Ideals and Reality: Projective Modules and Number of Generators of Ideals

Springer Monographs in Mathematics

Friedrich Ischebeck Ravi A. Rao

Ideals and Reality Projective Modules and Number of Generators of Ideals

GI - Springer

Friedrich Ischebeck Institut fur Mathematik Universitat Munster Einsteinstr. 62 48149 Munster, Germany e-mail: [email protected]

Ravi A. Rao School of Mathematics Tata Institute of Fundamental Research Dr Homi Bhabha Road 400005 Mumbai, India e-mail: [email protected]

Library of Congress Control Number: 2004114476

Mathematics Subject Classification (2000): 11C99,13A99,13Cio,13C40,13D15,55R25 ISSN 1439-7382 ISBN 3-540-23032-7 Springer Berlin Heidelberg New York This workis subject to copyright. All rights are reserved, whether the whole or part ofthe material is concerned,specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9,1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable for prosecution under the German Copyright Law. Springer is a part of Springer Science+Business Media springeronline.com O Springer-Verlag Berlin Heidelberg 2005 Printed in Germany The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Typeset by the authors using a Springer !+VEXmacro package Production: LE-TEX Jelonek, Schmidt & Vockler GbR, Leipzig Cover design: Erich Kirchner, Heidelberg Printed on acid-free paper

4113142YL - 5 4 3 2 1 0

Dedicated t o our close friend and former colleague Hartmut Lindel who left us much too early

Preface

Besides giving an introduction to Commutative Algebra - the theory of commutative rings - this book is devoted to the study of projective modules and the minimal number of generators of modules and ideals. The notion of a module over a ring R is a generalization of that of a vector space over a field k. The axioms are identical. But whereas every vector space possesses a basis, a module need not always have one. Modules possessing a basis are called free. So a finitely generated free R-module is of the form Rn for some n E IN,equipped with the usual operations. A module is called projective, iff it is a direct summand of a free one. Especially a finitely generated R-module P is projective iff there is an R-module Q with P @ Q S Rn for some n. Remarkably enough there do exist nonfree projective modules. Even there are nonfree P such that P @ Rm S Rn for some m and n. Modules P having the latter property are called stably free. On the other hand there are many rings, all of whose projective modules are free, e.g. local rings and principal ideal domains. (A commutative ring is called local iff it has exactly one maximal ideal.) For two decades it was a challenging problem whether every projective module over the polynomial ring k[X1,. . . ,X,] with a field k was free. It was known from the beginning that such a module had to be stably free. The statement that it should be actually free was called Serre's Conjecture. This was proved independently by D. Quillen and A. Suslin in 1976. We give several proofs of it. Later we show how vector bundles over a compact Hausdorff space X (more generally vector bundles of special type over any topological space) can be interpreted as projective modules over the ring of (real, complex or quaternion) continuous functions on X . This gives the concept of projective modules an intuitive meaning. For instance it is no surprise that nontrivial vector bundles exist - at least once one has seen the Mijbius band. In the second half of the book we study the question what one can say about the minimal number of generators of certain ideals. This often - but not

viii

Preface

always - is connected with the theory of projective modules. We begin with dimension theory on commutative so named Noetherian rings, i.e. such whose ideals are finitely generated. (This property of rings was first identified and studied by E. Noether, a student of D. Hilbert.) Its fundamental theorem states the equality of two numbers: Let R be a local Noetherian ring with maximal ideal m. Then the minimal number n of elements a l , . . . , a n such that m is a minimal prime over-ideal of (al, . . . ,an) equals the number of steps of a maximal chain of prime ideals in R, i.e.

with maximal n. This number n is called the (Krull) dimension of R. If m itself can be generated by n = dim(R) elements then R is called a regular local ring. A not necessarily local ring whose ideals are finitely generated is called regular, if all its localizations are so. (To every commutative ring R and every prime ideal p of it one can associate in a canonical way a local ring, the localization of R in p.) An example is k[X1,. . . ,X,] with a field k. All of its maximal ideals can be generated by n elements. Other examples are the domains all of whose ideals are projective modules. Namely these are the regular Noetherian domains of dimension 5 1, i.e. whose nonzero prime ideals are maximal. These rings are called Dedekind rings. We give a complete classification of the finitely generated modules, especially of the projective ones over those rings. Also we prove the theorem of the finiteness of the class number, which says that over certain 'classical' Dedekind rings there are only finitely many projective modules upto 'stable isomorphy', i.e. upto 'adding free modules7. The Forster-Swan Theorem gives upper bounds to the number of generators of modules if one knows these numbers for the localizations. As a consequence one sees that an ideal I of R = XI,. . . ,Xn] can be generated by n 1 generators if R / I is regular.

+

Improving an old theorem of Kronecker, in 1972 Eisenbud and Evans and independently Storch have shown that every prime ideal in k[X1, . . . ,X,] is a minimal prime overideal of an ideal generated by n elements. (In the book we will give a version of this theorem which is not restricted to prime ideals.) Geometrically this means that every set defined by polynomial equations can already be defined by n equations. The last chapter is dedicated to the question: Under what hypotheses can one describe an algebraic curve in the affine n-space by n - 1 equations? i.e. let p be a prime ideal of R = k[Xl,. . . ,Xn] with dim(R/p) = 1. When is p a minimal prime overideal of an ideal, generated by n - 1 elements? (Also here the good formulation does not restrict to prime ideals.) We show that the answer is positive in the following two cases: 1. R/p is regular, i.e. a Dedekind ring. (Mohan Kumar) 2. k is of positive characteristic. (Cowsik and Nori)

Preface

ix

The general answer is still not known. That is the reality today. The 'book' started as a M. Phil. project of the junior author with Selby Jose, in the course of which they found the expository article [I061 of Valla. The initiative and determination of the senior author, and the urging of colleagues, led to Setby's thesis becoming into a book which could be used as a graduate course or for an intensive workshop. The whole book tells the story of a philosophy of J-P. Serre and his vision of relating that philosophy to problems in Affine geometry. A thorough development of this subject till the end of 1980 is done in this book. The intermix of Classical Algebraic K-theory and Complete Intersection problems is emphasized in this text for the first time. The results of Eisenbud-Evans, Swan's connection between vector bundles and projective modules, Lindel's proof of Serre's conjecture appear for the first time in a student text form. The book is almost self-contained, and serves as an introduction to basic Commutative algebra and its applications in problems of &ne algebraic geometry. In a first reading, the student could skip Chapter 5; but a better understanding of the subject and its interconnection with other parts of Mathematics can only be had by eventually perusing this chapter. The reader could also skip the Eisenbud-Evans theorem in Chapter 9, 59.4; though he should know that this is the best achieved via general position arguments, the recurrent theme. The material in this text has been crystallized from various books, research and expository articles. Earlier works on this subject are the notes of Geyer, Ohm, BadesEu; survey articles by Valla, Lyubeznik, Murthy, Bass, Suslin; and the books of Lam, Szpiro, Kunz, Mandal respectively. Following the tradition in the books written by J-P. Serre, we have not let the exercises interrupt the flow of reading. We have placed them chapter-wise at the very end. The exercises are challenging; the student who wades through them will be very proficient to work in this active research area. This is because most of the exercises are culled from the research papers of experts. Interesting developments, not found here, include- Effective methods of Sturmfels, etc.; Bose on Serre's conjecture via Groebner bases, and its relation to problems in Electrical Engineering; numerous applications of the Local-Global techniques due to Bass, Suslin, Asanuma, etc; generalizations of Serre's conjecture: the Bass conjectures, the Bass-Quillen conjecture by Lindel, Popescu, and the Anderson conjecture over monoidal rings due to Gubeladze; surprising developments in the orthogonal groups due to Parimala, Sridharan, Ojanguren; analogous results of Suslin and his students in the classical groups; G. Lyubeznik's work in higher dimension which uses the Cowsik-Nori Theorem as the inductive start; Murthy's results on complete intersection questions; and Mohan Kumar's on Eisenbud-Evans conjectures over affine algebras; later developments, inspired by Nori, by Bhatwadekar, Mandal, Raja Sridharan on complete interestion and Euler classes; parallel development on

x

Preface

complete intersections over real algebras by Ojanguren, Ischebeck, etc.; Herzog on monomial curves; results of Warfield, Stafford, etc. on development of Serre's conjecture and the Eisenbud-Evans theorem in non-commutative noetherian rings; examples of Macaulay, Moh; projective varieties, and the corresponding problems there; connections with local cohomology started by Faltings, and developed in Lyubeznik's thesis; scheme theoretic and ideal theoretic questions studied by Hartshorne, Abhyankar amongst others; local algebra - 'symbolic primes' results of Cowsik, Huneke, etc.; motivations from algebraic topology; principal G-bundles; . . . We would like to thank many colleagues for their valuable suggestions which encouraged us, and have helped to improve the material presented. We thank our wives for their support. The senior author acknowledges the help from Deutsche Forschungsgemeinschaft (DFG) and T.I.F.R. which made it possible for him to interact with the junior author. The book was finally completed in the idyllic setting a t the I.C.T.P. this summer, whom we thank for their hospitality. Finally, we look forward to hearing from you, to improve the material and its presentation. September 13, 2004

Friedrich Ischebeck Ravi A. Rao

Contents

1

............................. The Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Modules ................................................

Basic Commutative Algebra 1

1

1.1

1

1.2

6

............................................15 Multiplicatively Closed Subsets ..................... 15 Rings and Modules of Fractions ..................... 17 Localization Technique ............................. 20

1.3 Localization 1.3.1 1.3.2 1.3.3

1.3.4 Prime Ideals of a Localized Ring . . . . . . . . . . . . . . . . . . . . 21 1.4 Integral Ring Extensions

................................. 23

1.4.1 Integral Elements

.................................23

1.4.2 Integrality and Primes 1.5 Direct Sums and Products 1.6 The Tensor Product

1.6.3 1.6.4 1.6.5

................................ 30

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

........................................ 38 Functoriality ...................................... 42 Exactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 Flat Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 Exterior Powers ................................... 46

1.6.1 Definition 1.6.2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

xii

2

Contents

Introduction to Projective Modules . . . . . . . . . . . . . . . . . . . . . . . . 49

........................ 49 Rank ..................................................52 Special Residue Class Rings ...............................56

2.1 Generalities on Projective Modules 2.2 2.3

2.4 Projective Modules of Rank 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 3

Stably Free Modules .......................................69 3.1 Generalities .............................................69 3.2 Localized Polynomial Rings ............................... 74 3.3 Action of GLn(R) on Umn(R)

............................

75

3.4 Elementary Action on Unimodular Rows ................... 76 3.5 Examples of Completable Rows

. . . . . . . . . . . . . . . . . . . . . . . . . . . 80

3.6 Direct sums of a stably free module ........................ 82 3.7 Stable Freeness over Polynomial Rings ..................... 83 3.7.1 Schanuel's Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 3.7.2 Proof of Stable Freeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 4

Serre's Conjecture .........................................87 4.1 Elementary Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 4.2 Horrocks' Theorem

...................................... 89

4.3 Quillen's Local Global Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

........................................... 98 Vaserstein's Proof .......................................101

4.4 Suslin's Proof 4.5 5

Continuous Vector Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 5.1 Categories and Functors ..................................105 5.2 Vector Bundles ..........................................108

....................114 Examples ...............................................121

5.3 Vector Bundles and Projective Modules 5.4

5.5 Vector Bundles and Grassmannians . . . . . . . . . . . . . . . . . . . . . . . .126 5.5.1 The Direct Limit and Infinite Matrices

...............126

Contents

xiii

5.5.2 Metrization of the Set of Continuous Maps . . . . . . . . . . . 128 5.5.3 Correspondence of Vector Bundles and Classes of Maps 129 5.6 Projective Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 5.7 Algebraization of Vector Bundles

. . . . . . . . . . . . . . . . . . . . . . . . . .135

...........136 Projective Modules as Pull-Backs ....................140 Construction of a Noetherian Subalgebra . . . . . . . . . . . . .143

5.7.1 Projective Modules over Topological Rings 5.7.2 5.7.3

6

Basic Commutative Algebra I1 .............................149 6.1 Noetherian Rings and Modules ............................ 149

......................................... 152 Dimension of Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .154

6.2 Irreducible Sets 6.3

..........................................156 Small Dimension Theorem ................................158 Noether Normalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .162

6.4 Artinian Rings 6.5 6.6

. 6.7 Affine Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .164 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 6.9 Dimension of a Polynomial ring ........................... 171

6.8 Hilbert's Nullstellensatz

7

Splitting Theorem and Lindel's Proof ......................175 7.1 Serre's Splitting Theorem .................................175 7.2 Lindel's Proof ...........................................180

8

.............................................185 ..............................................185

Regular Rings 8.1 Definition

8.2 Regular Residue Class Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .189 Associated Prime Ideals .................................. 194

8.3 Homological Dimension 8.4

.............................196 DedekindRings .........................................200 Examples ...............................................203 Modules over Dedekind Rings .............................205 Finiteness of Class Numbers ..............................209

8.5 Homological Characterization 8.6 8.7 8.8 8.9

xiv

9

Contents Number of Generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .213 9.1 The Problems ........................................... 213 9.2 Regular sequences

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .216

...................................221 Basic elements ....................................226

9.3 Forster-Swan Theorem 9.3.1

9.3.2 Basic elements and the Forster-Swan theorem

......... 227

. . . . . . . . . . . . . . . . . 228 . 9.4 Eisenbud-Evans theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 9.5 Varieties as Intersections of n Hypersurfaces . . . . . . . . . . . . . . . . 236 9.6 The Eisenbud-Evans conjectures . . . . . . . . . . . . . . . . . . . . . . . . . . .238 9.3.3

Forster-Swan theorem via K-theory

10 Curves as Complete Intersection ...........................241

10.1 A Motivation of Serre's Conjecture 10.2 The Conormal Module

........................241

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .243

10.3 Local Complete Intersection Curves ........................245 10.4 Cowsik-Nori Theorem ....................................248 10.4.1 A Projection Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .248 10.4.2 Proof of Cowsik-Nori

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .253

A

Normality of En in GL.

B

Some Homological Algebra B.l B.2

..............................251

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .257 Extensions and ~ x t ..................................... ' 257 Derived functors .........................................264

C

Complete intersections and Connectedness

D

Odds and Ends

E

Exercises

. . . . . . . . . . . . . . . .269

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .271 .

..................................................289

. References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .325 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .333 .

Basic Commutative Algebra 1

Here and in Chapter 6 we give an introduction to Commutative Algebra. This is the theory of commutative rings and their modules, i.e. abelian groups on which a ring operates as a field on a vector space. A ring in our sense is associative, but not always commutative. Upto one or two exercises, where it is expressively mentioned, a ring always is supposed to have a 1. (This does not excluude the zero ring, where 1 = 0.) Every ring homomorphism A -+ B has to map the 1 of A to the 1 of B. Consistently with this, we require a subring A of a ring B to have the same 1.

1.1 The Spectrum Throughout this section all rings are supposed to be commutative. We introduce the Zariski topology on the set of all prime ideals of the ring. Remember the notion of a prime ideal, which is fundamental for Commutative Algebra. A prime ideal p of a ring A is an ideal such that Alp is a domain, i.e. has no zero-divisors and is not the zero ring. (So A itself is not a prime ideal of A.)

Lemma 1.1.1 Let p be a n ideal of a ring A, different from A.

a) The following properties are equivalent: (1) p is a prime ideal; (2) f o r a l l a , b ~ Aw e h a v e a b ~ p * a € p o r b E p ; (3) for all ideals a, b of A one has ab c p

* a c p or b C p,

1 Basic Commutative Algebra 1

2

b) If p is a prime ideal, then a n b of A.

c p +a cp

or b

c p, for

all ideals a, b

c

Proof. a) (3) + (2) e (1) is clear. To prove (2) + (3), assume a p and b $ p.Then there are a E a \ p and b E b \ p. By (2) we have ab # p, whence ab 8.

c

b) follows from a) and the fact that ab

c a n b.

0

(Recall that by ab we mean the ideal generated by the set {ab I a E a, b E b), i.e. the set of all finite sums aibi with ai E a, bi E 6.)

xi

The converse of b) is not true: Every ideal of Z which is generated by a power of a prime number with exponent > 1 is a counterexample.

Remarks 1.1.2 a) Every maximal ideal of a ring A, i.e. a maximal one under all ideals different from A, is a prime ideal, since its residue class ring has only the 2 obvious ideals and hence is a field! It is an easy application of Zorn's Lemma, that there exist maximal ideals in every ring A p (0). More generally every ideal # A is contained in a maximal one. To apply Zorn's Lemma, it is enough to show that the union I of a chain of proper ideals (i.e. a set of ideals # A which is totally ordered by inclusion) is also a proper ideal. To see that it is proper, note that 1 # I. Make sure, you have understood the following: The union of a chain of ideals is an ideal, whereas the union of a general set of ideals need not be an ideal! b) By a minimal prime ideal we mean one which is minimal under all prime ideals. So in a domain there is exactly one minimal prime ideal, namely (0). Using Zorn's Lemma (in an 'order reversing' kind) we also easily see that every prime ideal q contains a minimal one. Namely, let {pi I i E I) be a chain of prime ideals, contained in q. Then pi is a prime ideal, since the complements of the pi form a chain of multiplicatively closed sets. So the set of all prime sub-ideals of q is inductively ordered by '<'='>'.

ni

As a corollary we get: If q is a prime over-ideal of an ideal I, i.e. q there is a minimal prime over-ideal of I, contained in q.

> I, then

1.1.3 Let X be the set of all prime ideals of a ring A. We have X # 0, if A 9 0. We define a topology on the set X called the Zariski topology. X together with this topology is denoted by Spec(A) and called the spectrum of A. The Zariski topology of Spec(A) is defined as follows:

If E c A, let

V ( E ):= {p E X l p

> E).

3

1.1 The Spectrum

(We write V(fi,. . . ,fT) := V({fl,. . . , fT)).) The subsets V(E) are defined to be the closed subsets of Spec(A). Note that (a) V(0) = V(0) = X, V(A) = V(1) = 0.

(c)

UyZl V(Ei) = Uy=l V(ai) = V(ny=, ai) = V(ny=, ai), where a; = (E;) denotes the ideal, generated by Ei.

Hence the collection {V(E) I E c A) - as the collection of closed set a topology (the Zariski toplogy) on X .

-

gives

I p = 0 or a prime number). V((0)) = X , V(mZ) = {pZ plm), for m E Z. So a subset of Spec(Z) is closed, if and only if it is either the whole set or a finite set of maximal ideals. Example 1.1.4 Let A = Z. Then X = {pZ

/

In every ring A a prime ideal is a closed point of Spec(A) if and only if it is maximal. But also the set of maximal ideals of a ring, equipped with the induced topology, in most cases is not Hausdorff, for instance in the case A=Z Remarks 1.1.5 Let us study some simple properties of Spec(A).

a) For f E A, let D(f) := {p E X I f $ p ) = Spec(A)\V(f). Then D ( f ) is an open subset of Spec(A), called the principal open subset associated with f . The collection {D(f) I f E A) is a base of Spec(A) because

b) With the help of these basic open sets it is easy to see that X = Spec(A) is quasi-compact: For, if X = UaEID(fa), then CaEIAfa = A. Clearly there are finitely many al,.. . ,a, E I and XI,. . . ,A, E A such that 1 = Cy=,Xi fai . And it is easily verified that X = U:=l D(fai). Now let (Ui)iEI be an open covering of X . Since every Ui is a union of principal open subsets, and finitely many of the latter cover X , also finitely many of the U; will cover X . c) Given a subset E of A we have associated with it a closed set V(E) c Spec(A). Conversely, given a subset Y c Spec(A) we can associate with it the ideal I ( Y ) = p. Both the correspondences

npEy

4

1 Basic Commutative Algebra 1

are inclusion reversing. One has V(I(Y)) = P the closure of Y: For, if K = V(J) > Y for some ideal J, then I(Y) c I ( K ) and V(I(K)) > V(I(Y)). But V(I(K)) = V(I(V(J))) = V(J). Hence K > V(I(Y)), i.e. V(I(Y)) is the smallest closed set containing Y and so is its closure. Later we will see that I(V(E)) = fi = {a E A 1 a n E a for some n) where a is the ideal, generated by E.

Definition 1.1.6 T h e maximal spectrum of a ring A i s the set of maximal ideals of A equipped with the induced topology of Spec(A). (Recall that the maximal ideals are prime.) T h e maximal spectrum of A i s denoted by Spmax(A) . 1.1.7 The next lemma (and not only that) will be better understood, if we interprete elements of a (commutative) ring A as 'functions' on its spectrum, namely we regard f E A as the 'function'

Especially f (p) = 0 means f E p. This is motivated by the following observations: Let k be a field and A := k[X1,. . . ,X,] the polynomial ring in n indeterminates (=variables) over k. Then there is a canonical injection

Namely the ideal (XI - al)A+. k-algebra homomorphism A

.+(X, - an)A is the kernel of the (surjective)

-+ k defined by Xi * ai,

hence maximal. And if, say, ai # bi, then Xi - bi clearly is not in the above kernel. So different n-tuples are mapped to different maximal ideals.

+ +

If a = (al,. . . ,an), write ma := (XI - al)A . . . (X, - a,)A. We will see later that for algebraically closed k every maximal ideal of A is of the form ma for some a E kn. (Hilbert's Nullstellensatz). But already here (for a general field k) we observe the following: For any f E A and a E kn we have:

Moreover f (a) maps to ( f mod ma) via the only possible k-algebra isomorphism k + Alma.

1.1 The Spectrum

5

Compare the following lemma to Urysohn's Lemma, which says that two disjoint closed subsets of a normal topological space can be 'separated' by a continuous real function on the whole space in the way that this takes the value 0 on one and the value 1 on the other subset. By the following lemma one can extend the zero function on a closed set V(I) of the spectrum of a ring to the whole space in such a way that it takes non-zero values on finitely many (not necessarily closed) given points outside V(I). (More generally a function on V(I) which is extendable to Spec(A) can also be extended in the way that it takes non-zero values on finitely many given points outside V(I).) The exact analogue of Urysohn's Lemma on two disjoint closed set is a direct consequence of the Chinese Remainder Theorem, which we will prove later in Section 1.5.

Lemma 1.1.8 (Prime Avoidance Lemma) Let I be an ideal and f an element of a ring A. a) Let pi, Pa, . . . ,Pr E Spec(A), such that f for some io.

+ I c Ui=, pi, then (f, I ) c pi,

b) Let pl, . . . ,pr be ideals of A, such that at most two of them are not prime. If I c Ui='=,pi, then I c pi, for some io. Proof. a) Choose a counterexample with minimal r. We must have r pi (t pj for i # j. We claim that f E n p i .

> 1 and

Suppose f 6 pi for some i. Then f +pi1 is disjoint from pi, so f +pi1 c Ujgi pj. By the minimality of r, we infer that (f, pi I ) c pj for some j # i. Since pi @ pj and pj is a prime ideal, this gives (f, I ) C pj, a contradiction. So f E n i p i . So also I c UI='=,pi. Hence the rest of a) follows, once we have shown b). b) Assume that I,pl, . . . ,p, form a counterexample with minimal r and that p3,. . . ,pr are prime ideals. Then we have I n pi (f Ujf pi, for every i = 1,... , r . Choose f i E ( I n p i ) \ n j f i p j . Then

since either p, is a prime ideal or r = 2. This contradiction proves b).

0

Remarks 1.1.9 a) In this book we will need part a) of the lemma as well as part b). There is an example of ideals al, a2, I and an element f in a ring, such that f I c al U a2, but f I (t ai for i = 1,2. (See Exercise 1.)

+

+

b) The simple looking Prime Avoidance Lemma will be useful in the context of several important theorems proved later. We will come across at least five important applications - the Serre's Splitting Theorem, the Bass Cancellation Theorem, the Forster-Swan Theorem, the Eisenbud-Evans Theorem, and H.

1 Basic Commutative Algebra 3

6

Lindel's ring theoretic proof of the Quillen-Suslin Theorem i.e. Serre's Conjecture. c) The Prime Avoidance Lemma is a typical argument known as a general position argument. By refining the method used in proving this lemma we will later derive the Forster-Swan bounds (on the number of generators of a module) and also the Eisenbud-Evans Theorems, from which all the previously mentioned theorems can be independently deduced. d) Let V be a vector space (of finite or infinite dimension) over an infinite field, U, Ul, . . . , U, be subspaces and X I , . . . ,x, be elements of V with U c U~=,((x; Ui). Then U c xi U; = Uifor some i. (You will find hints, how to prove this, in Exercise 2.) So if a ring A contains an infinite subfield, then Lemma 1.1.8 holds for arbitrary ideals pl,. . . , p,, not necessarily prime ones.

+

+

e) Lemma 1.1.8 frequently is applied in the following situation: Let al, . . . , a, be ideals and pl, . . . ,p, prime ideals of a ring such that a; @ pj for all i ,j . T h e n there is a n f E (a1 n . . . f l a,) \ (p1 U . . . Up,). Indeed set I := al n . . . n a, and use Lemma 1.1.1 b).

1.2 Modules Modules are 'vector spaces over rings', i.e. the concept of a module generalizes that of a vector space, replacing the underlying field by a general ring. The reader should not be deterred by the lot of - easy - concepts we will introduce here. It makes sense to get only a rough impression by the first reading and to repeat the details later, when they really are used. Also, since our deeper results concern modules only over commutative rings, the reader may well restrict his attention to this case, and so avoid several complications, appearing in the non-commutative case.

Definition 1.2.1 Let R be a (not necessarily commutative) ring. A left Rmodule M is a (here always additively written) abelian group M together with a map R x M -+ M, (a, x) I+ ax (the scalar multiplication), satisfying the following conditions.

+ y ) = ax + a y ; (a + b)x = ax + bx;

(1) a(x

a E R,

(2)

a, b E R, y E M

x, y E M

1.2 Modules

7

Remark 1.2.2 A right R-module is defined in nearly the same way. The only difference is that (c) is replaced by (3') (ab)x = b(ax).

In this case everyone writes the 'scalars' on the right side, i.e. xu instead of ax with x E M, a E R, so that (3') gets the form of an associativity law: (3') x (ab) = (xa)b. (But the essential difference between right and left modules is not the place where the scalars are written!) Over commutative rings there is no difference between left and right modules.

Examples 1.2.3 a) A vector space V over a field K is a K-module. b) An abelian group G is a Z-module. (For n E Z and x E G one defines nx := f(x . - . x) where the sum has In1 summands and the sign is that of n. Here we have used additive notation in G.)

+ +

c) Let V be a K-vector space together with an endomorphism a. Then (V, a ) becomes a module over the polynomial ring K[X] by f - v := f (a)(v). More explicitely let f = a i x i , then f v = ai - ai(v), where ai := a0 . . . o a , i-times.

xi

xi

d) The set Rn = R x . . . x R of n-tuples of elements of a ring R is an R-module in the same manner as K n is a vector space over a field K . Modules of this type, i.e. isomorphic to Rn will be called finitely generated free modules. (The meaning of 'isomorphic' should be clear, and is anyhow defined below.)

Especially R can be regarded as a left or as a right R-module. Its submodules (see below) are its left, resp. right ideals. The zero module which consists only of the zero element 0 is the special case n = 0 of the Rn. We will denote it by 0, not worrying that this symbol also denotes the zero element of every ring and module. e) If R and S are rings and cp : R + S is a ring homomorphism then S can be considered as an R-module as follows. Define the scalar multiplication by r s = cp(r)s for r E R, s E S. In particular we do so if R is a subring of S. f) In the situation of e) let M be an S-module. Then M becomes an R-module by r x := cp(r)x. g) Let I be a two-sided ideal of R and M an R-module. Assume I M = 0, i.e. cx = 0 for all c E I, x E M. Then M can also be regarded as an RII-module by hx := ax, since ax = a'x if a al(mod I ) . We often will do this tacitly! (Here and also elsewhere h denotes the residue class a I.

=

+

h) Let M be an R-module and R[x] the polynomial ring over R. By M[x] we mean the analogously defined 'polynomial module'. Its elements are formal sums

8

1 Basic Commutative Algebra 1

x 00

mixi, mi E M, mi = 0 for almost every ?:. i=O (By 'almost every' resp. 'almost all' in this book we always mean 'all but finitely many'.) The addition in M[x] is defined componentwise:

The scalar multiplication is defined analogously to that in R[x]:

Remarks 1.2.4 The most useful and peculiar property of vector spaces is the existence of bases and the uniqueness of their cardinality. So we have a very easy classification of vector spaces. But this is not at all so for general modules. On the other hand there are many definitions, theorems and proofs for modules which are verbatim the same as for the special case of vector spaces. Here is a list. a) The definition of a submodule is clear. The intersection of any family of submodules of a module is again one. The s u m of finitely many submodules Ul, . . . ,Un of a module M is

Of course one can also define the sum of an infinite family of submodules:

x {x / Ui :=

xi

Xi

E Ui, $6 = 0 for almost all i}.

Any sum of submodules clearly is a submodule (whether the sum is finite or infinite). b) If E is any subset of an A-module M , the submodule generated by E, is the intersection of all submodules which contain E. It is also the set of all linear combinations of elements of E, viz.

{x

nee / a, E A, a, = 0 for almost all e )

eEE

E is called a generating set (or a s e t of generators) of M , if the submodule generated by E is M itself.

1.2 Modules

9

c) A module M is said to be finitely generated if it has a finite set of generators {ml,. . . ,m,); i.e. M=

{x

aimi I ai E A).

M is called monogene or cyclic, if it has a generating set consisting of one element. If this element is x, we call x a generator. This is equivalent to M=Ax:={axla~A). d) A module M is called simple, if M # (0) and M and (0) are the only submodules of M . Clearly every simple module is monogene, every nonzero element of M being a generator. Note that the Z-modules 12 and Z / 4 both are monogene but not simple. e) A subset E of a module M is called linearly independent, if

x

a,e = 0 implies a, = 0 for all e E E.

eEE

(Here we assume tacitly a, = 0 for almost all e E E.) Otherwise we call it linearly dependent. The linear independence of E is equivalent to: xaee= eEE

x

bee

a, = b e for all e E E.

eEE

Definitions 1.2.5 a) A basis of a module is a linearly independent generating set of it. If E is a basis of M , then every x E M can uniquely be written as 2 = a,e, a, E A, a, = 0 for almost all c .

x

eEE

b) An R-module F is called free, if it admits a basis. Remarks 1.2.6 a) If R is neither the O-ring nor a skew-field (= division ring), then there will always exist nonfree R-modules. b) The module Rn is indeed a free R-module. A basis is formed by the elements el := (l,O, . . . ,O), ez := (0, l , O , . . . ,0), . . .

, en := (0,. . . ,O, 1)

as in the case of vector spaces. Conversely every free R-module with a finite basis is isomorphic to some Rn. c) If a non-zero ideal of a commutative ring R is free as an R-module, then necessarily it is a principal ideal, generated by a non-zero-divisor. Namely every pair (a, b) in R is linearly dependent, since ba - ab = 0.

10

1 Basic Commutative Algebra 1

Definition 1.2.7 If U is a submodule of an R-module M , then the factor group M/U is also an R-module, the so called factor module M modulo U. Namely if x E M and Z denotes the residue class of any z E M , then a 5 := a. This is clearly well defined. Remarks 1.2.8 a) Especially R / I is a left R-module if I is a left ideal of the ring R.

Now assume, R is commutative and I different from R and (0). Then R / I is not free, since no subset of R / I can be linearly independent because x E R I I , a E I \ (0) implies a x = 0. The case of non-commutative R is a trifle more difficult. b) Let M, U be as in the definition. The submodules of M/U are exactly those of the form V/U, where V is a module 'between' U and M , i.e. a submodule of M , containing U. c) Modules over a ring R often reflect the properties of the ring. For instance, if R is a principal (ideal) domain (a domain where every ideal is of type R a for some a E R), then also the finitely generated modules over R have a nice structure. For more details see the Section 4.1 on elementary divisors. Homomorphisms of modules, i.e. linear maps are defined as for vector spaces: Definitions 1.2.9 a) Let M and N be R-modules. A map f : M -+ N is called a n R-module homomorphism or R-linear i f

where x , y

E

M and a E R

b) The kernel ker(f) and the image im(f) of a module homomorphism f : M -+ N are defined to be the kernel ( f P 1 ( 0 ) ) ,resp. the image (f (M)) of f as a group homomorphism. (Clearly they are submodules of M , resp. N.) The cokernel coker(f) of f is defined to be N/im(f).

c) A n injective (i.e. one-to-one) linear map is also called a monomorphism, a surjective (i.e. onto) linear map an epimorphism. A bijective linear map is called an isomorphism. This is justified by the fact that the inverse map of a bijective linear map is linear as well. (Prove this! Continuous maps behave differently. The inverse map of a bijective continuous map of topological spaces need not be continuous!) Modules M , M' are called isomorphic i f there exists an isomorphism f : M -+ M'. d) A n endomorphism of a module M is a homomorphism M automorphism of M is a bijective endomorphism of M .

-+

M. A n

1.2 Modules

11

Examples 1.2.10 a) If R is a commutative ring, then for any a E R, the map ha : M -i M , given by scalar multiplication by a , i.e. ha(x) = ax is an Rmodule homomorphism. This map is often pompously called the homothesy of a; but really it is a simple thing!

Note that for a non-commutative R the homothesy of an element is not necessarily R-linear; for e.g. let M = R and a not commute with all elements of R. But, clearly if a belongs to the center of R, i.e. a commutes with all elements of R, then the homothesy of a on every R-module is R-linear. b) On the other hand, regard R as a left R-module and let a E R (not R, iz t+ xa is R-linear. necessarily in the center of R). Then the map R Further all R-linear maps of the left module R into itself are of this kind. c) More generally, if we consider Rm, Rn as left R-modules and write their elements as rows, then every R-linear map Rm -i Rn is given by v e va, where a E Rmxnis a suitable matrix with m rows and n columns. Conversely every such matrix gives us an R-linear map Rm -i Rn. (If you consider Rm, Rn as right R-modules and write their elements as columns, then every R-linear map Rm -i Rn is given by v e av, where cr is a matrix with n rows and m columns.) d) Another generalization of b) is the following: Let M be a left R-module and consider R as a left R-module. The R-linear maps R -i M are in bijective correspondence with the elements of M. Namely to x E M associate the map R -i M, a e ax, and to f : R -i M associate f (1) E M. The usual Homomorphism Theorem and Isomorphism Theorems hold. We state them here without proofs, since these are the same as for (abelian) groups. Proposition 1.2.11 (Homomorphism Theorem) Let f : M -i N be a module homomorphism and U c ker(f) be a submodule. T h e n f factorizes through the canonical map ts : M -i M/U, i.e. f is the compositions of linear maps

where f' is (well-)defined by fl(x mod U) := f (x). This is a typical case where we say that f ' is induced by f . Especially, f induces a n isomorphism M/ ker( f )

-i im(f).

Proposition 1.2.12 (First Isomorphism Theorem) Let U c V be submodules of a module M . T h e n canonical map M -+ M/V induces a homomorphism M/U -i M/V. And this induces a n isomorphism

12

1 Basic Commutative Algebra 1

Proposition 1.2.13 (Second, i.e. Noether Isomorphism Theorem ) Let U, V be submodules of a module M . Then the inclusion map V L, U V induces a n isomorphism v / ( u n v) 2 (U v)/U.

+

+

Remarks 1.2.14 a) Let M, N be R-modules. The set of all R-module homomorphisms from M to N is denoted by HomR(M,N). . It is an abelian group, addition defined by (f g)(x) := f (x) g(x).

+

+

If R is commutative, HomR(M,N ) has the structure of an R-module: (af ) (x) := a (f (x)). (If R is not commutative, the map a f : M -+ N defined by (af)(x) := a(f(x)) is not always R-linear. Also the trial to define (af ) (x) := f (ax) does not succeed.) Especially HomR(Rm,Rn) can be identified with the additive group of all m x n-matrices, addition defined componentwise. We write M * := HomR(M, R) and call it the dual of M. If M is a left module, then M * is a right module and vice versa. Namely for a E R, f E M*, x E M define (fa)(x) := f(x)a. For commutative R there is an isomorphism HomR(Rn,R) 2 Rn. Namely if one writes the elements of Rn as columns, then as usual the elements of HomR(Rn,R are written as one row matrices. This isomorphism is a special case of the map constructed in b). This isomorphism is not canonical in the following sense: Let F be a finitely generated free R-module. Then by different bases of F different isomorphisms Rn 2 F are defined. And these induce diflerent isomorphisms F*4 F. b) Note that if R is commutative and the R-module M finitely generated, say by ml, . . . ,m,, then every homomorphism f E M* is determined by the values f (mi). In this way M * can be regarded as an R-submodule of Rn via the injective R-homomorphism:

c ) Let M** := (M*)*, the bidual. The evaluation map M x M * -+ R, defined by (x, f ) ++ f(x) induces an R-linear map a : M -+ M**. If M is a free finitely generated R-module, a is an isomorphism. Note that here and also in b) the finite generation is essential.

d) Let f : M -+ MI be an R-module homomorphism and N another R-module. Then f induces group (resp. module) homomorphisms f * : HomR(M1,N )

+ HomR(M,N) and f,

: H o ~ R ( NM , ) -+ H O ~ R ( N MI) ,

1.2 Modules

13

by a I-, t r o f , and a ++f o a . Note the 'arrow reversing' in the first case. See also in the section on the Tensor Product for more properties of Hom. e) The endomorphism of an R-module M form a ring, the multiplication giving by composing homomorphisms: f g := fog. This ring is denoted by EndR(M). The ring EndR(Rn) can be identified with the ring Mn(R) of n x n-matrices over R. One must be cautious with the order of factors: Usually fog is defined by (fog)(%) := f(g(x)) (i.e. 'first g, then f'); now if R n is regarded as a left module of n-tuples, written in rows, and the homomorphisms are given by multiplying a row on the right side by a matrix, then Endn(M) is isomorphic to the opposite ring - i.e. where the multiplication is reversed - of Mn(R). 1.2.15 Now we go a step into the field of Homological Algebra. At first sight its results do not look of great importance, and its proofs are simple, or even automatic. (Therefore S. Lang in the first edition of his 'Algebra' suggested to 'take any book on Homological Algebra and prove all its theorems without looking a t the proofs given in that book.') But this first view is deceiving. Homological Algebra is a strong tool in many mathematical disciplines, as Algebraic Topology, Algebraic Geometry, Class Field Theory, Group Theory, . . . Definitions 1.2.16 a) A finite or infinite sequence of linear maps (i.e. a complex)

. - .+ Mn+l fqMn 3 Mn-l

-f

.-.

is called an exact sequence if im(fn+l) = ker(fn) for every n where this statement makes sense.

4

9

b) An exact sequence of the form 0 -+ M N -+ L-+ 0 is called a short exact sequence . Here exactness means: f is injective, g is surjective and im(f) = ker(g), i.e. M can be identified with a submodule of N via f , and L via g with the factor module of N modulo that submodule. Examples 1.2.17 a) Let U be a submodule of a module M then the sequence

is exact, if i , p are the inclusion map, resp. the canonical projection. b) Let m E Z \ (0) and h, : Z -+ Z be the multiplication by m, i.e. the homothesy of m, further n : Z -+ Z l m Z the canonical projection. Then the sequence

O + Z ~ Z A Z / ~ Z + O is exact.

14

1 Basic Commutative Algebra 1

f c) Let Ml -+ M2 4 M1 be an exact sequence of modules.

If M1 = 0 or more generally f = 0, then g is injective. If M3 = 0 or more generally g = 0, then f is surjective. Especially, the sequence

is exact if and only if f is an isomorphism. An example of a homological result is the so called Snake Lemma.

Lemma 1.2.18 Let

be a commutative diagram with exact rows. Then there is a homomorphism S : ker(al') -+ coker(al) such that the sequence

where the maps between the kernels and cokernels are the induced ones- is exact.

-

Moreover, if f 1 is injective, so is the induced map ker(al) -+ ker(a). And i f g is surjective, so is coker(a) -+ coker(aU). (We leave the explanation of the lemma's name to the imagination of the reader.)

Proof. We only define the connecting homomorphism S and leave the rest to the reader. So let x E ker(aU) c MI'. Choose a preimage y E M. Then a(y) E ker(g) has a unique preimage y' E N'. Let S(x) be the class of y' in 0 coker(al). It is well defined.

Remark 1.2.19 Let M be a finitely generated R-module, say generated by 21,. . . ,xn. Define fo : Rn -+ M by fo(a1,. . . , a n ) := Cy=,aixi. Then fo is surjective. Hence we get an exact sequence

1.3 Localization

15

Now assume that also ker fo is finitely generated, say by yl, . . . ,ym and define fl : Rm -+ Rn by fl(a1,. . . , a m ) := Ccl aiyi. Then clearly the sequence

is exact. One can continue this construction, provided the kernel of the last constructed map is finitely generated. If one allows to use infinitely generated free modules, then for any module M over every ring R one can construct an infinite exact sequence

with free R-modules Fi, a free resolution of M. It may happen that this process will "stop" with the zero module after some stage, in which case we say that the module M has a finite free resolution

We will see in Chapter 3 that this is the case for all finitely generated modules over a polynomial ring Ic[xl,. . . ,x,] with a field Ic. It is also true for modules over this ring which are not finitely generated. But we will not prove it.

1.3 Localization In this section every ring is supposed to be commutative. We duplicate the construction of the rational numbers from the integers. The rational number field Q is the field got from the ring of integers % 'by inverting all non-zero integers.' 1.3.1 Multiplicatively Closed Subsets Definition 1.3.1 Let R be a ring. A subset S of R is said to be multiplicatively closed or multiplicative if (1) 1 E S. (2)

sl,s2

E S implies sls2 E S.

Examples 1.3.2 a) If s E R, take S = {sn ( n E IN).

b) If p is a prime ideal of R, take S = R \ p.

16

1 Basic Commutative Algebra 1

c) If p, are prime ideals of R, take S = R \

U,

p,.

d) Let S be the set of all non-zero-divisors of R. e) If I is an ideal of R, take S = 1

+I.

f) Let (Si)iEx be a family of multiplicative subsets of R, the the intersection of the Si as well as their product Si, i.e. the set, generated multiplicatively by the union UiEx Si are multiplicative.

niEx

Lemma 1.3.3 (Krull) Let S be a multiplicatively closed subset and I an ideal of a ring R with S n I = 0. Then in the set of ideals a with I c a c R \ S exist maximal elements and these are prime ideals.

Proof. The existence of maximal elements in the above set of ideals follows directly by Zorn's Lemma. Let p be one of these. If x # p, y # p then there are sl E (x,p) n S, sz E ( y ,p) n S by the maximality of p. Let sl = Ax pl, s2 = py +p2, for A,p E R, pl,p2 E p. If x y E p then sls2 E p. But p r l S = 0 and so x y #p. 0

+

Definitions 1.3.4 a) The radical of an ideal I is the set

a:=

{a E R

I an E I for some n E N}.

It is easy to veri;fy - using the Binomial Theorem - that (This is not generally so for non-commutative rings.)

Jf is an ideal of R.

b) A n ideal is called a radical ideal i f it is its own radical. (Of course, = Jf and so the radical of any ideal is a radical ideal.)

fi

c)

is also called the nilradical of R and denoted by Nil(R).

d) A n element a of a ring R is called nilpotent if it belongs to Nil(R), i.e. an = 0 for a suitable n E IN. e) A ring R is called reduced if Nil(R) = (0).

Corollary 1.3.5 Let I be an ideal of a ring R. Then f i =

npEv(r) p.

Proof. Clearly is a subset of the right hand side ideal. In view of the canonical bijection between prime ideals containing I and prime ideals of R I I we may as well work in R I I . Hence we can assume that I = (0). Let a belong to the right hand side and let S = {an I n 2 0). If a were not nilpotent, i.e. ( 0 ) n S = 0, by Lemma 1.3.3 there would be a prime ideal p c R \ S. But since a belongs to every prime ideal, we would get a E p, a contradiction to a E S. Therefore, an = 0 for some m. Cl

1.3 Localization

Using Remark 1.1.2 b) we conclude that minimal prime over-ideals of I.

z/T

17

is also the intersection of the

Corollary 1.3.6 The correspondence a e V(a) gives a bijective correspondence between radical ideals of R and closed subsets of Spec(R). More explicitly, the map I from the set of closed sets of Spec(R) to the set of radical ideals which assigns to the closed set X the intersection I(X) := p provides the inverse map to the above correspondence as is evident from:

np,,

Corollary 1.3.7 If a is an ideal of R then I(V(a)) =

A.

1.3.2 Rings a n d Modules of Fractions

We now construct the ring of fractions (or localization) of R w.r.t. S and the module of fractions (or localization) of an R-module M w.r.t. a multiplicative set S: Define an equivalence relation on M x S by ( m )

-

there is an s" E S with s"slm = d'sm'.

( I ,s )

(If S consists of non-zero-divisors, this is equivalent to s'm = sm'. But the latter does not define an equivalence relation in general. It may lack transitivity. We will not restrict ourselves to the case where there are no zero-divisors in S. The important multiplicative complements of prime ideals often contain zero-divisors.) Define the analogous equivalence relation on R x S.

-,

Let S-'M M x S/

-

or Ms, resp. S-lR or Rs denote the set of equivalence classes resp. R x S/ under this equivalence relation. x The equivalence class of (x, s), x E M, resp. R, s E S will be denoted by S or fairly often - for reasons of typographical convenience - by x/s. We can make S-' M , resp. S-lR an additive group by defining addition m m' -+-:= S

s'

s'm

+ sm' ss'

for m, m' E M or in R, s, s' E S. This is easily checked to be well defined. We define a (commutative) ring structure on S-'R by defining multiplication as follows: r .r'_ . - _ rr' _ .S s' ss'

18

1 Basic Commutative Algebra 1

for r, r1 E R, s, st E S. The ring axioms can easily be checked. We make SP1M an S-I R-module analogously: r x rx .S sf SS' for r E R, x E M, s, sf E S. Note that, if M is a finitely generated R-module then S-I M is a finitely generated S-I R-module. 1.3.8 Notations: a) For f E R we write Rf := SelR and MI = SP1M where S = (1, f , f 2 , . . .). Also the notation R[l/f] is in use.

b) If S = R \ p, where p is a prime ideal of R, then S-I M is denoted by Mp, and S-lR by Rp. (We do not worry that this notation is not quite consistent with that of a) and c).) c) We often write Rs and Ms instead of S-lR, resp. S-lM. Especially, if S = 1 I we use the notations MI+^ and R I + ~ .

+

d) If S is the set of all non-zero-divisors in R, SP1R is called the total quotient ring of R or its full ring of fractions. It is denoted by Q(R). If R is a domain, then Q(R) is the quotient field of R. 1.3.9 Remarks and Definitions. a) Let M be an R-module and S c R be S-lR i and iM,S : M -+ multiplicative.There are canonical maps iR,S: R SP1M , defined by x H 7 for x E R or in M. In general these maps are neither injective nor surjective. We often write xs instead of :, especially when there are several multiplicative sets involved.

The map i,,, is a ring-homomorphism. Therefore S-lM is also an R-module. (See Example 1.2.3 f).) And iM,sis an R-module homomorphism. The kernel of i,,, consists of the x E M with sx = 0 for some s E S, analogously if M = R. So if 0 E S (or more generally, if s M = (0) fore some s E S), then S-I M E 0. b) Let f : M -+ N be an R-module homomorphism. We define the (clearly well defined) SP1R-module homomorphism

This is the unique R-module homomorphism making the following diagram commutative: M-N iWS1

f

s-lf

/iw

S-lM S-IN c) The property to be commutative for the above diagram means S-l f oiM,s = i N , Sf~. We think the reader will understand by this example the meaning of commutativity of any diagram!

1.3 Localization

19

Remarks 1.3.10 a) The assignments M t, S-'M and f t, S-'f form a so-called functor, i.e. S-'idM = idS-lM and S-'(gof) = (S-'g)o(S-' f ) , whenever g o f is defined.

b ) This functor is R-linear; which means that the map f

t+

S-'f

is R-linear. (Note that the 'Hom-group' on the right is an S-'R-module, hence an R-module via iRIs.) c) Every R-linear map S-'M -+ S-lN is also S-lR-linear, therefore H O ~ ~ - I ~ ( S - ' M , S - '= N )HornR(,!?-'M,S-IN), both as sets and S-'Rmodules. d ) There is a canonical SIR-linear map S-'HomR(M,N) -+ H O ~ ~ - ~ ~ ( S - ' M , S - ' which N ) , in important cases, but not always, is an isomorphism. We will discuss this later, when we introduce the notion of a finitely presented module. Lemma 1.3.11 (Exactness of Localization) If the sequence of R-module ho-

momorphisms M' f, M

3 M"

is exact, then the sequence

is exact as well. Proof. S-lgoS-' f = S - ' ( p f ) = 0 and so i m ( S - ' f ) c ker(S-lg). If x = m / s E ker(S-'g) then stlg(m)= 0 for some st' E S , i.e. g(st'm) = 0 and so 0 strm E ker(g) = i m ( f ) . Let stlm = f (m'). Then S-'f ( m l / s t t )= m . Corollary 1.3.12 Let S be a multiplicatively closed subset of R. Let

be an exact sequence of R-modules. Then

is exact. In particular if N is sub-module of M then S-lN can be regarded as a submodule of S-' M and S-' ( M / N ) E S-' M/S-'N canonically. Localization and taking quotients commute. In particular, if I is an ideal of R then S - ' ( R / I ) Z S-' R/S-'I as R-modules and also as rings canonically. firther S - l ( R / I ) S-'(RII), if 3 denotes 0 the image of S in R I I .

20

1 Basic Commutative Algebra 1

Corollary 1.3.13 Let S be a multiplicatively closed subset of R. Let M, N be R-modules and f : M + N be a R-linear map. Then ker(S-l f ) = S-I ker(f ), and im(S-' f) = S l i m (f ) . Proof. Apply Lemma 1.3.11 to the exact sequences O-+ker(f)-+M-$N

and

ker(f)+M-$im(f)+O.

Lemma 1.3.14 Let Ui for i E I be submodules of an R-module M . a) S-l

CiEIUi = CiEIUi.

b) If I is finite, then S-'

niEIUi = flier

S-I Ui.

If I = S is the set of positive integers and Ui = i Z , M = Z, then s-lui = Q and S-l ui = o.

niEI

niEI

Proof. a) Both sides of the claimed equality are submodules of S-I M . Note ui with ui that the elements of Xi,, S-lUi are essentially finite sums CiEI in S-'Ui, i.e. all but finitely many summands are 0. So there is a common denominator. Hence they belong to S-I Xi,, Ui. The converse inclusion is trivial. b) We may assume I = {1,2). Clearly S-l(U1 n U2) C S-lUi for i = 1,2, whence S-'(Ul n U2) c (S-I Ul) n (S-I U2). Conversely, let ul/sl = u2/s2 in l (S-lU1) n (S-lU2) with ui in Ui, si E S. Then there is an s E S with s s ~ u = ss1u2 =: v. Since v E Ul n U2, we have u l l s l = v/(ssls2) E S-l(U1 n 772). 0 1.3.3 Localization Technique

There are many properties of rings, modules and homomorphisms which can be checked 'locally'. This means, some property holds, say, for M if and only if it holds for Mp for all prime (or only all maximal) ideals p.

Lemma 1.3.15 Let M be an R-module. Then M = 0 if and only if Mp = 0 for all maximal ideals p of R. Proof. One direction is trivial. So assume M # 0 and take an x E M \ (0). The ideal Ann($) := {a E R I ax = 0), called the annhilator of x, is different from R, since l x # 0. So by Lemma 1.3.3 there is a maximal ideal p > Ann(x). It is enough to disprove Mp = 0. If it were so, in Mp we would have the equality x / l = 011, i.e. there would be an s E R \ p with sx = 0, contradicting Ann(x) c p. 0

1.3 Localization

21

Another illustration: Given an R-linear map f : M -+ N of R-modules M, N one can test whether it is injective (resp. surjective, resp. 0) by checking this locally. Corollary 1.3.16 An R-linear map f : M -+ N is injective (resp. surjective, resp. the zero map) if and only if f p : Mp -+ Np is injective (resp. surjective, resp. zero) for all maximal ideals p of R.

Proof. Let K = ker(f). One has the exact sequence

From Corollary 1.3.13 we get K p E ker(fp). Therfore we have the equivalences fp injective for all p. K p = 0 for all p f injective a K = 0 0 The other assertions follow in the same way. Corollary 1.3.17 Let U, V be submodules of an R-module M such that Up = Vp for every maximal ideal p of R. Then U = V.

Proof. U = V is equivalent to (U 1.3.14 and 1.3.15.

+ V)/U = (U + V)/V = 0. Use the Lemmas 0

Proposition 1.3.18 Let M be a finitely generated R-module and let p E Spec(R) such that Mp = 0. Then there is an s E R \ p with M, = 0.

Proof. Let 2 1 , . . . ,x, generate M . Since Mp = 0, there are s l , . . . ,s, E R \ p with six1 for i = 1 , . . . ,n. The element s := sl . . - s, fulfills the claim. 0 1.3.4 Prime Ideals of a Localized Ring

We next discuss the structure of prime ideals in a localized ring. One has the natural ring homomorphism i = iR,S : R -+ SP1R given by i(r) = r/1. Remember that i need not be injective and that ker(i) = {r E R I sr = 0 for some s E S). (The study of ker(i) plays an important role in Commutative Algebra - see [I] Theorem 10.17, Corollary 10.21). Lemma 1.3.19 Let p : R -+ S be a ring homomorphism. If p is a prime ideal of S then p-'(p) is a prime ideal of R.

Proof. S/p is a domain, and p induces an injective homomorphism 0 R/p-'(p) -+ S/p, whence Rip-'(p) is a domain, too.

22

1 Basic Commutative Algebra 1

Corollary 1.3.20 Let cp : R -i S be a ring homomorphism. Then cp induces a continuous map cp* : Spec(S) -i Spec(R) given by cp*(p) := cp-'(p) (= p n R if R c S).

Proof. I t is easy to check that c p * - ' ( ~ ( f ) ) = D(cp(f)),for any f E R. Theorem 1.3.21 There is a bijective correspondence between the prime ideals of S - l R and the prime ideals of R which do not intersect S , given by i* : p I-+ ii-'(p) = p n R.

Proof. By Lemma 1.3.19 the preimage i - ' ( p ) is a prime ideal of R. Since p is a proper ideal of S-'R, i-'(p) n S must be empty. I f p' is a prime ideal of R , then S-'p' is a prime ideal of S-'R as S - l ( R / p l ) is a domain, provided S rl p' = 0. One defines S-lR/S-'p' the inverse of the above map i* viz. i, : p' c, S-lp'. To show i* and i , are inverses of each other we need to show i-'(S-'p') = p', and S-li-'(p) = p, for primes p' of R , p of S-'R with p' n S = 0. LLi-lS-lp' = p'": x = a / s E S - l R , a E p', s E S implies st(sx-a) = 0 E p' for some st E S , and this implies sx - a E p' as st # p', whence x E p' (as x E R and s # p l ) .

<

"S-'i-'(p) = p": x = P E p, a E R , s E S implies sl(sx - a ) = 0 E p for some st E S , which implies sx - a E p as s' # p. Clearly sx E i - ' ( p ) and so a E i - I (8).Hence x = f E S-li-l ( p ) . 0 Corollary 1.3.22 Let R be a ring and f E R . Let further i : R Ri f be the natural map. Then the map i* : Spec(Rf) -i Spec(R) induces a homeomorphism Spec(Rf) D ( f ) .

) D ( f ) . The map j* : p I-+ p f Proof. By Theorem 1.3.21 we see i e ( S p e c ( R f )= from D ( f ) to Spec(Rf) is a continuous map as j * - ' ( ~ ( fg ) ) = D( f g ) for all g E R. Then j* is the inverse of i* : Spec(Rf)-i D ( f ) by Theorem 1.3.21 Corollary 1.3.23 Let R be a ring and let p, p l , .. . ,p, be prime ideals of R. Let S = R \ ULlp i . Then R p has a unique maximal ideal and S-I R has only finitely many maximal ideals, namely the S-'pi, 1 5 i 5 n. 0 Corollary 1.3.24 Let R be reduced. Then the set of zero divisors of R equals the union of minimal prime ideals.

Proof. Let p be a minimal prime ideal and a E p. Then pRp is the only prime ideal of R p , and so a l l in nilpotent in this ring. This means, there is an s E R \ p and an n 1 with sun = 0, but sun-' # 0. SO a is a zero divisor.

>

Conversely let a $ p for every minimal prime ideal p and ab = 0. But then b E p for every minimal prime ideal. Since R is reduced, this implies b = 0.

1.4 Integral Ring Extensions

23

Definition 1.3.25 A ring R which has a unique maximal ideal m is called a local ring and is sometimes denoted by (R,m). A ring which has only finitely m a n y maximal ideals is called a semilocal ring.

1.4 Integral Ring Extensions In this section all rings are supposed to be commutative. The ring of fractions is an over-ring of a ring which is intimately related to it. We introduce some over-rings, or ring extensions, which enjoy nice properties related to the ring. It is found that their study can often give back interesting information of the parent ring. 1.4.1 Integral Elements Definition 1.4.1 Let f : R + A be a ring homomorphism (by which we consider A as a n R-algebra). ( I n most cases R will be a subring of A and f the inclusion.)

a) A is called finite over R (or f is called finite), if A is finitely generated as a n R-module. b) A is called of finite type over R (or f is called of finite type), if A is finitely generated as a n R-algebra, i.e. zf and only i f f factorizes as

where K is the canonical imbedding into a polynomial algebra in finitely many indeterminates and g is surjective. Remark 1.4.2 If A is finite over R, it is clearly also of finite type, a generating set of A as an R-module being also a generating set of A as an R-algebra. On the other hand any polynomial R-algebra in finitely many (and at least one) indeterminates is of finite type, but not finite over R if R 0. Definitions 1.4.3 a) A polynomial f E R[X] is called monic, if it is not zero and its leading coeficient is 1, i.e. i f f is of the form

b) The annihilator of a n x E M is the set

1 Basic Commutative Algebra 1

24

Clearly Ann(M) and Ann(%) are ideals of R. (In the case where R is not necessarily commutative, Ann(M) is a two sided ideal, whereas the annihilator of an element x E M , i.e. Ann(%) is a left ideal.)

Proposition 1.4.4 Let R C A be a ring extension and a E A. The following statements are equivalent: (i) a is a zero of a monic polynomial f E R[X]. (ii) R[a], i.e. the subring of A, generated by R U {a), (which clearly consists of all polynomials i n a with coeficients in R) is finite over R. (iii) There is a ring B between R[a] and A, which is finite over R. (iv) There exists an R[a]-module M , which has a trivial annihilator and is finitely generated as an R-module.

Proof. (i)+ (ii). Any element of R[a] is of the form g(a) where g E R[X] is a polynomial. But since f is rnonic, in R[X] we can write g = qf g1 with deg(gl) < n. And then g(a) = gl(a), which shows that R[a] is generated by 1, a,. . . ,an-l as an R-module.

+

+ (iii). Set B = R[a]. (iii) + (iv). Set M = B. (ii)

+

(iv) (i). Let X I , . . . ,x, be a generating set for the R-module M. Let aij E R be so that axi = Cj aijxj. Then for the matrix ,B := (aij - Gijtr) we get

= 0 for in the module Mn. (Recall the meaning of Kronecker's Symbol # j , 6ii = 1.)By Cramer's Rule adj(P)P = det(P)I,. (By I, we denote the n x n unit matrix.) Therefore det(,B)xi = 0, whence det(P)M = 0 since the xi generate M. So det(P) itself is zero, M having a trivial annihilator. By the Leibniz Formula this gives an equation of the form i

the a, being combinations of certain aij, so lying in R.

Definition 1.4.5 In the above situation a is called integral over R, i f it fu&lls the equivalent conditions (i) to (iv). The ring A is called integral over R, i f every a E A is integral over R.

1.4 Integral Ring Extensions

25

Proposition 1.4.6 a) Every finite ring extension is integral. b) Let A C B and B C C be finite ring extensions. Then so is A

c C.

c) Let R c A be any ring extension. The set B of elements of A which are integral over R is a subring of A (clearly containing R).

Proof. a) This is clear by condition (iii). b) Let $1,. . . ,x, and y l , . . . ,yn be finite generating sets of the A-module B , resp. the B-module C. Then the xiyj with 1 5 i 5 m, 1 5 j 5 n generate C as an A-module. The reader will easily see this, if he remembers the proof of it in the case where A, B , C are fields. c) Let a, ,B E A be integral over R. We have to show that a f p, ap are integral too. They are so by a), since they are elements of R[a,P], which is finite over R by b). (Note that p is integral over R[a], if it is so over R.)

Examples 1.4.7 a) The ring Z + Z i of Gauss numbers is finite, hence integral over Z. b) More generally, let I be an ideal of the polynomial algebra R[X] over a ring R. If I contains a monic polynomial, then the ring extension R / ( I n R) c R[X]/I is finite, hence integral. Namely as an R-module it is generated by 1,x , . . . ,xnP1 where x denotes the residue class of X and n is the degree of a monic polynomial in I. 1.4.8 Note that a ring extension is finite if and only if it is integral and of finite type.

There are non-finite integral extensions, for e.g. Z c A, where A is the subring of C consisting of all complex numbers which are integral over Z . (If it were finite, A would lie in a finite dimensional subspace of the $-vector space C. But there are linearly independent subsets of arbitrarily big finite cardinality. Namely if a is a root of the polynomial X n - 2, which is irreducible by Eisenstein's Criterion, then a,a2. . . ,anare linearly independent over $.)

A ring extension A subextensions.

c

B is integral, if and only if B is a union of finite

Definitions 1.4.9 a) Let A c B be a ring extension. The ring of all elements of B which are integral over A is called the integral closure of A in B . b) Let A be a domain. The integral closure of A i n &(A) is simply called the integral closure of A. c) A domain which coincides with its integral closure is said to be integrally closed.

26

1 Basic Commutative Algebra 1

Examples 1.4.10 a) Every factorial domain A is integrally closed. Especially the polynomial ring Ic[xl, . . . ,x,] is so, if Ic is a field or a principal domain.

Namely let alb E Q(A) be integral over A with a, b mutually prime. We have an equation of the form

Multiplying by bn we get

So every prime factor of b is one of an, hence of a , since A is factorial. But b is prime to a; therefore b has no prime factor, i.e. alb E A.

+

b) The ring Z Zi of Gauss numbers is integral over Z and integrally closed, since it is a Euclidean domain, as one knows. Hence it is the integral closure of Z in $(i). c) In Algebraic Number Theory the integral closures of Z in general finite field extensions of $ are considered. These are not always principal domains. They are so called Dedekind rings . We refer our readers to Chapter 8 on this class of rings. To check the integrality of elements the following proposition is highly useful. Proposition 1.4.11 Let A be a n integrally closed domain and Q(A) c L a field extension. A n element a E L is integral over A if and only if it is algebraic and its minimal polynomial f over Q(A) has coefficients i n A.

By definition the minimal polynomial of a over &(A) is the - uniquely determined - monic polynomial f E Q(A)[X] of minimal degree with f (a) = 0.

Proof. If f fulfills the above condition, a is integral over A by Proposition 1.4.4 (i). Conversely, if a is integral over A, so clearly are its conjugates in the algebraic closure of L. And the (non-leading) coefficients of f are (the elementary symmetric) polynomials in the conjugates of a, so are integral over A and belonging to Q(A). Since A is integrally closed they even belong to A. The following two propositions will be of great use for us. Proposition 1.4.12 Let R c A be an integral ring extension and r E R be a unit in A. Then r is already a unit in R.

1.4 Integral Ring Extensions

27

Proof. Let a E A be an inverse of r. We have to show a E R. But a fulfills an equation of the form

Multiplying this equation by rn-' we derive a E R.

0

Proposition 1.4.13 Let A c B be an integral extension of domains. Then A is a field if and only if B is one.

Proof. Assume, B is a field. Then A is clearly one by Proposition 1.4.12. Now assume, A is a field. It is enough to show that A[/?] is a field for every ,b E B. But B' = A[/?]is a finite dimensional A-vector space and a domain. Multiplying by a non-zero element b of B' is an injective A-linear endomor0 phism of B', hence a bijective one. So there is a b' E B' with bb' = 1. 1.4.2 Integral Extensions and Prime Ideals

Proposition 1.4.14 Let A C B be an integral extension of rings and S a multiplicative subset of A, then S-lB is integral over S-lA.

Proof. Let x / s E S P 1 Bwhere x E B and s x is a zero of a monic polynomial, say,

E

S. But as B is integral over A,

with ai E A. But then

( x / s )+ ~ ( a l / s ) ( ~ / s ) ~+--'- - + (a,/sn) = 0. This shows that x / s is integral over S-'A.

0

Proposition 1.4.15 Let A c B be an integral extension of rings, I an ideal I7 of B and J = A n I. Then B / I is integral over A/J. Proposition 1.4.16 Let A c B be an integral extension. Let p be a prime ideal of B and p' = p n A. Then p is maximal if and only if p' is maximal.

Proof. By Proposition 1.4.15, B/p is integral over Alp' and both of these rings are integral domains. So by Proposition 1.4.13, Alp is a field if and only if B/pt is a field, i.e. p is a maximal ideal if and only if p' is one. 0 Corollary 1.4.17 Let A c B be an integral extension. Let p prime ideals of B. Then p n A # q n A.

5q

be diferent

28

1 Basic Commutative Algebra 1

If two different prime ideals of B do not contain each other they may have the same intersection with A.

Proof. Let S = A \ q nA. By Proposition 1.4.14, S-'A c SP1B is an integral extension. Clearly, S-lp 5 S-lq, and S-lp n A = S-l(p n A), S-lq n A = S-l (q nA). Therefore, it suffices to assume that A is a local ring with maximal ideal q n A. If p n A = q n A, then p would be maximal by Proposition 1.4.16; 0 hence p = q, a contradiction. Proposition 1.4.18 (Lying Over Theorem) Let A C B be a n integral extension and let p be a prime ideal of A. Then there exists a prime ideal p' such that p' n A = p.

Proof. By Propositon 1.4.14, Bp := (A \ p)-lB is integral over Ap. Now consider the diagram,

~i-B

where j is induced by the inclusion map i and a and P are the canonical homomorphisms. Let m be a maximal ideal of Bp, then n = m n Ap is a maximal ideal by Proposition 1.4.16, hence is the unique maximal ideal of the local ring Ap, i.e. m n A, = pAp. If p' = PP1(m), then p' is a prime ideal and 0 we have p' r7 A = a-l (n) = p.

Theorem 1.4.19 (Going-Up Theorem) Let A c B be a n integral ring extension,

PI 5 ... Pn be a chain of prime ideals of A and pi E Spec(B) lie over p1, i.e. pi n A = pl T h e n there is a chain

P:

S S -..S P:,

of prime ideals in B, with p: lying over pi. Proof. Using induction, we need only construct pi. The induced ring extension Alpl c, Blpi is integral and hence there is prime ideal of B/p: over p2/p1. This is of the form pi/p: with a prime ideal pi E Spec(B). That pi is the prime ideal, searched for. 0 Corollary 1.4.20 Let A c B be as above. Then: a) dim A = dim B . b) Let p E Spec(A).

1.4 Integral Ring Extensions

29

1. For every prime ideal p' of B over p one has dim(B/pl) = dim(A/p) and ht p' 5 ht p. 2. If ht p < m, there exists a prime ideal p' of B over p with ht p = ht p'.

Proof. a) By Corollary 1.4.17 a chain of n different prime ideals in B gives such a chain in A. Hence dim A 2 dim B. The converse inequality follows from the Going-Up Theorem. b) 1. is a consequence of Corollary 1.4.17 too, whereas 2. will follow from 0 Theorem 1.4.19. Theorem 1.4.21 (Going-Down Theorem) Let A C B be a n integral extension of domains, A integrally closed,

a chain of prime ideals of A and p; chain

E

Spec(B) lying over p,. Then there is a

s . - .s

s ~k

~k-1

of prime ideals i n B , with pi lying over pi. Proof. Going down step by step, we need only construct p;-,. Set So := A \ pn-1, Sl := B \ pk and S := So.S1. These are multiplicatively closed. From this the theorem follows. Namely by Krull's Lemma 1.3.3 there is a prime ideal pk-, E Spec(B) with p,-,B C pLp1 C B \ S . By n S1 = 0 we see pk-, c pk, and by pk-, n So= 0 we see pk-l n A = p,-l.

PROOF OF

THE

CLAIM: Assume there were an a E pnPlB f?S and let

be the minimal polynomial of cr over Q(A). (Since a is integral over A, it is algebraic over Q(A) .) We will show that ai E pn-l. Namely these are polynomials in the conjugates of a . And if B' arises from B by adjoining the conjugates of a, then the latter belong to p,-1 B', and so ai E A npn-, B'. Since B' is integral over A we have A n pn-,B' = pn-1; for pn-lB' is contained in any prime ideal of B' over Pn-1. Since a E S = So.S1, wecan write a = r s with r E So The minimal polynomial of s = a/r then is

c A,

s E

SI C B .

with bi = ai/ri Since s is integral over A we have bi E A. Then ribi = ai E pn-l and r E So = A \ pn-1 imply bi E pn-1. So Equation (1.1) says sn E pn-IB, which contradicts Sl n pn-,B = 0 as Sl is multiplicative. 17

30

1 Basic Commutative Algebra 1

Corollary 1.4.22 Under the assumptions of the Going-Down Theorem for 0 every prime ideal p of B we have ht p = ht(p n A).

1.5 Direct Sums and Products To study modules over a ring, the notion of addition and product of modules over a ring R is necessary and useful. Definition 1.5.1 a) Let M I , . . . ,Mn be finitely many A-modules. Their direct s u m (or direct product), written in two manners

is defined to be their cartesian product with the obvious operations making it an A-module:

a(xl,. . . ,x,) := ax^, . . . ,axn).

-

b) For an A-module M and n E N we write

M n : = M @ ...@ M = @ M ~ n-times i=l with Mi = M for all i. The latter is in accordance with the notation Rn above. Nevertheless there is a notational ambiguity. If I is a (left, right or two-sided) ideal of A, by In usually we denote the additive subgroup of A generated by all products bl . . . . .bn with bi E I (which will be a left, right, resp. two-sided ideal again). 1.5.2 Given a direct sum M = map' j : Mi

+ M,

x

I-,

eiMi for every i we have an 'inclusion

(0,. . . ,0, x, 0,. . . ,O), the x placed in the i-th position,

and a 'projection'

Clearly prioji = id^;. Write Mi' := ji(Mi). It is a submodule of M , isomorphic to Mi. We have M,! = M and Mi' n Cjfi M j = 0 for every i . The converse statement (in some sense) is:

xi

1.5 Direct Sums and Products

31

Proposition 1.5.3 Let Mi, .. . ,MA be submodules of a module M . The following statements are equivalent.

(2) The map n

is an isomorphism. Indeed (1) a) is equivalent to the surjectivity of the map in (Z), whereas b) is equivalent to its injectivity. 0 In this situation we say that M is the direct sum of the submodules MI, ... , M n and write

There is an easy but useful criterion for a module to be a direct summand of another module. Lemma 1.5.4 Let

N

~

M

~

N

be module homomorphisms with poj = idlv. Then j is injective, p is surjective and M = $(N) @ kerb). Proof. Since poj is injective, j is so. Analogously p is surjective. Let x E M , then x = jop(x)+(x- jop(x)). But jop(x) E j ( N ) and x- jop(x) E kerb), since ) = p(x) -p(x) = 0. So M = j(N) +kerb). At p(x - jop(x)) = ~ ( x-pojop(x) last let x = j(y) E j(N)nker(p), then 0 = p(x) = poj(y) = y. S o x = j(0) = 0, 0 whence j(N) n ker(p) = 0. Corollary 1.5.5 Let

be an exact sequence of modules. The following statements are equivalent: (1) There is a homomorphism q : M

+ M'

with q o i = id^,;

(2) There is a homomorphism j : M" +- M with poj = idM" ;

32

1 Basic Commutative Algebra 1

(3) There is a commutative diagram

where the middle vertical arrow denotes an isomorphism.

0

Definition 1.5.6 The exact sequence of the Corollary 1.5.4 is called split exact i f the equivalent properties (i) to (iii) hold. Also i n this situation we say that the sequence splits. Further we say that an epimorphism p (resp. a monomorphism i ) splits i f it fits i n a split exact sequence. Now let (Mi)i,g be a (not necessarily finite) family of modules. We generalize the above definition in two different ways.

Definitions 1.5.7 a) The direct sum

is defined to be the set of those families (xi)iEI where xi = 0 for nearly all, i.e. for all but finitely many i E I . b) The direct product

is defined to be the set of all families ( x ~ ) ~ , = I I n both cases the module structure is defined as above. c) Let I be some set and M a module. Then

M(') := @ Mi

and

iEI

where Mi = M for all i E I . Note that an A-module M clearly is free, if and only if there is a set I , such that M r A(').

1.5 Direct Sums and Products

33

R e m a r k s 1.5.8 a) Proposition 1.5.3 generalizes without difficulties to infinitely many direct summands. So if (M,!)iE~is a family of submodules of a module M which fulfills the equivalent statements of Proposition 1.5.3, we say that M is the direct sum of its submodules Mi, i E I and write

b) A direct summand of a free module, i.e. a module P such that P @ Q is free for some module Q , will be called p r o j e c t i v e . The study of projective modules - beginning in Chapter 2 - is one of the themes of our book. There do exist nonfree projective modules, but our central theorem will be that projective modules over polynomial rings XI,. . . ,Xn] over a field k are actually free. This was called "Serre's Conjecture" before it was proved, and is sometimes called so today.

D e f i n i t i o n 1.5.9 Let (Ai)iEI be a finite or infinite family of rings. Their direct product

is defined to be, as a set their Cartesian product, with component-wise addition and multiplication - analogously to the addition in Definition 1.5.1. I n the case of a finite I = (1,. . . ,n) we also write

R e m a r k s 1.5.10 a) Let I be infinite and Ai P 0 for all i . Then the set of families {(ai)iEII ai = 0 for almost all i ) lacks the 1-element and so is no ring in our sense. b) Some people call - for finite I - the direct product of rings a direct sum and use the notations $ or @. But one of the authors heavily dislikes this, since the direct product of rings is indeed what usually is called the direct product in the category of rings, whereas the direct sum in the category of rings would be the tensor product (over Z).

D e f i n i t i o n 1.5.11 Two-sided ideals I, J of a ring A are called c o m a x i m a l , if I + J = A or equivalently i f there are a E I , b E J with a + b = 1. R e m a r k s 1.5.12 a) For commutative A this means that I,J are not contained in a common prime ideal, in other words that V(I) n V ( J ) = 0.

34

1 Basic Commutative Algebra 1

b ) Let I , Ji be comaximal for i = 1,2, then I , J1J2 are comaximal. Namely let ai E I , bi E Ji such that a1 bl = a2 b2 = 1. Then 1 = (al + bl)(a2+ b2) = (ala2 +alb2 blaz) blb2 E I + J1J2.

+

+

+

+

Consequently, i f I , J are comaximal, so are I n , Jm. Lemma 1.5.13 (Chinese Remainder Theorem) Let R be a ring, Il , . . . , In

+

be two-sided ideals of R and q : R

n n

R / I i be the natural homomorphism i=l defined by q(r) = ( r + 11,. . . ,r + I,). Then q is surjective if and only if the 11,.. . ,In are mutually pairwise comaximal. In this case we have

If the Ii pair-wise commute, especially if R is commutative, then we have further that Il . . .In = Il n - - . n I, in the above case. Proof. Assume Ii

+ Ij

i # 1, and let rl =

= R for i

nyzl

# j. Let 1 = ui + vi, ui

n

= n ( l - ui) = 1

E I l , vi E Ii for

+

u for some u E 1 1 . Then i=2 q(r1) = el := (1 + I l , 0 + 12,.. . ,0 + I,). Analogously, ei E im(q) for all i , and so q is onto. Vi

Conversely, assume q is onto. I f q(ai) = ei then 1 - ai E Ii and ai E I j i f j # i. Therefore, 1 = (1 - ai) + ai E Ii + I j . Clearly ker(q) = Il n . . - n In. We prove the last assertion by induction on n , the case n = 1 being trivial. I f I + J = R we compute

~n J =

( I + J ) ( I nJ ) = I ( I n J )

+ J ( I n J ) c IJ,

whence I n J = I J . The induction step is

Compare this lemma to Urysohn's Lemma, which says that for any pair V ,W of disjoint closed subsets of a normal topological space X , there is a continuous real function being 0 on V and 1 on W . By the above lemma (in the commutative case) for closed sets V ( I ) ,V ( J ) with V ( I )n V ( J ) = 0 we find an f E A with f 1 mod I , f 0 mod J .

35

1.5 Direct Sums and Products

Definition 1.5.14 An element e of a (not necessarily commutative ring) is called idempotent if e2 = e. Remark 1.5.15 Let A = Al x .. . x An be a finite direct product of rings and ei := (0,. . . ,0,1,0,. . . , O), where the i-th component is the 1. Then n

(i) e: = ei

(ii) eiej = 0 for i

#j

(*)

e =1

(iii) i= 1

Further the ei belong to the center of A. There is a converse.

Proposition 1.5.16 Let e l , . . . , e n in the center of a ring R be finitely many elements fulfilling (*). Then the ideals Rei are also rings with I-element e; and the map

is an isomorphism of rings. Proof. Since ei is idempotent, (aei)ei = aei. So ei is the 1-element of Rei and Rei is a ring. It is not hard to show that the map qi : R t Rei, a + aei is a surjective ring homomorphism. Using (ii) and Ciae; = a, which holds by (iii), one shows ker(qi) = CjitiRej. Therefore the different ker(qi) are mutually comaximal. On the other hand ker(q1) n . . . n ker(qn) = (0), since a E ker(qi) for all i implies aei = 0 for all aei = 0. So by Lemma 1.5.13 the proposition follows. i, hence a =

xi

1.5.17 An analogous results holds for modules. Let M = M1 a finite direct sum. Let

ei : M t M,

.- .

Mn be

( X I , .. . ,xn) +-+(0,. . . ,0, xi, 0 , . . . , 0),

the xi standing in the i-th place. Clearly the e l , . . . ,en are endomorphisms satisfying (*), where (iii) is to be read e; = idM. (The ei need not belong to the center of the endomorphism ring of M.) Conversely,

xi

Proposition 1.5.18 Let M be a module and e l , . . . ,en be endomorphisms of M satisfying (*). Then n

M =@ei(~). i=l

36

1 Basic Commutative Algebra 1

xi

Proof. From (iii) we get e i ( M ) = M . Let x E e i ( M ) n ( C j f e i ( M ) . We have to show x = 0. From x = ei(y) we get e i ( x ) = e:(y) = e i ( y ) = x . From x = CjZie j ( y j ) we get e i ( x ) = C j f ieiej(yj) = 0. Hence x = 0. 0 The theory of modules over a finite direct product of rings can be completely reduced to the theories of those over the individual factors. This is the content of the following theorem. Theorem 1.5.19 Let A1 , . . . ,A, be rings and A = A1 x

. . . x A,. Let further e l , . . . ,en E A be the idempotents defined by this splitting as above. Then any A-module M admits the splitting M = MI

@

- .- $ M,, where Mi

:= eiM.

Mi is an Ai-module in an obvious way. Conversely if Mi is an Ai-module for i = 1 , . . . ,n then the group Ml @ . . . $ M, is an A-module in a canonical and obvious way.

ei

eiU. Conversely, if Ui is an For any A-submodule U of M we have U = Ai-submodule of Mi, then U := Ul+- - .+U, = Ul @. - .@U, is an A-submodule of M . Further M / U Z Ml/Ul @ - . .@ Mn/Un. I f f : M + N is an A-module homomorphism and Ni := e i N , then f ( M i ) c Ni, so that

Proof. Note that the ei belong to the center of A, so that multiplication by them are endomorphisms. Apply Proposition 1.5.18. Let a E Ai and x E Mi c M . Then the only sensible definition of ax is ax := ( 0 , . . . ,0, a, 0 , . . . , 0 ) x , a in the ith place. Since eiU c U by the definition of a submodule we get for U the analogous splitting as for M . We leave the rest to the reader. Note 1.5.20 Consider any non-zero module E. Then the diagonal submodule D := { ( x ,x ) I x E E ) of E @ E is not of the form e l D $ e2D, where ei are the endomorphisms of E @ E , defined by el ( X I , 2 2 ) = ( X I , O), e2 ( X I , 2 2 ) = (0,x2). In general the submodules of direct sums of modules cannot easily be derived from submodules of the individual summands. This is different in the case of ideals in a direct product of rings. Corollary 1.5.21 Let A = A1 x - .. x A, as in Theorem 1.5.19 and I a left (resp. right, resp. two-sided) ideal of A, then I = I1 x . . - x I,, where I j is a suitable left (resp. right, resp. two-sided) ideal of A j for j = 1 , . . . ,n. Conversely such a product is an ideal of A . In this situation there is a canonical 0 isomorphism A / I E ( A 1 / I l )x . . . x (A,/I,).

1.5 Direct Sums and Products

37

Proposition 1.5.22 let A = Al x ... x A, be a commutative ring. Then every prime ideal of A is of the form

p = Al x - . . x Ai-1 x pi x Ai+1 x - . -x A,,

for some i E { I , . . . ,n ) and some pi E Spec(Ai). Conversely every ideal of this form is prime. So Spec(A) may be identified with the disjoint union of the Spec(A;). Moreover it is the 'coproduct' of the Spec(Ai) i n the category of topological spaces; i.e. a subset U c Spec(A) = Ur=, Spec(Ai) is closed i n Spec(A), if and only i f U n Spec(Ai) is closed i n Spec(Ai) for every i. Proof. Note that every product of rings, where more than one factor is nontrivial, has zero divisors # 0. (In B x C we have (1,0)(0,1) = (0, O).) Let I = Il x . . . x Inbe anideal ofA. Then A / I E (A1/Il) x - - . x (An/In) is a domain, if and only if Aj = Ij for all but one j , say i and Ai/Ii is a domain, i.e. Ii is a prime ideal of Ai. We prove the last assertion. A subset T of Spec(A) is closed, if and only if it is of the form T = V(I) for some ideal I = Il x - .. x I, of A. But clearly T n Spec(Aj) = V(Ij) in Spec(Aj). 0 So every decomposition of a commutative ring A = B x C gives rise to a partition Spec(A) = Spec(B)USpec(C) where Spec(B) and Spec(C) are closed and open in Spec(A) and mutually disjoint. There is a converse. Proposition 1.5.23 Let A be a commutative ring and Spec(A) be the disjoint union of two closed (hence open) subsets Ul and U2. Then A can be decomposed: A = A1 x A2 with Spec(Ai) = Ui.

