MONOMIAL ALGEBRAS
PURE AND APPLIED MATHEMATICS A Program of Monographs, Textbooks, and Lecture Notes
EXECUTIVE EDITORS Zuhair Nashed University of Delaware Newark, Delaware
Earl J . Taft Rutgers University New Brunswick, New Jersey
EDITORIAL BOARD M. S. Baouendi University of California, San Diego Jane Cronin Rutgers University Jack K. Hale Georgia Institute of Technology
AniI Nerode Cornell University Donald Passman University of Wisconsin, Madison Fred S. Roberts Rutgers University
S. Kobayashi University of California, Berkeley
David L. Russell Virginia Polytechnic Institute and State University
Marvin Marcus University of California, Santa Barbara
Walter Schempp Universitat Siegen
W. S. Massey Yale University
Mark Teply University of Wisconsin, Milwaukee
MONOGRAPHS AND TEXTBOOKS IN PURE AND APPLIED MATHEMATICS 1. K. Yano, Integral Formulas in Riemannian Geometry (1970) 2. S. Kobayashi, Hyperbolic Manifolds and Holomorphic Mappings (1970) 3. V. S. Vladimirov, Equations of Mathematical Physics (A. Jeffrey, ed.; A. Littlewood, trans.) (1970) 4. B. N. Pshenichnyi, Necessary Conditions for an Extremum (L. Neustadt, translation ed.; K. Makowski, trans.) (1971) 5. L. Narici et a/., Functional Analysis and Valuation Theory (1971) 6. S. S. Passman, Infinite Group Rings (1971) 7. L. Dornhoff, Group Representation Theory. Part A: Ordinary Representation Theory. Part B: Modular Representation Theory (1971, 1972) 8. W. Boothby and G. L. Weiss, eds., Symmetric Spaces (1972) 9. Y. Mafsushima, Differentiable Manifolds (E. T. Kobayashi, trans.) (1972) 10. L. E. Ward, Jr., Topology (1972) 11. A. Babakhanian, Cohomological Methods in Group Theory (1972) 12. R. Gilmer, Multiplicative Ideal Theory (1972) 13. J. Yeh, Stochastic Processes and the Wiener Integral (1973) 14. J. Barns-Nefo, Introductionto the Theory of Distributions (1973) 15. R. Larsen, Functional Analysis (1973) 16. K. Yano and S. lshihara, Tangent and Cotangent Bundles (1973) 17. C. Procesi, Rings with Polynomial Identities (1973) 18. R. Hermann, Geometry, Physics, and Systems (1973) 19. N. R. Wallach, Harmonic Analysis on Homogeneous Spaces ( I 973) 20. J. Dieudonnd, Introduction to the Theory of Formal Groups (1973) 21. 1. Vaisman, Cohomology and Differential Forms (1973) 22. B.-Y. Chen, Geometry of Submanifolds (1973) 23. M. Marcus, Finite Dimensional Multilinear Algebra (in two parts) (1973, 1975) 24. R. Larsen, Banach Algebras (1973) 25. R. 0. Kujala and A. L. Vilfer, eds., Value Distribution Theory: Part A; Part B: Deficit and Bezout Estimates by Wilhelm Stoll (1973) 26. K. B. Stolarsky, Algebraic Numbers and Diophantine Approximation (1974) 27. A. R. Magid, The Separable Galois Theory of Commutative Rings (1974) 28. B. R. McDonald, Finite Rings with Identity (1974) 29. J. Safake, Linear Algebra (S. Koh et al., trans.) (1975) 30. J. S.Golan, Localizationof Noncommutative Rings (1975) 31. G. Klambauer, Mathematical Analysis (1975) 32. M.K. Agosfon, Algebraic Topology ( I976) 33. K. R. Goodearl, Ring Theory (1976) 34. L. E. Mansfield, Linear Algebra with Geometric Applications (1976) 35. N. J. Pullman, Matrix Theory and Its Applications (1976) 36. B. R. McDonald, Geometric Algebra Over Local Rings (1976) 37. C. W. Groefsch, Generalized Inverses of Linear Operators (1977) 38. J. E. Kuczkowski and J. L. Gersfing, Abstract Algebra (1977) 39. C. 0. Chrisfensonand W. L. Voxman, Aspects of Topology (1977) 40. M. Nagata, Field Theory (1977) 41. R. L. Long, Algebraic Number Theory (1977) 42. W. F. Pfeffer, Integrals and Measures (1977) 43. R. L. Wheeden and A. Zygmund, Measure and Integral (1977) 44. J. H. Cuttiss, Introduction to Functions of a Complex Variable (1978) 45. K. Hrbacek and T. Jech, Introduction to Set Theory (1978) 46. W. S. Massey, Homology and Cohomology Theory (1978) 47. M. Marcus, Introduction to Modern Algebra (1978) 48. E. C. Young, Vector and Tensor Analysis (1978) 49. S. €3. Nadler, Jr., Hyperspaces of Sets (1978) 50. S. K. Segal, Topics in Group Kings (1978) 51. A. C. M. van Roog, Nondrchimedean Functional Analysis ( I978) 52. L. Corwin and R. Szczarba, Calculus in Vector Spaces (1979) 53. C. Sadosky, Interpolationof Operators and Singular Integrals (1979) 54. J. Cmnin, Differential Equations (1980) 55. C. W. Groefsch, Elements of Applicable Functional Analysis (1980)
56. 1. Vaisman, Foundations of Three-Dimensional Euclidean Geometry (1980) 57. H. 1. Freedan, Deterministic Mathematical Models in Population Ecology (1980) 58. S. B. Chae, Lebesgue Integration (1980) 59. C.S. Rees et a/., Theory and Applications of Fourier Analysis (1981) 60. L. Nachbin, Introduction to Functional Analysis (R.M. Aron, trans.) (1981) 61. G. Onech and M. Onech, Plane Algebraic Curves (1981) 62. R. Johnsonbaugh and W. E. Pfaffenberger, FoundationsofMathematicalAnalysis (1981) 63. W. L. Voxman and R. H. Goetschel,Advanced Calculus (1981) 64. L. J. Cowin and R. H. Szczarba, Multivariable Calculus (1982) 65. K 1. lsfratescu, Introduction to Linear Operator Theory (1981) 66. R. D. Jawinen, Finite and Infinite Dimensional Linear Spaces (1981) 67. J. K. Beem and P. E. Ehrlich, Global Lorentzian Geometry (1981) 68. D. L. Annacost, The Structure of Locally Compact Abelian Groups (1981) 69. J. W Brewer and M. K. Smith, eds., Emmy Noether: A Tribute (1981) 70. K. H.Kim, Boolean Matrix Theory and Applications (1982) 71. T. W. Wieting, The Mathematical Theoryof Chromatic Plane Ornaments (1982) 72. 0.6. Gauld, Differential Topology (1 982) 73. R. L. Faber, Foundations of Euclidean and Non-Euclidean Geometry (1983) 74. M. Canneli, Statistical Theory and Random Matrices (1983) 75. J. H. Carmth et a/., The Theoryof Topological Semigroups (1983) 76. R. L. Faber, Differential Geometry and Relativity Theory (1983) 77. S. Barnett, Polynomials and Linear Control Systems (1983) 78. G. Karpilovsky, Commutative Group Algebras (1983) 79. F. Van Oystaeyenand A. Verschoren, Relative Invariants of Rings (1983) 80. 1. Vaisman, A First Coursein Differential Geometry (1984) 81. G. W. Swan, Applications of Optimal Control Theory in Biomedicine (1984) 82. T. Petrie and J. D. Randall, Transformation Groups on Manifolds (1 984) 83. K. Goebel and S. Reich, Uniform Convexity, Hyperbolic Geometry, and Nonexpansive Mappings (1 984) 84. T. Albu and C.Ndsfdsescu, Relative Finiteness in Module Theory (1984) 85. K. Hrbacek and T. Jech, Introduction to Set Theory: Second Edition (1984) 86. F. Van Oysfaeyen and A. Verschoren, Relative Invariants of Rings (1984) 87. B. R. McDonald, Linear Algebra Over Commutative Rings (1984) 88. M. Namba, Geometry of Projective Algebraic Curves (1984) 89. G. F, Webb, Theory of Nonlinear Age-Dependent Population Dynamics(1 985) 90. M. R. Bremner et a/., Tables of Dominant Weight Multiplicities for Representations of Simple Lie Algebras(1985) 91. A. E. Fekefe, Real Linear Algebra(1985) 92. S. B. Chae, Holomorphy and Calculusin Normed Spaces (1985) 93. A. J. Jerri, Introduction to Integral Equations with Applications (1985) 94. G. Karpilovsky, Projective Representations of Finite Groups (1985) 95. L. Narici and E. Beckenstein, Topological Vector Spaces( I985) 96. J. Weeks, The Shape of Space (1985) 97. P. R. Gribik and K. 0.Kortanek, Extrema1 Methods of Operations Research (1985) 98. J.-A. Chao and W A.Woyczynski, eds., Probability Theory and Harmonic Analysis (1 986) 99. 0. D. Crown et a/., Abstract Algebra (1986) 100. J. H. Carmth et a/.,The Theory of Topological Semigroups, Volume 2 (1986) 101. R. S. Doran and V. A. Belfi, Characterizationsof C*-Algebras (1 986) 102. M. W Jeter, Mathematical Programming (1 986) 103. M. Alfman, AUnifiedTheoryofNonlinearOperatorandEvolutionEquationswith Applications (1986) 104. A. Verschoren, Relative Invariants of Sheaves (1987) 105. R. A. Usmani, Applied Linear Algebra (1987) 106. P. Blass and J. Lang, Zariskl Surfaces and Differential Equations in Characteristic p > 0 (1 987) 107. J. A. Reneke eta/., Structured Hereditary Systems (1 987) 108. H. Busemann and B. B. fhadke, Spaces with Distinguished Geodesics ( I 987) 109. R. Hade, lnvertibility and Singularity for Bounded Linear Operators (1988) 110. G. S. Laddeet a/., Oscillation Theory of Differential Equations with Deviating Arguments (1987) 111. L. Dudkin et a/., Iterative Aggregation Theory (1987) 112. T. Okubo, Differential Geometry (1987)
113. D. L. Stancl and M. L. Stancl, Real Analysis with Point-Set Topology(1987) 114. T. C. Gard, Introduction to Stochastic Differential Equations (1988) 1 15. S. S. Abhyankar, Enu,merative Combinatoriqs of,Young TabJeaux(1988) 11 6. H. Wade and R. Farnsfeher, Ido#ui%i-ie AIgebGS and'Their eepreseniations (1988) 117. J. A. Huckaba, Commutative Rings with Zero Divisors (1988) 118. W. D. Wallis, Combinatorial Designs (1988) 11 9. W. Wips/aw,Topological Fields (1 988) 120. G.Karpilovsky, Field Theory (1988) 121. S.Caenepeel and F, Van Oystaeyen, Brauer Groups and the Cohomology of Graded Rings (1989) ' 122. W. Kozlowski, ModularFunctionSpaces(1988) 123. E. Lowen-Colebunders, Function Classes of Cauchy Continuous Maps (1989) 124. M. Pave/, Fundamentals of Pattern Recognition (1989) 125. V. Lakshmikantham et a/., Stability Analysisof Nonlinear Systems (1989) 126. R. Sivaramaktishnan, The Classical Theory of Arithmetic Functions (1 989) 127. N. A. Watson, Parabolic Equations on an Infinite Strip (1989) 128. K. J. Hastings, Introduction to the Mathematics of Operations Research (1989) 129. B. Fine, Algebraic Theory of the Bianchi Groups (1989) 130. D. N. Dikranjan et a/., Topological Groups (1989) 131. J. C. Morgan /I, Point Set Theory (1 990) 132. P. Biler andA. Witkowski, Problems in Mathematical Analysis (1 990) 133. H. J. Sussmann, Nonlinear Controllability and Optimal Control (1 990) 134. J.-P. Florens et a/., Elements of Bayesian Statistics (1990) 135. N. Shell, Topological Fields and Near Valuations (1 990) 136. B. F. Doolin and C. F. Martin, IntroductiontoDifferentialGeometryforEngineers (1990) 137. S.S.Holland, Jr., Applied Analysis by the Hilbert Space Method (1990) 138. J. Okninski, Semigroup Algebras (1990) 139. K. Zhu, OperatorTheoryinFunctionSpaces(1990) ' 140. G. B. Price, An Introduction to Multicomplex Spaces and Functions (1991) 141. R. B. Darst, Introduction to Linear Programming (1991) 142. P. L. Sachdev, Nonlinear Ordinary Differential Equations and Their Applications (1991) 143. T. Husain, Orthogonal Schauder Bases (1991) 144. J. Foran, Fundamentals of Real Analysis (1991) 145. W. C. Brown, Matrices and Vector Spaces (1991) 146. M. M. Rao and Z. D. Ren, Theory of Orlicz Spaces (1991) 147. J. S. Golan and T. Head, Modules and the Structures of Rings (1991) 148. C. Small, Arithmetic of Finite Fields (1991) 149. K. Yang, Complex Algebraic Geometry (1991) 150. D. G. Hoffman et a/., Coding Theory (1991) 151. M. 0. GonZ&lez, Classical Complex Analysis (1992) 152. M. 0. Gonzdlez, Complex Analysis(1992) 153. L. W. Baggeff, Functional Analysis (1992) 154. M. Sniedovich, Dynamic Programming (1992) 155. R. P.Agawal, Difference Equations and Inequalities (1992) 156. C. Brezinski, Biorthogonality and Its Applications to Numerical Analysis (1992) 157. C, Swartz,An Introduction to Functional Analysis (1992) 158. S.8.Nadler, h . , Continuum Theory (1992) 159. M. A. A/-Gwaiz,Theory of Distributions (1992) 160. E. Perry, Geometry: Axiomatic Developments with Problem Solving (1992) 161. E. Castillo and M. R. Ruiz-Cobo, Functional Equations and Modelling in Science and Engineering (1992) 162. A. J. Jeri, IntegralandDiscreteTransformswithApplicationsandErrorAnalysis (1992) 163. A. Chariier et a/.,Tensors and the Clifford Algebra (1992) 164. P. Bilerand T. Nadzieja, Problems and Examples in Differential Equations (1992) 165. E. Hansen, Global Optimization Using Interval Analysis (1992) 166. S. Guerre-Delabriere,Classical Sequences in Banach Spaces (1 992) 167. Y. C. Wong, Introductory Theoryof Topological Vector Spaces(1992) 168. S. H. Kulkami and B. V. Limaye, Real Function Algebras (1992) 169. W. C. Brown, Matrices Over Commutative Rings (1993) 170. J. Loustau and M. Dillon, Linear Geometry with Computer Graphics (1 993) 171, W, V. Petryshyn, Approximation-Solvability of NonlinearFunctionatandDifferential Equations (1993)
172. 173. 174. 175. 176. 177. 178. 179. 180. 181. 182. 183. 184. 185. 186. 187. 188. 189. 190. 191. 192. 193. 194. 195. 196. 197. 198. 199. 200. 201. 202. 203. 204. 205. 206. 207. 208. 209. 210. 21 1. 212. 213. 214. 215. 216. 217. 218. 219. 220. 221. 222. 223. 224. 225.
d
E. C.Young,Vector and Tensor Analysis: Second Edition(1993) T. A. Bick, Elementary Boundary Value Problems(1993) M. Pave/, Fundamentals of Pattern Recognition: Second Edition(1993) S. A. AIbeverio et a/., Noncommutative Distributions(1993) W Fulks, Complex Variables (1 993) M. M. Rao, Conditional Measures and Applications(1 993) A.JanickiandA. Wemn, SimulationandChaoticBehaviorofa-StableStochastic Processes (1 994) P. Neittaanmaki andD. Tiba, Optimal Control of Nonlinear Parabolic Systems(1994) J. Cmnin, Differential Equations: Introduction and Qualitative Theory, Second Edition (1994) S. Heikkila and V. Lakshrnikantham, Monotone Iterative Techniques for Discontinuous Nonlinear Dlfferential Equations (1994) X. Mao, Exponential Stability of Stochastic Differential Equations(1994) B. S. Thornson,Symmetric Properties of Real Functions(1994) J. E, Rubio, Optimization and Nonstandard Analysis(1994) J. L. Bueso et a/.,Compatibility, Stability, and Sheaves(1995) A. N. Michel and K. Wang, Qualitative Theory of Dynamical Systems(1995) M. R. Darnel, Theory of Lattice-Ordered Groups(1995) Z. Naniewiczand P. D. Panagiotopoulos, MathematicalTheoryofHemivariational Inequalities and Applications(1995) L. J. Cotwin and R. H. Szczanba, Calculus In Vector Spaces: Second Edition(1995) L. H. €he et a/.,Oscillation Theory for Functional Differential Equations (1995) S. Agaian et a/., Binary Polynomial Transforms and Nonlinear Digital Filters(1995) M. 1. Gilt Norm Estimations for Operation-Valued Functions and Applications(1995) P. A. Grillet, Semigroups: An Introduction to the Structure Theory (1995) S. Kichenassamy,Nonlinear Wave Equations (1996) V. F. Kmtov, Global Methods in Optimal Control Theory (1996) K. 1. Beidaret a/., Rings with Generalized Identities(1996) V. 1. Amautovet a/., Introduction to theTheory of TopologicalRingsandModules (1 996) G. Sierksma, Linear and Integer Programming (1996) R. Lasser, Introduction to Fourier Series(1996) V. Sima, Algorithms for Linear-Quadratic Optimization (1996) D. Redmond, Number Theory ( I 996) J. K. Beem et a/., Global Lorentzian Geometry: Second Edition(1996) M. Fontana et a/., Priifer Domains (1997) H. Tanabe, Functional Analytic Methods for Partial Differential Equations(1997) C.Q. Zhang, Integer Flows and Cycle Covers of Graphs(1997) E. Spiegel and C.J. O’Donnell, Incidence Algebras (1997) €3.Jakubczyk and W. Respondek, Geometry of Feedback and Optimal Control(1998) T. W. Haynes et a/., Fundamentals of Domination in Graphs (1998) T. W. Haynes et a/., Domination in Graphs: Advanced Topics(1 998) L. A. D’Aloffo et a/., A Unified Signal Algebra Approach to Two-Dimensional Parallel (1 998) Digital Signal Processing F. Halter-Koch, Ideal Systems(1 998) N. K. Govil et a/., Approximation Theory (1998) R. Cross, Multivalued Linear Operators (1998) A.A.Madynyuk, StabilitybyLiapunov’sMatrixFunctionMethodwithApplications (1 998) A. Favini and A. Yagi, Degenerate Differential Equationsin Banach Spaces (1999) A. Manesand S. Nadler, Jr., Hyperspaces:FundamentalsandRecentAdvances (1 999) G.Kat0 and D. Struppa, Fundamentals of Algebraic Microlocal Analysis(1999) G.X.-Z. Yuan, KKM Theory and Applicationsin Nonlinear Analysis (1999) D. Motmanuand N. H. Pave/, Tangency, Flow Invariance for Differential Equations, and Optimization Problems(1999) K. Hrbacek and T. Jech, Introduction to Set Theory, Third Edition(1999) G. E. Kolosov, Optimal Design of Control Systems(1999) N. L. Johnson, Subplane Covered Nets(2000) B. Fine and G. Rosenberger,Algebraic Generalizations of Discrete Groups(1999) M. Vath,Volterra and Integral Equations of Vector Functions(2000) S. S. Miller and P. T. Mocanu, Differential Subordinations (2000)
t t
226. R. Li et a/., GeneralizedDifferenceMethodsforDifferentialEquations:Numerical AnalysisofFiniteVolumeMethods(2000) l-,'.-s~,. ;',,? 227. H. Li and F. Van Oystaeyen, A Primer of Algebraic Geometj;' (2dOOj . 228. R. P. Aganval, Difference Equations and Inequalities: Theory, Methods, and Applications, Second Edition (2000) 229. A. B. Kharazishvili, Strange Functions in Real Analysis (2000) 230. J. M. Appell et a/., Partial Integral Operators and Integro-Differential Equations (2000) in Mathematical Physics 231. A. 1. Prilepkoet a/., MethodsforSolvingInverseProblems (2000) 232. F. Van Oystgeyen,Algebraic Geometry for Associative Algebras (2000) 233. 0.L. Jagerman, Difference Equations with Applications to Queues (2000) 234. 0.R. Hankerson et a/., Coding Theory and Cryptography: The Essentials, Second Edition, Revised and Expanded(2000) 235. S. Ddscdescu et a/., Hopf Algebras: An Introduction (2001) 236. R. Hagen et a/., C-Algebras and Numerical Analysis (2001) 237. Y. Talpaerf, DifferentialGeometry:WithApplicationstoMechanicsandPhysics (2001) 238. R. H. Villaneal,Monomial Algebras (2001) - ~ ~ ~ ~ 2 t , - s - ' ~ , ~
Additional Volumes in Preparation
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MONOMIAL ALG EBRAS Rafael H. Villarreal Departamento de Matematicas, Centro de Investigacidn y de €studios Avanzados del lnstituto Politecnico Nacional (IPN) Mexico Civ, Mexico
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Preface Let R = k [ x ] = k [ x l , .. . ,xn] be a polynomial ring in the indeterminates . . ,xn,over the field k. Let
21,.
be a finite set of monomials of R. We are interested in studying several algebras and ideals associated to these monomials, Some of these are:
. . ,f q ] c k [ x ] , k [ x ,f i t , . . . ,fqt] c k [ x ,t ] ,which is also a monomial
the monomial subring: the Rees algebra: subring, 0
0
lc[fi,.
the face ring or Stanley-Reisner ring: k [ x ]/ (fi mials are square-free, and
, . . . , fq), if the mono-
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 occur in this text. Rees algebra
Face ring
Monomial subring
J
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.
...
111
iv
Preface
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 [28] and [139]. 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 first chapter on commutative algebra 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 [250] and [311]. The first goal is to present basic properties of monomial algebras. For this purpose in Chapter 2 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 Grobner bases theory and minimal resolutions. In Chapter 3 a thorough presentation of complete and normal ideals is given. Here the systematic use of Rees algebras and associated graded algebras makes clear their importance for the area. Chapter 4 deals with Hilbert series of graded modules and algebras, a topic that is quite useful in Stanley's proof of the upper bound conjecture for simplicia1 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 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 5. The first version of this chapter was some notes originally prepared to teach a short course during the X X V I I Congreso Nacional de la Sociedad Matema'tica Mexicana in October of 1994. In this course we pre-
Preface
V
<,! -,,,:'
,y-,
.:":
I \ . !' ,
.
'h+&&$.?~lZ '
'
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 two sections of this chapter, discussing Hilbert series of face rings and the upper bound conjectures, was inspired by [25, 441 and [271]. 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 theory. We have included three chapters that focus on monomial algebras associated to graphs. One of them is Chapter 6 , where we present connections between graphs and ideals and study theCohen-Macaulay property of the face ring. Another is Chapter 8, where we present a combinatorial description of the integral closure of the corresponding monomial subring and give some applications to graph theory. In Chapter 9 we consider monomial subrings and toricideals of complete graphs withthe aimof computing their Hilbert series, Noether normalizations and Grobner bases. ' The central topic of Chapter 7 is the normality of monomial subrings and ideals; some features of toric ideals are presented here. At the end Chapter 10 is devoted to studying 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 toricvarietiesand their toric ideals are studied in Chapter 11. Most material in this textbook hasbeen 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 algebraicgeometry and combinatoricshastaken place. Excellent references for computationalandcombinatorialaspects that complement some of the material included here are [74, 92, 2981, [44, 274, 2781 and [39, 3151. 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 inadvertent 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 givenat the University of Messina covering Chapter 6 to Chapter 9 with the support of the Istituto Naxionale Di Alta Maternatica Francesco Severi. It is a pleasure to thank VittoriaBonanzinga, Marilena Crupi, Gaetana Restuccia, Rossana Utano, Maurizio Imbesi, Giancarlo Rinaldi, Fabio I
vi
Preface
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 specially grateful to Adrian Alciintar, Joe Brennan, Albert0Corso, Josh Martinez-Bernal, Susan Morey, Carlos Renteria, Enrique Reyes and Aron Simis. We are also grateful to Laura Valencia for her competent secretarial assistance. The ConsejoNacional de Ciencia y Tecnologz’a (CONACyT)andthe Sistema Nacional de Investigadores (SNI) deserve special acknowledgement 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 279313. In the homepage “http://www.math.cinvestav.mx/profesores/vila”we will maintain an updated list of corrections.
RAFAELH . VILLARREAL
I
Contents Preface
iii
11 Commutative Algebra 1.1 Module theory .......................... 1.2 Graded modules . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Cohen-Macaulay modules . . . . . . . . . . . . . . . . . . . . 1.4 Normal rings . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Koszul homology . . . . . . . . . . . . . . . . . . . . . . . . .
1 12 16 23 28
31 Algebras Graded and2 Affine 2.1 Noether normalizations ..................... 2.2 Cohen-Macaulay gradedalgebras . . . . . . . . . . . . . . 2.3 Hilbert Nullstellensatz . . . . . . . . . . . . . . . . . . . . . . 2.4 Grobner bases . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Minimal resolutions . . . . . . . . . . . . . . . . . . . . . . .
31 35 45 48 58
..
3 Rees andAlgebras Normality 65 3.1 Symmetricalgebras . . . . . . . . . . . . . . . . . . . . . . . . 65 3.2 Rees algebras and syzygetic ideals . . . . . . . . . . . . . . . 66 3.3 Complete and normal ideals . . . . . . . . . . . . . . . . . . . 69 3.4 A criterion of Jiirgen Herzog . . . . . . . . . . . . . . . . . . 82 3.5 Jacobiancriterion . . . . . . . . . . . . . . . . . . . . . . . . . 89 4 Hilbert Series 97 4.1 Hilbert-Serre’s Theorem . . . . . . . . . . . . . . . . . . . . . 97 4.2 a-invariants and h-vectors . . . . . . . . . . . . . . . . . . . . 106 4.3 Extrema1algebras . . . . . . . . . . . . . . . . . . . . . . . . 110 4.4 Initial degrees of Gorenstein ideals . . . . . . . . . . . . . . . 118 4.5 A symbolic study of Koszul homology . . . . . . . . . . . . . 125
vii
viii
Contents
5 Monomial Ideals and St anley-Reisner Rings 5.1 Primary decomposition . . . . . . . . . . . . . . . . . . . . . . 5.2 Simplicia1 complexes and homology 5.3 Face rings . . . . . . . . . . . . . . .
129 129
. . . . . . . . . . . . . . .138
.............. .................... ....................
150 154
6 Edge Ideals 6.1 Graphs and ideals . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Cohen-Macaulay graphs . . . . . . . . . . . . . . . . . . . . . 6.3 Trees. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Bipartite graphs . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Links of some edge ideals . . . . . . . . . . . . . . . . . . . .
161 161 172 178 182 185
5.4 Hilbert series of face rings 5.5 Upper bound conjectures
6.6 First syzygy module of an edge ideal . . . . . . . . 6.7 Edge rings with linear resolutions . . . . . . . . .
142
. . . . . . . 188 . . . . . . . 192
7 Monomial Subrings 201 7.1 Basic properties . . . . . . . . . . . . . . . . . . . . . . . . . . 201 7.2 Integral closure of subrings . . . . . . . . . . . . . . . . . . . 209 7.3 Integral closure of monomial ideals . . . . . . . . . . . . . . . 233 7.4 Normality of some Rees algebras . . . . . . . . . . . . . . . . 239 7.5 Ideals of mixed products . . . . . . . . . . . . . . . . . . . . . 248 7.6 Degree bounds for some integral closures . . . . . . . . . . . . 259 7.7 Degree bounds in the square-free case . . . . . . . . . . . . . 261 7.8 Some lexicographical Grobner bases . . . . . . . . . . . . . .276 8 Monomial Subrings of Graphs 8.1 The subring associated to a graph . . . 8.2 Rees algebras of edge ideals . . . . . . . 8.3 Incidence matrix of a graph . . . . . . . 8.4 Circuits of a graph and Grobner bases . 8.5 Edge subrings of bipartite planar graphs 8.6 Normality of bipartite graphs . . . . . . 8.7 The integral closure of an edge subring . 8.8 The equations of the edge cone . . . . .
9 Semigroup Rings of Complete Graphs 9.1 Monomial subrings of bipartite graphs . 9.2 Monomial subrings of complete graphs . 9.3 Noether normalizations of edge subrings
281
. . . . . . . . . . . .282 . . . . . . . . . . . . 286 . . . . . . . . . . . . 294
. . . . . . . . . . . . 296 . . . . . . . . . . . . 303 . . . . . . . . . . . . 311 . . . . . . . . . . . . 315
. . . . . . . . . . . . 325 335
. . . . . . . . . . . . 335 . . . . . . . . . . . . 350 . . . . . . . . . . . .359
ix
Contents
Diagrams
10 Monomial Curves 10.1 Defining equations of monomialcurves 10.2 Symmetricsemigroups . . . . . . . . . 10.3Idealsgenerated by criticalbinomials . 10.4 An algorithm for criticalbinomials . .
...,...... .......,.. .......... .......,.,
11 Affine Toric Varieties and Toric Ideals 11.1 Systems of binomialsintoricideals .. 11.2 Affine toricvarieties . . . . . . . . . . 11.3Curvesinpositivecharacteristic ...,
. . . . . . . , . . . , . 403 . . . . . . . , . . . . . 409 . . . . . . . . . . . , . 414
A Graph A . l Cohen-Macaulaygraphs A.2 Unmixed graphs . . .
367
. . . 367 . . . 377 . , . 385 . . . 399
403
421
. . . . . . . . , , 421 . . . . . . . . . . . . . . . . . . . . . . 424 .......,.
, ,
Bibliography
425
Notation
447
Index
449
This Page Intentionally Left Blank
Chapter 1
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 works of W. Bruns andJ. Herzog [44], D.Eisenbud [92], H. Matsumura [218] and W. Vasconcelos [298]. Some of the results presented below are just stated without giving proofs, if need be, the reader may locate the missing proofs in one of those references.
1.1 Module theory Noetherian modules and localizations 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 f l , . . . ,f q such that
Let R be a commutative ring with unit and let Ad be an R-module. Recall that M is called Noetherian if every submodule N of M is finitely generated, that is, N = Rfi Rfq, for some ff,. . . , f q in N .
+ +
If M is a finitely generated R-module overa Noetherian ring R, then M i s a Noetherian module.
Proposition 1 . i .1
Corollary 1.1.2 If R is a Noetherian ring and I i s a n ideal of R, then R I I and Rn are Noetherian R-modules. In particular any submodule of R" is finitely generated.
1
Chapter 1
2
Theorem 1.1.3 (Hilbert basis theorem) A polynomial ring a Noetherian ring A is Noetherian.
A[z] over
One of the important examples of a Noetherian ring is a polynomial ring over a field IC. Often we will denote a polynomial ring in several variables by k[x] and a polynomial ring in one variable by k [ x ] . Unless otherwise stated the letters IC and K will always denote fields. 1
Theorem 1.1.4 An R-module M is Noetherian if and only if M satisfies the ascending chain condition for submodules, that is, M is Noetherian if and only if for every ascending chain of submodules of M
there exists an integer IC such that Ni = Nk for every i 2
IC.
In this book by a ring (resp. module) we shall always meana Noetherian ring (resp. finitely generated module). The spectrum of a ring R will be denoted by Spec(R), it is the set of prime ideals of R. The minimal primes of R are the minimalelements 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 X = Spec(R). Given an ideal I of R, the set of all prime ideals of R containing I will be denoted by V ( I ) . It is not hard to verify that the pair ( X ,2) is a topological space, where 2 is the family of open sets of X , and where U is in 2 iff U = X \ V ( I ) ,for some ideal I . This topology is called the Zariski topology of the prime spectrum of R.
A local ring (R, m, k ) is a Noetherian ring R with exactly one maximal ideal m, the field IC = R/m is called the residue field of R. A homomorphism of rings is a map 9: R
+ S such that:
(i) cp(a + b) = cpb) + cp(b), vu, b E A , (ii) cp(ab) = cp(a)cp(b),V u , b E A , and (iii) cp(1) = 1. Let R be a ring and let 9: Z + R be the canonical homomorphism
then ker(cp) = nZ,for some n 2 0. The integer n is called the characteristic of R and is denoted by char(R). Proposition 1.1.5 Let (R, m) be alocal ring, then either char(R) = 0 or char(R) = p”, for some prime number p and some integer IC 2 1.
Commutative Algebra
3
Proof. Let n = char(R) 2 2. If n = pqm, where p , q are prime numbers with q # p , then p IR and q IR are both in m because they are non invertible, but this is impossible because 1 = ap bq and from the equality a
e
+
one derives 1~ E m,a contradiction.
0
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 E S. Then the module of fractions of M with respect to S, or the ZocaZization of M with respect to S, is defined by
S-'(M) = { m / s l m E M ,s E S ) , where m / s = ml/sl if and only if t(s1m - s m l ) = 0 for some t E S. In particular S-lR has a ring structure given by the usual rules of addition andmultiplication,and is a module over the ring S - l R with the operations:
There is a canonical map 9:M -+ S-l M ,where ~ ( m=) m/l. If f : M + N is a homomorphism of R-modules, then there is an induced homomorphism S-lf: S - l M "+ 5"'N
of S-lR-modules given by f ( m / s ) = f(m)/s. Example 1.1.6 If f is in a ring R and S = { f " l n 2 0)) then S"lR is usually written Rf . For instance if R = C [z] is a polynomial ring in one variable over the field C of complex numbers, then
is the ring of Laurent polynomials. Definition 1.1.7 Let p be a prime ideal of a ring R and S = R \ p. In this case S - l R is written R, and is called the localization of R at p. Example 1.1.8 Let p be a prime ideal of a ring R. The local ring
is the prototype of a local ring, where k ( p ) = R,/pRp denotes the residue field of R,.
Chapter 1
4
Krull dimension and height By a chain of prime ideals of a ring R we mean a finite strictly increasing sequence of prime ideals
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
Po
c p1 c ..' c Pn = p
which end at p . Note dim(R,) = ht ( p ) . If I is an ideal of R, then ht ( I ) , the height of I , is defined as ht(I) = min{ht(p)l I
c p and p E Spec(R)}.
+
In general dim(R/I) ht ( I ) 5 dim(R). The difference dim(R) - dim(R/I) is called the codimension of I . Let M be an R-module. The annihilator of
M is given by
annR(M) = {x E R ( x M = 0 } ,
4
if rn E M the annihilator of m is ann (m)= ann (Rrn).It is convenient to generalize the notion of annihilator to ideals and submodules. Let N1 and N2 be two submodules of M , their ideal quotient or colon is defined as
( N ~ : R N= z )(x E RlxN2
c Nl].
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 1.1.9 If A [ x ] is a polynomialringover then dim A[x] = dim(A) + 1.
a Noetherianring A ,
Primary decomposition of modules Let I be an ideal of a ring R. The radical of I is rad ( I ) = {x E RI x n E I for some n
> 0},
m,
the radical is also denoted by fi. In particular denoted by ' 3 2 ~is ~ the set of nilpotent elements of R and is called the nilradical of R. A ring is reduced if its nilradical is zero.
5
Commutative Algebra
Proposition 1.1.10 If I is a proper ideal of a ring R, then rad ( I ) is the intersection of all prime ideals containing I . Definition 1.1 . l l Let M be a module over a ring R, the setof associated primes of A d , denoted by A s s ~ ( l W )is, the set of all prime idealsp of R with the property that there is a monomorphism q5 of R-modules:
Note that p = ann (q5( 1)).
If A4 = 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) = A s s ( R / I ) . Proposition 1.1.12 Let R be a ring and S a multiplicatively closed subset of R. If M is an R-module and p is a prime ideal of R with S np = 8, then p is an associated prime of M if and only if S - l p is an associated prime of S-lM. Proof. If p is in Ass(M), then R / p C ) M . Hence
S-’R/S”p c3 S”M
5”’
and S - l p is an associated prime of ill. For the converse assume S-lp = ann(rn/l). Since R is Noetherian p is generated by a finite set a l , . . . ,a,, hence for each i there is si E S such that Siaim = 0. Set s = s1 sn. We claim p = ann(sm). Clearlyone has p c ann(sm). To show the other containment take x E ann(srn), then xsrn = 0 and x/1 E ann(rn/l) = S-lp. Hence z E p. 0 a
Definition 1.1.13 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 M p # 0.
4
A sequence’o ”+ M’ M 4 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:
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.
-+ M’ -+ M modules over a ring R, then
Lemma 1.1.14 If 0
-+ M” -+ 0 is a short exact sequence
Supp(M) = Supp(M‘) u Supp(M“).
of
6
Chapter 1
Proof. Let p be a prime ideal of R. It suffices to observe that from the exact sequence O+M~-+Mp+M~-+O, we get M p # 0 if and only if
kt; # 0 or M: # 0.
0
Theorem 1.1.15 If M is an R-module, then there is a filtration modules
of sub-
(O)=M~CM~C~..CM~=M and prime ideals p i , . . . ,Pn of R such that
MiIMi-1 N
Rlpi for all i.
In general the primes p 1 , . . . ,Pn that occur in a filtration of the type described in the previous result are not associated primes of the module M ; see [87] for a careful discussion of filtrations and some of their applications to combinatorics. Lemma 1.1.16 If 0 +=MI -+ M -+ M" + 0 is a short exact sequence modules over a ring R, then
% A s s ( Mc) Ass(")
of
U Ass(M").
Corollary 1.l.17 If M is an R-module, then ASSR( M )is a finite set. Proof. Let P I , . . . , p n be prime ideals as in Theorem 1.1.15. By a repeated 0 use of Lemma 1.1.16 onehas A s s R ( M )C { P I , . . . ,pn} C Supp(M). Let M be an R-module. An element x E R is a zero divisor of M if there is 0 # m E M such that xna = 0. The set of zero divisors of M is denoted by 2 ( M ) . If x is not a zero divisor on M we say that x is a regular element of M . Lemma 1.1.18 If M is an R-module, then
Proof. See [218, Theorem 6.11.
0
Definition 1.1.19 Let M be an R-module. A submodule N of M is said to be a primary submodule if ASSR(M / N ) = { p } . An ideal q of a ring R is a primary ideal if AssR(R/q) = { p } , thus q is a primary ideal iff zy E q and x 4 q implies yn E q for some n 2 1. Definition 1.1.20 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 .
gebra
7
Commutative Proposition 1.1.21 Let M be anR-module. If Q submodule of M ,then Q is a primary submodule.
# M is an irreducible
Proof. Assume there are plandp 2 distinct associated prime ideals of M / Q and pick ro E p1 \ p 2 (or vice versa).There is xi in 114 \ Q such that p i = ann(Zi), where Z i = zi + Q. We claim that
+
+
If x is in the intersection, then x = Xlsl 41 = X 2 2 2 42, for some X i E R and qi E Q . Note that rox E Q , hence roX222 isin Q and consequently roX2 E p 2 . Thus X 2 E p 2 and we get X 2 2 2 E Q . This shows z E Q and completes the proof of the claim. As Q is irreducible one has Q = Rzl Q or Q = Rx2 Q , which is a contra.dictionbecause xi # Q for i = 1 , 2 . 0
+
+
Definition 1.1.22 Let M be an R-module and N , N1, . . , ,N,. submodules of M .A decomposition N = N1 n - n N,. of N is said to be irredundant if
Theorem 1.1.23 Let M be an R-module. If N is a submodule of M ,then N has an irredundant primary decomposition N = Nl n - . n N , so that: (a) AssR(M/N~)= { p i } for all i.
# N1 n n Ni-1 n Ni+l n - n N,. for all i. (c) pi # ~j if Ni # N j Remark 1.1.24 If N # M and N = NI n. nN,. is an irredundant primary (b) N
decomposition of N with A s s R ( M / N ~=) { p i } , then
ann ( M / N i ) is a pi primary ideal, qnd rad (ann (M/Ni)) = pi for all i. Corollary 1.1.25 If R is a Noetherian ring and I a proper ideal of R, then I has an irredundant primary decomposition
such that qi is a
p i -primary
ideal and AssR(R/I) = { p l , . . . ,p,. } ,
Proof. Let (0) = I/I = (ql/I) n .r) (qr/I) be an irredundant decomposi- - n qr and qi/Iis pi-primary, tion of the zero ideal of R/I. Then I = q1 that is, ASSR((R/I)/(qi/I)) = Ass(R/qi) = {pi}. Let us show that qi is a primary ideal. If xy E qi and x !$ q i , then y is a zero-divisor of R/qi, but Z ( R / q i ) = p i , hence y E pi = rad (ann (R/qi)) = rad (qi) and yn is in qi for 0 some n > 0. n
e
Chapter 1
8
Definition 1.1.26 Let R be a ring and let S be the set of nonzero divisors of R. The ring S-lR is called the total ring of fractions of R. If R is a domain, S-lR is the field of fractions of R. Proposition 1.1.27 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. p1, . . . ,pT be the minimal primes of R and S = R \ Ur=,pi. Since R is reduced one has (0) = p1 n n p,- and K = S'lR. Define
Proof. Let
0
+ S-'R/S-lpl by 4 ( x ) = (X + S-lp,, . . . , x + S-'p,). 4: K
0
X
e
*
*
X
S-'R/S-'p,
As S-lpl,. . . ,S-lp, are maximal ideals and its intersection is zero, it follows from the Chinese remainder theorem that 4 is an isomorphism (see Exercise 1.1.38). 0 Lemma 1.1.28 Let M be an R-module and L a n ideal of R. If L M = M , then there is z l(mod L ) such that X M = ( 0 ) .
+ +
Proof. Let M = R a l Ran, ai E M. As LM = M , there are b, in L such that ai = Cj"=,b i j a i . Set a = (ai,.. . )a,) and H = ( b i j ) - I , where I is the identity matrix. Since Hat = 0 and Hadj(H) = det(H)I, one concludes det (H)ai = 0 for all i. Hence xlM = (0) and x 1(mod L ) where x = det(H). 0 a
)
Lemma 1.1.29 (Nakayama) Let R be a ring and N a submodule of a n R-module M . If I is an ideal of R contained in the intersection of all the maximal ideals of R such that M = I M iN , then M = N . Proof. Note I ( M / N ) = M / N . By Lemma 1.1.28 there exists an element x E l(modI) 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 because contradiction I c m. 0 Let M be an R-module. The minimum number of generators of M will be denoted by v ( M ) . A consequence of Nakayama's lemma is an expression for v ( M ) ,when the ring R is local (cf. Corollary 2.5.2). Corollary 1.1.30 If
M is a module over a local ring (R, m), then
Proof. Let ~ 1 , ... be a minimal generating setfor M and Zi = ai+mM. After a permutation of the ai one may assume that El,.. , ,E, is a basis for + Ra,. M/mM as a k-vector space, for some r 5 q. Set N = R a l + Note the equality M = N m M , then by Nakayama's Lemma N = M . Therefore required. r = q, as 0 ) cyQ
+
a
a
1
Commutative Algebra
9
Modules of finite length An R-module M has finite length if there is a composition series
where Mi/Mi-1 is a nonzero simple module (that is, Mi/Mi-l has no proper submodules other than (0)) for all i. Note that Mi/Mi-l 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 ( Mor) simply l ( M ) .
M is an R-module, then
Proposition 1.1.31 If
Ass(M)
c 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 v M and thus 0 # (R/p),-M,. Hence p is in the support of M ,this shows the first containment. Let p E Supp(M) and let x E ann(M). If x jZ p, then x m = 0 for all m E M and hfp = (0), which is absurd. Therefore p 'is in V(ann (M)). Conversely let p be in V(ann ( M ) ) and let ml, . . . ,mr be a finite set of generators of M . If Mp = (0), then for each i there is si jZ p so that Simi = 0, therefore s1 sr is in ann (M) c 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 M p # (0) there is an associated prime p1 R, of M p ,where p1 is a prime ideal of R contained in p. Since Mpl (Mp)pl# (0), we get that p1 is in the support of M and p = p1. Therefore using Proposition 1.1.12 one concludes p E Ass(M). 0 a
0
-
Proposition 1.1.32 If 0 + M' + M modules over a ring R, then
-+ MI' + 0 is an exact sequence of
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 1.1.31 we obtain Mp # (0). Therefore using Lemma 1.1.14 one has M i # (0) or M: # (0), thus either p contains ann ( M I ) or p contains ann(M"). This proves d 5 max{d', d"}, On the other hand ann (M) is contained in 0 ann (MI) n ann (M") and consequently max{d', d") 5 d. 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
10
Chapter 1
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 minimalprimes of I are precisely the minimal primes of A s s R ( R / I ) .In particular the minimal primes of R are precisely the minimal primes of AssR(R). Proposition 1.1.33 If M is an R-module, then M has finite length ifand only if every prime ideal in Supp(M) is a maximal ideal. Definition 1.1.34 A ring R is Artinian if ~ ? R ( R <)00. Proposition 1.1.35 Let R be a ring. Then R is Artinian ifandonlyif is Noetherian and every prime ideal of R is maximal.
R
Definition 1.1.36 Let R be a ring with total ring of fractions Q. An Rmodule M is said to have rank r if M @ R Q is a free &-module of rank equal to r. Lemma 1.1.37 If M is an R-module of positive rank r , then dim(M) = dim(&). Proof. Let p 1 , . . . ,pn be the associated primes of R and S = R \ U?=,pi. By hypothesis S - l M I I(5"'R)". Since
Mpi
( S - l M ) p i N [(S-'R)"Jpi N [(S-lR)pi]rN (Bpi)'# (0),
we obtain that all the minimal primes of R are in the support of M and consequently one dim(M) has = dim(R). 0
Exercises
1.1.39 Let I1,. . . ,I n , p be ideals of a ring R. If p is prime and n?=,Ii c p , prove that Ii C p for some i . In particular if n?!,Ii = p , then p = Ii for some i. 1.1.40 Let I,P I , .. . ,P,. be ideals of a ring R. If I is contained in Ur=, Pi and Pi is a prime ideal for i 2 3, prove that I c Pi for some i (cf. Lemma 2.2.18). 1.1.41 Let R be a ring and I an ideal. If P is a prime ideal such that c P, prove that ht(1) 5 ht(Ip) and show an example where the strict inequality holds.
I
Commutative Algebra
11
1.1.42 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 c P , prove that ht(1) = ht(Ip). 1.1.43 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 E R. 1.1.44 Let 111 be an R-module and I an ideal of R contained in annR(M). Note that 111 inherits a structure of R/I-module. Prove that p e AssR(M) if and only if p / I E ASSRII(M). 1J.45 Let R' = k [ x z , . . . ,xn] and R = R'[x1 J be polynomial rings over a field IC. If I' is an ideal of R' and p E Ass~(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'. 1.1.46 Let A be a ring and q a proper ideal of A. Prove that q is a primary , ideal if and only if 2 ( A / q ) C ( Y Z A / ~ where
%A/q
is the nilradical of A/q.
1.1.47 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. 1.1.48 Let A be a ring and q an ideal of A such that rad (4) = m is a
maximal ideal. Show that q is a primary ideal.
Hint Note Spec(A/q) = {m/q) and
%A/q
= m/q.
1.1.49 If B = A[x] is a polynomial ring over a ring A and q a primary ideal of A , then q B is a primary ideal of B.
Hint Note (A/q)[x]
E B/qB.
Use induction together with the next exercise.
1.1.50 Let A[z] be a polynomial ring in one variable over a ring A . Prove
that a polynomial f(x) E A[x] is a zero divisor iff there exists a such that a f ( z ) = 0.
# 0 in A
1.1.51 Let M be an R-module and S c T two multiplicatively closed subsets of R. Prove 2"lM N Z " l ( S - ' M ) as T-lR modules.
Hint Consider 4(m/t) = ( m / l ) / t ,which is induced by the canonical map from A4 to S-lM. 1.1.52 Let M be an R-module and S a multiplicatively closed subset of R. If p is a prime ideal such that S n p = 8, prove M p N (S-'
Chapter 1
12
1.1.53 Let R be a ring and I an ideal. If x E R \ I , then there is an exact sequence of R-modules:
0 "+ R / ( I : x )
-% R / I -% R / ( I , x ) -+
where @(T) = 2;' is multiplication by z and
0,
4 ( F ) = 7.
4
1.1.54 If 0 + M' M -%- M" -+ 0 is an exact sequence of R-modules, prove that the following sequence is also exact
0 + S-lM'
+ S-lM $3S-lM" +0.
S"f
1J.55 Let N1 and N2 be submodules of an R-module M . If S is a multiplicatively closed subset of R, then
(a) S-l(N1
+ N2) = S-l(N1) + S-'(N2),
(b) S-l(N1 n N2) = S-l(N1) n S - l ( N z ) , and
( c ) S-l (N1/&)
N
S-l (Nl)/S-l (N2), as S-l (R)-modules.
1.1.56 Let N1 and N2 be submodules of an R-module M . If S is a multiplicatively closed subset of R, then
(a) S-'(ann (iV2))= ann ( S - l ( N Z ) ) ,and (b) S-1(N1:N2)= (S-'(N1):S- (N2)).
Hint Use that N2 is a finitely generated R-module. 1.1.57 If I and J are ideals of a ring R and S is a multiplicatively closed
subset of R, then (a)
d
m = S"l (d), and
(b) S - l ( I J ) = S-'(I)S-'(J).
1.2
Graded modules
Let (H,+) be an abelian semigroup. An H-graded ring is a ring R together with a decomposition R=
@ Ra
(as a Z-module),
aEH
such that RaRb C Z-graded ring.
Ra+b
for all a, b E H . A graded ring is by definition a
Commutative Algebra
13
If R is an H-graded ring and M is an R-module with a decomposition
such that RaMb C Ma+b for all a, b E H, we say that is an H-graded module. An element f E M is said to be homogeneous of degree a if f E .Ma, in this case we set deg( f) = a. The elements in R, are also called f o r m s of degree a. A map p: M -+ N between H-graded modules is graded if p(Ma) c Na for all a E H. If M = $acHMa is an H-graded module and N = @aEHNa is a graded submodule, that is, Na c Ma for all a , then M / N is an H-graded R-module with ( M / N ) , = Ma/Na. Example 1.2.1 Let R = k [ z l ,. . . ,xn] be a polynomial ring over a field k . A natural way to grade R is to assign degree di to the variable xi, where d l , . . . ,dn are positive integers. For a = ( a l , . . . , a n ) in RP we set xa = X;' - and1 . 1 = aldl - + and,. The induced N-grading is:
+
one may extend this grading to a Z-grading by setting Ri = 0 for i
< 0.
Let R = k [ x l , . . . ,xn] be polynomial ring over a field IC endowed with an induced positive grading as above. In this case the homogeneous elements of R are sometimes called quasi-homogeneous polynomials. An ideal I of R is homogeneous, or graded, if there are homogeneous polynomials f 1 , . . . , fr such that I = (fl , . . . ,f r ) . Set deg(fi) = &. The ideal I is graded by
Hence R / I inherits a structure of N-graded R-module whose components are given by ( R / I ) i= Ri/Ii. Definition 1.2.2 The standard grading or usual grading of a polynomial ring k [ x l ,. . . ,xn] is the N-grading induced by setting deg(zi) = 1 for all i. Let us now recall some good properties connection with primary decompositions.
of graded modules mainly in
Lemma 1.2.3 If M is a n N-graded R-module and p E A s s ( M ) , t h e n p is a graded ideal and there i s m E M homogeneous such that p = ann(m). Proof. See [44, Lemma 1.5.61.
0
14
Chapter 1
Proposition 1.2.4 Let M be a n N-graded R-module and Q 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. Proof. See [217, Chapter 41.
0
Theorem 1.2.5 Let M be a n N-graded R-module. If N is a gradedsubmodule of M , then N has an irredundant primary decomposition
such that: (a) Ni is a graded submodule and AssR(M/N~)= {pi} f o r all i, (b) N
# N1 n - - - n Ni-1 n Ni+l n
f l Nr
f o r all i, and
Proof. Use Lemma 1.2.3, Proposition 1.2.4 andTheorem
1.1.23.
0
Finding primary decompositions of graded ideals in polynomial rings over fields is a difficult task, the punch line is that there is not yet a best general strategy for computing primary decompositions[77]; for some of the main algorithms that can been used see [95, 127, 255, 3131. For a specially nice treatment of the principles of primary decomposition consult the book of Wolmer V. Vasconcelos [298, Chapter 31. Lemma 1.2.6 (Graded Nakayama lemma) Let M be a n N-graded Rmodule and let m = R+. If N is a graded submodule of M and I c m is a graded ideal of R such that M = N I M , then N = M .
+
Proof. Since M / N = I ( M / N ) , one may assume N = (0). If x E M is a homogeneous element of degree r , then a recursive use of the equality M = I M yields that x E I r + l M and x must be zero. 0 Hilbert polynomial Let R = $goR;bean N-graded ring. We recall that R is a Noetherian ring if and only if R o is a Noetherian ring and R = &[XI,. . . ,xn] for some X I , .. . ,x, in R. If M is a finitely generated N-graded R-module and Ro is Artinian define the Hilbert function of M as
H ( M ,i) = l ? ~ ~ ( M i () ! R ~= lengthw.r.t R o ) . Definition 1.2.7 An N-graded ring R = $ g o R i is called a homogeneous ring if R = R o [ z l , . . . ,xn],where deg(xi) = 1.
Commutative Algebra
15
Theorem 1.2.8 (Hilbert) Let R = $EoRi be a homogeneousring and let M be a finitely generated N-graded R-module with d = dim(M). If Ro is Artinian, then there is a unique polynomial PM(t) E Q[t] of degree d - 1 such that P M ( ~=) H ( M ,i) for i >) 0. The polynomial PM(t) is called the Hilbert polynomial of M . Proof. See [44, Theorem 4.1.31.
0
+
Definition 1.2.9 If P M ( t ) = a&"td-' (10 is the Hilbert polynomial of M , the integer (d - l)!ad-l is called the multiplicity of and is denoted by + . e
Lemma 1.2.10 Let f ( t ) E Q[t] be a polynomial of degree d - 1 such, that f (n)E Zfor all n E Z, then there are unique integers ao, . . . ,a&." such that d-1
f ( t ) = x a i f i ( t ) , where fi(t) = i=O
Proof. The polynomials f i ( t ) , i E space. Hence
(": i).
N,are a basis for Q[t] as a @vector
d- 1
i=O
for some ai E Q. Using the Pascal triangle we get
thus by induction on the degree it follows that ai E Z for all i.
Exercises 1.2.11 Let I be a graded ideal of a polynomial ring R. Prove that the radical of I is also graded. 1.2.12 Let d,m E
W. Prove the equality
1.2.13 Let f:Z + Z be a numerical function, f is said to be a polynomial function of degree d if there is a polynomial P ( t ) E Q[t] such that P ( i ) = f ( i ) for i >> 0. Prove that f is a polynomial function of degree d if and only if the numerical function g: Z + Z given by g ( i ) = f ( i ) - f ( i - 1 ) is a polynomial function of degree d - 1.
Chapter 1
16
1.2.14 Let R = Ic[xl,.. .,X,] be a polynomial ring over a field k and let d l , . . . ,d, be a sequence of integers. Prove that R is a graded ring with the Z-grading: Ri = @ kxa (i E Z), lal=i
where x u = x:'
*
x$ and la1 = aldl
+ . + andn for a = ( a i ) E P.
1.2.15 Let R = @iEzRi be a graded ring. Prove that Ro is a subring of R with 1 E Ro. A graded ring R with Ri = 0 for i < 0 is called a graded Ro-algebra. 1.2.16 Let R = Ic[xl,1 2 1 be a polynomial ringover a field k with the grading induced by deg(zi) = (-l)i. Determine the subring Ro.
1.3
Cohen-Macaulay modules
Here we introduce some special types of rings and modules and present the following fundamental result of dimension theory. Theorem 1.3.1 (Dimension theorem) Let (R, m) be a local ring and let M be an R-module. Set
S ( M ) = min(r1 there are
21,.
. . ,Xr
E m with t ~ ( M / ( x l.,..x r ) M ) < 00},
then dim(M) = d(M). Proof. See [218, Theorem 13.41.
0
Definition 1.3.2 Let (R, m) be a local ring and let M be an R-module of dimension d. A system of parameters (s.0.p) of M is a set 81,. . . ,8d E m such that tR(MI(01,' ,O d ) M ) < 00*
Corollary 1.3.3 Let (R, m) be a local ring and let M be an R-module of dimension d . If hl, . . . ,h d E m is a system of parameters of M , then
Definition 1.3.4 Let M be an R-module. A sequence 8 = 81,. . . ,On in R is called a regular sequence of M or an M-regular sequence if ( 8 ) M # M and 1 9 ~4 Z(M/(81,. . . ,8i-1)M) for all i.
Commutative Algebra
17
Proposition 1.3.5 Let M be an R-module and I a n ideal of R such that I M # M. If 8 = 81, . . . ,Or is an "regular sequence in I, then 8 can be extended to a maximal "regular sequence in I . Proof. By induction assume there is an "regular I for some i 2 r. Set = M / ( 8 1 , . . . ,&)M.If I which is regular on M.Since
sequence 81, . . . ,Bi in 2 ( M ) pick 8i+l in I
is an increasing sequence of ideals in a Noetherian ring R, this inductive sequence in I . 0 construction must stop at a maximal "regular Lemma 1.3.6 Let M be a module over a local ring (R, m), If 81, . . . ,e,. is an M-regular sequence in m, then r 5 dim(M) .
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. Using the second part of the proof of Lemma 1.3.10 one has dim(M/&M) < dim(M). Since 0 2 , . . . ,e, is a regular sequence on M/&M by induction one derives r 5 dim(&!). 0 Proposition 1.3.7 Let M be an R-module and let I be a n ideal of R.
( a ) HomR(R/I, M)= (0) if there is x E I which is regular o n M. ( b ) Extk(R/I, M ) N HomR(R/I, M / e M ) , where "regular sequence in I.
e = 81,. . . , O r
is any
Proof. ( u ) *) Assume I c 2 ( M ) . Using Lemma 1.1.18 one has I c p for some p E AssR(M). Hence there is a monomorphism 111: R / p + M . To derive a contradiction note that the composition R / I 'P, R/p
-%M
is a nonzero map, where cp is the canonical map from R/I to R/p. The converse is left as an exercise. (b) Consider the exact sequence
According to [250, Theorem 7.31 there is a long exact sequence with natural connecting homomorphisms
18
Chapter 1
Since 81 is in I , using [250, Theorem 7.161 it follows that in the last exact sequence the maps given by multiplication by 81 are zero. Hence
the and
proofby follows
induction on
r.
0
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 1.3.7 one derives depth(M) = inf(r1 Extk(R/m, M)# (0)). In general by Lemma 1.3.6 we have depth(M) 5 dim(M). Definition 1.3.8 An R-module M is called Cohen-Macaulay (C-M for short) if depth(M) = dim(M), or if M = (0). Lemma 1.3.9 (Depth lemma) If 0 + N + M sequence of modules over a local ring R, then
+ L 3 0 is a short exact
( a ) If depth(M) < depth(l), then depth(N) = depth(").
( b ) If depth(M) = depth(l), then depth(N) 2 depth("). ( c ) If depth(A4)
> depth(l), then depth(N) = depth(l) + 1.
Proof. See [298, Corollary A.6.31. Lemma 1.3.10 If M is a module over a element of M ,then
0
local ring R and x is aregular
(a) depth(M/xM) = depth(M) - 1, and
(b) dim(M/zM) = dim(M) - 1. Proof. As depth M sequence
> depth M / x M applying the
depth lemma to the exact
O-+M~M"+M/zM"+O yields depth(M) = depth(M/xM)+l. To prove the second equality first observe the validity of the inequality dim(M) > dim(M/zM), otherwise there is a saturated chain of prime ideals ann ( M ) c ann ( k f / x M )c po
c
a
cpd,
where d is the dimension of M and po is minimal over ann (M). According to Proposition 1.1.31 the ideal po consists of zero divisors, a contradiction since z E ann ( M / z M ) c po. On the other hand the reverse inequality dim(M) 5 dim(M/zM) 1 follows from Theorem 1.3.1. 0
+
Commutative Algebra
,
. . I.'.
I
, L
-
19
L
Proposition 1.3.11 If M # (0) is a Cohen-MacaulayR-moduleover local ring (R, m) and p E A s s ( M ) , then
a
dim(R/p) = depth(A4). Definition 1.3.12 Let (R,m) be a local ring of dimension d. A system of parameters (s.0.p) of R is a set 81, . . . ,6'd generating an m-primary ideal. Theorem 1.3.13 Let (R,m, k ) be alocal ring and let S(R) = min{v(I) = dimk(I/mI) I I is an ideal of R with rad(I) = m}.
Then dim( R)= S( R ). Proof. This result
is a particular instance of Theorem 1.3.1.
0
Theorem 1.3.14 (Krull principal ideal theorem) Let I be a n ideal of a ring R generated by a sequence hl, . . . , h,, T h e n
5 r for any minimal primep of I . (b) If hl, . . . , h, is a regular sequence, then ht (p) (a) ht (p)
= r for any minimal
prime p of I.
ht
Proof. (a) Since = pR, from Theorem 1.3.13 one hasthe inequality ht (p) = dim(Rp) 5 r. (b)Set J = ( h l , ...,h i - l ) a n d K = ( h l , ...,hi). A s s u m e h t ( P ) = i - 1 for any minimal prime P over J . Since hi is regular on R / J and K / J is a principal ideal one has ht ( K / J ) = 1, thus there is a prime ideal po minimal over J such that J C po c p and consequently ht (p) 2 i. Using (a) one gets (p) = i. 0 Corollary 1.3.15 If ( R ,m,k ) is a local ring, then dim(R) 5 dimk(m/m2) and R has finite I h l l dimension. Proof. Let 21, . . . ,xq be a set of elements in m whose images in m/m2 form a basis of this vector space. Then m = (21,.. . ,xq) m2. Hence by Nakayama's lemma we get m = ( ~ 1 ,. .. ,xq). Hence dim R 5 Q by 0 Theorem 1.3.13.
+
Remark If R is a Noetherian ring, then dim R, < 00 for all p E Spec(R). However, there areexamples of Noetherian rings of infinite Krull dimension. Definition 1.3.16 A local ring ( R ,m,k ) is called regular if dim( R ) = .dimk(m/m2).
A ring R is regular if R, is a regular local ring for every p
E Spec(R).
20
Chapter 1
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 R, is a C-M local ring for all p E Spec(R), thenwe 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-l (R) is a Cohen-Macaulay ring (see [44, Theorem 2.1.31). Proposition 1.3.17 Let M be a module of dimension d over alocal ring (R, m) and let 8 = 81, . . . ,8 d be a system of parameters of M . T h e n M is Cohen-Macaulay if and only if 8 is an M-regular sequence. Proof. =+) Let p bean associatedprime of M . By Proposition 1.3.11 one has dim(R/p) = d. We claim that 81 is not in p. If 81 E p, then by Nakayama’s lemma one has (M/Ol M ) , # (0). Hence p is in the support of M/B1M, a contradiction because by Corollary 1.3.3 one has the equality dim(M/BIM) = d - 1. Therefore 81 $! p and 81 is regular on M .Thus the proof follows by induction, because accordingto Lemma 1.3.10one has that MI81 M is a C-M module of dimension d - 1 and the images of 02, . . . ,8 d in R/& form a system of parameters of M/(&)M. e)By Corollary 1.3.3 one has dim(M/eM) = 0 and M / ( @ ) Mis C-M. Therefore M is Cohen-Macaulay by Lemma 1.3.10. 0 Lemma 1.3.18 Let (R, m) be a local ring and let (f1,. . . , f,.) be a n ideal of height equal t o r. Then there are f,.+l,. , , ,fd in m such that f l , . . . , fd is a system of parameters of R. Proof. Set d = dim(R) and I = (f1, . . . ,f,.). One may assume r < d, otherwise there is nothing to prove. Let p1,. , . , p s be the minimal primes of I . Note that ht (pi) = r for all i by Theorem 1.3.14. Hence ifwe pick f r + l in m \ U;=,pi, one has the equality ht ( I ,f r + l ) = r 1, and the result 0 follows induction.
+
by
Definition 1.3.19 LetRbe 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 1.3.20 An ideal I of a ring R is called a set-theoretic complete intersection if there are f1,. . . ,f,. in I such that rad ( I ) = rad ( f i , . . . ,f,.), where r = ht ( I ) . Definition 1.3.21 An ideal I of a ring R is height unmixed or unmixed if ht ( I ) = ht (p) for all p in ass^ ( R / I ). Proposition 1.3.22 Let ( R ,m) be a Cohen-Macaulay local ring and let I be a n ideal of R. If I is a complete intersection, then R / I is Cohen-Macaulay and I is unmixed.
lgebra
Commutative
21 ,’
111.1
~
I,
,I
,
,
~ ~
1..
_I
,
Proof. Set d = dim(R) and r = ht ( I ) . By Lemma 1.3.10 R / I is CohenMacaulay and dim(R/I) = d- r. Let p be an associated prime of R / I , then using Proposition 1.3.11 yields dim(R) - ht (p) As a consequence ht (p)
2 dim(R/p) = depth(R/I)
= d - r.
5 r and thus ht (p) = r. Hence I is unmixed.
0
Theorem 1.3.23 Let R be a Cohen-Macaulay ring and I a proper ideal of R of height r . If I is generated by r elements f l , . . . , f r , then I is unmixed. Proof. It is enough to prove that I has no embeddedprimes,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 c q , then I R , c pR, c qR,. Since IR, has height r and ‘is generated by r elements, one obtains (using Lemma 1.3.18) that f1/1,. . . ,fr/1 is part of a system of parameters of R,. Therefore by Proposition 1.3.17 the sequence f 1/1, . . . , f p / lis a regular sequence and thus Proposition 1.3.22 proves that IR, is unmixed. Noticing that pR, and qR, are bothassociated primes of IR,, one derives pR, = qR, 0 and hence p = q. Proposition 1.3.24 If I is an ideal of height r in a Cohen-Macaulay ring R, then there is a regular sequence f!, . . . , f r in I. Proof. By induction assume that f l , . . . , fs is a regular sequence in I such that s < r. Note that (f1, . . . ,fs) is unmixed by Theorem 1.3.23. If I c 2 ( R / ( f l , . . . , fs)), then ht ( I ) = s, which is a contradiction. Therefore there is an element fs+l in I which is regularmodulo (f1, . . . ,fs). 0 Let A be a (Noetherian) ring one says that A is a catenary ring if for every pair p c 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 (4) - ht (p) for every pair of prime ideals p c q. Theorem 1.3.25 If R is a Cohen-Macaulay ring, then R is catenary. Proof. See [44,Theorem 2.1.121.
0
Gorenstein rings Let 211 # (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) ~ = {z E
MI mx = ( O ) } ,
and the type of A4 is defined as type(M) = dimk Soc(M/gA4),
22
Chapter 1
where g is a maximal "sequence in m. Observe that the typeof M is well defined because by Proposition 1.3.7 one has: Extk(k, M)N HomR(k, M I E M ) N Soc(M/gM), where T = depth(M). The ring R is said to be Gorenstein if R is a CohenMacaulay ring of type 1, and an ideal I c R is called Gorenstein if R / I is a Gorenstein ring. For a thorough studyof Gorenstein rings see [14,44]; a vivid presentation of the ubiquity of Gorenstein rings is given in [184].
Exercises 1.3.26 Let 4: A + B be an epimorphism of rings. If dim(A) = dim(B) < 00 and A is a domain, prove that 4 is injective. 1.3.27 Let R be a catenary ring. If S c R is a multiplicatively closed set and I is an ideal of R, prove that R / I and S-lR are both catenary rings. 1.3.28 Let M = Zrx ZP;l X - x Z p g m ,where p1, . . . ,pm are prime numbers. Prove the following: (a) M has finite length iff T = 0. (b) l ( M ) = a1
+ - + a,,
(c) Ass(M) # Supp(M), if
if l ( M ) T
< 00.
# 0.
1.3.29 If M , N are R-modules, prove Ass(M @ N ) = A s s ( M ) U Ass(N). t
1.3.30 (Cayley-Hamilton) Let cp: M -+ M be an endomorphism of an R-module M . If I is an ideal of R such that cp(M) c IM, then cp satisfies an equation:
Hint If M = Rml
+ + Rmn, write a
j=1
and consider the matrix C = cpI - ( a i j ) . Use that the entries of C are endomorphisms of M that commute with eachother toconclude the equality Cadj(C) = adj(C)C = det(C)I. Note that C operates in Mn by matrix multiplication and C(m1,. . . ,mn)t=O.
1.3.31 Let I be an ideal of a ring R and let z E R \ I . If I is unmixed and I = rad (I),then (I:z) is unmixed and ht(I) = ht(I: x).
Commutative Algebra
23
1.3.32 Let A4 be a module over a local ring (R,m). If 8 = 81, . . . ,8, is an “regular sequence in m, prove that can be extended to a system of parameters of M .
e
1.3.33 Let M be an R-module and let I be an ideal of R. If there is z E I which is regular on M ,show that HomR (R/I, M ) = (0). 1.3.34 Let A4 be an R-module and let I be an ideal of R. If J
c I , then:
HomR(R/I, M / J M ) N ( J M : MI ) / J M .
1.4
Normal rings
Let A and B be two rings. One says that B is an A-algebra if there is a homomorphism of rings cp:A ”+ B . Note that B has an A-module structure (compatible with its ring structure) given by u - b = p(a)b for all a E A and b E B . Thus if A c 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 cp is injective, in this case one may always assume that K c B is a ring extension. , Given F = {fl, . . . ,fq) a finite subset of B , we denote the subring of B generated by F and cp(A)by A[F]= A[ff,. . . ,fq]. Sometimes A[F]is 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 cp is finite or that B is a finite A-algebra if B is a finitely generated A-module. A homomorphism 4: B + C of A-algebras is a map 4 which is both a homomorphism of rings and a homomorphism of A-modules. Note that 4(a 1) = a 1 for all a E A. Definition 1.4.1 Let A be a subring of B . An element b E B is integral over A , if there is a monic polynomial 0 # f(z)E A[z] such that f ( b ) = 0. The set 71 of all elements b E 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 c B is an integral extension of rings or that B is integral over A , if b is integral over A for all b E B.
A homomorphism of rings cp: A + B is called integral if B is integral over cp(A), in this case one also says that B is integral over A.
Chapter 1
24
Proposition 1.4.2 Let A be a subring of B . If B is a finitely generated A-module, then B is integral over A . Proof. Let p E B. There are al, . . . ,an E B such that
One can write
n j=l
where mij E A. Set M = ( r n i j ) , N = M - PI and a = (a1,. . . ,an). Here I denotes the identity matrix. Since Na' = 0, onecanuse the formula Nadj(N) = det(N)I toconclude ai det(N) = 0 for all i. Hence det(N) = 0. To complete theproof note that
f(z)= (-l)ndet(M - X I ) is a monicpolynomial
in A[z] and
f(8)= (-l)ndet(N) = 0.
Corollary 1.4.3 If A is a subring of B , then the integral closure in B is a subring of B .
0
2 of
A
z.
Proof. Let a,/3 E If a and p satisfy monic polynomials withcoefficients in A of degree m and n respectively, then A[a,p] is a finitely generated Amodule with basis
Hence by Proposition 1.4.2 A[a,p ] is integral over A. In particular a f p integral and ap are over A. 0 Corollary 1.4.4 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. *) This implication follows using thearguments given in the proof of Corollary 1.4.3. +=) It follows Proposition from 1.4.2. 0 Lemma 1.4.5 Let A c B be a ring extension. If B is a domain and b is an element of B which is integral over A, then A[b] isa field if and only if A is a field.
Proof. +) Let a E A \ (0) and c = integral over A , there is an equation
its inverse in A[b]. Since A[b]is
'
Commutative Algebra
25
multiplying by a" it follows rapidly that c E A. e)Let A [ x ] be a polynomial ring and let
+: A[x]-+ A[b] be the epimorphism given by + ( f ( x ) )= f ( b ) . Note that ker(+) is a nonzero prime ideal. As A [ x ] is a principal ideal domain, ker($) is a maximal ideal and consequently is A[b] a field. 0 Proposition 1.4.6 Let A c 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 1.4.5.
0
Corollary 1.4.7 Let A c B be anintegral extension of rings. If P is a prime ideal of B and p = P n 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 Proposition1.4.6,because Alp c B / P is an integral extension of rings. 0 Corollary 1.4.8 Let A c B be anintegral extension of rings. If P are two prime ideals of B such that P n A = Q n A , then P = Q . Proof. Set p = P
cQ
n A . As B, is integral over A, and PBp n Ap = PA,,
using Corollary 1.4.7 we get that P B , is maximal and hence P = Q.
0
Lemma 1.4.9 If A c B is an integral extension of rings and p is a maximal ideal of A, then pB # B and p = P n A for any maximal ideal P of B containing pB .
+
+
Proof. If pB = B , one canwrite 1 = albl a,b, with ai E p and bi E B . Set S = A [ b l , .. . ,bq],the subring of B generated by b l , . . . , b,. Since S is a finitely generated A-module and pS = S, then by Lemma 1.1.28 there is x G l(modp) such that xS = (0). As 1 E B , one derives x = 0 and 1 E p, which is impossible. Hence pB # B. Note that this part of proof holds if p is a proper ideal of A. If pB c P is a maximal ideal of B , then p c P n A and thus p = A n P by the maximality of p. 0 Proposition 1.4.10 If A c B is an integral extension of rings and p is a prime ideal of A , then there is a prime ideal P of B such that p = P n A .
26
Chapter 1
by Lemma 1.4.9 Proof. Since A, c B, is anintegralextension,then pB, # B, . Note that P n A = p for any prime ideal P of B such that PB, iscontains prime and pB, (see Exercise 1.4.23). 0 Theorem 1.4.1 1 (Going-up) Let A C B be anintegralring extension. If p c q are two prime ideals of A and p = P n A for some P in Spec(B), then there is Q E Spec(B) such that q = Q n A and P c Q .
Proof. Since A l p c B/P is an integral extension, by Proposition 1.4.10 one has q/p = ( Q / P )n (Alp) for some prime ideal Q of B containing P. It follows readily that q = A n Q. 0 Proposition 1.4.12 Let A c B be an integral extension of rings. If B is integral over A, then dim(A) = dim(B). Proof. The formula follows usingTheorem 1.4.11, Corollary 1.4.8 and Proposition 1.4.10. 0 Let A be an integral domain and let K A be its field of fractions. The integral closure or normalization of A, denoted X,is the set of all f E K A satisfying an equation of the form
f"
+ un-1fn-' + + alf + a0 = 0
(ai E A and n 2 1).
-
By Corollary 1.4.3 the integral closure of A is a subring of K A . If A = A we say that A is integrally closed or normal. If A is not a domain we say that A is normal ifA, is a normal domain for every prime ideal p of A. A useful fact is that a normal ring is a direct product of finitely many normal domains (see [92]).
Proposition 1.4.13 Let A c B be an integral extension of rings. If B is a domain and A is a normal domain, then (a) (going-down) if p c q are two prime ideals of A and q = Q n A for some Q in Spec(B), then there is P E Spec(B) such that p = P n A and P c
Q.
(b) ht ( I ) = ht ( I n A) for any ideal I of B. Proof. See [217, Theorem 51 and [217, Theorem 201.
0
Proposition 1.4.14 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. Theorem 1.4.15 (Serre's criterion) A ring A is normal ifandonlyif (25'2) depth(A,) 2 inf(2, ht ( p ) ] for all p E Spec(A), and (R1)A, is regular for all p E Spec(A) with ht ( p ) 5 1. Theorem 1.4.16 (Auslander-Buchsbaum) If (A,m) is a regular local ring, then A is a unique factorization domain.
Commutative Algebra
27
Flat and faithfully flat algebras Let 4: A -+ B be a homomorphism of rings and consider an exact sequence of A-modules
If the sequence
is exact for any N,, B is called a flat A-algebra and q!~ is said to be flat. The homomorphism 4 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. Inparticular a polynomialring A[x] inseveral variables with coefficients in A is a faithfully flat A-algebra. Theorem 1.4.17 Let 4: A following holds:
+B
be a flat homomorphism of rings. The
(a) (I1 n I2)B = I1 B n I2B, for any two ideals I1,I2 of A. (b) T h e going down theorem holds for
4.
(c) If g is a p-primary ideal of A such that pB is prime, then pB-primary ideal.
qB is a
Proof. See [217, Sections 3.H, 5.D and 9.C].
0
Theorem 1.4.18 Let 4; A -+ B be a faithfully flat homomorphismof rings. The following holds: (a) The map $*: Spec(B) ”+ Spec(A), $ * ( P )= &-‘(P) is surjective. (b) I B n A = I and ht ( I ) = ht ( I B ) for any ideal I of A . Proof. [217, Sections 4.C and 13.B].
0
Exercises 1.4.19 Let I< be a field and ‘p: A + B an isomorphism of K-algebras, note ‘p(r)= r for all r E K . If A = @i?oAi is a graded K-algebra, then B is also a graded K-algebra graded by B = @iyoV(Ai). 1.4.20 Let ‘p: A -+ B be an isomorphism between two integral domains. If
I
‘p
extends to
Chapter 1
28
1.4.21 If A is a unique factorization domain, show that A is normal.
c B , such that A is a field and B is not an integral domain (cf. Lemma 1.4.5). 1.4.22 Show an example of an integral extensionof rings A
c B be a ring extension and p a prime ideal of A. If S = A\p and B, = S-lB, prove that the map P ePB, gives a bijection between the set of prime ideals of P of B such that P n A = p and the set of prime ideals of B, containing pB,. 1.4.23 Let A
1.4.24 Let B = A [ x ] be a polynomial ring over a ring A and let I be an
ideal of A . Prove ( A / I ) [ x ]N B / I B , where the left hand sideis a polynomial ring with coefficients in A / I . 1.4.25 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 @primary ideal of B.
Hint Use Theorem 1.4.17. 1.4.26 Let B = A[x] be a polynomial ring over a ring A and letI be anideal of A . If I = nT=lqi is a primary decomposition of I , then I B = n;=,qiB is a primary decomposition of I B .
1.5
Koszul homology
Let a be an element of the ring R and K,(a) be the complex defined as
Ki =
{
R 0
for i = O , l , otherwise,
with dl: K l ( a ) + &(a) being multiplication by a. Let I be an ideal of R generated by the sequence g = (21,, . , , xn}. The ordinary Koszul complex associated to a: is defined as
K*(x; R) = K*(x1) 8
* *
8 K*(xn).
For an R-module M we shall write Kk(g;M ) for K*(g;R) 8 M . The Koszul complex associatedto E = Rn complex K*(:; R) is then the exterior algebra and the map 8 : E -+ R, defined as e(xl,. . . , x n ) = xlzl ... xnzn. That is, 8 defines a differential a = d8 on the exterior algebra A ( E ) of E given in degree r by
+ +
d(e1 A
-
r e
A e,.) = x ( - l ) i - l e ( e i ) e l A i= 1
A
Ei
A
- .. A e,..
b
Commutative Algebra
29
A consequence of the definition of the differential of K*(g;R) is that if w and w' are homogeneous elements of A ( E ) ,of degrees p and q respectively, then ~ ( U AJ w') = ( - 1 ) " ~ A a ( ~ ' ) a(w)A w'.
+
This implies that the cycles 2(K,)lform a subalgebra of A ( E ) , and that the boundaries B(K,) form a two-sided ideal of 2(K,). As a consequence the homology of the Koszul complex, H,(g),inherits a skew commutative R-algebra structure. One can also see that H*(g) is annihilated by I = (1). Indeed, if e E E and w E 2r(K*), we have from the last formula a ( e A w)= O(e)w. The ordinary Koszul complex K,(g) = K,(g;R ) is simply the complex of free modules
where AkRn is the lcth exterior power of Rn; thus A kRn is a free R-module of rank (L) with basis ,
,
Proposition 1.5.1 If g = X I , . . . ,x n is a regular sequence in R, then the Kosxul complex is acyclic, that is, the complex IC,(:) is exact. Proof. See [290, Theorem 2.31.
0
Sliding depth Let ( R ,m) be a Cohen-Macaulay local ring and let I be an ideal of R generated by 21,. . . ,x,, denote by H*(g) the homology of the ordinary Koszul complex built on the sequence E = { 2 1 , . . . , x n } a
Definition 1.5.2 (i) (SD)I satisfies sliding depth if depth Hi(:) 2 dim(R) - n + i, V i 2 0. (ii) ( S C M ) I is strongEy Cohen-Macaulay if Hi(:) are C-M, V i 2 0. (Depths are computed with respect to maximal ideals. It is usual to set depth(0) equal to 00.) Remark 1.5.3 (a) The (SD) condition localizes [156], (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:
Chapter 1
30
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 = ( x 1 , .. . ,xn} in I n J such that
I = ((g): J ) and J = ((2): I), if in addition I and J are unmixed ideals of the same height n without common components and such that I n J = (g),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 11,. . . ,Im such that N
The ideal J is said to be in the even linkage class of I if rn 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 [233] showed that J is CohenMacaulay. A dramatic resultof Huneke [179]proves that (SCM) is preserved under even linkage, his method can be adapted to prove the following. Proposition 1.5.4 Let I and J be two ideals in a Gorenstein local ring R of dimension d, and let g = { X I ., . . ,xn) be a generating set for I . Assume that J is evenly linked to I . If I satisfies the condition
( S D k ) depth Hi(:;R) 2 d - n + i, 0 5 i 5 k, then J satisfies the ( S D k ) condition as well.
Exercises 1.5.5 Let R = k [ x l , . . . ,xg] be a polynomial ring over a field k and let I = I 2 ( X ) be the ideal of 2 X 2-minors of the symmetric matrix:
Prove that I is linked to I z ( X ' ) , where X' is the symmetric matrix
In particularif char(k) = 2 the ideal I is self-linked. For other characteristics this is an open question.
t
Chapter 2
Affine and Graded Algebras A few topics connected with affine and graded algebras are studied in this chapter, e.g., Grobner bases, Hilbert Nullstellensatz, minimal resolutions and Betti numbers. We present the affine and graded versions of the Noether normalization lemma and some of their applications to affine and CohenMacaulay graded algebras. As before all base rings considered here are Noetherian and modules are finitely generated.
2.1
Noether normalizations
Affine algebras occur naturally in algebraic combinatorics and geometry. One of the key results to understand their behavior is the famous Noether normalization lemma, first we present its affine version.
Definition 2.1.1 Let k be a field and let S be a k-algebra. We say that S is an afine k-algebra if S = Ic[yl,. . . , y,.] for some 91,. . . ,y,. E S. Definition 2.1.2 Let a and is obtained by declaring
p be in W . The lexicographical order in W /?>a
if the last nonzero entry of ,8 - a is positive. Notation The set of positive integers will be denoted by N+. If a,/?E Rn, here a - /? will denote the usual inner product of a and p.
31
32
Chapter
Lemma 2.1.3 Let a1 > > am be asequence of m distinct points in W ordered lexicographically. Then there is w = (w1,. . . ,wn) E W+ such that ~1 = 1 and w > w .ai for i 2 2. Proof. Let ai = ( a i l , .. . ,ai,) and pi = (ail,.. . ,aqn-l)). We proceed by induction on n 1 2. There is k so that o l n = - .= a k n and a k n > CXin for i > k. Onemayassume k < m; for otherwise > > Dm and onecan use induction. Since p1 > - > p k by induction there is w' = (1,w2,. , . , W n - 1 ) such that w' p1 > w' pi for 2 5 i 5 k. On the other hand for every i > k one can choose Si E N+ so that e
e . .
8
W'
*
p1
+ alndi > W'
pi
+ ainSi.
To finish the proof set
wn = max(6il k and observe w
sal
< i 5 m)
and w = (w',wn)
> w - ai for i 12.
0
Proposition 2.1.4 Let R = k [ x l , .. . , ~ n be] a polynomial ring over a field k and let f be a polynomial in R \ k . Then there is a change of variables
xi = 2:'
+ yi
(i 2 2)
such that
f ( 2 1 , x y 2 + 32, . . . ,xyn + g n ) = cox; + c1x;-l + ' ' + cr-151 where 0 # co E k, r > 0, and C i E k[y2,.. . ,yn] for i 2 1. Proof. The polynomial f can be written as *
+
CT,
m
i=1
where 0 # bi E k for all i. One may assume that a1 > . - - > am are ordered lexicographically. By Lemma 2.1.3 there is w E W+ such that xi = x"' + yi satisfies the required properties. Note r = w ol. 0 Lemma 2.1.5 If R = k [ x l , .. . ,X,] then R is a polynomial ring.
is an afine k-algebra of dimension n,
Proof. Onemayassume R N B / I , where B is a polynomialring in n variables with coefficients in the field k and I is an ideal of B. Let
r
Graded
and
Affine
33
Theorem 2.1.6 Let R = k[x1, . . . ,x,] be a polynomial ring over a field k and let I # R be a n ideal of R. Then there are z1,. . . ,x, in R such that (a) k[zl,. . . , x,] is integral over k[zl,. . . , z,J, and (b) I
n k [ ~ l ,... , x,]
= zlk[zl,. . . ,zn]
+ + ~ , k [ ~ l ,. , z,]. . * * *
Proof. The proof is by induction on n. If n = 1 and I = (f(z1))# 0, then one sets x1 = f(z1). Assume n 2 2 and I # 0. One may assume that I contains a monic polynomial in 21. Otherwise take a nonzero polynomial g in I and apply Proposition 2.1.4 to get an isomorphism of k-algebras
induced by cp(x1) = x1 and cp(x:i) = xyi monic in x1. Let
+
for i 2 2, such that cp(g) is
be a polynomial in I with ci E k [ x 2 , .. . ,x,] and
T
> 0.
Set
By induction there are 2 2 , . . . , z, such that: (i) k [ ~. . ., ,zn]is integral over k [ z z , .. . ,x,], and
(ii)
I1
n k [ z 2 , .. . , zn] = ( ~ 2 , .. . ,x,).
Set z1 = f. It is not hard to see that R is integral over k[zl,. . . , x,] and k[zl,. . . ,z,] n I = ( z l , . . . , z g ) , as required
0
Corollary 2.1.7 If R = k[zl, . . . ,x,] 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,. . . ,z, in R and an integer g such that the conditions(a) and (b) of Theorem 2.1.6 are satisfied. We will show that g is equal to the height of I. By Lemma 2.1.5 the zi’s are algebraically independent. Hence Proposition 1.4.13 yields
ht ( I ) = ht ( I n k[zl,. . . , x,]) = 9. Note that there is an integral extension cp
k[xg+l,...,x,]~k[zl,...,x,]/(xl,...,xg)~k.[xl,..~)~,]/I.
Therefore n - g = dim(R/I).
0
34
Chapter 2
Corollary 2.1.8 (Noether Normalization Lemma) If R = k[x] is a polynomialringover afield k and I # R isanideal,thenthereisan integral extension k[hl, 9 hd]-R/I,
- -
where hl ,. . . ,hd are in R and d = dim(R / I ) . Proof. By Theorem 2.1.6 there is an integral extension
To conclude theargumentset
hi = xg+i for i = 1,. . . ,d.
Corollary 2.1.9 If R = k[xl,. . then R is a catenary ring.
,~
n is]
0
a polynomial ring over a field k ,
Proof. Note ht (q/p) = dim(R/p) - dim(R/q) for any two prime ideals p C q in R. Hence by Corollary 2.1.7 we get ht (q/p) = ht (4) - ht (p). 0 Definition 2.1.10 Let k c 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 2.1.ll If k c L is a field extension, then any two transcendence basis have the same cardinality.
Proof. See [191, Theorem 8.351.
0
Definition 2.1.12 Let k c L be a field extension.The transcendence degree of L over k, denoted trdegk( L ) ,is the cardinalityof any transcendence basis of L over k. Corollary 2.1.13 Let k be afieldandlet A be a finitelygenerated algebra. If A is a domain with field of fractions L, then dim(A) = trdegk(l).
Exercises 2.1.14 If A and B are affine algebras over a field k, then dim(A @k B) = dim(A)
Hint Use the normalization lemma.
+ dim(B).
k-
Graded
and
Affine
2.2
Cohen-Macaulaygraded algebras
In this section we will emphasize the relationship between graded CohenMacaulay rings and their homogeneous Noether normalizations. Then some useful characterizations of those rings will be given. Definition 2.2.1 Let k be a field. A standard algebra or homogeneous algebra is a finitely generated N-graded k-algebra 00
s = $Si '
=k[ar1,...,9,]
i=O
such that yi E SIfor all i and So = k. If we only require y i homogeneous and deg(yi) > 0 for all i , we say that S is a positively graded k-algebra. The irrelevant maximal ideal m of S is defined by 0
0
'
m=S+=$~i. i=l
Definition 2.2.2 Let IC be a field and S = k[yl . . ; ,yr] a positively graded k-graded with yi homogeneous of degree di. There is a graded epimorphism p:R=k[sl,
...,x,] "+ s
given by f ( ~ l , " * , ~Ir 4 ~ f(Yl,",YT),
where R is a polynomial ring graded by deg(zi) = d i , the presentation of S is the k-algebra R / ker(cp). The graded ideal ker(cp) is called the ideal of relations or the presentation ideal of S. Proposition 2.2.3 If S is a Cohen-Macaulay standqrd 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 c p and g = ht (p/I). There is a saturated chain of prime ideals
I c po
c p1 c ' . ' c pg
= p.
Since po is a minimal prime ideal of I using Proposition 1.3.11one obtains dim(R/I) = dim(R/po), that is, r = ht (PO). Hence there is a saturated chain of prime ideals
36
Chapter
+
As R is a catenary domain one obtains ht (p) = r g and consequently ht ( J ) 5 r + g. It is not hard to verify that this inequality is equivalent to the inequality dim(S) 5 ht ( L ) dim(S/L).
+
To conclude the proof recall that the inequality
+
dim(S) 2 ht (L) dim(S/L) in
holds
0
Corollary 2.2.4 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 fromProposition
1.3.11 andProposition
2.2.3.
0
Definition 2.2.5 Let k be a field and letS be a positively graded k-algebra. A set of homogeneous elements = (81, . . . , 8 d ) is called a homogeneous system of parameters (h.s.0.p for short) if d = dim(S) and rad = S+.
e
(e)
Corollary 2.2.6 Let S be apositively gradedalgebra overafield h l , . . . ,hd a homogeneous system of parameters for S. Then
k and
dim S / ( h l , .. . ,hi) = dim(S) - i,
f o r 1 5 i 5 dim(S). Proof. Set 3 = S / ( h l , .. . ) h i ) and d = dim(S). First note that the set of images of hi+l,. . , hd in 3 generate an %-primary ideal, where m = S+ is the irrelevant maximal ideal of S and H = m??. Hence )
dim@) ,< d - i by the graded version of -the dimension theorem (see Theorem 1.3.13). On the other hand if 31, . . . ,e,. is an homogeneous system of parameters for 3, then the sequence 81,. . . ,8, h i , . . ,hi generates an m-primary once again one has dim@)
Thus
= d - i.
ideal of S, and applying the dimension theorem
d 5 r + i = dim@)
+ i. 0
Proposition 2.2.7 Let S be a positively graded algebra over a field k and
e = 8 1 , . . . , @ d a n h.s.0.p for S. Then S is Cohen-Macaulay if and only if 8 is a regular sequence.
