10)) this matrix reduces to [Ir , 0], as required. (b): By part (a) we have Δr (A) = 1. Consider the matrix B with column vectors (v1 , 1), . . . , (vq , 1). Since |vi | = d for all i, one has Δr (B) = 1 (see Exercise 14.4.23). Then by Theorem 9.3.25 this condition is equivalent to the equality K[F t] = A(P ). Hence K[F t] = A(P) because K[F t] is a normal domain. By Lehman theorem (Theorem 14.3.17), Q(A) is integral. Thus C satisﬁes max-ﬂow min-cut by Proposition 14.3.30 and Theorem 14.3.6. (c): xv1 t, . . . , xvq t are algebraically independent and K[F t] is normal because it is a polynomial ring. Hence the result follows from (b). 2 Recall that I is called minimally non-normal if I is not normal and all its proper minors are normal. Theorem 14.4.16 Let C be a d-uniform clutter and {Xi }di=1 a partition of X such that Xi = {x2i−1 , x2i } is a minimal vertex cover of C for all i. Then (a) rank(A) ≤ d + 1. (b) If C is a minimal vertex cover of C, then 2 ≤ |C| ≤ d. (c) If C satisﬁes the K¨ onig property and there is a minimal vertex cover C of C with |C| = d ≥ 3, then rank(A) = d + 1. (d) If I = I(C) is minimally non-normal and C has the packing property, then rank(A) = d + 1. Proof. (a): For each odd integer k the sum of rows k and k + 1 of the matrix A is equal to 1. Thus the rank of A is bounded by d + 1. (b): By the pigeon hole principle, any C ∈ C ∨ satisﬁes 2 ≤ |C| ≤ d. (c): Notice that C contains exactly one vertex of each Xj because Xj ⊂ C. Thus we may assume C = {x1 , . . . , x2d−1 }. Consider xα := x2 · · · x2d and notice that xk xα ∈ I for each xk ∈ C because xk xα is in every minimal prime of I. Writing xk = x2i−1 with 1 ≤ i ≤ d we conclude that the monomial xαi = x2 · · · x2(i−1) x2i−1 x2(i+1) · · · x2d is a minimal generator of I. Thus we may assume xαi = xvi for i = 1, . . . , d. The vector 1 belongs to QA because C has the K¨ onig property. It follows that the matrix with rows v1 , . . . , vd , 1 has rank d + 1, as required.

Combinatorial Optimization and Blowup Algebras

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(d): Let xα tb be a minimal generator of R[It] not in R[It] and let m = x . Using Proposition 14.3.29, one has deg(m) = bd. By hypothesis m is the only associated prime of N = I b /I b . Hence, by Corollary 2.1.29, we have rad(ann(N )) = p = m = (x1 , . . . , xn ) α

p∈Ass(N )

and m ⊂ ann(N ) for some r > 0. Thus for i odd we can write r

xri xα = (xv1 )a1 · · · (xvq )aq xδ , 2 where a1 + · · · + aq = b and deg(xδ ) = r. If we write xδ = xsi 1 xsi+1 xγ with γ xi , xi+1 not in the support of x , making xj = 1 for j ∈ / {i, i + 1}, it is not hard to see that r = s1 + s2 and γ = 0. Thus we get an equation: 2 xsi 2 xα = (xv1 )a1 · · · (xvq )aq xsi+1

with s2 > 0. Using a similar argument we obtain an equation: α v1 b1 vq bq w1 1 xw with w1 > 0. i+1 x = (x ) · · · (x ) xi 2 +w1 (xv1 )a1 · · · (xvq )aq = xsi 2 +w1 (xv1 )b1 · · · (xvq )bq . As Zn /ZA is Hence, xsi+1 torsion-free (Theorem 14.4.15), we get ei − ei+1 ∈ ZA for i odd. Finally to conclude that rank(A) = d + 1 notice that 1 ∈ ZA. 2

The next theorem gives an interesting class of uniform clutters, coming from combinatorial optimization, that satisfy Conjecture 9.6.22. Theorem 14.4.17 If A is a balanced matrix and C is a d-uniform clutter, then any regular triangulation of the cone R+ A is weakly unimodular. Proof. Let A1 , . . . , Am be the elements of a regular triangulation of R+ A. Then dim R+ Ai = dim R+ A and Ai is linearly independent for all i. Let Ci be the subclutter of C whose edges correspond to the vectors in Ai , and let Ai be the incidence matrix of Ci . As Ai is a balanced matrix, using Corollary 14.3.11 and Theorem 14.3.6, we get that the clutter Ci has the max-ﬂow min-cut property. Hence by Theorem 14.4.11 one has Δr (Ai ) = 1, where r is the rank of A. Thus the invariant factors of Ai are all equal to 1 (see Theorem 1.3.15). Therefore by the fundamental structure theorem for ﬁnitely generated abelian groups (see Theorem 1.3.16) the quotient group Zn /ZAi is torsion-free for all i. Notice that dim R+ A = rank ZA and dim R+ Ai = rank ZAi for all i. Since r is equal to dim R+ A. It follows that the quotient group ZA/ZAi is torsion-free and has rank 0 for all i, consequently ZA = ZAi for all i, i.e., the triangulation is weakly unimodular. 2

608

Chapter 14

Deﬁnition 14.4.18 Let K[F t] = K[xv1 t, . . . , xvq t] be a homogeneous subring with the standard grading induced by deg(xa tb ) = b. Let K[t1 , . . . , tq ]/IA K[F t],

ti → xvi t,

be a presentation of K[F t]. The regularity of K[F t], denoted reg(K[F t]), is deﬁned as the regularity of K[t1 , . . . , tq ]/IA as an K[t1 , . . . , tq ]-module. Theorem 14.4.19 Let C be a d-uniform unmixed clutter with g = α0 (C). If C has the max-ﬂow min-cut property, then K[F t] = A(P), the a-invariant of A(P) is bounded from above by −g and reg(A(P)) ≤ (d − 1)(g − 1). Proof. Setting B = {(vi , 1)}qi=1 and A = B ∪ {ei }ni=1 , one has the equality R+ B = R B ∩ R+ A .

(14.11)

By Theorem 14.3.6, R[I(C)t] is normal and Q(A) is integral. Hence, using Theorem 9.3.31, we obtain K[F t] = A(P) and A(P) becomes a standard graded K-algebra. By Theorems 14.1.1 and 14.2.7, we obtain the equality + + R+ (I) = He+1 ∩ · · · ∩ He+n+1 ∩ H(u ∩ · · · ∩ H(u , 1 ,−1) s ,−1)

(14.12)

where u1 , . . . , us are the characteristic vectors of the minimal vertex covers of C. Hence, by Proposition 14.4.4, there are X1 , . . . , Xd mutually disjoint minimal vertex covers of C of size g such that X = ∪di=1 Xi . Notice that |Xi ∩ f | = 1 for 1 ≤ i ≤ d and f ∈ E(C). We may assume that Xi = Ci for 1 ≤ i ≤ d. Therefore, using Eqs. (14.11) and (14.12), we get ! + R+ B = R B ∩ He+1 ∩ · · · ∩ He+n ∩ He+n+1 ∩ H(u , (14.13) i ,−1) i∈I

where i ∈ I if and only if deﬁnes a proper face of R+ B. The Ehrhart ring A(P) is Cohen–Macaulay (Theorem 9.1.6). Then, by Theorem 9.1.5 and Eq. (14.13), the canonical module of A(P) is the ideal given by + H(u i ,−1)

ωA(P) = ((xt)a | a ∈ R B; ai ≥ 1 ∀ i; (uk , −1), a ≥ 1 for k ∈ I), (14.14) where (xt)a = xa1 1 · · · xann tan+1 . The a-invariant of A(P) is given by a(A(P)) = −min{ i | (ωA(P) )i = 0},

(14.15)

see Proposition 5.2.3. Take an arbitrary monomial xa tb = xa1 1 · · · xann tb in the ideal ωA(P) . By Eqs. (14.13) and (14.14), the qvector (a, b) is in R+ B and ai ≥ 1 for all i. Thus we can write (a, b) = i=1 λi (vi , 1) with λi ≥ 0 for all i. Since vi , uk = 1 for all i, k, we obtain g = |uk | ≤

xi ∈Ck

ai = a, uk =

q i=1

λi vi , uk =

q i=1

λi = b

Combinatorial Optimization and Blowup Algebras

609

for 1 ≤ k ≤ d. This means that deg(xa tb ) ≥ g. Thus −a(A(P)) ≥ g, as required. Next we show that reg(A(P)) ≤ (d − 1)(g − 1). Since A(P) is Cohen–Macaulay, by Theorem 6.4.1, we have reg(A(P)) = dim(A(P)) + a(A(P)) ≤ dim(A(P)) − g.

(14.16)

Using that vi , uk = 1 for all i, k, by induction on d, it is seen that rank(A) ≤ g+(d−1)(g−1). Thus, using the fact that dim(A(P)) = rank(A) and Eq. (14.16), we get reg(A(P)) ≤ (d − 1)(g − 1). 2

Exercises 14.4.20 Let C be the clutter with vertex set X = {x1 , . . . , x9 } and edges: f1 = {x1 , x2 }, f2 = {x3 , x4 , x5 , x6 }, f3 = {x7 , x8 , x9 }, f6 = {x5 , x7 }, f7 = {x6 , x8 }. f4 = {x1 , x3 }, f5 = {x2 , x4 }, Prove the following: the incidence matrix of C is a balanced matrix, Q(A) is integral, |C ∩ fi | ≥ 2 for any C ∈ C ∨ and for any i, C has a perfect matching, α0 (C \ {x9 }) = α0 (C) = 4, C is not vertex critical, and the uniformity hypothesis is essential in Propositions 14.4.4 and 14.4.9. 14.4.21 Let A be the incidence matrix of a cycle of length 3 and let B be the matrix obtained from A by adjoining the row 1. Prove that det(A) = 2 and Δ3 (B) = 1. In particular A and B are not equivalent over Z. 14.4.22 Consider the clutter C whose incidence matrix is ⎡ ⎤ 1 0 0 1 ⎢ 0 1 0 1 ⎥ ⎢ ⎥ ⎢ 0 0 1 1 ⎥. ⎢ ⎥ ⎣ 0 1 1 0 ⎦ 1 0 1 0 Prove that C satisﬁes max-ﬂow min-cut, A is not equivalent over Z to an identity matrix, and the columns of A do not form a Hilbert basis for the cone they generate. Thus the uniformity hypothesis is essential in the two statements of Theorem 14.4.11. 14.4.23 Let A = {v1 , . . . , vq } ⊂ Zn . If |vi | = d = 0 for all i and Zn /ZA is a torsion-free Z-module, then Zn+1 /Z{(v1 , 1)), . . . , (vq , 1)} is torsion-free. 14.4.24 Let G be an unmixed bipartite graph, let A∨ = {u1 , . . . , us } be the set of column vectors of the incidence matrix of G∨ , and let P ∨ = conv(A∨ ). Then K[xu1 t, . . . , xus t] = A(P ∨ ) and reg(A(P ∨ )) ≤ (|V (G)|/2) − 1.

