For our wives,
Ulrike and Maja
Contents
Preface to the revised edition Preface to the rst edition
xi xii
I Basic concepts
1
1 Regular sequences and depth 1.1 Regular sequences : : : : : : : : : : : : : : : : : : : 1.2 Grade and depth : : : : : : : : : : : : : : : : : : : : 1.3 Depth and projective dimension : : : : : : : : : : : 1.4 Some linear algebra : : : : : : : : : : : : : : : : : : 1.5 Graded rings and modules : : : : : : : : : : : : : : 1.6 The Koszul complex : : : : : : : : : : : : : : : : : : Notes : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2 Cohen{Macaulay rings 2.1 Cohen{Macaulay rings and modules : : : : : : : : 2.2 Regular rings and normal rings : : : : : : : : : : : 2.3 Complete intersections : : : : : : : : : : : : : : : : : Notes : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 3 The canonical module. Gorenstein rings 3.1 Finite modules of nite injective dimension : : : : 3.2 Injective hulls. Matlis duality : : : : : : : : : : : : : 3.3 The canonical module : : : : : : : : : : : : : : : : : 3.4 Gorenstein ideals of grade 3. Poincare duality : : : 3.5 Local cohomology. The local duality theorem : : : 3.6 The canonical module of a graded ring : : : : : : : Notes : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 4 Hilbert functions and multiplicities 4.1 Hilbert functions over homogeneous rings : : : : : 4.2 Macaulay's theorem on Hilbert functions : : : : : 4.3 Gotzmann's regularity and persistence theorem : : 4.4 Hilbert functions over graded rings : : : : : : : : : 4.5 Filtered rings : : : : : : : : : : : : : : : : : : : : : : 4.6 The Hilbert{Samuel function and reduction ideals 4.7 The multiplicity symbol : : : : : : : : : : : : : : : : Notes : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : vii
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3 3 8 16 18 27 40 55 57 57 65 73 85 88 88 97 107 120 127 136 144 147 147 156 168 173 182 188 192 201
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II Classes of Cohen{Macaulay rings 5 Stanley{Reisner rings 5.1 Simplicial complexes : : : : : : : : : : : : : : : : : 5.2 Polytopes : : : : : : : : : : : : : : : : : : : : : : : 5.3 Local cohomology of Stanley{Reisner rings : : : 5.4 The upper bound theorem : : : : : : : : : : : : : 5.5 Betti numbers of Stanley-Reisner rings : : : : : : 5.6 Gorenstein complexes : : : : : : : : : : : : : : : : 5.7 The canonical module of a Stanley{Reisner ring Notes : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 6 Semigroup rings and invariant theory 6.1 Ane semigroup rings : : : : : : : : : : : : : : : : 6.2 Local cohomology of ane semigroup rings : : : 6.3 Normal semigroup rings : : : : : : : : : : : : : : : 6.4 Invariants of tori and nite groups : : : : : : : : 6.5 Invariants of linearly reductive groups : : : : : : Notes : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 7 Determinantal rings 7.1 Graded Hodge algebras : : : : : : : : : : : : : : : 7.2 Straightening laws on posets of minors : : : : : : 7.3 Properties of determinantal rings : : : : : : : : : Notes : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
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III Characteristic p methods 8 Big Cohen{Macaulay modules 8.1 The annihilators of local cohomology : : : : : : : : 8.2 The Frobenius functor : : : : : : : : : : : : : : : : : 8.3 Modi cations and non-degeneracy : : : : : : : : : 8.4 Hochster's niteness theorem : : : : : : : : : : : : : 8.5 Balanced big Cohen{Macaulay modules : : : : : : Notes : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 9 Homological theorems 9.1 Grade and acyclicity : : : : : : : : : : : : : : : : : : 9.2 Regular rings as direct summands : : : : : : : : : : 9.3 Canonical elements in local cohomology modules 9.4 Intersection theorems : : : : : : : : : : : : : : : : : 9.5 Ranks of syzygies : : : : : : : : : : : : : : : : : : : 9.6 Bass numbers : : : : : : : : : : : : : : : : : : : : : : Notes : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
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ix
Contents
10 Tight closure 10.1 The tight closure of an ideal : : : 10.2 The Briancon{Skoda theorem : : 10.3 F -rational rings : : : : : : : : : : : 10.4 Direct summands of regular rings Notes : : : : : : : : : : : : : : : : : : : : Appendix: A summary of dimension theory References Notation Index
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Preface to the revised edition
The main change in the revised edition is the new Chapter 10 on tight closure. This theory was created by Mel Hochster and Craig Huneke about ten years ago and is still strongly expanding. We treat the basic ideas, F -regular rings, and F -rational rings, including Smith's theorem by which F -rationality implies pseudo-rationality. Among the numerous applications of tight closure we have selected the Briancon{Skoda theorem and the theorem of Hochster and Huneke saying that equicharacteristic direct summands of regular rings are Cohen{Macaulay. To cover these applications, Section 8.4, which develops the technique of reduction to characteristic p, had to be rewritten. The title of Part III, no longer appropriate, has been changed. Another noteworthy addition are the theorems of Gotzmann in the new Section 4.3. We believe that Chapter 4 now treats all the basic theorems on Hilbert functions. Moreover, this chapter has been slightly reorganized. The new Section 5.5 contains a proof of Hochster's formula for the Betti numbers of a Stanley{Reisner ring since the free resolutions of such rings have recently received much attention. In the rst edition the formula was used without proof. We are grateful to all the readers of the rst edition who have suggested corrections and improvements. Our special thanks go to L. Avramov, A. Conca, S. Iyengar, R. Y. Sharp, B. Ulrich, and K.-i. Watanabe. Osnabruck and Essen, October 1997
Winfried Bruns Jurgen Herzog
xi
Preface to the rst edition