Graded
and
Affine
37
Proof. By Corollary 2.2.6 dim S/(81, . , . ,e,) = d - i. If S is C-M, then by Proposition 2.2.3 one has ht (01, . . . ,8i) = i. Assume el, . . . ,8+" is a regular sequence. Next we show that 0i is regular on A = S/(&,. . . ,8i-l), Otherwise if Bi is a zero divisor of A, then 8i belongs to someminimal prime p of (81,. . . ,Oi-1). Since ht (p) 5 i - 1 (see Theorem 1.3.14) we obtain ht (81, . . . ,S i ) 5 i - 1, which is impossible. Conversely if 8 is a regular sequence, then depth(S) = d and S is C-M. 0 Proposition 2.2.8 Let S be a positively graded algebra over a field k . If S is Cohen-Macaulay and 8 = 81,. . . 8, is a regular sequence of homogeneous elements in S+ then S/(@)is Cohen-Macaulay. )
)
Proof. It suffices to prove the case q = 1, which is a direct consequence of Lemma 1.3.10. 0 Lemma 2.2.9 Let V # (0) be a vector space over an infinite field K . T h e n V is not a finite union of proper subspaces of V .
Proof. We proceed by contradiction. Assume that there are proper subspaces VI, . . . ,Vm of V such that
where m is the least positive integer with this property. Let
Pick rn + 1 distinct nonzero scalars ko, . . . ,k, in IC. Consider the vectors
the
for i = 0 , . . . ,m. By the pigeon-hole principle there are distinct vectors p,., ,Bs in vj for some j . Since ,Br - ps E vj we get 212 E vj. Thus j = 2 by the choice of 212. To finish the proof observe that p,. E V2 imply v1 E V2, which 0 contradicts choice of 211. Proposition 2.2.10 Let R = k[zl,. . . ,xn] be a polynomialring over a field k with a positive grading and let I be a gradedideal of R. Then there are homogeneous polynomials h l , . . , , hd in R+ such that dim R / ( I , h l , .. . ,hi) = d - i, for i = 1,., . , d ,
where dim(R/I) = d. If k is an infinite field and deg(xi) = 1 f o r all i, then h l , . . . ,hd can be chosen in R1.
38
Chapter 2
Proof. Assume d > 0. Let p1,. . . ,p,. be the set of minimal primes of I of height g = ht (I).We claim that there is a homogeneous polynomial hl not in u:=,pi. To show it we use induction on r. Since p1 pr-l 4 p,. and because p i is graded we can pick f E p1 . pr-l n Rdl and f 6 p,., dl > 0. By induction there is g E Rdz and g !$ UIl;pi, d2 > 0. Assume Ri C U:=,pi for all i > 0, hence g E p,.. To complete the proof of the claim consider h = f d z - gdl to derive a contradiction. Note dim(R/I) > dim R/(I, hl), by the choice of hl. Hence a repeated application of the claim rapidly yields a sequence hl , . . . , h, of homogeneous polynomials in R+ with s 5 d and such that ht (I,hl, . . . ,h,) = dim(R). Therefore by Theorem 1.3.13 one concludes d = s, as required. If k is infinite and deg(xi) = 1 for all i, then there is hl in R1 and not in Ur=,pi; for otherwise one has R1 = U:=l(pi)l and since R1 cannot be a finite union of proper subspaces (see Lemma 2.2.9) we derive R+ = pi for some i, which is impossible. Thus one may proceed as above to get the required
-
Theorem 2.2.1 1 (Noether normalization lemma) Let k be a field and let R be a polynomialring with coeficientsin k . If R is positively graded and I is a graded ideal of R, then there are homogeneous polynomials hl, . . . ,hd in R+, with d = dim(R/I), and a natural embedding
A = k[hl,. . . ,hd] 4 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 hl, . . . ,hd can be chosen in R1. Proof. Let R = k[xl, . . . ,x,] and set Bi = {xal deg(xa) = i}. According to Proposition 2.2.10there arehomogeneous polynomialshl, . . . , hd in m = R+ with rad (I,hl, . . . ,hd) = m. Hence there is r such that mr C ( I , h l , . . . ,hd).
Note that Bi C ( I ,hl, . . . ,hd) for i 2 s = r 6 , where 6 is the maximum of the degrees of X I , . . . ,x,. Set B = U;=,Bi. Using induction on i wenow show that & C J = I Bk[hl,. ..,&I, v i 2 S.
+
The containment is clear for i 5 s. Assume i > s and take xa in Bi. Let ff ,. . .,fq be a homogeneous generating set for I. Since (I,h l , . . . ,hd) is graded we can write d
Q
xu =
b jf j j=l
+Ccjhj, j=1
Graded
and
39
Affine
where bl , . . . ,b, and c1, . . . , cd are homogeneous and deg(cjhj) = i. Hence all the monomials in the support of cj have degree less than i. Thus by induction one gets cj in J for all j and consequently xa E J . Observe that the image of l3 in R / I generates R / I as an A-module. Next we verify that the canonical map cp form A t o R / I is injective. Since R / I is integral over cp(A) by Proposition 1.4.12 one obtains that d is equal to dimcp(A). Using A / ker(cp) N cp(A) and the fact thatA has dimension at most d we get ker(cp) = 0. If k is infinite and deg(xi) = 1, then according to Proposition 2.2.10 one can choose hl, . . . , h d of degree one. 0 Definition 2.2.12 Let R be a positively graded polynomial ring over a field IC and I # R a graded ideal. A homogeneous Noether normalization of S = R / I is an integral extension
where hl , . . . , h d are homogeneous polynomials in R+ and d = dim(S). Proposition 2.2.13 (Stanley) Let R = k [ x l , .. . ,~ n be] a polynomial ring over a field k and let I be a monomial ideal with d = dim(R/I) . T h e n
is a Noether normalization, where
cri
is the ith symmetric polynomial.
Proof. It suffices to prove that S is integral over A. Let ~i = xi + I . Since d = dim(R/rad(I)), one has Oi E rad(I) for i > d. Hence from the equality (Z- 2 1 )
* * *
one has "n Zi
(Z- X n )
- i7;q-l
= xn - 01xn-l
+
* *
+ +(-I>~o, *
+ (-1)d&q-d
'
=h,
where is nilpotent in S. Hence if hm E I , raising the last equality to the mth yields power that ?& is integral over A. 0 Proposition 2.2.14 Let R = k [ x l , .. . ,~ n be] a positively graded polynomial ring over a field IC and let I # R be a graded ideal of R. If S = R / I is a Cohen-Macaulay ring and
is a homogeneous Noether normalization of S , then S = A ~ b 1@ . . . @ A ~ b m f o r any set of monomials B = (xal,. . . , ,Om 1 whose image in the Artinian ring 3 = R / ( I , h l , . . . ,hd) is a k-vector space basis of
s,
40
Chapter 2
Proof. Set Mi = {x" E RJdeg(z") = i}. First we show that the image of B generate S as an A-module. It suffices to prove Mi C J = I + AB for all i 2 0. Let x" E Mi. There are homogeneous polynomials p1, . . . ,p d in R
. . . ,,X
and XI,
in k such that
where f E R i n I and deg(pjhj) = i if pj # 0. Since degpj < i, by induction one obtains that pj E J for all j and consequently x" E J . We claim that if c1xO1+ - .+crnx'm belongs to ( I ,hl , . . . , hi-1) for some 2 5 i 5 d and c1,. . . ,Cm in k[hi,, . , , hd], then C j = 0 for all j . To prove the claim note that hl, . . . ,hd is a regular sequence (see Proposition 2.2.7) and use descending induction on i starting with i = d. Next we show S = AZp1 @ATfim. 9
Assume
m
i= 1
for some al, . , . ,am in A. If at
# 0 for some e, write j=O
where aij are in k[h2) . . . hd]. One may assumea,o # 0 for some r ; otherwise we may factor outhl tosome power and applythat hl is regular onS. Hence amOxPm is in ( I ,h l ) , and applying the claim with i = 2 we u1oxP1 derive a,o = 0, which is a contradiction.Therefore a j = 0 for all j , as required. 0
+ + 9
9
Lemma 2.2.15 Let k be an infinite field and let
be a standard k-algebra. If S is Cohen-Macaulay, then there exists a h.s.0.p
-9 = (el, . . . ,ed} such that 8 is a regular sequence and 9i E S1 for all i.
Proof. Let Ass(S) = {PI,. . . ,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 c 2 ( S ) = Uf=,pi, then S1 = Uf=l(pi)l. Since k is infinite, by Lemma 2.2.9, we obtain that s = 1 and p1 = S+,that is, dim(S) = 0, which is a contradiction. Hence there is 81 E S1 whichis regularon S.
Graded
and
Affine
41
By Proposition 2.2.3 dimS/(hl) = d - 1. Applying the depth lemma (see Cemma 1.3.9) to the exact sequence
o -+~ ( - 1 )-% s -+ s/(el)-+ o yields depth S/(81) = d - 1. Altogether S / ( & )is C-M and the result follows by induction. 0 Lemma 2.2.16 Let R be a n 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 = ( f l , . . . , f,>. Proof. Let hl, . , . , h, be a 'regular sequence generating the ideal I . Using the graded version of Corollary 1.1.30 one has
r = v(I) = dimk(I/mI) 5 g , where m = R+. On the other hand,since I is graded, there are homogeneous polynomials f i , . . . , f r generating I and such that {fi m I } b l is a k-basis for I / m I . By Theorem 1.3.14 one concludes r = g. 0
+
Proposition 2.2.17 Let R be a positively graded polynomial ring over a field IC and let I be a graded ideal of R. If I is a complete intersection, then R/ I is Cohen-Macaula y. Proof. By Lemma 2.2.16 the ideal I is generated by a sequence f 1 , . . . , fg consisting of homogeneous polynomials, where g denotes the height of I . According to Proposition 2.2.10 there is a sequence hl, , . . ,hnWgconsisting of homogeneous polynomials such that rad (fl,. , . , f,,hl, . . . ,hn-9) = m = R+. Therefore {fl, . . . , f g , h l , . . . ,hn-g} is a homogeneous system of parameters for R and by Proposition 2.2.7 we derive that f 1 , . . . , fg is a regular sequence. Finally Lemma 1.3.10 yields that R / I is Cohen-Macaulay. 0 Lemma 2.2.18 Let I , P I , . . . ,p m be graded ideals of a polynomial ring R over a field k such that p1, . . . , pm are prime ideals. If f E U E , p i f o r a n y f E I homogeneous, then I C pi for some i . Proof. By induction on m. If m = 2 and I (f pi for i = 1,2, then for each i pick a homogeneous polynomial fi E I \ pi of degree ui. Since fF2 f2u1 is homogeneous we readily derive a contradiction. Hence I C pi for some i. Let G be the setof homogeneous elements in I and suppose G c U Z , p i . One may assume pi pj for i # j . If I @ pi for all i, then by induction there is f l in G \ U K l ' p i . On the other hand since Ipl - . p m - l (f p m (see Exercise 1.1.39) there is a homogeneous polynomial f2 in Ipl - - p m - l and f2 not in p m . Since + f;', deg(fi) = ui, ishomogeneous we rapidly derive a contradiction. Thus I c p i for some i. 0
+
fr2
a
42
Chapter 2
Proposition 2.2.19 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 f 1 , . . . ,f,. in I of homogeneous polynomials. Proof. We will proceed by induction. Assume that f l , . . . , fs is a regular sequence of homogeneous polynomials in I , where s < r . Note that the ideal ( f 1 ,. . . ,fs) is Cohen-Macaulay by Proposition 2.2.17, and hence it is unmixed by Corollary 2.2.4. If all the homogeneouspolynomials in I belong to Z(Rl(f1,.. . ,fs)), then by Lemma 2.2.18 the ideal I must be contained in an associated prime of ( f 1 , . . . ,fs) and consequently ht ( I ) 5 s, which is impossible. Thus there is a homogeneous polynomial f s + l in I such that f s + l is regular modulo ( f 1 , . . . ,fs). 0 Tensor product ofaffinealgebras Let A , B be two affine algebras R2/I2, where over a field k and considerpresentations A N R1 /I1, B 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
+
induces a k-algebra homomorphism:
On the other hand thereis a k-bilinear map
-
givenby multiplication $(J,g) = fg. By the universal property of the tensor product there is a map $ that makes commutative the following diagram
-
where 4 is the canonical map and t,b = $4. As a consequence $ is the inverse of cp. Thus we have proved: Proposition 2.2.20 If A and B are two afine algebras over a field k with presentations A N k[x]/I1and B N k[y]/I2,then
A @k
2
+ 12).
k[x,Y ] / ( I l
,
Graded
and
Affine
43
Theorem 2.2.21 If A and B are two standard algebras over a field depth(A @k B ) = depth(A)
IC, then
+ depth(B).
Proof. Pick a regular sequence g = 91,.. . ,gr (resp. h = h l , , . . , h,) on A = k[x]/11(resp. B = k[y]/I2) with g c (x) (resp. c (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
gives a natural commutative diagram
such that the map in the first row is injective. Since the vertical arrows are natural ismorphisms by Proposition 2.2.20, one concludes that the map in the second row is also injective. As a consequence g is a regular sequence on IC[x,y]/(11,12). Similarly if we tensor the injective map
with the faithfully flat k-algebra k[x]/(Il,g) - it follows rapidly that h is a regular sequence on k [x,y)/ (11,12,2). Altogether 9,h is a regular sequence on k[x,y]/(11,1 2 ) . To finish the proof note that-(x, y) is an associated prime ideal of k [ x ,y]/(Il,I2,g,h) by Exercise 2.2.27, thus T s is the length of a maximal regular sequence in (x, and y) by Proposition 1.3.7 one obtains depth(A @k B ) = r s. For an alternative proof see [115]. 0
+
+
Corollary 2.2.22 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 greateror equal than the depth by Lemma 1.3.6, the result follows at once using Theorem 2.2.21 together with 0 Exercise 2.1.14.
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 [115].
Chapter 2
44
Exercises 2.2.23 Let V be a vector space over an infinite field k and W,V I , .. . , Vm vector subspaces of V . If W (f K for all i, prove W $ U E l v i .
Hint If W c uEIK,note W = Uzl(W r lK) and use Lemma 2.2.9.
.
2.2.24 Let I , I 1 , .. , Im be ideals of a ring R and let k be an infinite field.
If I
c uZl Ii
and k is a subring of R, then I
c Ii
for some i.
Hint Note that I , I 1 ,... ,Im are vector spaces over k . 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 2 1, prove that there are forms fi, . . . ,fm of degree p in I such that f1,.. . , f,. is a regular sequence and I is minimally generated by f l , . . . ,f m .
2.2.25 Let
Hint Find a regular sequence sequence to a k-basis of Ip.
fl,
. . . ,f r
in I p = Rp n I and extend this
R = k[x] be a polynomial ring over a field k and I c ( x ) an ideal. Note that k is an R-module with multiplication f(z) a = f(O)a,for f(z)E R and a E k. When is the canonical map
2.2.26 Let
-
a homomorphism of R-modules?. 2.2.27 Let A = k [ x ] / I l and B = k [ y ] / I 2be affine algebras over a field k . If (x)(resp. (y)) is an associated prime of k[x]/I1(resp. k [ y ] / I 2 ) ,then (x,y )
is as associated prime of k [ x , y ] / ( I l , I z ) .
Hint there are embeddings
2.2.28 Let k [ x ]be a polynomial ring and I an ideal (resp. graded). If k
cK
is a field extension, then there is a natural (resp. graded) isomorphism of K-algebras: K [ x ] / I K [ x 21 ] k [ x ] / I@k K. 2.2.29 Let A be an affine algebra over a field k and let k extension.Prove: (a) dim(A) = dim(A @k K ) .
cK
be a field
(b) depth(A) = depth(A @k K ) if A is a standard algebra. 2.2.30 Let K be a field and let A be an affine K-algebra. Prove that A is
Artinian if and only if dimK(A) < 00.
Graded
and
Affine
2.3
Algebras
45
Hilbert Nullstellensatz
Let k be a field and let R = k [ z l , .. . ,z n ] be a polynomial ring, we define the a f i n e space of dimension n over k , denoted by , !A to be the Cartesian x k. product kn = k x Given an ideal I of R define the zero set or variety of I as
V ( 1 )= (a E knl f ( a )= 0 , Vf
E
I},
by the Hilbert basis theorem V ( I ) is the zero locus of a finite collection of polynomials. Conversely for any X c bn define I ( X ) ,the ideal of X, as the set of polynomials that vanish on X. An afine variety is the zero set of an ideal. The dimension of a variety X is the Krull dimension of its coordinate ring R/ I ( X ) . Proposition 2.3.1 If R = k [ x ] is a polynomial ring over a field
k, then
(a) V ( I n J ) = V ( I )U V ( J )= V ( I J ) ,for all ideals I and J of R.
(b) n v ( I , ) = V ( u I , ) ) where {I,} is any family of ideals of R. (c) V ( l )= 8 and V ( 0 )= A:; where n is the number of variables of R. Proof. It is left as an exercise.
0
By the previous result the sets in the family
r = {A:
\ V ( I ) I( C R ]
are the open sets of a topology on A!, called the Zariski topology. Definition 2.3.2 An affine variety X c kn is reducible if there are affine varieties X1 # X and X2 # X such that X = X1 U Xa, otherwise X is irreducible. Theorem 2.3.3 Let k be a field and X an afine variety of A :) irreducible if and only if I ( X ) is a prime ideal. Proof. See [120, Chapter 11.
then X is 0
Proposition 2.3.4 If X c A: is an afine variety overa field k , then there are unique irreducible afine varieties X I , . , . , X,. in A: such that
and Xi $ X j for all i # j .
Chapter 2
46
0
Proof. See [141, Proposition 1.31.
The irreducibleaffine varieties X I ,. . . ,X,. of Proposition 2.3.4 are called the irreducible components of X . Theorem 2.3.5 Let R = k [ x l , . . . xn] be a polynomial ring over a field k and let m be a maximal ideal of R. If IC is algebraically closed, then there are a l , . . . ,a, E k such that )
m = (21 - ~
. . ,xn - an).
1 , .
Proof. According to Corollary 2.1.8 there is an integral extension
k [ h ,' ' > hd]-R/m, where d = dim(R/m) = 0. Hence the canonical map 9:k -+ R / m is an isomorphism because k is algebraically closed. To completethe proof choose ai E k so that ai) = iii = ?& = cp(xi). 0 Proposition 2.3.6 If I is an ideal of a ring R and f E R, then f E fl if and only if ( I ,1 - tf) = R[t],where t is a new variable. Proof. +) Let f be an element in fl. If ( I , 1 - t f ) # R[t],take a prime ideal p containing ( I ,1 - tf). Since f must be in p, because f is in I for some n > 0, one concludes 1 E p which is impossible. e)If 1 = a l f ~ . aqfq aq+l( 1 - t f ) , where fi E I , f E R and ai E R[t].Set z = l / t and note 1 - tf = ( z - f ) / t , multiplying the first equation by zm, with m large enough, one derives an equality
+ +
+
zm = b l f l
+ -+
a
'*
bqfq
+
bq+l(z
- f),
where bi E R[z].As R[z]is a polynomial ring over R, making z = f gives f" E I and f E fl. 0 Theorem 2.3.7 (Hilbert Nullstellensatz) Let R be a polynomial ring over an algebraically closed field k and let I be a n ideal of R, then
I ( V ( I ) )=
a.
Proof. One clearly has I c I ( V ( I ) ) ,hence fi c I ( V ( I ) )because I ( V ( I ) ) is a radical ideal. For the other containment take f E I ( V ( I ) )and consider the ideal J = ( I ,1 - f t ) of the ring R[t],where R = k [ z l ,. . . ,X,] and t is a new variable. Next we show the equality J = R[t]. If J # R[t]by Theorem 2.3.5 one has
( I )1 - f t ) c
(21
- al, -
, x , - an, t
an+l), ai E
k.
Hence p E V ( I ) ,where /3 = ( a l , . . . ,an). Using that f E I ( V ( I ) gives f(P) = 0, but this is impossible because 1- f(P)a,+l = 0. Hence J = R[t]. Thus f E fl by Proposition 2.3.6. 0
Graded Affine and
47
Algebras ,
I
Corollary 2.3.8 Let X = V (f l , . . . fq) be a n a&ne variety, defined by polynomials f1,. . . ,f, in n variables over an algebraically closed field k . T h e n q 2 n - dim(X). )
Proof. By the Nullstellensatz I ( X ) = rad (fi, . , . , f,). Let p be a minimal prime of ( f l ) . . . fq) of height n - dim(X). Applying Theorem 1.3.14 we get n - dim(X) 5 q. cl )
Proposition 2.3.9 Let R = k [ x l ) .. . ,x,] be a polynomial ring over a field k and let m be a maximal ideal of R. If k is uncountable, then R/m is a finite extension of k . Proof. Assume Ti is not algebraic over k , where, as usual, 5; denotes T + m for any T in R. For simplicity we set i = 1. As R/m has a countable basis consisting of monomials, one has that the (uncountable) setdefined by
is linearly dependent. Thus 3 c, c l , . . . , cq in k distinct and XI,, k \ (0) such that: Q 1 xi -
. , , X,
in
"
2=
1
Consider the polynomial
f(x) = l"J(x - Ci) - (x - E )
-
rI(x - E i ) X i . j=1 i # j
i=l
Since f(z1)= 0 and is notalgebraic over k we conclude that f is the zero polynomial, which is impossible because f(E) = (c - ci) # 0. 0
nf==,
Exercises 2.3.10 Let I and J be two ideals in a polynomial ring R over a field k . If rad ( I ) = rad ( J ) , prove that V ( I )= V ( J ) . 2.3.11 If k is a field and X an affine variety.
c A;)
then X c V ( I ( X ) )with equality if X is
2.3.12 Let J be an ideal of a polynomial ring R over a field k. Prove that J c I ( V (J ) ) with equality if J is the ideal of an affine variety. 2.3.13 Let k be analgebraically closed field. Prove that thereis a one t o one correspondence between affine varieties (resp. irreduciblevarieties) in k" and radical ideals (resp. prime ideals) in the polynomial ring k[sl,. . . 3 4 . )
48
Chapter 2
2.4
Grobner bases
In this sectionwe review some basic facts and definitions on Grobner bases. The reader may consult [l, 74, 2431 for a detailed discussion of Grobner bases and for the missing proofs of this section. Let k be a field and let R = k [ x l , .. . ,xn] be a polynomial ring. We set
The monomials in
Pn={xaI.EW) are called the terms. Definition 2.4.1 A total order > of Pn is called a monomial order or term order if (a) p 2 1 for all p E Pn, and (b) for all p , q, r E Pn, q
> p implies qr > pr.
Two examples of monomial orders in Pn are the lexicographical order or lex order defined as 'x > x a iff the last nonzero entry of b - a is positive, and the reverselexicographicalorder or revlex order givenby x b > xa iff the last nonzero entry of b - a is negative. In the sequel we assume that a monomial order for Pn has been fixed. Let f be a nonzero polynomial in R. One can write
> M r . The leading t e r m with ai E k* = k \ {0}, Mi E Pn and M I > M I of f is denoted by It(f). The leading coeficient a1 of f and its leading monomial a1 M I are denoted by lc(f) and Im(f) respectively. Definition 2.4.2 Let I be an ideal of R. The initial ideal of I, denoted by in(1) , is given by i d 4 = ( { W l f E I}). Lemma 2.4.3 (Dickson) If {Mi}& is a sequence in Pn, then there is an integer k so that Mi is a multiple of some term in the set { M l , . . . ,A l k } for every i > k. Proof. We consider the ideal I c k [ x l ,. . . ,~ n generated ] by {Mi}gl. By the Hilbert basis theorem I is finitely generated. It is readily seen that I can be generated by a finite set of terms MI, . . . ,Mk. Hence for each i > k, there is 1 5 j 5 k such that Mi is a multiple of M j . 0
Graded
and
Affine
for some fi E F , u E coefficient a.
Pn,a
E k* such that
Proposition 2.4.5 The reduction relation any sequence of reductions
a - u lt(fi) occurs in f with
“+F
”
is Noetherian, that is,
is stationary. Proof. Notice that at the ith step of the reductionsome term of gi is replaced by terms of lower degree. Therefore if the sequence above is not stationary, then thereis a never ending decreasing sequence of terms in P n , this but is impossible according to Dickson’s lemma. 0 Proposition 2.4.6 (Division algorithm) Iff,f1, . . . , f q are polynomials in R, then f can be written as
where ai, r E R and either r = 0 or r # 0 and no term of r is divisible b y one of lt(fi), . . . , I t ( f q ) . Furthermore ifaifi # 0, then lt(f) 2 lt(aifi), Definition 2.4.7 The polynomial r in the division algorithm is called a remainder of f with respect to F = { f l , . . . , f,). Definition 2.4.8 Let I # (0) be an ideal of R and let F = { f l , . . . ,f,.) be a subset of I . The set F is called a Grobner basis of , I if Definition 2.4.9 A Grobner basis F = { f l , . . . , f,.} of an ideal I is called a reduced Grobner basis for I if (i) IC( fi) = 1 V i , and (ii) none of the terms occurring in
fi
belongs to in(F\
{fi})
Vi.
Note that by the Hilbert basis theorem a Grobner basis of I always exist.
A nice feature of a reduced Grobner basis is its uniqueness. Definition 2.4.10 Let f,g E R. The S-polynomial of f and g is given by
Chapter 2
50
Given a set of generators of a polynomial ideal one can determine Grobner basis using the next fundamental procedure:
a
Theorem 2.4.11 (Buchberger) If F = (f1, . . . ,fq} is a set of generators of an ideal I of R, then one can construct a Grobner basis for I using the following algorithm: Input: F Output: a Grobner basis G for I Initialization: G := F, B := {{fi, fj}l fi # f j E G} while B # 0 do pick any (f, 91 E B B := B \ ( { f , g H r := remainder of S(f,g ) with respect to G if r # 0 then B := B u { { r ,h}l h E G} G := G u { r }
Proof. See [l,53, 2431.
0
Proposition 2.4.12 Let I be an ideal of R and let F = (f 1 , . generating set for I . If
. . ,fq} be
a
B = {’ii I u E Pn and u @ (lt(fl), . . . ,It(&))}, then B is a basis for the k-vector space R I I . First we show that B is a generating set for R / I . For that take f E R/I. Because “+F” is Noetherian we can write
-Proof.
4
r
i=l
i= 1
where X i E k* and such that every ui is a term which is not a multiple of any of the terms It (fj). Accordingly zii is in l3 for all i and 7 is a linear combination of the Ei’s. To prove that B is linearly independent assume h = Xiui E I, where ui E B and X i E k. We must show h = 0. If h # 0, then we can label the ui’s so that u1 > > us and X1 # 0. Hence lt(h) = Xlul E in(l), but this is a clear contradiction because in(I ) = (lt(fi), . . . ,It( fq)). Therefore 0 h = 0, as
x&,
required.
Corollary 2.4.13 (Macaulay) If I is a graded ideal of R, then R I I and R/ in(I) have the same Hilbert function.
Graded
and
Affine Lemma 2.4.14 Let f and g be two polynomials in R and let F = { f,g}. If lm( f) and l m ( g ) are relatively prime, then S(f,g ) "+F 0.
+ +
Proof. Let f = MI - M r and g = N1 are monomials such that MI > - > Mr , N1 hi = S(f,9) - Mrg - *
+ + N s , where Mi and Ni > . - - > Ns. Set - Alr-ig.
The monomials
we obtain S(f,g) +F ho + F hl -+F notice that MI Ni+l > Mk Nm, for m 2 i s.
hr-2. To complete the proof
+F
+ 1. Hence using the identity
hr-2+N2f +.*.+Nif= - MlNi+l - MrNi+l
-
e
*
*
...
Ml Ns - Mr Ns
we obtain
that is, S(f , g)
-+F
0.
0
The next criterion uf Buchberger will be used at several places to prove that a given set of polynomials form a Grobner basis. Observe that this criterion is important in Buchberger algorithm. Theorem 2.4.15 ([53]) Let I be a n ideal of R and let F = { f l , . . . , fp) be a set of generators of I, then F is a Grobner basis f o r I if and only if S(f i , f') -+F
0 f u r all i
-
# j.
Definition 2.4.16 Let I be an ideal of R generated by F = { f1,. . . , fr) and consider the homomorphism of R-modules (p:R'
+I ,
ei
where ei is the ith unit vector. The kernel of the syzygy module of I with respect to F .
fi,
(p, denoted by Z ( F ) ,is called
Chapter 2
52
Theorem 2.4.17 Let G = (91,. . . ,g,.} be a Grobner basis and write
where a i j k E R and lt(S(gi,gj)) 2 It(aijkgk) for all i , j , k . Then the set
generates the syzygy module of I = (91,. . . ,g,.) with respect to G. Proof. See [I, Theorem 3.4.11 and [75, Theorem 3.21.
l
0
Elimination of variables Let k[zl,. . . ,z,] be a polynomial ring over a field k. A useful monomial order is the elimination order with respect to the variables X I , . . . ,zi. It is defined as
z b > Za if and only if the total degree of zb in the variables X I , . . . ,xi is greater than that of za, or both degrees are equal, and the last nonzero component of b - a is negative.
Theorem 2.4.18 Let S = k[xl,. . . ,xn, t l , . . . ,t q ]be a polynomial ring ouer a field k with a term order such that terms in the X i 's are greater than terms in the ti 's. If I is an ideal of S with a Grobner basis G, then G n k [ t l , .. . , tq] is a Grobner basis of I n k [ t l ,. . . ,tq]. Proof. Set B = Ic[tl,.. . ,tqJand Ic= I n B. Assume Ad E in(Ic). There is f E I c with lt(f) = M . Hence M = Alt(g) for some g E G, because G is a Grobner basis. Since M E B and x" > tB for all Q and p we obtain g E G n B , that is, M E (in(G n B ) ) . Thus in(Ic) = (in(G n B ) ) , its
required.
0
Next we illustrate the previous result using the elimination order on S with respect to 51, . . . ,2,. Thus this order is given by xatc > xbtd
if and only if deg(za) > deg(xb), or both degrees are equal and the last nonzero entry of (a,c) - ( b , d ) is negative.
Graded
and
53
Affine
Example 2.4.19 Let us assign deg(ti) = 2 and deg(xj) = 1 for all i, j . Using Macaulay [15]one rapidly obtains that the reduced Grobner basis of the ideal
Proof. left It is
as an exercise.
0
Definition 2.4.21 Let I and J be two ideals of a ring R. One defines the saturation of I with respect to J as:
( I :J") = U ( I : RJi). i> 1
Proposition 2.4.22 If I is an ideal of a ring R and f E R, then
( I :f") = U ( I : Rf i ) = ( I ,1 - t f )n R, i> 1
where t is a new variable. Proof. Let g E ( I ,1 - t f ) n R. One can write 9 =d
l
+ + uqfq + U , + l ( l ' * *
- if),
where fi E I and ai E R[t]. Making t = 1/ f in the last equation and multiplying by f", with rn large enough, one derives an equality
sf"
= h f 1
+..*+b,f,,
'
where bi E R. Hence gf" E I and g E ( I :f"). Conversely let g E ( I :f"), hence there is rn 2 1 such that Since one can write g = ( 1 - t"f")g
for some
+ t"f"g
sf"
E I.
and 1 - tmf" = ( 1 - t f ) b ,
b E R [ t ] one , derives g E ( I ,1 - t f ) n R.
0
54
Chapter 2
Lemma 2.4.23 (1981) Let R be a ring and let ideal of R, then the radical of I satisfies
d7 = &:
(x1
. . . ,x n
x1,
. - .x n ) m ) n d m n
nd
E R. If I i s a n
G).
Proof. The left hand side is clearly a subset of the right hand side. To prove the other containment takea prime p containing I , it suffices to prove that p contains one of the ideals on the right hand side. If
~ ( I( 2 1I
*
xn>m>
there is f E R \ p such that f ( x 1 -2,)' E I Hence xi E p for some i and p contains
P,
c p for some integer s 2
d m , as required.
1. 0
Proposition 2.4.24 Let p be a prime ideal of a ring R and let x l , . . . ,x , be a sequence in R \ p. If I c p i s a n ideal, then f l = p if and only if
d(I:(x1 (b) d m =
(a) p =
9
x ~ ) ~and) , I
f o r all i.
Proof. If J is any ideal of R and y1, . . . ,yn E R, then by Lemma 2.4.23 the radical of J satisfies:
n rad(J, y1) n
rad(J) = rad(J: (y1 . -y,)")
's.
n rad(J, yn).
Hence the result follows by applying this formula to I and p, together with x1 . - x n is regular on R/p. 0 that fact the The homogenization of an ideal Let R = k [ x l , .. , ,x,] and S = R[w] be two polynomial rings over a field k . For f E R of degree e define f h
= w e f ($.,"-), W
that is, f is the homogenization of the polynomial f with respect to w. The homogenization of an ideal I c R is the ideal I h of R[w]given by
I h = ({fhlf E
11).
Proposition 2.4.25 If I i s a n ideal of R, then Proof. For the inclusion
fi
C
f i , we need only observe
(fh>"
for f E R and m 2 0.
h fl= d .
=
the equality
Affine Algebras and Graded
55
Conversely let g E f i . As flis a graded ideal one may assume that g is homogeneous. Write g = w'g?, where g1 E R. Since g m isin Ih,it follows that g1 E fi, as required. 0 Let to 21,
21,.
> be
the elimination order on the monomials of R[w]with respect to on the monomials of R.
. . ,X n ,ut, this order extends the elimination order with respect
. . . ,X n
Proposition 2.4.26 If I is an ideal of R generated by F = { ff, , , . ,fr}, then { . . . , f:) is a Grobner basis if and only if F is a Grobner basis.
fp,
.
Proof. +) Assume { f f , . . . ,f,h} is a Grobner basis. Set J = (ff, . . ,f,"). It suffices to prove in(I) C (lt(fl), . . . ,lt(fr)). Let MI be a monomial in the initial ideal of I . There is f E I c R of degree e such that It(f) = M I , We
write
I
f =91f1 't-.**+gr.fr
'
and consider
multiplying both sides by w c for t >> 0, one has W S f h=
hlf:
+ . + hrf,h ' *
for some s. Thus w s f hE J . Since lt(fh) = It( f) we have lt(w"fh) = w"M1. As {fp, . . . ,f,"} is a Grobner basis one obtains
w"M1 E (Wfih), ' - ' 7 It(f,h)). Using lt(fi) = lt(f:) yields M1 E (lt(jl), . . . ,lt(fr)). +) Assume { f1, . . . , f r } is a Grobner basis. Set 7 = , . . . ,f,"). We now show that in(7) c (It(f,"), . . . ,lt(f,h)). Let M E in(r), there is f E I such that M = It( f ). We may assume that f is a homogeneous polynomial. We can write f = wP(M1 M 2 W e 2 - .* Mswe,),
(ft
+
where MI, . . . ,
+ +
are monomials in R such that
Using that Miwe; all have the same degree we obtain It follows that f ' = f (21, . . . ,xn, 1) belongs to I and It(f ') = MI. Therefore MI belongs to in(I) = (lt(fl), . . . ,It(fr)). Since It( ff) = lt(f:)we obtain to ideal (lt(ft), . . . ,lt(f:)). ' 0 that M belongsthe
Chapter 2
56
Corollary 2.4.27 Let I = (f1, ... ,f m ) be a n ideal of R and let
fi is a Griibner basis,
be the homogenization of I . If fp, . . . ,
Proof. It is enough to show { f h l f E I } c (f,",. . , , is a Grobner basis any f E I can be written as
then
fk). Since f i , . , . , f m
hence from:
Projective closure Let k be a field. We define the projective space of dimension n over k, denoted by Pi,to be the quotient space
where two points a,p in kn+l \ {O} are equivalent if a = cp for some c E k. It is usual to denote the equivalence class of a by [a]. Given a homogeneous ideal I of S = k [ z l , .. . ,xn,w]define the zero set O f I a s
V ( I )= {[a] E PEI f (a)= 0, V homogeneous polynomials f E I } , conversely for any X c PF define I ( X ) the ideal of X , as theideal generated by the homogeneous polynomialsin S that vanish on X . A projective variety is the zero set of a homogeneous ideal. It is not difficult to see that the members of the family )
7=
{PE \ V(I)II
cS}
are the open setsof a topology on P?,called the Zariski topology.
Graded
and
57
Affine
Definition 2.4.28 Let Y c At. The projective closure of Y is defined as Y := p(Y), where cp is the map
"
9:At "-+
IF';, a
I"+
[(a,l)],
and cp(Y)is the closure of cp(Y)in the Zariski topology of Pz, Proposition 2.4.29 If Y is a subset of the afine space A$ over a field k , then the projective closure 7of Y is given by
Proof. It is left as an exercise, see Proposition 7.1.14 for the affine case. 0 Proposition 2.4.30 Let Y c : A and let -5 c Pg be its projective closure. If f l , . . . , fr is a Grobner basis of I ( Y ) , then
I(L)= (ft,. ..J;). Proof. Note that I(y)= I ( Y ) h and use Corollary 2.4.27.
0
I E D ;
Corollary 2.4.31 Let Y c A; andlet C be itsprojectiveclosure. Then the height of I ( Y ) in R is equal to the height of I ( 7 ) in S .
Exercises 2.4.32 Let I be an ideal of a ring R and f a nonzero element of R, then
2.4.33 Let R = k[x] be a polynomial ring, where k is a field. Recall that f is a binomial in R if f = x" - x0 for some cy, p in W . Use Buchberger's
algorithm to prove that an ideal of R generated by a finite set of binomials has a Grobner basis consisting of binomials. 2.4.34 Let I be an ideal of a ring R and f E R. Prove that the following
statements are equivalent: (a) f is regular on R/ I .
(b) ( I : Rf) = I . (c) I = ( I ,1 - t f ) n R, where t is a new variable. 2.4.35 If I is a prime ideal of a polynomial ring R, prove that I h is also a
prime ideal.
58
Chapter
2.4.36 Let I V be the setof n-tuples of non-negative integers endowed with the partial order given by
if and only if ai 2 bi for all i. If A c PP , prove that A has only a finite number of minimal elements.
Hint Use Dickson’s lemma. 2.4.37 Let Y = { ( t 3 ,t2,t)I t E k } c Ai and let If IC is an infinite field prove that
2.5
be its projective closure.
Minimal resolutions
The aim of this section is to study homogeneous resolutions of positively graded modules over polynomial rings. In the sequel we shall be interested in the numerical data of those resolutions and in particular in the Betti numbers of those modules.
We begin with a result which is a consequence of the gradedNakayama’s lemma (see Lemma 1.2.6). Proposition 2.5.1 Let R = k[x] be a polynomial ring over a field k and let M be a n N-graded R-module. If F = { fi, . . . ,fp) 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 MImM is a k-basis of M l m M . Corollary 2.5.2 Let R = k [ x ] be a polynomial ring over a field m = (x). If M is a n N-graded R-module, then
k and let I
u ( M ) = dimk(M/rnM), where u ( M ) is the minimum number of generators of M . Lemma 2.5.3 Let R = k[x] be a polynomial ring over a field k and let M be a n W-graded module over R with a presentation 0
-+K
= ker(cp)
-+
Rq ’P, M
”+ 0,
ei
I-%
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 K c mRQ.
59
Algebras Graded Affine and
Proof. e)Assume I< c mR4. By Proposition 2.5.1 we need only show that the image of F in M/mM is linearly independent over IC. If
i= 1
for some X i E k, then
for some a1 , . . . ,u, in m. Hence (X,, . . . , X,) is in mRQand X j = 0 for all j . +) If K mRQ,pick z = ( X I , . . . ,z,) E K such that z 4 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. 0 Definition 2.5.4 Let R = @EoRibe a positively graded ring and a E N. The gradedR-module obtained by a shift in the graduation of R is given by
where the ith graded component of R ( - a ) is R(-a)i = R-a+i. Proposition 2.5.5 Let R = @EoRibe a positively graded polynomial ring over a field k with maximal irrelevant ideal m = R+ and M a n N-graded R-module. Then there is an exact sequence of graded free modules
where (
~ is k
a degree preserving map and im(cpk) c mRbb-1 f o r k
2 1.
Proof. Let fl,. , . ,f, be a set of homogeneous elements that minimally generate M . Set di = deg( f i ) . There is an exact sequence
where 90 is a degree preserving homomorphism such that qo(ei) = f i for all i. Note that 21 c mRq by Lemma 2.5.3. Since 21 is once again a finitely generated graded R-module one may iterate the process to obtain exactthe required sequence IF. 0
Chapter
60
Theorem 2.5.6 (Hilbert syzygy theorem) Let R = k [ q , . . . , xn] be a polynomial ring over a field k and let M be a n N-graded R-module. Then M has a graded free resolution of length at most n. Proof. Set m = R+. First notice that TorF(M, R / m ) = 0 for i 2 n + 1, because the ordinary Koszul complex & ( E ) is a graded free resolution of R / m = k of length n. On the other hand assume that
is a graded free resolution of M as in Proposition 2.5.5. Applying the functor (.) 8~R/m yields the complex
Using im(cpk) c mFk-1 for all k 2 1 one obtains that all the maps are zero. Hence TorF(M, R / m ) 21 Fi 8 R / m
cpk @
1
for i 2 1. In particular Fi 8 R/m 21 Fi/mFi = 0 for i 2 n + 1 and by Nakayama’s lemma one obtains Fi = 0 for all i 2 n 1. This proof was adapted from [92]. 0
+
Corollary 2.5.7 Let R = $ZoRi be a polynomial ring of dimension n over a field IC and let M be a n N-graded R-module. Then there is a unique (up to complex isomorphism) exact sequence of graded modules
i= 1
i= 1
such that (a) g = sup{i I TorF(M, k) # 0} 5 n, and (b) im(cpk) C mRbk-l for all k 2 1, where m = R+. Proof. Notice the following isomorphisms: bj
Tor?(”,
k) N @ k(-dii)
21 Fj/mFj,
i= 1
where Fj is the j t h 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. 0
I ~
..
,
Lil
Graded
and
Affine
61
Remark 2.5.8 The entries of the matrices cpk are in m. This condition is equivalent to require that at each stage we use a minimal generating set, see Lemma 2.5.3. Remark 2.5.9 If
dkl
5
5 d k b b for all k , then
Definition 2.5.10 The integers bo,. . . ,bg are theBetti numbers of M . The dji’s are the twists, they indicate a shift in the graduation. In general the Betti numbers and twists of A4 may depend on the base field k , see for instance [108, Example 2.101 and [237, Remark 31. Definition 2.5.11 The R-module Zk = ker(pk-1) is called the k-syzygy module of M . Definition 2.5.12 The homogeneous resolution or minimal resolution of M is the unique graded resolution of M by free R-modules described in Corollary 2.5.7.
If M has a minimal free resolution as above, then note thatpd,(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 [ x l , . . . ,x,]-modules are free [250]. Theorem 2.5.13 (Auslander-Buchsbaum) Let M be an R-module. If R is a regular local ring, then pd,(M)
+ depth(M) = dim(R).
Proof. See [log, Theorem 3.11 or [193].
0
Corollary 2.5.14 If R is a polynomial ring over a field k and I is a graded ideal,then Pd,(RII) L ht (0, with equality if and only if R / I is a Cohen-Macaulay R-module. formula.
Proof. The result follows from anappropriategraded Auslander-Buchsbaum
version of the 0
Chapter 2
62
Theorem 2.5.15 (Hilbert-Burch) Let R be apolynomial ring overa field k and let I be a graded Cohen-Macaulay ideal of height two. If
0 -+ RQ-’ ’P, Rq + R
-+ R / I -+0
is the minimal resolution of RII, then I is generated by all the minors of size q - 1 of the matrix cp. Example Let R = Q[x,y, z , w] and let I = ( f 1 , f2, fl
= li2 - xz, f 2 = x3 - yzw,
f3
f3),
where
= x2y - 2221).
Let us construct theminimal resolution of R / I . Consider the exactsequence ,. . .
0 -+ ker(cp1) = 21 id, R(-2) @ R2(-3) 3 I
+ 0,
ei
I-%
fi.
Here 2 1 is the module of syzygies of I . Using Macaulay we find that 21 is generated as an R-modulo by the column vectors, $ 1 , 3b2, of the matrix:
Notice that R(-2) @ R2(-3) is graded by
Hence
el,
$2
E
[R(-2)@ R2(-3)I4. Next consider the exact sequence
+
Since $1R 3b2R is a free R-module, we have minimal homogeneous resolution of R / I is:
22
= (0). Altogether the
0 + R2(-4) CPZ, R(-2) @ R2(-3) -% R ”+ R / I ”+ 0. In this example R / I is Cohen-Macaulay because pd,(R/I)
= 2 = ht ( I ) .
Pure and linear resolutions Let R = $goRi be a polynomial ring over a field k with its usual graduation and I a graded ideal of R. Let b,
0 -+
@ R(-dga) -+
bl
* *
-+ @ R(-dli) + R + S = R / I + 0
(2.1)
i=1
i= 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 dl
< d2 <
< dg
.”.
.
.
-
...
Graded
and
Affine
63
such that
dli = d l ,
. . . , d,i
= d,
for all i. If in addition
di=dl+i-l for 2 5 i 5 g the resolution is said to be dl-linear . Theorem 2.5.16 (Herzog-Kuhl) If S is a Cohen-Macuulay algebra with a pure resolution, then
Proof. See [154] for the proof and for a generalized version of this result modules. valid for graded 0 Theorem 2.5.17 (Huneke-Miller) Let S be a Cohen-Macaulay algebra. If S has a pure resolution, then the multiplicity e ( S ) of S is given by
Proof. See [185] for the proof and for a generalized version of this result valid not only for cyclic modules. For another simple proof see [158]. 0
Since the Betti numbers and the twists are positive integers note that these formulae impose severe restrictions on the numbers d l , . . . ,d , that can occur as the resolution type of a Cohen-Macaulay standard algebra with a pure resolution.
Exercises 2.5.18 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 ) ) -+ Rb'(-d)
-+
I -+ 0,
where a , b 2 1. Assume bl = 6 and b = 1. Show that x = a y = d a 1 is a solution of the diophantine equation
+ +
+ 1 and
- 1) = y ( y - 1).
~X(Z
Then prove that the integral solutions of this equation are given by
x = (X+ 1)/2 and y = (Y + 1)/2, where &X
+ Y = f ( 1f &)(5 + 2fi)n,
n E Z.
,
Chapter 2
64
If char(k) # 2, prove the equality
I = ( 2 2 + y2 - x 2 , x 3 + y3 - x 3 , z 2 y 2 ) . 2.5.21 Let
R = k[z,y, x] and let I = (zn+ yn - znl n 2 2).
If char(k) # 2, prove that
is a primary decomposition of I , where 41
= (Y2,S - 4 ,
q2
= ( x 2 ,y - x), and
q3
= (y3,x3,22y2,22
+ y2 - 22).
Chapter 3
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 algebras will make clear their importance for the area. One of the outstanding references for blowup algebras is (2961. An important reference for the normality of Rees algebras is [40].
3.1
Symmetric algebras
-
Let M be an R-module. Given n 2 0 we define
T n ( M )= M 8
8 M and T o ( M )= R.
n-times
Recall that the tensor algebra T ( M ) of M is the non commutative graded a1geb r a 00
T ( M )= $ T n ( M ) , n=O
where the product is induced by juxtaposition. The symmetric algebra of M , denoted by Sym,(M) or simply Sym(M), is defined as the quotient algebra
where 3 is the two sided ideal generated by the elements
zy
- yz = z
8y - y 8z E T 2 ( M )
65
Chapter 3
66
with x and y running through M. Observe that Sym(M) is commutative. Since 3 is a graded ideal generated by homogeneous elements of degree two, the symmetric algebra is graded by Sym,(M) = T n ( M ) / J n T n ( M ) andSym,(M)
= R.
with x and y running through M .
Exercises 3.1.1 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.
3.2
Rees algebrasand syzygetic ideals be an ideal of a ring R generated by fi , . . . ,f,. The Rees algebra of
Let I I , denoted by R ( I ) or R[It],is the subring of R[t]given by
where t is a new variable. Note
There is an epimorphism of R-algebras
where B is a polynomial ring in the indeterminates t l , . . . , t, over the ring R. The kernel of ‘p, denoted by J , is the presentation ideal or toric ideal of R ( I ) with respect to fi, . . . ,f,. Notice that J is a graded ideal in the ti-variables: i= 1
where B has the standard grading induced by deg(ti) = 1. The mapping
$ : R q -+
given by
I
and Rees Algebras
67
Normality
induces an R-algebra epimorphism
P: R[tl,.. . ,tq]-+ Sym,(I). Thus the symmetric algebra of I is:
where ker(P) is an ideal of R[t] generated by linear forms:
On the other hand thekernel of cp is generated by all forms F(t1,. , . ,t 4 ) such that F ( f 1 , . . . ,fq) = 0. Inparticular, onemay factor cp through SymR(I) and obtain the commutative diagram:
R[tl,.. . ,tq]‘ ’p, R ( I )
SYmR ( I ) We say that I is an ideal of Zinear type if
Q
is an isomorphism.
An important module-theoretic ,obstructionto“SymR(I) given by the following result.
n ( I ) ” is
Proposition 3.2.1 (Herzog-Simis-Vasconcelos)Let I be an ideal of a ring R. If SYmR(q 2 R ( I )>
then for each prime p containing I , I p can be generated by ht (p) elements. Proof. See [155].
0
Syzygetic ideals Let I be an ideal of a ring R andlet H l ( I ) be the first homology module of the Koszul complex H*(g,R) associated to a set x = 21, . . . xn of generators of I . In E2601 it is pointed out that H1 ( I ) is related to 1/12,the conormal module of I , by the following exact sequence:
Here f([z]) = z @ 1 and h(ei 8 1) = xi 8 1. Set S ( I ) = ker(f). Definition 3.2.2 The ideal I is called syzygetic if S ( I ) = 0.
Chapter 3
68
Although H1 ( I ) may depend on the setof generators for I , S ( I ) depends only on I . Indeed A. Simis and W. Vasconcelos [260] proved that 6(I) N ker(Sym,(I)
-+ 12),
where Sym2(I) -+ I2 is the surjection induced by the multiplication map. Definition 3.2.3 An ideal I of a ring R is said to begenerically a complete intersection if I R p is a complete intersection for all p E A S S R ( R / I ) . Remark 3.2.4 If I is unmixed and generically a complete intersection, then the R/I-torsion of H I ( I ) equals S ( I ) . To prove this, notice that I p is syzygetic for all p E ass^ (R/I), consequently S ( I ) is a torsion module. Computing the toric ideal of a Rees algebra Let I be a graded ideal of a polynomial ring R over a field IC and f1, . . . ,f q a generating set for I. Consider the presentation of the Rees algebra:
* R ( I ) +0,
9:B = R [ t l , .. . ,tq]
(ti
* tfi).
The kernel J of cp can be obtained as follows:
J = (tl - t f i , . . . ,t, - t f q )fI B. Since J is a graded ideal in the ti-variables, J = $izl 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/BlJ1
= S ( I ) -+ H 1 ( I ) + (R/I)Q_.) 1/12"+ 0.
In particular one can decide whether I is syzygetic - that is, Jz = BlJI-or of linear type-that is, J = JIB.
Exercises 3.2.5 Let R = A[x]be a polynomial ring over a ring A and fl ,. . . ,f q forms of degree d 2 1. Prove that there is a graded isomorphism of A-algebras 9:A [ f i , ..
,fq]+ A[&. . . ,t f q ] , with cp(fi) = t h ,
where t is a new variable and both rings have an appropriate grading such that deg(fi) = 1 and deg(tfi) = 1.
3.2.6 Let I = 1 2 ( X ) be the ideal of 2 x 2-minors of the symmetric matrix:
[ :; :; :I x2x1
X =
x3 )
where the entries of X are indeterminates over a field k. Prove that I is a syzygetic ideal that satisfies sliding depth. See [161].
6
and Rees Algebras
3.3
Normality
69
Complete andnormal ideals
Let us recall the notion of integrality of ideals. Let R be a ring and I an ideal of R, an element z E R is integral over I if z satisfies an equation
the integral closure of I is the set of all elements x E R which are integral over I . This set will be denoted by 7 or I,. Definition 3.3.1 If I = 7, 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. Lemma 3.3.2 Let R be a ring and I a n ideal of R. An element a E R i s in the integral closure of I if and only if there is an ideal L of R such that (i) aL c I L , and (ii) a'ann(l) = (0) for some integer r
2 0.
Proof. =+) As a is in integral closure of I there is an equation:
an = alan-'
+ - + an-lct + an, *
where ai E I i ,
and n 2 1 is some integer. The required L is obtained by setting
L = Ran-l + Ian-2 + . . . + In"2a + In-1 Indeed from the equation above
and if i
2 1 one has
-
a(Iian-i-l) = Iian-i - I ( I ~ - W - ~c )IL. Hence a L c IL. To finish this part of the proof note an-lann(l) = (0). e)Since R is Noetherian L = (f1, . . . ,fn). One can write
where bij E I . Set B = (bij),C = B - aI,, and f = (ff , . , . , f n ) . Here In denotes the identity matrix. As C f 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 a' det(C) = 0 for some r. Expanding in powers of a shows that Q! is integral over I . c3
70
ChaDter 3
Proposition 3.3.3 If I is an ideal of a ring R, then 7 is an ideal of R. Proof. Let Qi E ?, i = 1,2. By the proof of Lemma 3.3.2, there is an ideal Li such that (i) aiLi c ILi, (ii) cr:ann(li)
= (0) and Iri C Li for some ri 2 0.
Since ai is integral over I , one has a:' E I for some ni 2 1. Therefore for n large enough ,f3" is in L1 La, where p = a1 c l 2 , thus
+
pnann(L1L2) = (0). On theotherhand one clearly hasthe inclusion pL1L2 c Hence by Lemma 3.3.2 one concludes p E 7, To finish proving that 7 is an ideal note X Q ~E 7 for x E R, this follows from at once the definition of f. 0 Proposition 3.3.4 If I is an ideal of a ring R and S is a multiplicatively closed subset of R, then S - l ( I ) = S-l(T). Proof. To prove S - l ( I ) c S"(7) take f E ,S"l(I). Onemayassume f = x / l with x E R because the integral closure of an ideal is again an ideal. There is an equation
where ai/si E S-' ( I ) i = S-' (Ii).Thus one may assume ai E I i and si E S for all i . Clearing denominators and multiplyingby an appropriate element of S one has an equality (in the ring R) of the form
multiplying both sides of this equation by sn-' yields sx is in 7. Therefore f = ( x s ) / sis in S-'(T). The other containment follows rapidly using that 0 localizations commute with powers of ideals. Lemma 3.3.5 Let A be a domain and x E A \ (0) such that A, is normal. Then A is normal ifand only if (x) is a complete ideal. Proof. Assume A is normal. Let z E (.). equation
Since z satisfies a polynomial
Rees Algebras and Normality
71
dividing by x m yields that x / x is integral over A and x E (x). Conversely assume (x)is complete. Let x be an element of the quotient field of A whichis integral over A . As A and A , have the same field of fractions and A , is normal one may assume x = a / x r , for some a E A and r 2 1. To show x E A it suffices to verify a E (x). There is an equation Xm
+ b 1 . P - l + * ' .+ bm-lz + b,
= 0,
bi E
A,
multiplying by x", yields the identity
am
+ (b1xr)um-l + - + (bm-12r(m-l))a+ '
'
(bm2T,)
hence a E (.) = ( x ) ,as required. Proposition 3.3.6 If A is a domain and x E A
= 0, 0
\ (O},
then
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 x E A, and x E A,, for all p E Ass A / ( z ) . Write x = a / P , a E A and n ,> 1. It is enough to prove that a E (x). If a is not in (x)note that W : a )c
Z(A/(x))* Hence ((x):a ) c p, for some p E A s s A / ( x ) . Since x E A , one may write x = u / x n = b/s, 6 E A and s $ p . Therefore s E ((x):a ) c p, which yields a contradiction. 0 Proposition 3.3.7 If A is an integral domain and closed subset of A , then
S is a multiplicatively
S-l(A) = S - ' ( Z ) . Proof. Note that A and S - l ( A ) have the same field of fractions K . First we prove S-l ( A ) c S-l Take any x in K integra1 over S-l ( A ) . There is an equation
(x).
where ai E A and si E S for all i. Set s = s1 - . sn. If we multiply by sn, it follows that sx E and x E S " l Conversely take x E There is s E S such that sx is integral over A. Hence sx satisfies
(x). $"(x).
72
Chapter 3
for some a1 , . . . ,a, in A , dividing by sn immediately yields that x is integral 0 over S-l ( A ) , as required. Corollary 3.3.8 If A is a normal domain and is S a multiplicatively closed subset of A , then ,S”l(A) is a normal domain. Corollary 3.3.9 Let A be a domain and x E A \ (0). Then A is normal if and only if A, and A, are normal for every p E Ass A / ( x ) .
quotient same
Proof. If A is normal, then by the corollary above A , and A , are normal for every 12: E A \ (0) and p E Spec(A). For the converse use Proposition 3.3.6 and observe that A, and A, have the field as A .
To state a useful criterion of normality of Rees algebras, itis convenient to introduce the extended Bees algebra of an ideal I in a ring R: A = R [ I t ,u ] , u = t-’. Part of its usefulness derives from the equality A , = R[t, t-l]. Proposition 3.3.10 Let I be an ideal of a ring R and A = R[It,t-l] its extendedReesalgebra. If R is a normal domain, then A is normal ifand only if A , is normal for each associated prime p of t-lA. Corollary
Proof. Note At-1 = R[t,t”3. Hence At-l is a normal domain and one can use 3.3.9. Lemma 3.3.11 Let A be a ring and x a regular element of A . If A / ( x ) is reduced and p E ASSAA / ( x ) , then A , is a normal domain. Proof. Let ASSAA / ( z ) = {PI,.
. . ,p,.).
Since R / ( x ) is reduced
hence ( x ) A P i= piApi 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 A,; is a domainnormal by Theorem 1.4.16. 0 Proposition 3.3.12 Let A be a ring and x a regular element of A . If A , is a normal domain and A / ( x ) is reduced, then A is a normal domain. Proof. As A , is a domain and x is a regular element on A one obtains that A is a domain. By Lemma 3.3.11 A, is normal for all p E ASSAA / ( z ) , hence A must normal be according Corollary to 3.3.9. 0
,
.
._._ , , ._.
_.
.-
Algebras
Rees
and Normality
73
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 algebra of I and is defined as:
with multiplication
Given a generating set f 1 , . . . ,f, 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 l = fl i- 12,. . . , f q = f q 4- 12. Thus one has gr,(R) = ( R / I ) [ J , ', ' Lemma 3.3.13
9
f,].
If I is an ideal of a ring R , then
R [ I t ] / I R [ I tN] grI(R) and A/t-lA
3
gr,(R),
where A = R[It,t-'] is the extended Rees algebra of I . Proof. It is left as an exercise.
0
T h e o r e m 3.3.14 Let I be a n ideal of a ring R. If I is generated by a regular sequence f l , . . . ,f q , then the epimorphism of graded algebras: 'p: @ / I ) [tl, '
- - &I + gI-1( R )= ( R / I )[ 71 7
7 *
cp(ti)=
'
9
fql
I
= fi
+ 12,
is an isomorphism, where t l , . . . ,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 ( f 1 , . . . , f , - l ) and consider the epimorphism:
-
cp'(ti) = f =
+
fi J 2 . Note ( J :fq) = J because fq is regular on R/ J , moreover since p' is an isomorphism it follows {by induction on m) that one has the following equalities
( P :f,) = Jm for all m >, I. Let F E R[t]= R[tl,.. . , t,] be a homogeneous polynomial of degree d such that its image in ( R / I ) [ t ]is in the kernel of cp, that is,
F(f)= F(f1,.* ., f q ) E Id+l.
Chapter 3
74
As cp is graded it suffices to prove F E IR[t]. To show F E IR[t] we proceed by induction on d, the case d = 0 is clear. There is W E R[t] of degree d 1 with F ( f ) = W ( f ) write ,
+
P
w =CtiWi, i=l
where Wi = 0 or Wi is a homogeneous polynomialin R[t] of degree d. There are polynomials G in R [ t l , .. . ,t,-1] and H in R[t] of degrees d and d - 1 respectively such that Q
F'=F-CfiWi=G+t,H. i= 1
Observe that F ' ( f ) = 0, hence H ( f ) E ( J d :f,) = J d c 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' E R [ t l ,. . . ,t,-1] of degree d with H ( f ) = H'(f). Set
F" = G + f,H' E R[tl,. . . , noting F"(f)= F ' ( f ) = 0 and using that cp' is an isomorphism one derives that F" has coefficients in J , which implies that G has coefficients in I as 0 required. Theorem 3.3.15 Let I be anideal of a ring R and A = R [ I t , t - l ] its extended Rees algebra. If R is a normal domain and grI(R) is reduced, then A is normal. Proof. As A/t-'A Proposition use may one domain,
2
grI(R) is reduced and At-l = R[t,t-l] is a normal 3.3.12. 0
Definition 3.3.16 Let A be an integral domain andK its field of fractions. An element x E K is almost integral over A if there is 0 # a E A such that axn E A for all n 2 0. Proposition 3.3.17 Let A be an integral domain and let K be its field of fractions. Then anelement x E 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
+ b1xm-' + + bm-lx + bm = 0, bi E A.
Setting a = dm one has ad" E A for all n > 0. Conversely assume there is 0 # a E A such that axn E A for all n > 0. Since a-lA is a Noetherian A-module andA [ x ] c a-lA one has that A[x]is a finitely generated A-module. As A is a subring of 4x1 by Proposition 1.4.2 one derives that x is integral over A . 0
k
Rees Algebras and Normality
75
Theorem 3.3.18 ([157]) Let I be a n ideal of a normal domain R. T h e n the following are equivalent: (a) I is a normal ideal of R. (b) The Rees algebra R[It]is normal. (c) T h e ideal I R [ I t ]c R[It] is complete. (d) The ideal (t-') C R[It,t-l] is complete. (e) The extended Rees algebra R[It,t-'1 is normal. Proof. (a) =+ (b) Set A = R[It ] .Let x E
c R[t]and write
5
X
=
biti. i=O
It suffices to prove bsts E A. First we prove that b,ts E 2. As x is almost integral over A there is 0 # f E A such that f x n E A for all n > 0. Hence there is 0 # fin E I m such that ( f m t m ) ( b s t s ) nis in A for all n > 0, that is, b,ts is almost integral over A . As R is a Noetherian integral domain using Proposition 3.3.17 one derives that b,ts is integral over A . Thus there is an equation (bst")m+ al(bsts)m"l
+ - + am-l(b,t') + am = 0, *
*
aijtj and a,j E I j . Grouping all the terms of t-degree where ai = equal to s m one has the equation m i=l
Thus b, is integral over Is and consequently b, E Is,as required. b,ts be an element of R[It]which is (b) 3 (c) Let z = bo + blt integral over IR[It].Then x satisfies an equation of the form
+ + e .
multiplying by tm one obtainsthat t x is integral over R(It]. Therefore t x E R[lt],which proves that z E I R [ l t ] . (c) 3 (d) Set B = R [ I t ,t-l]. Let x be an element of B integral over t-lB. As the negative part of the Laurent expansion of z is already in t-lB one may assume x = biti, s 2 0 and bi E Ii for all i 2 0. By descending induction on s it suffices to prove that b, E Is+'. There is an equation
x:=,
Chanter 3
76
hence x t is almost integral over B . It follows rapidly that b,ts+l is almost integral over B and thus b s t s f l is integral over B. A straightforward calculation yields that b,ts+l is integral over I R [ I t ] ,which shows that b, is in P+l.
(d) =$ (e) Set u = t-' and note R [ I t , u ] , = R[t,u]is a normal domain. Therefore by Lemma 3.3.5 one concludes that R [ I t , u ]is normal. (a) If z E F , then x satisfies a polynomial equation (e)
+
multiplying by trm yields that ztr is integral over the ring R [ I t ,t"]. Hence z E Ir. 0 Corollary 3.3.19 Let I be a n ideal of a normal domain R and R [ I t ] its Rees algebra. If gr,(R) is reduced, then R [ I t ]is normal. Proof. Use Theorem 3.3.15 and Theorem 3.3.18. An ideal I of a ring R is a radical ideal if I = rad ( I ) . Note: (i) a proper ideal I i s radical if and only if I is an intersection of finitely many primes, and (ii) 'P c rad ( I ) and equality occurs if I is e radical ideal, Corollary 3.3.20 Let I be a radicalideal of a normal domain R. If I is generatedby a regular sequence, then gr,(R) is reduced and R [ I t ] i s a normal domain.
Proof. It follows from Theorem
3.3.14 and Corollary 3.3.19.
0
Example 3.3.21 Let R = Q[x,y]be a polynomial ring in two variables over the field of rational numbers and I = (x2,y2). Observe the equality
thus R[It]is not normal because I is not even complete. Definition 3.3.22 Let I be an ideal of a ring R and p1,. . , ,p r the minimal primes of I . Given an integer n 2 1, the nth symbolic power of I is defined to be the ideal I ( ~=) 41 n n q r , where q i 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 m I , although the answer is negative in general, theproblem remains openin characteristic zero. For some insight into this problem see [97, 1861.
ality and
77
Algebras Rees
S - 9 " = s-l
((i
q i ) = f)
i= 1
s-lqi
i= 1
To finish the argument note that S-lqi n R = qi and intersect with R the equality above. 0 Proposition 3.3.24 Let I be a radical ideal of a ring R and minimal primes of I . T h e n
InRpi =
p1,. ,
. ,p T
the
= ( ~ i R p=~pYRpi. )~
Thus prR,, = qiRpi andcontractingonehas
pin) = qi, as required.
0
An aspect of symbolic powers that has attracted attention is to describe when the symbolic and ordinary powers of a given ideal I coincide, see [173] and [230] for a detailed discussion. Next we present a few cases where equality of symbolic and ordinary can be described in terms of properties of the associated graded ring. Definition 3.3.25 An ideal I of a ring R is called normally torsion free if A s s ( R / I i )is contained in A s s ( R / I ) for all i 2 1 and I # R.
In [38] M. Brodman showed that when 8 is a Noetherian ring and I is an ideal of R the sets Ass(R/In) 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 2 1, that is, in this case the notion of normally torsion free is a strong form of stability.
78
Chapter 3
Proposition 3.3.26 Let I be a n ideal of a ring R . If I has no embedded primes, then I is normally torsion freeif and only if I n = I ( n )f o r all n 2 1. Proof. +) Note Ass( R / I n ) = A s s ( R / I ) for n 2 1, because any associated prime p of I is a minimal prime of I n and thus p E Ass(R/In). As I n has no embedded primes one concludes that I n has a unique irredundant minimal primary decomposition and I n = I(") for n 2 1. e)Let P I , . . . ,p,. be the associated primes of I . Since pi is a minimal prime of I for all i, one derives
A S S ( R / P )= A S S ( R / I ( ~=) ){ p l , . . . ,p r } , as required.
0
Proposition 3.3.27 Let I be a n ideal of a Cohen-Macaulay ring is generated b y a regular sequence, then I n = I ( n )for n 2 1.
R. If I
Proof. Let f 1 , . . . ,f,. be an R-regular sequence such that I = ( f 1 , . . . , f r ) . By Krull's principal ideal theorem ht ( I ) = r , hence I is an unmixed ideal by Theorem 1.3.23. From Theorem 3.3.14 there is an isomorphism of graded rings 00
9:B = ( R / I ) [ t l ,... , trJ+ @ Ii/Ii'',
ti ~3 fi
+ 12,
i=O
where B is a polynomial ring with coefficients in R / I . Therefore Ii/Ii+I is a free R/I-modulo because it is isomorphic to Bi, the i th graded component of B. Thus AssR(Ii/Ii+l)= A s s R ( R / I ) . Using the exact sequences
o -+ P p i + l
"+ R
/I~+ "+ ~ R / I ~"+ 0,
it follows by induction that ASSR(R / P ) c A s s ~ ( R / lfor ) n 2 1. To finish the proof one may apply Proposition 3.3.26 to conclude the equalitybetween ordinary and the symbolic powers of I . 0 Definition 3.3.28 A proper ideal I of a ring R is said to be Eocally a complete intersection if IR, is a complete intersection for all p E V ( I ) . Proposition 3.3.29 Let R be a Cohen-Macaulay ring and I a prime ideal. If I is locally a complete intersection, then In = I ( n )f o r all n 2 1. Proof. By Proposition 3.3.26 it suffices to prove A s s ( R / I n ) c A s s ( R / I ) for n 2 1. If p is an associated prime of R / I n , then pR, is an associated prime of R p / I r . Using Proposition 3.3.27 yields that I, is normally torsion free, that is, the only associated prime of R, / I ; is I,, Therefore pR, = I, and p = I . 0
n
and Rees Algebras
Normality
79
Proposition 3.3.30 Let p be a prime ideal of a ring R such that pR, is a complete intersection, then p(n) = p" f o r all n 2 1 if and only if gr, ( R ) is a domain. Proof. +) Since gr,(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 + . - - + artr be an element in R[pt] and x' = (ao/l) ( a l / l ) t * ( a r / l ) t rthe image of x in Rp[pRpt]. Note that ai is in pi+' if and only if (sill) is in pi+lRp, because the ordinary and symbolic powers of p coincide, thus z is in pR[pt] if and only if x' is in pR, [pR,t]. As a consequence pR[pt]is prime if and only if pR,[pR,t] is prime. To finish the argument use Theorem 3.3.14 and the hypothesis that pRp is generated by a regular sequence to get that grPRp (R,) is a domain, e )First we prove A s s ~ ( p ~ / p ~ +=' )(p} for i 2 1. Let p1 be an associated prime of p i / p i + ' , that is, p1 = ann (z + pi+l), for some x in pi \ pi+1. Note a E p1 iff ax E pi+', and since gr,(R) is a domain, one readily concludes that a E p1 iff a E p . Hence p1 = p. There is an exact sequence:
+ +
+
0 -+ pi/pi+l
_3
+
R/p"l
"+ R / p i
_.)
0.
Hence using induction on i 2 1 and Lemma 1.1.16, one rapidly derives, from the exact sequence above, that Ass(R/pi) c Ass(R/p) = (p}. Thus p is normallytorsion free and by Proposition 3.3.26 onehas p n = for all n ,> 1. 0 Theorem 3.3.31 ([262]) 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 ( I ) = R[It]is a normal domain. Proof. By Corollary 3.3.19 we only haveto show that theassociated graded t ] reduced. Let algebra grI(R) = R [ l t ] / I R [ l is f = a0 a1t ' a,ts IR[It]
+ + +
+
be a nilpotent element of grI(R). Hence ( U ~ P is) in~ I R [ I t ]for some rn 2 1, by induction it suffices to verify that a,ts is in IR[It].Let p 1 , . . . ,p,. be the minimalprimes of I . Since IR,, = p i l l p , is a complete intersection one derives that grpiR, ( R p i )is reduced (see Corollary 3.3.20). Therefore the image of a,ts in grPiRpi(Rpi)is zero for all i, and one readily concludes that a, belongs to the following intersection: I
( Rn ~ Y + ~ Rn, ,o)- n ( Rn p ; + l ~ , , ) = pis+') n a
- - n p?+l).
As I is normally torsion free onecanwrite In = q 1 n n q,., where qi is pi-primary, localizing yields InRpl = prRpi = q i R p i and consequently qi = pi"). Making n equal to s + 1 proves that a, is in Is+'. 0
For ideals, in polynomial rings, with small data there are effective computational tests of normality, see Section 3.5, [40] and [296, Section 10.61.
. ...".
.-
".
.
Chapter 3
80
Descent of normality Let A c B is an extension of rings such that B is normal, in general the normality of A is not inherit from B. We discuss some sufficient conditions for A to be normal. Lemma 3.3.32 Let A c B be an extension of rings. If B = A A-modules), then I B n A = I for every ideal I of A.
@
C (as
x:==,
Proof. Let z E I B n A and write z = bifi, where bi f B and f i € I . By hypothesis bi = ai + ci, ai E A and ci E C. Since x E A it follows that X = aifiE I . This proves thecontainment I B n A C I , the reverse containment is clear. 0 Proposition 3.3.33 Let A c B be two integral domains with field of fractions K A and K B respectively. If B = A @ C (as A-modules), then
In particular if B is normal, then A is normal. Proof. By Lemma 3.3.32 one has I B r lA = I for every ideal I of A. Let b = a / c E B , a , c E A, then a E ( c ) Bn A = ( c ) , hence a = Xc = bc, with X E A . Therefore b = X E A. 0
Exercises 3.3.34 Let R = k [ z l , .. . ,x,] be a polynomial ring over a field k and KR its field of fractions. If R' = k[ztl,. . . , x;'] c KR is the ring of Laurent polynomials, prove that R' is a normal domain. Hint R' is the localization of R at the multiplicative set of monomials.
3.3.35 Let R be a ring and I , J ideals of R. If I and J are integrally closed, prove that I n J is integrally closed. 3.3.36 Let p be a prime ideal of a ring R. Prove that the symbolic power p(") is a primary ideal for n 2 1. Hint Note that pnR, is a primary ideal because its radical is maximal.
3.3.37 Let R be a ring and S a multiplicatively closed subset of R. If p is a prime ideal such that S n p = 8, then
3.3.38 Let I be an ideal of a Cohen-Macaulay ring R. If I is generated by a regular sequence, then
Ass(R/In)= Min(R/I) for n
2 1.
and
Algebras
Rees
Normality
81
3.3.39 Let I be an ideal of a ring R generated by a regular sequence. Prove that Ii/Ii+l is a free R/I-module for all i.
Hint Use Theorem 3.3.14. 3.3.40 Let I be a radical ideal of a ring R. Prove that I is normally torsion free if and only if grI(R) is torsion-free as an R/I-module (see [188] for other equivalent conditions).
Hint If g r I ( R ) is torsion-free show
AssR(I~/I~+ C 'A ) ss(R/I) for i
2 1 and use the exact sequence
o -+P / P + l + R / I ~ +-+ ~ R / I ~-+ 0. 3.3.41 Let R be a ring and F = {li)i,=Na family of ideals in R. It is said that F is a filtration of R, if li+l c Ii, Io = R, and I& c Ii+j for all i, j E N. If I is an ideal of R, prove that the following are filtrations of R:
(a) Ii = Ii,the ordinary powers of I , (b) Ii = F, the integral closure of I i , and (c) Ii = I ( i ) ,the symbolic powers of I .
Hint For (b) use the method of proof of Proposition 3.3.3.
3.3.42 Let R be a normal domain and F = {Ii}iEN a filtration of R. If Ii is integrally closed for all i, prove that the Rees algebra
of the filtration .F is integrally closed. See [131] for a study of the CohenMacaulay and Gorenstein property of R(F). 3.3.43 Let R be a domain and I an ideal, then 00
R a ( I ) = @ Fti c R[ltl i=O
with equality if R is a normal domain.
Hint To show equality use the very first part of the proof of Theorem 3.3.18.
Chanter 3
82
3.3.44 Let R = k [ z l , .. . ,x,] be a polynomial ring over a field k and m
the maximal ideal ( X I , .. . ,x,). If I = md and J = (x?,. . , ,x$)for some positive integer d, then R(J)=R(I).
3.3.45 Let I and J be two ideals of a normal domain R. Define the multi
Rees algebra of I and J as
R ( I @ J ) = R[UI)V J ] ) here u, v are new variables. Prove that
-
1
-
where I: = I n and Jam = J m .
3.4
A criterion of Jurgen Herzog
In this section we present an elegant and useful Cohen-Macaulay criterion (see Theorem 3.4.12). Once more we recall that all rings considered in this book are Noetherian and modules are finitely generated. Quasi-regular sequences It isuseful to generalize Theorem 3.3.14 to modules by introducing polynomial with coefficients in a module. Let M be an R-module and R[tl,. . . ,t q ]a polynomial ring over the ring R. Set
and note that anelement 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
.
cp: ( M / I I M ) [. .~ ~ ,tql ,
00
-+ grI(M) = @ P M / P + ' M , i=O
such that
cp(F(t)) = F ( f 1 , . . . ,fq) + Ii+lM)
for all F(t) in M[t] homogeneous of degree i, where the notation F(t) means reducing the coefficients of F(t) modulo I M . It is not hard to verify the equivalence between the following two conditions: (a) cp is injective.
(b) For every homogeneous polynomialF in M[t] of positive degree n such that F(f) E I"+lM, one has F E IM[t].
and Rees Algebras
Normality
83
Definition 3.4.1 If the map cp is an isomorphism and I M quence f 1 , . . . ,fq is called an h4-quasi-regular sequence.
# M , the se-
Theorem 3.4.2 Let M be an R-module and f = f 1 , . . . , fq a sequence in R. Iff is an "regular sequence, then f is an M-quasi-regular sequence. Proof. It follows adapting the
proof of Theorem 3.3.14.
0
Theorem 3.4.3 (Krull Intersection Theorem) Let Ad be an R-module and I an ideal of R. Then there is a E R such that a
E
1mod(1)and
a
( fi
I " M ) = 0.
i= 1
Proof. See [218, Theorem 8.91 or [298, Appendix A].
0
Lemma 3.4.4 ' L e t ( R ,m) be a local ring (resp. N-graded ring) and M a n R-module (resp. N-graded module). If N is a submodule of &I (resp. graded submodule) and I c m is an ideal of R (resp. graded ideal), then 00
N = n ( N + IiM). i= 1
Proof. Using Krull's intersection theorem one has
i=l
(this equality holds also in the graded case), where required equality follows at once.
M' = M / N . Hence the El
Proposition 3.4.5 Let ( R ,m) be a local ring(resp. N-graded ring)and A4 an R-module (resp. N-graded module). If f = f l , . . . ,fq is an M-quasi regular sequence of elements in m (resp. homogeneous elementsin m = R+), then f is an M-regular sequence. Proof. Fix an integer 1 ,< T 5 q and set I = ( f 1 , . . . ,fq), To begin with we split the ideal I as I = J + L , where J = ( f 1 , . . . ? f T - l ) and L = (fr,. , . ,fq), it is convenient to set J = (0) if T = 1. It suffices to show the equality ( J M : mfr) = J M , because this is equivalent to prove that fr is not a zero divisor of M/J M . Let rn E ( J M : M f,.), according to Lemma 3.4.4 one has
84
Chapter 3
thus the proof reduces t o proving by induction on n that m E J M for n 2 1. Since m E ( J M : M f,),there are mi in M such that
+ LnM
Consider the polynomial
regular sequence we Note deg,(F) = 1 and F(f) = 0. As f is an "quasi get rn E I M = J M L M . By induction assume m E J M L n M , that is, there is G E M [ t r ,. . . ,tP] homogeneous of degree n such that
+
+
Combining Eq. (3.1) and Eq. (3.2) yields
Using that f is an "quasi regular sequenceonemayassume(after an induction argument) that ai E I n M , and consequently t,G(t,, . . . , t q )is in I M [ t ] ,that is, G(t,, . , t q )is in I M [ t ] .One can write G = G I +Gz,where GI, Gz are homogeneous polynomials in M [ t ]of degree n with G1 E J M [ t ] and Gz E L M [ t ] . Since G1 ( f r ,. . . ,fq) is in JIM and Gz (f,, . ,fp) is in Ln'+%, from Eq. (3.2)we get that m is in JM Ln+lM, as required, 0
..
.
+
Multiplicities Let (R,m)be a local ring, M a finitely generated Rmodule and q an ideal with rad ( 4 ) = m. As grq( M ) is a finitely generated grq(R)-module and A0 = R/q is Artinian, by Theorem 1.2.8 there exists a polynomial P(t) E Q[t] of degree equal to d - 1, where d = dim(grg(M)), such that ~ ( i=)tAo(qiM/qi+lM), for i 2 no. By Lemma 1.2.10 there are integers ao,. . . , a d e l such that
From the short exact sequences
one derives t(M/qi+lM) - t ( M / q i M ) = P ( i ) for i identity
> no. Hence using the
Rees Algebras and Normality
85
we get
d-1
j+i+l
j=O
for i 2 no. Altogether the function
is a polynomial function of degree d, called the Samuel function of M with respect to q. The integer a d - 1 , is the multiplicity of M with respect to q and is denoted by e(q, M ) . Note that e(q, M ) / d ! is the leading coefficient of xk(i). We recall the following result in dimension theory that relates thedegree of the Samuel function with the dimension of the module. Theorem 3.4.6 The Samuel function
( i )= t(M/qi+’M) is a polynomial function for i >> 0 of degree equal to the dimension of M . Proof. See [218, Theorem 13.41.
0
Definition 3.4.7 Let (23,m) be a local ring of dimension d, M an R-module and q an ideal of R with rad (9) = m. We define
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 3.4.8 (Artin-Rees lemma) 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 N = In-c(IcM n N ) , ’dn> C. Proof. See [218, Theorem 8.51.
0
86
Chapter 3
Proposition 3.4.9 Let (R,m) be a local ring of dimension d and let q be an m-primary ideal of R. If
is an exact sequence of R-modules, then
Proof. By Proposition 1.1.32 one has dim(M) = max{dim(N), dim(N')}, henceonemayassume d = dim(M). TensoringwithR/q"+' sequence above yields an exact sequence
theexact
Taking lengths with respect to R/q gives
By the Artin-Rees lemma N n q"+lM C qn+l-CN,for some integer c Hence
> 0.
as x& (n) - x& ( 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. 0 Proposition 3.4.10 Let (R,m) be a local ring of dimension d and q an m-primary ideal. If M i s a finitely generated R-module and A is the set of all prime ideals p of R with dim(R/p) = d, then
Proof. Set B = A n Supp(M). If B = 8, then dim(M) < d and ed(q, M ) is equal to 0, thus in this case the identity above holds. Hence we may assume f3 # 8, this yields the equality dim(M) = d. By Theorem 1.1.15 there are prime ideals p 1 , . . . ,pn of R and a filtration of submodules:
such that Mi/Mi-1
N
R/pi for all i. Observe
,
87
Rees Algebras and Normality to show this equality note
see Corollary 1.1.17 and its proof, and recall that the minimal elements of Supp(M) are in Ass(M). Using Lemma 3.4.9 and the exact sequences
we get
4%w =
e(%RlPi),
where the sum is taken over all pi in the multiset {pl,, dim(R/pi) = d. Let p E B,then
. . ,pn}
such that
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, . . . ,p n ) . Therefore
as required. This
proof was adapted from [44].
17
Proposition 3.4.11 Let (S,m) be a local ringand M an S-module with a positive rank r = rank(M). If q is an m-primary ideul, then e(q, M )= e(q, S)rank(M). Proof. First note d = dim(M) = dim(S) by Lemma 1.1.37. 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 N (S,,)rfor any associated prime p of S, thus Mp 31 (Sp)rfor any p E A. Therefore using Proposition 3.4.10 one obtains
AS e(q, M )= ed(q, M )and
e(q,
S) = ed(q, S),the proof is complete.
0
In Chapter 4 we use the following important Cohen-Macaulay criterion of Jurgen Herzog to study the Koszul homology of graded ideals.
Chapter 3
88
Theorem 3.4.12 ([148]) Let (S,m) be a Cohen-Macaulay local ring and let M be a finitely generated S-module witha well defined and positive rank. If y = y1,. . .,Y d is a system of parameters of S , then
Furthermore equality holds if and only if M is Cohen-Macaulay. Proof. ByLemma 1.1.37 onehas d = dim(M) = dim(S). Let q be the m-primary ideal generated by y. There is a (graded) epimorphism
I
+
such that cp(F) = F ( y 1 , . . . ,~ d ) qiflM, for all F in M[t] homogeneous of degree i, where means reducing the coefficients of F modulo qM. By restriction of cp to the ith graded component of (M/qM)[t] gives
observe that both sides of the inequality are polynomial functions of degree d - 1 and consequently l(M/qM) 2 e(q, M). On the other hand y is a regularsequence,because S is Cohen-Macaulay (see Proposition 1.3.17), this forces cp to be an isomorphism when M = S (see Theorem 3.4.2). It follows rapidly that l(S/q) = e(q,S). To finish the first part of the proof note that making use of Proposition 3.4.11 one obtains the required inequality. At thispointoneshouldobserve that, by the arguments above, the proof reduces to show that C(M/qM) = e(q, M )if and only if M is CohenMacaulay. Next we show both implications. +) Taking into account Proposition 3.4.5 and Proposition 1.3.17 it suffices to prove that cp is injective; because this implies that y is an "regular sequence and hence M is Cohen-Macaulay. Using Eq. (3.4) we obtain an exact sequence
hence we get l(ker(cp)i) = C(M/qM)
(i + d-- ') - l(qiM/qi+'M)
for i
2 0.
As the polynomial functions on the right hand side have both degree d - 1 and their leading terms cancel out, one concludes that l(ker(cp)i) grows as a polynomial function of degree at most d - 2.
I
Rees Algebras and Normality
89
Let F = mat" be a homogeneous polynomial in M[t] of degree p , with m, in M , anddenote E , = m, qM. Assume is in ker(cp) and # 0. Since mk c q , there is an s 2 1 such that m9-lF # 0 and m s F = 0, thus one may assume m F = 0 and # 0. Hence there is a graded epimorphism $ induced by multiplication by
+
We now show that $ is injective. Let g = goto be an element in the kernel of $, where bp E S and z p = bp m. Thus (Cxptfl)(Cm,ta) = 0, where bp = bp + q. One may assume that &ptP and mat" are the leading terms of g and F respectively, w.r.t the lexicographical ordering of the ti variables. Thus ipm,ta+p is the leading term of g F and b p m , E qM. Hence bp q is a zero divisor of M/q M and since AssRjq(M/qM) = (m/q 1, one has that bp + q is in m/q, which proves b, E m and zp = 0. Altogether one derives g = 0 and $ is injective. Using that II) is a graded isomorphism yields
u
+
+
which is a contradiction because the left hand side is a polynomial function of degree d - 1. Therefore F = 0 and cp is injective, as required. e)As M is Cohen-Macaulay, the sequence y is an "regular sequence, an application of Theorem 3.4.2 yields that cp is an isomorphism. Therefore Eq. (3.5)becomes an equalityand we get l(M/qM) = e(q,M). 0
3.5
Jacobian criterion
Here we introduce the Jacobiancriterion and present some applications and examples to illustrate its use. Along the way some facts about regular local rings are discussed. Let ( R ,m) be a local ring. By Nakayama's Lemma it follows rapidly 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 over a field k, then R is a regular ring, that is, R, is a regular local ring for all p in $pec(R) or equivalently R, is a regular local ring for any maximal ideal m of R (see [218, Theorem 19.51). Proposition 3.5.1 If R is a regular local ring, then R is Cohen-Macaulay. Proof. Let m be the maximal ideal of R and let IC = R / m be its residue field. Assume 2 1 , . . . ,xd is a set of generators of m, where d = dim(R). If
Chapter 3
90
k [ t l , . , . ,t d ] is a ring of polynomial over the field k, there is an epimorphism of graded k-algebras
+
induced by cp(ti)= xi m2 for all i. It suffices to prove that the map cp is injective. Indeed if cp is injective, then 2 1 , . . . ,z d is a regular sequence by Proposition 3.4.5,and hence R is Cohen-Macaulay by Proposition 1.3.17. If I = ker(cp) # 0, pick a homogeneous polynomialf in I of degree s 2 1. Using the isomorphism
W l C " = fW1
together with Eq. (3.6)one concludes: dimk(mi/rni+l) = dimk k[t]i
and zero
- dimk Ii 5
(i+d-1) d-1
- (i-s+d-l) d-1
'
By Theorem 3.4.6the left hand side is a polynomial function of degree d - 1 while the right handside is a polynomial function of degree d - 2. Therefore beI must cp is injective. 0 Corollary 3.5.2 If (R,m) is a regular local ring, then R is a domain.
Proof. By Proposition 3.5.1m is generated by a regular system of parameters. Thus gr,(R) is a domain. Let z,y E R such that zy = 0. If x # 0 and y # 0, then by Theorem 3.4.3 there are T , s E N with z e mr \ mTfl and y E m8 \ m5+l. Multiplying the images 3,p of x , y in gr,(R), one obtains zy = 0. Hence 3 = 0 or 5 = 0, whichis impossible. This proves R is a domain. 0 Proposition 3.5.3 Let (R,m) be a regular local ring and I # R a n ideal of R. If R I I is a regular local ring, then I is a complete intersection.
+
Proof. Let 2 1 , . . . ,Zd be a generating set of iii= m/I, where = zi I and d is the dimension of R I I . Note that the set of images of 2 1 , . . . ,Xd in Tii/iiT2 is a basis for iii/Ei2 as a vector space over k = R/m. Hence z1+ m 2 , . . . , x d
+ m2
are linearly independent in m/m2. Set n = dim(R) and m' = ( X I ,. . . , xd). As R is a regular local ring n = dimk(m/m2), hence using the proof of Corollary 1.1.30and the equality m = m' I one derives m = m' J , for some ideal J = (xd+l,. . . ,z), contained in I . By Proposition 3.5.121,. . . ,x, is a regular sequence, thus it suffices to show J = I , to prove this equality observe that R/ J is a regular local ring
+
+
and Rees Algebras
Normality
91
of dimension d, and therefore R / J is a domain by Corollary 3.5.2. Since R / I has also dimension d, we see that the canonical homomorphism
is an isomorphism, thus I =asserted. J , as
0
Definition 3.5.4 Let R = Ic[zl,.. . Zn] be a polynomial ring over a field k and I = (fi, . . . ,fq) an ideal. The,Jacobian matrix of I is the matrix
We denote the Jacobian matrixof I taken modulo an ideal P by
Theorem 3.5.5 (Jacobian criterion) Let B = R / I be a quotient ring, where R = IC[xl,.. . ,~ n is] a polynomial ring in n variables over a field IC, and let I = ( f l , . . . ,f q ) c R be an ideal. If P c R is a prime ideal containing I and p = P / I , one has: (a) rank(afi/azj)(P) 5 ht(Ip).
(b) If rank(afi/aaj)(P) = ht(Ip)) then Bp is a regular ring. ( c ) If IC is a perfect field and
B, is regular, then
Proof. See [217, p. 2131.
0
Lemma 3.5.6 Let R be a ringand I an ideal. If I is height unmixed and P is a prime idealsuch that I c P , then Ip is also heightunmixed and ht(I) = ht(Ip).
Proof. Let
I = q 1 n * ' .n qr be a primary decomposition of I , we may assume Localizing at P gives a primary decomposition
fi c P for i = 1,. . .s.
9
I R p = (7 qiRP. i= 1
Hence the associated primes of Ip are f i R p , where i proof note h t ( f i R p ) = h t ( 6 ) = ht(1) for i I s.
5
s. To finish the 0
92
Chapter 3
Definition 3.5.7 Let k be a field and I = (f1 , . . . ,fq) an ideal of height
g of k[xl,.. . ,xn].The Jacobian ideal J of I is the ideal generated by the g x g minors of the Jacobian matrix3 = (li)fi/&j).