610

Chapter 14

14.4.25 If we do not require that |vi | = d for all i in Theorem 14.4.17, give an example to show that this theorem is false even if K[F ] is homogeneous. 14.4.26 [119] Let {u1 , . . . , ur } be the set of all characteristic vectors of the collection of bases B of a transversal matroid M. Then the polyhedral cone R+ {u1 , . . . , ur } has a weakly unimodular regular triangulation.

14.5

Clique clutters of comparability graphs

In this section we prove that the clique clutter of a comparability graph satisﬁes the max-ﬂow min-cut property. We also present some classical combinatorial results on posets, e.g., Dilworth decomposition theorem and a min-max theorem of Menger. Let P = (X, ≺) be a partially ordered set (poset for short) on the ﬁnite vertex set X = {x1 , . . . , xn } and let G be its comparability graph. Recall that the vertex set of G is X and the edge set of G is the set of all unordered pairs {xi , xj } such that xi and xj are comparable. A clique of G is a subset of the set of vertices that induces a complete subgraph. Lemma 14.5.1 If G1 (resp. cl(G)1 ) is the graph (resp. clutter) obtained from G (resp. cl(G)) by duplicating the vertex x1 , then cl(G)1 = cl(G1 ). Proof. Let y1 be the duplication of x1 . We now prove the inclusion E(cl(G)1 ) ⊂ E(cl(G1 )). The other inclusion follows readily using similar arguments. Take e ∈ E(cl(G)1 ). Case (i): Assume y1 ∈ / e. Then e ∈ E(cl(G)). Clearly e is a clique of / E(cl(G1 )), then e can be extended to a maximal clique of G1 . G1 . If e ∈ Hence e ∪ {y1 } must be a clique of G1 . Note that x1 ∈ / e because {x1 , y1 } is not an edge of G1 . Then e ∪ {x1 } is a clique of G, a contradiction. Case (ii): Assume y1 ∈ e. Then there is f ∈ E(cl(G)), with x1 ∈ f , such that e = (f \ {x1 }) ∪ {y1 }. Since {x, x1 } ∈ E(G) for any x in f \ {x1 }, one has that {x, y1 } ∈ E(G1 ) for any x in f \ {x1 }. Then e is a clique of G1 . / e which is adjacent in G to If e is not a maximal clique of G1 , there is x ∈ any vertex of f \ {x1 } and x is adjacent to y1 in G1 . In particular x = x1 . Then x is adjacent in G to x1 and consequently x is adjacent in G to any vertex of f , a contradiction because f is a maximal clique of G. 2 Let D be a digraph; that is, D consists of a ﬁnite set V (D) of vertices and a set E(D) of ordered pairs of distinct vertices called edges. Let A, B be two sets of vertices of D. For use below recall that a (directed) path of D is called an A–B path if it runs from a vertex in A to a vertex in B. A common vertex of A and B is also an A–B path. A set C of vertices is called an A–B disconnecting set if C intersects each A–B path. Menger [316] gave a min-max theorem for the maximum number of disjoint A–B paths in an undirected graph. This theorem also holds for digraphs.

Combinatorial Optimization and Blowup Algebras

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Theorem 14.5.2 (Menger’s theorem, see [373, Theorem 9.1]) Let D be a digraph and let A, B be two subsets of V (D). Then the maximum number of vertex-disjoint A–B paths is equal to the minimum size of an A–B disconnecting vertex set. Proof. Let k be the minimum size of an A–B disconnecting vertex set and let be the maximum number of vertex-disjoint A–B paths. Clearly ≤ k. Equality will be shown by induction on E(D), the number of edges of D. The case E(D) = ∅ is trivial. Pick an edge a = (u, v) of D. If each A–B disconnecting vertex set of D \ a has size at least k, then inductively there are k vertex-disjoint A–B paths in D \ a, hence in D. Thus we may assume that there exists an A–B disconnecting vertex set C of D \ a of size less than or equal to k − 1. Then C ∪ {u} and C ∪ {v} are A–B disconnecting vertex sets of D of size k. Now, each A–(C ∪ {u}) disconnecting vertex set A of D \ a has size at least k, as A is an A–B disconnecting vertex set of D. Indeed, if P is an A–B path of D, then P intersects C ∪ {u}, and hence P contains an A– (C ∪ {u}) path. So P intersects A. Therefore by induction D \ a contains k vertex-disjoint A–(C ∪ {u}) paths P0 , . . . , Pk−1 . Similarly D \ a contains k vertex-disjoint (C ∪ {v})–B paths Q0 , . . . , Qk−1 . If pi is the last vertex of Pi , we may assume—by reducing the size of Pi if necessary—that any other vertex of Pi is not in C ∪ {u}. Similarly if qi is the ﬁrst vertex of Qi , we may assume that all other vertices of Qi are not in C ∪ {v}. Since |C| = k − 1, we have P0 Pi

r - r q q q r a0 r - r q q q r ai

- r u - r ci

Q0 r - r q q q r v Qi r - r q q q r ci

- r b0 - r bi

where C = {c1 , . . . , ck−1 }, ai ∈ A, bi ∈ B for i = 0, . . . , k − 1. Any path in the ﬁrst collection intersects any path in the second collection only in C, since otherwise D \ a contains an A–B path avoiding C. Therefore we can pairwise concatenate the paths to obtain k vertex-disjoint A–B paths. This proof was adapted from [195]. 2 Theorem 14.5.3 [118] Let P = (X, ≺) be a poset on the vertex set X and let G be its comparability graph. If C = cl(G) is the clutter of maximal cliques of G, then C satisﬁes the max-ﬂow min-cut property. Proof. We can regard P as a transitive digraph without cycles of length two with vertex set X and edge set E(P ), i.e., the edges of P are ordered pairs (a, b) of distinct vertices with a ≺ b such that: (i) (a, b) ∈ E(P ) and (b, c) ∈ E(P ) ⇒ (a, c) ∈ E(P ), and (ii) (a, b) ∈ E(P ) ⇒ (b, a) ∈ / E(P ).

612

Chapter 14

Note that by (i) P is acyclic; that is, it has no directed cycles. We set X = {x1 , . . . , xn }. Let x1 be a vertex of P and let y1 be a new vertex. Consider the digraph P 1 with vertex set X 1 = X ∪ {y1 } and edge set E(P 1 ) = E(P ) ∪ {(y1 , x)| (x1 , x) ∈ E(P )} ∪ {(x, y1 )| (x, x1 ) ∈ E(P )}. The digraph P 1 is transitive. Indeed let (a, b) and (b, c) be two edges of P 1 . If y1 ∈ / {a, b, c}, then (a, c) ∈ E(P ) ⊂ E(P 1 ) because P is transitive. If y1 = a, then (x1 , b) and (b, c) are in E(P ). Hence (x1 , c) ∈ E(P ) and (y1 , c) ∈ E(P 1 ). The cases y1 = b and y1 = c are treated similarly. Thus P 1 deﬁnes a poset P 1 = (V (P 1 ), ≺1 ). The comparability graph H of P 1 is precisely the graph G1 obtained from G by duplicating the vertex x1 by the vertex y1 . From Lemma 14.5.1 we get that cl(G)1 = cl(G1 ), where cl(G)1 is the clutter obtained from cl(G) by duplicating the vertex x1 by the vertex y1 . Altogether we obtain that the clutter cl(G)1 is the clique clutter of the comparability graph G1 of the poset P 1 . By Theorem 14.3.6(vii) it suﬃces to prove that cl(G)w has the K¨onig property for all w ∈ Nn . Since duplications commute with deletions we may assume that w = (w1 , . . . , wr , 0, . . . , 0), where wi ≥ 1 for i = 1, . . . , r. Consider the clutter C1 obtained from cl(G) by duplicating wi − 1 times the vertex xi for i = 1, . . . , r. We denote the vertex set of C1 by X1 . By successively applying the fact that cl(G)1 = cl(G1 ), we conclude that there is a poset P1 with comparability graph G1 and vertex set X1 such that C1 = cl(G1 ). As before we regard P1 as a transitive acyclic digraph. Let A and B be the set of minimal and maximal elements of the poset P1 , i.e., the elements of A and B are the sources and sinks of P1 , respectively. We set S = {xr+1 , . . . , xn }. Consider the digraph D whose vertex set is V (D) = X1 \ S and whose edge set is deﬁned as follows. A pair (x, y) in V (D) × V (D) is in E(D) if and only if (x, y) ∈ E(P1 ) and there is no vertex z in X1 with x ≺ z ≺ y. Notice that D is a sub-digraph of P1 which is not necessarily the digraph of a poset. We set A1 = A \ S and B1 = B \ S. Note that C w = C1 \ S, the clutter obtained from C1 by removing all vertices of S and all edges sharing a vertex with S. If every edge of C1 intersects S, then E(C w ) = ∅ and there is nothing to prove. Thus we may assume that there is a maximal clique K of G1 disjoint form S. Note that by the maximality of K and by the transitivity of P1 we get that K contains at least one source and one sink of P1 , i.e., A1 = ∅ and B1 = ∅ (see argument below). The maximal cliques of G1 not containing any vertex of S correspond exactly to the A1 –B1 paths of D. Indeed let c = {v1 , . . . , vs } be a maximal clique of G1 disjoint from S. Consider the sub-poset Pc of P1 induced by c. Note that Pc is a tournament, i.e., Pc is an oriented graph (nocycles of length two) such that any two vertices of Pc are comparable. By Exercise 7.1.21 any tournament has a Hamiltonian path, i.e., a spanning