Proof. Since the Uj are closed there are ideals 11,I2of A with Uj = V(Ij). Since UI r l U:! = 0, we have I1 I 2 = A. So there are a j E Ij with a1 a2 = 1. Now, since Ul U U2 = Spec(A) we have Il n I2c Nil(A) = Therefore ~ suitable n. There is no prime ideal from a1a2 E I1 n I 2 we get ( ~ 1 ~ =2 0)for containing both a1 and a2, because a1 a2 = 1. Hence there is also no prime ideal containing a? and a;. So the ideals (a?) and (a;) are comaximal, whence V(a;l) n V(a;) = 0. Since (ay) c I j , whence V(a7) 2 Uj, we get V(a7) = Uj. Now we apply the Chinese Remainder Theorem, to see

+

+

m.

+

(A/(ay)) x (A/(a,")

E A/(aT)

n (a,") = A/(aTa,") = A. 0

Corollary 1.5.24 Let A be a commutative ring. The following statements are equivalent:

(i)

Spec(A) is connected.

(ii) If A S A1 x A2, one of the rings Al, A2 is a zero ring. (iii) A possesses no idempotent elements but 0 and 1.

0

38

1 Basic Commutative Algebra 1

Corollary 1.5.25 Let A have at least one of the following properties: (i) A is local, (ii) A has only one minimal prime ideal.

Then A is not directly decomposable, i.e. there are n o non-zero rings Al, A2 with A A1 x A2. 0

1.6 The Tensor Product The category of modules over a ring has a much more natural concept of product, in a categorical sense, called the tensor product.

1.6.1 Definition Let R be a ring (not necessarily commutative), M a right and N a left Rmodule. A map f : M x N -+ P where P is an abelian group is called a balanced m a p if

where a E R, m, ml ,ma E M and n, n l , n2 E N . Given a right R-module M and a left R-module N , there exists a unique (upto a unique isomorphism) abelian group M @R N , (sometimes also written M @ N), called the tensor product of M and N over R, and a balanced map $ : M x N + M @ R N with the 'universal property' that for any balanced map f : M x N + P with any abelian group P, there exists a unique group homomorphism g : M @ N + P such that go$ = f , i.e. the following diagram commutes:

M x N

c P

1.6 The Tensor Product

39

We show now that a tensor product of a right R-module M and a left Rmodule N over R exists. Let F be the free abelian group (i.e. free Z-module) with the set M x N as basis (i.e. F E B ( ~ ~and ~ let ) H ) be the subgroup of F generated by all elements of the types (x + Y, 2) - (2, z) - (Y,z), (x,z + t) - (x, z ) - (x, t), (xu, Y)- (x, ay), w h e r e x , y ~M, z , t € N a n d a € R . L e t M @ R N : = F / H a n d c p : M x N + M @R N be the composite M x N + F -+ M @R N , where F -+ M @R N is the canonical epimorphism. Then, it is easily seen that M @R N is a tensor product of M and N over R. The uniqueness is a formal consequence of the universal property: Let t) : M x N -+ M @R N and $' : M x N + M @& N be two tensor products with the accompanying balanced maps. By the universal properties of .II, and $I there are unique group homomorphisms a : M @R N + M @& N and @ : M @ & N-+ M @ R Nwith a o $ = $' and Po$' = $. Then (@oa)o$ = idM8N~t). By the uniqueness claim in the universal property we get @ o a = id^^^. Analogously we have a o @ = idM@,N.SO a,@are isomorphisms and unique. The tensor product with the accompanying balanced map is unique up to a unique isomorphism. For (x, y) E M x N,we denote its image under cp in M @R N by x @ y. (These are called the p u r e tensors). Note that any element of M @R N is of the form E x i @ yi (which are called mixed tensors, with xi E M, y; E N). But i

this description is not at all unique! This non-uniqueness is the main source of difficulties, when using the tensor product. Therefore we give here recipes and rules how to work with it. R e m a r k 1.6.1 How to show

Answer: Use successively identities of the following types

to convert one side in (*) into the other. Conversely, the existence of such a series of transformations is necessary for the equation (*). But since it is not easy to show that such transformations do not exist, here is another recipe: How to show

40

1 Basic Commutative Algebra 1

Answer: Find an abelian group G and a balanced map p : M x N that

+ G, such

+

Examples 1.6.2 a) Always x @ 0 = 0 @ y = 0. Namely x @ 0 x @ 0 = x @ (O+O) = x @ O . Consequently M @ 0 O 0 8 N N 0. (Note that the 0 above has six different meanings: the 0-element of M, N, M @ N, the trivial left, right module or group.)

Another consequence is (-2) @ y = -(z @ Y) = x @ (-Y). b) $/Z @a $ / Z

0.

Namely let x, y E $/Z. There is an n E Z \ (0) with nx = 0 and a y' E $ / Z withny'=y.Thenx@y=x@nyl=n~@yl=O@y'=O. c) The balanced map R x M -+ M given by (a, x) e ax induces a canonical M. isomorphism R @ R M

>

d) More generally, let I be a right ideal of R. Then the balanced map

induces a group isomorphism (R/I) @ R M M/IM. (Here the overbar denotes the residue class, and I M is the subgroup of M generated by all ax with a E I and x E M. If I is a two-sided ideal, then I M and hence M / I M are R-modules.)

>

e) Let R be commutative and S c R be a multiplicative subset. Consider the map a ax S-'R x M +S-'M, (;, x) e -. S

It is well defined and balanced, and hence induces a homomorphism f : ( S I R ) @ R M -+ S-'M.

CLAIM: f is an isomorphism. An inverse map is g : S-I M -+ (SP1R) @ R M , defined by m/s e (11s) @ m. This is well defined. Namely if m/s = m1/s', there is a t E S with tslm = tsm'. So (11s) @ m = (llsts') @ ts'm = (l/slts) @ tsm' = (l/sl) @ m'. Further gof ((als) @ m) = (11s) @ am = (als) @ m. Clearly fog = ids-,,. Since (SP1R) @ R M is generated by elements of the form (als) @ m, we see that g is inverse to f.

1.6 The Tensor Product

41

Remarks 1.6.3 a) Let f : R -+ S be a ring homomorphism and M a left R-module. Then S is a right R-module by s r := sf (r). Hence we can form the tensor product S @R M , and this is an S-module by a(b @ m) := (ab) @ m for a, b E S. Note that for a general element C si@mi E S @ R Mthis means a ( C si@rni)= C(asi) @mi.Using the above remark, one sees easily, that this is well-defined. If I is a twosided ideal of R, the isomorphism (R/I) @R M 7 M / I M of example d) above is R/I-linear. The same holds for the isomorphism f in example e). b) Now let R be commutative and M, N any R-modules - right or left having no different meaning here. Then M @R N is an R-module by r(m @ n) := (rm) @ n = m @ (rn). (Note, that this does not work for noncommutative R. Namely the attempt to define r(m @ n) := m @ (rn) does not make much sense, since for r, s E R this would imply m @ rsn = r(m @ sn) = r(ms @ n) = ms @ r n = m @ srn. But in general the equation m @ rsn = n @ srn does not hold in M @R N.) For commutative R the canonical map $ : M x N -+ M @R N is not only balanced, but even bilinear, i.e. $(rm, n) = $(m,rn) = r$(m, n). And we may define the tensor product by a universal property with respect to bilinear instead of balanced maps. Also we may prove its existence - over a commutative ring R - to be the factor module F/U, where F is a free R-module with basis M x N and U is generated by all elements of the form (m m', n) - (m, n) (m1,n), (m,n n') - (m,n) - (m,nl), (rm,n) - r(m,n), (m,rn) - r(m,n).

+

Finally we clearly have a canonical isomorphism M @ N is commutative.

+

E

N @ M , when R

c) The module structures on S @R N , resp. M @R N in a) and b) are special cases of the following: Let M be an S-R-bimodule, i.e. an abelian group, which has as well a left S- as a right R-module structure, with (sm)r = s(mr). Then M @R N is an S-module by s(m @ n) := (sm) @ n. d) The balanced map (M @ N) x P -+ (M @R P ) @ (N @R P ) given ky ((x, y ), z) t-, (x @z,y @z)induces a canonical R-isomorphism (M @ N ) @ P -+ (M@P)@(N@P). This holds also for infinite direct sums: we have a canonical isomorphism

(The analogue for infinite direct products does not generally hold.) Especially we have: Let F be a free R-module with basis {eijiG1and let R -+ S be a ring homomorphism. Then S @R F is a free S-module with (1 @ ei)iE1 as a basis.

42

1 Basic Commutative Algebra 1

e) Let R[x] be the polynomial R-algebra. Then we have a canonical R[x]-1'inear isomorphism R[x] @R M E M[x] for every R-module M. 1.6.2 Functoriality

Definition 1.6.4 Let R be a ring, M, M' right, N, N' left R-modules, further f : M + M' and g : N + N' be R-linear maps. These induce a diagram

Since @o(f x g) clearly is balanced, by definition of the tensor product there is a unique group homomorphism f @ g : M @R N -+ M' @R N which makes the above diagram commutative - i.e. (f @ g)o$ = @o(f x g). We say, f

@g

is induced by f and g.

Sometimes we write M

@ g :=

id^

@ g.

Remarks 1.6.5 a) We have id^ @ idN = idMBNand idM @ (g'og) = (idM @ gt)o(idM@g),whenever g'og is defined. This means, for a given right R-module M , the assignments N H M @R N, g H idM @ g form a so called functor from the category of left R-modules and R-linear maps to that of abelian groups and group homomorphisms. One may write this functor as M@R?. Further the map HomR(N, N')

+ Homz(M @ N, M @ N'),

g H idM @ g

is a group homomorphism. This is expressed by: the tensor product is an 'additive functor'. If R is commutative the above map is even R-linear: The tensor product M@R?is an 'R-linear functor' for commutative R. Analogously these statements hold if one interchanges the roles of M and N. b) For f : M tive:

+ M',

g :N

+ N'

as above, the following square is commutaf id^

M@N-M1@N

This is expressed by the phrase, that the tensor product is a 'bifunctor'.

1.6 The Tensor Product

43

1.6.3 To what Extent do Horn and @ Preserve Exactness

The assumption that M'

'M 'M"

is an exact sequence of modules, does not generally imply the exactness of either of the following induced sequences:

Hom(MU,N ) -+ Hom(M, N ) -+ Hom(M1,N),

N@M1-+ N @ M -+ N@M"

For example let the ring be Z , N := Z / 2 Z and consider the exact sequence Z -+ Z / 2 Z -+ 0 in the first, resp. 0 -+ Z 3 Z in the second and third case. (By -2 we mean the homothesy of 2.) Here we will show that partial results do hold.

Proposition 1.6.6 a) Let MI+

M -+ M" 0'

resp. O + N 1 + N

+N"

be exact sequences of R-modules. Then the induced sequences

resp.

0 -+ Hom(M, N') tHom(M, N ) tHom(M, N")

are exact as well. This is the so called left exactness of the Horn-functor. b) Moreover, if for a (not necessarily exact) sequence

the induced sequence

is exact for every module N , then (*) is exact. Proof. a) If one understands the meaning of the hypotheses and the assertions, the proof is automatic. b) First set N := M1'/im(g). Then the canonical map n : M" -+ N belongs to the kernel of g*. So the injectivity of g* implies the surjectivity of g. Secondly set N := MI'. Then f *og* = 0 implies gof = f *og*(idM") = 0. At last set N := M/im(f) and let n : M -+ N be the canonical map. Then f*(n) = nof = 0, i.e. n E ker(f*). So by hypothesis there is a linear map X : MI1 -+ N with n = Xog. Now if x E ker(g), then n(x) = 0, i.e. x E im(f).O

44

1 Basic Commutative Algebra 1

There is a useful connection between the tensor product and the Hom-functor. Namely there is a canonical isomorphism Homn(M

@R N,

G)

7HomR(N,

Homz(M, G)).

(Here G is an abelian group, M is a right, N a left R-module.) The principal idea how to construct this, is that a balanced map M x N -+ G means: every x E N gives - in an R-linear way - a B-linear map M -+ G. We leave the details to the reader. Note that HomZ(M, G) becomes a left R-module, if M is a right one. The isomorphism f is 'natural', which means that it is compatible with maps M -+ M', N -+ N' and G -+ G': The induced squares are commutative; for e.g. the induced square

is commutative. Let f : N -+ N' be an epimorphism, then one easily shows that M @ f is so too. One needs a bit more effort to prove the following. Proposition 1.6.7 Let

be exact. Then the induced sequence M@N'+M@N+M@N1'-+O

is exact too. This is called the right exactness of the tensor product. Proof. Assume that M1+M-+M''+O is exact. By the left exactness of the Hom-functor we get that

is exact, and hence

is exact for every abelian group G. So the sequence M'@~N--+M@~N---+M"@RN+O is exact.

1.6 The Tensor Product

45

Let us give a first application.

D e f i n i t i o n 1.6.8 A n R-module M is called finitely p r e s e n t e d , i f and only if there are m , n E IN and an exact sequence of module homomorphisms

The right exactness of the tensor product enables us to describe M @ R N for a finitely presented right R-module M as a cokernel of a group homomorphism which can easily be described. Namely from an exact sequence as (*) we get the exact sequence of abelian groups

where a' is given by the same matrix as a. (Note that the matrix operates from the left, since M is a right module and so Rm, Rn are regarded as right modules.) If one allows infinite matrices, one can do the same for general modules. 1.6.4 Flat A l g e b r a s

D e f i n i t i o n 1.6.9 A module E is called flat, if and only if tensoring with it is exact. This means: M'

+ M + MI'

exact

M'

@

E

+M

@E

+ M"

@E

exact,

the analogue for right modules. A n R-algebra A (for commutative R) is called flat, if and only if it is flat as an R-module. E x a m p l e s 1.6.10 S - l R and R [ X ]are flat R-algebras. The first follows from Example 1.6.2 e) combined with Lemma 1.3.11, the second by Example 1.6.2 c) combined with Remark 1.6.3 e). P r o p o s i t i o n 1.6.11 Let R be a commutative ring, M a finitely presented, N any R-module and A a flat R-algebra. Then the canonical map

is an isomorphism of groups.

46

1 Basic Commutative Algebra 1

Proof. By hypothesis there is an exact sequence

By the right exactness of the tensor product this gives an exact sequence

Since A is flat over R the left exactness of Hom gives a commutative diagram with exact sequences

Since the second and the third vertical arrow are isomorphisms, so is the first 0 one. 1.6.5 Exterior Powers

Let R be a commutative ring, M an R-module and n 2 1 an integer. Let BnM denote the tensor product of M with itself, n times. An Rhomomorphism f of 8"M into an R-module N is said to be alternating, if for X I , . . . ,x, E M , we have f (XI 8 x2 @ - - 8x,) = 0, whenever xi = xj for some i , j, i # j, 1 5 i, j 5 n. An n-fold exterior power of M is a pair (E,, $), where E, is an R-module and $ : 8"M + E, is an alternating R-homomorphism, such that for any alternating R-homomorphism f of 8"M into an R-module N , there exists a unique R-homomorphism f' : E, -+ N which makes the following diagram commute:

It is easy - and analogously to the uniqueness of the tensor product - to see that the n-fold exterior power of M is unique, if it exists. We now show that

1.6 The Tensor Product

47

n-fold exterior power of M exists. Let P be the submodule of 8"M generated by all elements ofthe for XI @ x 2 @ . . . @ x nwhere x1,x2, ... , x n E M and xi = x j for some i, j with i # j. Let An M = ( 8 " M ) / P and : 8"M --+ An M be the canonical homomorphism. Then it is easy to see that the pair (An M, +) (or An M ) is the n-fold exterior power of M. For xl, . . . ,xn E M , the image of XI @ . - .@ xn in An M under $ is denoted by xl A . . A x,.

+

We set M.

M = R so that

An M is defined for all n E IN.Note that

M=

1.6.12 Some properties of exterior powers: a) Let R + S be a ring homomorphism and let M be an R-module. Composing the S-isomorphism

~ --+ S @ R ( AM), ~ We have an with the S-homomorphism I s @ $ : S @ R ( @M) alternating S-homomorphism B n ( S @R M ) + S @R (An M ) , which induces an S-isomorphism n

A(s

@R

M)

7s @R ( A M ) .

--

b) Let M, N be R-modules. Then for 0 5 i 5 n, the map n

MX-.-XMXNX...XN-+A(M~N)

given, for X I , .

m-times

(n-m)times

. . ,xm E M

and yl, . . . ,yn-,

induces an R-homomorphism (Am M ) have an isomorphism n

n

@

E N , by

(An-m N ) -+ An(M @ N ) and we

m

n-m

c) If F is a free R-module with a basis of n elements, then l\i F = 0 for i > n.

An F % R

and

Introduction to Projective Modules

Projective modules are one of the main themes in our book.

2.1 Generalities on Projective Modules Proposition 2.1.1 Let A be a ring and P an A-module. The following properties are equivalent: (1) for every diagram of A-modules

with exact line, there is a homomorphism g : P f - making the diagram commutative;

-+

M

-

a so-called lift of

(2) to every epimorphism f : M + P exists a cross section, i.e. a homomorphism s : P Mi , such that fos = i d p - this means that P is isomorphic to a direct summand of M ; (3) P is a direct summand of some free A-module.

Proof. (1) + (2). Apply (1) to the situation N = P and f = idp. (2) + (3). This follows, since there is a surjective homomorphism F with a free module F.

+P

(3) + (1). First assume, P is free and B a basis of P. The restriction f lB can be lifted to a map y : B -+ M, since M + N is surjective. The linear extension g : P --+ M of y fulfills the assertion.

50

2 Introduction to Projective Modules

Now let P @ Q be free. Then f extends to a linear map f ' : P $ Q -+ N , (P, 9 ) e f(p). Let g' : P @Q -+ M be a lift off'. Then g := glP is a lift 0 off. Remark 2.1.2 If the reader has accustomed himself to "Hom" as a functor and its left exactness, it will be clear to him that the properties of the proposition are also equivalent to

( 4 ) the functor H o m (P, ~ -) is exact;

( 5 ) the functor HomA(P,-) maps epimorphisms to epimorphisms. Definition 2.1.3 A module with the equivalent properties of the proposition is called projective. Proposition 2.1.4 For a projective A-module P the following properties are equivalent:

(1) P is finitely generated; (2) P is a direct summand of a finitely generated free module; (3) For some n E IN there is an exact sequence

(So P is even finitely presented.) Proof. (1) 3 (2). Since P is finitely generated, there is a surjective homomorphism f : An P for some n E IN. By Proposition 2.1.1 (2) there is a section s of f , which identifies P with a direct summand of An. (2) 3 (3). If An = P$Q, the projection An -+Q and the canonical embedding Q -+ An compose to a map whose image Q is the kernel of the projection A" -+ P. (3) 3 (1) is trivial.

0

Remarks 2.1.5 a) A direct sum of (finitely or infinitely many) projective modules is projective. (This need not be so for direct products.)

b) If A is commutative and P and Q are projective A-modules, then so is P @ A Q. Namely remember the "distributivity" of the tensor product w.r.t. direct sums. By this F @ G is a free A-module, if F and G are so. And if Pep' = F and Q@Q1= G weget ( P @ Q ) $ ( P @ Q 1 ) @ ( P ' @ Q ) @ ( P ' @ Q ' ) F @ G , whence P @ Q is a direct summand of a free module.

2.1 Generalities on Projective Modules

51

c) If P is projective and A commutative, then for n E N the module An P is also projective. This is clearly true for free P, hence for a direct summand P of a free module by Paragraph 1.6.12. d) Every projective A-module is flat. This is true first for P = A, then for any free P and finally for a general projective P. e) Let P be a finitely generated projective left A-module, then P* = HomA(P,A) is a finitely generated projective right A-module. This can be derived from the following property of direct sums. Let be a family of A-modules. The families (fi : Mi + N)iEr of A-module homomorphisms are in bijective correspondence to the A-module homomorphisms Mi + N, which for xi E Mi is defined by

eiEI

Let P @ Q = An. Then Hom(P, A) @ Hom(Q, A) Hom(An, A) r Hom(A, A)" S An.

S

Hom(P

$

Q, A) =

f) Every finitely generated projective module is reflexive, i.e. the canonical map P + P**is bijective. Namely with the above notations clearly P &, Q = An is reflexive and ( P @ Q)** = P**@ Q**. The canonical isomorphism P @ Q + ( P @ Q)**maps P to P**and Q to Q**. So P + P**must be an isomorphism. Corollary 2.1.6 Let A be a ring. The isomorphism classes of finitely generated projective A-modules form a commutative monoid (semigroup with a neutral element O), the addition being defined by

([PI denotes the isomorphism class of P.) Moreover, if A is commutative, they form a commutative semiring, the multiplication defined by [PI . [Q] := [ P @ A Q]. (A semiring fu&lls the axioms of a ring with the exception that its additive structure only is a commutative monoid, not necessarily a commutative group.) This semiring admits a canonical involution, given by [PI -t [P*]. Proof. The only question is, if the isomorphism classes of finitely generated projective A-modules form a set. But they all can be realized as quotients of I7 one A-module, namely of A("). The rest is clear. Definition 2.1.7 The set (monoid, semiring) of isomorphism classes of finitely generated projective A-modules will be denoted by IP(A).

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2 Introduction to Projective Modules

Remark 2.1.8 Let f : A -+B be a ring homomorphism. We definc

In this way IP becomes a so called 'functor' from the 'category' of rings to that of monoids, resp. semirings. (See Section 5.1.)

2.2 Projective Modules over Local Rings, Rank Here all rings are supposed to bc commutative; except in the last example. We next show that all finitely generated projective modules over a local ring are free. For instance this is clear when the local ring is a field. Thc general case is reduced to this case by means of 'Nakayama's Lemma', i.e. Corollary 2.2.4.

Definition 2.2.1 The Jacobson radical Jac(A) of a ring A is the intersection of all maximal ideals of A. Lemma 2.2.2 An element x unit for all a E A.

E

A belongs to Jac(A) if and only if 1- ax i s a

Proof. Let x E Jac(A) . So ax E Jac(A) , whence 1- ax does not belong to any maximal ideal. Hence A(1 - ax) = A. i.e. there is a b with b(l - ax) = 1. Conversely if x $! Jac(A), there is a maximal ideal m of A with x $! m. Since A/m is a field, there is an a E A with ax 1 (mod m), i.e. 1-ax t m, whence A(1- ax) # A.

Lemma 2.2.3 Let M be a finitely generated R-module and I an ideal of R such that I M = M . Then there exists an element i E I such that ( l + i ) M = 0. Proof. (See [39].) We prove the result by induction on the minimal number p := p(M) of generators of the module M . If p ( M ) = 0 take i = 0. Let m l , . . . ,m, be generators of M . Let M' = M/Rm,. By induction there is an x E I such that (1 + x)M1 = 0, i.e. (1 x ) M c Rm,. Therefore, as M = I M , (1+ x ) I M = I ( 1 x)M c Im,. Hence (1 x)m, = ym, for some y E I , i.e. (1 + x - y)m, = 0. Thus, (1 x - y ) ( l x)M = 0 and clearly ( l + x - y)(l + x ) = 1 + i for some i E I.

+

+

+

+ +

Corollary 2.2.4 (Nakayama's Lemma) Let M be a finitely generated Rmodule and N a sub-module of M , further I c Jac(A) an ideal of A. a) If I M = M then M = 0. b) I f N + I M = M t h e n N = M .

2.2 Rank

53

Proof. a) Clear from Lemma 2.2.3.

+

b) This will follow by noting that I . (MIN) = ( N I M ) / N = M I N . Now apply (i) to the finitely generated R-module M / N and I to get M / N = 0, i.e. 0 M = N. Proposition 2.2.5 Let A be local (see Definition 1.3.25) and P a finitely generated projective A-module. Then P is free.

Proof. Let m be the maximal ideal, k := Alm and XI,. . . ,xn elements of P , whose residue classes form a basis of the k-vector space PImP. Then P is generated by 1x1, . . . ,x,) U mP, hence by Nakayama's Lemma it is already generated by 1x1,. . . ,xn). So one gets an exact sequence:

where f is defined by $1, . . . ,xn and Q the kernel. This sequence splits, because P is projective. Hence first Q is also finitely generated and secondly the sequence remains exact after tensoring. Especially one gets the exact sequence

By construction k B f is an isomorphism, whence Q/mQ = k @ Q = 0. Again 0 by "Nakayama" we get Q = 0, which proves the proposition. Remark 2.2.6 The proposition holds also for non finitely generated projective modules. See [35]. It also holds in the non-commutative case, if A has exactly one maximal left ideal, which then is automatically two-sided. Lemma 2.2.7 Let M be a finitely presented, N a finitely generated A-module and f : N -+ M an epimorphism, then K := ker(f) is finitely generated

Proof. By hypothesis there are natural numbers m, n and an exact sequence

Let g : AT -+ M be any linear map. Since AT is projective and f surjective, there is a homomorphism g' : AT -+ An with fog' = g. We get an exact sequence

(3M +0, where a(%,y) := (x - gl(y), y) and (f, g)(x, y) := f (x) + g(y). Therefore, for 0 +ker(f) @ A T4 An @ A T

big enough r, we may construct a a commutative diagram with exact lines and surjective p

2 Introduction to Projective Modules

54

Since by the Snake Lemma a! is surjective too, the lemma is proven.

0

Proposition 2.2.8 Let A be a ring and P be finitely presented A-module. The following properties are equivalent:

(1) P is projective; (2) for every m E Spmax A the A, -module Pmis free;

(3) there are f i , . . . ,fn E A, such that A = C Afi and Pfi is free over Afi for every i. Proof. (1) + (2) is clear. (2) + (3). Let m be a maximal ideal of A. The Am-module Pmhas a finite basis of the form x1/1,. . . ,xn/l. Let An + P be defined by XI,.. . ,xn and C its cokernel. We have C, = 0. Since C is finitely generated, there exists already an f E A \ m with Cf = 0. So we have an epimorphism A; + Pf . For its kernel K we have K, = 0. Since K is finitely generated by the above lemma, we get Kg = 0 for some g E A \ m. So Pf, is free. In this way we get a - maybe infinite - family (fi)iEr in A, such that Pfi is free over A and to every maximal ideal m there is an i with fi 4 m. The latter means

i.e. there are gi E A, gi = 0 for nearly all i with replace I by a finite subset.

+

xifigi = 1. So we can

(3) (2). Let m E Spmax(A). Since X A f i = A, there is an i with fi So Pmas a localization of Pf, is free over A,.

4 m.

(2) + (1). Let M + M" be an epimorphism. We have to show that an epimorphism HomA(P,M ) + HomA(P,Mu) is induced. It is enough to see that HomA(P, M), + Hom(P, M"), is surjective for every maximal ideal m. Since P is finitely presented, H o ~ A ( PN), , H o ~ A(Pm, , Nm) in a natural 0 way. So (i) follows, since P, is projective over A,. Lemma 2.2.9 Let A 0 be a (commutative) ring. Then Am 2 An implies m = n for a l l m , n E IN.

Proof. Since A possesses maximal ideals, there exist homomorphisms A + k, where k is a field. So Am E An implies km E ,En. (Justify!) Hence m = n.

2.2 Rank

55

Remark 2.2.10 If A is the non-commutative (!) endomorphism ring of an infinite dimensional vector space, then A r A2 as A-modules, hence Am r An for all m, n > 0. (Exercise!) Definitions 2.2.11 a) Let A 9 0 be a ring. The rank of a finitely generated free A-module Am is defined by rkA(Am)= rk(Am) = m. b) Let P be a finitely generated projective A-module. T h e n we define the map rk(P) : Spec(A) + N by rk(P)(p) = r k ~(,P p ) W e write also rkv(P) instead of rk(P) (p) . rk(P) is called the rank (map). c) B y a projective module of rank 5 n, resp. of rank n we mean a finitely generated projective module whose rank is bounded by n, resp. whose rank equals n everywhere. I t is clear what we mean by a projective module of constant rank. (Especially it has t o be finitely generated)

Proposition 2.2.12 Let P be a finitely generated projective A-module. The map rk(P) : Spec(A) + N is locally constant. Especially if Spec(A) is connected, i.e. if A is not a direct product of nontrivial rings, then rk(P) is constant.

Proof. Let p E Spec(A). By Proposition 2.2.8 there is an f E A \ p, such that Pf is free over Af . So rkp(P) = rk, ( P ) for every q E D(f). (Remember 0 D ( f ) = {P E Spec(A)I f 4 PI.) Note 2.2.13 Let A be a direct product of infinitely many fields ki, i E I . For k j is a projective A-module. If J any subset J of I the direct sum P = is infinite, P is not finitely generated, and the proposition may happen to be not valid.

ejEJ

Proposition 2.2.14 Let A be a semilocal ring and P a (finitely generated) projective A-module of constant rank. T h e n P is free.

Proof. Let J be the intersection of the maximal ideals of A. Then PIJP is 0 free over A I J . The rest is as in the proof of Proposition 2.2.8. Proposition 2.2.15 Let P be a projective A-module of constant rank n which can be generated by n elements pl , . . . ,p, . T h e n P is free with basis pl , . . . ,p, .

Proof. The generating system PI,. . . ,p,_defines an epimorphism f : An + P. This induces an isomorphism fv : A; + Pv for every prime ideal p, since Ppr A;. For A; r A; @ ker(fv). So f must be an isomorphism. 0

56

2 Introduction to Projective Modules

2.3 Special Residue Class Rings First we extend some definitions, already used in the commutative case, to not necessarily commutative rings.

Definitions 2.3.1 Let A be any ring. a) The Jacobson radical Jac(A) of A is the intersection of all its maximal left ideals. b) A n element x E A is called strongly nilpotent if there is an n E IN, such that every product XI - .x, in A vanishes, provided that at least n of its factors are equal to x. c) The nilradical Nil(A) of A is the set of all its strongly nilpotent elements.

Remarks 2.3.2 a) The simple left A-modules are up to isomorphism the modules A/m with a maximal left ideal m. Namely the latter clearly are simple. Now let E be simple and x E E \ (0). Then the map R + E, 1 I+ x is surjective, its image being a non-zero submodule of the simple E, and its kernel must be a maximal left ideal. (Note that over a noncommutative ring A modules A / I and A/ J with different left ideals I,J may happen to be isomorphic. For example let A = M,(k) be the n x n-matrix ring over a field k with n > 1 and I, resp. J consist of the matrices whose first, resp. second column is zero. Then A / I and A / J both are isomorphic to kn with its natural A-module structure. This left A-module is simple and I and J are maximal left ideals.) b) J := Jac(A) is a two-sided ideal of A. Namely let a E A. We only have to show J a c m for every maximal left ideal m. This is clear for a E m. Otherwise let n := {x E A I xa E m}, which clearly is a left ideal. Once we have shown that n is maximal, J a c na c m, and we are done. Consider the left linear map f : A + A/m, x I-, (xu mod m). Then ker(f) = n. Further f is surjective, because a @ m and Alm is simple, i.e. has only the two obvious submodules. Since therefore A/n must also be simple, there is no left ideal properly contained between n and A. c) Every proper left ideal is contained in a maximal one, as one easily sees with the help of Zorn's Lemma. We use this to show:

A n element x E A belongs to Jac(A) if and only if 1 - ax is left invertible for every a E A. Namely let x E Jac(A). So ax E Jac(A), whence 1-ax does not belong to any maximal left ideal. Hence A ( l - ax) = A, i.e. there is a b with b(1- ax) = 1.

2.3 Special Residue Class Rings

57

Conversely if x f Jac(A), there is a maximal left ideal m of A with x f m. Since the left A-module A/m is simple, there is an a E A with ax E 1 (mod m), i.e. 1- ax E m, whence A ( l - ax) # A. d) The Lemma of Nakayama and its proof are word-by-word the same as in the commutative case. See Corollary 2.2.4. e) If a'a = aa" = 1 then a' = a'aa" = a". (In this case we write a-I := a'(= a") and call a invertible.) We use this to show: Jac(A) is also the intersection of the maximal right ideals. By symmetry it is enough to show that the latter contains Jac(A), i.e. that 1 - xu is right invertible for all x E Jac(A), a E A, i.e. - Jac(A) being also a right ideal - that 1 - x is right invertible for all x E Jac(A). Since it is left invertible there is a y with y(l - x) = 1. So 1 - y = -yx E Jac(A). Therefore there is an y' with 1 = yl(l - (1 - y)) = y'y. So y is invertible, hence 1- x = y-l too. f) In a commutative ring the the concepts of nilpotent and strongly nilpotent elements coincide. If x, y E A are strongly nilpotent, so are x y, ax, xu for every a E A. Hence Nil(A) is a two-sided ideal of A.

+

If I is a nilpotent left or right ideal of A, all elements of I are strongly nilpotent. So I c Nil(A). If all entries of an n x n-matrix a are strongly nilpotent elements of A, so is a in the ring M,(A). g) Always we have Nil(A)

c Jac(A).

Namely let x E Nil(A) and a E A; then (ax)n = 0 for some n. Therefore (ax)i = 1 - (ax)n = 1, and the criterion of b) is fulfilled. (1 - ax) Co
Proof. We have two obvious epimorphisms

P --+ P1/IP' and P'

-+

P'/IP1.

By the projectivity of P we get a commutative diagram

2 Introduction to Projective Modules

58

+

From this we get f (P) IP' = P, hence f (P)= P', by Nakayama's Lemma. By the surjectivity of f and the projectivity of P' we see, ker(f) is a direct summand of P (hence also finitely generated). Since ker(f) is mapped to zero 0 in P/IP, we conclude - again by Nakayama's Lemma - ker(f) = 0. This means that P(A) -+ P ( A / I ) is injective, if I c Jac(A). It is not always surjective. (The isomorphism class of every so called stably free (A/I)-module is in the image, as we will see later.) 2.3.4 Lifting idempotents. Recall that the decompositions of a module into direct sums of two submodules are in bijective correspondence with its idempotent endomorphisms. Namely to the decomposition M = MI $ M2 ~ ;to the idempotent enattach the idempotent endomorphism id^^ $ 0 ~ and domorphism e attach the decomposition M = e(M) $ ker(e). (Check that one gets a direct decomposition.) Therefore evey (finitely generated) projective module is isomorphic to the image of an idempotent endomorphism of a (finitely generated) free module.

Now, let cp : A -+ B be a ring homomorphism. If P is finitely generated projective over A, say P g e(An) for some idempotent e E Mn(A), then B @I P g cp(e)Bn. So, to show the surjectivity of P(cp), it is sufficient to show, that every idempotent e E Mn(B) for any n has an idempotent preimage in Mn (A)Proposition 2.3.5 Let I c Nil(A) be a two-sided ideal of a ring A and cp : A -+ A / I the canonical surjection. Then P ( p ) is bijective.

Proof. The surjectivity follows from the last proposition and Nil(A)

c Jac(A).

Now to the surjectivity of P(cp). We show that every idempotent i? E Mn(A/I) has an idempotent preimage in Mn(A). Let e be any preimage of Z. Then e2 -e E Mn(I). Let J be the two-sided ideal, generated by all entries of e2 -e. , Note that (e mod M,(J)) (Let cij be these entries, then J = C iAc,A.) already is idempotent. Now J, generated by finitely many strongly nilpotent elements, is nilpotent. Since we may go stepwise, from A/ J to A/ J~to A/ J~ to . . . , we may assume J~= (0). Then also M,(J)~ = (0).

+

We search for an x E Mn(J), such that e x is idempotent. Since (e e2 ex xe x2 = e2 ex xe, we have to solve the equation

+ + +

+ +

+ x ) =~

by an element x E Mn(I).

+

The element 1 - 2e is a unit in M,(A). For, (1 - 2e)2 = 1 - 4e 4e2 = 1 4(e2 - e) is a unit, since e2 - e is nilpotent. The element (1 - 2e)-l(e2 - e) belongs to Mn(I), and since it commutes with e, it solves Equation (2.1). Cl

+

2.4 Projective Modules of Rank 1

59

2.4 Projective Modules of Rank 1 In this section all rings are supposed to be commutative. We will study projective modules of rank 1. These are isomorphic to ideals in many cases. And we are able to find many nonfree examples of them.

Proposition 2.4.1 Let A be a ring and M be a finitely presented A-module. The following statements are equivalent: ( 1 ) The "evaluation" map M

@

M*

-+

A, ( x ,a ) e a ( x ) is a n isomorphism.

( 2 ) There is a n A-module N with M @ N (3) M,

EA,

( 4 ) Mb

E Ap

as A,-modules

E A.

for every maximal ideal m of A .

as Ap-modules for every p E SpecA.

( 5 ) M is projective of rank 1. Proof. (1) + ( 2 ) is clear. ( 2 ) + (3). For every maximal ideal m we have Mm @A,,,N m E A,. SOwithout restriction of generality we may assume that A is local with maximal ideal m and residue class field k . We get ( M l m M )@ ( N l m N ) E k , hence M / m M E k . So by Nakayama's Lemma M is monogene, i.e. M E A / I for some ideal I . This implies I - ( M 18 N ) = 0. Since M @ N S A, we get I = 0, i.e. M E A. (3) # ( 4 ) is clear.

( 3 ) + (5) follows from Proposition 2.2.8. ( 5 ) + ( 1 ) :As M is finitely presented ( M m ) * E (M*),. Therefore M @ AM * 2 A is equivalent to M , @A, MG G A, for all maximal ideals m. But the latter 0 is so if M , r A,. Corollary 2.4.2 The isomorphism classes of projective A-modules of rank 1 form an abelian group, the multiplication being defined by [ P ]- [&I := [ P @ A & ] . The neutral element is [A],and the inverse of [PI is [PI-' := [ P * ] . 0 Definitions 2.4.3 a) The group of isomorphism classes of projective A modules of rank 1 is called the Picard group of A and denoted by Pic A . b) Let Q ( A ) be the total ring of fractions of A, i.e. Q ( A ) = S-'A where S is the set of the non-zero-divisors of A. A fractional ideal of A is a n A-submodule I of &(A)for which there exists an s E S, such that s A c I c s-lA. (Note that s A c I c tP1Aimplies s t A c I c (st)-'A. Therefore we made the definition with only one s.) Ideals in the usual sense are sometimes called integral ideals, i n order t o distinguish them from fractional ideals.

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2 Introduction to Projective Modules

c) Let I , J be fractional ideals of A. Then we define: I : J := { x E S - ' A I x J c I )

and I-'

:= A : I .

As for integral ideals, the product I J of fractional ideals I , J is the additive group generated by all products ab with a E I , b E J. Note 2.4.4 Every fractional ideal as an A-module is isomorphic to an integral fractional ideal. Namely i f sA c I c s-lA, Then I sI and s2A C S I C A. Lemma 2.4.5 If I , J are fractional ideals, then so are I : J and I J .

Proof. There are non-zero-divisors s, t E A, such that sA c I c s-'A and t A c J c t-lA. From J c t-'A, i.e. tJ C A and sA C I , we get stJ C I , hence st E I : J. Let x E I : J , i.e. X J c I c s-lA. So xtA c X J C I C s-lA, whence x E (st)-lA. For IJ we get immediately stA C IJ C (&)-'A.

0

Proposition 2.4.6 Let I , J be fractional ideals of a ring. The map cp: I:J+HomA(J,I)

xH(jHxj)

is an isomorphism of A-modules. Proof. Clearly cp is a homomorphism. Since J contains a non-zero-divisor,cp is injective. To show surjectivity, let a : J -+ I be an A-linear map and t E S with t A c J c t-'A. Set x := t-la(t). Now for any j E J we get (using t j E A): a ( j ) = t-2(t2a(j))= t-2a(t2j)= t-2a(tj-t)= t-2-tj.a(t)= t-la(t).j = xj.

2.4.7 By the definition of I-I we see immediately:

b) If there is a fractional ideal J with IJ = A, then J c I-l, whence II-' = A. Aso we have J = I-' in this case, as one sees by multiplying the equation IJ = A by I-'.

Definition 2.4.8 An invertible ideal of a ring A is a fractional ideal I such that II-I = A. Proposition 2.4.9 The following properties of a fractional ideal I of a commutative ring A are equivalent:

2.4 Projective Modules of Rank 1

61

(1) I is invertible; (2) I is projective as an A-module; (3) I is projective of rank 1. (Especially I is finitely generated). Proof. (1) + (2). Since 1 E I I - I by (i), there are a l , . . . , a n E I , bl, . . . ,bn E I-' with C aibi = 1. We define homomorphisms: f :An

+I

ei e a i ,

and

(Here e l , . . . ,en is the canonical basis of An. And g maps I to An, since bi E 1-1.) We get fg(x) = C aibix = x, i.e. f g = idI. So I is isomorphic to a direct summand of An, hence projective and finitely generated. (2) + (1). I is isomorphic to a direct summand of some ~ ( ~SO 1 let.

g

:

x t - , x g~i ( x ) e i

~

be homomorphisms with f g = id^. To every i E E by Proposition 2.4.6 there is a bi E I-' E Hom(I,A) with gi(x) = biz for x E I. For x E I we have gi(x) = 0 hence biz = 0 for nearly every i E E. Since there is a unit of S-'A in I, we have bi = 0 for nearly every i. Now x = fg(x) = C aibix for x E I . Again, since some of these x are units in S-lA, we see Caibi = 1. This implies 1 E I I - l , hence (1). (3) + (1) is weaker than (2) + (1).

(1) + (3). By the proof of (1) + (2) we know already that I is projective and finitely generated. So it suffices to show I @ I - I E A. This is implied Now I c s-lA + s I c A. Hence I is isomorphic to an by I @ I-' E IIP1. integral ideal in A. Since further I-' is invertible, whence projective, whence flat, the proof will be finished by the following observation. Lemma 2.4.10 Let J be an ideal of a ring A and E be an A-module.

a) By j @ e t-, je a surjective homomorphism p : J

@E

-+ JE

is defined.

b) If E is flat (especially if E is projective), p is bijective. (This lemma holds also for a noncommutative ring, if J is a right ideal and E a left module.) Proof of the lemma. a) is trivial.

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2 Introduction to Projective Modules

b) Consider the commutative diagram

The lines are exact, the upper by the flatness of E. Since the middle vertical arrow and the right one are isomorphisms, so is the left one. (Use the Snake Lemma). 0 Definitions 2.4.11 a) Let Inv(A) denote the group of invertible ideals of A. b) Let T be some multiplicative subset of A consisting of non-zero-divisors. Let Inv(A,T) denote the subgroup of Inv(A) consisting of such invertible ideals I of A, that tA c I c t-'A for some t E T. c) Let Prin(A) denote the group of invertible principal ideals of A, i.e. of those fractional ideals of A, which are generated as A-modules by a single unit of &(A). d) Prin(A, T) := Inv(A, T ) n Prin(A). I t is easily seen that we indeed have groups in the above situations. By Proposition 2.4.9 there is a canonical map

It is a homomorphism by Lemma 2.4.10. Its kernel is Prin(A, T). Proposition 2.4.12 With the above notations there is a canonical exact sequence

Proof. Only the exactness a t PicA is not trivial. 'im c ker'. For I E I ~ v ( A , T there ) is a t implies

E

T with At c I

c At-l. This

'ker c im'. Let [PI E Pic(A) with [T-'PI = 0 in Pic(T-'A). Then there is an isomorphism g : T-'P + T-'A. Let p E P, t E T be such that g(p/t) = 1 in T-'A. Non-zero-divisors of A are such of P, the latter being a direct summand of a free module. So we get a monomorphism P + T-'P E T-'A. Its image is an A-submodule I. It suffices to show sA c I c s-lA for some s E T. But g(p/l) = (t/l), whence t E I, with the above t. On the other hand I 5 P is finitely generated, say generated by a l l t l , . . . ,an/tn. Then s := t - tl - . .tn 0 will do.

2.4 Projective Modules of Rank 1

63

R e m a r k 2.4.13 Assume, there is a multiplicative set T, consisting of non zero divisors of A such that T-lA is semilocal. Then for such a T by Proposition 2.4.12 we have Pic(A) EZ' Inv(A, T)/Prin(A, T). Namely, if B is semilocal, then Pic(B) = 0 by Proposition 2.2.14. This means that evry projective A-module of rank 1 is isomorphic to some integral invertible ideal. Since Inv(A, T) c Inv(A) we have then also Pic(A)/Inv(A) =Pic(A). Let now T be the set of all non-zero-divisors. T-lA then is semilocal, if for instance A is reduced with only finitely many minimal prime ideals, for e.g. a domain. See Corollary 1.3.24. The same holds for rings whose ideals are finitely generated, which were first studied 'abstractly' by E. Noether, who showed that they also satisfy the property that ascending chains of ideals will terminate. For more details we refer the reader to Chapter 6 . We will see there that in such a ring, the socalled Noetherian ring, the set of zero-divisors is a union of finitely many prime ideals. Corollary 2.4.14 If A is a factorial domain, Pic(A) = 0. The converse does not hold, since there are many non-factorial local domains.

Proof. We show, that any invertible ideal I c A is principal. Let x be the greatest common divisor of its elements. It is sufficient to show J := x-I I = A. But since the greatest common divisor of J is 1, one sees easily that J-' = A, 0 whence J = A. Examples 2.4.15 We give examples of nonfree projective modules of rank one. a) Let A := H field of A is $(-) (2--,I+-)ofA.

+ H-

c C. It

=$

is clearly a subring of C. The quotient

+ $a. Consider the ideal a := (3, 1 + -)

=

CLAIM:a is invertible! Proof: An element a + bE $(-) belongs to a-' if and only if 3(a + b-) E A and (1 + -)(a bE A. The first condition means If we write a = a1/3, b = b'3 with a', b' E Z,the second condition 3a, 3b E means a1 b' = 0 (3) and a' - 5b1 = 0 (3); but that means a' b' = 0 (3). So for e.g. (-1 + -)/3, (2 + -)/3 E a-l. And we compute (2 - -)(2 G ) / 3 + (1 + G)(-l + G ) / 3 = 3 - 2 = 1.

+.

+

+

On the other hand a is not principal. Consider the norm

+

+

64

2 Introduction to Projective Modules

We know that it is multiplicative, i.e. N(aP) = N(a)N(@).By this, one easily sees that a E A is a unit of A if and only if N(a) = f1. Therefore 3 and 1 fiare irreducible, since the factors 2 and 3 of their norms 9, resp. 6 are not norms themselves. So the only principal ideal of A which contains a is A. And a # A, since a-I # A.

+

b) The following examples are similar to the above one. Let k be a field with char(k) # 2 and f E k[X] be a polynomial of degree > 1which has only simple roots in an algebraic closure of k. Our ring is A := k[X, Y]/(Y2 - f ) which clearly can also be written as A = k[X] &, k[X]y, where the multiplication is defined by y2 = f . (The ring A is the "coordinate ring" of a conic (deg(f) = 2), resp. of an elliptic curve (deg(f) = 3 or 4), resp. a hyperelliptic curve (deg(f) 2 51.1 Now let a , b E k with b2 = f (a). (This means the point (a, b) E k2 belongs to the curve.) For the sake of simplicity we further suppose b # 0 and f'(a) # 0. Consider the ideal a = (X - a , y - b) of A. We have

and so Ala r k[X, Y]/(X - a, Y - b) 5 k. Especially a that a is invertible and a-' = (a, P) with

a=

y+b 2b(X - a)

and

,B=--

#

A. We will show

1 2b'

- f (X) - f (a) Wehavea.(X-a) E Aandalsoa-(y-b) = y2 - b2 2b(X-a) 2b(X-a) This gives a a c A. The inclusion pa c A is trivial. Further we have

A.

This proves our statement. Now we will show that in many cases a is not principal. Again we do this using the norm

>

+

i) First we consider the case where deg(f) is odd and 3. An element g(X) h(X)y is a unit of A, if its norm is in kX. This implies first that h(X) = 0 and then g(X) E kX.So kX is the unit group of A. Now N(X - a) = (X - a)2 and one sees, by arguing with degrees, that c (X - a) is not a norm for any c E k X .SOX - a is irreducible, i.e. (X - a) is maximal amongst all principal ideals # A.

2.4 Projective Modules of Rank 1

65

Now N(y - b) = b2 - f (X) = f (a) - f (X), which is divisible by X - a, but not by (X - a)2 = N ( X - a), since we supposed fl(a) # 0. Therefore y - b is not contained in (X - a), whence a cannot be principal. ii) Secondly let k be an ordered field, e.g. any subfield of R and f (X) of even degree (> 2) with negative leading coefficient. Then for all non zero polynomi~ are als g(X), h(X) E k[X] the leading coefficients of g(X)2 and - h ( ~ f) (X) positive. Hence deg(N(g(X) h(X)y) = Max12 deg(g(X)), 2 deg(h(X)) 2). By the same arguments as in i) one sees that the invertible ideal a is not principal.

+

+

iii) As a special case of ii) let us consider the field k = IR and the polynomial 1. Then A = R[X, Y]/(X2 Y2 - 1) is the coordinate ring f (X) = -X2 of the circle. The projective module a may be called Mobius-module, since it is related to the Mobius band. (Note that we have excluded the 4 points ( f 1,0), (0, f1) of the circle for simplicity. Any orthogonal transformation different from f id of R2 will put these points to ones we have considered.)

+

+

Let us see, what happens in this case, if we extend the groundfield to C. Then the ideal I := (X - a, Y - b) of C[X, Y] coincides with the ideal J := (X2 Y2 - 1, X Yi - a - bi), i.e. I/(X2 Y2 - 1) is principal. Namely remember a2 b2 = 1 -

+

+

+

+

whence J c I .

(X - Y i ) ( X + Y i ) - (a - &)(a+ bi) - (X -Yi)(X + Y i - a -bi)

+ bi) + (X - Yi)(a + bi) = (X - Yi - (a - bi))(a + bi). So J contains X - a + (Y - b)i and X - a - (Y - b)i as well. By that one easily gets I c J. Therefore in A = C[X, Y]/(X2 + Y2 - 1) the ideal a = I / ( X 2 + Y2 - 1) is generated by the residue class of X + Yi - a - bi. -(a - bi)(a

C) Let A = R[X] with R = k[g,z](,,,)/(y3 - z2) S k[t2,t3](tz,t3),where k is any field. Then Q(R) = k(t). Now tn E R for all n 2 2. By Proposition 2.4.18 - see below - we conclude that there is a projective A-module P of rank 1, which is not free. Lemma 2.4.16 Let R be a domain. Then R =

n

Rp.

p prime

66

2 Introduction to Projective Modules

np

Proof. Let a E Rp and consider the ideal I := {s E R I sa E R}. For every prime ideal p there is an s E I \ p, since a E Rp. SOI is not contained in any 0 prime ideal, hence I = R. Therefore Ra c R, i.e. a E R. Corollary 2.4.17 Let R be an integral domain with quotient field K and let a E K \ R. Then there is a prime ideal p of R with a $! Rp. Proposition 2.4.18 Let R be an integral domain with quotient field K . Suppose that there is an element a E K \ R such that an E R for all suficiently large n. Then there is a rank 1 projective R[X]-module which is not extended from R, i.e. not of the form P[X] for a projective R-module P.

Proof. Replace a by a suitable power of it and assume that an E R for all 2 2. Let f = a x E K[X]. Let P = (1 f)R[X] (1 f f 2 ) ~ [ xC] K[X]. We show that P is an invertible ideal, hence a projective R[X]-module of rank 1:

+

n

Let Q = (1 - f)R[X]

+ + +

+ (1 - f + f 2)R[X]. Then

as f 2, f 3, f ER[X]. We show P Q = R[X]. If the overbar is meant modulo 1 f 2 + f 4 =O, and so P Q , then Z = (I - f 3 ) + ( 1 + f3) =i5, - - 1= 1 . 1 = ( J ~ = ) ~ = i5. Thus, P Q = R[X]. Then P is a projective module of rank 1 by Proposition 2.4.9.

=-+

7

+

If P were extended from R, the Pp would be extended from Rpwhence Pp would be free of rank 1, i.e. Pp = Rp[X]c for some c E K[X]. Now 1 = ( 1 + f + f 2 ) - f ( l + f ) E K[X]Pp = cK[X]. Hence c E K X . Then Pp is generated by 1 a x , 1 a x a2X2, whence it is generated by 1 modulo X . This means c E RpX. Therefore, 1 a x E Pp = cRp[X] = Rp[X] whence 0 a E RD. A contradiction.

+

+

+

+

+ + +

d) Let H = {a bi c j dkla, b, c, d E 1W) be the division ring of the real quaternions. It can be shown that all the projective JH[x]-modules are free. However, there are nonfree projective H[x, y]-modules. This is a special case of the 1971 theorem of M. Ojanguren-RSridharan [69] that there are nonfree rank one projective modules over a polynomial extension D[xl,. . . ,x,], n > 1,of a division ring D, which is not a field. The reader can also find a proof of this theorem in ([44], Chapter 2, $3). Here we give the example in the special case when D is the division ring H of the real quaternions, our argument expands on the one found in [85]. History is witness to the fact that in the decade after this counter-example there were a spate of papers on the Serre's conjecture in the early to midseventies, with several partially successful results reconfirming that all should

2.4 Projective Modules of Rank 1

67

be well in the case of the base ring being a field! In 1980, four years after Serre's conjecture was proved correct by Quillen-Suslin, J. StafTord proved in [go] that every projective module of rank > 2 over D[xl, .. . ,x,], where D is a division ring, is free. (In [go] the reader can also find a simple proof in the case when D is finite dimensional over its center.) We now construct the counter-example mentioned above. Let F = M[x, y] @ H[x, y] be the free left M[x, y]-module of rank 2, with el = (1, O), e2 = (0,l) as basis. Define a H[x, y]-module homomorphism cp : F + H[x, y] by sending el to x i, and e2 to y j. Note that cp is a surjective map as cp((y j)el - (x i)e2) = -2k. Hence, ker(cp) = P will be a left projective H[x, y]-module of rank 1. We show that P is not free.

+

+

+

+

+

+

+

+ +

Let tl = (y j)(x - i)el - (x2 l)e2, t2 = -(y2 1)el (x i)(y - j)e2. It is easily verified that cp(tl) = 0 = cp(t2), i.e. tl, t2 E ker(cp) = P. If P were free of rank 1, then it would be generated by a single element. It is easy to check that if P = H[x, yltl, then t2 $! H[x, yltl, and if P = H[x, ylta, then tl $! H[x, y]t2. Thus, tl, t2 will not generate P. For instance, if t2 E H[x, yltl, say t2 = p(x, Y){(Y j)(x - i)el - (x2 l)e21, for some P(X,Y) E M[x, y]. Let deg denote the total degree. Then, since deg(fg) = deg f degg, for f , g E H[x, y], we conclude that degp(x, y) = 0, i.e. p(x, y) = p(0,O) = k. But this cannot be. Similarly, show that tl 4 H[x, y]t2.

+

+

+

So let ael +be2 E P be a generator, with a, b non-zero. Since ael +be2 E ker(cp), a(x i) b(y j ) = 0. Since tl E P, there exists p(= p(x, y)), q(= q(x, y)) E H[x, y] such that tl = p(ael be2) tz = q(ael be2). These equations show that degp deg a = 2, deg deg b = 2, deg q deg a = 2, etc. They also allow us to conclude that a E H[y], b E H[x], p(x, y) E M[x], q E H[y].

+ + + +

+

+

+

+

Consequently, degp, degq 5 2, deg a = deg b. But degp # 2, as then deg a = 0, which is is not possible as a(x+i) b(y j) = 0. Similarly degq # 2. Again degp # 0, as then dega = 2, deg b = 2, and one cannot get a(x+i)+b(y+ j ) = 0. (How can the degree two term corresponding to y2 in a(x i) be cancelled by terms of b(y j)? Don't forget that b E H[x].)

+ +

+

+

The only case that remains is that degp = 1 = dega = degb. So let a = aoly, b = boo 6102, and study the relation (aoo aoly)(x i) (boo blox)(y j) = 0. By comparing the xy, constant, y, and x coefficients, we get a01 = 4 1 0 , aooi = -booj, aoli = -boo, aoo = -bloj. a00

+ +

+

+

+

+ +

Therefore, aooi = -booj = aolij = aolk. However, a00 = - b l o j = sol j. Hence, aooi = a01ji = -aolk!! Hence, a01 = 0. But then b = 0, a contradiction.

Stably Free Modules

A module P is projective, if and only if there is a module Q such that P@Q is free. We know that this does not imply that P itself is free. It is a fascinating fact that even if Q is finitely generated and free and P CB Q is free, P need not be free.

3.1 Generalities In this section R will denote an arbitrary ring, unless otherwise specified. We are considering right R-modules instead of left R-modules.

A finitely generated R-module P is projective if and only if there exists a finitely generated R-module Q such that P CB Q is free. Here we introduce a smaller class of finitely generated modules, the so called stably free modules P, which have the property that there exists a finitely generated free module F such that P @ F is free. Historically it was so that one knew 'from the beginning' that projective modules over a polynomial ring k[xl,. . . ,x,], where k is a field, are stably free. This will be shown here in Theorem 3.7.6. And in Proposition 3.1.10 we will give an example of a nonfree, but stably free module, which 'in geometric terms' is the tangent bundle on the 2 dimensional sphere. In Chapter 4 we will give several proofs of the fact that projective modules over k[xl,. . . ,x,] are actually free. Three of them use the fact that these modules are already known to be stably free. Definition 3.1.1 A (right) R-module P is said to be stably free of type m with (0 5 m < 00) if P @ Rm is free. A module is said t o be stably free if it is stably free of type m for some m E IN. (Such a module is of course projective.)

70

3 Stably Free Modules

A projective module P is stably free of type m if and only if

for a suitable epimorphism f , which automatically splits. If M is the m x n matrix associated with f , then M is right invertible, i.e, there exists an n x mmatrix N such that M N = I,. (By I, we denote the m x m unit matrix.) Conversely any right invertible m x n-matrix M defines a finitely generated stably free (right R-module) P of type m, namely the solution space of M,

In this way the study of finitely generated stably free right R modules is equivalent to the study of right invertible rectangular matrices over R. Proposition 3.1.2 The kernel P of an epimorphism f : Rn -+ Rm is free if and only i f f can be lifted to an isomorphism f' : Rn -+ Rm @ RT for some r, ~ = f. such that p r , f'

Proof. Suppose P is free, then there exists an r such that P -% RT is an isomorphism. One may write Rn = Q @ P in such a way that the restriction of f to Q gives an isomorphism fo : Q -+ Rm. Then fo @ g : Rn -+ Rm @ RT clearly gives the desired isomorphism. Conversely suppose that an isomorphism f ' with pr,ofl = f exists. Then 17 P = kerf 2 ker(pr,) = RT is free.

+

Note that in the situation above Rm @ RT E Rn need not imply m r = n , in general. A well-known example of P.M. Cohn [14] is: Let V be an infinite dimensional vector space over a field k, and R = Endk(V) the endomorphism ring of R. Then R E [email protected], since V is infinite dimensional, V@V E V and so

Definition 3.1.3 We say that a ring R satisfies the (right) invariant basis property (IBP) if for any s,t 2 0, RS Z Rt + s = t . Examples 3.1.4 a) A division ring (= skew field) satisfies IBP.

b) A commutative ring

(0) satisfies IBP; see Lemma 2.2.9.

c) Any (right) Noetherian ring satisfies (right) IBP. Otherwise an isomorphism Rm 7Rn with m < n , composed with the canonical projection Rn -+ Rm

3.1 Generalities

71

would give a surjective endomorphism f : Rn --+ Rn with nontrivial kernel. Then the series ker(f) 5 ker(f2) 5 ker(f3) 5 ... would be an infinite strictly increasing series of submodules of a finitely generated module over a Noetherian ring. This is impossible as we will see in Chapter 6.

Remark 3.1.5 Consider f , f' in Proposition 3.1.2. Let M denote the m x n matrix corresponding to f and let N denote the (m+r) x n matrix corresponding to f', i f f ' exists. The condition " prlo f' = f " says that M is a submatrix of N , consisting of its first n rows. The condition "f is an isomorphism" says that N is a (not necessarily square) invertible matrix. i.e. there exists another matrix N' of size n x (m+r) such that N N ' = I,+,, N ' N = I,. The following is the matrix theoretic version of Proposition 3.1.2. Proposition 3.1.6 For any right invertible m x n matrix M , m < n, the (stably free) solution space of M is free if and only if M can be completed to an invertible matrix by adding a suitable number of new rows. Definition 3.1.7 We say that v = (vl, . . . ,v,) is a (right) unimodular viR = R. row if

x:.=,

The set of all (right) unimodular rows of length n with entries in R is denoted by Umn (R),

Remarks 3.1.8 a) A row (vl , . . . ,v,) is unimodular, if and only if there are viwi = 1. Therefore, if w = (wl,. . . ,w,), then vwt = 1. Thus, v E Um,(R) if and only if there is a row w of length n such that (v, w) = vwt = 1. For example, the ithco-ordinate vector

wi E R such that

x

is a unimodular vector. b) A row, considered as a 1xn-matrix, describes a linear map of right modules Rn -+ R, which is surjective if and only if the row is unimodular. It follows that there is a surjective map from the set of unimodular rows of length n to the set of isomorphism classes of stably free modules of rank n - 1 and type 1.

Example3.1.9 LetA=IR[xo, ... , X , ] / ( X ~ + - . - + x ~ - l ) n, > 1 . L e t P b e the projective module corresponding to the unimodular row (xo,.. . ,x,) E Umn+1(A). Then P is free for n = 1,3,7.

72

3 Stably Free Modules

Proof. Let eo, . . . ,en be a basis for the complex numbers (n = l ) , quaternions (n = 3), and Cayley numbers (n = 7) over IR such that eo = 1 and the norm n

n

Consider the matrix Hz of the linear transformation y ++yx. If x # 0 then Hz E GLn+l(A). Moreover, elHz = (TO,.. . ,x,) and so H, is a completion of (xo,. . . ,x,) in the sense of Proposition 3.1.6. 0 Here is an interesting example of a stably free projective module which is not free. Proposition 3.1.10 Let

Then the projective A-module P , corresponding to the unimodular row (To,TI, T2), (where Zi denotes the residue class of x i ) is not free. Proof. If P where free then by Proposition 3.1.6 the row (To,?El, a)can be completed to an invertible matrix a with e l a = (z0,zl,C2). We shall think of aij, det(a) as "functions" on the real Zsphere

below. For any point on S2 we can define a tangent vector as follows: If v E S2 consider p(v) = (a,' (v) (v) ,a;' (v)) E R3.

,a2

Clearly, (v, cp(v)) = 0 and so cp(v) is a tangent vector to S2 at the point v. are polynomials, the map cp : S2t R3 is a differentiable function. Since a;' Since ~ 7 - lE~GL3(A) the vector cp(v) can never be the zero vector. Thus cp is a nowhere zero vector field on S2. This is well-known to be impossible by elementary topology (cf. [56]). (We will give a short sketch of the argument 0 in Proposition 5.4.4.) Thus, P cannot be a free module. Proposition 3.1.11 For any ring R, the fo2lowing statements are equivalent.

( 1 ) Any finitely generated stably free right R-module is free. (2) Any finitely generated stably free right R-module of type 1 is free.

(3) Any right unimodular row over R can be completed to a n invertible matrix (by adding a suitable number of new rows).

3.1 Generalities

73

Proof. (2) @ (3) follows from Proposition 3.1.6. (I) j(2) is obvious. (2)+ (1). We prove the result by induction on m. This is clear for m = 1 by (2). Assume the result for m - 1, and let P be finitely generated stably free of type m. Then P $ Rm r Rn, i.e. ( P Bi Rm-l) @ R % Rn. This implies P ~i Rm-l is &ably free of type 1. By (2) it is free. Thus P @ RnL-l R q o r some k . By induction P is free.

"

Definitions 3.1.12 a) A ring which satisfies the above properties is called a (right) Hermite ring. b) A (right unimodular) row which satisfies condition (iii) above is called completable. More generally, the reader can establish

Corollary 3.1.13 T h e following properties of a ring R are equivalent:

2 t is free. rank 2 t and of

(1) A n y finitely generated stably free right R-module of rank (2) A n y finitely generated stably free right R-module of 1 is free. (3) A n y unimodular row of length

type

> t + 1 i s completable.

Lemma 3.1.14 Let R be a commutative ring. Suppose P $ Rn-' = Rn, i.e. P is stably free of rank 1. T h e n P r R. Proof. R.epresent P as the solution space of a right invertible (n- 1 xn)-matrix M . We show that the maximal minors bl, . . . ,b, of M generate the unit ideal. Suppose not; then there exists a maximal ideal m containing b l , . . . ,b,. Since M E Mn-l,n(R/m) is right invertible, it has rank n-1. By linear algebra it can be completed to an invertible matrix M* E GL,(R/m). But by Laplace expansion of det(M*) w.r.t. the last row we get det(M*) = 0 as all the maximal minors are in m. A contradiction.

+

+

. . . a, b, = 1. By the above there are a l , . . . , a, E R such that a l bl Hence we can complete M to a matrix of determinant 1 by adding a last row a l , . . . ,a, with appropriate signs. Then by Proposition 3.1.6 this implies that P is free. Hence P E R.

Alternatively the above argument can be written neatly using exterior powers as follows:

P m o f A n ( P Bi Rn-') = Ei+j=,(/\" @ A' Rn-l) = /\I P @ An-' R"-' as Ai P = 0 if i > 1, and A' Rn-' = 0 if j # n-1. Therefore, as A"-' Rn-l % R, A n ( P @ Rn-l) r A' P F P . But P $ Rn-' = R71 and so A n ( P Bi RnP1) = An Rn r R. Hence P R.

74

3 Stably Free Modules

3.2 Non- Free over Localized Polynomial Rings Here we give a general method to find polynomials f E k[X1,. . . ,X,] and nonfree, stably free modules over k[X1, . . . ,X,] f . If I is an ideal, contained in the nilradical of a ring R, then by Proposition 2.3.5 every projective module over R I I can be lifted to one over R. This does not generally hold if I is only assumed to be contained in the Jacobson radical of R. But for stably free modules the situation is different. Proposition 3.2.1 Let A be a not necessarily commutative ring and I c Jac(A) a two-sided ideal. Every (finitely generated) stably free AII-module can be lified to a stably free A-module.

For simplicity we give here a short proof of this special case of Theorem 5.7.12

Proof. If Q is stably free over AII, there is a split exact sequence

with a matrix ,4 over AII. Let a : Am + An be any lift of P. Since I c Jac(A) and a(Am) IAn = An, by Nakayama's Lemma a is surjective as well. So with P := ker(a) we get an exact sequence

+

which splits, because An is free. So

is exact as well, which proves Q E P 8 AII.

0

Proposition 3.2.2 Let A be a (commutative) ring and I a n ideal such that there exists a non-free, stably free AII-module. T h e n there is a n f E 1 I and a non-free stably free Af -module.

+

Proof. Let S := 1+I.then S-'1 c Jac(SP1A) and S-'AIS-lI r AII. So a stably free, nonfree AII-module lifts to one, say P', over S-lA by the above proposition. Since the description of P' as a stably free module can be done by finitely many relations, there is already a stably free module P over Af for 0 some f E S with S-lP E P'. One may apply this to A = XI, . . . ,X,] to get a lot of examples of non-free projective modules over some Af which are not extended from A, since every projective A-module is free, as we will see in the next chapter.

3.3 Action of GL,(R) on Um,(R)

75

3.3 Action of GLn(R) on Umn(R) The group GL,(R) of invertible n x n square matrices over R acts on the set Um,(R) of unimodular rows in the following natural manner: If v E Um,(R), a E GL,(R), then

Note that if w E Ml,,(R) is such that vwt = 1, then va(wa-lt)t = 1, and so va E Um,(R). Thus, the above map defines an action of GLn(R) on Um,(R). If v' = va for some a E GL,(R), then we write this as v v' or v NGL,(R) v'. We will use this notation sometimes also if v, v' are not necessarily unimodular. N

Let G be a subgroup of GL,(R). We write v va = v'.

N~

v', if there is a a E G with

Proposition 3.3.1 The orbits of Um,(R) under the GL,(R)-action are in

one to one corespocorrespondencehe isomorphism classes of right R modules P for which P @ R g Rn. Under this correspondence orbit of (1,0,. . . ,0) corresponds to the free module Rn-l. Proof. To any (bl,. . . ,b,) E Um,(R) we can associate P = P(b1,. . . , b,), the solution space (i.e. kernel) of (bl, . . . ,b,) : Rn -+R. Such a P is a typical module for which P $ R E Rn. Suppose

for another (el,. . . ,en) E Um,(R). Then we can complete the following commutative diagram,

with a suitable isomorphism Rn A Rn. (Note that the rows are split exact). If M E GL,(R) denote the matrix of this isomorphism cr we will have

Conversely suppose (bl,. . . ,b,) = (el,. . . ,cn)M for some M E GL,(R). Then the automorphism defined by M induces an isomorphism of the two kernels P ( c ,~. . . ,en). P(bl, . . . ,b,)

0

76

3 Stably Free Modules

Corollary 3.3.2 Let (bl, . . . ,b,) E Um,(R). The following statements are equivalent: (1) (bl, . . . ,b,) is completable; (2) P(bl,. .. , b,)

Rn-l.

(3) (bl,... lbn) w ( l , O , + . .10). Proof. (2)

e (3) follows from Proposition 3.3.1.

(1) =. (3). (bl,. . . ,b,) E Um,(R) is completable to an invertible matrix M' E GL,(R). If M'M = I,, then elM1M = (bl,. . . ,b,)M = elln = el, i.e. (bl,... ,bn) - e l . (3) + (1): Suppose (bl, . . . ,b,) = (1,0,. . . ,O)M. Then M is a completion of (bl, . . . ,b,) to a square invertible matrix. 0

3.4 Elementary Action on Unimodular Rows Definitions 3.4.1 Let A be a ring, a E A and r

> 0 an integer.

<

a) For i, j r let eij denote the r x r-matrix whose only non-zero entry is 1 on the (i, j)-th place, and define E$ := I, +aeij for a E A and i # j. Such E$ are called elementary matrices. They are clearly invertible, since E$E%;" = I,. (The exponential place of the a is justified by the rule E$E$ = E;+~.) b) Let E, (A) denote the subgroup of GL, (A), generated by all E$ (where i and a E A). E,(A) is called the elementary group.

#j

c) By diag(al, . . . ,a,) we denote the diagonal matrix with entries a l l . . . ,a, along the diagonal. R e m a r k s 3.4.2 a) Clearly any surjective ring homomorphism A + B induces a surjective group homomorphism E,(A) + E,(B). This so called 'lifting property' will be extremely useful for us. (Note that it does not hold generally for GL, or SL,.) b) If A is commutative - such that SL, (A) is defined - E,(A) c SL,(A). These two groups do not always coincide, even not for all principal domains. Nevertheless they are equal in important cases, so for semilocal or Euclidean domains, as we will see below.

c) Multiplying a matrix on the right (resp. left) side by E$ means to change its j-th column (resp. i-th row) by adding the i-th column times a (resp. a times the j-th row) to it.

3.4 Elementary Action on Unimodular Rows

77

We call these transformations elementary (column, resp. row) transformations. d) By three elementary column transformations one can interchange two columns - up to a change of a sign. Namely let v, w denote the two columns, forget about the others and carry out the following elementary transformations (v, w) C) (2, W v) 1--) (-w, W v) 1--) (-20, v).

+

+

The same holds for rows. This clearly implies e.g.

Proposition 3.4.3 If R is commutative, any diagonal matrix in SL,(R) belongs to E,(R).

Proof. If D = diag(dl, . . . ,d,) where di7sare unit, we can factorize

D = diag(dl, dl-', 1,.. . , 1) - diag(1, dl, da, d3,. . . ,d,). By induction on n we are reduced to proving that diag(d, d-l) E E,(R). This is shown by bringing diag(d, d-l) to the identity matrix by elementary row and column transformations. 0 d-I

( d l ) ( - d

0)-

(-d

1 d-l 0 ) +

(idIL)1--)~2

Remark 3.4.4 The above is a special case of Whitehead's Lemma:

If

a!

a!

E GL,(R) then (O

0

t E2,(R). This is proved analogously:

Proposition 3.4.5 If (bl, . . . ,b,) E Um,(R) contains a right unimodular sub-row of shorter length (especially if one of the bi is a unit or if one is 0), then (b1, . . . ,bn) -E, (R)el.

Proof. Let (bi,, . . . ,bi,) E Um,(R) be contained in (bl,. . . , b,) E Um,(R). Let i 4 {il, . . . ,i,). Write bilail . . . bi,ai, = 1 for some bij E R 1 I:j 5 m. After a sequence of elementary transformations we may change bi to bi (bil ail . . . bim aim)(bi - 1) = 1. Further elementary transformations change

+ +

+ +

the other bj's to 0. Then we have (bl, . . . ,b,) -E,(R) ei. Since E2(R), we can show that ei -E,(R) el and

SO

(-y i) 0

(bl, . . . ,b,) -E,(R) el.

78

3 Stably Free Modules

Proposition 3.4.6 Let R be a Euclidean domain and (al, . . . ,a,) E Um,(R) with n > 1. Then a) (al,. . . ,a,) is elementarily equivalent to el ( a i , . ,an) -E,(R) el.

=

(l,O,. . . ,o), i.e.

b) Consequently SL,(R) = E,(R). Proof. a) Let 6 : R + IN be the Euclidean norm (also called Euclidean degree.) In the set of all (a:, . . . ,a;), elementarily equivalent to (al,. . . ,a,), choose one with minimal 6(a:). If a: = 0, we are through by Proposition 3.4.5. Otherwise aila; for j > 1. Namely if, say, a', 1 aa, write a; = aiq r with 6(r) < 6 ( 4 ) . Then (a:, a;, . . . ,a;) -En (R) (a:, r, a$, . . . ,a;) -E, (R) (r, -a:, a$, . . . ,a;), which contradicts the minimality of &(a:). But if a', la:, then clearly (a:, . . . ,a;) -E,(R) (a:, 0, . . . ,0). Again we use Proposition 3.4.5.

+

b) As R is commutative ring E,(R) c SL,(R). For the converse inclusion above let M E SL,(R). By (i) we can perform suitable elementary transformations to bring M to MI with first row (1,0,. . . ,0). Now a sequence of row transformations brings Ml to

where M' E SL,-1(R). The proof now proceeds by induction on n.

0

(Note that a field is a special case of a Euclidean domain. But of course, if R is a field, then the proposition nearly is trivial. Namely (bl, . . . ,b,) over a field is unimodular, if and only if one of the bi is non-zero, i.e. a unit. To transform (bl, . . . ,b,) to el = (1,0,. . . ,0) in this case by elementary transformations, is the classical 'Gauss Algorithm'.)

Corollary 3.4.7 Let R = n:='=,ki be a finite direct product of fields (or more generally Euclidean domains) ki. Then the statements of the Proposition 3.4.6) also hold.

niZl

GLn(ki), i.e. one can regard Proof. Clearly one can identify: GL,(R) = an a E GL,(R) as an r-tuple ( a l , . . . ,a,) with ai E GLn(ki). SO let a = (a1,. . . , a r ) E GL,(R) = GL,(~~).

n;,

CLAIM:a E E,(R)

cri E En(ki) for i = 1,. . . ,r.

The implication '*'being clear, let us show 'w'.So assume that every is a product of elementary matrices, say

ai

79

3.4 Elementary Action on Unimodular Rows

Every r-tuple E : ~:= (I,, . . . ,I,, ~ i jI,,, . . . ,I,) (with ~ an elementary matrix over nl=l Ici. Then

i in j

the i-th place) is

Now let ( b l , . . . ,b,) be unimodular over R and regard it as an r-tuple ( ( b l , . . . ,b , ) ~ ,. . . ,(bl,. . . , , , where (bl,. . . ,b,)i is unimodular over b ) ) ki. By Proposition 3.4.6 there are ai E E,(lci) with (bl,. . . ,bn)iai = (1,0,. . . ,O)i. For a := ( a l ,. . . ,a,) we get (bl, . . . ,b,)a = el and a E E n ( R ) by the claim. 17

.

Corollary 3.4.8 Let R be a semilocal ring and v E Um,(R) be a unimodular vector. Then v -E,(R) el. Proof. Let m l , . . . ,m, be the maximal ideals of R and J = nLZlmi its Jacobson radical. By the Chinese Remainder Theorem 1.5.13

is a finite direct product of fields. Let 'overbar' denote 'modulo J'. By Corol= q.We can lift t o an E E E n ( R ) , lary 3.4.7 there is an Z E E,(R with and we get V E = ( a l , . . . ,a,) with a1 l ( m o d J ) . So a1 E RX and therefore ( a l , . . . ,a,) -E,(R) e l , which implies the corollary. 0 Clearly the same holds, if R/Jac(R) is a finite product of Euclidean domains.

Proposition 3.4.9 Let R be a commutative ring with ( a l , . . . ,a,, bl,. . . ,b,) E Um,+,(R); s 2 1. Let I = C Rai and R = R / I . If -

(h,.. . ,bs)

-

(i,D,.. . ,o),

then ( a l , . . . ,aT,bl,.-- ,bs) -E,+,(R) ( 1 , 0 , . . . 70). Proof. Lifting the elementary transformations which bring (T,G ... ,0) we get

-

(51,

. . . ,b,) to

( l , O , . . . ,0) (mod I ) . Performing further elwhere (bll,.. . ,b,') to ementary transformations we can bring ( a , . . . ,, , . . . ) ( a l , . . . ,a., 1,0,. . . ,0). So the row ( a l , . . . , a T ,bl,. . . ,b,) contains a right unimodular subrow of shorter length. Hence by Proposition 3.4.5 we get

80

3 Stably Free Modules

3.5 Interesting Examples of Completable Rows R. G. Swan and J. Towber in [loll, while studying "cancellative" properties of projective modules over &ne algebras, stumbled upon the following remarkable fact: Given any unimodular bvector (a, b, c) E Um3(R) over a commutative ring R, then the unimodular 3-vector (a2,b,c) can always be completed to an invertible matrix! Later on in [41] M. Krusemeyer gave a delightful explanation as to why (a2,b, c) E Um3(R) is always completable, and exhibited the identity b

-c

+ 2ab' -a'

= (aa' - b'c'

+ bb' +

=1

bt2

(Recall that the determinant of an alternating matrix is a square - now figure out M. Krusemeyer's explanation!) The above theorem of Swan-Towber was also explained and dramatically generalized by A. Suslin in his doctoral thesis - see [92] - wherein he shows that if (ao, a l , . . . ,a,) E Urn,+l(R), then the unimodular vector (ao,a l , a$, . . . ,a:) can always be completed to an invertible matrix! Before proving this we need the following Lemma 3.5.1 Let R be a commutative ring and let a, b E R such that R a Rb = R. Then

a(;

with some Proof. Let 3.4.4

E

E

R

In)l:

+

= a&+ b,

En(R) and p E Mn(R). = RIRb, then G := (a mod b) E

0 where 8 E En@). Let

E

In-1

a

E En(R) be a lift of 8. Then

(a; for some p E Mn (R) .

EX.By Whitehead's Lemma

In)l:

=

+ b,

3.5 Examples of Completable Rows

81

Proposition 3.5.2 ( M . P. Murthy) [63] Let R be a commutative ring and ($0, X I , . . . ,x,) E Umn+l (R). Suppose (Zo,t, . . . ,Zn-l) is completable over R = R/Rxn. Then ($0, . . . , xn-1, x:) is completable. Proof. Since (Z0, . . . ,z ~ - ~is )completable over R = R/Rxn, we have a y? E GL,(R) such that the first column of y? is (30, . . . ,Zn-l)t. Let cp be a lift of y? in ~ . b := det (cp) is a unit modulo Mn (R), whose first column is (xo, . . . ,x ~ )Since x,, there are a , c E R with ab - exn = 1. Let cp' be the n x n-matrix which arises from cp by multiplying its second column by a. Then for the adjoint matrix adj (cp') of cp' we get cp' . adj (9') = det(cpl)In= a det (cp)In = abIn. So

Hence

(xnIn " adj 'In ) cp'

t GL2,(R). Now by Lemma 3.5.1, there exist

E

t

En(R), p t Mn(R) such that

Let y = padj(cpf). Then ycp' = padj(cp')cpl = abp. Now consider the matrix

:( :)

E GL2.(R). We have

xnIn adj cp'

0

In-1

It is now easy to make elementary row and column operations and transform this matrix to a matrix of the form

(EIn\l)

with a E GLn+l(R) and aef =

n t

(xo,... ,xn-l,xn) .

Theorem 3.5.3 (A. Suslin) [92] Let R be a commutative ring and let (xo,. . . ,x,) (xa,X I , x;, . . . ,xz) is completable. Proof. It follows by induction on n. If n = 1 and xoyo

0 E

Um,+l(R).

Then

+ xl yl = 1, then

is a completion. n-1 If n > 1, by induction assumption (xo, a1,x i , . . . ,xn-,) is completable mod0 ulo x,. So Proposition 3.5.2 performs the induction step.

Note that this construction of A.A. Suslin leads to the same matrix as M. Krusemeyer's in the case n = 2.

82

3 Stably Free Modules

3.6 Direct sums of a stably free module In this section we shall prove an interesting theorem of M. Gabel, which was sharpened by T. Y. Lam.

Lemma 3.6.1 [26] (M. Gabel) Let cp : Rn -+ Rm be a n onto homomorphism. Let P = kercp. Assume that there is a basis { d l , . . . ,d,} on Rn such that, for some k 5 n - m, the images {cp(dl), . . . ,cp(dk)) generate Rm. Then P 7RnPnL.

Proof. Let F = C=,Rdi 2j Rk. Since (cp(d1),... , c p ( d k ) } generate Rm, P + F = Rm. If K denotes the kernel of cp I P : P + Rn, then K = P n F . Let n - m = k + s. We have two exact sequences.

where P / K

Y

Rn/F Y Rn-k

Y

Rm+".Thus,

~ ~ K $ R ~ ~ ~ + ~ ~ K $ R " @ R " E F @ R " E R ~ + + " Y R as claimcd.

Lemma 3.6.2 (Whitehead's Lemma for Rectangular Matrices) Let A E Mm,,(R),B E Ms,t(R).Assume that B has a right inverse. Let A = (M,V ) with M E Mmxs(R), V E Mm,,-,(R). Let P, Q denote the solution spaces of A, B. Then P @ Q is isomorphic t o the solution space of ( M B ,V ) .

Proof. Let S E Mt,,(R) with BC = I,. Q @ P is the solution spacc of By elementary transformations:

The last matrix clearly has solution space isomorphic to that of ( M ,B, V ) .

Corollary 3.6.3 Let A E Mm,,(R), B E Mn,t( R ) ,and B have a right inverse. Then the solution space of AB is isomorphic to the direct s u m of the solution spaces of A and B.

3.7 Stable Freeness over Polynomial Rings

83

Theorem 3.6.4 (T. Y.Lam) Let R be a commutative ring, and P be a stably free R-module with P $ Rm 21 Rn, with n > m. Then, the r-fold sum rP = P $ . . . $ P (r times) is free for T 2 m+m/(n-m).