Corollary 3.5.8 Let R = k[xl,.. . ,xn] be a polynomial ring over a perfect field b and I = (f1,. . . ,fp) c R an unmixed ideal. If P is a prime ideal of R containing I and p = P / I , then (fill),is regular if and only if J / I @ p) where J is the Jacobian ideal of I . Proof. *) By Theorem 3.5.5 and Lemma 3.5.6 one has rank(af@q)(P) = ht(I), 1
hence there is f E J \ P and J/I $ P/ I . e)If ( R / I ) , is not regular, thenby Theorem 3.5.5 one has the inequality rank(afi/aaj)(P) < ht(I), hence J/P = (0) and J/I c P / I , which is impossible. Note that this part of the proofisvalidfor any field k. 0 Proposition 3.5.9 Let R be a polynomial ring over a field k and I a n unmixed ideal of height g. If J is the Jacobian ideal of I and ht (J,I ) 2 g+2, then ( R / l ) pis regular for ang prime p of RII such that ht (p) 5 1, Proof. Let B = R / I and p = P / I , where P is a prime ideal of R. If B, is not regular, then according to Corollary 3.5.8 one has J/I c p and hence ( J ,I ) c 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 ) 5 ht (P) 5 g 1, which is impossible. Thus B, must be regular, w required. 0
+
+
Another consequence of the Jacobian criterion is the following radical test of [294]. Let k be a field and I = (fl,.. . ,f q ) c R = k [ x l , .. . ,xn]an ideal of height g. Let J be the Jacobian ideal of I generated by the g x g minors of the Jacobian matrix 3 = (afi/li)zj).
Proof. +) Let P I , .. . ,P,. be the associated primes of I . If J c 2 ( R / I ) , then J c Pi for some i because 2 ( R / I ) = U:=,Pi. Hence by Corollary 3.5.8 one derives that ( R / I ) p i is not a regular ring, whichis a contradiction because ( R / I ) p i = Rpi /Pi Rpi is a field. e)Let I = q 1 n n 4,. be an irredundant primary decomposition of I and Pi = fi an associated prime ideal of I . Note (R / I ) p i 2 ( R /I),; and Ipi = ( q i ) p i , where pi = Pi/I. First we observe that (R/I>,,, is a regular ring; otherwise by Corollary 3.5.8 one has J / I C P i / I , hence f E Pi, which
-
k
Rees Algebras and Normality
93
is impossible because f is regular modulo I . Therefore ( R / I ) p zis a regular local ring and thus an integral domain by Theorem 1.4.16. As Ipi = (qi)pi must be prime, oneconcludes that qi is also prime and consequentlyqi = Pi. Note that here the hypothesis k perfect is not being used. 0
Example 3.5.11 LetR=k[x1, ...,x 7 ] a n d I = ( ~ 1 ~ 7 - x 2 ~ 6 , ~ 3 ~ 5 - ~ 4 ~ 7 ) , where k is a field. We now show that B = R / I is a normal ring using Theorem 1.4.15. The Jacobian matrix of I is:
Note that the monomials
are in the Jacobian idealJ of I . Hence ht ( J ,I ) 2 g+2, where g = ht ( I ) = 2. By Proposition 3.5.9one has that B satisfies (R1)and since I is a complete intersection it follows by Proposition 2.2.17 that B satisfies ( 5 ’ 2 ) . Thus by Serre’s normality criterion the ring B is normal. Lemma 3.5.12 Let R be a ring and I a prime ideal. I?’ = IF for some n 2 1, then r ( n )=
If f E R
\I
and
(r”:f”).
Proof. If z E ( I n :f”), then x f k E I n for some k 2 1, hence z E I(n). Conversely if z E I ( n ) ,then sz E I n for some s 4 I . Since s/l # I f and (s/l)(z/l) E IF, then z / l is in the nth symbolic power of I f . Thus by hypothesis x / l E I? and this means that x E ( I n :f“). 0 There are a few methods to compute symbolic powers of prime ideals in polynomial rings, see [257] and [298, Chapter 31. The following is a subtle application of the Jacobian criterion due to Wolmer Vasconcelos. Proposition 3.5.13 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(”) = (rn:f”)for n 2 I, and such element f exist if k is a perfect field. Proof. Since the localization Rf is Cohen-Macaulay, by Proposition 3.3.29 and Lemma 3.5.12 it is enough to prove that I f is locally a complete intersection. Let P be a prime ideal such that I c P and f @ P. Set p = P / I . One has
94
..
Chapter 3
As J / I $ p, using Corollary 3.5.8 we conclude that R p / I p is a regular ring and consequently ( I f ) p , is a complete intersection (see Proposition 3.5.3). assertion lastThe followsTheorem from 3.5.10. 0 Example 3.5.14 Let R = Q[x, y,z] and
I = (x3 - yz, y2 - x z , z 2 - x2y). The ideal I is prime because it is the kernel of the homomorphism:
Using COCOA[60] and Proposition 3.5.13 we get = (12:fm) = (12,A),
where f = 2y2
+ zz and A = -x5 + 3x2yz - xy3 - z 3 . See [199].
Exercises 3.5.15 Use the following procedure in COCOA[60] to verify the formula for the second symbolic power of Example 3.5.14
Fl := xn3 - y * z ; - - variables must begin with capital letters F2 := yA2 - X * Z ; F3 := ~ " 2- x2 * y; Jac := Jacobian([Fl, F2,F3]); J := Minors(2, Mat(Jac)); - - 2 X 2 minors of the Jacobian matrix I := Ideal(F1,F2, F3); H := E l i m ( t , 1 2 + Ideal(1 - t ( 2 * y*2 + x t 2))); Print (H);
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.
3.5.16 Let
Hint Use Lemma 1.3.18. 3-5-17 Let
R =
Q [ ~ , u , zand ] I the prime ideal generated by the set of
binomials fi
= x3 - yz, f2 = y2 - xz,
f3
= z2 - 2 y .
Prove: (a) I, = (fi,
f2),
that is, Ix is a complete intersection, and
(b) I ( n )= (In:xm)for n 2 1.
Rees Algebras and Normality
95
3.5.18 Let R be a Cohen-Macaulay ring and I a prime ideal. If f E R \ I and I f is a complete intersection, then
3.5.19 Let R = Q[Q , . . . ,x71 and let I be the principal ideal generated by the binomial f = xlx3X5x7 - 2 2 2 42 x 6 . Prove that R / I is a normal domain. Give a description of the irreducible binomials f such that R / ( f )is normal.
3.5.20 Let R be a Noetherian ring and a an element in the Jacobsonradical of R. If a is regular and R / ( a ) is an integral domain, prove that R is an integral domain. Recall that the Jacobson radical of R is the intersection of all the maximal ideals of R. 3.5.21 Let I = I z ( X ) be the ideal of 2 x 2-minors of the symmetric matrix:
where the entries of X are indeterminates over a field k . Prove that Izl is a complete intersection. Find a set of genera.tors for I ( 2 ) and prove that depth(H1) = 1, where HI is the first Koszul homology module of I .
This Page Intentionally Left Blank
Chapter 4
Hilbert Series Hilbert series of graded modules and algebras are introduced and studied in this chapter. The h-vector and a-invariant of a graded algebra S are defined through the Hilbert-Serre’s theorem, when S is Cohen-Macaulay we present some of their properties. Some characteristics of Cohen-Macaulay and Gorenstein extremal algebras will be presented.
4.1
Hilbert-Serre’s Theorem
Unless otherwise stated we shall always assume that modules are finitely generated and N-graded. Let R = k [ q ,. . . ,a,] be a polynomial ring over a field k with the grading induced by deg(ai) = d i , where di is a positive integer. If 00
M=@Mi i=O
is a finitely generated N-graded module over R, its Hilbert function and Hilbert series are defined by ,
00
H ( M , i ) = [ ( M i ) and F ( M , t ) = x H ( M , i ) t i i=O
respectively, where !(Mi)denotes the length of Mi as a k-module, in our case [(Mi)= dimk(Mi). If d E N, then M ( - d ) is the regrading of M obtained by a shift of the graduation of M , more precisely 00
M(-d) = $M(-d)i, i=O
97
"
.
98
Chapter 4
where M(-d)i = " d + i . Note that we are assuming Mi = 0 for i this way M ( - d ) becomes an N-graded R-module.
< 0. In
Lemma 4.1.1 F ( M ( - d ) , t ) = t d F ( M , t ) . Proof. Since M(-d)i = Mi-d one has:
where the first equality follows using that Mi-d = 0 for i = 0, , . , ,d - 1. 0 Lemma 4.1.2 Let M be a graded R-module. If z E degree preserving exact sequence
0 "+ (M/(O:z ) ) ( - d ) "% M 'P, M / z M
_I)
Rd,
then thereis a
0 (cp(m)= m
+ zM),
where (0:z ) = ( m E MI z m = 0 } and the first map is multiplication b y z . Proof. As the map $: M ( - d ) + M , given by $(m)= z m , is a degree zero homomorphism one has that (0: z)(-d) is a graded submodule of M(-d). The exactness of the sequence above follows because $ induces an exact sequence
0 + (0: z ) ( 4
) 2\M(-d) -% M -% M / x M -+ 0,
where 2 is an inclusion.
0
Theorem 4.1.3 (Hilbert-Serre) Let b be a field and R = k [ z l , ,. . ,an] a ' polynomial ring graded by deg(ei) = di E N+ . If M is a finitely generated N-graded R-module, then the Hilbert series of M is a rational function that can be written as
J'(M,t) =
h(t) for some h(t) E rl[(l - tdi) n
z[t].
i= 1
In particular if di = 1 for all i, then there is a unique polynomial h(t) E Z[t] such that F(M,t)= h(t) and h(1)# 0. (1 - t ) d "
Proof. If n = 0, then
Hilbert Series
99
Hence 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
Since the ends of this exact sequence are finitely generated modules over the polynomial ring k [ z l , .. . ,z,-1] the proof follows readily by induction on n. 0 Remark 4.1.4 The conclusion of the Hilbert-Serre theorem is valid ifwe assume that R = k [ z l , .. . ,z], is an N-graded algebra over an Artinian ring k. See [9, Chapter 111 for details. The integer d in the Hilbert-Serre's theorem will be denoted by d ( M ) ; thisinteger will turn out to be equal to the Krulldimension of M , see Proposition 4.1.8 and its proof.
Definition 4.1.5 The degree of F ( M ,t ) as a rational function is denoted by a ( M ) ,it is called the a-invariant of M .
The Hilbert polynomial One interesting invariant of a standard algebra is its degree or multiplicity, here we introduce multiplicities via Hilbertpolynomials and explain how the a-invariant is useful to measure the difference between the Hilbert polynomial and the Hilbert function. For the restof this section we shall always assume that R is a polynomial ring over a field IC with the usual grading. Lemma 4.1.6 Let n 2 1 be an integer, then
1 (1 - t)"
m=O
m
Proof. As the case n = 1 is clear, by induction on n one has:
m=O
Using the identity
m + l m+n rn+n n ( m + 1 ) = ( m the induction process is complete.
)
100
Chapter 4
Observe that Lemma 4.1.6 also follows rapidly from the classical binomial expansion:
(1
+
(;),n,
=
r
E R.
k20
Indeed recall that if r E R and k E N, then
Using the binomial expansion for (1 - t)-" one has 00
(1 - t)"
m
m=l
m=l
Lemma 4.1.7 If R is a polynomial ring in n variables with coeficients in a field k, then
1 F ( R ) t )= - and ( 1 - t)" Proof. Let z1 be one of the variables. From the exact sequence
0
"+ R(-1)
-% R -+ R / ( z ~3) 0,
I
we obtain
F ( R , t=) W q - l ) , t ) + F ( R / ( X l ) , t ) Hence the formula for the Hilbert series follows by induction on n. To show 4.1.6. 0 Lemma useequality the second Proposition 4.1.8 Let R be a polynomialringover a field k withthe R . Then there are integers usualgrading, and Ad a gradedmoduleover a - d , . . . ,a-1 so that d-1
H ( M ,i) =
U-(d-j)
j=O
d-j-1
where d = d ( M ) is the dimension of M and a ( M ) is the degree of F ( M , t ) as a rational function. Proof. By the Hilbert-Serre's Theorem there is a polynomial h(t) E Z[t] such that
Hilbert Series
101
and h ( 1 ) # 0. If d = 0, then a ( M ) is equal to deg(h,) and N ( M , i ) = 0 for i 2 a ( M ) 1. Thus one may assume d > 0. Observe that by the division algorithm we can find e ( t ) E Z [ t ]so that the Laurent expansion of F ( M ,t ) - e ( t ) , in negative powers of ( 1 - t ) ,is equal to
+
Next we expand ( 1 - t ) d - j in powers of t to obtain
observe that cp(i)= H ( M , i ) for i 2 deg e ( t ) + l , 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 1.2.10, and that d = d ( M ) is the dimension of M by Theorem 1.2.8. 0 Corollary 4.1.9 If M is a graded R-module of dimension d , then there is a unique polynomial ( P M( t ) E Z [ t ] of degree d - 1 so that H ( M ,i ) = c p ( i~) for i 2 a ( M ) 1.
+
Proof. Is an immediate consequence of Proposition 4.1.8.
0
is the Hilbert polynomial of M and the integer cd-l(d - l)!is the degree or multiplicity of M and is denoted by e ( M ) . If d = 0, we set e ( M ) = l ( M ) , where l(M)is the length M. Remark 4.1.11 The leading coefficient of v M ( t ) is equal to h ( l ) / ( d - l ) ! , where h(t) is the polynomial F ( M ,t ) ( l - t ) d .If d = 0, then
h ( 1 ) = l ( M ) = dimk(M). Corollary 4.1.12 Let M be agraded R-module. Set TO
= min{r E N l H ( M , i ) = y ~ ( i 'd) ,i
2r } .
T h e n ro = 0 if a ( M ) < 0 and ro = a ( M ) + 1 otherwise. Corollary 4.1.13 Let S = R / I be a standard graded algebra of dimension d ouer a field k . If S is Cohen-Macaulay and h = { h l ,, . . ,h d ) a system of parameters of S consisting of linear forms, then
e ( S ) = C(S/&').
Chanter 4
102
Proof. By Proposition 2.2.7 ( h l ,. . . ,hd} is a regular sequence. Thus there are exact sequences of k-vector spaces 0 + (S[-l])i = si-1
hl, si
4(S/hlS)i
-+ 0.
Therefore one has the relation cpg(i) = cps(i)- cps(i - 1) between Hilbert polynomials for i >> 0, where 3 = S/hlS. Let
I
Now, observe that thanks to cpg(i) = cps(i)- ps(i - 1) one derives:
consequently e ( 3 ) = e(S). Since 3 is Cohen-Macaulay of dimension d - 1, see Lemma 1.3.10, result the follows by induction. 0 Proposition 4.1.14 If A and B are two standard algebras over a field k , then dim(A @ k B ) = dim(A) + dim(B).
Proof. Since
00
A @ k B = $C,, r=O
where
@
Cr=
Ai~kBj,
i+ j = r
one has the next equality between Hilbert series:
00
= x ( d i m k Ci)ti = F ( A @k B , t ) . i=O
Theorem Thus
4.1.3 yields asserted the
equality.
0
Making use of the Noether normalization lemma it is not difficult to prove that the result above holdsfor arbitrary affine algebras overa field k .
Hilbert Series
103
Computation of Hilbert series Let R be a polynomial ring over a field k and I c R a graded ideal. Since R / I and R/ in(1) have the same Hilbert function (see Corollary 2.4.13), the actual computation of the Hilbert series of R / I is a two step process: 0
0
first one finds a Grobner basis of I using Buchberger's algorithm, and second one computes the Hilbert series of R/ in(1) using elimination of variables (see [16, 211).
Another approaches to the computation of Hilbert series are through minimal resolutions and Stanley decompositions (see [276]). An alternative method to compute Hilbert series of Cohen-Macaulay N-graded algebras is given by Proposition 2.2.14. Some exemplifications are given below. Example 4.1.15 Let R = ] C [ X ~ , X ~ , Xbe ~ ] a polynomial ring and
Let us compute the Hilbert series of R / I using elimination. Pick any monomial involving more than one variable, say z;x3. The idea is to eliminate x; from the monomials containing more than one variable. From the exact sequence of graded modules:
and applying Exercise 4.1.21, we get
+ 2t2 + 2t + 1
(1- t2)2 -t4 (1 - t)3
(1- t)
Example 4.1.16 Let R = Q[X,~J, z,w] and I = ( f l yf 2 , fl
= y2 - xz, f2 = x3 - yzw,
f3
f3),
where
= x2y - 2220.
Using Macaulay one finds that the minimal resolution of R / I is:
O
-+
R2(-4)
CPZ, R("2)
CB 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 ) G3 R 2 ( - 3 ) ,t ) F(R2(-4>, t ) 2t4 1 " t2-2t3 - " (1 - t)4 (1- t y (14 4 (1-t)4 1 + 2t + 2t2 * (1 - t ) 2
+-
Chapter 4
104
Example 4.1.17 Let R = k [ z l , .. . ,253 and I = (2122,232425). Note that I is a complete intersection, hence by Exercise 4.1.21 one has
F ( R / I ,t ) =
(1- t 2 ) ( 1 - t 3 ) - 1 (1 - t)5
The Laurent expansion of F ( R / I ,t )
5t2 - t + 2 (1 - t ) 3
+ 2t + 2t2 + t3 = - 1
+2
(1 - t ) 3
'
+ 1, in negative powers of (1 - t ) is
a-3 (1 - t ) 3
a-3
5t2 - t
(1- t ) 3
a-2 +-+(1 -
"
where a-1 = 5 , a-2 = -9, S = R / I is
+
t)2
a-1 (1 - t ) '
= 6. Hence theHilbert polynomial of
)
)
97s(i)=6(i f 2 - 9 ( i + l + 5 ( : ) = 3 i 2 + 2 .
Exercises 4.1.18 Let k be a field and R = k [ z l , .. . ,zn]a polynomial ring graded by deg(zi) = di E N+ . Then the Hilbert series of R is 1
W , t )=
n
U(1- t d i ) i= 1
4.1.19 Let fi , . . . ,fg be a regular sequence of forms of degree p in a polynomial ring R in n variables and I = ( f l , . . . ,fg). If 0 i < p show:
<
4.1.20 Let R = k [ 2 ,y, z] be a polynomial ring over a field
k and
I = (xa,yb,x c , 2 a l y b l z c ' ) . If a 2 a l , b 2 bl and c 2 c1, then e ( R / I )= dimk(R/I) = abcl
+ (c - c l ) ( a l b+ ( a - a1)bl).
4.1.21 Let R be a polynomial ring in n variables over a field and f l , , . . , f r a homogeneous regular sequence in R. If ai = deg(fi), then r
Hilbert
105
4.1.22 If R = k[xo,.. , , xn] is a polynomial ring over a field k and
then B is a Grobner basis for I = (B)w.r.t the reverse lex ordering. 4.1.23 In Exercise 4.1.22 show that the a-invariant of R / I is n(q - 1) by proving that the Hilbert series F ( t ) = F ( R / I ,t ) can be written as:
where
20
+ + - +t4-I.
=1 t
4.1.24 Let
a
e
R = k[xl,
x31 x2, be a polynomial ring and
I = (x?, Prove that the a-invariant of R / I is 3 and H ( R / I ,i) = 4 for i 2 4. 4.1.25 Let A, B be two standard algebras over a field k and define their Segre product as the graded algebra
Use Hilbert functions to prove dim(S) = dim(A)
+ ditn(B) - 1.
Note dim(Ai @k B,) = dim(Ai) dim(&). 4.1.26 Use the following procedure in COCOA[60] to verify the formulas
for the Hilbert series and Hilbert polynomial found in Example 4.1.17 Use R ::= Q[x[1..5]],Lex; 9 := Ideal(x[l] * x[2],x[3] * x[4] I x[5]); Poincare(R/Q); Hilbert (R/Q);
106
4.2
Chapter 4
a-invariantsandh-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 non negative. We include Stanley's characterization of CohenMacaulay algebras in terms of Hilbert series. We begin with a general inequality relating the a-invariant of a graded algebra with the shiftsof its minimal resolution. Proposition 4.2.1 Let R = k [ x l , .. . ,xn] be a positively graded polynomial ring over a field k and I a graded ideal. If b,
bl
bk
IF* : 0 -+ @ R(-dgi)
+ @ R(-dki) 3 - + @ R(-dli) 3 R 0
i=l
*
i= 1
i= 1
is the minimal resolution of S = R / I , t h e n a(S) 5 max{dk;llL k
5 g and 15 i 5 b k ) + a ( R ) .
Proof. From the resolution of S one can write the Hilbert series of S as:
i=l
hence a(S) 5 max{dki}
+ a(R).
0
Definition 4.2.2 Let R = k[z1,.. . ,zn]be a positively graded polynomial ring over a field k and I a Cohen-Macaulay graded ideal of height g. The canonical module of S = R / I is defined as US
= EXtgR(R/I,WR),
where U R = R(-S) and S = Cy==, deg(xi). Proposition 4.2.3 Let R = k [ z l , .. . ,xn] be a positively graded polynomial ring over a field k and I a Cohen-Macaulay graded ideal. If b,
IF*:
61
bb
o + @ R(-dgi) 3 - - -+ @ ~ ( - d k i )2
*.
+ @ ~ ( - d l i )3 R
i= 1
i= 1
i= 1
is the minimal resolution of S = R / I , then maxi{dli} and the a-invariant of S is given by a(S) = max { d g i )
lszsb,
+ a(R)= -min(i I
US)^
<-
e
a
< maxi{d,i}
# 0).
Hilbert Series
.
,*
I
.
*".
>,,, ,,,, .
-,.
"
.?*.+>.
107
Proof. One may order the shifts as d k l 5 5 d k b k for k = 1,.. . , g . From the resolution above one can express the Hilbert series of S as: 0
IJ(1 - t d e d z i ) ) i= 1
In order to make use of this formula we first need to prove the assertion about the behaviour of the shifts d k i . Since S is Cohen-Macaulay, the height of I is equal 9,this follows from Corollary 2.5.14. Recall that by Proposition 2.2.19 there exists a regular sequence f 1 , . . . , fg inside I , hence using Proposition 1.3.7 yields: Extk(S,wR) = 0 for i
< 9.
Set S = Cy=ldeg(zi)and W R = R(-6). Thereforedualizing IF*, with respect to the canonical module of R, one obtains the (exact) complex:
Horn@'*, L J R ); 0 -+Hom(R, W R )
-+ - -
)
@ R ( - d g i ) , W R -+
"+ Horn l:i(
0.
0bserve
Since Hom(IF,, W R ) is a minimal resolution of Extg(S, W R ) , by Remark 2.5.9, onehas d l b l < d2b2 < < d g b g . As a consequence, from the expression for F ( S ,t ) given above, one concludes the equality a(S) = dgb, + a ( R ) . In particular one has a surjection e .
To complete the proof note that since 6 - d g b g 5 equalities min(i1 ( w s ) ~# 0) = S - dSb, = -a(S).
- 5 6 - dgl,one has the 0
Chapter 4
108
h-vectors Next we introduce the notion of h-vector of standard algebras which is based on the knowledge of the Hilbert-Serre theorem for Hilbert series (see Theorem 4.1.3). Definition 4.2.4 Let S be a standard algebra and
h ( t ) = ho
+ hit + + h,t' * *
the (unique) polynomial with integral coefficients such that h(1) # 0 and satisfying h(t) F ( S , t )= (1 - t)d' where d = dim(S). The h-vector of S is defined by h ( S ) = (ho, Theorem 4.2.5 ([269]) Let S be a standard algebra and 81,. . . , e d a homogeneous system of parameters for S with ai = deg(0i). Let A be the quotient ring S/(t?l,. . . ,e d ) . Then is Cohen-Macaulay if and only if
s
rJ(1 - tUi) i=1
Proof. =+) Assume S is Cohen-Macaulay. It is enough to prove the equality
for 1 5 s 5 d. We proceed by induction on s. Assume s = 1. From the exact sequence of graded S-modules
we obtain Since
F(S["Ul],t) = C S [ - a l ] i t i =
i=al
= t"'F(S, t )
S[-U1]iti
Hilbert Series
109
we conclude
F(S/(Ol),t ) = (1 - t"l)F(S,t ) , which proves the case s = 1. Next assume s > 1 and that the equality (4.3) is true for s - 1. From the exact sequence
o -+ (s/(e,, . . . ,e,-,))
% sip,, . . . ,e,-,) -+s/(e,,. . . ,es) -+o
we get
F ( s / ( e l , Y & ) , t=) (1 -t"~)F(s/(el,...,e,-l),t), and the induction hypothesis yields the required equality. e)Let f(t) = aiti and g ( t ) = biti be two formal power series, we say that f ( t ) 2 g ( t ) if ai 2 bi for all i. From the exact sequence
xilo
we obtain
Hence the power series F ( S / ( & ) t, ) is greater or equal than(1- t a l ) F ( St, ) , and by induction it follows that r
F(S/(Bl,.. . O,),t)
2 n(1- t u i ) F ( S , t ) for 1 5 T 5 d. i= 1
Set St = S/(B1,.. . ,t9*-l). Using the exact sequence
we get
F(s'/(ed),t ) = (1 - t)""F(S',t ) + tadF(ann(ed),t ) .
On the other hand using the hypothesis we have
Therefore
-.
(1 - t)""F(S',t) - (1 - t"') * . *(1 - t""-')F(S,t)) 20
/
v
20
= "tudF(ann ( B d ) , t ) , and F(ann (e,), t ) = 0, which shows that t9d is a regular element on S'. By induction it follows readily that 81, . . . , 1 3 d is a regular sequence. This proof was adapted from [269]. 0 /
a
110
Chapter 4
Theorem 4.2.6 Let S be a Cohen-Macaulay standard algebra over a field k and h ( S ) = (hi) the h-vector of S , then hi 2 0 f o r all i. Proof. One may assumeIC infinite, otherwise one may changethe coefficient field k using the functor (.) @k K , where K is an infinite field extension of k . There is a h.s.0.p y = {yl, . . . ,yd} for S, where each yi is a form in S of degree one. Set 3 = S/(y)S. From Theorem 4.2.5 we derive
for all i and consequently hi
2 0.
0
Exercises 4.2.7 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)= ho hlt h,t' the polynomial in Z [ t ] so that h(1) # 0 and
+
+ +
F(M,t)= h(t) where d = dim(M). (1 - t ) d ' Prove that hi 2 0 for all i.
4.3 Extrema1 algebras The concept of an extrema1 Cohen-Macaulayor Gorenstein standard algebra (resp. ideal) came up in the works of J. Sally [251] and P. Schenzel [252]. Those algebras (resp. ideals) have the smallest possible reduction number (resp. smallest possible number of generators of least degree). We will study with certain detail those two distinguished classes of algebras.
The Cohen-Macaulay case Let R be a polynomial ring over a field k with its usual grading and
a graded ideal with Ip # (0). The integer p is the initial degree of I . As usual v ( 1 ) denotes the minimal number of generators of I , and v ( I p )stands for the minimal number of generators of I in degree p . The next result, when applied to a Cohen-Macaulay ideal I generated by forms of the same degree, provides a bound for the minimum number of generators of I .
Hilbert Series
111
Proposition 4.3.1 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 1 is Cohen-Macaulay, then
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 coefficientfield 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 2.2.15, there is a homogeneous system of parameters y = ( ~ 1 , .. . ,yd) for S , where each yi is a linear form in R. If we tensor the minimal resolution of S with = R / ( y ) it follows that
where 9 = 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.
0
Definition 4.3.2 If S is an Artinian positively graded algebra the socle of S is given by Soc(S) = (0:sS,). Lemma 4.3.3 ([118]) Let S = R / I be a quotient ring of a polynomial ring R over a field k modulo a graded ideal I . Let
be the minimal resolution of S . If S is Artinian, then there is a degree zero isomorphism of graded k-vector spaces: b*
Soc(S) &
@ k[g - d,J. i= 1
On the other hand b,
Tor: (S,k) N
@ k [-d,i] i= 1
--
-
1 " "
---
"
Chapter 4
112
(see the proof of Corollary 2.5.7). To complete the argument note
Hg (K* @ R s)[g]
N
2, (K* @ R
s)[g]
Soc(S).
Altogether one has
as required. Proposition 4.3.4 If S a Cohen-Macaulay positively graded k-algebra over a field k with presentation RII, 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 makinga 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 2.2.11 there existsa system of parameters y = {yl, . . . ,yd} for S = R / I with eachyi a form of degree oneof R. Since Tori(S, R/(y)) = 0 for i 2 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 4.3.3. Corollary 4.3.5 If R is a polynomial ring over a field k, then a graded ideal I of R is Gorenstein iff RII is Cohen-Macaulay and the last Betti number in the minimal graded resolution of R / I is equal to 1. Corollary 4.3.6 Let S be aCohen-Macaulaypositively over a field k and ws its canonical module. Then
graded k-algebra
where u(ws) is the minimum number of generators of u s . Proof. Use the proof of Proposition 4.2.3 andProposition 4.3.4.
0
Theorem 4.3.7 ([Sl]) 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 S = R / I is a CohenMacaulay ring, then S has a p-linear resolution if and only if the following equality holds
Hilbert
113
Proof. Let
be the minimal resolution of S . We order the shifts so that d k l 5 - 5 d& for k = 1 , . . . ,g; by the minimality of the resolution dl1 < < . < dgl, Observe that if the resolution of S is linear then the Herzog-Kuhl formulas of Theorem 2.5.16 give the required equality. Conversely assume the equality above. Since we may assume that k is infinite, there is a h.s.0.p y = {yl, . . . ,yd} for S, where each y i is a form of degree one in R. Set 3 = S / ( y ) S and = R / ( y ) . From Theorem 4.2.5 we derive that the h-vector of S satisfies h, = 0, and also get hi = H ( $ i). Therefore hi = 0 for all i 2 p . Using Lemma 4.3.3 one obtains a degree zero isomorphism
-
bf7
So@)
c2
@ k[g - dgi] i= 1
of graded k-vector spaces, in particular the socle of can only live in degrees d,i - g, hence we conclude the inequality dgi - g 5 p - 1. On the other (g - 1) 5 dgi for hand the minimality of the resolution of 3 gives p all i. Altogether we have dgi = p g - 1. Because I is C-M, then by Proposition 4.2.3 we must have p 5 dlbl < d2bz < . < d g b g = p + g - 1, which implies that the resolution is p-linear, as required. 0
+
+
-
9
Example 4.3.8 ([237])Let R = k[a,. . . , j] be a polynomial ring over a field k and
I = (abc, abd, ace, adf, ae f , bcf, bde,be f , 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 examples of Cohen-Macaulay algebras with linear resolutions include rings of minimal multiplicity [251], the coordinate ring of a variety defined by the submaximal minors of a generic symmetric matrix [194], the coordinate ring of a variety defined by the maximal minors of a generic matrix [89], and some face rings [114]. Cohen-Macaulay rings with linear resolutions have been studied by Sally [251] for the case p = 2, and by Schenzel [252] for the general case; more general rings with linear resolutions have been examined in [93, 1551.
114
Chapter 4
The Gorenstein case To give a sharp bound for the number of generators in 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 hereis to introduce and characterize extrema1 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.
+ +
Proposition 4.3.9 If S = So S, isapositively graded Artinian Gorenstein algebra over a field k , then the multiplication maps
Si
X
S,-i
-+
S, = SOC(S)21 k
are a perfect pairing, that is, the homomorphisms:
are isomorphisms of k-vector spaces. Proof. To show that cpi is one to one take r E ker(cp) and assume r # 0 and i < s. Since r is not in the socle of S, one may pick z E S homogeneous of maximal positive degree such that zr # 0. Note that 0 < deg(x) < s - i, because r z = 0 for x E S34. Hence zr has degree less than s, using that the socle of S lives only in degree s one obtains a homogeneous element wwithdeg(w) > 0 such that wxr = 0, which contradictsthe choice of x . Therefore r = 0 and cp is injective. Using the injectivity of pi one has 0 dimk Si 5 dimk Ss-i5 dimk Si, andthus cpi is an isomorphism.
Corollary 4.3.10 Let h = (ho,. . . ,h,) be the h e c t o r of a standard Gorenstein k-algebra, where k is a field. Then h is symmetric, that is,
Proposition 4.3.11 Let R be apolynomial ring overafield k with the standard grading and I agraded Gorenstein ideal of initial degree p and height g . If the minimal resolution of S = R / I by free R-modules is:
i= 1
i=l
i= 1
then bg- b
bb
@ R(-(d, - d(++))
@R(-dki)
i=l
i= 1
for 15 k 5 g
- 1.
(4.6)
Hilbert
115
Proof. One may always order the shifts so that
As b, = 1 and d,i = d,, by Proposition 4.2.3 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*, W R ) is a minimal resolution of
and using ws
N
S(d, - n ) ,
one obtains that IF* is self dual. With the same assumptions and notation of Proposition 4.3.11 one has: Proposition 4.3.12 Let h ( S ) = (ho,.. . ,h,) be the h-vector of S . Then s 2 2(p - 1) with equality if and only if the minimal resolution of S is of the form 0 + R(-(2p
+g - 2))
-$ l i b , - ] 0
*
+
( - ( p g - 2)) -+ - -+ R6' (- ( p 1))-+ Rbl( - p )
+
+ R.
Proof. As in the proof of Theorem 4.2.6 one may assume S Artinian and hi equal to H ( S , i ) . Note that the socle of S lives indegree dg - g by Lemma 4.3.3, hence s 2 dg - g. By Proposition 4.3.11 and the minimality of the resolution one concludes
+ -
2 p k 1 for 1 5 k 5 g - 1. For where thelast inequality uses the second assertion use the chain of inequalities above and compute the 0 h-vector of S using its resolution. Definition 4.3.13 If s = 2 ( p - l),the ring S is called an extremal Gorenstein ring and its resolution is called pure and almost linear. Proposition 4.3.14 ([252]) Let R be a polynomial ring over a field IC and I a graded ideal of initial degree p and height g . If S = R / I is an extremal Gorenstein algebra, then for 1 5 i < g the ith Betti number of S is given b y p+g-1 p+i= ( p + i - I) i - 1 2, +(P+y)(P+g-iP-1
(
2,
-
(3(P;:y2)*
116
Chapter 4
Proof. As in the proof of Theorem 4.3.7 one may assume S Artinian and R a polynomial ring in g variables. Set s = 2 ( p - 1). From the minimal resolution of S
'
p+m-1
bm =
(-1)P-l-i i=O
p+m-1-i
)hi, l < r n L g - l
Using the identity
and the symmetry of the h-vector of S one has p+m-1
+m-1
h,-i.
i=p
Note p+m-l
(-1)P-1-2
( p + m g- l - i
i=p
i=O
,
hS-i
1,
rn-1-i
since the terms in the right hand side are zero for i mation reduces to
> p - 1 the last
sum-
m-1-i
i=O P-1
g-l+i
gfp-2
i=o
thelastequality
follows fromEq. (4.7) making m equal to g
- m.
0
Hilbert Series
117
Exercises 4.3.15 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. Let (ho,. . . ,h,) be the h-vector of R / I . Prove that s 2 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. 4.3.16 Let R be a polynomial ring over a field k and I 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 P-1
(y; -
F ( S )t ) =
i=O
,
(1 - t ) d
where d = dim S,
if and only if S has a p-linear resolution.
4.3.17 Let R be a polynomial ring over a field IC and I a Cohen-Macaulay graded ideal in R of height g and initialdegree p. If S has a linear resolution, show that the Betti numbers of S = R / I and its multiplicity are given by h = ( p +k -kl-" > . (
and e ( S ) = ( p + g - '>.
p +g 9- k- l )
4.3.18 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:
with equality if and only if S is an extremal algebra.
Hint The h-vector h ( S ) = (ho,. . . h,) is symmetric and s compute explicitly )
2 2 ( p - 1). Then
9-2
i=O
4.3.19 Let R be a polynomial ring over a field IC and I a graded Gorenstein ideal of height 4 generated by forms of degree p . Let h ( S ) = (ho,. : . , h,) be the h-vector of S = R / I . Prove that the minimal resolution of S has the form: ~-2p+3 0 + R ( - ( s + ~ )+ ) R b ( - ( s - p + 4 ) ) -+ @ R a a ( - ( p + i ) )3 R b ( - p ) -+ R, '
i= 1
where
T
= s - 2p
+ 3 and ai = a,"i+l
for i 2 1.
118
Chapter 4
4.3.20 If z > 1 is an integer, prove that z(z + 8) is never a perfect square. 4.3.21 Let I be a graded Cohen-Macaulay ideal of a polynomial ring R with a minimal resolution: 0-,~~9(-(p+a+b))~~~a(-(p+a))-,~~l(-p)-)~3~/1.
If
bl
= 9 , use the Herzog-Kuhl formulas to prove that I is Gorenstein.
+
Hint Note b 3 - b2 bl - 1 = 0 and prove bl b2b3 = 9 b 3 ( b 3 square, then use Exercise 4.3.20.
4.4
+ 8) is a perfect
Initial degrees of Gorenstein ideals
Assume R is a polynomial ring over a field and I a homogeneous Gorenstein ideal of codimension g 1 3 and initial degree p 2 2. Let e(9,d=
(9 + p -
l)+(S+P“2
) and vo= (p + s -
9-1
l)-(
P+P” g-l
).
A consequence of the symmetry of the h-vector of R / I is that the multiplicity of R / I satisfies e ( R I I ) 1 e(g,PI, see Exercise 4.3.18. Given codimension g andinitialdegree p, a graded Gorensteinalgebra R/J with multiplicity e ( R / J ) = e(g,p) is extremal. This has strong structuralimplications: the minimal free resolution of R/ J must be pure and almost linear; consequently all the Betti numbersb i ( R / J ) are determined, and in particular J is generated by vo forms of degree p (see Section 4.3). Results in the literature dealing with Cohen-Macaulay ideals (such as [106,2491) give upper bounds for the minimal number of generators v(1) 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 4.4.3). 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 4.4.1 Let h, j be positive integers. Then h can be uniquely written h=
(y) + (7:;)+ * . . +(:),
Hilbert Series
119
From the inequality h=
(7)+ (-.;)
+,..+
(7) < ( ) (7)
we obtain (9";')< and this implies uniqueness. Assume h can be written as h= where bj
aj-1
(Y)+ (]!yJ + . . e +
> bj-1 > - > bk 2 k 2 1.
-
+ (3ajl)
aj j+ 1
<
aj. Let us show the
(;),
By induction on j one has
Since aj is unique we derive aj = b j . Hence by induction hypothesis k = i and aj-1 = b j - 1 , . . . ,ai = bi. 0 The unique expression for h in Eq. (4.8) is called the binomial expansion of h in base j . We set
h(j) =
( )+ ' j
j+l
+ ... +
(i +ai l)'
(4.9)
sometimes h(j)is refer to as the Macaulay symbol. Theorem 4.4.2 (Macaulay) Let h:N -+N be a numerical function and k a field. The following are equivalent (a) there exists a homogeneous k-algebra S with H ( S ,i ) = h(i)f o r i 2 0. ( b ) h(0) = 1 and h(i
+ 1) ,< h(i)(i)f o r all i 2 1.
Proof. See [298, Appendix B].
0
Theorem 4.4.3 ([222]) If I is a graded Gorenstein ideal of codimension 2 3 and initial degree p 2 2, then
g
and I is itself extrema1 if equality holds.
Chapter 4
120
Proof. If either ~ ( 1>~VO, ) or v ( I P )= vo and I is not extremal, then by the symmetry of the h-vector h ( R / I )there is some j 2 p so that
where
The ideaof the argument is to use the Macaulay bound h y ) for hj+l to see that such a small value of hj cannot grow to such a large value of
Recall that this estimate is calculated from the binomial expansion for hj:
(4.11) where aj
> aj-1 >
> ai 2 i 2 1. Then
a
We may assume that hj > j , for if not, then hj 5 j would imply that ae = for all C, and hence hj 2 hj+l, which contradicts our assumption. Notice that by grouping the terms of (4.11) according to the value of ae - C the binomial expansion for hj can be written as
(4.13)
it follows that k 5 g - 2. From (4.11), (4.12), and (4.13), together with Pascal’s identity and
a+b+l b+l
(
a+l
)=bfl(
a+b+l b
)’ I
Hilbert
121
we have
On the other hand from the upper bound on hj and hj+l = h,-l we
+
get
+
Since ( g - l)/(p - 1) > ( k l ) / ( j 1) it follows from (4.13) and the last two inequalities that FO < 0, where for 0 5 s 5 T we set
To derive a contradiction we are going to show the following inequalities
Assume 1 5 s
+ 15
T.
Notice that
Therefore
Let A , denote the first summation and B, the second in this last inequality; clearly A, - B, > 0. Also note that
ChaDter 4
122
Putting these together,we compute
+
Hence Fo
S+I
+1
-
"> +
B,
js 1
> FT > 0, which contradictstheobservation
that FO < 0.
0
One might hopefor even stronger estimates than suggestedby the result above, for instance that v(1) 5 VO. Next we present a counterexample showing that Theorem 4.4.3 cannot be extended to bound the number of generators in all degrees.
Example 4.4.4 Let
Using Macaulay one readily obtains that I is given by:
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,l),whereas the h-vector for an extrema1 Gorenstein algebraof codimension four and initialdegree twois (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 valueof 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 bl = bz = 2p 1.
+
Theorem 4.4.5 Let R be a polynomial ring over a field and I be a homogeneous Gorenstein ideal of height three. If p is the initial degree of I , then v(1) 5 2p 1 and b2(R/I) 5 2p + 1.
+
Hilbert Series
123
Proof. (sketch) We may assume, without loss of generality, that R is equal to Ic[zl,2 2 , 2 3 1 and A = R / I is Artinian and local with socle(A) = A,. Then by [54] the minimal free resolution of A has the form
where Y is an alternating matrix, and the generators f 1 , . . . , f v 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 degree p. 0
+
has
Conjecture 4.4.6 (Miller-Villarreal) Let I c R be a graded Gorenstein ideal of initial degree p and height g . If 1 5 i < 9 and yi is the ith initial Betti number of R I I , then
7'5
p+ii-1"> +(P+;-l)(P+g-iP-1
p+g-1 (p+i-l)(
"> (4) -
Proposition 4.4.7 Let n be a given positive integer and f ,g: N numerical functions given by g(z) = n(')
and
+N
the
f(z)=
where h(3) is the Macaulay symbol. Then is strictly increasing.
g(x) is non increasing and f (x)
Proof. See [22] and [44, Lemma 4.2.131 respectively.
0
There is a short and elegant argument, based on the behaviour of the combinatorial functions g(z) = n(")and f(z)= z ( ~ )to, show the inequality of Theorem 4.4.3. We include the details of this argumentin the proof below. Corollary 4.4.8 If R is a polynomial ring over a field IC and I agraded Gorenstein ideal of R of codimension g 2 3 and initial degree p 2 2, then
4 l , ) < v o = ( p +g-ll)-( g-
P gf g- -l 3
).
Proof. First we make a change of coefficients using the functor @k K , where I< 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)
Chapter 4
124
where h = hl,. . . ,h d is a system of parameters of S with hi a form of degree 1 for all i. Note that one can write A = R/T, where i? = R / @ ) is a polynomialring in g variables and 'i; = I z is a graded Gorenstein ideal ofof codimension g and initial degree p . By Proposition 4.3.9 the Hilbert function of A is symmetric. Since the h-vector h ( S ) = (hi) satisfy hi = dimk ( A i ) there is some j 2p so that
thus
The idea of the argument is to use the Macaulay bound
< h!),
hj+l
see Theorem 4.4.2. From the inequalities:
we conclude P f 9 - 3
( p-2 )
(P--2)
P+P" p
[(
) -44
(P--2) *
Using that f(z) is non decreasing yields the required inequality.
0
Exercises 4.4.9 Let I be a graded Gorenstein ideal of height 4 and initial degree p ,
prove v ( I p ) 5 ( p occur.
+ 1)2. If p = 4 show the values 21 ,< ~ ( 1 4 )5 24 do not
4.4.10 Let I be a graded Gorenstein ideal of height 4 generated by forms of degree p and R / J an extrema1 algebra of the same codimension and initial
degree, prove that their Betti numbers satisfy
Hilbert 4.4.11 Let R = k [ q , .. . ,x9] be a polynomial ring over a field k and I a graded ideal of height g generated by forms of degree p. Prove, that the
Hilbert function of I satisfies:
+
H ( I , p i)
4.5
2H(I,p)+ i
(3
for all i
2 0.
A symbolic study of Koszulhomology
The first Koszul homology module of a Cohen-Macaulay graded ideal is studied here using the Cohen-Macaulay criterion of Jurgen Herzog introduced in Section 3.4, our approach is symbolic and makes use of Hilbert functions. Lemma 4.5.1 Let p and g be positive integers and let
If p
1 2 and g 2 6 then $ ( p ) > 0.
Proof. Notice the equality
which is certainly positive for g 2 6. We proceed by induction on p . Assume $ ( p ) > 0. It is easy to check that $(p 1) is greater than
+
(P
step induction
+ 1)(4(9 - 2)p2 + (3g2 - 39 - 8)p + g2 - 39 - 2) (P + + + 2)
Since the right hand side of this inequality is positive for p the is complete.
1 2 and g >_ 3, 0
Theorem 4.5.2 ([240]) Let R = $goRi be a polynomialringwith its usual graduation over an infinite field k and I a Cohen-Macaulay graded ideal of R of height g with Q p-linear resolution
t ( H l ( I ) / y & ( I ) ) > rank(Hl(1)) t ( S / y S ) .
Chapter 4
126
We start by making a specialization to the case of a polynomial ring of dimension g. Since k is infinite according to Theorem 2.2.11, there exists a system of parameters y = ( y 1 . . . 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 2 1, it is clear thatthe minimal resolution of 3 = S/(yS) asan Ti: (= R/(y))-module has the same twists and Betti numbers as the minimal resolution of S over R. (ii) With H I Cohen-Macaulay it is easy to see that the first Koszulhomologymodule of I 8 over is precisely H1 ( I ) 8 We may then in the sequel assume that S is zero-dimensional. We will compare the integer r=rank(original H 1 ( I ) ) . l ( S ) ,with partial Hilbert sums contributing to ! ( H I ) . First notice (see Exercise 4.3.17) that the length of S and its Betti numbers can be calculated using:
z.
bk
= ( p +k k- 1- 2 )
1
(""'> 9-k
and
e(s)=
(
p + 9 - '>.
(4.14)
From the Koszul complex we obtain the exact sequences
0 3 22
-+
R(*2)(-2p) "+ B1 3 0,
where Zi and Bi are the modules of cycles and boundaries defining Hi(1). To simplify notation we set C(M)i = H ( M , i ) 7the dimension of the ith component of the graded module M . One may write
and therefore 2P
2P
2P W l )
2 C t ( H 1 ) i 2C C ( Z 1 ) i i=O
i=O
(t)
CC(R(-2p))i.
(4.15)
i=O
From the minimal resolution of S we obtain
Hence e(&), 2 b ~ t ( R ( - p ) ) i l(R)i and leads to 2P
2P
i=O
i=p+l
e(&)i
= 0 for 0
5 i 5 p , which
22)
(4.16) Finally by (4.15) and (4.16) we have
i=p+l
127
Hilbert
The proof now reduces to showing that the right hand side of (4.17) is greater than rank(H1( I ) )' !(S) = (bl - g ) !(S), that is, we must show that the following inequality holds for p 2 2 and g 2 3 e
It is not hard to see that f(p,9 ) simplifies to
Observe that from this equality we obtain the inequality:
It is easy to check that f(p,g ) > 0 for g E (3,4,5) and p inequality is now a direct consequence of Lemma 4.5.1.
1 2.
The required 0
The result above complements one of Bernd Ulrich [286] in which the twisted conormal module U S @R I = U S 8~ 1/12 was shown not to be Cohen-Macaulay; either result implies that I is not in the linkage class of a complete intersection. Corollary 4.5.3 Let I be a graded Cohen-Macaulay ideal of height g 2 3 in polynomial ring R. If I is generically a complete intersection and has 2 2, then I is not in the linkage class of a ap-linearresolutionwithp 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 CohenMacaulay. See [179] and 1296, Corollary 4.2.41. 0
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Chapter 5
Monomial Ideals and St anley-Reisner Rings 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 [25, 2711. The main objects of study in this chapter are monomial ideals, face rings and simplicial complexes. Here we emphasize the connection between algebraic properties of face rings and the reduced simplicia1 homology of the corresponding Stanley-R,eisner complex. An understated goal here is to highlight some of the work of M. Hochster, G. Reisner and R. Stanley. It turns out that some notions encountered while studying graded algebras, such as Cohen-Macaulay rings, Gorenstein rings and Hilbert series are useful to solve combinatorial problems. Some key results leading to the proofs of the upper bound conjectures for convex polytopes and simplicial spheres are presented, most of them without detailed proofs but giving the appropriate references. Our hope is to give a flavor of the methods and ideas of the area and to trace some routes for further reading.
5.1
Primary decomposition
General monomial ideals will be examined first, subsequently we specialize to the case of square-free monomial ideals and present some of their relevant properties. Let R = k [ x ] = k [ z l , .. . , x,] be a polynomial ring over a field k . To make notation simpler we set
129
Chapter 5
130
Definition 5.1.1 An ideal I of R is called a monomial ideal if there is c W such that I is generated by {xala E A). If I is a monomial ideal the quotient ring R / I is called a monomial ring.
A
Note that a monomial ideal is always generated by a finite set of monomials by Dickson’s lemma. Definition 5.1.2 A face ideal is an ideal p of R generated by a subset of the set of variables, that is, p = ( x i l , .. . ,xik) for some variables xij. Proposition 5.1.3 Let R = k[xl,. . . ,xn] be a polynomial ring over a field k . If I c R is a monomial ideal, then every associated prime of I is a face ideal. Proof. By induction on the number of variables that occur in a minimal generating set of I consisting of monomials. Set m = (21, . . . ,2,). 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
Since R is Noetherian one has I k = (Ik: X I ) for some k. There are two cases to consider. If p is an associated prime of ( I n , x l ) for some n, then by induction p is a face ideal because one can write ( I n , X I ) = (I;, X I ) , where I; is an ideal minimally generated by a finite set of monomials in the variables 22,. . . ,Xn (cf. Exercise 1.1.45). Assume we are in the opposite case. By Lemma 4.1.2 for each n there is an exact sequence
hence making a recursive application of Lemma 1.1.16 one obtains that p is an associatedprime of I n for all n. Since 51 is regularonR/Ik one concludes that I k is an ideal minimallygenerated by monomials in the variables q , . . . ,x n , thus by induction p is a face ideal. 0 Definition 5.1.4 A monomial f in R is called square-free if
for some 1 5 il
<
< i, 5 n.
Corollary 5.1.5 Let R = k[xl,. . . ,x ~ beJ a polynomial ring over a field k. If I is a n ideal of R generated by square-free monomials and p1,. . . ,ps are the associated primes of I , then
I = p1 n -rips.
Monomial Ideals and Stanley-Reisner Rings ,
I
.
,
I I /
131 .
Proof. Set J = n:=lpi and note J = rad ( I ) . We only have to check J c I because I is contained in any of its associated primes. Take a monomial f in J = rad(I) and write f = x:: - - - x:;, where il < - < i, and ai > 0 for all i. Since f k E I for some k 2 1, and using that I is generated by square-free monomials, we obtain xil - . xir E I . Hence f E I . To finish the proof observe that J is a monomial ideal because the intersection of 0 monomial ideals is again a monomial ideal (see Exercise 5.1.24). e
a
Definition 5.1.6 Let R = k [ z l , .. . ,x,] be a polynomial ring, the support of a monomial x a = 2;' - x? in R is given by supp(za) = {xi I ai > 0). e
Proposition 5.1.7 ([113]) Let I be a monomial ideal in a polynomial ring R over a field IC and 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 R = k[x] and let F = {fl,. . . ,fr)be a set of monomials that minimally generate I . Assume that one of the variables, say x l , occurs in at least one of the monomials in F with multiplicity greater than 1. Thus one may write f l = X Y ' 9 1 , . . * ,fs = X Y 3 9 S , where ai 2 1 for all i, a1 2 2, z1 4 supp(gi) for all i 5 s and z1 supp(fi) for i > s. Set
where 20 is a new variable. We claim that x0 - x1 is a nonzero divisor of S' = R'/I'. On the contrary assume that 20 - x1 belongs to an associated prime p of I' and write p = (1':h ) , for somemonomial h. Since p is a face ideal, one has x i h E I' for i = 0 , l . Hence using h 4 I' gives x1 h = zozy"-lgihl for some i and consequently h = zoz:"2gihl. Hence
In both cases one obtains 20 E supp(M) and h E 1', a contradiction. Since S'/(zo - 2 1 ) =
s,
one can repeat the construction to obtain the asserted monomial ideal I'. isThis proof to due R. Froberg. 0 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).
Chapter 5
132
Proposition 5.1.8 Let q be a monomial ideal of R = k[xl,. . . ,X,]. Then q is a primary ideal if and only if, after permutation of the variables, q has the form: q = (x;l,. . x;', x?. . . ,xba), a ,
where ai 2 1 and Uf=lsupp(xbi)C
(21,.
. . ,x,.},
Proof. If Ass(R/q) = {p), then by Proposition 5.1.3 one has that p is equal to (21, . . . ,x,.). Since rad (4) = p the ideal q is minimally generated by a set of the form: (x1"' 9 . , . ,x;?,xbl,. * . ,xba). Let x j E supp(xb'), then xbi= XjN, where N is a monomial not in q. Since q is primary, a power of x j is in q. Thus x j E ( X I , . . . ,x,-) and consequently 1 5 j 5 T , as required. For the converse note that any associated prime p of R/q can be written as p = (4: f), for some monomial f. It follows rapidly that ( X I , . . . ,x,.) is prime associated the only of q. 0
Corollary 5.1.9 If p is a face ideal, then pn is a primary ideal for all n. Proposition 5.1.10 If I is a monomial ideal of R = k [ x l , .. . ,xn], then I has an irredundant primary decomposition I = q1 n n qr, where qi is a primary monomial ideal for all i and rad ( s i ) # rad (qj) if i # j .
-
Proof. Let fi, . . . ,f, be a 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 fi ,. . . ,f,. One may assume that one of the variables in supp(f i ) , say x,, satisfy x: 4 I for all i, otherwise I is a primary ideal and there is nothing toprove. Next we permutethe fi in orderto find integers 0 5 a1 5 5 a,, with a, 2 1, such that fi is divisible by x: but by not higher power of xn. If we apply this procedure to ( I ,x$), instead of I , note that one must choose a variable different from xn. As ( I :x?) is generated by monomials in less than n variables and because of the equality I = ( I ,X$) n ( I :~ 2 )
@=,
one may apply the argument above recursively to both monomial ideals occurring in the intersection and use induction to obtain a decomposition of I into primary monomial ideals qi . . . ,qL. Finally we remove redundant primary ideals from qi . . . ,q: and group those primaryideals with the same radical. 0 Even for monomial ideals a minimal irredundant primary decomposition is not unique, what unique is is the number of terms insuch a decomposition and also the primary components thatcorrespond to minimal primes.
Monomial Ideals and St anley-Reisner Rings .-
133
. ,.
Example 5.1.11 If I = (z2,zy) c k[x,y], then
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 5.1.10. Now we illustrate this procedure with a specific ideal. Example 5.1.12 If R = k[z,y, z] and I = (yz2,x2z,z3y2).
/
J = ( I ,x3) = ( x 3 ,z x 2 ,z2y)
( J :2 ) = ( x 2 ,y)
( J ,2 ) = ( 2 , 2 3 , 2 2 4 . Thus I = (z,y2) n (z2,y) n ( z ~ , z ~ , z ~ z ) . Corollary 5.1.13 Let R = k [ x l , .. . ,xn] be a polynomial ring over a field IC. If I is a monomial ideal, then I has a primary decomposition
such that q i is generated b y powers of variables for all i. Proof. Let q be a primary ideal minimally generated by the set of mono. . , fl, . . . ,fs and such that mials
where ai > 0 for all i. Note that if and one has a decomposition:
f1
=
e
.
xp' and
bl
> 0, then
al
> bl
where in the first member of the intersection we have lower the degree of and have eliminated f i , while in the second member we have eliminated the variable z1 from f1. Applying the same argument repeatedly itfollows 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 z1 are pure powers of z1. Therefore by induction q is the intersection of ideals generated by powers of variables. Hence the result follows from Proposition 5.1.10 and Proposition 5.1.8. D
Chapter 5
134
Proposition 5.1.14 Let I be an ideal of k[xl,. . . ,xn] generated b y monomials in the first variables 2 1 , . . . ,x r . If
is an irredundant decomposition of I into monomial ideals, then none of the Ii can contain a monomial in k[x,+l, . , . ,x,]. Proof. Set X = ( X I ) .. . xr} and X' = X \ { X I , . . . ,x r ) . Assume some of the Ii contain monomials in k [ X ' ] , One may split the li's into two sets SO that 1 1 , . . . , Im do not contain monomials in k [ X ' ] (note m could be zero), while I m + l , . . . ,Is contain monomials in the set of variables X ' . For i 2 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 E nE, Ii and f g' ng,+, Ii, where we set f = 1 if rn = 0. To derive a contradiction consider f1 = fgm+l - - g8. As f l E I , we get f E I which is absurd. 0 )
Corollary 5.1.15 Let I C k [ x l , .. . ,x,] be an ideal generated b y monomials in the first variables 2 1 , . . . ,x,. If
is an irredundant primary decomposition into monomialideals, then q i is generated b y monomials in k [ x l ,. . ,x T ] . Proof. It follows fromProposition
5.1.8 andProposition
5.1.14.
0
Proposition 5.1.16 If R = k[xl,.. . x,] is a polynomial rang over a field k and I a monomial ideal, then I is irreducible if and only if up to permutation of the variables I can be written as )
I = ( x ; 1 ) .. . , x ; r ) , where ai
> 0 for all i.
Proof. =+) Since I must be primary (see Proposition 1.1.21) from the proof of Corollary 5.1.13 one derives that I is generated by powers of variables. e)If I is reducible, then by Corollary 5.1.13 there is an irredundant decomposition I = q1 n n qs, such that q i # I and qi is generated by some powers of the variables X I ,. . . ,x r , for all i. Let xi: E q i \ I , where 1 5 ji 5 r and ci < a i i . Set f = lcm{x:i
, . . . ,x:; } .
Note that one can write f = x z l x z k , where ei < a,; and r n l , . . . distinct,butthis is impossible since f E I , thus I is irreducible. 0
)
m k
0
Monomial Ideals and Stanley-Reisner Rings
135 , .
Theorem 5.1.I 7 If I is a monomial ideal in a polynomial ring R over a field k , then there is a unique irredundant decomposition
r = q1 n-nq, such that
qi
is an irreducible monomial ideal.
Proof. The existence follows from Corollary 5.1.13. For the uniqueness assume one has two irredundant decompositions: q1 n-nq,
= 4: n-.nq/,,
where q i and qi are irreducible for all i, j. Using the arguments given in the proof of Proposition 5.1.16 one concludes that for each i, there is g i such that qui c q i and vice versa for each j there is nj such that qkj c q j . Therefore T = s and qi = qbi for some permutation p. 0 There are various routines for Macaulay2 developed by S. Popescu [235] for dealing with monomial ideals, one of them computes the irredundant decomposition into irreducible monomial ideals of any monomial ideal. 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 188, 143, 144) and [307]. Example 5.1.18 Let I = (xy2,x2y) c k [ x ,y ] and V ( 1 )the affine variety defined by I . From the primary decomposition
I = ( 4 n( Y ) nb 2 ? Y 2 ) one has:
V ( 1 )= { (0,O)) U V ( x )U V ( y ) = V ( z )U V(y)
c Ai
where (z2,y2) corresponds to ((0,O)) which is embedded in V ( x )U V(y). For this reason ( x 2 ,y2) is said to be an embedded primary component. If k is infinite, then the coordinate axes V(z) and V(y) are the irreducible components of V ( 1 ) . Monomial ideals form a lattice Let L be a lattice, that is, L is a partially ordered setor poset, in which any two elements z, y have a greatest lower bound or meet II: A y and a lowest upper bound or j o i n 2 V y . The lattice L is called distributive if (a) z A ( y V z ) = (x A y ) V (z A z ) , and (b) z V ( y A z ) = ( x V y ) A ( z V z ) , V ’ z , y , x i n I ; . Proposition 5.1.19 If L is the family of monomial ideals, order by inclusion, of a polynomial ring R over a field k , then L is a distributive lattice under the operations I A J = I n J and I V J = I + J . Proof. Use Exercise 5.1.24 and Exercise 5.1.25.
0
136
Chapter
An interpretationof the minimal primes Let F be a set of monomials in the ring k[yl,. . . ,yn] and I = I ( F ) the ideal generated by F . Consider the hypergraph
whose vertex set is
u
SUPP(f > *
fE F An element in Y ( F ) is called a minimal vertex cover of F . In order to relate the ofset minimal primesof I with the setof monomials occurring in a certain geometric link of I , consider a copy F1 of F in the polynomial ring k [ x l , . . . ,xn]in a new set of variables 2 1 , . . . ,xn. Proposition 5.1.20 ([3]) Let Y 1 be the set of minimal vertez covers of I$. If L is the ideal
then ((z):I ( F ) )= (z, L ) , where z = (xlyl,.
. . ,xnyn}.
Proof. Take ( x i l , . . . , xi,,}E 'TI and y" E F . Hence xij divides x" for some 1 5 j 5 r and we obtain xil - * - Zi,y" E (z). This shows (z, L ) C ((z): I ( F ) ) . Conversely, assume M is a monomial in ((2): I ( F ) ) \ (z). Let y" be any monomial in F . Since My" = ztytM1 and M is not in (z), one has that xt divides M and yt divides y". Hence supp(M) n supp(xa) # 8 for any 2" in P I , and M E L. 0
Exercises 5.1.21 If f = fi , . . . , f,. is a sequence of monomials in R, then f is a regular sequence if and only if supp(fi) n supp(fj) = 8 for i # j . 5.1.22 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. 5.1.23 If L is a lattice, then the following conditions are equivalent:
(a) x A (y V z ) = (xA y) V (xA
X),
V x ,y, z in L.
(b) x V (Y A X ) = (ZV y) A (x V x ) , Vx, y, z in L. 5.1.24 Let I and J be two ideals generated by finite sets of monomials F and G respectively, prove that the intersection I n J is generated by the set
Monomial Ideals and St anley-Reisner Rings
137
5.1.25 If I , J , IC are monomial ideals, prove
I n ( J -I- Ii) = ( I n J ) + ( I nK ) . 5.1.26 If I and J are two monomial ideals, then ( I :J ) is a monomial ideal. 5.1.27 Let q be a primary ideal of k [ x l , .. . ,xn] generated by
and ai 2 1 for all i. If xc = xfl - - x:? is a monomial of least degree in the variables 21,. . . , such that zc # q, then (x1, . . . ,x T ) = (9: xc). a
5.1.28 If q is a primary monomial ideal, then qn is a primary ideal and qn = q(n) for n 2 1.
Hint Use Proposition 5.1.8. 5.1.29 If 4 1 , . . . ,q,. are primary monomial ideals of R with non comparable radicals and I = ql n ... n q r , then ~ ( n= ) qy
Hint Proceed
n . . . n 4;.
as in the proof of Proposition 3.3.24.
5.1.30 Let I = (x;,z ~ x ~ , x 1 x ~ xnote 3 ) ,that I = q1 n q 2 is a primary decomposition of I, where q1 = (xi,2;) and q 2 = (x:, 21x3, xi). Prove I2 = I ( 2 ) and I 3 #
Hint If x = x ~ x and ~ xa ~= x1 + x; + 23, then uz E I 3 and x # 13. 5.1.31 Use the next procedure in COCOAto verify Exercise 5.1.30. Use R ::= Q[x[1..3]];- - DegRevLex is the default order Q1 := Ideal(x[l]3, x[2]^2); 42 := Ideal(x[l]3, x[l] * x[3], x[3]^2); I := Intersection(Q1, Q2); S2 := Minimalized(Intersection(Qi?2, Q2-2)); I2 := Minimalized(I2); Len(S2); - - checking number of generators S3 := Minirnalized(1ntersect i o n ( Q i 2 , Q2^3)); EqSet(Gens(S2)) Gens(I2)); - - checking equality G := Gens(S3); G[4] IsIn I^3; I
Chapter 5
138
5.1.33 Use Proposition 5.1.20 to find the minimal primes of the ideal I generated by
5.1.34 Show that I = (y;, y1 y2) is a non primary ideal with a prime radical. 5.1.35 Let I be a monomial ideal in Ic[xl,.. . ,s], and X = V(I) c An(Ic) the associated monomial variety. If IC is infinite, prove that X is irreducible if and only if X = V(xi,, . . . ,xi,,). Hint Note I(V(x1,. . . ,x,.))= (XI,.. . ,x,.).
5.2
Simplicial complexes and homology
A finite simplicial complex consists of a finite set V of vertices and a collection A of subsets of V called faces or simplices such that (i) If u E V, then ( u } E A. (ii) If F E A and G
c F , then G E A.
Let A be a simplicial complex and F a face of A. Define the dimensions of F and A by dim(F) = IF1
- 1 and dim(A)
= sup(dim(F)I F f A)
respectively. A face of dimension q is sometimes refer to as a q-face or as a q-simplex.
Example 5.2.1 Take the following triangulation of the disk with a whisker attached.
\
A x2
0-simplices: X I , x2,23, x4 1-simplices: x1 x2, x2 2 3 , x 3 2 4 , ~ 2 ~ 4 2-simplices: 22 x3x4 x4
Let F be a q-simplex in A with vertices vo,u1, . , . ,uq. '#e say that two total orderings of the vertices
Monomial Ideals and
Stanley-Reisner Rings
139
are equivalent if ( i o , . . . ,i,) is an even permutation of (~'0, . . . ,jq). This is an equivalence relation because the set of even permutations form a group, and for q > 1, it partitions the total orderings of vo,. . . ,v, into two equivalence classes. An oriented q-simplex of A is a q-simplex F with a choice of one of these equivalence classes, If vo,. . . ,vq are the vertices of F, the oriented simplex determined by the ordering vo < - < vq will be denoted by b o , . . . ,vq]. Suppose A is a commutative ring with unit. Let Cq(A) be the free Amodule with basis consisting of the oriented q-simplices in A, modulo the relations [vo,v1, v2,. . .,vql + [m,0 0 , w 2 , . . * ,vq]. 0
e
In particular Cq(A) is defined for any field k and dimk Cq(A) equals the number of q-simplices of A. For q 2 1 we define the homomorphism
induced by
where iji means that the symbol vi is to be deleted. Since aqOq+l = 0 we obtain the chain complex C(A) = (C,(A), aq],which is called the oriented chain complex of A. The augmentedorientedchain cornpie2 of A is the complex
) 1 for every vertex u of A. This augmented where d = dim(A) and ~ ( v = chain complex is denoted by (C(A),e ) . Set 80 = e and C-1 (A) = A. Let Z,(A,A) = ker(aq), B, = im(8,+1) and
The elements of Zq(A, A ) and B, (A, A ) are called cycles and boundaries respectively and a q ( A ;A ) is the qth reduced simplicial homology group of A with coefficients in A . Remark 5.2.2 Note that if A
# 8, then Ei(A; A ) = 0 for i < 0.
Proposition 5.2.3 If A is a nonempty simplicial complex, then &(A; A ) is a free A-module of rank r - 1, where r is the number of connected components of A .
Chapter 5
140
Proof. Let V = {xl). . . x , } be the vertex set of A and A,). . AT its connected components. Denote the vertex setof Ai by & and pick a vertex in each one of the &, one - may assume xi E for i = 1,.. . r . Set pi = xi - x,. and p i = pi im(&) for i = 1,.. . ,r - 1. We claim that the set ti = {pl,.. . ,pT-l}is an A-basis for ao(A; A) = ker(&)/im(dl). First we prove that is linearly independent. Assume Clz: aipi is in im(81) for some ai in A , by making an appropriate grouping of terms one can write )
)
+
)
for some bij in A. Since Co(A) is a free A-module with basis V and the are mutually disjoint one has
applying 80 to both sides of this equality we obtain ak = 0 for all IC. Next we show that is a set of generators.Let xil and x j l be two vertices of A, it suffices to prove that x i l - xjl +im(&) is in the submodule generated by because ker(d0) is generated by the elements of the form xi - x j . Assume xil E Vm and xj1 E Vs. There is a path xil . . . ,xit x m in A m and a path xjl . . . xj,, x , in A, such that every two consecutive vertices of those paths form a 1-face of A. From the equalities )
)
Definition 5.2.4 Let A and A, be simplicial complexes on disjoint vertex sets V and W respectively. The join
is the simplicial complex on the vertex set V U W with faces F U G where F E A and G E A,. The cone cn(A) = w * A of A is the join of a point {w} with A. Proposition 5.2.5 If A is a simplicialcomplex and cn(A) = w cone, then i i , ( c n ( ~ )A; ) = o for all p .
*A
its
Monomial and Ideals
Stanley-Reisner Rings
141
Proof. If CT = [VO,. . . ,' u p ] is an oriented p-simplex, set [w, CT]equal to [w, V O ,. . ,v,]. In general if cp = ni0i is a p-chain, set [w, cp] equal ni [ w ,oil. This bracketoperation is a homomorphism from C,(A) to into Cp+l(w* A). If CT is an oriented p-simplex from the definition of the boundary operator one obtains
.
a[w,a]=
if dim 0 = 0, [w, 801 if dim c > 0.
0-W
0-
As a consequence one has the more general formulas:
By Proposition 5.2.3 we obtain &(w * A) = 0, because w * A is connected. Assume p > 0. Let z p E Z ( w * A) be a p-cycle of w * A, we will show that X, is a boundary. Write
where cp consist of those termsof z p with support in A, and dp-l is a chain. It is enough to show that zp
- a[w, c,] = 0.
Using the identities (5.1) and (5.2) we obtain
where ep-l = d,-1 + dc, is a chain in A. Applying d to the equality above and using that X, is a cycle yields d[w,e,-1] = 0. By identities (5.1) and (5.2) one has
Now, the part of this chain with support in A is ep-l. Therefore ep-l = 0 and [w,ae,-,] = 0, which shows X, = ~ [ w , c , ] . 0
Definition 5.2.6 The reduced Euler characteristic Z(A) of a simplicial complex A of dimension d is given by d
Z(A) = -1
+ C(-l)ifi = -1 + x(A), i=O
where
fi
is the number of i-faces in A and x(A) is its Euler characteristic.
Chapter 5
142
Exercises 5.2.7 Let A be a simplicial complex of dimension d and
fi
the number of
i-faces in A. If k is a field, then d
(-l)%ankEi(A; k) = -1
c(-l)ifi. d
i=O
iz-1
5.2.8 Let
+
A be a simplicial complex, prove that there is an exact sequence
and conclude
Ho(A;A) N Eo(A, A ) @ A . 5.2.9 Let
A be a simplicial complex with vertex set V = { X I , .. . ,x n > and
the right end of the augmented chain complex over a ring A . Prove that ker(&) is generated by the cycles of the form zi - xj. 5.2.10 Let
A be a simplicial complex and fi the number of i-faces of A. If
is the augmented chain complex of A over a field k , then dimk (21)is equal to f1 - fo r; where r is the number of connected components of A and 21 = ker(&).
+
5.3
Face rings
Let R = k[xl,.. . ,zn] be a polynomial ring over a field IC. If I is an ideal of R generated by square-free monomials, the Stanley-Reisner simplicial complex AI associated to I has vertex set V = {xiI xi 4 I } and its faces are defined by
Conversely if A is a simplicial complex with vertex set V contained in
(21,. . . ,x,}, the Stanley-Reisner ideal I A is defined as
and its Stanley-Reisner ring k[A]is defined as the quotient ring R / I A .
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143
Definition 5.3.1 Let I be an ideal of R generated by square-free monomials. The quotient ring R / I is called a face ring. Example 5.3.2 A simplicial complex A of dimension 1 and its StanleyReisner ideal:
Definition 5.3.3 Let A be a silnplicial complex and F E A, define lk(F), the link of F , as
Definition 5.3.4 Let k be a field. A simplicial complex A is said to be Cuhen-Macaulay over IC if the Stanley-Reisnerring k[A] is a CohenMacaulay ring. Theorem 5.3.5 (Reisner) Let A be a simplicial complex. If k is a field, then the following conditions are equivalent: (a) A is Cohen-Macaulay over (b) gi(1k F ; k ) = 0 for F
IC.
A and i < dim lk F.
Proof. See the originalpaper [237], or [44] for a detailed proof that combines a vanishing theorem of Grothendieck [137] with a formula of M. Hochster for the Hilbert series of the local cohomology modules of k[A] respect with the to fine grading. 0 We clarify that in particular a qohen-Macaulay complex A, being the link of its empty face, must satisfy Hi(A; k ) = 0 for i < dim A. In general the Cohen-Macaulay property of a face ring may depend on the characteristicof the base field, classical examples of this dependence are triangulations of the projective plane (see Exercise 5.3.31). Example 5.3.6 If 4 is a discrete set with n vertices
then A is Cohen-Macaulay and I A = (zizj 11 5 i
< j 5 n),
144
Chapter 5
Corollary 5.3.7 If A is a 1-dimensional simplicial complex, then A is connected if and only if A is Cohen-Macaulay. Proof. By Reisner theorem A is Cohen-Macaulay iff &(A; IC) = 0, because the link of a vertex of A consists of a discrete set of vertices. To complete Proposition apply argument the 5.2.3. 0 Proposition 5.3.8 ([174]) If k is a field, then a simplicial complex A is Cohen-Macaulay over k if and only if lk(F) is Cohen-Macaulay over IC f o r all F E A. Proof. +) Let F be a fixed face in A and A' = lk(F). If G E A', using the equality 1kAl (G) = 1kA (G U F ) and Reisner criterion one infers gi(lkAt (G), k ) = 0 for i < dim lkAl (G), thus A' is Cohen-Macaulay. e)Since all the links are Cohen-Macaulay, it suffices to take F = 0 and 0 observe A = lk(0) to conclude that A is Cohen-Macaulay. Definition 5.3.9 A face F of a simplicial complex A is said to be a facet if F is not properly contained in any other face of A. Proposition 5.3.10 If A is a simplicial complex with vertices 21,.. .,Xn, then the primary decomposition of the Stanley-Reisner ideal of A is:
where the intersection is taken over all facets F of A, and PF denotes the face ideal generated b y all xi such that xi 4 F . Proof. Let F be a face of A and PF the face ideal generated by the xi such that xi 4 F . It is not hard to prove that PF is a minimal prime of IA if and only if F is a facet. Therefore the result follow from Corollary 5.1.5. 0
Corollary 5.3.11 Let A be asimplicialcomplex vertex set V = ( X I , . . . ,xn} and k a field. Then
+
of dimension d on the
Proof. Let F be a facet of A with d 1 vertices. Note that by Proposition 5.3.10 the face ideal PF generated by the variables not in F has height equal to the height of IA. Hence ht(IA) = n - d - 1, taking into account 0 Proposition 2.2.3 one has dim k[A] = d 1.
+
Monomial Ideals and Stanley-Reisner Rings
145
Corollary 5.3.12 A Cohen-Macaulaysimplicialcomplex is, all its maximal faces have the same dimension.
A is pure, that
Proof. Use Proposition 5.3.10 and Corollary 2.2.4.
cl
Definition 5.3.13 The q-skeleton of a simplicial complex A is the simplicial complex AQconsisting of all p-simplices of A with p 5 q. Proposition 5.3.14 Let A be a simplicial complex of dimension d and AQ its q-skeleton. If A is Cohen-Macaulay over a field k , then AQ is CohenMacaulay for q 5 d. Proof. Set A' = AQand take F E A' with dim(F)
< q 5 d, Note
Since A is pure there is a facet F' of A of dimension d containing F , hence F' \ F is a face of lkA(F) of dimension d - IFI. It follows that thedimension of lkAr(F) is q IFI. Using Proposition 5.3.8we get
-
and consequently
Observe that theequality between the homology modules follows simply be~ and 1kA ( F ) cause the augmented oriented chain complexes of ( 1 1 (F))q-lF1 are equal up to degree q - IF1 and thus their homologies have to agree up to degree one less. Hence by Reisner criterion AQis Cohen-Macaulay. 0 Example 5.3.15 ([114]) Let R = k [ x 1 , .. . ,xn] be a polynomial ring over a field k and A the ( p - 2)-skeleton of a simplex of dimension n - 1, where p 2 2. Then IAis a Cohen-Macaulay ideal and the face ring
has a p-linear resolution by Theorem 4.3.7. Face rings with pure and linear resolutionshave been studied in [46]and [114, 1161; special results areobtained when the correspondingStanleyReisner ideal is generated in degree two. Proposition 5.3.16 If A, and A, are simplicial complexes, then their join A, *A2 is Cohen-Macaulay if and only if Ai is Cohen-Macaulay for i = 1 , 2 .
Chapter
146
Proof. Since one has the equality
there is a graded isomorphism of k-algebras
see Proposition 2.2.20. To complete the proof useCorollary 2.2.22. Given a simplicial complex A and A,, define their union
. . ,A,
0
sub complexes of A, we
6
Ai i=l as the simplicial complex with vertex set Uf=lVi and such that F is a face of U:=,Aiif and only if F is a face of Ai for some i. The intersection can be defined similarly. Definition 5.3.17 A pure simplicial complex A of dimension d is shellable if the facets of A can be order F1,.. . , F, such that
is pure of dimension d - 1 for all i 2 2. Here
If A is shellable,
F1,
. .. ,Fs is called a shelling.
Theorem 5.3.18 Let A be a simplicial complex. If A is shellable, then A is Cohen-Macaulay over any field k. Proof. Our proofis basedon a refined argument of T. Hibi [162]. Let V = {XI,. . . , a , } be the vertex set of A and R = k[x] a polynomial ring over a field k . Assume that F1,.. . ,F, is a linear order of the facets of A such that Fi n (UiEiFj) is pure of dimension d - 1 for all i 2 2, where d is the dimension of A. Set
one may assume
(r
= { X I , . . . ,x,.}. There is a short exact sequence
0 -+ R / ( I :f) f, R / I
+ I t / ( & f) ”-+ 0,
(*>
Monomial Ideals and Stanley-Reisner Rings
147
where f = x1 - - x,. and I = IA is the Stanley-Reisner ideal of A. First we show the equality
R / ( I :f ) = note that k [ F s ]is a polynomial ring in d+ 1variables. By Proposition 5.3.10 one can write I = &pi, where pi is generated by V \ Fi for all i. Hence
Observe that if o c Fi for some i, then i = s; otherwise if i < s, then o belongs to F, n ( U3j i 1l F j ) , but since this simplicial complex is pure of dimension d - 1, there is v E o such that o c F, \ {v), a contradiction. Hence i = s and ( I :f ) = (p,: f ) = p,, as asserted. Next let us show
R / ( I ,f) = k[Fll J ' . - u Fs-11. Note that f E pi for i = 1,.. . ,s - 1; otherwise if f 4 pi for some i < s, then u c Fi and the previous argument yields a contradiction. If p is a minimal prime of ( I ,f ) different from p 1 , . . . ,ps-l, then p, C p and xi E p for some xi E 0,thus by construction of 0 one has Fa \ {xi) c F k for some k < s. Therefore V \ F k c (xi) U (V \ F,) c p, and by the minimality of p one derives p k = ps, whichis impossible. Altogether we conclude that p 1 , . . . ,ps-l are the minimal primes of ( I ,f ) and
as required. Using induction on s and the exact sequence (*) it follows that R / I is Cohen-Macaulay by a direct application of the depth lemma. To apply the depth lemma one uses that the extremesof the exact sequence (*) are both 0 Cohen-Macaulay R-modules of dimension d 1.
+
Exercises 5.3.19 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. 5.3.20 A monomial ideal I , with coefficientsin
a field, is generated by square-free monomials iff any of the following conditions hold: (a) I is an intersection of prime ideals.
Chapter 5
148
(b) rad(I) = I . (c) A monomial f is in I iff x1
- - x,- E I , where supp(f) =
5.3.21 If A is the simplicial complex
prove the equality
IA = (x1,x2) n ( 2 1 ,2 3 ) n ( 2 1 , 2 4 ) n ( 2 2 , 2 3 ) n ( x 3 ,x4) by using the correspondence between facets and minimal primes. 5.3.22 Let A be a Cohen-Macaulay simplicial complex and F E A. If lk(F)
is not a discrete set of vertices, prove that lk(F) is connected. 5.3.23 Let A be a simplicial complex on the vertex set V and I its Stanley-
Reisner ideal. Take x E V and set J = ( I :.). (a) Prove that J = nSepipi, where Ass(1) = ( P I , . (b) Find a relation between A and A J .
. . ,pT},
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 2 is a variable not in I such that 2 E U;=,supp(fi), then R/(I: x) is Cohen-Macaulay.
5.3.24 Let
5.3.25 Let A be a pure simplicial complex on the vertex set V and I = In.
Assume that x is a variable in U:=lsupp(fa), where the fi 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: z) = ht(I,z). 5.3.26 Let A a simplicial complex and o E A, define the star of o as
star(o) = {G E A ( o U G E A}. Prove that the facets of star(o) are the facets of A that contain o. 5.3.27 Let R = k(x]be a polynomial ring over a field k and A a simplicial
complex with vertex set x. If
CT
= (21,
. . . ,xr} E A, prove the equality
k[star(o)] = R/(IA:f), where f =
nL=,zi and star(o) is the star of
o.
Monomial Ideals and Stanley-Reisner Rings
149
5.3.28 If A is a Cohen-Macaulay complex and F a face of A, then star(F) is a Cohen-Macaulay complex. Hint star(F) = F * lk(F). 5.3.29 A pure simplicial complex A is shellable if and only if the facets of A can be listed F1,. . . , F, such that for all 1 5 j < i 5 s, there exist some v E Fi \ Fj and some k E ( I , . . . , i - I} with Fi \ F k = {v}. 5.3.30 Let A be a pure simplicial complex. Prove that A is shellable if and only if the facets of A can be ordered PI,. . . ,F, such that for all 1 5 j < i 5 S, there is v E Fi\Fj and k < i with FinFj C FinFk = Fi\{v}. The next exercise is a subdivision of the minimal triangulation of the projective plane given by G. Reisner [237, Remark 31.
5.3.31 (N. Terai) If k is a field and A is the following triangulation of the real projective plane P2 2.5 x1
then the link of any vertex is a cycle and k[A] is Cohen-Macaulay if and only if char(k) # 2.
Hint
2 1(A; k ) N
(P2;k) = k / 2 k and use Theorem 5.3.5.
5.3.32 Let R be a ring and sequence of R-modules:
1 1 , 12
ideals of R. Prove that there is an exact
5.3.33 ([172]) Let R = k[x] be a polynomial ring over a field k and A,, A2 simplicial complexes whose vertex sets are contained in x. Prove that there is an exact sequence of R-modules:
Chapter 5
150
5.4
Hilbert series of face rings
For an Stanley-Reisner ring k[A] there are explicit formulas for its Hilbert function and Hilbert series in terms of the combinatorial data of the complex, we begin with a direct approach to compute these formulae. Definition 5.4.1 Let k and n be two positive integers. A k-partition of n is an element a = ( 0 1 , . . . ,arc) in N$ so that n = a1 +arc.
+
Lemma 5.4.2 The number c(n,k ) of k-partitions of n is equal to
Proof. If a = (01,. . . ,arc) is a k-partition of n define $ ( a ) by
To finish the proof note that the map $ induces a bijection between the set of k-partitions of n and the set of subsets of { 1,.. . ,n - l} with k - 1 elements. 0 Definition 5.4.3 Given a simplicial complex A of dimension d its f-vector is the (d 1)-tuple: f (A) = ( f 0 , 9 f d ) , where fi is the number of i-faces of A. Note f-1 = 1.
+
-
Proposition 5.4.4 If A is a simplicial complex of dimension d and (fi) its f -vector, then the Hilbert function of k[A] is: H(k[A],j) =
2 (j i=o
a
I)
fi,
f o r j 2 1 and H(k[A],O) = 1.
-
Proof. Let 21,.. . ,x, be the vertices of A. Given x" = xyl . xEn we denote its image in k[A] by P . The face ideal I A is generated by squarefree monomials, hence a monomial z" is in I A if and only if supp(za) = (xi1ai > 0) is not a face of A. Therefore the set of all monomials 3" such that supp(za) is a face in A form a basis for k[A] as a vector space over k . If {xil , . . . ,xi, } is a face of A of dimension r - 1, we can use Lemma 5.4.2 to count the number of monomials of the form ztl . . such that (al, . . . ,a,) is an r-partition of j . The desired formula follows at once. 0 ax:
Corollary 5.4.5 If A is a simplicial complex and f (A) = (fo) f -vector, then the Hilbert series of k[A] is given by
., .
)
fd)
its
Monomial and Ideals
Stanley-Reisner Rings
151
Proof. By the proposition above we have
Hence using the identities
the formula is established.
0
A result of Kruskal-Katona Kruskal and Katona [202] showed that (fo,f1, . , . ,fd) E Zd+l is the f-vector of some &dimensional simplicial complex if and only if 0 < fi+l
where
fji+')
5 f / i + l ) , 0 5 i 5 d - 1,
is defined according to Equation 4.9.
Simplicial complexes and their h-vectors As an application we derive formulas for h ( A ) , the h-vector of k[A]. If A is Cohen-Macaulay, we present an important numerical constraint for h ( A ) which is basic for some applications of Commutative Algebra to Combinatorics. Theorem 5.4.6 If A is a simplicial complex of dimension d and (hi) the h-vector of k[A], then h k = 0 for k > d 1 and,
+
Proof. The idea is to write the Hilbert series of k[A] intwo ways. By Corollary 5.4.5 we have:
On the other hand by the Hilbert-Serre Theorem 4.1.3 there is a (unique) polynomial h ( z ) = ho hlz - + h,zr with integral coefficients so that h(1) # 0 and satisfying
+
+
e
F ( k [ A ]z, ) =
h(4 (1 -
*
Comparing (5.4) and (5.5) yields the asserted equalities.
(5.5) 0
152
Chapter 5
Definition 5.4.7 The h-vector of a simplicial complex A, denoted by h(A), is defined as the h-vector of the standard graded algebra k [ A ] .
Computation of the h-vector of face rings To compute theh-vector (of face rings) the following procedure of R. Stanley [271]can be used. Consider the 3-dimensional convex polytope P consisting of two solid tetrahedrons joined by a 2-face:
one has f(P)= (5,9,6). Write down the f-vector on a diagonal, and put a 1 to the left of fo.
1
5 9
6 Complete this array constructinga table, by placing below a pair of consecutive entries their difference, and placing a 1 on the left edge:
1
1 1 h ( P ) = (1
5
4 3
2
9 5
2
6 1)
After completing the table the next row of differences will be the h-vector. Hence the boundary simplicial complex A = A(P) is 2-dimensional, k[A] has multiplicity 6, and I A = ( ~ 1 x 5~, 2 ~ 3 ~ 4 ) .
Now we present a result that plays a role in the proof of the upper bound conjecture for spheres. Theorem 5.4.8 Let A be a simplicial complex of dimension d with n vertices. If k[A] is Cohen-Macaulay and k is an infinite field, then the h-vector of A satisfies
ifn-d-2
) f o rO < i < d + l .
Monomial Ideals and Stanley-Reisner Rings
153
Proof. Set S = k[A] = R / I , where R = k [ x l , . . . ,xn] and I = IA. By Lemma 2.2.15 there exists a regular homogeneous system of parameters -8 = 81, . . . , 8d+l for S such that each 19i has degree one. Since A = S/(e)S is Artinian, in this case Theorem 4.2.5 says that hi(A) = H ( A , i ) . Note that S/(e)S N R/T, where = Rfig) is a polynomialringin n - d - 1 variables and 7 is the image of I in R. Therefore we have
as required.
0
Example 5.4.9 Let A be a. 2-dimensional complex with f ( A ) = (5,6,2). Thus k[A] is not Cohen-Macaulay because its h vector has a negative entry:
1 1 1
h ( A ) = (1
5 4
3
2
6 2
-1
2 0).
A shellable complex Let R = k [ x l , . . . ,xn] be a polynomial ring over a field k and A a simplicial complex with vertices X I , . . . x,. Associated to the Stanley-Reisner ideal IA we define another ideal )
I'(A) = (IA,xlYl)-..,xnYn)c
IG[21,...rxn,Yl)...,Yn],
and denote by S(A) the Stanley-Reisner complex corresponding to this ideal 1'(A). Proposition 5.4.10 The simplicial complex S(A) is shellable. Proof. To prove that S(A) is shellable we will use the criterion of Exercise 5.3.29. Let us begin by describing the facets of S ( A ) . It is not hard to show that S ( A ) is pure. Notice that the facets of S ( A ) are of the form F' = {y,., , . . . ,yrr,1 U F, where F = { x s l ) .. . ,x,,) is a face of A so that k + t = n , { r l )...,r ~ ) n { s l..., , s e ) = 0 . ThereforeFGF'definesaone to one correspondence between the faces of A and the facets of S ( A ) . To construct a shelling for S(A) we first linearly order the faces of A according to its dimension, and then linearly order the facets of S ( A ) in such a way that if F , G are faces of A and F < G then F' < G'. Let F', G' be facets of S(A), with F' < G', there is xj E G' \ F', setting H' = (G' U {yj}) \ (xj} we obtain H' < G' and G' \ H' = (zj}, as required. 0 Corollary 5.4.11 S ( A ) is a Cohen-Macaulay complex and the A is related to the h-vector of S(A) by
f i ( A ) = hi+l(S(A))for all i.
f -vector of
Chapter 5
154
Proof. The first assertionfollows from Theorem5.3.18 and theprevious result. For the other claim set J = ( I A ,x:, . . . ,x:) and note that theface ring k[S(A)] = k [ x ,y]/Ii reduces to k [ x ] / Jmodulo z = (91 - 2 1 , . . . ,pn - xn}. Thus z is a system of parameters for IC[S(A)]. Since S(A) is CohenMacaulay, z must be a regular sequence on k [ S ( A ) ]according to Proposition 2.2.7. Using Theorem 4.2.5 one has that the h-vector of S(A) is then equal to the h-vector of b [ x ] / J ,which is nothing other than f ( A ) shifted 0 by one.
Exercises 5.4.12 Let A be a simplicial complex of dimension d and h(A) = (hi) its
h-vector. Prove
hd+l = ( -qdZ(A), where Z(A) is the reduced Euler characteristic of A. 5.4.13 Let S = k[A] be the Stanley-Reisner ring of a simplicial complex A
over a field IC and cp(t)the Hilbert polynomial of S. Prove: (a) cp(i)= dimk(Si) for all i
2 1, and
(b) cps(0) = 1 if and only if %(A)= 0. 5.4.14 Let A be a simplicial complex of dimension d and (fo,
. . . ,fd)
its
, multiplicity of k [ A ] . f-vector. Prove that fd is equal to e ( k [ A ] ) the 5.4.15 Let S = k[A] be the Stanley-Reisner ring of a simplicial complex A
with vertices 2 1 , . . . ,xn and A = S / ( x ; , . . . ,x:). Prove that A is pure if and only if all the nonzero elements of the socle of A have the same degree.
5.5
Upper bound conjectures
It is our intention here to give flashes of some of the steps in the proof of the upper bound conjecture for simplicial spheres. An excellent reference for a detailed proof of this conjecture is [44]. Convex polytopes A convex polytope P c Rs,or simply a polytope, is the convex hull of a finite set of points 2 1 , . . . , x , in Rs, that is, P is the set of all convex combinations of those points:
P = (alxl
+ + anxnl ai 2 0 , a1 + + a, 0
.
= 1, ai E R}.