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oriented path. Therefore we may assume that v1 ≺ v2 ≺ · · · ≺ vs−1 ≺ vs By the maximality of c we get that v1 is a source of P1 , vs is a sink of P1 , and (vi , vi+1 ) is an edge of D for i = 1, . . . , s − 1. Thus c is an A1 –B1 path of D, as required. Conversely let c = {v1 , . . . , vs } be an A1 –B1 path of D. Clearly c is a clique of P1 because P1 is a poset. Assume that c is not a maximal clique of G1 . Then there is a vertex v ∈ X1 \ c such that v is related to every vertex of c. Since v1 , vs are a source and a sink of P1 , respectively, we get v1 ≺ v ≺ vs . We claim that vi ≺ v for i = 1, . . . , s. By induction assume that vi ≺ v for some 1 ≤ i < s. If v ≺ vi+1 , then vi ≺ v ≺ vi+1 , a contradiction to the fact that (vi , vi+1 ) is an edge of D. Thus vi+1 ≺ v. Making i = s we get that vs ≺ v, a contradiction. This proves that c is a maximal clique of G1 . Therefore, since the maximal cliques of G1 not containing any vertex in S are exactly the edges of C w = C1 \ S, by Menger’s theorem (see Theorem 14.5.2) we obtain that C w satisﬁes the K¨onig property. 2 Corollary 14.5.4 Let P = (X, ≺) be a poset on the vertex X and let G be its comparability graph. Then the maximum number of disjoint maximal cliques of G is equal to the minimum size of a set intersecting all maximal cliques of G. Proof. By Theorem 14.5.3, the clique clutter of G satisﬁes the max-ﬂow min-cut property. Then by Theorem 14.3.6, the clique clutter of G satisﬁes the K¨onig property. 2 Let P = (X, ≺) be a poset with vertex set X. A subset L of X is called a chain of P if x ≺ y or y ≺ x for all x = y in L. A set A of X is called an anti-chain of P if x and y are not related for all x = y in A. Proposition 14.5.5 Let P = (X, ≺) be a poset with vertex set X. Then the maximum size of a chain of P equals the minimum number of disjoint anti-chains into which X can be decomposed (covered). Proof. Let L be a chain of maximum size and let m = |L|. Clearly max ≤ min, because if X1 , . . . , Xk is a decomposition of X into disjoint anti-chains, then |L ∩ Xi | ≤ i for all i, so |L| ≤ k. Let x ∈ X and let h(x) be the size of the longest chain of P with end vertex x. Since any two vertices x, y with h(x) = h(y) are incomparable, we get a decomposition X = {x| h(x) = 1} ∪ {x| h(x) = 2} ∪ · · · ∪ {x| h(x) = m} into at most m disjoint anti-chains, so min ≤ max.

2

614

Chapter 14

Corollary 14.5.6 If G is the comparability graph of a poset P = (X, ≺), then G is perfect. Proof. Let ω(G) and χ(G) be the clique number and the chromatic number of G, respectively. By Proposition 14.5.5, we have ω(G) = χ(G). As the class of comparability graphs is closed under taking induced subgraphs, we get that G is a perfect graph. 2 Theorem 14.5.7 (Dilworth decomposition theorem) Let P = (X, ≺) be a poset on X. Then the maximum size of an anti-chain is equal to the minimum number of disjoint chains into which X can be decomposed. Proof. Let G be the comparability graph of P . By Corollary 14.5.6, G is perfect. Hence by the weak perfect graph theorem (see Theorem 13.6.1), the complement G of G is also perfect. In particular ω(G) = χ(G) and the result follows readily. 2 Recall that the vertex-clique matrix of a graph G is the incidence matrix of the clique clutter of G. Corollary 14.5.8 Let G be a comparability graph and let A be the vertexclique matrix of G. Then the following polytopes are integral : P(A) = {x| x ≥ 0; xA ≤ 1} and Q(A) = {x| x ≥ 0; xA ≥ 1} Proof. G is a perfect graph by Corollary 14.5.6. Hence the polytope P(A) is equal to the independence polytope of G (Theorem 13.6.1). In particular P(A) is integral. By Theorem 14.5.3 the clique clutter cl(G) has the maxﬂow min-cut property. Thus applying Theorem 14.3.6, we get that Q(A) has only integral vertices. 2 Conjecture 14.5.9 Let G be a perfect graph and let A be the vertex-clique matrix of G. If the polyhedron Q(A) = {x| x ≥ 0; xA ≥ 1} is integral, then the system x ≥ 0; xA ≥ 1 is TDI. The conjecture holds when the clique clutter of a perfect graph G is uniform (see Corollary 14.3.32). Corollary 14.5.10 If G is a comparability graph and cl(G) is its clique clutter, then the edge ideal of cl(G) is normally torsion-free and normal. Proof. It follows from Theorems 14.5.3 and 14.3.6.

2

Deﬁnition 14.5.11 Let d ≥ 2, g ≥ 2 be two integers and let X 1 = {x11 , . . . , x1g }, X 2 = {x21 , . . . , x2g }, . . . , X d = {xd1 , . . . , xdg } be disjoint sets of variables. A clutter C with vertex set X = X 1 ∪ · · · ∪ X d and edge set E(C) = {{x1i1 , x2i2 , . . . , xdid }| 1 ≤ i1 ≤ i2 ≤ · · · ≤ id ≤ g} is called a complete admissible d-uniform clutter (cf. Deﬁnition 6.6.4).

Combinatorial Optimization and Blowup Algebras

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Complete admissible uniform clutters were ﬁrst studied in [160] and more recently in [205, 325] (see Section 6.6). Theorem 14.5.12 If C is a complete admissible d-uniform clutter, then its edge ideal I(C) is normally torsion-free and normal. Proof. Let P = (X, ≺) be the poset with vertex set X and partial order given by xk ≺ xm p if and only if 1 ≤ < m ≤ d and 1 ≤ k ≤ p ≤ g. We denote the comparability graph of P by G. We claim that E(C) = E(cl(G)), where cl(G) is the clique clutter of G. Let f = {x1i1 , x2i2 , . . . , xdid } be an edge of C, i.e., we have 1 ≤ i1 ≤ i2 ≤ · · · ≤ id ≤ g. Clearly f is a clique of G. If f is not maximal, then there is a vertex xk not in f which is adjacent in G to every vertex of f . In particular xk must be comparable to xi , which is impossible. Thus f is an edge of cl(G). Conversely let f be an edge of cl(G). We can write f = {xki11 , xki22 , . . . , xkiss }, where k1 < · · · < ks and i1 ≤ · · · ≤ is . By the maximality of f we get that s = d and ki = i for i = 1, . . . , d. Thus f is an edge of C. Hence by Corollary 14.5.10 we obtain that I(C) is normally torsion-free and normal. 2

Exercises 14.5.13 Let G be a graph. Let G1 = G \ x1 (resp. cl(G)1 = cl(G) \ x1 ) be the graph (resp. clutter) obtained from G (resp. cl(G)) by deleting the vertex x1 . If G is a cycle of length three, prove that cl(G)1 = cl(G1 ). 14.5.14 Let cl(G) be the clique clutter of a perfect graph G. If cl(G) is d-uniform and satisﬁes the packing property, prove that cl(G) satisﬁes the max-ﬂow min-cut property.

14.6

Duality and integer rounding problems

Throughout this section we keep the notation of Section 14.1 but allow I to be a monomial ideal and A to be a matrix with entries in N. In this section we characterize the normality of a monomial ideal in terms of integer rounding properties of linear systems and show a duality theorem for monomial subrings. We show that the Rees algebra of the ideal of covers of a perfect graph is normal. Let I be a monomial ideal of R generated by xv1 , . . . , xvq , and let A be the n × q matrix with column vectors v1 , . . . , vq . In what follows consider the homogeneous monomial subring S = K[xw1 t, . . . , xwr t] ⊂ R[t], where {w1 , . . . , wr } is the set of all α ∈ Nn such that 0 ≤ α ≤ vi for some i.

616

Chapter 14

Deﬁnition 14.6.1 Given a polyhedron Q in Rn , its blocking polyhedron, denoted by B(Q), is deﬁned as: B(Q) := {z ∈ Rn | z ≥ 0; z, x ≥ 1 for all x in Q}. Lemma 14.6.2 If Q = Q(A), then B(Q) = Rn+ + conv(v1 , . . . , vq ). Proof. “⊂”: Take z in B(Q), then z, x ≥ 1 for all x ∈ Q and z ≥ 0. Let u1 , . . . , ur be the vertex set of Q. In particular z, ui ≥ 1 for all i. Then (z, 1), (ui , −1) ≥ 0 for all i. From Theorem 1.4.2, we get that (z, 1) is in the cone generated by A . Thus z is in Rn+ + conv(v1 , . . . , vq ). 2 “⊃ : This inclusion is clear. Deﬁnition 14.6.3 A rational polyhedron Q has the integer decomposition property if for each natural number k and for each integer vector a in kQ, a is the sum of k integer vectors in Q, where kQ is equal to {ka| a ∈ Q}. Theorem 14.6.4 I is normal if and only if the blocking polyhedron B(Q) of Q = Q(A) has the integer decomposition property and all minimal integer vectors of B(Q) are columns of A (minimal with respect to ≤). Proof. By Lemma 14.6.2 we get B(Q) ∩ Qn = Qn+ + convQ (v1 , . . . , vq ), because B(Q) is a rational polyhedron. From this equality we readily obtain I k = ({xa | a ∈ kB(Q) ∩ Zn }) for 0 = k ∈ N.

(14.17)

⇒) Assume that I is normal, i.e., I k = I k for k ≥ 1. Let a be an integral vector in kB(Q). Then, by Eq. (14.17), xa ∈ I k and a is the sum of k integral vectors in B(Q); that is, B(Q) has the integer decomposition property. Take a minimal integer vector a in B(Q). Then xa ∈ I = I and we can write a = δ + vi for some vi and for some δ ∈ Nn . Thus a = vi by the minimality of a. ⇐) Assume that B(Q) has the integer decomposition property and all minimal integer vectors of B(Q) are columns of A. Take xa ∈ I k , i.e., a is an integral vector of kB(Q). Hence a is the sum of k integral vectors α1 , . . . , αk in B(Q). Since any minimal vector of B(Q) is a column of A we 2 may assume that αi = ci + vi for i = 1, . . . , k. Hence xa ∈ I k . Theorem 14.6.5 If I = I and Q = Q(A), then I is normal if and only if the blocking polyhedron B(Q) has the integer decomposition property. Proof. ⇒) If I is normal, by Theorem 14.6.4, the blocking polyhedron B(Q) has the integer decomposition property. ⇐) Take xa ∈ I k . By Eq. (14.17), a is an integral vector of kB(Q). Hence a is the sum of k integral vectors α1 , . . . , αk in B(Q). By Eq. (14.17), 2 with k = 1, we get that α1 , . . . , αk are in I = I. Hence xa ∈ I k .