The notion of a Cohen{Macaulay ring marks the cross-roads of two powerful lines of research in present-day commutative algebra. While its main development belongs to the homological theory of commutative rings, it nds surprising and fruitful applications in the realm of algebraic combinatorics. Consequently this book is an introduction to the homological and combinatorial aspects of commutative algebra. We have tried to keep the text self-contained. However, it has not proved possible, and would perhaps not have been appropriate, to develop commutative ring theory from scratch. Instead we assume the reader has acquired some uency in the language of rings, ideals, and modules by working through an introductory text like Atiyah and Macdonald 15] or Sharp 344]. Nevertheless, to ease the access for the non-expert, the essentials of dimension theory have been collected in an appendix. As exempli ed by Matsumura's standard textbook 270], it is natural to have the notions of grade and depth follow dimension theory, and so Chapter 1 opens with the introduction of regular sequences on which their de nition is based. From the very beginning we stress their connection with homological and linear algebra, and in particular with the Koszul complex. Chapter 2 introduces Cohen{Macaulay rings and modules, our main subjects. Next we study regular local rings. They form the most special class of Cohen{Macaulay rings their theory culminates in the Auslander{ Buchsbaum{Serre and Auslander{Buchsbaum{Nagata theorems. Unlike the Cohen{Macaulay property in general, regularity has a very clear geometric interpretation: it is the algebraic counterpart of the notion of a non-singular point. Similarly the third class of rings introduced in Chapter 2, that of complete intersections, is of geometric signi cance. In Chapter 3 a new homological aspect determines the development of the theory, namely the existence of injective resolutions. It leads us to the study of Gorenstein rings which in several respects are distinguished by their duality properties. When a Cohen{Macaulay local ring is not Gorenstein, then (almost always) it has at least a canonical module which, so to speak, acts as its natural partner in duality theorems, a decisive fact for many combinatorial applications. We then introduce local cohomology and prove Grothendieck's vanishing and local duality theorems. xii
Preface to the rst edition
xiii
Chapter 4 contains the combinatorial theory of commutative rings which mainly consists in the study of the Hilbert function of a graded module and the numerical invariants derived from it. A central point is Macaulay's theorem describing all possible Hilbert functions of homogeneous rings by a numerical condition. The intimate connection between homological and combinatorial data is displayed by several theorems, among them Stanley's characterization of Gorenstein domains. In the second part of this chapter the method of associated rings and modules is developed and used for assigning numerical invariants to modules over local rings. Chapters 1{4 form the rst part of the book. We consider this material as basic. The second part consists of Chapters 5{7 each of which is devoted to a special class of rings. Chapter 5 contains the theory of Stanley{Reisner rings of simplicial complexes. Its main goal is the proof of Stanley's upper bound theorem for simplicial spheres. The transformation of this topological notion into an algebraic condition is through Hochster's theorem which relates simplicial homology and local cohomology. Furthermore we study the Gorenstein property for simplicialcomplexes and their canonical modules. In Chapter 6 we investigate normal semigroup rings. The combinatorial object represented by a normal semigroup ring is the set of lattice points within a convex cone. According to a theorem of Hochster, normal semigroup rings are Cohen{Macaulay. Again the crucial point is the interplay between cellular homology on the geometric side and local cohomology on the algebraic. The fact that the ring of invariants of a linear torus action on a polynomial ring is a normal semigroup ring leads us naturally to the study of invariant rings, in particular those of nite groups. The chapter closes with the Hochster{Roberts theorem by which a ring of invariants of a linearly reductive group is Cohen{Macaulay. Chapter 7 is devoted to determinantal rings. They are discussed in the framework of Hodge algebras and algebras with straightening laws. We establish the straightening laws of Hodge and of Doubilet, Rota, and Stein, prove that determinantal rings are Cohen{Macaulay, compute their canonical module, and determine the Gorenstein rings among them. In view of the extensive treatment available in 61], we have restricted this chapter to the absolutely essential. The third part of the book is constituted by Chapters 8 and 9. They owe their existence to the fact that a Noetherian local ring is in general not Cohen{Macaulay. But Hochster has shown that such a ring possesses a (not necessarily nite) Cohen{Macaulay module, at least when it contains a eld. The construction of these `big' Cohen{Macaulay modules in Chapter 8 is a paradigm of characteristic p methods in commutative algebra, and we hope that it will prepare the reader for the more recent developments in this area which are centered around the
xiv
Preface to the rst edition
notion of tight closure introduced by Hochster and Huneke 190]. In Chapter 9 we deduce the consequences of the existence of big Cohen{Macaulay modules, for example the intersection theorems of Peskine and Szpiro and Roberts, the Evans{Grith syzygy theorem, and bounds for the Bass numbers of a module. Chapters 8 and 9 are completely independent of Chapters 4{7, and the reader who is only interested in the homological theory may proceed from the end of Section 3.5 directly to Chapter 8. It is only to be expected that the basic notions of homological algebra are ubiquitous in our book. But most of the time we will only use the long exact sequences for Ext and Tor, and the behaviour of these functors under at extensions. Where we go beyond that, we have inserted a reference to Rotman 318]. One may regard it as paradoxical that we freely use the Ext functors while Chapter 3 contains a complete treatment of injective modules. However, their theory has several peculiar aspects so that we thought such a treatment would be welcomed by many readers. The book contains numerous exercises. Some of them will be used in the main text. For these we have provided hints or even references to the literature, unless their solutions are completely straightforward. A reference of type A.n points to a result in the appendix. Parts of this book were planned while we were guests of the Mathematisches Forschungsinstitut Oberwolfach. We thank the Forschungsinstitut for its generous hospitality. We are grateful to all our friends, colleagues, and students, among them L. Avramov, C. Baetica, M. Barile, A. Conca, H.{B. Foxby, C. Huneke, D. Popescu, P. Schenzel, and W. Vasconcelos who helped us by providing valuable information and by pointing out mistakes in preliminary versions. Our sincere thanks go to H. Matsumura and R. Sharp for their support in the early stages of this project. We are deeply indebted to our friend Udo Vetter for reading a large part of the manuscript and for his unfailing criticism. Vechta and Essen, February 1993