Proof. Let P = ker(p), where p : Rn A E Mm,,(R) be the matrix of p.

+ Rm + 0 is an epimorphism. Let

Let A = (M, V), M E Mm,,(R), V E Mm,,-,(R). 3.6.2 (1) P (2) P

$P

By Whitehead's Lemma

corresponds to (M(M,V), V) = (M2,MV, V). corresponds to (M3,M2V, MV, V), etc.

$P @P

Let Ni denote the submodule of Rm generated by the column vectors of the matrix (Mi-', Mi-2V1,. . . ,MV, V). By Cayley-Hamilton theorem M satisfies its characteristic polynomial of degree m. Consequently, Ni = NmP1 for all i m - 1.

>

+

> +

Let N = r ( n - m) m for some r m m/(n - m). Let a = RN -+ Rm be , a is given by given by the matrix (M', MT-'V,. . . ,MV, V) E M r , , ~ ( R )i.e. a e j = j-th column of this matrix. Clearly, ker(a) 21 T P .Now,

where N-p = (m- 1)(n-m). Thus Rm is generated by images of m+ (N-p) 5 N - m basis vectors of Rn. By Gabel's lemma, ker a E r P is free. 0

3.7 Projectives over k [ x l ,.

.. ,x,]

are Stably Free

In the remaining part of this chapter we will show that finitely generated projective modules over the polynomial ring k[xl,. . . ,x,], where Ic is a field, are stably free. This is a preparation for some of the proofs of Serre's Conjecture, which says that they are indeed free. 3.7.1 Schanuel's Lemma

We begin with Schanuel's Lemma:

3 Stably Free Modules

84

Lemma 3.7.1 (Schanuel) If P, P' are projective A-modules and we have exact sequences of A-modules,

then P'

@K S

P @ K'.

Proof. Since P, P' are projective one has R-linear maps f , g so that

POf

= a , aog = P.

Let X

c P @ P'

be the pull-back of a , P, i.e.

x = { b , ~ ' E) P @ P' I a b ) = P(P')). One has the diagram

Here pl,p2 are the natural projections which are surjective as a , ,B are so. Their kernels are K', K respectively. Since P, P' are projective the crossed 0 exact sequences splits. Hence P @ K' E X S P' @ K . 3.7.2 Proof of the Stable Freeness Lemma 3.7.2 (R.G. Swan) Let R be a subring of A. Let P, Q be projective A-modules and let a : Q -i P , ,Ll : P -i Q be monomorphisms. If

3.7 Stable Freeness over Polynomial Rings

85

A, P / a P P , QlPaiQ are projective over R, then P / a Q and Q / P P are also R-projective. Proof. Clearly, P and Q are R-projective as they are direct summands o f free A-modules and A is a direct summand o f a free R-module. W e have exact sequences

B y Schanuel's Lemma, P / a P P @ Q Similarly, P / a Q is R-projective.

Q / P P @ P . Hence, Q / P P is R-projective.

Corollary 3.7.3 Let R be a subring of A and s Let P , Q be projective A-modules with s P c Q R-projective then so is P / Q .

0 E

c

A be a non-zero-divisor. P . If A and A / s A are

Proof. Take ai t o b e t h e inclusion map Q v P and /? t o b e multiplication b y s from P + Q . T h e n P / a P P = P / s P and Q/&Q = Q / s Q are A / s A projective, and so R-projective. By Lemma 3.7.2 t h e module P / Q ( = P/aiQ) is R-projective. 0 Definition 3.7.4 R-modules P, Q are called stably isomorphic, if P @ Rm E Q $ Rm for some m. Theorem 3.7.5 (M. P. Murthy - C . Pedrini) Let P , Q be finitely generated projective R[x]-modules. Suppose that f P c Q c P for some monic polynomial f E R [ x ] .then P and Q are stably isomorphic. I n particular, if P f E Q f , then P and Q are stably isomorphic.

Proof. Let M = P / Q . Since f is rnonic, R [ x ] / ( f )is free as an R-module. Therefore P / f P is R-projective, whence M is R-projective by Corollary 3.7.3. W e have t h e exact sequences o f R[x]-modules

86

3 Stably Free Modules

where M[x] = M @ R R[x] is defined as in Example 1.2.3 h) and .x denotes the homothesy of x. By Schanuel's Lemma, Q @ M[x] E P@M[x]. Since M is R-projective, M[x] is R[x]-projective, and so the above isomorphism enables 0 us to conclude that Q, P are stably isomorphic. Theorem 3.7.6 (D. Hilbert) Let P be a finitely generated projective k[xl, . . . ,x,]-module, where k is a field. Then P is stably free.

Proof. We prove the result by induction on n, it being clear for n = 0. Let S = k[x,] \ (0). Then S-' k[xl, . . . ,x,] = k(x,)[xl,. . . , x,-11, where k(x,) is the quotient field of k[x,]. By the inductive hypothesis S-'P is stably free. Therefore, Pf(,n)is stably free for some monic f (x,) E k[x,]. By Remark 0 3.7.5 the module P is stably free. Remark 3.7.7 Let A = k[xl, . . . ,x,], k a field and M be a finitely generated A-module. D.Hilbert showed that every finitely generated A-module M has a finite free resolution of length 5 n

For projective M this implies M

$ A P ~E $ AD^

@ i

odd

i

even

Also it implies that for every multiplicative subset S of A every finitely generated projective S-lA-module is stably free. We will show this. For a short and independent proof we refer the reader to Zariski and Samuel's book on Commutative Algebra [110].

Serre's Conjecture

In his paper [84] J-P. Serre posed the question whether any projective modules over the polynomial ring k[xl,. . . ,x,] in several variables over a field is actually free. Later this was called Serre's conjecture. It was proved in 1976 independently by D. Quillen and A. Suslin by different methods. In this chapter we present four proofs, and present one more proof in Chapter 7. The reader can also find other interesting proofs in [94], [go]. In the whole chapter all rings are supposed to be commutative.

4.1 Elementary Divisors over Principal Domains Theorem 4.1.1 Let A be a principal domain and a E MTxs(A). Then there = diag(a1, . . . , at) with ai lai+l, are E E SL, (A), E' E SL, (A) ) such that &a&' for 1 i 5 t - 1, where t := Min{r, s).

<

= (We do not exclude the case that there is a j 2 1 with aj = aj+l = at = 0. It should be clear what we mean by diag(a1,. . . , a t ) also if r # s. It's not the geometric diagonal.) The elements ai are called the elementary divisors of a. We will apply this theorem primarily in the case of a Euclidean ring A, where E,(A) = SL,(A) by Proposition 3.4.6. The proof of the theorem in the case of a Euclidean A is simpler and shows directly that one can choose E,E' in the elementary groups then. Therefore we will give this proof first. Proof. a) Let first A be Euclidean and 6 the Euclidean norm. The claim is clear, if cr = 0 or r = s = 1. Otherwise consider the set S := {oar I a E E,(A), T E E,(A)). In S choose a matrix a' = (a!,j) such that ail # 0 and d(a:,) 5 d(aL) for all a" = (a:;) E S with :a: # 0.

88

4 Serre's Conjecture

Then ai,laij and aillail for all i, j . For otherwise, using division with remainder, by an elementary transformation we could replace say a i j by r = a'13. - qa',, with a suitable q E A, such that r # 0 and 6 ( r ) < &(ail) and again - as shown in Remark 3.4.2 d) - by three elementary transformations we could replace a',, by r . This would contradict the minimality of h(ai1).

+

Now, using at most r s - 2 elementary transformations, we may assume a& = ail = 0 for i , j > 1, i.e. that a' is of the form

By induction on M a x { r , s ) we see that ,B can be transformed to the form diag(a2,. . . ,at). It only remains to show that a1 la,. By an elementary transformation we get - looking only at the top left 2 x 2-matrix

Now we see a, la2 by the same argument as above. So we are done for Euclidean rings. b) Now let A be a general principal domain. The proof in this general case is similar. Consider the set S := { c a I~(T E SLT(A),T E SLs(A)}. In S choose a matrix a' = (aij) such that Aai, is maximal among all ideals Aa:, for a" = (arj) E S. Again we claim that a',, divides all entries in the first row and first column of a'. Assume e.g., it does not divide a',,, so that Ad 3 Aa:,, if d is a g.c.d. of a',, and a',,. Let d = bla;, b2ai2. Then for

+

This contradicts the maximality condition on Aai, . Then we proceed as done previously. Corollary 4.1.2 Let A be a principal domain.

a) Every submodule U of AT i s free of some rank m 5 r , and there are bases X I , .. . ,x, of AT and y l , . . . ,y, of U , such that yi = aixi for suitable U i E A, i 5 m.

If A i s even Euclidean, one m a y pass by elementary transformations from a given pair of bases of AT and U t o one which fulfills the above property.

89

4.2 Horrocks' Theorem

b) Every finitely generated A-module is of the form

for some ai E A. (ai = 0 is not excluded.) Proof. Choosing a finite generating system of U one gets a linear map AS + AT given by a matrix a. Changing bases means multiplying a on both sides by invertible matrices. The 0 rest is clear.

Remark 4.1.3 Things become much more difficult for A-modules which are not finitely generated For example $ is not a direct sum of cyclic groups. Namely show Hornz($, Z / m ) = 0 for every m E Z. If m # 0, the group Hom is as well a $-vector space, as annihilated by m. For m = 0, if n were the smallest positive integer in the image of some non-zero map f : $ + Z , say n = f (a), what should be the image of a12 ? 4.1.4 In Section 8.8 we will prove an analogous structure theorem for finitely generated modules over so called Dedekind rings, which were studied by Dedekind in connection with Fermat's Last Theorem and other Problems of Number Theory.

4.2 Horrocks' Theorem We first prove some preparatory lemmas, which show that if (R, m) is a local ring then Spec(R[x]) can be written as a union of two principal open sets:

where f E R[x] is an arbitrary monic polynomial and g E R[x] is any polynomial all of whose coefficients, except for its constant term, are lying in m.

Lemma 4.2.1 Let R be a ring, a E R and I an ideal of the polynomial ring R[x]. Assume that I contains a monic polynomial as well as an element of the form 1 + a h with h E R[x]. Then I contains an element of the form 1 ar, for some r E R .

+

Proof. The ring R[x]/I is integral over R / ( I n R), since I contains a monic polynomial. And 1 ah E I means that the residue class of a is invertible in R[x]/I, hence by Proposition 1.4.12 it is also invertible in R / I n R, which proves the statement. 0

+

90

4 Serre's Conjecture

Corollary 4.2.2 Let (R,n) be a local ring and let m be a maximal ideal of R[x] possessing a monic polynomial. T h e n m n R = n. Consequently, A = R[x]/(f) i s semilocal for any monic f . Proof. Assume, there were an a E n\(mnR). Since m is maximal (m, a) = R[x], and so 1 ah E m for some h E R[x]. By Lemma 4.2.1 we have 1 aX E m for some X E R. But 1 aX is a unit as R is local. Therefore, m n R = n.

+

+

+

We note that (~/n)[x]/(J)has only finitely many prime ideals corresponding to the irreducible factors of f E (R/n)[x]. Since every maximal ideal of R[x]/(f) contracts to n in R there is a bijective correspondence between ) The latter maximal ideals of R[x]/(f ) and maximal ideals of ( ~ l n[x]/(T). correspond to the irreducible factors of 7 . This settles the rest. 0

Corollary 4.2.3 Let (R,n) be a local ring and let f E R[x] be a monic polynomial. If h E 1+ n[x] then f , h are comaximal. Proof. Let m be a maximal ideal of R[x] containing f . Then by Corollary 4.2.2 m n R = n, i.e. n[x] c m. Therefore, any h E 1 nix] cannot lie in m. Thus, f , h are comaximal. 0

+

We now prove the main theorem of this section, due to G. Horrocks - see [33]. The monic inversion principle is one of the chief techniques in this subject.

Theorem 4.2.4 (Horrocks) Let R be a local ring, A = R[x] and S be the set of monic polynomials of A. If P i s a finitely generated projective A-module such that Ps i s free over As, then P is a free A-module. Proof. Since A has no non-trivial idempotents, P has a constant rank; say rk P = n. First we show by induction on n that P splits as a direct sum L @ Rn-' for some rank-1-projective A-module L. This is clear for n = 1, so let n 2 2. Let m be the maximal ideal of R and k = Rim. Let pl, . . . ,p, E P make up an As-basis of Ps. (Since S consists of non-zero-divisors of A we may regard P as a submodule of Ps and further may multiply any basis of Ps by a common = P/m[x]P = P @ A k[x] denominator, thereby pushing it into P.) Now is a projective, hence a free k[x]-module of rank n. So let 91,. . . ,q, E P be chosen in such a way that their residue classes . . . , form a basis of P = P / m P = P @ A k[x] over k[x]. By the Elementary Divisors Theorem 4.1.1 we can change pl, . . . ,p, and z , . . . ,q, by elementary transformations such that - giving the new elements the old names - p1 = hq2, for some a E R[x]. Since elementary transformations can be lifted, the new E, . . . ,p, have representatives pl, . . . ,p, which form again a base of Ps.

z,

4.3 Quillen's Local Global Principle

+

91

+

Set p = ql xTpl, with r to be specified later. Then p = ij, xrZij2. SO p,ij2,. - .,i& is a basis of P. We have sql = Cl,i5, aipi for some s E S , a l , . . - ,a, E A. Thus sp = (a1 +sxT)pl C2,i,n alpi. Choose r large enough, such that al sx' is rnonic. (Recall that is a rnonic polynomial.) Then p, p2, . - - ,p, is an As-basis of Ps.

+

+

s

+

Let T = 1 m[x]. Then m is contained in Jac(AT). Note that P = P T . Since p, ij2, - . - ,ij, generate P = P T , it follows from Nakayama's Lemma that p,q2, . - -,q, generate PT. Since PT is projective of rank n, by Proposition 2.2.15 we conclude that PT is actually free with basis p, q2,. . . ,q,. Now consider P' = PIAp. Note that P;, P&are free of rank n-1. By Corollary 4.2.3 any maximal ideal n of A either avoids S or avoids T. Thus P A is free of rank n - 1. By Proposition 2.2.8 the module P' is projective. Hence the exact sequence O+Ap+P+P'+O splits, i.e. P r [email protected] induction P' A-module L. Hence P L @ An-'.

=

= L@An-2 for some rank 1 projective

It remains to show that L is free. We know that after inverting some rnonic f we have Rn-1 @ L f %PfE R ? , i.e. Lf is stably free hence free by Lemma 3.1.14. The rest is done by the following lemma. 13

Lemma 4.2.5 Let L be a rankl-projective R[x]-module such that Lf is stably free for some monic polynomial f E R[x]. Then L is free. Proof. L is stably free by Theorem 3.7.5. Since rk L = 1 we know by Lemma 3.1.14 that L is free. 0

4.3 Quillen's Local Global Principle This principle is the crux of the matter.

Definition 4.3.1 Let R + A be a ring homomorphism. An A-module M is called extended from R, if there is an R-module N with M N @ R A. Especially, if R + R[x] is the canonical injection, an R[x]-module M is extended from R, if and only if M N[x] for some R-module N .

=

In the latter case N M is not possible.

= M I x M . In general such a simple description of N by

92

4 Serre's Conjecture

Lemma 4.3.2 Let A be a ring, s, t E A be comaximal, i.e. As M an A-module. Then the sequence

+ At = A, and

is exact. Note that module homomorphisms M + Nl @ Nl, resp. Ml @ M2 + N are in one to one correspondence with pairs (fl, f 2 ) where fi : M + Ni, resp. fi : Mi -+ N are homomorphisms.

+

Proof. The equality As At = A means that for every maximal ideal m of A one has s $ m or t 6 m. Therefore for every m both of (iM,,),, ( i ~ , , , ) , or , isomorphisms. So one easily checks the exactness both of (iM,t)rn, ( i ~ . , ~ )are locally. Corollary 4.3.3 Let A be a commutative ring and let s, t E A be comaximal. Let M , M' be two A-modules such that there are isomorphisms a! : Ms 7Mi and ,B : Mt Mi with at = ,Bs (as maps Mst + Mi,). Then M M'.

If you wish, you may see D.0.2. Corollary 4.3.4 (Patching Local Isomorphisms) Let A be a ring and s, t E A comaximal. Let M, M' be A-modules and a : Ms 7 Mi and ,B : Mt 7 M,' be isomorphisms. If (ao,B-'),t = for suitable yl E Aut ML, y2 E Aut M,', then M E MI.

Proof. Let a' = yl-loa 4.3.3.

, ,B'

= y20,L!?. Then a: =

Pi.

Now apply Corollary 0

The above corollary leads one to search for automorphism of Mst which factorize as a product of two automorphism as above. D. Quillen found an important class of such automorphism of an extended finitely generated module Mst[x]: Any automorphisms a(%)E Hom(Mst,Mst)[x] of M8t[XI with a(0) = idMSt! Lemma 4.3.5 Let P be a finitely generated projective R-module. Then

and EndR, ( P , ) E EndR( P )s. This is a special case of Proposition 1.6.11. We will give here an extra proof.

4.3 Quillen's Local Global Principle

93

Proof. If P S Rn is a free module then

In general one has a natural map

and this can be checked to be "locally" an isomorphism due to the above case. Hence by Corollary 1.3.16 it is an isomorphism. 0

Analogously one proves the second statement.

Remark 4.3.6 The above lemma also extends to finitely presented Rmodules M . Namely for such modules M , by Proposition 1.6.11, we know that H O m ~ [ x ](M [XI, [XI) H ~ mM,~N(, [XI

Lemma 4.3.7 Let S be a multiplicative subset of a ring R, P a finitely generated projective R-module and a($) E Aut(Ps[x]) with a(0) = idps. Then there is an s E S such that a(sx) is in the image of the canonical homomorphism Aut (Pix]) + Aut (Ps[x]). Consequently a(bx) belongs to this image for every b E Rs.

proof. EndRs[xl(Ps[x]) = end^, (Ps)[x] by Lemma 4.3.5. Since a($) E EndRs (Psi Ps)[x] is an automorphism with a(0) = idp,, we can write

for some ai,a: E End(Ps) = End(P)s. As P is finitely generated, for each ai,a:, there are si, s!, E S such that siai, s:ai are in the image H of End(P) -+ End(Ps). Let s' be the product of all si, s!,, then slai, s'ai E H for all i. This implies that a(slx),a' (s'x) are images of say P(x), P'(x) under the map End(P[x]) + End(Ps[x]), where we may assume P(0) = P'(0) = idp. Let y(x) := P(x)P1(x). - .+ymxm with Then y(0) = idp and y ( x ) ~= idPs[xl,i.e. y(x) = idp +ylx yi E EndR(P) such that tiyi = 0 for suitable ti E S. Therefore y(tx) = idp[,] 0 for t = ti. Then s = s't fulfills the statement of the lemma.

+

n

94

4 Serre's Conjecture

Lemma 4.3.8 (D. Quillen) (Splitting Lemma) Let P be a finitely generated projective R-module and sl,s2 be comaximal elements of R . Let a(x) E A U ~ ~ [ ~ , ~,[XI) ( P ,be, an automorphism with a(0) = idpsls2. Then there are automorphisms a l ( x ) E Aut(Psl [x]), a 2 E Aut(P,, [x]), with al(0) = idpsl , a2 (0) = idps2, and with a(x) = a1(x),, a 2 (x),, . Proof. Consider P(x, y, z) := a((z+y)x)a(zx)-I E Aut(P,,,, [x,y, z]). Clearly P(x, 0, z) = idpsls2[+. Therefore by Lemma 4.3.7, there exists an m E IN such that,

+

Note that there are X,p E R with Xslm pszm = 1. Also note that P(x, sTy, z) = a ( ( z syy)x)a(zx)-l. Now specialize z e Xslm, y I-+ p to get

+

and let

a1(x)

:= Pl(x, p, Xsy) E Aut(Ps2[x]). Then

as desired.

Theorem 4.3.9 (Quillen's Local Global Principle) Let R be a ring and P a finitely generated projective module over R[x]. Suppose that Pmis extended from R,,,for all m E Spmax R, then P is extended from R. Clearly by Pmwe mean the Rm[x]-moduleSP1P, where S := R \ m. Proof. Define Q(P) := {s E R I P, is extended from R,). The crucial point is:

CLAIM:Q(P) is an ideal of R. (It is called the Quillen ideal of P.) PROOF:Clearly, if s E Q(P), X E R, then Xs E Q(P). So we need only prove that s l , s 2 E Q(P) implies sl s2 E Q(P). i.e. we have to show that PSI+,, is extended from R,,+,, , provided PSI,P,, are extended from R,, , resp. R,, . But the latter imply that P,,~,,+,,~ and P,,~,,+,,~ are extended as well. So we may rename R,,+,, by R and P,,+,, by P and we have to show that P is extended, provided P,, and P,, are so for comaximal sl, s 2 .

+

Consider the 'covering' diagram,

-

4.3 Quillen's Local Global Principle

R[xl

95

RS, [XI

Let F = P / x P and generally the overbar mean 'modulo x'. We are given isomorphisms:

= idF = We may assume that by replacing pl by plo(?pl-l 8 idR,, and ~2 by cP2~(?p2-1 8 idR,2[xl)respectively. (E:= pi @ R [ ~R) ] We are interested in patching these 'local' data to an isomorphism

By the patching technique, described in Corollary 4.3.4, this will follow if the automorphism 6(x) = (p2)s,o(cpl)s,l E Aut(FS,,,[x]) splits as a product

for 19i E ~ u t ( F , [x]), ; i = 1,2. Since 6(0) = idasls2, this will be the case by Lemma 4.3.8. The claim suffices to establish the theorem: Let m be a maximal ideal of R. As P,,, is extended from &,, there is an s !$ m such that Ps is extended from Rs . Hence s E Q (P) . Thus Q (P) If m for all m E Spmax(R) . Therefore Q (P) , being an ideal, must be the whole ring, i.e. 1 E Q(P), i.e. P is extended.

96

4 Serre's Conjecture

Remark 4.3.10 The theorem also holds for any finitely presented R[x]module P , since Lemma 4.3.5 does so. Theorem 4.3.11 Let S = So $ S1 $ S2 @ . - - be a commutative graded ring and let M be a finitely presented S-module. Assume that for every maxzmal ideal m of So. Mm is extended from ( S o ) , . Then M is extended from So.

Proof. We recall C. Weibel's homotopy trick: One has a ring homomorphism S + S [ T ]defined by

and let P = S [ T ]@,#, M be extension of M and P via 9. By assumption, for every maximal ideal m of So, we have Mm = S @soNm, where, obviously, N = MIS+ M with S+ = S1 @ S2 $ . . . This implies that

Thus P is locally extended fom S via the canonical inclusions Sm + Sm[T].By Quillen's Local Global Principle, P = S [ T ]8 s Q where Q E P / ( T - a ) P , for any a E S . Since the composite homomorphism S + S [ T ]+ S [ T ] / ( T -1) = S is the identity of S , we have P / ( T - l ) P = S [ T ] / ( T- 1) 8 s S [ T ]8 , M = S M = M . On the other hand, the composite homomorphism S -% S [ T ]+ S[T]/(T= ) S is the same as S + S/S+ = So + S , hence

and this shows that M

CY S @soN .

0

Theorem 4.3.12 (Quillen-SuslinMonic Inversion Principle) Let R be a ring, P a finitely generated projective R[x]-modulesuch that Pf is free over R[x] for some monic polynomial f in R [ x ] .Then P is extended from R .

By Horrocks' Proof. Let m E Spmax R. Then ( P m ) fis a free R,[~]~-rnodule. Theorem 4.2.4, the Rp-module Pm is free, hence extended from Rm. Therefore P is extended from R by Quillen's Local-Global Theorem 4.3.9. 0 Lemma 4.3.13 (Nagata Transformation) Let k be a field and f a polynomial in k [ x l ,.. . ,xd]. There exists a change of variables

such that

f ( t i ,t 2 -I-t i T 2 , . .) = c . h(t1,. . . ,t d )

where c E k X and h is monic as a polynomial in tl over k[t2,... ,t d ] .

4.3 Quillen's Local Global Principle

Proof. Let f(t1, ... , t d ) =

97

aitlil - - - t d i d SO ,

i

+ terms of lower degree in tl such that the integers il + r2i2 + . - .+ rdidare

We can choose r2,. . . ,rd distinct for all the intervening d-tuples i = (il,. . . ,id). In fact if m is an integer greater than all involved ij, we may choose r j = mj-l, since then the integers il +r2i2 . - .+rdid will have different rn-adic expansions. Having thus .'+did in f ( t l , t 2 + t l r 2 , ...) chosen the 7-2, . . . ,r d , the monomials aitli1+T2i2+ will not cancel out each other, and the one with the highest degree and with ai # 0 will emerge as the leading term in f (tl ,t2 tlT2,. . .), regarded as a polynomial in tl.

+

+

We now give two proofs of Serre's Conjecture on the freeness of projective modules over a polynomial extension of a field. Theorem 4.3.14 (Quillen - Suslin) If k is a field, every finitely generated projective k[xl,. . . ,xn]-module is free.

Proof. (Via the Monic Inversion Principle) We prove the result by induction on the number n of variables. When n = 1 the result is known by Corollary 4.1.2. So let n > 1. Assume that all finitely generated projective k[xl,. . . ,x,-11-modules are free and let P be a finitely generated projective module over k[xl,. . . ,x,]. Let S := k[xn] \ (0). Then S-lk[xl,. . . ,x,] = k(xn)[xl,. . . ,x,-I], where k(x,) = S-lk[x,] is the quotient field of k[x,]. Now S-lP is a projective module over a polynomial extension of n - 1 variables over the field k(x,). By induction S-'P is free. Since P is finitely generated there exists a monic f E S = k[xn] \ (0), such that Pf is free. Therefore P is extended from k[xl,. . . ,%,-I] by the Monic Inversion Principle 4.3.12, i.e. there exists a P' E IP(k[zl,. . . ,xn-I]) such that P E P' 8 k[xl, . . . ,x,]. By induction P' is free. Hence P is free. Proof. (Via Quillen's Local-Global Principle) By Theorem 3.7.6 we know that P is stably free. It therefore suffices to show that k[x1, . . . ,xn] is an Hermite ring.

98

4 Serre's Conjecture

Let v(x,) = ( v l ( x n ) ., . . ,v,(x,)) E Um,(k[xl,. . . ,x,]). We can make a Nagata transformation of variables, fixing x,, by Lemma 4.3.13 so that vl(x,) is a monic polynomial in x, with coefficients in B := k [ x i , .. . ,X L - ~ ] for , some new variables xi, 1 5 i 5 n - 1. We show that v(x,) NSL, ( q X , l ) v(0). By Quillen's Local-Global Principle 4.3.9 it suffices to show that v(x,), N ~ ~ , ( ~ m [ x ,v(0)m7 l) for every maximal ideal m of B. This will follow from:

Lemma 4.3.15 Let ( R ,m) be a local ring and v ( x ) = ( v l ( x ) ,. . . ,v,(x)) E Um,(R[x]),r 2 3. If v l ( x ) is a monic polynomial, then v ( x ) can be completed to an elementary matrix.

Proof: Write S = R[x],let the 'overbar' denote 'modulo (vl(x))'and el := (1,O,... ,O).

-

-

-

S, is a semilocal ring by Corollary 4.2.2. And (vz( x ), . . . ,v,(x)) NEr-l

-

( s m ) el

by Corollary 3.4.8. Let Z E E , - ~ (3) be such that (vz(s), . . . ,v,(s))= ~Z i , and let E E ETP1( S ) be a lift of Z. Write (212 ( x ) ,. . . ,v,(x))E = (vi( x ) ,. . . ,V: ( 2 ) ) . Now ~ ( 5NE,(R[r]) ) (211( 2 1 1 vh ( x ) ,. - . ,v:(x)) N E p ( R [ ~ ] )el 0 the last assertion following as v;(x) 1 (mod vl ( x ) ) .

-

4.4 Suslin's Proof of Serre's Conjecture He finds a fascinating method to analyse the action of invertible matrices on unimodular polynomial rows with a monic entry.

Lemma 4.4.1 Let R be a ring, I an ideal of R[t] which contains a monic polynomial and J an ideal of R such that I+ J[t]= R[t].Then ( R f l I )+ J = R. Write R := R / ( R n I ) and 7 the image of J in fE. The ring extension R [ t ] / Iis integral, since I contains a monic polynomial.

Proof. R

+

If ( R n I ) J # R , there would be a maximal ideal m of R containing 7.By the Lying Over Theorem 1.4.18 and Proposition 1.4.16 there were a maximal ideal m' of R [ t ] / Iover m. Its preimage in R[t]would contain I J[t],contradicting 0 the assumption.

+

+

Lemma 4.4.2 Let R be a ring and f = ( f l , f 2 ) E ~ [ t Let ] ~ c. E R f l ( f l R[t] fzR[t]).Then for any commutative R-algebra A for which c is a non-zerodivisor and any b, b1 E A,

b

b1 (mod cA)

=+

f (b) - s L z ( ~ ) f (bl).

4.4 Suslin's Proof

Proof. We can write c = fig1 matrix M=-

+ f292 with g1,g2 E R[t]. Over A,

99

define the

It belongs to SL2(A). Namely first det(M) = (l/c2) . c . c = 1. Then modulo cA, the product of the two matrices on the right hand side is congruent to 0. So the entries of M belong to A. Now we have

Remark 4.4.3 The condition that c is a non-zero-divisor for A the only case we need here may be dropped. For this aim one can carry out a formal consideration, which is not quite obvious. (See Exercise 2.) -

-

Lemma 4.4.4 Let R be a ring and f E R[tIn. Then for any commutative R-algebra A and any subgroup G C GL,(A)

I = If,A,G := {C E R

Ib

b' (mod Ac)

+ f (b) -G

f (b'))

is always an ideal in R. Proof. Let c, c' E I , r, r' E R. To see that rc+r'cl E I , let b- b' = a(rc+rlc') where b, b', a E A. Since b - a r c = b' ar'c', we have f (b) -G f (b - (ar)c) = f (b' (ar')cl) -c f (b'). SO if,^,^ is an ideal in R.

+

+

Theorem 4.4.5 (A. Suslin) Let R be a ring and f = (fl, . . . , f,) E 2, with fi monic. Then for any commutative R-algebra Um,(R[t]), n A which is a domain, and any b, b' E A we have f (b) -G f (b') where G is the subgroup of GL,(A) generated by E,(A) and SL2(A).

>

Proof. For f and G as in the statement of the theorem, we want to prove that the ideal If,A,G in Lemma 4.4.4 is the unit ideal in R. So for any given maximal ideal m c R , it suffices to find an element c E I \ m.

100

4 Serre's Conjecture -

Note that (T2,.. . ,f ), E um,-l(R[tl) where R[tl = ( R / m ) [ t ] / ( J , ) which , is a semilocal ring, since it is a residue class ring of a principal domain by a nonzero ideal. Thus-byCorollary 3.4.8 there is an 2 E E n - l ( R [ t ] / ( f l ) m [ t ] ) such that (T,, . . . ,f n ) M = (i,Ti,. .. ,iS). ~ i f 2 t to M E ~ , - ~ ( R [ tand l ) let

+

=

+

+

+

1 (mod ( f s ) m[t]),whence f l R[t] g2R[t] m[t]= R [ t ] .Since Then g2 f l is monic, we can infer from Lemma 4.4.1 that R n ( f l , g2) + m = R. In particular there exists an element c E R n ( f l , g2), c $ m. We will be done if . check this let b G b1(mod cA). For i 2 2 we we can show that c E I f , ~ , GTO have gi (b) - gi(bl) E (b - b')A c C Ac f l (b)A g2 (b)A.

+

Thus via Lemma 4.4.2

>

Corollary 4.4.6 Let R be a ring, f ( t )= ( f l ( t ).,. . ,f,(t)), ( n 2) a unimodular row over R[t]and f l ( t ) be monic. Then f (t)NG f (0) where G is the subgroup of GLn(R[t])generated by En(R[t])and SL2(R[t]).

Proof. Apply Theorem 4.4.5 to A = R[t]and b = t , b' = 0.

0

Corollary 4.4.7 If k is a field then every finitely generated projective module P over the polynomial ring A := k[ts,. .. ,td]is free.

Proof. Since P is stably free by Theorem 3.7.6, it is enough to show that every unimodular row ( f s , . . . ,f,) over k [ t l , . .. ,td]can be completed to an invertible matrix. (See Corollary 3.1.11.) This is clear, if fl = 0 , by Proposition 3.4.5. If f l # 0 , using some Nagata transformation we may assume that fl is monic in ts (upto a factor in k X ) .Then by Corollary 4.4.6

We finish by induction on d.

0

In Appendix A we give a proof of a deep theorem of A. Suslin that the elementary subgroup En(A) of G L n ( A ) is a normal subgroup when n 2 3. We use this fact to deduce sharper results.

4.5 Vaserstein's Proof

101

Corollary 4.4.8 (R.A. Rao) Let R be a ring, f (x) = (fl (x), - . - ,fn(x)) E Umn(R[x]) with n 2 3 and f ~ ( x )monic. Then f (x) can be completed to an elementary matrix. We use a trick of S. Mandal [53]. (See the last Exercise on Appendix A for a Horrock's like argument.)

+

Proof. Consider f *(x,t) = (td"(x t - t-l)) E Umn(R[x, t]), where di = deg fi(x). (See Lemma 4.5.2). By Corollary 4.4.6 there is a a ( x , t ) in the group generated by SL2(R[x,t]) and En(R[x,t]) such that f *(x,t)a = el. Since En(R[x,t]) is a normal subgroup of GLn(R[x,t]) we may write a = E(X,t)(In-2 I 6(x,t)) for some ~ ( x , t )E En(R[x,t]), 6(x,t) E SL2(R[x,t]). Hence f * (x, t ) ~ ( xt), (In-2 I6(x, t)) = el, whence

f *(x,t ) ~ ( xt), = el (In-2 16(x, t)-l) = el. Now put t = 1 to recover f (x)e(x, 1) = el.

Theorem 4.4.9 (Quillen - Suslin) If k is a field and A = k[tl, . .. ,td] then any f = (fi, . . . ,fn) E Umn(A) with n 2 3 can be completed to an elementary matrix. The proof is analogous to that of Corollary 4.4.7. We only use Corollary 4.4.8 to achieve the stronger result.

4.5 Vaserstein's Proof of Serre's Conjecture Quillen's proof of Serre's conjecture does not use the fact that projective modules over K[xl, . . . ,xn] are already stably free. But using this fact and Quillen's ideas, one obtains a remarkably short proof of Serre's conjecture, as L.N. Vaserstein wrote in a letter to H. Bass. The next lemma is a special case of Lemma 4.3.7 when P = R[xIn. Also the proof is the same as there. The only difference is that is is written in terms of matrices. Note that there is an obvious ring isomorphism Mn(R[x]) 7 Mn (R)1x1-

Lemma 4.5.1 Let R be a commutative ring and S be a multiplicatively closed subset of R. Let r(x) E GLn(Rs[x]) be such that r(0) = In. Then there exists ) some s E S, a matrix ?(x) E GLn(R[x]) such that ?(x) localizes to ~ ( s x for (i.e. ?(x), = ~ ( s x)) and ?(0) = I,.

102

4 Serre's Conjecture

Proof. Since T ( X ) E GLn(Rs[x]),there exists p(x) E GLn(Rs[x])such that r ( x ) p ( x )= I,. Also as ~ ( 0=) I,, we have p(0) = I,.

+

+

Thus T ( X ) = (6ij x f i j ( x ) )and p(x) = (6ij xgij ( 2 ) )where f i j ( x ) ,g;j ( x ) E Rs[x].Since there are only a finite number of denominators, we can find an sl E S such that ~ ( s l xand ) p(s1x) are images of matrices r1( x ) ,resp. pl ( x ) over R[x]with 7 1 (0) = In = pl(0).

+

Let r l ( x ) p l ( x )= (bij x h i j ( x ) ) ,then, since ( ~ ~= I,,p there ~ ) is ~an $2 E S with = 0 for all i ,j. So ?(x) := rl(s2x) E GL,(R) and localizes to ~(s1~2~). Lemma 4.5.2 (S. Mandal) a) Let f ( x ) E R[x] be a rnonic polynomial of degree d. Then f * ( x ,t ) = t df ( x t - t-l) E ( R [ x ] ) [ tis] a bimonic polynomial in t (i.e. the leading coeficient and the constant term of f * ( x ,t ) , regarded as a polynomial in t over R[x] are 1).

+

b) Let f ( x ) = ( f l ( x ) ,.. . ,f,(x)) E Um,(R[x]) with f l ( x ) rnonic. Then f * ( x t, ) = (tdlf l ( x t - t - l ) , . . . ,tdrf r ( x t - t - l ) E Um,(R[x,t ] ) ,where di = deg fi. Moreover, clearly f * ( x ,1) = f ( x ) .

+

+

Proof. a) Let f ( x ) = xd

+ alxd-I + -

- +ad, for ai

E

R, 1

< i < d. From

it is clear that tdf ( x + t - t - l ) = t2d+ - . - + 1. b ) For this part note that f * ( x ) E ~ m , ( ~ [ x , t , t - ' 1 )Hence . t1 E EL1tdif,*(x,t ) ) ~ [t]x ,for some I 2 0. But tdlf,*(x,t) = 1 mod t R [ x ,t ] , whence t"[x,t] + tdlf,*(x,t ) R [ x t] , = R [ x ,t]. Theorem 4.5.3 If k is a field, any finitely generated projective module over A = k [ x l ,. . . ,x,] is free.

Proof. (L.N. Vaserstein) W e know by Theorem 3.7.6 that every projective module over Ic[xl,.. . ,x,] is stably free. By Lemma 3.1.11, it is enough to show that every unimodular row of length r 2 3 over k [ x l ,... ,x,] is completable. Let R = k [ x l ,. . . ,x,-l] and f (x,) = ( f l ( z , ) , .. . ,f,(x,)) E Um,(R[x,]). After making a Nagata transformationof variables (see Lemma 4.3.13), fixing x,, the polynomial f l ( x n ) becomes rnonic. W e claim that f (x,) N ~ ~ , ( ~ [ z , fl )(0). Consider f * (x,, t ) E Um, ( R [ x n t, ] )defined in Lemma 4.5.2, Since f * (x,, 1) = f (x,) it suffices to show that f *(x,, t ) is completable. In particular we may assume (after renaming f c ( x n , t ) as f ( t ) E Um,(A[t]))that we have a unimodular row f ( t ) = ( f l ( t ) ,. . . ,f , (t))with f l (t)rnonic and fl(0) = 1. Using elementary transformations we may further ensure that f;(O) = 0 for i > 1. By Lemma 4.3.15, i f m E Spmax(A) then f (t), NE,(A,[t]) el. Therefore,there is an s E A \ m such that f (t),N ~ , ( ~ , [ t l el. )

4.5 Vaserstein's Proof

103

Let Q(f (t)) = {s E A I f (t), - S L , ( A . [ ~ ~ ) el) be the "Quillen ideal" of f (t). As in Quillen's proof it is enough to show show that Q(f (t)) is indeed an ideal: For then it will follow from above that Q(f (t)) = A[t], i.e. 1 E Q(f (t)), i.e. f (t) -SL, (A[t]) el as required.

+

Clearly we need only show that s l , sa E Q(f (t)) implies sl sa E Q(f (t)). Further, after inverting sl s2, we may assume sl sa = 1, and also s l , sa E Q(f (t)) \ (0). Let a(t) E SL,(As, [t]) with f (t)a(t) = el. Since f (0) = el we may modify a(t) and assume a(0) = I,. Let y(x, y, t) = a((% y)t)a(yt)-l E SL, (Asl [x,y, t]) and P(t) E SL, (As2[t]) such that f (t)P(t) = el and P(0) = I,. By Lemma 4.5.1 there is an m > 0 such that for all b E (ST), c E (sy),

+

+

+

(We consider A,, as subrings of the quotient field Q(A), the si being nonzero in a domain.) Note that

+

Since (sl, s2) = A there are b E (SF), c E (SF) with b c = 1. Now, put x = b, y = c, then y(b, c,t) E SL,(A[t]), P(ct) E SL,(A[t]), and

as required.

0

Continuous Vector Bundles

There are a large number of useful analogies and relations between algebra and topology. In this chapter we will describe one such relationship: that existing between projective modules and vector bundles. The main application of this so far have been to the construction of non-trivial examples of projective modules, the non-triviality being proved by passing to the associated vector bundle and using topological methods.

5.1 Categories and Functors To give an adequate formulation of the statements in this chapter, we introduce here the language of categories and functors, which is extremely important in many fields in modern mathematics. 5.1.1 We begin with examples, one of a category and one of a functor. The class of all R-modules over a fixed ring R together with the sets HomR(M,N ) for all pairs of R-modules M, N and the compositions

make up a category. Let R + S be a ring homomorphism. The map which assigns to every R-module M the S-module S @R M and to every R-linear map f : M the S-linear map ids @R f : S @R M + S @R N is a functor. Definitions 5.1.2 a) A category C consists of ( 1 ) a class of objects (sometimes denoted by Ob(C) ),

+N

106

5 Continuous Vector Bundles

(2) for every pair M, N of objects a set of morphisms Homc(M, N ) (sometimes one writes Morc(M, N ) or C(M, N) for this set),

(3) a specified element id^ E Homc(M, M ) for every object M ,

( 4 ) for every triple L, M, N of objects a map

such that ho(gof ) = (hog). f , and idNof = f , f oidM = f ,

whenever the left hand sides are defined. b) A n isomorphism i n a category is a morphism f E Hom(M, N ) for which there is a morphism g E Hom(N, M) with gof = id^ and fog = id^. Objects M, N are called isomorphic i f there exists a n isomorphism i n Hom(M, N ) . W e write M E N i n this case. Clearly the relation ' E ' is a n equivalence relation i n the class of objects. Examples 5.1.3 a) Sets and maps. Note that Lord Russel's famous antinomy tells us that we cannot speak about the set of all sets. But to speak of the class of all sets does not lead to a contradiction. The inconsistency in the definition of the barber of the village, to be the man who shaves every man who doesn't shave himself, totally disappears, if the barber is a woman. b) Topological spaces and continuous maps. c) Rings (resp. groups, resp. R-modules) and ring (resp. group, resp. Rmodule) homomorphisms. d) Vector bundles over a topological space X and bundle homomorphisms over X (which will be defined in the next section). And so on. e) If G is a group (or more generally a monoid) one can form a category with exactly one object, say X , and Hom(X,X) = G, '0' being the product in G. Definition 5.1.4 Let C, D be categories. A functor F : C-+D is a map which associates to every object M i n C an object F ( M ) in D and to every morphism f E Homc(M, N ) i n C a morphism F ( f ) E HomD(F(M),F ( N ) ) in D , such that F ( i d ~= ) i d q q and F(g0f) = F(g)oF(f) i f gof is defined. N o t e 5.1.5 One often writes f : M + N for f E Home (M, N). The meaning of a commutative diagram in a category then is clear. A functor transforms isomorphisms into isomorphisms and commutative diagrams into commutative diagrams.

5.1 Categories and Functors

107

Example 5.1.6 Let A be a ring and E an A-module. Then the assignments H o ~ A ( E , M ) f, I-, f,, (resp. M M @ A E, f f E ) make up M a functor from the category of A-modules to that of abelian groups. (If A is commutative, these are also functors from the category of A-modules to itself.)

*

*

*

Definitions 5.1.7 Let F : C -+ D be a functor. a) F is called faithful if F maps Homc(M, N ) Hom,(F(M), F ( N ) ) for every pair M, N of objects in C. b) F is called full if F maps Homc(M,N) HomD(F(M),F ( N ) ) for every pair M, N of objects in C .

injectively

to

surjectively

to

c) F is called a n equivalence if it is full and faithful and for every object P in D there is a n object M in C with P S F ( M ) .

Remark 5.1.8 If F is full and faithful and M, N are objects in C , then F ( M ) r F ( N ) implies M r N. So a full and faithful functor gives an injective map of the class of isomorphism classes of objects in C into that in D. An equivalence gives a bijective map of the class of isomorphism classes of objects in C to that in D. Further, using the Axiom of Choice for classes, to an equivalence F : C -+ D one can construct a functor G : D -+ C such that G(F(M)) r M and F(G(P)) E! P for every object M in C and P in D. Moreover, these isomorphisms can be chosen as so called natural isomorphisms. This means that for morphisms f : M + N, g : P + Q in C , resp. D one gets commutative diagrams

Remark 5.1.9 A functor as defined above is often called covariant, since it preserves the "direction of arrows". A contravariant functor reverses it. This means it gives maps Homc (M, N ) + HomD(F(N) , F(M)) . Clearly one must require F(g0f) = F(f)oF(G). An example is the following: Fix an R-module E. Then one gets a contravariant functor F from the category of R-modules to that of abelian groups by F ( M ) := HomR(M,E ) and F ( f ) := f * , where f * is defined as in Chapter 1 Section 4. If C is a category one defines the opposite category COP as follows: Cop has the same objects as C and Homc,,(M, N ) := Homc(N, M). A contravariant

108

5 Continuous Vector Bundles

functor C ---+ D is the same as a covariant functor COP +D or C +DOP. So in abstract category theory one may restrict oneself to covariant functors, and call them simply 'functors'.

5.2 Vector Bundles 5.2.1 We asked DALEHUSEMOLLER to give us some information on the historical development of vector bundles theory. He wrote us a short note, which we - thankfully - reproduce here without change.

"The general concept of fibre bundles has its origins already in the 1930's with Seifert, Ehresmann and Whitney. Then in the 1940's came the book of Steenrod, and in 1949-1950 the Seminar of Henri Cartan containing a general introduction to fibre bundles through principal bundles. In the 1950's differential topology, that is the topology of smooth, piecewise linear, and topological manifolds, had a great development. The first impulse came from Thom's 1953 thesis which reconsidered the theory of characteristic classes and his 1954 article on cobordism and transversality and the second impulse came from Milnor's 1956 discovery of exotic differential structures on the 7-sphere followed by his work on the Hauptvermutung. Fibre bundles and especially vector bundles which are fibre bundles with a vector space V as fibre and structure group GL(V) played a role everywhere in this development. Milnor gave courses at Princeton and wrote up notes, which were mimeographed and distributed in a small circle of people - sometimes supported by the Government Grants of the time. It was just before the Xerox machine and long before electronic servers. There were two sets of notes of Milnor which had a wider influence, especially when the new Xerox machine made their reproduction easier. They were Differential topology and Characteristic classes. The second were notes of Milnor's lectures which were taken by Stasheff and which were eventually published with additions by Milnor and Stasheff in the Annals of Mathematics series. In Differential topology Milnor studies vector bundles directly without a reference to a general theory of fibre bundles, and in these and another set of notes he proved the homotopy classification theorem. Then in 1957 came the K-theory in Grothendieck's general Riemann-Roch theorem. This was followed a year later by Bott's periodicity theorem which was very soon after interpreted as giving a new cohomology theory, called topological K-theory. This was introduced by Atiyah and Hirzebruch. The analytic implications were soon after discovered by Atiyah and Singer in the general index theorem. The affine version of Grothendieck's K-theory lead to the study of the class groups of projective modules over a ring. Both Serre and Swan showed that

5.2 Vector Bundles

109

finitely generated projective modules over the ring C(X) of continuous complex (or real) valued functions on a compact space X came as modules of cross sections of vector bundles. So the vector bundles were related to finitely generated projective modules by an equivalence of categories." In [I031 L.N. Vaserstein gave a new very general version as well of this correspondence as well as that of a correspondence to homotopy classes of maps to Grassmannians, which we will present here. 5.2.2 We denote by IF one of the skew-fields IR, C or H (of the real or complex numbers or quaternions). The reader may restrict his attention to the fields IR and C.

Recall that a finite dimensional IF-vector space V possesses a canonical topology. (IFn has the product topology, and since the automorphisms of IF" clearly are continuous all isomorphisms IFn 7V induce the same toplogy on V.) Definitions 5.2.3 Let X be a topological space.

a) A quasi vector bundle J over X consists of a topological space E(J), a continuous map pc : E(J) -+ X , and the structure of a finite dimensional IF-vector space o n every fibre F'(J) := p ~ ' ( x ) (where x E X), i n such a way, that the two topologies on every F,(J), one coming from the vector space structure, one induced by the topology of E(J), coincide. E(J) is called the total space of J.

The quasi vector bundle, given by p : E -+ X will sometimes be denoted by Ip : E -+ XI, sometimes, i f possible, by E b) A homomorphism, f : J -+ Q of quasi vector bundles over X is a continuous map (denoted by the same letter) f : E(J) -+ E(J1), such that

commutes, and for every x E X the induced map F, (J) + F, (5') is IF-linear. Consequently an isomorphism is a homomorphism which has an inverse, or equivalently which is bijective. It is then clear, what we mean by the phrase "J and are isomorphic" or "J is isomorphic to q ". c) A quasi vector bundle over X is called a trivial vector bundle, i f it is isomorphic to the quasi vector bundle pr, : V x X -+ X , where V is a finite dimensional IF-vector space. (The vector space structure o n the fibres of pr, should be clear.)

110

5 Continuous Vector Bundles

d) Let U be a subspace of X . The restriction [Iu of a quasi vector bundle 5 over X t o U is obviowly defined, namely by the restriction of pg t o p,'(u) + U. e) A vector bundle over X is a quasi vector bundle [ which is locally trivial. B y this we mean that every x E X has a neighbourhood U, such that the restriction [lu is a trivial vector bundle. Homomorphisms of vector bundles are those of quasi vector bundles. Remarks 5.2.4 a) Let X be a topological space. The vector bundles over X together with the vector bundle homomorphisms make up a category. b) The rank or dimensionrk : X + IN, x e rk~F,([) = dim^^ F,(<) clearly is locally constant for a vector bundle. (This does not hold for quasi vector bundles.) Definition 5.2.5 A vector bundle of constant rank 1 is called a line bundle.

5.2.6 A vector bundle gives us data of the following kind. There is a covering (Ui)iE~of X , such that := [IUi is trivial for every i , i.e. we may assume E(ti) = & x Ui, where & is an IF-vector space and pci = pr,. If Ui n Uj # 0 clearly V , Z Vj. For every x E Ui n Uj we get an isomorphism a, : V , + Vj. Choosing bases of every V , we may assume a, E Gl,(IF). So we have a map Ui n Uj -+ Gl,(F), x e a,.

ci

CLAIM: This map is continuous. To see this, it is enough to show, that every column of a, depends continuously on x. Fix a basis in V,, take its r-th vector e, and consider

The maps of the first line are the the isomorphisms according to (5.2.3) b) and c). They are continuous. Therefore (e,, x) e (a,(e,), x) is continuous, hence x H a,(e,) is so too. 5.2.7 On the other hand let X be a topological space, (Ui)iEr a covering of it, & an IF-vector space for every i E I and continuous maps

with aii(x) = idv, for x E Ui and dk(x)oaij (x) = ai"x) for x E Ui nUj n Uk. (Here we wrote a? := a G (x).)

5.2 Vector Bundles

From such data we construct a vector bundle

111

5 in the following way.

On the 'disjoint union' (or 'coproduct') &,(I4 x Ui) we define the equivalence relation by (v, x) (w,y) tj there are i, j such that x = y E Ui r l Uj, v E &, w E V,, aij(x)(v) = w. "N"

Then E(5) := U(V, x Ui)/ be the obvious one.

N

and the projection pc : E(5) -+ X is defined to

Clearly every fibre has a well defined vector space structure. Further every map

is a homeomorphism. Namely it factorizes through & x Isom(V,, Vj) x (Ui fIUj) in the following way:

And both maps are continuous. Since aii(x) = idv, and aji(x)oaij(x) = aii(x) , the inverse is continuous too. So, since the glueing of the & x Ui is done by homeomorphism, one sees, that the canonical maps 6 x Ui -+ E(5) are embeddings. By this one easily derives the property of being a vector bundle.

Definition 5.2.8 Kernel and image of a vector bundle homomorphism f : 5 -+ 7 are sensibly defined. Namely the kernel ker(f) is the union of the kernels ker, (f) of the induced maps F, (5) -+ F, (7) together with the restricted map PC : Uzexkerx (f -+ X The image is the image of the map f : E(J) -+ E(7) together with the restriction of p,, to f (E(5)). Clearly kernel and image are quasi vector bundles, but in general they need not be vector bundles. Consider e.g. the (trivial) bundle 5' := [pr2 : F x F -+ IF] and the bundle homomorphism 5 -+ 5, given by F x F + F x IF, (y, x) I-) (xy, x) - which is a notorious counterexample in mathematics. But in important cases kernel and image are vector bundles, as the following lemma shows.

Lemma 5.2.9 Let f : 5 -+ 11 be a vector bundle homomorphism (over a topological space X). The following assertions are equivalent: (1) The rank of ker(f) is locally constant; (2) the rank of im( f ) is locally constant;

112

5 Continuous Vector Bundles

(3) ker(f) is a vector bundle; (4) im(f) is a vector bundle; Proof. (1)

(2) is trivial, and (3) + (1) and (4) + (2) are so as well.

(I), (2) + (4) and (3). Let x E X and U be a neighbourhood of x over which J and q are trivial, say k and y E U' near x we can write f (si, y) in the form

c:=,

a,(y)sj for i > k belongs to the kernel of f for every So sj(y) := si y E U'. The s',+,(y), . . . , sk(y) are linearly independent and are in the right 0 number. So they generate the kernel for y E U'. We will use this especially in the case when f is surjective hence (2) is fulfilled.

Definitions 5.2.10 a) The ring of continuous functions f : X -+ F will be denoted by C(X). b) Let J be a vector bundle over the topological space X . A section s of J is defined to be a continuous map X -+ E(J) with pcos = idx. (This means S(X)E Fz(J) for every x E X.) c) The set of sections of J is denoted by r ( J ) . I t is a C(X)-module in a canonical way. d) Let f : J -+ q be a vector bundle homomorphism over X. This is given by a ffibrepreserving' map f : E(J) -+ E(q). Therefore f o s E T(q) if s E r ( J ) . Define r ( f ) ( s ) := f 0s. Since f is fibrewise linear, r ( f ) is a C(X)-module homomorphism. e) Let U c X. By a section over U we mean a section of the restricted bundle Elu. We use the notion ' global section: if we want to stress that we mean a section over all of X.

5.2 Vector Bundles

113

Remarks 5.2.11 a) A vector bundle J over X is trivial, if and only if there are sl, .. . ,s, E r ( J ) such that for every x E X the n-tuple sl(x), . . . ,sn(x) is a basis of F, (5). b) Clearly r is a functor from the category of vector bundles over X to the category of C(X)-modules. c) If X = [pr, : IFn x X + XI is a trivial vector bundle, then r(X) E C(X),. And a global basis of J is a basis of r ( J ) as a C(X)-module.

Definition 5.2.12 Let J1, . . . ,Jn be finitely many vector bundles over the same space X . Their direct sum @;=l Jj = Jl @ - .- @ &is defined by the total space E(@yZl Jj) :=

and the obvious projection: pcle...ecn(al, . . . ,an) := pel ( a ) (= pgj (aj) for any j). For every fibre we have F,(51@. - .@In) = Fx(J) x . x Fx(Jn)- Define a quasi vector bundle structure by equipping the latter cartesian product with the direct sum structure i n the usual way. Remarks 5.2.13 a) If Jj is one.

51,. . . ,J,

are all trivial vector bundles, then clearly

b) If (Ui)iEl1,.. . , (Ui)iE~n(where the Ij are assumed to be pairwise disjoint) be trivializing open coverings of J, resp. q. Then all Jj are trivial over every of the following open sets: Uil n . . . fl Ui,, (il, . . . ,in) E Il x , . . . x I,. These Jj is a vector bundle. sets make up a covering of X. This means that c) If Ji, . . . ,Jn are subbundles of a vector bundle ( such that over every x E X one has F,(() = Fz(E1) @ . . . @ F z ( J n )then ( E 51 CB...CBJn in a canonical way. d) A trivial vector bundle of rank n is a direct sum of n trivial vector bundles of rank 1. e) r(@j"=l Jj)

S

r ( & ) in a canonical way.

5.2.14 The direct sum of vector bundles has some formal properties in common with the direct sum of modules. For every i = 1,.. . ,n there is a canonical projection pi : @;=, Jj + ti.There is a bijective correspondence between the vector bundle homomorphisms

with the n-tuples of homomorphisms

114

5 Continuous Vector Bundles

given by f

+ ppf

@Yzl

For every i = 1, . . . ,n there is a canonical embedding e i : ti + .There is a canonical bijective correspondence of the vector bundle homomorphisms

with the n-tuples of homomorphisms

given by fi = f oei. 5.2.15 Therefore every vector bundle homomorphism

is uniquely described by an m x n-matrix ( f i j ) whose entries are homomorphisms fij :Sj

And it is clear that

n

+ Vim

is uniquely described by the m x n-matrix (r(fij)) whose entries are the homomorphisms r(fij) : r(Sj) +r(qi)A special case is that all bundle homomorphism

S j , qi

are trivial bundles of rank 1. Every vector

is uniquely described by an m x n-matrix whose entries are elements of C ( X ) . And r ( g ) is described by the same matrix.

5.3 Vector Bundles and Projective Modules Definitions 5.3.1 a) A finite partition of unity on a topological space X is a finite family (or set) of continuous functions f i : X + [O, 11 with fi(x) = 1 for e v e y x E X . ( B y [0,11 we denote the closed unit interval.)

xi

5.3 Vector Bundles and Projective Modules

115

b ) finite envelope of unity on a topological space X is a finite family of continuous functions fi : X + [O, 11 with Maxi{fi(x))= 1 for every x E X . c) The support Supp(f) of a function f : X { x E X I f ( x ) # 0).

-+ IR is the closure of

the set

d ) Let U = (Ui)lSiLn be a finite open covering of X . A partition of unity, resp. envelope of unity is called subordinate to U , i f and only if it is of the f iUi. ) form ( f i ) l l i l n with S u ~ ~ ( C e) A locally finite partition of unity on X is a family of continuous functions ( f i : X + [0,l ] ) i G 1such that for every x E X there are only finitely f i ( x ) = 1. many i E I with f i ( x ) # 0 and

xi,-*

Remark 5.3.2 Let g l , . . . ,gn be finitely many nonnegative continuous real functions on a topological space X such that for every x

+

E

X there is an i with gi(x) > 0.

(5.1)

+

Define s := gl - .- gn, which clearly has no zero. Then apparently { g l / s , . . . ,gn/s) is a partition of unity on X . Analogously define s f ( x ) := Maxi{gi(x)) for every x Max{gl/st,. . . ,gn/st) is an envelope of unity.

E

X . Then

Clearly any partition and any envelope of unity fulfil Equation (5.1). So the three concepts of a partition of unity, an envelope of unity and a family of nonnegative functions, fulfilling Equation (5.1) are equivalent in some sense. Lemma 5.3.3 Let ( f i ) l l i < _ nbe a finite partition of unity on a topological space X . Then U := (Ui)l
a

Proof. For every x E X there is at least one i with fi(x) > 0. ThereforeU is a covering. But note, i f Ui # (E)for some i, then (fi)i=l,...,, is not subordinate to [U]. Define

(Remember: n is the number of the fi.) Then h ( t ) > 0 for t 2 l l n . So the gi := hofi have the property (5.1) since for every x E X at least for one

i E ( 1 , . .. ,n ) we have fi(x) 2 l l n , whence gi(x) > 0. Further Supp(gi) C { x E X I fi 2 l / ( n 1 ) ) c Ui. Remark 5.3.2 then proves the lemma. 0

+

116

5 Continuous Vector Bundles

P r o p o s i t i o n 5.3.4 Let J be a vector bundle o n a topological space X . The following two properties are equivalent: (i) There is a finite partition of unity (fi)l 0), is trivial. (ii) There is a finite open covering U = (Ui)lliln of X , such that J restricted to any Ui is trivial and there exists a partition of unity subordinate to U .

Proof. (i)

(ii) is Lemma 5.3.3.

(i). Let (fi)lSiln be a partition of unity, subordinate to U . Define V , (ii) as in (i). Then (V,)lliln is a finite open covering of X , as we know. Since 0 V, C C Ui, the bundle J restricted to any V , is trivial.

D e f i n i t i o n 5.3.5 A vector bundle J o n a topological space X is called s o f t (strongly o f finite type, if and only if it has the properties (i), (ii) of the Proposition 5.3.4. R e m a r k s 5.3.6 a) If X is a normal space and U = (Ui)i any finite open covering of X , then there exists a partition of unity, subordinate to U . See [38] Chapter 5, Problem W. (In our notion of 'normal' the Hausdorff condition is included.) Therefore, if J is a vector bundle on a normal space X which is of finite t y p e , i.e. if there is a finite open covering (Ui) of X , such that is trivial for every i, then it is soft. b) If X is a compact space, then every vector bundle over X is soft. (In our notion of 'compact' the Hausdorff condition is included.) 5.3.7 In this chapter we will use several times the possibility to equip IFn with an inner product ( , ) : IFn x IFn + IF by

where $jequals y if IF = IR and is the complex, resp. quaternionic conjugate of y if IF = C, resp. H. (Note that in the last case jjZ = 3.g.) It should be clear what we mean by the conjugate of a matrix. As usual, vectors v, w E Fnare called o r t h o g o n a l , if (v, w) = 0. In this case we write v Iw

m.

We define the norm Ivl of any v E Fn by Ivl := (This is the usual Euclidean 'length' of v, if one identifies IFn with R k n , with k = 1,2 or 4.) It is well known and easy to show that the topology, defined by this norm, on IFn coincides with the product topology.

5.3 Vector Bundles and Projective Modules

117

An orthonormal basis of Fnis an n-tuple of vectors vl, . . . ,V, in Fn,such that (vi, vj) = dij. (Kronecker's 6.) It is automatically a basis. The canonical basis of IFn is an orthonormal basis. If wl, . . . ,wn is any basis of Fn, then there is a canonical way to transform this into a basis, the G r a m Schmidt orthogonalization. Namely set first vl := Iwll-lwl. Then assume that vl, . . . ,v, is already an orthonormal basis of the space, generated by wl,. . . ,w,. If r < n, set first v;+, := w,+l - Z)~=l(w,+l,vi)vi. Then .;+I is orthogonal to all the vl, . . . ,v,. So, setting v,+l := IV:+~ l-' ~ :+~, we get the orthonormal basis vl,.. . ,v,+l of the space, generated by wl,. . . ,w,, w,+l.We will use the fact that the Gram - Schmidt orthogonalization is a continuous map from the set of all bases of IFn to itself. Here the topology on the set of bases is given by considering a base as an n2-tuple over F. Check this! 5.3.8 To every endomorphism a of Fnthere is a unique adjoint a*,defined by the property (av, w) = (v, o*w) for all v, w E Fn.With respect to the canonical basis and the above described canonical inner product, a* as a matrix is the transposed conjugate of a. One has (Boa)* = a*o/3*.

Following a suggestion of S. Lang ([45] VIII, 7 'Terminology') we will call an n x n-matrix a over F unitary, if it describes an automorphism of Fnas an inner product space, i.e. if besides linearity it fulfills (o(x),o(y)) = (x, y) for all x, y E IFn in any of the three cases F = IR, C or M.A matrix a is unitary if and only if it is invertible and a-' = a*. One calls a hermitian, if and only if a = a*.By the Spectral Theorem ([45] XV Theorem 6.7) an endomorphism a is hermitian if and only if there is a unitary a such that a a o * = diag(A1,. . . ,A,)

with A j E

R

(5.3)

5.3.9 The space Hom(Fm,F n ) of all homomorphisms Fm-+ Fncarries the so called operator n o r m I 1, defined by

It defines a topology on Hom(Fm,F n ) which coincides with the product topology if Hom(Fm,Fn) is interpreted as the set of n x m-matrices. It fulfills lao/?I IaIIPI. In case of a hermitian endomorphism, la1 equals the maximum of the absolute values of the eigenvalues of a.

<

If a is unitary, IaJ= 1. 5.3.10 Every linear subspace U of IFn is the image of a unique orthogonal projection e E Mn(F).

CLAIM:E E Mn(F) is an orthogonal projection if and only if E = c2 = E*.

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5 Contirluous Vector Bundles

A projection E is an endomorphism of Fn such that there is a splitting Fn= U@U' with e = idu@Ou,. We know by Paragraph 1.5.17 and Proposition 1.5.18 that E is a projection, if and only if e = e2. An orthogonal projection is a projection such that for the U, U' as above the condition U' = U' holds. (U' := {v E JFn I (v,u) = 0 for all u E U} means the orthogonal complement of U.) If E is an orthogonal projection with image U, then orthogonal bases of U and U' together make up an orthogonal base of F n . With respect t o such a base, E is described by the matrix diag(1,. . . , l , 0 , . . . , 0 ) and so clearly fulfills E =&*.

Conversely from equivalences

= E* we derive ( ~ ( x )y), = (x,c(y)) and hence have the

E

x E ker(&)e ( ~ ( x )y), = 0 for all y e (x, ~ ( y )for ) all y H x E i m ( ~ ) '

(5.4) (5.5)

So the claim holds. Therefore the set G;(IF) of linear subspaces of dimension k of Fn,the so called Grassmannian is in bijective correspondence to the set of idempoterlt hermitian n x n-matrices over F of rank k. We equip G;(IF') with the topology given by the operator norm on Mn(IF'). Arid the set G n ( F ) of all sub vector spaces of IF7', i.e. the disjoint union of the G;(F), gets the disjoint union topology. Note that J E = ~ 1 if E = E' = E* and E Hausdorff space since M n ( F ) is one.)

# 0. Hence G7"(F is compact. (It is a

Lemma 5.3.11 Let ( be a subbundle of a trivial vector bundle X = [Fnx X 3 XI. The orthogonal projection ~ ( x :)Fn -+ F, (E) is continuosly depending on x. Proof. Fix x E X . There is an open neighbourhood U' of x such that ( l u ~is trivial. Let t l , . . . , t k be a basis of over U' and s l , . . . ,sn be the canonical basis of X over X , hence over U'. There are n - k sections under the sl, . . . ,s,, say sk+l, . . . , s, such that t l (x), . . . ,t k(x), sk+l(x), . . . , sn(x) is a basis of F, (A). By the determinant argument of the proof of Lemma 5.2.9 we see that there is a neighbourhood U c U' of x over which t l , . . . ,tk, sk+l,. . . , s n is a basis. The Gram-Schmidt orthogonalization, which depends continuously on y E U, transforms the latter basis into a basis t i , . . . ,t',, s&+,,. . . ,s;, which is an orthonormal basis of IFn over every y E U. With respect t o this basis the orthogonal projection to ( is given by diag(1,. . . , l , O , . . . , 0 ) where k is the number of the 1's. With respect to the canonical basis this projection is given by adiag(1,. . . , l , 0 , . . . ,O)a-' where the columns of Q are I the t i , . . . ,t,, sI, + ~.,. . ,S; which are continuously dependent on y E U.

5.3 Vector Bundles and Projective Modules

119

Theorem 5.3.12 If J is a soft vector bundle over the topological space X , then there are a trivial vector bundle X and a vector bundle q over X with XEJCBq.

Proof. Let (Ui)i=l,..., be a finite covering of X together with a subordinate partition of unity (fi)i=l,...,, such that ( I u i is trivial and isomorphic to

with vector spaces V,. For any i we define a map from the trivial bundle V ,x X over X to J by (v,x)

for x $! Ui

V,) x X -+ E(J), which clearly is a Together these define a map 7r : surjective vector bundle homomorphism. Its kernel q is a vector bundle since its image is a vector bundle, namely J. By Lemma 5.3.11 the orthogonal projection E : X -+ q is a vector bundle ) q is a vector bundle, its kernel is so, too. And homomorphism. Since i m ( ~ = ) q. Finally ker(&)clearly is mapped isomorphically to J we have X = k e r ( ~ CB 0 by 7r.

Corollary 5.3.13 If X is a topological space and J a soft vector bundle over X , then r ( J ) is a finitely generated projective C(X)-module . We will show the converse:

Proposition 5.3.14 For every finitely generated projective C(X)-module P there is a soft vector bundle J over X with r ( J ) E P

Proof. There is an n and an idempotent endomorphism a of C(X)" with a(1'2(X)~)E P. (Compare this with Paragraph 1.5.17 and Proposition 1.5.18.) We interprete a as a matrix whose entries are continuous F-valued functions on X. So a may be considered as a vector bundle endomorphism of the ndimensional trivial vector bundle X := [pr, : Fn x X + XI. Let J denote its image (which a priori only is a quasi vector bundle).

CLAIM:J is a vector bundle. We have to show its local triviality.

+

Let x, y E X and set y(x, y) := a(x)a(y) (I - a(x))(I - a(y)), where I denotes the identity. Since a(%), a(y) are idempotent, we see immediately

120

5 Continuous Vector Bundles

Now fix x. Since y(x,x) = I, there is an open neighbourhood U of x such that y(x, y) is invertible for every y E U. So for y E U we have a(y) = y(x, y)-la(x)y (x, 9). So over U we get the vector bundle isomorphism

) ( a(y)(Fn) F") (Ifv E a(x)(pn), then ?'(X,Y)-'(V) E ~ ( x , Y ) - ~ ~ ( x ) ~ ( x , Y = and conversely.)

CLAIM:5 is a soft vector bundle. By Lemma 5.3.11, [ is the image of an orthogonal projection, i.e. we have an E E Mn(C(X)) with E = E~ = E* and &(A)= 5, i.e. E(X)= Fx([), where E(X)= E ( x ) ~= E(X)* E Mn(IF) depends continuously on x. The set Gn := {a E Mn(F) I a = a2 = a*)is compact. Clearly it is closed and la[= 1 or 0. So on Gn there is a finite partition of unity fl, . . . , f, such that la - PI < 113 whenever fi ( a )fi(P) # 0 for some i. Considering E as a continuous map X -+ Gn, the fioa make up a finite partition of unity on X. We will show that

5 is trivial on every Ui := {x E X I

fio&(x)# 0).

+ +

. For x, y E X set as above y(x, y) := E(X)E(Y) ( I - E(x))(I - ~ ( y ) ) We compute y(x, y) = ~ E ( x ) E ( ~I) - E(X)- ~ ( y = ) I ( ~ ( y) ~ ( x ) ) (l 24~).

+

<

1 Now, if x, y E Ui by the definition of fi and Ui we derive: ly(x, y) - I le(y) - E(X)I . I I - 249) I < $ . 3 = 1 From this we conclude that y(x, y) is invertible for x, y E Ui. Now, for x, y E Ui we have ~ ( y = ) ~ ( x9)-l~(x)y(x, , 9). AS above we have the triviality on Ui. 0

Theorem 5.3.15 The functor r from the category of soft vector bundles over X and the category of finitely generated projective C(X)-modules is an equivalence.

Proof. Since we know already that r induces a bijective map of the isomorphism classes of the objects in these categories, it will be enough to show that it is full and faithful, i.e. r induces an isomorphism

where Homx ( 5 , ~denotes ) the group of vector bundle homomorphisms 5 -+ 77. CLAIM:Let XI, X2 be trivial vector bundles. Then the canonical map Homx (XI, X2) -+ Homc(x) (r(X1),r(X2)) is bijective. Indeed homomorphisms XI -+ A2 and r(X1) -+ r(X2) both are described by matrices of the same shape whose entries are in C(X).

5.4 Examples

Then let f : 5 -+ q be any vector bundle homomorphism and q @ qt = X2 be trivial bundles.

5 @ 5'

121

= XI,

We may describe f as the composition

where the first and the last homomorphism are the canonical ones. Applying r we get

(5.6) By the claim r ( f ) # 0, if f # 0. Also by the claim we see that to every module homomorphism g : r ( t ) + r ( q ) there is a bundle homomorphism

with r(hl1) = g,r(hlz) = r(h21) = r(h22) = 0. We know already that this implies h12 = hzl = hz2 = 0. So the surjectivity of the map Homx(t, q) -+ Hornccx)( r ( t ) , r(q)) is also 0 clear.

5.4 Examples We construct several vector bundles, whose nontriviality can be proven by topological means. In certain cases we can give here complete proofs of the nontriviality. And we use these topological examples to find nonfree projective modules over Noetherian rings.

+

5.4.1 The Miibius Bundle. Let A := R[X,Y]/(X2 Y2 - 1) be the real coordinate ring of the circle and x, y the residue classes of X, Y respectively. . is generConsider the maximal ideal I associated to the point (1, 0) E I R ~ It ated e.g. by x + y - 1, x - y - 1. It is an invertible non-principal ideal as we have seen in Example 2.4.15 b) iii). Here we give a topological proof of the

CLAIM:I is not free, i.e. not principal. Consider A as a subring of C(S1) and set J := (x+y-l)C(S1)+(x-y-l)C(S1). Let P+ := (0,l) and P- := (0, -1). Then on S1 - {P+) the module J is

122

5 Continuous Vector Bundles

+

generated by x y - 1, hence corresponds there to a trivial line bundle. Analogously on S1 - {P-) it is generated by x - y - 1. The glueing of these trivial line bundles is given by S1 - {P+,P-) + GL1(R) = R X, (x, y) I+ (x y - l)/(x - y - 1). This function is < 0 on {(x, y) E S1I x > 01, and > 0 on {(a, y) E S1I x < 0). So we obtain the Mobius bundle, which is not trivial.

+

If it were so, it would have a continuous section without zeros. This would lead to continuous functions S1 - P+ -+ R Xand S1 - P- -+ RX,both of which could not change their sign. But, according to the glueing, one of them must 5.4.2 A projective module over A/Jac(A), which is not induced from one over A. Let X be any subset of Rn and define Sx c IR[tl,. . . ,t,] to be the set of all polynomials which have no zeros along X. Then the maximal ideals of ~ ; l R [ t l , . . . ,t,] belong to the points of X. (i.e. they are generated by n-tuples of the form tl - X I , . . . ,t, - x, with (XI,.. . ,x,) E X.)

Namely let the ideal I = (fl, . . . ,f,) of R[tl,. . . ,t,] be not contained in any of the maximal ideals attached to the points of X. Then f - - f: E Sfl I. SOS-'I = S-lR[tl, - . . ,tn]-

+ +

+

Apply this to X := S1 = {(x, y) E R2 1 x2 y2 = 1). Let A := ~ ; l R [ t ~t2]. , Then Jac(A) is generated by t: +t; - 1,and A/Jac(A) E R[tl, tz]/(t! +tz - 1). Since A - as a localization of a factorial ring - is factorial, every projective A-module of rank 1 is free. Hence the Mobius module over A/Jac(A) is not induced by any projective A-module. 5.4.3 Tangent bundles on spheres. Let Sn := { a E Rnfl the n-sphere - embedded in Rn+lin the usual way.

I

la( = 1) be

Consider the section o of the trivial real vector bundle X := [Ktn+' x Sn -+ Sn] defined by a($) := (x, a). This generates a trivial subbundle v of rank 1 of X, the normal bundle of Sn in R n f . (Recall that the vector x is normal to the sphere in the point x.) The tangent bundle of Snis the orthogonal complement r of v in A, i.e.

E ( r ) = {(y, x) E Rnfl x Sn I (y, x) = 0) So X = v

@r

(5.7)

and F,(x) is orthogonal to F,(x) in Fx(x) = Rn+'.

In spite of the fact that X and v are always trivial, the tangent bundle r is only trivial in the cases n = 1,3,7. This is a deep theorem in topology, whose proof is out of our scope. But at least - due to Milnor - we are able to prove the nontriviality of r if n > 0 is even - using only higher dimensional calculus. (Also see [8] for another accessible argument.)

If r is a trivial vector bundle, there are n elements el,. . . ,en E r ( r ) , such that el (x),. . . ,en(x) are linear independent for every x E Sn.

5.4 Examples

123

Therefore the following proposition is stronger than the nontriviality of the tangent bundle of an even dimensional sphere.

Proposition 5.4.4 Let r be the tangent bundle of an even dimensional sphere S2n.There does not exist any v E r(7)(a SO called vector field) such that v(x) # 0 for every x E S2". The hypothesis that the dimension of the sphere is even, is essential, as otherwise

gives a differentiable nowhere vanishing vector field on

S2"-I

Proof. (Milnor) Assume there is such a vector field, Using Weierstrass' Approximation Theorem we see that there exists even a continuously differentiable vector field v on S2"- see [57]. Replacing v by Ivl-'v, we may and will assume that Iv(x)l = 1 for every x E S2n.We can extend v throughout the region A := {x E R ~ ~ I +a ' 1x1 5 b), where 0 < a < 1 < b, by setting v(ru) = rv(u) for a 5 r 5 b and u E S2" = { U E I'R2"+' I IuI = 1)

<

For t E IR consider the function ft : A

-+ IR2"+',

ft(x) := x

+ tv(z).

We will show: a) ft is injective for small values of t, b) ft, for small values of t, transforms the annulus A onto a nearby region ft(A) whose volume can be expressed as a polynomial function of t. c) ft, for small values of t, transforms A onto the region

< <

{x E R ~ ~ I + aJ ' S1x1 b JS), whose volume equals ( d m ) 2 n + 1 ~ o(A). l This implies the proposition. Namely ( d m ) 2 n + ' is not a polynomial function of t, and so c) contradicts b). Proof the above items: a) Since v is continuously differentiable on the compact region A it satisfies a Lipschitz condition , i.e. there is a constant c > 0 so that Iv(x) - v(y)I 5 clx - yl for all x, y E A. In particular if It1 < c-' then ft is injective: Namely let x ft(x) = ft(y). Then x - y = t(v(y) - v(x)), hence

a contradiction.

# y, but

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5 Continuous Vector Bundles

b) We have

+

+

aift bvi det ((x)) = det (I t -(x)) = 1 tcl (x) axj axj

where the t. So

ak

+ . -. + t2"+l c2n+1(x),

are continuous functions of x. The determinant is Vol ft(A) = a0

> 0 for small

+ a l t + .- - + a2n+lt2n+1,

c) For small values of t, the sphere S of radius r around 0 is mapped by ft around 0.Namely, the matrix of first onto the sphere T of radius rdderivatives of ft is non-singular on A. By the Inverse Function Theorem ft is an open mapping on A. Hence ft(S2") is a relatively open subset of the sphere of radius r ( 2 / m ) . But the sphere S is compact and so ft(S) is compact and so closed. Since T is connected, any nonempty open and closed subset of T equals T. 0 Applying the functor r we see that r ( r ) is a stably free C(Sn)-module for all n > 0. It is nonfree for n # 1,3,7. We have shown this only for even n. Further we have seen that T(T) has a free direct summand of rank 1 if n is odd by Formula (5.8) and that it is free if n = 1 , 3 or 7 in Example 3.1.9. 5.4.5 One can apply this to rings, which are more of 'algebraic kind' than C(Sn). Consider e.g. A := IR[Xl,. . . ,Xn+l]/(X,2 . . . X:+, - I), which is the ring of polynomial (real) functions on Sn.This is a Noetherian subring of C(Sn). (See Chapter 6, Section 1.) Let xi be the residue classes of the Xi in A and consider the row (XI,. . . ,xn+l), which is unimodular, since C x: = 1. It defines a stably free module P , which, tensoring with C(Sn), becomes r ( r ) by Equation (5.7). So P is nonfree, if r is nontrivial, especially for even n.

+ +

5.4.6 Let B = IR[Xl,. . . ,Xn+l] and f = E X > Then (XI,. . . ,Xn+,) is unimodular over Bf and defines a projective Af-module P , which is stably free but not free for even n. (More general for n # 1,3,7.)

Namely the elements of Bf define continuous maps on IRnfl \ (0) and so on Sn. Therefore we get an obvious ring homomorphism Bf -+ C(Sn). Via this map P B B f C(s2) is the projective C(Sn)-module that corresponds to the tangent bundle on Sn. 5.4.7 Let B, f be as in the last example with even n, and m = BX1

-.+ BXn+l. Then

again - the stably free (B,) unimodular row (XI,. . . ,Xn+l), is not free. -

+

-module P , defined by the

Assume P is free. Then it is extended from B,, Hence there is a g E B \ m such that P is extended from the subring Bfg of (B,,,)f. (To describe P, only

a finite set of denominators of B \ m is required.) Let Z(g) c ELnf1 be the set of real zeros of g. It is closed and does not contain 0. So there is an nsphere S round 0 of some (small) radius with S n Z(g) = 0. The elements of B f g are continuous real functions on S (without poles). Therefore there is a ) well known ring embedding B f g L, C(S). And in C(S) by (XI,. . . ,X n + ~our nonfree, stably free module is defined. In these examples one may replace IR by any subring of IR. 5.4.8 Extension Problem. In the last example f belongs to the square of the maximal ideal mA, of the regular local ring A,. (For the definition of 'regular' see Chapter 8.) It is an open question, whether for a regular local ring (R, m) and an f E m \ m2 every projective Rf-module is free. This would imply that for every regular ring A every projective A[X]-module is extended from A - a conjecture of H. Bass, and now generally refered to as the BassQuillen conjecture, due to the above question raised by D. Quillen. Quillen's question was raised in the paper [71] where he solved Serre's conjecture, and is also known as Quillen's question! Some known cases of Quillen's question can be found in [7], [loo]. The known cases of the Bass-Quillen conjecture can be found in 1471, [74]. 5.4.9 We will give here an example of a nontrivial complex vector bundle on each sphere of odd dimension 5. It would take us afar to prove here the topological facts we will use.

>

Let S2"+l be interpreted as the subset of the

en+'defined by

Consider the bundle with total space

and with the obvious projection to S2n+1. Every fibre is an n-dimensional subspace of (En+' and the fibre over z is orthogonal to z, also if z and w are interpreted as real vectors. Namely the real part of zjwjis the usual ~ the ' ~vector . bundle E = real inner product if one interpretes en+ as 1 ~ ~ SO [E -+ S2n+1] is a subbundle of the real tangent bundle over S2n+1 of real rank 2n. For z E S2nf1the vector iz also is orthogonal to z over IR,i.e. a tangent vector to S2n+1 in z. (The vector field z iz is the same as that in Formula 5.8 in Proposition 5.4.4!) The vector bundle {(z, riz) E S2n+'x en+'I r E IR) Over every z E S2n+1 it is a trivial real vector bundle of rank 1 over S2n+1. is orthogonal to F,(t). The direct sum of this and the complex vector bundle

Cj

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5 Continuous Vector Bundles

E is the full tangent bundle T on S2,+'. As we already have mentioned - but

>

<

not proven - the latter is not trivial for n = 2, n 4. So a fortiori cannot be trivial as an IR-vector bundle since it has a trivial complement in T. So cannot be trivial as a C-vector bundle. By other topological means on can even show that it is nontrivial for all n 2.