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Corollary 14.6.6 If I is the edge ideal of a clutter, then I is normal if and only if the blocking polyhedron B(Q) has the integer decomposition property. Proof. Recall that I is an intersection of prime ideals. Thus it is seen that I = I and the result follows from Theorem 14.6.5. 2 Deﬁnition 14.6.7 Let A be a matrix with entries in N. The linear system x ≥ 0; xA ≥ 1 has the integer rounding property if max{y, 1| Ay ≤ w; y ∈ Nq } = (max{y, 1| y ≥ 0; Ay ≤ w})

(14.18)

for each integral vector w for which the right-hand side is ﬁnite. Systems with the integer rounding property have been widely studied; see [372, Chapter 22], [373, Chapter 5], and the references therein. Theorem 14.6.8 ([19], [373, p. 82]) The system x ≥ 0; xA ≥ 1 has the integer rounding property if and only if the blocking polyhedron B(Q) of Q = Q(A) has the integer decomposition property and all minimal integer vectors of B(Q) are columns of A (minimal with respect to ≤). Corollary 14.6.9 Let I = (xv1 , . . . , xvq ) be a monomial ideal and let A be the matrix with column vectors v1 , . . . , vq . Then I is a normal ideal if and only if the system xA ≥ 1; x ≥ 0 has the integer rounding property. Proof. According to Theorem 14.6.8, the system xA ≥ 1; x ≥ 0 has the integer rounding property if and only if the blocking polyhedron B(Q) of Q = Q(A) has the integer decomposition property and all minimal integer vectors of B(Q) are columns of A (minimal with respect to ≤). Thus the result follows at once from Theorem 14.6.4. 2 This corollary was ﬁrst observed by N. V. Trung when I is the edge ideal of a hypergraph. Lemma 14.6.10 The following equation holds: conv(w1 , . . . , wr ) = Rn+ ∩ (conv(w1 , . . . , wr ) + R+ {−e1 , . . . , −en }). Proof. The inclusion “⊂” is clear. To show the inclusion “⊃” take a vector z such that z ≥ 0 and z = λ1 w1 + · · · + λr wr − δ1 e1 − · · · − δn en ,

(14.19)

where λi ≥ 0, λ1 + · · · + λr = 1, and δj ≥ 0 for all i, j. Consider the vector z = λ1 w1 + · · · + λr wr − δ1 e1 . We set T = conv(w1 , . . . , wr ) and wi = (wi1 , . . . , win ). We claim that z is in T . We may assume that δ1 > 0,

618

Chapter 14

λi > 0 for all i, and that the ﬁrst entry wi1 of wi is positive for 1 ≤ i ≤ s and is equal to zero for i > s. From Eq. (14.19) we get λ1 w11 + · · · + λs ws1 ≥ δ1 . Case (I): λ1 w11 ≥ δ1 . Then we can write δ1 δ1 z = (w1 − w11 e1 ) + λ1 − w1 + λ2 w2 + · · · + λr wr . w11 w11 Notice that w1 − w11 e1 is again in {w1 , . . . , wr }. Thus z is a convex combination of w1 , . . . , wr , i.e., z ∈ T . Case (II): λ1 w11 < δ1 . Let m be the largest integer less than or equal to s such that λ1 w11 + · · · + λm−1 w(m−1)1 < δ1 ≤ λ1 w11 + · · · + λm wm1 . Then !: 9 m−1 m−1 λi wi1 δ 1 (wm − wm1 e1 ) + z = λi (wi − wi1 e1 ) + − wm1 wm1 i=1 i=1 !: 9 m−1 r λi wi1 δ1 wm + λm − + λi wi . wm1 wm1 i=1 i=m+1 Notice that wi − wi1 e1 is again in {w1 , . . . , wr } for i = 1, . . . , m. Thus z is a convex combination of w1 , . . . , wr , i.e., z ∈ T . This proves the claim. We can apply the argument above to any entry of z or z , thus we obtain that z − δ2 e2 ∈ T . Thus by induction we obtain that z ∈ T , as required. 2 Deﬁnition 14.6.11 [372, p. 117] Let P be a rational polyhedron in Rn . The antiblocking polyhedron of P is deﬁned as: T (P) := {z| z ≥ 0; z, x ≤ 1 for all x ∈ P}. Lemma 14.6.12 If P = {x| x ≥ 0; xA ≤ 1}, then T (P) = conv(w1 , . . . , wr ). Proof. One has the equality P = {z| z ≥ 0; z, wi ≤ 1 ∀i} because for each wi there is vj such that wi ≤ vj . Hence, by Corollary 1.1.34, we can write P = {z| z ≥ 0; z, wi ≤ 1 ∀i} = conv(β0 , β1 , . . . , βm )

(14.20)

for some β1 , . . . , βm in Qn+ and β0 = 0. From Eq. (14.20) we get {z| z ≥ 0; z, βi ≤ 1 ∀i} = T (P).

(14.21)

Using Eq. (14.20) and noticing that βi , wj ≤ 1 for all i, j, we get Rn+ ∩ (conv(β0 , . . . , βm ) + R+ {−e1 , . . . , −en }) = {z| z ≥ 0; z, wi ≤ 1 ∀i}. Hence using this equality and [372, Theorem 9.4] we obtain Rn+ ∩ (conv(w1 , . . . , wr ) + R+ {−e1 , . . . , −en }) = {z| z ≥ 0; z, βi ≤ 1 ∀i}. (14.22)

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Therefore, by Lemma 14.6.10 together with Eqs. (14.21) and (14.22), we 2 conclude that T (P) is equal to conv(w1 , . . . , wr ), as required. Deﬁnition 14.6.13 Let A be a matrix with entries in N. The linear system x ≥ 0; xA ≤ 1 has the integer rounding property if min{y, 1| y ≥ 0; Ay ≥ a} = min{y, 1| Ay ≥ a; y ∈ Nq } for each integral vector a for which min{y, 1| y ≥ 0; Ay ≥ a} is ﬁnite. Remark 14.6.14 Let A be a matrix with entries in N. Then the linear system x ≥ 0; xA ≤ 1 has the integer rounding property if and only if min{y, 1| y ≥ 0; Ay ≥ a} = min{y, 1| Ay ≥ a; y ∈ Nq } for each vector a ∈ Nn for which min{y, 1| y ≥ 0; Ay ≥ a} is ﬁnite. This follows decomposing an integral vector a as a = a+ − a− and noticing that for y ≥ 0 we have that Ay ≥ a if and only if Ay ≥ a+ Theorem 14.6.15 ([19], [373, p. 82]) If P = {x| x ≥ 0; xA ≤ 1}, then the system xA ≤ 1; x ≥ 0 has the integer rounding property if and only if T (P) has the integer decomposition property and all maximal integer vectors of T (P) are columns of A (maximal with respect to ≤). Theorem 14.6.16 The system x ≥ 0; xA ≤ 1 has the integer rounding property if and only if the subring K[xw1 t, . . . , xwr t] is normal. Proof. Let P = {x| x ≥ 0; xA ≤ 1} and let T (P) be its antiblocking polyhedron. According to Lemma 14.6.12 one has T (P) = conv(w1 , . . . , wr ).

(14.23)

Setting B = {(wi , 1)}ri=1 , by Theorem 14.6.15 and Corollary 9.1.3, it suﬃces to prove that the equality R+ B ∩ ZB = NB holds true if and only if T (P) has the integer decomposition property and all maximal integer vectors of T (P) are columns of A. Assume that R+ B ∩ ZB = NB. Let b be a natural number and let a be an integer vector in bT (P). Then using Eq. (14.23) it is seen that (a, b) is in R+ B. In our situation one has ZB = Zn+1 . Hence (a, b) ∈ NB and a is the sum of b integer vectors in T (P). Thus T (P) has the integer decomposition property. Assume that a is a maximal integer vector of T (P). It is not hard to see that (a, 1) is in R+ B, i.e., (a, 1) ∈ NB. Thus (a, 1) is a linear combination of vectors in B with coeﬃcients in N. Hence (a, 1) is equal to (wj , 1) for some j. There exists vi such that a = wj ≤ vi . Therefore by the maximality of a, we get a = vi for some i. Thus a is a column of A.

620

Chapter 14

Conversely assume that T (P) has the integer decomposition property and that all maximal integer vectors of T (P) are columns of A. Let (a, b) be an integral vector in R+ B with a ∈ Nn and b ∈ N. Hence, using Eq. (14.23), we get a ∈ bT (P). Thus a = α1 + · · · + αb , where αi is an integral vector of T (P) for all i. Since each αi is less than or equal to a maximal integer vector of T (P), we get that αi ∈ {w1 , . . . , wr }. Then (a, b) ∈ NB. 2 Corollary 14.6.17 Let A be the incidence matrix of a uniform clutter. If either linear system x ≥ 0; xA ≤ 1 or x ≥ 0; xA ≥ 1 has the integer rounding property and P = conv(v1 , . . . , vq ), then K[xv1 t, . . . , xvq t] = A(P). Proof. In general the inclusion “⊂” holds. Assume that x ≥ 0; xA ≤ 1 has the integer rounding property and that every edge of C has d elements. Let w1 , . . . , wr be the set of all α ∈ Nn such that α ≤ vi for some i. Then by Theorem 14.6.16 the subring K[xw1 t, . . . , xwr t] is normal. Using that v1 , . . . , vq is the set of wi with |wi | = d, it is not hard to see that A(P) is contained in K[xv1 t, . . . , xvq t]. Assume that x ≥ 0; xA ≥ 1 has the integer rounding property. Let I = I(C) be the edge ideal of C and let R[It] be its Rees algebra. By Corollary 14.6.9, R[It] is a normal domain. Since the clutter C is uniform the required equality follows at once from Theorem 9.3.31. 2 Corollary 14.6.18 If G is a perfect unmixed graph and v1 , . . . , vq are the characteristic vectors of the maximal stable sets of G, then the subring K[xv1 t, . . . , xvq t] is normal. Proof. The minimal vertex covers of G are exactly the complements of the maximal stable sets of G. Thus |vi | = d for all i, where d = dim(R/I(G)). On the other hand the maximal stable sets of G are exactly the maximal cliques of G. Thus, by Theorem 13.6.8 and Corollary 14.6.17, the subring K[xv1 t, . . . , xvq t] is an Ehrhart ring, and consequently it is normal. 2 Proposition 14.6.19 Let I = (xv1 , . . . , xvq ) be a monomial ideal and let vi∗ = 1 − vi . Then R[It] is normal if and only if the set Γ = {−e1 , . . . , −en , (v1∗ , 1), . . . , (vq∗ , 1)} is a Hilbert basis. Proof. Let A = {e1 , . . . , en , (v1 , 1), . . . , (vq , 1)}. Assume that R[It] is normal. Then A is a Hilbert basis. Let (a, b) be an integral vector in R+ Γ, with a ∈ Zn and b ∈ Z. Then we can write (a, b) = μ1 (−e1 ) + · · · + μn (−en ) + λ1 (v1∗ , 1) + · · · + λq (vq∗ , 1),