Winfried Bruns Jurgen Herzog
Part I
Basic concepts
1
1 Regular sequences and depth
After dimension, depth is the most fundamental numerical invariant of a Noetherian local ring R or a nite R -module M . While depth is de ned in terms of regular sequences, it can be measured by the (non-)vanishing of certain Ext modules. This connection opens commutative algebra to the application of homological methods. Depth is connected with projective dimension and several notions of linear algebra over Noetherian rings. Equally important is the description of depth (and its global relative grade) in terms of the Koszul complex which, in a sense, holds an intermediate position between arithmetic and homological algebra. This introductory chapter also contains a section on graded rings and modules. These allow a decomposition of their elements into homogeneous components and therefore have a more accessible structure than rings and modules in general. 1.1 Regular sequences Let M be a module over a ring R . We say that x 2 R is an M-regular element if xz = 0 for z 2 M implies z = 0, in other words, if x is not a zero-divisor on M . Regular sequences are composed of successively regular elements: De nition 1.1.1. A sequence x = x1 . . . xn of elements of R is called an M -regular sequence or simply an M -sequence if the following conditions are satis ed: (i) xi is an M=(x1 . . . xi;1)M -regular element for i = 1 . . . n, and (ii) M=xM 6= 0. In this situation we shall sometimes say that M is an x-regular module. A regular sequence is an R -sequence. A weak M-sequence is only required to satisfy condition (i). Very often R will be a local ring with maximal ideal , and M 6= 0 a nite R -module. If x , then condition (ii) is satis ed automatically because of Nakayama's lemma. The classical example of a regular sequence is the sequence X1 . . . Xn of indeterminates in a polynomial ring R = S X1 . . . Xn]. Conversely we shall see below that an M -sequence behaves to some extent like a sequence of indeterminates this will be made precise in 1.1.8. The next proposition contains a condition under which a regular sequence stays regular when the module or the ring is extended. m
m
3
4
1. Regular sequences and depth
Proposition 1.1.2. Let R be a ring, M an R-module, and x R a weak M-sequence. Suppose ' : R ! S is a ring homomorphism, and N an R-at S-module. Then x R and '(x) S are weak (M R N )-sequences. If x(M R N ) 6= M R N, then x and '(x) are (M R N )-sequences. Proof. Multiplication by xi is the same operation on M N as multiplication by '(xi) so it suces to consider x. The homothety x1 : M ! M is injective, and x1 N is injective too, because N is at. Now x1 N is just multiplication by x1 on M N . So x1 is an (M N )-regular element. Next we have (M N )=x1 (M N ) = (M=x1M ) N an inductive argument will therefore complete the proof. The most important special cases of 1.1.2 are given in the following corollary. In its part (b) we use M^ to denote the -adic completion of a module M over a local ring (R k) (by this notation we indicate that R has maximal ideal and residue class eld k = R= ). Corollary 1.1.3. Let R be a Noetherian ring, M a nite R-module, and x m
m
m
m
an M-sequence. (a) Suppose that a prime ideal Supp M contains x. Then x (as a sequence in R ) is an M -sequence. (b) Suppose that R is local with maximal ideal . Then x (as a sequence ^ in R^ ) is an M-sequence.
2
p
p
p
m
and R ! R^ are at. (a) By hypothesis M 6= 0, and Nakayama's lemma implies M 6= M . A fortiori we have xM 6= M . (b) It suces to note that M^ = M R^ is a nite R^ -module. Proof. Both the extensions R
!R
p
p
p
p
p
p
p
The interplay between regular sequences and homological invariants is a major theme of this book, and numerous arguments will be based on the next proposition. Proposition 1.1.4. Let R be a ring, M an R-module, and x a weak Msequence. Then an exact sequence N2
' ' ' ;! N1 ;! N0 ;! M ;! 0 2
1
0
of R-modules induces an exact sequence N2 =xN2
;! N1=xN1 ;! N0=xN0 ;! M=xM ;! 0:
By induction it is enough to consider the case in which x consists of a single M -regular element x. We obtain the induced sequence if we tensor the original one by R=(x). Since tensor product is a right exact functor, we only need to verify exactness at N1 =xN1 . Let denote residue classes modulo x. If ' 1( y) = 0, then '1 (y) = xz for some z 2 N0 and Proof.
5
1.1. Regular sequences
2
x'0(z ) = 0. By hypothesis we have '0(z ) = 0 hence there is y0 N1 with z = '1 (y0 ). It follows that '1 (y xy0 ) = 0. So y xy0 '2(N2 ), and y ' 2(N 2 ) as desired.
;
2
; 2
If we want to preserve the exactness of a longer sequence, then we need a stronger hypothesis. Proposition 1.1.5. Let R be a ring and 'm ' N : ;! Nm ;! Nm;1 ;! ;! N0 ;! N;1 ;! 0 0
.
an exact complex of R-modules. If x is weakly Ni -regular for all i, then N. R=(x) is exact again.
Proof. Once more one uses induction on the length of the sequence x. So it is enough to treat the case x = x. Since x is regular on Ni , it is regular on Im 'i+1 too. Therefore we can apply 1.1.4 to each exact sequence Ni+3 ! Ni+2 ! Ni+1 ! Im 'i+1 ! 0. Easy examples show that a permutation of a regular sequence need not be a regular sequence see 1.1.13. Nevertheless there are natural conditions under which regular sequences can be permuted. Let x1 x2 be an M -sequence, and denote the kernel of the multiplication by x2 on M by K . Suppose that z 2 K . Then we must have z 2 x1M , z = x1 z 0 , and x1 (x2z 0 ) = 0, whence x2 z 0 = 0 and z 0 2 K , too. This shows K = x1 K so that K = 0 if Nakayama's lemma is applicable. Somewhat surprisingly, x1 is always regular on M=x2M the reader may check this easily. Proposition 1.1.6. Let R be a Noetherian local ring, M a nite R-module, and x = x1 . . . xn an M-sequence. Then every permutation of x is an
M-sequence.