<

>

5.4.10 We will exploit this in order to construct nontrivial projective modules. Define the ring B, := C[zo,. . . ,z,, WO, . . . ,wn]/(Cj zjwj - 1) and a ring homomorphism B, + C(S2"+') by zj c, zj, wj c, zj.Then the stably free Bn-module defined by the unimodular row (20,. . . ,z,) extends to the and therefore is not free. C(S2n+')-module r([) Clearly one can replace C by any subring of it.

5.5 Vector Bundles and Grassmannians We are going to construct a correspondence between isomorphism classes of soft vector bundles over X and the homotopy classes of continuous maps from X to certain spaces.

But first we need some preparations. 5.5.1 The Direct Limit and Infinite Matrices We define the so called direct limit in a special situation. Definition 5.5.1 Let

G~ 3~~ % G% ~-

(5-9)

be a sequence of maps of sets (or group homomorphisms). Define gii := idGi and gik := gk-10. . - ogi for i < k . O n the disjoint union UiEwGi define the following equivalence relation for a E Gi, b E Gj:

The direct limit of the above sequence is then defined to be

Canonical maps tcj : Gj

+ limG, -+

(Gj + UncW Gn + UnEw Gn/

N.)

are defined i n the obvious way.

5.5 Vector Bundles and Grassmannians

127

Remarks 5.5.2 a) The canonical maps commute with the g,. b) To get a good feeling of the meaning of the direct limit: An element of lim G, is one which appears in some G,. Elements are equal in lim G, if and ----+ + only if they 'become' equal in some G,. c) If the Gi are groups and the gi homomorphisms, lim Gi is also a group. + d) If the gi are injective and one regards Gi as a subset of Gitl due to the map gi, then limGi = U iGi and every G, can be interpreted as a subset of 4 lim G, . ---+ e) As an exercise the reader should try to show that the additive group of $ is isomorphic to the direct limit of the following sequence:

where the maps are the homotheties of the integers 2 2. Proposition 5.5.3 In the situation of the definition let H be a set (or group) and h, : G, -+ H be maps (homomorphisms) which commute with the g,. Then there is a unique map (homomorphism) limG, -+ H , which commutes + with the g,.

Since we do not need this proposition, we leave the proof to the reader. 5.5.4 Recall that M,(R) denotes the ring of n x n-matrices over a ring R, and GL,(R) the group of invertible n x n-matrices. We have injective maps

(Note the difference!) According to these maps we define M,(R) := lim Mn(R) and GL, (R) := lim GL,(R). + + The map M,(R) + Mn+l(R) is a homomorphism of additive groups and respects multiplication, but does not map I, to In+l. Insofar it is not a ring homomorphism. So the direct limit lim, M,(R) is nearly a ring, but lacks a 1. The maps GL,(R) -+ GLn+l(R) are group homomorphisms. So GL,(R) is a group. The elements of M,(R) may be interpreteted as infinite matrices (aij)i,j>o with the property that there is an n with aij = 0 if i or j > n.

128

5 Continuous Vector Bundles

Analogously the elements of GL,(R) may be interpreteted as infinite matrices (aij)i,j with the properties that aii = 1 for all i and that there is an n such that the submatrix (aij)i,jl, is invertible and that aij = 0, if i # j and i or j > n. One may perform the product a/?if a , as well as /? belongs to any of M,(R) or GL,(R). If both belong to GL,(R) then a/?E GL,(R); in the three other cases a/?E M, (R). For example for a E GL, (R), a E M, (R) we may define the conjugated or similar element aaa-l. This induces on M,(R) the equivalence relation, to be similar, i.e. conjugated a

N

,8

there is a a E GL,(R)

with /? = ma-'

(5.13)

It is clear that every matrix, similar to an idempotent one, is again idempotent. The equivalence classes of this relation are called the conjugation classes of M , (R). 5.5.2 Metrization of the Set of Continuous Maps By C(C, D) we denote the set of continuous maps f : C -+ D if C, D are topological spaces. It is not always possible to equip C(C, D) with a useful topology. But if C is compact and D a metric space, we even get a good metric on C(C, D). 5.5.5 In the following let C be a compact topological space and D be a metric space with distance function d. The set C(C, D) of continuous maps f : C -+ D then inherits a metric according to the definition:

, is continuous and so the maximum Namely the map C -+ R,t e d(f ( t ) g(t)) exists by the compactness of C. The axioms for a metric easily follow. Let X be an arbitrary topological space. Every (not necessarily continuous) map f : X x C -+ D induces a map X -+ Map(C, D), x e f (x, where Map(C, D) denotes the set of all (not necessarily continuous) maps C -+ D. a ) ,

Lemma 5.5.6 Under the above conditions on C and D a map f : X x C -+ D is continuous i f and only if every map f (x, -) : C -+ D and the map X -+ C (C, D), x e f (x, -) are continuous. The map f (x, .) is the composition off with i, : C -+ X x C, c Proof. '3''. (x, c), hence is continuous. Let further x 6 X and E > 0. We have to show:

5.5 Vector Bundles and Grassmannians

129

there is a neighbourhood U of x in X such that d(f (x, .) , f (sf,.)) < E for all x' E U. Since f is continuous, for every c E C there are open neighbourhoods V, of x in X and W, of c in C such that f (V, x W,) is contained in a11 Eneighbourhood of f (x, c). Since C is compact, there are finitely many of the Wj = C. Let K , . . . , V, be the corresponding W,, say Wl, . . . , W, with neighbourhoods of x. We claim that U := 5 h a the desired property. We have to show: d(f (x', c), f (x, c)) < E for all x' E U, c E C. Now there is an i with c E Wi. Then f ( U x Wi) c f ( & x Wi) is contained in an Eneighbourhood of f (x, c) .

UEl

n>

"e". Let (x,c) E X x C and E > 0. Since the map X -+ C(C, D ) , x l H f(x,.) is continuous, there is a neighbourhood V of x in X such that d(f (x', c'), f (x, c') < ~ / 2V (x', c') E V x C. Since f (x, .) : C + D is continuous, there is a neighbourhood W of c in C such that d(f (z, e'), f (x, e)) < ~ / 2 for every c' E W. Therefore d(f (x', c'), f (x, c)) < E V (x', c') E V x W. 5.5.3 Correspondence of Vector Bundles and Classes of Maps Lemma 5.5.7 Let R be an arbitrary ring. a) If M is an R-module and M = P @ Q = P' @ Q' are splittings with P 2 P',

then: Q @M r Q ' @ M . b) If E, E' E Mn(R) are idempotents with im(e) im(Ef). Then there is a y E GL2,(R) with y ~ y - ' = E'. Here E,E' are regarded as elements of Ma, by filling up the matrices with zeros. (One may consider all matrices as elements of Mm(R).) Proof. a) If M is projective which is the only case we need this is a special case of Schanuel's Lemma 3.7.1. (Note that sirice there exist stably free nonfree modules one cannot expect Q S Q'.) -

But the proof of the general case is easy: Since P @ Q' Q@MrQ@P@Q'EM@Q1.

-

E M E P$Q

b) Let P = i m ( ~ ) P' , = i m ( ~ ) ,Q = ker(e), Q' = ker(el). Then P hypothesis and and Q @ Rn 2 Q' @ R7' by a). So we have splittings

we have

r P' by

R~~ = P @(Q @ Rn) = P' @ (Q' 63 Rn). Let r1 : P -+ PI, rz : Q @ Rn -+ Q' $ Rn be isomorphisms. Then a := TI $7-2 clearly fulfills the claim. Namely E is zero on Q and the identity on P. So ma-' is = 0 on Q' and r1idprc1 = idp, on P'. We want to show the following:

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5 Continuous Vector Bundles

Theorem 5.5.8 The soft vector bundles on X correspond upto isomorphism bijectively to the elements of lim r ( X ,G n ( F ) ) . where r ( X ,G n ( F ) ) denotes 4 the set of homotopy classes of continuous maps X + G n ( F ) .

Proof. Let a , P E M,(C(X)) be hermitian and idempotent. They can also be interpreted as elements of lim C ( X ,Gn(lF))We have to show that the following 4 equivalence is true: a , p are conjugated H a , P are homotopic. Assume

p

= crag-'. Then

By Whitehead's Lemma the matrix

6

is a product of elementary ma-

trices, hence homotopic to the 1-matrix. Therefore also a and ,D are homotopic. Conversely assume that a , /? E G n ( F ) are homotopic. Then there is a continuous map y : X x [O,l] + Gn(F) with y(., 0 ) = a, y ( - ,1) = p. Regard y as a map from X to the space r G n ( F ) of d l continuous maps [O,1]+ G n ( F ) . This space is metric by the distance function d ( f ,g ) := Maxt{d(f (t),d t ) )I 0 L t I 1 ) Therefore there is a locally finite partition of unity (e;)iEron r G n ( F ) with d(u,v ) < 119 if e;(u)ei(v) # 0 for some i E I. By the continuous map X + r G n ( F ) , x c+ y ( x , - ) this can be lifted to a locally finite partition of unity (ei)iE1on X. So for x , y E X with ei(x)ei(g) # 0 we have d(y ( x ,t ),y (y ,t ) ) < 119 for all t E [0,1].For every i E I we choose a point xi E Ui := { z E X / e i ( z ) # 0 ) and a positive number ~i such that d ( y ( ~ ti),y ( x i ,t')) < 119 whenever It - t'I < ~ i Then . d ( y ( z ,t ) ,y ( z , t')) 5 d ( y ( z ,t ) ,?'(xi, t ) ) + d ( y ( x i ,t ) ,y(xi, tl))+d(y(xi,t f ) y, ( z , t') < 1/9+1/9+1/9 = 113 for every z E Ui whenever It - t'l < ~ i .

&,

Define 6 ( z ) := eiei ( z ) for z E X . Then clearly 6 is a continuous positive function on X such that d ( y ( z ,t ) ,$2, t f ) < 113 whenever It - t'l < 6(x). We define continuous maps tk : X integer k 2 0 by

+ [O, 11 and yk

:X

+ G n ( F ) for every

Then yo = cr and yk ( z ) = P(z) for k 2 l / 6 ( z ) ,further d(yk ( z ) ,yk+i ( z ) ) < 113 for d l z and k.

5.6 Projective Spaces

131

Interprete the a , ,f?,yk as idempotent hermitian matrices in Mn(IF) for big enough n, let I = In be the unit matrix and set

a := aoalaa -.. , i.e. a ( z ) = a o ( z ) a l ( z .). . c k (

Z)

for k

> 1/6(z).

Then o is an invertible matrix over C ( X ) and a ( z ) c ( z )= y o ( z ) .- - y k ( z )= c ( z ) P ( z )for k > 1/6(z).This means a o = a p , hence a = apa-'. 0 5.5.9 A topological space X is called contractible if and only if there is a point p E X such that the constant map X + X , x ++ p is homotopic to idx. For instance Rn is contractible by the homotopy: h : Rn x [0,11 -+ IRn, ( 2 ,t ) tx.

*

This was the reason why Serre's Conjecture was born! Namely C ( R n ] is the topological counterpart of k [ x l ,. . . ,x,]. It is clear that every continuous map X -+ Y is homotopic to a constant map, if X is contractible. Therefore, by the theorem, on Rn there are only trivial soft vector bundles over Rn.Indeed there are only trivial vector bundles on Rn at all.

5.6 Projective Spaces 5.6.1 The n-dimensional projective space IFPn over IF is the Grassmannian of one dimensional linear subspaces of IFn+', i.e. F P n = G?+' ( I F ) .

Clearly we have a surjective map

which assigns to any nonzero x E IFn+' the subspace, generated by x. Elements x , y E IFnt1 \ ( 0 ) belong to the same fibre of p, if and only if there is an a E I F X with ax = y. This means that F P n is the orbit space of the obvious operation of I F X on IFn+' \ ( 0 ) . The map p is continuous. Namely the topology on IFPn is given by assigning to any one dimensional subspace of I F n f 1 the orthogonal projection onto it, and then by using the operator norm on the space of endomorphisms of IFn+'. But the orthogonal projection onto the subspace, generated by v is the map w ++ ( v ,v)-l ( v ,w)v, which clearly depends continuously on v.

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5 Continuous Vector Bundles

By [xo: . . . : x,] we denote the point p(xo,. . . ,x,). (Of course [xo : . . . : x,] is only defined for ( x o ,. . . ,x,) # (0,. . . ,O).) The set Sn'+'-' := { ( x o ., . . ,x,) E F n f l I C lxi12 = 1 ) is the (real) unit sphere in lRnT+' if this is identified with Fn+'. Here r = 1,2,4, depending on whether F = IR, C or M.By p this sphere is mapped surjectively to F P . Elements x, y E Snr+'-' have the same image under p if and only if there is a u E F of absolute value 1 with ux = y. The elements u E F with lul = 1 make up a subgroup of the multiplicative group of F and form the unit sphere 3'-1 in F . Therefore F P n is a quotient of the sphere Snr+'-' group ST-' = { x E IF 1 1x1 = 1).

by the operation of the

Note that Ui := {[xo: . . . ,x,] ( xi # 0 ) for every i = 0,. . . ,n is an open subset of FPn. It is homeomorphic to F n . We describe the maps F n + Ui + F n only for i = 0.

(91,.. . ,9,) e [l : y'

: . . . : y,],

[xo: X l

: . . . : x,]

Xn e (-,2 1 . . . , -) 20

20

(It is clear, how to do this for general i.)

Definition 5.6.2 The universal line bundle on F P n is the one dimensional subbundle of the trivial bundle [pr, : Fn+' x F P n + FPn] whose total space is { ( v ,x ) E IFn+' x F P n I v E x ) . (Note that every x E lFPn is a one dimensional subspace of Fn+' so that the relation v E x makes sense.) The universal bundle is also called the tautological bundle since the fibre over x is x itself, considered as vector space. Also the name canonical line bundle is i n use. 5.6.3 The universal line bundle y is trivial on every one of the Ui, which we defined in Paragraph 5.6.1. Namely (for i = 0 ) we have the isomorphism

5.6.4 The complex projective line CP1 is the so called Riemann sphere. Indeed it is homeomorphic to the 2-sphere by the stereographic projection.

We will explain this in detail.

CP' is covered by the two open sets: Uj := {[zo: zl] I zj # 0). And there are homeomorphisms Uo + C, [zo: zl] e zl/zo and Ul -+ C, [zo: zl] I+ zo/zl.

+

Now let S2 := { ( z , t ) E C x lR 1 1 . ~ 1 ~ t 2 = 1) and let n := ( 0 , l ) E S 2 , s := (0,-1) E S 2 denote the 'north pole', resp. the 'south pole' of S 2 of the 2-sphere. We have two maps cp,, cp, : C + S2 given by

5.6 Projective Spaces

133

(Here and some lines later denotes the complex conjugate of z.) We will show that these are homeomorphisms from C with S2\ {n), resp. with S2\ {s). We do this by giving the inverse maps.

( zl )/ z We leave the computations to the reader. Also one computes $ ~ ~ o ( p ~= for z E 61'. But this means that S2 arises from two copies of C by the same glueing as CP1 does. Therefore they are homeomorphic to each other and can well be identified. The above described map p : S3 + CP1 can now be viewed as a map p : S3+ S2.This is the famous Hopf map. Note 5.6.5 One may do the same in the cases IF = IR and IF = M. The first case is contained in the above computations. Only one has to restrict z to real numbers. One gets a map S1 -+ S1 of degree 2. This means it can be when one identifies S1 with the set of complex numbers described by 5 cs of absolute value 1.

c2

In the second case one proves as above that MP1 is homeomorphic to S4and one gets a Hopf map S7 + S4. 5.6.6 Let us consider the universal C-line bundle y. On Uo we have the trivialization C x Uo -+ E(ylUo), (a, [ZO: 211) (a,a2), zo

-

To compute the glueing map aol we have to find the preimage of (1, z1/zo) under the second map. It is (zl /zo, [zo : zl]), since it clearly maps to (1'21 120). Therefore aol(z) = z if we identify [zo : zl] with z = zl/zo on Uo n U1. Fore later use, now we replace the Uj by the closed halfspheres Yo := {[zo : zl] 1 Izl/zol 5 I), Yl := {[zo : zl] ) Jzo/zl\ 5 1). Then Yo n Y1 = {z I IzI = 1) where z = zo/zl Of course in this situation we also have aO1(z) = z. 5.6.7 Now let us describe the tangent bundle of the S2in the same manner. It is the tangent Cline bundle T on CP1, considered as an IR-bundle of rank 2.

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5 Continuous Vector Bundles

On each of the two halfspheres Yj the tangent bundle is trivial and the glueing map Pol of the bundle is the derivative w.r.t. z := zl/zo of the glueing map of the two charts Yo n K -+ Yo n Yl. The latter is z I+ 2-l. So the glueing map of the tangent bundle is Pol(z) = - z - ~ . Since we may change the coordinate of Yo from z to iz, we may take as glueing map of the bundle the map @'O1 (z) = z2. (The overbar here denotes the complex conjugation.) This means that yg2 := y @ c y = 7. Since 7 as a real vector bundle is not trivial, a fortiori it is not trivial over C. So the universal bundle 7 on C P cannot be trivial. 5.6.8 Note, one has a similar situation in the case IF = R . The onedimensional real projective space is the 1-sphere S1, the universal bundle is the Mobius bundle, its tensor square is the tangent bundle, which is trivial in this case. 5.6.9 Let F = IR. On RPn we consider the universal bundle y. The map p : Sn -+ RPn induces a map p* : C(IRPn) -+ C(Sn), which is injective and identifies C(IRPn) with the subring Co of C(S2) consisting of those f E C(Sn) which satisfy f (-x) = f (x).

We want to interprete r ( y ) in a similar manner. Let x E S n , p(x) E RPn.An element of the fibre of y over p(x) is of the form (ax,p(x)) = ((-a) (-x), p(x)) with a E R . So a section over RPn is given by a continuous real function g E C(Sn) with g(-x) = -g(x). This clearly gives a good description of r ( y ) as a module over Ca = C(RPn). Namely, I f f , g E C(Sn) satisfy f (-2) = f (x) and g(-2) = -g(x), then (fg)(-2) = -(fg)(x). We will now study the minimal number of generators of this projective Co module of rank 1. Unfortunately we only can do that by using some facts on Stiefel-Whitney classes we now cite here. Stiefel-whitney classes (wo,wl, w2,. . .) assign to a real vector bundle J on a topological space X cohomology classes wi(<) E Hi(X, Z/2Z) of X with wo(J) = 1. The total Stiefel-Whitney class of J is W (J) = 1 wl(J) w2 (5) . . . E H*(X, ZI2Z). One has the following properties:

+

+

+

a) W(X) = 1 if X is trivial. C) wi(J) = 0 for i

> rk(5).

The cohomology of the real projective space is of the form H * ( R P n , 2 / 2 2 ) = ( Z I 2 Z )[TI/(Tn+l) and, letting t denote the residue class of T , we have t i E H ~ ( R P ~and ) W(Y) = 1 + t. Proposition 5.6.10 Let y be as above and A'' the trivial bundle of rank r o n R P n . T h e n the C(RPn)-module r ( y AT) needs n r 1 generators.

+ +

5.7 Algebraization of Vector Bundles

+

135

Proof. Assume that r ( y @ AT) could be generated by n r generators. Then y @ AT is a direct summand of An+' the trivial bundle of rank n r, i.e. there is a bundle q with y $ AT $ = An+'. Clearly rk(q) = n - 1. So tS for some ai E Z / 2 Z and some s < n. W(q) = 1 a l t - . - a,-lts-l Therefore W(y @ AT @ q) = 1 . - . tSfl # 1, since s 1 n and therefore 0 tS+l # 0. On the other hand W(An+') = 1. A contradiction.

+

+ +

+ + +

+

+ <

5.6.11 Now we consider the ring A := RIXo,. . . , X n ] / ( C Xp - 1). It is the subring of C(Sn), consisting of the polynomial functions on Sn. By xi we denote the residue class of Xi. Then A. := Go n A consists of the residue classes of those polynomials, all of whose monomials have even (total) degree. We have a splitting A = A. $ A1 of the additive group of A, where Al is the subgroup of the residue classes of those polynomials, all of whose monomials have odd (total) degree. This splitting is a so called Z/2Z-grading of A, which means AOAOc Ao, A0 A1 c Al, A1A1 c Ao. So Al is an Ao-module, and we will see now that it is a finitely generated projective one. One has a surjective linear map f : A:+' + Al, given by f (ao,.. . ,a,) = Cy=o xiai and a section to this map s : A1 + A;;"+', b (xob,xlb,. . . ,x,b). Note that f 0s = id follows directly from Cy=o xz = 1. According to the above description of r ( y ) as a Go-module we have parallely Go-linear maps f i : -+ r ( ~ ) (ao, , . . . ,an) t-, Ckoxiai and si : r ( y ) + b ++ (zob,. . . ,xnb) with f'osi = id. A1. And therefore Al is a projective Ao-module of rank 1, So r ( y ) = Co such that A1 $ A; cannot be generated by less than n 1 r elements.

+ +

5.7 Algebraization of Vector Bundles Let X be a compact subset of Rn. We have seen how the continuous vector bundles on X may be interpreted as finitely generated projective C(X)modules. Algebraists might prefer to have examples of projective modules over rings which are more algebraic, i.e. 'smaller' in some good sense. At least they should be Noetherian. A ring is called (left) Noetherian if its (left) ideals are finitely generated. This property will be introduced and studied in the first section of Chapter 6. We will even show that the rings we construct, are essentially of finite type over IF. Definition 5.7.1 A ring A is called essentially of finite type over a ring R, if it is of the form A = S-lR[al,. . . ,on].

If R is Noetherian, say a field, then every R-algebra, essentially of finite type, is Noetherian, too.

136

5 Continuous Vector Bundles

It is clear that for a single finitely generated projective C(X)-module P (or finitely many ones) there are finitely many functions f i , . . . , fn E C(X) such that P is extended from Z[fl,. . . , f,]. Namely P is the image of an idempotent endomorphism of a finitely generated free C(X)-module, i.e. P is given by a finite matrix. Let f l , . . . ,f n be the entries of this matrix. But clearly one prefers, to find Noetherian rings or algebras A, essentially of finite type, with interesting behaviour of the totality of their projective modules. At first K. L~nstedin 1481 gave a method to produce Noetherian such A. Later R.G. Swan in [99] constructed such A, essentially of finite type over IF. M. Carral in 1161 then was able to replace a part of Swan's construction by a more efficient one. We reproduce here the method of Swan and Carral and Milnor's famous construction of projective modules, which is used for that.

5.7.1 Projective Modules over Topological Rings Here, following Swan, we will study the behaviour of P under a ring homomorphism A +-B with certain topological properties.

5.7.2 Let A be a topological ring and P, Q finitely generated projective Amodules. We define topologies on P and Hom(P, Q) as follows. First identify Hom(AT,AS) with ATS and give it the product topology. If a , p are automorphisms of As and A' respectively, then a H a a p is a homeomorphism of Hom(AT,AS)with itself, since it is continuous as well as /3 ea-lpp-l. Therefore the topology on Hom(AT,AS) is independent of the chosen bases. Now let P I , Q1 be so that P @ P' E AT and Q $ Q' E AS. Then a H a $ 0 embeds Hom(P, Q) into Hom(AT,AS). The induced topology on Hom(P, Q) is then independent of the choices of P' and Q'. To see this note first that replacing P' by P' @ An and Q' by Q' $ Am does not change the topology. Let P $ P r l E A n a n d Q @ Q " E A m . Then P $ P 1 $ A n E P $ P " $ A T and analogously Q $ Q1 $ Am E Q $ Q1' $ AS. Hence the independence follows. The topology on P can be defined via the identification P = HomA(A,P ) or equivalently via the inclusion P r P $ P' = AT. The following hypothesis is not always true, but in important cases it is: (0) The units of A form an open subset and u H up' is continuous on it. This is equivalent to saying that some neighbourhood of 1 consists of units and up' is near 1 if u is near enough to 1. This is easy to show for A = C(X) with compact X. (Exercise.) More generally it holds for any Banach algebra.

5.7 Algebraization of Vector Bundles

137

The matrix ring Mn(A) regarded as Hom(An, An) has the above defined topology, namely the product topology, if one regards Mn(A) as A"'. With this topology Mn(A) is a topological ring. Lemma 5.7.3 Let A satisfy (0). There are a neighbourhood U of Inin Mn(A) and continuous maps ei, fi,d : U -+ Mn(A), i = 1,. . . ,r = n(n - 1)/2, such that

a) all ei (x) ,fi (x) are elementary and the d(x) are invertible diagonal matrices for x E U which reduce to Infor x = In and

Proof. Let x = (aij) be near enough In. Then all is near 1, hence a unit in A. So by n - 1elementary row and n - 1 column operations we can transform x to a matrix of the form a l l N0 ) . Since a:; is near 1 and the ail, a l j for

(

i, j # 1 were near 0, the matrix N is near on n we get through.

In-1 in

GLn-1(A). By induction

Proposition 5.7.4 If A satisfies (O), so does End(P) for any finitely generated projective module P.

Proof. At first we assume P = An to be free. By Lemma 5.7.3 there is an open neighbourhood U of In in Mn(A) which is contained in the general linear group and on which the map x 4 xP1 is continuous. Namely, for x near In we have x = el(%)- ..e,(x)d(x) f l (x) . - fT(x), hence x-I = fT(x)-' ... fl(x)-ld(x)-leT(x)-l .. .e1(x)-l. In the general case let P @ Q = An. Then by the above definition of the topology on End(P) = Hom(P, P ) an endomorphism g of P is near idp if and 0 only if g @ idQ is near In in Mn(A). Proposition 5.7.5 Let A satisfy (0) and f , g : P + Q be linear maps between finitely generated projective A-modules. If f is surjective and g suficiently near f , then g is surjective too, and ker(g) ? ker(f).

Proof. Let s be a section of f , i.e. f s = idQ. If g is sufficiently near f then gs is near f s = idQ and so by Hypothesis (0), gs is an automorphism of Q. So g is surjective. Let t = s(gs)-' coker s E ker f .

:

Q

+ P . Then gt

= idQ and therefore kerg E cokert =

0

Proposition 5.7.6 Let A satisfy (0) and F be a finitely generated projective A-module, further e : F -+ F be an idempotent. I f f : F -+ F is idempotent too and suficiently near e, then e ( F ) S' f (F).

138

5 Continuous Vector Bundles

Proof. Let P = e(F), Q = f (F) and Q' = id^ - f )(F). If f is sufficiently id^ - f ) will be near enough e id^ - e) = id^, hence near e, then e an automorphism by Proposition 5.7.4. Therefore F = P Q' and so the composition a of the canonical maps P -+ F -+ Q is surjective. We have a(x) = f (x) for x E P. Define ,l:? Q -+ P by P(y) = e(y). If f is sufficiently near e, then pa : P -+ P will be very near to the map x e e2(x) = x. Hence again pa is an automorphism and so a will also be injective.

+

+

+

5.7.7 Now we consider a ring homomorphism cp : A -+ B and conditions

(1) B is a topological ring which satisfies (0) above; (2) cp(A) is dense i n B ; (3) there is an open neighbourhood U of 1 i n B with cp-I (U) c AX.

Remarks 5.7.8 a) Let I c Jac(A) be an ideal and B := A/I equipped with the discrete topology. Then (1) and (2) are obvious, and (3) means 1+I c A X , which follows easily from I c Jac(A). b) If cp : A -+ B satisfies (I), (2) and (3') cp-l(BX) c A X ,i.e. cp-l(BX) = A X , then cp satisfies (1) to (3). Especially, if cp satisfies (1) and (2), and S := cp-'(BX) is contained in the center of A, then the induced homomorphism S-'A -+ B satisfies (1) to (3). (If A is a noncommutative ring and S is a multiplicative subset of the center of A, then S-'A can be defined in the same way as in the commutative case.) c) Let cp : A -+ B satisfy (1). Define a topology on A by taking the neighbourhoods of 0 to be the sets cp-l(U) where U is a neighbourhood of 0 in B. Then we have the equivalence: AX is open in A

&

cp satisfies (3).

Since cp(u-l) = cp(u)-' for units in A we also have the equivalence A satisfies (0) a cp satisfies (3). d) Let X be a compact subset of some IRn. The ring R(X) of regular functions on X is defined by n ( x ) := s i l l ~ [ t l ,... ,~,]/Ix where Ix is the zero ideal of X , i.e. consists of the polynomials vanishing along X , whereas Sx consists of those polynomials which are nowhere zero on X . Then the inclusion cp : R ( X ) -+ C(X) satisfies (1) to (3). Namely (1) is trivial, whereas (2) follows by Weierstrass' Approximation Theorem, and (3) is clear, since we even have cp-l (C(X) X , = R(X) X .

5.7 Algebraization of Vector Bundles

Lemma 5.7.9 If cp : A Mn(A) -+ Mn(B)

-

+

139

B satisfies (1) to (3), then so does Mn(cp) :

Proof. (1) has already been proved in Proposition 5.7.4, and (2) is trivial. For (3), define a topology on A resp. Mn(A) as in Remark c) above by taking the neighbourhoods of 0 to be the sets cp-'(U) where U is a neighbourhood of 0 in B, resp. of 0 in Mn(B). The resulting topology on M,(A) is the same as the topology, induced by that on A. Now by Remark c), condition (3) for cp implies Hypothesis (0) for A, hence for M,(A) by Proposition 5.7.4. Again 0 by Remark c) we obtain (3) for Mn(cp).

Lemma 5.7.10 Let cp : A -+ B satisfy (2) and f : B 8 P + B 8 Q be B linear, where P, Q are finitely generated and projective over A. Then we can find g : P -+ Q such that idB 8 g is arbitrarily close to f . Proof. This is clear, if P, Q are free. Let P @ P' = F and Q @ Q' = G be free and find h : F -+ G such that h approximates f @ 0. Then the composition g of P f F G 4 Q will be near f , where i, p are the canonical inclusion 0 resp. projection.

1

Lemma 5.7.11 (Generalized Nakayarna Lemma) Let cp : A -+ B satisfy (1) to (3) and M be a finitely generated A-module with B 8 M = 0. Then M = 0. Proof. Let F' F + M -+ 0 be an exact sequence of A-modules where F, F' are free and F is finitely generated By hypothesis B 8 f is surjective. Hence there is a section s of it: ( B 8 f)os = idF. There is a finite subset of a base of F', generating a free submodule G, such that s(B 8 F ) is contained in B 8 G. By Lemma 5.7.10 find t : F + G with B 8 t near s. So B 8 (f ot) is near (B 8 f)os = idp. Since Mn(cp) satisfies (3), the endomorphism f ot will be an automorphism, if B 8 t is sufficiently close to s. Therefore f is surjective. 0

Theorem 5.7.12 Let cp : A + B satisfy (1) to (3). a) IP(cp) : IP(A) -+ P ( B ) is injective. b) The isomorphism class of a finitely generated projective B-module belongs to the image of P ( p ) if it is stably isomorphic to another finitely generated projective B-modzlle whose isomorphism class is in the image. Especially the isomorphism class of any stably free module is in the image of IP(cp). Proof. a) Let P, Q be finitely generated projective A-modules and let f : B 8 P -+ B @ Q be an isomorphism. Find a g : P + Q with i d s 8 g so close to f , that B 8 g is an isomorphism, using Lemma 5.7.10 and Proposition 5.7.4. Then B 8 coker(g) = 0, whence coker(g) = 0 by the Generalized Nakayama

140

5 Continuous Vector Bundles

Lemma. So the sequence 0 + ker(g) + P + Q + 0 is split exact, hence ker(g) as a factor of P is finitely generated. Therefore from B @ ker(g) = 0 we obtain ker(g) = 0. Hence P E Q. b) Let P be a finitely generated projective B-module, Q a finitely generated projective A-module with P @ BT r (B @A Q) @ BT for some r. With Q' := Q @ AT we get

the projection P @ BT + BT composed with this isomorphism gives us a homomorphism p : B @A Q' + BT. Using Lemma 5.7.10, we find g : Q' + AT with idB @ gso close top, that idg @ gis surjective by Proposition 5.7.5. Then g is also surjective by the Generalized Nakayama Lemma. So P' := ker(g) is a direct summand of Q' and hence projective. Finally we get B @A P' = ker(idB @ g) r ker(p) = P by Proposition 5.7.5 if only B @ g was chosen close 0 enough to p.

5.7.2 Projective Modules as Pull-Backs We present here Milnor's construction of projective modules.

Definition 5.7.13 A diagram of the form

of ring or group homomo~phismsis called a pull-back diagram or Cartesian, if A r {(al, a2) E A1 x A2 I jl(al) = j2(a2)) and il, i2 are induced by the projections. (From this clearly the commutativity of the diagram follows.) In this case one calls A, or more exactly (A, il, i2) the pull back of the pair of maps (jl ,j2). (Compare Appendix B.)

Examples 5.7.14 a) Let Il,I 2 be (two-sided) ideals of a ring R and let A = Rl(I1 n I,), A, = R/Ik, k = 1,2, A' = R/(Il I,) and il, i2, jl, j2 be canonical maps.

+

For our purpose this will be the most important case. Namely let Vl, V2 be closed subsets of a normal topological space X and set A := C(Vl UV2), Ak := C(Vk) and A' := C(& f l V2) Then the restriction maps A + Ak, Ak + A' are surjective by Tietze's Extension Theorem. And if Ik= ker[A + Ak], then I 1 I 2 = ker[A + A']. Namely if f : fi U & + F is 0 on Vl or on V2, then

+

5.7 Algebraization of Vector Bundles

141

clearly it is 0 on Vl n V2, whence Ikc ker[A + A'] for k = 1,2. So we only have to show Il I2> ker[A + A']. But if f : Vl U V2 + IF is 0 on Vl n h, then define

+

Since f k is continuous on the closed sets Vl and V2, it is continuous on Vl UVz. Further f k E I k and f = f l f2.

+

b) Let A c B be a ring extension and I c A be an ideal of B (so that I is a common ideal of A and B). Then the diagram with the obvious maps A-B

is Cartesian. We note that this situation for example arises in the case where A is a domain, B its integral closure and B is finite over A. Then the so called conductor &(B/A) := AnnA(B/A), which is the maximal common ideal of A and B, is not zero. (We will use the concept of the conductor in Chapter 10, see Definition 10.4.6.) c) Let A' be any ring, A1, A2 subrings of A' and A their intersection.

5.7.15 Construction. In the following let

be a pull back diagram and assume additionally that jl or j2 is surjective. (The last requirement is fulfilled in Examples a) and b) above, but not generally in Example c).) If f : A + B is a ring homomorphism and P a projective A-module, we denote by f # P the projective B-module A @ A P and by f, : P + f # P the canonical A-linear map m t, 18 m. From a pair of projective Ak-modules Pk:for k = 1,2 and an isomorphism h : jl# PI + j 2 # P 2 we construct M = M(Pl, P2,h) to be the pull back of the pair of group homomorphisms (hojl,, j2,). So the diagram

142

5 Continuous Vector Bundles

is Cartesian. i.e. M consists of the pairs (pl,p2) E PI x P 2 which fulfill hojl, (pl) = j2* (p2). It is a subgroup of Pl x P2and becomes an A-module by a . (pl,m) = (il(a)pl,i2(a)m).

Theorem 5.7.16 In the above situation the following holds: a) M = M(Pl, P2,h) is a projective A-module, and it is finitely generated if Pl and P2 are so. b) Every (finitely generated) projective A-module is isomorphic to such an M(Pl ,P2,h) for suitably chosen (finitely generated) PI,P2,h.

In the following proof we will restrict ourselves to the case where all rings have Invariant Basis Property (see Definition 3.1.3), and all modules are finitely generated. We leave the general case to the reader, who may consult Milnor's book [58]. Proof. The first and biggest part of the proof will be to show a). We do this in several steps: 1) Let PI,P 2 be free (of the same rank). Thereover we assume that PI has a basis 21,. . . ,x, and P2 has a basis yl, . . . ,yn such that hojl,(xk) = jz,(yk) for k = 1,. . . ,n. Then clearly M is free with basis (XI,yl), . . . , (x,, y,). 2) From 1) we immediately derive

Lemma 5.7.17 Let PI,P 2 be free with bases X I , . . . ,x,, resp. 91,. . . ,y, and let the matrix a = (a,,) over A' describe h with respect to these bases. We assume that a is the image under j2 of an invertible matrix y over A2. Then M is also free of rankn. Namely, using the matrix a, one may change the basis yl, . . . ,y, in such a way that one arrives at the situation of 1). 3) Let still PI,P 2 be free and j 2 be surjective. Let again a describe h. Then let Q1, Qz be free over A1, resp. A2 of rank n. Then one has an isomorphism

PI @ Q1 --+ P 2 @ Q2 given by the matrix /3 = By Whitehead's Lemma 3.4.4 we know that /? is the product of elementary matrices, which are images of invertible matrices over A2, since j 2 is surjective. Therefore M (PI $ Q1, P2$ Q2, P) is free over A, according to Lemma 5.7.17. On the other hand let g : jl#Q1

+j2#Q2

be described by a-l. Then

5.7 Algebraization of Vector Bundles

143

Therefore M(Pl, P2,h) is a finitely generated projective A-module. 4) Now consider the general case (only with the restrictions indicated before the proof).

CLAIM:There are finitely generated projective modules Q1, Q2 over A1, resp. A2, such that P k $ Qk are free for k = 1,2 and fl#Ql Z f2#Q2. Namely there are Nl, N2 with PI @ Nl P' = jl#PI E j2#P2, we get

Z

A; and P2@ N2

E

A;. Writing

5) To show that M(P1, P 2 , h) is projective, let Qk be as under 4) and choose some isomorphism hl : Q1 + Q2. Then

Now the module on the right hand side is finitely generated and projective by 3) and therefore M (PI, P 2 , h) is so. So we have proven a). We prove b). Let P be a finitely generated projective A-module. Then set P k := i k # P and let h :jl#oil#P + jz#oiz#P be the canonical isomorphism. Then clearly P = M(Pl ,P2,h). We prove c). There is a natural A-linear map M = M(P1, P 2 , h) + PI. This induces an Al-module homomorphism f : il# M -+ PI.We have to show that f is an isomorphism.

If the conditions of Lemma 5.7.17 are fulfilled, the statement clearly is true. In the general case during the proof of a) we constructed projective modules Qk such that Pk@ Qk are free and the conditions of Lemma 5.7.17 are satisfied for some isomorphism h @ h' : jl#P1 @ jl#Q1 + j2#P2@ j2#Q2. Therefore, setting M' := M(Q1, Q2, h'), the canonical map f' : jl#M' + Q1 has the property that f @ f' : jl#M @ jl#Mt -+ PI $ Q1 is an isomorphism. This implies that f is one. 0

Remark 5.7.18 Note for later use that is f is an automorphism of f' : A' @ P2+ A' @ P 2 the induced automorphism, then M(Pl,Pz,h) E M(P1,P2,f'oh).

P2

and

5.7.3 Construction of a Noetherian Subalgebra 5.7.19 We will now consider the following situation. Let F = R or C. (We exclude the case F = H.) Let A c B be an extension of F-algebras with the following properties:

144

5 Continuous Vector Bundles

(1) B = C(X) where X the maximum norm.

c IRn is a compact set. B is a topological ring via

(2) Every polynomial function on X belongs to A.

(Later we will consider A of the form A = S-lIF[xl,. . . ,x,, 91,. . . ,g,], where . . ,x, are the coordinate functions, restricted to X and 91,. . . ,g, are finitely many continuous functions on X , and finally S consists of all s E IF[xl,. . . ,x,, 91,. . . ,g,] which have no zero on X . For certain X by a skilful choice of the gj we will achieve that the finitely generated B-modules - i.e. vector bundles on X - are extended from A.) XI,.

5.7.20 We will draw some conclusions from the assumptions (I), (2), (3).

The extension of rings A c B satisfies (I), (2), (3') and (3) of Paragraph 5.7.7 and Remark 5.7.8 b). Note that A = B is not excluded. Therefore true statements for general A are true for B. To every point x E X we associate a maximal ideal m, of A by m, := {f E A I f (x) = 0), which is the kernel of the ring homomorphism A -+ IF, f ++ f (x). This homomorphism is surjective, since IF c A. We define for every x E X a maximal ideal n, of B analogously. We have m, = A n n , . If x, y E X are different, then m, f E A with f (x) = 0, f (y) # 0.

# m,. Namely there is a polynomial function

CLAIM:Every maximal ideal of A is of the form m,. The same holds for B, since A = B is not excluded . . .

PROOF: Let m be any maximal ideal of A. It is enough to show that m c m, for some x E X. Assume this were not the case. Then for every x E X there were an f E m with f (x) # 0. Since, for a fixed f , the set of the x E X with f ( x ) # 0 is open, and since X is compact there were finitely many fi, . . . ,f, E m without common zero. But then f = f l E . - - f p Z would have no zero on X , i.e. it would belong to B X ,hence to A X .But on the other hand f belongs to the proper ideal m, a contradiction. (Here 5 denotes the complex conjugate of fk. If IF = IR, then 5 = f k.)

+ +

In all, there are canonical bijective maps X -+ Spmax(B) -+ Spmax(A). If one equips Spmax(B) and Spmax(A) with their Zariski topology, these maps become continuous.

5.7 Algebraization of Vector Bundles

145

The first one, X -+ Spmax(B) then even is a homeomorphism. Namely if Y c X is a closed subset, the distance function f (x) = dist(x, Y) is continuous and has the property that f ( x ) = 0, if and only if x E Y, since Y is a compact subset of IRn. So V ( f ) n Spmax(B) is the set of the maximal ideals m, of B with x E Y. 5.7.21 For an ideal I of a commutative ring, by rad(I) we will denote the intersection of all maximal ideals which contain I. (This means that rad(I)/I = Jac(A/I) .) Clearly = rad(I) and f i c rad(I).

If I is an ideal of A, then rad(IB) fl A = rad(1). Namely for x E X we have the equivalences

Further we know already n, n A = m,. Proposition 5.7.22 In the above situation let I be a n ideal of A with rad(I) = I . Set J := rad(IB). T h e n the extension of IF-algebras A / I c B I J has the properties (i), (ii) and (iii).

Proof. Let Y := {y E X I f (y) = 0 for all f E I) = {y E X I my 3 I) = {y E X 1 ny > I B ) . Then Y is a closed subset of X. By Tietze's extension theorem, every continuous function on Y can be extended to X . This means that the restriction honlornorphism C(X) + C(Y) is surjective. Its kernel clearly is rad(IC(X)) = J . Therefore B / J = C(Y) in a canonical way. This is (i). The polynomial functions on Y extend to polynomial functions on X . Polynomial functions on X restrict to equal functions on Y, if and only if they are congruent modulo I . Therefore A / I contains the polynomial functions on Y. This is (ii). We have seen at the beginning of the proof that the maximal ideals of A / I are the m,/I with y E Y. Let a E A such that its residue class a belongs to (B/J)'. This means that a Bf n, for all y E Y. But then a Bf my for all y E Y. So a is a unit in A/I. This is (iii). Lemma 5.7.23 Let B = C(X) be as above (X c IRn compact) and J a n ideal of B . Let a E GL,(B/J). There i s a n E > 0 such that for any m x mmatrix 6 = (dij) over B with Idij I < E for all i, j there is a E GL,(B) with pa = a + 6. (Here denotes the residue class modulo J.)

p

Proof. Let a' be an m x m-matrix over B such that its residue class modulo J is a p l , and set ,B = I, + 6a1, which is invertible if E is small enough.

146

5 Continuous Vector Bundles

Theorem 5.7.24 (Carral) Let A c B be a ring extension which fulfills (i), (22) and (iii) above. Let further Q be a finitely generated projective B-module for which there are finitely many ideals 11,. . . ,I d of A with rad(Ik) = I k and Ik= (0) such that Q/IkQ is free of constant rank over B/IkB for k = 1 , . . . ,d. Then there is a projective A-module P with B @ P P Q.

niI1

Proof. For k = 1,. . . , d set Jk:= rad(IkB). There are free A/Ik-modules P k with B @ Pk P &/Jk& (= &/IkQ). We will glue these by Milnor's method step by step.

n;~:

I k , J := rad(IB) and assume, there is a So let 1 < r 5 d, set I = projective A/I-module Pl with an isomorphism cp : Q / J Q 7B 8 PI.

-

We have a Cartesian diagram of rings with surjective homomorphisms B / ( J n J,)

B/J

Write Q' := Q I J Q , Q, := Q/ J,Q. Then there is an isomorphism h : Q'/JrQ1 + QT/JQr with Q / ( J n J,)Q = M(Qt,Q,,h). Note that Q, E (B/ Jr)"Hence Q / ( J n J,)Q Ei M ((B/ J) @ PI, (B/ J,)", on the overbar means 'modulo J J,.'

+

+

hov), where here and later

+

Since Q / ( J J,)Q is free, by Theorem 5.7.12 also P l / ( J J,)Pl is free over B / ( J J,). Hence hocp is given by a matrix a E GL,(B). There is a matrix a' E GLm(A/(I + I,)), arbitrarily near to a. By Lemma 5.7.23 there is a matrix p E GL,(B/J,) such that & = a'. By Remark 5.7.18 we see that

+

Corollary 5.7.25 Let X C Rn be a compact subset which is the union of finitely many contractible closed subsets. Then there is an IF-subalgebra A of B = C(X), which is essentially of finite type over IF such that the induced map IP(A) + P ( B ) is bijective. Proof. Let X = u;=, Yk with closed contractible Yk. Further for k = 1,. . . ,d let f k E C(X) be the distance function fk(x) := dist(x,Yk). Then set A := S-lIF[xl,. . . ,xn, fi, . . . ,fd], where the xj are the coordinate functions restricted to X . For k = 1 , . . . ,d set I k := rad(fkA). Then the hypotheses of Theorem 5.7.24 are fulfilled for every finitely generated projective B-module 0 Q. So the corollary follows.

5.7 Algebraization of Vector Bundles

147

Remarks 5.7.26 a) This corollary clearly applies to the spheres, but also to the infinite union X of the following n-spheres: the k-th of them has the radius l / k and the midpoint (l/k,0,0,. . . ,0). Then X = Yl U Y2 with Yl = ( ( ~ 0 ,- .- xn) E X I X n 2 01, Y2 = {(xO,.- . ,x,) E X 1 x
>

This can be generalized to finite 'CW complexes', see [16].

Remark 5.7.27 R.G. Swan in his paper [99] constructs rings A, essentially of finite type over IF with an interesting behaviour of their projective modules. For instance for every m =: 2 mod 4 he constructs a Noetherian ring A of Krull dimension m (see Chapter 6) such that there is a nonfree projective A-module of rank m, but of no other rank. To show this, he uses deep results on homotopy theory we cannot present here. Therefore we recommend the interested reader to read his nicely written original paper.

Basic Commutative Algebra I1

In the rest of the book our main interest will be, to study the minimal numbers of generators of ideals. A natural beginning is to turn the attention to those rings whose ideals are all finitely generated. Apart from the very beginning in this chapter all rings are supposed to be commutative, unless stated otherwise.

6.1 Noetherian Rings and Modules Proposition 6.1.1 Let R be a (not necessarily commutative) ring and M an R-module. The following are equivalent statements. (1) Finite generation: Every submodule N of M (including M itself) is finitely generated. (2) Ascending chain condition: E v e y ascending chain of submodules

becomes stationary, i.e. there is a k such that M n = M k for all n

> k.

(3) Maximal condition: Every non-empty subset of the set of all submodules of M has a maximal element.

The proof is a nice exercise.

Definition 6.1.2 A module satisfying the above conditions is called a Noetherian R-module. A ring R is called a (left) Noetherian ring i f it is a Noetherian as a (left) R-module.

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6 Basic Commutative Algebra I1

Remarks 6.1.3 a) Every sub-module and factor module of a Noetherian module clearly is Noetherian. And every residue class ring of a Noetherian ring is so, too. But, as we will see under c), a subring A of a Noetherian ring B need not be Noetherian. Conversely, if M is a module and U a submodule of M such that U and M/U are Noetherian, then M is so. Namely, let E be any submodule of M. If X I , . . . ,xn generate E n U and residue . . ,?/, of yl, . .. , ym E E generate E / ( E n U) G' ( E U)/U, then classes E,. X I , .. . ,x,, y1,. . . ,ym generate E, as one easily shows.

+

b) Consequently, if R is a Noetherian ring, then any finitely generated free module Rn is a Noetherian module. (Induction on n.) And therefore any finitely generated module over a Noetherian ring R - being a factor module of some Rn is Noetherian. -

By this we see that every finitely generated module M over a Noetherian ring R has a resolution (see Remark 1.2.19) by finitely generated free modules -

c) Consider ring extensions A c B. Since an ideal of A need not be one of B, there are cases, where B is Noetherian but A not. One may take A to be any non-Noetherian domain - e.g. a polynomial ring in infinitely many indeterminates over a field, or the ring of holomorphic functions on a connected open non-empty subset of @ - and B to be the quotient field of A. (We indicate why the ring A of holomorphic functions on a non-empty connected open subset U of C is a domain and not Noetherian. First let f g = 0 on U, then at least one of the functions f , g is zero on a non-discrete subset of U and must therefore be identically zero by the Identity Theorem. Secondly let (z,) be a sequence of points which is discrete in U. We define an ascending chain of ideals I, := {f E A I f (2,) = 0 for all n 2 m). By Weierstrass' Product Theorem for every m there is an f E A whose zeros are exactly the zn with n 2 m. Therefore the I, form an infinite strictly ascending chain of ideals.)

Theorem 6.1.4 (Hilbert's Basissatz) Let R be a Noetherian ring. Then R[x] is Noetherian ring. Proof. If not, then we can find an ideal I of R[x] which is not finitely generated Choose f l E I \ ( 0 ) of least degree, say nl. Inductively choose f k E I \ (fi,. . . ,fk-1) of least degree nk, for all k 2 1. This is possible since I is not finitely generated Clearly ni 5 ni+l for all i. Let ai be the leading coefficient of fi, and Ji := Ral .. . Rai for all i. Since R is Noetherian we have J k = J k + 1 for some k. But then ak+l = c:=, Xiai for some Xi E R, and

+ +

6.1 Noetherian Rings and Modules

This contradicts the choice of

fk+l

.

151

0

Corollary 6.1.5 A n algebra of finite type XI, . . . ,xn]/I over a field or prin0 cipal domain k i s a Noetherian ring. We now give I.N. Herstein's proof ([37]) of Krull's Intersection Theorem and begin with the following

Lemma 6.1.6 Let R be a Noetherian ring, I a n ideal of R, further M a finitely generated R-module and N a submodule of M . Let N' be a submodule of M which i s maximal with respect t o the property N' n N = I N . T h e n I n M C N' for some n. Proof. Since I is finitely generated it suffices to show that for any a E I there is an m with a m M c N'. (If I = (al, . . . ,a,) and ":a c N', then, with m = Maxi{mi), we have Im'M C N'.) Define an ascending chain of submodules D, of M by: D, := {x E M

I arx E N').

By the Noetherian property the chain D, becomes stationary, say at r = m.

CLAIM:

( a m M + N')

nN

= IN.

+

+

Clearly, I N = N ' n N C (amM N') n N. Conversely, if t = a m x y E l.h.s., for some x E M , y E N', then at = am+'x ay E a N C I N c N' and so am+'$ E N'. Hence amx E N', because Dm+' = Dm. And so t E N' n N = IN.

+

By the maximal property of N' we see a m M c N' !

0

Theorem 6.1.7 (Krull's Intersection Theorem) Let R be a Noetherian ring, I a n ideal of R, M a finitely generated R-module, and Mo = InM. Then IMo = Mo. In particular, if R i s a domain or a local ring and I # R, then Mo = 0.

nn2,

Proof. Let S := { N submodule of M I N > IMo, and N n Mo = IMo). By the Noetherian property S has a maximal element N'. By Lemma 6.1.6 I n M C N', for some n; and so Mo C I n M C N' C IMo and so Mo = N' n Mo = IMo. The last assertion follows via Nakayama's Lemma. 0 Noetherianity behaves well under localization.

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6 Basic Commutative Algebra I1

Proposition 6.1.8 Let R be a commutative ring and S c R a multiplicative subset. If M is a Noetherian R-module, then S-'M is a Noetherian S I R module. Especially, if R is Noetherian, then so is SP1R.

Proof. Let El c E2 be S-lR-submodules of S-'M. Let E$ := i G S ( ~ j ) . Then S-I E$ = Ej and Ei c E i . So we have an injective order preserving map from the set of all S-'R-submodules of S-'M to the set of all R-submodules of M . From this the statement follows. Note that as an R-module, S - l M need not be Noetherian. For example $ is finitely generated over $; but it is not so over Z.

6.2 Irreducible Sets 6.2.1 A topological space X is called irreducible if it is nonempty and either of the following equivalent statements holds

(1) every non-empty open set is dense in it,

(2) any two non-empty open sets intersect non-trivially,

(3) X = Fl U F2, Fl , F2closed, implies X = Fl or X = F2. Proposition 6.2.2 Let A be a ring. The irreducible closed subsets of Spec(A) are those of the form V(p) with p E Spec(A). Namely, let V(p) = V ( I ) u V ( J ) . Thenp = JiS= V(I) > V(p) or V(J) > Vp).

m.So I c p or J C p, i.e.

Conversely, assume that V(I) is irreducible for a radical ideal I. If I were not prime, there would be al,a2 4 I with alas E I. Then V(I,ai) 5 V(I) for i = 1,2, since I = and also V(I) = V(I, a l ) U V(I, az), contradicting the irreducibility. 0

a,

Thus, the map V gives an inclusion reversing bijection between prime ideals and irreducible closed subsets of Spec(R). 6.2.3 An irreducible component of a topological space X is a maximal irreducible subset of X. If Y is an irreducible subset of X then so is its closure L. Therefore, any irreducible component of X is a closed set.

A topological space X is called Noetherian if it satisfies the descending chain condition on closed subsets, i.e. any sequence

of closed subsets Y,, i

> 1,is eventually stationary.

6.2 Irreducible Sets

153

Example 6.2.4 If R is a Noetherian ring then Spec(R) is a Noetherian space. The converse is not true. See Exercise 22. Proposition 6.2.5 I n a Noetherian topological space X every non-empty closed subset Y can be expressed as a finite union Y = YI U Y2 U . - U Y, of irreducible closed subsets Y,. If in addition we assume that Y , $ Y,, for all i # j , then the Yi are uniquely determined. These are the irreducible components of Y. Proof. Let S be the set of non-empty closed subsets of X which cannot be written as a finite union of irreducible closed subsets. If S is non-empty, then, since X is Noetherian, it must contain a minimal element, say Y. Since Y, belonging to S, is not irreducible we may write Y = Yl U Y2 where Yl and Y2 are proper closed subsets of Y. By the minimality of Y both Y1,Y2 can be expressed as finite union of closed irreducible subsets, hence also Y, a contradiction. Hence S = 0 and so every closed set Y has an expression as a finite union of irreducible closed subsets Yi. By throwing away a few we may $ Y j if i # j . assume that

x

Now let Y = UL=, of Y, by

x such that x $ Yj for i # j . If Z is any irreducible subset T

z =U ( z n x ) i= 1 we get Z c for some i, since Z is irreducible. So Yl, . . . ,YT are the maximal irreducible subsets of Y, i.e. the irreducible components of Y. From this the 0 uniqueness follows directly.

Corollary 6.2.6 (Finiteness of minimal prime over-ideals) Let R be a ring with a Noetherian spectrum (e.g. R Noetherian) and a a n ideal of R. T h e n there are finitely mang prime ideals PI, . . . ,p, such that 1/? = n;='=, pi.

x

Proof. Spec(R) is a Noetherian topological space and V(I) = ULz1 with Y , irreducible closed sets. Therefore Yi = V(pi) for some prime ideals pi of R. Pi. 0 Hence, I(V(I)) = fi = I(UL='=,)V(pi)= n;='=, I(V(Pi)) =

nLzl

154

6 Basic Commutative Algebra I1

6.3 Dimension of Topological Spaces and Rings 6.3.1 If X is a topological space we define the dimension of X (denoted by dim X) to be the supremum of all integers n such that there is a chain

of distinct irreducible closed subsets of X. Note that this definition of dimension is only suited to spaces with a 'Zariski'like topology. Consider e.g. Cn with its usual (Euclidean) topology. Then its irreducible sets are the one-point sets. So its dimension, as defined above, would be 0. The 'Zariski topology' on Cn is defined as follows: The closed subsets are defined to be the zero sets of systems of polynomials (in n variables). Cn equipped with this topology is homeomorphic to the maximal spectrum of CIX1,. . . ,X,] and has dimension n. (See Section 8. on Hilbert's Nullstellensatz . This section also will explain the geometric meaning of 'dimension'.) We define the Krull dimension of a ring R to be dim(Spec(R)). Thus, the Krull dimension of R is the length n of a chain of maximal length, if any, of prime ideals in R: PO 5 PI 5 . . . 5 Pn. If no such chain exists (and R dimension.

0) then we say that R has infinite Krull

The dimension of a vector space V over a field K is denoted by dim^ V R e m a r k 6.3.2 If a ring B is integral over a subring A, then dim A = dim B. This follows from Proposition 1.4.18 and Theorem 1.4.19. The polynomial ring in infinitely many variables is an example of a ring with infinite dimension. It is, however, not Noetherian. The following example of M. Nagata shows that there are Noetherian rings of infinite dimension. Example 6.3.3 Let R = k[xl, 2 2 , . . .] be the polynomial ring over a field k in infinitely, but countably many indeterminates and pl = (xl), p2 = (22, 23), p3 = (24,25, XG),. . . . Let S = R\Ui21 pi, which is a multiplicatively closed subset of R. Set A := S-lR and mi := S-lpi First we show that the mi are the maximal ideals of A. Let a # (0) be an ideal of R which is contained in Ui21 pi. It suffices to show that a is already contained in one of the pi. Assume it were not so. For any n we have a C pl U . -- U p, U (xi,, pi). So by Proposition 1.1.8 we get a C x i > , Pi, since we assumed a @ pi for all i. But clearly nn>l(&npi) = (0), which leads to a contradiction. (We owe this argument t o 3 . J . Fendrich (Mainz).)

6.3 Dimension of Rings

155

Note that any non-zero element in A belongs to only finitely many mi as f (xl,. . . ,x,) E R \ (0) is not in pi for large i. So every non-zero ideal I of A is contained in only finitely many mi. Now every A,; = Rpi is a localization of a polynomial ring over some field (of infinite transcendence degree over L), hence Noetherian. Let Ei := {ail,. . . ,air,) with aij E I generate I,,,,in A,, , and let J be the ideal of A which is generated by the union of those Ei with I c mi. Then J is finitely generated J C I and I, = J, for all maximal ideals of A, whence I = J is finitely generated. An important theorem in Commutative Algebra is that a Noetherian semilocal ring has finite dimension. This will be established in the next two sections.

Definition 6.3.4 W e say that a prime ideal p has height r if there is a chain of prime ideals

P = PT 2 PT-1 2 . . . 2 PO and there is n o such chain of longer length. Notation: ht(p) or htR(p) if the ring R is t o be specified. It may happen that there are maximal (i.e. non refinable and non elongable) such chains of different length! In view of the bijective correspondence between prime ideals of R contained in p and prime ideals of Rp we have ht(p) = dim(Rp) = dim(Spec(Rp)). For example if R is a domain then (0) has height 0. If R is principal ideal domain and not a field then dim R = 1. In general a principal prime ideal of Noetherian ring has height 5 1.

PROOF: Let p := XR and q be different prime ideals with XR > q then q = xq. Namely every element of q is of the form xr with r E q, since x 4 q and q is prime. In the local ring Rp Nakayama's Lemma implies then q,, = (0), whence q is minimal. (We will soon generalize this to prime ideals which are not necessarily principal themselves, but only minimal prime over-ideals of a principal ideal.) The chain of prime ideals

in k[xl,. . . ,xn] shows that dim(k[xl, . . . ,xn]) 2 n. (Later we show equality.)

Definition 6.3.5 If I i s a n ideal of R we define its height t o be ht(I) = inf ht(p). Especially if I = R then ht(I) = CQ. PEV(I)

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If S is a multiplicatively closed subset of R and I is an ideal of R then ht(I) and ht(Is) = ht(I) if and only if some minimal prime overht(Is) ideal of I which has the same height as I does not intersect S.

>

Later on we will show how the concept of dimension suffices to establish a valuable bound on the number p(M) of generators of a finitely generated Rmodule M in terms of the spectrum of R. Further we present a proof of Serre's Conjecture due to H. Lindel, which only needs the concept of dimension of a Noetherian ring and its main properties!

6.4 Artinian Rings: 0-Dimensional Noetherian Rings We first recall the Jordan

- HGlder theorem (cf. [110], [45]).

A Jordan-HGlder series for an R-module M is a chain

< <

of submodules Mi of M, 0 i n, such that for every i the factor Mi/Mi+l is simple, i.e. has no proper submodule and is not isomorphic to 0. We say that (*) is a Jordan-Holder series of length n. Theorem 6.4.1 (Jordan - Holder) If M has a Jordan-Holder series then any

chain of submodules of M can be refined to a Jordan-Holder series, and any two Jordan-Holder series have the same length. Definition 6.4.2 W e say, an R-module M has finite length I i f it has a

Jordan-Holder series of length 1. Examples 6.4.3 A finite dimensional vector space has a finite length; any finite abelian group G has finite length; for any ring R and finitely generated maximal ideal m of R the ring R/mn has finite length as

is a chain of ideals where the i-th factor is isomorphic to milmi+', which is a finite dimensional vector space over R/m as m is finitely generated The series can therefore be refined to a Jordan-Holder series for R/mn. The next proposition shows that this is a typical prototype of rings of finite length: Proposition 6.4.4 Let R be a ring. The following statements are equivalent:

6.4 ArtinianRings

157

(1) R is a Noetherian ring with every prime ideal a maximal ideal. (2) Every finitely generated R-module has finite length. (3) R has finite length as a R-module. (4) Descending chain condition on ideals: Every descending chain

of ideals Ii of R is stationary after some stage, i.e. there is a k that Ik = Ik+, for all r 2 0.

> 0 such

(5) Minimal condition: A n y non-empty subset of the set of all ideals of R has a minimal element. (6) (0) = mF1 . . - m? for finitely m a n y maximal ideals mi of R and ni E IN. To show this, we suggest (1) + (2) + (3) + (4) u (5) + (6) + (1). For details we refer to 1110, 52, 53 see pages 203 - 2081. It is much easier, and totally sufficient for our purposes, to prove the equivalence of (1) to (6) under the general hypothesis, that R is Noetherian. The proof that (4) or ( 5 ) imply the Noetherian property of R is not at all automatic. (Note that modules, satisfying the analogue of (4) or (5) need not be Noetherian! For example let p be a prime number and A c $ the additive group consisting of the fractions of the form m / p n . Then A / Z is an Artinian not Noetherian Z-module.) Definition 6.4.5 A ring satisfying any, hence all, of the above six condition is called a n Artinian ring. Thus an Artinian ring is a zero-dimensional Noetherian ring and conversely. mi = ml . . em, is the Jacobson radical of an Artinian ring If Jac(R) = R then due to (vi) Jac(R) is a nilpotent ideal, i.e. Jac(R)l = 0 for some 1 2 0.

nL=,

Note that by the Chinese Remainder Theorem for an Artinian ring R we have

R E n;=l R/mni with the notation of (vi) above. Corollary 6.4.6 A reduced Artinian ring is a finite direct product of fields.

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6.5 Small Dimension Theorem A central theorem of Commutative Algebra says that the dimension of a local and hence of a semilocal Noetherian ring is finite. We recall a proof of Kaplansky (cf. [37]) here. A preliminary lemma: Lemma 6.5.1 Let R be a domain and let x, y E R \ (0). Assume that the R-module (2, ?/)/(x2)has finite length and that ((y) : x) = ((y) : x2). Then ( 2 , ~= ) (x2,y).

Proof. Since R is a domain, multiplication by x induces an isomorphism

Since ((y) : x) = ((y) : x2) it follows that the kernel of the map R (x2,y)/(x2, xy), 1 ++ / is (2). Hence,

+

Proposition 6.5.2 Let R be a Noetherian local domain with maximal ideal m. Assume that there is a z E m such that m is a minimal prime over-ideal of (z). Then ht(m) 5 1. In fact Spec(R) = {(0),m).

Proof. Note that by Proposition 6.4.4 (i) the ring R/(z) is Artinian since Spec(R/(x)) = {m) and R/(z) is Noetherian. Let y E m \ (0). The ascending chain {((y) : z")

becomes stationary say

((y) : z") = ( ( 9 ) : zn+l) = . . . = ((y) : 22") = . . Let x = zn. Consider the R/x2-module (x, y)/(x2). Since R/(x2) = R/(z2") is clearly Artinian, (x, y)/(x2) has finite length. Now (x, y) = (x2,y). By Lemma 6.5.1. Therefore, x = Ax2 +py, for some A, p E R. Hence, x(1- Ax) E (y) + x E (y) as 1 - Ax is a unit in R. If there were a prime ideal p different from (0) and m, we may choose y E p \ (0). From zn = x E (y) E p we get z E p, contradicting the hypothesis 0 that m should be a minimal prime over-ideal of (z).

6.5 Small Dimension Theorem

159

Definition 6.5.3 a) If M is a finitely generated R-module, for example a finitely generated ideal, we denote by pR(M) or p(M) the 'minimal number of generators' i.e. the minimal cardinality of all generating sets of M . b) If further p is a prime ideal we write pp(M) := pR, (Mp) . To find bounds of this number will be one of our aims!

Remark 6.5.4 It may happen that there is a minimal set of generators of M with more then p ( M ) elements. For example 2, 3 generate the Z-module %, but not 2 or 3 alone. On the other hand, if (R, m) is local and X I , . . . ,x, form a minimal generating set (with different xi), then their residue classes :I, .. . , modulo mM form a basis over Rlm, whence p(M) = n in this case. Theorem 6.5.5 (Small Dimension Theorem) Let R be a Noetherian ring and let p be a minimal prime over-ideal of an ideal I generated by r elements. Then ht(p) 5 r . In particular, ht(p) p(p) .

<

The case r = 1 is also called the Principal Ideal Theorem.

Proof. We prove the result by induction on r. Since ht(p) = ht(pRp) and pRp is minimal prime over-ideal over I R p we may assume that R is a local ring with maximal ideal p minimal over I = (al,. . . ,a,). If r = 1 apply Proposition 6.5.2. Assume the result for r - 1. Suppose now that you have a strictly ascending chain of prime ideals

We may further ensure (since R is Noetherian) that there is no prime ideal properly between p, and @,+I. We may also assume w.1.g. that a l $ p,. Note that Spec(Rl(al,p,)) = {p/(al,p,)), whence by R/(al,p,) is an Artinian local ring by Proposition 6.4.4 (2). Hence its maximal ideal is nilpotent, due to which we can find a t > 0 such that

i = 2,. . . , r , for some Xi E R, bi E p,. Let J = (b2,. . . ,b,)

c pT and let q be a minimal prime over-ideal of J contained in p,. By induction hypothesis ht(q) 5 r - 1. Since ht(p,) 2 r, q Zp,.

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Note that I1 c (al, J ) for some 1. Therefore any prime over-ideal of (al, J) contains I and therefore p is the unique minimal prime over-ideal of (al, J ) . Now in the ring R / J the prime ideal p / J is of height a t least 2 and also minimal over the principal ideal generated by (a1 mod J ) . A contradiction to Proposition 6.5.2. The above theorem has an important converse.

Proposition 6.5.6 Let p be a prime ideal of height r in a Noetherian ring R . Then there are a l , . . . , a , E R such that p is a minimal prime over-ideal of ( a l , . . . ,a,). Proof. Induction on r , the case r = 0 being obvious. Let p l , . . . ,p, be the finitely many minimal prime ideals of R. If r = ht(p) 2 1 then p $ U:=l pi as by Lemma 1.1.8 it would follow that p C pi for some i . Let a1 E p \ Ul='=,Pi. In the ring R/(al) the prime ideal p is of height less than r , since the minimal prime ideals pl, . . . ,p, have disappeared. Now use induction. Corollary 6.5.7 Let R be a Noetherian local ring with maximal ideal m. Then dim R = ht(m) equals the minimal number r of elements a l , . . . ,a,, such that m is a minimal prime over-ideal of (al, . . . , a,). In particular dim R is finite. Also if p is any prime ideal in any Noetherian ring then ht(p) is finite.

6.5 Small Dimension Theorem

161

Proposition 6.5.8 Let R = k[xl,. . . ,x,] be a polynomial ring in n variables over a field k. Then dim(R) = n. Proof. We prove the result by induction on n, it being well known for n 5 1. The chain of prime ideals

>

shows that dim(R) n. Let m be a maximal ideal of R of height 2 n, and let S = k[xl] \ (0). If S n m = 0, then ht(m) = ht(SP1m) 5 n - 1 by the induction hypothesis; hence assume S n m # 0. Let f ($1) E m be an irreducible polynomial. Then R/(f (XI)) E (k[xl]/(f (xl)))[x2,.. . ,x,]. Since k [ ~ l ] / ( (XI)) f is a field we get by induction that dim(R/(f (XI)) = n - 1. By Proposition 6.5.6 the ideal m/(f (XI)) is minimal over an ideal generated by n - 1 elements, hence m is minimal over an ideal generated by n elements. 0 Hence ht(m) 5 n by 6.5.5. Another consequence of Theorem 6.5.5 is

Corollary 6.5.9 A Noetherian domain R is factorial if and only if every prime ideal of height 1 of R is principal. Proof. Let R be factorial and p a prime ideal of height 1. There is an a E p\ (0). Let a = pl -. .p, with prime elements pi. Then pi E p for a t least one i. But plR is a non zero prime ideal, contained in p. Since the latter is of height 1, we see piR = p. Conversely, let a E R be neither 0 nor a unit. Then there is a prime ideal p of height 1, containing a. By assumption p = plR for some element pl E R. This, generating a prime ideal, is a so called prime element. (We define it by this property.) a E p l R means a = p l a ~for some a1 E R. If a1 is not yet a unit, then by the same argument a1 = pzaz for some prime element pz and some a2 E R. One gets a strictly ascending chain of ideals

By Noetherianity this must stop with (a,) = R. So a = pl - - . ~ , a , with prime elements pi and a unit a,. A decomposition of an element into prime elements always is essentially unique. This the reader should either know or prove as 0 an exercise.

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6.6 Noether Normalization Definitions 6.6.1 Let K

> k be a field

extension.

a) A n n-tuple $ 1 , . . . ,x , of elements i n K is called algebraically independent over k , i f there is no non-trivial polynomial relation between the x i , i.e. if the k-algebra homomorphism

where the X I , . . . ,X , are indeterminates, is injective. I n this case we also use the phrase: the X I , . . . ,x , are algebraically independent. (This is unprecise insofar, as algebraic independence is indeed a property of the n-tuple, and not of the individual xi.)

>

b) W e say that the transcendence degree of the field extension K k is n and write trdgkK = n , if the following holds: There are algebraically independent X I , . . . ,x , E K such that K is algebraic over k ( x l , . . . ,x,). Note that one can easily replace the above n by any - also transfinite cardinal number. It is shown in ([45],VIII, 1) that trdgkK is uniquely defined. -

>

Remark 6.6.2 If K t K 2 k are field extensions and K t is algebraic over K , then clearly trdgkK = trdgkK1. Definition 6.6.3 A n affine algebra is an algebra of finite type over a field. If we want to specify this field, we say affine k-algebra. It is clear what we mean by an affine domain. The name is explained by the fact that such algebras are function algebras on so called affine varieties at least if k is algebraically closed and the algebra reduced - as we will see later. -

Theorem 6.6.4 (Noether Normalization Theorem) Let A be a k-algebra of finite type over a field k . Let I be an ideal of A . Then there are natural numbers 6 d and elements 91, . . . ,yd E A such that

<

(1) the y l , . . . ,yd are algebraically independent over k ; (2) A is finite over k [ y l ,. . . , yd] (i.e. it is finitely generated as a k [ y l ,. . . ,yd]module). (3) I n k [ y l , - . - ,

~ d= ]

( ~ 6 + l , . . .7 ~ d ) -

6.6 Noether Normalization

163

Proof. We consider three cases :

CASE1: Let A be the polynomial ring k[xl, . . . ,x,] and I = ( f ) a principal ideal, where f is a non-constant polynomial.

<

, (1 i < n). AS in Lemma 4.3.13 we may Set yn = f , and yi = xi - xri transform f , for suitable Ti, to an element of the form f = ax:

+ alx:-'

+ . . . + am

where a E k* and ai E k[yl,. . . ,y,-l] for 1 5 i 5 n - 1. Clearly A = k [ ~ i.,. . ,yn][xn]. Since

x, is integral over k[yl, . . . ,y,], and therefore A = k[yl, . . . ,y,][x,] is finite over k[yl, . . . ,y,] by Proposition 1.4.4. If the yl, . . . ,y, were not algebraic independent over k, then n trdgkk(yl,.. . ,yn) = trdgkk(xl,.. . ,x,) = n, a contradiction.

>

We now prove that I n k[yl, . . . ,y,] = (y,). Clearly (y,) c I n k[yl, . . . ,y,]. Let g E I f l k[yl, . . . ,y,], then g = hy, for some h E A. Since A is finite, hence integral over k[yl,. . . ,y,], the polynomial h satisfies an equation of the form h S + b l h s - l + - . . + b s = O with bi E k[yl, ... ,y,], from which we get

This implies that y, divides gS in the factorial ring k[yl,. . . ,y,], whence y, divides g, i.e. g = h'y, for some h' E k[yl,. . . ,y,]. So g E (9,) as required.

CASE2: Now let I be an arbitrary ideal in A = k[xl,. . . ,x,]. The result is clear for I = (0). So we may assume that there is a non-constant polynomial f E I. We now proceed by induction on n, the case n = 1 being contained in Case 1. So assume n > 1. Let k[yl, . . . ,y,] with y, = f be constructed as in Case 1. By induction we may assume that there are elements tl, . . . , tdPl E k[yl,. . . ,yn-I] algebraically independent over k such that k[yl,. . . ,y,-l] is finite over k[tl, . . . ,td-l] and I n k[tl, . . . ,td-I] = (t6+1,.. . ,tdPl) with some S < d. Then k[yl, . . . ,y,] is finite over k[tl,. .. , td-1, y,], whence also A, being finite over k[yl, - . . ,y,], is finite over k[tl , . . . ,td-1, yn]. Further n - 1 = dim k[yl,. . . ,y,-l] = dim k[tl,. . . ,td-l] = d - 1, and so n = d. Therefore and since k(x1,. . . ,x,) is algebraic over k(tl, . . . ,t,-1, y,), the tl , . . . , t,-1, y, are algebraically independent over k.

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Since y, = f E I , clearly I r l k[tl, . . . ,t,-I, y,] = (t&+l,. . . ,td-1, yn). Conversely let g E I n k[tl,. . . ,t,-1, y,] then g = g* hy, for some g* E I n k[tl, . . . ,t,-I] = (t&+l,.. . ,td-I), and h E k[tl,. . . ,t,-1, yn]. So I n k[tl,. . . ,td-I, y,] is generated by ts+l, . . . ,td-I, 9,-

+

CASE 3: For the general case let A = k[xl, . . . ,x,]/J. As in Case 2 determine a subalgebra k[yl,. . . ,y,] of k[xl,. . . ,x,] with J n k[yl,. . . ,y,] = (yd+l,. . . ,yn). The image of k[yl,. . . ,y,] in A can be identified with the polynomial algebra k[yl,. . . ,yd]. And A is then finite over this ring. Again apply Case 2 to I' = I n k[yl,. . . ,yd] we get a polynomial subalgebra k[tl,. . . ,td] c k[yl,. . . ,yd] over which k[yl,. . . ,yd] is finite and I I I k[tl,. . . ,td] = I' n k[tl,. . . ,td] = (t6+1,. . . ,td) with some 6 5 d. Since A is finite over k[tl,. . . ,td] the elements tl, . . . ,td satisfies all the requirements of the theorem. 0

Corollary 6.6.5 Let A be a k-algebra of finite type of dimension d. Assume that A is a domain. Then the transcendence degree of &(A), the quotient field of A, over k is d. Proof. By the Noether Normalization Theorem there is an integral extension

with algebraically independent yl, . . . ,y,. By Remark 6.3.2 we see dim A = dim k[y1, . . . ,yn]. By Remark6.5.8 we have dim k[yl, . . . ,y,] = n. Hence n = d. Also Q ( A ) is an algebraic extension of k[yl, . . . ,yd] and so trdgk Q (A) = trdgkk(yl,.. . ,yd) = d by Remark 6.6.2. 0

6.7 Affine Algebras Theorem 6.7.1 (Algebraic form of Hilbert's Nullstellensatz) Let k be a field. A an afine k-algebra and m a maxima1 ideal of A. Then the canonical inclusion k L) Alm is a finite field extension. Proof. The field Alm is a k-algebra of finite type. By Noether's Normalization Lemma 6.6.5 we have trdg,(A/m) = dim(A/m) = 0. So Alm is algebraic and of finite type, hence finite over k. 0

Corollary 6.7.2 Let k be a field, f : A -+ B a homomorphism of afine k-algebras and m a maximal ideal of B. Then fP1(m) is a maximal ideal of A.

6.7 Affine Algebras

165

Proof. We have inclusions k c A/ f -'(m) c Blm. By the Theorem B/m is finite over k, hence finite over A/ f -'(m). Therefore the latter must be a field 0 by Proposition 1.4.13. This is a remarkable fact for affine k-algebras, and certainly not true for general rings.

Corollary 6.7.3 Every maximal ideal m of the polynomial ring XI,. . . ,X,] over a field k is of height n and generated by n and not by less elements. (Symbolically ht m = p(m) = n.) Proof. Induction on n , the case n 5 1 being clear. So let n > 1, B' := k[X1,. . . ,X,-11 C B := k[X1,. . . ,X,], m E Spmax B and m' := B1 n m. By Corollary 6.7.2 mf is a maximal ideal of B'. So by induction m' is generated by n - 1 elements and of height n - 1. Then p := m'B = ml[X,] is also generated by n - 1 elements and of height 2 n - 1. Namely a chain

of prime ideals in B' induces the chain

in B . Now B/p = B'[X,]/ml[X,] (B'/mf)[X,]. So m/p is a maximal ideal in a principal domain and therefore generated by 1 element. Therefore m can be generated by n - 1 1 = n elements. Further ht(m) = ht(mf[X,]) 1 n - 1 1 = n. So we have n ht(m), p(m) n and, by dimension theory, ht(m) 5 p(m), which finishes the proof. 0

+

+

<

+ >

<

This shows once more, but - using Noether's Normalization with more effort, that dim(k[X1,. . . ,X,]) = n. The reader will realize from the above proof that m is of the form (fl(X1), f2(X1, X2), . . . ,fn(X1, . . . ,Xn))-

<

Corollary 6.7.4 Let 1 6 5 n . Then the prime ideal (X6+1, . . . ,X,) of k[Xl,. . . ,X,] is of height n - 6 . Proof. Let S := k[X1,. . . ,Xs] \ (0). Then S-'k[xl,. . . ,X,] = k(X1,. . . ,Xa)[X6+1,.. . ,X,] is a polynomial ring over a field in n - 6 indeterminate~Xb+',. . . ,Xn, where the ideal generated by these is of height n - 6. The analogous prime ideal (Xs+',. . . ,X,) in the non localized ring k[X1,. . . ,X,] then must be of the same height. 0 Proposition 6.7.5 Let A be an afine domain and p

E

Spec(A). Then

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Proof. There is a Noether Normalization k[yl,. . . , y,] c A with n = dim A, such that for some 6 5 n we have p' := p n k[yl, . . . ,yn] = ( Y ~ +. .~. ,,yn). Since A is finite over the integrally closed k[yl, . . . ,y,] we can apply "going down" and see ht p = ht p' = n - 6. On the other hand Alp is finite over k[yl, . . . ,yn]/p' 2 k[yl, . . . , yg] and hence of dimension 6. Since (n - 6)

+ 6 = n we are through.

0

Corollary 6.7.6 Let A be as above. Then all maximal chains of prime ideals in A (i.e. those which are neither refinable nor extendable) have the length dim A. Proof. Let A be a counterexample of minimal dimension and be a maximal chain of prime ideals in A with m < n := dim A. Clearly ht pl = 1. Since, by the minimality of dim A, the ring Alp1 fulfills the corollary, every maximal chain of prime ideals in Alpl has length m - 1. Therefore dim(Alp1) = m - 1. But by Proposition 6.7.5 we get dim(A/pl) = n - 1 > m - 1. Contradiction. 0

Examples 6.7.7 a) If A is not a domain, Proposition 6.7.5 and its corollary may fail to hold. For example let A := k[X, Y, Z]/(XZ, YZ). Denote the residue classes of X , Y, Z by x, y, z. Then A has the minimal prime ideals (x, y) and (z), and dim(A/(x, y)) = 1,whereas dim(A/(z)) = 2. One may also replace A by its localization in the maximal ideal (x, y, z) to get a local ring B with a minimal prime ideal p such that dim(B/p) < dim(B). (Geometrically speaking, A is the ring of the union of a plane and a line, intersecting in one point. After having read the section on Hilbert's Nullstellensatz, you will understand this better.) b) Let A be a discrete valuation ring, i.e. a principal domain with exactly one non-zero maximal ideal (p). (Up to associates then p is the only prime element of A.) Consider the ideal (pX - l)A[X]. As a principal ideal it is of height 1. On the other hand it is maximal, since the residue class ring A[X]/(pX - 1) is the field &(A). So the Zdimensional domain A[X] has maximal ideals of height 1. (Other maximal ideals of height 1 are (pXn - 1). The maximal ideals of height 2 are those of the form 03,f ) , where f is irreducible modulo p.)

Definition 6.7.8 A (Noetherian) ring A is called catenary if for every pair of prime ideals q c p in A any two maximal (i.e. non refinable) chains of prime ideals starting with q and ending with p:

4 =Po have the same length.

5 . - -5 p n = p

6.7 Affine Algebras

167

Note that the examples above are catenary. Remarkably enough there do exist non catenary Noetherian rings. But: Corollary 6.7.9 E v e r y afine algebra A is catenary. Proof. Let q c p be as in the definition above. We may assume that q = (0) and so that A is a domain. If there were two maximal chains of prime ideals

with r # s , they could be prolonged by one maximal chain beginning in p and ending in any maximal ideal m p. In this way we get two maximal chains of prime ideals in A of different length, contradicting Corollary 6.7.6. 0

>

The following proposition shows that the spectrum and the maximal spectrum of an f i n e algebra are highly related. We prepare for this by two lemmas. Lemma 6.7.10 L e t A be a Noetherian d o m a i n of dimension a t least 2. T h e n A possesses infinitely m a n y p r i m e ideals of height 1, a n d (0) i s t h e i r intersection. Proof. By the Small Dimension Theorem in the case r = 1, every nonunit lies in some prime ideal of height 1. So any maximal ideal m is contained in the union of the prime ideals of height 1. If there were only finitely many of them, by the Prime Avoidance Lemma 1.1.8, at least one of them would contain m, whence ht m 5 1. But by hypothesis there are maximal ideals of height 2.