For A C Rs, let aff( A ) be the affine space generated by A . Recall that aff(A) is the set of all ufine combinations of points in A:
Stanley-Reisner Rings
Monomial Ideals and
155
+
One can represent aff(A) as aff(A) = x0 V , where 20 E Rs and V is a (unique) linear subspace of Rs. The dimension of A is defined as dim(A) = dimw ( V ). Given y E Rs \ (0) and a E R, define the hyperplane
H ( Y ,4 = { x E where ( , ) is the usual innerproduct bounded by H ( y , a) are
I
IRs (2, Y)
in
= a},
R3. The two closedhalfspaces
H f ( y ,a ) = {x E R3)(x,y ) 2 a) and H - ( y , a ) = H+(-y, -a). Definition 5.5.1 A proper face of a convex polytope P is a subset F c P such that there is a supporting hyperplane H ( y , a ) satisfying:
( 4 F = p n H(P,a># 0, (b) P $? H ( y , a ) and P
c H-(y,a).
The improper faces of a polytope P are P itself and 8. It is us1la1 to call vertex to a face of P of dimension zero; it is a result of polyhedral geometry that P is in fact the convex hull of its vertices, see [39, 3151. Proposition 5.5.2 If P C Iw" is a convex polytope, then P is a compact convex subset of Rs with a finitely many faces and any faceof P is a convex polytope. Proof. See [39].
A set of points y1, , . . , y m
0 E
R3 is afineky independent if the relation
m
i= 1
can hold only if a, = 0 for all i. Definition 5.5.3 A q-simplex is a polytope generated by q + 1 affinely independent points. A polytope is called a simplicial if every proper face is a simplex.
+
Let P be a polytope of dimension d 1 and denote by fi the number of its faces of dimension i, thus fo is the number of vertices of P . The f-vector of P is the vector f(P)= (fo,. . . ,fd) and we set f-1 = 1.
Chanter 5
156
Simplicial spheres Let A be a finite simplicial complexwith vertices . . . ,x n and ei the ith unit vector in IW". Given a face F E A set
21,
18'1
= conv(ei1 xi E F'I >
where "conv" denotes convex hull. Define the geometric realization A as
PI=
u
IFL
FEA
thus A is a topological space with the induced usual topology of
IW",
The boundary complex A ( P ) of a simplicial polytope of dimension d + 1 isby 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 A ( P ) has dimension d. Note that P and A ( P ) have the same f-vector and accordingly one defines the h-vector of P as the h-vector of A ( P ) . The geometric realization IA(P)I is therefore homeomorphic to a d-sphere S d . This suggest the following more general concept. Definition 5.5.4 A simplicial complex A of dimension d is called a simplicial sphere if lAl 21 Sd and in this case A is said to be a triangulation of Sd. Theorem 5.5.5 Let A be a simplicial complex of dimension d. If IA I 2 Sd, then IC for i = dim(lk(F)), Proof. See [44,Chapter 51.
0
Proposition 5.5.6 If A is a simplicial sphere, then A is Cohen-Macaulay and the following equality holds z(lk F ) = (-l)dim Ik Proof. Use Reisner's criterion Theorem and
f o r F E A. 5.5.5.
0
Remark 5.5.7 If P is a simplicial d-polytope, then A ( P ) is a simplicial sphere of dimension d- 1. Hence from Proposition5.5.6 we obtain theEuler formula
where f = (fo, . . . ,fd-1) is the f-vector of P . The Euler formula is valid for any d-polytope, not just the simplicial ones (see [39, Theorem 16.11).
157
Monomial Ideals and St anley-Reisner Rings
Corollary 5.5.8 If A is a simplicial sphere, then k[A] is a Gorenstein ring. Proof. It follows from (274, Theorem 5.11 and Theorem 5.5.5.
c1
Theorem 5.5.9 If A is asimplicialsphere of dimension d , then its hvector satisfies hi = hd+l-i f o r i = 0 , . . . , d 1.
+
Proof. Since k[A] is a Gorenstein ring, its h-vector is symmetric according 0 to Exercise 4.3.9. Remark 5.5.10 If A is a simplicial sphere, then the linear relations
of its h-vector contain the Euler formula as a particular case (make i = 0). Those relations are called the Dehn-Sommerville equations. Corollary 5.5.11 If h = (ho,. . . ,hd+l) is the h-vector of a simplicial polytope of dimension d + 1, then h satisfies
Let us mention two conjectures raised around 1993 in connection with the problem of bounding the Betti numbers of graded Gorenstein ideals [222]. Conjecture 5.5.12 (Miller-Villarreal) Let A be a simplicial sphere and I the Stanley-Reisner ideal of A. If I has initial degree p and height g , then f o r 1 _< i < g the ith Betti number Pi of k[A] satisfies: pfg-1 (p+i-1)(
p+iG I 2 ) +(P+q-l)(P+g-iP-1
”>
-
(:)
( p t a ” T ” ) *
Conjecture 5.5.13 (Miller-Villarreal) Let A be a simplicial sphere and I its Stanley-Reisner ideal. If I has initial degree p and height g, then
Chapter 5
158
Cyclic polytopes In order to formulate the upper bound conjecture for simplicial spheres and the upper boundconjectures for convex polytopes we need to introduce some results oncyclic polytopes, the readeris referred to [39] for a detailed discussion of this distinguished class of polytopes. Consider the monomial curve I' c Rd+l given parametrically by
r = {@,. . . , # + l ) )
E R}.
A cyclic polytope, denoted by C ( n ,d+ l ) ,is the convex hull of any n distinct of points in I' such that n > d 1. The combinatorial type or f-vector C ( n ,d 1) depends only on n and d and not on the points chosen and dim C(n,d 1) = d 1. One of the relevant properties of cyclic polytope is that they are simplicial, thus the boundaryd C ( n ,d 1) defines an abstract simplicial complex A(n,d 1) = A ( C ( n ,d 1)) so that lA(n,d 1)1 S Sd. Notice that d C ( n ,d 1 ) is a union of proper faces (in fact of facets) of C(n,d 1). A polytope P has the highest possible number of i-faces when every set of i 1 vertices is the vertex set of a proper face of P . In this case it is said that P is (i 1)-neighbourly.
+
+
+
+
+
+
+
+
+
+
+
+
Theorem 5.5.14 A cyclic polytope C ( n ,d In particular
+ 1) is L(d + 1)/2J -neighbourly.
Proof. See [39].
0
Corollary 5.5.15 If h = (ho,. . . ,hd+l) is the h vector of C ( n , d + l), then
Proof. Let 0 5 k
x(-I)k-i ( k
hk
=
5 [ ( d + 1 ) / 2 J . One has d+l-i
k-i
i=O
k )fi-1
(
)( J
= x(-1)2(k-i) k -kd--i 2 n i=O
The second equality uses Eq. (4.1) and the last equality follows using induction on k, or setting a = n and c = k - d - 2 in:
For a proof of this formula see [39, Appendix 31.
0
159
Monomial Ideals and Stanley-Reisner Rings ,I,
,
Upper bound conjecture for simplicial spheres Next we present the upperboundconjecture for simplicial spheres. It was conjectured by V. Klee and proved in 1975 by R. Stanley [268]. It was motivated by the performance of the simplex algorithm in linear programming. Conjecture 5.5.16 (UBCS) Let A be simplicial complex of dimension d with n vertices. If (A(S Sd, then
Proof. (sketch) Consider the following four conditions: (a) hi(A) 5 hi(A(C(n,d
+
+ 1)))for 0 5 i 5 [(d + 1)/2J.
(b) hi(A(C(n,d 1)))= hd+l-i(A(C(n, d (c) hi(A) 5 (i+nyd-2) for all i. (d) hi(A) = hd+l-i(A) for 0 5 i 5 d
+ 1)))for 0 5 i 5 d + 1.
+ 1.
Observe that (a), (b)and(d) imply theupper boundconjecture. To show it notice that if [(d+l)/ZJ < i 5 d + l , then 0 5 d + l - i 5 [(d+ 1)/2J and
+
hence h ( A ) 5 h ( A ( C ( n , d 1)))for 0 5 i
5 d + 1. Therefore
= fk-l(A(C(n, d + 1))) for 1 5 k 5 d
+ 1,
and the proof of the observation is complete. By Corollary 5.5.15 (c) + (a), and by Corollary 5.5.11 (b) is satisfied because C(n,d 1) is a simplicial polytope. As a consequence the upper bound conjecture is reduced to prove (c) and (d) . To complete the proof notice that (c) follows from and follow from Theorem 5.5.9. 0 Theorem 5.4.8,(d)
+
As a particular case of the UBCS one obtains the upper bound conjecture for convex polytopes, this conjecture was posed by Motzkin in 1957 and proved in 1970 by McMullen [221].
Corollary 5.5.17 (UBCP) If P is a convex polytope of dimension d with n vertices, then
Chapter 5
160
Proof. (sketch) By a process known as “pulling the vertices” P canbe transformed into a simplicial polytope with the same number of vertices as P , and at least as many faces of higher dimension. Hence we may assume that P is simplicial. Since P is simplicial the boundary complex A ( P ) is a simplicial sphere, that is, IA(P)I EZ Sd-l. Using the UBCS we obtain
for i = 0,.
. . ,d.
0
Remark Not every triangulation of a sphere is the boundary complex of some simplicial polytope. There are examples, foundby Grunbaum in 1965, with d = 4 and n = 8 satisfying ( A (S Sd-l and A not isomorphic to A ( P ) for every d-polytope. Moreover it follows from a result of Steinitz (a graph is the 1-skeleton of a 3-polytope in R3 if and only if it is a 3-connected planar graph) that such an example does not exist for d = 3.
Exercises 5.5.18 If A is a connected simplicial complex havingall its vertices of degree two, then A is a cycle. 5.5.19 If A is a simplicial sphere on n vertices whose geometric realization
[AI is isomorphic to the unit circle S’, then (a) A is a cycle of length.
(b) v ( I A ) =
(g) - n for n 2 4.
5.5.20 ([lS]) Let R = k [ z l , .. . ,xn] be a polynomial ring over a field k . If I is a monomial Gorenstein ideal and rad ( I ) = ( X I ,. . . ,zn),prove that
I = (x;’,. . . ,x>)
Hint Prove that I is irreducible. See [43] and [218, Theorem 18.11. 5.5.21 Give an example of a monomial Gorenstein ideal whichis not a complete intersection.
I
Chapter 6
Edge Ideals Edge ideals are the most simple polynomial ideals after face ideals, they are generated by square-free monomials of degreetwo and they can be associated to graphs and to Stanley-Reisner complexes. At first sight the use of graph theory to study edge ideals is convenient because there are common objectsof study, but in fact some remarkable applicationsof graph theoretical results will take place in this chapter. There are many works in the literature relating graph theory with other branches of algebra, see for instance [17, 24, 135, 136, 164, 262, 2631. In the first section the connection between ideals and graphs is established through the notion of minimal vertex cover. Some graph theory results and terminology are recollected and various versions of a theorem due to Konig are discussed. As we learned graph theory from [28] and [139], those remain as our main references for the subject and they are highly recommended. We study some properties of edge ideals, such as the Cohen-Macaulay property, and stress thatsome of those properties can be better understood using graph theoretical notions. A structure theorem for Cohen-Macaulay trees is presented in Section 6.3; this leads to a deeper result showing that Cohen-Macaulay bipartite graphs can be defined recursively. The notion of “ideal of covers” of a monomial ideal is introduced in Section 6.7, and we study the connection between properties of an edge ideal and properties of its ideal of covers, the notion of triangulated graph is used to link properties of those two ideals.
6.1
Graphs and ideals
A simple graph G consists of a finite set V of vertices and a collection E of subsets of V called edges such that every edge of G is a pair {wi, v j ] for
161
162
ChaDter 6
some vi, 1-3 in V . Note that a simple graph G has no loops, that is, we are requiring vi # vj for all edges (vi,Vj} of G. 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. Let H and G be two graphs, it is said that H is a subgraph of G if V ( H ) c V ( G ) and E ( H ) c E(G). A spanning subgraph is a subgraph H of G containing all the vertices of G. Let G be a graph on the vertex setV . If z = {vi,Vi} is an edge of G one says that the vertices vi and vj are adjacent or connected by z , in this case it is also usual to say that the edge z is incident with the vertex vi. The degree of a vertex v in V, denoted by deg(v), is the number of edges incident with v. A vertex of degree one (resp. degree zero) is called an end vertex (resp. isolated vertex). If all the vertices of G are isolated, G is called a discrete graph. A walk of length n in G is an alternating sequence of vertices and edges
where zi = {vi-l, vi} is the edge joining Vi-1 and vi. A walk may also be written {vo,. . . , v,} with the edges understood, or z1,z2,. . . ,xn with the vertices understood. If vo = vn, the walk w is called a closed walk. A path is a walk with all its vertices distinct. and v2 there We say that G is connected if for every pair of vertices is a path from 211 to 212. Note that G has a vertex disjoint decomposition
where G I , . . . ,G, are the maximal (with respect to inclusion) connected subgraphs of G, the Gi are called the connected components of G. If x is a edge, we denote by G \ { z } the spanning subgraphof G obtained by deleting z and keeping all the vertices of G. The removal of a vertex v from a graph G results in that subgraph G \ {v) of G consisting of all the vertices in G except v and all the edges not incident with v. For instance:
If A is a set of vertices of G one can define G \ A recursively by removing one vertex at a time.
Edge
Ideals 163
A cycle of length n is a closed walk (vo,. . . ,vn} in which n 2 3 and the vertices 211,. . . ,v n are distinct. 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. A tree is a connected graph without cycles, and a forest is an acyclic graph. The complete graph I C , has every pair of its n vertices adjacent. For instance:
A graph G is bipartite if its vertex set V can be partitioned into disjoint subsets VI and V2 such that every edge of G joins VI with V,. 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) = 00. Proposition 6.1.1 A graph G is bipartite if and only if all the cycles of G are even. Proof. +) Let VI and V2 be a partition of V ( G )such that every edge of G joins 6with fi. If (VO, . . . ,v n 1 is a cycle of G, one may assume vo E V I . Note E V2, it follows at once that vi is in VI if and only if i is even, thus n must be even. e)It suffices to prove that each connected component of G is bipartite, thus one may assume G connected. Pick a vertex vo E V . Set
& = {v E V(G)ld(v,VO) is even} and fi = V ( G )\ V I . It follows that no two vertices of are adjacent for i = 1,2, otherwise G would contain an odd cycle. Therefore G is bipartite. 0 Corollary 6.1.2 If T is a forest, then T is a bipartite graph. Definition 6.1.3 A set of edges in a graph G is independent if no two of them have a vertex in common. Let G be a graph on the vertex set V ( G ) . Given a subset U c V ( G ) , the neighbor set of U , denote NG(U)or simply N ( U ) ,is defined as
N ( U ) = {v E V ( G )1 v is adjacent to some vertex in U). Notice that if C,. is a cycle, then C,.
c N(C,).
164
Chapter 6
Proposition 6.1.4 Let G be a bipartite graph with vertex sets V I ,V2 having m and n vertices respectively. If
for all A c VI, 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 V I )thus there is one independent edge. Assume m 2 2. Case (I): First assume IAI < IN(A)( for all A c & with IAI < m. Let x1 E VI and y1 E V2 be two adjacent vertices. Consider the subgraph
obtained from G by removing $1 and y1. Note that for every A c Vl \ {x1) one has NH(A) = N ( A ) if y1 4 N ( A ) and N H ( A )= N ( A ) \ (91) otherwise. Hence by induction there are 22,. . . xm independent edges in H , which together with z1 = { X I ,31) yield the required number of independent edges. Case (11): Nextassume IAI = IN(A)( for some A c with 1AI < m. Set r = 1AI. Consider the subgraph )
H = G \ (VI \ A ) . Since N H ( S )= N ( S ) for all S c A, by induction there are independent edges 21,. . . , x, in H . On the other hand consider F = G \ ( A U N ( A ) ) . If S c VI \ A, then
since
N ( S uA ) = N ( A )U N F ( S ) and r = (N(A)I one derives (NF(S)J,> IS[. Applying induction there are w1), . , wm+ independent edges in F . Altogether it follows that there are m independent edges in G, as required. This proof is due to Halmos and 0 Vaughn. )
Corollary 6.1.5 Let G be a bipartite graph with vertex sets V I ,V2. If there is an integer p 2 0 such that IA( - p 5 IN(A)I for all A c K , then there are m - p independent edges in G, where m = I& I. Proof. By Proposition 6.1.4 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 VI to y. Let A c V I ,since NF( A ) = N ( A ) U { y } one concludes INF(A)I 2 IAI - ( p - 1). Hence by induction there arern -p+ 1independent edgesin F , it follows that there are m - p independent edges in G. 0
Edge Ideals
165 ,
e
Let G be a graph with vertex set V . A subset A c V is a minimal vertex cover for G if (i) every edge of G is incident with one vertex in A, and (ii) there is no proper subset of A with the first property. If A satisfies condition (i) only, then A is called a vertex cover of G and A is said to cover all the lines of G.
It is convenient to regard the empty setas a minimal vertex cover for a graph with all its vertices isolated. A set of vertices in G is independent if no two of them are adjacent. Notice that a set in G is a maximal independent setiff its complement is a minimal vertex cover for G. The vertex covering number of G, ao(G),is the smallest number of vertices in any minimal vertex cover. Definition 6.1.6 The independencenumber of a graph G, denoted by the maximum number of independent edges in G.
p 1 ( G ) ,is
Theorem 6.1.7 (Kiinig) If G is a bipartite graph, then ,&(G) = ao(G). Proof. Let V = VI U V2 be a partition of the vertex set of G such that every edge of G joins VI with Vz. Set r = ,&(G) and m = IV1 I. Since there are r independent edges in G one obtains ao(G) 2 r. Note there is A c VI such that IN(A)I 5 IAI - ( m - r ) , otherwise if IN(A)I > IAI - (m - r ) for all A c V I ,then by Corollary 6.1.5 one would have r + 1 independent edges which is impossible. Therefore N ( A )U (VI\A) is a vertex cover of G with at most r elements, thus ao(G) 5 r. 0
A pairing off of all the vertices of a graph G is called a perfect matching. Thus G has a perfect matching if and only if G has an even number of vertices and there is a set of independent lines "covering" (containing) all the vertices. Thereare severalresults equivalent to Konig theorem (see [28, 139, 223]), next we present its connection with perfect matchings; see Proposition 8.8.11 for a generalization of the marriage problem. Theorem 6.1.8 (Marriage problem) If G is a bipartite graph with vertex set V, then the following are equivalent (a) G has a perfect matching. (b) IAI 5 IN(A)(for all A c V independent set of vertices.
166
Chapter6
Proof. Let VI, V2 be a partition of V such that every edge of G joins VI with h.Set r = pl(G). It is clear that (a) implies (b). (b) 3 (a): Since 6 is an independent set for i = 1,2, one readily derives IV1 I = IV2 I. By Konig theorem there arez1, . . . ,zr independent edges and a minimal vertex cover A of G with r elements, hence xi n A has exactly one vertex for any i and A is an independent set. Using that also V \ A is an independent set of vertices and the equality N ( V \ A ) = A we conclude
IV \ AI I IN(V \ All = I It follows that IV1 I = r , thus
21,.
4
. . , xT yields a perfect
matching.
0
The cover complexity Given a graph G , the family of its minimal vertex covers will be denoted by T ( G ) . One may call the integer IT(G)l the cover complexity of the graph G. We are interested in algebraic ways of grasping this important number. The best known estimate is due to E. Sperner (cf. [112, Theorem 3.11): Theorem 6.1.9 ([265]) If G is a graph with n vertices, then
Proposition 6.1.10 Let
= sup (IY(G)(I G has n vertices)
.
If {Fn) denotes the Fibonacci sequence given by Fo = F1 = 1 and Fn = Fn-1 then Tn
+ Fn-2,
5 Fn for n 2 1.
Proof. The proof is by induction on n. The required inequality is clearly satisfied if n I 4. Assume yk 5 Fk for k < n and n 2 5 . Let G be a graph with n vertices. We write G = G1 uG2 where G1 is a connected component of G with at least two vertices. Let x be a vertex in G1 and 2 1 , . . . ,xT the vertices of G adjacent to x. Let Y l be the set of minimal covers of G that contain x and "2 = 'Y ( G ) \ Y l . If A E Y l notice that A \ { p ) is a (possible empty) minimal vertex cover for G \ { x } , hence I'Y1 I 5 Fn-1. If A E Y 2 , then A \ {XI,. . . ,X,,} is a (possible empty) minimal vertex cover , implies IT2 I 5 Fn-r-l. Therefore yn ,< Fn as for G \ { x ,2 1 , . . . ,x ~ }which required. 0
Edge Ideals
167
Proof. To show it note: (i) The minimal vertex covers of Pn not containing x1 are in one to one correspondence with the minimal vertex covers of the graph Pn \ {x1,x2}) and (ii) the minimal vertex covers of Pn containing 2 1 are in one to one correspondence with the minimal vertex covers of the 0 graph pn \ {xl,X2)x3). Proposition 6.1.12 If T is a forest, then IT(T)(2 ao(T)+ 1. Proof. The proof is by induction on the number of vertices. Assume the inequality for all forests with less than n vertices and assume T is a forest with n vertices. Let x be a vertex of degree one in T and x the vertex in T adjacent to z. Set T, = T \ { z ) . Let A E T(T,). If z E A one has A E T(T),and if x 4 A, then A U { z } E T(T). Hence IT(T)I 2 IY(Tz)I. If cro(T)= ao(T,) the inequality follows by induction hypothesis. If
then c r O ( T ' ' ) = ao(T)- 1. Therefore there is a minimal vertex cover A1 of T so that lAll = ao(T) and z E A I . Notice that
is not a minimal vertex cover for T, because A1 \ (2) is a minimal cover for T,.As a consequence A2 is a cover of T which is not a lift of a cover of Tz. The required inequality nowfollows using theinductionhypothesis.
0
Some types of graphs For convenience let usrecall the notion of CohenMacaulay standard algebra. Let R be a polynomial ring over a field k with irrelevant maximal ideal m and I a graded ideal of R. The depth of R / I , denoted by depth R / I , is the largestinteger r so that thereis a homogeneous sequence f 1 ) . . . ,f,. in m with fi not a zero-divisor of R / ( I ,f1) . . . , fi-1) for all 1 5 i 5 r , here we set fo = 0. This number r is bounded by dim R / I according Lemma 1.3.6. Definition 6.1.13 The quotient ring R / I is called Cohen-Macaulay (C-M for short) if depth(R/I) = dim(R/I), and the ideal I is said to be Cohen-Macaulay if R / I is Cohen-Macaulay. Let G be a graph with vertices V I , .. . ,v n and
a polynomial ring over a field IC, with one variable xi for each vertex vi, we will often identify the vertex vi with the variable xi.
168
Chapter6
Definition 6.1.14 The edgeideal I(G) associated to the graph G is the ideal of R generated by the set of square-free monomials x p j such that V i is adjacent to vj, that is,
If all the vertices of G are isolated we set I ( G ) = (0). Note that thenon zero edgeideals are precisely the ideals of R generated by square-free monomials of degree two. Definition 6.1.15 The graph G is said to be Cohen-Mucaulay over the field k (C-M graph for short) if R / I ( G ) is a Cohen-Macaulay ring. According to Theorem 6.4.7 the Cohen-Macaulay property of a bipartite graph is independent of the field k, but in general this property may depend on the field k (see Example 5.3.31). The next resultestablishes a one to one correspondence betweenthe minimal vertex coversof a graph and the minimal primes of the corresponding edge ideal. Proposition 6.1.16 Let R = k [ x l , .. . ,xn] be a polynomial ring over a field k and G a graph with vertices X I ,. . . xn. If p is an ideal of R generated b y A = ( x i 1 , .. . ,xi,.}, then p is a minimal prime of I ( G ) if and only if A is a minimal vertex cover of G. )
Proof. Note that I ( G ) c p if and only if A is a vertex cover of G. Assume that A is a minimal vertex cover of G, hence I ( G ) c p. By Proposition 5.1.3 any minimal primeof I ( G ) is a face ideal, thus p is a minimal prime of I ( G ) . The converse is clear. 0 Definition 6.1.17 A graph G is said to be an unmixed graph if any two minimal vertex covers of G have the same cardinality. Corollary 6.1.18 If G is a graph and I ( G ) its edge ideal, then the vertex covering number ao(G) is equal to the height of the ideal I(G). Proof. It follows at Proposition from once
6.1.16.
0
Definition 6.1.19 An edge x of a graph G is critical if
If all the edges in G are critical G is called edge-critical . A vertex v of G is critical if
ao(G \ V ) < ao(G), and G is called vertex-critical if all its vertices are critical.
serted
Edge Ideals
169
Proposition 6.1.20 Let G be agraph. If ao(G \ v) vertex v of G , then ao(G \ v) = aO(G)- 1.
< ao(G) for some
+
Proof. Set r = ao(G \ v) and note r 1 5 ao(G). Pick a minimal vertex cover A of G \ v with r vertices. Since v cannot be an isolatedvertex A U {v) is a minimal vertex cover of G. Hence ao(G) 5 T- 1 and one has 0 the equality.
+
If G is a graph without isolated vertices, then the relationships between the various notions defined above are the following: C-NI =+ unmixed + ,vertex-critical e= edge-critical, where none of the arrows can bereversed in general. We now prove the first two implications and leave the third implication as an exercise. Proposition 6.1.21 If G is a Cohen-Macaulay graph, then G is unmixed. Proof. By Corollary 5.1.5 I ( G ) is the intersection of its minimal primes. Therefore I ( G ) is an unmixed ideal by Corollary 2.2.4, thus G is unmixed by the correspondence between minimal covers and minimal primes. 0 Proposition 6.1.22 If G is an unmixed graph without isolated vertices, then G is vertex-critical. Proof. Let v E V ( G ) .Since G is unmixed there is a minimal vertex cover A of G containing v and such that IAI = ao(G). As A \ {v) is a minimal vertex cover of G \ v, one has ao(G \ v) < aO(G). 0 Definition 6.1.23 A partially ordered set or poset is a pair P = (V,L), where V is a finite set of vertices and 5 is a binary relation on V satisfying: (a) u 5 u , Vu E V (reflexivity). (b) u 5 v and v 5 u , imply u = v (antisymmetry). (c) u
5 v and 21 5 w,imply u 5 w (transitivity).
A poset P = (V,5 ) with vertex set V = {XI,. . . ,x,) can be displayed by an inclusion diagram, as with the diagram
.8 4
3
2
of the divisors of 12. In such a diagram xi is joined to if Xi < Xj and there is no other vertex in between.
xj
by raising a line
170
ChaDter 6
Given a poset P with vertex set V = (21.. . . ,xn} its order complex, denoted by A(P), is the simplicial complex on V whose faces are the chains (linearly ordered sets) in P , thus
+
k[A(P)] = k[V]/(xixjlxi xi) is the Stanley-Reisnerring of A(P). Here xi x.j means that xi is not comparable to xi. The Cohen-Macaulay property of those rings associated to posets has beenextensively studied, see [12, 26,44, 123, 162, 2741 and the references there. Those simplicial complexes that occur as order complexes have been nicely characterized by Stanley [270], see also [162].
+
Example 6.1.24 Let P = (V,5 ) be a partially ordered set with vertex set V = { X I ,. . . ,.a} ordered according to the next inclusion diagram. The Stanley-Reisner ring of the order complex A(P) is k[V]/(x6x7, 22x3).
x3+x2
Simplicialcomplexes of edge ideals Given a graph on thevertex set V one defines a simplicial complex Simp(G) on the same set of vertices: Simp@) = {FI 3 H a subgraph of
such that
F = V ( H ) and H
2
X,.},
where K , denotes a complete graph on r vertices. Conversely a simplicial complex A defines a graph Skel(A), the 1-skeleton of A, on the same set of vertices: Skel(A) = { F E AI dim(F) 5 1). In general one has A c Simp(Skel(A)). Proposition 6.1.25 Let A be asimplicialcomplexand Reisnerideal.Then A = Simp(Skel(A)) if andonly b y square-free monomials of degree two.
IA its Stanleyif I A isgenerated
Proof. Let V = { X I ,. . . ,x,) be the vertex set of A . Assume I A is generated by square-free monomials of degree two. Take F E Simp(Skel(A)), for simplicity set F = {XI,.. . ,x,.}, that is, (xi,xj} for all 1 5 i < j 5 r. If F A, then x1 - x? is in I A and by assumption xixj is in I A for some 1 5 i < j 5 r. Hence {xi,xj} @ A, whichis a contradiction. This shows A = Simp(Skel(A)). The converse also follows rapidly. 0 )
Simplicial complexes associated to edge ideals have been studied in the literature, they are called flagcomplexes, see [274, Chapter 1111 and the references there for a series of amusing related problems.
Edge
Ideals
Exercises 6.1.26 Prove that
c.
= Skel(Simp(c)), for anygraph
6.1.27 Let G be a graph with vertex set V = identity:
. . , x n } , prove Euler’s
(~1,.
n
i= 1
6.1.28 (P. Hall) Let X be a finite set and SI,. . . , S, a collection of subsets of X. If
for all i l , . . . ,ik distinct, prove that there are 2 1 , . that xi E Si for all i.
,
. ,x,
in X distinct such
Hint Consider the bipartite graph with vertex sets VI = {SI,. . . ,Sm} and Vz = X , where x E X is adjacent to Sj if x E Sj.. 6.1.29 Let H be a subgroup of G of index n and G I H (resp. ( G / H ) L )the set of right (resp. left) cosets. Then there are 91,. . . ,gn in G such that
Hint If G / H = { H I ,. . . , Hn 1 and (G/ H ) L = { H i , . . . , HA}, define Si as the set of j such that Hi n Hj # 8 and write
where B, is equal to T
s= 1
with
il,
. . . ,i,
distinct. Note that at least r of the B, n Hl are non empty.
6.1.30 If G is a connected graph and d the distance function, then:
6.1.31 Let G be a graph. If CYO(G\X) < aO(G)for some edge x of G , prove that ao(G \x) = ao(G) - 1.
172
Chapter 6
6.1.32 If G is an edge-critical graph without isolated vertices, prove that G is vertex-critical. 6.1.33 In [234] there is a list of all the known edge-critical graphs with fewer than nine vertices, prove that all these graphs are unmixed. Can you give an example of an edge-critical graph which is not unmixed?.
6.2
Cohen-Macaulay graphs
Let G be a graph on the vertex set V = { X I , .. . ,z n } . We define the complementary simplicial complex AG of a graph G by
AG = { A c VI A is an independent set in G}, notice that AG is precisely the Stanley-Reisner simplicial complex of I(G). According to Reisner’s Theorem 5.3.5 the Cohen-Macaulay property of a given graph can be decided oncethe reduced homology of the links of AG is computed. Unfortunately the simpler the graph G the more complicated the complex AG. Following [262, 3023 we will study C-M graphs by looking at both the complementary simplicial complex and the graph itself.
Construction of Cohen-Macaulay graphs One of the purposes hereis to show how large classes of C-M graphs can be produced and to show some obstructions for a graph to be C-M.
A
First construction Let H be a graph with vertex set V(H)= {XI,. . . ,zn,x,w) and J its edge ideal. Assume that x is adjacent to w with de&) 2 2 and deg(w) = 1. We label the vertices of H such that X1 22 x k 21,.. . , z k , w arethe vertices of H adjacent to x, as shown in thefigure. The next two resultsdescribe how the Cohen-Macaulay property of H relates to that of the two subgraphs G = H \ ( x , w} and F = G \ {z1,. . . ,xk}. One has the equalities
J = (I,z~z,. . .,z k z , zw) and ( I ,X I , .. . ,xk) = (L,z1. . . , z ~ ) , 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 x is not isolated, there is a minimal prime p over I containing (21,., . ,zk} and such that ht(1) = ht(p) = 9. It is not difficult to prove that k < n and deg(zi) 2 2 for i = 1,.. . ,IC.
+
Proposition 6.2.1 If H is a C-M graph, then F and G are C-M graphs.
Edge Ideals
173
Proof. Set A = k[zl,. . . ,xn]and R = A[z,w].By Proposition 9.3.2 there exists a homogeneous system of parameters {fi,. . . , fd} for A / I , where fi E A+ for all i. Because of the hypothesis and the equalities
z ( z - w)+ zw = z2 and w(w - z j + zw = w 2 , the set { fi, . . . , fd, z - w} is a regular system of parameters for R/ J . Hence f l , . . . f d is a regular sequence on R / I , that is, G is C-M. To show the corresponding property for F we consider the sequence
where the first map is multiplication by z and $ is induced by a projection. The exactness of this sequence follows from Corollary 5.1.5. Taking depths w .r.t the maximal ideal of R, by the depth lemma one has n -9
+ 1 5 depth R / ( 1 7 x 1 , .. . , x k , w ) , where g = ht(1).
Since ( I ,q , . . . ,xk, w)= (L,x1,. . . , x ~ , w )F,is C-M.
0
Proposition 6.2.2 If F and G are C-M graphs and { X I , . . . ,xk} form a part of a minimal vertex cover for G, then H is a C-M graph.
graph.
Proof. Consider the exact sequence of Proposition 6.2.1. Since the ends of this sequence have R-depth equal to dim(R/J), by the depth lemma H is a Cohen-Macaulay 0 Corollary 6.2.3 If G is C-M and for G, then H is C-M.
{XI,.
. . , xk}
is a minimal vertex cover
Proof. Note that in this case F is Cohen-Macaulay because I ( F ) = (0). 0 Second construction For the discussion of the secH be ond construction we change our notation. Let a graph on the vertex set V ( H ) = { X I , . . . ,xn,z}. Let { X I , . . . , X I ; } be the vertices of H adjacent to z , as shown in the figure. Taking into account the first construction one may assume deg(xi) 2 2 for i = 1,.. . ,k and deg(z) 2 2. Setting G = H \ { x } and F = G \ {XI,. . . ,x k } , notice that the ideals J, I and L associated to H , G and F respectively are related by the equalities J = ( I , x l z , . . .xkz) and ( I , x 1 ,. . . , x k ) = ( L , x 1 , .. . , x k ) . X1
x2
xk
Proposition 6.2.4 If H is C-M, then F is C-M.
+
Proof. We set R = k[zl,. . . ,x n , 23, A = k [ x l ?. . . ,X,] and ht(J) = g 1. The polynomial f = z - x1 - * - xk is regularon R / J because it is
174
Chapter 6
clearly not contained in any associated prime of J . Therefore there is a sequence (f, f1, . . . ,fm) regular on R / J so that { f l , . . . , fm) c A + , where rn = n - g - 1. Observe that {fl,. . . , fm} is in fact a regular sequence on A/I, which gives depth(A/I) 2 n - g - 1. Next, we use the exact sequence
. . . ,xk) = g + 1 to conclude that F is C-M.
andht(I,x1,
0
Proposition 6.2.5 Assume X I , .. . ,xk do not form a part of a minimal vertex cover for G and ht(I,x1,. . . ,z k ) = ht(I) 1. If F and G are C-M, then H is C-M.
+
Proof. The assumption on { q ,. . . ,xk) forces ht(J) = ht(I) exact sequence
+ 1. From the
H is C-M.
that we obtain
0
Corollary 6.2.6 If G is C-M and for G, then H is C-M.
{XI,.
. . ,xb-1)
is a minimal vertex cover
Connected components of Cohen-Macaulay graphs A goodproperty of Cohen-Macaulay graphs is its additivity with respect to connected components. Lemma 6.2.7 Let R1 = IC[x] and R2 = k[y] be two polynomial rings over a field IC and R = k [ x ,y]. If I1 and I 2 aregradedideals in R1 and Rz respectively, then depth(RI/Il)
+ depth(&/&)
= depth R / ( h
+
12).
Proof. We give two proofs. First proof the formula is a direct consequence of Proposition 2.2.20 and Theorem 2.2.21. Second proof by a change of base ( R ~ / I ~ ) @ ft: ~R I R/I2R is a flat R1-module. The equality now follows from a general property of tensorproducts [217, Theorem 501. 0 Proposition 6.2.8 If G is a graphand G I , .. . ,G, its connected components, then G is C-M if and only if Gi is C-M for all i. Proof. result The
follows Lemma from
6.2.7.
0
Edge
Ideals
Complementary complex versus the complement
c
The complement of a graph G has vertex set V ( G )and two vertices are in G. Along the chapterwe adjacent in if and only if they are not adjacent will encounter the complement E, and witness its usefulness to encapsulate information of the edge ideal I ( G ) .
c
Proposition 6.2.9 Let G be a graph and AG its complementary simplicial complex.Then c A c , withequality if andonly if has no triangles.
c,
Proof. If z = {vl ,Q } is an edge of then { V I , v2} is an independent c AG. Assume hasnotriangles. If set of G and z E AG. Hence A = {VI ,. . . ,vn} is an independent set of G, then n = 1 or n = 2; otherwise if n > 2, then would contain thetriangle {wl,v2,213). Therefore A is either a pointor an edge of G, thus = AG. The converse also follows rapidly. 0 In general a Cohen-Macaulay graph G can have all the subgraphsG\ {v} failing to be C-M, for w in G; see Example 6.2.24. Under special circumstances we will study the existence and distribution of the vertices of G whose removal preserve the C-M property, the case of bipartite graphs is analyzed in Section 6.4. Definition 6.2.10 A cutpoint of a graph is a vertex whose removal increases the number of connected components. Proposition 6.2.11 If u, w are two vertices at maximum distance in a n o n discrete connected graph G , then u,v are not cutpoints. Proof. Assume 21 is a cutpoint and write G \ {w) = G1 U - U G,, where GI, . . . ,G, are the connected components of G \ {v}. One may assume u E G1, since s 2 2 there is w E Gi for some i 2 2. Note that v is in any path joining u and w,thus d(u,w)> d(u,w), a contradiction. This 0 argument is due to Harary. e
Definition 6.2.12 A set of vertices D of a graph G is called dominant if every vertex of G is adjacent with at least one vertex in D. Lemma 6.2.13 Let G be a connectedgraph. If
is connected and has no
triangles, then
D = {w E V(@l is a dominant set of vertices of G.
\ {v} is connected}
Chapter 6
176
Proof. Assume E \ {w} is disconnected for some w E V(G). We set
where 2 1 , . . . ,xk are the vertices of G adjacent to v. Since has no triangles d ( x i ,xj) = 2 for k 1 5 i < j 5 r n , where d is the distance in Notice that at least one of the vertices ( ~ k + l , .. . ,x,} has degree greater or equal than two in G, say xm. We order the vertex set so that xk and x, are adjacent in Let G I ,. . . ,Ge be the connected components of \ {IJ}. We may also assume that zk and x m are in V(G1).Notice that d ( ~ kb,) 2 3 for all b in V(G2). Using Proposition 6.2.11 one has that any pair of vertices at maximum distance in must be in D. Therefore \ {xi} is connected for some 1 5 i 5 IC, that is, xi is a vertex in D adjacent to IJ in G.
c.
+
c.
c
If
Proposition 6.2.14 Let G be a connected graph and its complement. -
G is connected and has no triangles, then
D=
{ z E V(G)I G\ { z >
is C - M )
is a dominant set.
Proof. Let-AG be the complementary simplicial complex of G. Observe that AG = G, hence by Reisner’s Theorem G is C-M. The rest of the proof immediate is consequence an of Lemma 6.2.13. 0
Exercises 6.2.15 Show that a cycle n
Cn of length n 2 3 is unmixed if and only if
= 3,4,5,7.
5 is connected and has no triangles. Show that I ( G ) is Gorenstein if and only if is a cycle.
6.2.16 Let G be a graph such that
c
Hint It follows from [274, Theorem 5.11 that I ( G ) is Gorenstein iff is a pseudo manifold. Since is a graph it is a pseudo manifold iff it is a cycle.
6.2.17 ([309]) Let GI, G2 be two graphs with vertex set x. If their complements are triangle free and they have the same number of edges, then the Hilbert series of k[x]/I(GI) and k[x]/I(Ga) are identical. 6.2.18 Let G be a graph on the vertex set V = ht(I(G)) = p - 2. Prove the following:
{ X I ,.
. . ,xP}
such that
(a) G is Cohen-Macaulay if and only if AG is connected, and (b) if I ( G ) is not Cohen-Macaulay, then I ( G ) is unmixed if and only if AG has no isolated vertices.
Edge
Ideals 177
6.2.19 If G is a C-M graph with q edges, prove that
where g is the height of I ( G ) . 6.2.20 A graph G with q edges is said to be suturutedif q = g(g+1)/2, where
g is the height of I ( G ) , and G is C-M. Let G be a graph with connected components G I , . . . ,G,. Prove that G is saturated if and only if Gi is saturated for all i. 6.2.21 If G is a saturated graph, prove that at most one of its connected
components has more than one vertex. 6.2.22 Let G be a Cohen-Macaulay graph over a field k . Prove that G is
saturated if and only if R / I ( G ) has a 2-linear resolution. 6.2.23 Prove that the following graph is Cohen-Macaulay.
6.2.24 Prove that the ideal I = I A of Example 5.3.31 is generated by
Let Ri = XI,. . . , x i , , . . , and Gi = G \ { x i } . Note I = I ( G ) for some graph G , Prove that the f-vector of A c i is (10,25,15) or (10,24,14). Thus the h-vector of R i / I ( G i ) is (1,7,8,-1) or (1,7,7, -1). Use Theorem 4.2.6 to conclude that Gi is not a’C-M graph for all i. /c
6.2.25 If G is a Cohen-Macaulay graph andX I , x2 are two vertices of degree
1, prove that they cannot have an adjacent vertex in common. 6.2.26 Let G be a connected graph and (21,. . . ,zn} a path with n 2 4. If deg(z1) = 1, deg(xi) = 2 for i = 2,. . . n - 1 and deg(z,) 2 3, prove that G
cannot be unmixed.
178
6.3
Chapter 6
Trees
In this section we give an effective description of Cohen-Macaulaytrees and show an interesting family of graphs containing all Cohen-Macaulay trees. The first Koszul homology module of a Cohen-Macaulay forest is also studied. Proposition 6.3.1 Let S be a positively graded algebra of dimension d over a field k . If A = k [ h l ,. . . ,hd] C ) 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.
+
Proof. By Theorem 2.5.13 pdA(S) depth(S) = dim(A), where the depth of S is taken withrespectto 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. Converselyassume S Cohen-Macaulay.Note that A is a polynomial ring. Let p E Spec S be a minimal (graded) prime over A+S. There is an integral extension k = A/A+ L) S/p, hence S / p is a zero dimensional domain and consequently p is a maximal ideal. Therefore rad (A+S)= S+, that is, hl, . . .,h d is a h.s.0.p for S. Now use Proposition 2.2.7 to conclude that hl, . . . , 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. 0 Next we apply the above characterization of Cohen-Macaulay graded algebras to give a proof, due to Wolmer Vasconcelos, of the following key result: Proposition 6.3.2 Let R = k [ x l ,. . . ,x,, y1, . . . ,yn] be a polynomial ring over a field k and M c ((i, j )I 1 5 i < j 5 n). Then the ideal
is a Cohen-Macaulay ideal. Proof. We start by doing a change of variables xi = xi - yi for i = 1,. . . ,n. Its effect is that A=k[zl,...,zn,yl,...,y,]/l becomes an integral graded algebra over k [ z l ,. . . , z n ] , where
Edge
Ideals 179
Observe that the images of yil ' yi, (1 5 i l < . < i,. 5 n ) not divisible by any of the monomials yjye form a generating set for the k [ z l , . . . ,zn]module A. To complete the proof we now show that they are in fact a basis. We set z = (21, . . . , xn) and y = (31,.. . ,yn). Assume the following equality e
= 0. First Using induction we willprove that fi = filiz = * * = fil.,,i, we show fi = 0 for 1 5 i 5 n. Making y j = 0 for j # i in the equation above we obtain f f ( x ) = gi(yi, x) (pi zz), which forces fi = 0. We set the next induction step by assuming fi = filiz = . . . - fil..,i,.-l = 0. The substitutions y j = 0 for y j 4 {pil,. . . ,yi, } and yip = -zip for 1 < p 5 r into the first equality gives
+
therefore fil,.,i, (x) '= 0. Altogether A is a free module over k [ z l , ., . ,xn] and this implies that I is C-M by Proposition 6.3.1. Let us give a quite different second proof pointed out by Jurgen Herzog. Note that the polarization of the ideal
J = ({x:,xjxe(1 5 i 5 n and ( j , C ) E M}) is equal to I andProposition apply
5.1.7.
0
Lemma 6.3.3 Let T be atree. If v and w are twoadjacentvertices of degree at least two, then there is a minimal vertex cover for T containing both v and w.
for
Proof. Let x be the edge joining v and w. Since T \ {x) has exactly two components, say TI and T2, there are minimal vertex covers A and B for 1'2 and 7" respectively so that v E A and w E B. Therefore A U B is the T. 0 required cover Theorem 6.3.4 Let T be a tree with vertex set V and edge set E . T h e n T is C-M if and only if IVl 5 2 or 2 < IV) = 2r and there are vertices x1,.. . ,x,.,y1,. . . , y,. so that deg(zi) = 1, deg(yi) 2 2, and {xi,yi} E E f o r i = 1,...,r. Proof. *) As T is a bipartite graph there are disjoint sets VI and V2 such that V = VI U V2 and every edge of T joins VI with V2. Since VI and V2 are minimal vertex covers for T , and T is unmixed, we have
Chapter 6
180
By Theorem 6.1.7 there are r independent edges. Therefore we may assume VI = (21,. . . ,z,.), V2 = (w1,. . . ,w,.}, and {zi,wi} E E for i = 1 , . . . , r . To complete the proofis enough to show that for each i either x i or wi has degree one. Assume deg(zi) 2 2 anddeg(wi) 2 2 for some i. By Lemma 6.3.3 there is a minimal vertex cover for T,say A , containing zi and wi. Since { z j , wj } A # 0 for j # i we conclude IAI 2 r + 1, which is a contradiction. +=) The sufficiency follows from Proposition 6.3.2. 0
n
Corollary 6.3.5 If G is a tree, then G is C-M if and only if G is unmixed. Proof. If G is unmixed, using the proof above it follows that I ( G ) is a C-M ideal of kind the described in Proposition 6.3.2. 0 Corollary 6.3.6 The only C-M cycles are a triangle and a pentagon. Proof. Let Cn = {xo,2 1 , . . . ,Xn = ZO} be a C-M cycle of length n. The cases n = 4, 6, and 7 can be treated separately, so we assume n 2 8. By Proposition 6.2.4 Cn \ { x o , x ~ , ~ - is1 a) C-M path of length n - 3 which is impossible by Theorem 6.3.4 0 Proposition 6.3.7 If I = I ( T ) c R is the edge ideal of a Cohen-Macaulay forest with 2n vertices and without isolated vertices, then depth(R/12) 2 n - 1. Proof. By Theorem 6.3.4 there are vertices 2 1 , . . . ,x,, y1,. . . ,y , with deg(zi) = 1, deg(yi) 2 1 for all i, such that I has a generating set:
Set R = k[x,y], where k is a field. We proceed by induction on n. If I is a complete intersection or 1 5 n 5 2 the result follows from Theorem 3.3.14 together with the exact sequence 0 + 1/12_I) R/12
-+R / I + 0.
Notice that if 1is not a complete intersection, thenT has a vertex of degree two, its existence follows from Theorem 6.3.4 and the fact the sum of the degrees of the vertices of a graph is twice the number of its edges. We order the variables so that de&,) = 2 and Yn-19, E I and assume the result true for CM forest with less than 2n vertices. The ideal of R obtained by evaluation of I at y , = 0 will be denoted by 3, while L will denote the ideal in R obtained by evaluation of J at yn-l = 0. We also set K = ( J 2 , y n - 1 y n , ~ n y n ) .
Edge Ideals
181
In the sequel we shall also make use of the fact that, if I is a monomial ideal, and f is a monomial, then ( I :f ) is a monomial ideal. Consider the complex (easily shown to be exact by the preceding remark):
By an appropriate induction hypothesis
Hence from (1) depth(R/(J2,y,-1y,)) exact sequence
>, n - 1, which together with the
gives depth(R/K) 2 n - 1. Because of the equality K = (I2,yn-lyn, zy,), we shift our attention to the exact sequence
To estimate the depth of the module on the left, we use the exact sequence
Taking into account that L and J are attached toCM forests, we have that the left end of the last sequence has depth equal to n 1 while its right end has depth equal to n. Hence a combination of (3) and (4) gives
+
depth(R/(12,x,yn)) 2 n - 1. To complete the proof observe that from the sequence
we get depth(R/12) 2 n - 1.
0
Exercises 6.3.8 Let T be a tree with 2r vertices. If r 2 2, prove that T is CohenMacaulay iff ht I ( T ) = P and T has exactly lr vertices of degree 1. 6.3.9 Let G be a graph with vertex set V = ( 2 1 , . . . ,x,, 31,. . . ,y), and e d g e s e t X = ( { z k , y k ) , { y i , y j ) I I c = l , ...,n a n d l i i < j < n ) . Showthat G is a saturated graph over any field k.
182
6.4
Chapter 6
Bipartite graphs
Let G be a Cohen-Macaulay bipartite graph and I = I ( G ) its edge ideal. We prove that G \ {v} is Cohen-Macaulay for some vertex v in G. Then we show that the Stanley-Reisner simplicial complex of I is shellable. Cohen-Macaulay Bipartite graphs A graph G is a complete bipartite graph if its vertex set V can be partitioned into disjoint subsets VI and V2 such that G contains every edge joining VI and V2. If VI and V2 have rn and n vertices respectively, we denote such a complete bipartite graph by Km,n. In the following illustration VI (resp. V2) is the set of vertices in the top (resp. bottom) row.
K3,3
Lemma 6.4.1 The complete bipartite graph and only if m = n = 1.
ICm,n
is Cohen-Macaulay if
Proof. By Exercise 5.3.22 it is enough to observe that the complementary complex of ICm,n is disconnected for m n >_ 3. 0
+
Lemma 6.4.2 Let G be an unmixed bipartite graph and I ( G ) its edge ideal. If I ( G ) has height g , then there are disjoint sets of vertices VI = { X I , . . . ,xg} and V2 = (91 , . . . ,yg) such that: (i) {xi,yi} is an edge of G for all i, and (ii) every edge of G joins VI with Vz. Proof. The statement follows rapidly if one uses that g is equal to the maximumnumber of independentedges of G, see Theorem 6.1.7. 0 Definition 6.4.3 Let G be a graph and 21 E V ( G ) . The neighbor set N ( v ) of 21 is the set of vertices of G adjacent to v.
Theorem 6.4.4 Let G be a Cohen-Macaulay bipartite graph. If G is n o t a discrete graph, then there is a vertex v E V ( G ) such that deg(v) = 1. Proof. We may harmlessly assume that G has no isolated vertices. Set g = ht I ( G ) . Let VI and V2 be as in Lemma 6.4.2. We proceed by contradiction. Assumedeg(v) 2 2 for all v E V I . Let v E VI be a vertex of minimal degree. We may assume v = 2 1 . If deg(w) = 9,then G = X,,, is a complete bipartite graph, but this is impossible by Lemma 6.4.1. Therefore we may also assume 2 5 deg(x1) 5 g - 1. Set r = deg(z1). Let N ( x 1 ) be the set of vertices in V adjacent to q , say N(x1) = { y1,. . . ,yr}.
Edge Ideals
183 8.
,:
.
I
"
I
, A - . J >
, :.-
'7
We claim that deg(zi) = r , for all i = 1,. . . ,r. For simplicity we assume i = 2. If deg(x2) > r , then x2 is adjacent to y j for some r 1 5 j 5 g. Let G' = G \ { X I , y1,. . . ,y,.}. By Proposition 6.2.4 we derive that G' is C-M, and hence G' is unmixed. Notice that {yr+1, , . . , yg) is a minimal vertex cover for G' and there is also a minimal vertex cover for G' containing { x ~z,+1 , . . . ,zg},which is a contradiction since G' is unmixed. Thus deg(z2) = r. The previous argument also shows that N(xi) = { 91, , . . ,yr 1, for all 1 i 5 r. Hence IV(yi) is a subset of {z,.+l,.. . , x g } , for all r 1 5 i 5 g. Consider the graph
+
+
<
To complete the proof observe that a repeated use of Proposition 6.2.4shows that H is a C-M bipartite graph. This is impossible since H = KT,,.is a and graph bipartite complete r 2 2. 0 Corollary 6.4.5 If G is a Cohen-Macaulay bipartite graph, then is C-M for some vertex 2) in G.
G \ {v}
Proof. Let {w, z } be an edge of the graph G such that deg(v) = 1. Then 0 by Proposition 6.2.1 we obtain that G \ { z } is C-M.
The results above yield a recursive algorithm to construct all the C-M bipartite graphs with a given number of vertices, see exercises. Shellability The vertexsetand edge set of a graph G are denoted by V ( G ) and E(G) respectively. Recall that the faces of AG, the complementary simplicial complex of G, are precisely the independent sets of G (together with the empty set). Notice that the facets (faces of maximal dimension) of A c are the maximal independent sets of G. Let G be a graph on the vertex set V = (21, . . . , zn,x,y ) . Assume that z is adjacent to y and deg(y) = 1. We label G so that 'y,21,. . . ,z k are the vertices of G adjacent to x. Lemma 6.4.6 Let H = G \ {x, y) and A c V . If G is Cohen-Macaulay, then A is a facet of A, if and only if A can be written as
c V ( H ) is a facet
of A H ) or
u {x}, where F c V(H)is a facet
of AH and
(a) A = F U {y}, where F (b) A = F in F .
21,.
. . , z b are not
Proof. Let A be a facet of AG. Suppose x A, then y E A . Set W = V(H) and F = A \ {y] c 14'. We must show that F is a facet of AH. First notice that F is an independent set of H , because E ( H ) c E(G).Next let us show that F is not contained in any other independent set of H . Let v E W \ F .
Chapter 6
184
Since A is a facet of AG, we obtain that there is an edge L in E ( G ) so that L c A U { v } . As z 4 A and de&) = 1, we have L E E ( H ) and L c F U {v}. Hence F is a facet of AH and we are in case (a). On the other hand if z E A , then we have y, z1,. . . ,z k $! A . Set F = A \ (x}. Let C be a facet of AH containing F . Since C U {y) is an independent set of G, and G is C-M, we obtain that IC1 1 5 IAI = IF1 1. Hence C = F and we are in case (b), as required. The converse also follows readily using similar arguments. 0
+
+
Theorem 6.4.7 Let G be a Cohen-Macaulay graph. If G is bipartite, then the Stanley-Reisner simplicial complex A c of I ( G ) is shellable.
Proof. Wewill use the shellability criterion of Exercise 5.3.29. One may assume that G has at least one vertex of positive degree, for otherwise AG is a simplex with a unique facet. Set g = ht I ( G ) . We argue by induction on g. Let VI and V2 beas in Lemma 6.4.2. By Theorem 6.4.4 we may assume that deg(y1) = 1. Let us label the vertices of G so that y 2 , . . . ,yt be the vertices of G (if any) adjacent to 2 1 . According to Proposition 6.2.1 the graph H = G \ {11,y1} is C-M, hence by induction AH is shellable. Let Fl,. . . ,Fm be a shelling of AH. By Lemma 6.4.6there are integers 15 r l < < r 8 5 rn so that the facets of AG can be linearly ordered as follows 9
9
F 1
U {PI}, -
- ,Frn u {PI}, Fr1 u {XI}, - Fr5u {XI}, *
where Frl ,. . . ,Fr, are the facets of H not containing 9 2 , . . , gt (or equivalently containing 2 2 , . . . ,z t ) . We claim that the subcomplex A' of AG generated by Frl U {XI}, . . . ,Frs U { q )is shellable, with shelling given by Frl U {XI}, . . . , Fr, U {XI}. TOshow the claim assume 1 5 j < i 5 s. Since AH is shellable there is v E Fr; \ Frj and l! < ri so that Fri \ Fe = {v}. It follows that z2, . . . ,xt E Fe. Hence Fe = Frk , for some 1 5 k < i. Therefore A' is shellable. It now follows readily that the linear ordering of the facets of AGabove given is a shelling for AG. 0 Computing the type Let G be a C-M bipartitegraph on thevertex set V . Assume G has no isolated vertices. There are disjoint sets VI and V2 so that V = h U V2 and everyedge of G joins VI with V2. Since VI and V2 are minimalvertex coversfor G, and G is unmixed, we have g = IV1 I = IV2 I = height I ( G ) . By Konig's theorem there are g independent edges. Therefore we may assume
Vi = {yl,. . . ,yg}, V2 = {wl,. . . , t u g } , and yiwi
E I ( G ) for all
i.
Notice that w = {yi - ' ~ i } : = ~is a regular system of parameters for R / I ( G ) , where R = k [ y ,w], and R / I ( G ) modulo (w) reduces to:
A = k[Yl/(PT,
* ' '
,
$7
I(H)),
Edge
Ideals ' II 1
.
,w
.:;.. . ,
,
.,
where H is a graph on the vertex set VI. The algebra A is generated, as a k-vector space, by the image of 1 and the images of all the monomials yil yi, such that { p i l , , . . ,yiT1 is an independent set of H . Observe that the type of R / I ( G ) is equal to ( T ( H ) (because , Soc(A) is generated by the yi, such that {vi,, . . . ,yi, 1 is a maximal images of all the monomials yil independent set of H . 9
e
e
-
0
a
I
Example 6.4.8 Let G be the C-M bipartite graph whose edge ideal is:
The graph H obtained by reduction has edges y 1 y 2 , ~ 1 ~ 3 , ~ 1 ~ 4 Since this graph has only three minimal vertex covers, the type of R / I ( G ) is equal to three.
Exercises 6.4.9 Let G be a C-M bipartite graph. Show that the h-vector of R / I ( G ) has length at most g = ht I ( G ) . Can we replace C-M by unmixed? 6.4.10 Prove that thefollowing is the full list of Cohen-Macaulay connected bipartite graphswith eight vertices, in each case determine the type and the multiplicity of the edge ring.
6.4.11 Let I = (ziyjll 5 i 5 j 5 n), draw a picture of the bipartite graph G such that I = I(G) and prove that I is a Cohen-Macaulay ideal. 6.4.12 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.
6.5
Links of some edge ideals
In this section we give graphtheoreticalinterpretations certain ideals attached to graphs.
for the links of
, ' ~ 2 ~ 3 , ~ 2 ~ 4 .
Chapter 6
186
Let Go be a graph on the vertex set V = ( ~ 1 )... Yn} and I = I(G0). Take a new set of variables X I , . . . ,xn and define S(G0) to be the new graph obtained by attaching toeach vertexyi a new vertex xi and theedge {zi,yi}. We set K = I(S(G0)) = (ZlYl, * ' ' 7 XnYn, I ) c R = k[x,Y], where k is a field. The S inthenotation is intended for suspension or attaching whiskers to the base graph. The significance of this construction lies partly in the following restatement of Theorem 6.3.4: )
Theorem 6.5.1 If G is a graph, then G is a Cohen-Macaulay tree only if G = S(G0) for some tree Go.
if and
Of course, all the structureof S(G0) lies in the graph Go. The following simple computation conforms to this view. Proposition 6.5.2 Let Go be a graph and G = S(G0) its suspension. Then the multiplicity of R / I ( G ) is d i=- 1
where
fi
is the number of i-faces of the Stanley-Reisner complex of I(G0).
Proof. Let 91,. . . ,Yn be the vertices of Go. By Proposition 6.3.2, the set of linear forms {yi - x i } t l is a regular system of parameters for R / I ( G ) . The reduction of R / I ( G ) modulo these forms is JVYl/(YL
*
'9
Y2, W o ) ) ,
whose k-vector space basis is formed by the standard monomials, that is, 0 the facesattached of the simplicial complex. The next proposition describes the first link of S(G0) relative to the regular sequence z = {x1y1, . . . ,xnyn}. Proposition 6.5.3 Let Go and G1 be twocopies of thesamegraphon disjoint sets of vertices (y1, . . . ,gn} and { X I , . . ,x n } respectively. Let Y be the set of minimal vertex covers of G I and
T h e n ((z): K ) = (z, L ) , where K = I(S(G0)). Proof. Take {xil,.. . ,x i r } E T and ykyl E I . Since i , E {k,e} for some s we obtain xil .xi,ykye E (z). This shows (z, L ) c ((z): K ) . Conversely, assume M is a monomial and M E ((z): K ) )\ (2;).Let { yk, yt } be any edge in Go. Since Mykyl = xtytM1, either x k divides M or xe divides M ; in either case, M E L. 0
--
187
Edge Ideals
Corollary 6.5.4 The type of R/I(S(Go)) is equal to IY(G0)l. Proof. Since R / K is a Cohen-Macaulay ring and z is a maximal regular sequence in IC, where K = I(S(Go)),using Proposition 1.3.7 one has: ExtE(R/K, R )
HomR(R/K, R / ( z ) )N
((2): K ) / ( z ) .
Thus by Corollary 4.3.6 one obtains type(R/K) =,~(((2):K ) / ( z ) )and the assertion Proposition follows from 6.5.3. 0 Observe that these results can be used to compute the primary decomposition of an edge ideal. Example 605.5 Let I(G0) = (YIP39y1y4,y2y4, !/2y5, 9395,Y3Y6r y4yS). Note that the ideal ( ( a ) :I(S(C0))is generated by z = (xlyl, . . . ,XSyS} together with: x2x3x4, 2324x5, xlx2X5xf3,xlx2X3x6, xlx4X5x6,
and these monomials correspond to the minimal primes of I(G0). The type of R/I(S(Go))is 5 and the multiplicity of the face ring of I(G0) is 2. Definition 6.5.6 A star on n vertices is a complete bipartite graph of the form Kl,n-land is denoted by star(n). Corollary 6.5.7 If Go = star(n) is the star on n vertices, then I(S(G0)) lies in the linkage class of a complete intersection. Proof. Let y l , . . . ,yn be the vertices ofGo and y1 its center; then the only minimal vertex covers of G1 are (x1 ') and (x2, . . . ,xn}, so that, in the notation above, the first link is
= ( ( z ) : I ( S ( G o ) )=) ( X Z Y ~ , . . . , X ~ Y ~ , X ~ , X ~ " ' Z ~ ) = (x17 ~ 2 ~ 2 - ,x n ~ n r2 2 .xn). Take now its link with respect to the regular sequence formed by the first n elements: J
1
*
(x1,~2Y2,.'.,xnvn): J = (Xl,Y2,...,Yn),
which intersection. is a complete
0
Exercises 6.5.8 Let Go and GI be two copies of the same graph on disjoint sets of vertices (yl, . . . , gn) and {XI,.. . ,xn} respectively. Let Y be the set of
minimal vertex covers of G1 and
L = ({xil -xisI {xil j . . ,xi8} E T(G1))). *
Then ( K ,z) and
(2,L )
a
are geometrically linked, where K = I(S(G0)).
(xiyjll 5 i 5 j 5 n) C R = k[x,y]. Prove that R / I has type n and its a-invariant is - (n - 1).
6.5.9 Let I =
Chapter 6
188
6.6
First syzygy module of an edge ideal
Let G be a graph and I(G) c R its edge ideal. An interesting problem is to express some of the first initial Betti numbers of I ( G ) in terms of the graph theoretical data of G; below we give an indication that this might be possible, see [lo31 for further information. Let R be a polynomial ring over a field K with the usual grading and Ia graded ideal of R. The minimal graded resolution of R / I by free R-modules can be expressed as:
where all the maps are degree preserving and the dij are positive integers. < d j c j for all j = 1,.. , ,g . The integer g is equal One may assume djl < to pd,(R/I), the projective dimension of R / I . To simplify notation set
-
9
The rank of Fj is the jth Betti number of I . R o m the minimality of the resolution dl1 < < d,,) see Remark 2.5.9. We define the jth initial Betti number of I as Tj = bjl and the jth initial virtual Betti number of I as 9
--
Note that some of the initial virtual Betti numbersmay be zero. For a general monomial ideal f there is an explicit and elegant description of a minimal set of generators for the first module of syzygies of I due to S. Eliahou [loll; see also [43].
Definition 6.6.1 The edge graph of G, denoted by L(G), has vertex set equal to E = E(G)with two vertices of L(G)adjacent whenever the corresponding edges of G have exactly one common vertex. Proposition 6.6.2 If G is a graph with vertices x1,.. . ,x n and edge set E(G), then the number of edges of the edge graph L(G) is given b y
Proof. To prove the first equality note that each vertex xi of G of degree di contributes with ($) edges to the number of edges of L(G).
Edge
Ideals
The second equality follows from the Euler's identity n
2lJw)l=C d e g h ) , i= 1
see Exercise 6.1.27.
0
Let us now present a formula, in graph theoretical terms, for the second initial Betti number of I(G). Proposition 6.6.3 ([103]) Let I = I ( G ) c R be the edgeideal of a graph G and V the vertex set of G. If 0
.
.
+ RC(-4) @ R b ( - 3 )__$ R q ( - 2 )
R
-+
R/I
+0
is the minimal graded resolution of R / I . Then b = IE(L(G))I- N t , where Nt is the number of triangles of G and L ( G ) is the edge graph of G. Proof. Assume I = (fi, . . . , f q ) and $ ( e i ) = fi. Let 2; be the set of elements in ker($) of degree 3. We may regard the fi as the vertices of L(G). Every edge e = {fi, fj} in L ( G ) determines the syzygy syz(e) = xjei - xiej, where f i = xiz and f j = z j z for some X , xi,xj in V . According to Theorem 2.4.17 the set of those syzygies generate 2:. Given a triangle C 3 = {z1,z2,z3} in G we set
4(c3)= {zlej - z3ei9 zlej - Z S e k , r2ek - z3ei}, where fi = ~ 1 x 2 f, j = 2 2 x 3 , and fk = ~ 1 ~ Notice 3 . that 4 ( C 3 ) f~ 4(Ci)= 8 if C3 # Ci. For every triangle C3 in G choose an element p ( C 3 ) E 4 ( C 3 ) . It is not hard to show that the set
B = ,{syz(e)I e E L ( G ) )\ (p(C3)I C 3 is a triangle in G } is a minimal generating set for 2:.
0
It is a fact that thesecond Betti number of a Stanley-Reisner ideal (resp. monomial ideal) is independent of the field [169] (resp. [43, 1011))moreover for edge ideals the third and fourth Betti numbers are also independent of the field [168]. An applicationtoHilbertseries Let A be a simplicial complex of dimension d and f = (fo,. . . , f d ) its f-vector, where fi is the number of faces of dimension i in A and f - 1 = 1. By Corollary 5.4.5 the Hilbert Series of the Stanley-Reisner ring S = R / I A can be expressed as
We now use this formula to compute a particular instance.
190
Chapter 6
Corollary 6.6.4 Let G be a graph with q edges, V = (11 , . . . ,x,} its vertex set and F ( x ) the Hilbert series of S = R/I(G). If S has dimension 3, then F(z)(lis equal to l+gz+((9:1)-q)22+(
29
+ 3g2 + g3 - 6q6 - 6gq - 6 N t + 3~ ) 2 3 ,
where g is the height of I(G), Nt is the number of triangles of G, and n
v=
deg2xi. i=l
Proof. Let A = AG be the Stanley-Reisner complex of I = I ( G ) . The first entries of the f-vector of A are given by
To compute
f2
notice that the resolution above yields dim(R/I)3 =
(: t) - q n + b
and
where the last expression for f 2 uses Proposition 5.4.4.The desired formula 0 followsby substitution of the fi in Eq. (6.2) and usingEq. (6.1). The number of triangles of a graph Let G be a graph with q edges and vertex set V = 1x1,. . . , xn}. As the number of edges of the edge graph L(G) of G is given by the formula:
JE(L(G))J=
2 (degixi))+ 2 7, = -q
i=l
deg2 xi
i= 1
observe that Proposition 6.6.3 provides a method to compute the number of triangles of a graph G by computing syzygies with Macaulay or COCOA. An alternative method using linear algebra is recalled next. The adjacency matrix of G is the n X n matrix A = ( a i j ) whose entries are given by 1 if xi is adjacent xj, Uij = 0 otherwise.
It follows directly that A is symmetric and that its trace is equal zero. If f(z) = xn
+ c 1 P - l + C 2 x n - 2 + C 3 x n - 3 + . * ' + Cn7
Edge Ideals
191
is the characteristicpolynomial 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 ti's consult [24]. Let G be the graphwhose complement is the union of two disjoint copies of a path of length two. The edge ideal of G is equal to:
and part of its minimal resolution is:
+ R24(-3) "+ R11(-2) -+
R "+ R / I ( G ) -+ 0.
Hence the number of triangles Nt of G is equal to 6. Of course the same number is obtained noticing that
f(x) = x6 - 1lx4 - 12x3 + 3x2 + 4x
-1
is the characteristic polynomial of the adjacency matrix of G.
Exercises FG(x)the Hilbert series of k [ x ] / l ( G )where , k is a field. For any vertex x E V ( G )one has:
6.6.5 Let G be a graph with vertex set x and
6.6.6 If
is a complete bipartitegraph,then J?K,Jz)
=
1
'
1
+- 1. (1 - x)' (1 x)8
6.6.7 ([309]) Let Pn be a path graph with vertices x = 2 1 , . . . ,xn. Prove that the Hilbert series of k [ x ] / l ( P , ) can be express as:
6.6.8 Let Pn be a path graph with vertices x = X I , . Hilbert function of IC[x]/I(P,)is given by
Notice that for m 2
. . , xn. Prove that the
L q J this is the Hilbert polynomial.
Chapter 6
192 6.6.9 Let G = C7 be an heptagon. Prove the formula:
What is the number of minimal primes of I(G)?
6.6.10 Let G = C, be an heptagon and AG its complementary simplicial complex. Prove that AG is a triangulation of a Mobius band whose reduced Euler characteristic is ~ ( A G=) - 1. 6.6.11 Let I = I ( G ) c R be the edge ideal of a graph G. If 0
.
.
+ RC(-4) @ Rb(-3) + Rq(-2)
R
-+
R/I
+0
is the minimal graded resolution of R / I . Prove that c is equal to thenumber of unordered pairs of lines { fi, fj) such that fi and fj are independent lines that cannot be joined by an edge.
6.7
Edge rings with linear resolutions
A noteworthy use of graph theoretical methods in commutative algebra is implicit in Lyubeznik thesis [212] and explicit in the work of R. Froberg [116]. In both cases the central notion is that of a triangulated graph. It will turn out that this is the right notionto express some homological properties of edge ideals and theCohen-Macaulay property of unmixed ideals of height two generated by square-free monomials. the ideal Given a graph G an interesting duality betweenI ( G ) and IC (G), of covers of G, will take place in this section. Under special circumstances I ( G ) reflects properties of I,(G) and viceversa. The Cohen-Macaulay case of dimension two Let us show a family of flag complexes where shellability is equivalent to Cohen-Macaulayness, We begin with: Lemma 6.7.1 Let A be a simplicial complex of dimension 2 on the vertex set V . If A is Cohen-Macaulay, then there are facets F1,.. . , Fs such that
Edge
Ideals 193
Proof. We will proceed by induction on IC. Assume that there are facets . . , F k in A with IF1 U U Fit = 2 i for 1 5 i 5 IC, and such that (F1 U . - U Fi-1) n Fi has dimension 1 for 2 5 i 5 IC. Set Ui = F1 u Fi. Assume u k # V , otherwise the proof is complete, Choose F $ U k so that IF n UkI has maximal dimension. Set G = F n U k and r = IGl. We will show T = 2. Assume r = 1, i.e., G = { p } , Notice that p E F!c Uk for some i 5 k. We pick zo E F \ uk and x0 E Fi, x0 # p . Observe that zo and x0 are both in lk (G) and that lk (G) is connected, hence there is a path
+
F1,.
u
{zo = wo,.
. , w, *
s
e
= xo)
in Ik (G) joining zo and XO. We pick j so that wj @ U k and wj+l E u k . Since {wj, w j + l ) E lk (G) we obtain that F' = {wj, wj+l,p)is a facet of A so that F' (t Uk and F' n U k = { p , wj+l},which contradicts the choice of F . Thus r = 2, and we can set F = Fk+1 to complete the induction. 0 Proposition 6.7.2 (Ills])Let A be asimplicialcomplex of dimension 2. If theStanley-Reisner ring k[A] has a 2-linearresolution,then A is a Cohen-Macaulay complex if and only if A is shellable. Proof. Assume that k[A] is a Cohen-Macaulay ring with a Z-linear resolution. The f-vector of A is given by fo = n, f i = 2n - 3 and fi = n - 2, where n is the number of vertices of A (see Exercise 6.7.17). By Lemma 6.7.1 there are facets PI, . . , ,Fs in A so that V = u;=,Fi, IF1 U U Fil = 2 i for 1 5 i 5 s , and (F1 U U F k - 1 ) n F k is a face of dimension 1 of A for 2 5 IC 5 s. Notice that in our case s = n - 2 , hence F1,. . . , F, are precisely the facets of A . Set Ui = F1 U * U Fi and Ai = F1 U U Fi,
-
a
+
-
e
-
where Fi = {u E A l u c Fi). We claim that Ai = U i . Since Ai is a subcomplex of Vi,it suffices to show that they have the same f-vector. 0bserve that f l (Pi) + 2 L f l (&+I) and 2i + 1 I f l ( A i ) I f l ( T f i ) . Using f 1 (V9) = 2n - 3 yields fi ( U i ) = f1 (Ai) = 2i 1 for all i. A similar for all i. argument shows f2 ( A i ) = f:! To complete the proof notice that
(vi)
+
G = Fk n Uk-1 E F k n Ai = F k n u k - 1 , accordingly G c
Fk
n Fj for some j < k , using that IGl = 2 we get F k n Fj = F k \ {X}
for some x E V . Thus A is shellable by Exercise 5.3.29. The converse is quite true generally (see Theorem 5.3.18). 0
194
Chapter 6
Corollary 6.7.3 If k[A] is a Cohen-Macaulay ring of dimension 3 with a 2-linear resolution over a field k, then K [ A ] is a Cohen-Macaulay ring with a 2-linear resolution for any other field K . Example 6.7.4 Let R = k[a,.. .,f ] be a polynomial ring over a field k and A the following triangulation of the real projective plane P2:
Note that IA = (abc, abd, ace, adf, bc aef, f , bde, bef, cde, cdf). The face ring k[A] was studied by Reisner [237]. One has that k[A] 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 (compare with [114, Example 31). Hence A is not shellable and dim A = 2. The general case Let G be a graph and its complement. The graph is said to be triangulated or chordal if every cycle C n in of length n 2 4 has a chord in ??. A chord of C n is an edge joiningtwo non adjacent vertices of C n . The following result provides a graph theoretical characterization of the rings k[A] having a 2-linear resolution. Theorem 6.7.5 ([lis]) Let G be a graph and A the Stanley-Reisner comif andonly if isa plex of I(G). Then k[A] hasa2-linearresolution triangulated graph. Next we introduce the notion of d-tree and state an interesting consequence due to R. Froberg. Definition 6.7.6 A d-tree is defined inductively: (i) a complete graph K d + 1 is a d-tree, (ii) let G be a d-tree and H a subgraph of G with H N Kd, if v is a new vertex connected to all vertices in H , then G U (v} is a d-tree. Theorem 6.7.7 ([lis]) Let G be a graph and A the Stanley-Reisner complex of I ( G ) . If G is Cohen-Macaulay over a fieldk , then k[A]has a 2-linear resolution if and only if is a d-tree, where d = dim A .
Edge Ideals
195
Corollary 6.7.8 Let G be a connectedCohen-Macaulaybipartitegraph with q edges. If q is equal t o g(g 1)/2, where g = ht I ( G ) , then
+
Proof. First notice that R / I ( G ) has a linear resolution by Theorem 4.3.7. Hence, according to Theorem 6.7.7, the I-skeleton of AG is a ( g - 1)-tree, which implies that there is a vertex w of G of degree g. We now use the notation (and simiIar arguments) of the proof of Theorem 6.4.4 to conclude that deg(zi) = 1 and deg(yj) = 1, for some i ,j . To finish the proof consider G \ {w}and use induction. 0 graphthe The more general problem of determining the simplicial complexes A whose Stanley-Reisner ideal I A has a pure resolution has been addressed by W. Bruns and T. Hibi [46, 471, where a very concrete solution can be found if I A is generated by square-free monomials of degree two. Edge ideals and their ideals of vertex covers Triangulatedgraphs have been extensively studied, they can be constructed according to a result of G. A. Dirac, see [84, 2851. Those graphs have appeared in the commutative algebra literature in the works of R. Fkoberg [116] and G.Lyubeznik [213], here we point out a connection between these two works that leads to an interaction between some properties of an edge ideal and some properties of the ideal of minimal covers of the graph. Lemma 6.7.9 Let G be a connected graph with vertex set
V ( G ) and
d = min{deg(v)l v E V ( G ) } .
If G is not a complete graph, then there is disconnected and d = ISl.
S
c V ( G ) such that
G \ S is
Proof. Let v be a vertex of G of degree d and S = N ( v ) the neighbor set of v. Since G is not a complete graph and v has minimum degree there is a vertex w # S U {v}. Thus G \ S has at leasttwocomponents. 0 Definition 6.7.10 Let G be a graph and S a set of vertices, the induced subgraph <S> is the maximal subgraph of G with vertex set S. Thus two vertices of S are adjacent in <S> if and only if they are adjacent in G. Proposition 6.7.11 ([84]) If H is a triangulated non complete graph, then it can be constructed out of two smaller disjoint triangulated graphs H I and H2 by identifying two (possibly empty) complete subgraphs of the same size in HI and H2.
Chapter 6
196
Proof. One may assumethat H is a connected graph with n vertices. Since H has at least one vertexof degree less than n - 1, by Lemma 6.7.9 there is a minimal set of vertices S such that H \ S is disconnected and IS1 5 n - 2. Hence we can write H \ S = F1 UF2, where Fl and F2 are disjoint subgraphs of H . Set
H I = < V ( F l ) U S > and Hz = , where < A > is the induced subgraph with vertex setA. Note HI U Hz = H and HI n H2 =<S>. Next we show that < S > is a complete graph. Assume there arevertices z1 and x2 in S not connected by an edge of H . Since H \ (S \ {xi)) is connected for i = 1,2,there are paths pl and p2 in HI and Hz respectively joining z1 with 2 2 , whose only vertices in S are x1 and x2. If pl and p2 are the shortest paths with this property, thenp l U p2 is a cycle of length greater or equal than four without a chord, a contradiction. a
Lemma 6.7.12 ([285]) Let H be a triangulated graph and K a complete subgraph of H . If K # H , then there is x # V ( K ) such that the subgraph < N ( x )> induced by the neighbor set N(x) of x is a complete subgraph. Proof. By inductionon V ( H ) . If H is complete,any x V ( K )would satisfy the requirement. Otherwise, using Proposition 6.7.11 H = H I U Hz, where H I , H2 are triangulated graphs smaller than H such that H1 n H2 is a complete subgraph. It follows that I< is a subgraph of Hi for some i, say i = 1. As H1 nH2 is a proper complete subgraph of HZ,by induction x can in chosen be H2 \ ( H I n Hz). 0 Let R = Ic[xl, . . .xn] be a polynomial ringover a field IC and I a monomial ideal of I . The ideal of minimal covers or simply the ideal of covers of I , denoted IC , is the ideal of R generated by all the monomials xil - xik such that the ideal (xil,.. . ,xik) is an associated prime ideal of I . This terminology is intended to emphasize that a minimalprime"covers" or contains all the monomials in I . If I is a Stanley-Reisner ring one has a duality:
-
I
=
(xil -
... , )
. ~ i ~ ,
$ IC = (xil,.. . , x i , ) n
m .
= ( x j l , .. . , x j a )n . . '
=
3
(xil * x i a ,,.. , ) * *
Let G be a graph and E = E(G) its edge set, in this case we denote the ideal of minimal covers of I ( G ) by I, (G). One has
W)
=
n (GY)EE
Thus Ic(G)is an unmixed ideal of height two. The Cohen-Macaulay property of this ideal was successfully studied by G. Lyubeznik.
Edge
Ideals 197 <
L
..
'
,:
Theorem 6.7.13 ([213]) Let G be agraphanditscomplement.Then theideal of minimal covers Ic(G)isCohen-Macaulay if and only if triangulated graph. Proposition 6.7.14 Let G be agraphand Ic(G) isCohen-Macaulay,then
is a
Ic(G) its ideal of covers, If
where u(Ic(G))is the minimum number of generators of I,(G) and g is the height of I ( G ) . Proof. By the Hilbert-Burch Theorem 2.5.15, Ic(G) is generated by the m - 1 minors of an m x ( m - 1) matrix A with homogeneous entries, where rn = v( IC(G)). Take f in IC(G) of degree g, since any m - 1 minor of A has obtains deg(f) = g 2 m - 1. 0 degree at least m - 1, one
Proposition 6.7.15 Let G be a graph whose complement ?? is a triangulated graph. Then I ( G ) is Cohen-Macaulay if and only if Ic(G) has a linear resolution. Proof. +) By the previous results one has that Ic(G)is Cohen-Macaulay and the following inequality holds
according to Theorem 4.3.7 it suffices to prove that one has equality. Note that v(I,(G)) is equal to the multiplicity of R / I ( G ) , because all the minimal primes of I ( G ) are of height g = ht I ( G ) . On the other hand the multiplicity of R / I ( G ) is equal to g 1 (see Exercise 6.7.17). -+) By induction on the height of I ( G ) . One may assume
+
g = ht I ( G ) >_ 1, otherwise I ( G ) = 0. Note that the hypotheses imply that G is an unmixed graph such that the multiplicity of R / I ( G )is equal to e = g + 1. Hence any two maximalcompletesubgraphs of are isomorphic. Note d+l=n-g, where n is the number of vertices of G and d+ 1 is the dimension of R / I ( G ) . Let K be a proper complete subgraphof E on d + l vertices, by Lemma 6.7.12 there is z V ( K )such that the subgraph < N ( x ) > induced by the neighbor set N ( z ) of z is a complete subgraph. Since z belongs to a complete subgraph L of E on d + 1 vertices, it follows rapidly that degc(z) = d or equivalently degG(z) F 9. Hence by induction I(G \ {x}) is C-M of height g - 1. Using Proposition 6.2.5 one concludes that I ( G ) is C-M. 0
Chapter 6
198
One can use a duality criterion described next to see that this result is valid in a more general setting and without the hypothesis that the graph G is triangulated (see Exercise 6.7.21).
Definition 6.7.16 Let A be a simplicial complex on the vertex set V, the Alexander dual A* of A is the simplicial complex given by
According to [go] the Stanley-Reisner ideal of a simplicial complex has a linear resolution if and only if its Alexander dual is Cohen-Macaulay. This has been recently generalized [151] replacing linear resolution by the notion of componentwise linear ideal, and Cohen-Macaulay by sequentially Cohen-Macaulay.
Exercises 6.7.17 Let A be a simplicial complex of dimension d with n vertices and f = (fo, . . . ,fd) its f-vector. If k is a field and k[A] is a Cohen-Macaulay ring with a 2-linear resolution, then
6.7.18 Let A be a simplicial complex and k a field. If the ring k[A] is Cohen-Macaulay and has a 2-linear resolution, then thereduced Euler characteristic g(A) is equal to 0 and fid(A) = 0, where d = dim A . 6.7.19 Let A be a simplicial complex. If k[A] is a Cohen-Macaulay ring with a 2-linear resolution over a field k , prove that A is shellable.
Hint Use Theorem 6.7.7. 6.7.20 Show that the complement of a triangulated graph cannot have a chordless cycle with five or more vertices. 6.7.21 Let A be a simplicial complex on the vertex set V = (11,. . . ,x,} and A* its Alexander dual. Prove that the Stanley-Reisner ideal of A* is equal to the ideal of minimal covers of I = I A , that is:
6.7.22 Prove that the third Betti number of the ideal I = (abc,abd,
ace,adf, ae f,bcf, bde, bef , cde, cdf)
depends on the base field, and show that I is equal to its ideal of vertex covers. See Example 6.7.4.
Edge Ideals
199 - 2
6.7.23 Let k be a field of characteristic zero or two. Usekacaulay to verify , that the initial Betti numbers and initialdegrees of the Stanley-Reisner ideal I of Exercise 6.2.24 are as follows:
lo:& 78
8:
9:& Y7
6:& 76
Ys
5:& Y4
4:& Y3
3:& Y2
2:& 71
where the number before the colon indicates initial degree. What are the initial virtual Betti numbers of I ?
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Chapter 7
Monomial Subrings Two of the main objects of study here are subalgebras generated by monomials and the corresponding Rees algebras. In general one would like to relate properties of those two algebras, such as the normality property or the Cohen-Macaulay property. Some of the relevant topics of the chapter are the descriptions of the integral closure of ideals and monomial subrings and how the various notions of normality for algebras and ideals relate to each other. Special attention is given to showing the relation between the normality of the Rees algebra and properties of other related objects. We present some degree bounds for the generators of the integral closure of Rees algebras and monomial subrings. Upper bounds for the a-invariant of some subrings will also be given.
7.1
Basic properties
Let R = k[x] = k[zl). . . x , ~be ] a polynomial ring over a field k in the indeterminates 21).. . ,x, and F = {fl, . . , ,fq} a finite set of distinct monomials in R such that fi # 1 for all i . For c E Nn we set zc = xfl * * * x 2 .The monomial subring spanned by F is the k-subalgebra )
kEF1 = q j l , . . . , fq]
c R.
The exponent vector of fi = zai is denoted by log(fi) = ai and log(F) denotes the set of exponent vectors of the monomials in F . To try t o avoid cumbersome notation we set f" = f,"' figif c E Nq . Note that k [ F ]is equal to the semigroup ring
- -
20 1
Chapter 7
202
where C = Wlog(fl)+. .+Nlog(,fq) is the subsemigroup of W generated by log(F). Thus asa k-vector space k [ F ]is generated by the set of monomials of the form x", with a E C. An important featureof k [ F ]is that it is a graded subring of R with the grading k [ F ] i= k [ F ]fl Ria There is a graded epimorphism of k-algebras: 'p: B
= k [ t l , . . . ,tqJ-+ k [ F ]+0, induced by 'p(ti)= fi,
where B is a polynomial ring graded by de&) map 'p is given by
= deg(fi). Note that the
v(h(t1,.. . ,tq))= h ( f 1 , .. . ,f q ) , for all h E B. The kernel PF of 'p is called the presentation ideal or toric ideal of k [ F ] with respect to f1, . . , ,f q . Definition 7.1.1 A binomial f in B = k [ t l , ,. . , tq]is a difference of two monomials, that is, f = t" - to, for some a,,@in Ng,An ideal generated by
binomials is called a binomial ideal. Proposition 7.1.2 The presentation ideal PF of k [ F ] is a graded prime ideal generated by a finite set of binomials. Proof. Observe B/PF 21 k [ F ]and that k [ F ]is a Noetherian domain. If a polynomial ho + + h, is in PF with hi E Bi, then cp(hi) = 0 because cp is graded. Hence hi E ( P F ) i for all i, and PF is a graded ideal. Consider the ideal L of B generated by the binomials in PF. Since PF has a finite basis of homogeneous polynomials, to show L = PF it suffices to prove B d fl P ' C L for ail d. Let h E B d fl P', write
-
T
h =CaYitY' i= 1
with aYi E k and tYi E B d . We show that h E L by induction on r. Assume 2 3. Making the substitution zi = tl in the equality aYif'i = 0 one obtains a,,)tt = 0 and uYi = 0. One may assume f" = f Y j for some i # j , otherwise aTi = 0 for all i. For simplicity assume i = 1 and j = 2. Since uY2= -aT1 ,8, with p E k one has r
xi=,
xi=1
(E;=,
+
h = UT, (trl - t")
+ pt" +
r
aTit" = uy1(trl
- t'2) + 9.
i=3
By induction one gets g E L and h E L,as required,
0
Definition 7.1.3 If F = { f ! .,. ., f q ) is a set of monomials in k[x], the associated matrix of k [ F ] is the n X q matrix M whose columns are the exponent vectors log(fi), . . . ,log(fq).
Monomial Subrings
203
~
~~
~
Recall that the support of a vector ,L? E Rq is supp(p) = { i I pi # 0) , Note that any vector Q in Rn can be written as Q = a+ - a", where a+ and a- are two non negative vectors with disjoint support. Corollary 7.1.4 If PF is the toric ideal of the monomial subring k[F] and M is the associated matrix of k [ F ] ,then
PF = ( { t a t - t"- I a
Z q and M a = 0)).
Proof. Let t" - to be a binomial in PF. If supp(a) n supp(p) # 0, then we can write f" - ffi = f7(fa' - f p ' ) = 0, where ~ y i= min{ ai,pi). Note that a' and p' have disjoint support and f a - ffi = 0 iff M ( a - /I) = 0. To complete theargument use that PF is generated by binomials. 0
Corollary 7.1.5 Thetoricideal PF of a monomialsubring k[F] has a Grobner basis consisting of binomials with respect to any monomial ordering of the polynomial ring B . Proof. Let E be a finite set of generators of PF consisting of binomials and let f , g E E. Since the S-polynomial S(f,g ) is again a binomial and the remainder of S(f,g ) with respect to E is also a binomial, it follows that the output of the Buchberger algorithm (see Theorem 2.4.11) is a Grobner basis of PF consisting of binomials. cl Definition 7.1.6 Let R = K [ z l , .. . ,3 4 be a polynomial ring over a field K and k 2 1 an integer. The monomial subring of k-products or simply the subring of k-products is:
K[V,] c R, where V , = (xil
- - x i k I 1 5 il < *
The kth Veronese subring of R, denoted by I?('),
*
*
< ix: 5 n}.
is by definition:
i=O
where an element in Rik is said to have normalized degree i. Thus R(,) is again a standard algebra generated by elements of normalized degree 1. Sometimes the subring K[Vk] is called the square-free kth Veronese subring of R, because it is contained in I$('"). Corollary 7.1.7 ([lll])Let R = K [ z l , .. . ,'xn] be a polynomial ring over a field I( and let P be the toric ideal of the subring of k-products K[Vk]. If k 2 2, then P is generated by homogeneous binomials of degree two.
Chapter 7
204
Proof. Let v k = {xi, x i k I 1 5 il < < ik 5 n } be the set of squarefree monomials of degree k in R. To begin with consider a polynomial ring: a
B = K [ { t i,...ikI 1 5 il <
''
with one variable ti , . . . i s for each monomial xil graded homomorphism cp: B
+ K[V',],
< n)],
< ik
xik in
9
++cp
induced by ti,
xil
Vk.
--
Let cp be the
xik,
where B has the usual grading andK[Vk]has the normalized grading. Thus P = ker(cp) is the toric ideal of K[Vk]. By Proposition 7.1.2 the toric ideal P = Ps is a graded ideal generated by homogeneous binomials, hence it suffices to show that all the binomials in Psare in (P2)for s 2 2. We proceed by induction on s. Assume s 2 3 and that all the binomials in P2, . . . , P,-1 are in P2. Let
@z,
where ti, ti E 'T for all i. If cp(ti)= f i and (p(t!,)= fi, then after reordering the variables we may write f l = x1 x,zp+l . xk, where x l , . . . ,x, are in nglsupp(fi) = ng,supp(fil) and x,+l,. . . ,xk are not in nglsupp(fi). Set h equal to x1 - 'x, and write fi = hair+, ' x i k , If xp+l 4 supp(fl), then = ' x p k for some p 2 2. By our assumption there is & SO that xi,+, is not in supp(fj) for some l 2 2. First consider the case x,+l E supp(fj).Write f j = hz,+lxe,+, x e k . Note that fffj = fi'fj', where ff' = ( x , - + ~ f i ) / x i , .and +~ = (xi,+Ji)/z,+1. Since fi'and fi' are in V k ) there are t: and ty in 'J mapping under cp to fi' and respectively. Let G = titi - tyt;. From the equality
-
a
-
fi
fi'
F 1
fi'
= (tl t, - t'l t ; ) + (tit; - t':t!)(tb = tl . . * t , - tyt; . . .t'e-1 t"t' e t, = F e + l 9
.