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where μi ≥ 0 and λj ≥ 0 for all i, j. Therefore −(a, −b) + b1 = μ1 e1 + · · · + μn en + λ1 (v1 , 1) + · · · + λq (vq , 1), where 1 = e1 + · · · + en . As A is a Hilbert basis we can write −(a, −b) + b1 = μ1 e1 + · · · + μn en + λ1 (v1 , 1) + · · · + λq (vq , 1), where μi ∈ N and λj ∈ N for all i, j. Thus (a, b) ∈ NΓ. This proves that Γ is a Hilbert basis. The converse can be shown using similar arguments. 2 Deﬁnition 14.6.20 Let A = (aij ) be a matrix with entries in {0, 1}. Its dual is the matrix A∗ = (a∗ij ), where a∗ij = 1 − aij . The following duality is valid for incidence matrices of clutters. It will be used later to establish a duality theorem for monomial subrings. Theorem 14.6.21 If A is the incidence matrix of a clutter and A∗ is its dual matrix, then x ≥ 0; xA ≥ 1 has the integer rounding property if and only if x ≥ 0; xA∗ ≤ 1 has the integer rounding property. Proof. We set Q = {x|x ≥ 0; xA ≥ 1} and P ∗ = {x|x ≥ 0; xA∗ ≤ 1}. If vi∗ = 1 − vi for all i, then A∗ is the matrix with column vectors v1∗ , . . . , vq∗ . Let w1∗ , . . . , ws∗ be the set of all α ∈ Nn such that α ≤ vi∗ for some i. Then, using Lemmas 14.6.2 and 14.6.12, we obtain B(Q) = Rn+ + conv(v1 , . . . , vq ) and T (P ∗ ) = conv(w1∗ , . . . , ws∗ ). ⇒) Thanks to Theorem 14.6.15 we need only show that T (P ∗ ) has the integer decomposition property and all maximal integer vectors of T (P ∗ ) ∗ are columns of A∗ . Let b ∈ N+ and let a be an integer vector in bT (P ). ∗ ∗ Then we can write a = b(λ1 w1 + · · · + λs ws ) with i λi = 1 and λi ≥ 0. For each 1 ≤ i ≤ s there is vj∗i in {v1∗ , . . . , vq∗ } such that wi∗ ≤ vj∗i . Thus for each i we can write 1 − wi∗ = vji + δi , where δi ∈ Nn . Therefore 1 − a/b = λ1 (vj1 + δ1 ) + · · · + λs (vjs + δs ). This means that 1 − a/b ∈ B(Q), i.e., b1 − a is an integer vector in bB(Q). Hence by Theorem 14.6.8 we can write b1 − a = α1 + · · · + αb for some α1 , . . . , αb integer vectors in B(Q), and for each αi there is vki in {v1 , . . . , vq } such that vki ≤ αi . Thus αi = vki + i for some i ∈ Nn and consequently: a = (1 − vk1 ) + · · · + (1 − vkb ) − c = vk∗1 + · · · + vk∗b − c, where c = (c1 , . . . , cn ) ∈ Nn . Notice that vk∗1 + · · · + vk∗b ≥ c because a ≥ 0. If c1 ≥ 1, then the ﬁrst entry of vk∗i is non-zero for some i and we can write a = vk∗1 + · · · + vk∗i−1 + (vk∗i − e1 ) + vk∗i+1 + · · · + vk∗b − (c − e1 ).

622

Chapter 14

Since vk∗i − e1 is again in {w1∗ , . . . , ws∗ }, we can apply this argument recursively to obtain that a is the sum of b integer vectors in {w1∗ , . . . , ws∗ }. This proves that T (P ∗ ) has the integer decomposition property. Let a be a maximal integer vector of T (P ∗ ). As the vectors w1∗ , . . . , ws∗ have entries in {0, 1}, we get T (P ∗ ) ∩ Zn = {w1∗ , . . . , ws∗ }. Then a = wi∗ for some i. As wi∗ ≤ vj∗ for some j, we get that a = vj∗ , i.e., a is a column of A∗ . ⇐) Thanks to Corollary 14.6.9, the system x ≥ 0; xA ≥ 1 has the integer rounding property if and only if R[It] is normal. Thus by Proposition 14.6.19 we need only show that the set Γ = {−e1 , . . . , −en , (v1∗ , 1), . . . , (vq∗ , 1)} is a Hilbert basis. Let (a, b) be an integral vector in R+ Γ, with a ∈ Zn and b ∈ Z. Then we can write (a, b) = μ1 (−e1 ) + · · · + μn (−en ) + λ1 (v1∗ , 1) + · · · + λq (vq∗ , 1), where μi ≥ 0, λj ≥ 0 for all i, j. Hence A∗ λ ≥ a, where λ = (λi ). By hypothesis the system x ≥ 0; xA∗ ≤ 1 has the integer rounding property. Then one has b ≥ min{y, 1| y ≥ 0; A∗ y ≥ a} = min{y, 1| A∗ y ≥ a; y ∈ Nq } = y0 , 1 for some y0 = (yi ) ∈ Nq such that |y0 | = y0 , 1 ≤ b and a ≤ A∗ y0 . Then a = y1 v1∗ + · · · + yq vq∗ − δ1 e1 − · · · − δn en , where δ1 , . . . , δn are in N. Hence we can write ∗ , 1) + (yq + b − |y0 |)(vq∗ , 1) − (b − |y0 |)vq∗ − δ, (a, b) = y1 (v1∗ , 1) + · · · + yq−1 (vq−1

where δ = (δi ). As the entries of A∗ are in N, the vector −vq∗ can be written as a nonnegative integer combination of −e1 , . . . , −en . Thus (a, b) ∈ NΓ. This proves that Γ is a Hilbert basis. 2 The following result is a duality theorem for monomial subrings. Theorem 14.6.22 [52] Let A be the incidence matrix of a clutter and let vi∗ be the vector 1 − vi . If w1∗ , . . . , ws∗ is the set of all α ∈ Nn such that α ≤ vi∗ for some i, then the following conditions are equivalent: (a) R[It] is normal, where I = (xv1 , . . . , xvq ). ∗

∗

(b) S ∗ = K[xw1 t, . . . , xws t] is normal. (c) {−e1 , . . . , −en , (v1∗ , 1), . . . , (vq∗ , 1)} is a Hilbert basis. (d) x ≥ 0; xA ≥ 1 has the integer rounding property. (e) x ≥ 0; xA∗ ≤ 1 has the integer rounding property. Proof. (a) ⇔ (c): This was shown in Proposition 14.6.19. (a) ⇔ (d): This was shown in Corollary 14.6.9. (b) ⇔ (e): This part was shown in Theorem 14.6.16. (d) ⇔ (e): This follows from Theorem 14.6.21. 2

Combinatorial Optimization and Blowup Algebras

623

Deﬁnition 14.6.23 Let I be the edge ideal of a clutter C. The dual of I, denoted by I ∗ , is the ideal of R generated ; by all monomials x1 · · · xn /xe such that e is an edge of C, where xe = xi ∈e xi . Corollary 14.6.24 Let C be a clutter and let A be its incidence matrix. If P = {x|x ≥ 0; xA ≤ 1} is an integral polytope and I = I(C), then R[I ∗ t] and S = K[xw1 t, . . . , xwr t] are normal. Proof. Since P has only integral vertices, by a theorem of Lov´asz (see Theorem 13.6.8) the system x ≥ 0; xA ≤ 1 is totally dual integral, i.e., the minimum in the LP-duality equation max{a, x| x ≥ 0; xA ≤ 1} = min{y, 1| y ≥ 0; Ay ≥ a}

(14.24)

has an integral optimum solution y for each integral vector a with ﬁnite minimum. In particular the system x ≥ 0; xA ≤ 1 satisﬁes the integer rounding property. Therefore R[I ∗ t] and K[xw1 t, . . . , xwr t] are normal by Theorem 14.6.22. 2 Corollary 14.6.25 [423] If G is a perfect graph, then R[Ic (G)t] is normal. Proof. Let C be the clique clutter of G, the complement of G, and let A be its incidence matrix. The graph G is perfect by the weak perfect graph theorem (see Theorem 13.6.1). Hence, by Theorem 13.6.8, the polytope P(A) = {x| x ≥ 0; xA ≤ 1} is integral. If I is the edge ideal of C, by Corollary 14.6.24, the ideal I ∗ is normal. To complete the proof notice that 2 I ∗ is equal to Ic (G). Corollary 14.6.26 [165, Corollary 5.11] If Ic (G) is the ideal of covers of a perfect graph G, then Ass(R/Ic (G)k ) form an ascending chain. Proof. We set J = Ic (G). By Corollary 14.6.25, R[Jt] is normal. Thus J k = J k for all k, so by Theorem 7.7.3, J has the persistence property. 2 Proposition 14.6.27 Let C be a balanced clutter and let I be its edge ideal. Then the Rees algebra R[I ∗ t] of I ∗ is normal. Proof. Let A be the incidence matrix of C. This matrix is balanced by hypothesis. By Theorem 13.6.7 A is balanced if and only if every submatrix of A is perfect. Thus, the result follows from Corollary 14.6.24. 2 Corollary 14.6.28 If G is a perfect and unmixed graph, then R[Ic (G)t] is a Gorenstein standard graded K-algebra. Proof. We set g = ht(I(G)) and J = Ic (G). By assigning deg(xi ) = 1 and deg(t) = −(g − 1), the Rees algebra R[Jt] becomes a standard graded