>

Now assume there were an x # 0 contained in every prime ideal of height 1. Then the Noetherian ring A/(%) would possess infinitely many minimal prime 0 ideals, contradicting Corollary 6.2.6. Lemma 6.7.11 L e t A be a n afine d o m a i n of dimension 1. It possesses infinitely m a n y m a x i m a l ideals a n d t h e i r intersection is (0). Proof. There is a finite ring extension k[X] c A with a field k. First we show that k[X]has infinitely many maximal ideals. These are in bijective correspondence to the irreducible monic polynomials. If there were only finitely many, say f i , . . . ,f, of them, following Euclid form g := 1+ fi . - - f,. Since g is not a unit, it is divisible by a monic irreducible f , which clearly is different from all f i .

Over every maximal ideal of k[X] lies one of A. And these are all of height 1. By the same argument as in the proof above, their intersection is (0). 0

168

6 Basic Commutative Algebra I1

Proposition 6.7.12 Every radical ideal I of an a f i n e algebra A is the intersection of the maximal ideals containing I .

Proof. Since I is the intersection of prime ideals, it is enough to prove the proposition for prime ideals I. Let p be a maximal counterexample in A. Then p is no maximal ideal. If dim(A/p) = 1, then p is the intersection of the maximal ideals containing it by Lemma 6.7.11. If dim(A/p) > 1, then p is an intersection of strictly bigger prime ideals by Lemma 6.7.10, the latter being intersections of maximal ideals, since p was a 0 maximal counterexample. So p is an intersection of maximal ideals. Corollary 6.7.13 Let A be a n a f i n e algebra. The relation Z H Z n Spmax(A) gives a bijection of the set of closed subsets of Spec(A) to the set of those of Spmax(A). This map is compatible with the forming of unions and intersections and with the relation of inclusion. Therefore a closed subset Z of Spec(A) is irreducible i f and only if ZnSpmax(A) is irreducible in Spmax(A). Especially dim(A) = dim(Spmax A). Note that, if A is semilocal of dimension > 0, things are totally different.

Proof. By definition a subset Z' c Spmax(A) is closed if and only if it is of the form Z n Spmax(A) with Z closed in Spec(A). So we just have to show that different closed subsets of Spec(A) have different intersections with Spmax(A). But if V(a) is closed in Spec(A) we know by Proposition 6.7.12 that J;f is the intersection of the maximal ideals m E V(a) n Spmax(A). So we recover V(a) = V(&) from V(a) n Spmax(A). The rest is then clear.

El

6.8 Hilbert's Nullstellensatz It is well-known that the maximal ideals in the ring C[O, 11of continuous functions [O,l] + B are the 'points' m, = { f I f (x) = 0) for x E [O, 11. The algebraic analogue of this is the 'Zero-point Theorem' or 'Nullstellensatz' of D. Hilbert, which classifies the maximal ideals in a polynomial ring k[X1, . . . ,X,] over an algebraically closed field k (for example C,! the field of complex numbers) as 'points' (XI - a1, . .. ,X, - a,). Note that the maximal ideal (X2 1) in B[X] is not a 'point' - so the condition that the field is algebraically closed is essential.

+

6.8 Hilbert's Nullstellensatz

169

Proposition 6.8.1 (Weak Hilbert's Nullstellensatz) Let k be an algebraically

closed field. Then the maximal ideals of k[Xl,. . . ,X,] are those of the form (XI - al, . . . ,X, - a,) where (al, . . . ,a,) E kn. This means that the injective map

which we know from Paragraph 1.1.7, is bijective i f k is algebraically closed. (The reader should not be confused by the notational ambiguity: As usual by (XI -al, . . . ,X, -a,) we mean an ideal of k[Xl, . . . ,X,], and by (al, . . . ,a,) an n-tuple in kn.)

Proof. First note that the ideal (XI,. .. ,X,) is a maximal ideal of A := k[X1,. . . ,X,], since the residue class ring is the field k. So the ideal (XI - a l , . . . ,X, - a,) for (al,. . . ,a,) E kn is also maximal, since it is the image of (XI,. . . ,X,) under the k-algebra automorphism A -+ A defined by Xi H Xi - ai. For the converse we need the hypothesis that k is algebraically closed. Let m be any maximal ideal of A. Then the inclusion k L-, Alm is a finite field extension by Theorem 6.7.1, hence an isomorphism, since k is algebraically closed. So, for every i = 1 , . . . , n there is an ai E k such that Xi G ai (mod m). Therefore (XI -al,. . . ,Xn-a,) c m. Equality holds as (XI-al, . . . ,X,-a,) 0 is maximal. Note 6.8.2 Let f E A := k[X1, . . . ,X,]. Recall from Paragraph 1.1.7 that then f E (XI - a1, . . . ,Xn - an) f (a1, - .. ,an) = 0,

*

(also if k is not necessarily algebraically closed). We will repeat the argument. Here and in the sequel let us write g := ( a l , . . . ,a,) E kn and ma := (XI - a l , . . . ,X, - a,) E Spmax(A), The composition of canonical maps

is an isomorphism, by which we identify k with Alma. Now &(Xi)= &(ai)the latter being equal to ai under the above identification- So, iff E k[X1,. . . ,X,] is arbitrary, ~ ( f =) f (a).This implies the above statement. This note enables us to derive from the proposition the following corollary, which deserves as well to be called 'Nullstellensatz'.

170

6 Basic Commutative Algebra I1

Corollary 6.8.3 Let k be algebraically closed and E c A generate a proper ideal (E) 5 A. Then there is an a E kn with f (a) = 0 for all f E E. Proof. By hypothesis there is a maximal ideal m of A containing the ideal ,. So, by the above (E). By Proposition 6.8.1 there is an a E kn with m = m 0 note, f (a) = 0 for all f E (E). Definition 6.8.4 Let E be an arbitrary subset of A = k[Xl,. . . ,X,].Then define V(E) := { a E kn I f (a) = 0 for all f E E ) . It is called the algebraic (sub)set (of kn)defined by E. (Sometimes one calls it the set of zeros of E.) The letter 'V' stands for 'variety'. And indeed the algebraic sets - equipped with suitable rings of functions are the so called 'affine varieties' of Algebraic Geometry. -

On the other hand the notion 'V(E)' is also used to denote the set of prime ideals containing E. This makes sense, since under the correspondence of points in kn and maximal ideals of A the set V(E) corresponds to the set of maximal ideals containing E . This follows directly from the above note.

Definition 6.8.5 Let M be an arbitrary subset of kn. Then define I(M):={f E A I f ( a ) = O for a l l a E M).

It is called the zero ideal of M Remarks 6.8.6 a) Clearly I(M) is an ideal, even a radical ideal of A. b) If a := (E) is the ideal generated by E , then clearly

# 0, if a # A by the above corollary. C) E c I(V(E)) and M c V(I(M)). d) El c E 2 ==+= V(E1) 2 V(E2) and MI c M2 ==+ I(M1) > I(M2). V(E)

Theorem 6.8.7 (Strong Hilbert's Nullstellensatz) Let E be any subset of A = k[X1, .. . ,X,] and a = (E) the ideal generated by E. Then

6.9 Dimension of a Polynomial ring

171

Proof. By Remark b) we see that V(E) consists of those points 9 E kn for which f (a) = 0 for all f E fi.The maximal ideals ma of A corresponding to the points 9 E V(E) are therefore exactly those which contain &. And by Proposition 6.7.12 their intersection is fi.Therefore the set of f E A with f (a) = 0, i.e. with f E ma - for all a E V(E) equals fi. 0 Corollary 6.8.8 B y the maps V and I we get a bijective correspondence between the radical ideals of A and the algebraic subsets of kn, provided k is algebraically closed.

Proof. If a = fi is a radical ideal then I(V(a)) = a by Theorem 6.8.7. Conversely let M = V(E) be an algebraic subset of k". Then E c I(M) and so M = V(E) 2 V(I(M)), hence M = V(I(M)), since trivially M c V(I(M)) as we already know. 0 Definition 6.8.9 A subset of kn is called (Zariski) closed if it is a n algebraic subset. Clearly a subset of kn, with an algebraically closed field k, is closed if and only if it corresponds to a closed subset of Spmax(A) where Spmax(A) c Spec(A) is equipped with the induced topology. So k" by the definition of closed subsets gets a topology, also called Zariski topology, with respect to which it is homeomorphic to Spmax(A). Note that a subset of k = k1 is Zariski closed if and only if it is either finite or equal to k. Therefore (for n > 1) on kn the product topology is different from the Zariski topology. The latter is finer. (Exercise!) Further, in the case k = C, the Euclidean topology on Cn is strictly finer than the Zariski topology. (Exercise!) The Euclidean topology is given by the Euclidean metric on Cn = R2".)In the following we only consider Zariski topology. 6.8.10 Recall the definition of irreducible subsets and the dimension of a topological space in Section 6.3.1.

The closed irreducible subsets of kn are in bijective correspondence to the prime ideals of A. This is easily shown with the help of Corollary 6.8.8. So the dimension of V(a) equals dim(A/a) for every ideal a of A.

6.9 Dimension of a Polynomial Ring It is fundamental to expect that the dimension of a polynomial ring R [ X ]is one more than that of the base ring R, when R is a Noetherian ring!

172

6 Basic Commutative Algebra I1

The contents of this section will be needed in Lindel's proof of Serre's Conjecture in Chapter 7, and in the proof of the Theorem of Cowsik and Nori in Chapter 10. We will study the height of prime ideals in a polynomial ring R[x] over a Noetherian ring R and, as a consequence, show that dim(R[x]) = dim(R) 1. This generalizes the fact that dim(k[Xl, . . . ,X,]) = n for a field k, which we already know. See Proposition 6.5.8.

+

Let p be a (prime) ideal of R . Then p[x] = pR[x] is a (prime) ideal of R[x]. Here p[x] is defined to be the set of polynomials whose coefficients belong to p. Clearly p[x] n R = p and (p' n R)[x] c p' for p' E Spec(R[x]). In the following lemmas R need not necessarily be Noetherian. Lemma 6.9.1 Let p' p1 = (pl n R) [XI.

5 p"

be prime ideals of R[x] with p'n R = p" nR. Then

Proof. Set p := p' n R. The inclusion p[x] c p' is clear. Were p[x] # p', the chain p[x] c p' c p" would strictly increase. We may assume p = (0), since we have the same situation for R/p c (R/p)[x] = R[x]/p[x]. Now set S := R \ (0). Since by hypothesis p' n R = p" n R = (0), we would have p' n S = p" n S = 0 and so get a strictly increasing chain of prime ideals

>

in S-'R[x], whence dim(S-lR[x] 2. But the latter ring is a principal domain, since SP1R is a field. A contradiction. 0 Lemma 6.9.2 Let p be a minimal prime over-ideal of an ideal I in R. Then p[x] is a minimal prime over-ideal of I[$]in R[x].

Proof. Assume there were a p' E Spec(R[x]) with I[$]c p' 5 p[x]. SO p = p[x] n R 2 p' n R. Since p is minimal over I, we would get p = p' n R, and by Lemma 6.9.1 this would imply p' = pix], a contradiction. 0 Theorem 6.9.3 Let R be Noetherian, p' E Spec(R[x]) and p := p'n R. Then

a) htR[,]pl = htRp in the case p' = p[x];

+ 1 in the case p' # p[x]. C) dim(R[x]) = dim(R) + 1.

b) htRjx1p1 = htRp

6.9 Dimension of a Polynomial ring

173

Proof. Let htRp = n. There is a chain

of prime ideals in R, from which we derive the following chain of prime ideals in R[x] PO[X] 5 . . . 5 ~ n [ x= ] P[x]. Therefore htR[,]pl

> h t ~ ifp p[x] = p' and htR[,]pl 2 h t ~ +p 1 if not.

Since htRp = n, there are n elements a l , . . . , a n E R such that p is a minimal prime over-ideal of Ral + - . - + Ran in R. By Lemma 6.9.2, p[x] is a minimal prime over-ideal of alR[x] + . . .+a,R[x]. Hence htR[zlp' n if p' = p[x]. This proves a).

<

Now assume p' # p[x] and htR[,]p1= r. In R[x] there is a chain of prime ideals

Set qi = pi n R. Then q, = p' If

n R = pn = p. And we have to show r 5 n + 1. 40

5 q l 5 . . . 5 %-,

then even r 5 n. SOlet j be maximal such that q j = q j + l . By Lemma 6.9.1 hence h t ~jq= htR[,]p 2 j. By the choice of j we have this implies pi = q j q j = q j + l 5 qj+z 5 ... 5 q,. Therefore n = h t ~ q ,2 r - ( j 1) h t ~ q 2 j r - j - 1 j = r - 1. (One might think, these arguments are unnecessarily complicated. But it is not clear from the beginning that there is a chain of prime ideals of length r in R[x] ending with p' and passing through p[x].)

[XI,

+

+ + <

+

We are done with a) and b). And these clearly imply dim(R[x] dim(R) 1. To show the converse inequality assume dim(R) = n < oo. Then a chain of prime ideals of length n in R like Equation (6.3) produces in R[x] the following chain of prime ideals of length n + 1

Serre's Splitting Theorem and Lindel's Proof of Serre's Conjecture

In this chapter all rings are supposed t o be commutative. W e shall present a proof of the Quillen-Suslin theorem due t o H . Lindel which only uses the concept of the Krull dimension o f a noetherian ring, and its property that it increases by one for a polynomial extension in one variable, and that it decreases by one i f we go modulo a non-zero divisor.

7.1 Serre's Splitting Theorem In Section 1.1 we had proved the Prime Avoidance Lemma: Let R be a ring, p 1 , p 2 , . . . ,pr E Spec(R), I an ideal of R and x E R. If x I c p i , then ( x ,I ) C pi, for some io.

+

The above statement implies Lemma 7.1.1 Let ( a l ,. . . ,a,) @ p i , for prime ideals p l , . . . ,p,. Then there pi. exist b2,. . . ,b, E R such that c = a1 b2a2 + . . - bnan $

+

+

UiZl

Indeed, set x = a1 and I = (aa,. . . ,a,). Proposition 7.1.2 Let A be a Noetherian ring and I c A an ideal of height n generated by n elements a l , . . . ,a, E A. Then there is a matrix E E E,(A) such that the n-tuple (bl,.. . , b,) := ( a l ,. .. ,a,)& generates I , and satisfies

>

In particular, if n 2 dim(A) + 2, then ( a l ,. . . ,a,) is completable to a matrix of E n ( A ).

176

7 Splitting Theorem and Lindel's Proof

Proof. By the above lemma we can find cl2 . . . ,cl, in A, such that the element bl = a1 c12a2 . . . clnan does not belong to any of the minimal prime ideals of A. Applying the above lemma again we can find c21, ~ 2 3 ,... ,can in A, such that the element bh = a2 c21bl ~ 2 3 ~ .3- . canan does not belong to any of the minimal prime overideals of (bl). But then bz := bb - czlbl = a2 c23a3 . . . cznan also does not belong to any of the minimal prime overideals of (bl). In particular ht(b1, b2) = 2. Proceeding as above we can obtain a set of generators as required.

+

+ +

+

+

+

+ +

+ +

(bl, . . . ,b,). Let Now to the last sentence. Clearly ( a l , . . . , a n ) -E,(A) d := dim(A). Then d 1 < n. Since ht(b1,. . . ,bd+l) > dim(A), already (bl,. . . ,bd+1) is unimodular. So (bl, . . . ,bd+l,. . . ,6,) is 'elementarily' completable, since a proper subrow is unimodular. 0

+

Remark 7.1.3 From Proposition 7.1.2 we can conclude that if (al, . . . , a,, s) is in Um,+l(A) then there are elements cl, ... ,c, E A such that ht(a1 f s c l , . . . ,a, +sc,) 2 r . Let us illustate this in the special case when r = 2; the general case is similarly done. By Proposition 7.1.2 there are A, p l , p2 E A such that the ideal (al Xa2 pis, a2 p2s) has height 2. Note that we can write a1 Xu2 p l s = a1 X(a2 p2s) + p i s . But then ht(a1 + p i s , a2 +pas) 2.

+ +

+

>

+

>

+

+

+

<

Lemma 7.1.4 Let a l , . . . , a n , s E A. Then there are elements s E A, 1 i n such that ht(al + scl,. . . , ai + sci)A, 2 i for 1 5 i 5 n in the ring A,.

<

(If I is an ideal of A, by IA, we clearly denote the ideal of A, which is generated by the image of I.) Proof. Let pl, . . . ,p, be those minimal prime ideals of A which do not contain pi, i.e. there is a s. Since s @ pi for all i, we conclude that a1 sA (f cl E A such that (a1 scl)A, has height 2 1.

+

Uzl

+

+

Let p', , . . . ,ph, be those minimal prime overideals of (al scl) which do not contain s. Again, as above, a2 sA $ Up:, i.e. there is a c2 E A such that a2 sc2 @ U p: whence ht(a1 S C ,~a2 SCZ)A, 2.

+ + + + > Continuing thus we get ht(a1 + SCI,. . . ,ai + sci)A, > i for all 1 < i 5 n.

0

We shall 'globalize' the above lemma next.

Definitions 7.1.5 Let M be an A-module and z

E

M.

a) z is called unimodular if Az is a direct summand of M and Ann(z) = (0). The set of all unimodular elements of M is denoted by Um(M). b) The order ideal C 3 ~ ( z )of z is defined by

7.1 Serre's Splitting Theorem

177

Clearly OM(z) is an ideal. And z is unimodular, if and only if 0 M ( 2 ) = A.

Remarks 7.1.6 a) If S c A is multiplicative and M finitely presented, then S-lOM(z) = O ~ - ~ M ( Z ~ ) . This follows from the fact that S - l H o r n A ( ~A) , cally, if M is finitely presented.

H o m ~ (Ms, , As) canoni-

b) Let f = (fl, ... ,fn) € A n = : F. Then OF(^) =CY&Afi. Lemma 7.1.7 Let P be a projective A-module of rank r and p E P . Let further s E A such that P, is free. Then there exists a q E P such that htOp(p sq)A, 2 r. In particular, if jj E Um(P/sP) over A/(s), then htOp(p sq) 2 r .

+

+

-

Proof. Let pl, . . . ,p, E P such that their images in Ps make up a basis of P, over A,, and write ps =

a . PZ . - with a, E A, n E N. sn 1 i=l

We may assume n > 0. By Lemma 7.1.4 there are elements ci E A such that r . Let q := sn-l C cipi. ht(a1 clsn, . . . ,a, c,sn)A,

+

+

>

By the above remarks OP@

+ sq)A, = Ope((p + sq),) 3 (a1 +

CI

sn, . . . ,a,

+ c,sn)A,,

+ sq)A, 2 r. If 1, E Um(P/sP) then 1 + s d E Op(p) for some x' E A, whence 1 + sx E Op(p + sq) for some x E A. Therefore, if a prime ideal p > Op(p + sq) then 0 s # p. Hence htOp(p + sq) = htOp(p + sq)A, 2 r. which implies htOp(p

Theorem 7.1.8 (Serre's Splitting Theorem) Let A be a commutative Noethe-

rian ring of finite Krull dimension. Let P be a finitely generated projective A-module of rank > dim A. Then P has a unimodular element. Moreover, if s E A such that Ps is free and (p, s) E Um(P $ A) then there is a q E P such that p sq E Um(P).

+

Proof 1. (H. Lindel): We may assume that A is a reduced ring with connected spectrum, whence rk P is constant. We prove the result by induction on dim A. If dim A = 0, then A is a finite direct product of fields and P is free. The second assertion is then easily verified. If S is the multiplicatively closed subset of all non-zero-divisors of A then by Corollary 6.4.6 the ring SP1A is a finite direct product of fields. Hence S-'P is

178

7 Splitting Theorem and Lindel's Proof

free; so P, is free for some s E S. Since dim(A/(s)) < dim A, by the induction hypothesis Um(P/sP) # 0.Let p E Um(P/sP) where p is the residue class of p E P. By Lemma 7.1.7 there is a q E P such that ht Op(p+ sq) 2 rk P . By 0 rk P > dim A we see p sq E Um(P).

+

Proof 2. (R. A. Rao): We sketch the idea of the proof. As before we may assume that the ring A is reduced. View the ring A as a fibre product

+

where T = 1 sA. (It is an easy check that the fibre product is equal to A.) Similarly, the projective module P is the fibre product

We use induction on the dimension of A. We choose a non-zero-divisor s so that P, is free. Let pl E Um(P,), and let p2 E Um(PT) which will exist by induction. Suppose that there is an a E Au~(P,T)such that

Then ( p l a ~ ) , = (p2al),, and so p = (plal,p2a;1) p is "locally unimodular" it is unimodular.

E

P,

XI=,

PT = P . Since

To check the existence of an a with the desired properties one observes that by the general position arguments there is an elementary matrix E E E,(A,T) such that (pl)T&= (p2),. So we only have to verify that the second property holds. It is well known that the elementary matrices have the desired 'splitting property'. More generally, one can show that

Proposition 7.1.9 Let s , t E R with (s,t) = 1. Let P be a projective Rmodule such that Psi is free of rank n, and let E E En(RSt) (regarded as a ) , , some subset of Aut(PSt) r Aut(RFt) = GLn(R,t).) Then E = ( E I ) ~ ( E ~for 0 ~1 E Aut(P,), ~2 E Aut(Pt). This follows from the fact that elementary matrices are homotopic to the identity. For then one can apply Quillen's Splitting Lemma 4.3.8 to a homotopy.

Corollary 7.1.10 Let R be a one-dimensional Noetherian ring and P a finitely generated projective R-module of rank r . Then P 5 AT P @ RT-l.

7.1 Serre's Splitting Theorem

179

Proof. By repeatedly using Serre's Splitting Theorem we get P r Q @ RT-I with Q projective of rank 1. Taking the r-th exterior power will now give us A ' P Z Q. Let P be a projective R-module. We first describe some autornorphisms of P $ R called Flips which correspond to the elementary transformations of , a E R. The following a free module. Let p,q E P , cp E P* = H o m ~ ( pR), automorphisms of P @ R are called Flips:

These have the following nice property: If I is an ideal of A, then every Flip of ( P / I P ) $ (R/I) over the ring R / I can be lifted to one of P @ R. As a consequence we now derive the famous Cancellation Theorem of Hyman Bass, which was proved in the early sixties: Theorem 7.1.11 (Bass' Cancellatiorl Theorem) Let R be a Noetherian ring of dimension d and P a finitely generated projective R-module of rank > d.

"

Then P is "cancellative", i.e. P $ & P' @ Q, for some finitely generated projective module Q implies that P r PI. In fact, if (p, a ) E Um(P @ R) then there is a product r of Flips such that (p, a ) r = ( 0 , l ) . Proof. We may assume that R is a reduced ring. Let Q Q' = F be free; then P @ F r P' @ F. So we may assume Q is free above. Therefore, it will suffice to show that P @ R 5 P' @ R implies that P P'. Let (p,a ) E P @ R denote the image of (0,l) E P' @ R. Then (p,a ) E Um(P 8 R); i.e. Op(p) Ra = R. To show that P is cancellative, it suffices to show that there is an automorphism r E Aut(P $ R) such that (p, a ) r = (0,l). For then P' r coker(p,a) 2 coker(0,l) = P. We show that there is a flip r satisfying this, by induction on d. If d = 0 then R is a finite direct product of fields, and the assertion is easy.

"

+

To perform the induction step, let us first recall that by the Splitting Theorem there is a p~ E Um(P). Further there is a non-zero-divisor s E R such that P, is free. Since dim(R/(s)) < dim R, by induction hypothesis by applying a product of flips we can map @,a)to (-6, 1). By further flips we map (-6, i ) first ) finally to @p,-6). (The 'overline' denotes 'modulo (s)'.) Then to ( ~ , iand we lift the composition of these flips to get an automorphism r' such that, if (p, a)r' = (p',a') then E Um(P), and a' = ST E sR.

>

By Lemma 7.1.7 there are y E P , n E IN such that ht(Op,, (p'+(sr)"q) dfl. Note that if a prime ideal p > Op(pr ST)^^) then ST 4 p. This is due to the ST)^^, ST)! Therefore, any minimal prime overideal of unimodularity of (p'

+

+

180

7 Splitting Theorem and Lindel's Proof

Op(p'+ ( ~ r ) ~does q ) not contain sr. Consequently, ht(Op(p'+(sr)nq) whence p' ( ~ r ) E~ Um(P). q

+

> d+ 1;

Now we can perform the following sequence of flips (p', s r ) I+

+ ( ~ r ) ~ST) q ,H (p' +

ST)^^, 1) I--+ ( 0 , l ) .

7.2 Lindel's Proof of Serre's Conjecture Theorem 7.2.1 Let R be a Noetherian ring and P a finitely genrated projective module over the polynomial ring A := R[Xl,. . . ,X,] of rank > dim R. Then P possesses a unimodular element. (cf. Definition 7.1.5.) This means that P admits a free direct summand of rank 1. By induction on rk P this implies that P is free, if R is a field. If dim R = l and Spec(R) is connected, then P splits into a direct sum of a free module and a rank-lprojective one. So it is free, if R is a principal domain. (It is extended from R, if R is a Dedekind ring, more generally, if R is a so called seminormal Noetherian ring of dimension 1. We will not go into this.) It was asked in the early seventies by H. Bass, whether Theorem 7.2.1 might be true. It was established by S. M. Bhatwadekar and A. Roy in [6],where they used the Quillen-Suslin Theorem to start the induction process on dim(R). Later H. Lindel could do without this, and so gave a new proof of Serre's Conjecture. Here we outline a variant of his argument. Remarks 7.2.2 a) Let P be a projective A-module and I an ideal of A. Consider P := P/IP as an A/I-module and let a E OF@), where z denotes the residue class of z. Then there is a b E Op(z) whose residue class in A / I is a.

Namely, since P is projective, one can lift every homomorphism P/IP A / I to a homomorphism P -+ A.

+

b) Recall that, if f = (fl, . . . , f,) E An =: F , then OF(f) = xy=l Afi. Especially if q : A -+ A is a ring endomorphism, f as above and f, := ( ~ ( f l )- ,. - ,q(fn)). Then OF(^,) = r l ( O ~ ( f ) ) Awhere , we denote by the latter term the ideal of A, generated by ~ ( C ) Ff)). ( Let us apply Remark b) to prove

Lemma 7.2.3 Let f (X) = (fl(X), . . . ,fn(X)) E RIXln, a E R n OF(f). If there is a g(X) E R[X] with 1 + Xg(X) E OF( f ) , then f (abX) := (fl(abX), . . . ,fn(abX)) is unimodular in F := R[XIn for every b E R.

7.2 Lindel's Proof

181

Proof. Clearly the map Q : REX] -+ R [ X ] , h ( X ) e h(abX) is an endomorphism of the ring R [ X ] .So by Remark b) we get l+abXg(abX) = q ( l + X g ( X ) )E OF(^ ( a b X ) and also a = ~ ( aE ) OF(^ ( a b X ) But a and 1

+ abXg(abX) generate the unit ideal of R [ X ] .

0

Proposition 7.2.4 Let M be a finitely presented A[X]-module,m E M 7 s E A such that

Then there is a unimodular m' E M with m'

f m (mod

sX M ) .

Proof. By ( 3 ) there is an r E A such that 1 - sr E OM(m). We may rename sr by s. Namely, if ( 1 ) holds for s it holds for r s ; and if the assertion holds for r s it holds for s. So we may assume

Fix an identification M, = A,[XIn and let m, = ( f i , . . . ,f,) with fi E A,[X]. For N E N, consider the endomorphism

of A,[X]. We can write

with some v E Ms. We choose N big enough such that there is a w E M with s N X v = sXW,. We set m' := m s X w . Then

+

m I m'

(mod s X M ) and mi = ( f l ( ( -l s N ) x ) , . . .

, f n ( ( l- s N ) x ) )

Since M is finitely presented, we see ( 1- s), E OM=(m,) by Remark 7.1.6 and Relation (7.1). So, using (ii), the lemma, applied to the ring A,, to a = (1- s ) , and ab = ( 1 - sN),, gives that m$ is unimodular in M , over A,[X]. But this means that sT E O M ( m l )for some r E N. Finally by (7.1) there is a linear map a : M -+ A with a(m) = 1 - s. Therefore a ( m l ) = a ( m - s X w ) E 1 sA. Together with sT E O M ( m l )this 0 implies 1 E OM(^'), i.e. that m' is unimodular.

+

Definition 7.2.5 Let I be an ideal of a polynomial ring A [ X ] (in one indeterminate). By l(1) we denote the set consisting of 0 and all leading coeficients o f f E I \ (0). Obviously l ( 1 ) is an ideal of A.

182

7 Splitting Theorem and Lindel's Proof

Lemma 7.2.6 (H. Bass, A. Suslin) Let A be a Noetherian ring and I an ideal of A[X]. Then htal(I) 2 htaix1I. Proof. Let first I be a prime ideal and p = I n A. If I = p[X], then clearly l(1) = p. With ht(p) = ht(p[X]) the lemma follows in this case. If on the other hand I # p[X], i.e. I 2 p[X], let g E I \ p[X]. Then there is an h E p[X], such that the leading coefficient of f = g - h does not belong to p. Since f E I and p c l(I), we have ht l(I) ht(p) 1 = ht I. (The latter equation is Theorem 6.9.3 b)).

>

+

Now let I be arbitrary and ql, . . . ,q, the prime ideals of A[X] which are minimal over I. (Their number is finite, since A[X] is Noetherian.) The case I = A[X] being obvious, we may assume r > 0. Then ( n i qi)N c I for some N E IN.Since apparently 1(I) . I ( J ) c 1( IJ ) , one derives 1(qi) c 1(I).

n

Let p > l(I) be a prime ideal of A with ht p = ht l(1). Then l(qi) C p for some I, and so ht I < h t q i < h t p = h t l ( I ) . 0

Proposition 7.2.7 Let I be an ideal of a polynomial ring R[X1,. . . ,Xn] with ht I > dim R. Then there exists a so called Nagata transformation of variables, fixing Xn, and sending Xi I+ Xi: = Xi X,: for suitable ri E N, 1 5 i 5 n - 1, such that I contains a polynomial that is monic in Xn with coeficients in R[Xi, . . . ,Xk-,].

+

Proof. Induction on n. The case n = 1 follows directly from the Lemma 7.2.6. So assume n > 1. Set B := RIX1, . . . ,Xn-l] and view A as a polynomial ring in one indeterminate Xn over B. Then l(1) c B is of height > dim R. By induction hypothesis we may assume, that there is a g E l(I) which is rnonic in XI, i.e. of the form g

= x T + ~ T - ~ x T - +~. - . + g o

with gi E R[X2 ,... ,Xn-11.

By the definition of 1(I), in I there is a polynomial f E A of the form

Let M be the highest power of X1 occurring in the bi and let K E lN be specified later. Then set :=Xi

for i < n and Y, :=X,-x:.

gT-l~T-l

Now g . X : = (qT+ + - .- + gO)(Yn+ Y ~ is monic ~ ) of degree ~ T + K N in Yl, whereas Yl occurs in b~-lX,N-l + . .. + bo with exponent M+K(N-1).IfwechooseKsufficientlylarge,T+KN > M + K ( N - I ) , 0 and so f is monic in Yl .

<

7.2 Lindel's Proof

183

Proof of the theorem: Induction on on n (the number of variables). If n = 0, this is Serre's Splitting Theorem 7.1.8. We may assume that R hence R[Xl, . . . ,X,] is a reduced ring with connected spectrum. Assume first that R = k is a field. Then R[Xl, . . . ,X,] = k[X1][X2,. . . ,Xn]. Then by the induction hypothesis P E L @ RIX1,. . . ,XnIT-l where L is a projective k[Xl,. . . ,X,]-module of rank 1. Since /?[XI,. . . ,X,] is a factorial, L, and hence P is free. For general R let S be the set of all non-zero-divisors in R. Since R is reduced and Noetherian, S-lR is a finite direct product of fields. By the case, handled above, S-'P is free. Since P is finitely generated. there is a non-zero-divisor s E R such that P, is free. Consider P/sX,P over (R[x,]/(sx,)) [XI, . . . ,X,-I]. And note that P,x, is free over R[Xl,. .. ,Xn],xn. Since the rank of P/sX,P over R[Xl,. . . ,X,]/(sX,) equals that of P and so is bigger than dim R dim R[X]/(sX,) (cf. Theorem 6.9.3 c)), by induction hypothesis there is a p E P, such that its residue class I, is unimodular in P/sX,P over R[Xl,. . . ,X,]/(sX,). Therefore for every q E P there is an h E R[Xl, . . . ,X,] with

>

Now we write t := sX,. By Lemma 7.1.7 there is a q E P such that ht(Op, (P + t d t ) rk(P).

>

+

+

But p tq = p, and so again by Lemma 7.1.7 we have ht(Op(p tq)) 2 rk(P) > dim(R). Applying Proposition 7.2.7, we see that OpO)+ tq) contains a monic polynomial f (X,) E R[X{, . . . ,XA-,] [X,] with coefficients in A := R[Xi, . . . ,XL-,], for some suitable variables X i , . . . ,Xk-, . This implies that A[X,]/Op (p+tq) is integral over A/An OpO)+ tq). SO,since s by the relation (7.2) is invertible in the first ring, it is invertible in the second one. This means (1 sA) n OpO) tq) # 0. SOthe hypotheses of Proposition 7.2.4 are fulfilled for M = P and m = p tq. Therefore Um(P) # 0. 0

+

+

+

Regular Rings

8.1 Definition and Jacobian Criterion Let (R,m) be a Noetherian local ring. By the Small Dimension Theorem dim(R) = htm 5 p(m) = dimR/,(m/m2) by Nakayama's Lemma. The case when there is equality is an event of much algebraic geometric significance and one isolates it as Definition 8.1.1 A local ring (R,m), which is Noetherian and for which dim R = dimRlm(m/m2), is called a regular local ring. A ring R is said to be be regular if it is Noetherian and R, is regular for every maximal ideal m of R. Examples 8.1.2 a) A regular local ring of dimension 0 is a field and vice versa.

b) k[x](,), k[[x]] are regular local rings of dimension 1 if k is a field. More generally, a local principal domain is a regular local ring of dimension 5 1. The converse is also true, but will be proved later as Corollary 8.2.3 Onedimensional regular local rings are called discrete valuation rings. Namely one can define a discrete valuation v :R

\ (0) -+ Z by v(a) = n where

a E mn \ mn+'

This is a consequence of Krull's Intersection Theorem. The map v extends to a valuation of Q(R) by v(a/b) := v(a) - v(b). (The reader who has read Exercise 18 of Chapter 1 will know about this already.) Every principal domain is a (not necessarily local) regular ring. But the class 1 is bigger than that of the principal of regular domains of dimension domains. It is the class of the so called Dedekind rings, which we will study later in this chapter.

<

186

8 Regular Rings

c) Also k[xl, . . . ,x,], is a regular local ring of dimension n, if k is a field and m = (xl, . . . ,x,). More generally this holds for every maximal ideal m of the polynomial ring k[xl,. . . ,x,], since m is of height n and generated by n elements by Corollary 6.7.3. Proposition 8.1.3 (Jacobian Criterion) Let I = ( f l , . . . ,f,) be an ideal of A = k[xl ,... ,x,] with afieldk. Letfurtherm= (XI-a1 ,... ,%,-a,) > I be a maximal ideal, and write a := (al, . . . ,a,). Then (AII), is a regular local ring if and only if

afi

rk (-(a)) axj

= n - dim(A/I),

Proof. Note that the canonical map k -+ A/m is an isomorphism, due to the special form of m. For every a E kn one has a k-linear map e:a+kn,

:;(-(

6(f)=

Clearly, 6(xi - ai) = ei, 6(m2) = {O), and

af a), ... ,G(a)) SO

6 induces an isomorphism

We get rk (%(a)) axj

= dimk 6(1) = dimr (I

+ m2)/m2.

Further

Therefore

and the statement follows.

0

J-P. Serre and also Auslander and Buchsbaum proved the following homological characterization of regular local rings: A Noetherian local ring (R, m) is regular, if and only if every R-module M has finite homological dimension, if and only if m has finite homological dimension. We give a proof of this later in this chapter after defining the requisite notions.

8.2 Regular Residue Class Rings

187

8.2 Regular Residue Class Rings Proposition 8.2.1 ( J . Ohm) Let (A, m) be a Noetherian local ring in which there is an element x, such that (x) is a prime ideal of height 1. Then A is a domain. Proof. Since (x) is a prime ideal of height 1, there is a minimal prime ideal p of A, properly contained in (x). We will show that p = (0). Let y E p c (x). Then y = ax for some a. Since p is prime, a E p. So p = px. Using x E m by Nakayama's Lemma we get p = (0), whence A is a domain. 0 Proposition 8.2.2 Every regular local ring is a domain. Proof. We proceed by induction on dim(A). If dim A = 0, then A is a field, and so a domain. So let dim(A) = n > 0. Let m be the maximal ideal and p ~. ., . ,p, the minimal prime ideals of A, and set k := Aim. We have m @ m2 U p~U - . - Up,. Otherwise, since m # m2, we have m C pi for some i by the Prime Avoidance Lemma 1.1.8. This contradicts dim(A) > 0. C h o o s e x ~ m \ ( m ~ ~ p ~ ~ - . - u p , ),..., a n dxx, ~2 m s o t h a t x , x 2, . . . ,x, represent a basis of m/m2 over k. We know dim(A/(x)) 2 n - 1. The maximal ideal n := m/(x) of the ring A/(x) is generated by n - 1 elements, namely by the classes of 22, . . . ,x,. Since generally dim(A/(x)) 5 p(n) we see that A/(%) is regular of dimension n - 1. By induction A/(x) is a domain, whence (x) is a prime ideal. By the choice of x the ideal (x) is not minimal. Proposition 0 8.2.1 now says that A is a domain.

Corollary 8.2.3 A regular local ring of dimension 1 is a principal domain. Proof. It is a domain by the proposition. And it is factorial by Corollary 6.5.9. Let p be a generator of the maximal ideal. Since @) is the only prime ideal of height one, p is essentially (i.e. up to multiplication with a unit) the only prime element. Every non-zero element is of the form upn with a unit u. From this one easily derives that every non-zero ideal is generated by some pm. 0 Definition 8.2.4 Let A be a ring, M an A-module. An r-tuple XI,.. . ,x, is called an M-regular sequence (or simply an M-sequence), if the xi belong to m and every xi is a non-zero-divisor of M/(xl M+. . .+xi-l M ) . (Especially XI has to be a non-zero-divisor of M . ) Also this applies to M = A. Proposition 8.2.5 a) Let ( A ,m) be a local Noetherian ring, and X I , . . . ,x, an A-regular sequence in m. Then dim(Al(x1,. . . ,x,)) = dim(A) - r. b) Let A be a regular local ring. Every minimal generating system XI,.. . ,a, of its maximal ideal m is an A-regular sequence.

188

8 Regular Rings

c) Let A be a Noetherian local ring whose maximal ideal m is generated by a n A-regular sequence 21,. . . ,x,. T h e n A is regular, and $1,. . . ,x, is a minimal generating system of m.

Proof. a) Since xl is a non-zero-divisor, it lies in no minimal prime ideal of A. Hence s := dim(Al(x1)) dim(A) - 1. On the other hand, there are s residue . . ,& modulo (xl) such that m/(xl) is a minimal prime over-ideal classes E,. of (5,. . . ,?/,). But then m is a minimal prime over ideal of (xl, 91,. . . ,y,), whence dim(A) s 1.

<

< +

In this way one shows a) inductively. b) Induction on n, the case n = 0 being obvious. Let n > 0. As in the proof of Proposition 8.2.2 we see that A/(xl) is regular of dimension n - 1. By induction hypothesis the residue classes 2,.. . ,% of x2, . . . ,x, form an A/(xl)-regular sequence, and since A is a domain, x1 is a non-zero-divisor of A. Together we see that X I , x2,. . . ,x, is an A-regular sequence. c) Since xl E m is a non-zero-divisor, xl is not contained in any minimal prime ideal. Hence dim(Al(x1)) = dim(A) - 1. Inductively we see dim(Al(x1,. . . ,xi)) = dim(A) - i , especially 0 = dim(Al(x1,. . . ,x,)) = dim(A) - n, i.e. dim(A) = n, which proves that A is regular. Since in an n-dimensional local ring the maximal ideal cannot generated by less then n 0 elements, the n-tuple $1,. . . ,x, is a minimal generating system. Corollary 8.2.6 Let (A, m) be a regular local ring, zl , . . . ,x, E m. The following statements are equivalent: (i) 21,. . . ,x, is a part of a minimal generating system of m; (ii) the residue ciasses of

XI,

. . . ,x, are linear independent modulo m2;

(iii) Al(x1,. . . ,x,) is regular of dimension dim(A) - r .

Proof. A subset of a finitely generated module M over a local ring (A,m) is a minimal generating set if and only if it represents a basis of M/mM over A/m. Therefore (i) (ii) is clear. (ii)==+(iii). The ring B := A/(xl,. . . ,x,) is of dimension n - r where n = dim A, since 21,. . . ,x, is an A-regular sequence. And the maximal ideal of B is generated by n - r elements, since dimA/, m/(m2, 21,. . . ,x,) = n - r . Therefore B is regular. (iii) ==+ (i). Let B and n be as above. By hypothesis the maximal ideal of B is generated by n - r elements, say ?/,+I,. . . , y,. Let $,+I,. . . ,x, E A represent them. Then X I , . . . ,x, generate m. Since dim A = n, this is a minimal generating system of m and X I , .. . ,x, is a part of it. 0

8.3 Homological Dimension

189

Proposition 8.2.7 Let I be a n ideal of a re.gular local ring (A,m) such that A / I is regular. T h e n I i s generated by a part of a minimal generating system of m. Proof. Choose y l , . . . , y, E m so that their residue classes modulo I form a minimal generating system of m/I. They form a basis of m/(m2 I)over Aim. So we have (yl, . . . ,y,) I = m. Then choose X I , .. . ,x, E I so that their classes form a basis of (m2 I)/m2 over Aim. Then X I , .. . ,x,, y l , . . . ,y, is a minimal generating set of m. Set I' := ( z l , . . . ,x,). The proposition now follows from the

+

+

+

Clearly I > I'. Further A / I and AII' both are regular of the same dimension s. So I and I' are prime ideals with I > I' and dim(A/I) = dim(A/I1). This implies I = I'.

Corollary 8.2.8 Let A be a regular local ring. T h e n the numbers p(p) for the ideals p of A with regular Alp are bounded by dim(A).

8.3 Homological Dimension The concept of homological dimension h d R M of an R-module M is a measure of how far an R-module is from being projective. Therefore one takes that a projective module is the simplest type, i.e. hdRM = 0 if and only if M is R-projective. One studies other R-modules in terms of projective R-modules. Any module M is the image of free module F, i.e. there is an exact sequence

The next simplest modules are those for which K is projective. Note that if for some free module F and some epimorphism F -+ M -+ 0 the kernel K is projective, then for any other free module F' and surjection F' -+ M , with kernel K', we will have K ' projective! This is a consequence of Schanuel's Lemma: 0 - K - F - M - 0

One has F @ K ' 2 F'@ K , and so K ' is projective. We will say that h d R M = 1 in the above case.

190

8 Regular Rings

Definition 8.3.1 More generally we say that hdRM = n if there is an exact sequence (called projective resolution of length n ) with projective R-modules Pi, 0 i n, and maps

< <

and no such sequence of smaller length. B y Schanuel's Lemma it u~illfollow = n, and one had a sequence ass above with Pi R-projective for that if h d ~ M 0 i n - 1, then necessarily P, must be R-projective.

< <

If no such inte,qer n exists we say that hdRM is infinite. Examples 8.3.2 (1) hdRM = 0 if and only if M is R-projective.

(2) hdRM = 1 if and only if M is not R-projective and M projective.

-. P / Q , wit,h P, Q

(3) Let a E R be neither a zero-divisor nor a unit. Theri hdRR/(a) = 1 as one has a free resolution

Moreover since Ann(R/(a)) not projective.

=

(a), and a is a non-zero-divisor, R/(a) is

"

(4) Let R = 2 / ( 4 ) and I = (2)R Rl(2). Then h d R I = co as one has an exact sequence O--+I--+RtI-+O, which does not split. The statement follows by splicing this sequence with itself. Note that we have seen

(5) More generally, let R be an Artinian local ring. Then for any finitely generated R-module M either h d R M = 0 or hdRM = oo. (Details are left to the reader). The key lemma, used actually to compute homological dimension of an Rmodule M , is Lemma 8.3.3 Let 0

+ M' + M

modules. Then

t 0 be an exact sequence of R -

< Max(hd M', hd M") when hd M" = hd M' + 1.

hd M

with equality except possibly

-+ MI'

8.3 Homological Dimension

191

Proof. We first claim that if any two of hd M, hd MI, hd MI1 are finite then so is the third: If M or MI1 is projective, the result is easy to see. So assume hd M > 0, hd M" > 0, and let M = P/Q, with P projective. Therefore, M' = Q'/Q with Q c Q' c P, and M" .v PI&'. Now hd Q' = hd M" - 1, hd Q = hd M - 1. One has the exact sequence

By induction on the sum of the two finite homological dimensions in an exact sequence then we prove the claim. Now proceed by induction on the sum of all three finite homological dimension! By induction one gets hd Q' 5 Max(hd Q, hd M') with equality except possibly when hd M' = hd Q

+ 1. Therefore,

hd M" - 1 = Max(hd M - 1,hd M') with equality except possibly when hd M' = hd M. Thus,

< Max(hd M', hd MU) with equality except possibly when hd M" = hd M' + 1 (by considering the hd M

various possibilities which may occur).

0

Remark 8.3.4 The above lemma is most useful to compute hd M" from hd M' and hd M. This is successful except in the case when hd M' = hd M; then it is ambiguous as one only gets an inequality. We note that homological dimension of a finitely generated module over a Noetherian ring is a 'local property' i.e. to compute hdRM one may assume that R is a local ring to begin with. This is due to Lemma 8.3.5 Let R be a Noetherian ring and M a finitely generated R-

module. Then hd M = sup hdR, Mp, p E Spec(R).

Proof. Let r.h.s. = d. Since localization preserves exact sequences, hdR, Mp 5 hdRM. Hence d hdRM. So if d is infinite then hdRM is infinite. Let us consider an exact sequence

<

with finitely generated free R-modules Pi. Then (Kd)mis a finitely generated free Rm-module for every maximal ideal m of R. Now K is projective by Proposition 2.2.8, and so hd M 5 d. 0

192

8 Regular Rings

Lemma 8.3.6 If 2n a commutat2ve diagram of module homomorphisms of the form 0 0 0

the columns and the first two rows are exact, then so is the third row. Proof. To show the surjectivity at the south east corner is easy. For the rest 0 use the Snake Lemma 1.2.18. Proposition 8.3.7 Let ( R , m ) be a local noetherian commutative ring, M a finitely generated R-module, and x E m a non-zero-divisor of R and M . Then

Proof. Choose an exact sequence

with a free R-module F. We get a commutative diagram whose columns and first two rows are exact:

8.3 Homological Dimension

193

By Lemma 8.3.6 also the last row is exact. Applying the same procedure to an exact sequence

with a free R-module Fl , we get again an exact sequence

'Splicing' this with the previous one, we get the exact sequence

Iterating this procedure, from an R-free resolution of M

we get the R/(x)-free resolution of MIxM:

This clearly implies hdRl(,)(M/xM) 5 hdRM. (Note that upto this point we did not need that M is finitely generated, R is local noetherian and x in its maximal ideal.) To show the inverse inequality assume hdRl(,)(M/xM) = n < oo and set L := ker(Fn-1 + F,-z) in a free resolution of M . By assumption LIxL is free over R/(x). So we need prove the following:

CLAIM:If M l x M is finitely generated and free over R/(x), then M is free over R.

PROOF: There is an isomorphism f' : Rn/xRn 7 MIxM, which gives an epimorphism Rn -+ MIxM. This lifts to a linear map f : Rn -+ M , so that writing F := Rn - we have a commutative square: F-M

f

+

We show that f is an isomorphism. Since f (Rn) X M = M and x E Jac(R), by Nakayama's Lemma, we see that f is surjective. Now the square fits into a diagram as above with K = ker(f). The column and the first two rows are again exact. Since this implies the exactness of the third row we get K I x K = 0. Now K is finitely generated, since F = Rn is so and R is noetherian. Again 0 by Nakayama's Lemma we see K = 0, i.e. that f is an isomorphism.

194

8 Regular Rings

8.4 Associated Prime Ideals We will need the following device in the proof the homological characterization of regular local rings.

Definition 8.4.1 Let M be an A-module. A prime ideal p of A is called associated to M , i f there is an injective A-linear map Alp + M . The set of associated prime ideals is denoted by ASSAMor Ass(M). Equivalently p is associated to M , if and only if there is an x E M with p = Ann(x). (Namely, for such an x define Alp + M by i I-, x, and conversely define x as the image of i under the injection Alp + M.) This definition formally makes sense for any ring and module. But it is of significance only if the ring A is Noetherian!

c Ass(M), if U is a submodule of M. b) If p is a prime ideal of A, then AssA(A/p) = {p). More generally, if N # 0 Remarks 8.4.2 a) Clearly Ass(U)

is a submodule of Alp, then ASSAN = {p). Namely Ann(x) = p for any x E Alp \ PI. c) If p E Ass(M), then p is the annihilator of a submodule of M, hence p > Ann(M).

Proposition 8.4.3 Let A be Noetherian and M any (not necessarily finitely generated) A-module. The set of the zero-divisors of M , i.e. of the a E A with am = 0 for some m E M \ {0), is the union of the p E AssA(M).

Proof. Every p E Ass(M) consists of zero-divisors. Namely, if Alp + M is injective, let x be the image of ( 1 mod p). Then x # 0 and px = (0). Conversely let ax = 0 with a E A, x E M \ (0). Let p be maximal in those ideals J > (a) with J y = (0) for some y E M \ (0). We claim that p is prime. First, p Then bcy = 0.

# A , since l y

=y

# 0. Now let bc

E

p.

+

If cy = 0 then p (c) annihilates y. From the maximality condition on p we derive that c E p. Now let cx # 0. Then p

+ (b) annihilates cy. As above we derive b E p.

At last define A + M by 1 I-, y. Its kernel contains p, hence equals p by its CI maximality property. So p E Ass(M).

Corollary 8.4.4 Under the assumptions of the proposition Noetherian, M any A-module - we have:

-

namely A

8.4 Associated Prime Ideals

Proof. I f M = 0 , there is no zero-divisor. I f M zero-divisor o f M . So ASSAM cannot be empty.

# 0 at least 0

E

195

A is a

Proposition 8.4.5 Let S be a multiplicative subset of the Noetherian ring A and M an A-module. Then ASSA,Ms = {ps I p E Ass(M), p n S = 0).

Proof. '2'. An injection A l p v M induces one As/ps

L,

Ms.

' c ' . Let ps E AssA,M,, say ps = A n n ( x / t ) with x E M , t E S and { a l , .. . ,an) be a generating set o f p. Then ( a i / l ) ( x / t )= 0 , i.e. there are si E S with siaix = 0. So, i f s := sl - . .s,, then sax = 0 for all a E p, whence p c Ann(sx). Conversely, i f b E A annihilates sx, then b l l annihilates x l t E Ms, i.e. b / l E ps. This implies b E p, the latter being prime. So we have shown that p = Ann(sx). Most important is the finiteness o f Ass(M) for finitely generated M , which we will show in the next proposition. Lemma 8.4.6 Let U be a submodule of a module M . Then Ass(M)

c

Ass(U) U Ass(M/U). Proof. Let p E Ass(M) and E c M be the image o f an injection Alp v M . I f E n U # 0 , then { p ) = Ass(E n U ) c Ass(U). I f E n U = 0, then E E ( E U ) / U c M / U , hence p E Ass(M/U).

+

Proposition 8.4.7 Let M be a finitely generated module over a Noetherian ring A.

a) There are finitely many prime ideals p l , . .. , p , and a composition series, i. e. a finite filtration of M :

with MilMi-1

2 Alpi.

(There is no uniqueness statement!)

b ) For any such composition series and prime ideals as in a) we have Ass(M) C { P I ., .. ,p,). Especially Ass(M) is finite. c) In the above situation for p E Spec(A) the following statements are equivalent: ( 1 ) p is a minimal prime over-ideal of A n n ( M ) ; (2) p is minimal in Ass(M); (3) p is minimal in { P I , .. . ,pn).

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8 Regular Rings

Proof. a) I f M = 0 then n = 0 and all is trivial. If M # 0, let pl E Ass(M) and MI be the image of some injection Alpl c+ M. Then apply the same procedure to MIMl to find Mz, and so on. The sequence Mo 5 MI 5 . . . will stop, since M is Noetherian. But it does not stop as long as M/Mi # 0. So a) is established. b) follows by the Lemma 8.4.6. c) Let {XI,.. . ,x,) be a generating set of M. Then Ann(M) = ni=l Ann(xi). So, if p is a minimal prime over-ideal of Ann(M) it is one of some Ann(xi) =: a. To show that p E Ass(M), it is enough to show that p is associated to the submodule Axi E Ala. Now p,, is the only prime ideal of A,,/a,,, hence associated to A,,. By Lemma 8.4.5 we conclude that p E Ass(A/a) c Ass(M). Now let pl, . . . ,p, and Mo c - .. c M, be as above, then Ann(M) clearly annihilates every Mi/Mi-l, and so pi > Ann(M). Therefore we have inclusions

Since we have shown that the minimal elements of V(Ann(M)) belong to Ass(M), statement c) is true. 0 Corollary 8.4.8 Let A be a Noetherian ring. There is a finite set of prime ideals, including the minimal ones, whose union is the set of zero-divisors of A. 0

8.5 Homological Characterization of Regular Local Rings The characterization is: A local Noetherian ring R is regular if and only if every finitely generated R-module has a finite homological dimension, 2.e. it possesses a finite free resolution. Definition 8.5.1 Let M be a finitely generated module of finite homological dimension over a local Noetherian ring R. Then there is an exact sequence

with finite pi. The Euler characteristic of M is denoted by x(M) and is n

defined to be x ( - l ) i ~ , i=l Using Schanuel's Lemma one can show that this number does not depend on the particular free resolution of M that is chosen.

8.5 Homological Characterization

197

Lemma 8.5.2 Let (R,m) be a Noetherian local ring and M a finitely generated R-module with a finite free resolution

Suppose that m has a non-zero annihilator. Then M is free. Proof. Using induction on n we easily see that it suffices to prove the result when n = 1. So assume n = 1 and that one has an exact sequence

We may assume that f maps the canonical basis of RPO to a minimal generating set of M i.e. g ( ~ P 1 c ) mRP0. By the hypothesis there is a z E R \ (0) such that zmRP0 = 0, and so zRPl = 0. Hence RBI = (0) and M 2 RPO is free. 0

Remarks 8.5.3 a) Localizing does not change the Euler characteristic: XR, (Mp) = xR(M). It follows that x(M) 2 0 (provided it is defined). Namely localize in a minimal prime ideal p. Then Mp is free over Rp by Lemma 8.5.2, and XR, (Mp) is equal to the rank of Mp over Rp. b) Let I be an ideal of finite homological dimension over a Noetherian local ring R. Then also the R-module R / I has finite homological dimension and x(I) x(R/I) = 1. Namely from a finite free resolution of I

+

by finitely generated free modules we get the finite free resolution

of R / I , and one directly observes that xR(R/I) = 1- xR(I).

Lemma 8.5.4 Let R be a Noetherian ring, M a finitely generated R-module of finite homological dimension. If x(M) = 0 then Ann(M) contains a nonzero-divisor.

Proof. Let Ass(R) = (91, . . . ,p,), then Mpi is an Rpi-module having a finite free resolution. Now Mpi is free by Lemma 8.5.2. But x(Mpi) = x(M) = 0 and SO Mpi = 0. Hence Ann(M) (t Pi for any pi E Ass(R). But then Ann(M) (t Upi and so must contain a non-zero-divisor. 0 Theorem 8.5.5 (Ferrand, Vasconcelos) Let (R, m) be a local Noetherian ring and I c m be an ideal of R. Assume that hd I < oo. Then 1/12is a free R I I module if and only i f I can be generated by a n R-sequence.

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8 Regular Rings

Proof. Here we prove only one implication of the equivalence and postpone the other one to Lemma 10.3.2. is free over RII. If I = (0), it is generated by the empty So assume that 1/12 regular sequence. So assume I # (0). Then also 1/12# 0 by Nakayama's Lemma, and - since 1/12is free over R / I - we have Ann(I/12) = I, hence Ann(I) c I. By hypothesis I is of finite homological dimension. Hence R / I over R is so, too, and x(I) x(R/I) = 1. We distinguish two cases.

+

CASE1: x(R/I) = 1 and x(I) = 0. Then for every minimal prime ideal p we have Ip= (0). This implies first I C p. Otherwise there is an a E I \ p and for this sa # 0 for every s E R \ p, and hence Ip# (0). Secondly Ip= (0) implies Ann(I) p. Namely every a E I is annihilated by some s E R \ p, and so does a whole finite generating set. Together this contradicts Ann(I)

c I.

CASE2: x(R/I) = 0 and x(I) = 1. By Lemma 8.5.4 we see that I has a non-zero-divisor. By the Prime Avoidance Lemma 1.1.8 - applied to {mI) U Ass(R) - we can deduce that I \ m I has a non-zero-divisor, say a. One can by Nakayama's Lemma complete { a ) to a set {a, b2,. . . ,b,} of elements of I which represent a basis of I / m I over Rim, hence a basis of the free R/Imodule I/12.

+

It is easily verified that the conormal module ( I l ( ~ ) ) / ( I l ( a ) = ) ~1/(12 (a)) is a free R / ( I (a))-module of rank r - 1. Note that hdRl(,)(I/aI) < co by Proposition 8.3.7.

+

Therefore by induction I/(a) can be generated by an R/(a) sequence 0 ca, . . . ,c,; whence I = (a, c2, . . . ,c,) is generated by an R-sequence.

-

Corollary 8.5.6 A local Noetherian ring R whose maximal ideal m has finite 0 homological dimension is a regular local ring. Conversely,

Proposition 8.5.7 Let (R,m) be a regular local ring of dimension d. Then any finitely generated R-module M has finite homological dimension 5 d. Proof. We may assume d > 0. Let x E m \ m2. Then R/(x) is regular of dimension d - 1. Consider an exact sequence

with F a finitely generated free R-module. Since .rr is not a zero-divisor on K , by induction hypothesis hdRl(,)K/xK = hdRK 5 d - 1. Hence hd M = hdK+lSd. 0

8.5 Homological Characterization

199

We put together: Theorem 8.5.8 (J-P. Serre) Let (R,m) be a Noetherian local ring. The following properties are equivalent: (1) R is regular. (2) Rlm is of finite homological dimension.

(3) Every finitely generated R-module of finite homological dimension.

0

Corollary 8.5.9 I f R is a regular local ring and p E Spec(R) then Rp is regular.

Proof. Let M be a (finitely generated) Rp-module, generated by say . . , x n , and N be the R-submodule, generated by the same XI,.. . ,x,. Then M E Np. Any finite free resolution of N over R produces one of M over 0 Rp. Hence the Rp-module M has finite homological dimension. XI,.

Remarkably, this corollary had not been proved in total generality before one knew the homological characterization of regular local rings. Corollary 8.5.10 A not necessarily local ring R is regular i f it is Noetherian I7 and every localization Rp i n any prime ideal p is regular.

The finiteness of the homological dimension of modules over regular local rings permits one to deduce that regular local rings are factorial, i.e. Unique Factorization Domains! Lemma 8.5.11 (Nagata) Let R be a Noetherian domain and p prime element. Rp is factorial i f and only i f R is factorial.

E

R be a

Proof. It is easy to see that if R is factorial. then Rp is so. For the converse, first note that all the units of Rp are of the type upn, for some n E Z and u E R*. Moreover, any prime element in Rp is of the type uq, where u is a unit in Rp and q prime in R: If n- E Rp is prime and n- = q/pn, then n-Rp = q R , is a prime ideal and so (qRp) n R = qR is a prime ideal, i.e. q is a prime element. 0 From this it is easy to deduce that R is factorial. Theorem 8.5.12 (Auslander - Buchsbaum) A regular local ring (R, m) is factorial.

200

8 Regular Rings

Proof. We use induction on dim(R) , the case dim(R) = 0, i.e. R a field, being obvious. Let w E m \ m2. Then R/(w) is a regular local ring, and hence by Proposition 8.2.2 it is a domain. Thus w is a prime element. By Nagata's Lemma 8.5.11 it is enough to show that R, is factorial. We have dim(&) < dim R, since m does not 'survive' in R,. So, by induction, (R,)q is factorial for every prime ideal q of R. If p is a prime ideal of & of height 1 then pq is principal for every prime ideal q of R,. Therefore, by Proposition 2.4.1 we see that p is a projective R,-module of rank 1. Now p has a finite free resolution; namely every finitely generated R-module, hence every finitely generated R,-module has one. (See the proof of Lemma 8.3.5.) Therefore p is stably free, and so p is free of rank 1 by Lemma 3.1.14. Hence 0 p is principal. So R, is factorial by Corollary 6.5.9.

Corollary 8.5.13 A regular local ring is an integrally closed domain. This is clear by Example 1.4.10 a).

A domain is integrally closed if the localization in every maximal ideal is integrally closed. Therefore a regular domain is integrally closed.

I7

Corollary 8.5.14 A regular domain A is factorial, if and only i f Pic(A) = 0.

Proof. Recall that by our definition a regular ring is Noetherian. So A is factorial, if and only if its prime ideals of height 1 are principal. They are locally principal, hence invertible by Theorem 8.5.12. So they are principal if 0 Pic(A) = 0. The converse is Corollary 2.4.14.

8.6 Dedekind Rings Dedekind rings are certain (commutative) domains which generalize slightly principal domains. They appear as integral closures of Z in finite extensions of $ and also as 'coordinate rings of nonsingular a&e curves'. There are several formally different, but actually equivalent possibilities to define them.

Theorem 8.6.1 Let A be a domain. The following properties are equivalent: (1) A is Noetherian, integrally closed and of dimension

5 1;

(2) every non-zero ideal of A is invertible; (2') A is regular of dimension 5 1, i.e. A is Noetherian and A, is a principal ideal domain for every maximal ideal m; (3) evey non-zero ideal of A is a unique product of maximal ideals. (Of course the uniqueness is meant upto the order of the factors.)

8.6 Dedekind Rings

201

Definition 8.6.2 A domain A which fulfills the above equivalent properties is called a Dedekind ring. Clearly all principal domains - fields included - fulfill all above properties, so they are Dedekind rings. The equivalence (ii) (ii') follows directly from the fact that invertible ideals are those which are finitely presented and locally principal. We will prove the theorem during this section. Without loss of generality we will assume that A is not a field. At first we prove two lemmas.

Lemma 8.6.3 Every minimal prime ideal p of a (not necessarily Noetherian) ring A consists of zero-divisors. If additionally p is finitely generated, there is a b E A \ (0) with bp = (0). Proof. a) Let a E p \ (0) and up be its canonical image in pAp. Since pAp is the only prime ideal of Ap, we have a; = 0 and a;-' # 0 for some n 1, hence sun-'a = 0 for some s E A \ p and sun-I # 0.

>

b) Since p is finitely generated, there is an n 2 1 with PA,,)^ = (0). Let n be minimal with this property. Let b E A have an image in Ap which is non-zero and lies in pn-'Ap . If a l , . . . ,a,. generate p there are si E A \ p with sibai = 0. 0 For s := sl .--s,.we have sb # 0 and sbp = (0).

Lemma 8.6.4 Let A be a local Noetherian domain whose maximal ideal m is non-zero and principal, say m = (p). Then every element of A \ (0) is of the form pnu with n E IN and u E A X . Consequently m is the only non-zero prime ideal of A. Proof. Let a E A \ (0). Since, by Krull's Intersection Theorem, n,(pn) = n n m n = (0), there is an n E IN such that a E (pn), but a $! (pn+l). This means a = pnu with some u E A \ (p) = A X . Now if q is any non-zero prime !J ideal of A and pnu E q \ (0), then u 4 q, whence p 6 q, whence (p) = q. 8.6.5 We prove (1)

+ (2),

(3) of Theorem 8.6.1

First we show that every maximal ideal m is invertible. Since A is not a field, m # (0). Let a E m \ (0) and consider m-l. We have A c m-' c a-lA and we will show that A # m-l. Since m/(a) is a minimal prime ideal of A/(a) and since it is finitely generated, by Lemma 8.6.3 there is a b E A \ (a) with bm c (a). Now clearly ba-lm c A and ba-l $! A, whence ba-l E m - l \ A. We have m c mm-' C A. The &st inclusion follows from A C m-' and the second one from the very definition of m-l. We have to exclude mm-l = m.

202

8 Regular Rings

Assume it were so and x 6 m-l. Then xm C m, hence xn+lm c xnm, hence by induction - xnm C m for all n E IN. So xn E m-I C a-lA for all n. Therefore every x E m-I is integral over A. Since A is integrally closed, we get m-I = A, which we already have disproved.

-

The only possibility is mm-I = A, i.e. m is invertible. Now we show that every ideal I f (0) in A is the product of maximal ideals. I is the empty product, if I = A. If not, I is contained in some maximal ideal m. Clearly I c m-lI. But also I # m-lI. Namely otherwise m I = rnmW1I= I. Localizing in m and using Nakayama's Lemma, we would get I, = (0), hence I = (O), since I c, I,. So we derive a chain of ideals in A

with maximal ideals mi, which must eventually stop by Noetherian property, but which can be continued as long as the last ideal differs from A. Therefore

with suitable maximal ideals ml , . . . ,m,. This implies that I is invertible, since the mi are so. Now to the uniqueness. Let

>

with different maximal ideals ml ,. . . ,m,, ni 1 and m any maximal ideal of A. Then Im=mn4,,,, i f m = m i , and Im= A i f m # m i fori = 1,... , r . Since A is Noetherian, mnAm = mmAmimplies n = m by Nakayama's Lemma. So the mi and ni are unique. 8.6.6 We prove (3) + (2). Let I be a non-zero ideal of A and a E I \ (0). We can write I=m:l..-mr and (a) = m y l . . . m p *

>

0. Since (a), C I, for every maximal with maximal ideals mi and ni,mi ideal m, we get ni 5 mi for every i. (Note that we do not need the Noetherian property for this argument.) Define

Then IJ = (a), whence I(a-'J) = A. So I is invertible. 8.6.7 At last we show (2) erated, A is Noetherian.

+ (1).Since every invertible ideal is finitely gen-

8.7 Examples

203

Now let p be a non-zero prime idcal of A. Since it is invertible, its localization pAp is principal and so by Lemma 8.6.4 is the only non-zero prime ideal of Ap. Therefore, in view of Theorem 1.3.21, there does not exist any non-zero prime ideal properly corlttained in p. Still we have to show that A is integrally closed. So assume that x E Q ( A ) is integral over A, i.e. the ring I := A[$] is a finitely generated A-module. Therefore I is also a fractional ideal, isomorphic as a module to an ideal, contained in A. So I is invertible and, since I is a ring, 12= I. The invertible ideals form a group, whence I = 12implies that I is the identity element of this group, i.e. I = A. So x E A.

Proposition 8.6.8 Every ideal of a Dedekind ring A is generated by 2 elements. More precisely, i f I # (0) is an ideal and c E I \ (0), then there is an a E I with I = (a, c). Proof. Let ml, . . . ,m, be the maximal ideals containing c and S = A \ U imi. Let SP1Iin SP1A be generated by a E I. Then (a, c), = I, for cvery maximal ideal m of A, as one sees by considering the two cases c E m and c 4 m.

8.7 Examples Proposition 8.7.1 Let R be a Dedelcind ring (e.g. R = Z or R = k [ X ]with a field k ) , K := Q(R), K C L a finite field extension and A the integral closure of R i n L. Assume further that A is finite over R. Then A is a Dedekind ring.

<

Proof. We have dim(A) = dim(R) 1by Remark 1.4.4. Since R is Noetherian and A is a finitely generated R-module, every ideal of A is a finitely generated R-module as well. So, a fortiori, it is finitely generated as an A-module. Finally A, being an integral closure in L, is integrally closed.

Remarks 8.7.2 a) The so called Theorem of Krull-Akizuki assures that the assumptiorl of the finiteness of A over R is not necessary. (See [lo] VII.2.5.) Though its proof is much longer, and we will see that the finiteness holds in important cases. b) In the above situation, L is the quotient field of A. Moreover L = {alr I a E A, r E R \ (0)). Namely every x E L fulfills an equation of the form

with ai E R, a, # 0. Multiplying this equation by a:-', is integral over R, i.e. b E A, whence x = bla,.

we see that b := anx

c) From b) we see that there exists a basis of L over K consisting of elements of A.

204

8 Regular Rings

Proposition 8.7.3 Let R be a Noetherian integrally closed domain with quotient field K and A the integral closure of R i n a finite extension field L > K . Assume that there exists a non-zero K-linear map T : L -+ K with r(A) c R. Then A is finite over R. Especially this holds i f L is separable over K . Proof. For every R-submodule M of L define

M~

:= {b E L

I +a)

E

R for all a E A).

Then M c N implies M~ > N". Especially we get AD c ED if E = Ral + - . - + R a n where a l , ... , a n is a basis of L over K with ai E A. Now T(A) C R implies A C AD. Further we have an R-linear map E~ -+ Rn by b t-, ( ~ ( b a l ).,. . ,r(ban)), which is injective. Otherwise r(bal) = - . -= ~(ba,) = 0 for some b E L X .This would imply T = 0, since hal,. . . , ban is a basis of L over K. In all, we have an R-linear embedding A v Rn, which gives the result, since R is Noetherian. If L > K is separable, the trace nLIK is known to be different from zero (see ([110], Vol I, pg. 92)) and it maps A into R. Namely, if a E A, all conjugates of a also are integral over R, and so is their sum TrLIK(a). Since R is integrally 0 closed, TrLIK(a) E R.

Corollary 8.7.4 Let k be a field, R a k-algebra of finite type and a domain and L > Q(R) a finite field extension. Then the integral closure A of R in L is finite over R. Proof. According t o Noether's Normalization Theorem 6.6.4, the ring R is finite over some polynomial ring k [ x l , .. . ,xn]. So we may assume R = k[xl, . . . ,x,], in which case R is integrally closed. If char k = 0 then L is separable over K and the statement follows directly from Proposition 8.7.3. So assume char k = p # 0. Let L' be the normal closure of L over K . Then - R being Noetherian - it is enough to show that the integral closure of R in L' is finite over R. So we assume the extension L > K to be normal. In this case L is separable over the maximal purely inseparable sub-extension Li of K . Once we have shown that the integral closure A' of R in Li is finite over R, the corollary follows. For, then A' is integrally closed and Noetherian and L > Li = &(A') is separable, whence A is finite over A' and so over R by Proposition 1.4.6 b). So finally we may assume R = k[xl, . . . ,x,] and L purely inseparable over K = Q(R). We will construct a T : L -+ R which fulfills the conditions on T in Proposition 8.7.3.

8.8 Modules over Dedekind Rings

205

There is a power q of p with L c ~ ~ 1Then ' . = F ~ / Q [ x ,' /. .~. ,x;Iq] is integrally closed (being isomorphic to R) and integral over R, so is the integral ' . A = B1/q n L. closure of R in ~ ~ 1Then

CLAIM:R1/q is a free R-module. Namely B := F[X:/~,. . . ,x;'~] is free over R, a basis being the family of all products x ~ ' ./- .~x 2 I q with 0 ai < q. Further R1/' = F1/~[x'/' 1 ,... , X L/q] n l/q is free (but not necessarily finitely generated) over B = k[x:lq,. . . ,xn 1, a basis being any basis of F1/q over F. Then all the products of one basis element of R'/' over B by one basis element of B over R form a basis of ~ ' 1 4over R.

<

Since R ' / ~ is free over R there is an R-linear projection p p(A) # 0. Define then r :L

+K

by

:

~ l / qt R with

alb ++p(a)/b for a E A, b E R.

This has the required properties.

Examples 8.7.5 Let R be a principal domain and d E R be neither a square of a unit, nor divisible by a square of a prime demcnt. Consider L := K(&) where K = Q(R). Assume char(K) # 2, in which case especially L is separable over K . We will compute the integral closure A of R in L. Let a = a b& with a, b E K . If a t A clearly the trace t = 2a and the norm n = a2 - b2d of a must belong to R. Conversely, if they do, a is integral over R, since then a2 - t a n = 0. We consider two special cases, and the reader is invited to try other ones.

+

+

1) Suppose 21d. Then one easily derives A = R

+ R&

= R $ R&.

This includes the case that 2 is a unit, especially the case where A is an algebra over a field of characteristic # 2. So the rings, considered in Examples 2.4.15 b) are Dedekind rings, but often not principal domains. 2) Suppose R = Z.Then one computes

Especially the ring Z $ Zfi, considered in Examples 2.4.15 a) is a Dedekind ring, but not a principal domain.

8.8 Modules over Dedekind Rings The following proposition will be improved later:

206

8 Regular Rings

Proposition 8.8.1 Let An be a finitely generated free module over the Dedekind ring A and U C An be a submodule. T h e n U is the direct s u m of at most n projective A-modules of rank 1.

Proof. We show this by induction on n, the statement being true for n 5 1. Let pr, be the projection onto the last factor and denote its kernel by An-'. Then we get an exact sequence

Since I , := pr,(U) c A is an ideal, hence projective (maybe O), the sequence splits. So U (U f l An-') @ I,. By the induction hypothesis we are done.

Remark 8.8.2 Using transfinite induction or Zorn's Lemma in a clever way, one can show the same also in the case that U is a submodule of any free module, not necessarily finitely generated. Corollary 8.8.3 Every finitely generated projective module over a Dedekind ring i s a direct s u m of projective modules of rank 1. As already mentioned, one can do better as we will see in the sequel.

Definitions 8.8.4 Let R be any domain and M any R-module. a) Define its torsion (or torsion submodule) t o be T(M) := {a E M

I there is a n a E R \ (0) with ax = 0)

b) M is called a torsion module, if T(M) = M c) M is called torsion free, if T ( M ) = 0.

Remark 8.8.5 Clearly T(M) is a submodule of M , and T(T(M)) = T(M). So T(M) is a torsion module. And M/T(M) is torsion free. Proposition 8.8.6 a) Every finitely generated torsion-free module over a Dedekind ring is projective. b) Consequently every finitely generated module M over a Dedekind ring is of the form M

E T $ P,

T = T(M) a torsion module , P E M/T(M) projective.

Note that the torsion free, but not finitely generated Z-module $ is not projective. The projective summand P above is not a canonically defined submodule of M.

8.8 Modules over Dedekind Rings

207

Proof. a) Let M be a finitely generated torsion-free module over the Dedekind ring A and S := A\(O). Since M is torsion-free, iM,s: M + S-lM is injective. We regard M as a submodule of S-I M. The latter is a finitely generated vector space over K := Q(A), and we can choose a basis XI,. . . ,x, with xi E M . Since M is finitely generated, there is a b E A \ (0), with bM c C A x i . n a,. Namely if M is generated by y1, . . . ,y, and yj = A x i then one may set

n.

sij

b = w. sij. The map M + bM, x I-, bx is an isbmorphism. Since bM is projective by Corollary 8.8.3, so is M.

0

b) is clear, since M/T(M) is projective.

For a complete classification of all finitely generated modules over a Dedekind ring we have to analyze the finitely generated torsion modules and the finitely generated projective ones. The following proposition is essentially a special case of Serre's Splitting Theorem. Here we give an independent elementary proof.

Proposition 8.8.7 Let P , Q be projective R-modules of rant 1 where R is a Noetherian ring of dimension 1. Then

Proof. P and Q are isomorphic to non-zero (invertible and integral) ideals I, resp. J of R. CLAIM.There is an (integral) ideal I' r I with I' + J = R. Namely there are only finitely many prime ideals containing J, say p1, . . . ,pn. p. Then SP1(I-l) is a principal fractional Set S := R \ (Uy=, pi U UpcAss(R) ideal of the semilocal ring S-'R. We know that every invertible fractional ideal of a semilocal ring is principal by Proposition 2.2.14. Let x E I-l be a generator of S-l(I-l). Then I' := zI c R and S-lI' = S-IR. Therefore I' is not contained in any of the prime ideals containing J, whence I' J = R.

+

Therefore the map I' @ J -+ R , defined by (x, y) I-, x - y, is surjective. And its kernel consists of the pairs (x, x) with x E I' n J. So we have an exact sequence 0--+I'~J-&I'@J-%R-+o, with f (a) = (a, a), g(a, b) = a - b which clearly splits. By the Chinese Remainder Theorem and Lemma 2.4.10 we get I' n J = I'J P @ Q. And so ( P @ Q ) @ R EP @ Q . 0

"

Corollary 8.8.8 Any finitely generated projective module P of rank n over a Dedekind ring is of the form

208

8 Regular Rings

where L is a rank 1 projective module. L is unique upto isomorphism. Proof. We know already P Z L1 $ - ..$ L, with Li of rank 1. By Proposition 8.8.7 we get P E (L1 8 . . . @ L,) $ An-'. The uniqueness of L follows from An(L @ An-') E L. 0

Remark 8.8.9 Again we see that every ideal of a Dedekind ring is generated by at most 2 elements. Namely let I # (0) be an ideal. Then I $ I - I r 11-I 63 R % R2. Therefore there is a surjective homomorphism R2 -+ I. Consequently, any projective module of rank n over a Dedekind ring is generated by a set of at most n 1 elements.

+

Proposition 8.8.10 (Elementary Divisors for Dedekind Rings) A finitely generated torsion module M over a Dedekind ring A is of the form

M

(A/Il) ED - .@ (A/In),

where Il 3 - .. 3 I, is a descending series of non-zero ideals of A. The Ij are unique. Proof. Let e l , . . . , e m form a generating set of M. For every ei there is a cl E A \ (0) with ciei = 0. So c M = 0 for the non-zero c = c l . - - e m . There are only finitely many maximal ideals of A containing c, say ml, . . . ,m,. Let S = A \ Ui mi. Then S-lA is a semilocal Dedekind ring, hence a principal domain. The elements of S are units modulo (c), since their residue classes are outside every maximal ideal of A/(c). Therefore S-lA/(S-lA)c

S

A/(c).

So M as an A/(c)-module is an S-lA-module. By the Elementary Divisors Theorem 4.1.1 for principal domains we get where Ij := aj n A. By an easy consideration we see that it is enough to show the uniqueness of the Ij locally. So assume that A is a local principal domain with the maximal ideal m and k = Alm. Then Ij = mTj where rl 5 7-2 5 . .. r,. The numbers rj are determined by the vector space dimensions di := rkk(miF1M/miM) in the following way:

<

< <

for 0 < i rl di = n di = n - 1 for rl < i rz

. .. .. .

.. . . . . . . .. . .

di = 1 di = 0

for rn-1 < i for r, < i

< r,

8.9 Finiteness of Class Numbers

209

8.9 The Finiteness of the Class Number Our goal is to prove the following important classical result.

+

Theorem 8.9.1 Let R = or R = F[X] with a finite field F . Let further A 1 R be a finite ring extension, A a domain. Then there are only finitely many isomorphism classes of ideals of A. Especially, Pic(A) is a finite group.

(&(A) is a so named global field and A an order in it.) Remarks 8.9.2 a) Recall the following facts on ideals, fractional ideals and rank 1 projective modules of a domain A. Ideals I, J are isomorphic, if and only if there is an x E Q(A)X with XI= J. By Remark 2.4.13 every rank 1 projective module over the domain A is isomorphic to an invertible ideal of A. Especially Pic(A) E Inv(A)/Prin(A), where Inv(A) denotes the group of invertible fractional ideals, Prin(A) that of principal fractional ideals of A. Note that every fractional ideal is isomorphic to an 'integral' one, i.e. one contained in A. (If I c s-lA, then sI c A.)

b) If A is the integral closure of R in a finite field extension of Q(R), then A is finite over R according to Proposition 8.7.3 and Corollary 8.7.4. 8.9.3 If f E R \ (0), then R/ f R is a finite ring. We define ( f 1 := #(R/R f ) if f # 0 and 101 = 0. If R = Z this is the usual absolute value. If R = F[X], q := #F and f E R \ (0) we have If 1 = qdeg(f). In both cases:

l f g l = I f 1 .I91 and

(Even

I f + gl

I f + gl II f 1 + 191.

(8.1)

5 Max{l f l,lgI) in the case R = F[X].)

As an R-module the ring A is torsion-free and finitely generated, hence free. Since K = &(A) = {als I a E A, s E R \ (0)) by Remark 8.7.2 b), clearly A is of rank n := [&(A) : Q(R)]. Fix an R-basis al, . . . ,anof A. This is also a basis of Q (A) as a vector space over Q (R) . Now let I # (0) be an ideal of A. It is also a finitely generated torsion-free R-module, hence free. Let a E I \ (0). Then the homothesy of a on the Q(R)vector-space &(A) is an automorphism. Therefore a a l , . . . ,aa, are linearly independent over Q(R), hence over R. It follows that I, contained in A and containing cual,. . . ,aa,, is a free R-module of rank n and that A / I is a finite ring. We write IlIll := #(A/I). Lemma 8.9.4 Let a E A and h, denote the homothesy of a on A, regarded as a free R-module. Then:

210

8 Regular Rings

Proof. The case a = 0 being clear, assume a # 0. By the Elementary Divisor Theorem 4.1.1 there are an R-basis a:, . . . ,a; of A and d l , . . . ,dn E R with aa: = dial. So det(h,) = d := dl . . - d n E Aa. Further both sides of b) are equal to Id/. As a consequence of a) we see that I n R # (0) for every non-zero ideal I of A, in other words, that A / I is a torsion module over R. Lemma 8.9.5 There is an integer C > 0 such that in every non-zero ideal I ofA there is a y # 0 with I(Ayll CllIll, i.e. #(I/Ay) C.

<

<

Proof. To determine y E I, we distinguish two cases.

CASE1: R = 22. Let m E IN be such that mn of the following (m l)n elements

+

must be congruent modulo I, since #(A/I) be our y.

< IlIll < (m + l ) n . At least two

< (m + l ) n . Their difference will

CASE2: R = F[X]. Let q = #F and s E IN, such that qsn Two of the following q(s+l)n elements

< llIll < q(Sfl)n.

must be congruent modulo I. Again call their difference y. In both cases y has the properties:

The latter means Imj I 5 r , where we set r :=

m.