Proof. Every permutation is a product of transpositions of adjacent elements. Therefore it is enough to show that x1 . . . xi+1 xi . . . xn is = an M -sequence. The hypothesis of the proposition is satis ed for M M=(x1 . . . xi;1)M and the M -sequence xi . . . xn . So it suces to treat the case i = 1 and to show that x2 x1 is an M -sequence. In view of the discussion above we only need to appeal to Nakayama's lemma.
Quasi-regular sequences. Let R be a ring, M an R -module, and X = X1 . . . Xn be indeterminates over R . Then we write M X ] for M R X ] and call its elements polynomials with coecients in M . If x = x1 . . . xn is a sequence of elements of R , then the substitution Xi xi induces an R -algebra homomorphism R X ] R and also an R -module homomorphism M X ] M . We write F (x) for the image of F M X ] under this map. (Since the monomials form a basis of the free R -module R X ], we may speak of the coecients and the degree of an element of M X ].)
!
!
2
7!
6
1. Regular sequences and depth
Theorem 1.1.7 (Rees). Let R be a ring, M an R-module, x = x1 . . . xn an M-sequence, and I = (x1 . . . xn). Let X = X1 . . . Xn be indeterminates over R. If F 2 M X ] is homogeneous of (total) degree d and F (x) 2 I d +1M, then the coecients of F are in IM.
We use induction on n. The case n = 1 is easy. Let n > 1 and suppose that the theorem holds for regular sequences of length at most n ; 1. We must rst prove an auxiliary result which is an interesting fact in itself: let J = (x1 . . . xn;1) then xn is regular on M=J j M for all j 1. In fact, suppose that xny 2 J j M for some j > 1. Arguing by induction we have y 2 J j ;1M so y = G(x1 . . . xn;1) where G 2 M X1 . . . Xn;1] is homogeneous of degree j ; 1. Set G0 = xn G. Then the theorem applied to G0 2 M X1 . . . Xn;1] yields that the coecients of G0 are in JM . Since xn is regular modulo JM , it follows that the coecients of G are in JM too, and therefore y 2 J j M . The proof of the theorem for sequences of length n requires induction on d . The case d = 0 is trivial. Assume that d > 0. First we reduce to the case in which F (x) = 0. Since F (x) 2 I d +1P M , one has F (x) = G(x) with G homogeneous of degree d + 1. Then GP= ni=1 Xi Gi with Gi homogeneous of degree d . Set G0i = xi Gi and G0 = ni=1 G0i. So F ; G0 is homogeneous of degree d , and (F ; G0 )(x) = 0. Furthermore, F ; G0 has coecients in IM if and only if this holds for F . Thus assume that F (x) = 0. Then we write F = G + Xn H with G 2 M X1 . . . Xn;1]. The auxiliary claim above implies that H (x) 2 J d M I d M . By induction on d the coecients of H are in IM . On the other hand H (x) = H 0 (x1 . . . xn;1) with H 0 2 M X1 . . . Xn;1] homogeneous of degree d . As Proof.
(G + xn H 0 )(x1 . . . xn;1) = F (x) = 0 it follows by induction on n that G + xn H 0 has coecients in JM . Since xnH 0 has its coecients in IM , the coecients of G must be in IM too. Let I be an ideal in R . One de nes the associated graded ring of R with respect to I by grI (R ) =
1 M i=0
I i =I i+1:
The multiplication in grI (R ) is induced by the multiplication I i I j ! I i+j , and grI (R ) is a graded ring with (grI (R ))0 = R=I . If M is an R -module, one similarly constructs the associated graded module grI (M ) =
1 M i=0
I i M=I i+1M:
7
1.1. Regular sequences
It is straightforward to verify that grI (M ) is a graded grI (R )-module. (Graded rings and modules will be discussed in Section 1.5. The reader not familiar with the basic terminology may wish to consult 1.5.) Let I be generated by x1 . . . xn. Then one has a natural surjection R X ] = R X1 . . . Xn ] ! grI (R ) which is induced by the natural homomorphism R ! R=I and the substitution Xi 7! x i 2 I=I 2 . Similarly there is an epimorphism : M X ] ! grI (M ). One rst de nes on the homogeneous components by assigning to a homogeneous polynomial F 2 M X ] of degree d the residue class of F (x) in I d M=I d +1M then is extended additively. As the reader may check, is an epimorphism of graded R X ]-modules. Obviously IM X ] Ker via the identi cation M X ]=IM X ] = (M=IM )X ], we therefore get an induced epimorphism ' : (M=IM )X ] ! grI (M ). The kernel of is generated by the homogeneous polynomials F 2 M X ] of degree d , d 2 N, such that F (x) 2 I d +1M . So we obtain as a reformulation of 1.1.7 Theorem 1.1.8. Let R be a ring, M an R-module, x = x1 . . . xn an M-sequence, and I = (x). Then the map (M=IM )X1 . . . Xn ] ! grI (M ) induced by the substitution Xi 7! x i 2 I=I 2 is an isomorphism. This theorem says very precisely to what extent a regular sequence resembles a sequence of indeterminates: the residue classes x i 2 I=I 2 operate on grI (M ) exactly like indeterminates. Since a regular sequence may lose regularity under a permutation, whereas 1.1.8 is independent of the order in which x is given, it is not possible to reverse 1.1.8 see however 1.1.15. Later on it will be useful to have a name for sequences x satisfying the conclusion of 1.1.8 we call them M -quasi-regular if, in addition, xM 6= M . Exercises
Let 0 ! U ! M ! N ! 0 be an exact sequence of R -modules, and x a sequence which is weakly U -regular and (weakly) N -regular. Prove that x is (weakly) M -regular too. 1.1.10. (a) Let x1 . . . xi . . . xn and x1 . . . xi . . . xn be (weakly) M -regular. Show that x1 . . . xi xi . . . xn is (weakly) M -regular. (Hint: In the essential case i = 1 one nds an exact sequence as in 1.1.9 with M=x1 x1 M as the middle term.) (b) Prove that xe1 . . . xenn is (weakly) M -regular for all ei 1. 1.1.11. Prove that the converse of 1.1.2 holds if, in the situation of 1.1.2, N is faithfully at over R . 1.1.12. (a) Prove that if x is a weak M -sequence, then TorR1 (M R=(x)) = 0. (b) Prove that if, in addition, x is a weak R -sequence, then TorRi (M R=(x)) = 0 for all i 1. 1.1.13. Let R = K X Y Z ], k a eld. Show that X , Y (1 ; X ), Z (1 ; X ) is an R -sequence, but Y (1 ; X ), Z (1 ; X ), X is not. 1.1.9.