*
+
ti&+, ' .t ; ) Gt', . ti-lt;+l a
a
t',
we get F E (P2) if and only if F1 E (Pz). On theotherhand if zr+l is not in supp(fj),write fi = hxe,+, Note that (ze,+,,. . . , q k ] is not contained in A = { I ~ , +. .~. ,xPk1, say xt,+, is not in A. Setting !f = ( ~ + l f l ) / ~ t , +and , f: = (zer+1fA)/z,+1,we get G2 = titb-tytg E P2, where ty , t; map under cp to f/, f: respectively. From the equality
-
F2
+
= F G2(t: *ti-lt;+l * t,-ltp+l I t * * * t',) = tl * . . t s - tyt; . . .t'1-1 t'e+1 * t;-1t;t;+1
t',
we obtain F E (Pz) if and only if F2 E ( P 2 ) . Set fjt' = ( $ ~ ~ , + ~ ) / x , . + ~ , fi' = ( f ~ ~ ~ + 1 ) / zand i , +G3 , = tit: - tyty'. From the identity
F3
+
= F2 G3(th - * ti-,ti+l - tb-ltit;+l * * t:) = tl . . t , - t:'t; . * . t'1-1 p't' e c+1 * ' * t;-lt;t;+l * ' ' t', *
a
Monomial Subrings
205
we obtain F2 E (P2) if and only if F3 E (P2). Hence from the outset one may assume
Using the same argument as above with x,+2 and xi,+z playing the role of xT+l and xi,.+l, and so on, we derive that there is a binomial Fm E P, of the form Fm = tl t , - tlu2 - u s ,with 212, - us E 'T, so that F E (P2) if and only if Fm E (P2). To complete the proof notice that Fm is in (P2) by the hypothesis. 0 e .
induction
e
,
Corollary 7.1.8 Let R = K[xl,. . . ,x,] be a polynomial ring over a field K and P the toric ideal of the Veronese subring R ( k ) .If k 2 2, then P is generated by homogeneous binomials of degree two. Proof. Consider a polynomial ring:
with one variable t , for each monomial x" of degree IC. Let cp be the graded homomorphism cp: B
-+
induced by t, I"% x",
where B hasthe usualgradingandhasthenormalizedgrading. By Proposition 7.1.2 the toric ideal P = ker(cp) is a graded ideal P = Ps generated by homogeneous binomials. Hence we need only show that all the binomials in Ps are in (P2) for s 2 2. We proceed by induction on s. Let
@z2
where t i , ui E for all i. If cp(ti) = fi and cp(ui) = g i , then onehas the equality f l f, = g1 - 9,. Write f 1 = . . - x? and g1 = . . . , To show that F belongs to (P2) we consider a few simple cases, Case(a):Assume ci < k and xi E supp(gj) for some j # 1, For simplicity of notation set i = 1 and j = 2. Pick xe # x l in the support of 91. Note that g l g 2 = hl ha, where hl = xlgl/ze and h2 = zeg2/xl. Since hl and h2 have degree IC, there are u\ and ui in 7- mapping under cp to hl and ha respectively. From the equality
-
a
a
we conclude that F E (P2) if and only if F1 E (P2). Case (b): x; divides g1 g, €or some i , say i = 1. By a repeated use of case (a) one may assume 91 = x!. Hence by symmetry one may also assume f 1 =x:, that is, F = tl...t, -t1u2".uS, which by induction is in (P2). '
0
Chapter 7
206
Case (c): x: does not divide g1
- - - gs for all i. Let
Note cij < k for all j. Thus using case (a) one may assume g1 = x::* - . xcir %r ' where xil 4 Ui+lsupp(gi). A repeated application of case (a) yields that C' C' supp(gi) for all j . By onemayassume g1 = xifl xi:' , where xij 4 symmetry one may write f1 = x:; . x:;, where xij 4 U+lsupp(fi), that is, one can reduce to the case F = t l - t, - tlu2 us, which by induction 0 gives F E (P2).
-
-
a
Proposition 7.1.9 If R is a polynomial ring over a field k and are in R, then the kernel of the homomorphism of k-algebras cp:
B = k[tl,. . . ,tq] -+ k[fl,.
is the ideal (tl - f1,. . . ,t,
f1,.
. . ,f,
. . ,f,], induced by cp(ti) = fi,
- f,) n B .
Proof. Let F E ker(cp). Making ti = (ti - fi)
+ fi
we can write F as:
Since F(f1,. . . ,frr)= 0 one derives & ag f 0 = 0, and consequently F is in (tl - f 1 , . . . t, - fq)n B. The other containment also follows readily. Let us illustrate a technique used to compute the toric ideal PF of a monomial subring k [ F ]using Grobner bases and elimination of variables. Example 7.1.10 Let F = (zixjll 5 i < j 5 4) and 9:k[tij] + k[F], the graded epimorphism of k-algebras with cp(tij) = xixj and deg(tij) = 2. By Proposition 7.1.9 one has
Using Macaulay [15] one obtains that the reduced Grobner basis of L w.r.t the elimination orderingin the first variables 21,. . . ,x4 is equal to:
207
Monomial Subrings
Another method to compute PF can be found in [83]; this method uses Buchberger's algorithm [53] adapted for the specialized situation of toric ideals. In [23] some algorithms are devised, that in many respects improve existing algorithms for the computation of toric ideals. For a general study of binomial ideals we refer the' reader to [98]. Lemma 7.1.1 1 Let f be a polynomial in IC[zl,. . . , xn]. If IC is an infinite field and f ( a ) = 0 for all a E IC", then f is the zero polynomial.
Proof. By induction on n. The case n = 1 is clear because any nonzero polynomial in k [ q ]of degree r has at most r roots. If n 2 2 write
where f i E k [ z l , . . . , zn-l] for all i. Fix c = (c1,. . . , cn-l) in kn-l and consider the polynomial
As g vanishes on k and g is in IC[z,] one obtains thatg is the zero polynomial. Hence fi(c) = 0 for all i. Since c was an arbitrary point i n the affine space kn-l by induction one has that fi is the zero polynomial for all i, thus f is U the zero polynomial. Corollary 7.1.12 Let R = k [ z l ,. . . ,xn] be a polynomial ring over a field IC and X c asetgivenparametricallyby ti = fi(x1,.. . ,xn)) where F = { f l , . . . ,fp) c R. If IC is infinite, then the ideal I ( X ) of polynomials f ( t l ,. . . ,t q )that vanish on X is equal to PF, the presentation ideal of k [ F ] . ICQ
Proof. As PF = (tl - f l , . . . ,t, - f q ) n lc[tl,.. . ,t4J,one has PF c I ( X ) . Conversely take f E I ( X ) and consider the revlex ordering such that ti > xj for all i, j. By the division algorithm f ( t l ,. . . ,t q )= gi(ti - f i ) f r , where T- is a polynomial in R. Hence for any a E kn one has
Ci
This proves r ( a ) = 0 for any Q E k n , thus T is the zero polynomial because k is an infinite (seefield Lemma 7.1.11). 0 Definition 7.1.13 Given a subset X of the affine space A; over a field k , the Zariski closure of X ,denoted by is the closure of X in the Zariski topology of A;. Thus is the smallest affine variety of At containing X .
x,
x
Proposition 7.1.14 If IC is a field and X a subset of the ufine space then = V ( I ( X ) ) ,where is the Zariski closure of X.
x
x
A;)
Chapter 7
208
x
Proof. As V ( I ( X ) )is closed and X c V ( I ( X ) ) one , has c V ( I ( X ) ) .For the other containment note = V ( J ) ,where J is an ideal of a polynomial ring R over k. From X c V ( J )one derives I ( V ( J ) )c I ( X ) , and since one also has the containment J c I ( V ( J ) )by transitivity J c I ( X ) . Therefore V ( I ( X ) )c V ( J )and consequently V ( I ( X ) )c 0
x
x.
Definition 7.1.15 An afine toric variety V is defined as the zero set of a toric ideal, that is, V = V ( P p )for some toric ideal PF. Corollary 7.1.16 Let R = k [ s l , . . . ,~ n be ] a polynomial ring over a field k and X c kq a set given parametrically b y ti = fi(x1, . . . ,xn), where F = { f 1 , . . . , fq) is a subset of R. If k is infinite, then the Zariski closure of X is equal to V(P*).
Proof. By Corollary 7.1.12 and Proposition 7.1.14 one has the equalities I ( X ) = PF and X = V ( I ( X ) ) . 0 Proposition 7.1.17 If k[F] is a monomial subring over a field k generated by a finite set F of monomials and M is its associated matrix, then dim k [ F ]= rank(M),
where dim k [ F ] is the Krull dimension of k[F]. Proof. Let F = {fi,. . . ,fq} and f i = P i , where ai = (ail,. . . ,ai,) for i = 1 , . . Q. Let M = (aji)be the n X q matrix of k[F]. Set r = rank(&?). By the Noether normalization lemma dim k [ F ]= trdegk k[F], therefore it suffices to prove the equality r = trdegk k [ F ] . We may assume a l , . . . ,a,. is a basis for the column space of M . Wenow show that f 1 , . . . ,f r are # ,
algebraically independent over k. If then there are
f1,
. . . ,f r
satisfy a polynomial relation,
such that f a = f". Hence M ( a ) = M ( c ) and ? l a 1 + - + ?;.a,. = 0, where Ti = ai - Cia Therefore ai = C i , that is f a = f is a trivial relation. To show that f,.+i is integral over the quotient field k ( f 1 , . . . ,f,.) pick s E N such that SQT+i = Ala1 - - . X,.a,., where A j E Z for all j . Therefore f:+i = fix1 - f,?', as required. 0 e
+ +
Exercises 7.1.18 Let k = Z2and X c k3 the space curve given parametrically by tl = X:, t2 = sf,t 3 = x:. Prove that and PF # I ( X ) .
gs
209
Monomial
7.1.19 If P is the toric ideal of the subring of Ic-products s,,k = K[Vk]and n 2 2Ic 2 4, then the minimal number of generators of P is given by: V ( P )=
E (”>( i=O
) [(2@ -!’- 1) - 11 .
2(k - 2 )
2
i*
IC-2-1
7.1.20 Use Theorem 2.4.18 together with Proposition7.1.9 to give a second proof of the fact that toricideals of monomial subrings have Grobner bases consisting of binomials. 7.1.21 Let R = K [ x l , 2 2 y1, , . . . ,ys] be a polynomial ring over a field K and S = K [ F ]the monomial subring spanned by F = { x I Y ~ Y ~ , x ~1YI ~i Y<~j (L 5 ) . Prove that
( 2 1 ~ 1 ~ 2z 2: 9 s3 9 4 )
=(~1~192,
213ay192y&
xqy1~2~52).
7.1.22 Let R = k [ z l , 2 2 , 2 3 1 and R(2)the second Veronese subring. If the homomorphism of Ic-algebras
-+
‘p
is
R(~)
‘p: k [ t i j ~
induced by ‘ p ( t i j ) = x i x j , prove that the toricideal P = ker(’p) is generated by the 2-minors of the symmetric matrix
x=
( ii:
tll
tt1132 t22
t23
tt3233
)-
7.1.23 Let R =
k [ 2 1 , x 2 , x 3 ] andlet S = R(2) bethe subring endowed with the normalized grading
secondVeronese
s = @Si, i=O
where Si = R 2 i . Prove that the Hilbert series of R(2)is equal to (1
+3 4
F ( S , Z )= (1 - 4 3 ‘
7.2
Integral closure of subrings
Here we present a description of the integral closure of a monomial subring, which links polyhedral geometry and integer programming techniques with the problem of finding integral closures of monomial subrings. The problem of the effective determination of the integral closure of a monomial subring has a satisfactory answer given by an algorithm of W. Bruns and R. Koch; see [48] for details on this algorithm.
Chapter 7
210
Polyhedral sets and cones AnufJinespace or linear variety in Rn is by definition a translation of a linear subspace of Iw't. Let A c Rn . Let us recall that aff ( A ) , the afine space generated by A, is the set of all ufJine combinations of points in A:
There is a unique linear subspace V of Rn such that
aff(A) = x.
+ V,
for some 20 E Rn . The dimension of A is defined as dim A = dim W( V ) . A point x E Rn is called a conwex combination of p l ) . . . ,p r E Rn if there are non negative real numbers al, . . . ,a, such that
x = alp1
+ - + arpr and a1 + - + a, = 1.
Let A c Rn. The conwex 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. Definition 7.2.1 A conwex polytope P C Rn is the convex hull of a finite set of points P I , . . . ,p, in Rn, that is, P = conv(p1,. . . , p , ) . The inner product of two vector x = in Rn is defined by (5,~= ) ~ I Y+ I Given a f Rn
\ (0)
(21).
. . ,xn) and y = (yl, . . . ,yn)
+ XnYn.
and c E R, define the hyperplane
H ( a ,c) = (2 E Rn 1 (2,a ) = c } . Note that every hyperplane of IW" is of this form. The two closed halfspaces boundedby H ( a , c ) are
H+(a,c) = {z E Rnl (z,a) 2 c} and H - ( a , c ) = H+(-a, - c ) . Definition 7.2.2 A polyhedral set or convex polyhedron is a set Q in Rn which is the intersection of a finite number of closed halfspaces of Rn. Since the intersection of an arbitrary family of convex sets in Rn is convex, one derives that any polyhedral set is convex and closed. It is a fact that P is a convex polytope in Rn if and only if P is a bounded polyhedral set; see [310,Theorem 3.2.51. Definition 7.2.3 If C C Rn is closed under linear combinations with non negative real coefficients, we say that C is a conwex cone or simply a cone. A polyhedral cone is a convex cone which is also a polyhedral set.
211
Monomial Subrings
If d is a set of points in R" , the cone generated by A, denoted by Et+ d or cone(d), is defined as
where R+ is the set of non negative real numbers. We say that C is a finitely generated cone in Rn if C = &A, for some finite set of points d c R". Theorem 7.2.4 If C c Etn, then C is a polyhedral cone in Rn if and only if C is a finitely generated cone.
Proof. See [310, Theorem 4.1.11.
0
One of the fundamental results on polyhedral geometry is the following remarkable representation theorem. Theorem 7.2.5 (Finite basis theorem) If Q is a set in Rn, then Q is a polyhedral set if and only if Q can be expressed as Q = P + C, where P is a convex polytope and C is a finitely generated cone. Proof. See [64, Theorem 16.21 and [310, Theorem 4.1.31.
U
Thus according to the finite basis theorem a convex polyhedron has two representations. We have found it very convenient to use thecomputer program PolyhedronRepresentationTransformationAlgorithms (PORTA for short) to switch between the two representations. See [63]; for the case of polyhedral cones see [48]. Definition 7.2.6 Let Q be a closed convex set in IIB". 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 n H # 8. Definition 7.2.7 A proper face of a polyhedral set Q is a set F c Q such that there is a supporting hyperplane H ( y , c) satisfying the conditions: ( 4 F = Q n W y ,4 #
0,
The improper faces of a polyhedral set Q ase Q itself and
0.
Proposition 7.2.8 If Q is a polyhedral set in Rn and Fl, Q , then their intersection F = Fl n F2 is a face of Q .
F2
are faces of
Chapter 7
212
Proof. Let Hi = H ( a i ,ci) be supporting hyperplanes of Q , where ci E R and 0 # ai E Rn, such that Fi = Q n Hi and Q c HZ for i = 1,2. Let US prove the equality F = Q n H(a1 04, ~1 ~ 2 ) .
+
+
The left hand side is clearly contained in the right hand side. On the other hand if z E Q n H(a1 a2, c1 c2), then using (x,ai) 2 ci one has:
+
+
c1 + c2 = (2,a1 3- a2) = (x,a1)
+ (x,a2) 2 c1 + c2,
hence (x,ai) = ci for i = 1 , 2 and x E F . Since
set
the
F is a face of Q.
0
Definition 7.2.9 Let Q be a polyhedral set and x0 E Q. The point x0 is called a vertex or an extreme point of Q if (20) is a proper face of Q. Definition 7.2.10 Given z = (
. . ,xn) E Rn its Euclidean norm
~ 1 , .
Lemma 7.2.1 1 Let Q be a convex polyhedron in Rn and is a vertex of Q if and only if Q \ {xo} is a convex set.
x0 E
is
Q. T h e n x0
Proof. *) Let H ( a , c) be a proper supporting hyperplane of Q such that (10) = Q n H ( a ,c) and Q c H+(a,c). Take x,y E Q \ {zo) and consider x = t y (1 - t ) z with 0 < t < 1. Note that x,y are not in H ( a ,c), thus
+
(x,a) = t ( y , a) + (1 - t ) ( z a, ) > tc + (1- t ) c = c. Hence ( x , a) > c, that is, x # zo and x E Q, as required. e)Let Q = n;=,H+(ai, ci) be a decomposition of Q as an intersection of closed halfspaces, where 0 # ai E Rn and ci E R for all i. First observe that 20 is in H(ai,ci) for some i, otherwise if (ZO,ai) > C i for all i, there is an open ball Bs(x0) in Etn, of radius 6 centered at $ 0 , whose closure lies in Q. Hence taking two antipodal points 21, 2 2 in the boundary of Bd(z0) one obtains zo E Q \ (zo}, a contradiction. One may now assume there is IC 2 1 such that 2 0 E H(ai, ci)for i 5 k and (zo, ai) > ci for i > k. Set
A = H ( a l , c ln)
+
- - n H ( u ~c ,~ ) .
We claim that A = { zo). If there is z1 E A \ (zo}, pick l36 (zo) whose closure is contained in H+(ai)c i ) for all i > k . Since x = tzo (1- t)zl is in A for all t E R, making t = 1 S/llzl - x0 11 one derives z E A and IIx - zoll = S, thus x E Q . Note that if k = r , then x1 is already in Q
+
+
gs
Monomial
213
+
and in this case we set z = 2 1 . Making t = -1 in z1 = t z (1 - t ) z o one concludes z1 E A and llzl - x011 = 6. Altogether z,z1 are in Q \ { E O } and zo= ( z 21)/2, a contradiction because Q \ (20)is a convex set. From the equality (20) = A = AnQ one has that (20)is an intersection of faces. Using Proposition 7.2.8 yields that {x01 is a face, as required. 0
+
Proposition 7.2.12 Let Q be a polyhedral set in Rn ulhzch is n o t a n a f i n e space and
nH+(ai, T
Q=
ci),
i=l
where 0 # ai E Rn and ci E R f o r all i. If Q if and only if (a) {xo} = ni,cIH(ai,ci) for some I (b)
(ZO, ai)
x0
E R", then x0 is a vertex of
c { I , .. . ,T } ,
and
2 ci f o r all i = 1,.. . ,r .
Proof. =+) This direction follows at once from the proof of Lemma 7.2.11. e)Note 20 E Q. Since the intersection of faces of Q is a face, see Proposition 7.2.8, onethat has (50)is a face. 0
If a E R", a # 0, then the set Ha will denote the hyperplane of Rn through the origin with normal vector a , that is, Ha = {x E Rn I (x,a ) = 0). This hyperplane determines two closed halfspaces denoted by
H z = {x E R"] ( z , a )2 0} and H; = {x E RnI (z,a) 5 0). Proposition 7.2.13 If C is a closed convex cone in Rn and H a supporting hyperplane of C , then H is a hyperplane passing through the origin. Proof. Let H = H(a,c). Since 0 E C c H f , one has c 5 0. If CnH = {0}, then c = 0 as required. Assume C n H # (0) and pick 0 # z E C such that ( z ,a ) = c. Using t z E C c H+(a, c ) for all t 2 0, one derives
tc = t ( z , a ) = ( t z ,a ) 2
c,
for every t 2 0. Thus c = 0.
0
Proposition 7.2.14 Let C be a cone in Rn and let
be a representation of C. If ai E Q" generated rational cone, that is, there are
\
(0) f o r all i, then C is a finitely pi E Qn such that
214
Chapter7
Proof. Consider the polyhedral set
where
A = {Elel
+ - + Enenl
&i
E { 1,-1) for all
i).
Note that P is a bounded set and theorigin is an interior point of P. Hence P is a polytope and consequently the set Q = C n P is also a polytope. Thus by [39, Theorem 7.21 one can represent Q as Q = conv(p1,. . . ,pS), where PI,. . . ps are the vertices of Q. By Proposition 7.2.12 each vertex of Q is the unique solution of a system of linear equations with rational coefficients. Thus all the pi have rational entries. E&.ps c C. Conversely if Since pi E C for all i one has E&.p 1 + x E C, there is an scalar t > 0 such that t x E Q, and one rapidly obtains in cone generated by p 1 , . . . ,pS. 0 that x isthe )
-- + a
Definition 7.2.15 A proper face F of a polyhedral set Q facet of Q if dim(F) =dim(&) - 1.
c Rn is called a
Theorem 7.2.16 Let Q be a poZyhedraZ set in Rn which is not an afine space such that
where a i is in Bn\ (0)) ci E B for alZ i and none of the halfspaces H+ (ai,c i ) can be omitted in the intersection. Then the facets of Q are precisely the sets F1, . . . ,Fr where Fi = Q n H(ai, C i ) ) and each proper face of Q is the intersection of those facets of Q that contain it. )
Proof. See [310, Theorem 3.2.11. Definition 7.2.17 If a polyhedron Q in Rn is represented as
and satisfies
for all j, we say that (*) is an irreducible representation of Q.
0
Monomial Subrings
215
We shall always assume that an affine space in Rn has the topology induced by the usual topology of Itn. By the relative interior of a set X in Etn, denoted by ri(X), we mean the interior of X in aff (X). Theorem 7.2.18 Let Q be apolyhedralset such that Q # Rn. Let
in Rn with dim(&) = n and
be a representation of Q with H+(al,cl), . . . ,H+(a,, c,) distinct, where ai is in Rn \ (0) for all i. Set Fi = Q n H ( a i ,ci), for i = 1,.. . , r e T h e n (a) ri(Q) = {x E
RnI (x,a l ) > c1,. . . , (x,a,) > c,).
(b) Each facet F of Q is of the form F = Fi for some i. (c) Each Fi is a facet of Q if and only (*) is irreducible. Proof. See [39, Theorem 8.21 and [310, Theorem 3.2.11. Corollary 7.2.19 If Q # Rn is a polyhedral set has a unique irreducible representation
0
of dimension n, then Q
as an intersection of closed hayspaces. Proof. Follows from parts (b) and (c) of Theorem 7.2.18. Note that if the set F = Q n H ( a , c) is a facet of &, thenaff(F) = H ( a ,c). 0 Corollary 7.2.20 If C # Rn isapolyhedralcone there is a unique irreducible representation
of dimension n, then
Proof. Note that according to Proposition 7.2.13 a set F is a proper face of C if there is a supporting hyperplane Ha of C such that
In particular the facets of C are defined by hyperplanes through the origin, therefore the irreduciblerepresentation of C has the requiredform. 0 The following two results are quite useful for determining the facets of a polyhedral cone. Proposition 7.2.21 Let A be a finite set of points in En. If F is a nonzero face of &A, then F = IR+ A' for some A' c A.
216
Chapter
Proof. Let F = & A n Ha with & A c H l . Then F is equal to the cone A’ = { a E -41 ( a ,a) = 0). 0 generated by the set Corollary 7.2.22 Let A be a finite set in Z n and F a face of &A. (a) Ifdim(F) = 1 and A
c W , then F = &a for some cy
E A.
(b) If dim(R+A) = n and F is a facet defined b y the supporting hyperplane Ha, then Ha is generated b y a linearly independent subset of A. Definition 7.2.23 Let Q be a polyhedral cone in Rn with dim(&) = n and such that Q # Itn. Let r
Q=nH: i= 1
be the irreducible representation of the cone Q. If ai = (ail,,, . ,ai,) we shall call ai151 * * * a i n x n = O (i = I , . . ,r )
+ +
a set of
equations of the cone Q.
Proposition 7.2.24 If Q is a proper cone an Rn that can be written as
with ai for all i and dim(&) = n, thenthere are unique(uptosign) a l , . . . , a, in Z n with relatively prime entries and such that
is the irreducible representation of Q. Proof. First note that if Hb is a supporting hyperplane of Q generated by a set of n - 1 linearly independent vectors in { a l ,. . . , ag},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 E Qn and Ha = Hb. Hencemultiplying a by a suitable integer andthen dividing by the greatest common divisor of the entries, one may assume Ha = Hb, where a is in Z nand has relative prime entries. Observe that a, b are linearly dependent because the orthogonal complement of Hais one dimensional, It is readily seen that a is uniquely determined up to sign. To complete the proof use Corollary 7.2.22 (b) to see that any facet of Q is defined by a supporting hyperplane Hb as above. 0
Monomial Subrings
217
Definition 7.2.25 Given a = (ul,. . . , a n ) E Rn,we set
Theorem 7.2.26 Let A be a finite set of nonzero points in Nn and A' the set of Q E A such that &a is a face of A of dimension 1. Then
Proof. Set 4 = {a1, . . . ,Qg} and S = conv(e1, . . . , e,). Consider the convex polytope P = S n Iw+ A. Note that P = conv(p1, . . . ,pg), where pi = ai/]~il. Therefore by [39, Theorem 7.21 the set of vertices V ( P ) of P is contained in { , . . . ,pg1 and P is the convex hull of V ( P ) . Since Iw+ A = I&- V ( P )it suffices to show that lit-+,B is a face of the cone spanned by A for every /? in V ( P ) . Let p E V (P ) . There is a hyperplane H = { z E Rn I (z, u ) = c) such that P n H = { p } and P c H-. Set a = (ul, . . . ,a,). To complete the proof we now show that
where b = (a1 - c, . . . ,an
- c ) . Since ai/lail E P c H - for all i one has
( fI q
=c=
ai
(",(.
,"., .)).
1% I
reverse inclusion take 0 ,c)) = cIz(
* (;,a)
# x E lR+ A n Hb. = c.
Thus z/lzI E H . Observe that x/lzl E S and z/lzl E &A, hence X _.
1.1 Altogether
X E PnH
1x1
this proves z E JR+ B.
ESn&A=P.
= (P), 0
218
Chapter
Notation Given x = ( 2 1 , . . . x,) E Rn, we write x 2 0 if xi >, 0 for all i. )
Theorem 7.2.27 (Farkas) Let A an n x q matrix with entries in a field K and z E Kn.Assume K = Q or K = R. Then either there exists x E K Q with Axt = xt and x 2 0 or there exists v E K n with vA 2 0 and vxt < 0 , but not both. Here xt denotes the transpose of x. Proof. See [315,Proposition 1.81 for the case I( = R and [207, 2161 for the case K = Q. 0 Normality of monomial subrings One of the issues to bediscussed in first place is a description of the normalization of a monomial subring. For convenience recall that if A is a domain with field of fractions K A , the normalization or integral closure of A is the subring ';LT consisting of all the elements of K A which are integral over A. Theorem 7.2.28 Let R = k [ x l , .. . ,x,] be a polynomial ring over a field k and let F be a finite set of monomials in R. T h e n
-
k[F]= k [ { x aa( E zdn rw+A}],
where A = log(F) is the set of exponent vectors of the monomials in F . Proof. Set
-
-
B = k[{x"I a E ZAn R+A)].
-
First we show k [ F ]C B. Take x E k [ F ] ,since k [ F ]c R and R is normal we get k [ F ]c R and x E R. Hence one can write
where di E k \ (0) and 71,.. . 7,. distinctnonzeropoints in Bin. Write "yi = n&, with ni equal to the gcd of theentries of ~i (observe that PI,. . . ,pr are not necessarily distinct). Consider the cone C spanned by A and (PI,. . . ,P,.}. Onemust show Pi E A for all i. Assume on the contrary that C is not equal to &A. By Theorem 7.2.26 one may assume that lR+pp,is a face of C not contained in It+ A. Let Ha be a hyperplane so that Ha n C = R+p1 and C c H;. There is 1 5 C 5 r such that
Since z is integral over k [ F ]it satisfies a monic polynomial f of degree rn with coefficients in k [ F ] .To derive a contradiction we claim that the term
gs
Monomial xmre occurs only once in the expansion of f ( x ) as a sum of monomials.
Assume the equality ( X y m
= Za(x7"
)mil
...
) W,
where mij > 0 for all j , m 2 mij and ct E &A. As ~t = n&, one has (mrl,a ) = 0 and from this equality one rapidly derives Hence a , & , . . . , p i t E Ha f~C = Iw+p1. Note p1 4 &A. Thus a = 0 and = for all j . Therefore t = 1 and yl = 7i1, as claimed.Altogether we get f ( z ) # 0, which is impossible. This proves that "yi E rw+dfor all i. On the other hand using that z belongs to the field of fractions of k [ F ]it follows that yi E Z A for all i. Let us now show the other containment B c k[F]. Note the equality
pi,
ZdnK+d=ZAnQ+d, which follows from Theorem 7.2.27. An straightforward calculation rapidly shows that x" is in the field of fractions of k [ F ]if a E ZA, and z" is an 0 integral element over k[F]if a E Q+ A. Hence B c k [ F ] .
-
Corollary 7.2.29 Let R be a polynomial ring over a field k and F a finite set of monomials in R. T h e n k [ F ]is normal if and only if
Nd=ZdnIIB,A, where d = log(F) . Proof. It suffices to observe the equality
k [ F ]= k [ { z a ( aE Nd)] toand
-
use the description of k [ F ]given above.
0
Theorem 7.2.30 (E. Noether) Let B = k [ t l , ., . t q ]be a polynomial ring over a field k and p a prime ideal of B . Then the integral closure of B / p is a finitely generated B/p module. Proof. See [92, Corollary 13.131.
0
In general it is hard to compute the integral closure of an affine domain; see [295] and [298, Chapter 61. Corollary 7.2.31 Let R = k [ x l , .. . ,x,] be a polynomial ring over a field k and F a finite set of monomials in R. T h e n
-
+ + xPmk [ F ] ,
k [ F ]= zP1 k[F]
-
for some ~ 0 1 .,. . ,xBm in k [ F ] .
0
* *
,
220
Chapter 7
Proof. Use Theorem 7.2.28 and the finiteness of the integral closure of an affine domain, see Theorem 7.2.30. This result also follows from the next lemma. 0 Lemma 7.2.32 Let A = ( a l ,. . . , a q ) be a finite set of points in Zn. Then there exist 7 1 , . . . ,ym in Z n such that
ln
r
Proof. Set M = q max la: I and N = -q max la; 1Szla
l
Let
p
1.
Recall the equality
E C. One can write
where xi E N and 0 # yi E N. By the division algorithm there are non negative integers T i , ni such that xi = niyi ri and 0 5 ri < yi. Therefore one can write
+
a
U
i=1
i=1
where ni E N and ai E [0,1] n Q. Since aiai is in C n [ N ,MIn one rapidly obtains that A U (C n [ N ,MIn) is a generating set for C with the required property. Our argument based on the proof of Gordan's Lemma given in [44].
was 0
Definition 7.2.33 Let R be a polynomial ring over a field k and F a finite set of monomials in R. A decomposition
of the k-vector space k [ F ] is an admissiblegrading if k [ F ] is a positively graded k-algebra with respect to this decomposition and each component k[F]ihas a finite k-basis consisting of monomials.
Monomial Subrings
221
Theorem 7.2.34 (Danilov, Stanley) Let R = k [ z l , .. . , xn] be a polynomial ring over a field k and F a finite set of monomials in R. If k[F] is normal, then the canonical module w k [ F ] of k [ F ] ,with respect to an arbitrary admissible grading, can be expressed as '
where A = log(F) and ri(IW+A) denotes the relative interior of IF& A. The formula above represents the canonical module of k[F] as an ideal of k[F] generated by monomials. Soon it will become clear how to apply ittoestimate somea-invariants. For a comprehensive treatment of the Danilov-Stanley formula see [44,Theorem 6.3.51 and [76]. Theorem 7.2.35 (Hochster) Let R be a polynomial ring over a field k and F a finiteset of monomials in R. If k [ F ] is normal,then L[F]is Cohen- Macaula y. Proof. See [30] and [172].
0
Theorem 7.2.36 Let R = k [ x l , .. . ,xn] be a polynomial ring over a field IC and F a finite set of monomials in R. If
A = k [ h l , ...,ha] L) k [ F ]
-
is a homogeneous Noether normalization of k [ F ] , then k[F] is a free Ad modulewhosegeneratorshave degree atmost deg(hi). Proof. Let S = k [ F ]endowed with the induced grading Si = S n Ri. Note that the composition
A q S v S
s
is a homogeneous Noether normalization. Since is Cohen-Macaulay by Theorem 7.2.35,one can then use Proposition 2.2.14 to obtain a direct sum decomposition
S=A#1$...$ATfim, Therefore the Hilbertseries of
3 can be expressed as m
m
F ( A x o l ,t ) =
F ( S ,t ) = i= 1
c
tdeg
i=l
n(1- tdep",) i= 1
ApplyingTheorem 7.2.34 one derives that a@), the a-invariant of 3, is negative. Using that a@) is equal to the degree of F(S,t ) as a rational function (see Proposition 4.2.3) one concludes deg xfii 5 EL1deg hi. 0
ChaDter 7
222
An algorithm to compute normalizations Let us describe some steps of an algorithm of Winfried Bruns and Robert Koch [48] that effectively computes normalizations. Let d be a finite set of points in Nn. First one finds w1, . . . ,wd in Z n such that
There exists an isomorphism of Z-modules Zd
T:Zd-+
with T(wi) = ei. Because of Lemma 7.2.32 there areP I , . . . p,. in a bounding box [ N ,MId such that
The determinationof the pi’s can be accomplishedby finding the irreducible representation of the cone JWtT(d) as an intersection of closed half spaces (see Corollary 7.2.20), and then testing which elements in
[ N ,MId f l Z d lie in the cone spanned by T ( d ) . To finish the description note
Observe that the computationof the irreducible representation of &T(A) can be performed using Corollary 7.2.22; in several specific examples we have used [48]or PORTA [63] to rapidly determine such representation. Example 7.2.37 Let A C
be the set of points
Note
Zd=ZwlCB*.*@Zw7, where the wi’s are the row vectors of the matrix: - 1 0 0 P= 0 0 0 0
1 1 0 0 0 0 0
0 1 1 0 0 0 0
0 0 1 1 0 0 0
0 0 0 1 1 0 0
0 0 0 0 1 1 0
0 0 0 0 0 1 2
Monomial Subrings
223 '
,
L
:
Next we consider the affine transformation T determined by T ( w ; ) = ei. An straightforward computation shows that the points in T ( A )correspond to the rows of the matrix: - 1 0 0 1 1 -1 0 0 &= 0 0 0 0 0 0 0 0
0 0 0 0 2 -2 1 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 2 -2 1 0 0 0 0 0 0 1 0 0 0 1 0 1 - 1 1
Using [48] or PORTA [63] we obtain that a point x = ( X I ) .. . ,x7) is in & T ( d ) if and only if x satisfies the system of linear inequalities
-25
-x4 -x4
"2x5
-23
-x3
-x4 "X5
-22
-x2 -222
-21 -221 -21
-25
+x3
-x4 -22
It follows that where ,8 = ( 1 , 0 , 1 , - 1 , 2 , - 1 , l ) .Hence
Z A n & A = NA + N(1, I , 1 , 0 , 1 , 1, 1). The linear inequalities shown above define the facets of the cone spanned by T ( 4 . Homogeneous subrings The class of homogeneous monomial subrings has been studied with detail in [278]. Examples that fall into this class are Rees algebras of ideals generated by monomials of the samedegree, subrings generated by monomials of the same degree, and polytopal subrings.
Chapter
224
Definition 7.2.38 Let F = (x"', . . . ,z"q} be a set of monomials in the ring of polynomials R = K[zl, . . . ,X,], where K is a field. The monomial subring K [ F ]is said to be homogeneous if there is v E such that ( a i , u ) = 1 for i = 1,.. . , g . Proposition 7.2.39 Let F = ( x a 1, . . . ,x"q} be a set of monomials in the ring R = K [ x ~. . , ,~ n ]where , K is a field. Then K [ F ]is a homogeneous subring if and only if K [ F ] is a standard graded algebra with the grading
Proof. =+) Let v E Qn such that (ai,u ) = 1 for all i. It suffices to prove
because one clearly has
K [ F ] i K [ F ]cj K[F]i+j for all i, j . First let us prove that K [ F ] in K [ F ] j= (0) for i not zero, one has an equality
with1 . 1
# j . If this intersection is
= i and Icl = j. Hence
(
i =1 . 1 = &ai,u) i=l
=
(
= (cl = j ,
i=l
+
a contradiction. To finish assume fil + f i r = 0, with fij E K [ F ] i jand < < i,.. If fi, # 0, using that the monomials of R form a K-basis, one has K[F]i,n K[F]i,# (0) for some j < T , a contradiction. Thus f i r = 0. By induction all the fij must be zero, as required. e)Set A = affQ(a1,.. . , and T = dim@). To begin with we claim that 0 is not in A . Otherwise if 0 E A , write 0 = ala1 + aqaq with ai E Q for all i and ai = 1. There is 0 # p E Z such that pai E Z for all i. One may assume, by reordering the ai, that pui 2 0 for i 5 s and pai 5 0 for i > s. As K [ F ]is graded, from the equality il
-
e
cyQ)
one derives p(a1 + *
- + a s ) = -p(as+l + *
+
*
*
+ up),
t
Monomial Subrings
225
x!=,
that is, p ai = 0, a contradiction. Thus 0 4 A, and consequently r < n. It is well known that a linear variety A in Qnof dimension r can be written as an intersection of hyperplanes
n WA,
n- r
A=
i= 1
where 0 # yi E Qn and ci E Q for all i; see [310, Corollary 1.4.21. Since 0 4 A , one has cj # 0 for some j . To finish it suffices to set w = y j / c j . 0 Corollary 7.2.40 Let F c K [ x ] be a finite set ‘of monomials and PF the toric ideal of K [ F ] .Then PF is a graded ideal with respect to the standard grading if and only if K [ F ]is a homogeneous ,subring. Proof. As PF is generated by a finite set of binomials, the result follows rapidly. 0 Proposition 7.2.41 Let F = { fi, . . , ,f,} be a set of monomials in the polynomial ring R = K [ x ] ower a field K and T I ,. . . T,, T a new set of variables. If cp, $ are the maps of I<-algebras defined b y the diagram
K [ T l , .. . ,T,]
K[F]
Ti
‘
4
fi
then there is a unique an epimorphism $ such $J = $9.In addition $ is an isomorphism if and only if K [ F ]and K [TF]have the same Krull dimension.
5
Proof. To show the existence of it suffices to prove ker(cp) c ker($). This containment follows at once because any binomial in ker(cp) clearly belongs to ker($), and ker(9) is a binomiaI ideal. Note that ker(cp) and ker($) areboth primeideals. Themap is an isomorphism if and only if ker(cp) = ker($), ‘thus to finish we need only observe that the last equality holds if and only if KIF] and K [ T F ]have the same dimension. 0 With the notation above one has Corollary 7.2.42 Themap q:K [ T F ] + K [ F ] is a isomorphism ifand only if K [ F ]is a homogeneous subring. Proof. If is an isomorphism, then by the proof above ker(q) = ker(cp) and K [ F Jis homogeneous because ker(cp) is homogeneous with the standard grading of K [TI,. . . , T q ]. Conversely assume ker($) homogeneous with the standard grading.Note that any homogeneous binomial in ker($) is also in ker(cp) , thus one has 0 ker($) = ker(cp) and q is an isomorphism.
Chapter 7
226
Ehrhart function In this paragraph we consider a fixed set of distinct monomials F = ( ~ ~ 1 .,. ,. z a g } in a polynomial ring R = K[z1 . . . ,z n ] over a field K . Let P = conv(d) be the convex hull of the set A = ring of P is the graded algebra
(01,.
. . ,a,}.
The normalized Ehrhart
where the ith component is given by
Note that because of the convexity of P one has
(AP)i(AP)jc (AP)i+j and ( A p ) i n ( A p ) j = (0) for i # j . Thus Ap is in fact a graded K-algebra. Proposition 7.2.43 K [ F T ]c A p , with equality if K [ F ]is homogeneous. Proof. Set
d'= ((al,l),. . . , (a,,1)) c Zn+l. First we show:
Take z = (a,i) in Z A' n & A' and write
where nj E Z and A j 2 0 for all j . Note a = 0 if i = 0, and a / i E P if i 2 1. Hence z is in the right hand side of (*). By Theorem 7.2.28 one concludes the required inclusion. Assume K [ F ]is a homogeneoussubring.Let v be a vectorinsuch that ( a j ,v) = 1 for all j . It suffices to prove that equality holds in (*). Take z = (a, i) in the right hand side of (*) and write
where n j E Z, A j 2 0, and z
(a,v) = (a$)
+ . + A, = 1. Hence = n1 + - . - + n ,= i , and = n1(a1,1) + ... + n q ( a q1) , = iAl(al,l)+ . * .+ iX,(a,, 1).
A1
*
Thus z E Z A' n & A', as required.
gs
Monomial Corollary 7.2.44 If K [ F ] ishomogeneous, then K [ F ] is normal ifand only if K [ F T ]= Ap. Proof. Use Corollary 7.2.42 and Proposition 7.2.43.
a
The normaljzed Ehrhart function of P is defined as
E p ( i )= dimK(Ap)i = JZAniPJ. Corollary 7.2.45 If K [ F ] is Hilbert function of K [ F ] ,then (a) h(i) 5 Ep(i) for all i
2 0,
a
homogeneoussubring of R and h isthe
,
and
(b) h ( i ) = Ep(i) for all i 2 0 ifandonlyif
K [ F ]is normal
Proposition 7.2.46 If K [ F ] is a homogeneoussubring of dimension d, then Ep is a polynomial function of degree d - 1.
-
Proof. As the Ehrhart ring Ap is equal to K [ F T ] ,by Corollary 7.2.31 one has that A p is a finitely generated I(-algebra. Hence Ap is a finitely generated graded K[FT]-module. Using Theorem 1.2.8 andProposition 1.4.12 one concludes that the Hilbertfunction of Ap is a polynomial function of degree d - 1. 0
Corollary 7.2.47 If K [ F ] isnormal and homogeneous, then the Hilbert polynomial of K [ F ]is equal to the normalized Ehrhart polynomial of P . Example 7.2.48 Let F = {x4, x3y, xy3, y4} and P = conv(A), where A = log(F). The normalized Ehrhart ring is
A p = K[x4T,x3yT, xy3T, y4T, x 2 g 2 T ] . The subrings K [ F ]and A p have different Hilbert functions, but they have the same Hilbert polynomial, which is equal to 4t 1. ,
+
Proposition 7.2.49 ([278]) If K [ F ]is a homogeneous monomial subring whose toric ideal has a square-free initial idealwithrespect to some term order, then K [ F ]is normal. Proof. See [278, Proposition 13.151.
0
Normal polytopes Some special types of monomial subrings associated to polytopes are introduced next.Our exposition has been influenced by the beautiful paper [41],that presents an appealing study of normal polytopal subrings.
228
Chapter 7
Let P be a lattice polytope in Rn, that is, P has all its vertices in Zn. For simplicity we assume that the vertices of P are in W . Given a field K the monomial subring
is called the polytopal subring of P. The polytope P is said to be normal if K[P] is normal. Proposition 7.2.50 If P is a lattice polytope in Rn and K a field, then dimK[P] = dim(P) + 1. Proof. Let al, . . . ,aq E observe the equality
Nn such that P = conv(a1,. . . ,aq). First we
aff(P) = aff(a1,. . . ,aq)= a1
+ V,
where V = R(a2 -al)+.."tR(cx, -a1). Thus one has dim(P) = dimw(V). On the other hand using the identity
the asserted formula
7.1.17.
follows Proposition from
0
Proposition 7.2.51 Let Ao, . . . ,An be a set of afinely independent points in Rn and A = conv(A0,. . . ,An). Then the volume of the simplex A is
det vol(A) =
A1
- A0 J
)
An - A 0
. -
n!
(
e
n!
Proof. The result follows using linear algebra or applying the change variables formula. See [206, Chapter XI11 or [310, Chapter 61.
of 0
Let A be an n-dimensional lattice simplex in Rn, that is, A is the convex hull of a set of n+l affinely independent pointin Zn.The normalized volume of A is defined as n!vol(A). From the result above one has vol(A) 2
1 n! '
recall that A is called a unimodular simplex if equality holds.
Monomial Subrings
229
Proposition 7.2.52 Let A be an n-dimensional lattice simplex in Rn with vertices Ao, . . . , An in Zn. Then A is unimodular if and only if either of the following two equivalent conditions hold
+ + Z(An, 1). (b) Z n = Z(A1- Ao) + + Z(An - Ao). (a) Zn+' = Z(A0,l)
*
Proof. Recall that a set of vectors a l , . . . , Qn in Z n generate Z n as a Z-module if and only if the n x n matrix A whose rows are the ai has determinant f 1. Thusthe result follows from Proposition 7.2.51. 0 Proposition 7.2.53 If A is a unimodularsimplex in Rn with vertex set A = {ao,. . . ,a,), then A n Z n = A. Proof. It suffices to show that A n Z n is contained in A. Let a E A n Zn. There are X i in I& such that Q = X o a o - Xnan and Cy==, X i = 1. If X0 > 0, from the equalities
+ + *
one derives A0 = 1, and consequently X i = 0 for all i Similarly if Ai > 0 for some i > 0, then a = ai.
2 1. Hence ct = QO. 0
Corollary 7.2.54 Let I< be a field and P a unimodular simplex in Rn with vertex set A = { QO, . . . ,a,} Then
K [ P ]= K [ X " O T ,
e . .
,X""T]
and K [ P ]is a polynomial ring. Proof. The asserted equality is a direct consequence of the result above. As K [ P ] has dimension n 1, using Lemma 2.1.5 gives that K [ P ]is a polynomial ring. 0
+
Proposition 7.2.55 ([41]) Let {C,Ci) be a family of finitely generated subsemigroups of I V suchthat C = UiCi. If ZG = ZCi and K[Ci] is normal for all i, then K [ C ]is normal. Proof. By Corollary 7.2.29 it suffices to show that ZCn I&C is contained in C. Take z E ZCnR+C. There is an integer s 2 1 such that sz E C, thus sz E Ci for some i. As ZC = ZCi and K[Ci]is normal we get z E Ci c C , as required. 0
Chapter 7
230
Definition 7.2.56 Let P be a convex polytope in Rn. A collection {Ai} of unimodular lattice simplices in Rn is called a unimodular covering of P if P = UiAi. Theorem 7.2.57 ([41]) Let P be a lattice polytope in Rn of dimension n with vertices in RP . If P has a unimodular covering and K is a field, then K [ P ]is normal.
Proof. Let {At} be a unimodular coveringof P and a l , . . . ,aq the vertices of P. By Proposition 7.2.55 we need only show that C = UtC!, where C and Ce are the subsemigroups of W+l defined as:
C = N{(a.,1)1CY. E ZnnP } and Ct = N{(a,1)1 a
E
Znnne}.
One clearly has the containment UtC! c C, because P = &A,. Conversely take z E C 2 = (P,s> =nl(Pl,l) +-*+nr(Pr,l), where ni E
N and Pi E Z nn P for all i. There are scalars Pij 2 0 such that
xi
Note that p / s is in P , where /3 = Ean$i and s = ni. Hence P / s is in At for some e. We can write At = conv(y0,. . . ,m ) , with {+yo,. . . , +yn}the vertex set of Ai. Thus
for some pi 2 0 satisfying
xi
pi
= 1. Hence
by the unimodularity of At one has
Therefore using that K[Ae] is normal (see Corollary 7.2.54) yields z E Ct, as required. This proof is due to W. Bruns, J. Gubeladze and N. Trung. 0
Exercises 7.2.58 Let A c I t n . Prove that A is an affine variety in X1A X2A c A for all XI, X2 in JR such that X1 X2 = 1.
+
+
Rn if and only if
7.2.59 If K [ F ]is a homogeneous monomial subring over a field K , use the
) 0. Danilov-Stanley formula to show a ( K [ F ] <
Monomial Subrings
7.2.60 Let a l , .. . ,a, E W and C = (&a1
231
+ + &a,)
n W . Then
+
+Ny,.,for some 71,.. . ,yrin W . (b) k[ C ]= k [ { ~ " (Ea C}] c k[x] is a normal domain, where k is a field. (a) C = Ny1
7.2.61 Let A c R"'. Prove that A is an affine variety in Rn if and only if AIA + AzA c A for all X I , A 2 in IR such that A1 + A 2 = 1. 7.2.62 Determine the facets of the convex polytope P = conv(fe1,. . . ,he,)
c IW".
+
+
7.2.63 Let d = {al,. . . ,a q } c w" and C = No1 . . No,. Prove that the semigroup ring k[C]is normal if and only if any element a E Zd satisfying T Q E C for some integer r 2 1 belongs to C. 7.2.64 If P is a unimodular lattice simplex in IRn with vertices ao, . . . ,an, prove that the polytopal subring K [ P ]is not isomorphic (as a ring) to the monomial subring K [ z " O , . . . , ~ " n ] c K [ x ] ,where K is a field. 7.2.65 Let 1 5 IC 5 n be two integers and
where ei is the ith unit vector in Rn. If P = conv(a1,. . . ,aq),for Some a l , .. . , a, E A, then P n Z n = ( ~ 1 , ... ,a q ) .
7.2.66 Let F = {x"',. . . ,x " q } be a finite set of square-free monomials of degree k in a polynomial ring R = K [ z l , .. . , zn3 over a field K . If P = conv(a1,. . . ,a,), then K [ P ]N K [ F ] . 7.2.67 Let F = { x a 1 ., . . ,x"q } be a set of monomials in a polynomial ring R = K [ z ~. ,. . , z,] over a field I<. Prove that K [ F ]is a homogeneous subring of R if and only if 0 # A = affq(a1,. . . , a q )c Q n ,
Hint A is an intersection of n - r rational hyperplanes, where T = dim(A). 7.2.68 Let { a l ,. . . ,a,) be a set of vectors in Qnand A = affQ(a1,.. . ,aq). Prove that 0 # A if and only if
7.2.69 Is the integral closure of a homogeneous subring homogeneous?
ChaDter 7
232
7.2.70 Is the monomial subring K[z2,zy,y3]C K [ z ,y] homogeneous? 7.2.71 Let F = ( x a 1 , .. . , x a q } c K [ x ] . Prove that K [ F ]is homogeneous if and only if
7.2.72 Let F = { P I , . . . ,x"q} be a set of monomials in R = K [ s l , .. . ,x,] and consider P = conv(log(F)). If A' = { ( p , 1)1p E ;Zn n P ) , then
and K [ F T ]c K [ P ]c A ( P ) . The ring A ( P )is called the Ehrhart ring of P. In the following problems we keep the notations of the previous exercise. For more details on Ehrhart rings consult [44, 163, 2721. See [48] for related algorithms.
7.2.73 Prove that the Ehrhart ring A ( P ) is a finitely generated K-algebra and a normal domain. 7.2.74 Prove that 00
A(P)= @A(P)i i=O
is a graded K-algebra with its ith component given by
7.2.75 Prove that the Ehrhart ring A ( P ) is integral over KEFT],and that the Hilbert function of A ( P ) is a polynomial function of degree d = dim(P). The corresponding polynomial is called the Ehrhart polynomial of P . 7.2.76 If d = dim(P), then according to [272,Proposition 4.6.301 one has: vol(P) = lim
i+oo
IZn n i P J id
*
Thus vol(P) is the leading coefficient of the Ehrhart polynomial of P. Prove this formula for the case n = dim(P). In practice one can compute volumes of lattice polytopes using [48].
Monomial Subrings
7.3
233
Integral closure of monomial ideals
As noted earlier the simplest kind of integrally closed monomial ideals are the Stanley-Reisner ideals and the powers of face ideals. Below we give a few basic properties of the'integral closure of monomial ideals. Lemma 7.3.1 Let R be a polynomial ring over a field k and let I be a n ideal of R. Assume that MI,. . . , Mq is a set of linearly independent monomials over k such that
N and rji E N f o r j = 1, . . . ,q. If bmj is a monomial in > 0 and bmj E k f o r mj = 0, then Ad: E I t for some t > 0.
mj E mj
Imj
Proof. We argue by induction on q. The case q = 1 is clear. Assume q For j 2 2 we substitute M:' into M;"' to obtain:
for
1 2.
'
AIJsjsl-rljrjl
= bzjb%:
IT
Mtslrji+rlirJ1
l#:i#j
Notice that in these equations M I has been eliminated. Hence applying the inductionhypothesis to M2, . . . ,Mq yields Me E I t for some > 0. 0
As a consequence of Lemma 7.3.1 we derive the following result using a short algebraic argument. Proposition 7.3.2 ([198]) Let R be a polynomial ring over a field k and let I be a monomial ideal in R. T h e n ?, theintegralclosure of I , is a monomial ideal.
+
Proof. Let z E 7 and let z = A l l + M q , where the Mi's are linearly independent monomials over k . ,There is an equation 0
a
By induction it suffices to verify that M i E 7 for some i. If M r E I", then Mi E 7. Hence we may assume M r 4 I n for all i. Since I i is a monomial ideal, all the monomials in the supportof ai belong to Ii. Using the equation above we obtain (after canceIling out common terms) equations of the form
N and rji e N for j = 1,. . . ,q, and such that b,, are monomials in for mj > 0, and bmj E k for mj = 0. As a consequence Lemma 7.3.1 yields Mq E 7. 0 mj
Irn3
E
234
ChaDter 7
Proposition 7.3.3 Let R = Ic[xl,.. . X,] be a polynomial ring over a field k and I c R a monomial ideal. Then the integral closure of I is given b y I = (x"(x"" E Im for some m 2 1). )
+
Proof. If xma E I m , then x" satisfies a polynomial of the form zm a, with a, E I", hence xa is in 7. On the other hand if x = x" E 7, then there is an equation
Since I is a monomial ideal one obtains zm E Imfor some m 2 1. Noting asserted equality follows. 0 that 7 is a monomial ideal the
A geometric description of the integral closure Let Q E C&",where Q+ is the set of non negative rational numbers. We define the upper right corner or ceiling of a as the vector [a1 whose entries are given by
where [ail stands far the integral part of ai. Let ai = (ail,. . . ,ai,) E W and let conv(a1,.. . , a,.) be its convex hull (over the rationals), that is,
is the set of all convex combinations of a1, . . . ,arm Proposition 7.3.4 Let R = Ic[xl,.. , X,] be a polynomial ring over a field k. If 1 c R is an ideal generated by monomials xal,.. . ,xar, then its integral closure is: I = xra1 E conv(a1,. . . ,U T ) } ) .
I F = { xla1 I a ({
Proof. We set
E conv(a1).
. . , a,)} .
Let xr"1 E F and let a = Xiai be a convex combination of a1 , . . . , a,. Since [a] 2 a , there is p E Qn) with p 2 0, such that [a]= p + a. Hence there exists p > 0 so that p p E Nn and pXi E N for all i. Therefore
showing zr"1 E
and ( F ) c 7.
gs
Monomial ConverseIy let zr E 7, that is, xP7 E IP for some p non-negative integers si satisfying
Hence
>
0. There are
, = - s+ q ;r) u i . P
x:==,
i=l
Set a = (si/p)ai. By dividing the entries of S by p , we canwrite y = O+P+a, where 0 5 pi < 1 for all i and 8 E W . Notice P+a E W . To complete the argument we claim that fa1 = /3 + a. If pi = 0, then ai E N and [ali = pi + ai. Assume 0 < pi < 1, since pi + ai E N we have ai N and fa1i = IaiJ + 1. If [ai]+ 1 < ai + pi < ai + 1, then LaiJ + 1 < ai, which is impossible, hence ai < ai + pi 5 [ai]-t-1 and pi + ai = [aiJ +l. Consequently x7 required. E ( F ) , as 0 Corollary 7.3.5 Let R = k [ q , .. . ,x,] be a polynomial ring over a field k and I c R a n idealgeneratedbymonomials x a l , ., . ,xar . Then is generated by monomials of degree at most d S n - 1, where d is the maximum of the degrees of x a l , .. . ,xar , Proof. Let a E conv(a1, . . . , a,) and p = [ a ] .As in [0, 1),deg(zb) has one < d n,as required.
+
pi = ai + &,for some Si 0
Example 7.3.6 Let I = (x3,y4) c k[x,y]. The integral closure of I is generated by the monomials of the points marked in black dots in the figure:
"t
Therefore I = ( 1 3 , y4)xy3)xy4)x2y2,x2y3,x3y,x3y2) = (x3, y4,xy3,x2y2).
236
Chapter
Proposition 7.3.7 Let k be a field and R = k [ x l ,x21 a polynomial ring in -two variables. If I is generated by m o n o m i a l s x a l , .. . ,xaq of degree d, then I is also generated by monomials of degree d . Proof. Onemayassume all < - .. < a q l , where ai = (ail,ai2). We claim that 7 is generated by the set A of all x c such that c1 c2 = d and all I c1 L aq1. If x' E A, then c1 = tall (1 - t)a,l for some 0 5 t 5 1. Hence c = tal (1 - t ) a , and c E conv(a1, u q ) , that is, x c E 7. Conversely let xp E 7, by Proposition 7.3.4 one may assume p = [a1 for some a in conv(a1,. . . ,a*). Set c = ( [ a l l d, - [all),by the previous argument x c is in A. Since p 2 c one has xfl E (A). 0
+
+
+
Proposition 7.3.8 Let R = k [ x l ,x21 and I an ideal of R generated by monomials of the same degree d. If I is integrally closed, then I is normal. Proof. The ideal I is of the form I = z J , where z is the greatest common divisor of the monomials in I. Since I is integrally closed if and only if J is integrally closed, one may assume I is m-primary, where m = ( 8 1 , x2). As x$ and x$ belong to I , one rapidly derives I = m. In polynomial rings in two variables any integrally closed ideal is normal; more generally one has thefollowing result of Zariski. Theorem 7.3.9 If R is a regular local ring of dimension two and I l , . . . ,I,. are integrally closed ideals, then I1 IT is integrally closed.
-
0
Proof. See [314, Appendix 5 , Theorem 2'1.
0
Example 7.3.10 Let R = k [ x l ,2 2 , x31 and I = ( x ; , x2.i). Using the program Normalix [48] onecancomputetheintegral closure of I . The input and part of the output file are: example. in
example.out
integral closure of I 2
3 3 0 3
0 1
0 2
(number of generators) (dimension of ambient lattice) (generators)
3 0
0 1
0 2 2
mode) (option or
Definition 7.3.11 If x a is a monomial in R = k [ x ]we set
log(x") = a. Given a set F of monomials, the log set of F , denoted log(F), consistsof all log(aa)with x a E F .
1
1
Monomial Subrings
237
Proposition 7.3.12 Let R = k [ x l , . , . , X,] be a polynomial ring overa field IC and I C R a monomial ideal. Then the integral closure of I is given by
-
I = ( x Q /a E conv(log(I)) n Zn).
Proof. Set
J = (xa/a! E conv(1og I)). Let a E conv(1og I ) . One can write 4
a=
PiPi,
i= 1
x:==,
where pi 2 0, pi = 1 and x! E I . Note that Apt = (1, where A is the (n 1) X q matrix whose columns are (1,pi)" i = 1,. . . ,q , hence by Farkas's Lemma one may assume pi E Q. By choosing m E N+ such that mpi - E N for all i one obtains xmQ E I" and x" E 7. Conversely let us show I c J . Since 7 is generated by monomials for any x* E 7 there is m E N+ such that xma E I m . It follows readily that a! E conv(1og I ) . 0
+
Proposition 7.3.13 Let F be a finite set of monomials of degree k in a polynomial ring R over a field K . If I = ( F ) is complete, then
(P)
conv(log(F)) n Z n = log(F).
Proof. Set F = (x"*,. . . ,zag). Let p E conv(1ogF) n Z", then 4 i= 1
x:.-,
where pi 2 0 and pi = 1. Since = k and x0 is in Proposition 7.3.12), then x0 = x"' for some i and /3 = ai.
7 = I (see 0
Condition ( P ) says that log(F) is the set of all lattice points in the convex polytope spanned by itself. We may interpret it as asserting that the integral closedness of I is not violated in the lowest possible degree. If F consists of square-free monomials, then P is certainly satisfied. Symbolic Rees algebras of monomial ideals For a monomial ideal I generated by square-free monomials it is possible to compare the integral closure of the powers of I with the symbolic powers of I , to make the comparison we begin with the following description of the symbolic powers. Proposition 7.3.14 Let R be a polynomial ring over a field K and let I be a n ideal of R generated by square-free monomials. Then
f o r all n >_ 1, where
p1,
. . . ,p,.
are the minimal primes of I .
Chapter 7
238
Proof. From Proposition 5.1.8 one has pp = pin). Since I is a radical ideal result the follows Proposition from 3.3.24. 0 Corollary 7.3.15 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 f o r n 2 1. Proof. Set J = I(n). Let p l y .. . ,p,. be the associated primes of I . If f E 5, then there is m 2 1 so that f" E p?" for all i. By Corollary 3.3.20 p? is complete, thus f E p? for all i. Hence Proposition 7.3.14 shows f E J . 0 Corollary 7.3.16 Let R be a polynomial ring over a field K and let I be a monomial ideal. If I is a radical ideal, then In c I ( n ) f o r n 2 1. Proof. Let x E I", there is k 2 1 so that 'x E (In)' c thus x is in E I(n), Note that I(n) is integrally closed by Corollary 7.3.15, hence
x E I(n).
0
Let us present the monomial version of Theorem 3.3.31, which can be proved directly in the monomial case. Proposition 7.3.17 Let R be a polynomial ring with coeficients 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 3.3.26 I n = I(n) for n 2 1. Hence thanksto Corollary 7.3.16 I n is integrally closed for n ,> 1, and consequently R[It]is normal. 0 Proposition 7.3.18 ([214]) Let R = k [ q , .. . ,xn] be a polynomial ring over a field k and I a n ideal generated b y square-free monomials. Then the symbolic Rees algebra
R , ( I ) = R + I t + I(2)t2+
+ I(")tn +
0
. -
c R[t]
is a finitely generated R-algebra. Proof. We follow Lyubeznik's original argument. Let p 1 , . . . ps be the minimal primes of I . Given a monomial y = x;' - - x p , set degri (3) = ri and degpi( 3 ) = C z j Edeg,, p i (y). To every monomial y in I ( m ) we associate the vector: )
By Proposition 7.3.14 a monomial y is in I(") if and only if degpi(y) for all i. Hence w (y) E Nn+s for all y E I(").
2 rn
Monomial Subrings
ai
239
Note that the set Nn+' is partially ordered by: ( a i ) 2 (bi) if and only if 2 bi for all i. Denote by A the set of all w(y) such that y is a monomial
whichis part of a minimal generating set of I ( m ) for some m 2 1. By Dickson's lemma A has only a finite number of minimal elements, and for each c in A there is a minimal element d of A with c = b + d for some b E P P s . Let y1,. . . ,yp be the monomials in R such that u ~ ( y l .) ., . , w(yp) are the minimal elements of A. Thus yi is part of a minimal generating set of I ( ~ Xfor ) ,some mi 2 1. We claim R,( I ) = R [ y lt m l ,. . . , yptmp]. Let y t m E I(m)tm,where y is a monomial in I ( m ) ,note that one can write w(y) = b + w(yj) for some j and some b E R P S . One has: (i) y = xayj, and (3) - rn >_ deg,, (yj ) - mj for all i. (ii) degpZ
Combining (i) and (ii) yields:
Exercises 7.3.19 Let R = K[zl,.. . , x,] be a polynomial ring over a field K and m its irrelevant maximalideal, then md is the integral closure of the ideal (x!, . . . ,x;>. 7.3.20 Prove I n J
# T n T , where I = (x5,y3)and J
= (y5, z 2 ) .
+
7.3.21 Let R = k [ x l , .. . , x m , y l , . . . , yn] and L = 1 1 Jr IrJ1, where Jt (resp. I s ) denote the ideal of R generated by the square-free monomials of degree t (resp. s ) in the y i (resp. xi) variables. If r 2 3, prove that L3 is not integrally closed.
Hint Set f = x ~ ~ . . x ~ y a~n d. p. r~o yv e~f E D \ L 3 .
7.4
Normality of some Rees algebras
There areseveral interesting examples of normal ideals generated by squarefree monomials, here we present some of them. Since the normality of an ideal I , in a polynomial ring R over a field k , is equivalent to thenormality of the Rees algebra of I (see Theorem 3.3.18), it is often the case that is much easier to verify the normality of the ideal than the normality of its Rees algebra.
ChaDter 7
240
An advantage of proving normalityof the Rees algebra of I occurs when the ideal is generated in a single degree p , in this case one automatically obtains the normality of the subring k [ I p ] ,by the next general result on descent of normality. See Example 8.7.14 for a counter-example if I is generated in different degrees. Proposition 7.4.1 Let R be a polynomial ring over a field k and I an ideal of R generated b y homogeneous polynomials f 1 , . . . ,f q , Assume deg(fi) = d for all i. If R(1)is normal, then k [ f l , .. . ,f q ] is also normal. Proof. Let m = R+ be the irrelevant maximal ideal of the polynomial ring R and A = k[Tf1,.. . ,T f q ] . Observe that there is a decomposition of A-modules:
As A
N
k [ f i , .. . ,f q ] (as rings) the result follows from Proposition 3.3.33. 0
Proposition 7.4.2 Let R = k [ x l , .. . ,x,] and R [ x ] be polynomialrings over a field k . If I is a normal ideal of R generated by square-free monomials of the same degree t 3 2, then the ideal J = I (2x1,. . . ,x x n ) is normal.
+
Proof. Set V = ( ~ 1 ,. . ,x n ) and J = I + (2x1,. . . ,xxn). Here to make the notation simpler we denote the integral closure of JP by J,P. By induction on p we will show that J i = JP for all p 2 1. If p = 1 then J , being an ideal generated by square-free monomials, is complete (see Corollary 5.1.5). Assume J i = Ji for i < p and p 2 2. Using Proposition 7.3.3 we have
J," = ((91 y is a monomial in R [ x ]and ym E Jmp for some m 2 1)). Let y be a monomial in J,P, then ym E J m P , m Since J l C J,"-l = JP-l we can write y = zW(zw1) ' ' ' (zw,)f1 ' '
> 0.
Let
US
show y E Jp.
fp-8-1,
for some 0 5 s 5 p - 1, where M is a monomial with x 4 supp(M ), the fi's are degree t monomials in J with x 4 supp(fi) for all i, and wi E V for 1 5 i 5 5. Likewise we can write
for some 0 5 r 5 mp, where N is a monomial, the gi's are monomials of degree t in J with z 4 supp(gi) for all i , and xi E V for all 1 5 i 5 r. To clarify notation set xwo = zzo = 1 and fo = go = 1. Altogether we obtain:
Monomial Subrings
241
making x = 1 and taking degrees one readily derives the inequality where r 5 min{mp, ml + m s ) . (a) If deg(M) = 0, the inequality above gives 4 2 2 since r 5 m ( l s), and s 5 p - 2 since T 5 mp. Therefore x'fl E J2and y E JP. (b) Assume deg(A4) > 0, notice that = 0 and T 5 m s , otherwise y E J P . Let y1 be the result of evaluating y at x = 1. From Eq.(7.2) we derive y? E Imp-r C Im(P-S),and y1 E = IP-s. Using Eq.(7.2) and Eq. (7.3) we obtain deg(y) 2 2s + NP - 4 , it follows that y E J P . 0 as y = x9y1 by a degree argument
+
The next example shows that Proposition 7.4.2 does not extend to monomial ideals with minimal generators in different degrees.
If I is the ideal of R generated by whereas the ideal
J =I
f1,
. . . ,f 6 ,
+ (XXl,. .
then R(I) is a normal domain, zx8)
is not a normal ideal of the polynomial ring R[x]. Let us briefly explain how to verify the assertions in Example 7.4.3. Let P be the toric ideal of R ( I ) with respect to f l , . . . , f 6 and let L be its Jacobian ideal, that is, L is generated by the g x g minors of the Jacobian matrix of P , where g is the height of P (in our case g = 5 ) . Using the computer algebra system Macaulay [15] we readily obtain the equality pd, ( B / P )= height(P), where B = R[Tl, . . . ,T']. Therefore R ( I ) is Cohen-Macaulay by Corollary 2.5.14. Again making use of Macaulay we derive that ( L ,P ) is an ideal of height 7. Therefore the normality of R ( I ) follows from the Serre's normality criterion together with the Jacobian criterion for regularity in polynomial rings; see Example 3.5.11 for details. On the other hand observe that J 3 is not complete because xf1 f2 $ J 3 and (xfif2)2 = (xx5)(xz6).f3f4f5f6 E J 6 . Proposition 7.4.4 Let R = k[xl,. . . ,x,] and R[x] 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 ,xxl) is a normal ideal of R [ x ] .
Chapter 7
242
Proof. By induction on p we will show J z = JP for all p 2 1. If p = 1, then JP is integrally closed by Corollary 5.I .5. Assume JZ = J i for i < p and p 2 2. Let y be a monomial in J,P, then ym E P m , rn > 0. Since J: c Jz-l = Jp-l we can write y = z t ( x z l ) P M f l' ' ' Pf"
= zt(zzl)%,
where M is a monomial with z 4 supp(M) and the fi's are monomials in J with z 4 supp(fi) for all i. By Proposition 7.3.2 it suffices to show y E JP. Since ym E JPm we have
3P . = xtm(xzl)Tmz" = N ( z z l ) S g l' ' gmp"s, *
(7.4)
where N is a monomial, x 4 supp(gi) for all i, and the si's are monomials in J . We distinguish two cases (a) Assume t = 0, then s ,< rrn. Since xim divides ym from Eq. (7.4) we obtain zm E I(P-P)m,hence z E = IP"' and y = ( z q ) ' z E Jp. (b) If t > 0, we may assume z1 4 supp(M), otherwise y E JP. We may also assume z1 4 supp(fi) for all i, otherwise it is not hard to see that we are back in case (a). Notice that s 5 rm, because z1 supp(z). Since xim divides ym it follows that z E = I*-" and y = Z ~ ( X Z ~ E) JP. ~ Z Proposition 7.4.5 Let R = IC[zl,.. .,xn] be a polynomial ring over a field IC. If It is the ideal of R generated by the square-free monomials of degree t , then It is a normal ideal. Proof. We will proceed by induction on t. Assume that Ii is a normal ideal for i < t and let I = It. We set m = ( X I , . . . , x n ) . It is enough to show (by induction on p 2 0) that IP is integrally closed for all p . Assume p > 0 and I' is integrally closed for i < p . In our case the integral closure of Ip is a monomial ideal given by
see Proposition 7.3.2. Assume M = I,f/IP is non-zero. We claim that M has finite length. If p E ass^ M\ (m}, say 2 1 4 p, then Ip= ( I t - l ) p ,where It-1 is theideal of R generated by the square-free monomials of degree t - 1 in 2 2 , . . . ,zn. By the first induction hypothesis we get Mp = (0), which is impossible. Hence ASSR M = {m). Let y be a monomial in I: \ IP which is conducted by m into IP. Note that, by the above description of I!, any monomial in I: has degree at least tp. We have y E I: c = P - l , so we can write y = 21 * * ' zp-lu, where z1, . . . ,zp-l are monomials of degree t in I and u is a monomial with deg(u) 1 t . Notice that Jsupp(u))5 t - 1, otherwise y E IP. We can write,
Monomial Subrings
243
after permutation of variables, u = x!1 - x?, bi 2 1 all i. If x1 4 supp(zi) for some i, then there is xjl E supp(zi) \ supp(u), and we can re-write a
y = z1 "'zi-1z;zi+l
*
e
*zp-1x h1 - l x262 . , , x bxx x
j1,
where z: = ( z i / x j l ) x l .We have three cases to consider. (i) The number of zi's not containing x1 is greater or equal than b l , then we can re-write
where wi are monomials in I \ mI, the variables x2, . . . ,x x , x j l.,. . ,xjbl distinct, and X bl 5 t. (ii) The number q of zi's not containing x1 is less than b l , then we can write
+
where x1 ,x2, . . . ,zx, xjl , . . . ,X j q are distinct, wi are monomials in I \ m1 and x1 in supp(wi) for all i. (iii) All the zi's contain X I , which reduces to case (ii). Applying the same arguments to 2 2 , . . . ,xx yields that we can write y = y 1 ' " ypwhere yi are monomials in I distinct and
1 X:;
*
x:,, xjl
*
xj8,
\ m I , ai 2 1 for all i, x i l , .. . ,xi,, x j l , . , . ,x j s
n
P-1
xil
, . . . ,xi, E
supp(yi).
i= 1
Note that T + s 5 t - 1, otherwisey E Ip. We set x j s = 1 if s = 0. We substitute xi1 = . = xi, = 1 in xily and denote by g the monomial obtained from zily after evaluation. To derive a contradiction observe that
while on the other hand, since xil y E Ip we have deg(g) 2p ( t - T ) . Hence s2 is t , which impossible. CI
T
+
Some of the arguments given in Proposition 7.4.5 may apply for other groups of square-free monomials whose elements are of degree t or less. Next we show a result in this direction. Theorem 7.4.6 Let R = k [ x l , .. , xn] and R [ x ]be polynomial rings over a field IC and let Lt = It XI"
+
where It is the ideal of R generated b y the square-free monomials of degree t 2 2. Then Lt is a normal ideal of R [ x ].
ChaDter 7
244
Proof. We use induction on t. If t = 2, then Proposition 7.4.5 yields that L2 = 1 2 XR is normal. Assume that Li is a normal ideal for i < t and set m = (21,. . . ,xn,x), L = Lt , I = I t - 2 , and J = It. By induction on p 2 1 we now show that LP is integrally closed for all p . Recall that L is an intersection of prime ideals because L is generated by square-free monomials (see Corollary 5.1.5), hence L = La. Assume that Li is integrally closed for i < p and p 2 2. Let us show Lfl = Lp. If on the contrary Lp # Lfl, we set (0) # M = Lz/Lp. Observe that for any p E ASSAM \ {m} one has L, = (&-I), if xi 4 p for some i and L, = (It-z),, if x 4 p, therefore by induction hypothesis and Proposition7.4.5 follows it that theonly associated prime of M is m. Hence there is a monomial y E LP, \ Lp so that my c LP, since LP,c Lg-l = Lp-l we can write
+
where the hi's are monomials in I of degree t - 2 and thefj 's are monomials in J of degree t. Observe that if x E supp(u), then we may re-write
for some xi, xj E supp(f1) and i # j, a repeated use of this rearrangement of the factors of y yields that either y = u(xh1) - - (xh,)f1 fp"s-l with x 4 supp(u),or y = u(xh1) (xhp-l) with x E supp(u).Inthe second case using that ym E LmP for some m 2 1, together with Proposition 7.4.5, rapidly gives y E Lp. As a consequencefrom the start we mayassume x 4 supp(u); notice that deg(u) 2 t. Let u = xrl . x : x , 1 5 X 5 t - 1, a1 2 2. There are three cases to consider: (a) Assume (supp('1~) I = t - 1. In this case
-
-
otherwise y E LP. We may assume x1 4 supp(hc) for some e, otherwise by induction on t we obtain y E LP. We claim that
If the first containment is false, then pick xr E supp(hi) \ supp(u). If not in supp(hi), then we re-write
-
x1
is
where f = f1 f p - s - l , which contradicts the choice of y. On the other hand if x1 E supp(hi), then supp(h1) $ supp(hi) \ (21) and we may pick
Monomial Subrings z E supp(he) and z
.,>
,
-. -
.
.
245
..
4 supp( hi).Therefore
where f = f l . . . fp”S-l, hence y E LP, and again this is a contradiction. To show the second containment note that we can write he = 22 . . xt-1. If supp(u) $? supp(fj), pick x, in supp(u) \ supp(fj) and re-write
where hi = (zlhe)/z, and f j = (s,fj)/z1, since x1 is not in supp(fj) and supp(fj) $ supp(u) we get y E LP, and the proof of the claim is complete. Letg be the result of evaluating y at z1 = = ~ t - 1 = 1, notice that deg(g) = p - 1, while by evaluating sly E LP at the same points we obtain deg(g) >_ p , which is a contradiction. Altogether in this case y E LP. (b) Assume Isupp(u)I = t - 2. If
then by induction hypothesis on t we derive y E LP, otherwise we reduce to a case similar to (a). (c) If Isupp(u)) 5 t - 3. Since deg(hi) = t - 2, supp(hi) @ supp(u) and supp(fj) @ supp(u) for all i , j . Therefore we may apply the method of proof of Proposition 7.4.5 to conclude that (after at most a1 - ax rearrangements) we reduce to a case similar to (a), (b), or there is some xj in the support of u , hi, and f j for all i, j , which implies (by induction on t ) that y E LP. 0
+ + e
Definition 7.4.7 Let I and J be two monomial ideals of the polynomial rings k[xo, . . . ,x,] and k[yo, . . . ,yn] respectively. The join of I and J is:
I*J=I+J+K,
Theorem 7.4.8 ([72])Let R = k[xo,.. . ,x m ] and S = k[yo,.. . , gn] be polynomial rings over a field k , andlet I , J be two monomial ideals of R and S respectively. If I and J are normalidealsgenerated b y square-free monomials of the same degree t 2 2, then their join I * J is normal.
Chapter 7
246
+ +
Proof. Set X = (10, . . . ,xm}, Y = {yo, . . . ,yn} and L = I J K , where K = ( X )( Y ). By induction on p we will show that LP, = LP for all p 2 1, where Lz denotes the integral closure of LP. If p = 1 then L is a radical ideal (see Corollary 5.1.5), hence L is integrally closed. Assume L6 = Li for i < p and p 2 2. Using Proposition 7.3.2, we have
Lz = ( { X I z is a monomial in k [ X ,Y ]and zq E LqPfor some q 2 1)). Let x be a monomial in LP,, then zQ E LQP,q Since LP,c Lz-' = we canwrite
> 0. Let us show z
E
LP.
where M is a monomial, the hi's are monomials of degree two in K ,the gi's and fi's are degree t monomials in J and I respectively. Likewise we can write z Q = N h ' , . . . h ' , , g ~ . ' . g : , f : ' . 'f:P-rl-sl) where N is a monomial, deg(hi) = 2 and hi is a monomial in K for all i, and fj are degree t monomials in J and I respectively for all i, j . From the last two equalities we have
From Eq. (7.5) one readily derives the inequality
We may assume M = x" or M = y", otherwise x E LP. By symmetry we may assume M = x", where CY 1 0. (a) If t 2 3, then r = 0 or p - r - s - 1 = 0, otherwise g1 f i E L3 and hence x E LP. First we treat the case r = 0. By taking degrees in Eq. (7.5) w.r.t. the variables yo,. . . ,Yn one has s1 trl 5 qs, which together with Eq. (7.6) yields deg M 2 t. Let z1 be the result of evaluating z at yi = 1. From Eq. (7.5) we derive E ~ ~ P " s l " r 1c p ( P - 4 ,
+
and x1 E = IP-'. We maywrite z = yezland x1 = xpw, where deg(ye) = s and w is a monomial in I*-3 of degree t ( p - s). Since deg(z1) = deg(M)
+
+ s + t ( p - s - 1)
we obtain deg(x1) 2 s t ( p - s), hence deg(x8) 2 s. Altogether we derive z = yezflw E LP. Next we consider the case r = p - s - 1 >, 1, observe that
Monomial Subrings
247
deg(k1) 5 1,otherwise x E LP. Therefore either x = hl h,gl may rewrite X = Y'hl *--hshg+lgl **-gr-1, e
a
-
a
g,., or we
where deg(h,+l) = 2 and hs+l E K , interchanging the xi and yi variables we may apply the arguments above to conclude X LP. (b) Assume t = 2. Using x 9 E LqP onerapidly obtains deg(M) 2 2, hence we may assume r = 0 (otherwise x E LP) and the arguments of case (a) can be applied to conclude x E LP. 0 Corollary 7.4.9 ([256]) Let X = (20, . . . ,,xm) and {yo,. . . ,yn) be two disjoint sets of indeterminates 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 + K , where K = ( X ) ( Y ) . T h e n L i s a normal ideal. Proof. Proceed as in the proof of Theorem 7.4.8 and notice that in this caser=p-s-1. 0
Exercises 7.4.10 Let R = k[zl, . . . ,zn] be a polynomial ring and I = (fl, , . . ,fq), where k is a fieId. If fi is homogeneous of degree d 2 1 for a11 i. Prove (a) R ( I ) 21 R ( I ) / m R ( I )@ m R ( I ) as R(1)modules, where m = R+, (b) k[f1,. . . ,fq]
= R ( I ) / m E ( I )as k-algebras,
(c) if F = { f l , . . . ,f q ) is a' set of monomials, then R ( I ) k-algebras, where x = (11, . . . ,xn).
Use Proposition 7.1.2 to show that 4 and
k[F,tx] as
+ have the same kernel.
7.4.11 Let R = K [ z l ,. . . ,z m ,y1, . . . ,yn] and L = IsJt , where Is (resp. J t ) 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.
7.4.12 Let R = K [ x ]be a polynomial ring over a field K . If I is an ideal of R generated by square-free monomials of degree 2, then I and I 2 are both integrally closed.
Chapter 7
248
7.5
Ideals of mixed products
Let R = K [ x ,y] be a polynomial ring in two disjoint sets of variables x, y over a field K . We will study ideals of mixed products
+
such that k r = s + t , where Ik (resp. J,.) denotes theideal of R generated by the square-free monomials of degree k (resp. r ) in the x (resp. y) variables. The notion of ideal of mixed products was introduced in [241]. The aim is to present a complete classification of the normal ideals of mixed products, the main result (see Theorem 7.5.8 and [241])generalizes the normality of various previously known normal ideals and contain as a particular case the normality of generalized graph ideals associated to complete bipartite graphs. The dimension of the monomial subring K [ F ]will be examined, where F denotes the minimal generating set of L consisting of monomials.
Powers of ideals of mixed products Let R = K [ q , .. . , x m , y1,. . . ,yn] a polynomial ring over a field K . Given non negativeintegers k,r, s, t such that k + r = s + t consider the square-free monomial ideal of mixed products given by
where Ik (resp. J,.) is the ideal of R generated by the square-free monomials of degree IC in the variables X I , . . . ,x, (resp. 91) . . . ,yn) . Note that L has a d u d ideal L' obtained by an interchange of the two sets of variables:
It is convenient to set IO= JO = R. By symmetry there are essentially two cases: (i) L = IkJ,.
+ Is& with 0 ,< k < s, or
(ii) L = Ik Jr with k
2 1 or r 2 1.
One of the results that will be used throughout this section is the fact that the powers of the ideals Ik and Jr are complete, see Proposition 7.4.5. Remark 7.5.1 (a) If A = R [ y i - l ] is the Laurent polynomial ring for some yi, then J t A = J,",A,
Monomial Subrings
249
where Jt'-l is the ideal of R generated by the square-freemonomials of degree t - 1 in the variables y1, . . . , yi-1, yi+l, . . . ,Yn. In fact one clearly has Jt c Ji- 1, thus Jt A c Ji- A . On the other hand consider a monomial f in Jt"l, then yif E Jt and f E JtA. Hence JI_,A c JtA. (b) If a variable yi is not in a prime ideal p c R, then the localization of Jt at p is the same as the localization of Ji-l at p. A similar statement holds if a variable xi is not in p andone consider 11,(resp.instead of Jt (resp. Jt-l). We need this observation only when p is a face ideal. For use below recall that products of ideals and quotients of modules commutewith localizations. We also recall that the integral closure of ideals commute with localizations, see Proposition 3.3.4. For basic facts on monomial ideals see Chapter 5 . As usual a monomial f = x;1 . .x$y;l y> will be denoted simply by f = zayp, where a = (al, . . . am) and ,f3 = ( b l , . . . ,bn). The support of f , denoted by supp(f), is: e
e
-
We begin by treating some special cases. Lemma 7.5.2 Let R = K[xl, . . . ,X m , y1, . . . , yn] be a polynomial ring over a field IC. If
L = Jr+Im, then L is a normal ideal for any r. Proof. Let us prove directly that = LP for all p 2 1. Take f = P y p in F , where z" = x:' - x%. Thus fi is in LPi for some i 2 1. If aj = 0 for some j , then yip E J,? and yp E = JF. Hence one may assume that a1 2 1 is the smallest value of the ai. Note that p 5 a1 implies x" E I&. Assume p > al. It follows that yip E J$P-al). Using that J,. is normal 0 yields yo E J,P-"l, which together with x" e I$ proves f E LP. a
+
Example 7.5.3 If m = n = 3, then L = 52 12 is not a normal ideal \ L3, where f is the product of all the variables. because f E Proposition 7.5.4 Let R = K[xl, . . . ,Xm, y1, . . . ,yn] be a polynomial ring over a field K . If L = Jr+ImJt
with r = m -k t , then L is a normal ideal. Proof. By induction on r. If r = 1 or t = 0 the result is clear because of Lemma 7.5.2. Next we show by induction on p that = LP for all p 2 1; the case p = 1 is clear because L is a radical ideal. Assume Liis integrally
ChaDter 7
250
-
closed for 1 5 i < p and set M = LP/LP. If M prime ideal p of M.As M L) R / L P ,
#
(0), take an associated
the prime ideal p is a face ideal. Note that yi is in p for all i; for otherwise if yi 4 p , then by Remark 7.5.1 one concludes
and
Ji-l
is the ideal generated by the square-free monomials of degree
r - 1 in the variables y1, . . . ,yi-1, yi+l,. . . ,yn. Thus by induction on r , the immediate consequence is that the localization of M at p is zero, that is,
Mp = (LP/LP), = (O), a contradiction because p is in the support of the R-module M . Hence yi is in p for all i. - As p is generated by a subset of the variables, there is a monomial f in LP \ LP such that p = (LP:f). Since gif is in LP and p 2 2, one can pick a variable y1 in supp(f) such that deg,, (f) 2 degYi(f) for all i. Therefore one can write Y l f = gw1 ' wp, (1) where 201, . . . ,w p are monomials of degree T in L and g is some monomial of R. Note deg(g) > 0, because f E LPi for some i 2 1, and y1 # supp(g), because f # LP. Case (a): First we assume y j divides g for some j, note j # 1 because f 4 LP. Set c = deg,, (f). Since yFsl divides y1f one has that yf+' divides wl...w,,thusonemayassumeylisinthesupportofwifori=l, ..., c + l . In order to derive a contradiction note that y j E wi for all i = 1,. . , , c + 1, this follows by observing that if y j 4 wi, then from the equality
one derives f E L,. Therefore yjcsl divides f , which contradicts the choice of c. Case (b): Next we assume g = x". Onemayorder the w i such that wt E Im Jt for l 5 j and wc E J,. for l > j, where j is some fixed integer between 0 and p . Therefore from Eq. (1) one has
251
Monomial Subrings
On the other hand since f i E LPi for some positive integer i, one can find monomials 21, . . . , zpi in L of degree r such that fi
= hz1 *
(3)
zpi,
*
for some monomial h in R. We order the zi such that X[ E I m Jt for l 5 j1 and ze E J r for k' > j1, where j1 is some integer between 0 and pi. Using Eq.(2) and taking degrees in Eq.(3) gives i(jt
+ ( p - j ) r - 1) 2 i deg,(f)
2j1t + r(pi - j l )
e=+( r - t ) ( j l - ij) 2 i.
(4)
Hence j < p and consequently Isupp(z")( < m, because otherwise y E LP. Pick a variable x1 $ supp(x") and note that Eq. (1) gives degZl( f i ) = ij, which together with Eq. (3) yields j 1 5 ij, but this contradicts Eq. (4). 0 Proposition 7.5.5 Let R = K[zl,.. . , x,, y1,. . . ,yn] and let
L = Jr
+
I 8
Jt.
If t 2 1, r = s + t and 2 5 s < m, then L2 is not integrally closed. Proof. Since s
< m it makes sense to set f=(x1z2...~,tl)(y12...~t2_1yt"'~r).
Inorder to prove that f E equalities:
g , set
yo = 1 andobserve
that from the
belongs to I:+'
T
S+I
terms in J~
we obtain f E L28,as required. By counting degrees, and using s 2 2, it follows that f 4 L2. 0 Proposition 7.5.6 Let R = K[s.l,. . . ,x,, 91,. . . ,yn] and let
If s > k f 1 2 2, t 2 1 and k + r = s + t , then L3 is not integrally closed.
Chapter 7
252
Proof. Set
where zo = yo = 1. To show that f is in the integral closure of L3, itsuffices to observe that from the equalities.
it follows that f 2 is in L6. As deg, (f)= 2s + k - 1 and deg,(f) = 2r by a countingdegreeargumentit follows that f is not in L3. Proposition 7.5.7 Let R = K [ q ,. . . ,x m , 91, . . . ,~ L = Is
+ t - 1, 0
n and ] let
+ Js.
If 2 5 s < inf(m, n), then LS+l is not integrally closed.
-
Proof. Let f = (xf" zzJ:)(yl . y,+l). To show that f lies inthe integral closure of L5+l note that from the equality: e
a
it follows that f is in L(s+l)s. As s LS+l. in f is not follows that
2 2 , by a counting degree argument, it 0
Theorem 7.5.8 ([241]) Let R = K[xl,. . . ,xm,y1,. ring over a field K . If L = I k J r IsJt # R
. . ,yn] be a polynomial
+
is an ideal of mixed products with k ir = s + t , then L is normal ifand only if L can be written (up to permutation of k , s and r , t )in one of the following forms: (a) L = I k J r
(b) L =
+ Ik+lJr-l,
IkJr,
k
2 0 and r 2 1.
k 2 1 o r r 2 1.
(c) L = I k J , . + I s J t , O = k < s = r n ,
orO=t
Monomial Subrings
253
Proof. =+) If k = s, then r = t and we are in case (b). By symmetry one may assume 0 5 k < s 5 m. Taking into account the previous results one can readily conclude the following classification. If k = t = 0, then
J1 Js
+ 11, s = 1,
case (c). case (c) . + I s , 2 5 s < inf{ m,n } , not normal.
If k = 0 and t 2 1, then
L= If k
{
+ +
Jr IlJt, s = 1, JT 17nJt, s = m, Jr + Is J t , 2 5 s < m,
case (a). case (c). not normal.
2 1 and t = 0, then L=
If k 2 1 and t
{
+
IkJl Ik+l, P = 1, IkJn iIs, r = n, Ik Jr + I s , 2 5 r < n,
case (a). case (c). not normal.
2 1, then
L={
IkJ, 4- Ik+l Jr-1,
s =k
I k Jr
s
+ Is Jt
+ 1,
case (a).
2 IC + 2, not'normal.
Altogether one obtains that L has the required form.