624

Chapter 14

K-algebra. The ring R[Jt] is normal by Theorem 14.6.25. Then, according to Theorem 9.1.5, its canonical module is the ideal of R[Jt] given by ωR[Jt] = ({xa1 1 · · · xann tan+1 | a = (ai ) ∈ R+ (J)o ∩ Zn+1 }). By Theorem 9.1.6 the ring R[Jt] is Cohen–Macaulay. Using Eq. (13.10) of Proposition 13.6.3 it is seen that x1 · · · xn t belongs to ωR[Jt] . Take any monomial xa tb = xa1 1 · · · xann tb in the ideal ωR[Jt] , that is (a, b) ∈ R+ (J)o . Hence the vector (a, b) has positive integer entries and satisﬁes (14.25) xi ∈Kr ai ≥ (r − 1)b + 1 for every complete subgraph Kr of G. If b = 1, clearly xa tb is a multiple of x1 · · · xn t. Now assume b ≥ 2. Using the normality of R[Jt] and Eqs. (13.10) and (14.25) it follows that the monomial m = xa1 1 −1 · · · xann −1 tb−1 belongs to R[Jt]. Since xa tb = mx1 · · · xn t, we obtain that ωR[Jt] is generated by x1 · · · xn t and thus R[Jt] is a Gorenstein ring. 2 Corollary 14.6.29 Let B1 , . . . , Bq be the collection of basis of a matroid M with vertex set X and let v1 , . . . , vq be their characteristic vectors. If A is the matrix with column vectors v1 , . . . , vq , then all systems x ≥ 0; xA ≥ 1,

x ≥ 0; xA∗ ≥ 1,

x ≥ 0; xA ≤ 1,

x ≥ 0; xA∗ ≤ 1

have the integer rounding property. Proof. Consider the basis monomial ideal I = (xv1 , . . . , xvq ) of the matroid M . By [338, Theorem 2.1.1], the collection of basis of the dual matroid M ∗ of M is given by X \ B1 , . . . , X \ Bq . Now, the basis monomial ideal of a matroid is normal (Corollary 12.3.12). Thus the result follows at once from the duality of Theorem 14.6.22. 2 Linear systems of the form xA ≤ 1 We now turn our attention to study the integer rounding property of linear systems of inequalities of the form xA ≤ 1 (see Deﬁnition 1.3.23). Proposition 14.6.30 Let v1 , . . . , vq be the column vectors of a nonnegative integer matrix A and let A(P) be the Ehrhart ring of P = conv(0, v1 , . . . , vq ). Then the system xA ≤ 1 has the integer rounding property if and only if K[xv1 t, . . . , xvq t, t] = A(P). Proof. By Theorem 1.3.24, we have that the system xA ≤ 1 has the integer rounding property if and only if the set B = {(v1 , 1), . . . , (vq , 1), (0, 1)} is an integral Hilbert basis. Thus the proposition follows by noticing the equality A(P) = K[{xa tb |(a, b) ∈ R+ B ∩ Zn+1 }] and the inclusion K[xv1 t, . . . , xvq t, t] ⊂ A(P).

2

Combinatorial Optimization and Blowup Algebras

625

Theorem 14.6.31 Let A = {v1 , . . . , vq } be the set of column vectors of a matrix A with entries in N. If the system xA ≤ 1 has the integer rounding property, then (a) K[F ] is normal, where F = {xv1 , . . . , xvq }, and (b) Zn /ZA is a torsion-free group. The converse holds if |vi | = d for all i. Proof. If xA ≤ 1 has the integer rounding property, then (a) and (b) follow from Corollary 1.3.32. Conversely assume that |vi | = d for all i and that (a) and (b) hold. We need only show that B = {(vi , 1)}qi=1 ∪ {en+1 } is a Hilbert basis. Let (a, b) be an integral vector in R+ B, where a ∈ Nn and b ∈ N. Then we can write (a, b) = λ1 (v1 , 1) + · · · + λq (vq , 1) + μ(0, 1),

(14.26)

for some λ1 , . . . , λq , μ in Q+ . Hence using (b) gives that a is in R+ A ∩ ZA. q Hence xa ∈ K[F ] = K[F ], i.e., a ∈ NA. Then we can write a = i=1 ηi vi for some η1 , . . . , ηq in Nn . Since |vi | = d for all i, one has i λi = i ηi . Therefore using Eq. (14.26), we get μ ∈ N. Therefore we get (a, b) = η1 (v1 , 1) + · · · + ηq (vq , 1) + μ(0, 1) ⇒ (a, b) ∈ NB.

2

Corollary 14.6.32 Let A be the incidence matrix of a connected graph G. Then the system xA ≤ 1 has the integer rounding property if and only if G is a bipartite graph. Proof. ⇒) Let A = {v1 , . . . , vq } be the set of columns of A. If G is not bipartite, then according to Corollary 10.2.11 one has Zn /ZA Z2 , a contradiction to Theorem 14.6.31(b). ⇐) By Corollary 10.2.11 and Proposition 10.3.1 the ring K[xv1 , . . . , xvq ] is normal and Zn /ZA Z. Thus, by Theorem 14.6.31, the system xA ≤ 1 has the integer rounding property, as required. 2 Integer rounding properties in simple graphs Let G be a simple graph with vertex set X = {x1 , . . . , xn }, let v1 , . . . , vq be the column vectors of the incidence matrix A of G, let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and let I = I(G) be the edge ideal of G. Recall that the extended Rees algebra of I is the subring R[It, t−1 ] := R[It][t−1 ] ⊂ R[t, t−1 ], where R[It] is the Rees algebra of I. Lemma 14.6.33 R[It, t−1 ] K[t, x1 t, . . . , xn t, xv1 t, . . . , xvq t].

626

Chapter 14

Proof. We set S = K[t, x1 t, . . . , xn t, xv1 t, . . . , xvq t]. Note that S and R[It, t−1 ] are integral domains of the same dimension. This follows from the dimension formula given in Corollary 8.2.21. Thus it suﬃces to prove that there is an epimorphism ψ : S → R[It, t−1 ] of K-algebras. Let u0 , u1 , . . . , un , t1 , . . . , tq be a new set of variables and let ϕ, ψ be the maps of K-algebras deﬁned by the diagram K[u0 , u1 , . . . , un , t1 , . . . , tq ]

ψ

ψ k k k k k k k Sk

/ R[It, t−1 ] k5

ϕ

ϕ

u0 −→ t, ϕ ui −→ xi t, ϕ ti −→ xvi t,

ψ

u0 −→ t−1 , ψ

ui −→ xi , ψ

ti −→ xvi t.

As ker(ϕ) is a binomial ideal (Corollary 8.2.18) we get ker(ϕ) ⊂ ker(ψ). Hence there is an epimorphism ψ of K-algebras that makes the diagram 2 commutative, i.e., ψ = ψϕ. Theorem 14.6.34 Let G be a connected graph and let A be its incidence matrix. Then K[G] := K[xv1 , . . . , xvq ] is normal if and only if the system x ≥ 0; xA ≤ 1 has the integer rounding property. Proof. Let I = I(G) be the edge ideal of G. By Corollary 10.5.6 the subring K[G] is normal if and only if R[It] is normal. Using Theorem 4.3.17, we get that R[It] is normal if and only if R[It, t−1 ] is normal. By Lemma 14.6.33, R[It, t−1 ] is normal if and only if the subring S = K[t, x1 t, . . . , xn t, xv1 t, . . . , xvq t] is normal. Finally applying Theorem 14.6.16 we get that S is normal if and only if the system x ≥ 0; xA ≤ 1 has the integer rounding property. 2 Theorem 14.6.35 Let G be a connected graph and let A be its incidence matrix. Then the system x ≥ 0; xA ≤ 1 has the integer rounding property if and only if any of the following equivalent conditions hold. (a) x ≥ 0; xA ≥ 1 is a system with the integer rounding property. (b) R[It] is a normal domain, where I = I(G) is the edge ideal of G. (c) K[xv1 t, . . . , xvq t] is normal, where v1 , . . . , vq are the columns of A. (d) K[t, x1 t, . . . , xn t, xv1 t, . . . , xvq t] is normal. Proof. According to Corollary 14.6.9, the system x ≥ 0; xA ≥ 1 has the integer rounding property if and only if the Rees algebra R[It] is normal. Thus the result follows from the proof of Theorem 14.6.34. 2 Corollary 14.6.36 Let G be a graph and let I = I(G) be its edge ideal. Then R[It] is normal if and only if R[I ∗ t] is normal.

Combinatorial Optimization and Blowup Algebras

627

Proof. Recall that: (i) the Rees algebra R[It] is normal if and only if the extended Rees algebra R[It, t−1 ] is normal (Theorem 4.3.17), and (ii) R[It, t−1 ] is isomorphic to K[t, x1 t, . . . , xn t, xv1 t, . . . , xvq t], where I is the edge ideal of G (Lemma 14.6.33). Then the result follows applying Theorem 14.6.22. 2 The next example shows that Corollary 14.6.36 does not extend to arbitrary uniform clutters. Example 14.6.37 Consider the clutter C whose incidence matrix A is the transpose of the matrix: ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

0 0 0 1 0 1 1 1 1 1

0 0 1 1 1 1 1 1 1 1

1 1 1 0 1 1 1 1 1 1

1 0 0 0 0 1 1 1 1 1

0 1 0 0 1 1 1 1 1 0

1 1 1 1 0 0 0 0 1 0

1 1 1 1 1 0 0 1 1 1

1 1 1 1 1 1 1 1 0 1

1 1 1 1 1 1 0 0 0 0

1 1 1 1 1 0 1 0 0 1

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Let I = I(C) be the edge ideal of C. Note that C is uniform. Using Normaliz [68] it is seen that R[It] is normal and that R[I ∗ t] is not normal. Proposition 14.6.38 Let G be a graph without isolated vertices and let G be its complement. Then I(G)∨ = I(G)∗ if and only if G is triangle free. Proof. ⇒) Assume that G has a triangle C3 = {x1 , x2 , x3 }. We may assume n ≥ 4. Notice that C = {x4 , . . . , xn } is a vertex cover of G, i.e., x4 · · · xn belongs to I(G)∨ and consequently it belongs to I(G)∗ , a contradiction because I(G)∗ is generated by monomials of degree n − 2. ⇐) Let xa = x1 · · · xr be a minimal generator of I(G)∨ . Then the set C = {x1 , . . . , xr } is a minimal vertex cover of G. Hence X \ C is a maximal complete subgraph of G. Thus by hypothesis X \ C is an edge of G, i.e., xa ∈ I(G)∗ . This proves the inclusion “⊂”. Conversely, let xa be a minimal generator of I(G)∗ . There is an edge {x1 , x2 } of G such that xa = x3 · · · xn . Every edge of G must intersect C = {x3 , . . . , xn }, i.e., xa ∈ I(G)∨ . 2 Corollary 14.6.39 Let G be a triangle free graph without isolated vertices. Then R[I(G)t] is normal if and only if R[I(G)∨ t] is normal. Proof. It follows from Corollary 14.6.36 and Proposition 14.6.38.