0
0
0
1
8
1. Regular sequences and depth
Prove that x1 . . . xn is M -quasi-regular if and only if x1 . . . xn 2 I=I 2 is a grI (M )-regular sequence where I = (x1 . . . xn ). 1.1.15. Suppose that x is M -quasi-regular, and let I = (x1 . . . xn ). Prove (a) if x1 z 2 I i M for z 2 M , then z 2 I i 1 M , (b) x2 . . . xn is (M=x1 M )-quasi-regular, (c) if R is Noetherian local and M is nite, then x is an M -sequence.
1.1.14.
;
1.2 Grade and depth Let R be a Noetherian ring and M an R -module. If x = x1 . . . xn is an M -sequence, then the sequence (x1) (x1 x2) (x1 . . . xn ) ascends strictly for obvious reasons. Therefore an M -sequence can be extended to a maximal such sequence: an M -sequence x (contained in an ideal I ) is maximal (in I ), if x1 . . . xn+1 is not an M -sequence for any xn+1 2 R (xn+1 2 I ). We will prove that all maximal M -sequences in an ideal I with IM 6= M have the same length if M is nite. This allows us to introduce the fundamental notions of grade and depth. In connection with regular sequences, nite modules over Noetherian rings are distinguished for two reasons: rst, every zero-divisor of M is contained in an associated prime ideal, and, second, the number of these prime ideals is nite. Both facts together imply the following proposition that is `among the most useful in the theory of commutative rings' (Kaplansky 231], p. 56). Proposition 1.2.1. Let R be a Noetherian ring, and M a nite R-module.
If an ideal I R consists of zero-divisors of M, then I for some Ass M. Proof. If I for all Ass M , then there exists a I with a = for all Ass M . This follows immediately from 1.2.2.
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The following lemma, which we have just used in its simplest form, is the standard argument of `prime avoidance'. Lemma 1.2.2. Let R be a ring, P 1 . . . m prime ideals, M an R-module, and x1 . . . xn 2 M. Set N = ni=1 Rxi. If NPj 6 j M j for j = 1 . . . m, then there exist a2 . . . an 2 R such that x1 + ni=2 ai xi 2= j M j for j = 1 . . . m. Proof. We use induction on m, and so suppose that there are a02 . . . a0n 2 R P n 0 for which x1 = x1 + i=2 a0i xi 2= j M j for j = 1 . . . m ; 1. Moreover, it is no restriction to assume that the i are pairwise distinct and that T m;1 m is a minimal member of 1 . . . m . So Pthere exists r 2 ( j=1 j ) n m. Put x0i = rxi for i = 2 . . . n and N 0 = ni=1 Rx0i. Since r 2= m we have N 0 m = N m . On the other hand, as r 2 j for j = 1 . . . m ; 1, it follows that x01 + x0i 2= j M j for i = 2 . . . n and j = 1 . . . m ; 1. If x01 2= m M m , then x01 is the element desired otherwise x01 + x0i 2= m M m for some i 2 f2 . . . ng, and we choose x01 + x0i. p
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9
1.2. Grade and depth
Note that if M = R and N = I R , then the condition N j 6 j M j simpli es to I 6 j . Suppose that an ideal I is contained in 2 Ass M . By de nition, there exists z 2 M with = Ann z . Hence the assignment 1 7! z induces a monomorphism '0 : R= ! M , and thus a non-zero homomorphism ' : R=I ! M . This simple observation allows us to describe in homological terms that a certain ideal consists of zero-divisors: Proposition 1.2.3. Let R be a ring, and M N R-modules. Set I = Ann N. (a) If I contains an M-regular element, then HomR (N M ) = 0. (b) Conversely, if R is Noetherian, and M, N are nite, HomR (N M ) = 0 p
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implies that I contains an M-regular element.
(a) is evident. (b) Assume that I consists of zero-divisors of M , and apply 1.2.1 to nd a 2 Ass M such that I . By hypothesis, 2 Supp N so N k( ) 6= 0 by Nakayama's lemma, and since N k( ) is just a direct sum of copies of k( ), one has an epimorphism N ! k( ). (By k( ) we denote the residue class eld R = R of R .) Note that R 2 Ass M . Hence the observation above yields a non-zero '0 2 HomR (N M ). Since HomR (N M ) = HomR (N M ) , it follows that HomR (N M ) 6= 0. (See 318], Theorem 3.84 for the isomorphism just applied.) Lemma 1.2.4. Let R be a ring, M, N be R-modules, and x = x1 . . . xn a weak M-sequence in Ann N. Then HomR (N M=xM ) = ExtnR (N M ): Proof.
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We use induction on n, starting from the vacuous case n = 0. Let n 1, and set x0 = x1 . . . xn;1. Then the induction hypothesis implies that ExtnR;1(N M ) = HomR (N M=x0 M ). As xn is (M=x0 M )n ; 1 regular, ExtR (N M ) = 0 by 1.2.3. Therefore the exact sequence Proof.