Let as above al, . . . , a nbe a basis of A over R. Then one can write aiaj = xi,j,kaijkak with aijk E R. For y = x j m j a j (with m j E R) we have mjaijkak. So det(h,) is a homogeneous polynomial of degree n in yai = the m j over R. Therefore by facts (8.1) there is a C E IN such that ImjI r 1 1 . 0 implies I det(h,) 1 5 Crn = CII 1

xj,k

<

8.9 Finiteness of Class Numbers

211

Proof of Theorem 8.9.1: Let c E R be the product of all a E R \ ( 0 ) with la1 5 C. (There are only finitely many of them.) We will show that every ideal I # ( 0 ) of A is isomorphic to one between A and Ac. Since A / A c is a finite ring there are only finitely many of the latter. Choose y as in Lemma 8.9.5. Then I y - ' / A E I / A y is of order 5 C. There is an R-module isomorphism I y - ' / A S R / ( d l ) @ - @ R/(d,) with suitable di E R \ ( 0 ) . Then ldil C , whence dilc. So c ( I y - l / A ) = 0 , i.e. cy-'I c A. 0 On the other hand c A c c y l I , and we are done.

<

Corollary 8.9.6 Let A be as above. For every n E lN there are only finitely many isomorphism classes of projective A-modules P of rank n.

Proof. Since dim(A) = dim(R) = 1,by Proposition 8.8.7 we have an invertible ideal I with P E I @ An-'. 0

A strong generalization of the theorem and its corollary is the JordanZassenhaus Theorem. See ([97]Theorem 3.9). The finiteness of class number has other interesting consequences:

Proposition 8.9.7 Let A be a Dedekind ring with finite Picard group. (Or more general, let A be any domain with only finitely many isomorphism classes of ideals.) Then there is an f E A \ ( 0 ) , such that Af is principal.

Proof. Let the ideals I 1 , . . . ,In represent the isomorphism classes of the nonzero ideals and choose f E Il . - . In \ ( 0 ) . Then every non-zero ideal of A is 0 isomorphic to one of the ( I j ) = Af , hence principal. Proposition 8.9.8 Let A be a Dedekind ring whose Picard group is a torsion group. Then for every ideal I of A there is an f E I with J? = In the terminology of the next chapter, every ideal is a set theoretical complete intersection.

m.

Proof. For I # ( 0 ) there is an r with I T = A f for some f E A. This f has the 0 required property.

Bounds on the Number of Generators

The theory of the numbers of generators is the second main theme of our book. All rings in this chapter will be supposed to be commutative.

9.1 The Problems Recall some facts of Linear Geometry (or Vector Space Theory). Let k be a field, U a linear subspace of kn of dimension n -r and c E kn. Sets of the form c U are sometimes called affine linear subsets of kn. We know that c U is the set of solutions of a system of r linear equations

+

+

has rank r. With the notations we used where the matrix A' := (aij)l ji in the section on Hilbert's Nullstellensatz

is an algebraic subset of kn The dimension of c+ U as an algebraic subset of kn also is n -r, the dimension of U as a vector space. This can for e.g be seen as follows: The matrix A' can be completed to an invertible n x n-matrix A := (aij)lli,jln. Performing the following change of variables:

for i = 1 , . . . ,n with bi = 0 for i

> r , we see

I(c + U) = (Yl, . . . ,YT) and k[Xl,. . . ,Xn]/I(c + U)

k[Y,+i,. . . ,%I.

214

9 Number of Generators

So, in algebraic terms, we get the following: Let k be a field, A := k[X1,. . . ,X,] and I an ideal of A which is generated by polynomials of degree 1. Then ( I is a prime ideal and)

We ask to what extent these equalities will hold if I is not necessarily generated by linear polynomials, or if A is a more general Noetherian ring. The second equality, by Proposition 6.7.5, holds for every affine domain. We will concentrate our attention on the first one. Generally for any Noetherian ring A and any ideal I # A we only have the inequality p(I) 2 ht(I). The case of equality deserves a name:

Definitions 9.1.1 a) A n ideal I of a Noetherian ring A is called a n ideal theoretic complete intersection or simply a complete intersection if p(I) = ht(I). b) T h e ideal I is called a local complete intersection if I, is a complete intersection in A, for every maximal ideal m containing I. In the next section we will show that a complete intersection in the polynomial ring A = k[Xl,. . . ,X,] (more generally in any regular or Tohen-Macaulay' ring) can be generated by an A-regular sequence. (Note: If I c A of height r is generated by r elements a l , . . . , a r then a l , . . . , a r need not be a regular sequence itself. See Remark 9.2.4.)

Examples 9.1.2 a) Every maximal ideal of A := k[X1,. . . ,X,], where k is a (not necessarily algebraically closed) field, is of height n and generated by n elements, as we know from Corollary 6.7.3. So it is a complete intersection. (As one easily sees from the construction in the proof of Corollary 6.7.3, the maximal ideals are even generated by regular sequences.) b) Also every prime ideal of height 1 of A is a complete intersection, since A is factorial. See Corollary 6.5.9 c) If A is a regular local ring and I an ideal with regular residue class ring AII. Then I is a complete intersection (and prime) by Corollary 8.2.6. d) On the other hand there are trivial counterexamples. Let for e.g. A = k[X,Y] and m := (X,Y). Then mn is of height 2 for every n, but X n , Xn-'Y, . . . ,XYn-l, Yn is a generating system of mn whose residue classes form a basis of mn/mn+l over k = Alm. So mn cannot be generated by less than n 1 elements.

+

e) There are also prime ideals in polynomial rings which are not complete intersections. Macaulay gave examples which show that p(p) is not bounded

9.1 The Problems

215

for prime ideals of height 2 in k[X, Y, Z], where k is an infinite field. cf. ([52] p. 36), ([27], $4) and [59]. (Moh gave examples in the formal power series ring k[[X, Y, Z]], in which Macaulay's examples do not remain prime ideals.) In the exercises we will give an example of a curve in 3-space (i.e. a prime ideal of height 2 in k[X, Y,21) which is no complete intersection, even not locally. f) An invertible ideal is locally a principal ideal, generated by a non-zerodivisor, hence a locally complete intersection. But in Examples 2.4.15 we found many invertible ideals which are not principal and hence not (global) complete intersections. In these cases the rings were Dedekind rings and hence the ideals generated by 2 elements. In the Exercises of Chapter 10 we also will give an example of a regular curve in 3-space which is not a (ideal-theoretical) complete intersection, though it locally clearly is so by c). 9.1.3 Also we are interested to what extent we can bound the number of polynomials which are needed to describe an algebraic subset E of kn, where k is an algebraically closed field. 1.e. what can we say about the minimal r such that E = V(fl,. . . ,f,) for suitable polynomials fi? We may ask the same question if E is a closed subset of Spec(A) where A is any Noetherian ring.

Since we ask, how to describe a set E = V(I), and we have (in both cases) we define: V(I) = v(.J?), Definitions 9.1.4 a) An ideal I of a ring A is called set-theoretically rgenerated if there are r elements a l , . . . ,a, E A with & = J-. b) An ideal I of a Noetherian ring A is called a set-theoretic complete intersection if I is set theoretically r-generated where r = ht(I). The latter means V(I) = V(fl) n - . . n V(f,) with r = ht(I). In geometric terms: 'V(I) is the intersection of the right number of hypersurfaces.' Dimension Theory tells us that the maximal ideal of a local Noetherian ring is a set-theoretic complete intersection. Using prime avoiding techniques, one can show this for Jac(A) in a semilocal Noetherian ring A. But in general one only knows the following: Given a prime ideal p of height r in a Noetherian ring A, then there are r elements a l , . . . ,a, and finitely many prime ideals p2,. . . ,ps (f p, such that

One can achieve that the p2,. . . ,p, are different from finitely many given ones, but not always get rid of them.

216

9 Number of Generators

If (A,m) is a local ring then it is often easy to compute p(I) of a finitely generated ideal I or more general p(M) of a finitely generated module M , since it equals the vector space dimension of M/mM over Aim. So the question, whether some ideal is a locally complete intersection, is easy in some sense (may the answer be negative in most cases). But the question, whether some ideal might be a set-theoretic complete intersection, often is troublesome. In Chapter 10 we will show the following: Let I be an 'ideal of a curve', i.e. an ideal of A := k[X1,. . . ,X,] with dim(A/I) = 1. a) If k is an arbitrary field and I locally a complete intersection, then I is a set-theoretic complete intersection. b) If k is a field of positive characteristic, then I is a set theoretic complete intersection. It is still open, whether the ideal of a curve is a set-theoretic complete intersection in every case.

9.2 Regular Sequences in Regular Rings Proposition 9.2.1 Let A be a Noetherian ring, a l , . . . ,a, be an A-regular sequence, p a minimal prime over-ideal of I := (al, . . . ,a,). Then h t ~ ( p = ) r. Consequently also ht(I) = r .

Proof. We know already ht(p) 5 r by the Small Dimension Theorem. To show the converse, we may assume by induction that htAlca,,(p/(al)) = r - 1. Therefore there is a series of r prime ideals

with (al) c pl. Now a l , as a non-zero-divisor, does not belong to any minmal prime ideal, since these are associated to A. Hence pl is not a minimal prime ideal of A. So there is a minimal prime ideal po 5 pl, by which we conclude ht(p) 2 r. The goal of this section is, to show some kind of converse, namely: Let A be a (not necessarily local) regular (or more generally a 'Cohen-Macaulay') ring, and I = (al,. . . ,a,) with r = ht(I) (< m). Then there is an A-regular sequence b l , . . . , b, , generating I . (The definition of a Cohen-Macaulay ring will be given below.)

9.2 Regular sequences

217

Remarks 9.2.2 a) Let A be a Noetherian ring, M finitely generated and I an ideal with I M # M . Then for every M-regular sequence a l , . . . ,an in I we have (al,. . . , a i ) M # (al,. . . ,ai+l)M for 0 5 i 5 n - 1 since ai+l operates injectively on the non-zero module M/(al, . . . ,ai)M . Consequently every M-regular sequence in I can be extended to a maximal (finite) one. b) If a l , . . . ,a, is an M-regular sequence then for every multiplicative S C A the sequence a l / l , . . . ,a n / l is SP1M-regular. This follows from the exactness of the localization functor and the fact that localization commutes with the forming of factor modules.

Lemma 9.2.3 a) Let a, b be an M-regular sequence and b not a zero-divisor of M . Then b, a is an M-regular sequence, too. b) If A is Noetherian, M finitely generated and a l , . . . , a n an M-regular sequence contained in the Jacobson radical of A, then any permutation of it is also M -regular. Proof. a) Assume, a were a zero-divisor modulo bM, say ax = by with x, y E M, x $ bM. We have y E a M , say y = az, since b is no zero divisor modulo a M . So ax = abz, whence x = bz ( a being no zero divisor of M ) in spite of x 6 bM. b) It is enough to show that one can interchange ai with ai+l for every i . Arguing modulo ( a l , . . . ,ai-l)M, one sees that one may assume n = 2. So we have to show that b, a is an M-regular sequence, if a, b is one. By a) it remains to show that b is a non-zero-divisor of M . So assume, b were a zero-divisor of M , and choose x E M \ (0) in such a way that Ax is maximal under those monogene submodules with bx = 0. Since b is a non-zero-divisor modulo a M , we have x E a M , i.e. x = ay for some y E M . So bay = 0, which implies by = 0, since a is no zero-divisor of M . From y # 0, a E Jac(A), x = ay we see with Nakayarna's Lemma that Ax 5 Ay, 0 contradicting the maximality condition on Ax.

Remark 9.2.4 In general, regular sequences need not remain regular under permutation. Consider for e.g. the sequence 1, 0, which is A-regular. If you think this is too primitive, then take a product A = B x C , a B-regular sequence bl, b2 of B , and consider the sequence of pairs (bl, 1), (b2,0) . And if you are not yet content, begin with an a1 E A such that the ring A/(al) splits. For example (if k is a field) in the ring k[X, Y, Z] the sequence X ( X + I ) , ( X 1)(Z 1) - 1, Y(X 1) is regular, whereas X ( X l ) , Y(X l ) , ( X + 1 ) ( Z+ 1) - 1 is not so. Moreover, the latter sequence is not regular, but it is of length 3 and generates an ideal of height 3, namely (X, Y, 2 ) .

+

+

+

+

+

218

9 Number of Generators

Proposition 9.2.5 Let I be an ideal of a Noetherian ring A and M a finitely generated A-module with I M # M. Then any two maximal M-regular sequences in I have the same length. Proof. Assume, n is the minimal length of maximal M-regular sequences in I, and let b1,. .. ,bn such one. We use induction on n, the case n = 0 being obvious. If a l , . . . , a n is another regular sequence in I , we have to show that it is maximal too, i.e. that I consists of zero-divisors of M/(al,. . . ,an)M. Let first n = 1. Then I is contained in the finite union U p of primes p E Ass(M/blM), hence I c p for some p E Ass(M/blM). By the definition of Ass there is some x E M - blM with px c blM, hence for such z we have I z c blM, h e n c e a l x ~blM,say a l x = bly.

CLAIM: y @ a1M but Iy c al M. This shows that I consists of zero-divisors of M/alM, whence also a1 is a maximal regular sequence in I.

PROOF OF THE CLAIM: The assumption y E a l M would imply alz = albly' for some y', hence z = blyl E blM, which was excluded. For c E I we have cbl y = al(cx) E a1(bl M), which implies cy E a1M since bl is no zero-divisor of M. Now assume n > 1. Form the - finite - union U of all prime ideals which are associated to M/(al,. . . ,a k ) Mor M/(bl,. . . ,bk)M for some k < n. Then I $ U, since for every prime ideal p of the above kind an 4 p or bn 4 p. Choose c E I \ U . Then c is no zero-divisor of M/(al,. .. ,ak)M nor of M/(bl,. . . ,ba)M for every k = 0 , . . . ,n - 1. Therefore, by Lemma 9.2.3 a), we may permute c successively with an-1,. . . ,a l , resp. bn-I,. . . ,bl to see that c, al, . . . ,an-1 and c, bl, . . . ,bn-l are regular sequences. The latter is a maximal one. Otherwise there were a non-zero-divisor b E I of M/(bl, . . . ,bn-1, c)M , whence bl, . . . ,bn-l, c, b would be regular. But bl, . . . ,bn-1, c is already maximal by the case n = 1 applied to M/(bl,. . . ,bn-1)M. So, arguing modulo cM and using the induction hypothesis, one sees that c, al, . . . ,an-1 is a maximal Mregular sequence in I. So c, a1,. . . ,a n - ~hence also a ~. .,. ,an-l, c is maximal. Applying the case n = 1 again, we finally see that a l , . . . , a n is a maximal 0 A-regular sequence in I. Definition 9.2.6 If I is an ideal of a Noetherian ring A and M a finitely generated A-module with I M # M , we denote by depth(I, M) the common length of the maximal M-regular sequences in I and call it the I-depth of M. If additionally (A, m) is local we call depth(m, M) simply the d e p t h of M and denote it by depth(M). (depth(1, M) is also called grade of I on M.)

9.2 Regular sequences

219

Corollary 9.2.7 In the above situation let a l , . . . ,a, be a n M-regular sequence in I. Then depth(1, M/(al,. . . ,a,)M) = depth(1, M) - r . Corollary 9.2.8 Let A be Noetherian, M finitely generated, the ideal I with I M # M generated by n elements and depth(1, M ) = m. T h e n m 5 n and there is a generating system a l , . . . , a n of I such that its first m elements a l , . . . ,a, make u p a n M-regular sequence.

Proof. Induction on m, the case m = 0 being obvious. So assume m > 0. Then clearly n > 0. Let I = (bl,. . . ,b,). We will see that there is a non-zero-divisor a1 of M with I = (al, 212,. . . , b,). The corollary will then follow by induction, arguing modulo a1M.

+

The condition m > 0 means I $ Up, for p E Ass(M). Hence also bl (b2,. . . ,b,) 6 Up, for p E Ass(M), by Lemma 1.1.8 a). Then there is a nonzero-divisor of M of the form bl EL2Xibi with Xi E A. Clearly this may 0 be chosen to be al.

+

Definition 9.2.9 The dimension dim(M) of a finitely generated A-module M is defined t o be dim(A/AnnM). Proposition 9.2.10 Let M be a non-zero finitely generated module over a local Noetherian ring A. Then

Proof. Since p E Ass(M) implies p > Ann(M), the second inequality is clear. We prove the first inequality by induction on n := depth(M), the case n = 0 being obvious. So let n > 0 and a be a non-zero-divisor of M. Since depth(M/aM) = n - 1 we have by induction hypothesis that n - 1 5 MinpEAss(MlaM) dim(A/p). Therefore it will be enough to show that for every p E Ass(M) there is a p' E Ass(M/aM) with p (a) c p'. Namely then n - 1 5 dim(A/pl) 5 dim(A/p) - 1.

+

Consider the submodules N' := {x E M ( px = 0) and N := {x E M I px

c aM).

We have N' c N, a M c N and N' # 0, the latter since p E Ass(M). Any p' E Ass(N/aM) will fulfil the above statements. So we only have to show that a M # N. Assume a M = N and let y E N'(c N). Then y = ax for some x E M and pax = py = 0, whence px = 0 since a is a non-zero-divisor of M . This would mean aN' = N', contradicting N' # 0 by Nakayama's Lemma. 0

220

9 Number of Generators

Definition 9.2.11 Let A be a Noetherian ring. A Cohen-Macaulay Am o d u l e is a finitely generated module M with depthAmM, = dim(M,,,) for every maximal ideal m. A ring A is called a Cohen-Macaulay r i n g i f it is Noetherian and Cohen-Macaulay as an A-module. For us, the most interesting examples of Cohen-Macaulay rings are the regular ones, especially the polynomial rings over fields. Corollary 9.2.12 Let M be a Cohen-Macaulay module over a Noetherian ring. a) Every element of Ass(M) is minimal in Ass(M). In other words: p,pl E Ass(M) and p c p' imply p = p'. b) If i n addition A is local, then dim(M) = dim(A/p) for every minimal prime over-ideal p of Ann(M).

Proof. a) Since Ass behaves well under localization according to Proposition 8.4.5, we may assume that A is local. Suppose p # p'. Then dim(M) 2 dim(A/p) > dim(A/pl) 2 depth(M), contrary to dim(M) = depth(M). b) dim(M) Ann(M).

2 dim(A/p) 2

depth(M) for every minimal prime over-ideal of 0

Corollary 9.2.13 Let (A,m) be a local Noetherian ring, M a finitely generated Cohen-Macaulay A-module and a l , . . . ,a, an M-regular sequence in m. Then M/(al, . . . ,a,) M is a Cohen-Macaulay module, too.

Proof. By induction, we may assume r = 1. Clearly depth(M/alM) = depth(M) - 1. On the other hand, dim(M/alM) 5 dim(M) - 1. This is because Ann(M) c Ann(M) (al) C Ann(M/al M ) . Moreover no minimal prime over-ideal p of Ann(M) contains a l , as p E Ass(M) by Proposition 8.4.7 c) 0

+

T h e o r e m 9.2.14 Let A be a Cohen-Macaulay ring, I # A an ideal of height r , generated by r elements. Then there is an A-regular sequence a l , . . . ,a,, generating I .

Proof. Let a l , . . . ,a, be a maximal A-regular sequence in I and J := ( a l ,. . . ,a,). Then I consists of zero-divisors of A / J and so is contained in some p E Ass(A/J). But since A / J is a Cohen-Macaulay module, its associated prime ideals are minimal prime over-ideals of J . By Proposition 9.2.1 we get ht(p) = v. So r 5 v, whence r = v. Thus depth(I,A) = r . Hence by Corollary 9.2.8 the r generators of I can be replaced by a regular sequence 0 which generates I.

9.3 Forster-Swan Theorem

221

9.3 Forster-Swan Theorem Let R be a commutative noetherian ring with 1, and let M be a finitely generated R-module. The Forster-Swan theorem gives a bound on the number of generators of a finitely generated R-module M in terms of its local number of generators ,up(M). (Note that the local number of generators of M is always less than the number of generators of M. So the theorem gives a hold in the reverse direction.) By Nakayama's lemma, to find the minimal local number of generators ,up(M) of M is the same as to calculate the dimension of the vector space Mp/pMp over the field k ( p ) = RP/pRP. Thus the calculation of a bound on p(M) is reduced to a problem in linear algebra. Let us describe this bound. Given a prime ideal p define the local p-bound as the number bp (M) = ,up(M) dim RIP.

+

The Forster-Swan theorem asserts that

The theorem was discovered by 0. Forster in 1964, and shortly generalised by R. G. Swan. We give four proofs of this theorem. 0

0. Forster's proof via the Kaplansky ideals. R.G. Swan's proof via basic elements, a concept which was introduced by him.

R. A. Rao's proof based on classical K-theory. (Alternatively, use Quillen's Splitting Lemma as done by Plumstead.) The advantage is one can prove similar bounds, conjectured by Eisenbud-Evans, for finitely generated modules over polynomial extensions of noetherian rings. 0

We deduce the Forster-Swan theorem from the Eisenbud-Evans theorem.

Schematically, the different viewpoints to approach the Forster-Swan theorem are given by the following diagram:

222

9 Number of Generators

Basic Elements

(

K-THEORY

1

9.3.1 Let M be a finitely generated R-module. The following ideals I,(M) for every r 2 0 were defined by S. Kaplansky:

where the summation is done over all possible collections of r elements ml, . . . ,m, in M . We call them Kaplansky ideals. They have the following properties: a) Io(M) = Ann(M), I k ( M ) c Ik+1 (M) for all k, and I,(M) = R if r 2 4M). b) For any multiplicatively closed subset S of R

(where on the left hand side S - l M is considered as an S-'A-module).

PROOF: One has

a x Anna, (S-' (M/

Rmi))

To see the equalities (1)and (3), show that for every family (Ni) of submodules S-I Ni = s - l ( C i Ni). For equality (2) we need of any module M we have that M , hence M I Rmi is finitely generated.

z xi

9.3 Forster-Swan Theorem

223

c) Recall the definition pp( M ) := p ~ (Mp) , = dimR,/pRp(Mp/pMp).

>

We have pp(M) k if and only if p > Ik-1(M). In particular if p > Ik-l(M) and p $ Ik(M) then pp(M) = k. (This is a topological characterization of PP(M)-

PROOF: Suppose that pp(M)

> k then

and so IkPl(Mp) c pRp. Since Ik-l(Mp) = I ~ - I ( M )it~ follows that Ik-l(M) C p, i.e. p E V(IkP1(M)). Conversely, if pp(M) < k then . Ik-l(Mv) = Rp, whence Ik-l(M)p = Rp, and so pRp $ I ~ - I ( M ) ~Consequently, p $ Ik-I (M). The third property allows us to induce a 'stratification' of Spmax(R), resp. Spec(R), with respect to a finitely generated R-module M: Let Xk(M) = {m E Spmax(R) I pm(M) then Spmax(R) = Xo(M) > Xl(M)

> k)

= V(Ik-l(M))

> .- . > X,+l(M)

n Spmax(R)

= 0,

where r = p(M), the minimal number of generators of M.

Definition 9.3.2 The Forster - Swan bound (on p(M)) is the number b(M) = max{k k

+ dim Xk(M) I k > 1,

Xk(M)

# 0).

We set b(M) = 0 if Xl(M) = 0, i.e. M = 0. Note that if N is a factor module of M then b(N) 5 b(M) as pp(N) 5 pp(M) for all primes p. Remark 9.3.3 An alternative description of the Forster - Swan bound is the following: Let

where we shortly write S(p) := dim(Spmax(R/p)) = dim(V(p) n Spmax(R)). (Clearly S(p) 5 dim(R/p).)

PROOF:Denote the right hand number by f ( M ) . Since Xk(M) = V(Ik-1(M)) n Spmax(R) there is a prime ideal po > (M) such that &(PO)= dim(Xk(M)). From po > &-1(M) we get pp,(M) 2 k. Hence k dim Xk(M) ppo(M) S(p0). Consequently b(M) 5 f (M)

+

<

+

To prove the converse inequality, let p E Spec(R) and pp(M) = k. Then p > (M), whence S(p) 5 dim s p m a ~ ( R / I ~ (M)) - ~ = dim(Xk(M)). Hence pp(M) + 6(p) k + dim(Xk(M)). Hence taking the supremum on both sides f (MI 5 b(M).

1k-l

<

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9 Number of Generators

Theorem 9.3.4 (Forster - Swan) Let R be a noetherian ring and M a finitely generated R-module. Then p ( M ) 5 b(M).

Proof. We may assume b(M) < oo and argue by induction on b(M). If b(M) = 0 then Ann(M) = R and M = 0 is generated by 0 elements! Let Xkj(M) be the finitely many irreducible components of XI,(M). Let J be the finite set {(k,j) I k 1 and k dim(Xkj(M)) = b(M)).

>

+

+ +

If (k, j ) E J the Xkj(M) @ Xk+l(M). Otherwise we would have k 1 dim(Xk+l(M)) > k dim(Xkj(M)) = b(M), which contradicts the definition of b(M).

+

Let mkj E Xkj(M) \ Xk+l(M). Then pmkj(M) = k

> 0 for (k, j )

EJ

CLAIM: There is an x E M such that x 6 mkjM for all (k, j ) E J. This is equivalent to saying that there is an element x E M whose image in every R/mkj-vector space M/mkjM belongs to some basis. Let J' be the set of all mkj with (k, j ) E J. (It may happen that mkj = mkj, for different j, j'.) By the Chinese Remainder Theorem the canonical map

is surjective. Hence there is an x E M whose image under component.

6

has no zero

Let N = M/Rx. By the choice of x, pmkj( N ) = k - 1 for all (k, j ) E J, i.e. mkj 6Xk+l(N). Therefore k + dim(Xk(N)) < b(M), hence b(N) < b(M). By induction hypothesis p(N) 5 b(N). Since b(N) < b(M), p ( M ) p ( N ) 1 b(M). 0

<

+ <

Corollary 9.3.5 Let P be a finitely generated projective R-module of rank r over a noetherian ring whose maximal spectrum has dimension d. Then P can be generated by d r elements.

+

Corollary 9.3.6 Let A be a Noetherian ring of dimension n and I an ideal which is a local complete intersection. Then p ( I ) 5 n 1.

+

Proof. Let p E Spec(A). We have to show p p ( I ) distinguish two cases.

+ dim(A/p) < n + 1. We

CASE 1: p 2 I. Then Ip = Ap, whence p p ( I ) = 1. But clearly 6(p) dim(A/p) 5 n.

< <

CASE 2: By hypothesis Ipis generated by a regular sequence of length dim(Ap) = ht(p). So p p ( I ) dim(A/p) 5 ht(p) dim(A/p) 5 n. 0

+

+

9.3 Forster-Swan Theorem

225

Corollary 9.3.7 Let A be a regular ring of dimension n, for example A := k[X1,. . . ,X,] with a field k , and I a n ideal such that A I I is regular. T h e n p ( I ) 5 n + 1. Indeed in this case I is a local complete intersection by Corollary 8.2.6. An immediate corollary is Corollary 9.3.8 Let A be a residue class ring of a finite dimensional regular ring, for e.g. a n a f i n e algebra, t h e n t h e set of the p ( I ) where I ranges over all ideals with regular A / I is bounded. Especially the set {p(m) ( m E Spmax(A)) is bounded. Note that an algebra A of finite type over any regular ring R (for e.g. R = Z) is of the above kind, namely a residue class ring of R[X1, . . . ,X,], which is regular as well.

+

Another special case is that I is invertible. Then b(I) = 1 dimSpmax(A). We get a new proof that ideals of a Dedekind ring can be generated by 2 elements. We know Dedekind rings which are not principal domains. For these the Forster-Swan bound is sharp. Also if A is a regular domain and I a prime ideal of height 1. Then by Theorem 8.5.12 I is locally l-generated, i.e. invertible. So the above inequality holds. The bound b ( M ) in the Forster-Swan Theorem is the best possible in general also in higher dimension. We recall an example of R.G. Swan which shows this:

+

+

Let A be the subring of R[xo,. . . ,xd]/(xi . . . x; - 1) consisting of all polynomials all of whose terms have even (total) degree. Let P be the Asubmodule of B[xo, . . . ,x ~ ] / ( x ; - . . x; - 1) generated by x ~. ., . ,xd. Thus P consists of the classes of those polynomials all of whose terms have odd (total) degree. It was shown by R.G. Swan in [96] that P is projective (of rank 1) and that if P, = P @ AT-' then rank P, = r and P, cannot be generated by fewer than d r elements, cf. Proposition 5.6.10.

+ +

+

We end this section by a weak, but very general generalization of Corollary 6.7.3. There is a wide class of Noetherian rings which have the following property. (Closedness of t h e singular locus) Definition 9.3.9 W e say t h a t a Noetherian ring A has a closed singular locus, if the set {p E Spec(A) I Ap is n o t regular) i s closed in Spec(A).

9 Number of Generators

226

One can derive from the Jacobian Criterion 8.1.3 that every afFine algebra has a closed singular locus. But of course the following proposition is of value, only if the class of rings with this property is much wider. This is the case, but unfortunately we cannot show it here.

Proposition 9.3.10 Let A be a Noetherian ring of dimension n < ca. Assume that Alp has a closed singular locus for all p E Spec(A). The the set {p(m) I m E Spmax(A)) is bounded. Proof. Induction on n = dim(A). If n = 0 the ring A has only finitely many prime ideals. So let n > 0. Since there are only finitely many minimal prime ideals, by reduction modulo these, we may assume that A is a domain. By hypothesis, the singular locus of A is closed, say it is V(I). Since the localization A(o) of A in the zero ideal is a field, (0) $ V(I), i.e. I # (0). So dim(A/I) < n. By induction hypothesis the set of the p(m/I) is bounded where m ranges over the maximal ideals containing I. Hence also the set of the p(m) is bounded for these m. The other maximal ideals m $ V(I) are locally complete intersections, hence, by Corollary 9.3.6 can be generated by 0 5 n 1 elements.

+

9.3.1 Basic elements

Let R be a commutative ring with 1, and let M be a finitely generated Rmodule. An element m E M is said to be unimodular in M if Rm is a direct summand of M , i.e. there is a (finitely generated) module M' such that M 21 Rm @ M'. For instance, any non-zero vector v in a vector space V is a unimodular element. For another example, Serre's Splitting Theorem can be restated as asserting that a finitely generated projective A-modules of rank > dim(A) has a unimodular element. One can attach an ideal OM(m) of R with an element m E M , called the order ideal of m, viz., o ~ ( m= ) { f (m) I f E M* = HomR(M, R)). Clearly, m is unimodular if and only if OR(m) = R. UmR(M) will denote the set of all unimodular elements of M. How far does the analogy of being "part of a basis" go in the ring-theoretic setup? Let us consider the (commutative) case closest to the vector space situation: This is the case when R is a local ring with maximal ideal m. Are you surprised? Recall Nakayama's lemma, and its consequence,

.. . , m l , . . . ,m, E M generate M i f and only i f El, space MImM.

is a basis of the Rlm-

9.3 Forster-Swan Theorem

227

This gives a clue. We can expect an element m E M to be "good" if r r ~6 mM; for then m E M/mM is part of a basis. Of course, in the case when R is not local, we would wish to have this property at every localization of R at its maximal ideals. Such an element is called a basic element of M . Formally,

--

Let m E M . We say rrl is basic at p if m 4 Mp m is part of basis of the k(p) (= Rp/pRp)-vector space Mp/pMp m # 0 in M @ R k(p). We say m is a basic element of M if rn is basic at p, for every prime ideal p of R.

--

Let us relate this definition to the "local number of generators of M" pp( M ) = the minimal number of generators of the Rp-module Mp. By Nakayama's lemma, pp(M) = dim Mp/pMp. Observe that

if and only if m PP(M) - 1.

6 pMp. Thus, m is basic in M at p if and only if pD(M/Rm)=

Of course, any unimodular element m E M is a basic element in M. The converse is true when M is projective. There are several ways to see this. For instance, consider the case when M is free of rank n, say M .v Rn. If m = (al,. . . ,an). Then m basic mcans that the ideal ( a l , . . . , a n ) $ m, for any maximal ideal m of M . Hence, (al,. . . , a,) = R, whence m is unimodular. For the general case, note that any finitely generated projective module is a summand of a free module Rn, for some n. So m is unimodular in An, by above; whence there is a linear map $J : An -+ R for which $(m) = 1. Composing with the inclusion map P -+ An yields a linear map 6 : P -+ R, with 8(m) = 1. This mcans that m is unimodular in M.

9.3.2 Basic elements and the Forster-Swan theorem

We begin with a basic lemma. Lemma 9.3.11 Let M be a non-zero finitely generated R-module. Let PI,. . . ,p, E Supp(M). Then there exists m E M which is basic at P I , . . . ,p,.

PROOF: By a reordering we may assume that pi is maximal among { ~ i t ~ i +.. l , ,pT), . for each i . We induct on r , the result being clear for r = 1. Let m' be basic at p1,. . . ,p,-1. If m' is basic at p,, then there is nothing more to be done. If not, choose n' E M which is basic a t p,. Let r E pl . . .pT-l \ p,. (Such a r clearly exists.) Let m = m' is basic at pl , - - . ,p,.

+ rn'.

Then m

228

9 Number of Generators

Swan's proof of Forster-Swan theorem:

Let f ( M ) = sup p{bp(M) I p prime ,Mp # 0). CLAIM: We show that the bound f (M) can be attained at atmost finitely many prime ideals. Note that U, = {p I pp(M) = r ) is an open subset of Spec(R). This is because if the images of ml, . . . ,m, E M , generate Mp, then the images of ml, .. . ,m, E M will also generate Mq, for all q E D(s), for some s 4 p. (The existence of the ml, .. . ,m, E M , as above, will determine a s.) Let X, = {p I pp(M) 2 n). Then X, = (U:;U~)" is a closed set. Hence, there is an ideal I, such that X, = V(I,). Let JI, = nYglpTi be its primary decomposition, so that the irreducible decomposition of the closed set :l V(p,i). is V(I,) = Ug Let E = {p,i I r 2 0 , l 5 i 5 n,). Note that E is a finite set as X, = 0, for r > p(M). We show that if f (M) = bp(M) then p E E. Let pp(M) = n. Then p E X,. Hence JI, c p. Hence p,i c p, for some i. Note that pp(M) = ppni(M). Therefore, if coht (p) < coht (pni), then bp(M) < bpni(M)= bp(M)!. Therefore, p = p,i E E. This settles the claim. above lemma, choose a m E M which is basic at all primes of E. Let M = MIRm. Then f ( x ) 5 f (M) - 1: This is clear if p 6 E, and also if p E E as then m is basicat E. By -

By induction, p(M) 5 f ( x ) . Hence p(M) 5 p ( z ) to be shown.

+ 1 5 f(M), as required 0

9.3.3 Forster-Swan theorem via K-theory

We first explain the notion of generalized dimension of a commutative noetherian ring R. This notion evolved from the work of J-P. Serre-H.Bass (who began by concentrating on the Maximum spectrum), then in the work of R.G.Swan (who started working with j-primes), and then by B. Plumstead who axiomatised the notion in his doctoral thesis [70]. Given a set P of prime ideals of R and a function S : P + N U {0), define a partial order << on P by setting p << q if p c q and S(p) > S(q). Definition 1. The function 6 : P + NU (0) is called a generalised dimension function on P (g.d.f. in short) if for any ideal I of R, V(I) n P has only a finite number of minimal elements with respect to <<.

9.3 Forster-Swan Theorem

229

We say R has generalised dimension S(R) if S(R) = min( max S(p)). 6

pESpec R

For instance, the standard dimension function S(p) = coheight P is a g.d.f. Thus g.d.R dim R.

<

We recollect an example of B. Plumstead in [70] of a ring having g.d. A dim A.

<

First observe that if s is an element of R such that R/(s) and R, have genare g.d.f's on R/(s), eralised dimension 5 d, then g.d. R 5 d. In fact, if R, respectively withh Si d for i = 1,2, then define 6 : Spec(R) -+ N U (0) as follows

<

Clearly, 6 is a g.d.f. on R with 6(p)

< d for all p E Spec(R).

Plumstead's Example: Take A = R[z], where R is a ring having an element s E radical of R with dim(R/(s)) < dim R. The above reasoning shows g.d. A < dim A. We begin a proof of the Forster-Swan theorem via K-theory due to R.A. Rao.

Theorem 9.3.12 (Forster-Swan) Let A be a noetherian ring of dimension d, and let 6 be a generalized dimension function on A. Then an A-module M is generated by

elements.

PROOF: Let f = f (M); we show that there is a free A-module of rank f mapping onto M. We prove this by induction on (dim(A),G(A)); it being clear when d = 0. Consider a surjection An --+ M this map. Find a (n.z.d.) s E R, such that

--+ 0, with n > f , and let K be the kernel of

230

9 Number of Generators

1. s $ p, if p is minimal with respect to the generalized dimension function 6. (Note that there are only finitely many primes, including the minimal ones, for which p is minimal with respect to 6.

2. such that K, is free. Choose XI E K,, with XI is unimodular in A.! Let T = 1+sR. Then g.d.AT 5 6(A), if 6(A) > 0, and so ( d , 6 ( A ~ ) < ) (d,S(A)). For any p E Spec(&),

Hence KT is (S(p) induction there is a

+ 1)-fold basic in A? 22

at p for every p E Spec(A~);by

E KT which is unimodular in A?.

Since n > f , n 2 S(A)+l. Note that &(AT,) 5 S(A) - 1, and so n 2 6(AT,) +2. By Proposition 7.1.2 one can find E E En(AT,) such that ~ ( $ 1 =) ~ (x2),. Let x = xi x, x2 E A! x, A?. Since x is "locally" unimodular, it is unimodular.

n

Any elementary matrix is homotopic to the identity matrix: if E = Eh then E(T) = EhT is a homotopy. By Quillen's Splitting Lemma 4.3.8, E splits as (&l)s(&2)t,for Some E l E SLn(As), E2 E SLn(AT).

n

In view of the fact that elementary matrices split, A! x, A? them. Let Ax @ P = An,

7An; identify

We are now in the situation

Since x, = SI E K, and XT = x2 E KT, the second projection map M is "locally" surjective, and so surjective.

r 2

:P

-+

But P is stably free of rank n - 1 2 f 2 S(A). By the Bass Cancellation Theorem 7.1.11, P is free. 0

9.4 Eisenbud-Evans theorem

231

9.4 Eisenbud-Evans theorem The Eisenbud-Evans theorem is the ultimate result known today by general position arguments regarding the number of generators of a finitely generated module M over a commutative noetherian ring R. (There are results also know over non-commutative noetherian rings due to R. Warfield, etc., but we shall not dwell on these here.) It can be used to derive all the earlier theorems: Serre's Splitting Theorem, Bass Cancellation Theorem, and the Forster-Swan Theorem; and in fact provides more information on the latter. These theorems play an important role in showing that curves in n-space are set-theoretic complete intersections. Today, one has more refined arguments, which combine general position arguments, and some Quillen-Suslin patching theory, (via Ktheory), to derive these consequences for curves directly; without recourse to the Eisenbud-Evans theorem per se. A schematic picture is given by

I

EISENBUD-

I

UNIMODULAR THEOREM

THEOREM

What are the Eisenbud-Evans theorem all about? One can ask under what condition can you guarantee that a submodule M' of a module M has a basic element? Moreover, in this case how will you go about finding such an element? Essentially, the condition should depend on M' being large enough, rather than on M. For instance, consider the case when M' = Q is a projective R-module. If rank(Q) > dim(R), then by Serre's Splitting Theorem Q will have a unimodular element q E Q. q will also be basic in Q, as Q is projective. Here ,up(&) > dim(R), for all primes ideals p of R. A similar situation holds in the general case. We have to first figure out a notion of "largeness" of a module. For this we define the notion of when a submodule MI is t-fold basic in M . If M' is a sub-module of M we say MI is t-fold basic in M at p if pp(M/M1) 5 pp(M) - t. MI is t-fold basic in M if M' is t-fold basic in M at p for every prime ideal p of R. Clearly, basic is 1-basic. Let us do a little linear algebra to understand this definition. One has

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9 Number of Generators

This is now a vector space dimension calculation. We are essentially doing linear algebra in commutative rings - the door being opened by Nakayama's Lemma. Thus M' = ~ f =Rmi , is t-fold basic in M at p if and only if dim{mi+pMp) 2 t. In particular, note that M' is t-fold basic in M a t p implies M' is t'-fold basic in M at p for t ' s t. Having defined the appropriate notion of "largeness" let us describe what Eisenbud-Evans found: The Eisenbud-Evans theorem is that if M' is (coht(p) 1)-fold basic in M at p, for every prime ideal p of R, then M' has a basic element.

+

The proof of the theorem is in the spirit in which the key starting Prime Avoidance Lemma 1.1.8 was proved. The technique used to prove it is called general position argument. One takes a set of generators m l , . . . ,mt of M', = +aimt) ~ of M', for ai,. . . ,at-1 E R. One and consider submodules ~ i R(mi tries to ensures that the choice of the submodule is such that it has similar properties as the original M' one started with. This requires a suitable choice of a l , . . . ,a,. The situation is then inductive and leads to a basic element having a specific type, as was expected by the proof of our key lemma. We schematically describe the different types of Mathematical thinking which help in general position arguments of different types:

-

ALGEBRAIC GEOMETRY AELD THEORY QUADRAI lC P O N S

CATEGORYTHEORY

FUNC'IION'SHBORY

BASS CANCPI LATION

THEORY

LISENBUD-EVANS BERTNI-KWNER SUSLIN CANCELLATlON

9.4 Eisenbud-Evans theorem

233

W e now go t o the proof o f the Eisenbud-Evans theorem. Let us, as EisenbudEvans did, first just state the two general position argument lemmas which will be needed, and prove the theorem assuming them. Lemma 9.4.1 Let R be a commutative noetherian ring with 1. Let M be a finitely generated R-module, and let M' be a submodule of M Let P be a subset of Spec(R) having the following property: For every prime ideal p E Spec(R), for which there i s a prime ideal q E P with q c p , q # p , one has p q ( M / M 1 ) pp( M ) - w , i.e. M' is w-fold basic i n M at p.

<

Then M i is w-fold basic in M at q , for all but finitely many prime ideals q E P . (These are utmost the prime ideals which are minimal over the Kaplansky ideals I , ( M / M 1 ) , for some r , and which belong to P.) Lemma 9.4.2 Let R be a commutative noetherian ring, M a finitely generated R- module, M' a submodule of M . Let p l , . . . , p , E Spec(R) be a finite set of prime ideals. Assume that M i is wi-fold basic in M at p i , with wi > 0 , for all i . Let M' = ~ t Rmi, = for ~ some m i E M .

If a R is such that ( a ,ml) E R @ M is basic at p l , . . . , p,, then there exist elements ai E R such that ( a ,rnl a a l m t ) E R @ M is basic at p 1 , . . . ,p, . Moreoaer, the submodule R ( m l + aalrnt) + R(mi + aimt) is minit - 1, wi)-fold basic i n M at p i , for all i .

+

tit

Let us now proceed t o formally state and prove the famous Eisenbud-Evans theorem. Theorem 9.4.3 Let R be a commutative noetherian ring of dimension d , M a finitely generated R-module, M' a sumbodule of M . Let 6 be a generalized dimension function on R.

If M' is (S(p)+ 1)-fold basic i n M at p , for every prime ideal p of R, then M' has a basic element. Furthermore, let ml, . . . ,mt generate M'. Let a E R such that ( a ,m l ) E R@M is basic. Then there is a basic element of M of tvpe ml + a aimi.

xIZa

PROOF:A subset { n l , . . . ,n k ) is called a basic set i f the R-submodule N' = c:=, Rni is min{k, 6(p) + 1)-fold basic in M at p , for all prime ideals p o f R. (Note that i f { m } is a basic set, then m is 1-basic in M.) By hypothesis, { m l , . . . ,m t ) is a basic set. T h e idea employed by EiscnbudEvans is t o "cut down" on number o f generators occuring in a basic set, retaining its basicity. This is done by means o f "elementary operations": More precisely, given a basic set { n l , . . . , n t ) , t > 1, one finds a l , . . . ,at-1 E R, such that the set

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9 Number of Generators

is a basic set. If we do this process (t - 1) times to {ml, . . . ,mt) we will land up with a basic element of the required type. CLAIM:There are atmost finitely many prime ideals for which N' is not {k, S(p) 2)-fold basic in M at p. We prove this.

+

Since R has a finite (generalized) dimension, it suffices to show, for fixed 6(p) = s, say, that there are only finitely many primes for which N' is not {k, S(p) 2)-fold basic in M at p.

+

Let P = {q I 6(q) = s). If p c q E P , p # q, then S(p) > S(q) = s. Hence, min{lc, S(p) + 1) 2 min{k, S(q) + 2) = w. Therefore, M' is w-fold basic in M at p. By Lemma 9.4.1, M' is w-fold basic in M at q, for all but finitely many q E P. Let E be this finite set of primes.

+

By Lemma 9.4.2, there are a l , . . . ,at-1 E A such that N" = A ( n l + aatnt) A(ni aint) is min{t - 1,6(p) 1)-fold basic in M at p E E . However, if p $ E, the N" is minit - 1, S(p) 1)-fold basic in M at p, for any choice of 0 ai's. Thus N" is a basic set as required.

+

+

+

We now prove the two lemma which were mentioned before the theorem and used in its proof.

PROOF OF LEMMA 9.4.1: Let p E P. Let pp(M/M1) = r + l . This is if and only if I,(M/M1) c p, and I,+l(M/M') $ p, for all 1 > 0. Suppose, in addition, that p is not minimial over I,(M/M1). By hypothesis, pq(M/M1)= w (= r). Therefore, pP(M/M1)= pq(M/M1)5 p,(M) - w 5 pp(M) - W.Thus, MI is 0 w-fold basic in M at p.

+

PROOF OF LEMMA 9.4.2: Of course, (a, ml aatmt) is basic at pl, . . . ,p,, for any choice of a1 E R. This is because it is the image of the basic element (a,ml) by a flip of R @ M. If wi,

> t = p(M1), for some io,then for any choice of ai, the submodule

is (t - 1)-fold basic in M at pi,, as required. Therefore, we may assume that wi < t, for all i, by simply omitting all the primes ideals pi for which the corresponding wi 2 t. We will induct on n. We start the induction with the vacuous case n = 0. Reorder the pi, and assume pt is minimal amongst P I , . . . ,pi, i.e. pt By the induction hypothesis we can choose a:,

. . . ,a:-,

ER

4 &:pi.

so that

9.4 Eisenbud-Evans theorem

generate a sub-module o f M which is wi-fold basic at

CLAIM:W e can choose a',', . . . ,a:-'_,

is wt-fold basic at

E

pi,

for i

235

< t.

R , so that for any r E R \ pt,

pt.

(Therefore,i f r E (nj<&) \ pt, then my,. . . , my-, is WI-foldbasic at p i , for i 5 t! This is because, for i < t , r E p i , and m i , . . . ,mi-,, are wi-fold basic in M at p i . For i = t , this is guaranteed by the claim.)

PROOF OF CLAIM: W e may localize at pt. Therefore, assume R is a local ring with maximal ideal pt. Since "basicness" can be checked after "going modulo pt" we assume R is a field.

c:::

If Rmi is wt-fold basic, thcn choose ai = 0 , for all i. I f not, let k < t , be the largest integer such that m', E Rmi. (Note that k exists as C;:: R7r4 is a vector space o f dimension 5 wt - 1 < t - 1). If k

#

1, the choose a! = 0 , for i

#

k , and choose a: = 1. I f k = 1, then = 1, and a! = 0 , for all i > 1. Then we can check that, m',' = mi = aaimt, my = mi = mi almt. Therefore, c:=, Rm; = Raaimt Ci R ( m i a:mt) is wt-basic at p t , by hypothesis.

mi = 0. Since ( a ,mi) is basic in R $ M , a # 0. Therefore, take a:

+

+

+

T h e Eisenbud-Evan's Theorem can be used t o deduce the famous Serre Splitting Theorem, the Bass Cancellation Theorem, the Forster-Swan Theorem, amongst others. W e give these three applications. Corollary 9.4.4 (Serre's Unimodular Element Theorem)

Let P be a finitely generated projective R-module of rank 2 d i m ( R )+ 1. Then P has a unimodular element, i.e. P G P' $ R, for some finitely generated projective R-module PI. In fact, if P is generated by m l , . . . , m t , then one can find a2,. . . ,at E A, so that m = m l a2m2 . . . atmt is unimodular.

+

+ +

PROOF: Let M = P $ Q be a free R-module. Let M' = P , in the EisenbudEvans theorem. Then M' is ( d + 1)-fold basic in M . Now (1,m l ) E R @ M is basic. By the Eisenbud-Evans theorem there is a2,. . . ,at E R, so that m = ml a2mz + . . . + atmt is basic in M . But then, since M is projective, rn is unimodular in M . Since rn E P , m is unimodular in P , and splits o f f a summand.

+

Corollary 9.4.5 (Bass Cancellation Theorem)

Let P be a projective R-module of rank > d i m ( R ) . Then P is '%ancellative", i.e. P $ Q E P' $ Q implies P = P', or equivalently, P $ R E P' @ R implies that P E PI.

236

9 Number of Generators

PROOF: Let a : P' @ R P @ R be an isomorphism, and let a(0.1) = (p, a). Then Pi G coker(0,l) r coker(p,a). It suffices to show that (p, a) can be mapped t o (0,l) E P @ R by some automorphism of P @ R. For this would show that coker(p, a) G coker(0,l) E P. We shall actually find a product of flips which maps @.a) to (0,l).

+

Let pl, . . . ,pn generate P. P is (d 1)-fold basic in a finitely generated free module Rn, for some n. Therefore, by Eisenbud-Evans theorem, there is an element p = pl + a(CYz2 aipi) E P which is basic in Rn. This element will be unimodular in Rn, and thereby in P. Now one can construct flips which map

The penultimate arrow exists as p is unimodular.

0

Corollary 9.4.6 (Forster-Swan) p(M) 5 f (M).

PROOF: We have a surjection Rn + M , for some n, with kernel K . Choose n least. If n > f (M), then pp(K) 2 n - pp(M) > S(p), for all prime ideals p. By Eisenbud-Evans theorem, K has an element k, which is basic (hence unimodular) in Rn. Hence Rn G [email protected] rank(Pi) = n - 1 2 dim(R)+ 1. By Bass Cancellation Theorem, P' is cancellative, and so P' is free of rank n - 1. But one can construct an onto map from P' onto M , as k E K . A contradiction. 0

9.5 Any Variety in the n-Dimensional Space is an Intersection of n Hypersurfaces In 1882 Kronecker stated without proof that if I is an ideal in the polynomial ring k[xl,. . . ,xn] over a field k then = J(f1,. . . ,f,+l) for some fl, . .. , fn+l E k[x1, . .. ,x,]. If k is algebraically closed, by Hilbert's Nullstellensatz this is equivalent to saying that the variety V(I) = {x E kn I f (x) = 0 for all f E I) c kn = A;

+

is the "set-theoretic" intersection V(fi) of n 1 hypersurfaces V(fi). (This was implicitely proved by Konig, Perron and van der Waerden.) We give here an algebraic version for general Noetherian rings.

Theorem 9.5.1 (Kronecker) Let I be an ideal of a Noetherian ring A of dimension n. Then there are f l , . .. ,fn+l e A such that d = J(f1,. . . ,fn+l).

9.5 Varieties as Intersections of n Hypersurfaces

237

Proof. We prove, by induction on r, that there are fl,. . . ,f,, such that V(f1,. . . ,f,) = V(I) U Z,, where Z, is a closed subset of Spec(A) of codimension r. By the codimension of a closed subset Z of Spec(A) we mean The case r = 0 being trivial, assume the result for an r 2 0 and let Z, = Z: U . . . U Z,S be the decomposition of Z, into irreducible components. We may assume, Z: $ V(I). Let pi E Z: \ V(I) and so I $ pi for i = 1,.. . ,s. By Lemma 1.1.8 there is an f,+1 E I \ Ui p i Then V(f1,. . . ,f,+l) = V(I) U zr+l with some closed set Z,+1 everyone of whose irreducible components is properly contained in one of the irreducible components of 2,. Therefore 0 codim(Z,+l) 2 r 1, which means Z,+l = 0.

+

In 1972, the above result, restricted to polynomial rings, was improved upon (to the best possible in general) by U. Storch [91] and D. Eisenbud - E. Evans independently in [19] who showed that n hypersurfaces in the affine n-space suffice. Theorem 9.5.2 (U. Storch; D. Eisenbud-E. Evans) Let R be a Noetherian commutative ring of finite Krull dimension d and let I be an ideal of the Then there are d 1 elements fl, . . . ,fd+1 E I with

+

Proof. We may assume that R is a reduced ring, since radical ideals of R and Rred are in bijective, inclusion preserving correspondence. We use induction on dim(R), the case dim(R) = 0 being obvious, since then R - as a Noether i m reduced ring of dimension 0 - is a product of finitely many fields, hence R[x] is a finite direct product of principal domains, whence all of its ideals are monogene. Let S be the multiplicatively closed subset R \ Ui=, pi where pl, . . . ,p, are all the minimal prime ideals of R. By Corollary 6.4.6 S-lR is a finite product of fields. Hence - as above - every ideal of S-lR[x] is generated by a single element. Let S-lI = (fl (x)) for some fl E I. Then there is an a E S , such that a I c (fl) c I. Hence V(I) c V(fl) c V(aI) = V(a) U V(I). Since a is not in any minimal prime ideal of R, we have dim(R/(a)) < d. Let the 'overbar' denote 'modulo (a)'. By the induction hypothesis there are f 2 , . . . , f d + l E I with

238

9 Number of Generators

Since on the other hand (fl,. . . , fd+l) c I, we get V(I) = V(fl,. . . ,fd+1) 0 or equivalently = d ( f i , . . . ,fd+1) as required.

J?

Corollary 9.5.3 let I be an ideal of the polynomial ring k[xl,. . . ,xn] over for some f l , . . . ,f, E a field k (with n > 0). Then f i = k[xl, . . . ,x,] . Equivalently, any variety V(I) in kn is the set theoretic inter0 section of n hypersurfaces V(fi), 1 i n.

d m < <

,

Remark 9.5.4 The above theorem had the effect of raising expectations that estimates for p(M) would improve by 1 in the case of finitely generated modules M over a polynomial rings R[x]. Eisenbud - Evans conjectured that if M is a finitely generated module over a polynomial ring R[x] then p(M) sup{pp(M) +dim R[x]/p I ht(p) < dim R[x]) or in 0. Forster's notation, let

<

Then M can be generated by b*(M) = sup{bg(M) : k 2 1) elements. This conjecture of Eisenbud-Evans was settled by N. Mohan Kumar in [60]; we give a proof of it in the next section.

Remark 9.5.5 Note that one cannot conclude that every prime ideal p in m for k[xl,. . . ,x,] is generated by n elements! One only has p = d some polynomials fi. In fact Macaulay has constructed examples of a sequence of prime ideals p, in k[x, y, x] with p(p,) = r for every r E N.

9.6 The Eisenbud-Evans conjectures Eisenbud-Evans made some conjectures regarding a bound on the number of generators needed for a finitely generated module over a polynomial ring R[X], when R is a noetherian ring. They also proposed a plan of action by which these conjectures should be solved - essentially, establish that Serre's Unimodular Element Theorem, and Bass Cancellation Theorem hold over polynomial ring, for projective modules of rank bigger than the dimension of the base ring. Then deduce a similar estimate for the bound as that for the base ring. The EE-conjecture for the bound were established for affine domains by A. Sathaye in [82], and for a general noetherian ring by N. Mohan Kumar in [60]. Later, B. Plumstead gave an argument in [70] modeled on the suggested method. We give a variant of this below.

9.6 The Eisenbud-Evans conjectures

239

Theorem 9.6.1 Let R be a Noetherian ring of dimension d, and A = R[x]. Then an A-module M is generated by e(M) elements, where

e(M) =

sup

{pp(M)

p,dim A/p
+ dim Alp).

Proof. Let e = e(M); we show that there is a free A-module of rank e mapping onto M . Consider a surjection An this map.

-+M -+0, with n > e, and let K

be the kernel of

Find a non-zero-divisor s' E R, xl E K,, such that K, is free and x l is unimodular in A.: Let s = fas, T = 1 sR. Let 6 on Spec(&) be a g.d.f. 5 d, as defined by Plumstead, cf. 59.3.3.

+

For any p E S p e c ( A ~ ) , T p) p ( F ~ ) pp(MS) > dim Alp 2 S(p). p p ( F ~) ~ ~ ( F T I K=

+

Hence KT is (a(#) 1)-fold basic in A? at p for every p E Spec(AT); by Theorem 7.1.8 there is an $2 E KT which is unimodular in A?. By Proposition 7.1.2 one can find an as, clearly, e 2 dim A.

E

E En(AT,) such that E(XI)T= (x2),,

Let x = xi x, 2 2 E A: X, A?. By Theorem 7.1.11 we see A: x, A? 2j An; identify them, Let Ax @ P = An. We are now in the situation

Since x, = s l E K, and XT = x2 E KT, the second projection map M is 'locally' surjective, and so surjective.

7 ~ 2: P

+

But P is stably free of rank n - 1 2 e 2 dim(A). We show that P is free, by showing that the associated unimodular row v(x) = (VI(x), . . . ,vn(x)) can be completed to an invertible matrix. Using Prime Avoidance Lemma, the reader will be easily able to show that a unimodular row of length 2 &(A)+ 2, can always be completed to an elementary matrix. (Here S will denote a g.d.f. on Spec(A).) In view of this, for any p E Spec(Rp[x]),v(x) can be completed over Rp[x], whence the associated projective module Ppis free. By Quillen's Local Global Principle, P is extended from R, i.e. P .v P I x P 8 R[x]. By Theorem 7.1.11, the stably free module P I x P is actually free. Hence, P is free. 0

Curves as Complete Intersection

J-P. Serre studied the codimension 2 complete intersection problem; in particular, how does one recognize a height 2 complete intersection ideal in R = XI,. . . ,X,]. He pointed out necessary conditions are that it should locally be a complete intersection, all its associated primes should have the same height, have homological dimension 1, and also that the module ~ x t ' ( 1 ,R) should be monogene. These conditions are also sufficient. This is our initial objective. In the latter part we prove known results of (local complete intersection) curves in n-space which are set-theoretical complete intersections.

10.1 A Motivation of Serre's Conjecture Hilbert's epoch theorem established the finiteness of the number of generators needed to generate an ideal I in a polynomial ring XI,. . . ,x,], with lc = Z or lc a field. The novelty and beauty of his proof was in its existential content: A natural outgrowth was to persevere with the old approach of finding actual generators perhaps tempered by the new methods of homological algebra which had been so grandly started by Hilbert. An ideal I of R is generated by n elements if and only if there is a surjection Rn -+ I -+ 0. J-P. Serre's reasoning was that it would be possibly easier to show that I is locally generated by n elements, i.e. there is a cover

and surjections RYi

-+I&.

One could then endeavour to 'patch' this local information. The global information he expected was to get a finitely generated projective R-module P of rank n and a surjection

242

10 Curves as Complete Intersection

If, say R = k[xl, . . . ,x,] - with k being a field - then Serre's Conjecture was that all finitely generated projective R-module were free. Thus, P would be free and I would be generated by n elements! J-P. Serre proceeded to show how to 'patch the local information' in a special case : When I was an ideal of height 2 in k[xl, . . . ,x,]. We need here (but not in later sections) some interesting additional facts of homological algebra especially about the functors Exti, which remedy the non-exactness of the functor Hom. The group (for commutative R it is an R-module) Extl (M, N ) = E X ~ ~ (N) M ,parameterizes the extensions of M by N , i.e. the short exact sequences

This is shown in Appendix B. The homological dimension hdRM of an R-module M can also be defined as the supremum of the n such that Extl)l(M, N ) # 0 for at least one R-module N. If one knows the definition of the Extl)l(M, N ) as homology groups of a complex which is built, starting with a projective resolution of M , it is easy to show that the just given definition of the homological dimension coincides with our old one Definition 8.3.1. (Some details of this can be found in Appendix c.1 The crucial step was

Lemma 10.1.1 (J-P. Serre) Let R be a Noetherian ring and M a finitely generated R-module with hd M 5 1. Suppose that Extl(M, R) is a cyclic (i.e. monogene) R-module. Then there is an exact sequence

with projective P. Proof. Let q generate ~ x t(M, l R) and let

be the extension corresponding to q. Then hd P 5 Max(hd M,hd R) = 1. We show P is projective. Look at a part of the long exact homology sequence, derived from the sequence 10.1 Hom(R, R)

3 Extl (M, R) --+ Extl (P, R) -+Extl (R, R).

The last term of this sequence vanishes since R is projective. Now A is surjective, since by Appendix B.1.13 we see that A(idR) = 1). Therefore,

10.2 The Conormal Module

243

Extl(p, R) = 0, whence Extl(p, Rn) = 0 whence Extl(p, Q) = 0 for every finitely generated projective Q. Since hd P 5 1 we have a resolution

with PI,Pofinitely generated projective. Since ~ x t l ( PPI) , = 0, the sequence 10.2 splits by extension theory, i.e. P Cg PI = Po,whence P is projective.

10.2 Projective Generation of I Let R be a local ring with maximal ideal m. For any ideal I c m we have

since I2c mI. For a non local ring the numbers p(I) and p(I/12) can differ, but at most by 1, as the following lemma will show.

Lemma 10.2.1 Let R be a ring and I a proper finitely generated ideal. Let further a l , . . . ,a, E I, such that their residue classes hl, . . . ,h, generate 111'. Then there is an s E I with (1 - s)I C (a1, . . . ,a,). Hence I = (al,. . . ,a,,s) and so

Proof. The first inequality is obvious. Let J := (al, . . . ,a,). If S = 1+ I , the ideal Is is contained in the Jacobson radical of Rs. Therefore by Nakayama's Lemma, elements whose residue classes generate Is/Ig already generate Is. So Js = Is,whence there is an s E I with (1 - s ) I c J. For every a E I we derive a - sa = (a17 . - - ,am, 8).

CEl &ai

for some Xi E R, i.e. a E

The height of I can be checked in the localizations with respect to the prime ideals containing I. For such a prime ideal p we have p+,(I)= p+,(I/12),since I+, is in the Jacobson radical of Rp. SO

By Lemma 10.2.1 there are two possibilities for p(I) - either it equals p(I/12) or p(I/12) 1. Besides, the proof of Lemma 10.2.1 also shows that Spec(R) is covered by two open sets D(s) and D(t), on which I is generated by 2 p(I/12) elements. (Namely set t = 1- s.) In other words one has surjections

+

244

10 Curves as Complete Intersection

where p = ,u(I/12) and with

(3, t)

= R.

0

Lemma 10.2.1 shows that the study of p(I) is closely linked to the study of the minimal number of generators of the conormal R/I-module I/12. In view of Serre's motivation it is natural to query, whether there is a projective module P of rank p which maps surjectively onto I. (i.e. we are trying to 'globalize the local' information.) This was done by M. Boratynski (see [9]) and his proof provides also a non-homological proof of Serre's result mentioned in the introduction of the last section. Theorem 10.2.2 (M. Boratynski) Let I be a finitely generated ideal of a ring R. Then there is an ideal J c I with -\/j = fi and a projective R-module P of rank p = p(I/12) which maps surjectzvely onto J. Proof. As in the proof of Lemma 10.2.1 there are s, al, . . . ,a, E I, such that (1 - s)I c (a1, . . . ,a,) and there are surjections

Since I,(,-1)= R,(,-l), the vector v := (al/l, . . . ,a,/l) is unimodular over &(8-1)

-

Suppose that v is completable and let ( a ~.,. . ,a,) = e ~ a - Ifor some a E GL,(R,(,-l)). Then one has a commutative diagram

with f := (1,0, . . . ,O), g := (a1, . . . ,a,). Let P := Rf x, Ry-, = {(x, y) E Rf x Rf-, I axl-, = y,) be the so called 'fibre product' of Rf and Rr-l over a. We have P, E Rf and PI-, E Ry-,. So P is projective of rank p. We can define an epimorphism cp : P + I by

If p = 2 then v is always completable. So we have shown that, if p(I/12) = 2 there is a projective module P of rank 2 which maps epimorphically to I. (This provides a non-homological proof of Serre's theorem.)

10.3 Local Complete Intersection Curves

245

In general one has to modify the ideal I. Let

where a l , . . . ,a, are as above. Then fl = f i and J, = R,, the latter since V(J,) = V(I,) = 0. Further J1-, = ( a l , a i , . . . ,a;-')l-,; namely (1 - s)Ip-l c ( a l , a f , .. . ,a;-1), and P-1 so 11-,c (a1, a;, . . . ,a;-l)l-,. By Suslin's Theorem 3.5.3 the unimodular row (al /1, a i / l , . . . ,a;-'/l) over R,(l-,) is completable. Hence as above we get an epimorphism P -+ J for some projective module P of rank p. 0

Corollary 10.2.3 Let k be a field or a principal ideal domain and I an ideal of the polynomial ring k[xl,. . . ,x,] and p = p(I/12). Then

for suitable fl , . . . ,f,. Proof. By Theorem 10.2.2 there are an ideal J with fi = fi,a projective module P of rank p and an epimorphism P -+ J. But P is free by the Quillen0 Suslin Theorem.

10.3 Local Complete Intersection Curves in n-Space are Set-Theoretical Complete Intersections A classical open problem asks whether every curve in the affine n-space is a set-theoretical complete intersection. Initial progress in Serre's Conjecture led Ferrand and Szpiro to independently affirm that a curve in 3-space which is a local complete intersection is a set-theoretic complete intersection. Due to Serre's Conjecture's solution and M. Boratynski's results we can now establish this by a similar argument in n-space. (These results were conjectured by 0. Forster and were first settled for n > 3 by N. Mohan Kumar in [60]). We need few preliminary observations on rank 1 projective modules and on regular sequences. For a rank 1 projective R-module L we denote by LBn =

L @ - .. @ L n times, if n L* 8 . - .@ L* In1 times, if n

>1 < -1

ifn=O It is easily verified that L@("+") = L@" @ L@mfor all n, m E Z.

246

10 Curves as Complete Intersection

Lemma 10.3.1 Let R be a one-dimensional Noetherian ring and P a finitely generated projective R-module of constant rank r 2. Then every projective R-module Q of rank: I is a direct summand of P.

>

Proof. By Proposition 8.8.7 we see P Z P' @ Rr-I with some P' of rank 1. Further r - 1 1. Then

>

Lemma 10.3.2 Let R be a local Noetherian ring and I # R be an ideal, generated by an R-regular sequence X I , . . . ,x,. Then 1/12is a free RII-module.

Proof. We show that the residue classes G, .. . , form a basis. They clearly generate I/12.Assume they were linearly dependent, say Ciaixi E 12with some a j $! I. By Lemma 9.2.3 we may assume j = n, i.e. a, 4 I . Then a,x, E (XI,. . . ,$,-I) I2= (XI,. . . , x,-I) x,I, i.e. (a, b)x, E (XI,.. . ,$,-I) b $! I, whence a fortiori a, b for some b E I. Since a, $! I, also a, (XI,. . . ,%,-I). But this means that x, is a zero-divisor modulo (XI,. . . ,xnWl) - a contradiction. 0

+

+

+

+

+

Theorem 10.3.3 (Ferrand) Let I be a n ideal of a Noetherian ring R of Krull dimension d 2 3. Assume further that I is a local complete intersection ideal of height (d - 1), i.e. Im= (XI,.. . ,xd-1) for some regular sequence XI,. . . ,xd-1 in R,, for e v e y maximal ideal m of R containing I . Then there is an ideal J of R such that

b) J is a local complete intersection of height (d - 1) c) J/J2is a free R I J-module of rank (d - 1).

Proof. By Lemma 10.3.2 the R/I-module 1/12 is locally free, i.e. projective of rank ( d - 1). Since dim(R/I) = 1and Ad-' (1/12)* is a projective RII-module such that for of rank 1, by Lemma 10.3.1 there is an ideal J, containing 12, K := A~-'(I/I~)* we have

We show that the ideal J has the required properties. We only have to prove b) and c). b) To show that J is a local complete intersection, we may assume that R such that (hence also R I I ) is local. Then there is a basis X I , . . . ,xd-1 of 1/12 X I , . . . ,xd-2 is a basis of J/12and xd-1 one of K/12. Then

10.3 Local Complete Intersection Curves

247

-

2 (xl,... ,xd-2,xd-1)So p ( J ) 5 d - 1. On the other hand ht(J) = ht(I) = d - 1, since 12c J c I.

c) J/ J2is a projective R/ J-module of rank (d - 1). We show that J/J2is generated by (d - 1) elements; this will suffice to show that J/ J2 is free of rank (d - 1). Since ( I J / J 2 ) 2 = 0 in the ring R / J 2 , by Nakayama's Lemma we have p ( J / I J ) = p(J/ J 2 ) . So we need only show p ( J / I J ) = d - 1. Let us analyse the R/I-module J I I J . Notice that J/IJ = J @ R R / I = J @ R R / J B R / J R / I = J/J2@ R / J R / I and so is a projective RII-module of rank (d - 1). We will show that it is actually a free RII-module of rank (d - 1). By Serre's splitting Theorem 7.1.8 it suffices to show that I\~-'(J/IJ) 2 R/I. We have an exact sequence of RII-modules Since J/IJ and J/12both are projective over R I I , this sequence splits and 12/IJis also projective over R I I . By the equality (10.3) we have rk(J/12) = d - 2, and so rk(12/IJ) = 1 by the exactness of (10.4) . Since the sequence (10.4) splits, J/IJ 2 12/IJ$ J/12. Taking exterior powers gives

Recall that K = ~ o m ( I \ ~ (-I' / I ~ ) ,R/I) r I/J via the equality (10.3). In particular, since A~-'(I/I~) K@-l, we get K@-' r A~-'(K @ J/12) ( ~ 1 1 ~ ) A~-~(J/I8 ~ )K , hence ~ ~ ~ ~E KBp2. Multiplication in I gives us a natural (R/I)-bilinear map

I/J x I / J + I~/IJ, ( x + J , y + J ) e x y + IJ, which induces an R/I-homomorphism I/J BRlr I/J --+ 12/IJ.This map is locally an isomorphism as I/J Z K is projective of rank 1 over R/I. Hence it is an isomorphism and so 12/IJ2 K @ K . Thus as required, (J/I J) 7 A ~ - ~ ( J / I ~ ) @ I ~ / I J = K @2 - ~R/I. @K@~ 0 Corollary 10.3.4 (Ferrand - Szpiro, n = 3; Mohan Kumar ) Let I be a local complete intersection ideal of k[xl, . . . ,x,] of height (n - 1). Then Jf = J(f1,. . . ,fn-1) for suitable f i , . . . , fn-1. Proof. In view of Ferrand's construction Jf = for some local complete intersection J with J/ J2free of rank n - 1. By M. Boratynski's theorem, there = and a projective k[zl,. . . ,x,]-module P is an ideal J' c J with of rank n - 1,mapping onto J'. By Quillen-Suslin's theorem P is free of rank n - 1, and so J' is generated by n - 1 elements. 0

a' a

248

10 Curves as Complete Intersection

10.4 The Theorem of Cowsik and Nori The Theorem of Cowsik and Nori says that curves in the &ne n-space over a field of positive characteristic are set-theoretical complete intersections. We need a preparation. 10.4.1 A Projection Lemma

The following statement seems intuitively plausible: A curve C in n-space can be projected into a plane in such a way, that this projection maps C isomorphically to its image outside a finite set of points. And a proof along this intuitive idea can be given, if the ground field is algebraically closed (or at least infinite) (cf. [30], Chapter IV). But if this field is finite, there are only finitely many directions along which to project. And if the curve is the union of lines in every possible direction, the above statement fails to hold. So one has to admit "nonlinear projections", e.g. Nagata transformations. We will formulate and prove the correct statement in purely algebraic terms. We follow the ideas of [53]. Definition 10.4.1 A homomorphism A + B of reduced Noetherian rings is

called birational, if it is injective and induces an isomorphism of the total quotient rings &(A) + Q(B). We fix a prime number p and a perfect field k of characteristic p. Proposition 10.4.2 (Projection Lemma) Let I be a purely 1-codimensional radical ideal of A = Ic[Xl,. . . ,X,], i.e. I = p1 fl . - - n p, with dim(A/pi) = 1. Then there is a change of variables, such that after this change the ring

extension k[X1,X2]/(k[X1,X2]fl I ) v A / I is finite and birational. The proof requires some preparations. First recall that every f E A with d f /dXi = 0 for all i is of the form gp with some g E A, since k is perfect. So in this case f is not irreducible. As a consequence we get Lemma 10.4.3 Let f E k[X, Y] be irreducible. Then for big enough m, not

divisible by p, we have that f (X + Ym, Y) is monic i n Y upto a unit and

Proof. Write F := f (X + Ym, Y). We know that F is monic in Y upto a unit for large m. The chain rule for differentiation, valid also in this formal situation, yields

10.4 Cowsik-Nori Theorem

249

(Distinguish well between and -d(fX + Y m , Y ) . af(x+ym'y) dY dY The first means: first replace X by X Ym, then differentiate w.r.t. Y; the second means: first differentiate w.r.t. Y, then replace X by X Ym.)

+

+

+

If d f / d Y # 0, then d f / d Y # (Yr) for some r . Then also d f / d Y ( X Ym, Y) # (YT).Since for m > r the first summand of the right hand side of (10.5) belongs to (YT),the left hand side does not, and so even more does not vanish.

If on the other hand d f /dY = 0, then, according to the above remark, d f / d X f 0. And so if p ;(m.

The following lemma is well known.

Lemma 10.4.4 Let k c k(x, y) be a finite field extension and y separable over k. Then k(x, y) = k(x + ay) for all but finitely many a E k. Lemma 10.4.5 Let pl, p2 be two distinct maximal ideals of K [ X , Y]. Then pl n K [ X aY] # p2 n K [ X + aY] for all but finitely many a E K .

+

Proof. By Hilbert's Nullstellensatz the fields Li := K[X, Y]/pi are finite over K . Fix K-embeddings (i.e. K-algebra-homomorphisms) Li L, E, the algebraic closure of K , and call xi, yi the images of the residue classes of X , resp. Y modulo pi under these iembeddings The set E of all K-iembeddings L1 v K is finite. For every a E E there is at most one b E K with b(a(y1) - y2) = x2 - a(x1). Otherwise we had x2 = a ( x l ) and y2 = a(y2) for some a and hence pl = p2 - as kernels of 'isomorphic' homomorphisms. Therefore the set

S := {b E K

/

b(a(y1) - 92) = x2 - a(x1) for some a E E )

is finite.

+

+

We claim that pl n K [ X aY] # p2 n K [ X aY] for every a E K \ S . By definition of S we have a ( x l + ayl) # 2 2 aya V a E E, whence XI ayl and 22 ay2 are not conjugate to each other. So, if fi(T) is the minimal (monic) polynomial of xi ayi, we have f l # f2. Now it is clear that pi n K [ X aY] is generated by f i ( X aY). So pl n K [ X aY] # p2 n K [ X aY].

+

+

+

+

+

+

+

+

10 Curves as Complete Intersection

250

Proof of the Projection Lemma: First we handle the case n = 3 and call the indeterminates X , Y, Z. Let (as above) I = pl n . -- n p,, where pi are prime ideals with dim(k[X, Y, Z]/pi) = 1, i.e. ht pi = 2 by Proposition 6.7.5. According to Theorem 6.9.3 we have

The case ht(pi n k[X, Y]) = 1 is the more general one: A curve projects to a curve. But the other case may also happen, e.g. if pi = (X, Y). To escape this difficulty, let gi(X, Y) E pi n k[X, Y] \ (0). For big enough m, n-m all gi(X+Zm, Y +Zn) are unital in Z. Hence k[X, Y, Z]/pi is integral over k[X1,Y', Z]/pi nk[X1,Y'] for new variables X' = X And therefore dim(k[X1,Y1, Z]/pi n k[X1,Y']) = 1.

+ Zm, Y' = Y + Zn.

So we may assume that ht(pi n k[X,Y]) = 1. In this case pi n k[X,Y] = fik[X, Y] for certain polynomials fi E k[X, Y], since prime ideals of height 1 in factorial rings are principal. Using Lemma 10.4.3, after a change of variables, we may assume that the fi are monic upto a constant factor in Y and dfi/bY # 0. Set Ki := Q(k[X, Y, Z]/pi) and denote by xi,yi, xi the images of X , Y, Z in Ki. Then the field extensions k(X) L, Ki are finite and the yi are separable over k(X). Since k(X) is an infinite field, by Lemmas 10.4.4 and 10.4.5, there is an a E k(X) such that first Ki = k(X)(ayi xi) for i = 1,. . . ,r and secondly the k(X)[aY Z] n p i are distinct. Then k(X)[Y, Z]/Ik(X)[Y, Z] is generated by the class of aY Z over k(X).

+

+

+

Now write a = c/d with coprime c, d E k[X] and let y,S E k[X] be chosen such that yc - Sd = 1. Then k(X)[Y, Z]/Ik(X) is generated by the class of cY dZ over k(X).

+

+

+

Replacing cY dZ by Y and SY y Z by Z we reach the situation that the ring k(X)[Y, Z]/Ik(X)[Y, Z] is generated by Y over k(X). Hence

becomes an isomorphism after tensoring with k(X), whence it is birational. And it is finite since I contains a polynomial which is monic in Z. Now assume n > 3. Since, by the first case, after a change of variables the extension k[Xl, X2]/I n k[Xl, X2] L) k[Xl, Xz7X n ] / I n k[Xl, X2, X,] is finite and birational, the same holds for k[Xl,. . . ,Xn-1111 n k[Xl,. . . ,Xn-l] 9 k[X1, . . . ,Xn]/I. By induction we are through. 17

10.4 Cowsik-Nori Theorem

251

10.4.2 The Proof of the Theorem of Cowsik and Nori

We need the following notion:

Definition 10.4.6 Let A C B be a ring extension. The conductor C of this extension is defined by C := A ~ ~ A ( B / A ) , where B/A denotes the residue class module of the A-module B bg its submodule A.

Remarks 10.4.7 a) The conductor C is an ideal of B contained in A. Moreover it is the biggest common ideal of A and B. b) Assume that the extension A c B is a birational finite extension of reduced rings. Then C contains a non-zero-divisor. Namely write the elements of a finite generating system of the A-module B as fractions over A. Then every common denominator will belong to C.

Theorem 10.4.8 (Cowsik-Nori) Every curve in the afine n-space over a field k of positive characteristic p > 0, is a set-theoretic complete intersection. Proof. Let p

c

B := XI,. . . , X I ] be the radical ideal defining the curve

C. If the extension pe of p in kf [XI,;. . ,x,] is generated upto radical by N (n - 1) elements f l , . .. ,f,-l, then f,P ,. .. ,f,P-l E B for some N and p =

Jr

fn-,). Therefore, we may assume k = kllpm, i.e. that k is perfect.

By Proposition 10.4.2 after a change of variables we may assume that k [ x ~~, ~ ] / p n kxz] [~ , is a finite and birational extension. Therefore the L,~B/p conductor C of this extension contains a non-zero-divisor.Since dim(B/p) = 1, we have dim(B/p)/C = 0, and so (B/p)/C is Artinian. It therefore is a finite dimensional vector space over k by Lemma 6.5.1. r. p~+l Let Li,, (for 3 i n) be the k-vector space, generated by 3: ,Zi , . . . in (B/p)/C. Then Li,o > Li,1 > . . . > Li,, > . . . . This series must eventually stabilize: say Li,, = = . . . . Thus there exists an r such that ;z;< E Li,,+l for all i, i.e. p'+ 1 -pr+t" = 0, - SilZi - . .. -

< <

zr

with some sij E k. Thus

for suitable q, whose residue classes modulo p belong to C, hence to k[xl, xz]/p n k[xl,x2]. Therefore we may and will choose ci in k[xl,x2].

252

10 Curves as Complete Intersection

Let yi := x r , B' = k[xl,x2, y3,. . . ,y,] and p' = p n B'. Note that fi E p'. Since p'B c p and every element of p raised to its pr-th power belongs to p', = p. we see that

CLAIM:B" = B1/( f3, . . . , f,) is a regular ring.

>

Namely observe that dfi/dyj = Sij for i, j 3. By the Jacobian Criterion 8.1.3, B1/(f3, . . . ,f,) is regular of dimension 2. Note that p' = p1/(f3,. . . ,f,) has height one. For any maximal ideal m > p', B g is regular and local, whence factorial. Consequently being an intersection of finitely many prime ideals, is a principal ideal, i.e. p' is locally 1-generated. Therefore, p' is locally (n - 1)-generated. By Corollary 10.3.4 up 0 to radical p' is generated by (n - 1) elements.

Normality of the Elementary Subgroup

In 1964 P.M. Cohn had shown in [13] that E2(A) was not always a normal subgroup of SL2(A). For example, he showed that

It came as a surprise when A. Suslin announced in mid 1976 that En(A) is a normal subgroup of GLn(A) for n 3 and any commutative ring A.

>

We have to show that zf

if E

GLn(A) then i f ~ $ i f - l

Now if~;if-l

= ?(In

+ Xe:ej)if-l

Let vt = ife:, w = e j y P 1 . Then v , w yEij(X)y-l = In + vtw.

= In E

E

En(A)

+ Xye:ejy-l.

U m n ( A ) with wvt = 0 and

Yt 3

+ vtw is 1-stably elementary, i.e. ( I n En+ 1 ( A ) .In other words, in (n + 1)-space I, + vtw is elementary. First we will see that In

~ We prove this as follows: Let wvt = 0 and consider In +vtw. Since ( v ~ w=) 0, we see that In vtw is invertible with In - vtw as its inverse. Now

+

It follows that ( I n

+gVtW y)

E En+1(A).

Now we consider the special case that some coordinate vi of v = ( v l ,- .- ,v,), or some coordinate wj of w is zero. Then we will conclude from above that In vtw E E n ( A ) .

+

A Normality of En in GL,

254

Let, for example, wj = 0. By permuting coordinates we may assume w, = 0. Let v' = (vl, - - .,v,-I), w' = (wl, . - . ,w,-I), then w'v'~ = 0, and so In-1 vttw' E E,(A) by above. Now

+

(This observation was made by L.N. Vaserstein in [104] in 1973.) We are now ready to prove the general case by reducing it to the above case. For this we need some linear algebra :

Lemma A.O.l Let v = (vl, . . . ,v,) E Um,(A). Let further f : An + A be an A-linear map given by ei e vi where {ei)l
An. Then ker(f) = {(wl,. . . ,w,) E An elements (vjei - viej), 1 5 i

< j 5 n.