2

628

Chapter 14

Example 14.6.40 Let G be the graph consisting of two vertex disjoint odd cycles of length 5 and let G be its complement. By Proposition 10.5.8 and Corollary 14.6.39 the Rees algebras R[I(G)t] and R[I(G)∨ t] are not normal.

Exercises 14.6.41 Let G be a perfect graph with vertex set X = {x1 , . . . , xn } and let S be the subring generated by all xa t such that supp(xa ) is a clique of G. Prove that S is normal. 14.6.42 Let G be a bipartite graph without isolated vertices and let G be its complement. Then I(G)∨ = I(G)∗ . 14.6.43 If I ⊂ R is a Cohen–Macaulay square-free monomial ideal of height two, then R[It] is normal. 14.6.44 Let A be the vertex-clique matrix of a graph G and let v1 , . . . , vq ∗ ∗ be the column vectors of A. Consider the ideal I ∗ = (xv1 , . . . , xvq ), where vi∗ = 1 − vi for all i. Prove the duality: I ∗ is the ideal of covers of G. 14.6.45 If G is a pentagon, then the Rees algebra of Ic (G) is normal and G is not a perfect graph. 14.6.46 Let A be the incidence matrix of a clutter C. If C is uniform and has the max-ﬂow min-cut property, then the system xA ≤ 1 has the integer rounding property.

14.7

Canonical modules and integer rounding

Here we give a description of the canonical module and the a-invariant for subrings arising from systems with the integer rounding property. Let A be a matrix of size n×q with entries in N such that A has non-zero rows and non-zero columns. Let v1 , . . . , vq be the columns of A. For use below consider the set w1 , . . . , wr of all α ∈ Nn such that α ≤ vi for some i. Let R = K[x1 , . . . , xn ] be a polynomial ring over a ﬁeld K and let S = K[xw1 t, . . . , xwr t] ⊂ R[t] be the subring of R[t] generated by xw1 t, . . . , xwr t, where t is a new variable. The ring S is a standard graded K-algebra such that deg(xa tb ) = b for any monomial xa tb of S. If S is normal, then according to Theorem 9.1.5 the canonical module of S is the ideal given by ωS = ({xa tb | (a, b) ∈ NB ∩ (R+ B)o }),

(14.27)

Combinatorial Optimization and Blowup Algebras

629

where B = {(w1 , 1), . . . , (wr , 1)} and (R+ B)o is the interior of R+ B relative to aﬀ(R+ B), the aﬃne hull of R+ B. In our case aﬀ(R+ B) = Rn+1 . Let {β0 , β1 , . . . , βm } ⊂ Qn+ be the set of vertices of the polytope P = {x| x ≥ 0; xA ≤ 1}, where β0 = 0, and let β1 , . . . , βp be the set of all maximal elements of β0 , β1 , . . . , βm (maximal with respect to ≤). The following lemma is not hard to prove. Lemma 14.7.1 For each 1 ≤ i ≤ p there is a unique positive integer δi such that the non-zero entries of (−δi βi , δi ) are relatively prime. Notation In what follows {β1 , . . . , βp } is the set of maximal elements of {β0 , . . . , βm } and δ1 , . . . , δp are the unique positive integers in Lemma 14.7.1. Theorem 14.7.2 If the system x ≥ 0; xA ≤ 1 has the integer rounding property, then S is normal, the canonical module of S is given by −δ1 β1 · · · −δp βp e1 · · · en a b ωS = x t (a, b) ≥1 , δ1 · · · δp 0 · · · 0 (14.28) and the a-invariant of S is equal to − maxi {1/δi + |βi |}. Proof. Note that in Eq. (14.28) we regard (−δi βi , δi ) and ej as column vectors for all i, j. The normality of S follows from Theorem 14.6.16. Recall that we have the following duality: P = {x| x ≥ 0; x, wi ≤ 1 ∀i} = conv(β0 , β1 , . . . , βm ), T (P) = {x| x ≥ 0; x, βi ≤ 1∀i} = conv(w1 , . . . , wr ), see Lemma 14.6.12. Therefore, by the maximality of β1 , . . . , βp , we obtain conv(w1 , . . . , wr ) = {x| x ≥ 0; x, βi ≤ 1, ∀ i = 1, . . . , p}. Hence, setting B = {(wi , 1)}ri=1 and noticing ZB = Zn+1 , it is seen that + + R+ B = He+1 ∩ · · · ∩ He+n ∩ H(−δ ∩ · · · ∩ H(−δ . 1 β1 ,δ1 ) p βp ,δp )

(14.29)

Notice that Hei ∩ R+ B and H(−δj βj ,δj ) ∩ R+ B are proper faces of R+ B for any i, j. Hence from Eq. (14.29) we get that a vector (a, b), with a ∈ Zn , b ∈ Z, is in the relative interior of R+ B if and only if the entries of a are positive and (a, b), (−δi βi , δi ) ≥ 1 for all i. Thus the required expression for ωS follows using the normality of S and Eq. (14.27). It remains to prove the formula for a(S), the a-invariant of S. Consider the vector (1, b0 ), where b0 = maxi {1/δi + |βi |}. Using Eq. (14.28), it is

630

Chapter 14

not hard to see (by direct substitution of (1, b0 )), that the monomial x1 tb0 is in ωS . Thus, from Eq. (9.2) of Lemma 9.1.7, we get a(S) ≥ −b0 . Conversely if xa tb is in ωS , then again from Eq. (14.28) we get (−δi βi , δi ), (a, b) ≥ 1 for all i and ai ≥ 1 for all i, where a = (ai ). Hence bδi ≥ 1 + δi a, βi ≥ 1 + δi 1, βi = 1 + δi |βi |. Since b is an integer we obtain b ≥ 1/δi + |βi | for all i. Therefore b ≥ b0 , i.e., deg(xa tb ) = b ≥ b0 . As xa tb was an arbitrary monomial in ωS , by the formula for the a-invariant of S given in Eq. (9.2) of Lemma 9.1.7, we obtain that a(S) ≤ −b0 . Altogether one has a(S) = −b0 , as required. 2 Theorem 14.7.3 Assume that the system x ≥ 0; xA ≤ 1 has the integer rounding property. If S is Gorenstein and c0 = max{|βi | : 1 ≤ i ≤ p} is an integer, then |βk | = c0 for each 1 ≤ k ≤ p such that βk has integer entries. Proof. We proceed by contradiction. Assume that |βk | < c0 for some integer 1 ≤ k ≤ p such that βk is integral. We may assume that βk is (1, . . . , 1, 0, . . . , 0) and |βk | = s. From Eq. (14.29) xβk ts−1 cannot be in S + . Consider xa tb , where because (βk , s − 1) does not belong to H(−δ k βk ,δk ) a = βk + 1, b = b0 + s − 1 and b0 = −a(S). We claim that xa tb is in ωS . By Theorem 14.7.2 it suﬃces to show that (a, b), (−δj βj , δj ) ≥ 1 for 1 ≤ j ≤ p. Thus we need only show that (a, b), (−βj , 1) > 0 for 1 ≤ j ≤ p. From the proof of Theorem 14.7.2, it is seen that −a(S) = maxi {(|βi |)} + 1. Hence we get b0 = c0 + 1. One has the following equality (a, b), (−βj , 1) = −|βj | − βk , βj + c0 + s. Set βj = (βj1 , . . . , βjn ). From Eq. (14.29) we get that the entries of each βj are less than or equal to 1. Case (I): If βji < 1 for some 1 ≤ i ≤ s, then s − βk , βj > 0 and c0 ≥ |βj |. Case (II): βji = 1 for 1 ≤ i ≤ s. Then βj ≥ βk . Thus by the maximality of βk we obtain βj = βk . In both cases we obtain (a, b), (−βj , 1) > 0, as required. Hence the monomial xa tb is in ωS . Since S is Gorenstein and ωS is generated by x1 tb0 , we obtain that xa tb is a multiple of x1 tb0 , i.e., xβk ts−1 must be in S, a contradiction. 2 Theorem 14.7.4 Assume that x ≥ 0; xA ≤ 1 has the integer rounding property. If −a(S) = 1/δi + |βi | for i = 1, . . . , p, then S is Gorenstein. Proof. We set b0 = −a(S) and B = {(w1 , 1), . . . , (wr , 1)}. The ring S is normal by Theorem 14.6.16. Since the monomial x1 tb0 = x1 · · · xn tb0 is in ωS , we need only show that ωS = (x1 tb0 ). Take xa tb ∈ ωS . It suﬃces to prove that xa−1 tb−b0 is in S. Using Theorem 14.6.16, one has the equality R+ B ∩ Zn+1 = NB. Thus we need only show that (a − 1, b − b0 ) is in R+ B.

Combinatorial Optimization and Blowup Algebras

631

From Eq. (14.29), the proof reduces to showing that (a − 1, b − b0 ) is in + for i = 1, . . . , p. As (a, b) ∈ ωS , from Theorem 14.7.2, we get H(−β i ,1) (a, b), (−δi βi , δi ) = −a, δi βi + bδi ≥ 1 =⇒ −a, βi ≥ −b + 1/δi for i = 1, . . . , p. Therefore (a−1, b−b0), (−βi , 1) = −a, βi +|βi |+b−b0 ≥ −b+1/δi +|βi |+b−b0 = 0 for all i, as required.