0 ;! M ;! M ;! M=x1 M ;! 0 x1
yields an exact sequence
0 ;! ExtnR;1(N M=xM ) ;! ExtnR (N M ) ;! ExtnR (N M ):
'
The map ' is multiplication by x1 inherited from M , but multiplication by x1 on N also induces ' see 318], Theorem 7.16. Since x1 2 Ann N , one has ' = 0. Hence is an isomorphism, and a second application of the induction hypothesis yields the assertion. Let R be Noetherian, I an ideal, M a nite R -module with M 6= IM , and x = x1 . . . xn a maximal M -sequence in I . From 1.2.3 and 1.2.4
10
1. Regular sequences and depth
we have, since I contains an (M=(x1 . . . xi;1)M )-regular element for i = 1 . . . n, ; ExtiR;1(R=I M ) = HomR R=I M=(x1 . . . xi;1)M = 0: On the other hand, since IM 6= M and x is a maximal M -sequence in I , then I must consist of zero-divisors of M=xM , whence ExtnR (R=I M ) = HomR (R=I M=xM ) 6= 0: We have therefore proved Theorem 1.2.5 (Rees). Let R be a Noetherian ring, M a nite R-module, and I an ideal such that IM 6= M. Then all maximal M-sequences in I have the same length n given by
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n = min i : ExtiR (R=I M ) = 0 :
De nition 1.2.6. Let R be a Noetherian ring, M a nite R -module, and I an ideal such that IM 6= M . Then the common length of the maximal M -sequences in I is called the grade of I on M , denoted by grade(I M ): We complement this de nition by setting grade(I M ) = 1 if IM = M . This is consistent with 1.2.5: grade(I M ) = 1 () ExtiR (R=I M ) = 0 for all i: For, if IM = M , then Supp M \ Supp R=I = by Nakayama's lemma, hence (1) Supp ExtiR (R=I M ) Supp M \ Supp R=I = conversely, if ExtiR (R=I M ) = 0 for all i, then 1.2.5 gives IM = M . The inclusion in (1) results from the natural isomorphism ExtiR (N M ) = ExtiR (N M ) which holds if R is Noetherian, N a nite R -module, M an arbitrary R -module, and 2 Spec R see 318], Theorem 9.50. A special situation will occur so often that it merits a special notation: De nition 1.2.7. Let (R k) be a Noetherian local ring, and M a nite R -module. Then the grade of on M is called the depth of M , denoted depth M: p
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Because of its importance we repeat the most often used special case of 1.2.5:
11
1.2. Grade and depth
Theorem 1.2.8. Let (R k) be a Noetherian local ring, and M a nite non-zero R-module. Then depth M = min i : ExtiR (k M ) = 0 .
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Some formulas for grade. We now study the behaviour of grade(I M )
along exact sequences. Proposition 1.2.9. Let R be a Noetherian ring, I R an ideal, and 0 ! U ! M ! N ! 0 an exact sequence of nite R-modules. Then grade(I M ) minfgrade(I U ) grade(I N )g grade(I U ) minfgrade(I M ) grade(I N ) + 1g grade(I N ) minfgrade(I U ) ; 1 grade(I M )g: The given exact sequence induces a long exact sequence ! ExtiR;1(R=I N) ! ExtiR (R=I U) ! ExtiR (R=I M) ! ExtiR (R=I N) ! ExtiR+1(R=I U) !
Proof.
One observes that ExtiR (R=I M ) = 0 if ExtiR (R=I U ) and ExtiR (R=I N ) both vanish. Therefore the rst inequality follows from 1.2.5 and our discussion of the case grade(I ) = 1. Completely analogous arguments show the second and the third inequality. The next proposition collects some formulas which are useful in the computation of grades. (In the sequel V (I ) denotes the set of prime ideals containing I .) Proposition 1.2.10. Let R be a Noetherian ring, I J ideals of R, and M a
nite R-module. Then (a) grade(I M ) = inf depth M : V (I ) , (b) grade(I M ) = grade(Rad I M ), (c) grade(I J M ) = min grade(I M ), grade(J M ) , (d) if x = x1 . . . xn is an M-sequence in I, then grade(I=(x), M=xM ) = grade(I M=xM ) = grade(I M ) n, (e) if N is a nite R-module with Supp N = V (I ), then
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;
grade(I M ) = inf fi : ExtiR (N M ) 6= 0g:
grade( M) for 2 V (I ), and it follows from 1.1.3 that grade( M ) depth M . Furthermore, if grade(I M ) = 1, then Supp M \ V (I ) = so that depth M = 1 for all 2 V (I ). Thus suppose IM 6= M and choose a maximal M -sequence x in I . By 1.2.1 there exists 2 Ass(M=xM ) with I . Since R 2 Ass(M=xM ) and (M=xM ) = M =xM , the ideal Proof. (a) It is evident from the de nition that grade(I M ) p
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1. Regular sequences and depth
R consists of zero-divisors of M =xM , and x (as a sequence in R ) is a maximal M -sequence. (b) and (c) follow easily from (a). = M=xM . Elementary argu(d) Set R = R=(x), I = I=(x), and M () I M =M . Furthermore ments show that IM = M () I M = M y1 . . . yn 2 I form an M -sequence if and only if y 1 . . . y n 2 I form such a sequence. This proves the rst equation. The second equation results from 1.2.5. (e) The hypothesis entails that Rad Ann N = Rad I . By (b) we may therefore assume that I = Ann N . Now one repeats the proof of 1.2.5 (and the discussion of the case IM = M ) with R=I replaced by N . The name `grade' was originally used by Rees 303] for a di erent, though related invariant: De nition 1.2.11. Let R be a Noetherian ring and M 6= 0 a nite R module. Then the grade of M is given by grade M = minfi : ExtiR (M R ) 6= 0g: For systematic reasons the grade of the zero-module is in nity. It follows directly from 1.2.10(e) that grade M = grade(Ann M R ). It is customary to set grade I = grade R=I = grade(I R ) for an ideal I R , and we follow this convention. (Of course, grade I has two di erent meanings now, but we will never use it to denote the grade of the module I .) p
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Depth and dimension. Let (R ) be Noetherian local and M a nite R module. All the minimal elements of Supp M belong to Ass M . Therefore, if x is an M -regular element, then x = for all minimal elements of Supp M , and induction yields dim M=xM = dim M n if x = x1 . . . xn is an M -sequence. (Note that dim M=xM dim M n is automatic see m
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A.4.) We have proved: Proposition 1.2.12. Let (R ) be a Noetherian local ring and M = 6 0a m
nite R-module. Then every M-sequence is part of a system of parameters of M. In particular depth M dim M.