/

C w i v i = 0) is generated by the i= 1

Proof. Since v unimodular, f is surjective. Let K be the submodule of ker(f) generated by {(vjei - viej) : 1 5 i # j 5 n ) . We check "locally" that K p = (ker(f))p = ker(fp) for all p E Spec(A). This is easily done in the case v = e l ; and the general case can be easily reduced to this case as v can be completed to an invertible matrix over R,. 0

Proposition A.0.2 Let n 2 3, v = (vl,. . . ,v,) E Um,(A) and w = (wl,. . . ,w,) E Ml,,(A) be such that wvt = 0. Then I, + vtw E E,(A). Proof. Let (ei)lSisn be the canonical basis for An. Define an A-linear map f : An -+ A by f(ei) = vi. Since wvt = 0, we have w E ker(f) and hence by Lemma A.0.1, there exist aij E A such that

where

Wij

= aij(vjei - viej). Since wvt = O and w = C w i j , wijvt = 0. Note

i<j

that n 2 3. By Vaserstein's observation I,

+ vtwij E E,(A).

Now

We prove the claim by induction on the number of terms in the summation, say r. Suppose r = 1, then the result is clear. Suppose r = 2, then

A Normality of En in GL,

(I,

255

+ v t w1)(I, + v t w z ) = I, + v t W l + vtw2 + v L w 1 v t ~ =' 2I, + vtwl + vtw2

as wlvt = 0. Suppose r 2 3 and assume that the result is true for r - 1, then,

= (In

+ vt(wl + . . . + w,-l))(I, + vtw,)

(The middle step is by the case r = 2, and the last step is by the inductive assumption.) Thus by induction

as required. Open Question: Can one drop the condition that tion A.0.2?

71

is unimodular in Proposi-

This is possible when R is a regular ring containing a field.

Some Homological Algebra

In this Appendix we shall do the following:

* Extension theory. We introduce extensions and ~ x t and l their relationship.

* Derived functors, and higher Extn * Rees' homological characterization of grade. B . l Extensions and ~ x t l Definition B . l . l Let R be a ring, M , N , X be R-modules and f : M -+ X , g : N -+ X homomorphisms. A pull-back of ( f , g) is an R-module Y together with a pair of homomorphisms f' : Y -+ M , g' : Y -+ N such that f f' = gg' and the following universal property holds: If f" : Z -+ M , g" : Z -+ N are homomorphisms such that f f" = gg", then there exists a unique h : Z -+ Y such that f'h = f" and g'h = g".

258

B Some Homological Algebra

An obvious candidate for Y is Y = { ( m ,n ) E M @ N I f ( m ) = g(n)),i.e. the kernel of the map M @ N -+ X , ( m ,n) f ( m )- g(n) with the canonical maps f' : Y -+ M, g' : Y -+ N .

*

The following definition is dual to the above one.

Definition B.1.2 Let R be a ring, M, N , X be R-modules and f : X -+ M , g : X -+ N module homomorphisms. A push-out of ( f , g ) is a module Y together with a pair of homomorphisms f' : M -+ Y , g' : N -+ Y such that f ' f = g'g and the following universal property holds:

Iff'' : M -+ Z , g" : N -+ Z are homomorphisms such that f l ' f = gl'g, then there exists a unique h : Y -+ Z such that hf' = f" and hg' = g". Z

Here the obvious candidate is Y = M @ N / { ( f ( x ) ,-g(x)) I x E X ) , i.e. the cokernel of X -+ M @ N , x I+ ( f ( x ) ,-g(x)) with the canonical f' : M -+ Y, g ' : N + Y .

Proposition B.1.3 In the category of R-modules pull backs and push outs 0 do exist and are unique upto unique isomorphism. We leave this to the reader.

Lemma B.1.4 If the square

Y

f'

+ M

is a pull-back diagram, then g' maps every fibre fl-'(rn) o f f ' bzjectively to the corresponding fibre g-'( f ( m ) ) of g. Especially

B.l Extensions and ~ x t '

(1) g' induces ker(ft)

C

259

ker(g);

(2) If g is an epimorphism, then so is f'. Proof. We may assume Y = {(m, n ) E M @ N I f (m) = g(n) ) and f '(m, n ) = m, gt(m,n) = n. Then ft-'(m) = {(m,n) I n E g-l(f(m))), which clearly is mapped by g' bijectively to g-l(f (m))}. The consequences (1) and (2) are 0 immediate. (Note that surjectivity means: all fibres are nonempty.) Dually we have Lemma B.1.5 Let

X-M

f

be a push-out sqare. Then (1) f' induces an isomorphism coker(f) 2 coker(gt); (2) f injective

===$g'

injective.

This is also easily established Nl{(f (XI,4 ~ ) I x) E X}.

if

one identifies

Y

with

M

@

0

B.1.6 Let R be a ring and M, N be two R-modules. A short exact sequence

of R modules is called an extension of N by M . Two extensions O-+M-+E-+N-+O are said to be equivalent if there is an isomorphism the diagram

E

: E -+

Et such that

is commutative. (One need only require, E to be a homomorphism, since then bijectivity automatically holds.) This is an equivalence relation on the class

260

B Some Homological Algebra

of all extensions. The equivalence classes w.r.t. equivalence make up a set. (The scrupulous reader should convince himself that any E as above can be identified as a set with M x N - equipped with some 'skew' algebraic structure.) We denote the set of the equivalence classes of extensions by E ( N , M ) . Clearly E ( N , M ) # 0 since M $ N , together with the canonical inclusion map i and the canonical projection n-,yields an extension

Any extension equivalent to (*) is called a split extension of N by M . B.1.7 From homomorphisms cp : N' -+ N , resp. $ : M -+ MI we deduce maps cp* : E ( N , M ) -+ E(N1,M ) , resp. ,$I : E ( N , M ) -+ E ( N , MI). (Note the contravariance in the first and the covariance in the second variable!) We proceed as follows: Let a E E(N, M ) given by 0 -+ M -+ E -+ N -+ 0; then form the pull-back square

E-N resp. the push out square M-E

these fit into the following diagrams with exact rows

respectively 0-M-E-N-0

$1

O-M'-E"+N+O

i l

Then the first row in the next to the last diagram represents cp* ( a ) , whereas the second row in the last diagram represents $,(a). There are several properties which the reader is invited to prove:

B.l Extensions and ~ x t '

261

a) Functoriality: If N" 3 N' -f% N , respectively M 3 M' 3 MI1 are homomorphisms, then (cp20cp1)*( a ) N cp; (9; ( a ) ) , resp. ($20$1)* ( a ) % $~2*($~*(a)). Here and in the following we write 'E' for equivalence. b)

10*cp* ( a )

cp*$, (a).

c) If 0 is the zero endomorphism of N (resp. of M ) then O*(a) (resp. O,(a)) is equivalent, to the split extension of N by M .

B.1.8 We can now define the sum of two extensions in E ( N , M ) : Let 0 t M t E t N t 0, 0 t M t E' t N -+ 0 be representatives of two elements a, /3 in E ( M , N ) . First form the direct sum a @ P, i.e.

Define V : M $ M - + M by ( m , m 1 ) + + m + m 1and A : N + N @ N by n e ( n , n ) . Then define: a

+ /3 := A*(V, ( a CB P ) ) = V, (A*(a CE B)).

This composition is known as the Baer sum

.

It is related to the sum of homomorphisms in the following way. Let cpj : N' t N , resp. $j : M t MI with j = 1 , 2 be homomorphisms. Then (cpl+p2)*(a) = cp; ( a ) + cp; ( a ) and (101 + $2)* ( a ) = $1* ( a ) $2, ( a ) .

+

By the Bacr sum E ( N , M ) becomes an abelian group. The class of split extensions is the zero element of this group. The inverse of the class of

a : O - + M ~ E ~ N t O i s t , h e c l a s s o f O - + M - f E ~ N - + 0 , which is equivalent to 0 -+ M E 3 N -+ 0 and can also be described by (-l),(a), resp. (-l)*(a), where (-1) denotes the homothesy of -1 on M , resp. N .

4

The reader may prove these statements, using a), b), c) above. Lemma B.1.9 Let

be a commutative diagram with exact rows, then the right-hand square is a pull-back diagram.

262

B Some Homological Algebra

be a pull-back diagram. By (B.1.4) y is epimorphic and /3 induces an isomorphism ker(y) M . Hence we obtain an extension

By the universal property of Z there exists a map $ : E' -+ Z, such that ,B$ = E , y$ = gl. Since $ induces the identity both in M1 and M , it is an isomorphism. f N' 3 P -+ N + 0 B.l.10 Let R be a ring. A short exact sequence 0 -+ of R-modules with projective P is called a projective presentation of N .

Such a presentation induces for an R-module M an exact sequence

To the modules M and N and to the chosen projective presentation of N we can associate the abelian group E x t l ( ~M , ) = coker(f * : Hom(P, M )

+ Hom(N1,M)).

An element in Extl(N, M ) may be represented by an element cp in Hom(N1,M). The element represented by cp : N' -+ M will be denoted by [cp] E EX~'(M,N). Theorem B . l . l l There is a natural isomorphism : E(N, M )

7E x t l ( ~M, )

This shows by the way that Extl(N, M ) is independent from the chosen projective presentation of N . By the term 'natural' we especially mean 'natural' in M , i.e. a homomorphism f : M + M' induces a commutative diagram

If* The map q is also natural in N ; but to show this is a trifle more difficult.

B.l Extensions and ~ x t '

Proof. We first define an isomorphism of sets q : E(N, M )

Extl(N, M).

Given an element in E ( N , M ) , represented by the extension 0 N -+ 0 we may form a diagram 9 f , P - N - O N' 0

-

263

-+ M 4E 3

Namely cp exists (but is not unique) since P is projective; and d can be regarded as a restriction of cp since cp(kerg) c ker k. The homomorphism t) : N' -+ M defines an element [$] E Extl(N, M ) = coker(f * : Hom(P, N ) -+ Hom(N', N ) ) . We claim that this element does not depend on the particular cp : P + N chosen. Thus let cpi : P -+ N , i = 1,2 be two maps inducing $i : N' -+ M.Then cpl - cpa = h o for ~ some r : P -+ M . It follows that $1 - $2 = .of. Therefore [$I] = [$a 7 0 f ] = [dz] since TO f is mapped to 0 by f*. Since two representatives of the same element in E ( N , M ) obviously induce the same element in EX~'(N,M), we have defined a map q : E ( N , M ) -+ Ext' (N, M).

+

Conversely given an element in ~ x t ' ( N ,M ) we represent this element by a homomorphism II, : N' -+ M . Taking the push-out of (I), f ) we obtain the diagram

h M - E - M By Lemma B.1.5, the bottom row 0 -+ M

k

4E 3 N -+0 is an extension.

We claim that the equivalence class of this extension is independent of the particular representative $ : N' + M . Any other has the form $' = + of with some T : P + M . Then the diagram 9 N' f + P - N - o 0

+

-

+

~ commutative. It follows that the extension we arrive at with cp' = cp h o is does not depend on the representative. We thus have defined a map

264

B Some Homological Algebra

which is easily seen to be natural in M . By Lemma B.1.9 7 and 6 are inverse to each other. Thus we have an isomorphism 11 : E ( N , M ) Extl(N, M ) .

<

B.1.12 Let (*) 0 -+ A + B -+ C + 0 be a short exact sequence of R-modules and N another R-module. By functoriality we get scquenccs

and EX? (c, N)

-+ EX^^ (B, N) -+ EX^^ (A, N)

We know that the first of them is exact. It is left, to the reader to show that also the second one is so. If we want to splice these two exact sequences, there is an obvious way to do it provided Extl(C, N ) is interpreted as E(C, N ) . Namely to f : A -+ N assign f,(a) where a is given by the extension (*). This assignment defines a map A : Hom(A, N ) + Extl(C, N ) . We leave it to the reader to show that A is a homomorphism and that the resulting long sequence -

is exact. B.1.13 In the above situation assume that N = A. Then d ( i d ~ = ) a where cr is given by the extension (*).

B.2 Derived functors In this section we go to the heart of homological algebra, by briefly studying the theory of derived functors, in special cases, and which we shall use to introduce the derived functors of Ext(M, N). This theory historically arose as a generalization of the theory of Ext described in the previous Appendix C. The methods there to define Extl in terms of extensions has also been done by Yoneda for the higher Extn, which are equivalent to n-extensions. The reader will find more details of this, and about our brief sketch below, in ([28], Chapter 7) or ([32], Chapter IV). The classic foundational book in homological algebra is Cartan-Eilenberg's Homological algebra [12].

B.2 Derived functors

265

An additive contravariant (covariant) functor is a correspondence which associates to each R-module M a R-module T(M), and to each R-linear map f : M -+ M', a R-linear map T(f) : T(M1) -+ T(M) (resp. T ( f ) : T(M) -+ T(M1)) satisfying

Example: If N is a R-module then the functor T(M) = M @R N , and T ( f ) = f @ R IN,for f : M -+ M', is a covariant additive functor. The functor T(M) = Horn~(M,N) and T ( f ) = Hom(f, IN), for f : M -+ M', is an additive contravariant functor. We now define the n-th right derived functors RnT (n 2 0) of an additive contravariant functor. The reader should be able to similarly define the left derived functors Ln(T) (n 2 0) of an additive covariant functor. Let M be an R-module and P . -+ M be a projective resolution of M. Applying T to the complex P.:

d

-.--+Pn3Pn-1

-+

dn-1

.-.-+P1-+P0

we have the complex T(P)

The n-the homology of the complex T ( P ) is as usual the subquotients at the n-th stage, i.e. ker(T(dn+l)/S(T(dn)).This is defined to be RnT(M), the n-th right derived functor of T. To justify it is a functor, we must still define RnT(f), for any map f : M -+ M'. Let P, P' be projective resolutions of M , M' respectively. Then f can be lifted to a map F of complexes P -+ P'. This induces a map T ( f ) : T(P.) -+ T(P1.). Since T ( F ) is also a mapping of complexes it induces mappings on the homology RnT(f) : RnT(M') -+ RnT(M'), for n 2 0. It can be shown that (1) RnT(M), RnT(f) are well defined and independent of the projective resolution. (It is interesting to note that one cannot say this if one had only restricted oneself to free resolutions.) Moreover, RnT, n 2 0, are additive contravariant functors. (2) RnT(M) = 0, n

> 1, if M is projective.

266

B Some Homological Algebra

(3) RO(T)E T , if R is left exact.

(4) For any exact sequence

of R-modules, there is a long exact sequence

which satisfies the naturality conditions. The n-th right derived functor R n T of the contravariant additive functor T = Horn(-, N ) is denoted by Extn(-, N ) . The derived functors of Hom, @ R play a vital role in the homological dimensions (also called projective dimension) of rings and modules; which in turn plays a crucial role in the study of dimension of rings and modules. With this in mind, it is natural to first how Extn can be used to compute (theoretically) the grade of a module w.r.t. and ideal I . Theorem B.2.1 (Rees) Let A be a Noetherian ring, M a finitely generated A-module and I an ideal. Let n > 0 be an integer. Then the following are equivalent: (1) E x t ; ( ~ , M) = 0 for any finitely generated A-module N such that Supp(N) V(I) and for any i < n;

c

(2) EX~;(A/I, M ) = 0 for any i

< n;

(3) there exists an M-regular sequence f l , . . . ,f n in I .

+

(2) is trivial. (2) + (3): We have E&(A/I, M ) = Proof. (1) H o ~ A ( A / IM , ) = 0. If no elements of I are M-regular, then I UpEAss(M) p so that I p for some p E Ass(M). Then there exists an injection Alp -+ M , and by composing it with the natural map A I I -+ Alp we obtain a non-zero homomorphism A I I -+M , which is a contradiction.

c

c

Therefore there is an element fl of I which is M-regular. If n > 1, put f M -+ Ml -+ 0. We MI = M l f l M and consider the exact sequence 0 -+ M 1 then get a long exact sequence.

I , ) = 0 for i < n by assumption, we have E x t i ( A / ~MI) , =0 As E X ~ ~ ( A /M for i < n - 1. By induction on n there exists a MI-regular sequence f2, . . . ,fn in I. Then fl, f 2 , . . . ,f n is an M-regular sequence in I.

B.2 Derived functors

267

(3) + (1): Put MI = M/flM. By induction on n we have Exti(N, Ml) = 0 for i < n - 1. Considering the long cxact sequence

derived from 0 t M

9M t MI t 0, we sce that

is exact for each i < n. Sincc Supp(N) = V(Ann(N)) C_ V(I) we have I C_ radical of Ann(M). In particular, fir E Ann(N) for some r > 0. Therefore fl annihilates E X ~ ~ (M N ), also. Thus we have Exti(N, M ) = 0 for i < n.

Corollary B.2.2 The length of a maximal M-regular sequence in I depends on M and I only, and is equal to the number n such that EX~;(A/I, M ) = 0, for i < n , Extl(A/I, M ) # 0. Proof. It remains to prove that, if f ~. . .,, f, is a maximal M-regular sequence in I , then Ext;(A/I, M ) # 0. By induction on p, this is an easy consequence of the exact sequence 0 t M t M t M / f l M -+0.

Complete intersections and Connectedness

In this Appendix we shall prove a famous theorem of Robin Hartshorn. This theorem gives a topological characterisation of a set theoretic complete intersection ideal of dimension > 1 in a local ring in terms of the connectedness of the punctured spectrum SpecR \ {m). The notion and existence of a primary de2omposition of an ideal in a Noetherian ring, and the notion of a completion R of a local ring R w.r.t. its maximal ideal m, and the notion of a Cohen-Macaulay ring will be assumed in this section. The reader can refer to the standard texts like [I], [20], [54] for these topics. Theorem C.O.l (Hartshorne) Let R be a local ring with maximal ideal m. Assume that Spec(R) \ {m) is disconnected in the Zariski topology. T h e n depth(R) 5 1.

Proof. Since Spec(R)\{m) is disconnected all the minimal prime ideals cannot be in one connected component. Let (0) = nr=,qi be the primary decomposition of (0) in R, and let = pi be the associated prime ideals of R. If m is an associated prime then depth R = 0; so we exclude this possibility. Let P I , .. . ,pi be in one connected component of Spec(R) \ {m), and let pt+l,. . . ,p, be in the other. Let I = nf=lqi, J = nn 3=t+lqj. Then IJ = (0). = m, CLAIM:Spec(R) = V(I) U V(J), V(I) n V ( J ) = {m). In fact, 2/for i t , j > t : For if Spec(R) \ {m) = V(Il) U V(12) with V(Il) nV(12) = 0, and with V(I1) a connected component, then d m = m. Moreover, say qi >Illthen q j > I 2 , a n d p i + p j > I1+I2, whencem.

<

Thus, m = d m .By general position argument, choose a E I \ Ui<tpj so that a b $! pk, for 1 k mi whence, is a non-zero-divisor. ~ e n c e , depth(R) 2 1.

+

< <

270

C Complete intersections and Connectedness

+

+

Let R = R/(a b). Then a # 0: If not, a = x(a b), for some x E R, whence x E qi for 1 i t. Hence xb E IJ = 0, and so (1 - x)a = xb E J (1 - x)a = u + a = 0, as 1- x is a unit in the local ring R.

+

We show that m = m/(a

< <

+ b) consists of zero-divisors.

m,

>

for every x E m there is least integer n 1 such that Since m = zn = p + q, p E I, q E J. Now axn = up = p(a b), and so a f n = 0. Since azn-1 # 0, f is a zero-divisor in R. Thus, a + b is a maximal R-sequence, whence depth(R) = 1. 0

+

Corollary C.0.2 (Hartshorne's Connectedness Theorem ) Let (R, m) be a Cohen-Macaulay local ring. Let V = V(I) be a set-theoretical complete intersection in Spec(R) of dimension > 1. Then V \ {m) is a connected space.

Proof. Since V is a set-theoretic complete intersection, R I I is a CohenMacaulay local ring of dimension > 1. Hence depth(R/I) > 1. By the previous 0 theorem Spec(R/I) \ {m) is connected. The natural map i : R -+ k is a flat map, i.e. if one has an exact sequence 0 -+ M' -+ M -+ M" -+ 0 of R-modules, then the sequence 0 -+ M' @R R -+ MI' @R k -+ 0 is also exact. From this one can easily see that if a E R is a non-zero-divisor then its image 2 = i(a) is a non-zero-divisor in g. More generally, if I is an ideal of R, then one has ( T I ) .v Ell? Using this, one can show that if 21,. . . , a TE R is a regular sequence, then 4,.. . ,ZT is a regular sequence in R. The map i : R + k is, in fact faithfully flat, i.e. if 0 -+ M' -+ M -+ MI' -+ 0 the ind_uced sequence of is a sequence of finitely generated-R-mod~les~with 0i exact, then modules (and maps) 0 -+ M' @R R -+ M @R R -+ MI' @, R the first sequence is also exact. Using this, one can show that if R / I is a Cohen-Macaulay local ring then R / I .v R / I is also a Cohen-Macaulay ring. A

*

A,.

The proof of Corollary C.0.2 shows us that: Corollary C.0.3 (Formal connectedness ) ~f = sPec(k/F) then \ ($1 is also connected. ( W e say V \ {m) is formally connected.)

Proof. ( E l l ) E z/l^is Cohen-Macaulay of dimension R I I = depth(R/I) Apply Hartshorne's Connectendess theorem to deduce the result.

> 1. 0

Odds and Ends

Here is a 'local-global' property which can be proved using direct sums: Proposition D.O.l Let U, V be submodules of an A-module M .

a) For every multzplicatzve set S c A we have (S-I U) f l (S-I V) = S-I (U fl V) and S-I U + SP1V = SP1(U V) .

+

b) U = V

Urn = V, for all m E Spmax(A).

Proof. a) We have a commutative diagram with exact rows

Chapter 2 Let A be a principal ring, F a finitely generated free A-module and U c F a submodule. Remember that F possesses a basis XI,. . . ,x, such that U = Aalxl . . . Aanxn with ai E A and ailai+1 for i = 1,. . . , n - 1. Clearly U is free itself with basis a l x l , . . . ,a,x, where T is maximal with a, # 0.

+ +

Let A, F , U be as above and assume a1 # 0. Further let cp : F -+ A be an epimorphism with cp(U) = (al). Then every section (i.e. right inverse)

272

D Odds and Ends

a' : (al) -+ U of cp' := cplv : U + (al) extends uniquely to a section a : A -+ F of cp. If B is any domain, b E B \ (0) and x E bBn. Then there is exactly one y E Bn with by = x. (If there were another one, say y', we would have b(y - 9') = 0, which clearly cannot happen.) We denote this y by btlx.

Since al(al) E U c a l F , we may define a by a(1) := a ~ l d ( a 1 ) .And there is no other possibility to define a o which extends a', since a'(a1) = a(al) = ala(l). Now we have to show p a = ida. But alcpoa(1) = cpoa(a1) = 0 cp'oal(al) = a l , hence cpoa(1) = 1.

Chapter 4

+

Lemma D.0.2 Let A be a commutative ring and let s, t E A such that As At = A. Let M, M' be two A-modules such that a: : M, 7ML and /3 : Mt 7 Ml with at = p,. Then M M'.

Proof. Take any m E M. Then a t ( m / l ) = Ps(m/l). This gives a(m/l)t = /3(m/l),. If a ( m / l ) = x E ML and P(m/l) = y E Mi, then xt = y,. Since s and t are comaximal by the usual componendo and dividend0 trick we can find an element m' E M' such that m', = x and mi = y. Since sA tA = A the element m' is uniquely determined. Define 8 : M -+ M' by 6(m) = m'. Then .9 is A-linear and since 6, = a, 6t = P, 6 is locally an isomorphism. By 0 (1.3.16) 6 is an isomorphism.

+

Remark D.0.3 In view of Corollary 4.4.8, any unimodular vector v E Umr(k[xl, ,x,]), r 3, can be completed to an elementary matrix & E ET(k[xi,... ,%I). Consequently, if a E SLr(k[xl, - ,xn]), then using above and the normality of Er(k[xl,. . . ,xn]) one has E E ET(k[x1,.. . ,xn]) S U C ~that m = = I 6) for some 6 E SL2(k[~1,.. . ,xn]). Thus, SLr(k[xl,. . . SL2(k[xl, . - . ,xn])Er(k[xl,- . . ,xn]) for r 3, for all n0

.

.

>

>

In [93] A. Suslin established that SLr(k[xl, - .- ,xn]) = Er(k[xl,. . . ,xn]) for r 2 3, for all n. This has been posed as Exercise 15 to Appendix A. We have indicated a proof of this due to M. P. Murthy with Exercises 12-14 acting as a hint. A. Suslin's proves it by establishing a Monic Inversion Principle for the Elementary group Er(R[X]), for r 3, for any commutative ring R.

>

We give a quick bibliographic survey of the known applications of the L-G Principle in the literature, and some other directions which we feel are related.

* Extending [Kol[21 -

to linear and classical groups (A.Suslin, et. al. [93], [Su][lO],

D Odds and Ends

273

* Rigidity interpretation for Principle G-bundles (MS. Raghunathan [Ra][l]). * Constructive approach to Quillen-Suslin theorems ( Bose [BU[l], Laubenbacher [La][1], Fitchas [F][I], Sturmfels [St][l], etc.).

* Projective

generation of ideals in algebras, commutative rings (M. Boratfnski [Bt][l-lo],N. Mohan Kumar [MK][2, 4, g], S.M. Bhatwadekar, Raja Sridharan [Bh][7-101).

* Locally polynomial algebras (Bass-Connel-Wright [B]121, A. Suslin [Su][2]). * Normality of elementary subgroups, sandwich theorem for normal subgroups (L.N. Vaserstein [Val [6]).

* Solvability of the quotient by the elementary subgroup (A. Bak [Bk][l]),(cf. Appendix A, Exercise 4).

* Stabilization and Prestabilization for linear groups (van der Kallen [K][l-21, R.A. Rao [RR][8]).

*

Determining whether an element in a polynomial ring is a "variable" (A.Sathaye [Sa][2,3],T.Asanuma [A][l-31))S.M. Bhatwadekar [Bh][13-15, 211.

* Efficient generation upto radicals of ideals in polynomial rings (M. Boratfnski [Bt][l], N. Mohan Kumar [MK][7], S. Mandal [Ma][9]), (cf. Chapter 10, Exercise 16-19). * Lifting generators of the conormal module (S. Mandal [Ma,2, 31, S.M. Bhatwadekar & M. Keshari [Bt][lO]).

* Euler classes and complete intersections (S. Mandal [Ma][l, 21, S.M. Bhatwadekar [Bt][4], Raja Sridharan [RS]15, 71, M. K. Das [Dl121).

* Orbits of

linear and elementary subgroups (L. N. Vaserstein [Va][6], N. Mohan Kumar [MK 11, M. Roitman [Rt][l, 21, J. Gubeladze [GI[l,3, 41, R.A. Rao [RR][2]).

* Cancellation

technique for projective modules, monoid rings (A. Suslin [Su][6-91, R. A. Rao 1721, [RR][l], H. Lindel [Li][2],J. Gubeladze [G][6],R.G. Swan [Sw][4], S.M. Bhatwadekar [Bh][16, 17, 191).

*

Bass-Quillen-Suslin conjectures (M. P. Murthy [MP][11], H. Lindel [Li][1,3,4], N. Mohan Kumar [MK][6, 101, A. Suslin [Su][6, 71. M. Roitman [Rt][2], S.M. Bhatwadekar [Bh]116-201, D. Popescu [PI[2], R.G. Swan [Sw][2, 5, 61, R. A. Rao [RR][3, 4, 111).

* Defining new groups related to unimodular vectors, and establishing group structures on orbits of unimodular vectors (R.A. Rao & Selby Jose, in preparation).

* Establishing

Quillen-Suslin theory for Euler classes, other groups (M. K. Das [D][l], Selby Jose & R. A. Rao, preprint).

274

D Odds and Ends

Chapter 5

< be a soft vector bundle over X and x E X . there is a global section s of < with s ( x ) = a.

Lemma D.0.4 Let

a) For any a

E

F,

b) There are an open neighbourhood U of x and global sections s l , . .. ,sn such that s l ( y ) ,. . . ,sn(y) is a basis of Fy(<)for every y E U . (i.e. sllu,. . . , snlu is a basis over U . )

<

Proof. a) Let U l , . . . , Urn be a finite open covering of X , which trivializes and e l , . . . ,em a subordinate envelope of unity. There is an j with e j ( x ) = 1. Over U j there exist a section s' with si(x) = a. Then define a global section s : X + E ( [ ) by s ( y ) :=

for y

#

Uj

"

Since Supp(ej) c U j, this is continuous. Clearly s ( x ) = s l ( x ) = a. b) Let Ui, U j ,ei, ej be as above, and let s i , . . . ,sk be a base over U j . (i.e. s: ( y ) ,. . . ,& ( y ) is a base of Fy (<) for every y E Uj. Remember that
<

Corollary D.0.5 Let f , g : + q be homomorphisms of soft vector bundles over X with r(f)= r ( g ) . Then f = g.

Proof. Let a E E ( < ) ,say a E F,(<). B y Lemma D.0.4 there is a global section s with s ( x ) = a. Since by assumption r ( f ) ( s ) = T ( g ) ( s )we see f ( a ) = 0 r ( f) ( s ) ( x )= r ( g ) ( ~ ) ( =x )d a ) .

<

Lemma D.0.6 Let be a soft vector bundle over X and x E X , s E r(t) with s ( x ) = 0. Then there are so,. .. ,sn E r(<), ao,. . . ,an E C ( X ) with a i ( z ) = 0 and s = Cy=o aisi.

Proof. Let n = ( U j ) l L j 5 mbe a finite open covering of X and ( e j ) l l j l n be a subordinate envelope of unity such that
<

Let s l , . . . ,sn E r(<) be a basis of over Uj,. There are bi E C(Ujo)with s ( y ) = CZl bi(y)si(y)for all y E Ujo.Then bi(x) = 0 for i = 1,.. . ,n. Set

a i ( y ) :=

for y

# Ujo.

D Odds and Ends

275

The ai are continuous, since Supp(ejo)C Ujo.Set so := s - Cy=,aisi. Then so(y) = 0 for all y E X where e j o ( y )2 112. Therefore so = (goejo)- S O , since goejo(y)= 1 for all y E X where ejo(y)5 112. On the other hand goej(x) = 0. So finally s = CEO aisi with ai(x) = 0, if one defines a0 := goej,. 0

Corollary D.0.7 For x E X let I, := { a E C ( X ) I a ( x ) = 0). (It is a twosided ideal.) Then s I-, s ( ~ induces ) an isomorphism r ( E ) / I J ( E ) + F,(E). 0 Theorem D.0.8 Let E, q be soft vector bundles over X . To every C(X)-linear map F : r(c)+ r ( q ) there exists a unique bundle homomorphism f : 5 + q with r(f ) = F .

Proof. The uniqueness has been proved in Corollary D.0.5. The homomorphism F : r(5)+ r ( q ) induces for every x E X a map f , of fibres

These fit together to a map of the total spaces f : E(<) + E ( q ) over X , whose continuity will be proven later. Note first that r(f)would be equal to F . Namely by the definition of f , we have for s E r ( ( ) :

Now to the continuity of f . Let U c X be open such that
D.0.9 Now we will show that to every finitely generated projective C ( X ) module P there is a soft vector bundle 5 over X such that F ( t ) E P. P is a direct summand of some C ( X ) " , so it is the image of a (not necessarily 0 unique) idempotent endomorphism a of C ( X ) n . Lemma D.O.10 The map a

H

,/Z is continuous.

Proof. Let (Pk) be a sequence of semi-positive hermitian matrices with ,Bn = a. Then the sequence is bounded since Ifil = limk,, Let y be an accumulation point of the sequence (z/P;)and be a subsequence converging to y. Then y is also semi-positive hermitian and y2 = a. Therefore y is the only accumulation point of the bounded sequence 0 whence limk,, f i = 6.

(a

(am. (a),

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D Odds and Ends

Therefore, if an endomorphism a of C(X)n is semi-positive hermitian, i.e. a(x) is so for every x E X , then there exists a semi-positive hermitian endomorphism y of C(X)" with y2 = a.

Proposition D.O.11 Every finitely generated projective module P over C(X) is isomorphic to the image of a hermitian idempotent endomorphism E of C(X)n for some n. (That E is hermitian idempotent means E = E~ = E*.) From a2= a we derive (a*)2 = a* and

Since (1 - 2a)(l- 2a*) is positive hermitian, there is an y with y2 = I 2a)(I - 2a)*). As y2 is invertible, also y is so.

+(I -

We set E := y-lay. Then s2 = y-'a2y = y-lay And from y2a* = ay2 we infer E* = ya*y-l = y-lay = E. If s = 1, the A-module A I I has the composition series

The factors of this composition series all are isomorphic to RIP. By Proposition 8.4.7 b) the statement is true in this case. Now let the statement be true for some s < r and prove it for the case s. The A-module A I I has the compositon series O c ( X . . . X X k.+l--1 + ,Xs+2,. . . ,X,)/I c - -. c (x:', . . . , X ~ S , X , +.~ . .,,X,)/I c AII. All factors are isomorphic to E := A/(@, .. . ,x?, X,+',. . . ,X,). By induction assumption Ass(E) = p. Again use Proposition 8.4.7. 0

D.0.12 Construction of Vector Bundles We shall assume that the reader has some familiarity with the notion of a smooth manifold. We show how vector bundles arise in practice. The material here is standard, but we liked its presentation in [65], which we follow in this appendix. Smooth vector bundles formalize the notion of "smooth family of vector spaces". For example given a smooth manifold M and a vector space F we can think of the Cartesian product FM := F x M as a smooth family (Fx)xEMof vector spaces. This trivial example is not surprisingly called the trivial vector bundle with fibre F and base M. We can obtain more interesting examples by gluing these simple ones using gluing data. These consist of (i) an open cover (U,) of a smooth manifold M.

D Odds and Ends

B. a gluing cocycle i.e. a collection of smooth maps gp, : U,p (where U,p = U, n Up), such that

+ Aut

277

(F),

The open cover U, is also know as a trivializing cover. We will also say it is the support of the gp,. The map gp, describes the "transition from F, := Fuato Fup= Fp" in the sense that for every x E U,p the element (v,x) E F, is identified with the element (gp,(x)v, x) E Fp. Pasting together the trivial bundles F, following the instructions given by the gluing cocycle we obtain a smooth manifold E (called the total space), a smooth map 7r : E + M (called the canonical projection) and diffeomorphisms $, : 7r-I (U,) + Fa,(called local trivializations) such that for all x E U,p, v E V

E 4 M as above is called a vector bundle over M. The rank of E is by definition the dimension of the standard fibre F (over its field of scalars). Rank one bundles are of course known as line bundles. Example 1. Consider the projective space CPn defined as the set of onedimensional complex subspaces of P+'. There is natural projection

where ~ ( x := ) one-dimensional subspaces spanned by x. The fibres

are vector subspaces of P+l. The family x-lCp) is indeed a smooth family of vector spaces in the sense described above. It is called the tautological (or universal) line bundle over the projective space and is denoted by UJn. f Y is a smooth map and E + Y is a smooth vector bundle Suppose that X + given by a gluing cocyle gp, supported by an open cover (U,) of Y. Then f induces a vector bundle on X called the pull-back of E by f and denoted by f * E. It is given by the open cover (V, = f -l(U,)) and gluing cocycle hpa = g~cuof.

The following (exercise) describes a very general procedure of constructing smooth vector bundles. Consider a smooth map P from a compact connected smooth manifold X to the space End (V) of endomorphisms of a vector space V such that P2(2) = P(x) V x E X, i.e. P(x) is a smooth family of projectors of V. (a) Show that dim ker P(x) is independent of x E X. Denote by k this common dimension.

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D Odds and Ends

(b) Show that the assignment x H ker P(x) defines a rank k smooth vector bundle over X. (c) Provide a projective description of the tautological line bundle over Wn. (d) Show that any map X + V*\{O) defines in a canonical way a vector bundle over X of rank dim V - 1. Denote by Gk(V) the Grassmannian of k-dimensional subspaces of an ndimensional vector space V. Assume V is equipped with an inner product. For each k-dimensional subspace U C V denote by Pu the orthogonal pro jection onto U L . The smooth family Gk(V)3U H Pu defines according to the previous construction a rank k vector bundle over Gk(V) called the universal vector bundle and denoted by Uk,,. When k = 1 this is precisely the tautological line bundle over R3Pn-' or CPn-l . A smooth map s from a smooth manifold X to a vector space F is a smooth selection of an element s(x) in each fibre F x x of Fx. In other words, it is smooth map s : X + Fx such that n o s = lx w here n : Fx -+ X is the natural projection. Replacing Fx with any smooth vector bundle E 4 X we get the notion of smooth section of E. The space of smooth sections of E will be denoted by F(E) or CW(E). In terms of gluing cocycles we can describe a section as a collection of smooth maps

The functorial operations in linear algebra have a vector bundle counterpart. Suppose Ei ZI) X(i = 1,2) are two vector bundles over X with standard fibres Fi, i = 1,2, given by gluing cocycles Gp,(i) along the same support. For example the direct sum Fl $ F 2 corresponds to the direct (Whitney) sum El $ E2 given by the gluing cocycle Gp,(l) $ Gp,(2). The dual vector bundle Ef is defined by the gluing cocycle (G;,(l))-' "*" denotes the conjugate transpose.

where

We can form tensor products, symmetrix, exterior products of vector bundles, etc. In particular, the bundle E,' 63 E2 will be denoted by Hom (El, E,). Its sections are bundle morphisms, i.e. smooth maps T : El + E2 mapping the fibre El(x) of El linearly to the fibre Ez(x) of E2. When El = E2 = E we use the notation End(E). If the induced morphisms T(x) are all isomorphisms then T is called a bundle isomorphisms. A bundle automorphism of a vector bundle E is also called a gauge transformation. The group of bundle automorphisms of E is denoted by G(E) and is known as the gauge group of E. The line bundle A ' " " ~ ( ~ ~ ) E Iis called the determinant line bundle of El and is denoted by det El.

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279

In the exercises we will see how to use sections to prove that any complex line bundle over a compact manifold is the pullback of the universal line bundle over a complex projective space. For every smooth manifold M denote by Picw(M) the space of isomorphism classes of smooth complex line bundles over M and by [M, CPn], the set of (smooth) homotopy classes of smooth maps M + C?. This is an inductive family, [M, CP1], L, [M, CP2], L, . . . and we denote by [M,Cw], its inductive limit. If M is compact we have a bijection. Picm(M) [M, CPw],. The tensor product of line bundles induces a structure of Abelian group on Picw(M) This group has a cohomological interpretation (for the more advanced reader) viz.

=

Since the inductive limit CPw of the CPn 's is a K(Z, 2)-space we can conclude that we have isomorphism of groups

For any L E Picm(M) the element c?(L) C h e r n class of L.

is called the topological first

One is often led to study families of vector spaces satisfying additional properties such as vector spaces in which vectors have lengths and pairs of vectors have definite angles (as in Euclidean geometry). According to Felix Klein's philosophy, this is the same as looking at the symmetry group, i.e. the subgroup of linear maps which preserve these additional features. In the above case this is precisely the orthogonal group. If we want to deal with families of such spaces then we must impose restrictions on the gluing maps; they must be valued in the given symmetry group. Here is one way to formalize this discussion. Suppose we are given the following data. A Lie group G and a representation p : G + End (F). A smooth manifold X and open cover U,. 0

A G-valued gluing cocycle, i.e. a collection of smooth maps gp, : U,p such that g,,(x) = 1 E G V x E U, and

+ G,

Then the collection p(gp,) : U,p + End (F) defines a gluing cocycle for a vector bundle E with standard fibre F and symmetry group G. The vector bundle E is said to have a G-structure. Differential geometers usually phrase the above construction in terms of principal G-bundles. Given a gluing G-cocycle as above we can obtain a smooth

D Odds and Ends

280

manifold P as follows. Glue the product G x U, to G x Up along U,p using the following prescription: for each x E Uffpthe element ( g , x) in G x U, is identified with the element (g,p(x) .g.x) in G x Up. We obtain a smooth manifold P and a smooth map ?r : P + X whose fibres n-l(x) are diffeomorphic to the Lie group G. This is called the principal G-bundle determined by the gluing G-cocycle gp,. The above vector bundle E is said to be induced from P via the representation p and we write this as P x, F. 0

Chapter 6 Theorem D.0.13 (Finiteness of minimal prime over-ideals (cf. Corollary 6.2.6)

nkl

Let I be an idea2 of a Noetherian ring R , then J? = pi, for some prime ideals pi of R, 1 5 i n. Consequently, I has only a finite number of minimal prime over-ideals pl, .. . ,p,.

<

Proof. Let S be the set of all radical ideals which are not a finite intersection of prime ideals. Suppose S # 0. If I. E S is a maximal element, then I. is not a prime ideal and so there are a, b E R , with a # Io, b # Io, ab E lo.It is easy to verify that I. = But then, by the maximality of I. in S we get # S and # S , and so d(lo,b)are finite intersections of prime ideals. Hence I. too is such an intersection, which 0 contradicts that I. E S .

dm.

Corollary D.0.14 Let I be a n ideal of a Noetherian ring R and p be a prime ideal containing I then p contains a minimal prime over-ideal of I .

Proof. Since p is prime p > J? = nTZlpi. If no pi $ p and xi E pi \ p, 1 5 i 5 r , then x = 21x2 - . - x TE niZlpi \ p, a contradiction. Hence pi, c p for some io. 0

Chapter 7 Proposition D.0.15 Let P be a finitely generated projective A-module of rank r and s E A such that P, is free. For any p E P there is a q E P such that ht(ASOp& sq)) 2 r . I n particular, if the residue class ji of p is unimodular i n the A/s-module P / s P , then ht(Op(p + sq) 2 r.

>

+

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281

Proof. Let B be a ring and b = ( b ~ ,... , b,) E B T .Then the order ideal of b is generated by b l , . . . , b,. Let X I , . . . , x , E P such that ( X I ) , , . . . , (x,), form a base of P,. Then there is an n E N with snP E A x I + - , - + A x , . Let snP = alxl+...+a,x,, ai E A. By the usual prime avoiding procedure we find e l , . . . , c, E A, such that the ideal I := A,(aI sn+'cl) . . . A,(a, sn+lc,) of A, has height r. Set q := C cixi. Then A S 0 p ( p sq) = A,Op(snp sn+'q) = I.

+

+ + +

+

+

>

To show the 2nd statement, we use that P is projective. Since jj and hence

p is unimodular, there is a homomorphism P / s P -+ AIsA with f i e 1. By projectivity this lifts to homomorphism P + A with p + sq ++ 1 + as for some a E A. So no prime ideal containing s contains O p ( p+ sq), whence ht AsOp(p 89) = ht OP(P sq).

+

+

A corollary of this is a weak form of Serre's Splitting Theorem: If A, P are as above, rank P > dim A, then P has a unimodular element. Proof. Induction on dim A, the case dim A = 0 being clear. We may assume that A is reduced with connected spectrum. Then S-'P is free, if S is the set of all non-zero-divisors of A. This implies, P, is free for some non-zerodivisor s. By induction hypothesis there is a unimodular p in P/sP. Then apply Proposition D.0.15 to a representant p of p.

To see this we need to show that ( a )n ( a I ,ba, . . .,b,) = aI: This is easily checked by using the fact that 1/12has {a,&, . . . ,b,) as a basis. By Proposition 8.3.7 we have hdRl(,)I/(a)< m. Recall the definition of the norm

Let [K : Q(R)]= n and 0 be an algebraic closed field containing K. There are exactly n different field embeddings ai : K v f2 which leaves Q ( R )fixed. And the norm of cr E K is defined by N ( a ) := ai(a).One easily sees by Galois Theory that indeed N ( a ) E Q ( R ) ,and trivially one has N(a/3) = N(cr)N(/3). Further N(cr) = O u a = O andcr E A X N(a)E R X .

nyz1

Then a := N(cr) E I n R and # ( A / a A ) = # ( R n / a R n ) < m. Therefore also # ( A / I ) < m. To show Theorem 8.9.1 we prove some lemmas.

Lemma D.0.16 Let I be a non-zero ideal of a domain A of dimension 1 and P a (finitely generated) projective A-module of rank n. Then P / I P % as A-modules. In particular, if J # (0) is an invertible ideal of A, then J / I J E A / I as A-modules.

Consequently in this case # ( A / I J ) = # ( A / I ) # ( A / J ) ,2.e. IlIJll = IlIIl. IlJIl.

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Proof. Since I # (0) and A is a domain of dimension 1, there are only finitely many maximal ideals, say nl, . . . ,n, containing I . Set S := A \ Uz"_,ni. Then

since the elements of S act bijectively on PIIP. Thus we may assume A = S-'A, in which case A is semilocal, hence P free. Then the claim is trivial.

Chapter 8 Let A be a Dedekind ring, F a finitely generated free module and U C F a submodule. Assume that there is a c E A \ (0) with c F c U. Show that there are a basis 21,. . . ,x, of F and a decreasing series Il > . . - > I, of ideals, such that U = Ilxl . - . I,x,. (Consider all epimorphisms cp : F -+A and let cpl be one such that cpl(U) =: Il is maximal under all cp(U). First show that U c IIF. One can identify F with An, such that cp = pr,. Namely every stably free module over a Dedekind ring is free. If U @ IlF then pri(U) @ I1 for some i > 1, say i = 2. If not, let a : I, -+ U be a section (right inverse) of the epimorphism cplU : U t Il. The existence of a section follows from the projectivity of Il. We will extend this section to a section a' : A -+ F of cp. Let S := A\ (0) and consider F as a submodule of SP1Fin the canonical way. Choose some a E Il - (0) and define a' : A -+ S-'F by a'(1) := a-'a(a). Once we have shown a l ( l ) E F, we see immediately that c' : A -+ F is a section of 9.

+ +

Proposition D.0.17 (cf. Proposition 8.8.7) Let P, Q be rank 1 projective modules over a I-dimensional noetherian ring R. Then

"

L @ R for some projective Proof. By Serre's Splitting Theorem P @ Q module L of rank 1. Further P @ Q E A 2 ( p @ &) N (L @ R) L.

Proof. P and Q ase isomorphic to invcrtible ideals I, resp. J of R, contained in R.

CLAIM.There is an ideal I'

EI

with I'

+ J = R.

Namely since dim(R) = 1 and J contains a non-zero-divisor, there are only finitely many prime ideals containing J. Let P I , . . . ,p, be the prime ideals which either contain J or belong to Ass(R) and S := R \ l&pi. Then Spl(Ipl) is a principal fractional ideal of S-'R. Let z E I-' be a generator of Spl(I-I). Then I' := z I c R and S-'I' = SplR. Therefore I' is not

D Odds and Ends

contained in any of the prime ideals containing J , whence I' write I instead of I'.

283

+ J = R. Now

We have an exact sequence

(Justify!) which clearly splits. By part of the Chinese Remainder Theorem [email protected](P@Q)@R~P@Q. We now continue where we left off in Chapter 8, 58.3, by proving a famous formula of Auslander and Buchsbaum relating the depth and homological dimension as codimensions. We begin with the notion of a minimal free resolution of a module M over a local ring (R, m). A free resolution

is called a minimal free resolution of M if di : Li + ker(di_l) is minimal, for each i 2 0, i.e. di @ Rlm is an isomorphism, or equivalently, di is onto and ker(di) mLi, for all i .

c

Lemma 1. Let M be a finitely generated module over a local ring (R, m). Let 0 -+ K -+ F -+ M -+ 0 be an exact sequence with F a finitely generated free R-module.

I f depth(R)

> 0, depth(M) = 0, then depth(K) = 0.

Proof. Let x E m be a non-zero-divisor of R. Tensoring with the right exact functor @RR/(x)we get an exact sequence

Since depth(M) = 0, m E Ass(M), equivalently there is a m E M that m - m = 0.

\ {0), such

Clearly, m E M', whence m E Ass(M1). But then m E Ass(K/xK), whence depth(K1xK) = 0. Since x is not a zero-divisor of K (Why?) depth(K) = 1.

Theorem D.0.18 (Auslander-Buchsbaum [2]) Let M be a finitely generated module over a Noetherian local ring (R, m). If hd(M) < oo then hd(M) depth(M) = depth(A)

+

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Proof. Let n = hd(M), and

be a minimal free resolution. Let Ki = Irn(d+l), for i = 1,.. . ,n. We argue by induction on depth(R). If depth(R) = 0 there is a non-zero x E R with m . x = 0. If n > 0, then L, E mLn-1 implies that xLn g ~ r n L , - ~= (0), so = (0), a contradiction. Therefore, n = 0, and M is free, so depth(M) = depth(R) = 0. Now let depth(R) > 0. If depth(M) > 0 there is x E m that is not a zero-divisor of R or M by Prime Avoidance as m 2 Up, for p in the finite set Ass(M) UAss(R). Therefore, depth(R/(x)) = depth(R) - 1, and depth(M/xM) = depth(M) - 1. By Proposition 8.3.7 hdRl(z)(M/xM) = hdR(M). The Auslander Buchsbaum formula is established by induction in this case. Let depth(R) > 0. Then by Lemma 1 depth(K1) = 0. But hd(K1) = hd(M) - 1. Since hd(K1) depth(K1) = depth(R), by above, it follows that hd(M) = depth(R), as required. Induction establishes the Auslander Buchs0 baum formula.

+

Chapter 9 Let us give an interesting example, which shows that one can expect a certain class of ideals to have a bounded number of generators - namely the class of "local complete intersection ideals", i.e. those proper ideals I for which Ipis generated by a regular sequence for all prime ideals p > I. Proposition D.0.19 Let R be a noetherian r i n g of dimension d and I a local complete intersection ideal. Then ,u(I/12) d. In particular p(I) 5 d + 1.

<

Proof. Let p E V(I). Clearly pp(I/12) 5 ,up(I). Since I is a local complete intersection ideal, we have

,up(I) = W p )

5 ht(pRp) = ht(p).

Therefore, for the Forster-Swan bound

(Alternatively one can show that xk(1/12)= 0, if k > ht(I) and derive this.) By the Forster-Swan Theorem we get p(I/12) 5 dim(R), hence by (10.2.1)

D Odds and Ends

285

Chapter 10 Examples: Set-Theoretical C o m p l e t e Intersection C u r v e s

L e m m a D.0.21 We have the following equality of ideals in A: ( f l ,f 2 , f 3 )

n (x:",x;12)

= ( f l ,f 2 , ~ y ~ ~ f 3 , x ;=~(~f if ,f32 ))

>

>

Proof. First equality: Since a1 a21, a2 a12 we have f i , f 2 E So ( f l , f 2 , f 3 ) n (x:"',~;'" 3 ( f l , f 2 , ~ y ~ ~ f 3 , x ; ~ ~ f 3 )

03.3) (2:"

,x;12)

Let f = c:=, figi E ( f i ,f 2 , f 3 ) n (x?'l, x;12). Since f i , f 2 E (xy2l,x;12), we have f3g3 E (xY21,x;12). It is an exercise to show that A s s ( A / ( x y l ,a;'") = { ( x l , x 2 ) } . SO, since f 3 $! ( x I , x ~we ) see g3 E ( X ~ ~ ~ , X ;So ~ ~g3) f .3 E ( x Y 1 f 3 , x ; 1 2 f 3whence ) f E ( f ~f 2, , xY2lf3,x;'V3). Second equation: '3'is trivial. Using Equations (D.2) we see:

xF3'

fl

+ x3a13 + xyZ1 f2

f3

=

Analogously we compute

By these equations xT21f3 and x;12 f3 both belong to ( f l , f 2 ) , which establishes 0 the second equation.

Appendix B L e m m a 2. The square

Y

f'

b M

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is a pull-back diagram, then the sequence

is exact. Proof. Note that (f, -g) o { f', g') = f f' - gg' = 0. Hence only to show that ker{f,g) = 0. Since Y = {(m,n) E M x N : f(m) = g(n)), we have 0 (0,O) = {fl,g'Hm, n) = (fl(m, n)7g1(m7n)) = (m7n). (i) Note that ker f' = ((0, n) I n E N and g(b) = 0). So we can define a map 9 : kerf' + ker g by ~ ( 0n), = n. Clearly cp is an isomorphism. (ii) We have the exact sequence

Suppose m E M. Since g is epimorphic there exists n E N such that f (m) = g(n). Hence (m,n) E ker((f, -g)) = im({ f',gl) by exactness. Thus there exists y E Y with m = fl(y) (and n = g(y)). Hence f' is epimorphic. (i) Note that kerf' = ((0, n) I n E N and g(b) = 0). So we can define a map cp : kerf' + kerg by cp(0, n) = n. Clearly cp is an isomorphism. (ii) We have the exact sequence

Suppose m E M. Since g is epimorphic there exists n E N such that f(m) = g(n). Hence (m,n) E ker((f,-g)) = im({fl,g') by exactness. Thus there exists y E Y with m = fl(y) (and n = g(y)). Hence f ' is epimorphic.

Definition D.0.22 Let I be a finitely generated ideal of a commutative ring A with In= (0). The I-Boer submodule of AN consists of those ele-

,n,,

ments whech, considered as sequences, converge to 0 with respect to the I-adic topology.

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287

Lemma D.0.23 Let A, I be as in the Definition and H be the I-Baer subIH = H . module of A". Then A(") c H and A(")

+

Proof. The first claim being trivial, let a = (u,),,~ E H and let N be the and a; = a, maximum of the n with a, f I . Then a' E A(") if a' = (a;),," for n 5 N , a; = 0 for n > N. We have to show that a - a' E IH and know a, - a; E I. Let I be generated by x i , . . . ,x,. Then every element of IW1 can be written as a sum xlbl+- - .+x,b, with bj E Then write a,-a; = xlbln+.- -+x,b,, in this manner. Then the sequences b l := . . ,b, = (brn)nEN)belong 0 to H , whence a - a' = XI bl - .- x,b, E IH.

+ +

Theorem D.0.24 Let A be a ring, possessing two finitely generated ideals 11,4with I1 I2= A and Ijn = (0) for both j . Then A" and A(") are rejlexive A-modules.

+

,n,

Proof. For j = 1 , 2 let Hj be the Ij-Baer submodule of A".

+

We have HI H2 = A". Namely let (a,),," E A". Then, since I: +I?= A, b.: clearly (b,), E for every n there are b; E I?, b: E I; with a, = b; Hi, (b:) E H2. Now set Hi := H~/A("). Then Hi

+

+ Hi = A ~ / A ( " ) .

Claim: H O ~ ~ ( A " / A ( " ) ,A) = 0 Proof of this: Let f : A"/A(") + A be a homomorphism. Then f (Hi) = f (IjHi) = Ijf (Hi) by Lemma D.0.23. So also f (Hi) = Ijn f (Hj) c Ijn for every n. Therefore f ( H i ) = 0. But then im(f) = f (Hi + H i ) = f (Hi) + f (H;) = 0. This means that every homomorphism f : A" -+ A is already given by its restriction to A("). i.e. for f there is a sequence (b,), E A" with f ((a,),) = a,b,. But this defines a hononorphism, i.e. the homomorphism A(") + A, given by (b,), extends to A", if and only if b, = 0 for almost all n.

En

If A is a complete discrete valuation ring (see later), which is principal but not countable, then A" is a free A-module. 0

Exercises

Chapter 1 1. Let k be the field of 2 elements and A := k[X, Y]/ (X2,XY, Y2). Call x, y the residue classes of X , resp. Y and set I := (y). Then x+ I = {a, x+y) c (x) U (x + y), but neither I c (x) nor I C (x 9). Hence the assumption that the pi are prime ideals in Lemma 1.1.8 a) cannot be weakened to the extent that at most two of them need not be prime. We do not know, whether part a) of this lemma holds true under the assumption that at most one of the pi is not prime.

+

2. Prove the theorem in Remark 1.1.9 d). First reduce to the case U = V by intersecting the xi Ui with U. (If U n (xi Ui) # 0 one may assume X i E U.) Then assume, there is a counterexample. Take one with minimal r . Then xi Ui @ U;=,(xi Ui). If now a E (xi Ul) \ U;='=,(xi Ui) there is a line a Kv with v E V \ (0) through a not contained in $1 Ul. It intersects every xi Ui in a t most one point and so is not contained in the union of the xi Ui.

+

+

+

+

+

+

+

+

+

+

3. (M. Henriksen) A commutative ring R has no maximal ideal iff (i) it has no ideal I with R / I a field, (ii) R2 p R = R for every prime number p.

+

4. (Nagata) Let R be an integral domain. Let b # 0,and a be a non-zerodivisor modulo (b). If ( b , a) # R then show that (a + bX) is a prime ideal of R[X].

5. Let A be a commutative ring with 1. Let 4 I 2 = A. Then, as A-modules,

+

11, I 2

be two ideals of A with

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6. Let A be a reduced ring with only finitely many minimal prime ideals pl, . . . ,p, and set S := A \ Uy=, pi. Show S-lA E fly=, Q(A/pi).

7. If p, q are different prime ideals disjoint to a multiplicative set S then S-lp # S-lq. This does not hold for general ideals: If S := Z \ (2), then SP1(2) = SP1(6) in SPIZ. 8. Solve the equation

CHINA. CHINA = * * * * *CHINA, where the letters represent digits in the decimal system and the stars may be replaced by arbitrary digits. (Find all solutions!)

9. Let a , b, c, d E $. Show that the following system of linear equations always is solvable:

(Use a determinant and the fact that Z is integrally closed.) 10. Let f (X) E R[X] have leading coefficient a 4 p, for some p E Spec(R). Prove that there are only finitely many prime ideals P E Spec(R[X])such that f (X) E P and P n R = p. 11. Let I be an ideal of a polynomial ring containing a monic polynomial. Let J be an ideal of A such that I J[X] = A[X]. Then show that (InA)+J=A.

+

12. Let M be a R-module, N a submodule of M and I an ideal of R. If x, y E I and N :N (x) = N = N :N (y) then show that

13. Let C be the ring of continuous real valued functions on the unit interval lo, 11. a) Show that for every x E [O,1] the set of f E C with f (x) = 0 is a maximal ideal of C. (Here [0,1] may be replaced by any topological space.) (See e).) b) Show that the set of the f E C with f ([0, s]) = (0) for some s > 0 is an ideal but not a prime ideal of C.

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291

c) Show that there are non-maximal prime ideals in C. (Use Lemma 1.3.3. We do not know whether one can describe any non-maximal prime ideal explicitly in any sense.) d) One may also show the existence of an infinite strictly descending chain (sequence) of prime ideals

in C. Namely consider the following sequence of functions (fk), defined by

Now let po be the maximal ideal attached to the point 0 E [O,1] as in a). Now assume you have already found pk. Then define S to be the multiplicative set generated (multiplicatively) by C \ pk and f k and Cfi. Show I n S = 0 and apply Lemma 1.3.3 to find let I := Pk+l-

xi,,

e) Show that for every maximal ideal m of C there exists an x E [ O , l ] with m = { f E C I f (x) = 0). (Otherwise for every x E [O,1] there where an f E m with f (x) # 0. Since the set O(f) := {y E [O,1] I f (y) # 0) is open in [O,1] and [0,1] is compact, there would exist finitely many f l , . . . ,f n in m with O(f1) n - . .n O(fn) = 0. But then f; ... f: would belong to m but also be a unit in C. Here [O,1] may be replaced by any compact space.)

+ +

14. Let M be a left A-module. a) If x E A is a zero divisor of M , the ax is so for every a E A. b) If u E A operates bijectively (resp. injectively) on M and b E A nilpotently (i.e. there is an n E IN with bnx = 0 for all x E M ) , then u b operates bijectively (resp. injectively) on M . (Assume: For every x E M there is an n with bnx = 0. Does the above assertion hold also under this weaker condition?)

+

c) Let additionally A be commutative. Show that the set of zero divisors of M is the union of (suitable) prime ideals. (The non-zero-divisors make up a multiplicative set. If a is a zero divisor of M , say ax = 0, then the whole ideal Ann(%)consists of zero divisors of M. Use Lemma 1.3.3.) d) Give an example that not every multiplicative set is the complement of a union of prime ideals. e) Give examples which show that the set of zero divisors is not always an ideal.

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15. a) Show that the ideals I := (0) x Z and J := Z x (0) of the ring R := Z x Z are not isomorphic as R-modules. Show that also R I I 9 R I J as R-modules. (In spite of this, the latter are isomorphic as rings.) (Clearly Z can be replaced by any non-zero ring.) b) Let I,J be different ideals of a commutative ring R (or, more generally, two-sided ideals of any ring). Show that R I I and R I J are not isomorphic as R-modules. c) In the noncommutative ring R = M,(K) where n > 1, there are different left ideals I,J such that R I I and R I J are isomorphic Rmodules. Namely let for e.g. I (resp. J ) consist of the matrices all of whose columns but the first (resp. the second) vanish. 16. Infinite direct products may have surprising properties. a) Would you expect that a direct product of finite groups could have a factor group which is a non-zero $-vector space? Indeed this happens. Let P be the set of all prime numbers and let G := EP Z/(p). Show Z/(pY, i.e. consists of that the torsion subgroup of G is H := those families (a,),Ep, a, E Z/(p) with a, = 0 for almost all p E P. Then show that every n E Z \ (0) operates bijectively on GIH. This means that G I H is a $-vector space.

n

b) A consequence of a) is that G @z $ for every p.

# 0, whereas ( Z l b ) ) @n $ = 0

c) Show that a family (fi)icI of homomorphisms f i : M + Ni defines I Ni. This gives us a canoncanonically a homomorphism f : M + ical map cp : (niEI Ei) @A F + ni,--(Ei @A F) for A-modules Ei and F, which by b) is not always injective.

niG

d) The map cp in c) also need not be surjective. Indeed, let A be a non-zero ring and show that the canonical map ( A ~ ) (+~ () ~ ( ~ 1 ) ~ is not surjective. Then identify (A")(") with A " @ A ( ~ )and ( A ( ~ ) ) ~ with (A @ A("))". 17. Note that the above constructed G I H also is a ring A. Show that A has the following properties: a) An element of A is a unit if and only if an arbitrary representing element E G has only finitely many zero-components. Otherwise it is a zero-divisor. b) A has uncountably many zero-divisors, but no non-zero nilpotent element. c) AX has elements of every (finite or infinite) order.

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d) For every prime number p there is an injective group homomorphism of the additive group of the padic numbers into A X . For c) and d) one needs Dirichlet's theorem on prime numbers in arithmetic progressions or rather the weak form of it: For any integer m > 1 there are infinitely many prime numbers p l(modm). (This weak form can be proven much easier - without analytic tools - than the full Dirichlet's theorem.) Also one uses that any group (ZIP)' is cyclic, i.e. that for an arbitrary divisor m of p - 1 there is an element of order m in (ZIP) X .

=

a) An ordered abelian group is an abelian group G (here written additively) together with a structure as follows: there is given a subset P c G, a so called positivity domain, such that 1 . P u - P = G , 2.Pn-P={o), 3.P+PcP(-P:={-X~XE P), P + P : = { x + ~ I x , ~ E P ) . ) This defines a total ordering '5' on G by x the property 4. x y x z 5 y z.

<

+

+

5y

:

y - x E P, with

Conversely, given a total ordering on G with property 4., one may define a positivity domain P by P := {x E G I 0 5 x), which obeys I., 2., 3. Examples of ordered abelian groups are the subgroups of the additive group of IR. An ordered abelian group is isomorphic as an ordered group to a subgroup of IR if and only if it is Archimedean. This means that for every x, y E G with x > 0 there is an n E IN with nx > y. b) If G is an ordered abelian group, then by G U {oo) we denote G together with an extra element oo with oo 2 x and oo x = oo for every x E g U {oo).

+

Let K be a field. By a valuation of K one means a map v : K + G U {oo) obeying: 1. v(a) = oo e a = 0, 2. v(ab) = v(a) v(b), 3. v(a b) 2 Min(v(a), v(b)).

+

+

Show that the set 0, := {a E K I v(a) 2 0) is a local subring of K with maximal ideal m, = {a E K ( v(a) > 0). c) Show that 0, is a principal domain if and only if v ( K X ) ? Z as (ordered) groups. If you know already the definition of a Noetherian ring, show that this holds if and only if 0, is Noetherian. In this case v is called a discrete valuation. d) Let A be a domain and p E A a prime element, i.e. Ap is a non-zero prime ideal. Then every non-zero x E &(A) can be written in the

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E Exercises

form x = pnf with a , b E A, n E Z and p { ab. Here n is uniquely determined by x. The map Q(A) + Z U {oa) defined by v(0) = co, v(x) = n for x # 0 is a discrete valuation with 0, = AAp. e) A ring A is of the form 0, with a discrete valuation v, if and only if it is a local principal domain and not a field. These rings are called discrete valuation rings. 19. We will give an example of a discrete valuation ring A and a finite field extension &(A) c L such that the integral closure B of A in L is not finite over A, i.e. not a finitely generated A-module. (E. Artin, and independently 0. Zariski.) a) Let k be a field. Note that the quotient field of the formal power series ring k[[X]] can be understood as the ring of Laurent series with finite aiXi where principal part, i.e. as Laurent series of the form CEO=, n E Z is allowed to be negative. This field will be denoted by k((X)). b) Let K C L be a radicial (i.e. purely inseparable) field extension of degree p = char(K) (> 0). Show that for every a E K - L the minimal polynomial of a over K equals XP - cP. Consequently, if A c K is integrally closed and Q(A) = L, then B := { a E L I clip E A is the integral closure of A in L. c) Now we construct the announced example. Let k be a field of characteristic p > 0. The construction will take place within k((X)). Let := ciXi E k[[X]] be transcendental over k(X) . Such an 11 exists by a cardinality argument. (For simplicity you may assume that k is finite (or countable), hence k(X) and even its algebraic closure are countable. But k[[X]] is not countable.) Now, set Y := @ = cfXPi. Then define K := k(X, Y), L := k(X, 7) = K ( Y ~ / Pand ) A G ,%[[XI]n K . Show that Q(A) = K and A is a discrete valuation ring with prime element X . (Restrict the 'canonical' valuation of k((X)) to K.) Let C denote the integral closure of A in L. For every n E IN one can write Y in the form

Ci,,,

with a polynomial f, E k[X] of degree < n and Y, E k[[X]]. Then Yn E K , hence E A and further yl/p = X-,(V - f,) E L, hence in C.

+ + +

Finally let B := A A, - - - AqpP1, which is a subring of C with .. ., is a basis of L over K Q(B) = Q(C) = L. (Note that l , ~ as well as one of B over A.) If C were finite over A there would be a d E A with dC c B. Especially dy;lp E B. But Y;/P = X-n(q - f,) would imply Xnld for every n.

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295

Chapter 2 1. Let A be a Noetherian ring. Let I be an ideal of A such that I = 12.Then I is generated by an idempotent element e (i.e. e2 = e). 2. Let I be a finitely generated ideal of R. Prove that if 1/12is generated by r elements as an R/I-module then I is generated by (r 1) elements.

+

3. Let (R,m) be a local ring. Let x E m. Suppose that m is generated by L elements. When can you say that the least number of generators of the maximal ideal m/(x) of R/(x) is Ic - l?

4. (N. Mohan Kumar) Let A be a Noetherian ring. Let I be an ideal of A such that I = (al, . . . ,a,)+12, for some ai E A. Then I = (a,, . . . ,a,, e), with e(1- e ) E (al, . . . ,a,).

5. Let M be a finitely presented module over a reduced ring R. Show that M is a projective R-module if pp(M) is constant on Spec(R). (Here pv(M) denotes the minimal number of generators of the module Mv over the ring Rv. This equals the dimension of the vector space Mp/pMp over the field RplpRv.) 6. Let R be a principal (ideal) domain. Show that every submodule of any free R-module is also free. Consequently, every projective module over a principal domain is free. You may use the following hints.

a) Let f : M + R be R-linear. Since the image is free, hence projective, one obtains M E ker(f ) im(f ) .

+

b) Let M c Rn and p : Rn -i R be the projection to the first direct summand. Then ( M n kerb)) $ p(M) r M by a). So M is free by induction on n. c) Now let F be a - not necessarily finitely generated - free R-module with a basis B. Usind a well-ordering of B one may prove that any submodule of F is free analogously to the proof in the case where F r R,. d) But one may also use Zorn's Lemma. Consider pairs (C,D) where C c B and D is a basis of M n C,,, Rc.

7. Let M be a finitely generated torsion free module over a principal domain A. Show that M is free. 'Torsion free' means that am = 0 for a E A, m E M implies a = 0 or m = 0. Let S = A \ (0). By torsion freeness we get an imbedding M + SP1M. Since the latter is a (finite dimensional) S-lA-vector space, it has a basis, which we may assume to lie in M , i.e. . .. , Since M is finitely generated, there is an s E S of the form with s M c Am1 $ . - .$ Am,. Then use Exercise 6 and that M S sM.

y, y.

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8. Is Zm (i.e. the group of infinite sequences of integers) a free Z-module? The answer is: 'No'.

Show for every countable principal domain A which is not a field that Am is not a free A-module. Use that submodules of free modules over a principal domain are again free. (Exercise 1) Choose any prime element p of A and consider the submodule H of sequences (ao,a l , a2, . . . ) having the following property: For every m almost all a, are divisible by pm. Show that H is uncountable. If H were free it would have an uncountable basis and so would have the A/pA-vector space HipH. Then show that H / p H has a countable basis, consisting of the residue classes of the sequences (1,0,0,0,...), (0,1,0,0,...), (0,0,1,0,...) ,... . 9. (S.M. Bhatwadekar, Raja Sridharan) Let A be a Noetherian ring. Let I be an ideal of A. Let Il c I , I 2 g I2be two ideal of A such that I1 I 2 = A. Then I = Il (e), for some e E 12, and Il = I n K , with I 2 K = A.

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Chapter 3 1. Let v = (al,. . . ,a,) be a unimodular row over a commutative ring A. Suppose that there exists a unimodular row (el,.. . ,c,) with C aici = 0. Then show that the projective module Pv corresponding to v = (al,. . . ,a,) has a unimodular element, i.e. there is a p E P such that Ap @ Q = P , for some submodule Q of p. 2. Let v = (al,. . . ,a,) be a unimodular row over a commutative ring A. (i) (Bass) If n is even, prove that Pv has a unimodular element, (ii) (R.A. Rao-Raja Sridharan). Suppose that n is odd, and some ai is a square. Prove that Pv has a unimodular element. (iii) Give an example to show that the condition that some ai is a square is necessary. 3. There is a good reason why, in the definition of stably free, we require that the free summand is finitely generated Namely for every projective module P there is a - generally not finitely generated - free module F such that P 69 F is free, as we see by the following trick of Eilenberg:

Let P, Q be projective such that P @ Q = E is free and let F = E @ E @ E @..... E whichis also free. Then P @ Fr P @ E @ E @ .... r P@(QCBP)~(Q@P)B... r (P@Q)CB(P@Q)CE.... r E@E@E e... rF

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4. We know the following theorem of Gabel: If P is stably free, but not finitely generated, then P is actually free. Say P @ R mE F, where F is free with basis {ei}iE1. Since P is not finitely f Rm) for some generated, I must be an infinite set. View P as ker(F + epimorphism f. Let {xl, ...,x,} be a basis of Rm. Then there are yi E F such that f (yi) = xi. But every yi is a linear combination of {ei}iEI, where Ii is a finite subset of I. Take I. = UEIIli.Then I. is a finite subset of I . Define Fo = CiEIo(Rei) Then f : Fo + Rm is surjective. Thus F = P Fo.Put Q = P n Fo,we have two short exact sequences

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(2) is exact because kerh = {x E Fo : h(x) = 0) = {x E F o : x E P } =PnFo=Q=img.

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By the second isomorphism theorem Fo/Pn FoE P Fo/P,i.e. Fo/Q E F/P. So FIFO= P/Q. But FIFO= CiEIPIo(Rei).Since I \ I. is infinite we can write P / Q E Rm @ Fl for some free module Fl. Now the first exact sequence splits naturally. Also second one splits since Rm is free. So we have P g Q @ P / Q and Fo E Q @ Rm. ~0~ p E & @ P / Q E Q @ (Rm @ Fi) % (Q @ Rm) @ Fl = Fo@ Fl which is free. 5. (T.Y. Lam) Let R be a ring. let P be a stably free prjective R-module with P @ Rm Y Rn for n > m. Let P .v P1 @ Rk for some L > 0. Let r 2 {mlk) {m/(n - m)), where {a} denotes the least integer a. Show that .P = P @ . . . @ P (r times) is a free R-module.(Hint: By induction , /(n-m),andt>m/k, mi, i P @ R m . v ~ i ( n - m ) $ ~ m . I f r = s s+>tm then use the free summands coming from tp to 'liberate' s P then use the resulting free summands to 'liberate' tP.)

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6. (T.Y.Lam) Let R be right noetherian. Let {Pi,i 1) be a sequence of R-modules such that Pi $ R~~ Y Rni. Assume that ni > mi+l for all i. Then, for r sufficiently large, the partial sums PI @ . . . @ PTare all free (Hint: Imitate the proof of Theorem 4.4)

7. Let P, Q be R-modules. Assume that P @ Rm Y Rn, and Q @ RS E Rt. If n > s then show that Q @ P @ Rm is free. Deduce Whitehead's lemma for reactangular matrices from the proof of the above.

8. The following is a generalized version of Whitehead's lemma due to L.N. Vaserstein: Let m , n 2 1 be integers, y an m x n matrix over a ring A, x an n x m matrix over A. Assume that I, + xy E GLmA. Then

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E Exercises

)

I, In+yx E GLnA, ( I ~0 + x Y (In 0y ~ ) - l ' ( I m +0 x y O )

+

' ( ( l n + 0y ~ ) - l I0 m)

E Em+n(A).

9. For any a! E GLnA and P E GLnA, we have

E E2,(A). (Hint:

Take x = a(/?- 1,) E M,A, y = a-l E MnA in the previous exercise.)

Chapter 4 1. In Lemma 4.4.2 one may drop the hypothesis that c is a non-zero-divisor. We can still use the procedure of the proof to get explicit formulas for the entries of M , thus avoiding the use of c in the denominator.

We introduce new indeterminates y, z and write

We shall try to define M = (mij)2,2 E SL2A. The ( 1 , l ) entry in the product of the two matrices considered above is

Thus we can define in the expression in the bracket to be m l l . Similarly we can define ml2, m21, m22. We can conclude on formal grounds, that det(mij) = 1 and that f (b)(mij) = f (b'). 2. a) Let M be an A-module, I an ideal of A and z E Um(M). Then f E Um(M/IM), where f denotes the residue class of z in M / I M and the latter is regarded as an (A/I)-module. b) Conversely, if, with the above notations, I consists of nilpotent elements and M is projective, then f E Um(M/IM) implies z E Um(M).

3. (D. Quillen) Let R be an algebra (not necessarily comutative) over A, f E A, and let B E (1 TRf [TI)* (=the group of invertible elements in Rf [TI which are congruent to 1 modulo T). Then there exists an integer k 0 such that for any gl,g2 E A with gl - 92 E fkA, there exists $ E (1 + TR[T])* such that Gf (T) = B(g1T)B(g2T)-1.

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299

4. (D. Quillen) Let M be a finitely presented A[T]-module suppose that Mm is an extended Am[T]-modulefor each maximal ideal m of A, then M is extended from A.

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5. (Roitman) Let (20,. . . ,x,) E Um,+l(A),n 2. Let y,-l,yn E A such that xn-1 yn-1 xnyn is invertible modulo ($0, . . . , x,-2). Show that (20,.. . , x,) and (20,. . . ,x,-2, y n - ~ y,) , are in the same elementary orbit.

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6. (Roitman) Let (xo,. . . ,x,) E Umn+l(A),n 2. Define J(1) to be the intersection of all the maximal ideals containing I. Let Ij = (20,. . . ,xj-1, xj+l,. . . ,xn-1) for 0 5 j 5 n - 1. Show that if x x, modulo (I0 . . . In-1) then (20,.. . ,x,), (20,.. . ,xn-l, x) are in the same elementary orbit.

+ +

7. (Roitman) Let (xo, . . . ,x,) E Um,+l(A), n > 2. Let t be an elemnet of A which is invertible modulo ($0, . . . ,xnP2).Show that (xo, . . . ,x,) and (20, . . . ,~ ~ - t2xn) 1 , are in the same elementary orbit. 8. Open Questions: (Roitman) i) Can you replace 10+ . . . + In-1by J(Io + . . . + In-l) above ?. ii) Can you replace I. . . . In-1by t/(xo,. . . ,xn-1) above?.

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9. (Vaserstein) Let (xo, . . . ,x,), (yo,. . . ,y,) E Um,+I(A), n 2 be in the same elementary orbit. (1) For any m 1, ( x r , XI,.. . ,x,), (y?, y1, . . . ,y,) are in the same elementary orbit. (2) If xoxo = 1 modulo (XI,.. . ,x,), yoyo = 1 modulo (31,. . . ,yn), then (20,21,. . . ,x,), (yo,y1, . . . , y,) are in the same elementary orbit.

>

10. Let

(XYIl ,, -- -- ..,Xn)

E

M2,,(A) have right inverse. Let 6 E GL2(A). Show

,Yn

>

11. (Roitman) Let (xO,.. . ,x,) E Um,(A), n 2. Let k be an integer between 0 and n - 1. Choose yk,. .. ,yn E A, such that

1

( : : :)

K-1

+A

X

= A. Prove that (I.,

. . . ,xr-1,

st,.. . ,2,)

i

and (xo, . . . ,XK-1, y ~. ., . ,y,) are in the same elementary orbit. 12. Let a =

(." . . ' I") Ylt... '

,Yn

E

M2,,(A). Let &(A) be the ideal of A generated

by 2 x 2 minors of $, i.e. 12(A) is the ideal generated of A which is by {xiyi - xjyj I 1 5 i # j 5 n). Show that for any a E I2(D) there exists ,B such that a,B = dI2. 13. (Karoubi Squares or Patching diagrams) Let cp : B + A be a homomorphism of rings. Let s E B such that

300

E Exercises (i) s is a non-zero-divisor in B. (ii)p(s) is a non-zero-divisor in A. (iii) p indcues an isomorphism B / s B -+ A/p(s)A. Then the commutative diagram.

resulting from a situation as above will be called a patching diagram. One also says B + A is an analytic isomorphic along s. Show that (a) (ii) and (iii) imply B/snB -+ A / ~ ( s ) ~isAan isomorphism for all n. (b) B is the fibre product of B, and A over A,, i.e. the square is cartesian. (c) If B is noetherian then (i) follows from (ii) and (iii). This is not true in general; give an example. Examples. Show that the following are patching diagrams. (a) Let Rs R t = R. Then R -+ Rt is an analytic isomorphism along s. (Such diagrams are called covering diagrams.) (b) Let B = k[[tl, . - . ,t,-l]][t,], A = k[[tl, . . - ,t,]], where k is a field. Let f E B which is a distinguished monic in t,, i.e. it is a monic polynomial in t, with its lower degree coefficients belonging to the maximal ideal of k[[tl,. . . ,t,-I]]. Show that B -+ A is an analytic isomorphism along f . (c) Let (R, m) be a local ring. Let f E R[t] be a Weierstrass polynomial, i.e. f is a monic polynomial in t with its lower degree coefficents in m. Then R[t] + R[t](,,t) is an analytic isomorphism along f .

+

S

(d) Let B be a noetherian ring, and s a non-zero divisor in B. Let A =g denote the (s)-adic completion of B. (See remarks on completion of a ring, and I-adic completion of rings for the definition). Then B -+ A is an analytic isomorphism along s. (e) Let A be a flat Z-algebra. Show that applying to a patching diagram gives a new patching diagram.

Splitting property: Let B -+ A be an analytic isomorphism along s. Let a E E,(A,), (1) If n 2 3, then there exists a1 E En(A), a 2 E En(Bs) such that a =: (ai)s(a2 @ 1). (2) If n = 2, there exist a1 E S12(A), a 2 E En(Bs) such that a = ( ~ 1 ) (8 ~ 1). 2 Patching property Let IP(R) denote the category of all finitely generated projective R-modules show that if B to A is an analytic isomorphism along s then the corresponding square

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301

Chapter 5 1. An example of a nonsoft vector bundle.

+

Define the subsets D+, D- of S1 := {(x, y) E lR2 1 x2 y2 = 1) by D+ := {(x,y) E S1 I y > 0) and D- := {(x,y) E S2 1 y < 0. On S1 consider the topology given by the following definition of "open":

A subset of S1 is called open, iff it is of the form or D- - F , where F is a finite set.

0 or S1 - F

or D+ - F

It is easy to check the axioms of a general topological space. Now consider the two open sets U-1 := D+ U D- U {(-1,O)) = s1- {(1,0)) We can glue the trivial line bundles over UI and U-1 by the continuous function a : Ul

nKl

+lR,

with a(u)= 1 for u E Ul, a(U) = -1 for u E U-1

to get some kind of Mobius bundle. But there is no partition of unity subordinate to the open covering (Ul, U-1). Namely every continuous function Ul + lR, resp. U-1 -+ lR is constant. 2. Let R = Z[2i], S = Z[i]. Computer the conductor C(S/R).

3. Let n > 0 be an integer. Then Z[X]/(Xn - 1) is isomorphic to the so called group ring of the cyclic group of n elements over Z .

The exercises which follow are based on the material found in Odds and Ends, Chapter 5. 4. Describe a gluing cocycle for Un.

5. Suppose that x H P(x) is a smooth family of projectors of a vector space V parameterized by a connected smooth manifold X. Set k = dim ker P(x) and n = dim V and denote by f the map

Show that f is smooth and that the pull-back of Uk,,by f coincides with the vector bundle defined by the family of projections P(x).

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E Exercises

6 . Suppose L -+ X is a smooth complex line bundle over X. Show that G ( L ) E Coo(M.C*).

7. Suppose M is a smooth compact manifold and E + M is a complex line bundle. A subspace V c Cw(E) is said to be ample if for any x E M there exists u E V such that u(x) # 0. (a) show that there exist finite-dimensional ample subspaces V

c Cw(E).

(b) Let V be a finite-dimensional ample subspace of Cw(E). For each x E M set V, = {v E V : v(x) = 0). Equip V with a Hermitian metric and denote by P(x) : V + V the orthogonal projection onto V,. Show that dimkerP, = 1 and the family of projections {p(x) : x E M ) is smooth. we thus obtain a complex line bundle E, + V. (c) Show that the line bundle E is isomorphic to Ev. In particular, this shows that E is the pull-back of a universal line bundle over a projective space. are two (smoothly) homotopic maps. (d) Suppose that f.g : M -+ an Denote by Ef(resp. E,) the pull-backs of the universal line bundle Un via f (resp. 9). Show that Ef E E,. 8. Show that the manifold P described above comes with a natural free right G-action and the space of orbits can be naturally identified with X .