+
e)(a): By induction on k P we show that L is normal. If k + P = 1, then L = J1+ I; and L is normal. Assume k P _> 2. Next we use induction = LP for all p 2 1; the case p = 1 follows because on p to prove that L is a radical ideal. Assume Li is integrally closedfor 1 5 i < p and set M = Lp/ LP. If M # (0), take an associated prime ideal p of &f. Since M L) R I P , the prime ideal p is generated by a subset of the variables.Note that by Remark 7.5.1 one has L, = (Ll),, under the following conditions
+
L1 =
Here Ji-l and have thesame meaning as inRemark 7.5.1. Thus if p is different from the maximal ideal m = R+, using the induction hypothesis and the fact that I k and Jr-1 are normal, one qoncludes that M p = (O), a contradictionbecause p isin the support of M , Hence' m is the only associated prime ideal of M and there is a monomial f in -L?i \ LP such that
Chapter 7
254
m = (LP:f ) . Observe that the support of f contains one of the y i variables; because if f = z", then f E LPi for some i 2 1 but this can only take place c LP which if r = 1 and f i E ( I k + l ) P i ) as I k + 1 is normalonehas f E is a contradiction. Pick a variable y1 in supp(f) such that degVl( f ) >, deg,, (f) for all i, then one can write Y l f = gw1 ' ' wp, (1) where w1, . . . ,wPare monomials of degree k+r in L and gis some monomial of R. Note deg(g) > 0 because f i E LPi for some i 2 1, and y1 is not in supp(g) because f 4 LP. Case (I): First assume y j divides g for some j fixed, note j # 1 because f 4 LP. Set c = degyl(f). Since yf+' divides y1 f one has that yf+l divides w1- w P ,thus one may assumey1 is in the support of wi for i = 1,.. . ,c+ 1. Note that Y j is in supp(wi) for all i = 1,. . . ,c 1; because if y j 4 SUPp(wi), then from the equality *
+
91f = w1 * ' ' ( W Y j / Y l )
-
*
'
Wc+l
-
'*
wp(Ylg/Yj)
one derives f E L,. Hence yjC+l divides f , which contradicts the choice of 91 ' Case (11): Next assume g = x" and z j divides 9,for some fixed j , We divide this case into two subcases. Subcase (A): First assume that there is a term we of the form
we = (Xi1
. * * z i b ) ( y l ~ j*Z. * ~ j , ) , with
YI
E SUPP(WZ),
observe that z j is in supp(wl), otherwise write Y I =~
w *S.wt-l(zil * " z i k z j ) ( Y j Z
* * * ~ j , ) w e"'wp(Ylg/Zj), +l
to obtain f E LP. We claim that z j 4 n~'ls~pp(wi),because if z j is in the support ofwi for all i, then divides z j f , but since xj f E LP one such readily derives that f is also in LP, a contradiction. Then there is that (i)
w,={ (ii) If (i) occurs
4 suPP(wq), or 4 supp(w,)* say 4 {Xi1 ,
. * . z s b + l ) ( ~ t', * * ~ t , . - l ) , xj (281
{zsl 9
From
~ l =f gwewq
9
*
' * zSb)(Ytl
xsb+1}(xi1
a
i#h?
* *
z'j
!/t,), 9
9
},
251
*
wi = (Ylg/xj)(zslwo/Yl)(zjwq/zs*)
9
xik
}a
wi,
i#e,q
one derives f E LP. If (ii) occurs {gtl , . . . ,~ t } , pf {yj, , . . . , yj,. }, say gtl is not in {yjz, . . . ,yj, }. From
~ l =f gwewq wi
= (Y1g/zj)(Ytlwe/Y1)(zjwq/Ytl)wi, i#e,q
i#&q
Monomial Subrings
255 1
, I
one derives f E LP, again a contradiction. Subcase (B): Now assume all the terms wt that contain y1 in their support (and there is at least one of such terms) are of the form
. . , wq Consider the following two possibilities for the terms w1).
If (i) occurs and we is any term as in Eq. (*), pick
Note z j is in supp(w,) because otherwise from
one derives f E LP. Assume zj # supp(we) for some wl (that contains y1 in its support) and note that from
one derives f E LP. Thus the argument for case (i) is complete. Assume (ii), that is, all the wi have the form wi = z a g b , where the degree of za (resp. yb) is equal to IC + 1 (resp. r - 1). Using Eq.( 1) and the fact that g = x" one concludes degy ( f ) = p(r - 1) - 1. On the other
256
ChaDter 7
hand, since f j E LjP for some j 2 1, we get j deg,(f) 2 (r - 1)pj. Hence deg,(f) 2 (T - 1)p whichis impossible. This completes the proof of the normality in case (a). To finish the proof of the theorem note that the normality of case (b) follows readily using that Ik and 5,.are normal, and the normality of case (c) Proposition follows from 7.5.4. 0 Definition 7.5.9 Let G be a graph with vertices X I , . . . ,x n , the generalized graph ideal, denoted by Iq(G),is the ideal of K[xl,. . . x n ] generated by the square-free monomials xil xi, such that the xij is adjacent to xij+l for all 1 < j 5 q - 1. Corollary 7.5.10 If G is a complete bipartite graph, then the generalized graph ideal Iq(G)i s normal for all q 2 2.
Proof. Let X I , . . . ,x m , 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 1. Therefore =
resultthe and
+ It+l Jt { ItJt+l It Jt
follows Theorem from
if if
+1
q = 2t q = 2t,
7.5.8.
0
Segre products and a dimension formula Letusbeginwithtworesults that are useful. The first is the following dimension formula, and the second result states that certainmonomial subrings of mixed products are isomorphic to Segre products. Proposition 7.5.11 Let B and C be two standard algebras over a field I<, then their Segre product:
has Krull dimension equal to dim(B) + dim(C) - 1. Proof. See Exercise 4.1.25.
0
Definition 7.5.12 Let R = K [ x l , .. . ,xn] be a polynomial ring over a field K . The set of k-products in the Xi variables, denoted by vk, is defined as:
where k is a positive integer.
Monomial Subrings
257
Proposition 7.5.13 Let v k C K [ x ]and Vi C K [ y ] be the sets of k and r products respectively in disjoint sets of variables. Then the Segre product of K [ V k ] c K [ x ]and K[V,']C K [ y ) is isomorphic to
Proof. We grade K[Vk],K[V,'] and A with the normalized grading, thus their generators have normalized degree 1. It is no hard to see that there is a well defined graded epimorphism
such that ~ ( f g=) f @ g for all f E V #and g E V,', where
Since dimK(Ai) = d i m ~ ( K [ V &@ K K[V,']i) for all i one obtains that the map p is an isomorphism. Lemma 7.5.14 Let A be the matrix with rows
where ei is the ith unit vector in Rn. Then det(A) = r for n 2 2, Proof. Expanding by the first column of A and using induction on the size nmatrix, the of the result follows rapidly. 0 Proposition 7.5.15 If L is the ideal of mixed products in m and n variables and F the minimal generating set of L consisting of square-free monomials, then the Krull dimension d of K [ F ]is given b y
Proof. The first formula follows at once using Proposition 7.5.11 and Proposition 7.5.13. To prove the second formula consider the linear subspace W of Rm+n generated by the set of exponent vectors of the monomials that minimally generate L . Note that the exponent of xi is ei and the exponent of yi is e,+i, where ei is the ith unit vector. Recall that dim K [ F ]= dimK(W), that is, the proof reduces to showing dimK W = m + n.
.
"
Chapter 7
258
Give two integers 1 5 i < j 5 m, pick integers j 1 , . . . ,jk-1 between 1 and m such that i , j , j l , . . . ,jk-1 are all distinct, this choice is possible because 1 ,< IC < m. Hence subtracting the vectors
we get ei - ej in W for 1 5 i < j 5 m. A symmetricargumentusing 1 5 t < n shows that ei - ej is in W for m 1 5 i < j 5 rn n. Similarly subtracting
+
-+
yield that eik+l ei, - ejt+, or equivalently the vectors ei, all possible choices. By adding + e
a
.
+
- ej, are in W for all possible choices, - ejl - * * - ej,,-, are in W for
+ - - + ei,-k
s-k
r-t
L=2
e=2
to any of these vectors we conclude (s - Ic)ei, - ( r - t ) e j , E W , but by assumption s - k = lr - t , thus ei, - ej, are in W for 1 2 il 5 m and m<j&m-i-n. Altogether ei - e j are in W for all 1 5 i < j 5 rn + n. To finish the argument apply Lemma 7.5.14 to the matrix having first row
to conclude dirnK(W)
= n + m.
0
Remark 7.5.16 There is a notion of sequentially Koszul algebra that has been recently studied [7]and that is likely to hold for some members of the family of monomial subrings associated to mixed products. See also [152] for the related notion of strongly Koszul algebra.
Exercises 7.5.17 Let Vk c K [ x ] and Vi C K [ y ] be thesets of k and r products respectively, in disjoint sets of variables, and
A = K [ {f g {f
E v k and g E
Vi}].
If K[Vk],K[V,.]and A have the normalized grading, then for all d one has the equality dim&&) = d i m ~ ( K [ V &CIIK K[V&).
”.
.
Monomial Subrings
259
7.5.18 Let L = I k JT+ IsJt be an ideal of mixed products with k + T = s + t and F the minimal generating set of L consisting of monomials. Prove that dim K [ F ]= m n in the following three "extreme" cases:
+
(u) O < k < s < m (b) l < k < s < m (c) O = k < s < m ,
and and and
l
+
7.5.19 Let L = I1J 4 1451 be an ideal of mixed products in the polynomial ring K [ z l , . . . ,q , y1,. . . ,p4]. Prove that the integral closure of L3 is not generated by monomials of degree 15.
7.6
Degree bounds for some integral closures
Let S be an affine domain and 3 its integral closure or normalization. When S is graded, it is rarely possible to have estimates of the' degrees of the generators of 3 unless there are very circumscribedsituations. Here we discuss one of these when S is a monomial subring permitting a sharp regrading. In this section we use some techniques introduced by W. Brurls, J. Gubeladze and N. Trung [41]. Theorem 7.6.1 ([51]) Let R = K [ a l , .. . ,x,] be a polynomial ring over
an infinite field K and F a finite set of monomials in R of the same degree k . Let I be the ideal of R generatedby F and , 00
R = @ lit c R[t] i=O
its Rees algebras. Then the normalization R is generated as an R-module, and thus as an R-algebra, by elements g E R[t]of t-degree at most n. Proof. Let F = { f l , ... , f q } and R = K[zl, ...,x n , f l t , . . . , f q t ] . We introduce a new grading S on R[t] by setting
S(xi) = 1 and S(t) = -(k
- 1).
Note that R is generated as a K-algebra by elements of degree 1 because &(fit) = S(fi) 6(t) = 1. Consider the subsemigroup
+
of W+l generated by e l , . . . ,e,, log(f+t), . . . ,log(f,t). Since ZC = Z n f l , according to Theorem 7.2.28 the integral closure of R can be expressed as:
Chapter 7
260
where Iw+ C is the cone spanned by C in IWnfl. If M = xatb with (a,b) # 0 and (a, b) E Zn+l t l F C , then it is not hard to show that S ( M ) 2 b. Therefore is generated as a K-algebra by monomials of positive degree. There is a Noether normalization of R
A = K [ Z l , .. . ,&+I] where
$J-4R9 R,
. . ,xn+1 E R1.Note that
21,.
is a Noether normalization of R. By Theorem 7.2.35 R is Cohen-Macaulay. Hence by Proposition 2.2.14 R is a free module over A and one may write
where Mi = .Ditbi. Using that the length is additive one has the following expression for the Hilbert series
where hi = l { j I 6 ( M j ) = i}l. Note that a@), the a-invariant of E, is equal to s - (n 1). On the other hand by Proposition 4.2.3 one has
+
where WFis the canonical module of E. An application of Theorem 7.2.34 yields a ( R ) = s - (n 1) < o
+
and s 5 n. Altogether one has &(Mi)5 n. An straightforward computation shows that if Mi = .Ditbi, then the t-degree of Mi is less or equal than n, that is, bi 5 n. 0
For a more general version of Theorem 7.6.1 see [51]. We give a variant of the previous theorem for the subalgebra generated by monomials of the same degree k . A homogeneouspolynomial of degree ik is said to have normalized degree i. Theorem 7.6.2 ([51]) Let F be a finite set of monomials of degree IC and
s = K [ F ]c R = K [ q ,
a
n
.
,x,]
the monomial subring generated by F . Then the normalization 3 of S is g E R of generated as an S-module, and thus as an S-algebra, by elements normalized degree at most dim(S) - 1.
Monomial Subrings
261
Proof. Let
be the kth Veronese subalgebra of the polynomial ring R. Both S and 3 are contained in We grade 3 by setting = n R i k . Then we may n arguments the adapt of the previous proof.
si s
7.7
Degree bounds in the square-free case
Let R = K [ x l , .. . ,xn] be a polynomial ring over a field I< and F a finite set of square-free monomials of the same degree k . One of the goals is to present bounds for the generators of the normalization of the monomial subring K [ F ] .In this case there are embeddings
K [ F ]c S c R('), where S = K [ { x ~* ,
*
xik
I 1 5 i l < - < ik 5 n } ] ,
and R ( k )is the kth Veronese subring of R. All the constructions of K [ F ] take place inside of one of these algebras. Our exposition will follow closely that of [51]. Especially we want to present an explicit generating set for the canonical module ws of S that appears in [51]. In particular we will fully describe the algebras S with the Gorenstein property and compute thea-invariant of S. The analysis below will determine the Cohen-Macaulay type of S but also leads to the control of degrees in K [ F ] .
A description of the canonical module Let us fix some of the notation that will be used throughout the rest of the section. Let n 2 2k 2 4 be two integers (this is not an essential restriction; see Remark 7.7.12). We set
where el, . . . , e, are the canonical vectors in IW". The affine subsemigroup of Nn generated by d will be denoted by C, that is, we have
Chapter 7
262
The cone generated by C will be denoted by &C, thus
If 0 # a E Rn, then the setHa will denote the hyperplane of Rn through the origin with normal vector a. Thus Ha = (x E Rn I (x,a ) = 0 } , and Ha determines two closed half-spaces
Let R = K [ q ,. . . ,I C ~ ]be a polynomial ring over a field K . Observe that the K-subring of R spanned by the set (zala E A} is equal to the a f i n e semigroup ring K[C]. Remark 7.7.1 Let cp be the matrix whose columns are the vectors of A. By Proposition 7.1.17, we obtain rank(cp) = dimK[C] = n. Hence the vector subspace generatedby A is equal to Rn and dim I& C = n. The equations of the cone In order to describe the canonical module of S we will use the Danilov-Stanley formula. Thus the equations of the cone R+C should be determined first. Lemma 7.7.2 Set N1 = {-el,.
. . , -en})
N = Nl U N2 and
If H is a supporting hyperplane of the cone R+C such that H contains a set al,. . . ,an-1 of linearly independent vectors in A, then H = Ha for some vector a E N . Proof. Let 0 # a = (a1, . . . ,an), such that H = Ha. Let Ad be the (n - 1) x n matrix whose rows are the vectors 0 1 , . . . , an-l. First note that if the j-column of M is equal to zero, then (ai,e j ) = 0 for all i and H = H e j . On the other hand, if all the entries of the j-column of M areequalto 1, then H = Ha for some a N2. Therefore we may assume that all the columns of M have some zero entries and also some entries equal to 1. Set
without loss of generality we may assume
A = (ail i E Z} = { a l , . . . ,at},
Monomial Subrings
263
where e = lZl. By symmetry theproof can be easily reduced to thefollowing four cases. (a) Assume Z = 0. Using that all the columns of M have at least one entry equal to 1 we obtain that a = 0; hence this case cannot occur. (b) Assume 2 5 5 k - 1. We claim that there exists ai so that in the set of the first 4? entries of ai there is at least one entry equalto zero and at least one entry equalto 1. Otherwise, usingthat rank(lM) = n - 1 and rowreducing M to its normalform we obtain [ = 2, and hence a = ale1 - a1e2. Setting
p = el + e3 +
a
a
+ ek+l
and y = e2
+ e3 + . . - + ek+l
we get
(p,a) = a1 > 0 and (?,a) = -a1 < 0 , which is impossible because Ha is a supporting hyperplane of EL+C and the proof of the claim is complete. For simplicity we assume a1 = e i , +cia + . - + e i , +ej, +-+ej,, ei, = e l , rn r = k,
+
< . - .< i, 5 e < j 1 < < j r 5 n and r , m 2 1. Let us show (p,a) > 0, where ,b’ = el + + e l + e j , - + e i k - l . Note that (p,a) > a1 + a3 + ’ + at + aj, + - ’ + ajrome 3 5 i2
n . .
e
*
2
ai,
*
+ e
9
*
+ ai, + + ai, + aj, + + ajk-t 2 0, * * *
*
because (a1,a) = 0. As a consequence E A and (p,a) > 0, as required. If 2 5 1 31 5 k - 1, we may apply similar arguments to prove that there is a y E A so that ( y, a) < 0, which contradicts that H is a supporting hyperplane. Therefore we may further assume that 1 31is either equal to 1 or 1 312 k ; in this situation we can rapidly find y E A so that (y, a) < 0, which again yields a contradiction. (c)Assume = 1 and 131 2 k . Set r = 1 31and B = {aili E J } , say B = { a2, . . . ,a r + l } . If r = n - 1, then using (ai, a ) = 0 for all i we derive that all the entries of the first column of M are equal to 1, hence H = H a for some a E N2. Next we assume r 5 n - 2. Let ai be a vector with its first entry equal to zero. Since (a;,a) = 0 it follows that the first r 1 entries of ai are equal to zero and n > r + k. Setting
+
p = el + er+2 + -
+ er+k
and y = p
- el + e2
we obtain
(P,4 > 0
and (y,4 < 0, which is impossible. The case C = 1 and l J l = 1 can be treated similarly. (d) Assume t‘ 3 b and 1312 k. This case cannot occur because one can rapidly find vectors 0, y in A so that (p,a) > 0 and (7,a) < 0.
Chapter 7
264
Remark 7.7.3 The converse of Lemma 7.7.2 is also true because we are assuming n 2 2k 2 4, and this implies n 2 k 2. Note that if n = 3 and k = 2, then the cone IR+C has only three facets.
+
Proposition 7.7.4 (1511) A point x = ( X I , . . . ,x,) E Rn is in K C if and only if x is a feasible solution of the system of linear inequalities
rw+
-
Proof. Let C = H; n n H a , be the irreducible representation of R+C as an intersection of closed half spaces. By Theorem 7.2.18 the set Hbi fl& is a facet of &Note that Hb; is generated by a set of linearly independent vectors in A, see Corollary 7.2.22. Therefore by Lemma 7.7.2 one obtains Hbi = H a , for some a E N , here N denotes the set defined in 0 Lemma 7.7.2.
c
e
a
c.
A generating set for the canonical module Let X be an arbitrary subset of Etn. The relative interior of X , denoted ri(X), is the interior of X relative to aff ( X ) ,the affine hull of X . Lemma 7.7.5 Let a be a vector in Cnri(IW+C) and set A = (i I ai 2 2). If IAI 2 IC and il, . . . ,i k are distinct integers in A , then a’ = a - ei, - . - eik also belongs to C n ri(It.+C) . a
Proof. Without loss of generality onemayassume a1 2 a2 2 2 a,, ak 2 2 and a’ = a - el - . - .- ek. We claim that a‘ E ri(& C). First recall that C has dimension n; thus one has 1
rw+
by Proposition 7.7.4. Hence, using that a E ri(R+C) we readily obtain
On the other hand, if k
< i 5 n we have
Monomial Subrings As n 2 2IC
265
2 4 we obtain n
j=1
or equivalently, one has
Altogether making use of (7.7), (7.8), and (7.9) we obtain a' E ri(&C). By Proposition 7.4.5 the subring K [ C ]is a normal domain, and therefore C = Z C n Et+ C. Since a' E ZC we conclude a' E C. 0 Let R =
@E-,Ri be the standard grading of R and let
be the kth Veronese subring of R graded by
Notice that K [C ]is a graded subring of R(k)with the normalized grading:
where
(K[c]> =~K [ C ]n (R(")i. In the sequel we shall assume that K [ C ]has the normalized grading. Theorem 7.7.6 ([51]) Set S = K [ C ] .Let U S be the cunonical module of S and let 23 be the set of monomials M = x:' - . x 2 satisfying the following conditions:
If n 2 2k 2 4, then B is a generating set for u s .
Chapter 7
266 Proof. According to Theorem 7.2.34 we have ws =
({PI
aE
C n ri(&C)}).
Taking into account the arguments of the proof of Lemma 7.7.5 and by a repeated use of Lemma 7.7.5 it is enough to prove that !B c us. Let M E 2 3 ; without loss of generality we may assume M = xyl . xak-l k-l xk * x n , where a1 2 - 2 ak-1 2 1. The monomial N = xyl - xakdlcan be factored as
-
-
On the other hand, write
by the properties (a) and (b) it
where deg(Ni) =
n
k-1
i=k
i=1
follows that we can
(k - i)(ai - ai+l), if 1 ,< i 5 k - 2 ak-1)
ifi=k-1,
and deg(N') s 0 mod ( k ) . Hence
n
k-1
M = N'
(NiNi)
i= 1
is in K [ C ] which , readily implies M E us.
0
Applications of the description of the canonical module To find the presentation of K[C] as a quotient of a polynomialring B modulo a prime ideal P , consider the polynomial ring over the field K
with one variable Ti,...ik for each monomialX i l - . - x i k . Here B has the usual grading. There is a graded homomorphism of K-algebras
the ideal P = ker($) is the presentation ideal or toric ideal of K[C].Thus the Cohen-Macaulay type of the ring K [ C ] ,denoted by type(K[C]), is the last Betti number in the minimal free resolution of B / P as a B-module.
267
Monomial Subrings
Remark 7.7.7 Set S = K [C ]. By Corollary 4.3.6 the type of S is equal to the minimal number of generators of the canonical module U S of S. To compute the type of S notice that a monomial M = . . xab-l k-1 xk ' ' ' x n is in '123 if and only if for all 1 5 i 5 IC - 1 one has k-1
Caj=mk-n+k-1
forsome m 2 2 .
and 1 < a i < m - 1 ,
j=l
These two conditions imply '17.
-
Corollary 7.7.8 Let ws he thecanonicalmodule IC = 2 and n 2 2k. If n is odd, then
Corollary 7.7.9 If k = 2 and n 2 2k, then
type(V1) =
(n2- 4n 2
+ 2)
if
n
is even.
Proof. The details are left as an exercise. Corollary 7.7.10 If n = 2k
+ 1 2 5, then
Proof. Left as an exercise.
0
Let S be a Cohen-Macaulay positively graded K-algebra over a field K and U S the canonical module of S. We recall that a ( S ) , the a-invariant of S, is given by a(S) = -min{ i I ( w s ) ~ # 0). Corollary 7.7.11 ([51]) If n
2 2k 2 4, then the a-invariant
a ( K [ C ]= ) -
If1'
where [x1 is the least integer greater or equal than
x.
of K[C]is:
268
Chapter 7
Proof. Set S = K[C]and m = It follows fromRemark 7.7.7 that the degree of the generators in least degree of U S is at least m. To complete the proof we exhibit some generators of us living in degreem. Write n = qk+r, 0 ,< r < IC; note q 2 2. If r 2 1, observe that the monomials X: * * * x:-,Zk-p+l * Zk-1Zk * X n and belong to (us)m.In particular, S cannot be a Gorenstein ring in this case. If r = 0, then the monomial M = 2 1 2, satisfies M E 0 Remark 7.7.12 (Duality) Let R = K [ x l , ., . ,x,] be a polynomial ring over the field K , and IC an integer such that 1 5 IC <, n - I. Consider sn,k
= K [ { x ~ , xik I 1 5 il
< - < i k 5 n)],
the K-subring of R spanned by the xil - - . x i k ’ s . Observe that there is a graded isomorphism of K-algebras of degree zero: p:
Sn,k
+S n , n - k ,
where { j l , . . . ,j n - k } then
induced by
-
p ( ~ i ,* x i k ) = X j 1
*
‘Xj,,-k)
= { 1, . . . ,n } \ {il,. . . ,ik}. In particular if n
5 2k,
Because of this duality one may always assume that n 2 2k. The next corollary was shown by DeNegri and Hibi [Sl]using different methods. Corollary 7.7.13 ([51]) The subring S n , k is a Gorenstein ring if and only if k E (1,n - 1) or n = 2 k .
s
s
Proof. Bydualityonemayassume n 2 2k 2 4. Set = s n , k . If is Gorenstein, then by the proof of Corollary 7.7.11 we may assume n = qk. If q 2 3, then 2 1 * xn and x ~ x xiel ~ xk X , belong to (us)qand (os)q+l respectively, which is impossible. Therefore q = 2, as required. Converselyassume n = 2k. Let 93 beas in Theorem 7.7.6. Take a monomial M in 2 3 ; it suffices to verify that M is equal to 21 zn. One xak-’ x k * * Z n , where ai 2 ai+l 2 1. By may assume that M = x;1 hypothesis one has
-
-
9 .
1
-
a
e
k-1
j=1
for some m 2 2. On the other hand one has k-1
(7.11)
.
,
ngs
269
Monomial Next we combine (7.10) and (7.11) to obtain ai using (7.10) again we rapidly derive
5 m - 1 for all i. Therefore
k ( m - 1) - 1 5 ( k - l ) ( m - l),
which yields m 5 2. As a consequence ai = 1 for all i , as required.
0
Proposition 7.7.14 ([51]) Let F be a finite set of square-free monomials of degree k in R and let A = K [ F ]be the K-subring of R spanned by F . If dim(A) = n, then
a@) 5 a ( W I ) ,
where 2 denotes the integral closure of A and C is the subsemigroup of W generated b y A = { e i , + eikl 15 il < < ik 5 n } . . . e +
Proof. Let C and CF be the subsernigroups of I V generated by A and log(F) respectively. Set S = K[C].Since S is normal by Proposition 7.4.5, we obtain C S. Let
x
M I = {xal a E G
n r i ( R + G ) } and M2 = {xala E C f7 r i ( Q C)),
Notice Rt-G= R+ CF and aff (E%+C F )= Rn . Therefore the relative interior of Et+ equals its interior in Rn. For similar reasons we have ri(& C) = (IR+C)O. Hence r i ( & G ) c ri(IR+C). Altogether we obtain MI c Mz.Let xb be an element of minimal degree in MI so that deg(zb) = -a@). Set T = - a ( a ) . Since zb is in M2 and x b E AT c S,, we conclude
-a(S) = min{deg(za)la E M2) 5 T = -@). Hence a@)
5 a(S), as required.
0
Corollary 7.7.15 ([51]) Let R = K [ z l , .. . ,x,] be a polynomial ring over a field K and F a finite set of square-free monomials of the same degree k . If dim K [ F ]= n, then
Proof. Use Corollary 7.7.11, Proposition 7.7.14, and duality.
0
270
Chapter 7
Remark 7.7.16 It is not hard to verify the equality
where l3(&) is the k t h Veronese subring of R. Let K[C] be the subring generated by the square-free monomials of degree k. Notice that
a(K[C])5 a(R("). For n = 5 and b = 3 we have
a ( K [ C ] )= -3 < -2 = a(R(')). Because of this, to keep better control of the a-invariant, it is preferable to embed K[F] into K[C], instead of R ( k ) . Corollary 7.7.17 Let R = K[xl, . . ,xn] a polynomial ring over a field K and F a finite set of square-free monomials of degree k in R. If K[F] has dimension n and n 2 2k 2 4, then K[F] is generated as a K-algebra by elements of normalized degree less or equal than n -
Proof. Proceedalongthe Corollary 7.7.15.
lines of the proof of Theorem 7.6.2 and use 0
Example 7.7.18 Let K [ F ]be the subring of R = K[xl,. . . ,x81 spanned by the monomials of R defining the edges of the graph shown below.
-
The generators of K [ F ]can be computed using the program [48], see also [298, Section 7.31. A Noether normalization for K[F] is given by
ngs
Monomial
H ( K [ F ]x, ) =
271
+ + 4z2 + 8z3 + z4
1 32
(1 - 2)8
and a ( K [ F ] = ) -4.
Altogether K [ F ]is generated as anAo-module - by monomials of normalized degree less or equal than dim R u ( K [ F ] = ) 4. However, as a K-algebra it is generated by square-free monomials of normalized degree at most 3.
+
Next we show a genera1 statement on degrees valid when S and 3 are both Cohen-Macaulay. Proposition 7.7.19 Let S be a standard graded algebra and let M c N be finitely generated graded S-modules of the same dimensiond and multiplicity e. If M and N are Cohen-Macaulay and
are their Hilbert series, thendeg f ( t ) 2 deg g ( t ). Proof. Consider the exact sequence of graded modules. If M # N , P 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 1.3.9) implies that P is Cohen-Macaulay of dimension d - 1. We have the equality of Hilbert series,
and therefore
d t ) - f ( t >= (1 - t ) h ( t ) . The assertion follows since the h-vectors of these two modules are positive (see Chapter 4 and [44, Corollary 4.1.101). 0
272
Chapter 7
Lemma 7.7.20 If K [ F ]c K [ x ]is a monomial subring over a field K , then there is a positive integer no such that ( x " ) ~ O E K [ F ]for all x" E K [ F ] . Proof. By Lemma 7.2.32 there are 71, . . . ,yT E RP such that
K [ F ]= K [ x Y 1 ,... ,xrr]. For each i one can find a positive integer ni such that xniYi E K [ F ] . It is readily seen that the integer no = Icm(n1, . . . ,n,) satisfies the required condition, because any x" E K [ F ]can be written as x" = xalrl . x a ~ " r , for some a1, . . . ,a, E N. 0 Proposition 7.7.21 If A = K [ F ]c K [ x ] is a monomial subring over field K and I is the ideal
a
I = ( A : A X )= {U E A( a?l c A } , then there is a monomial x7 E I . In particular I
# 0.
Proof. Let 71, . . . ,7,. E RP such that 2 = K[xYl,. . . ,32771, One can write xri = &/xdi, where .pi and xdi arein A . By Lemma 7.7.20 there is 0 # no E N such that xnO" E A for all z" E X. Set x7 = xnodl xnodr, Take x" in andwrite x" = ( X Y ' ) ~. ~ where ai E N for all i. By the division algorithm there are integral equations ai = qino + ci, where 0 5 ci < no for all i. From the equality
--
x
"T
T
T
i=l
i= 1
i= 1
is in A
is in A
is in A
one concludes xYx" E A. Since 2" was an arbitrary monomial in 2, we get x Y x c A and x" E I . 0 Proposition 7.7.22 If A = K [ F ] c K [ x ] is a homogeneousmonomial is a positively graded K-algebra andthe subringover a field K , then multiplicity of 2 is equal to the multiplicity of A .
x
Proof. Set A = log(F) and P = conv(A). As A is homogeneous,using Proposition 7.2.43 one has that
i=O
is a graded K-algebra whose ith component is given by
Monomial Subrings
273
Recall that A is graded as in Proposition 7.2.39. Let h and E be the Hilbert functions of A and 2 respectively. Since A and 2 have the same dimension d, one can write
+
h(i) = aoid-’ terms of lower degree, ’ E ( i ) = coi““ + terms of lower degree,
-
for i >> 0. One clearly has a0 5 co because Ai c Ai for all i. On the other c A. hand by Proposition 7.7.21 there is 27 E A of degree nx such that Thus for each i there is a one to one map
Hence E ( i ) 5 h(i = CO. Therefore
+ m) for all i, and consequently
co
5 ao. Altogether
a0
e ( A ) = ao(d - I)! = co(d - I)! = e ( 2 ) , as required.
0
Corollary 7.7.23 ([51]) If K is a fieldand K [ F ] is a Cohen-Macaulay homogeneous monomial subring, then
Proof. Use Proposition 7.7.19 and Proposition 7.7.22.
0
Exercises 7.7.24 Let F be a finite set of square-free monomials of degree k in the polynomial ring K [ z l , .. . ,z n ] , I< a field. If n < 2k and K [ F ] is not a normal domain, prove that K [ F ]cannot be generated as a I< algebra by square-free monomials. 7.7.25 ([130, 2191) Let R(’”) be the k t h Veronese subring of a polynomial ring R in n variables over a field K . Prove that R(’) is Gorenstein if and only if k divides n. Hint The interior of the cone spanned by the exponent vectors in R(k)is the “first open quadrant’’ of Iw” and use the Danilov-Stanley formula.
7.7.26 Let R(’”)be the k t h Veronese subring of a polynomial ring R in n variables over a field I(. Prove thatthe n-invariant of is given by
Chapter 7
274
7.7.27 If K [ F ]is a two dimensional homogeneous monomialsubring, prove that the Hilbert polynomial of K [ F T ]is equal to the Hilbert polynomial of K [ F T ] ,where T is a new variable. 7.7.28 Let A = KIF] be a Cohen-Macaulay homogeneous monomial subring. If dim(A) = 2 and a ( A ) < 0, prove that A is normal.
7.7.29 Let S be the lcth square-free Veronese subring of a polynomial ring R = K[zl,.. . , x,] over a field IC. If KS is the field of fractions of S. Use the equality k+l h
xi-1xixi+l* *
21 i=2
(22
-
~ k + l
= x:,
’5k+l)k”1
to derive x! E K s . Using a similar equality proveK[x:, . . .,x:] c K s , then conclude tr.degK(S) = n.
7.7.30 Let F be the setof square-free monomials of degree k in the variables 5 1 , . . . ,x, and S = K [ F ]the monomial subring spannedby F endowed with the normalized grading. If n = 2k + 1 2 5, prove that the canonical module of S is given by US
= ({sal card({ila; = 2)) = k - 1 and card({ilai = 1)) = n - k
+ 1)).
Recall that a standard Artin algebraA over a field K is called a level algebra [27] if all thenonzero elementsof the socle of A have the samedegree. Prove that if hl, . . . ,h, is a system of parameters for S = K[C]with each hi a form of degree 1, then S/(hl,.
e . ,
hn)
is a level algebra. Consider the homogeneous subring A = K [ F ] where , F is the set of all monomials xixj such that X i and xj are connected by a line of the graph:
z,
Note that A is a graded subringof where both rings are endowed with the normalized grading. In the following problems we indicate how to compute some invariants of A and
x.
Monomial Subrings
275
7.7.31 Use CoCoA, and the procedures indicated below, to prove that the Hilbert series and Hilbert polynomial of A and are given by:
x
F(A,t)=
1+t+t2+t3+t4 (1 - t ) 7 1 144
cp(A,t ) = -t6
@, t ) = -t61 144
,
F ( Z , t )=
1 + t + t 2 + 2t3 (1 - t ) 7
1 55 + -t5 + -t4 +65t3 + E t 2 + Et + 1 , 16 144 48 12 18 + “ 174 5 + -t4 55 + -t3 21 + g t 2 + -t 157 + 1. 240
144
16
18
60
In particular e ( A ) = e@) = 5, a ( A ) = -3 and a@) = -4.
7.7.32 Use this CoCoA procedure to compute theHilbert series and Hilbert function of A. Note that we are first computing the toric ideal P of A , through the usual procedure of elimination of variables. Use R ::= Q[t[1..8],x[l..7]]; Q := Ideal(x[l] * x[2] - t[l],x[2] * x[3] - t[2], x[1] * x[3] - t[3], x[3] * x[4] - t[4], x[4] * x[5] - t[5], x[5] * x[6] - t[6], ~ [ 6*]~ [ 7-] t[7], ~ [ 5*]~ [ 7-] t[8]); P := Elim(x, Q); Use R ::= Q[t[1..8]]; P := Ideal(t[2] * t[3] * t[5]?2 * t[7] - t[l]* t[4]^2 * t[6] * t[8]); Poincare(R/P); Hilbert(R/P);
7.7.33 Use Normalix to verify that required for the next computation.
x = A [ x ~ x ~ x ~ x ~This x ~step x ~ ]is.
7.7.34 Use the following CoCoA procedure to compute the Hilbert series of 2. Note that we are also computing the toric ideal P of 2. Use R ::= Q[t[l..9],x[1..7]]; Q := Ideal(x[l] * x[2] - t[l],x[2] * x[3] - t[2], x[l] * x[3] - t[3], ~ [ 3*]~ [ 4-] t[4], ~ [ 4*]~ [ 5-] t[5], ~ [ 5*]~ [ 6-] t[6], ~ [ 6*]~ [ 7-] t[7], ~ [ 5*]~ [ 7-] t[8], ~ [ *l ~ ] [ 2*]~ [ 3* ]~ [ 5*]~ [ 6I]~ [ 7-] t[9]); P := Elim(x, Q); Use R ::= Q[t[l..9]], Weights([l, I, I, I,I, 1,1,I, 31); P := I d e a l ( t [ l ] * t[4] * t[6] * t[8] - t[5] * t[9], t[2] * t[3] * t[5] * t[7] - t[4] * t[9], t[l)* t[2] * t[3] * t[6] * t[7] * t[8] - t[9]^2); Poincare(R/P);
Chapter 7
276
2 using the following input file for the program Normalix, and running Normalix with the option -h.
7.7.35 Verify the assertions about
.
bowt ie in 8
7 1 0 1 0 0 0 0 0 2
1 O 1 1 0 1 0 1 0 0 0 0 0 0 0 0 mode)
O 0 0 1 1 0 0 0
O 0 0 0 1 1 0 1
O O 0 0 0 0 0 0 0 0 1 0 1 1 0 1 or
(number of generators) (dimensionlattice) of ambient (generators)
(option
Some lexicographical Grobner bases A lexicographical order of the terms in a polynomial ring K [ z l ). . . z,] 7.8
is an 2:; for a fixed permutation order obtained by representing terms as x;; 7r of the variables and then using the lexicographical order. The Grobner bases obtained by lexicographical orders are called lexicographical. There areseveral classical examples of graded primeideals that areminimally generated by a lexicographical Grobner basis [59,66, 159, 2761. Below we study some special linear orders and relate them to toric ideals of subrings of k-products. )
- -
24
Some linear orders of the variables Giventwo integers n 2 2k we consider a “label set”: and a set of variables:
T = {T[il,.. . ,ik]l [il,.. . ,ik] E 3:}. Let
A = ( a i j ) be a non-singular matrix of order k with entries in R and ui = (ail,.
s
aik)
its ith row. This matrix defines a linear ordering in Zk given by 5=(21,...,zk)>Ay=(ylc...)yk)iffAz>l,,Ay,
that is z > A y if the first non-zero entry of (z - u1 - y u1,. . . ,x uk - y is positive. Thus we can linearly order the variables in T by: T [ i l , . . ,ik] > T [ j l , .. . ) j k J iff (il, . . . ik) )
>A
(jl).. . , j k ) .
a
uk)
277
Monomial Subrings
Example 7.8.1 If n = 5 and k = 2, then the linear ordering induced by the matrix:
A = [ -1
'1
0
A distinguished group of matrices Consider the set A of matrices of order k of the form
such that the set { e i , , . . . e i k } is a permutation of the set of canonical vectors { e l , . . . e k } in Rk. Note that A is a group of order 2"!. As the rows of any matrix in A are orthonormal, we see that A is a subgroup of the orthogonal group 01,(R) . )
)
Finding lexicographical quadratic Griibner bases Let K be a field and R = K [ q ) .. . zn] a polynomial ring. Let V k be the set of square-free monomials of degree k in R. Consider the presentation: )
thus P = ker(p) is the toric ideal of K[Vk]. Remark 7.8.2 It has been shown by Bernd Sturmfels that there exist a quadratic Grobner basis, with respectto some term order, for the toric ideal P of K[Vk], see [278]. The examples below can be verified using Mucauluy. We thank Eduardo Viruefia Silva for writing some Pascal programs to generate all the matrices in A, and the corresponding linear orderings of the variables. Example 7.8.3 The matrices below are precisely the matrices in A that induce a quadratic Grobner bases for the toric ideal of Q [ V k J corresponding to k = 2 and n = 6.
A , = [ 01 1 01
A 3 = [ ;
-;]
A4=[
-;-;]
Chapter 7
278
Example 7.8.4 For n = 6 and k = 3 the full list of matrices in A that induce quadratic Grobner bases are:
;
A s = [ 1; 0
&I=
[
-HI
A g = [ 01 10 0 1 0 0 1
0 1 0 0 0 11 1 0 0
[ -%] 0 1
A45=
Example 7.8.5 For n = 8 and k = 4 the complete list of matrices in A that give quadratic Grobner bases is:
[ -!] -; : :] [ [ 4 -; ; -i] [ 8 -: ; -I] [ -;I 0 1 0
A25=
0 0 1 0
0
A367=
0 - 1
0
0
0
A79=[!
0 0 0
A89=
0
1 0 0 0
0 1 0 0
-;I
0 - 1
0
0 - 1
0
0 - 1 0 0
A351=
0
A377=
0 1 0 0 0 01
0
1 0 0 Conjecture 7.8.6 If P is the toric ideal of K[Vk] and n 2 2k 2 4, then there exists a linear order of the variables in T so that the reduced Grobner basis of P , with respect to the lexicographical order, is generated by quadrics.
gs
Monomial
279
Exercises 7.8.7 If we order g;, this induces an order on 3E-k. Prove that P ( v k ) has a lexicographical quadratic Grobner basis if and only if P(vn-k) does. Here P ( v k ) denotes the toric ideal of K[Vk]. 7.8.8 (V.Bonanzinga) Let n 2 d 2 1 be two integers. If 3; has the lex order and N [ i l ) .. . id] represents the position of [ i l , . . . ,id] with respect to this order. Prove the formula: )
See [29] for more details about this formula.
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Chapter 8
Monomial Subrings of Graphs Here we study the monomial subrings associated to graphs and their toric ideals, A full description of the integral closure of those subrings will be presented along with some connections with graph theory and polyhedral geometry. Let G be a graph. The following diagram gives an idea of the objects associated to G that will be, studied along this chapter. Graph: G
/\ \ /
Edge ideal: I ( G )
Edge subring: b[G]
1r
Rees algebra
Incidence matrix: MG
Toric ideal
Toric ideal
Circuits & Smith form
Grobner basis
The edge subring k[G]associated to a graph G is the monomial subring generated by the monomials corresponding to the edges of G, where k is a field. The description of the integral closure of k[G]will be given in
28 1
282
Chapter 8
terms of special circuits of the graph (see Theorem 8.7.9). This description links the normality property of k[G]with the combinatorics of the graph G (see Proposition 8.8.11). The incidence matrixof G plays an important role because it is related to thetoric ideal of k[G]and its rank can be interpreted in graph theoretical terms.
8.1
The subring associated to a graph
Let G be a graph on the vertex set V = { V I , . . . ,v n ) and 03
R = k[Zl,. . .,x,] = @ Ri i=O
a polynomial ring over a field k with the standard grading. To simplify notation sometimes one identifies the indeterminate xi with the vertex vi. The monomial subring or edge subring of the graph G is the k-subalgebra
k[G]= k[{zizjI vi is adjacent to v j ) ] c R. If F = { f 1 ,. . . ,f q } is the set of monomials zizj such vi is adjacent to vj, the elements in k[G]are polynomial expressions in F with coefficients in k. A monomial f = z i z j in R, is said to be an edge generator of G if vi is adjacent to vj. As k[G]is a standard k-algebra with the normalized grading
there is a graded epimorphism of k-algebras
where B is a polynomial ring graded by deg(Ti) = 1 for all i . The kernel of 9, denoted by P ( G ) , is a graded ideal of B called the toric ideal of k[G] with respect to fi, . . . ,f q .
Definition 8.1.1 Let w = ( V O , VI, . . . ,V r = 210) be an even closed walk in G such that f i = zi-lzi.As
the binomial
Tw = TI
*
TT--1
- Tz . ' * Tr
is in P(G). One says that Tw is the binomial associated to 20.
Subrings Monomial
of Graphs
283
Next we identify a set of generators for P ( G ) corresponding to the even closed walks of G. A closed walk of even length will be called a monomial walk.
Notation For use below Zs denotes the set of all non-decreasing sequences = (il,. ,is) of length s. Let a l , . . . ,ap be a sequence and a = ( i l , . ,is) in Zs, then we set a, = ai, . - - ai,. cy
-
Proposition 8.1.2 ([303])If G is a graph and P = P ( G ) the toric ideal of the edge subring k[G],then (a) P = ({T,I w is an even closed walk)), and (b) P = ({T, I w is an even cycle)) if G is bipartite.
Proof. Let f 1 , . . . , fq be the edge generators of G and P the kernel of the homomorphism of k-algebras 'p: B
-+b[G],
where B is the polynomial ring lc[T1,. . . ,Tq]and cp(Ti)= ff.Recall that P is graded and
P=B. (6P.J. s=2
First we set
x3 = {T,l w is an even closed walk}. One clearly has (B)c P. To prove the other containment we use induction on s. First recall that P is a binomial ideal by Proposition 7.1.2. If s = 2, it is enough to note that the binomials in P2 come from squares. Assume Ps-lc (B). To show Ps c (a), take T, - Tp in Ps,where a = (il, . . . ,is) and = ( j l , . . . ,jB). Define GI as the subgraph of G having vertex set
VI = (xi E VI xi divides
fa}
and edge set
El = {{x,y} E E(G)I ft = xy for some E
-
{il,..
. , i s , j l , . . . ,j s ) } .
Note that if fil - - fi, = fjl - - fjm for some m < s and for some ordering of the generators then T, - Tp E (B);to prove it, notice that T, - TB can be written as e
-
and use induction hypothesis. Nowwe can assume fil fi, # fjl . fj, for m < s and for any re-ordering of the fi's. Observe that this forces G1 a
284
Chapter8
to be connected.Take wo E V I . Since fa = f p , it is easy to check that after re-ordering we can write fil, = 212k-2v2k-1 and fjk = Z)2&1z]2k, where 1 5 k 5 s and wo = 212,. Therefore the monomial walk w = {WO, . . . ,' ~ 2 ~ satisfies Tw = T, - Tp and the induction is complete. This proves part (a). Assume G is bipartite. The second part of the proposition can be proved fi, # fjl - fjm similarly ifwe notice that this time the condition fil for m < s and for any re-ordering of the fi's implies that the graph GI is an even cycle. 0
-
a
-
Definition 8.1.3 Let F be a finite set of monomials in a polynomial ring k[x] and let P be the toric ideal of k[F]. A binomial
is called primitive if there is no other binomial divides T" and T 6 divides TP.
T Y
- T6 E P such that T r
According to a result of [278], the setof all primitive binomials in a toric ideal P is finite and is called a Graver basis of P. Corollary 8.1.4 If G is a graph and f = T" - TP 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 8.1.2.
0
Corollary 8.1.5 If G is a bipartite graph and f = T" - Tp 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 8.1.2.
0
Proposition 8.1.6 Let G be a graph and let P(G) be the toric ideal of the edge subring k[G]. If f = T" - TD is a primitive binomial in P(G), then the entries of Q satisfy ai 5 2 for all i. Proof. By Corollary 8.1.4 one can write
where l is even, and w is an even closed walk in G of length C:
w = ( w o , ~ ~.,..,we--1 we = wo}
)
Graphs Monomial of Subrings
285
(including the end vertices) in the closed walk w. If wo occurs more than three times in w, one can write
w = { V O , V l , . . . , ~ t l = v O , ~ t , + l , . . ' , v t l + t z= ~ O , ~ t l + t z + l , . " , v e = ~ o } , note that tl must be odd; otherwise using w1 = {vo, V I , . . . ,wtl = wo} one
has that T,, = T' - T', where T' divides TIT3 - - Te-1 and T6 divides . . Te, a contradiction because f is primitive. By a similar argument t 2 must be odd. Thus tl t 2 is even, to derive a contradiction note that the closed walk T2T4 ,
+
w' = ( U t l + t 2 = vo,V t l + t 2 + l , is of even length, and
'' '
9
ve = 210)
apply the previous argument.
0
Lemma 8.1.7 Let P be thetoricideal of a monomialsubring K [ F ] . If f = T" - TO is a binomial in the reduced Grobner basis of P with respect to some term order, then f is a primitive binomial. Proof. See [278, Lemma 4.61.
0
Lemma 8.1.8 Let G be a graphandlet P = P (G) be the toric ideal of k[G].If f is a polynomial in any reduced Grobner 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 = Twfor some even cycle w of the graph G. Proof. Let A be a reduced Grobner basis of P and take f E d. Using Corollary 7.1.5 together with the fact that normalized reducedGrobner bases are uniquely determined, we obtain f = 2'" - TP. By Lemma 8.1.7 f is primitive, thus by Corollary 8.1.4 and Corollary 8.1.5 f has the required form. 0 Proposition 8.1.9 If G is a graph and P = P(G) is the toric idealof k [ G ] , then the set
{T, I Tw is primitive and w is an even closed walk} is a universal Grobner basis of P . Proof. It follows from Lemma 8.1.8. Proposition 8.1.10 If G is a bipartite graph and ideal of k[ G ] ,then the set
0
P = P ( G ) is the toric
{T,I w is an even cycle)
is a universal Grobner basis of P . Proof. It follows from Lemma 8.1.8.
0
Chapter 8
286
Exercises 8.1.11 Let G be a graph with q edges and P ( G ) the toric ideal of the edge If f is a primitive binomial in P ( G ) prove: subring k[G]. q if q is even, and q - 1 otherwise.
8.1.12 Let G be a bipartite graph and k[G]its edge ideal. Prove that k[G] is a normal domain.
8.2
Rees algebras of edge ideals
We will look closely at the presentation ideal or toric ideal of the Rees algebra of an edge ideal, and then show a few applications.
Notation For f and g in a unique factorization domain we define
As before Zsdenotes the set of all non-decreasing sequences a = (il, ,is) of length s. Let a l , . . . ,upbe a sequence and let a = (il, . ,is) E Zs, then we set a, = ai, ai, and Z,, = ai, . . ai, ai,.
-
h
e
Theorem 8.2.1 Let I be the edge ideal of a graph G and F = { f 1 , . . . , fq} the set of edge generators of I . If B / J is the presentationof the Rees algebra of I with respect to F , then
J = B J l + B f s=2 jPs), where Ps= {T,- T’l
fa
= fp, for some a,/3 E Zs).
Proof. Let X I , . . . ,xn be the set of vertices of G and R a polynomial ring in n variables over a field k. Here we identify the vertices of G with the variables of R. Let R ( I ) = R[IT]be the Rees algebra of I . The toric ideal J of R ( I ) is the kernel of the graded epimorphism:
It is sufficient to show the recursive formula
Js = B1 Js-l
+ RP,
for s 2 2.
Monomial Subrings of Graphs
287
According to Theorem 2.4.17 J , is generated by polynomials of the form
where a = ( i l , . . . ,is), p = (jl,. . . ,j,) are in X,. It suffices to prove that B3-1J1 RP,. Since for fa = fp all those generators are in B1J3-1 expression (8.1) simplifies to T, - Tp E RP,, we may assume fa # f p . In this case there is a variable z E {XI,.. . ,xn} so that f a = zah and fp = zbg, with a # b. Assume b > a 2 0, it is not harg to see that there are integers m and t so that [fa!, fp] is a nlultiple of f j m f a t o Notice that the generators of J , can be re-written as
+
+
The proof now reduces to showing that there are polynomials X, p , A in R and integers rn and t so that
and
To complete the proof observe that such polynomials exist if multiple of f j m f a t for some integers m and t.
[fa,fp]
is a
h
0
The example below proves that Theorem 8.2.1 do not extends to ideals generated by square-free monomials of higher degree. Example 8.2.2 Let R = k [ x l , .. . ,x71 be a polynomial ring and let I be the ideal of R generated by
Then using Macaulay [15]we get that the toric ideal of R ( I ) with respect to F = { f l , . . . ,f 4 } is minimally generated by the binomials
Chapter 8
288
Corollary 8.2.3 Let I = I ( G ) be the edge ideal of a graph G . T h e n I is a syzygetic ideal if and only if G contains no squares. Proof. Notice that according to Proposition 8.2.1 I is syzygetic if and only if RP2 = (0). Hence it issufficient to observe that if I is generated by fi, . . , ,fq then a non trivial relation of the form f i f j = f k f t can only occur if f i f j is a square-free monomial. 0 Corollary 8.2.4 ([303]) Let G be a connected graph and let I be its edge ideal. T h e n I is an ideal of linear type if and only if G is a tree or G has a unique cycle of odd length. Proof. Let V = (21,.. . ,xn} be the vertex set of G and I = (f1, . . . , fq). If I is an ideal of linear type then by Proposition 3.2.1 G is a tree or G has a unique cycle of odd length. Conversely assume G is either a tree or G has a unique cycle of length. We claim that RP, = (0) for s 2 2, we proceed by induction on n. If n 5 3 this is easily verified, assume the claim true for graphs with less than n vertices. Using induction on s wenow show RP, = (0) for s 2 2, notice that by Corollary 8.2.3 RP2 = (0). Assume RP,-1 = (0) and RP, # (0), take a nontrivial relation fil fi, = f j l . . 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 { f i l , . . . , fi, } = (91,. . . ,g,}, where gcd(gi, g j ) = 1 for i # j and g1 *gk = x1 xn. Notice that this equality cannot occurif n is odd, hence RP, = (0) and theproof of the claim is complete. That I is an ideal of linear type is now a direct consequence of Proposition 8.2.1. 0
-
e
e
Let us show an application of Proposition 6.3.7 that lends some support to the following open question [291]: Conjecture 8.2.5 (Vasconcelos) Let I be a Cohen-Macaulay ideal in a regular local ring R. If I is syzygetic and 1/12 is Cohen-Macaulay, then I is a Gorenstein ideal. Corollary 8.2.6 Let T be a tree with vertex set V and let I be its edge ideal. If I is Cohen-Macaulay and m = IVl 2 3, then
m
depth(l/12) = - - 1. 2 Proof. First note that I has the description given in Theorem 6.3.4, here we use the notation of this theorem. Set
Subrings Monomial
289 of Graphs '
I
~
where rn = 2r. Since I is syzygetic (see Corollary 8.2.3), we have an exact sequence o "+ H ~ ( I ) "+ (R/I)"-~ -+I / I ~ "+ 0.
If 1/12is Cohen-Macaulay, then Torp(I/12, R / ( z ) )= 0. Therefore if we tensor this exact sequence with R / ( z )we get that theideal
({Y,~I
1I iL
+({Y~wJ
adjacent to ye)>
obtained from I by ma-king xi = yi is syzygetic, which is contradicted by the relations ( y j y ~ = ) ~y2y2 3 E' 0 Corollary 8.2.7 Let G = K , be the complete graph on n vertices and let I be its edge ideal. Then the toric ideal of R ( I ) is generated by binomials of degree two. Proof. Let P = ker(B = k[T1,.. . Tq]% k[G]), where cp(Ti) = fi. Recall that we can write
By Theorem 4.3.7 (cf. [251]) the edge ideal I has a linear resolution, hence using Theorem 8.2.1 the proof reduces to showing P = (P2). Using induction on s we will show Ps c P 2 for all s. Assume P.+" c P 2 . Let f (T) = 2'" - TP E Ps.By the induction hypothesis we may assume a and ,k? have disjoint support and that f(T)comes from a monomial walk. In particular we may assume that al, a2, and p3 are positive integers so that f l = 21x2, f 2 = 22x3 and f3 = 3,324, where $1, . . . ,x4 are distinct vertices. Note that f k = 21x4 for some k and g(T)= T1T2 - T3Tk E P 2 . Using
f ( T ) = TYl-l T2a2-1T'3 . . . T:qg(T)
the induction assumption shows f(T)E P,, as required.
0
The next result uses a formula of Huneke and Rossi [187, 2611 for the Krull dimension of the symmetric algebra of a module. For another proof of this formula based on slightly different ideas, as well as for a nice survey on symmetric algebras, see [293]. Theorem 8.2.8 ([303])Let G be a graph with n vertices and q edges and I = I ( G ) its edge ideal. If G is connected, then dim(Sym(1)) = sup{n
+ 1, q ) .
290
Chapter8
Proof. According to the Huneke-Rossi formula we have
dim Sym(1) = sup{v(l,)
+ dim (R/p)I p E Spec(R)} ,
where v(IP) denotes the minimal number of generators of the localization of I at p. By this formula it is clear that the Krull dimension of Sym(I) is at least the given bound. To prove the converse, take a prime ideal p of height n - i containing I and set B = {v E Vlv 4 p}. It follows that p contains a face ideal Q, that is, Q is an ideal generated by a subset of the vertices of G. We may assume Q = (A),where A is a minimal vertex cover for G. Define now C = {x E A1 z is adjacent to some vertex in B }
The formula is now a consequence of Lemma 8.2.9 below,
0
Using the notation above one has: Lemma 8.2.9 Let G be a connected (n, &graph with vertex set V = V ( G ) and edge set E = E ( G ) . Let A be a minimal vertex cover for G and B a subset of V\A. Then IC1 lYl+ lBl 5 sup{n 1,q } .
+
+
Proof, Let Y' betheset of edges coverby C, that is, Y' = E(G)\Y. Consider the subgraph G' of G with edge set equal to Y' and vertex set
V ( G ' ) = { x E V(G)I x lies in some edge in Y'}.
If Y = 0,then lBl+ IC1+ lYl 5 n. Assume Y # 0, and denote theconnected components of G' by GI,. . . ,G,. Set Bi = x3 n V(Gi)and Ci = C n V ( G i ) ; notice B = uBi and C = UCi. We claim that lBil + lCil 5 ni - 1, for all i, where ni = IV(Gi)l. For that, fix an edge { x , g } E Y. If z E V ( G i ) ,then x 4 Bi U Ci; using that and Ci are disjoint we get the asserted inequality. On the other hand, if E V ( G i ) at a minimum distance from x. This yields a path {x = xo,2 1 , . . . ,x,. = x). As x,-1 $ V ( G i ) ,x 4 Bi u Ci, which gives (&I ICi I 5 ni - 1. Altogether we have t3i
x
4 V ( G i ) ,we choose a vertex x
+
Monomial Subrings of Graphs
291 .
But Y' is the disjoint union of E(Gi) for i = 1,
, a
e . ,
m, and thus
m
i= 1
here qi = IE(Gi)l. This permits writing the last inequality as
Since Gi is a connected (ni,qi)-graph we have ni - 1 5 q i . therefore
establishing the claim.
0
A dimension formula Let us describe the cycle space of a graph G over the two element field F = Z2. We denote the edge set of G by
v = {XI,.
e
*
,xn}.
Let Co 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
of points and a 1-chain is a formal linear combination
of edges, where ai E F and bi E F . The boundary operator transformation defined by
a is the linear
The cycle space 2 ( G ) of G over F is equal to ker(a). The vectors in 2 ( G ) can be regarded as a set of edge-disjoint cycles. A cycle basis for G is a basis for 2(G) which consist entirely of cycles, such a basis can be constructed as follows:
Chapter 8
292
Remark 8.2.10 If G is connected, then G has an spanning tree T. The subgraph of G consisting of T and any edge in G has exactly onecycle, the collection of all the cycles obtained in this way form a cycle basis for G. See [139] for details. In particular dimF 2 ( G ) = q - n
+ 1,
see Exercise 5.2.10. Lemma 8.2.11 Let G be aconnectedgraphand 2 e ( G ) thesubspace 2 ( G ) of all cycle vectors of G with an even number of terms. Then q -n
dimF 2e(G) =
+1
of
if G is bipartite, and otherwise.
Proof. Let cl,. . . ,q , cl+l,. . . ,c, be a cycle basis for G, where c1,.. . , c1 are even cycles and cl+1, . . . ,c, odd cycles. It is clear that a bases for 2, (G) is given by (c1,.. . ,c ~c1+1 , cm, . . . ,c,-1 cm}. 0
+
+
The next result shows how the Krull dimension of k[G] is related to the cycle space of its graph. Proposition 8.2.12 ([303]) Let G be a connected graph and let toric ideal of the edge subring k[G]. T h e n
P be the
ht(P) = dimF 2 e (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 an spanning tree T of G so that (after re-ordering) I ( T ) = (f1, . . . ,fn-l). Since I ( T ) is an ideal of linear type dimk[G] = tr.degk k[G] 2 n - 1.
If G is bipartite notice that fk E k(f1,.. . , fn-l) for k 2 n; to prove it we write fk = x x and observe that the graph T U {x,x } has a unique cycle of even length. This shows the equality dim k[G]= n - 1. If G is not bipartite, thenfor some f k = xx, k 2 n. Note that the graph T U (x,z } has a unique cycle of odd length. By Corollary 8.2.4 the ideal (fi,. . . ,fn-l, fk) is of linear type, hence dim k[G] 2 n; to show equality recall that tr.degk k[G] 5 n. To finish the proof useLemma 8.2.11. 0 Corollary 8.2.13If edge subring, then
G is a connected graph with
n vertices and k[G] its
Proof. It follows from the proof of Proposition 8.2.12.
Monomial Subrings of Graphs
293
The role of the cone of a graph 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. Example 8.2.14 A pentagon and its cone:
U G
Proposition 8.2.15 Let G be a graph and let Then there is an isomorphism
C ( G ) be the cone over
G.
R ( I ( G ) )& k[C(G)]. Proof. Let V = ( 2 1 , . . . ,xT1}be the vertex set of G and
the Rees algebra of the ideal I ( G ) = (fi,. . . , fq). We assume that fi, are the monomials in the xi's corresponding to the edges of G. Let
. . . ,fq
be the monomial subring of the cone. As these two algebras are integral domains of the samedimension by Proposition 7.1.17 and Proposition8.2.12 (cf. [94]),it follows that there is an isomorphism
q:R(I(G)+ ) k[C(G)], induced by q ( x i ) = tzi and q ( t f i )= fi. That is R ( I ( G ) ) k[C(G)].
0
Exercises 8.2.16 Let G be a connected graph with vertices the edge generators of G. Prove that (a) k(f1,
- . . ,&, 2 1 )
= k ( z l , . . . , xn), and
= n. (b) dim k[G][x:]
Here k ( F ) denotes the field of fractions of k [ F ] .
. . ,xn and f l , . . . ,fq
~ 1 , .
Chapter 8
294
8.2.17 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.
+
8.2.18 Let G be a graph with vertices z1, . . . ,x, and I c R = XI,. . , ~ its edge ideal. If cp, $ are the maps of k-algebras defined by the diagram
where C(G) is the cone of G and ker(p) = ker(+).
8.3
f1,.
n ]
. . ,fq are the edge generators, then
Incidence matrix of a graph
One of the most interesting matrices that have been associated to a graph is its incidence matrix, here we present some properties of this important matrix. The reader is encouraged to consult [136] for a thorough study of the minors of an incidence matrix. Let G be a simple graph with vertex set V = { X I , . . . , xn} and edge set E = { z l , . . . ,xq}, where every edge zi is an unordered pair of distinct vertices xi = {xij, X i h } . The incidence matrix MG = [ b i j ] associated to G is the n x q matrix defined by bij
=
{
1 if xi E z j , and 0 if xi x j .
Note that each column of MG has exactly two 1’s and the rest of its entries equal to zero. If x i = {xij,xirc} define ai = eij e i k , where ei is the ith canonical vector in Rn. Thus the columns of MG are precisely the vectors a l , . . . ,aq regarded as column vectors. As a simple example consider a triangle G with vertices X I , ~ 2 ~ x In 3 .this case:
+
+
+
+
with the vectors a1 = el e2, a2 = e2 e3, and a3 = el e3 corresponding to the edges x1 = { z 1 , x 2 } , 22 = { x 2 , 2 3 } , and z3 = { x ~ , Q } .
Remark 8.3.1 Let A be a square submatrix of MG. In [136] it is shown that either det(A) = 0 ordet(A) = k2k, for some integer k such that 0 5 k 5 TO, where TO 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.
Monomial Subrings 295 of Graphs Recall that a graph G i s bipartite if all its cycles are of even length. Thus any tree and in particular any point is a bipartite graph. The number of bipartiteconnectedcomponents of G will be denoted by CO, and the number of non bipartite connected components will be denoted by c1. Thus c = co + c1 is the total number of components of G.
Proof. Let G I , . . . ,G, be the connected components of G, and ni the number of vertices of Gi. After permuting the vertices we may assurne that MG is a "diagonal" matrix
where M G is ~ the incidence matrix of Gi.By Proposition 8.2.12 the rank of MG~ is equal to ni - 1 if Gi is bipartite, and is equal to ni otherwise. Hence rankthe of MG is equal to n - CO. 0 Definition 8.3.3 Let U be a square matrix with entries in a commutative ring R. The matrix U is called unimodular if det(U) is a unit in R. Theorem 8.3.4 ([136]) If G is a graphwith n verticesand M G isthe incidence matrix of G, then there are unimodular integral matrices U, V such that
where
D = d i a g ( l , l , . . . , 1 , 2 , 2 , .. . , 2 ) ,
n - c is the number if 1's and c1 is the number of 2's. Proof. See [136, Theorem 3.31.
0
The matrix S is called the Smith normal form of MG and the numbers in the diagonal matrix D are the invariant factors of MG. Corollary 8.3.5 Let G be a graph with vertices 2 1 , . . . , x n and dG the set of all vectors ei e j such that {Xi, xj) is a n edge of G. T h e n
+
where r = n - co is the rank of MG. Proof. It follows at once from the fundamental structure theorem of finitely generated modules over a principal ideal domain. See [190, Chapter 31. 0
296
Chapter 8
Exercises 8.3.6 Let G be a cycle of length n and MG its incidence matrix. Prove det(Mc) =
f l if n is odd
0
if n is even.
8.3.7 Let U be a square matrixwith entries ina commutative ringR. Prove that U is unimodular if and only if U is invertible.
8.4
Circuits of a graph and Grobner bases
Let G be a graph. In this section we study the toric ideal P(G) of the edge subring k[G]from the view point of Grobner bases, and examine the elementary integral vectors of the incidence matrix of G. For general results on Grobner bases of toric ideals a standard reference is [278, Chapter 41. Theelementary integral vectors Thenotion of elementaryintegral vector occurs in convex analysis [247] and in the theory of toric ideals of graphs [278, 3031.
If a E Iwn, its support is defined as supp(cx) = (i I ai # 0). Note that a = a+ - a _ ,where a+ and a- are two non negative vectors with disjoint support. Remark 8.4.1 If a is a linear combination of
/31
, , . ,pT E Iwq , then
Definition 8.4.2 Let N be a linear subspace of (@.An elementary vector of N is a nonzero vector a in N whose support is minimal with respect to inclusion, i.e. supp(a) does not properly contain the support of any other nonzero vector in N . The concept of an elementary vector arises in graph theory when N is the kernel of the incidence matrix of a graph G. The reader is referred to [247, Section 221 for a precise interpretation of the vector in N as flows of G which are conservative at every vertex. Lemma 8.4.3 If N as a linear subspace of and a, /3 are two elementary a= for some X E Q. vectors of N with the same support, then
Monomial Subrings of Graphs
297
Proof. If i E supp(a), then one can write ai = Api for some scalar A. Since supp(a - AD) 5 supp(a), one concludes a - X@ = 0, as required. 0
v,
Definition 8.4.4 Let N be a linear subspace of An elementary integral vector or circuit of N is an elementary vector of N with relatively prime integral entries. Corollary 8.4.5 If N is a linear subspace of N is finite.
of
v ,then number of circuits
Proof. It follows from Lemma 8.4.3.
0
Definition 8.4.6 Two vectors a = (ai)and ,f3 = (pi) in if aipi 2 0 for every i.
p are in harmony
Lemma 8.4.7 ([246]) Let N be a linear subspace of Q4. If 0 # a E N , then there is an elementary vector y E N in harmony with a such that SUPPtY) c SUPP(4. Proof. Let ,8 = (PI, . . . ,&) be an elementary vector of N whose support is contained in supp(a). By replacing p by -p one may assume ai& > 0 for some i. Consider
Note that zi = (ai - Xjpi)ai 2 0 for all i. Indeed if a& 5 0, then clearly x i 2 0, and if a& > 0 by the minimality of X j one has xi 2 0. Thus the vector a - Xjp is in harmony with a and its supportis strictly contained in the support of a. If a - Ajp = 0 we are done, otherwise we apply the same argument with a - Ajp playing the role of a. Since being in harmony is an equivalence relation, the result follows by a recursive application of this procedure. 0 Theorem 8.4.8 ([246]) If N is a vector subspace of Q? and a E N then a can be written as
\ (0},
T
a=zpi i=l
for some elementary vectors P I , , . . ,,&. of N with r 5 dim N such that (i)
PI,.
. . ,pT are in harmony with
a,
(ii) supp(pi) c supp(a) f o r all i, and (iii) supp(pi) is not contained in the union of the supports of for all i 2 2.
PI,.
. . ,pi-1
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298
Proof. By induction on the numberof elements in the support of a. If a is not an elementaryvector of N , then by Lemma 8.4.7 there is an elementary vector a1 E N in harmony with a such that supp(a1) C supp(a). Using the proof of Lemma 8.4.7 it follows that there is a positive scalar X1 > 0 such that a - Ala1 is in harmony with a and the support of a - Ala1 is strictly contained in supp(a). Therefore the result follows applying induction. This proof uses the original argument given in [246]. 0 Corollary 8.4.9 If N is a linear subspace generate N as a Q-vector space. Theorem 8.4.10 Let homomorphism
M be a n n X
of
v, then the circuits
q integral matrix and
of N
$ the Z-linear
$:ZQ_3 Zn given by $(a) = M ( a ) . If M hasrank n, then ker(l(l) isgeneratedasa Z-module by the elementary integral vectors of ker($). Proof. Set N = ker($). Let al, . . . ,cyq be the column vectors of M and ei the ith unit vector of Zq. By the proof of [278, Lemma 4.91, the set of circuits of N is the set of vectors of the form
--
< i n + l 5 q, and h is the greatest common divisor of the where 1 5 il < entries of the vector v, if v is nonzero. For convenience we introduce the notation v = v(i1,. . . ,i n + l ) and
Set d equal to the greatestcommon divisor of all thenonzero integers having the special form [il,. . . ,ij, . . . ,i n + l ] and n
Note ZB c L c N , where L is the Z-module generated by the elementary integral vectors of N . We claim that ZB = N . First observe that L and N have both rank g - n, see Corollary 8.4.9. Since Z B and L have equal rank, one concludes that N / Z B is a finite group. Thus to show Z B = N it suffices to prove that N / Z B is torsion-free or equivalently that Z q / Z B is torsion-free of rank n. Consider the matrix A whoserows are the vectors of B. By the fundamental theorem of finitely generated abelian groups theproof reduces to
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showing that the ideal Iq+,(A) of Z generated by the (q - n)-minors of A is equal to Z. 9 Fix A = [ i l , .. . ,z j , . . . in+l]/d. For simplicity of notation one may consider the case A = [l,. . . ,n ] / d . Note that A occurs in ~ ( 1 .,.., n, i ) / d with a f sign for i = n 1, . . . ,q. Thus the matrix
+
~2~
a21
A'=[
:
0
A 0
. . . . . .
whose i t h row is the vector ~ ( 1 ,. . ,n, i ) / d is a submatrix of A. Therefore Aq-n belongs to I q - n ( A ) . If Iq-+(A) # Z, there is a primenumber p dividing for all A. Thus p divides A for all A, andthis implies that pd divides d, a contradiction. 0 On the circuits of a graph Let G be a graph on the vertex set V with edge set E and incidence matrix M . We set N = ker(M), the kernel of M in @. Note that a vector a E N n Zq determine the subgraph G, of G having vertex set V, = { x E V 12 divide f a + } and edge set
where f 1 , . . . , fq are the square-free monomials corresponding to the edges of G, that is, the edge generators of G. Note that we are using the following notation:
Notation For a = ( a l ,. . . ,aq)E NQ and f l , . . . ,f q in a commutative ring with identity we set f a = The support of f a is defined asthe set SUPP(f") = {fi I ai # 0).
fr' -
e
The next result gives a geometric description of the elementary vectors of the kernel of the incidence matrix of a graph G. Proposition 8.4.11 ([303]) Let G be a graph with incidence matrix M andlet N be thekernel of M in p. Thenavector a E ZQn N is an elementary vector of N if and only if (i) G, is an even cycle, or (ii) G, is a connected graph consisting of two edge disjoint odd cycles joined b y a path.
Chapter 8
300
Proof. Assume a E Zqn N is an elementary vector of N so that G, is not an even cycle. From the equation fa+ = fa- we obtain a walk in G,
. . . ,xe-1 are distinctvertices, n is odd and xl is in (21,. . . ,X C - ~ } . The walk w can be chosen so that zixi+l E supp(f"+) if i is even and xixi+l E supp(fa-) otherwise. We claim that xe is not in { q ,. . , ,xn-l}. Assume x[ = xm for some 1 5 rn 5 n - 1. If m and k' - n are odd then the monomial walk so that X I ,
2 ~ 1=
{xo,x1,.
, x m , x e - 1 , ~ ~ - 2 ,, x n = XO} e
a
yields a relation f s = f 7 with supp(6 - y) properly contained in supp(a), which is a contradiction. If rn is odd and e- m is even then we can consider the walk w1 = { x n , x n + l , . ~ * , x t = x r n , x r n + 1 , * * . , x n } and derive a contradiction, the remaining cases are treated similarly. We may now assume xe = xm for n 5 rn 5 k' - 2. The cycle
C1 = {xm,.. . ,xe = x,} must be odd, otherwise C1 would give a relation fs ='f with supp(d - y) a proper subset of supp(a). The walk w gives a relation fs = f? with
Hence supp(6 - 7) is a subset of supp(a), and since the support of a is minimal one has the equality supp(d- 7)= supp (a).Therefore
and G, is asrequired.The
converse follows fromCorollary 8.2.4.
0
Definition 8.4.12 A circuit of a graph G is either an evencycle or a subgraph consisting of two edge disjoint odd cycles joined by a path. Proposition 8.4.13 If G is a graph with incidence matrix M and P is the toric ideal of k[G], then
P = ({Ta+- T"- I a E Z q and M a = O}). Proof. The result
is a consequence of Corollary 7.1.4.
0
Let G be a graph with incidence matrix M and let N = ker(M) c Qpp . The next example was computed using Macaulay [15], it shows that the set of elementary integral vectors of N does not determine the presentation ideal of k[G].
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Example 8.4.14 Let R = k[a,. . . ,o]be a polynomial ring over a field IC and let G be the labeled graph a
b
f
IC
1
the toric ideal P of k[G] has height three and k[G] is not Cohen-Macaulay. The toric ideal P is minimally generated by
Remark 8.4.15 By Proposition 8.4.11 there are, up to sign, exactly nine elementary integral vectors of N . They are in one to one correspondence with the exponent vectors of the first nine generators given above. Therefore the set
{ T a t - Ta-I Q is an elementary integral vector of N} is not a generating set for the presentation ideal of k[G]. Corollary 8.4.16 Let G be a graph with incidence matrix M . If X = (Xi) is an elementary integral vector of N = ker(M), then lXil 5 2 for all i. Proof. Let X be an elementary integral vector of N . Since Gx is an even cycle or two edge disjoint cycles joined by a path we obtain an elementary integral vector a with supp(a) = supp(X) and so that ]ail 5 2 for all i. Because any two elementary vectors with the same support are dependent we obtain X = a. 0
302
Chapter 8
Theorem 8.4.17 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 Grobner basis of P is less or equal than 2q dim Ze(G). Proof. Let M be the incidence matrix of G and let
N={aEQqIMa=O}. Assume F is the reduced Grobner basis of P and take f = f ( T )E F . Using normalized Corollary 7.1.5 and Corollary 8.4.13, together with the fact that reduced Grobner basis are uniquely determined,we obtain f(T)= T" - TP. Notice that supp(a) n supp(p) = 8, otherwise take i in the intersection and write f(T)= Ti(T" - T b ) .Since T" - T breduces w.r.t F it follows that f(T)reduces w.r.t F \ {f} whichis impossible. We can now write f(T)= Ta+ Ta-for some a E N n Z? By Theorem 8.4,8we can write a= ai& for some ai E @, and for some elementary integral vectors X i each in harmony with a and such that Supp(Xi) c supp(a) for all i and r 5 dim N . Hence
xi=,
-
i= 1
i= 1
If ai > 1 for some i then ai+ and ai- are componentwise larger than Xi+ and Xi- respectively, note that thisis not possible by the previous argument. Hence 0 < ai 5 1. Using Corollary 8.4.16 it now follows that the entries of a satisfy lail 5 2 dim Ze(G)for all i, which gives the asserted inequality. 0 Let G be a graph with incidence matrix M and N = ker(M) the kernel of M in (@.Let XI, . . . , X, are the elementary integralvectors of N . Define the elementary zonotope of N as EN = [0, X,] [0, X,].
+ + a
Corollary 8.4.18 If N is the kernel of the incidence matrix of a graph G and P is the toric ideal of k[G], then
is a universal Grobner basis for P .
Exercises 8.4.19 Prove that theelementary integralvector of the kernel of the matrix
M are precisely the row vectors of the matrix A with a f sign M = ( ;
; ; ;)
-;
2 - 3
A = ( ;
1 0
-3O 21 ) -4
3
'
Monomial Subrings of Graphs
8.5
303
Edge subrings of bipartite planargraphs
It is the goal of this section to extract information about the edge subring k[G],where G is a bipartite planar graph, by studying the particular geometry of the graph G. Our exposition follows closelythat of [85],all the main results andproofs in this section are due to Luisa Doering and Tor Gunston. Atomic cycles Let G be a bipartite planar graph with vertex set
and
R = k [ z l , .. .,X,] a polynomial ring over a field k . If f1, . . . ,fn are the edge generators of G, that is, the monomials that correspond to theedges of G , then thepresentation ideal P(G) of the edge subring k[G]is the kernel of the homomorphism of k-algebras 5'1
= k[Tl,. . . ,Tq]+ k[G]
induced by the assignment Ti I+ fi. By abuse of notation, we will sometimes refer to "the edge Ti". Fix an embedding of the graph G in the plane. Let c be an even closed walk, c = Ti,, . . . ,Tia,. For any such walk, write
By Proposition 8.1.2, we have that the generators of P(G) correspond to cycles in G: P(G) = ({T,]c is a cycle in G}). Since G is embedded in the plane, G divides the plane into regions. If c is a closed walk that bounds a region (not including the unbounded region) we say that c is an atomic cycle (even though it need not have distinct vertices). Let A(G) c P ( G ) be the ideal generated by
{ T,I c is an atomic cycle in G }. As G is embedded in the plane, every cycle in G has a locus in the plane. We denote cycles by lower case letters, and use the same symbol for its locus; the corresponding upper case letter denotes the open region in the plane: a = dA. Similarly, for each atomic cycle ci, we denote the open region it bounds by Ci. If G can be written as G = G1 U G2, where GI and G2 are subgraphs with at most one vertex in common, then
Chapter 8
304
and
k[G]k[G1] @)lc k[G2]. Therefore, we will limit ourselves to graphs which are 2-connected (that is, connected graphs with no cutvertices). Under this assumption, atomic cycles will actually be cycles. For any vector w = ( ~ 1 , ... ,wq) we will write T ufor Tfl T:q. If u and v are vectors with non-negative components, then we define (u,w ) to be the vector whose ith component is given by (u,w ) i = max(0, ui - vi). 9
6
e
In the sequel, G isa2-connectedbipartiteplanargraphwithafixed embedding in the plane, n vertices, q edges and r regions. Polarizations The concept of polarization willallow us to define term orders which in some respects are the "rightones" to study P ( G ) and some homomorphic images of P(G).
Definition 8.5.1 Let c be an even closed walk of polarization of c is the set
G,c = Til,. . . ,Tit.A
or the set
{Ti2,Ti4,.",Tit}. Let
{CI, .
. . ,}.c
be the atomic cycles of G. A polarization P of G is a set
P = {PI,...)P r } where Pi is a polarization of ci for all i such that (a) for
i # j we have Pi n Pj = 8, and
(b) {TilTi 4 Pj for j = 1,.. . , r } is a polarization of w, where w is the boundary of the unbounded region. Proposition 8.5.2 ([85]) G has a polarization. Proof. Let { X I , .. . ,xn} be the vertex set of G. Form a new graph G' as follows. G' has vertex set
and edge set
{{xi,zj}( { Q , z ~ is } an edge in G } u { { x i , y j } l zi is in the atomic cycle c j } .
Monomial Graphs Subrings of
305
In other words, G' is obtained from G by adding a vertex in each bounded region and connecting it to each vertex in the boundary of that region. We view G as a subgraph of G'. It is clear that G has a polarization if the plane map determined by GI can be 2-colored. In order to verify this, we let GI' be the graph with vertex set
{viI vi is a bounded region of G') and edge set
{(vi, vj} I vi is adjacent to
vj
in G'}
We claim that GI' is bipartite. G" has an embedding in the plane induced from the embedding of G, so we need only check that every atomic cycle of GI' is even. The atomic cycles of G" are of two sorts: those corresponding to a vertex xi and those corresponding to a vertex yi. A cycle corresponding to a vertex yi is even because G is bipartite. It is easy to check that a cycle 0 corresponding to a vertex xi is even also. Thus GI' is bipartite. Lemma 8.5.3 If P = { P I , .. . , P r ) is a polarization of G, then by relabeling one may assume that Ti E Pi and that if 1 5 i < j 5 r , Tj does not appear in the cycle ci. Proof. We induct on r , the number of atomic cycles. If r = 1 the lemma is clear. Assume T > 1. It is not hard to see that there is a1 least one edge that appears in only one atomic cycle, that we label C r y and is contained in Pr. Label this edge T,.Let G' be the graph obtainedfrom G by deleting all edges and vertices that do not appear in any atomic cycle except cT. Then G' is planar and bipartite, and is oriented by
Apply the lemma,labeling with
T'1
through
2'?-1,
and we are done.
0
Proposition 8.5.4 If c1, . . . , cI.are the atomic cycles, then Tcl ,. . . ,Tcr is a regular sequence. Proof, Assume we have a labeling as in the previous lemma. Let lexicographical ordering on SI with
TI> T2 >
> be
the
> Tr > . . * > Tq
and let in(-) refer to the initial term or ideal with respect to this ordering. Let mi = in(Tci). Note that
so for i # j we have gcd(mi, mj) = 1. Hence the height of (Tcl ,. . . ,Tcr) is r , as required. 0
Chapter 8
306
Example 8.5.5 Let G be the following graph and consider the polarization obtained by taking pairs of dashed lines in each atomic cycle.
1
"-
2
"0-0-0
/ / Tc4 I
//
Thus with the lex ordering T I > > T4 > 2'5 > . > T I the ~ binomials T,,, . . . ,Tc4form a Grobner basis for the ideal they generate. e .
a
a
Fixing more notation Choose a polarization P = {PI,. , . ,Pr} of G. Assume that the edges are labeled such that Ti E Pi. We employ the following notation:
PC = { T l , . . . ,TJ \ U P i
9
Pt = (2'I Tj appears in ci} \ Pi and for any cycle a, with a = aA:
2'
Pa = (Tjl appears in a and Tj E Pi for some Ci c A } . Let
N = PCU {Ti- Tjl 1 5 i 5 r and Tj E Pi}, Set S2 = k[Ul, . . . ,UT]and define a homomorphism 6 of L-algebras by:
. Sl/(N). Note ker(6) = ( N ) and S2 = ~ [ U I. .,,UT] Termorders Define a termorder in 5'2 as follows: Assume thatthe atomic cycles of G are labeled c1, . . . ,c,. in such a way that for every atomic cycle ci , there is a sequence of atomic cycles
Monomial Subrings of Graphs such that il < i 2
307
< . . . < il and
and label that walk w = Cr. Let G' be obtained from G by deleting all those edges that appear in no atomic cycle other than cr. Inductively, label the cycles of G' with ~ 1 . ., . cr-1. Let >s2 be the reverse lexicographic ordering with respect to this ordering, with U1 >sz U2 >s2 - - - >s2 U T . Nowwe define a term order on S1 as follows: for monomials u and b in S I , say a + b if $(a) >s2 $(b). Let >sl be a term order on S1 that is a refinement of +. )
Remark. For a cycle a, we write T' = Ta+- Ta- where Ta+= in>,l (Ta). It is easy to see that Ta+= Ti. Ti E P
a
-
Grijbner Bases Inthispart we find Grobner bases for $(P(G)) and 4(A(G))with respect to the term orders introduced in the previous parah for the reduction of g to h with respect to a set of graph. We write g polynomials and a term order. For any two cycles a and b, we will write
Note that Ta,b is the S-polynomial of Ta and Tb with respect to >sl.If A is a region in the plane, we write for the closure of A .
x
Proposition 8.5.6 Let a andb be cycles.Supposetheconnectedcomponents of A \ B are A I ) . . . , A , and the connected components of B \ Z are B l , . . . ) B t . ThenTa,b E (Ta,)...)T',,Tbl,...,Tbt)., Proof. The proof as is left
an exercise.
0
Note that since the initialterms of the TUiand the Tbi are all relatively prime, they form a Grobner basis for the idealtheygenerate; see Lemma 2.4.14. Thus, it follows from Proposition 8.5.6 that Ta,b'L) 0 with respect to Tal) . . . T,, Tbl,.. . Tbi.
-
)
)
)
Lemma 8.5.7 Let A = k [ x l , .. . ,x,] andlet > be a term order on A. Suppose mg 0 with respect to f l y . . . , fs, where g , f l , . . . , fS E A and m is a monomial such that gcd(m, in>(fi)) = 1 for 1 h i 5 s. Then g 'L) 0 with respect to f l , . . . f,. )
Chapter 8
308 Proof. is The proof
left as exercise.
0
Theorem 8.5.8 ( [ 8 5 ] ) A4 = M ( G ) = {$(Tc)Ic is a cycle } is a Grobner basis f o r 4(P(G))with respect to >sa.
Let
By Buchberger's criterion, it suffices to show that Ua,a* 0 with respect to {Uc = 4(Tc)Ic is a cycle }. By Proposition 8.5.6, Ta,b * 0 with respect to T a l ,. . . ,Ta6 Tbl , . . . ' Tbt. Since every Tai and Tbi is a binomial or monomial, every step in the reduction Ta,b* 0 is a binomialormonomial. Further, since T U >sl T' implies U4(z")IS U 4 ( " ) , for ,anybinomial f E SI we have $(f)= 0 or ins, M f ) ) = $(ins, (f)).Thus #(T , = @ ( ( a + , b + ) ) + d ( b - ) - ~ 4 ( ( 6 + , ~ + ) ) + 4 ( a ?A -) 0 ab
with respect to U a l , ., . ,Ua6, ubi,. . . ,Ubt. It can be easily seen that Ua,b(#(Ta,b)and the quotient is a monomial U" such that UilU" implies Ci C A n B . Thus, by Lemma 8.5.7, U a ,b * 0 respect with to { Ucl c is a cycle }. 0 Note that it follows from this and from Buchberger's criterion that
{Tclc is a cycle in G} is a Grobner basis for P(G)with respect to >sl. In fact, it is actually a universal Grobner basis, even in the case where G is not planar according to Proposition 8.1.10. Theorem 8.5.9 ( [ 8 5 ] ) B = B(G) = ($(Tc)Ic is an atomic cycle Grobner basis for $(A(G))with respect to >sa.
1
isa
Proof.
UjP,I UCi
= 4(T,i) =
U,!pit - Ujl
if ci containsanyedges
in
PC,
. Ujk if ci doesn't contain any edges
in
PC,
where cjl,. . . ,cjk are the atomiccycles that share anedge with Pic, counting 0 multiplicities. Since in>sz (Uti) = U e j p i I , the result is clear.
Monomial Graphs Subrings of
309
Applications As applications of the results of the previous paragraph, we show that N is a system of parameters for k[G]. Corollary 8.5.10 height A ( G ) = r . Corollary 8.5.11 ([85]) The image of N in S,/A(G) forms a system Parameters.
of
Proof. Both corollaries follow from the fact that in>sz ( $ ( A ( G ) )= ) (uIp1I,. . . , u!?.') is (VI,. . , , UT)-primary, and the fact that N has the expected number of elements. From
I N I= IP'I
+ C(l~il1) = lpCl+
one obtains that
l~il-
r = 4 - r,
i
i
N has q - r elements.
0
Proposition 8.5.12 height P(G) = r . Proof. It follows from Proposition 8.2.12.
0
Corollary 8.5.13 ([85]) The image of IV in S l / P ( G ) S k[G]forms a syst e m of parameters. Proposition 8.5.14 If a is a cycle and T, E (TcIc is a cycle, c # a ) , then a has a chord. Proof. If T, E (TCJ c is a cycle, c # a ) , then each term of T, must be divisible by a term of T, for some c # a. For the purposes of computingHilbertfunctions, various rings involved are graded by
we clarify that the
deg(Ti) = deg(Ui) = 1 and deg(zi) = 1/2. Thus the isomorphism
is a graded isomorphism of standard k-algebras. Corollary 8.5 .I 5 The Hilbert series of k[G]can be written as
where Q(t) is a polynomial with positive integer coeficients and degQ(t) L
lPil - r. i
Chapter 8
310
-
+
Proof. Let k[G]= S I / ( N P ( G ) ) S 2 / 4 ( P ( G ) ) . Since the edge subring k[G]is Cohen-Macaulay, the image of N in k[G]is a regular sequence. Thus
We have that
. . ,ULprl, other monomials ). in>,,$(P(G)) = (U!’’l,. Then we have Q ( t ) = H m ( t ) with the desired properties.
0
Exercises 8.5.16 Let G be the bipartite graph
and P(G)the toricideal of k[G]obtained by mapping tij to xizj. Prove that P(G)is a Cohen-Macaulay ideal of height 5 . If we order the tij variables in reverse lexicographical order:
use this order to show that I = ( h l ,ha, h3, h4, h5) is a complete intersection inside P ( G ) , where
Find an appropriate embedding of G in the plane to explain how this order obtained andwhy this orderis useful to find a maximal regular sequence of binomials inside P(G).
WM
8.5.17 Let G be a graph that can be written asG = G1 uG2, where G1 and
G2 are subgraphs with at most one vertex in common. If G1 is bipartite, then P ( G ) = ( P ( G l ) , P ( G z )and )
Monomial Subrings of Graphs
311
8.5.18 Show that the following embedding in the plane of the bipartite graph consisting of two edge disjoint squaresthat meet at a point has atomic cycles which are not cycles.
8.6
Normality of bipartite graphs
In this section we characterize the torsion freeness of an edge ideal in graph theoretical terms and study the integral closure of the powers of an edge ideal. Two key notions to our examination are the even cycles and certain class of configurations of the graph closely related to the circuits of the graph.
Torsion freeness of bipartite graphs We shall prove here that edge ideals of bipartite graphs are normally torsion free. In particular, by Proposition 7.3.17 and Theorem 7.2.35, their Rees algebras are normal Cohen-Macaulay domains. The main technical device is the following: Lemma 8.6.1 Let G be a bipartite graph and let I = I ( G ) be its edge ideal. If z1 is a vertex in G, then x 1 I Sn (Is+':z l ) c Is+' for all s 2 0. Proof. The proof is by induction on s . The case s = 0 is easy to verify. Assume the containment true for all integers less than s. Assume M is a monomial belonging to x1 Isn zl) \ Is+1. There are square-free monomials of degree two MI,. . . , M,, N1, . . . ,N,+1 in I so that M = MI. * M s p and 2 1 Ad = N1 - . Ns+lLLfor some monomial L and for some monomial M divisible by 2 1 . We set M = z 1 z 1 . By an induction hypothesis we may assume Mi, - MiT # Nj, Njp for all increasing sequences {ik } , {jk 1 having values in { 1, . . . , s} and for all r _< s . For convenience, set M,+1 = N3+2 = 1. Given 1 5 k 5 s + 1 we claim that there are distinct vertices 2 1 , . . . ,x 2 k of G so that (after re-ordering the Ni's and Mi's) Mi = x2iz2i+l for i 5 k - 1, Ni = 22i-lz2i for i 5 k , and Mk Ms% =zzkNk+l N3+1H, for some monomial H . #
-
a
e
312
Chapter 8
The proof of the claim is by induction on IC. If k = 1, from the equation x1M = N1 N,+1H and using that ill 4 Is+1, we obtain Ni = x122 for some i. We reorder the Ni's such that N1 = 21x2, which shows that M = x2 N2 . Ns+lH . Assume that the claim is valid for k and consider the equality k ? k - M , M = x2kNk+l * * * Ns+l H.