The inequality in 1.2.12 can be somewhat re ned: Proposition 1.2.13. With the notation of 1.2.12 one has depth M dim R= for all 2 Ass M. Proof. We use induction on depth M . There is nothing to prove for depth M = 0. If depth M > 0, then there exists an M -regular x 2 . For p
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1.2. Grade and depth
2 Ass M we choose z 2 M such that Rz is maximal among the cyclic submodules of M annihilated by . If z 2 xM , then z = xy with y 2 M , and y = 0 since x is M -regular moreover, Rz is a proper submodule of Ry, contrary to the choice of z . Therefore consists of zero-divisors of M=xM , and is contained in some 2 Ass(M=xM ). As x 2= , we have 2= Supp(M=xM), and thus 6= . Now depth(M=xM) = depth M ; 1 by 1.2.10, whence, by induction, dim R= > dim R= depth(M=xM ) = depth M ; 1: A global variant of 1.2.12 says that height bounds grade. Proposition 1.2.14. Let R be a Noetherian ring and I R an ideal. Then grade I height I. Proof. Since grade I = inf fdepth R : 2 V (I )g by 1.2.10, and height I = inf fdim R : 2 V (I )g, the assertion follows from 1.2.12. p
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Depth, type, and at extensions. Finally we investigate how depth behaves
under at local extensions. As a by-product we obtain a result on the behaviour of the type of a module under such extensions. This is an invariant which re nes the information given by the depth: De nition 1.2.15. Let (R k) be a Noetherian local ring, and M a nite non-zero R -module of depth t. The number r(M ) = dimk ExttR (k M ) is called the type of M . Proposition 1.2.16. Let ' : (R k) ! (S l ) be a homomorphism of m
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Noetherian local rings. Suppose M is a nite R-module, and N is a nite S-module which is at over R. Then (a) depthS M R N = depthR M + depthS N= N, (b) rS (M R N ) = rR (M ) rS (N= N ).
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The proof of the proposition is by reduction to the case of depth 0. We collect the essential arguments in a lemma. Lemma 1.2.17. Under the hypotheses of 1.2.16 the following hold: (a) diml HomS (l M N ) = dimk HomR (k M ) diml HomS (l N= N ), (b) if y is an (N= N )-sequence in S, then y is an (M R N )-sequence, and m
N=y N is at over R. Proof. (a) Set T = S= S . There is a natural isomorphism m
;l Hom (T M N) = Hom (l M N) m
(2) HomS S S since the modules on both sides can be identi ed with the submodule U = fz 2 M N : z = 0g of M N . As N is at over R , we have a natural isomorphism HomS (T M N ) = HomS (k S M N ) = HomR (k M ) N: n
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1. Regular sequences and depth
(see 318], 3.82 and 3.83). Now HomR (k M ) = ks for some s 0, and s so HomR (k M ) N = (N= N ) . In conjunction with (2), this yields the equation asserted. (b) One has a natural isomorphism (M N )=J (M N ) = M (N=JN ) for an arbitrary ideal J S . Therefore we may use induction on the length n of y , and only the case n = 1, y = T y needs justi cation. By Krull's intersection theorem one has 1i=0 i (M N ) = 0. Suppose that yz = 0 for some z 2 M N . If z 6= 0, then there exists i such that z 2 i (M N ) n i+1(M N ), and y would be a zero-divisor on i (M N )= i+1(M N ). However, consider the embedding i M ! M . Since N is at, the induced map i M N ! M N is also injective, and its image is i (M N ). The same reasoning for i+1 and atness again then yield an isomorphism i (M N ) i+1(M N ) ( i M= i+1M ) N kt N (N= N )t = = = for some t 0. Since y is regular on N= N , it must be regular on i (M N )= i+1(M N ). In order to test atness of N=yN it suces to consider exact sequences 0 ;! M1 ;! M2 ;! M3 ;! 0 of nite R -modules (318], Theorem 3.53). By hypothesis 0 ;! M1 N ;! M2 N ;! M3 N ;! 0 is also exact. As has been shown previously, y is regular on M3 N , and (M3 N )=y(M3 N ) = M3 N=yN . Therefore 1.1.4 yields the exactness of 0 ;! M1 N=yN ;! M2 N=yN ;! M3 N=yN ;! 0: Proof of 1.2.16. Let x = x1 . . . xm be a maximal M -sequence, and y = y1 . . . yn a maximal (N= N )-sequence. First, '(x) = '(x1) . . . '(xm) N )is an (M N )-sequence see 1.1.2. Second, by 1.2.17, y is an (M = M=xM . Since M N sequence where M = (M N )='(x)(M N ), it follows that '(x) y is an M N -sequence. Set N 0 = N=y N . Then N 0 = N 0 = (N= N )=y (N= N ), and N0 : (M N ) ('(x) y )(M N ) =M An application of 1.2.4 therefore gives the isomorphisms ) HomR (k M = ExtmR (k M ) HomS (l N 0 = N 0 ) = ExtnS (l N= N ) N0 ) HomS (l M = ExtmS +n(l M N ): Part (a) of 1.2.17 implies that diml ExtmS +n(l M N ) has the dimension required for (b), and in particular is non-zero. Together with the fact that '(x) y is an (M N )-sequence this proves depth(M N ) = m + n. m
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1.2. Grade and depth
The type of a module of depth 0 is the dimension of its socle: De nition 1.2.18. Let M be a module over a local ring (R k). Then Soc M = (0 : )M = HomR (k M ) is called the socle of M . For ease of reference we formulate the following lemma which was already veri ed in the proof of 1.2.16. Lemma 1.2.19. Let (R k) be a Noetherian local ring, M a nite Rmodule and x a maximal M-sequence. Then r(M ) = dimk Soc(M=xM ). m
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Exercises
Let k be a eld and R = k