+

9. Regard S2"+las a real hypersurface in Cnfl given by the equation lzoI2 121l2 - - . 1znI2= 1. The group S1 = {eit;t E R) c G acts on S2n+1 by scalar multiplication. The quotient of this action is obvious an.

+ +

(a) Show that S2nt1+ an is a principal S1-bundle. (It is known as the Hopf bundle). (b) Show that the line bundle associated to it via the tautologival representation S1 -+ Aut(C1) is precisely the universal line bundle Un over

an. E + X is Bvector

10. If bundle then as metric on E is a section h of s y m r n 2 ( ~ *such ) that h(x) is positive definite for every x E X . If E is complex one defines similary Hermitian metrics on E. A Hermitian bundle is a vector bundle equipped with a Hermitian metric. Show that any metric on a rank n real vector bundle naturally defines an O(n)-structure.

Chapter 6 I. Let A = k[xl, X2,. . . ,Xd], B = k[X1, X2]/[X; - XZXl), and let C = k[X1, X2, . . . ,Xd+1]/(f ), where f is a homogeneous polynomial in

E Exercises

303

XI,. . . ,Xd+l, monic in Xd+1 and of degree n. (1) Show that A, B, C are graded rings with zero-th component = k. (2) Consider the following product of formal power series ...)(1 X2 X; . . .). .. (1 Xd + X i .. .) (a) (1 + X I +X; (b) (1 + X I +X,2 ...)( 1+X2) ...)(1+xd+l+.--+X;L~~) +Xf ...)...(l+Xd+X; (c) (1 (i) Show that each monomial of A (resp. B,C) occurs precisely once in the power series (a) (resp. (b), (c)). (ii) By specializing each X j = T , and comparing the coefficient of Ti in the resulting formal power series in T, find the dimension of each graded component of the rings A,B, C (Hint: The monomials (which survive) form a k-basis.) (iii) The Hilbert series of a graded ring A = A. @A1@ . . . , with A. = k, a field, is the power series CEO=, dimb Ai ti E Z[[t]]. Show that the Hilbert series of the rings A, B, C is a rational polynomial of type p(t)/(l - t)d, where d is the Krull dimension of the ring, and p(t) E Qt].

+ + +

+ +

+

+

+

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2. (Rees) Let A be a Noetherian ring, I = (al,. - . ,a,) be an ideal of A. The subring A[alt, . . . ,art, t-l] of A[t, t-'1 is called the Rees ring of A with respect to I, and denoted by R(A, I). Show that 4

c,tT / c, E IT, if r 2 0) is a graded Noetherian ring.

(i) R(A, I ) = { T=-p

(ii) (t-n)R(A, I ) n A = In,for all n

> 0.

3. (Krull's Intersection theorem) Let A be a Noetherian ring, I an ideal of A. Then x in if and only if x = ax for some a E I .

nr,lIn

(Hint: Prove this when I is principal. Deduce the general case by considering IR(A, I).) 4. (Artin-Rees Lemma) Let A be a Noetherian ring, and I,J be two ideals of A. Show that there is an integer k such that if n > k,

(Hint: Show that JR[t, t-'1 n R(A, I ) has a finite set of generators of the form j,tr. Take k to be the greatest exponent o f t occuring amongst these elements). 5. (The Principal Ideal Theorem) Let A be a local domain with maximal ideal m such that there exists an m-primary ideal generated by a single element x. Then m is the only non-zero prime ideal of A. (3) (Hint: Show that if y(# 0) E A then (y) is not primary i.e., x" for k chosen satisfying Artin-Rees lemma with I = (x), J = (9). It suffices to show (x" y) = (xk+l,y), which follows if e((xk,y)/(xb+l)) = e((y)/(xN1) n (y)). Prove this.)

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6. Any injective (resp. surjective) endormorphism cp of an Artinian (resply. Noetherian) R-module M is surjective (resply. injective).

7. Let A be an affine domain over a field k. If A is not a field then its quotient field Q(A) is never algebraically closed. 8. (Rings of Invariants) Let G be a finite subgroup of GL,(C). Let G act linearly on CIX1, . . . ,X,]= Any matrix a E GL,(C) acts on the variables by (XI,. . . ,X,) ++-+(XI,.. . ,Xn)T. Extend this action to CIX1, .. . ,X,] so that f r-, a ( f ) is an automorphism of C[X1, . . . ,X,].

Let @[XI,... ,xnIG = {f E C[X1, ... ,X,] I a ( f ) = f , for all f E CIX1,. . . ,X,]) CIX1,. . . ,XnIG is a subring of C[Xl,. . . ,X,], called the ring of invariants of G. (a) Show that dimension @[XI,.. . ,xnIG = n. (b) (Reynolds) Define the map p is called the Reynolds operator. Show =Identity. that p is k[X1,. . . ,XnIGlinear, and p I k[Xl,. .. ,xnIG (c) If I is an ideal of C[Xl, . .. ,xnIG, show that I q X 1 , . . . ,X,] n C[Xl,. . . ,xnIG = I. Deduce that CIX1,. . . ,xnIG is a (graded) noetherian ring. (d) Let fl, .. . ,fs be homogeneous elements of positive degree which generate the maximal graded ideal of positive degree. Show that R = C[fl, . . . , f,]. Deduce that C[X1, . . . ,XnjGis a f.g. Galgebra. (e) (Molien) Show that the Hilbert series of CIX1, . . . ,xnIG is 1 aEG

det(1- ta)

. (Hint: The Reynolds operator induces a linear map

Pi from XI,. . . ,Xn]i -+ k[Xi,. . . ,x ~ ] Q , for all i show that and rank (pi) = trace (pi) = dim(RG)i.) 9. Let R be a Noetherian ring. Let m be a maximal ideal of R[X1,. . . ,X,]. Show that m n R is a prime ideal of height equal to (height m - n).

10. (Zariski) Let k be a field. Assume that E is a f.g. k-algebra which is a field. Then E is an algebraic extension of k, and is a finite dimensional vector space over k. Deduce Hiblert's Nullstellensatz from this. 11. (Artin) Let R be a Noetherian ring, and m a maximal ideal of RIX1,. . . ,X,], n 2 1. Then Rlm n R is a semi-local ring of dimension 5 1. 12. (Davis-Geramita) Let R be a noetherian ring. Let m be a maximal ideal in R[X, Y]. Show that m n R[X - Ys] is a maximal ideal for large s.

13. (R.C. Cowsik) Let k be a field. Let A = k[T5,T6,T7,TS] k[T]. Let cp be the obvious homomorphism from k[X, Y, 2,W] onto A. Prove that ker cp is aprime ideal generated by XZ-Y2, YW-Z2, X3-ZW, W2 -X2Y and X W - YZ. Show that (ker cp)3+n # (ker cp)(3+n)for n 2 0. (Hint: It suffices to find f E ker cp, f 6(X, Y, Z, W)6+2n,with Yf E (ker P)~+".If

E Exercises

305

In= { X Z - y2)2(w2 - XY) + (XW - Y q 2 ( Y W - z 2 ) ) ( x w - Y q n then g, E (ker cp)3+n, (X,Y, 2,w ) ~ + ~ " ) .

and Y

I

g,. Let f = gnYP1. check that f

#

14. Let R be a ring such that every ideal of R is countably generated.

a) Show that this also holds for the polynomial ring R[X]. (Let J be an ideal of R[X] and Ikfor Ic E IN denote the ideal of R, consisting of 0 and the leading coefficients of the polynomials of degree Ic in J . Choose a countable generating set El, of every Ikand to every a E El, \ (0) a polynomial of degree k whose leading coefficient is a. The chosen polynomials make up a generating set of J.) b) Show that this also holds for the polynomial ring A := R[Xi I i E IN] in countably infinitely many indeterminates. (Consider the rings A, := R[Xi I i 5 n]. Since every polynomial envolves only finitely many indeterminates, we see A = U,, A,. Hence I = U ( I n A,). So the union of countable generating sets of the I n A, generates I.) c) On the other hand there are uncountable chains of ideals in the ring A of b). (One can take the rational numbers p as indices of the indeterminates instead of the natural ones. Having done this, for every real number r let I, be generated by the X, with p E $, p 5 r.) 15. Let A be a discrete valuation ring (i.e. a principal domain with only one maximal ideal # (0)). Or more generally let A be a commutative Noetherian ring of dimension 1 with only finitely many maximal ideals. Show that in the polynomial ring A[X] there are maximal ideals as well of height 2 as of height 1.

16. Let A c B be an extension of finite type of domains. Show that there are an s E A - (0) and 'variables' X I , . . . ,x, E B such that B, is integral over A,[xl,. . . ,x,]. (The meaning of X I , .. . ,x, being 'variables' is that the ring homomorphism from the polynomial ring AIX1,. . . ,X,] to A[zl,. . . ,x,], given by X j H xj, is an isomorphism.) (Use Noether normalization.)

17. Let A C K be a ring extension of finite type, where K is a field. Show: a) there is an s E A \ (0) with &(A) = A,; b) [K : &(A)] < 00; c) (0) is in A not the intersection of the non-zero prime ideals of A. 18. Let A be a Noetherian domain. Show that there is an x E Q(A) with Q(A) = A[%]if and only if A is semilocal of dimension 1.

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19. A ring is called a Hilbert ring (or a Jacobson ring) if everyone of its prime ideals is an intersection of maximal ideals. (If K is a field, every K-algebra of finite type is a Hilbert ring.) Let A be a Hilbert ring and cp : A + B be a homomorphism of finite type. Show: a) B is a Hilbert ring. b) If n is a maximal ideal of B, then m := cp-l(n) is a maximal ideal of A and the induced field extension Alm c B l n is finite. (Prove b) first. Then let p E Spec(B), s E A \ p and q > p be a prime ideal of B such that q, is maximal in B,. Note that (Blp), # 0. Since B, is of finite type over A, the counterimage cp-l(q) is maximal in A. By Hilbert's Nullstellensatz Blq is a Hilbert ring. Hence q is an intersection of maximal ideals. Since s $! q we have shown that no s E B \ p belongs to all maximal ideals that contain p.) 20. Let A be a principal domain with infinitely many maximal ideals. Show that every maximal ideal of AIX1, . . . ,X,] is of height n 1 and can be generated by n 1elements.

+

+

21. Let R be a discrete valuation ring (i.e. a local principal domain which is not a field) and let p be a generator of its maximal ideal. We consider the ring A := R[X]. It is factorial. So its prime ideals of height 1are principal. Some of them are maximal ideals, for e.g. (pX - 1). More general every a j X j with aj E R, p 1 ao, plaj for irreducible polynomial of the form CjnZo j 2 1 generates a maximal ideal. Special such polynomials are pmXn - 1 with gcd(m, n) = 1. (See [45] VI.9.) So there are infinitely many maximal ideals of height 1. If m is a maximal ideal of height 2 of A. Then m n R = pR. Otherwise there were a chain of three prime ideals lying over the ideal (0) of R. So m is of the form (p, f ) where f E A is irreducible modulo p. We see: All prime ideals p of A are so called complete intersections, i.e. they are generated by the 'right' number, namely ht(p), of generators. (See Section 9.1.) 22. Let k be a field and V be a k-vector space of infinite dimension. Define a multiplication on the additive group k83 V by (a,v)(b, w) := (ab, aw h). Show that A := Ic @ V together with this multiplication is a ring. Show further that A is not Noetherian, but has a Noetherian spectrum. Namely this consists of the single prime ideal 0 @ V.

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Chapter 7 1. Let A be a Noetherian ring. Let I = (al,. . . ,an, s) be an ideal of A, with s E 12.Show that there exist a l , . .. , a n E A such that (al + S C ~ ., . . ,a, sc,) = I n If,where I' As = A, and with height

I' 2 n.

+

+

Chapter 8 1. Make precise the following intuitive definitions of dimension of an &ne algebra r ( X ) of functions on a subset X c P. 0

Define dimX by induction on n; it being clear for n = 1. If X = P , then its dimension is n. If not, after a change of variables, show that p = (0,. . . ,1) 4 X, and q = (0,. . . , a ) E X , for some a E C. Show further that the closure Y of the image Y of X in P-l under the projection map, is an f i n e algebra. Define dimX = dimP. Show that there is a point p E X where r(X),, is a regular local ring. (Where m, denotes the maximal ideal corresponding to the point p). In this case X will be a manifold at p. Show that there exist local coordinates XI,. . . ,x, of p E P , such that X is locally defined by the vanishing of X I = $2 = - - - = xk = 0. Define dimX = n - k.

2. (Serre, Murthy) Let M be a finitely generated R-module with hd(M) = 1. Then the following natrual numbers are equal. (i) min { t 1 3 an exact sequence 0 + Rt + P + M --+ 0, with P finitely generated projective ). (ii) min {p(Pl) 1 3 an exact sequence 0 + PI + Po + M + 0 , with Po, PI f.g. projective ). (iii) p ( ~ x t (M, & R)).

3. Let I be an ideal of a commutative ring R. Suppose that 1/12is a free R / I module and that hdRI is finite. If p(I) = p(I/12) then I is generated by a regular sequence.

4. Let R be a Noetherian ring. Assume that all f.g. projective R-module are free. Show that the following conditions are equivalent. (a) I is generated by a regular sequence. ) oo. (b) (i) h d ~ ( I < (ii) for any set yl, . . . ,yn E I whose images gl,. . . ,g, E 1/12is a basis of 1/12 the associated unimodular row (yl,. . . ,y,) over R,,I can be "lifted" to a unimodular row over R,I, for some s, s' E R, (s, sf) = R. (i.e. the projective R,,I-module corresponding to (ul, . . . ,u,) is extended

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from a projective R,, -module corresponding to a unimodular row over Rsl. (c) (9 hd~(.iT)< (ii) I/I is a free R/I-module. (iii) there is a set yl, . . . ,y, E I with Gl,. . . ,jj, E I/12.a basis of 1/12and such that the associated unimodular row (yl, . . . ,y,) over R,,! (for some s and s') can be lifted to a unimodular row over R,, . Give an example to show that the conditions (b)(iii) and (c)(iii) are essential. 5. (Mount) Let R, denote the polynomial ring &Exl,.. . ,x,] over an algebraically closed field &. Let f = (fo,. . . ,fn) E R;+l. f is said to be a determinantal sequence if there is a n x (n+ 1) matrix M with maximal minors fi, for each i. Show that if the ideal generated by fo, . . . ,f n has homological dimension less than 2, then f is a determinantal sequence. 6. (Ischebeck)

a) We give a new example of a principal domain A with SKIA # 0. Consequently A is far from being a Euclidean ring. This example seems to be simpler than those in [Ba, 9.21 and [Is]. (SKIA := lim SL, (A)/ En(A) .) -+

+

Let A := S-lIR[x, y], where S is generated by x2 y2 and all polynomials f with f (0,O) # 0. ( S is not of the form Sx for some set X c IR2.) Then A is a principal domain with SKl A # 0. Idea of proof: Being a localization of a factorial ring, A is factorial. Further its dimension is 1, since it is the localization of the local ring of the origin in the real affine plane by x2 y2. So A is a principal ideal domain.

+

Now we consider the matrix

We argue as follows: If p became zero in SKIA, it would already become zero in SKI (IR[x,y](,z+,z). f ) for some f with f (0,O) # 0. Then there would be a 1-sphere around the origin which would not meet the real zero set of f . But if one restricts p to this 1-sphere, it would give a nontrivial element of SKI of this sphere.

Remark. There is even a principal ideal domain A such that SKIAf # 0 for every f E A - ( 0 ) . Namely let A := SP1IR[x,y], where S is generated by all (x - a)2 (y - b)2, a, b E IR and all polynomials without real zeros.

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b) If B is a discrete valuation ring, then it is Euclidean. Therefore SLn(B) = En(B) for every n. So the above matrix p is locally elementary but not globally.

7. In the Theorem on the Finiteness of Class Number, it is essential that the ring is a domain (or at least reduced). (Swan) Let A := Z[X]/(X2) r Z i2 Z where in the latter additive group multiplication is defined by (a, b) (a', b') = (aal,abl bat). For n E IN define ideals In := n Z $ Z c Z @ Z.Then (0, l)In = 0 $ n Z , hence In/(O, 1)InZ Z i2 (Z/(n)). So for n # m the ideals Inand I, cannot be isomorphic.

+

8. Show that that every regular, not necessarily local ring is reduced.

9. Let A be a commutative ring such that there are elements a , b E A with (a) = Ann@) and (b) = Ann(a). Show that A/(a) as an A-module has infinite projective dimension if neither A/(a) nor A/@) is projective (i.e. a direct factor of A). One can construct an explicite infinite free resolution. This is enough to classify those residue class rings of Z (or of any other principal domain) whose modules have finite projective dimension. One does not need Theorem 8.5.8. 10. a) Show that a left A-module F is flat if and only if for any (finitely generated) right ideal I of A the group homomorphism I @ A F + F, induced by the inclusion I v A is injective. b) Conclude that a module over a Dedekind ring is flat if and only if it is torsion free. (Show first that flatness localizes.) 11. a) Let k be a field. Show: If F is flat over k[Xl,. . . ,Xn], then Xi is a non-zero-divisor of F/(Xl,. . . ,Xi-1)F for every i = 1,.. . ,n. (If i = 1, then (XI,. .. ,Xi-l)F = 0.)

b) Show that the ring Cn of continuous functions f : lRn + lR is not flat over the polynomial ring IR[Xl,. . . ,X,] for n > 1. Clearly it is flat for n = 1. 12. Let IP be the set of prime numbers. The the torsion group of the abelian Z / p is T := Zip. We know already by Exercise group G := 1.9 that the residue class group G / T is a $-vector space. Derive that the canonical projection v; : G + G/T admits no cross section, i.e. there is no homomorphism a : G/T + G with ma = i d G p . So G is not a direct sum of its torsion and its torsion free part.

npEp

epEp

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+

13. Consider the ring A := Z [ a = H Z-. hence a Dedekind ring. of H in $(-),

It is the integral closure

a) Show that 1,-1 are the only units of this ring. (Use the norm N : $(-) + Q, a b f l H (a b-)(a - G) = a2 5b2 for a, b E $. It is multilicative, since the map (T : A + A, a b f l H a - bis an automorphism.

+

+

+

+

+

b) Show (using again the norm) that the elements 2 , 3 , 1 f6, 2 fl are irreducible, i.e. no nontrivial products. c) Show that A is not factorial by computing 2 . 3 , 32, (1 + -)(I -, (2 (2 -

+ G)G.

-

d) Prove the following equalities of ideals of A:

+ n)

are maximal ideals of A. e) Show that (3,2 , (2,l + -) What are the decompositions in prime ideals of (6) and (9)? 14. FINITEEXTENSIONS OF DEDEKIND RINGSLet A c B be a finite extension of Dedekind rings. If S := A \ (0), then S-'B is a finite extension of $-'A = Q(A), hence a field, hence S-lB = Q(B). Since B > A is finite, n := [Q(B) : &(A)] < co.As a finitely generated torsion-free A-module, B is projective and finitely generated. Further its rank is constant; for A is a domain and so Spec(A) is connected. Let rkAB = n. Now let m be any maximal ideal of A. Then B,(:= T-'B with T = A\m) is a free A,-module of rank n. Therefore

Let pl , . . . ,p, E Spec(B) be the prime ideals over m, i.e. those which contain mB. Then Alm L, B/pi are field extensions. Let fi denote their degrees: [Blpi : Alm] = fi. In the above situation let mB = p:' . . . p p be the factorization of mB in B. Then ei f i = n.

xi='=,

Idea of the proof. In B we have the following finite sequence of ideals:

Every factor module of subsequent members of this sequence is of the form I l p i I with some ideal I of B and some i 5 r. But I / p i I r B/pi. So for every i = 1,.. . ,r we have ei factors isomorphic to B/pi. And B/pi is a vector space of dimension fi over Aim. This finishes the proof.

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15. Let R be a principal domain and M a finitely generated R-module. Let E R/(fi) @ . . . @ R/(f,) with fiI . . . If,. Prove p ( M ) = r . (The divisibility condition is necessary, since for e.g. Z/(6) r Z/(2) @ Z/(3).) Show further that I,(M) = R, if and only if p ( M )

5 r.

16. Let R be a domain. In the polynomial ring A := RIX1, . . . ,X,] consider ): with positive integers r 5 n, kl, . . . , Ic,. the ideal I := (x:',. . . ,X Show that Ass(A/I) consists of exactly one element, namely p := ( X ~ , . ' .,XT)' (To see the general scheme you could first assume r = 2. Then consider first the case Ic2 = 1. You have the following composition series of A/I:

The factors of this composition series all are isomorphic to Rip. By Proposition 8.4.7 b) the statement is true in this case. For general

Ic2

2 1 you have the composition series

Every factor of this is isomorphic to A/(X;', X2), and you know already ASS(A/(X:', ~ 2 ) =) {p}. Again use (8.4.7). In the general case use induction on s for the claim: The statement is true in the case Ic,+l = . . - = Ic, = 1. The case s = 0 is obvious. The case s = 1 is proved as the first case above, the step s + s 1 as the second step above.)

+

17. a) Let A be a regular ring of dimension 5 2 and N a finitely generated A, is projective. Especially every module. Show that N * := H o r n ~ ( NA) finitely generated reflexive A-module is projective. (An A-module M is called reflexive if the canonical map M + M** is an isomorphism.) Note that from a projective resolution Q --+ P -+ N -+ 0 one derives an exact sequence 0 + N * + P*+ Q* + E + 0 with a suitable E. Since hd(E) 5 2 we get that N * is projective. b) Let A be as above and S c A be multiplicative. Show that then P ( A ) -+ P(S-'A) is surjective. (If P is finitely generated projective over S-'A there is a finitely generated A-submodule M E P with S-'M = P. Then also SP1M** = P.) 18. Let Ic be a field and S a multiplicative subset of the polynomial ring A := Ic[Xl, . . . ,X,]. Show that every finitely generated projective module P over SP1A is stably free. (There is a finitely generated A-module M with P projective resolution

E

S-lM. There is a

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with finitely generated Qj, whic is actually free. This induces a free resolution 0 + S-lQ, ---+. - . ---+ S - ~ Q+ ~ P +0. over SU1A.This 'splits' everywhere; so show that P is stably free.) 19. A ring R is called hereditary, if all of his ideals are projective R-modules. (So Dedekind rings are hereditary.) a) Show that R is hereditary if and only if every submodule of any projective (or free) R-module is projective. b) Show that every hereditary domain is a Dedekind ring. c) Show that every hereditary Noetherian commutative ring is a finite direct product of Dedekind rings. d) Let K be a field and A the subring of the infinite product K", consisting of all 'nearly constant' sequences. Show that A is hereditary. (An infinite sequence (a,) is called nearly constant, if there is an N E IN such that a, = a, for all n, m 2 N.)

Chapter 9 1. Let M be a finitely generated R-module. Let S be a multiplicatively closed subset in R, and I be an ideal of R. Assume S is disjoint from I and Ann(M). Show that the grade of M in I (i.e. the length of the maximal M-sequence in I ) is 5 grade of SP1M in SP1I.

2. (McRae) Let R be a Noetherian ring. Let I be an ideal of R such that each associated prime ideal of I has grade one. Let I-' = HomR(I, R). Assume that I has a finite free resolution.

(i) Show that is contained in a associated prime ideal p of I then Ipis both projective and non-invertible! (ii) If b E but not in any associated prime ideal of R then show that there a E I such that I = aR : bnR, for some n. (Hint: Ibis principal).

3. (McRae) Let R be a Noetherian domain. Let I be an ideal of R such that each associated prime ideal of I has grade one. If hdRI is finite then hdRI = 0. (Hint: Show thatif II-I is a proper ideal then it has grade

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atleast 2. Let p be a prime idealof R containing II-l. Show that Ip[XI is not an invertible ideal of R[X]. Deduce that Ip[X]Ip[X]-lhas an element p(x), not belonging to any associated prime ideal of Ip[XI, and which is a prime. Now use McRae's characterisation to show that Ip[XI is principal!) 4. Let R be a Noetherian domain. Let I be an ideal of finite homological dimension generated by two elements. Show that its homological dimension is atmost one.

5. (Auslander-Buchsbaum) Let (R, m) be a local ring, and M a R-module of finite homological dimension. Show that hdR(M) can atmost be grade of m, and that equality holds if and only if m is an associated prime of M.

6 . (Rees) Let M be a finitely generated R-module. If p is an associated prime ideal of M then grade of p is atmost hdRM.

7. (McRae) Let R be a Noetherian ring, and M be a finitely generated Rmodule. Let p be an associated prime ideal of M. Let S be a multiplicatively closed subset of R which is disjoint from p and Ann(M). Show that if hd(M) is finite then grade of p equals grade of S-lp. 8. Here is an example of a curve in affine 3-space which is not an ideal theoretic complete intersection, but is a set theoretic one (even not locally). Let k be an algebraically closed field, and let C := {(t3,t4,t5) I t E k). Set I := I ( C ) c k[X, Y, Z] and m := (X, Y, Z) c k[X, Y, Z]. Show that:

a) If fl := XZ-Y2,

f2

:= X3-YZ, f3 := X2Y-Z2, then fi, f2, f3 E I.

b) If f (X, Y.Z) E I, then the linear part of f is zero. So I c m2. c) fl, f2, f3 are linearly independent modulo m3, hence also modulo mI. So I is at the point, defined by m not a complete intersection, i.e. ~m(2 0 3. d) C is set-theoretically defined by X Z - Y2, X 5

+ Z3 - 2X2YZ.

9. Let I be generated by a R-sequence a l , a2. Consider the Koszul resolution K(al,a2) : 0 + R 3 R2 3 I + I + 0, where 4(1) = (a2, -al) E R2, +(el) = ai, i = 1,2. Analyse the induced Horn(-, I) sequence

(a) Show that the top row it is exact. (b) a(1)= [K(a1,a2)] generates Extl (I, R). (c) Show that the bottom row is exact.

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(d) Define a map y : ~xt'(1,R) -i R/I so that the above diagram is commutative (Hint: y([K(al, az)]) = I E R/I). (e) Deduce that Ext(1, R) E R/I. 10. Let R be a Noetherian ring, and I # R an ideal of R. Let p be a prime ideal of R containing I. Show that if I,,is generated by a regular sequence of length n, then there exists s 6 p such that I, is generated by a regular sequence of length n. 11. Show that R is Cohen-Macaulay if and only if every complete intersection ideal of R is generated by an R-sequence if and only if every complete intersection ideal of R is unmixed.

12. Show that I is a complete intersection ideal of R of hdRI = 1 if and only if I is generated by a R-sequence of length 2. 13. (Sally-Vasconcelos) Let R be an affine algebra over a field k of dimension 5 2. Then {p(p) I p E Spec(R)) is bounded. 14. (Fitting ideals) Let cp = Rn -i Rm be a linear map, and let M(cp) E Mm,,(R) denote the matrix of cp. F.(cp) denotes the ideal generated by t x t minors of M (9). a) Show that Ft(cp) only depends on cp, and independent of choice of bases. They are called the Fitting ideals of M . b) Let Rm 3 Rn -+ M -0, i Rn 3 Rp -+ M -+ 0 be finite free presentations of the R-module M. Show that F,-k(p) = Fp-k(t,6), for all k 2 0. (Hence In-k(cp) is also called the k-th Fitting invariant of MI. c) If S is an R-algebra then show that F t ( c p @S~) = Ft(cp)S. Let M have a finite free presentation Rn 4 Rm -i M 0.i Let p be a prime ideal of R. d) Show that the following are equivalent: 6) Ncp)

c P-

,

(ii) (im(p)) contains a unimodular element. (iii) p(M,,) 5 m - t. e) Show that the following are equivalent : (i) Ft(cp) $ P and F t + i ( ~ = ) ~0. (ii) (Imcp),, is a free direct summand of R r of rank 1. (iii) M,, is free of rank (m - t). Consequently, M is a R-projective of rank m-T if and only if F,(cp) = R and F,+l(cp) = 0.

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f) Prove that cp = Rn + Rm is an homomorphism then rank (cp) (=rank M(cp)) = r if and only if F,(cp) has a R-regular element, and F,.+l(cp) = 0. 15. Let M be a fintely generated R-module. Let Fi(M) be its Fitting ideals.

a) If R is a local ring with maximal ideal m then show that p(M) = r if and only if F,-1(M) c m, and F,(M) = R. b) M is projective of rank r if and only if Fo(M) =?, FrP1(M) = 0, Fr(M) = R. 16. (Geyer) Consider the subring S = k[a2,b, c, ab, ac] of k[a, b, c, 1. Show that S = ]C[Xl,X2, X3, X4, Xs]/p, where p is the prime ideal xi -xlx;, x5x2-x3x4, x5x4- x 1 x 2 x 3 ) . ( x i -xlX:,

Show that ,!? = S/SX2 is not a Cohen-Macaulay ring. (Hint: Ann(X4) = (x3, x4, x 5 ) c Ass(,!?), and contains the minimal prime ideal (X4). Deduce that S is not a Cohen-Macaulay ring. 17. (Geyer) Let char k = 2 in above example. Let f l = X$ - X1X: ,and X i - XIXi. Show that (i) fl ,f2 are relatively prime. (ii) V(fl, f2) C A; is a complete intersection. (iii) R = k[X1, . . . ,X5]/(fl, f2) is a Cohen-Macaulay ring. (iv) R / & e S.

f2

=

18. Let A be a Noetherian ring. Let I be an ideal of A. Then a l l .. . , a n E I generate I if (and only if) a l , . . . ,an generate I modulo I2and & = v ( I ) = v(f1,. . - ,fn).

d m 1

19. Let A be a Noetherian ring of dimension d. Let I be an ideal of A such that 1/12is generated by n 2 d + 1 elements. Then I can be generated by n elements. 20. Let A be a Noetherian ring. Let I be an ideal of A. If f l , . . . ,fn E I generate I modulo I2and if every maximal ideal which contains (fl, . . . ,fn) also contains I, then show that f l , . . . ,f, generate I. 21. (Northcott) Let M be a Noetherian R-module, I an ideal of R, and ro E R. Show that there is an integer q 0 such that ( I n M :M rO) = In-q(PM : M ro) (0 :M ro), for all n 2 q. Consequently, for n q,

+

>

>

22. (Northcott) Let MI,. . . ,M, be submodules of a Noetherian R- module M , and let I be an ideal of R. Show that there exists an integer q 2 0 such that n g l InMi = I n - 9 ( n g l I q Mi).

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23. Let a l , . . . ,a, be a R-sequence on M. Let s 2. Show that (al,. . . ,a,-2,as)M : M (a,-1) = (al,. . . ,a,-2,a,)M. Deduce is a R-sequence on M if and only if that a l , . . . ,a,-2,a,,a,-l (al,. . . ,aS-2)M : M (a,) = (al, . . . ,a,-2)M. 24. Let M be a R-module, and let I be an ideal of R. Let x, y E I which are not zero-divisors on M. Show that (xM : M I ) / x M R (yM : M I)/yM as R-modules. 25. Let M be an Noetherian R-module, I be an ideal of R such that I M # M. Let a l , . . . , a m , and bl . . . ,bn be R-sequence on M in I. If m < n then show that there exist a m + l . . . , a n so that al, . . . ,am,am+l,. . . , a n is a R-sequence on M in I. 26. Let M be a Noetherian R-module and I be an ideal of R with I M # M. Jf then show that J M # M and the If J c Jf is an ideal with grade of J w.r.t. M is equal to the grade of I w.r.t. M.

a=

27. (Cohen) Let k be a field. Show that A = k[X2,X3,Y,XY] not Cohen -Macaulay.

c k[X, Y] is

Let p = k[x, y, u, v] + A be the onto homomorphism which takes x, y, u, v, to X2,Y, X3, X Y respectively. Show that ker p is generated by 2x 2 minors

28. (Upper semicontinuity of p) Let I,p be ideals of a ring R, with p a prime ideal, and I f.g. Show that there is a s E R-p such that p(IR,) = p(IRp). 29. (Davis-Geramita) Let R be a commutative (not necessarily noetherian) ring. Let m be a maximal ideal of R[Xl,. . . ,Xn] and let p = m n R. Assume that p is f.g. Then (1) If p is maximal p(m) 5 p(p) (2) In general, p(m) L p(pRp)

+ n.

+n + 1

> 0, and p maximal, p(m) 5 p(pRp) + n (Hochster) Let t < r < s be integers, and let K be a field of characteristic

(3) If n 30.

zero. Let A = K[Xij] be the ring of polynomials in r s variables, and let It(X) be the ideal generated by the t x t minors of the r x s matrix (Xij). Then show that It(X) is not set theoretically a complete intersection. 31. Prove the following version of the Eisenbud-Evans theorem: Let R be a commutative ring, M be a R-module, P be a subset of Spec(R), and 6 : P + N U (0) be a generalized dimension function on P. Let M' c M be (6(p) + 1)-fold basic in M at all p E P. Then M' contains an element m' which is basic in Mp for every p E P.

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In particular, if (r, m) E R @ M is basic at all p E P, then there exists an element m1 E M such that m rml is basic at all primes of P.

+

32. Prove the Eisenbud-Evans estimates for a ring R having a g.d.f. S. EE I: A projective R-module P of rank > g.d. R has a unimodular element. EE 11: A projective R-module P of rank > g.d. R is cancellative. EE 111: A R-module M is generated by ea(M) = supp{pp(M) S(p)) elements.

+

33. (Plumstead) Prove the Eisenbud-Evans conjectures over a polynomial ring: Let R be a commutative noetherian ring of dimension d, A = R[X], P be a finitely generated projective A-module of rank 2 d + 1, and M be a finitely generated A-module. The show that P has a unimodular element, and is cancellative, and deduce from these two facts that

Chapter 10 1. Let C be the a%ine space curve with parametric equations X = t3, Y = t4, Z = t5. Show that the ideal I(C) of polynomial functions vanishing on C = {(t3,t4,t5), t E E) is generated by the maximal minors of the matrix

Let m = (X, Y, Z) and observe that I c m2, but I q! m3. Show that I/m3 is a k-vector space of dimension three. Deduce p(I) = 3, and that C is not a complete intersection (i.e. ht I # p(I)). Show that C = V(XZ - Y2) n V(X5 complete intersection.

+ Z3 - 2X2YZ) is a set-theoretic

2. Show that the ideal I = (Y2 - X3) in k[X, Y] is locally a complete intersection ideal at primes p > I, but that k[X, Y ) / I is not a regular ring. 3. (Scheja-Storch) Let C E A: be a non-singular curve. Show that after applying a suitable k-automorphism of EIX1,. . . ,X,] the canonical injection E [ x ~ , x ~ , x ~ ] / I ( cI)I E [ x ~ , x ~ , x ~v] E[x~, ... ,x,J/I(C) is an isomorphism. Deduce C is an complete intersection if C1 = V(I(C) n

XI, X2, X3]) is.

4. Let I be generated by a R-sequence a1,az. Show that h2(I/12) is a free R/I-module of rank 1 generated by Sil A ii2. Deduce that the evaluation I / I-+ ~ R/I ) , defined by e(h) = h(Gl A Si2) is an map e : H o ~ ~ ~ ( A ~ (R/I) R/I-module isomorphism.

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5. Let R be a Noetherian ring. Let I # R be an ideal of which is locally generated by a regular sequence of length 2 at every prime ideal containing I . Prove that Extl (I, R) E HomRp(A2(1/12),R/I).

6. Let X

c

Let J =

(C3

be a complete intersection of two equations, say f and g.

!")

be the Jacobian matrix. Using the Implicit function gx Qy Qz theorem, or otherwise, show that there is a point p E X at which atleast one of the 2 x 2 minors of J does not vanish. Deduce that there exist a, b, c E C[x, y, Z] such that aJ12 b J 2 3 cJ31 does not vanish anywhere on X (i.e. the 1-form w = adz bdy cdz does not vanish on X). Show that the existence of a nowhere vanishing 1 form on X implies that X is a complete intersection. Construct examples of non-complete intersection curves. (fr fg

+ + + +

7. Two surfaces (f) and (g) through a point a E A: are said to meet transversally at a if (1) a is a simple point on each,

(%%.g )

(&, ag

ag

ag

(i.e. f #.I 0, &, %) #.I 0. (2) The surfaces have distinct tangent planes at %. Show that a curve C E A: is a set-theoretic complete intersection of two surfaces f = 0, g = 0 which are transversal at every point of C if and only if C is non-singular and a complete intersection of f , g. 8. Let C E A%be the affine space curve with parametric equation x = t, y = t2, z = t3. Show that the ideal I(C) of polynomial functions vanishing on

, t3) I t E &) is generated by the maximal minors of C = { ( t t2, (Such ideals are called determinental ideals). 9. Show that the ideal of non-singular curve in A: has a generating set consisting of 2 x 2 subdeterminants of a 2 x 3 matrix over k[X, Y, Z].

10. Let I be a height 2 unmixed ideal of k[X1,. . . ,Xn] which is a local complete intersection ideal at every prime containing I. Then I is a complete intersection ideal if and only Extl(I, R) is cyclic. 11. Given an example of a complete intersection ideal I which is not generated by a R-sequence. 12. (J. Herzog) The curve C = {(tnl,tn\tn3)) t E I), with g.c.d. of (nl ,na, ns) = 1 is a complete intersection if and only if the subsemi+) generated by nl,nz,ns is "symmetric" (i.e. there is an group of (N, m ~ Z such that for all z E Z, z E H if andonly if m - z $ H). 13. (Hilbert-Burch) Let R be a Noetherian ring, and I an ideal with a free resolution.

o + R ~ ~ R ~ + ' + I + o

(*I

E Exercises

319

Then there exists an R-regular element a such that I = aFn(cp). If I is projective then I = (a). If hd I = 1, then Fn(cp) has a regular sequence of length 2. Conversely, if cp : Rn + Rn+l is an R-linear map, and Fn(cp) has a R) a free resolution (*). sequence of length 2 2, then I = F n ( ~has

14. Let M be a Noetherian R-module, and let I be an ideal of R. Let a l , . . . , a m , and b l , . . . ,b, be two R-sequences on M in I. Then show that there is an R-module isomorphism

15. (Geyer) Give an example in characteristic 0 of a complete intersection X whose co-ordinate ring is not Cohen-Macaulay. Question: Are there examples of this type in dimension 2? 16. Let A be a semilocal Noetherian ring. Let I be an ideal of A. Show that ~ ( =4P ( I / I ~ ) 17. Let A be a local Noetherian ring. Let I be an ideal of A[X]. If I has a monic polynomial then show that p(I) = p(I/12). 18. Let A be a Noetherian ring. Let I be an ideal of A. If p(I/12) 2 dim(A/J(A)) 1 then show that p(I) = p(I/12). (Here J(A) denotes the Jacobson radical of A).

+

19. (Forster's Conjecture) Let k be a field, and p c A = k[Xl,. . . ,Xn] be a prime ideal such that Alp is regular. Show that p/p2 is generated by n elements. Show that p is generated by n elements. (This was settled by N. Mohan Kumar in [60]). 20. Let A be a commutative ring with 1. Let II,I 2 be two ideals of A with I 1 I 2 = A. If I l / I z , 1211; are generated by n elements, then so is (11n I~)/(II n 1 ~ 1 ~ .

+

21. Show that the maximal ideal m = (X - I, P) of the coordinate ring of real circle A = R[X, Y]/(X2 + Y2 - 1) is not a principal ideal.

4

+

B = C[X,Y]/(X2 Y2 - 1) E C[u,v]/(uv - 1) z C[U,U-~], (Hint: A for u = X iY, v = X - iY. Check that mB = (u - 1). Show that Xun(u - 1) @ Im (i), for all X E C . Deduce that m is not principal.)

+

Let R be a commutative Noetherian ring. Let I be an ideal of R. Let S be the multiplicative closed subset of R consisting of all s E R such that (I, s) = R. Let n be the least number such that there exist X I , . . . ,x, E R with f i = If m denotes the least number such that there = J(yl,. . . ,y,), then show that exists y1, . . . ,g, E S-'R with m
320

E Exercises

+

23. Let I be an ideal of R. Let s,s' E R with R s Rs' = R, and with S' E I. Suppose that I, = (XI,.. . , x n ) where xi E R, for 1 5 i 5 n. If there exists a unimodular row (al,. . . , a n ) over R,! and an elementary matrix y E En(R,,t) such that y(x1,. . . ,xn) = ( a l , . . . , a n ) over R,,, then p(I) 5 n. 24. Let R be a Noetherian ring. Let R = A[X] and let m be a maximal ideal of R with htm = dim R, and with R, regular local. Show that there exists t E 1 + (m n A) such that p(mt) = dim R. Moreover, one can choose generators f l , . . . , fn of mt, n = p(mt), with f l , . . . ,fn-1 generating mt n At[X]. (Hint: Show that if p = m n A then Ap is regular local. Deduce p = (fl, . . . ,fn-1, e) where e(e - 1) E (fl, . . . ,fn-I), and m = (fl,. . . ,f n P l , e, f,). Take t = 1- e, and complete the argument. 25. Let R be a commutative ring which contains a field k. Let s E R such that R, 2 A[X] for some ring A. Let m be a rational maximal ideal of R (i.e. the natural map k + Rlm is an isomorphism). Assume that s # m. If R, is a regular local ring of dimension equal to dim R, then m is generated by a regular sequence (Hint: By a further localisation assume that A is a reduced ring. Now settle the case when dim R = 1. Using an earlier exercise show that (after a further localisation if necessary) m, = (fl,.. . ,fd), where d = dimR, and with f l , . . . , fd-1 generating m, r l A. Find s' E m, r l A so that ( f l , . . . ,fd) E elEd(R,,r).) n

26. Let R = &[XI,.. . ,x,, 91,. . . ,y n ] / ( ~ x i y i I), n 2 2. Show that each i=l is maximal ideal of R, other than the "origin" (51,. . . ,Zn, gl,. . . generated by a regular sequence.

,on)

Appendix A 1. (R. A. Rao & R. Khanna) For an almost commutative ring R, i.e. one which is finitely generated over its center, and an integern 2 3 the following properties are equivalent:

a) Normality: En(R) is a normal subgroup of SLn(R);

+

b) In vtw E En(R) if v E Umn(R) and (v, w) = 0; c) Local-Global Principle: If a ( X ) E SLn(R), a(0) = I n and a,(X) En(R,) [XI for all m E Spmax(R), then a ( X ) E E(R[X]);

E

d) Dilation Principle: If a ( X ) E Sln(RIX]), a(0) = IIn and a,(X) E En(R,[X]) for some non-zero s E R, then a(bX) E En(R[X]) for b E (sl) where L is large. Actually we mean there is some P(X) E En(R[X], (X)) (i.e. P in the normalizer of the the group generated by

E Exercises

321

the elementary generators which are congruent to In) with P,(X) = a(bX). But since there is no ambiguity, for simplicity we are using the notation a(bX) instead of P,(X).)

+

e) If a ( X ) = In Xdvtw, where v E En(R)el and (v, w) = 0, then a(X) E En(R[X]) and is a product of the form E:"~) for big d.

n

+

f) In vtw E En(R) if v E En(R)el and (v, W) = 0. g) In+vtw E En(R) if v E SLn(R)el and (v,w) = 0. 2. Give an example of a locally elementary polynomial matrix, but which is

not an elementary matrix. (Hint: As P.M. Cohn has shown,

is not elementary over Z[x, y]. Consider the matrix CE&C-I.) 3. Let R be a ring, s E R a non zero divisor and a E R. Using Quillen's Local Global Principle show for n 2 3 and m 2 2

4. Using the previous exercise, prove A. Bak's theorem that the group SLn(R)/En(R) is nilpotent of class d, where d is the dimension of R, for n 2 3. 5. Suppose a(X) E SLnR[X] such that a(0) = In and a(X), E [SLn(R,[X], SL(R,[X])] for every maximal ideal m of R. Then a(X) E [SLn(R[Xl, SLn(R[XI)IFor the following two exercises we need some definitions: Let M be a finitely generated module over a commutative ring R.

+

a) li3ansvection: An automorphism of M of the form I cp,, for some cp E M* = Hom(M, R), p E M, with cp(p) = 0, and with either p or cp unimodular. b) By Flip(M $ R) we denote the subgroup of Trans(M @ R) generated

6. (R.A. Rao) Dilation Principle for li3ansvections Let P be a projective R-module of rank r 2 2. If a E Aut(P[X] @ R[X]) with a(0) = id and cr, 6 E,+l(R,[X]) for a non zero divisor s, then there is an 1 such that u(bX) E Flip(P[X] @ R[X]) for all b E Rsz.

322

E Exercises

7. (R.A. Rao) Local Global Principle for Transvections Let P be a projective R-module of rank r 2 2. If u E Aut(P[X] @ R[X]) with u(0) = id and up E E,+l(Rp[X]) for all p E Spec(R), then a E Flip(P[X] @ R[X]). 8. (R.A. Rao) Let P be a projective R-module of rank 2 2. Then Trans(P[X] @ R[X]) = Flip(P[X] @ R[X]). 9. (Suslin) Let v = (v1,v2,-.. ,v,) E M1,,(R), n 2 3, with v l , v 2 , - - . ,v, a regular sequence. Let w E Ml,,(R) with wvt = 0. Show that I, + vtw E E n (R)10. (Vaserstein) Let A be an associative ring with unity, n 2 2, v = (vi) E An, u = (y)E (An)*, and uv = 0. Assume that 1 ukvk E GLIA for some index k. Then 1, + vu E E,(A). 11. (P.M. Cohn) The 2 x 2 elementary subgroup need not be a normal subgroup. Prove the following steps to get an example:

+

(1) Let E(a)

=

(-4 i). Show

that E2(A) is generated by

{E(a) l a E A). (2) Let a E D2(A)E2(A), where Da(A) is the subgroup of GL2(A) of diagonal matrices. Show that a = diag {a, b)E(al) . . . E(a,) such that ai # A* U {O), for 1 5 i < r. (3 ) Let A = k[X, Y] be a polynomial ring in 2 variables over a field k. Let a E D2(A)E2(A) have a representation as above. Assume further that a, # k. If e l a = (f, g), then deg > degg. (4) Let cr E GL2(k[X, Y]), with e l a = (f, g), with deg f = degg. Assume further that the leading forms of f and and g are linearly independent over k. Prove that a @ D2 (k[X, Y])E2 (k[X, Y]) . (5) Show that E2(k[X, Y]) is not a normal subgroup of S12(k[X, Y]). 12. (Mennicke symbols). Let A be a commutative ring and (a, b) E Um2(A). Let c,d E A such that ad - bc = 1. The Mennicke symbol ms (a, b) is defined to be the class of

(: i)in Sb(A)/ E3(A). This is independent of

e has the following properties. choice of c, d. Prove that the ~ e n n i c k symbol (MS2) ms (aa', b) = ms(a, b)ms(al, b) (MS1) ms (u, b) = 1 for u unit. (MS3) ms (a, b) = ms(b, a). (MS4) ms (a Xb, b) = ms(a, b), V X E A. 13. Let A be a local ring. Let f (X) E A[X] be a monic polynomial. Suppose that (f ( X ),g(X)) E Um2(A[X]) . Show that ms (f ( X ),g(X) = 1. 14. (Murthy) (Local Global Principle for Mennicke Symbols). Let (f ,g) E Um2 (A[X]). Suppose that ms (f, g) = ms(f (O), g(0)) over A,[X], for every maximal ideal m of A[X]. Then ms(f,g) = ms(f (O),g(O) over A[X]. 15. (Suslin) Let A = k[X1,. . . , X,] be a polynomial ring over a field k. Prove that S1,([X1, . . . ,X,]) = E, (k[Xl, . . . ,X,]), for all r 2 3.

+

E Exercises

323

>

16. (A. Suslin) Let a(x) E E,(R,[x]), n 3, with a(0) = I,. Show that there is a number of k such that for rl, 7-2 E R[x], with r l - r2 E skRR], a ( 7 ' 1 ~ ) - ~ ~ 2 E( ~En(R[x]). 2~) 17. Let v(x) E Um,(R[x]), n 2 3. Suppose that for all m E Spmax(R),v(x) 5 v(0) (mod E,(R,[x])). Then show that v(x) v(0) (mod E,(R[x])). 18. (R.A. Rao) Deduce Corollary 4.4.8 using the Local Global Principle for elementary action. (Hint: If v(x) E Um,(R[x]), then show v(x) v(1) (mod E,(R[x]). Consider w(x-l) = (X-~"~(X)),where v(x) = ( V ~ ( X.).,. ,v,(x)). Show w(x-') E Um,(R[x-I], and that w(0) el (mod E,(R)). Finish the proof.)

=

-

=

Appendix C 1. (Hartshorne) Let X be a connected topological -space. Let y be a closed subspace of X. Let X, = {x E X I y E {x)). Suppose that for each y E Y, X, \ {y) is a non-empty and connected space. Show that X \ Y is connected.

The concept of a topological space being connected in codimension k was made precise by R. Hartshorne, as follows: Let X be a Noetherian topological space, and Y be a clsoed subset of X. Let Y' be a closed irreducible subset of Y. The codimension of Y' in Y is the supremum of {r I there exists a sequence of closed irreducible subspaces Zi of X, Y' Zo Z1 E . . . E 2, X). The codimension of Y in X = inf{ codimension of Y' c Y in X, Y' closed and irreducible ). 2. (Hartshorne) Let X be a Noetherian topological space. We say that X is connected in codimension k if it satisfies any of the following equivalent conditons.

(i) If Y is a closed subset of X with codimension of Y in X X \ Y is connected.

> k then

(ii) If X', X" are irreducible components of X then there is a sequence X' = XI, X2,. . . ,Xr = X" of irreducible components of X , with Xi n Xi+l of codimension 5 k in X , for all 1 5 i 5 r - 1. Prove that (i) and (ii) are equivalent conditions.

3. Let X be a Noetherian topological space. Show that the following conditions are equivalent. (i) For any y E X , X, is connected in codimension k. (ii) Whenever y E X is such that dimX,

> k, then X, \ {y) is connected.

(A space X is said to be locally connected in codimension k if it satisfies any of the above conditions).

324

E Exercises

4. Let X = Spec(A) be a connected, locally Noetherian space. Let Y = V(I) be a closed subset of X such that A I I locally has depth 2 2. Show that X \ {y) is connected.

5. Let X = Spec(A) be a locally Noetherian space. For each p E X with dim(Ap) > k assume that depth(Ap) 2 2. Show that X is locally connected in codimension k.

References

Local Bibliography 1. Atiyah M.F. and I.G. Macdonald: Introduction to Commutative Algebra. Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont. (1969). 2. Auslander M., Buchsbaum D: Homological dimension in local rings, Trans. Am. Math. Soc. 85 (1957), 390-405. 3. BLdescu L.: On the Equations defining Affine curves. (Notes given in the School on Commutative Algebra and Interactions with Algebraic Geometry and Combinatorics at I.C.T.P. in May, 2004.) 4. Basu R.: On Forster's Conjecture and Related Results. M. Phil. dissertation (2002). 5. Bhatwadekar S.M., H. Lindel, R.A. Rao: The Bass - Murthy question : Serre dimension of Laurent polynomial extension. Invent. Math. 8 1 (1985), 189203. 6. Bhatwadekar S.M. and A. Roy: Some theorems about projective modules over polynomial rings. J. Algebra 86 (1984), 150-158. 7. Bhatwadekar, S. M., Rao, R. A.: On a question of Quillen. Trans. Amer. Math. Soc. 279 (1983), no. 2, 801-810. 8. Boothby W.: On two classical theorems of algebraic topology. Amer. Math. Monthly 78, (1971), 237-249. 9. Boratynski M.: A note on set-theoretic complete intersection ideals. J. Algebra, 54 (1978), 1-5. 10. Bourbaki, N.: Commutative algebra. Springer, (1988). 11. Bruske R., F. Ischebeck, F. Vogel: Kommutative Algebra. B.I.Wissenschaftsver1ag Mannheim, Wien Zurich, (1989). 12. Cartan H., Eilenberg, S.: Homological Algebra, Princeton (1956). With an appendix by David A. Buchsbaum. Reprint of the 1956 original. Princeton Landmarks in Mathematics. Princeton University Press, Princeton, NJ, (1999). ISBN: 0-691-04991-2. 13. Cohn P.M. : On the structure of the GL2 of a ring. Inst. Hautes Etudes Sci. Publ. Math. No.30 (1966), 365-413. 14. Cohn P.M. : Some remarks on the invariant basis property. Topology 5 (1966), 215-228.

326

References

15. Bruns W., Herzog J.: Cohen-Macaulay rings, Cambridge studies in advanced mathematics 39, Cambridge University Press, (1993, 1998). 16. Carral M.: Modules projectifs sur les anneaux de fonctions. (French) [Projective modules over rings of functions] J. Algebra 8 7 (1984), no. 1, 202-212. 17. Cowsik R.C and M.V. Nori: Curves in characteristic p are set theoretic complete intersections. Invent. Math. 4 5 (1978), 111-114. 18. Eisenbud D. and E.G. Evans Jr.: Generating module efficiently: Theorems from Algebraic K-Theory. J. Algebra 2 7 (1973), 278-315. 19. Eisenbud D. and E.G. Evans Jr.: Every Algebraic set in n-space is the intersection of n Hypersurfaces. Invent. Math. 1 9 (1973), 107-112. 20. Eisenbud D.: Commutative algebra with a view toward algebraic geometry. Graduate Texts in Mathematics 150, Berlin Heidelberg New York, Springer (1995). 21. Ferrand D.: Suite r6guli8re et intersection compl8te. C.R. Acad. Sci. Paris 264 (1967), 427-428. 22. Ferrand D.: Courbes gauches et fibres de rang 2, C. R. Acad. Sc. Paris 281, (1975), 345-347. 23. Flenner H., O'Carroll L., Vogel, W.: Joins and Intersections, Springer Monographs in Mathematics, ISBN 3-540-66319-3, (1999). r Anzahl der Erzeugenden eines Ideals in einem noether24. Forster 0.:~ b e die schen Ring. Math 2. 8 4 (1964), 80-87. 25. Biinic5 C., Forster 0.:Multiplicity structures on space curves. The Lefschetz centennial conference, Part I (Mexico City) (1984), 47-64, Contemp. Math., 58, Amer. Math. Soc., Providence, RI, 1986. 26. Gabel, M. R.: Generic orthogonal stably free projectives. J. Algebra 29 (1974), 477-488. 27. Geyer W.D.: On the Number of Equations which are Necessary to Describe an Algebraic Set in n-Space. Atas 3" Escola de Algebra. Brasilia (1976), 183-17. 28. Gopalakrishnan, N.: Commutative Algebra. Oxonian Press Pvt. Ltd. New Delhi, ISBN 81-7087-039-9, (1984, 1988). 29. Hartshorne, R.: Complete intersections and connectedness. Amer. J. Math. 8 4 (1962), 497-508. 30. Hartshorne R.: Algebraic Geometry. Springer New York etc. (1977). 31. Herzog J.: Generators and relations of abelian semigroups and semigroup rings. Manuscripta Math. 3 (1970) 175-193. 32. Hilton P., Stammbach U.: A Course in Homological Algebra, GTM 4, Springer Verlag, (1970). 33. Horrocks G.: Projective modules over an extension of a local ring. Proc. London Math. Soc. 1 4 (1964), 714-718. 34. Ischebeck F., Ojanguren M.: Another example of a projective module over a localized polynomial ring. Arch. Math. (Basel) 70 (1998), no. 1, 29-30. 35. Kaplansky I.: Projective modules. Ann. of Math. (2), 68 (1958), 372-377. 36. Kaplansky I.: Fields and Rings. Univ. of Chicago Press (1969). 37. Kaplansky I.: Commutative Rings. Allyn and Bacon, Boston (1970). 38. Kelley, J.L.: General topology. D. Van Nostrand Company, Inc., TorontoNew York-London, (1955). 39. Knus, M.A., Ojanguren, M.: Thhorie de la Descente et Algkbres d'Azumaya. LNM 389, Springer. 40. Kronecker L.: Grundziige einer arithmetischen Theorie der algebraischen Grijpen. J. reine angew. Math. 92 (1882), 1-123.

References

327

41. Krusemeyer M.: Skewly completable rows and a theorem of Swan and Towber. Communications of Algebra 4 (7) (1975), 657-663. 42. Kunz E.: Introduction to commutative algebra and algebraic geometry. Birkhauser, (1985). 43. Lam, T. Y. Series summation of stably free modules. Quart. J. Math. Oxford Series (2), 2 7 (1976), no. 105, 37-46. 44. Lam T.Y.: Serre's Conjecture. Springer Verlag, 635, New York, (1978). 45. Lang S.: Algebra 3'd ed. Addison-Wesley Reading, Massachusetts, (1993). 46. Lindel H.: Unimodular elements in projective module. J. Algebra 172 (1995), 301 - 319. 47. Lindel H.: On the Bass-Quillen conjecture concerning projective modules over polynomial rings. Invent. Math. 6 5 (1981/82), no. 2, 319-323. 48. Lonsted K.: Vector bundles over finite CW-complexes are algebraic. Proc. Amer. Math. Soc. 38 (1973), 27-31. 49. Lyubeznik G.: The number of defining equations of a f f i e algebraic sets. Amer. J. Math 1 1 4 (1992), 413463. 50. Lyubeznik G.: The number of defining equations of affine algebraic sets. Amer. J. Math 1 1 4 (1992), 413463. 51. Lyubeznik G.: A Survey of Problems and Results on the Number of Defining Equations. Commutative algebra, (Berkeley, CA, 1987), 375-390, Math. Sci. Res. Inst. Publ., 15, Springer, New York, (1989). 52. Macaulay, F.S.: Algebraic Theory of Modular Systems. Cambridge Tracts 1 9 (1916). 53. Mandal S.: On efficient generation of ideals. Invent. Math. 75 (1984), 59-67. 54. Matsumura H.: Commutative Algebra. Cambridge University Press, (1986). 55. McDonald B.: Linear Algebra over Commutative Rings. Marcel Dekker, Inc., New York and Basel, (1984). 56. Milnor J.: Analytic proofs of the "Hairy Ball Theorem" and the Brouwer Fixed Point Theorem. Bulletin AMS, (1978), 521-524. 57. Milnor J.: Topology from the differentiable viewpoint. Based on notes by David W. Weaver. Revised reprint of the 1965 original. Princeton Landmarks in Mathematics. Princeton University Press, Princeton, NJ. 58. J. Milnor; Introduction to Algebraic K-theory, Ann. of Mathematics studies 72, Princeton University Press. 59. Moh T.T.: On the unboundedness of generators of prime ideals in power series rings of three variables. J. Math. Soc. Japan 26 (1974), 722-734 60. Mohan Kumar N.: On two conjectures about polynomial rings. Invent. Math. 4 6 (1978), 225-236. 61. Mohan Kumar N.: Affine Geometry. appeared in Current Science. 62. Murthy M.P.: Complete intersections. In: Conf. Comm. Algebra; Kingston (1975), Queen's papers Pure and Appl. Math. 4 2 197-211. 63. Murthy M.P.: Suslin's work on linear groups over polynomial rings and Serre problem. (Notes by S.K. Gupta). IS1 Lecture Notes, 8. Macmillan Co. of India, Ltd., New Delhi, 1980. 64. Nashier B.S. and W. Nichols: Ideals containing monics. Proc. of the Amer. Math. Soc. 99 (1987), 634-636. 65. Nicolaescu L.: Notes on Seiberg-Witten Theory, Graduate Studies in Mathematics 28, Amer. Math. Soc. (2000).

328

References

66. Northcott D. G.: A first course of homological algebra. Cambridge University Press, Cambridge-New York, (1980), ISBN: 0-521-29976-4 18-01. 67. Northcott D. G.: Finite free resolutions. Cambridge Tracts in Mathematics, No. 71, Cambridge University Press, Cambridge-New York-Melbourne, (1976). 68. Ohm J.: Space curves as Ideal-Theoretic Complete Intersection. In: Studies i n Math. 20 Math. Assoc. Amer. (1980), Ed: A. Seidenberg, 47-115. 69. Ojanguren M. and R. Sridharan: Cancellation of Azumaya algebras. J.Algebra 1 8 (1971), 501-505. 70. Plumstead B.R. : The Conjectures of Eisenbud and Evans. Ph.D. Thesis (1979), Amer. J. Math. 1 0 (1983), 1417-1433. 71. Quillen D.: Projective modules over polynomial rings. Invent. Math. 36 (1976), 167-171. 72. Rao R.A.: Patching techniques in Algebra - - Stability theorems for overrings of Polynomial Rings and Extendability of Quadratic Modules with sufficient Witt Index: Doctoral thesis submitted in June, (1983) to the Bombay University, India. 73. Rao R.A.: An elementary transformation of a special unimodular vector to its top coefficient vector. Proc. Am. Math. 93 (1985), 21-24. 74. Rao R.A.: The Bass-Quillen conjecture in dimension three but characteristic # 2,3 via a question of A. Suslin. Invent. Math. 93 (1988), no. 3, 609-618. 75. Rees D.: Two classical theorems of ideal theory. Proc. Cambridge Philos. Soc. 52 (1956), 155-157. 76. Rees, D.: A theorem of homological algebra. Proc. Cambridge Philos. Soc. 52 (1956), 605-610. 77. Rotman M.: Introduction to Homological Algebra. Pure and Applied Mathematics, 85, Academic Press, Inc. [Harcourt Brace Jovanovich, Publishers], New York-London, (1979). 78. Roy A.: Application of patching diagrams to some questions about projective modules. J. of pure and Applied Alg 24, (1982), 313-319. 79. Roy A.: An approach to a question of Bass via patching diagram. article on a talk i n University of Poona, March (1981). 80. Roy A.: Remarks on a result of Roitman. J. Indian Math. Sac. (N.S) 44, (1980), 117-120. 81. Sarges H.: Ein Beweis des Hilbertschen Basissatzes. J. reine angew. Math. 283/284 (1976), 436 - 437. 82. Sathaye A.: On the Forster-Eisenbud-Evans Conjecture. Invent. Math. 46 (1978), 211-224. 83. Serre J-P.: Algebre locale. MultiplicitBs. (French) Cours au College de France, 1957-1958, rCdigC par Pierre Gabriel. Seconde Cdition, (1965). LNM 11, Springer-Verlag, Berlin-New York (1965). Local algebra. (English) Springer, 2000. 84. Serre J-P.: Faisceaux algbbriques cohCrents. (French) Ann. of Math. (2) 61, (1955), 197-278. 85. Sharma P.K.: Projective modules over group rings. J. Alg. 1 9 (1971), 303-314. 86. Raja Sridharan: Commutative algebra with applications to Combinatorics. M. Sc. thesis, University of Bombay, (1989). 87. Raja Sridharan: Non-vanishing sections of Algebraic Vector Bundles. J. Algebra 176, (1995), no. 3, 947-958.

References

329

88. Sally J. D.: Numbers of generators of ideals in local rings. Marcel Dekker, Inc., New York-Basel, (1978). ISBN: 0-8247-6645-8 89. Sally J. D.: Boundedness in two dimensional local rings. Amer. J . Math. 100 (1978), no. 3, 579-584. 90. Stafford J.: Projective modules over polynomial extensions of division rings. Invent. Math. 59 (1980), no. 2, 105-117. 91. Storch U.: Bemerkung zu einem Satz von Kneser. Arch. Math. 23 (1972), 403404. 92. Suslin A.A.: On Stably Free modules. Math. USSR Sb. 31 (1977), 479491. 93. A. Suslin; On the structure of special linear group over polynomial rings. Math. USSR. Izv 11 (1977), 221-238. 94. Suslin A.A.: On Projective modules over polynomial rings. Math. USSR Sb. 22 (1974), 5955602. (English translation). 95. Swan R.G.: Vector bundles and projective modules. Trans. AMS 105 (1962) 264-277 96. Swan R.G.: The number of generators of a module. Math. 2. 102 (1967), 318-322. 97. Swan R.G.: K-Theory of Finite Groups and Orders. LNM 149 SpringerVerlag Berlin, Heidelberg, New York (1970). 98. Swan R.G.: Cancellation theorem for the projective modules in metastable range. Invent. Math. 27 (1974), 23-43. 99. Swan R.G. Topological examples of projective modules. Trans. of A.M.S. 230 (1977), 201-234. 100. Swan R. G.: A simple proof of Gabber's theorem on projective modules over a localized local ring. Proc. Amer. Math. Soc. 103 (1988), no. 4, 1025-1030. 101. Swan R.G. and J. Towber: A class of projective modules which are nearly free. J. of Algebra 36 (1975), 427-434. 102. Szpiro L.: Lectures on equations defining space curves (Notes by N. Mohan Kumar). Tata Inst. Lecture Notes i n Math., Bombay (1979); by SpringerVerlag, Berlin-New York, 1979, ISBN: 3-540-09544-6. 103. Vaserstein L.N. : Vector Bundles and Projective Modules. Trans. Amer. Math. Soc. 294 (1986), no. 2, 749-755. 104. Vaserstein L.N. and Suslin A.: Serre's problem on projective modules over polynomial rings and Algebraic K-theory. Math. U.S.S.R. Izvestija 10 (1976), 937 - 1001. 105. Vasconcelos W.V. : Ideals generated by R-sequence. J. of Algebra 6 (1967), 309 - 316. 106. Valla G. : On set-theoretic complete intersections. Complete intersections (Acireale, 1983), 85-101, LNM 1092,Springer, Berlin, (1984). 107. Valla G.: On determinantal ideals which are set-theoretic complete intersections. Compositio Math. 42 (1980/81), no. 1, 3-11 108. Weil A. : Foundations of Algebraic Geometry. A.M.S Publication, Vol XXIX, (1962). 109. Zariski 0.: The concept of a simple point of an abstract Algebraic Variety. Trans. Am. Math. Soc. 62 (1947), 1-52. 110. Zariski 0. and P. Samuel : Commutative Algebra Vol 1-11. Van Nostrand, Princeton (1958, 1960).

Global Bibliography

The citations below list authors with the Math. Reviews number of the reviews of their articles. The citation [B][2]in the text will indicate that one is refering to the paper corresponding to the second Math Review reference given in [B], namely MR0432626. Further details can then be found via Math. Sci. Net. or Math Reviews. Asanuma, Teruo; MR0862714, MR0951192, MR0678522, MR0564418. Bass, Hyman; MR0527279, MR0432626, MR0527279, MR0564418. Bak, Anthony; MR1115826, MR1329456, MR1810843, MR1810843, MR1329456. Bhatwadekar, S.M.; MR1824883, MR1626485, MR1659957, MR1775418, MR1156463, MR0517136, MR0642336, MR0765772, MR1155426, MR2017619, MR0994128, MR1199695, MR1318088, MR1421074, MR1279265, MR1992040, MR1858341, MR1156463, MR1690788, MR0893595, MR1429296. Borat$nski, M.; MR0533101, MR0511453, MR0895462, MR0883967, MR0877531, MR0837812, MR0822425, MR0783093, MR0677709, MR0533101, MR0533113. Bose, N.K.; MR1861117, MR1803518. Cowsik, R.C.; MR0472835. Das, M. K.; MR1981423, MR2006420. Eisenbud, D.; MR0432627, MR0327742. Ferrand, D.; MR0444646 Fitchas, Noa; MR1124807. van der Kallen, W.; MR0987316, MR0704762. Kopeiko, V.I.; MR1815566, MR0497932, MR0978207. Gubeladze, Joseph; MR1824231, MR1360174, MR1206629, MR1161570, MR1079964, MR0937805. Lindel, Hartmut; MR0641133, MR1322406, MR0689374, MR0796196. Laubenbacher, Reinhard C.; MR1745576, MR1457848. MR1358266. Lam, T.Y.; MR1732042, MR0485842, MR0399212, MR0302739. Lyubeznik, G.; MR1156572, MR0954979, MR1015529, MR0940486, MR0857929, MR0954979. Mandal, Satya; MR1417820, MR1174904, MR1428789, MR1480172, MR1966529, MR1848970, MR1626712, MR0728138, MR0691806. McDonald, Bernard; MR0769104, MR0476639. N. Mohan Kumar; MR1444497, MR1155427, MR0977765, MR0749107, MR0828868, MR0681997, MR0499785, MR1155427, MR1044349, MR0815767. Murthy, Pavaman; MR0611151, MR1626712, MR1697952, MR0396591, MR0977765, MR0422276, MR0940495, MR1044346, MR1298718,MR1732046, MR0200289. Park, Hyungju; MR1358266. Parimala, R.; MR0517136, MR0712622, MR0694376, MR0644232, MR0636877, MR0732195, MR0627667, MR0610478. Popescu, D.; MR1060701, MR0986438, MR0868439, MR0818160. Ojanguren, M.; MR1487451, MR1048288, MR0802307, MR0776448,

References

331

[Ra] Raghunathan, MS.; MR0541022, MR0813075. [RR] Rao, R.A.; MR09555292, MR0766519, MR0952284, MR0991967, MR1317126, MR0709584, MR0784289, MR1317126, MR0784289, MR0709584, MR0919503. [Rt] Roitman, M.; MR0801320, MR0845969. [Ry] Roy, A.; MR0656854, MR0727374, MR0701360, MR0666638. MR0752648. [Sa] Sathaye, A.; MR0499784, MR0533106, MR0722731. [RS] Sridharan, Raja; MR1609901, MR1418947, MR1351372 MR1351370, MR1940666, MR1775418, MR1688449. [Se] Serre, J.-P.; MR0068874, MR0177011, MR0244257, MR0201468. [Su] Suslin, A.; MR0469905, MR0444647, MR0469914, MR0457525, MR0558957, MR0562623, MR0441949, MR0669571, MR0466104, MR0535484. [St] Stafford, J.; MR0617082, MR0577357, MR0662253. [Stm] Sturmfels, Bernd; MR1144671. [Sw] Swan, Richard; MR0469906, MR1697953, MR1256458, MR1144038, MR0954977, MR0921488, MR0396531, MR0218347. [Val Vaserstein, L.N.; MR0447245, MR0840588, MR1086811, MR0982288, MR0882801, MR0619316. MR0830354, MR0830354, MR1086811, MR0982288. [Vo] Vorst, Ton; MR0750693, MR0606650, MR0550060. [W] Warfield, R. B., Jr.; MR0593603, MR0550426, MR0548137.

Index

Artin-Rees Lemma, 303 Bass Cancellation Theorem, 179, 235 Boratynski's Theorem, 244 Carral's Theorem, 146 category, 105 opposite, 107 Cowsik-Nori Theorem, 248 curves, x complete intersection, 214, 241 connectedness, 269 formal connectedness, 270 Hartshorne's Theorem, 270 ideal theoretic, 214 local, 214 set-theoretic, 215 set-theoretical, 245 monomial, x diagram, 106 cartesian, 140 commutative, 14, 107, 140 in a category, 106 pull-back, 140 direct limit, 126 Direct Limits, 126 Eisenbud-Evans Theorem, 231 elementary divisors, 10, 87, 210 for Dedekind Rings, 208 for Principal Domains, 87, 90 Ferrand's construction, 246

fibre bundle, 108 Forster's Conjecture, 319 Forster-Swan Theorem, 221 functors, 105 n-th right derived functors, 265 contravariant, 107, 265 covariant, 107, 265 equivalence, 107 faithful, 107 full, 107 isomorphism, 106 morphism, 106 natural, 107 right derived functors Extn(-, N), 266 general position argument, 5, 231, 232 Prime Avoidance, 5, 167, 198, 232, 284 Gram-Schmidt orthogonalization, 118 Grassmannian, 118 Hilbert, viii, 4, 150 Basisatz, 150 Hilbert ring, 306 Hilbert-Burch Theorem, 318 Nullstellensatz, 4, 154, 168, 249, 306 Strong form, 170 Weak form, 169 Syzygy Theorem, 86 homotopy theory, 147 Hopf map, 133 Horrock's Theorem, 89 Manic Inversion Principle, 90, 96

334

Index

Hypothesis (O), 136 Banach algebras, 136 idempotent, 35 Integral Ring Extensions, 23 Going Up Theorem, 28 Going-Down Theorem, 29 Lying Over Theorem, 28 Inverse Function Theore~n,124 Jordan-Holder Theorem, 156 Jordan-Zassenhaus Theorem, 211 Krull's Intersection, 151, 185, 201, 303 Krull-Akizuki Theorem, 203 Lipschitz condition, 123 Local Global Principle, 91 Localization, 15 exactness, 19 module of fractions, 17 multiplicatively closed, 15 rings of fractions, 17 Macaulay's examples, 214 Modules, 6 M-sequence, 187 HomR(M,N), 12 bidual, 12 dual, 12 evaluation map, 12 left exactness, 43 Pic(A), 200 finite group, 209 torsion, 211 depth(M), 218 grade(1, M ) , 312 homological characterization, 266 Ass(M), 194 finite, 195 basic elements, 226 Cohen-Macaulay, 220 complex, 13 homology, 265 conormal module, 244 dimesnion, 219 direct product, 32 direct sum, 30, 31 Euler characteristic, 197

exact sequence, 13 free resolution, 15 minimal free resolution, 283 projective resolution, 190 short, 13 split, 32 extensions, 259 Baer sum, 261 equivalent, 259 split, 260 faithfully flat, 270 fibre product, 178, 244, 300 finitely generated, 9 finitely presented, 45 Fitting ideals, 314 flat, 45, 270, 309 flip, 321 Flips, 179 free, 9 grade(1, M ) , 218 homological dimension, 189, 242 homothesy, 11, 209 Invariant Basis Property, 142 monogene, 242 order ideals, 176, 226 projective, 33 Inv(A), 62 Prin(A), 62 Um,(R), 71 IPiA), 51 Pic A, 59 completable row, 73 Milnor's construction, 140 nonfree over M[x,y], 66 nonfree stably free, 59, 74 over Topological rings, 136 pull-back, 84 rank, 55 rank one, 59, 63 stably free, 69 unimodular, 226 unimodular row, 71, 176 pull-back, 257 push-out, 258 reflexive, 51, 311 resolution projective, 242 submodules, 8 torsion, 309

Index torsion submodule, 206 transvections, 321 Moh's examples, 215 Mohan Kumar's Theorem, 247 Nagata Transformation, 96 Nakayama's Lemma, 52, 53, 57-59, 74, 91, 151, 155, 185, 187, 193, 198, 202, 217, 219, 221, 226, 227, 232, 243, 247 Generalized, 139 Noether, viii Noether Isomorphism, 12 Noether Normalization, 204, 305 prime ideal, 1 associated, 241 minimal prime, 2, 280 symbolic primes, x Products, 30 direct, 30 infinite, 292 exterior power, 46 tensor, 33 R-linear, 42 additive functor, 42 balanced map, 38 bifunctor, 42 bilinear map, 41 mixed tensors, 39 pure tensors, 39 right exactness, 44 universal property, 38 Projection Lemma, 248 Projective Spaces, 131 canonical line bundle, 132 Riemann sphere, 132 universal line bundle, 132 rings, 1 affine algebra, 162 Ann(x), 20 catenary, 167 Chinese Remainder, 5, 34, 157, 207, 224, 283 Cohen-Macaulay, 220, 269 comaxmial ideals, 33 conductor, 141 Dedekind, 26, 185, 201

335

Class number, 209 Modules over, 205 Picard group, 211 Eisenstein's Criterion, 25 essentially of finite type, 135 factorial, 26, 199, 200 fractional ideals, 59 Gauss numbers, 25 global field, 209 order, 209 Hermite, 73 Hilbert series, 303 integral, 24 integral closure, 25 integrally closed, 25 invariant basis property, 70 invertible ideal, 60 Jacobson ring, 306 local, 23 monoidal, ix Noetherian, viii Artinian, 157 Krull dimension, 154 Small Dimension Theorem, 158, 185, 216 Principal Domain, 76 radical, 16 Jacobson, 52 nilradical, 16 reduced, 16 Rees ring, 303 regular, 185 discrete valuation, 185 factorial, 199 homological characterization, 186 jacobian criterion, 185, 252 local, 125 Rings of Invariants, 304 Molien series, 304 Reynold's operator, 304 semilocal, 23 semiring, 51 Schanuel's Lemma, 83, 85, 86, 129, 189, 196 Serre, 175 Splitting Theorem, 175, 235, 281 special case, 207 Serre's conjecture, vii, 87

336

Index

Lindel, 175 motivation, 241 Quillen Local Global Principle, 94 Quillen ideal, 94, 103 Quillen's conjecture, 125 Splitting Lemma, 94 Suslin, 98 Vaserstein, 101 Small Dimension Theorem, 159 Principal Ideal Theorem, 159, 303 Snake Lemma, 14, 54, 62, 192 Spectral Theorem, 117 hermitian, 117 unitary, 117 Spectrum, 1 maximal spectrum, 4 Stiefel-Whitney classes, 134 Swan's Theorem, 120 Szpiro's Theorem, 247 topological space, 131 C(X), 119 compact, 116 contractible, 131 CW-complex, 147 finite partition of unity subordinate to U ,115 Metric on C(C, D ) , 128 normal, 34, 116 partition of unity, 114 finite envelope of unity, 115 locally finite, 115 quasi-compact, 3 Tietze extension, 145 Urysohn's Lemma, 34 Zariski, 1, 171 Vector bundles, vii, 108 Algebrization, 135 construction, 277

direct sum, 113 dual, 278 finite type, 116 first Chern class, 279 gluing cocycle, 277 guage group, 278 guage tansformation, 278 homomorphism, 110 image, 111 kernel, 111 Hopf, 302 line bundle, 110 determinant, 278 tautological bundle, 132 locally trivial, 110 Mobius, 301 Mobius, vii, 121 metric, 302 Principal G-bundle, x, 279 projective modules, 114 pullback, 277 quasi vector bundle, 109 rank, 110 section, 112 soft, 116, 274, 301 subbundle, 113 tangent bundles, 122 trivial, 109 universal, 278 vector field, 123 Whitney sum, 278 Weierstrass Approximation, 123, 138 Weierstrass' Product Theorem, 150 Whitehead's Lemma, 77, 80, 83, 130, 142, 297 for Rectangular Matrices, 82 generalized, 297 Zorn's Lemma, 2, 56, 206, 295

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Abhyankar, S.S. Resolution of Singularities of Embedded Algebraic Surfaces 2nd enlarged ed. 1998 Andrievskii, V.V.; Blatt, H.-P. Discrepancy of Signed Measures and Polynomial Approximation 2002 Ara, P.; Mathieu, M. Local Multipliers of C*-Algebras 2003 Armitage, D.H.; Gardiner, S.J. Classical Potential Theory 2001 Arnold, I,. Random Dynamical Systems corr. 2nd printing 2003 (1st ed. 1998) Aubin, T. Some Nonlinear Problems in Riemannian Geometry 1998 Auslender, A.; Teboulle M. Asymptotic Cones and Functions in Optimization and Variational Inequalities 2003 Bang-Jensen, J.; Gutin, G. Digraphs 2001 Baues, H.-J. Combinatorial Foundation of Homology and Homotopy 1999 Brown, K.S. Buildings 3rd printing 2000 (1st ed. 1998) Cherry, W.; Ye, Z.Nevanlinna's Theory of Value Distribution 2001 Ching, W.K. Iterative Methods for Queuing and Manufacturing Systems 2001 Crabb, M.C.; James, I.M. Fibrewise Homotopy Theory 1998 Dineen, S. Complex Analysis on Infinite Dimensional Spaces 1999 Elstrodt, J.; Grunewald,F.Mennicke, J.Groups Actingon Hyperbolic Space 1998 Edmunds, D.E.; Evans, W.D.Hardy Operators,Function Spaces and Embeddings 2004 Fadell, E.R.; Husseini, S.Y. Geometry andTopology of Configuration Spaces 2001 Fedorov,Y.N.; Kozlov,V.V. A Memoir on Integrable Systems 2001 Flenner, H.; O'Carroll, L.Voge1, W. Joins and Intersections 1999 Gelfand, S.I.; Manin,Y.I. Methods of Homological Algebra 2nd ed. 2003 Griess,R.L.Jr. Twelve Sporadic Groups 1998 Gras, G. Class Field Theory 2003 Ischebeck, F.; Rao, R.A. Ideals and Reality 2005 Ivrii, V. Microlocal Analysis and Precise Spectral Asymptotics 1998 Jech, 'I: Set Theory (3rd revised edition 2002) Jorgenson, J.; Lang, S. Spherical Inversion on SLn (R) 2001 Kanamori, A.; The Higher Infinite (2nd edition 2003) Khoshnevisan, D. Multiparameter Processes 2002 Koch, H. Galois Theory of p-Extensions 2002 Kozlov, V.; Maz'ya, V. Differential Equations with Operator Coefficients 1999 Landsman, N.P. Mathematical Topics between Classical & Quantum Mechanics 1998 Lebedev, L.P.; Vorovich, 1.1. Functional Analysis in Mechanics 2002 Lemmermeyer, F. Reciprocity Laws: From Euler to Eisenstein 2000 Malle, G.; Matzat, B.H. Inverse Galois Theory 1999 Mardesic, S. Strong Shape and Homology 2000 Margulis, G.A. On Some Aspects of the Theory of Anosov Systems 2004 Murdock, J. Normal Forms and Unfoldings for Local Dynamical Systems 2002 Narkiewicz, W. The Development of Prime Number Theory 2000 Parker, C.; Rowley, P. Symplectic Amalgams 2002 Peller, V. (Ed.) Hankel Operators and Their Applications 2003 Prestel, A.; Delzell, C.N. Positive Polynomials 2001 Puig, L. Blocks of Finite Groups 2002 Ranicki, A. High-dimensional Knot Theory 1998 Ribenboim, P. The Theory of Classical Valuations 1999

Rowe, E.G.P. Geometrical Physics in Minkowski Spacetime 2001 Rudyak, Y.B. On Thom Spectra, Orientability and Cobordism 1998 Ryan, R.A. Introduction to Tensor Products of Banach Spaces 2002 Saranen, J.; Vainikko, G. Periodic Integral and Pseudodifferential Equations with Numerical Approximation 2002 Schneider, P. Nonarchimedean Functional Analysis 2002 Serre, J-P. Complex Semisimple Lie Algebras 2001 (reprint of first ed. 1987) Serre, J-P. Galois Cohomology corr. 2nd printing 2002 (1st ed. 1997) Serre, J-P. Local Algebra 2000 Serre, J-P.Trees corr. 2nd printing 2003 (1st ed. 1980) Smirnov, E. Hausdorff Spectra in Functional Analysis 2002 Springer, 'LA. Veldkamp, F.D. Octonions, Jordan Algebras, and Exceptional Groups 2000 Sznitman, A.4. Brownian Motion, Obstacles and Random Media 1998 Taira, K. Semigroups, Boundary Value Problems and Markov Processes 2003 'Tits, J.; Weiss, R.M. Moufang Polygons 2002 Uchiyama, A. Hardy Spaces on the Euclidean Space 2001 Ustiinel, A.-S.; Zakai, M. Transformation of Measure on Wiener Space 2000 Yang, Y. Solitons in Field Theory and Nonlinear Analysis 2001