--
-
*
Observe that if
Z2k
divides
m,then the equation
implies M E Is+1. Hence 52k divides Mi for some k 5 i 5 s ; as before we re-order the Mi's so that MI, = x2kz for some vertex x . If z E (81,. ,x ~ k } , then either z = xi with i odd and in this case we have an odd cycle, or i is even and we have a relation of the form Mi, Mi, = Nj, Nj, : using that G is a bipartite graph together with our previous assumptions we see that none of these cases can occur. Therefore z = x2k+l satisfies:
-
-
Since M $ IS+l the last equation shows that 22k+l divides Ni for some i, say N k + l = 22k+lw for some vertex w. We set w = x2k+2. Notice that by the last argument w $ (11, . . .x2k+l}, and the induction on k is complete. We now apply the claim for k = s 1 to obtain M = xZs+2H. To complete the induction on s, if k = s 1 then an application of Eq. (8.3) above leads 0 to M E P + l , which is a contradictiontotheinitial choice of M .
+
+
Lemma 8.6.2 Let G be a graph and U = {XI,.. . ,x,.} a subset of the vertex of G, then N G ( U ) A, where set of G. If A is a minimalvertexcover N G ( U ) is the neighbor set of U. Proof. Set N = N ~ ( x 1 ,... , z T ) . Assume A is a minimal vertex cover of G such that N c A. Take an edge e of G. If x1 E e, then e = (XI, z } with z E N , thus z E A \ (21) and e n ( A \ ( 2 1 ) ) # 8. If x1 e, then using e n A # 8, one has e n ( A \ (21)) # 8. Altogether A \ { X I } is a vertex cover of G, which contradicts the minimality of A . Therefore N cannot be contained in A. 0 Theorem 8.6.3 ([262]) Let G be agraphandlet ideal. The following conditions are equivalent: (i) G is bipartite. (ii) I is normally torsion free.
I = I ( G ) be its edge
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313
Proof. Let V = {XI,.. . ,xn> be the vertex set of G and let R = k [ ~=] I C E X I , .
. . ,x n ]
be a polynomial ring over a field IC. We set m = (x). (i) + (ii) Wewill proceed by inductionon n. The ideal I is clearly normally torsion free for n = 1 and n = 2. Assume I is normally torsion free for bipartite graphs with fewer than n vertices. Let P E Ass(R/IS) for some s. First notice that by Lemma 8.6.1 m # (Is:M )
for every monomial A4 in the x variables and for all s 2 1. As a consequence m is not an associated prime of R / I s for all s. Therefore, P is a face ideal properly contained in m, hence Ip can be written (after a permutation of the variables) as IP
= (x1,
--
9
zk,I(Gl))P,
where GI is a bipartite subgraph of G. Using the induction hypothesis we conclude that P R P E ASS(RP/IP), which gives P E A s s ( R / I ) and the proof is complete. (ii) + (i) Assume G is not bipartite and pick an odd cycle with vertices 2 1 , . .., Set IC = (T 1)/2 and f = 2 1 2,. Note f 4 I” becauseany monomial in Ik has degree at least 2k. Let
+
be the neighbor set of ( 2 1 , . . . ,x~}.Note that f E I(lC), because according to Lemma 8.6.2 the linear form s = x1 f . - - + x s is a nonzero divisor of R / I and sf E Ik.Therefore Ik # I ( k ) ,a contradiction to Proposition 3.3.26. 0 Let G be a non bipartite graph and I = I ( G ) its edge ideal. According to [38] the sets Ass(R/In) will stabilize for large n. A recent article [62] gives a constructive method to find this stable set and present an upper bound on where the stable set will occur. Corollary 8.6.4 ([262]) Let G be a graph and I = I(G) its edge ideal. If G is bipartite, then the Rees algebra R(I) is a normal domain. Proof. Note that I ( G ) is a radical ideal and intersection use Theorem 3.3.31.
which is generically a complete 0
Chapter 8
314
Integral closure We discuss here some typical behavior of the integral closure of the powers of an edge ideal. Proposition 8.6.5 If I is the edge ideal of a graph, then I 2 is integrally closed. Proof. an Is left as
exercise.
O
M. Hochster pointed out the next example showing that the proposition does not extend to the next power of I . Example 8.6.6 (Hochster’s configuration) Let G be the graph
Note that f = (112223)(252627)satisfies f E 13\ 13. This example leads naturally to thefollowing concept. Definition 8.6.7 Let G be a graph. A Hochster configuration of order t (H-configurationfor short) in G consists of two odd cycles C2,.+1 and C2s+l satisfying the following conditions: (i)
C2,.+1
n N(C2,+1) = 8 and t = r + s + 1.
(ii) No chord of either
C2r+l
or
C2s+1 is
an edge of G.
Given such a configuration, we consider the product of the monomials corresponding to the two cycles to produce new generators of the integral closure of the appropriate powers of I ( G ) . Proposition 8.6.8 If I is the edge ideal of a graph G admitting an H configuration of order t , then It # I t . Proof. The monomial obtained by multiplying all the variables in the two cycles of the configuration is an element of It \ I t . Indeed, let
be the monomials corresponding to the two cycles in the configuration. By definition, we have r s 1 = t. Therefore 6y E I t - l \ I t . On the other and y2 E 12,+1, thus ( 6 ~ E) 1~2 ( r + s + 1= ) 12t. This shows hand d2 E 0 that Sy E It.
+ +
Subrings Monomial
315
of Graphs
Conjecture 8.6.9 ([262]) The edgeideal of a graph G is normal if and only if G admits no H-configurations. As it will be seen in the nextsection this conjecture hasa positive answer, but first we need to find a good description for the integral cIosure of the edge subring of a graph.
Exercises
+
8.6.10 Let R = k [ 5 1 , 2 2 , 2 3 , ~ 1 , ~ 2 , ~ and 3 , ~ 4I ] = I1J3 I3J1, where IT (resp. J T ) denote the ideal of R generated by the square-free monomials of degree T- in the xi variables (resp. pi variables). If z = ZLz223$&/2Y3Y4, then z 2 E I 4 and z E p\ 12.
8.6.11 Let G be a graph. Explain the differences between a circuit of G and an H-configuration of G.
8.7
The integral closure of an edge subring
Let G be a graph and let k[G]be the edge subring of G over an arbitrary field IC. In general, it is difficult to certify whether k[G]has a given algebraic property or to obtain a measure of its numerical invariants. This arises because the number q of monomials is usually large. It is similarly difficult to carry out manipulationssuch as those required in the Noether normalization process. Our goal here is to unfold a construction for the integral closure or normalization k[G]of k[G],for this purpose the underlying graph theoretic aspects are very helpful. ,
,
A combinatorial description of normalizations Let us begin by proving the normality of edge subrings aasociated to graphs which are nearly bipartite. Proposition 8.7.1 Let G be a connected graph and k a field. If G has at most one odd cycle, then k[G]is normal. Proof. We may assume that G has a unique odd cycle, for otherwise G is a bipartite graph and by Theorem 8.6.3 k[G]is normal. Pick an edge e of G which is on the unique odd cycle and set H = G \ {e). Notice that H is a connected bipartite graph because e is not a bridge (it is on a cycle). Hence k [ H ]is a normal domain of dimension dim k[G]- 1 (see Proposition 8.2.12). Consider the epimorphism
k [ H ] [ t ]’P, k[G]”+ 0, induced by y ( t ) = fe,
316
Chapter8
where t is a new indeterminate and fe is the monomial corresponding to e. Since k [ H ] [ tand ] k[G]are both Noetheriandomains of the same dimension, cp is an isomorphism. Therefore k[G]is normal. 0 Lemma 8.7.2 Let G be a graph without even cycles. Then the intersection of any two cycles is either a point or the empty set. two distinct cycles of G such that 21 n 2 2 # 8. We may assume and w1 4 2 1 . Let j = min{i 2 11wi E &}, and write wj = zs.If 0 < s
x0
= 200
< n, considering the cycles
of lengths s -tj and j + n - s, respectively, we obtain that n is even, which is a contradiction. Therefore s = 0 or s = n. To concludeobserve that 2 2 = (WO, wl,. . . ,wj-1,ZO}, hence 2 1 n 2 2 = (zo}. 0 Definition 8.7.3 A graph G having just one cycle is said to be unicyclic. Definition 8.7.4 Let G be a graph without even cycles. A cycle 2 of G is said t o be a terminal cycle if N ( 2 ) = 2 U (v}, for some v 4 2. Lemma 8.7.5 Let G be aconnectedgraphwithoutevencycles.Assume that 2 1 nN ( Z 2 ) = 8, for any two cycles 2 1 , Z z of G. If G is not a unicyclic graph, and deg(v) 2 2 f o r all v E V ( G ) ,then G has at least two terminal cycles. Proof. Let 2 1 , . . . , 2,. be the cycles of the graph G. We now construct a new graph T having vertex set
Thus the vertices of T are the cycles of G, together with the vertices of G not lying on any cycle of G. Two points Yl,Yz of T are adjacent if and only if Y1 n N ( Y 2 ) # 8, here we regard a vertex v of G as the one point set {v}. Notice that T is a tree and that the terminal cycles of G correspond to the end points of T , Hence G must have at least two terminal cycles, a
Monomial Subrings of Graphs
317
The integral closure in terms of bow ties Let G be a graph with vertex set V = ( 2 1 , . . . ,x,} and let R = k[zl,. . . , x n ] be a polynomial ring over a field k . The set { f 1 , . . . ,fp) will denote the set of all monomials ziz:j in R such that { X i , z j } is an edge of G. Note
k[G]= k [ f l , . . . ,f*l c R In general it is difficult to predict which elements lie in the integral closure of an affine domain, butin the case of k[G]one has the following general description: k[G]is also a monomial subalgebra generated by monomials
f=zf1...zk =xp,
PEW,
with the following two properties (see Theorem 7.2.28):
frn=,fibi,
m,biEWandm?l.
The first condition asserts that f lies in the field of fractions of k[G],the other being the special form the integrality condition takes for such elements. TOexplain the description of k[G]we begin by exhibiting some monomials in the integral closure of k[G].A bow tie of a graph G is an induced subgraph w of G consisting of two edge disjoint odd cycles
joined by a path {x,., . . . ,zs}.In this case one sets Mu,= x1 xrxs+l 'xt. We observe that 21 and 2 2 are allowed to intersect and that only the varinot those in the path itself. ables in the cycles occur in M,,,, If w is a bow tie of a graph G, as above, then Mw is in the integral closure of k[G].Indeed if gi = zi- 1zi , then 0
-
Hence Mw E k[G].
a
Chapter 8
318
Lemma 8.7.6 Let G be a connected graph without even cycles, V the vertex set of G,and f 1 , . . . ,fq the monomials defining the edges of G. Assume the conditions:
(a) 2 1 n N(Z2) = 8, for any two cycles
Z1,22
of G,and
(b) deg(v) 2 2 f o r all v E V and G has an even number of vertices.
Then
nvEV v is in k[B],where B = {
f1,.
. . ,fq} U {M,J w is a
bow tie}.
Proof. The proof is by induction on the number of cycles of the graph G. Set x = v. If G has exactly two cycles, then G is a bow tie and we and 6 = MG is the product of the vertices can write z = f S , where f E k[G] in the two cycles. We may now assume that G has at least three cycles. By Lemma 8.7.5 there are two terminal cycles
nvEV 21
= (XO,Xl, * .- , X T = xo),
22
= {yo,y1,. - ., y s = yo).
+
There is a path { x r , x T + l , .. . , xT1},so that r 1 5 r l , deg(xj) = 2 for all r < j < T I , and deg(s,,) 2 3. Likewise there is a path { y s , y s + l , . . . ,ysl}, so that s 1 5 s1, deg(yj) = 2 for all s < j < S I , and deg(ys,) 2 3. We may assume rl and s1 even, for otherwise, if r1 is odd, we write
+
where VI = V show that
\ ($1, . . . ,sTl -1 1, and then
use induction. It is not hard to
Set
If zTl# ysl then we write
where Vl = V \ ( X I ,. . . ,xT1-l,y1, . . . ,ysl-l}, hence using induction we get z E k[B]. We may now assume that rl ,s1 are even and xTl = ysl . Consider the subgraph H = G \ { X I , . . . ,xTl-l,yl, * ‘ ,YS,-l). If deg(x,,) 2 4, then applying the induction hypothesis to H we obtain z E k [ B ] . Assume deg(xTl) = 3. Note that H has a terminal cycle a
Monomial Subrings of Graphs
319
+
and there is a path { z t ,zt+l,. . . ,zit 1, so that t 1 5 t l , deg(zj) = 2 for all t < j < t l , and deg(ztl) 2 3, by the previous arguments we are reduced to the case r1, S I , tl even and x,., = ysl = z t l . Observe that this forces the equality V = {~1,.~.,~rl~~1,...,~sl-l~~l,~..~~tl-l).
To complete the proof we use the identity r
z = (zlz2)(Z3z4)
''
(Ztl--lZtl)0102 i=l
9
xi
yi, i=l
to derive z E k[B].
0
Given a graph G we set AG = (log(fl), . , . ,log(f,)}, where f 1 , . . , ,fq are the monomials corresponding to the edges of G and log(ff)is the exponent vector of fi. The affine subsemigroup of RP generated by d G will be denoted by CG. Thus CG = k!. a€dc
The cone generated by CG will be denoted by
Note that k[G]is equal to the afine semigroup ring k [ c G ] . This cone will be studied closely in Section 8.8. Lemma 8.7.7 Let G1 and G2 be two finite sets of monomials in disjoint sets of indeterminates. Then k[G1 U Gz] N k[G1] @ k k[G2] and - ~ [ GUI G2] E k[G1] @lc k[G2] 2 k[G1] k[Gz], "
is the subring generated
b y the generators of k[G1] and k[Gz].
Proof. Set G = G1 U G2. By Theorem 7.2.28 we have
-
k[G]= k[(a"l a E ZCGn QCG}].
On the other hand there is a decomposition
~:CG~I~,CG=(~CG~~I~,CG~)~(~CG~~E+CG~). Using equality this
the assertion follows readily.
0
Corollary 8.7.8 Let G be a graphand G I , .. . G, its connected components. Then k[G]is normal if and only if k[Gi] is normal for each i. )
ChaDter 8
320
Theorem 8.7.9 ([263])Let G be a graph and k a field. Then the integral closure k[G]of k[G] is generated as a k-algebra by the set
B = { f l , . . . ,f q } where
fi,
. . . ,f q
U {M,(
w is a bow tie},
denote the monomials defining the edges of G .
Proof. Let V = (21,. . . ,x,) and E = {el,. . . , ep} be the vertex set and edge set of G, respectively. According to Theorem 7.2.28 one has:
The proof is by induction on q, the number of edges of G. The case q = 1 is clear. Assume q 2 2 and that the result is true for graphs with less than q edges. If ei = {Xj,Xk), We Set fi = XjXk. First we remove all the isolated vertices of G, without affecting the subring k[G].By Lemma 8.7.7 and the induction hypothesis we may assume that G is a connected graph. We may also assume that deg(w) 2 2 for all v E V ; because if de&) = 1 and fa = Wxj, then setting G' = G \ {ea) we the latter being a polynomial ring over k[G'],hence have k[G]= k[G'][fi],
and fi is not part of a bow tie, which by induction yields the required equality. Let z = xa E k[G] be a minimal generator of the integralclosure of k[G]. There is a positive integer m so that
Observe that if bi = X i = 0 for some i, then each bowtie of G \ {ea) is a bowtie of G andtheinduction hypothesis yields x E k [ B ] . Hence we may further assume that all the monomials f1, . . . ,f q must be present in Eq. (8.4). There are two cases to consider. Case (I): Assume that G contains an even cycle 2. We set vi = log(fa). Forconvenience we use the edges of G to represent the cycle 2 by the sequence (ei, ,. . . ,eak),where two consecutive edgesare adjacent. If bar = 0 for some 1 5 r ,< IC, then using the relation
that comes from the even cycle, we may rewrite Eq. (8.4) as
Monomial Subrings of Graphs
321
which by induction yields z E k[l?]. We may now assume bi, > 0 for all 1 5 r 5 k . We may harmlessly assume that bi, 5 5 bi,, Using Eq. (8.4) and Eq. (8.5) we obtain e
a
where the ci's are non negative integers. The net effect is that now we may rewrite Eq. (8.4) as
and di, = 6il = 0. Hence by induction z E k [ B ] . Case (11):Assume that all the cycles of G are odd. Let 21 and 2 2 be two distinct cycles of G, represented by the sequences of edges {ei, , . . . , e i k } and {eel , . . . , ets1 , respectively. If 21 f~2 2 # 8, then according to Lemma 8.7.2 21 n 2 2 = {x}, for some z E V , say x E SuPp(fi,) n supp(fe,). Using the identity
j ojdodd d
j even
j even
we may proceed as in Case (I) to derive z E k[B]. Therefore we are reduced to the case N(Z1)n 2 2 = 0,for any two cycles Z1, Zz. If xj is not in the support of f,"' , for some j , consider the set e
fg".
Then &- Xi = 0 and bi = 0, for all i E D. Note that, for any i E D,we may eliminate the monomial f i from Eq. (8.4), hence by induction we get x E k [ B ] .As a consequence we are reduced to the case supp(xm) = V , that x= - - - x> , with ai 2 1 for all i. By Corollary 7.7.17 we obtain that is, k [ G ]is generated as a k-algebra by monomials of degree at most 2(nAltogether we get
[:I).
hence n must be even and z = x1 as required. that x E k[B],
-
xn. Using Lemma 8.7.6 we conclude 0
Chapter 8
322
There is a version of Theorem 8.7.9, due to Hibi and Ohsugi [164], that allows loops in the graph G. Next we clarify that there are two special types of bowties that are irrelevant for the description of the integral closure of k[G]. Proposition 8.7.10 Let w be a bowtie consisting of two edge disjoint odd cycles Z1,Z2 that meet at a vertex or are connected by a n edge, that is, UJ has essentially one of the forms:
then Mw is in k[G]. anProof. Left as
exercise.
0
As an immediate consequence of Theorem 8.7.9 and Proposition 8.7.10 one has the following full characterization of when k[G] is integrally closed.
Corollary 8.7.11 Let G be a connected graph. T h e n k[G] is normal if and only if for any two edge disjoint odd cycles Z1,Zz either Z1 and 2 2 have a common vertex, or Z1 and 2 2 are connected by a n edge. Thus the smallest connected graph G such that k[G]is non normal is the Hochster configuration of Example 8.6.6, consisting of a graph G with two disjoint triangles joined by a 2-path. The corollary above prompts the following: Definition 8.7.12 A graph G is said to satisfy the odd cycle condition if every two vertex disjoint odd cycles of G can be joined by an edge in G. The odd cycEe condition has come up in the literature in connection with the normality of edge subrings [164, 262, 2631 and the description of the circuits of a graph [247, 3031; it also occurred before [119]. It could be said that this notion has been rediscovered during the study of monomial subrings of graphs. Corollary 8.7.13 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 ( I ) is normal.
Subrings Monomial
of Graphs
323
Proof. +) Let C ( G ) bethe cone over thegraph G; it is obtained by adding a new vertex t to G and joining every vertex of G to t . According to Proposition 8.2.15 there is a isomorphism
Let Mw be a bowtie of C ( G ) with edge disjoint cycles 21 and 2 2 joined by the path P , it is enough to verify that Mw E k[C(G)].If t 2 1 U 2 2 U P , then w is a bowtie of G and M, E k[G].Assume t E 21 U 2 2 , say t E 21, If 2 1 n 2 2 # 8, then M , E k[C(G)].On the other hand if 21 n 2 2 = 8, then hfw E k[C(G)] because in this case 21 and 2 2 are joined by the edge { t ,z } , where z is any vertex in 2 2 . It remains to consider the case t $ 21 U 2 2 and t E P , since G is connected there is a path in G joining 21 with 2 2 . Therefore Mw = M,, for some bowtie 201 of G and Mw E k[G]. e)Conversely if R ( I ( G ) )is normal, thenby Proposition 7.4.1 we obtain is normal. 0 that k[G] In the corollary above the hypothesis that G is connected is essential to prove that k[G]normal imply R(1)normal; for instance if G consists of two disjoint triangles, then k[G]is a polynomial ring and R ( I ) is non normal. We give the following counter-example to show that in general it is not true that normality of the Rees algebra implies normality of the monomial subring.
is clearly not normal but R(1)is normal by the Jacobian criterion. Proposition 8.7.15 Let G be a graph and let G I , . . . ,G, be the connected components of G. Then R ( I ( G ) )is normal ifand only if either: (i) G isbipartite, or
(ii) exactly one of the components, say GI, is non bipartite and a normal domain. Proof. One may
K[G1] is
proceed as in the proof of Corollary 8.7.13.
Corollary 8.7.16 ([262]) The edge ideal of a graph G isnormal ifand only if G admits no H-configurations.
0
324
Chapter 8
Remark 8.7.17 (a) Let G be a graph consisting of two disjoint triangles. Note that G itself is an H-configuration but G is not contained in a bowtie, since the two triangles cannot be joined by a path. (b) If two disjoint odd cycles 21, 2 2 of a graph G are in the same connected component, then the two cycles form an H-configuration if and only if 2 1 , 2 2 cannot be connected by an edge of G and 21, 2 2 have no chords in G Integral closure of Rees algebras Let G be a graph with vertices . . ,x n and I = I(G ) its edge ideal. Consider the endomorphism # of the field I C ( x 1 , . . . ,xn,t ) defined by x i xit, t e t - 2 . It induces an isomorphism
21,.
where R = k [ q ,. . . ,x,] is a polynomial ring over a field k and C(G) is the cone over G obtained by adding a new variable t. 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)J. If 2 1 = { X I , 2 2 , . . . ,x,.= x 1 1 and 2 2 = { z 1 , z 2 , . . . ,zs = 21) are two edge disjoint odd cycles in C(G). Then one has
for some bow tie w of C(G), this follows readily because C(G) is a connected graph. In particular M,,is in the integral closure of k[C(G)J.Observe that if t occurs in 21 or 2 2 , then Mw E k[C(G)]because in this case either 2 1 and 2 2 meet at a point, or 2 1 and Z 2 are joined by an edge in C(G);see Proposition 8.7.10. Thus in order to compute the integralclosure of k[C(G)] the only bow ties that matter are those defining the set:
x3 = (M,I
UI
is a bowtie in C(G) such that t
4 supp(M,)).
Therefore by Theorem 8.7.9 we have proved: Proposition 8.7.18 k[C(G)]= k[C(G)][B]. To obtain a description of the integral closure of the Rees algebra of I note that using t$ together with the equality
Monomial Subrings of Graphs
325
where xT+l = 2 1 , we see that Mw is mapped back in the Rees algebra in the element
M,t+ If
= 21x2 ' '
X T Z 1 Z 2 ' ' ' z,t*
*
B' is the set of all monomials of this form, then one obtains:
-
Proposition 8.7.19 R [ l t ]= R[lt][B'].
Exercises 8.7.20 Prove that k[G]is normal for every connected graph G with six vertices, where k is a field.
8.7.21 Let G be a graph on the vertex set V = { X I , .. . ,xn} and choose a new set of vertices y1, . . . ,Y n , define S(G), the suspension of G, to be the graph obtained by attaching to each vertex xi the new vertex yi and the edge (xi,yi}. If I ( G ) is a normal ideal, then I(C(G))and I ( S ( G ) )are normal ideals.
+
8.7.22 Let G be a graph and let rn = (T s)/2 be the least positive integer such that G has two vertex disjoint odd cycles of lengths T and s. If I = I ( G ) is the edge ideal of G, prove that Ii is integrally closed for i < rn.
8.8
The equations of the edge cone
Let G be a graph on the vertex set V = { V I , .. . ,Vn) and R = lc[sl,.. . ,x,] a polynomial ring over a field k. The edge cone of G is the cone A c Iw" spanned by the set A of all vectors ei ej such that vi is adjacent to vj, where ei denotes the ith unit vector. In this section we present a combinatorial descriptionof the facets (faces of maximal dimension) of Et+A when k[G]has dimension n. We show some estimates for the a-invariant of Ic[G],for certain G. The problem of finding upper bounds for the a-invariant of more general normal monomial subrings generated by square-free monomials of the same degree was addressed in Chapter 7.
+
rW+
The irreducible representation of the edge cone The computation of the equations defining the facets of the edge cone of a graph G is a crucial step toward theeffective determination of the canonical module and the a-invariant of k[G],whenever k[G]is a normal domain. For simplicity we have focus our attention on connected non bipartite graphs, but the methods can be applied to bipartite graphs as well.
326
Chapter 8
Definition 8.8.1 The support of a E Rn is defined as SUPP(Cu)
= {aiI ai # 0).
Lemma 8.8.2 Let G be a graph with connected components G I , . . . ,G, and M its incidence matrix. If G1 is a tree with at least two vertices and Ga, . . . ,G, are unicyclic n o n bipartite graphs, then ker(Mt) = (p) for some vector p whose support is contained in (0, 1, -1).
Proof. Let V = ( ~ 1 , .. . ,un) be the vertex set of G. If G is a tree, using induction on the number of vertices it is not hard to show that ker(lMt) = (p), where supp(/?) = {1,-1). On the other handif G = Gi for some i 2 2, then M t is non singular and ker(Mt)= (0). The general case follows readily by writing M t as a "diagonal" matrix
M t = diag(M&,, . . . , M & J , where MG, is the incidence matrix of the graph V(G1) = {vi E VI pi = f l ) .
Gi. Observe the equality 0
The facets of the edge cone Let us introduce some more terminology and fix the notation that will be used throughout. To ease the reading we recall some of the relevant notation on polyhedral geometry; see Chapter 7. We set AG (or simplyA if G is understood) equal to the set{ q ,. . . ,cyp) of row vectors of the transpose of the incidence matrix MG of G. The edge cone & A of G is defined as the cone generated by d,that is:
By Lemma 8.3.2 one has n - co(G) = dim k[G]= rank ( M G )= dim R+d,
where co(G) is the number of bipartite components of G. If a E Rn,a # 0, then the set Ha will denote the hyperplane of Rn through the origin with normal vector a, that is,
Ha = {X E Iw"I (X,U ) = 0). This hyperplane determines two closed half-spaces
H: = {x E Etn( (z,a) 2 0) and Hi = {X
E
RnI (qa) 5 0).
A subset F C Rn is a properface of the edgecone of G if there is a supporting hyperplane H a such that
Monomial Subrings of Graphs
327
Definition 8.8.3 A proper face F of the edge cone is a facet if dim F = dim R+d- 1. When the edge cone is properly embedded into Rn it will turn out that its facets are of two kinds: those defined by independent sets (see definition below) and those defined by supporting hyperplanes of the form He;. Lemma 8.8.4 If the incidence matrix of G has maximal rank n, then the set F = Hei n & A is a proper face of the edge cone. Proof. The proof isanleft as
exercise.
0
Let A be an independent set of vertices of G. The supporting hyperplane of the edge cone defined by
will be denoted by H A . Lemma 8.8.5 If the incidence matrix of G has maximal rank n and A is an independent set of vertices of G, then the set F = An H A is a proper face of the edge cone.
Proof. The proof is left as an exercise.
0
To determine whether H A defines a facet of It+ A consider the subgraph L = L1 U L2, where L1 is the subgraph of G with vertex set and edge set V ( L 1 )= A u N ( A ) and E(L1) = ( z E E(G)(x
n A # 0)
respectively, and L2 =<S> is the subgraph of G induced by S = The vectors in A n H A correspond precisely to the edges of L.
v\ v(L1).
Remark 8.8.6 Let F be a facet of & A defined by the hyperplane Ha.If the rank of MG is 72, then it is not hard to see that Ha is generated by a linearly independent subset of 4. See Corollary 7.2.22. Theorem 8.8.7 ([305]) Let G be a graph with uertex set V = (w1,. . . ,w,) and M its incidence matrix. If rank(M) = n, then F is a facet of It+ A if and only if either
( a ) F = Hei n & A for some i, where all the connected components G \ (vi> are n o n bipartite graphs, or
of
( b ) F = H A n & A for some independent set A c V such that L1 is a connected bipartite graph, and the connected components of L2 are n o n bipartite graphs.
Chapter 8
328
Proof. Let A = ( 0 1 , . . . ,a q }be the set of column vectors of the incidence matrix of G. Assume that F is a facet of &A. By Remark 8.8.6 we may assume that there is a E Rn such that
(i) F = R+AnH,, ( a , a i ) 5 0 V i E (1,...,q ] , and
. . . ,a,-1 are linearly independent vectors in Ha. (ii) al, Consider the subgraphD of G whose edges correspond to a1 , . . . , an-l and itsvertexset is the union of the vertices in the edges of D. Let I' be the transpose of the incidence matrix of D , and D l , . . . ,D, the connected components of D. Without loss of generality one may assume
r= . . , rank(r) = n - 1, o o ... r, *
.
where I'i 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 0 be the matrix with rows 0 1 , . . . ,an-l. We consider two cases. Case (I): Assume 0 has a zero column, hence it has exactly one zero column and D has vertex set V ( D )= V \ {vi>, for some i. This implies r
r
i= 1
i= 1
Since ni 5 qi for all i, one obtains ni = qi for all i. Note that co(D) = 0 , by Lemma 8.3.2. Therefore D l , . . . ,D, are unicyclic non bipartite graphs. As any component of G \ {Vi} contains a component of D , we are in case (a). Case (11): Assume that all the columns of 0 are non zero, that is, D has vertex set equal to V . Note that in this case r
r
i=1
i= 1
Since rank (I?)= n - ~ ( 0 = n) - 1, we get that D has exactly one bipartite component, say D l . Hence n1 - 1 5 q1 and ni 5 qi for all i 2 2, which togetherwith Eq. (8.6) yields q1 = nl - 1 and ni = qi for all i 2 2. Altogether Dl is a tree and Dz, . . . D, are unicyclic non bipartite graphs. By Lemma 8.8.2 there is p E Rn such that )
ker(r) = ( p ) and supp(P)
c {0,1, -1).
We may harmlessly assume a = p , because H , = Hp and a E (p). Set
!
Subrings Monomial
of Graphs
329
Note that A # 0, 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 {vi, v j 1 is an edge of G for some vi,vj in A , then a k = ei e j and by (i) we get (a,a k ) 5 0, which is impossible because ( a ,ak) = 2. This proves that A is an independent set. , Next we show N ( A ) = B. If wi E N ( A ) , then C L = ~ ei + e j for some wj in A, using (i) we obtain ( a , a k ) = ai 1 5 0 and ai = -1, hence vi E B. Conversely if vi E B , since D hits no isolated vertices, there is 1 5 k 5 n - 1 so that arc = ei e j , for some j , by (ii) we obtain (a,a k ) = -1 + aj = 0, which shows that v j E A and vi E N ( A ) . From the proof of Lemma 8.8.2 we obtain V ( D 1 ) = A U N ( A ) . Therefore al,. . . ,an-1 are in H A and Ha = HA. Observe that L1 (see notation before Remark 8.8.6) 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 Dl is an spanning tree of i;l. Note that every connected component of Lz is non bipartite because it contains some Di,with i 2 2. Conversely assume that F is as in (a). Since the vectors in A n Hei correspond precisely to the edges of G \ {vi) 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. 0
+
+
+
Corollary 8.8.8 ([305]) Let G be aconnected n o n bipartitegraphwith vertex set V = { V I , . . . ,Vn}. Then a vector x = ( X I ,. . . ,x,) E Rn is in & A if and only if x is a feasible solution of the system of linear inequalities
CviEA xi -
-xi 5 0 , i = 1,...,n xi 5 0 , for all independent sets A
x,,iEN(A)
c V.
Proof. Let
R + A = H p - n H ~ be theirreducible representation of the edge cone as anintersection of closed half-spaces. By Theorem 7.2.18 the set Hbi n & A is a facet of & A, and 8.8.7. Theorem we may apply Proposition 8.8.9 ([305]) Let G be a connected n o n bipartite graph with n vertices. If k[G]is a, normal domain, then a monomial x!' - .!x belongs to k[G]if and only if the following two conditions hold (i)
P = ( P I , . . . ,Pn) is in the edge cone of G, and
330
Chapter 8
Proof. Set d = d G = { a l ,. . .,q ) . Assume ,B E R+dand deg(xP) even. We proceed by induction on d e g ( d ) . Using Corollary 8.3.5 one has the isomorphism Z"l(Q1, ' ,aq) fs1 Z2, hence 2p E R+-d n Z d = N d and one may write
i=l
i= 1
by inductiononemayassume ~ i a= i 0. Thereforefrom theequality above one concludes that the subgraph whose edges are defined by the set {ai I ei = 1) is an edge disjoint union of cycles 21, . . . 2,. By induction one may further assume that all the Zi's are odd cycles. Note r ,> 2, because deg(zb) is even. As G is connected, using Corollary 8.7.11 it follows that zfi E k[G]. The converse is clear because k[G] is normal. 0 )
Remark 8.8.10 Let us give a simple example to see that the connectivity hypothesis is essential in Proposition 8.8.9. Consider the graph G = G1UGS with six vertices consisting of two disjoint triangles:
The vector ,B = (1,.. . 1) belongs to the edge cone because 2p E xfi is not in k[G] . )
NAY but
Let G be a connected non bipartite graph with k[G]normal. The last results can be used to give conditions for a graph tohave a perfect matching. Proposition 8.8.11 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 IAI 5 1N ( A ) (for every independent set of vertices A of G. e
Proof. +) Since b[G]is normal:
Hence a = (1,. . . ,1) = ( a l , . , . ,a,) is in I&+ dc. Using Corollary 8.8.8 we get that the vector a satisfies the inequalities:
Monomial Subrings 331 of Graphs for every independent set of vertices A of G, as required. +) First we use Corollary 8.8.8 to conclude that a is in the edge cone, apply Proposition then 8.8.9 to get a E k[G]. CJ Corollary 8.8.12 (Generalized marriage problem) Let G be a graph with an even number of vertices. If G isconnectedandsatisfiesthe odd cycle condition, theh the following are equivalent (a) G has a perfect matching. (b) IAI 5 ( N ( A ) [for all A independent set of vertices of G. Proof. The proof follows using Theorem 6.1.8 and Proposition 8.8.11.
0
Corollary 8.8.13 ([51]) Let G = K n be the complete graph on n vertices. Then a vector x E Rn is in I$+ A if and only if x = ( X I , .. . ,x,) satisfies
Example 8.8.14 If G is a triangle with vertices cone of G has three facets defined by
Note that
x1
2 0,
x2
2 0 and
x3
3. 0
{x1,22,x3
1, then the edge
define proper faces of dimension 1.
The a-invariant of some normal subrings Let X be an arbitrary subset of Rn. The relative interior of X , denoted ri(X), is the interior of X relative to aff(X), the affine hull of X. As usual the interior of X in Rn is denoted by X o . We grade the edge subring k[G]c R of a graph G with the normalized grading k[G];= k[G]n R2i, where
00
R=$R~ i=O
is a polynomial ring over the field k with the standard grading. Recall that k[G]is normal if and only if NAG = E%+ d G fl ZAG, this follows from Corollary 7.2.29. If k[G] is normal, then by Theorem 7.2.34, the canonical module # k [ G ] of k[G]is given by “k(G]
= ({xnla E NAG n r i ( b d G ) ) ) -
This formula will be used throughout to estimate some a-invariants.
(8.7)
Chapter 8
332
Remark 8.8.15 Let G be a connectednon bipartite graph and take a vector ,f3 in ri(IWt_&). If F is a proper face of It+&, then /3 $! F . This observation holds for all cones because a proper face is always contained in the boundary [39, Theorems 5.31. In particular pi > 0 for all i, because Hei n It+AG is a proper face. Definition 8.8.16 The j o i n GI * G2 of two disjoint graphs G1 and G2 consists of GI U G2 and all the lines joining GI with G2. Proposition 8.8.17 Let G I and G2 be two connected graphs o n n vertices. If k[G1]and k[G2]are normal, then
a(k[G1* Gz])= -n. Proof. Let fi = { V I , . . . ,un} and V2 = {v,+l, . . . , ~ 2 be~ the ) vertex set of GI and G2 respectively. One may assume n 2 2. Set G = GI * G2. Note that the vector ,f3 = (1,. . . , 1) E satisfies
for all independent sets A of G , because IAI 5 n - 1 < IN(A)(. Hence by Corollary 8.8.8 and Lemma 8.3.2 we get p E (Et+ JIG)" n NAG. Since k[G] is normal by Corollary 8.7.11, we apply Eq. (8.7) to get x p E wk[G]. This implies -n ,< a(lc[G]).On the other hand by Corollary 7.7.15, we conclude a(k[G])= -n. 0 Proposition 8.8.18 Let G be a connected n o n bipartite graph o n n vertices. If k[G]is normal, then
where [x] is the least integer greater or
equal than x.
Proof. Let w1 and L J ~bethe canonicalmodules of k[G] and k[C(G)J respectively. Set d = d G and V = V ( G ) .By Lemma 8.3.2 one has
As k[G]is normal and according to Eq.(8.7),we may pick p E (rw+d)"nNd so that deg(sb) is equal to -a(lc[G]). Set y = (PI 1,/?2,. . . ,pn,1). To prove the first inequality we will show that y is in (I&+ &(G))O f l NdC(G), that is, one must show that y is not on any facet of Rk d C ( G ) . Let F be a facet of and H the hyperplane defining F . By Theorem 8.8.7 therearethree cases to consider. If H = Hei for some
+
Monomial Subrings of Graphs
333
1 5 i 5 n,then y # H ; for otherwise p would be on the proper face of &d determined by H e i , which contradicts the choice of /?(see Remark 8.8.15). If H = Hen+l, note (y,en+l) = 1 and y 4 H . Assume that H = H A , for some independent set in C(G). If vn+l E A , then y satisfies
and y 6H A . Assume vn+l @ A, since p satisfies the inequality
it follows that y satisfies
Hence y $ H A . Observe that k[C(G)] is normal by Corollary 8.7.11. Thus using Eq.(8.7) one obtains ZY E C J ~and the proof of the first inequality is complete. The second inequality follows from Corollary 7.7.15. 0 Proposition 8.8.19 Let G1 and Ga be two connected non bipartite graphs. If k[Gl] and k[G2) are normal, then
Proof. Choose vectors a and b in
respectively, such that deg(z") = -a(k[GI]) and deg(zb) = -a(lc[G2]). Using arguments similar to those given in the proof of Proposition 8.8.18 it follows that (a,b) is an interior point of &-d G 1 * G 2 , which implies the 0 asserted inequality.
Exercises 8.8.20 Let G = GI UGz be a graph with two non bipartite connected com-
ponents G1 and G2. Find a relation between the irreducible representation of the cone R+ $ 2 and ~ the irreducible representations of the cones IR+ d G 1 and
k d G z .
Chapter 8
334
8.8.21 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. 8.8.22 Prove that the following graph does not have a perfect matching by exhibiting an independent set A such that IAI $ JN(A)I.
8.8.23 Prove that the following are the equations of the edge cone of the graph G shown on the left.
-21
-x4 -25
x3
x2
21
+ x4 + x5
Show that the a-invariant of k[G]is equal to -4.
0 0 0 x1 x1
x2
+ x2 + x4 + x5 + + x4 + x5 x3
+ x3
Chapter 9
Semigroup Rings of Complete Graphs The purpose of this chapter is to study some of the properties of edge subrings of complete graphs, including the bipartite case. Special attention is given to the computation of their Hilbert series, Grobner bases and Noether normalizations.
9.1
Monomial subrings of bipartite graphs
Here we treat two special types of monomial subrings attached to complete bipartite graphs and compute their Hilbert series and a-invariant. First we study the edge subring of a complete bipartite graph, and then we examine the Rees algebra of the edge ideal of a complete bipartite graph.
The monomial subring of a bipartite graph One of the results that simplify the study of an edge subring associated to a complete bipartite graphs is the fact that its toric ideal is a determinantal prime ideal. Let us begin with a classical formula for determinantal ideals.
Theorem 9.1.1 Let X = (xij) be an m x n matrix of indeterminates over a field I( and It(X) the ideal generated by the t-minors of X . If m 5 n and 15 t 5 m 1, then
+
height(It(X)) = (rn - t
+ l ) ( n- t + 1). 0
Proof. See [52, Theorem 2.51.
335
336
Chapter9
Proposition 9.1.2 Let K be a field and ICm,n the complete bipartite graph with vertex set V = (21, . . . ,x m , y1 . . . ,Yn} . Then the toric ideal P of K[ICm,n] is generated by the 2 X 2 minors of a generic rn X n matrix. Proof. Let
A = K[{TijJ1 5 i 5 m, 1 5 j 5 n}] be a polynomial ring over a field K in the indeterminates Tij and consider the graded homomorphism of K-algebras
is generated by the 2 x 2 minors of a We claim that the kernel of generic M x n matrix. Indeed let X = (Tij). It is clear that the ideal I’z(X), generated by the 2 x 2 minors of X , is contained in P = ker($). On the other hand, since the graph is bipartite one has $J
dim K[Km,,] = rn
+ n - 1,
see Proposition 8.2.12. Therefore
+
height(P) = (mn)- (rn n - 1) = (rn- l)(n - 1) = height(Iz(X)), the latter equality by Theorem 9.1.1. Since they are both prime ideals, we 0 have I 2 ( X ) = P. The Cohen-Macaulayness and Gorensteiness of algebras that include K[Ic,,n], has been dealt with in great detail in [50]. Proposition 9.1.3 Let Km,n be the complete bipartite graph and AIP the toric ideal of K[Km,,]. If n 2 rn, then the Hilbert series of AIP is given by
Proof. Let VI = (21). . . ,x,} and V - = (91, . . . ,g n } be two disjoint sets of vertices, and R = K[Vl U Vz]a polynomial ring over a field K . Let
be a polynomial ring in the indeterminates homomorphism of K-algebras
Tij
and consider the graded
By Proposition 9.1.2 the toric ideal P = ker($) is equal to 1 2 ( X ) , the ideal generated by the 2-minors of the m X n matrix X = (Tij). Note that the
Semigroup Rings of Complete Graphs
337
dimension of K[Km,n]is equal to rn+n - 1. The set of the 2 X 2 minors of X form a Grobner basis for I z ( X ) with respect to the lexicographical ordering induced by
>
T1'1'
* *
> T1n >
*
*
> Tm1 >
*
*
> Tmn,
see [59, 2761. Note that in(P), the initial ideal of P , is generated by the 2-diagonals of X, that is, one has
By Corollary 2.4.13 one has F(A/P, x) = F(A/in(P), z ) , thus the proof reduces to compute F(A/in(P), x). Consider the arrays:
and
(m-l)n*
Let It and Je be the ideals of A generated by the 2-diagonals of X e and Ye respectively. Note in(P ) = In. For 2 5 e 5 n there are graded exact sequences
where
Kt = (Tij(i < m, j < a) + Jt + (Tm(t+l),* - 1 T m n ) . As Tml,. . . ,Tme is a regular sequence on A / K t , it is not hard to see that one can write
where
Charher 9
338
that is, the polynomial h(m-l)(n-t+l) determines the h-vector of Ae/Je. Hence
F(A/In,z) =
( g p ( A / K t > z ) ) z + F(A/(TmZ,...,Tmn,Il),z)
- (1 + h(rn-1)2 +
identity.
+ h(m-l)(n-l))X (1 - z)rn+n-l
+ h(m-1)n
Solving this recursive formula for the Hilbert series of A / I n rapidly leads to the required 0 Corollary 9.1.4 Let Km,n be the complete bipartite graph. If n 2 m, then the a-invariant and the multiplicity of K[ICm,n]are given by:
Proof. Use Remark 4.1.11.
0
Another approach to the computation of the a-invariant is through the of [132]. theory of Segre products, and then appealing directly to the results Useful methods to compute the a-invariant of some interesting families of algebras, e.g. algebras with straightening lawsor Stanley-Reisner rings, are introduced in [42]. Letus state two generalresults on determinantal rings generalizing Proposition 9.1.3 and Corollary 9.1.4 respectively Theorem 9.1.5 ([69]) Let X = (zij) be an rn x n matrix of indeterminates over a field K with m 5 n and IT+l( X ) the ideal generated by the ( r 1)minors of X . Then the Hilbert series of R,+1 = K [ ~ i j ] / I r + l ( is: X)
+
Theorem 9.1.6 ([133,421) Let X = (zij) be an rn x n matrix of inde] by the terminates over a field K and & ( X ) theideal of K [ ~ i jgenerated r-minors of the matrix X . If rn 5 n, then
Proposition 9.1.7 Let G be anspanningconnectedsubgraph n 2 m, then the a-invariant of K[G]satisfies
of Km,n. If
Semigroup Rings of Complete Graphs
339
Proof. From Corollary 9.1.4 one has
Hence one may proceed as in the proof of Proposition 7.7.14 to rapidly derive the asserted inequality. 0 An application of Dedekind-Mertens formula If R is a commutative ring and f = f ( t ) E R[t]is a polynomial, say
f = no
+
e
+ amtm,
the content of f is the R-ideal (no,. . . , a,). It is denoted by c(f). Given another polynomial g, the Gaussian ideal of f and g is the R-ideal
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 4 f g ) = c(f)c(g>In general, these two ideals are very different but one aspect of their relationship is given by a formula originally due to Dedekind-Mertens (see [91] and [231]): c(fg>c(s>" = c(f)c(s)"+l. (9.1) We consider the ideal c(fg) in the case when f and 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(fg). One path to our analysis and its applications to Noether normalizations of some semigroup rings, starts by multiplying both sides of (9.1) by c(f)"; we get c(fg>[c(f>c(g)l" = c(f>c(s>[c(f)c(dlrn. (94 It will be shown that this last formula is sharp in terms of the exponent rn = 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 c I such that, for some non-negative integer T , the equality
holds. The smallest such integer is the reduction number T J ( I )of I relative to J , and the reduction n,umber T ( I )of I is the smallest reduction number amongst all reductions J of I .
ChaDter 9
340
Thus (9.2) says that J = c(fg) 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 9.1.8 ([91]) Let R = k[zo,. . . ,x m , yo, . . . ,yn] be a polynomial ring over a field k and let
i+j=q Then
is a Noether normalization of S . Proof. Let R[t]be a polynomial ring in a new variable t and f , g E R[t] the generic polynomials m
n
Set I = c(f)c(g) and J = c(fg). Note that J is a reduction of I by Eq. (9.2) and more precisely J I m = Im+l. Therefore
R[Jt]L) R [ I t ] 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:
where m = (20... , x m ,yo,. . . ,yn). To complete the proof note that the dimension of k [ ~ i y j ’is~ equal ] to n m 1, by Lemma 8.3.2. 0
+ +
Theorem 9.1.9 ([72]) Let R = k[xo,. . . ,xm,yo,. . . ,yn] be a polynomial ring over a field k and let f,g E R[t]be the generic polynomials m
n
i=O
j=O
with n 2 m. If I = c( f)c(g) and J = c( fg), then r J ( I )= m.
Semigroup Rings of Complete Graphs
341
Proof. By Theorem 9.1.8 there is a Noether normalization:
where h, =
~ i l ~ j . i+j=q
Set I = ( f l , . . . ,f 8 ) . We note that the ideal I = (xiyj's) is the edge ideal associated to a complete bipartite graph G which is the join of two discrete graphs, one with rn + 1 vertices and another with n + 1 vertices. Since k[G] is Cohen-Macaulay (see Corollary 8.7.11), using Proposition 2.2.14 we get
k[G]= AfP1 @
*
- CB A f B k , pi E Ns.
By Corollary 9.1.4 the a-invariant of k[G]is equal to -72- 1, we can compute the Hilbert series of k[G]using the decomposition above to conclude that max { deg(f P i ) } = m,
l<_z<_k
where the degree is taken with'respect to the normalized grading. On the other hand assume T = T J ( I )and JI' = ITS1,since we already know the inequality r 5 m. It suffices to prove r 2 rn. Note
where S
Z=(.
a = ( a l , .. . ,a8) E N8 and la1 = c a i ,< r i=l
If T < m this rapidly leads to a contradiction because there is 3 c Z such can be written as that k[G]
k[G]=
@ Af". "€3
See Exercise 9.1.18.
0
Multiproducts Inorder to see a different explanation of Eq. (9.2), we extend it to the product of 3 (or more) polynomials, but using the theory of Segre products as a tool. Let X = (20, . . . ,xm}, Y = {yo, . . . ,yn}, and 2 = { ZO,. . . ,z p ) be 3 sets of indeterminates. Defining the polynomials m
n
i=O
j=O
V
k=O
Chapter 9
342
one has that J = c ( f g h ) is a reduction of I = c ( f ) c ( g ) c ( h )by DedekindMertens formula. If rn 5 n 5 p , a simple calculation will show
r J ( I ) 5 rn
+ n.
We now resolve this inequality. Proposition 9.1.10 ([72]) Let X = ( 5 0 , .. . ,xm}?Y = {yo,. . . ,yn}, and 2 = {zo, . . . ,z p ) be sets of distinct indeterminates, let R = k [ X ,Y ,Z] be a polynomial ring over a field k , and let
I = (xiyjzal xi E x,y j E Y, Zk E 2 ) . T h e n I is a normal ideal of R. Proof. Wewillshow that Iq is complete for all q 2 1. Let I: bethe integral closure of Iq and let f E IaQ be a monomial. We write
Since f" E Isq for some w
> 0 we can write f" = x;; . . . x ; p ,
where M is a monomial whose support is contained in Y U 2. Therefore we obtain T x i=l
i=l
which implies i= 1
A similar argument shows 9
t
i= 1
i= 1
Therefore f E I Q .
0
By Theorem 7.2.35,the Rees algebra R[It]is a Cohen-Macaulay ring. Since F ( I ) = k[ziyjzkI xi E y j E Y, Zk E Z]
x,
is a normal domain by Proposition 7.4.1, and therefore Cohen-Macaulay by Theorem 7.2.35.We may thus more easily compute the reduction number of the subring F(1).
k
Semigroup Rings of Complete Graphs
343
Theorem 9.1.11 The reduction nurnberrJ(1) of the ideal I above is m+n. Proof. Since F ( I ) is a Cohen-Macaulay algebra,its reductionnumber can also be obtained from the degrees of the generators of its canonical module. But F(1) is a Segre product of standard Cohen-Macaulay algebras and the canonical module is given by an explicit formula from the canonical modules of the factors. Entering the data in [132, Theorem 4.3.11, we get r J ( I ) = m + n. a
Rees algebras of complete bipartite graphs The following result show that Rees algebras of edge ideals associated to complete bipartite graphs have some structure; its proof makes use of general facts about ladder ideals that can be found in the literature. Proposition 9.1.12 Let B/Q be the presentation of R ( I ) , the Rees algebra of the edge ideal I = ( x i y j l l 5 2' 5 m, 1 5 j 5 n). Then the toric ideal Q is generated b y the 2-minors of the ladder
Y=
X1
.
Xn,
Grobner basis w.r.t the lex ordering induced * > T17, > x2 > . ' ' > Tmn. Proof. Consider the presentation of the Rees algebra of I :
Note 12(Y)c Q = ker($). As they are both prime ideals of height mn - 1 by [67, Section 41 and [228] (cf. [65, Proposition 3.3.1]),one has equality. The assertion that the2-minors are a Grobner basis of I 2 ( Y ) is a general fact of ladder ideals [228]. 0 Proposition 9.1.13 Let B/Q be the presentation of R ( I ) , the Rees algebra of the edge ideal I = (xiyj 11 6 i 5 m, 1 5 j 5 n ) . If n 2 m, then the Hilbert series of B/Q can be written as
I n particular a ( B / Q ) = -(n
+ 1 ) and e ( B / Q ) = (*":)
- 1.
Chapter 9
344
Proof. Let
B = K[si,yj,TijI 15 i 5 m, 1 5 j 5 n] be a polynomial ring over a field K and consider the ladders
Let It and Jl be the ideals generated by the 2-diagonals of Me and Ne respectively, where a 2-diagonal is the product of the diagonal entries of a 2 x 2 submatrix. By Proposition 9.1.12 and Corollary 2.4.13 one has that the Hilbert series of B/Q is equal to the Hilbert series of B/I1. Hence the proof reduces to compute F(B/Il, z ) . For 1 5 5 rn - 1 consider the graded exact sequences
where K e = ( T i j ( t + l < i ) + ~ t + ( z ..., l , ze-1). Therefore
AS xe, . . . ,xm is a regular sequence on B/Ke using Proposition 9.1.3 one readily obtains
and consequently
Semigroup Rings of Complete345 Graphs Using the formula
yields the required equality. Theorem 9.1.14 ([306]) Let
be the monomial subring of the cone over a complete bipartite graph
Kmn.
If n 2 m 2 2, then the canonical module ws of S is minimally generated a8 follows: (a)
If n = m, then
+ 1, then ws is minimally generated b y the set of monomials: x1 - 'xmyl - - .y7Pxb,where n - m + 1 5 b 5 n + M - 1 and
(b) If n 2 m
(i)
a
n + m + b ~ 2 N , or (ii) x:' xkrny1 . y,x6, where m
i= 1
b = 2 , 3 ,...,n - m + l , a n d a i
21f o r a l l i .
Define
As S is normal of dimension n+m+ 1 (see Theorem8.7.9 and Lemma 8.3.2) one can use the Danilov-Stanley formula (see Theorem 7.2.34) to express its canonical module as:
where (&A)" is the interior of the cone generated by A and S has the normalized grading. Let
Chapter 9
346
be a minimal generatorof the ideal US. According to Theorem 8.8.7 a vector /3 E W+n+lis in (R+d)" if and only if p satisfies the inequalities:
m
mtn
i= 1
i=m+l
m+n
m
i=m+l
i= 1 m+n
Pm+n+l
L
pi,
-I+ i= 1
Therefore one has:
m
n
i= 1
i= 1
n
m
i= 1
i= 1
i= 1
i= 1
Using that a E N d one also obtains: m
n
i=1
i= 1
Observe that (9.4) and (9.5) yield b 2 2. We consider the following cases: I. Assume ai 2 2 for some i. We claim that bi = 1, for all i. For simplicity assume a1 2 2 and bl 2 2. Write: m
n
i= 1
i= 1
If 1 5 i 5 2, write = M'ylz. Since log M = log M'+logylz one has that log M' is in Z d . It is not hard to see that log M' satisfies all the equations in (9.3), thus log M' E (R+A)O. By thenormality of S oneconcludes
Semigroup Rings
of Complete Graphs
347
log M' E Nd,hence M' is in W S , which contradicts the minimality of M . If i 2 3, consider M = M'x1y1, as before one readily obtains M' E U S , which again contradicts the choice of M . 11. Assume n = m. First let us show ai = 1 and bj = 1 , for all i, j. By the previous case and symmetry if ai 2 2 for some i (resp. bi 2 2 ) , then bi = 1 for all i (resp. ai = 1 for all i). If ai 2 2 for some i , say i = 1, then M' E US, where M = M ' q x , and this is impossible by the choice of M . Next, using (9.4) and (9.5))it follows that b = 2i for some 1 <, i 5 n - 1 , and we are in case (a). 111. Assume n 2 m 1 2 3. We claim that b, = 1 , for all i. If bi 2 2, say i = 1, then by case I one has ai = 1 for all i. As before one readily derives a contradiction by writing M = M'ylz. Altogether one can write:
+
If ai = 1 for all i, then using (9.4) and (9.5) one has n-m+l_
and we are in case (b) . Next assume ai (9.4) and (9.5) one has
and m + n + b ~ 2 N ,
2 2 for some i, say i = 1. Combining
m
n<xai+b-2. i= 1
If the inequality
m
n<xai+b-3 i= 1
holds, one derives a contradiction by considering M = M ' x ~ x .Therefore m
m+l_<xai=n-b+2 i= 1
+
for b = 2 , . , n - rn 1, and we are in case (b). Thus any minimal generator of ws must be as in (a) or (b). To complete the argument observe that any of the monomials occurring in (a) or (b) are sare they and multiple of each other. 0 in unot #
,
Remark 9.1.15 If n > rn = I, then assertion (b) above still holds. To show it replace the very first set of inequalities in (9.3) by pi 2 1, i = 2 , . . . ,n+m. On the other hand if n = m = 1 , then u s = (z:y;z2). Corollary 9.1.16 Let S = R ( I ) be the Rees algebra of the edge ideal I of the complete bipartite graph Krnn. If n 2 m 2 2 or n > rn = 1, then
Chapter 9
348
Proof. If n = m 2 2 or if n > m = 1 the formula follows rapidly from Theorem 9.1.14(a) and Remark 9.1.15. Assume n 2 m 1 2 3. Observe that there arem - 1 generators of us as in Theorem 9.1.14(i), to count the remaining generators of us as in Theorem 9.1.14(ii) note that the sum from b = 2 to b = n - m 1 of the m-partitions of n - b 2 is equal to
+
+
+
n-m+l b=2 n-m
m-l+i
i=O
Hence type(S) = ( m - 1)
+);(
-1=
);(
+(m-2),
as required.
Exercises 9.1.17 Let R[t]be a polynomial ring over a ring R. Iff = f ( t )and g = g(t) are two polynomials in R[t],prove the Dedekind-Mertens formula
9.1.18 Let M be a graded free module over a polynomial ring A such that M = Af1 @ @ Af,, where fi is homogeneous for all i. If M is generated by homogeneous elements 91, . . . ,gs, then M = Agi, @ . Agi, , for some
-
21,.
. . ,a,..
a
9.1.19 Let A = k[zl,. . . ,X,] and R = k[yl,. . . ,yn] be two polynomial rings over a field k , prove that the Segre product
9.1.20 Let A and R be two standard algebras over a field k of dimensions m and n respectively. If S is the Segre product of A and R, prove the following relation between the multiplicities
Semigroup Rings of Complete Graphs
349
9.1.21 Let R = K [ x l ,. . . , x,] be a polynomial ring over a field K and m = (x1,. . . ,x,). Prove that the toricideal of the Rees algebra R(m)is the ideal generated by the 2-minors of the following matrix of indeterminates:
9.1.22 Let R = K[xl,. . . ,x41 be a polynomial ring over a field K and I the 3rd square-free Veronese subring of R generated by the monomials
Prove that the Rees algebra R ( I ) is a complete intersection. 9.1.23 Let R = K [ x l , .. . ,~ n be ] a polynomial ring over a field K and I = (21,.. . Set s = - [n/kl n, andwrite, n = qk q , for some q, r1 E N such that 0 5 IC < rl. By a "counting degrees" argument prove:
+
+
9.1.24 Let R = K[xl, . . . ,x,] be a polynomial ring over a field K and m = ( X I , . . . ,x,). If I = and J = (x?,. . . ,x:), then J is a reduction of I and the reduction number of I relative to J is given by
,IC
T J ( I )= -
+ n,
where [x1 is the c&ling'of x.
9.1.25 Let S be a standard algebra over an infinite field K and
A = K[hl,. . . ,h d ] cs S a Noether normalization of S with hi E 5'1 for all i. If bl, . . . ,bt is a minimal set of homogeneous generators of S and as an A-module, prove
where r = maxi{ deg(bi)} and that T 2 0 is the minirnum integer where equality occurs. The reduction number of S , denote by r ( S ) ,is the minimum T taken over all Noether normalizations. See [298, Chapter 93 for an study of several degrees of complexity associated to a graded module. 9.1.26 Let S be a standard algebra over an infinite field I< and a(S) its a-invariant. If S is Cohen-Macaulay prove: r ( S ) = a(S)
+ dim(S).
Chapter 9
350
9.1.27 Let 4: B 3 S be a graded epimorphism of standard K-algebras, where K is an infinite field. If dim(B) = dim(S) and
is a Noether normalization of B with hi E S I for all i, prove that S is integral over A' = K[q5(hl),. . . ,$(&)I and that the composition
is a Noether normalization of S.
9.1.28 Let 4: B -+ S be a graded epimorphism of standard K-algebras, where K is an infinite field. If dim(B) = dim(S), prove r ( B ) 2 r ( S ) .
9.2
Monomial subrings of complete graphs I
Lexicographical Grobner bases and Hilbert functions of toric ideals of edge subrings associated to complete graphs will be considered here. Let us recall some terminology and fix some of the notation thatwill be used throughout the section. The lettersi, j,k,e, n, r will denote positive integers. We denote by Ggn, the graph with vertex set
and edge set
where
on = ( ( i , j ) l I < i
< j 5 n}.
Given a set D c Vn, we denote by GD the maximal subgraph of Gg, with vertex set D. Let An = k[Vn] be a polynomial ring over a field IC and G a graph on the vertex set Vn. The edge-idea2 I ( G ) , of the graph G, is the ideal of An generated by the square free monomials TijTke so that Tij is adjacent to Tke in G. If all the vertices of a graph G are isolated we set I ( G ) = (0). The graph G is called Cohen-Macaulay (C-M for short) if I ( G ) is a C-M ideal. A subset C c Vn is a minimal vertex cover for G if every edge of G is incident with some vertex in C and there is no proper subset of C with this property.
Semigroup Rings
of Complete Graphs
351
Let R = k[xl,.. . ,xn] be a polynomial ring over the field k , and /cn the complete graph on the vertex set X = { x 1 ,. . . ,x,}. For i < j we set f i j = xixj. Let k [ G ]= k[{fijl1 I i < j 5 n}] be the k-subring of R spanned by the
fij.
Consider the homomorphism
The ideal Pn = ker(?+b)is the presentation ideal or toric ideal of k[lcn]. Proposition 9.2.1 ([304]) Let R = k [ z l , . . . ,zn] be a polynomial ring over a field k and Pn the presentation ideal of k[lCn]. Set
If the terms in A , = k [ { T i j l l 5 i < j 5 n)] are orderedlexicographically by T12 < < TIn < T23 < - - < T2n < . < T(n-lln, then B is a minimal generating set for Pn and x3 is a reduced Grobner basis f o r P,. 0
.
Proof. By Corollary 7.1.7 it follows that B is a minimal generating set for Pn. Let f,g E B. Taking into account Theorem 2.4.15, it suffices to verify that S(f,g ) , the S-polynomial of f and 9 , reduces to zero with respect to B. There aretwo sequences of positive integers i < j < k < l and p < q < T < s so that
Assume f and g as in (a) and (c) respectively, the otherscases can be treated similarly. We may assume that the leading terms of f and g have common factors, for otherwise S(f,g) reduces to zero w.r.t {f,g ) by Lemma 2.4.14. First we assume Tij = Tpq,in this case i = p and j = q , and the Spolynomial of f and g is equal to S(f,g) = -T,s~peTqr,+~keT,sTq,. Observe that
Hence S(f,g) --+B 0 if {k,e) n { T , s) in { T , s). If e > s and T > b then
# 8. We may then assume k and not
352
Chapter 9
On the other hand if k' > s and k
> r we have
In both cases S(f,9) +B 0. Assume s > k' and k
> r.
From the equality
we derive S ( f ,g ) +B 0. To complete the argument consider the case s and r > IC, using the equality
>C
we obtain S(f,g) -+B 0. Now consider the case Tij = T,,, note that the case Tkt = Tpqis symmetric. We have i = T , j = s and S(f,g ) = T'~~T,,T,,.- TpqTreTsk. Since
we obtain S ( f , g ) + B 0. Finally we show the case Tke = T,,, notice k = r ,
C = s and S(f,9 ) = TijTpsTq, - TpqTisTjr. If i 2 q then
and S(f,g ) +B 0. If i
< q, then
Example 9.2.2 If k[{TijI 1 5 i
< j 5 s)]
has the lex ordering induced by
Then the reduced Grobner basis for the toric ideal P of k[&] is equal to:
Semigroup Rings of Complete Graphs
353
Corollary 9.2.3 If Pn is the toric ideal of the subring k[Kn] and n 2 4, then Pn is the ideal 12(Y) generated by the 2-minors of the symmetric array
Lemma 9.2.4 Let n 2 4 be a fixed integer, and let r be positive integer so that 1 g r _ < n -1. Set
If I ( G a i )is a Cohen-Macaulay ideal of height i(i - 3) 2 f o r all i < n, then I(GD,,) is a Cohen-Macaulny ideal of height
Proof. We proceed by descending induction on k. If k = r we have D:T = 0 and Go,, N Gg,,-,. Assume IC 5 r - 1 and G D ( ~ +Cohen-Macaulay ~), of the appropriate height. Let
Since N is equal to
354
Chapter 9
it follows that
and that N is a minimal vertex cover for Go,,
. Therefore
) , , N can not be Observe that 2'(k+ljn is an isolated vertex in G D ( ~ + ~hence contained in a minimal vertex cover for G D ( , + ~ and )~
Consider the exact sequence
Since the endsof this sequence are bothCohen-Macaulay rings of dimension equal to dimAn/I(GDh,)we derive t h a t A n / I ( G ~ , , )is CM, as required. 0 Corollary 9.2.5 ( [ 2 8 0 ] ) Let
Proof. Take r = n - 1 and IC = 1 in Lemma 9.2.4. Theorem 9.2.6 The edge, ideal I(G,,,,) t o n(n - 3)/2.
0
is Cohen-Macaulay of height equal
Proof. The proof is by induction on n. The assertion is not hard to verify for n 5 5 . Assume the result for all graphs isomorphic to G,; for i < n. Set
We claim that Gc, is Cohen-Macaulay with ht I(Gc,) = ( r - 1)
+ (n - l ) ( n- 4)
9
for 1 5 r 5 n - 1. The proof of the claim is by induction on r . If r = 1 then I ( G c l ) = I(Ggn-l) and, by the first induction hypothesis, I(Gcl) is CM of height ( n - l ) ( n- 4)/2. Assume Gci CM with ht I(Gc,)= (i - 1 )
- 4) + (n - l)(n 2
Semigroup Rings of Complete Graphs for i
355
< r , Let U, = {Tij E C, I Tij is adjacent to T,,,},
and let Dl, be as in Lemma 9.2.4. Observe that
lU,I = ( r - l ) ( n- r - 1)
+ ( r - 2 )2( r - 1)
an application of Lemma 9.2.4 yields that L, is Cohen-Macaulay and
=
("-2"')+ ( r - - 2 ) + ( r - l ) ( n - r - l ) +
Hence ht (L,) = ( r - 1)
(Til).
+ ( n - l )2( n - 4)
On the other handby the second induction hypothesis ht (J,) = ht (L,) and J, is CM. Altogether we have that AIJ, and AIL, are Cohen-Macaulay rings with the same dimension. Since L, is also equal to ( I ( G c r W lU) r, ) we have ht ( U r , I(Gc,_l)) = ht I(Gcr-l) 1.
+
Therefore U, can not be a subset of a minimal cover for, G C , - ~and , thus
From the exact sequence
we obtain I(Gc,) Cohen-Macaulay. In particular for T = n - 1 we conclude a that I ( G g n )is Cohen-Macaulay of height equal to n(n - 3 ) / 2 . Proposition 9.2.7 Let Fn(X) = h , ( z ) ( l - X)" be the Hilbert series of the ring An/I(Gg,,). Then the polynomial h n ( x ) satisfies the diflerence equation
for n
2 5 with boundarg' conditionsh3(z) = 1 and hd(z),=
1
+ 22 + x 2 .
Chapter 9
356
Proof. Set A = A , and C, = V,-l U {TI,,T2,,. . . ,Tr,}. Let D l , and UT be as in Lemma 9.2.4 and Theorem 9.2.6 respectively. Consider the exact sequences
where J2 = (Tz,, . . . ,T(,-l),, I(Gg,,-l)). On the other hand, by induction on r , we readily obtain
Therefore
Lk=n-2
k=4
J
subtracting the last two equations we obtain
The boundary
conditions follow readily since I(Gg3)= (0).
0
Corollary 9.2.8 ([262]) The ring k[IC,] is a Cohen-Macaulay ring. Proof. Let A,/Pn be thepresentation of k[lc,]. Note that in(P), the initial ideal of P , is equal to I ( G u n ) .By Theorem 9.2.4 we obtain that A/in(P) is Cohen-Macaulay, and this implies (see [133, 2081) that A / P is Cohen-Macaulay as well. c1
357
Semigroup Rings of Complete Graphs Remark 9.2.9 Let
R ( I ) = k [ { ~ 1 , . , x n , T z i ~ j5l Ii < j 5 TZ}] be the Rees algebra of the edge-ideal I = I(Icn) and
Then by Proposition 8.2.15 there is an isomorphism
induced by c p ( ~ i= ) zizn+l
and ‘p(Tzizj)= zizj.
Therefore by Lemma 9.2.8 we obtain that the Rees algebra of the edgeideal of a completegraph is Cohen-Macaulay. In factsuchalgebras are even normal by Proposition 7.4.5. Theorem 9.2.10 ([304]) Let Icn be the complete graph o n n vertices and An/Pn be the presentation of k[Kn].If n 2 3, then the Hilbert series Fn(z) of An/ Pn satisfies:
Proof. Let Fn ( z ) = F(An/Pn,2 ) . Since in(Pn), the initial ideal of P,, equals I ( G g n )by , Corollary 2.4.13 the Hilbert series of An/I(Gg,) is equal to Fn(2). Let
and let
00
n=3
be the generating function of the sequence we have
( ~ m n ) n > 3By .
Proposition 9.2.7
Recall that 00
Go(z) = x z n , n=3
h3(2) =
1, and hd(z) = 1
+ 22 + z 2 .
358
Chapter 9
Therefore for m 2 3 we obtain
00
00
and consequently
for m 2 3. By a similar argument we derive
Hence a l n = n(n - 3)/2 €or n 2 3. To complete the proof observe that
-
Xm
Gm ( x ) - (1- x)2m+l for m 2 2. Hence amn =
(&)
n=2m
for m 2 2 and n 2 3.
0
Remark 9.2.11 The Hilbert function of k[ICn]can be computed using the fact that the Ehrhart polynomial of the second hypersimplex of order n is equal to the Hilbert function of k[rCn]. See Chapter 7 and [79, 2771. Corollary 9.2.12 The multiplicity e(k[lCn]),of the ring k[lCn],is equal to 2n-1 - n.
Exercises 9.2.13 Let k be a field and A = k [ { T i j ]1 5 i permutation of the variables:
<j 5
5 } ] . Consider the
Prove that the toric ideal P of k[&] is not invariant under the automorphism of A induced by 4.
Semigroup Rings of Complete Graphs
359
9.2.14 If Q is the toricideal of the Rees algebraR(1)of the monomial ideal I = (xixjp < i < j 5 n - 1), then Q is the ideal 12(Y) generated by the
2-minors of the symmetric array
and the 2-minors of Y form a Grobner basis for I,(Y) w.r.t the lex ordering TI2 < . * < Tl(n-1) < 2 1 < T23 < ' < T'(n-1) < 22 < * < ~ n - 1 . Hint Set B = K[T'jI 1 5 i < j 5 n] and zi = Tin for 1 5 i 5 n - 1. There is a commutative diagram *
*
Since a is a isomorphismonehas Q = ker($) = ker('p) , then use the description of ker(cp) given in Proposition 9.2.1..
9.3
Noether normalizations of edge subrings
In this section two explicit Noether normalization of the edge subring of a complete graph are presented, it will turn out that those normalizations can serve as a model to deal with some specific graphs. The results of this section are due to A. AlcAntar [2, 31. There are few classes of graded algebras where homogeneous Noether normalizations are known, such classes include quotientsof monomials ideals [298, Proposition 2.3.31, edge subringsof complete bipartite graphs [72] and edge subrings of bipartite planar graphs [85]. Systems of parameters of edge subrings Let G be a graph on the vertex set V = ( X I ,. . . ,z n ) and R = K [ z l ,. . . ,x,] a polynomial ring over a field K . One can express the toric ideal of the edge subring K[G]as the kernel of the epimorphism of K-algebras: $: A = I([(TijIi
<j
and xi is adjacent to z j } ] + K[G]
ChaDter 9
360
induced by +(Tij)= zixj)where {Tij) is a new set of variables in one to one correspondence with the edges of G. We shall always assume that A and R have the usual grading and that K[GJ has the normalized grading
thus $ is a gradedmap. In the sequel we denoteker($) by P(G), An standard Noether normalization of A / P ( G ) is an integral extension
where hl, . . . ,h d are homogeneous polynomials of degree one and d is the Krull dimension of K[G].Note that (*) is a Noether normalization if and only if rad(h1,. . . ,h d , P ( G ) )= m, where Dl is the maximal ideal of A generated by all the variables Tij.See Chapter 2. Edge subrings of complete graphs The completegraph with vertices 21,. . . ,xn, denoted by K n , has every pair of its n vertices adjacent. Let
be the edge subring of R associated to IC,. Thus the toric ideal P(ICn) of K[Kn]is the kernel of the homomorphism of K-algebras
induced by Tij -+ xixj.
Notation For use below we set Pn = P(Kn). If j < i, we set Tji := Tij.The homogeneous maximal ideal of An will be denoted by !Dln. Lemma 9.3.1 ([3]) Let Qn = (Pn)Jn) c An be the ideal of An generated b y Pn and Jn = (hl . . . hn), where )
)
Let P be an associated prime ideal of Qn. Assume Q p is an m,-primary then P = 9.lIn. ideal of A, for 4 5 p < n. If P contains a variable TTS, Proof. Let P be an associated prime ideal of Q n so that T,., E P , to simplify notation assume r = n - 1 and s = 72. Since Tk(n-l)Tjn- TkjT(n-1)n E Pn,
Semigroup Rings of Complete Graphs
361 ,~
,,
"
I
we obtain Tk(n-l)Tjn E P for j # k . If Tk(n.-l) 4 P for some k , then TjnE P for all j # k , j 5 n - 1. As hn E P we have
hence Qn-l c P. Using the hypothesis yields T k ( n - l ) E P , a contradiction. Therefore T k ( n - 1 ) E P for all k # n - 1. Let
Because Q c P , from the hypothesis we derive that P contains all variables not in { T k ( n - l ) ] k + n - l . Therefore P = ?Illn. 0 Theorem 9.3.2 ([3]) If K is either a field of characteristic zero or K is a field of characteristic p 2 2 and 2 , . , . , n - 1 are relatively prime to p , then the ideal Q n i s an DI,-primary ideal of An. Proof. We proceed by induction on n. It is not hard to verify that
i(i - 1 ) ht(Qi) = 2 for 3 5 i 5 6 (see arguments below). Assume n 2 7 and the result true for m < n. Let P be an associated prime ideal of Qn. By Lemma 9.3.1 it suffices to prove that P contains one of the Tij's variables. Set
Consider the equality
and define the following permutations of { 1,. . . ,n } 2
nn -2 -1
n
2
1 0 2 = ( nn - -- 31
2
01=(
1 n-2
1
)
nnn---321
2
1 2 nn-n-3-21 n n -n2- 1
Q=(
n n-1
n
2
1
1
n n-2 n n-3
) )
362
Chapter9
Observe that a permutation u of { I , ,. . ,n } induces an automorphism
we have a'(&,) Note that
c Qn.
Applying
0: to
Eq. (9.6) yields Ob(T2nLn) E Qn.
If T'k(k+l) $! P for all 2 5 k 5 n - 1, then Tli - Tlj E P for i # j. Using the identity n
hl
+ 7 3 1 2 - T l j ) = ( n- 1 y 1 2 j=3
we obtain
7'12
E
P.
Corollary 9.3.3 ([3]) If K is a field of characteristic zero, then
is a Noether normalization of K[Kn] Proof. Note that the monomial subring K[ICn]is integral over the subring K [ h l , .. . ,hn] by Theorem 9.3.2. 0 Corollary 9.3.4 If K is afield of characteristic zero, then finitely generated free module over K [ h l ,. . . ,hn]. Proof. Since K[ICn]is a Cohen-Macaulaygradedalgebra
An/Pn is a
of dimension
n, by Proposition 6.3.1 we obtain that An/Pn is a finitely generated free
module over K [ h l , .. . ,hn]. Note that { h l , . . .,hn} is a regular system of parameters for An/Pn. 0
Semigroup Rings of Complete Graphs
363
Another Noether normalization Let us present here a simpler Noether normalization of the monomial subring of the complete graph. Since any variable Tij occurs only in hi and hj, one has:
2h = hl
+ . - . +h,,
where h =
Tij. l
Setting
si =
-h
{
+ hl + hi+l
h+l
h-hl
for i = 1,2, for3
it is not hard tosee that F = (91 . . . g n } is a Grobner basis for (hl ,. . . hn) w.r.t. the lex ordering T12 > > Tln > 7'23 > > T(n-1)n. Thus one may try to eliminate some of the first variables from F with the intention of producing some "triangulated" Noether normalization whose first elements are variables. The next result proves that one may replace g1 by 2'12 and still get a normalization. )
)
)
Proposition 9.3.5 Let K be a field of characteristic p . If 2 , . . . ,n - 1 are relatively prime t o p , then
is a Noether normalization of K[ICn]. Proof. From the equalities
one has
K[hl,...)hn] =K[g1,*--)gn]. As K[Kn]is Cohen-Macaulay (see Proposition 7.4.5) by Theorem 9.3.2 one concludes that 9 1 ) .. . gn is a regular system of parameters for An/Fn. In particular 92). . . gn is a regular sequence on A,/Pn. To finish the proof it suffices to show that TI2 is a regular element on )
)
Let Q be an associated prime of S and assume TI2 is in Q. If hl E Q , then h3,. . . h,, h are in Q, which together with Eq. (9.7) gives h2 E Q . Hence g1 E Q ;a contradiction to the fact that g1 is regular on S. Therefore hl cannot be in Q and consequently there is k 2 3 so that T1k $ Q. )
If (1, IC} n { 2 , j } = 8 and j # 2, then TlkT2j - T12Tkj is in P, and thus is in Q for all j 4 (2,IC}. Note T 2 k is not in Q ; for otherwise ha,. . . ,h, and h - hl are in Q and from Eq. (9.7) h E Q , and hence hl E Q which is impossible. By considering T2kTij - T2jTik, one obtains Tij E Q for all ij so that i, j are not in { 2, k} and i # j . It follows readily that all the variables Tij not occurring in hk belong to &. Using Eq. (9.7) and noticing that T ik occurs exactly twice in hl ih,, one gets 2(h- hk) E &, but since hk is also in Q one derives h, h2 E Q and consequently g1 is in Q, which is again a contradiction. Altogether 7'12 is not in the union of the associated primes on S. 0 of S and is therefore a regular element T2j
Examples and one open question The minimalvertex covers of the complement of a graph G seem to be related to the problem of finding a Noether normalization of the edge subring K[G].
Example 9.3.6 LetGbe the following graph and its
complement. I
Using Macaulay one has P ( G ) = ( g 1 , 9 2 ,g 3 , g 4 ) , where
By Corollary 8.7.11 it follows that theedge subringK[G]is a normal CohenMacaulay domain. To find a Noether normalization for K[G]over a field K of characteristic zero consider the sequence:
unfortunately ht(P(G),8 1 , . . . ,135) = 7, but we can modify the coefficients of the variables in the support of I3i to obtain the required height 8, more precisely a Noether normalization for K [ G ]is:
I
Semigroup Rings
of Complete365 Graphs
where
Note that the coefficients of the variables occurring in t9i were modified by first finding a minimal vertex cover C = ( 1 4 ) for and then changing t i j by i t i j , if xj is in C. Remark 9.3.7 To verify the assertion that ht(P(G), hl, . . . , h 5 ) = 8 in the Example 9.3.6, note that hl, . . . , h 5 specializes nicely, that is,
hi, . . . ,hi-l, hi+, , . . . ,h; is a system of parameters of K[G \ {xi}] for each 1 L i 5 5 , where h> is obtained making til, = 0 in the sequence hl , . . . ,h 5 . Hence we may proceed essentially as in the proof of Proposition 9.3.5. Take an associated prime Q of ( P ( G ) hl, , . . . ,h 5 ) . If t 1 2 e Q , then t 1 5 t 2 3 E Q. Note one may assume t 2 3 E Q; otherwise t 1 5 and t 1 3 are in Q and from:
one concludes ht (Q) = 8. Using:
again one concludes ht(Q) = 8. Similar arguments show that if t 1 3 or t 1 5 are in Q, then ht(Q) = 8. Hence one may assume t 1 2 , t 1 3 , t 1 5 notin Q. Fkom the equalities:
one readily derives that Q is the irrelevant maximal ideal of
K[tij].
All the normalizations of monomial subrings of graphs that we have found, through the procedure of Example 9.3.6, suggest the following: Question 9.3.8 Let G be a graph with n vertices and A / P ( G )the presentation of K[G]over a field K of characteristic zero. Assume K[G]is a normal subring of dimension n such that deg(v) 2 2 and dim K[G \ {v)] = n - 1 for any vertex v of G. Is there a Noether normalization of the form:
366
Chapter 9
The next instanceof a graph is rather more complicated than the graph given in Example 9.3.6, but exhibits the same sensitivity to modifying the coefficients linked to subsets of minimal covers of the complement of the graph. Example 9.3.9 Let G bethe complement.
following graph on 10 vertices andits
By the description of the integral closure of K [ G ]given in Corollary 8.7.11 it follows that K[G]is a normal domain. It is not hard to verify that a Noether normalization for K [ G ]over a field K of characteristic zero is:
where
The coefficients of the variables occurring in hi were chosen usinga minimal vertex cover C for and then modifying the coefficient of tij, if xj is in C. Remark 9.3.10 The monomial subring of Example 9.3.9 was studied by Hibi and Ohsugi [165] to show a normal subring K[G]of a graph G such that none of the initial ideals of P ( G ) is squarefree.
L
Chapter 10
Monomial Curves In this chapter we present some material on toric ideals of monomial curves and symmetric semigroups. The emphasis will be on space curves and monomial curves in the affine space of dimension four whose toric ideals are generated by critical binomials. We should point out that Chapter 10 and Chapter 11 overlap at certain points. A few almost identical results will be proven in both chapters with slightly different proofs.
10.1 Defining equations of monomial curves Let k be a field and
S = dlN+
+dnN
the semigroup generated by a finite sequence d l ) . . . dn of relatively prime positive integers. A monomial curve I'in the affine space A: over k is given parametrically by xi = pi) )
that is, we have
r={(tdl, . . . )t d q E A F l t E k } . Let R = k[xl,. . . ~ n and ] k[t]be two polynomial rings over k graded by deg(xi) = di for all i and deg(t) = 1 respectively. Let q5 be the graded homomorphism of k-algebras
The image of 4 ) denoted by k [ S ]or k [ r ] ,is the semigroup ring associated with S and the homogeneous ideal P = ker(4) is the presentation ideal or toric ideal of k[S].
367
Chapter 10
368
By Proposition 7.1.2 the toric ideal P is generated by a finite set of binomials. Hence the exact sequence 0
+K
= ker(Q) -+
Zn -% Z "-+ 0,
$(ei)
= di
is related to the presentation ideal P of k [ S ]in the following manner: if - xf' is a binomial, then g E P if and only if a - /3 E ker($). Given a binomial g = xu - z b ,we set = a - b. Given a subset I c R we denote its zero set in A; by V ( I ) ,and given a subset X c A; we denote its va,nishing ideal in R by I ( X ) . If hl,.. . ,hr E En,we denote by
g = x"
the subgroup of Z n generated by these elements. Let G c Z n be a subgroup. We associate to G an equivalence relation on the monomials in R = k[xl,.. . ,xn], by x" -G xP if a - 0 E G. Note that this relation is compatible with the product, i.e., x" -G x0 implies x~z"-G Z'ZP for all y E W , Definition 10.1.1 A nonzero polynomial
in R is simple (with respect to - G ) if all its monomials, Le., those x" with nonzero coefficient A" , are equivalent under -G. Given any polynomial f E R \ (01, we can group together its monomials by equivalence classes under W G , thereby obtaining a decomposition
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 summandshi as the simple components of f (with respect to W G ) . Lemma 10.1.2 Let 91,. . . ,gr be binomials in R and G c Z n the subgroup generated by 31).. . ,g r . If f E (91,.. . ,gr), then every simple component of f with respect to NG also belongs to (91,. . . ,gr). Proof. First, each generator gi = 2"' -@ is simple, as its two constituting monomials x a i , zoi are equivalent under -G by construction.Moreover, zYgi remains simple for any y E W , since the relation -G on monomials
69
Monomial Curves is compatible with the product. Let now f beanyelement in the ideal (91, . . . ,g,.). Tben, f is a linear combination of polynomials of the form xygi, with i E { 1,. . . ,r ) and y E Nn , which are simple. Hence, every simple component of f is also a linear combination of some x'gi and therefore belongs to (91, . . . ,g,), ,
I
Theorem 10.1.3 ([104]) Let 91,. . . ,gr be a system of binomials in the toric ideal P and I = (91, . . . ,g,.). If char(k) = p # 0 (resp. char(k) = 0), then rad(I) = P if and only if
(a) pmker(@)c ($1,
. . . ,&)
for some m E N (resp. ker(@) = ($1, (b) V(g1,. . . , ST,xi) = (01, for all i .
. . . ,&)).
Proof. +) (a) First we consider the case char(k) = p # 0. By hypothesis, there exists m 2 0 such that f P m E I for all f E P. Let y E ker(+) and write y = a - /3 with a,p E Nn and g = x" - xo. Thus g belongs to P and ? = a - p = y. Set G = (&, . . . &) c Z". )
We know that gPm E I and gPm = zaprn- x p p " . Since the simple components ofgPm belong to I and hence to P (see Lemma 10.1.2), and since zapm does not belong to 1 because it does not belong to P , it follows that gPm is simple, Le., p m a -G pmp. Hence pma - p m p E G. This shows p"ker(+) c G. Next we treat the case char(k) = 0; this case was shown in [loo] using simplecomponents.Let G = (21, . , . ji,)and 0 # 7 ker(l(l). Note that one can write 7 = a - p, where a and ,6 are in N" and have disjoint support, thus f = x" - xp E P and f q is in I for some prime q >> 0. Hence )
m i= 1
where Ti
- Si are in G and
ci E k
\ {0}, for all i. Set
Observe that h equals the sum of all the monomials cixBi occurring in such that qa - 8i E G, because qa - yi E G iff qa - 6i E G. Expanding one has:
If qcll - qp $! G, thep
Chapter 10
370
for some 3 contained in { 1,. . . ,q - 1). As q is prime, q must divide ( f ) for 0 < i < q. Thus making xi = 1 yields l k = ( s q ) l k , s E Z, but since char(lc) = 0 one has 1 = qs, which is impossible. Hence q(a - p) E G. Pick a prime r # q such that 'f E I, by the previous argument r(a - p ) E G. Since gcd(q,r ) = 1, we get a - p E G. (b) Let a E V ( g 1 , .. . ,g r , xi),since V ( P )= V ( I )and x? - x;' are in P for all j # i, it follows that a = 0. e)First we assume char(lc) = p # 0. Let f = x c - xe be a binomial in P and write gi = xai - .pi. By (a) there are integers s i such that
one may assume Hence writing
si 2 0 by replacing zai/zBi by its inverse if necessary.
= ((x"' p
x"'
i
)
- 1)+ 1
and using the binomialtheorem it follows that MfPm is in I , for some monomial M . Since P is generated by a finite set of binomials, there is a monomial zr such that x7fp" E I, for all f E F. Let PI be a minimal prime of I , we will show that P1 contains P. As I is graded observe that PI is also graded (see Lemma 1.2.3). If none of the variables in the support of x7 are in P I , then P c P I . Otherwise, after a permutation of the variables and the binomials, one has the inclusions:
Ic
.. ~
(XI,.
mG') ,
c PI c ( ~ 1 , .. . ,xn),
where G' = ( 9 1 , .. . ,g q } is a subset of ( 9 1 , . . . ,g r } such that the set of variables occurring in any binomial in G' is disjoint from 2 1 , . . . ,x m and the binomials g q + l , . . . ,gr belong to the ideal ( 2 1 , . . . ,xm). Set ai = 0 for 1 5 i 5 m and ai = 1 for i > m. Since
one derives that G' must be the empty set. If m < n, then (ai)is again in V(g1, . . . ,g,., XI), which yield rn = n and thus P c PI. Hence P c rad(I) and rad(I) = P. Note that this part of the proof works if char(k) = 0 by making pm = 1. 0 Remark 10.1.4 The condition (a) above is equivalent to require that the group ker($)/G is a finite p-group, and this condition can be readily validated by noticing that ker($)/G is a finite pgroup if and only if there is an isomorphism Zn/G N Z x H , where H is a finite p-group.
i
Monomial Curves
371
Example 10.1.5 Consider the monomial curve
where k is a field of characteristic p
> 0. If g = x p - yP E k[z,y], then
9 = (a: - y)" and I' = V ( g ) .
Note that
5 does not generate the subgroup ker($)
= ((1,-1)) of Z2.
Proposition 10.1.6 If 91,. . . ,gr is a set of binomials generating the toric ideal P of k [ S] ,then 31,.. . , 3 r generate ker($). Proof. Let L = (&, . . . ,&) and 0 # 77 E ker($). Note 77 = Q - p, where a and ,G' are two non negative vectors with disjoint support, thus f = xa - zp E P and one can write
i= 1
where yi - Si are in L and ci E k, for all i. Set
Using zri - 2'; = (xe - x d i )- (xe - a : T i ) , with 8 = a or 8 = ,8 one may rewrite f as 9
m
i= 1
j=1
where a - ~ i /3, - Sj E L \ (01, for all i and j. One may assume P # a # Sj for all i, j; otherwise one has a - p E L. Hence 9
m
9
m
i= 1
j=1
i= 1
j=l
~i
and
Chapter 10
372 Lemma 10.1.7 Let
$3" _I) Z be the homomorphism induced by $(ei) = di, where dl,. . . ,dn is a sequence of relatively prime integers. If L is a submodule of ker($), then L = ker(+) if and only if Z n / L is a free Z-module of rank 1. Proof, Notice that ker(Q)/L is torsion free if and only if Zn/L is torsion free. Therefore one hasthe equality ker($) = L if and only if Z n / L is a free Z-module of rank 1. a Proposition 10.1.8 Let d l , . . , ,dn be a sequence of relatively prime inteE n + iz the linear map induced by $(ei) = di, then the set gers and
+:
is a generating set for ker(+). Here (di, dj)= gcd(di, d j ) and ei is the ith unit vector.
Proof. By Lemma 10.1.7 it suffices to prove that Z n / ( L ) is free of rank equal to 1. Let A be the matrix with rows
and let Ii be the ideal generated by the i-rowed minors of A . Let us show that In-l = Z. For a fix IC consider the ( n - 1) x n submatrix B of A with rows
i # k. It is not hard to see that the integers
occur as n - 1 minors of B. If a prime p divides ak, then it divides dk, therefore In-1 = Z. Since Ii C Ii-1 all the invariant factors of A must be equal to the1 and is proof complete. 0 Corollary 10.1.9 Let I be the ideal generated b y the set of binomials
Then rad ( I ) = P , where P is the toric ideal of k[S].
.
"
I
Monomial Curves
373
Corollary 10.1.10 If I = ({xfj - $1
15 i
< j 5 n)), then rad(1)
= P.
Proof. Using the method of proof of Proposition 10.1.8 it follows that ker($) is generated by
= P by Theorem 10.1.3.
rad(I)thus
0
Remark 10.1.11 Let 91, . . . ,gr be a finite set of binomials in the toric ideal P and I the ideal generated by 91, . . . ,gr. In general the algebraic equality rad(1) = P and the geometric equality
V ( P )= V ( 1 ) are not equivalent (see Exercise 10.1.20), but they are related as follows: 0 0
if k is any field and rad(1) = P , then V ( I )= V ( P ) , on the other hand if k is algebraically closed and V(1)= V ( P ) ,then by the Nullstellensatz rad(I) = P.