2

Corollary 14.7.5 If P = {x| x ≥ 0; xA ≤ 1} is an integral polytope, then S is Gorenstein if and only if a(S) = −(|βi | + 1) for i = 1, . . . , p. Proof. Notice that if P is integral, then βi has entries in {0, 1} for 1 ≤ i ≤ p and consequently δi = 1 for 1 ≤ i ≤ p. Thus the result follows from Theorems 14.7.3 and 14.7.4. 2 Conjecture 14.7.6 If A is the incidence matrix of a connected graph and the system x ≥ 0; xA ≤ 1 has the integer rounding property, then S is Gorenstein if and only if −a(S) = 1/δi + |βi | for i = 1, . . . , p. The answer to this conjecture is positive if A is the incidence matrix of a bipartite graph because in this case P is an integral polytope and we may apply Corollary 14.7.5. For use below we consider the empty set as a clique whose vertex set is empty. Note that supp(xa ) = ∅ if and only if a = 0. Theorem 14.7.7 Let G be a perfect graph and let S = K[xω1 t, . . . , xωr t] be the subring generated by all square-free monomials xa t such that supp(xa ) is a clique of G. Then the canonical module of S is given by −β1 · · · −βs e1 · · · en a b ωS = x t (a, b) ≥1 , 1 ··· 1 0 ··· 0 where β1 , . . . , βs are the characteristic vectors of the maximal independent sets of G, and the a-invariant of S is equal to −(maxi {|βi |} + 1). Proof. Let v1 , . . . , vq be the set of characteristic vectors of the maximal cliques of G. Note that w1 , . . . , wr is the set of all α ∈ Nn such that α ≤ vi for some i. Since G is a perfect graph, by Theorem 13.6.1, we have P = {x|x ≥ 0; xA ≤ 1} = conv(β0 , β1 , . . . , βp ), where β0 = 0 and β1 , . . . , βp are the characteristic vectors of the independent sets of G. We may assume that β1 , . . . , βs correspond to the maximal independent sets of G. Furthermore, since P has only integral vertices, by

632

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a result of Lov´ asz (see Theorem 13.6.8) the system x ≥ 0; xA ≤ 1 is totally dual integral, i.e., the minimum in the LP-duality equation max{α, x| x ≥ 0; xA ≤ 1} = min{y, 1| y ≥ 0; Ay ≥ α}

(14.30)

has an integral optimum solution y for each integral vector α with ﬁnite minimum. In particular the system x ≥ 0; xA ≤ 1 satisﬁes the integer rounding property. Therefore the result follows readily from Theorem 14.7.2. 2 Proposition 14.7.8 Let G be a perfect graph, let w1 , . . . , wr be the characteristic vectors of the cliques of G, and let β1 , . . . , βs be the characteristic vectors of the maximal independent sets of G. Then the set Γ = {(−β1 , 1), . . . , (−βs , 1), e1 , . . . , en } is a Hilbert basis of (R+ B)∗ , where B = {(w1 , 1), . . . , (wr , 1)}. Proof. The characteristic vector of the empty set is set to be equal to zero. As G is perfect we have conv(w1 , . . . , wr ) = {x| x ≥ 0; x(β1 · · · βs ) ≤ 1}. Therefore it is seen that R+ (w1 , 1) + · · · + R+ (wr , 1) = He+1 ∩ · · · ∩ He+n ∩ R+ (−β1 , 1) ∩ · · · R+ (−βs , 1). Thus, by duality (see Corollary 1.1.30), we obtain that (R+ B)∗ = R+ Γ. Using that the system x ≥ 0; x(β1 , . . . , βs ) ≤ 1 is TDI it follows that 2 Zn ∩ R+ Γ = NΓ, i.e., Γ is a Hilbert basis, as required. Corollary 14.7.9 Let G be a connected bipartite graph and let I = I(G) be its edge ideal. Then the extended Rees algebra R[It, t−1 ] is a Gorenstein standard K-algebra if and only if G is unmixed. Proof. Let ωS be the canonical module of S = R[It, t−1 ]. As S is Cohen– Macaulay, recall that S is Gorenstein if and only if ωS is a principal ideal. Since G is a perfect graph. The result follows using Lemma 14.6.33 together with the description of the canonical module given in Theorem 14.7.7. 2

Exercise 14.7.10 If A is the incidence matrix of a graph, then δi = 1 or δi = 1/2 for each i (see Lemma 14.7.1).

Combinatorial Optimization and Blowup Algebras

14.8

633

Clique clutters of Meyniel graphs

In this section it is shown that clique clutters of Meynel graphs are Ehrhart in the sense of [306] (see Deﬁnition 9.3.27). Recall that a Meyniel graph is a simple graph in which every odd cycle of length at least ﬁve has at least two chords. Theorem 14.8.1 [247] A graph G is Meyniel if and only if for each induced subgraph H and for each vertex v of H, there exists an independent set in H containing v and intersecting all maximal cliques of H (maximal with respect to inclusion). Theorem 14.8.2 Let G be a Meyniel graph and let A = {v1 , . . . , vq } be the set of columns of the vertex-clique matrix of G. If A(P) is the Ehrhart ring of P = conv(v1 , . . . , vq ), then K[xv1 t, . . . , xvq t] = A(P). Proof. The inclusion “⊂” is clear. To prove the other inclusion take xa tb ∈ A(P), i.e., a ∈ bP ∩ Zn and b ∈ N. Using that P is the convex hull of the vi ’s, it is not hard to see that we can write (a, b) = λ1 (v1 , 1) + · · · + λq (vq , 1), λi ≥ 0 ∀ i.

(14.31)

Let A be the vertex-clique matrix of G. Any Meyniel graph is perfect [373, Theorem 66.6]. Thus, by Theorem 13.6.8, the system x ≥ 0; xA ≤ 1 is TDI. Therefore, by Proposition 1.3.28, we get that the set H = {(v1 , 1), . . . , (vq , 1), −e1 , . . . , −en } is a Hilbert basis, i.e., R+ H ∩ Zn+1 = NH. As (a, b) is an integral vector in R+ H, we can write (a, b) = η1 (v1 , 1) + · · · + ηq (vq , 1) − μ1 e1 − · · · − μn en ,

(14.32)

ηi ∈ N, μj ∈ N for all i, j. By Theorem 14.8.1, for each xk in V (G) = {x1 , . . . , xn } there exists an independent set Bk of G containing xk and intersecting all maximal cliques of G. We set βk = xi ∈Bk ei for 1 ≤ k ≤ n. Notice that a clique of G and an independent set of G can meet in at most one vertex. Then β1 , . . . , βn are vectors in {0, 1}n such that vj , βi = 1 and ei , βi = 1 for all i, j. Hence, using Eqs. (14.31) and (14.32), we obtain a, βi a, βi

= λ1 v1 , βi + · · · + λq vq , βi = b, ∀ i. = η1 v1 , βi + · · · + ηq vq , βi − μ1 e1 , βi − · · · − μn en , βi

= b − μ1 e1 , βi − · · · − μn en , βi , ∀ i. n Therefore for i = 1, . . . , n, we get j=1 μj ej , βi = 0. Since ei , βi = 1 for all i, we get μi = 0 for all i. Hence the vector (a, b) belongs to the 2 semigroup NA. Thus xa tb ∈ K[xv1 t, . . . , xvq t].

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Corollary 14.8.3 Let G be a Meyniel graph, let A be the vertex-clique matrix of G, and let v1 , . . . , vq be the column vectors of A. The following conditions are equivalent: (i) x ≥ 0; xA ≥ 1 is a TDI system. (ii) I i = I (i) for i ≥ 1, where I = (xv1 , . . . , xvq ). (iii) Q(A) = {x| x ≥ 0; xA ≥ 1} is an integral polyhedron. Proof. (i) ⇒ (ii) and (ii) ⇒ (iii): Follow from Theorem 14.3.6. (iii) ⇒ (i): By Theorem 14.8.2 we get K[xv1 t, . . . , xvq t] = A(P). As Q(A) is integral, a direct application of Proposition 14.3.30 and Theorem 14.3.6 gives that the system x ≥ 0; xA ≥ 1 is TDI. 2

Exercises 14.8.4 Let v1 , . . . , v6 be the characteristic vectors of the maximal cliques of the graph below and let P be its convex hull. s s s s @ @ G @ @ s @s s @s Use Normaliz [68] to show that K[xv1 t, . . . , xv6 t] A(P). Show that the incidence matrix of cl(G), the clique clutter of G, is totally unimodular by showing that the incidence matrix of cl(G) is the transpose of the incidence matrix of the complete bipartite graph K2,4 . 14.8.5 Consider the following graph G: ppsp pp p p p s p ppp p p p p pp p p p pT p p p p pp p p s p p p p p ps pp T p p p p T pp p pp p p p pp p p p Tpsp pp s

s

s p p pp p pspp pp p p p p p p p p p ppT p p ppsppppppppp pp Ts pp T p pp p p p pp p p p pp p pppp p Ts p p ps

Prove that the edge ideal I = I(cl(G)) of the clique clutter of G is not normal. This graph is chordal, hence perfect. Thus edge ideals of clique clutters of perfect graphs are in general not normal.

Appendix Graph Diagrams For convenience we will display some Cohen–Macaulay graphs and unmixed graphs with small number of vertices.

A.1 Cohen–Macaulay graphs The complete list of Cohen–Macaulay connected graphs with at most six vertices is: •

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636

Appendix •

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Graph Diagrams

637

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638

Appendix

A.2 Unmixed graphs Next we display the set of all unmixed non-Cohen–Macaulay connected graphs with at most six vertices. ? / •/?/?? // ?? // ?? // ?• • // // / • •

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Mathematics

MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS

Monomial Algebras, Second Edition presents algebraic, combinatorial, and computational methods for studying monomial algebras and their ideals, including Stanley–Reisner rings, monomial subrings, Ehrhart rings, and blowup algebras. It emphasizes square-free monomials and the corresponding graphs, clutters, or hypergraphs.

Monomial Algebras

Bringing together several areas of pure and applied mathematics, this book shows how monomial algebras are related to polyhedral geometry, combinatorial optimization, and combinatorics of hypergraphs. It directly links the algebraic properties of monomial algebras to combinatorial structures (such as simplicial complexes, posets, digraphs, graphs, and clutters) and linear optimization problems.

Villarreal

Features • Presents computational and combinatorial methods in commutative algebra • Shows how to solve a variety of problems of monomial algebras • Covers various affine and graded rings, including Cohen–Macaulay, complete intersection, and normal • Examines their basic algebraic invariants, such as multiplicity, Betti numbers, projective dimension, and Hilbert polynomial • Contains more than 550 exercises and over 50 examples, many of which illustrate the use of computer algebra systems

Second Edition

New to the Second Edition • Four new chapters that focus on the algebraic properties of blowup algebras in combinatorial optimization problems of clutters and hypergraphs • Two new chapters that explore the algebraic and combinatorial properties of the edge ideal of clutters and hypergraphs • Full revisions of existing chapters to provide an up-to-date account of the subject

Monomial Algebras

MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS

Second Edition

Rafael H. Villarreal

K23008

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2/13/15 12:23 PM