Now, we want to investigate what it means geometrically for a set of binomials 91,. . . ,gr in P to satisfy only condition (a) in Theorem 10.1.3. The next result is a refinement of [loo, Lemma 31 in positive characteristic. Proposition 10.1.12 ([104]) Let G = (&, . . , ,z r ) , where 91,.. . , is a set of binomials in P . If k is a field of char(k) = p # 0 (resp. any field k) and pmker($) c G (resp. ker(llt) = G), then
Proof.
Assume char(k) # 0; for theother case see [loo]. Let a = be in V(g1, . . . ,gT), andsuppose a1 - a, # 0. One has
-
,. . . ,a,)
(a1
= a"' - .pi = 0
vi,
where gi = zai - zDi. Hence uai = upi. Since all the entries of u are distinct from zero, this can also be expressed as aai -pi = 1. Now, if g = x" - xp E P , then a - p E ker($) and p"(a - p) E G. Thus one can write p m ( a - p) = &(ai- pi), for some t i E Z. Hence
xi
ap"(0-P)
= aC;li("i"Pi)= ni(a"i - aBi)l* = 1
and upma = [email protected] uprna- apmB = (aa - a4)pm = 0 , which implies g ( a ) = a" - up = 0. As P is generated by binomials it follows that a is in claimed. V ( P ) = I?, as 0
374
Chapter 10
Monomial curvesareaffinetoric varieties An interestingproperty of monomial curves is that they are affine varieties defined by binomials. In Section 11.2 we give a criterion to decide when a set defined parametrically by monomials has this property. Lemma 10.1.13 Let k be a field and a, b E k\{ 0). If n, m are two relatively prime integers and an = bm, then the equation zm = a has a solution in k . Proof. Since 1 = ns
+ qm, one has
as required. Proposition 10.1.14 Let I be the ideal of R generated by the set
T h e n I? = V ( I ) ,where V ( I ) is the zero set of I . Proof. It is enough to prove the containment V ( I ) c I?, because clearly one has I' c V ( I ) . Let a = ( a l , . . . ,an) be an element in V(1). One may assume ai # 0 for all i, otherwise a = 0. Assume n = 2. As a? = at1 by Lemma 10.1.13 there is t E k such that a1 = t d l . Note that u = az/td2 satisfies ud1 = 1 and consequently gcd(o(u), dz) = 1. Thus there is an integer s so that sdz E 1 mod o(u), setting to = tu8 readily gives ai = tti, for i = 1 , 2 . Hence a E I?. Set h = gcd(d1,. . . ,&-I) and for 1 5 i 5 n - 1 write di = dih, where gcd(d\, . . . ,CZL-~)= 1. Let I' be the ideal generated by the binomials:
By induction assume V(I')C
r'= { ( t d i , . , . , t d i - l )tl E k } ,
Using the equalities:
one concludes that a' = ( a l ,. . . ,an-l) is in I?'.Hence there t E k such that < n. Set b = a: and x = b/&. F'rom the identities
ai = td:,for all i
t
375
Monomial Curves
onereadily derives z d : / ( d : j d n=) 1, thus o(z) divides d: for i < n, that is, o(z) = 1 and t d n = a;. By Lemma 10.1.13 there is tl E k satisfying t = t:. Set w = a n / t f n , since v h = 1 the order of w divides h and this implies gcd(o(w),d n ) = 1. Hence sd, E 1 mod o(w), for some integer s. To complete the proof set to = tlw3 and note
For an alternative shorter argument can use the next elementary fact.
to prove Proposition 10.1.14 one
Proposition 10.1.15 Let a l , . . . ,a, be a sequence of non zero elements of a field k . If d l , . . . , dn are relatively prime integers such that a t j = a;i for all i, j , then there is t E k with ai = tdi for all i. Proof. One can write equalities
1 = cldl
one derives that t = ai1
-
+ - + cndn, for some ci E Z.
a? is the required element.
From the
0
Remark 10.1.16 Assume k algebraically closed, then Proposition 10.1.14 is valid without assuming gcd(d1, . . . ,d n ) = 1. More generally, if X is the curve in At given parametrically by
where fi(t) E k [ t ] \ k for i = 1,.. . ,n, then by the extension theorem it follows that V ( P )=
x,
where P is the toric ideal of k [ f l ( t ) ,. . . , f n ( t ) ] .See [74, Chapter 31 for a beautiful exposition of the extension theorem. Proposition 10.1.17 Let I' = { ( t d l ,... ,tdn))l t E 6 ) be a monomial curve in the afine space A:. If dl > da > - > d,, and IC is the field of complex of I' is equal to numbers,thentheprojectiveclosure
Chapter 10
376
Proof. Let I" be the right hand side of Eq. (10.1). By Proposition 2.4.29 one has = V(l(p(r))),where cp is the map
given by p(a) = [(a,l)].Let
. . , x n ,w]such that then f (al,. . . ,an, b) = 0 for any polynomial f in k[zl,. f(tdl,. . . ,tdn, 1) = 0 for all t E IC and f homogeneous. One may assume b # 0, otherwise if b = 0, then f = - z f i w d l - d i is in I(cp(I')) for i 2 2, hencemaking u = 0 and t = we get a E r'. Since b # 0 we readily get f(ul/b,.. . ,an/b) = 0 for every binomial f in the toric ideal P of k [ t d l , .. . , t d n ] , hence ( a l / b , . . . ,an/b) E V ( P ) . As I' = V ( P ) ,there is t l E IC such that ai/b = t p for all i. Making t = bl/dltl and u = b 1 I d 1 one concludes a E I". Conversely let
be an element of I" and f a homogeneous polynomial of degree r such that f vanishes on p(I'). From the equality
one readily derives that f ( a )= 0, that is, a E y.
0
Exercises 10.1.18 Let dl,. . . ,d n be a sequence of relatively prime integers and
the linear map induced by $ ( e i ) = di . If G is a subgroup of ker($) , then Torsion(Zn/G)
c ker(@)/G,
with equality if ker($)/G is a finite group.
10.1.19 Let D = ( d i j ) be an m x n matrix with integer entries and
the linear map induced by $(ei) = Di, where Di is the ith column of D. If G is a subgroup of ker($), prove that G = ker($J)if and only if Zn/G is torsion free of rank equal to rank(D).
I
Monomial Curves
377
10.1.20 Let P be the toric ideal of the subring k [ t 6 ,t * , t g ]and I = (g2,g3), where g2 = x! - 2;) g3 = xi - 2;. Then (a) P = ( 9 2 , ~ :- xlxi) for any field k, (b) z3/(&)&> N z X z3) (c) rad(1) = P if char(k) = 3, and rad(1) (d)
# P if char(k) = 2,
= { ( t 6 , t 8 , t g ) l Et k) = { ( O ) O ) O ) ) ( l ) l J ) }
= V ( I ) ,if k = Z2.
10.1.21 Let P be the toric ideal of &[t6, t 8 ,t g ] and let
'
10.1.22 Let d l , . . . ,d, be a sequence of nonzero integers and $ the linear map from Z n to Z induced by $(ei) = d i . Prove that the set
is a generating set for ker($), where d = gcd(d1
.. .
dn).
10.1.23 Anaffine variety X in kn is said to be a set-theoretic complete intersection if its vanishing ideal I ( X ) is a set-theoretic complete intersection. If k is algebraically closed prove that X is a set-theoretic complete intersection if and only if X can be defined by n - dim(X) polynomials in n variables with coefficients in IC,
10.2
Symmetricsemigroups
Let S # (0) be a semigroup of N,that is, S is a subset of N which is closed < dn under addition and 0 E S. Note that there exists a sequence dl < of positive integers such that a .
If gcd(d1,. . , dn) = 1, it is said that S is a numerical semigroup. )
Lemma 10.2.1
If S c N is a numerical semigroup, then c+NcS
for some c E S.
-
378
Chapter 10
Proof. Let dl < - < d, be a sequence of positive integers generating S and such that gcd(d1,. . . ,dn) = 1. To determine c write 0
and set c = ()rlldl
+ + Irn)dn)d,. 9
*
It is not hard to see that c satisfies the required property.
0
Definition 10.2.2 Let S be a numerical semigroup and c the greatest inthis number is called the Robenius number of teger not in
s,
s.
Definition 10.2.3 Let S be a numerical semigroup and c its F'robenius number, the semigroup S is said to be symmetric if c - x E S for all x 4 S. Proposition 10.2.4 Let S be a numerical semigroup of N and let c be its Probenius number. Then S is symmetric if and only if (c
+ 1)/2 = IN \ SI.
Proof. Consider the map
given by cp(z) = c - z. Observe that S is symmetric if and only if
C ~ ( \NS ) = S n [o, c). Because of the decomposition
one obtains that S is symmetric if and only if c
+ 1 = 2)N \ SI.
Definition 10.2.5 A discrete valuation of a field K is a mapping
with the following properties. For all a, b E K : (a) v(ab) = v(a)
+ v(b),
(b) 4 2 + b) 1 min{v(a),+)), (c) v( a) = 00 if and only if a = 0. The valuation ring of v is the set of all a E K with v ( a ) >, 0.
0
Monomial Curves
379
Example 10.2.6 Let K = k ( t ) be the field of rational functions in one variable over a field k. For f ( t ) E k [ t ] write f ( t ) = t m g ( t ) , where m E N and g ( t ) is a polynomial with g(0) # 0. Define v(f) = rn, and extend 21 to a discrete valuation of K by setting
The valuation v is the t-adic valuation of K . Proposition 10.2.7 Let (A,m, IC) be Cohen-Macau2ay local ring with total ring of fractions Q and z E m a nonzero divisor. If dim(A) = 1, then there is an isomorphism of k-vector spaces
Soc(A/zA)
E
m-l/A,
where tn-l = { z E QIzm C A). Proof. The “multiplication by z” map p: m-’/A cp(z
-+
Soc(A/zA),
+ A) = z z + X Ayields the required isomorphism.
0
One of the motivations to study symmetric semigroups comes from the following result. Here we adapt a proof of E. Kunz to make it work for the graded case. Theorem 10.2.8 ([203])Let A = b [ t d l , .. . ,tdn]be the monomial subring of k [ t ] generated by t d l , .. . ,tdn over the field k and S = v(A), where v is
the t-adic valuation on K = k ( t ). If S c N is a numerical semigroup, then S is symmetric if and only if A is a Gorenstein ring. Proof. Assume S is symmetric. Set m = A+ and
Let h E rn-l \ A. If v(h) E S, then p = v(h) 2 1. Hence h E k [ t ] and we can write h = hl ha where hl E A and v(h2) +!S. We claim that v(h2) = c, where c is the greatest integer not in S . If v(h2) < c, then since S is symmetric 0 < p = c - v(h2) E S. Choose g E A with v(g) = p , then
+
therefore h2g 4 A, contradicting h2 E m-l. Thus v(h2) 2 c. To complete the proof of the claim notice that 4 h 2 ) 5 c, for otherwise v(h2) E S. Next we show that m-l = A + Af for any f E m - l \ A . We may assume v(f) # otherwise write f = f~ f2 with f1 E A and f2 # S and notice
s,
+
Chapter 10
380
A + A f = A + Af2. Hence v( f) = c and f is a polynomial in t. We may assume, by dividing f by an appropriate constant, that f ( t )= t c g ( t ) where g(0) = 1 and g ( t ) E k [ t ] . Take x E rn-l \ A, if v(x) E S write x = + x1 with 51 E A and 4 x 2 ) = c and notice x E A+Af if and only if x2 E A + A f . Hence we may assume v(x) 4 S, in this case v ( z ) = c and x is a polynomial in t of the form x ( t ) = t C g l( t )with g l ( 0 ) = 1. As a consequence v ( x-f) > c, which gives z - f E A and x E A Af. Therefore the length of rn-l/A as an A-module is 1 and according to Proposition 10.2.7 this implies that A is Gorenstein. Note that the proof shows rn-l c k[t]. Conversely assume A is a Gorenstein ring. Let c be the greatest integer not in S and
+
B = {ti + tc-lAIO 5 j 5 c, j E S } , C = (tc+l+i+ tcflA) 0 5 i 5 C, i 4 S } . First note dimk A/tC+'A = c A/tC+lA. Define
+ 1, because the set
B U C is a k-basis of
We now show that cp is onto. Let 0 5 i 5 c and i 4 S, it suffices to prove - i E S. If c - i is not in S, pick the largest i with c - i 4 S. Since the socle of A / t C + l Alives only in degree 2c 1, there is s E S \ (0) such that c
+
+
+
+
hence s i is not in S and thus i < s i 5 S and c - (i s) f S, which contradicts the choice of i. Hence c - i must be in S and cp is onto. Since cp is clearly injective we obtain
and S is symmetric by Prbposition 10.2.4.
0
A nice generalization of Theorem 10.2.8 due to R. F'roberg will be presented below; the idea is to find an appropriate expression for the CohenMacaulay type of A in terms of the semigroup using an Artinian reduction of A. Proposition 10.2.9 ([11'7]) Let
where S is a numerical semigroup of N with Fkobenius number c . Then S is symmetric if and only if T ( S )= { c} .
Monomial Curves
381
Proof. +) Let z E T ( S ) andassume z < c, then 0 < c - x E S and x + (c - x) = c is in S, which is impossible, hence x = c. e)Let x # S, z > 0, one must show c - x E S. If c - x 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 E S, s > 0 with ( c - z ) + s not in S, hence x - s > 0, which contradicts the choice of z. cl
Theorem 10.2.10 ([117]) Let A = k [ S ] be thesemigroupring merical semigroup S over a field k and
T ( S )= {x E N \ s I x + s E
of a nu-
s, v s E s, s > 0 ) .
Then ( T ( S ) is ( equal to the Cohen-Macaulay type A. Proof. Let dl < < d, be a minimal generating set of S with d l , , , . ,d, relatively prime and s E S, s > 0. Consider the map 9:T ( S )+ Soc(A/t"A),~ ( x=) tz+" + PA.
It is enough to verify that the image of T ( S ) is a k-vector space basis of Soc(A/t"A),because cp is clearly injective. Note that Soc(A/t"A)has a k-basis consisting of monomials of the form tW+ P A , with w - s # S and tdl(t" + t " A ) = t"A for all i, write di + w = s + si, where si is in S. We claim w - s E T ( S ) .It follows that (w - s) s' is in S for all s' E S, s' > 0. Observe that w - s > 0; otherwise from w - s = SI - d l = s2 - d2 we derive s1 = 0 and dz = s2 +dl , hence d2 = pdl , which is impossible because dl and d2 are part of a minimal generating set of S. Hence w - s E T ( S ) and cp(w - s) = tu. To finish note that cp(T(S))is linearly independent. 0
+
Arithmetical symmetric semigroups Let S be a numerical semigroup and c its F'robenius number, that is, c is the greatest integer not in S. The problem of Frobenius consists in determining the integer c; in non-mathematical terms 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 and estimating the Frobenius number has been examined by several authors [117, 248, 2541. Lemma 10.2.11 Let u , v E
N+ and assume gcd(u,v) = 1, then c=uv-u-v
is the largest integer not in U = uN + vN.
Chapter 10
382
Proof. It is easy to see that c is not in U . Indeed if c E U ,then uv - u - v = rlu
+ r2v
for some rl ,r2 in N,that is, u(v - rl
- 1) = v(r2 + 1)
and since gcd(u, v) = 1 one derives v - r1- 1 = qv 2 v, which is impossible. Next we show c i E U for i E N+ . One has 1 = plu p2v, hence
+ + c + i = (v - 1 + ip1)u + (-1 + ip2)v,
and by the division algorithm one may write 0 5 b2 < u. From the identity
one concludes
c
+i
= blu
bl 2 0 and consequently c + i E U.
+ b2v, where
0
Lemma 10.2.12 Let n 2 3 and v 2 1 be two integers and S the semigroup
+
+
where di = dl (i - l)v and dl = r k ( n - 1 ) for some k , r E N. If gcd(d1, v) = 1 and 2 5 r 5 n, then the greatest integer not in S is equal to (k v)dl - V.
+
+
Proof. Set c = (k v)dl - v, wewill show c assume 1 5 i 5 d l . Notice
+ i E S for i 2 1. One may
+ i = (k + 1)dl + X + i, where X = vdl - dl - v. Consider U = d l N + vN. By Lemma 10.2.11 X is the greatest integer not in U. Hence X + i = ad1 + bv, for some a, b in W. One has Wdl + i = ( a + l)dl + ( b + 1)v. Hence b 5 ( n- l)(k + a + 1) andwe can write b = j ( k + a + 1) +e, for some 15 k + a + 1 and 0 5 j 5 n - 2. Therefore the equality c
gives c
+ i in S. It remains to show c 4 S. If c E S, then
Monomial Curves
383
for some ai in W. Setting
n
s=C u i , i=l
note that one can rewrite c as: c = (IC
+ 1)dl + X =
n
aidi = sdl +pv, i= 1
for some p E N. Since X 4 U , one derives s ,< k and c ,< sdn 5 kdn, which 0 is a contradiction since c = kd, + ( r - 1)v.
N and let dl = r + k ( n - 1) and di = dl + (i - l)v be an arithmetical sequence. If 2 5 r 5 n # 2 and gcd(dl, v) = 1, then
Theorem 10.2.13 ([3, 107, 1951) Let r , k,n, v E
r = 2 if and only if the semigroup
S = dlN + - *
*
+ dnW
is symmetric.
+ (n - 1)k. Set c = ( k + v)dl - v = kdn + v
Proof. +-) Let dl = 2
+ vN. Assume z 4 S. First consider the case z E U , write z = rdl + sv, and U = dlN
T , s E N. By the Euclidean algorithm and Lemma 10.2.12 one can write z = ad, + bv, where a, b are integers so that 0 5 a 5 k and 1 5 b < dl. Writing a = k - i, where 0 5 i 5 k , we claim that 1 5 b 5 (n - 1)i 1. If b > ( n - 1)i 1, then z = zldl X ~ V ,with x1 = k v - i and 2 2 = b - (n - 1)i - 2. Since 0 5 2 2 5 (n - 1 ) q it is not hard to show that z E S, which is impossible. Hence 1 5 b 5 (n - 1)i 1. Set y1 = i and y 2 = ( n - 1)i- b 1. Since 0 5 y2 5 ( n - 1)yl from the equality c - z = yldl 92b, one gets c - z E S. Now consider the case z $! U . One can write z = ad1 + bv, with 1 5 b < d l and u < 0. Set w1 = k 1 and w2 = dl - b - 1. Using c - z = wldl - (1 + a ) & + w2v and 0 5 w2 5 ( n - l)wl, gives c - z E S. Therefore S is symmetric. e)Assume S is symmetric and keep the same notations as above. Let 2 5 r 5 n and d l = r k(n - 1). As v 4 S one has c - v E S and
+
+
+
+
+
+
+
+
+
n
c-v =
bidi, i= 1
where bi E N. Set a = Cy=,bi and notice that c - v = (k
+ 1)dl + X - v = ad1 + qv,
for some q E N. Therefore a 5 k, otherwise one obtains X E U . From the 0 inequality c - v 5 ad, 5 kdn, it follows that r = 2.
Chapter 10
384
Exercises 10.2.14 Let S equivalent:
# (0) be a semigroup of N. Prove that the following are
(a) S is a numerical semigroup. (b) c
+ N c S for some c E S.
( 4 IN \ SI < O 0 *
10.2 15 Let S be a numerical semigroup of N. Prove that there is a unique minimal set of generators { d l , . . . ,d,} of S with gcd(d1, . . . ,d,) = 1. 10.2.16 Prove that the semigroup
is symmetric.
10.2.17 Let S be the numerical semigroup
and IC a field. Prove that k [ S ]is a Gorenstein ring which is not a complete intersection.
10.2.18 Let S c N be a numerical semigroup with F'robenius number c and k[S]its semigroup ring over a field k. If
A = k[S]c k[t] hasthe inducedgrading, show that the A / t C + l A is 2c 1.
+
degree of theHilbert
series of
10.2.19 Let k be a field and let
where xi has degree di E N+ . If gcd(d1, d2) = 1 and c show that the Hilbert series of B is (1- ZC+1) (1 (1 - zdl)(l - Z d 2 )
+ 1 = aldl + a a d ~ ,
'
where c is the F'robenius number of the proper semigroup S = dlN Note B 2 A / t C + l A ,where A = k [ t d l ,t d 2 ] ] .
+ d2N.
Monomial Curves
10.3
385
Ideals generated by critical binomials
Let us begin by fixing some notation to be used throughout this section. We will always assume that
is a numerical semigroup generated by a sequence dl , . . . ,dn of relatively prime positive integers and that r is the corresponding monomial curve, The kernel of the homomorphism of k-algebras
induced by $(xi) = t d i , is the toric ideal of k[S]and will be denoted by P , where R is a polynomial ring in n variables over a field k. Definition 10.3.1 A binomial
is called a criticalbinomial with respect to xi if mi is the least positive integer such that midi E djN. j#i
Definition 10.3.2 A set {ff, . . . ,fn) is called a full set of critical binomials if fi is a critical binomial with respect to xi for all i . Given a full set F of critical binomials, a main problem is to determine when the ideal generated by F is equal to P. The interest in studying ideals generated by critical binomials comes from a result of J. Herzog [146] showing that theideal P of a monomial space curve is generated by a full set of critical binomials (see Theorem 10.3.10). This is not longer the case for toric ideals of monomial curves in higher dimensions: Example 10.3.3 ([99]) Let m be a positive integer and let (dl,d2,d3,d*) = (m2,2m2- 1,3m2 +m,4m2f m
- 1))
then v ( P ) 2 m, where v ( P ) is the minimum number of generators of P.
If S is symmetric and generated non redundantly by four integer, then P is minimally generated by 3 or 5 elements 1341; thus even in this case P is not generated by critical binomials. For simplicity of notation fi will stand for a critical binomialwith respect to xi, for i = 1,.. . ,n. Note that we do not assume &fi,. . . ,f f ndistinct.
386
Chapter 10
Lemma 10.3.4 ([146]) Let
be a full set of critical binomials. If bi > 0 , ci > 0 f o r all i , then ai > 0 f o r all i and a 1 = bl + c l , b 2 = a 2 + C ~ , C Q = a 3 b 3 .
+
Proof. If
a2
= 0, then
a3
2 c 3 . Using the equality
we derive a contradiction, hence a2 Define
+ +
> 0, by a similar argument a 3 > 0.
and ( V I ,~ 2 , 2 1 3 )= a p y. Next we show vi 2 0 for all i. Note a1 > bl , otherwise using x!"a1x2 f l f 2 one derives a contradiction, Hence using x$b f l f 2 , one derives a 3 b3 2 c 3 and 213 2 0. Similarly oneshows V I ,v 2 2 0. Since v 1 dl v2d2 v 3 d 3 = 0 we have vi = 0 for all i. 0
+
+
+ + +
Proposition 10.3.5 Let
be a full set of critical binomials and I = ( f 1 , f 2 , f 3 ) . If bi > 0 , ci > 0 f o r all i, then { f 1 , f 2 , f 3 ) is a Grobner basis f o r I with respect to the revlex ordering x 1 > x 2 > 2 3 . Proof. Use Lemma 10.3.4 and Buchberger's criterion Theorem 2.4.15. 0 Proposition 10.3.6 Let
be a full set of critical binomials. If bi > 0 , ci > 0 for all i, then the ideal equal to the toric ideal P of k [ S ] .
(f1, f 2 , f 3 ) is
Proof. Let f be a binomial in P and assume we orderthe monomials with the revlex ordering x 1 > x 2 > x 3 . As f 1 , f 2 , f 3are binomials, by the division algorithm Proposition 2.4.6 we can write
< a l , r 2 , 3 2 < b2 and such that none of the monomials occurring in r is divisible by the leading term of f 3 . To agree with Proposition 10.3.5 one must show r = 0. r1, s1
Monomial Curves
387
Assume r # 0. As x? or is a factor of r , one may assume that s3 = 0. 5 T I or s2 5 r2 we get a contradiction with the minimality of b2 or a1 respectively, hence s1 > T I and s2 > 7-2. From the equality T = xT1 x? h, - x ~ 1 - " 1 ~ ~ 2one - ~ 2concludes , h E P and 7-3 2 c 3 . Using the where h = identity h - x T - C 3 f3 = xr3-C3 x;l x;2 - x ; -7-1 ~ x s -7-2 ~
If
s1
2
3
and the minimality of b2 and a1 allow us to conclude s1 - T I > c1 and s2 - r2 > c 2 , a contradiction because xyl xX2 is not divisible by xElx7. 0 Remark 10.3.7 The proof of Proposition 10.3.6 does not use the hypothesis that the integers d l , 4 , d3 are relatively prime. Proposition 10.3.8
If
is a full set of critical binomials, then the ideal ideal P of k [ S ] .
( f l , f3)
is equal to the toric
Proof. Let f be a binomial in P and assume we order the monomials with the revlex ordering x3 > x1 > x2. As fl, f3 are binomials, by the division algorithm we can write
f = hlfl + h3f3 + r, where r = x:1x732g3 - xf1x;2x33E P , such that T I ,s1 < a1 and r 3 , s 3 < c 3 . Assume r # 0. As x? or x;2 is a factor of r , onemayassume that s2 = 0. Note s1 > T I and s 3 > 73. From the equality r = x;* h, where h= - x? -rl xs3-r3 , one concludes h E P and 7-2 2 u2. Therefore the identity h + ,.42-a2fl = x ; ~ x ? - a 2 - x ; l - r 1 x ; 3 - ~ 3 >
rapidly leads to contradiction with the minimality of c 3 because one has the 0 inequality a1 > SI - r1. Definition 10.3.9 The support of a binomial f = x" - xp, denoted by supp(f), is defined as supp(xa) Usupp(xfi),where supp(z") = { x i ]CY( > O } . Theorem 10.3.10 ([146]) If
is a full set of critical binomials, then the ideal toric ideal P of k[S].
(fl, f 2 , f3)
is equal
to
the
388
Chapter 10
Proof. By Proposition 10.3.6 one may assume supp(fi) # (21,2 2 , x 3 } for at least two fi. Hence one may assume a 3 = 0 and the cases to consider are the following: (i) bl = 0 , (ii) b 3 = 0, (iii) c1 = 0 and (iv) c 2 = 0. (i) Note b 3 2 c 3 , if b 3 = c 3 , then f 2 is a critical binomial w.r.t x 3 and by Proposition 10.3.8 one concludes that P is generated by { fl, f 2 ) . If b 3 > c 3 , from the identity
we get
c2
= 0. Thus
Applying Lemma 10.3.4 to f l , fl, f$ yield c 1 = a l . Hence f 3 is critical w.r.t 2 1 , using Proposition 10.3.8 one concludes that P is generated by {fi, f3}. (ii) As a 2 2 b2 and bl 2 a l , from fl - 5i2-b2f2= - x ; 2 - b 2 ~ ! 1 we get f1 = -f 2 and P = (f1, f 3 ) by Proposition 10.3.8. One may now assume bl > 0 and b 3 > 0, (iii) using x i 2 - b 2 f Z+ f 3 we readily see that this case cannot occur, (iv) usingxi2-bzf 2 + f l we derive a contradiction. This proof is to due A. Alchntar [3]. 17
Proposition 10.3.11 ([146]) If P is the toric ideal of k [ t d l t, d 2 td3] , over a field k and d l , d2, d 3 are relatively prime, then P is a set theoretic complete intersection.
Critical binomials in four variables Let I be an ideal generated by a full set of critical binomials f1, . . . ,fn, one fi for each variable xi. It would be desirable to findeffective means for determining when I is equal to the toricideal P. Here we address the case n = 4 and char(k) = 0 by presenting the elements of an algorithm, having fl, . , . ,f4 as input, that decides when I = P occurs. In the remaining of this Section we study ideals generated by critical binomials in four variables. Our aim is to determine when the ideal generated by a full set of critical binomials is prime and in this case determine its type. If the presentation ideal P of a monomial curve in Ai is generated by a full set of critical binomials, produced by any critical-binomial-algorithm, we predict how these generators should look like (see Theorem 10.3.15). Some of the resultspresented here can beshown using different methods introduced by W.Gastinger, E. Kunz and R. Waldi [124, 1251, and by H. Bresinsky [37]. A complete description of almost complete intersection monomial curves in A$ was given in [124],
Monomial Curves
389
Lemma 10.3.12 Let
Proof. (i) If bi
>, ai, the following equality implies that f e is not critical
Hence ai > bi. By symmetry be > at. (ii) The minimality of cj together with the identity x; fi - ~ q fe ~- x q a - -bi xt ~ be -~ x? +bix;e yields the inequality aj bj 2 c j . 0
+
Lemma 10.3.13 Let
f1,.
. . ,f4 be the full set of critical binomials:
If ai > 0 , bi > 0 for all i, then I = (fi, f 2 ,
f 3 , f4)
is not a prime ideal.
Proof. Assume I is prime. As in Lemma 10.3.12 we have al b2 > a2. From the identity
>
bl and
we derive g1 E I and there is a monomial M so that
Therefore c4 = 0 or then a1 > el and e4
we obtain
g2
e3
= 0. First we consider the case c4 = 0. If From the identity
> a4.
el
> 0,
E I and there is a monomial M so that
If (i) holds, then c2 = 0 and a1 -el 2 c1, since f l is critical and f3 = x i 3 -x:' we have a contradiction. If (ii) holds then, e2 = e3 = 0 and a1 - el 2 e l , which is impossible because f l is critical and f 4 = xi4 - x:1. Hence el = 0.
Chapter 10
390
xb~-e~x;4-b4 2
=
{ (ii)6 )
Mxi1xFx2 = Mxi'x? or, Mx;' x 2 x g 3 = Mxg2x i 3
(10.4)
If (i) occurs, thenc1 = 0 and b2 -e2 2 c2, which is a contradiction since f2 is critical and f3 = x? -x;'. If (ii) occurs, then el = e3 = 0 and b2 - e2 2 e2, which is also a contradiction since f2 is critical and f4 = x? - x i z . Hence e2 = 0. We claim that c1 = 0. If c1 > 0, then a1 > c1 and c3 > a3. Using the equality a4+c4 xF5$4 fl + x;l -c1 f3 = g 3 (x;l -c1 x y 3 - 2 az+cz 2 x4 ) = xg394 we see that x:l
94
E I and there is a monomial M so that
-c1,43-a3
=
{ (ii) (i)
M X ~ ~ =X M~ X~~ X~or,X~ ~~ ~(10.5) = Mxi3
MxPl
If (i) holds, then c2 = 0 and a1 > c1, which is impossible since f1 is critical and f3 = xg3 - xyl. If (ii) holds, then c3 > e3, which is impossible because f3 is critical and f4 = xz4 - x z 3 . Therefore c1 = 0. Altogether we have c1 = c4 = 0 and el = e2 = 0. We claim that b2 > c2, otherwise if c2 2 b2, then from the equality x?-b2f2 + f3 = - xbl x ~ ~ - b3b ~b4 1
2
x3 x4
and using the minimality of c3 we obtain a contradiction. To complete the proof for the case c4 = 0 notice that the inequality b2 > cg also yields a contradiction since f3 = x? - x? and f 2 is critical. It remains to consider the case e3 = 0. We may assume c4 > 0. We claim that c1 = 0. If c1 > 0, then by (10.5) either (i) or (ii) hold. Since c4 > 0 (ii) holds, as a result e2 = 0 and a1 > e l , which is a contradiction since f1 is critical and f4 = xz4 - x;l. Hence c1 = 0. Next we show el = 0. Assume el > 0, by (10.3) we have that (i) or (ii) hold. If (i) holds, then c2 = 0 and e4 > c4, which is a contradiction since f4 is critical and f3 = x? - x?. If (ii) holds, then e2 = 0 and a1 > e l , which is a contradiction since fi is critical and f4 = x? - x;l. Hence el = 0. Altogether we have c1 = 0 and el = e3 = 0. We claim that 152 > e2, if on the contrary e2 2 b2 from the identity x ; z - b z f 2 + f4 = - xb1xez-b2 b3 b4
xz
1
2
x3 x4
we contradictthe minimality of e4. To complete the proofnotice that b2 > e2 also leads to a contradiction since f 2 is critical and f4 = x:4 - x i 2 , Therefore I is not a prime ideal. a
Monomial Curves
If c3
391
> 0, then Lemma 10.3.12 and Lemma 10.3.13 yield > a3 and c4 > 0. Using the equality
c1
c2
= 0,
a1
> c1,
we obtain g E I and x~1-C1xg3-a3 = for some monomial M. Therefore e2 = 0, and by Lemma 10.3.12 and Lemma 10.3.13 we obtain el > 0 and e3 > 0. From the equation
we obtain g E I andthis is a contradiction since x;3-e3xi4-C4is not a multiple of any of the monomials that occur in f l , . . . ,f4. Therefore c1 = 0. Assume c4 > 0, note that Lemma 10.3.12 yields c2 = 0. If e3 > 0 , then el = e2 = 0 and it follows (see argument below) that f4 = -f 3 , as required. On the other hand if e3 = 0, we have e2 > 0 and from the equation
we obtain g E I and there is a monomial &I so that x;* " e 1 x ~ 4 - a=4 Mz:*. Hence el = 0 and we are in case (b). We may now assume c1 = c4 = 0. f 2 = 2g3 - xi2-b2 bs x3we obtain By minimality c2 2 b2 and since f3 + b3d3, which implies f3 = -f 2 . Write the identity c3d3 = (c2 - b2)& f4 = - z;'xi2xg3, we claim that el = 0. If el > 0 , then a1 > el and e4 > u4. Therefore from the equality
+
x:4-a4 = Mx7l xG2xg3, we have g E I . There is a monomial M so that it follows that e2 = e3 = 0 , which is a contradiction since f l is critical. Therefore el = the 0 and proof is complete. U Next we present a classification theorem.
392
Chapter
Proof. We set Si = supp(fi). Assume that lSil = 4 for some i, say IS11 = 4. Let f1 = xyl - X ~ ~ X ai~ >~ 0 Xfor ~all~ i. , By Lemma 10.3.13 we have lSil 5 3 for i = 2,3,4. Assume that lSil = 3 for i 2 2. If x1 !$ (S2U S 3 U S 4 ) , then (C)occurs, hence we may further assume x1 E Si for some i 2 2, say x1 E S2. We claim that x1 E S3 U S4. If x1 4 S3 u S4, then (after permutation of variables) we have
where ai
> 0, bi > 0, ci > 0, ei > 0 for all i.
Using the identity
we obtain g E I and this is a contradiction since ~ ~ 3 - ~ 3 3 $ 4 - ~ is 4 not a multiple of any of the monomials that occur in the fi's. Hence x1 E S3us4. ASsume that 2 1 E 5'3 \ s3 n5'4 (the case x1 E S4 \ 5'3 nS4 is similar), Notice that s2 = 5'3, otherwise after permutation of variables we have f1
= x;' - x;2xa33 xa4 4 ,f 2 = x$ -xllx;, 6 6
f3
= x? -xc1xc4 1 4 , f4 = x 2 - xe2xe3 2
39
and, as before, this leads to a contradiction, Thus S2 = S3 and we are in case (D).Nowwe may also assume x1 f S3 n S4, by similar arguments it
Monomial Curves
393
we obtain g E I and x;l-blx$"az = M x Y ' x ~ ~for x ~some monomial M . Hence a1 > c1, bz > c2 and c4 = 0, it follows readily that c1 > 0 and c2 > 0. A similar argument shows b4 = 0 and we are in case (C). The case c1 > 0
Chapter 10
394
is symmetric since we can interchange 22 and 2 3 . We may now assume bl = c1 = 0. If b4 = c4 = 0, then f3 = -f2 and (B) occurs. If b4 > 0 and c4 > 0, then k = 2 , 3 lead to case (B) and k = 1 lead to case (E). If b4 > 0 and c4 = 0, then k = 2 or k = 3 and (B) or (Bl) occur. On the other handif b4 = 0 and c4 > 0, then c2 = 0 and k = 1 cannot occur, observe that k = 2 and k = 3 lead to cases (Bl) and (B) respectively. To complete the proof of (iii) it remains toconsider the case j = 2. Assume j = 2, in this case we can ~ that x ~k # , 1. write f2 = x? - :x for some k and f4 = x? - x ~ ~ xnote First assumec1 > 0, using f1 and f3 we obtain x : 1 - C 1 x i 3 - a 3= M x e11 x2e 2 x3 e73 hence e2 = 0, a1 > el and c3 > e3. Notice that el > 0 and e3 > 0, otherwise we are in case j = 4. Therefore c4 = 0, k = 4 and we are in (E), We may now assume c1 = 0, notice k # 1. If k = 3, then (B) or (131) occur. If k = 4 and el = e3 = 0, then (B) occurs. If k = 4, el = 0 and e3 > 0, then e2 = c2 = 0 and (B) occurs. We are going to show, by contradiction, that the case k = 4, c1 = 0 and el > 0 cannot occur. If k = 4, c1 = 0 and el > 0, then using f2 and f4 gives e2 = 0 and we can write
where ai
> 0 for all i and el > 0.
Observe that
b4
2 e4 and consider
we obtain g E I and there is a monomial M so that
If (i) occurs, then c4 = 0 and b2 > c2 > 0, which contradicts the minimality of b2. On the other hand if (ii) occurs,then e3 = 0 and a1 > el > 0, which contradicts the minimality of a l . Therefore in both cases we obtain a contradiction the and proof is complete. 0 Remark 10.3.16 Here we keep the assumptions and notation of the theorem above. (i) In (E) a4 2 e4 and e2 = 0 is not possible. If this is the case we have
This gives the polynomial x4a4-e4f4 + f l = x:' - ~ i ~ - ~ ~ExI zfrom 3 which x;l - xcl xc2 x e 3 E I since e3 2 c3. Thinking about itas a diophantine equation,a1 > c1 since c2 > 0. Thus xT1-'l - 1 2 ~ 3 ~ ~ ~ E3 I and ~
c3~z4-e4
3
~
~
~
~
~
4
Monomial Curves
395 ,
.
,I, I
I
,
';
> 0 since c 2 > 0, a contradiction to a1 being minimal. Therefore in (E) we have bi > 0, ci > 0 for all i , e 3 > 0 and e 2 > 0. In (E) if a 4 > e 4 , then I = ( f { , f 2 ,f 3 ,f 4 ) , where fi = f l + x4a4"e4f4 = - xe2~e3xa4"e4 , that is, a1 - c1
I has a generating set as in (D). In (E) if a4 = e 4 , then I = (ff, f 2 , f 3 , f 4 ) , where fi = f l + f 4 = xpil - xe2 xe3and I has a generating set as in (C), (ii) Assume (D), then it follows from the proof of Proposition 10.3.23 that I has a generating set as in (C). (iii) If (B2) holds, then I = ( f l , f 2 , j ; , f 4 ) , where f$ = f 3 + f 4 = x?-x?l and I has a generating set as in (BI). If (B1) holds, then I = ( f l y f 2 , fi, f 4 ) , where fi = f l f3 = -f2, after permutation of x1 and x 4 we obtain that I has a generating set as in (B).
+
From this remark we obtain the following Corollary 10.3.17 Let I be a n ideal generated by four critical binomials, one for each variable. If I is a prime ideal of height 3, then there are critical binomials f l , f 2 , f 3 , f 4 with respect to x 1 , x 2 , x 3 , x 4 generating I so that, up to permutation of the variables, f1, f 2 , f 3 , f 4 can be written in one of the following forms.
- X bl 1 xb4 4 , f 3 = ei
- z C2 2 X4c 47
> 0 f o r all i.
Let f l , f2, f 3 ,f 4 be a full set of critical binomials having one of the forms (A)-(E) as in the theorem above, and I = ( f l , . . . , f 4 ) . The rest of the section is devoted to find the arithmeticalconditions, on the initial data, for I to be prime. Proposition 10.3.18 ([37, 1241) Let I = erated by a full set of critical binomials:
SO
that ai
d4
be a n ideal gen-
> 0, bi > 0, ci > 0, ei > 0 f o r all i. T h e n I is prime if and only
(a) al = bl (b)
(f1, f2, f 3 , f 4 )
=~
+ el, b2 = c 2 + e 2 , c 3 l b 2~ a3 s b 1 ~ 2 .
=a3
+
e3, e4
= a4 4- b4 -t- ~
4 and ,
if
396
Chapter 10
Proof. Assume I is prime. By Lemma 10.3.12 we have
From the equalities
we obtain 91,g2,g3 E I . Let M be the set of monomials that occur in fi for i = 1,2,3,4. Notice that any monomial p that occur in gi can be written as p = p1p2, where pl is a monomial and p2 is in M . As a consequence a1 = bl + el, b2 = c2 + e 2 , ~ 3= a3 e3. Set
+
and consider the exact sequence.
S i n c e v = a + / 3 + y + 6 ~ K w e o b t a i n e 4 = ~ 4 + b q + c 4 . ByTheorem10.1.3 K is generated by a , p, y, 6. Hence &(A) = Z, where A is the matrix with rows a,/3,y, S and &(A) is the ideal generated by the minors of order 3 of A. The equality -6 = a /3 y shows 13(A) = (A41, A 4 2 , A43, A 4 4 ) , where A4j are the cofactors of A corresponding to the last row. Since the column vectors C = (A41,A42, A43,A44) and q = (dl, dz, d3, d4) are in the kernel of A we have ker(A) = CZ = qZ.Therefore d4 = a1b 2 ~ 3- a3blcz and the first part of the proof is complete. Converselyassume (a) and(b) hold. Observe that C E ker(A) = $Z and write C = aq, a E Z. Using (b) gives1 . 1 = 1, that is, Z4/(a,P,7,6) is a torsion free Z-module. This readily implies ker($) = ( a , P , y ,S). An application of Theorem 10.1.3gives rad ( I ) = P. We claim that f1, f2, f3, f 4 form a Grobner basis with respect to the reverse lexicographical ordering 2 1 > x2 > x3 > x4, it is enough to show that the S-polynomial of fi and fj reduces to zero. As we may assume that the leading terms of fi and fj have common factors the claim follows from the identities
+ +
From the claim we derive that x4 is a regular element on R / I , i.e., R / I is a one dimensional Cohen-Macaulay ring. Altogether P is the only associated
397
Monomial Curves prime of I. Consider the cofactor given by
Since f
542
of the Jacobian matrix J of I , it is
4 P , Theorem 3.5.10 shows rad ( I ) = I. Therefore I = P.
0
In the following corollary we keep the same assumptions and notation as in the proposition above. Corollary 10.3.19 If I is prime, then R / I is a Cohen-Macaulay ring type 3.
of
Proof. Since I is prime a1 > el >, 1,b2 > e2 2 1,c3 > e3 2 1. Set S = R / I and A = S/x4S. Notice A -N B1/11,where R1 = k [ z l ,2 2 , 2 3 1 and 11 = (x:1, x?, x?, x:1 xi2 xg3). Set .
17 = Socle(A) = (I1:m)/Il.
It is readily seen that the socle of A is equal to
and the Therefore dimk ( V ) = 3 type
of S is 3.
0
Proposition 10.3.20 Let f1
= x;l - x 3 a 33 x a44 , f 2 = x: - 2 2 , f3
=,
4 2 , f4
= x;‘ - X i 2 X 5 3
be a set of critical binomials with respect to X I , x2,x3, x4 respectively. Let I = (f1, f 2 , f 4 ) . T h e n I is a prime ideal if and only if d3 = alb2e4. Proof. The proof is the same as that of Proposition 10.3.18. It is enough to make two observations: (i) f1, fi, f 4 form a Grobner bases with the rev. lex. ordering x1 >’ x4 > 23 > x2, hence x2 is regular on R / L (ii) The polynomial
f = a1b3e4xy”143-1x;4-1 is an element in theJacobian ideal of I which is regular on R/I.
0
Proposition 10.3.21 Let
be a set of critical binomials with respect to X I , x2, x3, x4 respectively. Let I = ( f l , f 2 , f3, f4). If ai > O,bi > 0,ci > 0 and ei 2 0 f o r all i , then I is a prime ideal if and only if dl = e 4 ( b 2 ~ 3- b 3 ~ 2 ) .
Chapter
398
Proof. We proceed as in the proof of Proposition 10.3.18. It is enough to make three observations: (i) By Remark 10.3.7 11 = (f1, fi,f3) is a prime ideal of height twoand f 4 is regular on R/I1. Hence R / I is Cohen-Macaulay. (ii) The polynomial
is an element in the Jacobian ideal of I which is regular on R / I . (iii) By the previouslemma a1 = bl c1, b2 = a2 c2 and c3 = a3 b3.
+
+
+
Let I be a prime ideal as in the proposition above, then we can also determine its type. We have Corollary 10.3.22 If I is prime, then R / I is a Cohen-Macaulay ring of type 2.
S is a Cohen-Macaulay algebra
of type 2.
0
be a set of critical binomials w.r.t X I , x2,23, x4 respectively. If a3 > 0, a4 > 0 and bi > 0, ci > 0 f o r all i, then I = ( f l , f 2 , f3, f4) is a prime ideal if and only if dl = ed(b2c3 - b 3 ~ 2 )and a4 = qe4 for some q E N. Proof. Assume I is prime. Consider the identity
Since I is prime g E I and there is a monomial M so that
+
Since (ii) can not occur we have a1 2 bl c1, hence Lemma 10.3.12 yields al = bl c1. Let ai be the least positive integers so that aid1 E d2N d3N and let fi‘ = - a’ By Lemma 10.3.4
+
+
in particular ai = a l . Notice ai 2 a2 and ab 2 a3 (see argument below). From f:’ - f1 = z;2X;3(x;4 - x2a i -a2 3 -a3 )
Monomial Curves
399
we obtain a4 2 e4. By the division algorithm we can write a4 = qe4 with 0 5 r < e4 and q 2 1. The equality
+ r,
shows I' = (f;, f2,
f3, f4)
= I . We now prove r = 0. Since
we conclude
To prove r = 0 it is enough to show that the coefficients of non-negative. From the equality
d2
and
d3
are
it follows that there is a monomial M so that
As (ii) can not occur we obtain ug - u3 - qe3 2 0. Similarly using fi and f2 yields ah - a2 - qe2 2 0. As in the proof of Theorem 10.3.18 we readily obtain dl = ed(bzc3 - b3c2). For the converse note that by Proposition 10.3.21 Lemma and 10.3.4 I' is a prime ideal. 0 Corollary 10.3.24 Let
be a set of critical binomials so that fi is critical w.r.t xi for all i . If bi > 0, > 0 for all i and e3 > 0, then I = (f1, f2, f3, f4) is a prime ideal if and only if dl = es(b2c3 - b 3 ~ 2 )and a4 = qe4 for some q E N.
ci
10.4
An algorithm for critical binomials
The critical binomials in the next example were computed using the algorithm below. As an application of the results in Section 3 we have
Chapter 10
400
is a set of critical binomials generating a height 3 prime ideal of type 2. (b) If dl = 204, d2 = 855, d3 = 1216, d4 = 1260, then
is a set of critical binomials generating a height 3 prime ideal of type 3.
(c) If dl = 9, d2 = 12, d3 = 18, dq = 19, then fl
= -f 3 = x: - 23, f 2 = x; - 23, f4 = x; - x 1 2 2 4
is a complete intersection prime ideal. Let us present an algorithm to compute a full set f1,. . . ,f n of critical d n N c W and P binomials, one fi for each variable xi. Let S = dl W the toric ideal of k[S], where gcd(d1,. . . ,d n ) = 1. For a = ( ~ 1 ,.., , a n )E I T we set deg(a) = Cy=lai. Let a,P E Wn. The graded reverse lexicographical ordering in W is defined by a < p if deg(a) < deg(p), or deg(a) = de@) and the first nonzero entry of p - a is negative. Since x? - x? E P for j # i, an upper bound for mi is given by d = min{ dl, . . . ,&, . . . ,d n } . For each i search for a E Nn-l so that
+ -+ ‘
a . (dl,.. . ,d i , . . . ,d n ) h
0 mod 4
0
,
(10.6)
-
where denotes the usual inner product, all this from degree zero up to degree d. For the a’s satisfying (10.6) compute ma = a
(dl,. . . ,d i , . . . ,d n ) / d i
and take mi = mmo= min{m,
n
Ia
satisfies (10.6)}.
If we search the a’s in increasing order with respect to < we can take a0 minimum with respect to this ordering. To generate the a’s in increasing order, from (0,. . . ,0) up to a certain degree g, can be accomplished with the following:
Monomial Curves
, _ I
>
,
,
2,.
401
*
Exercises 10.4.2 Let P be the toric ideal of k [ - t d l ,t d 2 , ids], where IC is a field and d l , d2, d3 are relatively prime positive integers. Prove that the following are equivalent: (a) P is a complete intersection. (b) P is Gorenstein. (c) S = dlN + d2N
+ d3N is symmetric.
10.4.3 If fi, f2, f3 is a full set of critical binomials, prove the equality
where $I: Z 3 + Z is the Z-linear map induced by $(ei) = di. Recall f = a-P iff=xa:-xO. h
+
10.4.4 Let f l = zyl f 2 = x? - z;'z$, f 3 = zg3 be a full set of critical binomials. If 6i > 0, ci > 0 for all i, prove the equalities dl = a263 + a3b2, d2 = alb3 ~ 3 6 1 d3 , = alba - a2b1.
+
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Chapter 11
Affine Toric Varieties and Toric Ideals We will study general toric sets and toric ideals over arbitrary fields. The aim is to examine when a given set of binomials define a toric ideal up to radical and to give a criterion to characterize when a given toric set is an affine toric variety. Here we generalize in a natural way some of the results of Chapter 10. I
11.1
Systems of binomials in toric ideals
Let k be any field and D a fixed rn x n matrix with non negative integer entries dij and with nonzero columns. The toric set determined by the matrix D is the set in the affine space A : given parametrically by xi = t f 1 i * * - t $ i
( i = 12...,n).
Thus one has
r= Let
((tf11
. . . t 2 1 , .. .
...t$n>
E
A;
1
~ 1 , ., . , t ,
Ek}.
and B = k[tl,. . . , t m ]
R = k[xl,. . . ,x,]
be two polynomial rings over k , graded by
+
deg(xi) = dli - - * + dmi for all i , and deg(tj) = 1 for all j ,
-
respectively. Let q$ be the graded homomorphism of k-algebras:
4: R -+ k [ t l , .. * ,t,],
403
xi
d'
t Di ,
Chapter 11
404
where Di = ( d l i , . . . ,d,i)
is the transpose of the ith column of
D and
tDi = tali . , . tkmi, 1
We denote the image of q5 by k[r]. The kernel of q5, denoted by P , is the toric ideal of k[r]. Closely related to the map q5 is the homomorphism
determined by the matrix D in the standard bases of Z n and Zm. Indeed, we have $(x") = for all a, E W . As a consequence, a binomial g = x" P = ker(q5) if and only if 5= a, - ,b' belongs to ker($).
- xp belongs to
To prove the following two results we will use the notion of simple component of a polynomial with respect to a subgroup of Zn; see Section 10.1. Proposition 11.1.1 If 91,. . . ,gr is a set of binomials generating the toric ideal P of Ic[I'], then 51,. . . ,& generate ker(@). Proof. Set G = ( $ 1 , . . . ,&) c Zn. Consider the equivalence relation on the monomials in R given by
x"
"G
-G
xP
if a - ,O E G. Let y E ker($) and write y = Q - p, where a,@ E W , Proving that y E G amounts toprove that x" -G xp, i.e., that the binomial g = x" - xp is simple. Now, g belongs to the toric ideal P= (91,. . . ,g T ) , hence the simple component h of g containing x" also belongs to P , by Lemma 10.1.2. Since x" 4 P , it follows that h = g, hence that g is simple, as required. 0 Definition 11.1.2 Let I be an ideal of a ring R and f E R, the saturation of I with respect to f is 00
U ( I : =~ ) E R Jr f i E I , for some i 2 I}.
(I:~o= o)
{T
i= 1
If R is a polynomial ring the saturationcan be computed using Grobner bases and the equality ( I :f") = ( I ,1 - t f ) n R, where t is a new variable. See Chapter 2. For toric ideals the condition (a) of Proposition 2.4.24 will be replaced by a condition which is much easier to verify.
Affine Toric
Ideals
Varieties Toric and
405
.,
Proposition 11.1.3 ([104]) Let 91,.. . ,g,. a set of binomials in the toric ideal P ojlc[r]and 1 = (91,.. . ,g,.). Ifchar(k) = p # 0 (resp. ,char(k) = 0), then the following two conditions are equivalent: (al) P = rad(]: zoo), where x = z1 (a2) p2"ker($) c
(GI,.. . ,&)
.
a
xn and ( I :zoo) = U Q ~( I :zi).
for some u E
N (resp. ker($) = (51,.. . ,&)).
Proof. (al) =+ ( a 2 ) : First we consider the case char(k) hypothesis, there exists u 2 0 such that
Let y E ker(@). Write y = a g E P a n d 3 = c r - p = r . Set
- ,8 with
a,P E W and g
G = (Gl, * .. ,&)
F
p
# 0. By
= xcy - xo, thus
c zn.
We know that dgPU E I for some S E PP and
Since the simplecomponents of xsgpu belong to I and hence to P (see Lemma l O . l . Z ) , and since xJz"Pu does not belong to I because it does not belong to P , it follows that x6gPu is simple, i.e.,
Hence pUa- p u p E G. This shows puker($J)c G. Nowwe consider the case char(k) = 0. Let G = ($1,. . . ,$,.) and let 0 # r) E ker($J). Note r) = a - p , where a and p are in I T , thus f = x" -x0 is in P and xrfQ is in I for some prime q >> 0 and for some y E W . Expanding f Q one has:
Let h be the simple component of x7fQwith respect to G containing zQ"+r. If qa - qP 4 G, then xqo+r does not occur in h. Hence
for some J contained in (1, , . . ,q - 1). Note that h E I by Lemma 10.1.2, As 4 is prime, q must divide (Y) for 0 < i < q. Thusmaking xi = 1 yields 1 k = (sq)lk, s E Z, but since char(k) = 0 one has 1 = qs, which is
Chapter 11
406
impossible. Hence q(a- p) E G. Pick a prime r # q such that x7 f E I, by the previous argument r ( a - p) E G. Since gcd(q,r ) = 1, we get a - p E G. (al) (ag): It suffices to prove
P
c rad(I: zoo),
because the other contention is clear. First we assume char(lc) = p
# 0. Let
be a binomial in P and write
By
(82)
there are integers si such that
where we may assumesi 2 0 by replacing z"'/ Hence writing Z"i
-=
d i
by its inverse if necessary.
((S)
-1) + I
and using the binomialtheorem it follows that x7fp' is in I, for some monomial ZT . Therefore f E rad(I: zoo). In thecase of characteristic zero, the same arguments as above will work, after replacing pu by 1 throughout. 0 Theorem 11.1.4 ([104]) Let I' be a toric set and 91, . . . ,gr a set of binomials in the toric ideal P of k[l?]. Set I = (91, . , . ,9,). If char(k) = p # 0 (resp. char(lc) = 0), then rad(I) = P if and only if
Proof. It is a consequence of Proposition 2.4.24 and Proposition 11.1.3. 0 Corollary 11.1.5 ([104])Let 91, . . . ,gr be a set of binomials in the ideal P of the monomial curve r. Set I = (91,. . . ,g T ) . If char(k) = p # 0 (resp. char(k) = 0), then rad(I) = P if and only if (a) pmker(+) c (&, . . . ,&) for some rn E N (resp. ker(@) = (&, . . . ,&.)) (b) V ( g 1 , . . . ,g r , zi)= (01, for all i.
Toric Affine
Varieties and Toric Ideals
407
Proof. +) By Theorem 11.1.4 one derives condition (a) and rad(1, xi) = rad(P, xi) for all i. On the other hand rad(P, xi) = ( x 1 , ,. , ,xn), because the height of P is n - 1 and xi is regular on RIP. Therefore
V ( I ,xz) = V(z1,.. . ,xn)= (0). +=) By Proposition 11.1.3 one has
P = rad(1: P ) , where x = x1
-
a
x,.
Hence there is monomial xd and an integer N such that x s P N c I . In order to prove that rad(1) = P , we will show that every prime ideal Q containing I also contains P. If Q contains no variable, then the inclusions
x 6 P Nc I
cQ
imply that P c Q, as desired. If Q contains at least one variable, then we claim that Q contains all the variables, implying
This is the point in the proof where we need a special argument, compensating for the fact that the condition (b) above is weaker than that in Theorem 11.1.4. Indeed, up to a renumbering of the variables 2 1 , . . . ,x,, one may assume that ( X I , . . . ,x.} is the list of all the variables contained in the prime ideal Q , for some integer 1 5 s 5 n. Let a = ( a i ) be the point in affine space given by
We claim that a E V ( I ) .Indeed, let g = x" - xp be any binomial belonging to I . Since g belongs to Q, it easily follows that x" E (21, . . . ,x,) if and only if x0 E (21,. . . , x,). As a consequence, a" = up, this common value being 0 if x" E (x1 . ,. . ,x$), or 1 otherwise. In either case, we have g ( a ) = 0. It follows that a E V ( I ) ,as claimed. But a1 = 0, hence a belongs to V ( I , X ~ ) , which is (0) by condition (b). Thus a = 0, which by definition of a implies that s = n. Hence, Q contains all the variables and consequently contains P , as desired. 0 Remark 11.1.6 In the proof of Corollary 11.1.5 the underlying hypothesis gcd(d1, . . . ,d,) = 1 is unnecessary.
408
Chapter 11
Corollary 11.I.? Let 91, . . ., be a set of binomials in a toric ideal P . a field of char(k) = p # 0 (resp. char(lc) = 0) and the containment p m ker($) C G = ($1,. . . ,&) holds for some rn 2 0 (resp. ker($) = G), then V ( g 1 , 9 . 3 gT) c V ( P )u V ( x 1 .* * E n ) .
If k i s
Proof. By Proposition 11.1.3, there is a monomial 'x and an integer N such that z'PN c I. It follows that
V ( I )c V ( P N )u V ( 2 )c V ( P )u V(Zl
*
xn),
as
required. 0
Exercises 11.1.8 Let I' be a toric set and P the toric ideal of k [ r ] .If k is an infinite field, prove that P = I ( r ) and V ( P )= r, where
is the Zariski closure I?.
11.1.9 Let P be the ideal of a toric set I?. If I is an ideal of R and prove that = PR',
fi = P ,
dEF
where R' = k[xfl, . . . ,x:']
is the ring of Laurent polynomials.
Hint Use that R' is the localization of R at the multiplicative setof monomials. 11.1.10 Let I be an ideal generated by a set of binomials in the ideal P of the toric set I'. Prove that the following conditions are equivalent: (al) rad(I R') = P .R', where R' = k[z?', . . . ,x:1] 9
(a2) zsPN c I for some monomial 'x E R and some integer N 2 1 11.1.11 Let D be an m x n integral matrix and $: Z n + Z m the homomorphism determined by D. If G is a subgroup of ker($), prove that
T ( Z n l G )= T(ker($)/G), where T denotes torsion. 11.1.12 Let I' be an arbitrary toric set and P the toric ideal of k[r]. If . . . , is a set of binomials in P such that
91,
v(g1,* ,ST, x i ) = (0)
for all i , prove that dim k[r]= dim(R/P) = 1.
Hint Use the proof of Corollary 11.1.5 to show rad (P,xi) = (
.
~ 1 , . , ,x,).
Affine Toric Varieties and Toric Ideals
409
11.1.13 Let D be an m x n integral matrix with nonzero columns and +:Zn -+ Z m the homomorphism determined by D. If G is a subgroup of ker($) and p 2 2 a prime number, prove that the following are equivalent:
(al) puker(+) (a2)
c G for some u E N.
ker(+)/G is a finite p-group or ker($)/G = (0).
(as) Zn/G 2 Z 3 x H , where H is a finite p-group or U = (0), and s is equal to rank(D). 11.1.14 (Rational quartic curve in P3) Let k be afield and P the toric ideal of k[r],where r is the toric set defined by the matrix:
11.2
Affine toric varieties
Our aim here is to use linear algebra to characterize when a toric set I? is an affine toric variety in terms of the existence of certain roots in the base field and a vanishing condition. We will make use of the fact that any integral matrix is equivalent to a diagonal matrix which is in Smith normal form. First we fix some notation. Let I? be a toric set defined by an rn x n matrix D = ( d i j ) . According to [229, Theorem 11.91, there are invertible integral matrices U = (uij) and Q = ( q i j ) of orders m and n respectively such that L = U D Q = diag(X1,. . . , X,,O,. . . ,O), where s is the rank of D and XI,. . . , A, are the invariant factors of D , that is, X i divides &+I and X i > 0 for all i.
Notation For use below set
In the sequel ei will denote the ith unit vector in Zn. Proposition 11.2.1 If $:Zn 3 Z m is the homomorphism determined b y D in the standard bases, then
where qi corresponds to the ith column of Q .
410
Chapter 11
Proof. Let z E Z n and make the change of variables y = Q-lz. As
L = UDQ it follows that D X = 0 if and only if Ly = 0. Set y = ( y 1 , . , . ,yn). First note qi E ker(q) for i 2 s 1, because LQ-lqi = Lei where ei is the ith unit vector in Zn.On the other hand if x is in ker($), then Xiyi = 0 for i = 1,.. . ,s. Thus
+
n
X
= QY =
yiqi.
i=a+l TOcomplete the proof observe that the columns of Q are a basis for Zn. Proposition 11.2.2 Let
0
+:zn+ zm
be the linear map determined by D and S
vj =
bij qi
,
i=l
where qi correspond to the ith column of Q . If e l , . . . ,en is the standard basis of Zn,then {vi - ei}y=l is a generating set for ker($). Proof. Set Q-l = ( b i j ) . Note
for all j , because QQ-l = I . Hence one can write n
S
and using Proposition 11.2.1 we obtain vj - ej E ker($) for j = 1,, . . ,n. On the other hand from the equality above:
n
and Ideals Toric
Varieties Toric Affine
411
Theorem 11.2.3 ([242]) Let k be afieldand P the toric ideal of k [ r ] , where I' c kn is the toric set defined by the matrix D. T h e n = V ( P ) if and only if the follqwing two conditions are satisfied: I?,
# 0 V i , then a:"
(a) If (ai) E V ( P ) and ai i = I, ...,s.
-..api has a
&-root in k
for
r f o r i = 1,.. . ,n. Proof. +=) One invariably has r C V ( P ) . To prove the other containment take a point a = (a1, . . . ,an) in V ( P ) ,by condition (b) one may assume ai # 0 for all i. Thus using (a) there are t i , . . . ,t$ in k such that (ti>X*= ay1i . . . aQni = aqi n (i = 1,.. . ,s). (11.1) (b) V ( P , z i )c
For convenience of notation we extend the definition of ti by putting ti = 1 for i = s + 1,. . . ,m and t' = ( t i , .. . ,t&).Set
where f i = ( f l i , . . . ,fmi) and d j = ( d l j , . . . , & j ) denote the ith and j t h columns of U-' and D respectively. Next we compare columns in
D = (U"L)Q-l to get: 8
dk
=
Xjbjk fj
( k = 192,. . ,TZ),
(11.4)
j= 1
where Q-l = ( b i j ) . Using UU-l = I and Eq.( 11.2) we rapidly conclude: tfk
= ti
( k = I , . . ., m ) .
(11.5)
From Proposition 11.2.2 we derive Dvj = Dej = dj for j = 1, . . . , n, where
i= 1
\e=
1
e= 1
Chapter 11
412
Therefore putting altogether
for k = 1 , . . . ,n. Thus a E I?, as required. +) It is clear that (b) holds because V ( P ,xi) c V ( P ). To prove (a) take (ai) in V ( P ) with ai # 0 for all i, then by definition of I? there are t l , . . . ,tm in k such that aj = t d j for j = 1 , . , . ,n. Therefore by Eq. (11.3) one has: t x i f i = t q l i d l . , , t q n i d n = a y l i , . . aqni n *
Thus ( t f i ) X= i aili
sei, as required.
0
Corollary 11.2.4 If k is an algebraically closed field, then
V ( P )c I? u V(x1
xn).
Proof. Let a = ( a i ) E V ( P ) such that ai # 0 for all i. Since k isalgebraically closed condition (a) above holds. Therefore one may proceed as of the proof of Theorem 11.2.3 to get a E I?. 0 inthefirstpart Next we present another consequence that can be used to prove that monomial curves over arbitrary fields are affine toric varieties and generalizes Proposition 10.1.14. Corollary 11.2.5 If the columns of D generate Z m as Z-module, then the equality I? = V (P ) holds if and only if V (P, xi) c I? for all i. Proof. Since
Zdl+
*
- + Zd,
= Zm,
one has Xi = 1 for all i , thus condition (a) holds. Therefore I? is an affine toric variety if and only if (b) holds. El
Proof. If k is algebraically closed, then (a) is satisfied, thus I? is a toric variety if and only if V ( P ,xi) c I' for all i. 0
As a more concrete application wenow show that Veronese toric sets are affine toric varieties. Corollary 11.2.7 Let d be a positive integer and
A = ((~1,. .,am) E W m I
+ . . *+ a, = 4
*
If k is an algebraically closed field and D the matrix whose columns are the vectors in A, then the toric set r determined by D is an afine toric variety.
413
Ideals Toric Varieties and Toric Affine Proof. Let
where d+m-1 s=( m-1
).
One can order the fi such that fi = t t for i = 1 , . . . ,rn and Isupp(fi)l for i > m, where supp(t") = {tilai > 01. Fix an integer 1 ,< i _< s, it suffices to prove
22
where P is the toric ideal associated with D. We use induction on m. Take a E V ( P ,xi). If i > rn and $(xi) = f i = ty' . - tz,note that the binomial
belongs to P , hence a E V ( P ,xj) for some 1 ,< j _< m. Therefore one may harmlessly assume 1 L i 5 m and $(xi) = t t , for simplicity of notation we with TI > 0 one has assume i = 1. Observe that for every fj = ty' aj = 0; indeed since x; - x;' . x 2 belongs to P and a E V ( P ,21) one has aj = 0. Let D' be the submatrix of D obtained by removing the first row and all the columns with nonzero first entry (from top to bottom), and P' the toric ideal of D'. The vector a' = (ai)tl 4 supp(fi)) is in V ( P ' ) , because P' c P. Since V ( P ' ) c I?'U V(z2 x,), where I" is the toric set associated with D', by induction one readily obtain a E I?. 0
tz
In the light of Exercise 11.2.10, a natural question is whether I? can be a variety but not a toric variety, to clarify consider: Example 11.2.8 Let k = Z3 and D = (2,4), then
On the other hand P = (x1 - xi) and (1,2) E V ( P ) .Thus i' # V ( P )
Exercises 11.2.9 Let D be an m x n integral matrix such that there are invertible integral matrices U and Q such that UDQ = diag(X1,. . , ,X,, 0 . . . ,O). If Q1 (resp. Q2) denote the submatrix of Q (resp. Q-') obtained by fixing the first s columns(resp. s rows) of Q(resp.Q-'), prove theequality DQ1Q2 =
Chapter 11
414 11.2.10 Let I' be a toric set and P the toric ideal of field and r is a variety defined by an ideal J , then J
cP
k[r].If k is an infinite
and I? = V ( J )= V ( P ) .
11.2.1 1 Let IC be an algebraically closed field and I? the toric setassociated
Prove that I' is an affine toric variety.
11.2.12 Prove that Corollary 11.2.4and Corollary 11.2.6 are valid assuming condition (a) of Theorem 11.2.3, instead of assuming k algebraically closed. 11.2.13 (Segre variety) Let m,nbe two positive integers. If
is the toric set associated with incidence the matrix D of a complete bipartite graph Km,n, prove that I? is an affine toric variety over any field k.
Hint Use that D is totally unimodular (Le., every minor is 0 or ztl).
11.3
Curvesin positive characteristic
Let I be an ideal of a ring R the arithmetical rank of I, denoted by
r = ara(I), is the least positive integer r such that there are f i , . , . ,f,. with rad(f1,. . . ,f,.) = rad(1). By Krull's principalideal theorem one hasara(I) 2 ht(I), note that equality occurs if I is a set-theoretic complete intersection. Let k be a field and P the associated toric ideal of a monomial curve I' in the affine space A;. In characteristic zero is an open problem whether P is a set theoretic complete intersections,several authors have contributed toward the solution of this problem, see [99, 215, 224, 2821, the case n = 3 is treated in [35]. If char(k) > 0, then by [73, Theorem 21 and [73, Remark 11 it follows that P is a set-theoretic complete intersection; a result of T. T. Moh E2241 shows that P is indeed generated up to radical by n - 1 binomials. Our aim is to give a different constructive proof of Moh's result due to AlcQntar, Reyes and ZQrate [3, 41 by identifying special generating sets for the solutions of linear diophantine equations.
Toric Affine
Varieties Toric and
Ideals
415
Lemma 11.3.1 Let D = (dl, . . . ,dn) be a 1 X n matrix with di E Z \ (01, then there is a unimodular matrix Q = (42, . . , qn, 41) with entries in Z whose first n - 1 columns are of the form:
qi =
I!],
such that (a) DQ = (0,. . . ,0, X), and
(b) ker(a) = 92z
+ + qnZ, where X is equul to gcd(d1,. . *
*
,
,dn).
Proof. The idea is to use an algorithm similarto theone used to diagonalize a matrix with entries in an Euclidean domain. To "diagonalize" D we use elementary operations on the columns of the following matrix:
where I is theidentitymatrix of order n. Note that using elementary operations one can obtain the following reductions: 0 421
b12
q22
b22
0
0
0 1
0
0
0
7-2
d3 0
--' * '
'''
*'. .. .. * .
."
f o
dn 0 0 0
1
O
r3
d4
d,
\
where r2 = gcd(d1, d2) and 7-3 = gcd(r2, d3). In the first reduction we have used elementary operations on the first two columns only, those operations come from the Euclidean algorithm used to compute the greatest common
Chapter 11
416
as
ker(D)
divisor of dl and d 2 . Similarly the second reduction can be obtainedby performing elementary operations on the second and third columns only, where those operations come from the Euclidean algorithm to compute gcd(7-2, d3). Thus by induction we rapidly get the required matrix &. To finish the proof use Proposition 11.2.1 to conclude that q 2 , . . . ,qn generate Z-module. 0 Proposition 11.3.2 Let D = ( d l , .. . , dn) be a matrix with di E Z \ (0). Then there are f 2 , . . . ,f n E Z n such that
(b) for each fi with i 2 2, it holds fii > 0 and fij 5 O for 2 5 j (c) ker(D) is generated b y f 2 , . . . , f n as Z -module.
5 i - 1,
Proof. Let 4 2 , . . . ,qn be as in Lemma 11.3.1 and let 2 5 k 5 n. It will be shown recursively that there are f 2 , . . . , f k satisfying conditions (a),(b), and ~ q 2 + " ' + ~ q k = Z f 2 + " ' + Z f k .
We proceed by induction on b. For k = 2, take f 2 = q 2 . Assume 2 5 k < n and that we have constructed f 2 , . . . ,f k satisfying (a), (b), and such that ~q2+"'+Z'Qk=~f2+"'+Zfk.
(*)
Next define f k + l , k + l = q k + l , k + l > 0 and f k + l , k = Q k + l , k - x k : f k , k , with large enough to have f k + l , k 5 0. In general for 2 5 i ,< k and haven chosen & + I , . . . ,X I , , define fk+l,i
= qk+l,i - Akfk,i - *
choosing x i >> 0 to have
a
*
- &+lfi+l,i - x if i , i ,
5 0. Finally define fk+l,l = qk+l,l - x k f k , l - ' ' ' - h f 2 , 1 , note that we are not requiring f k + l , l 5 0. If we set fk+l,i
'
fk+l
=
fk+l,l
fk+l,k+l
0
0
I
Affine Toric Varieties and Toric Ideals
417
clearly f 2 , . . . , f k + l satisfy the hypotheses (a), (b) of the proposition. On the other hand fk+l = Qk+1- X k f k - - ' * - X 2 f 2 , using Eq. (*) this implies:
(a) f i =
Then fii divides dl for all i 2 2. Proof. Since dlei - diel belong to ker(D) for i 2 2, one obtains
Clearly onehas dl = pnf n n if i = n and pn = 0 if i < p. By reverse induction follows that p n = pn-1 = * . = /li+l = 0 and dl = pi fii. 0 a
Theorem 11.3.4 ([224]) Let D = (dl, . . . ,dn) be an integral matrix with positive entries and P the toric ideal of k [ t d l ,.. . ,tdn].If char(k) = p > 0 , then P = rad(g1,. . , ,gn-1) for some binomials 91,. . . ,gn-1. Proof. Let f 2 , . . . , f n be as in Proposition 11.3.2. First we prove recursively that for 2 5 k 5 n there are w 2 , . . . ,w k E Z n and rk E N so that
(i) wi =
wii O I
, o
with
wij
5 0 for j = 1,.. . ,i - 1y
wii
> 0,
Chapter 11
418
To prove this we proceed by induction on IC, for IC = 2 it is enough to take w2 = f 2 and r2 = 0. Let 2 5 k < n and assume there are w2,. . . ,wk E Zn satisfying (i) and rk E N such that
By Lemma 11.3.3, we have fiildl for i = 2 , . . . ,n. Then dl = f k + l , k + l l for some e, let p a b + l be the greatestpower of p dividing e, that is, we can write
where p does not divide s k + l , hence using that the image jj of p in Z8b+lis non zero, one obtains that has finite order. Therefore there is P k + l E N+ such that f l k + l 1 modsk+l, hence p O k + l U z 1 mod S k + l , for all u E N, consequently there is c k + l E N (depending on u ) such that pPL+Iu
= sk+lck+l
choose u large enough to guarantee p a k + l the entries of w k + l as:
+ 1,
fk+l,l
=P f k + l , l - d k +C l k+l =pak+*fk+l,i i = 2,.. k Wk+l,k+l = P f k + l , k + l + dlCk-t.1 wk+l,j = 0 Wk+l,l Wk+l,i
Observe that equality
- d k + l c k + l < 0 and define
e ,
~ k+ l, i
5 0 for i = 1,...,k and
>k + l w k + l , k + 1 > 0.
Therefore
as a result
setting
wk+l
qk+l
E ker(D), hence
=ak+l
(2)
+ ,&+lu, one concludes Wk+l,k+l
= P qb+l f k + l , k + l .
for j
Note the
andIdeals Toric
Affine Varieties Toric
419
+
NOW,since only the first k 1 entries of wk+1 and f k + l can possible be different from zero (all the remaining entries are zero), we get from Eq. (3) that only the first k entries of wk+l - ( p q k + + l ) f i ~ + ~ can possible be different from zero (the rest are zero), hence using that ker(D) is generated by f:!, . . . , fn, one derives: Wk+l
- (pqk"+')fk+lE zf2 +
*
'
+ Zfk,
which together with Eq.(**) gives pTk(wk+l- (pqk+~)fk+~) E Z(wz,. . . ,w ~ ) , and consequently ( p q k + l +rk )fk+l E
setting r k + l = q k f l
Zb2,*
* *
,Wk,W f l )
+ r k , one concludes
Prk+1q(f2,'
9
fkfl)
c Z(w2, ' ' * , W k + l ) .
Altogether this terminates the induction argument and ends the first step of the proof. Making k = n, we obtain that there are w2,, . . ,w n such that the first i - 1 entries of wi are negative or zero, the ith entry of wi is positive, and for j > i the entries of wi are zero. Furthermore there is p r n such that prnqf2,
*
, fn) c Z(w2,
* * '
,wn).
Set gi = Gi+l, this means that gi is a binomial such that lji = ~ i + Let ~ . I = (91,. . . ,gn-1). Take a E V ( I ,x i ) , with a = ( a l , . . . ,an), and suppose that a # 0 and let aj be the first entry of a equal to zero (such entry exists because at least ai = 0). If j # 1, one would have and aj = 0, then some a, = 0 for T < j, which is not possible, thus j must be equal to 1. Let us see that a = (0,. . . ,0) by induction on the coordinates of a, assume a1 = = a k = 0 for 1 5 IC < n. One has:
and since a E V ( I ,x i ) C V ( I )and gk E I , one obtains ak+l = 0. Therefore a = (0,. . . ,0). As a result V ( 1 ,xi) = 0 for i = 1,.. . ,n, this fact together with prnker(D) c Z( ~ 2 , .. . ,wn) prove that the conditions (a) and (b) of Theorem 10.1.3 are satisfied, and consequently P = rad(g1, . . . ,gn-l). 0 Remark 11.3.5 Working over an algebraically closed field b of characteristic zero s. Eliahou [loo] proved that the toricideal P of a monomial curve in A: is the radical of an ideal generated by n binomials.
This Page Intentionally Left Blank
Appendix A
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:
T
42 1
Appendix A
422
0
r
Graph Diagrams
423
Appendix A
424
A.2
Unmixed graphs
Next we display the set of all unmixed non Cohen-Macaulayconnected graphs with at most six vertices.
17
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Not at ion ann R ( M), annihilator of M ,4 (G), independence number, 165 ~ R ( M )length , of a module, 9 gcd, greatestcommon divisor,218 ( , ), usual inner product, 210 1.r, ceiling of a , 234 lcm, least common multiple, 49 lk(F), link of a face, 143 log(F), log set of F ,236 log(xa), log of a monomial, 236 A:, affine space, 45 N+ , positive integers, 31 P;, projective space, 56 v ( M ) , number of generators, 8 U S , canonical module, 106 I or l a , integral closure, 69 fi or rad ( I ) ,radical of I , 4 ,1. diamond norm, 217 e ( M ) ,multiplicity of M , 101 e( q, M ) , relative multiplicity, 85 ei, unit vector, 156 j ( A ) , the f-vector of A, 150 f + F g , reduction w.r.t F , 49 h ( j ) ,Macaulay symbol, 119 k , a field, 2 k [ F ] ,monomial subring, 201 k[G],edge subring, 282 k [ A ] ,Stanley-Reisner ring, 142 k [ x ] ,polynomial ring, 2 C I' = k \ (01, 48 r ( I ) ,reduction number, 339 x 2 0, non negative vector, 218 xc, monomial, 201 Cn, cycle, 163
( I :J"), saturation of I , 53 ( N ~ : R N ideal ~ ) , quotient, 4 A ( P ) ,Ehrhart ring, 232 A [ F ] subring , generated by F , 23 A p , normalized Ehrhart ring, 226 C ( G ) ,cone over a graph G, 293 F ( M ,t ) , Hilbert series, 97 GI * G2, join of two graphs, 332 H ( M ,i), Hilbert function, 14 H ( y , c), hyperplane, 210 H+(y, c), halfspace, 210 I ( n ) ,symbolic power, 76 I A , Stanley-Reisner ideal, 142 I C ,ideal of covers, 196 K , a field, 2 K*(g),Koszul complex, 29 N G ( U ) ,neighbor set, 163 R ( - a ) , shift in the graduation, 59 R ( k ) kth , Veronese subring, 203 R,, localization at a prime, 3 S(f,g ) , the S-polynomial, 49 S-l ( M ) ,localization, 3 V (I ), prime ideals over I , 2 Ass(l), associated primes, 5 A s s R ( M ) ,associated primes, 5 Aq, q-skeleton of A, 145 AI, Stanley-Reisner complex, 142 Spec(R), spectrumof a ring, 2 Supp(M), support of a module, 5 T(G), minimal covers, 166 1 1 ~ 1 1 , Euclidean norm, 212 a+,positive part of a vector, 203 ao(G), covering number, 165 ht ( I ) ,height of an ideal, 4 447
448
X,, complete graph, 163 Km+, complete bipartite, 182 R ( I ) ,Rees algebra, 66 2(M) , zero divisors, 6 x(A), Euler characteristic, 141 I& = {x E R ( x 2 O}, 211 &A, cone generated by A, 211 Soc(M), socle of M ,21 SymB(M),symmetric algebra, 65 adj(H), adjoint of a matrix, 8 char(R), characteristic of R,2 conv(A), convex hull of A, 210 deg(v), degree of a vertex, 162 depth(M), depth of a module, 18 dim(M), Krull dimension, 4 gr, ( R ), associated algebra, 73 iff, if and only if, 2 in(I), initial ideal, 48 ri(X), relative interior, 215 s . o . ~system , of parameters, 16 star(a), star of a face, 148 supp(p), support of a vector, 203 supp(zu), support of xu, 131 trdeg, transcendence degree, 34 type(M), type of a module, 21 w.r.t, with respect to, 14 (A;A ) , reduced homology, 139
Notation
Index a-invariant, 99 adjacency matrix, 190 admissible grading, 220 affine k-algebra, 31 combination, 210 space, 45, 210 generated by a set, 210 algebra, 23 finite, 23 finitely generated, 23 of finite type, 23 almost integral, 74 annihilator, 4 arithmetical rank, 414 Artin-Rees lemma, 85 Artinian reduction, 123 ring, 10 ascending chain condition, 2 associated graded algebra, 73 matrix of a monomial subring, 202 prime of a module, 5 of an ideal, 5 atomic cycles, 303 Betti numbers, 61 initial, 188 initial virtual, 188 binomial, 202 expansion, 100, 119 ideal, 202 449
of a closed walk, 282 bipartite graph, 163 boundary complex, 156 bow tie, 317 Buchberger algorithm, 50 criterion, 5 1 canonical module, 106 catenary ring, 21 Cayley-Hamilton, 22 ceiling of a vector, 234 characteristic of a ring, 2 Chinese remainder theorem, 10 chord of a cycle, 194 chordal graph, 194 circuit, 297 of a graph, 300 COCOA see computer algebra, 94 codimension of a module, 4 of an ideal, 4 Cohen-Macaulay graph, 168 ideal, 20 module, 18 ring, 20 complement of a graph, 175 complementary complex, 172 complete bipartite graph, 182 graph, 163 ideal, 69 intersec,tion,20
Index
460
generically, 68 set theoretic, 20 composition series, 9 computer algebra systems COCOA,94 Macaulay, 53 Normaliz, 236 PORTA, 211 cone, 210 finitely generated, 211 of a complex, 140 over a graph, 293 conormal module, 67 content of a polynomial, 339 convex combination, 154, 210 cone, 210 hull, 154, 210 polyhedron, 210 set, 210 cover complexity, 166 critical binomials, 385 full set, 385 cutpoint, 175 cycle, 163 basis, 291 space, 291 cyclic polytope, 158 d-tree, 194 Danilov-Stanley formula, 221 Dedekind-Mertens formula, 339 degree of a module, 101 of a vertex, 162 Dehn-Sommerville equations, 157 depth lemma, 18 of a module, 18 Dickson’s lemma, 48 dimension of a set, 210 of a simplicial complex, 138 theorem, 16
discrete graph, 162 valuation, 378 distance, 163 division algorithm,, 49 dominant set, 175 duality in monomial subrings, 268 in Stanley-Reisner ideals, 196 edge, 161 cone, 326 critical graph, 168 generator, 282 graph, 188 ideal, 168 generalized, 256 subring, 282 Ehrhart polynomial, 232 normalized, 227 ring, 232 normalized, 226 elementary integral vector, 297 vector, 296 zonotope, 302 elimination order, 52 embedded prime, 10 end vertex, 162 equations of a cone, 216 Euler characteristic, 141 reduced, 141 formula, 156 even cycle, 163 exact sequence, 5 short, 5 extended Rees algebra, 72 extrema1 Cohen-Macaulay ring, 117 Gorenstein ring, 115 f-vector
Index
451
of a complex, 150 of a polytope, 155 face, 138, 211 ideal, 130 ring, 143 facets of a cone, 325 of a simplicial complex, 144 faithfully flat, 27 Farkas’s lemma, 218 field of fractions, 8 filtration of a ring, 81 finite homomorphism, 23 length,9 flag complexes, 170 flat homomorphism, 27 forest, 163 forms, 13 F’robenius number, 378 I
Gauss lemma, 339 geometrically linked, 30 going down, 26 UP,26 Gorenstein graded ideal, 112 ideal, 22 ring, 22 Grobner basis, 49 lexicographical, 276 of toric ideals, 203 reduced, 49 graded algebra, 16 ideal, 13 map, 13 module, 13 ring, 12 graph, 161 connected, 162 connected components, 162 Graver basis, 284
H-configuration, 314 h-vector of a C-M complex, 152 of a complex, 152 of a polytope, 156 of a standard algebra, 108 height of an ideal, 4 Herzog-Kuhl formulas, 63 Hilbert basis theorem, 2 function, 14, 97 polynomial, 101 of a face ring, 150 over an Artinian ring, 15 series, 97 of a face ring, 150 Hilbert-Bprch theorem, 62 Hochster configuration, 314 homogeneous element, 13 ideal, 13 resolution, 61 ring, 14 subring, 224 homogenization of an ideal, 54 homomorphism integral, 23 of algebras, 23 of rings, 2 ,
.
ideal of a subset, 45 of linear type, 67 of relations, 35 quotient, 4 improper’ face of a polyhedral set, 211 incidence matrix, 294 Smith form, 295 independence number, 165 independent set of edges, 163 of vertices, 165 induced subgraph, 195
Index
462
initial degree, 110 ideal, 48 integral closure, 23 commutewith localizations, 70, 71 of a domain, 26 of an edge subring, 320 of ideals, 69 of monomial ideals, 234 of monomial subrings, 218 integral extension, 23 integrally closed ideal, 69 ring, 26 irreducible representation, 214 of a cone, 215 submodule, 6 irredundant decomposition of modules, 7 irrelevant maximal ideal, 35
isolated prime, 10 vertex, 162 Jacobian criterion, 91 ideal, 92 matrix, 91 Jacobson radical, 95 join of graphs, 332 of ideals, 245 of simplicial complexes, 140 Konig Theorem, 165 Koszul complex, 28 Krull dimension, 4 intersection theorem, 83 principal ideal theorem, 19 Kruskal-Katona criterion, 151
lattice, 135 distributive, 135 of monomial ideals, 135 Laurent polynomials, 80 leading coefficient, 48 monomial, 48 term, 48 length of a module, 9 level algebra, 274 lex order, 31, 48 linear resolution, 63 variety, 210 link of a face, 143 linkage class, 30 even, 30 of ideals, 30 local ring, 2 localization, 3 at a prime, 3 locally a complete intersection,78 log set, 236 Macaulay see computer algebra, 53 symbol, 119 theorem, 119 marriage problem, 165 a generalization, 330, 331 minimal primes of a ring, 2 of a module, 9 resolution, 61 minimum number of gens, 58 module of finite length, 9 of fractions, 3 monomial curve, 367 ideal, 130 of minimal covers, 196
Index of mixed products, 248 order, 48 ring, 130 subring, 201 subring of a graph, 282 walk, 283 multiplicity of a face ring, 154 of a graded module, 15, 101 over a local ring, 85 Nakayama’s lemma general version, 8 graded version, 14 neighbor set of a subset, 163 of a vertex, 182 neighbourly polytope, 158 nilpotent element, 4 nilradical, 4 Noether normalization homogeneous, 39 lemma, 34 of a monomial ring, 39 of an edge subring, 309, 340, 362 Noetherian module, 1 ring, 1 normal ideal, 69 polytope, 228 ring, 26 normality descent, 80, 240 Normaliz, 236 normalization of a ring, 26 of an edge subring, 320 of monomial subrings, 218 normalized degree, 203 grading, 265 volume, 228
453
normally torsion free, 77 edge ideal, 312 versus normal, 79, 238 Nullstellensatz, 46 numerical semigroup, 377 odd cycle condition, 322 order complex, 170 partition, 150 Pascal identity, 15 path, 162 perfect matching, 165 polarization, 131 of a bipartite planar graph, 304 of an even closed walk, 304 polyhedral cone, 210 set, 210 face, 21 1 facet, 214 polynomial function, 15 quasi-homogeneous, 13 polytopal subring, 228 polytope, 154 lattice, 228 PORTA, 211 poset, 169 positively graded algebra, 35 presentation ideal, 35, 66, 202 of a graded algebra, 35 primary ideal, 6 extension of, 27 submodule, 6 primary decomposition irredundant , 7 of a graded module, 14 of a module, 7 of an ideal, 7 of monomial ideals, 132
Index
464
prime avoidance general version, 10 graded version, 41 primitive binomial, 284 projective closure, 57 of a monomial curve, 375 dimension, 61 space, 56 proper face of a polyhedral set, 211 of a polytope, 155 pure resolution, 62 pure simplicial complex, 145 quasi-regular sequence, 83 radical ideal, 76 radical of an ideal, 4 rank of a module, 10 reduced ring, 4 reduced simplicial homology, 139 reduction number of an algebra, 349 of an ideal, 339 reduction of a polynomial, 49 Rees algebra, 66 of a filtration, 81 regular element, 6 regular ring, 19 is Cohen-Macaulay, 89 regular sequence, 16 ReisKer Theorem, 143 relative interior, 2 15 remainder, 49 residue field, 2 revlex order, 48 ring extension, 23 p
p
_
p
p
_
S-polynomial, 49 Samuel function, 85 saturated graph, 177 saturation of an ideal, 53 Segre
p
p
-
p
product, 105 variety, 414 semigroup, 377 ring, 201, 367 Serre’s normality criterion, 26 set of k-products, 256 Shellable complex, 146 shelling, 146 shift in the graduation, 59 simple components, 368 simple module, 9 simplex in affine space, 155 lattice, 228 of a simplicial complex, 138 oriented, 139 unimodular, 228 simplicial complex, 138 Alexander dual, 198 Cohen-Macaulay, 143 geometric realization, 156 simplicial polytope, 155 simplicia1 sphere, 156 Gorenstein property, 157 skeleton of a complex, 145 sliding depth, 29 socle of a graded algebra, 111 of a module, 21 spanning subgraph, 162 spectrum of a ring, 2
psquare,pi63p square-free monomial, 130 standard algebra, 35 grading, 13 St anley-Reisner complex, 142 ideal, 142 ring, 142 star, 187 star of a face, 148 strongly Cohen-Macaulay, 29
Index subgraph, 162 subring of k-products, 203 support of a binomial, 387 of a module, 5 of a monomial, 131, 299 of a vector, 296 supporting hyperplane, 211 suspension of a graph, 325 symbolic power, 76 Rees algebra, 238 symmetric algebra, 65, 67 semigroup, 378 system of parameters, 19 for modules, 16 homogeneous, 36 syzygetic ideal, 67 syzygy module, 51, 61 computation of, 52 tensor algebra, 65 term order, 48 terminal cycle, 316 terms, 48 toric ideal, 202, 404 of an edge subring, 282 set, 403 variety, 208 total ring of fractions, 8 transcendence basis, 34 degree, 34 tree, 163 triangle, 163 triangulated graph, 194 twists of a graded module, 61 type, 266 Cohen-Macaulay, 112 of a module, 21 unicyclic graph, 316
455
unimodular covering, 230 unimodular matrix, 295 totally, 414 unmixed graph, 168 unmixed ideal, 20 upper bound conjecture, 159 usual grading, 13 valuation ring, 378 variety affine, 45 coordinate ring, 45 dimension of, 45 irreducible, 45 irreducible components, 46 projective, 56 toric, 208 Veronese subring, 203 square- fr ee , 203 Veronese variety, 412 vertex, 212 vertex cover, 165 minimal, 165 vertex covering number, 165 vertex critical graph, 168
walk, 162 closed, 162 Zar iski closure, 207 topology, 45 of the prime spectrum, 2 zero divisor, 6 zero set of an ideal, 45