X ]] Y ]. Deduce that X Y and 1 ; XY are maximal R -sequences. (This example shows that the condition IM 6= M in 1.2.5 is relevant.) 1.2.21. Let R be a Noetherian ring, I R an ideal, I = (x1 . . . xn ), and M a nite R -module with IM 6= M . Set g = grade(I M ). Prove (a) I can be generated by elements y1 . . . yn such that yi . . . yih form an M sequence for all i1 . . . ih with 1 i1 < < ih n, h g, (b) if y1 . . . yn satis es (a), then, in fact, every permutation of yi . . . yih is an M -sequence. P Hint: It is possible to choose yi = xi + j=i aj xj . Use the discussion above 1.1.6 for (b). 1.2.22. Let R be a Noetherian ring, I R an ideal, and M a nite R -module with IM 6= M . Set R = R= Ann M . (a) Prove that grade(I M ) height I R . (b) Give an example where grade(I M ) > height I . (c) Show that if I = (x1 . . . xn ), then grade(I M ) n. 1.2.23. Let R be a Noetherian local ring, and I R an ideal. Show grade I depth R ; dim R=I . (Hint: Use 1.2.13.) 1.2.24. Let R be a Noetherian ring, M a nite R -module, and I an ideal of R . Show that grade(I M ) 2 if and only if the natural homomorphism M ! HomR (I M ) is an isomorphism. 1.2.25. Let ' : (R ) ! (S ) be a homomorphism of local rings, and N an R - at S -module such that N= N has nite length over S . Show that for every nite length R -module M , `S (M N ) = `R (M ) `S (N= N ). (The symbol ` denotes length). Hint: use induction on `(M ). 1.2.26. Let ' : (R ) ! (S ) be a homomorphism of Noetherian local rings, and M an S -module which is nite as an R -module. (a) Suppose 2 AssS M , and let x 2 M with AnnS x = . Prove that ' induces an embedding R=( \ R ) ! S=
= Sx which makes S= a nite R=( \ R )-module. Conclude that \ R 6= , if 6= . (b) Show that depthR M = depthS M . (c) Suppose in addition that ' is surjective. Prove rR (M ) = rS (M ). 1.2.20.
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1. Regular sequences and depth
Let R be a Noetherian ring, M a nite R -module, and N an arbitrary R -module. Deduce that Ass HomR (M N ) = Supp M Ass N . 1.2.27.
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1.3 Depth and projective dimension Let R be a ring, and M an R -module M has an augmented projective resolution 'n ' ' P : ;! Pn ;! Pn;1 ;! ;! P1 ;! P0 ;! M ;! 0: (By de nition a projective resolution is non-augmented, i.e. M is replaced by 0 for the most part it is clear from the context whether one uses a nonaugmented resolution or an augmented one, so that one need not mention the attribute `augmented' explicitly.) Set M0 = M and Mi = Ker 'i;1 for i 1. The modules Mi depend obviously on P . However, M determines Mi up to projective equivalence (318], Theorem 9.4), and therefore it is justi ed to call Mi the i-th syzygy of M . The projective dimension of M , abbreviated proj dim M , is in nity if none of the modules Mi is projective. Otherwise proj dim M is the least integer n for which Mn is projective replacing Pn by Mn one gets a projective resolution of M of length n: 0 ;! Mn ;! Pn;1 ;! ;! P0 ;! M ;! 0: For a nite module M over a Noetherian local ring (R k) there is a very natural condition which, if satis ed by P , determines P uniquely. It is a consequence of Nakayama's lemma that x1 . . . xm 2 M form a minimal system of generators of M if and only if the residue classes x 1 . . . x m 2 M= M = M k are a k-basis of M k. Therefore m = dimk M k, and (M ) = dimk M k is the minimal number of generators of M . Set 0 = (M ). We choose a minimal system x1 . . . x of generators of M and specify an epimorphism '0 : R ! M by '0(ei ) = xi where e1 . . . e is the canonical basis of R . Next we set 1 = (Ker '0) and de ne similarly an epimorphism R ! Ker '0. Proceeding in this manner we construct a minimal free 1
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' ' ' ;! R ;! R ;! ;! R ;! R ;! M ;! 0: n
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It is left as an exercise for the reader to prove that F is determined by M up to an isomorphism of complexes. The number i (M ) = i is called the i-th Betti number of M . Proposition 1.3.1. Let (R k) be a Noetherian local ring, M a nite .
R-module, and F. :
m
' ' Fn;1 ;! ;! F1 ;! F0 ;! 0 ;! Fn ;! n
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1.3. Depth and projective dimension a free resolution of M. Then the following are equivalent: (a) F. is minimal (b) 'i (Fi) Fi;1 for all i 1 (c) rank Fi = dimk TorRi (M k) for all i 0, (d) rank Fi = dimk ExtiR (M k) for all i 0.
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The equivalence of (a) and (b) follows easily from Nakayama's lemma. Since TorRi (M k) = Hi (F k), (c) holds if and only if 'i k = 0 for all i 0. The latter condition is evidently equivalent to (b). To relate (b) to (d) one uses that ExtiR (M k) = H i (HomR (F k). Corollary 1.3.2. Let (R k) be a Noetherian local ring, and M a nite R-module. Then i(M ) = dimk TorRi (M k) for all i and proj dim M = supfi : TorRi (M k) 6= 0g: Proof.
.
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The following theorem, the `Auslander{Buchsbaum formula', is not only of theoretical importance, but also an e ective instrument for the computation of the depth of a module. Theorem 1.3.3 (Auslander{Buchsbaum). Let (R ) be a Noetherian local ring, and M 6= 0 a nite R-module. If proj dim M < 1, then proj dim M + depth M = depth R: m
The proof is by induction on depth R . We isolate the main arguments in two lemmas, the rst of which, in view of a later application, is more general than needed presently. Lemma 1.3.4. Let (R k) be a local ring, and ' : F ! G a homomorphism m
of nite R-modules. Suppose that F is free, and let M be an R-module with Ass M. Suppose that ' M is injective. Then (a) ' k is injective (b) if G is a free R-module, then ' is injective, and '(F ) is a free direct summand of G. Proof. Since Ass M , there exists an embedding : k M . As F is a free R -module, the map F is also injective. Furthermore we have a m
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!
commutative diagram
F 'k
F ? k ;;;;! F M ?
?y
?y'M
;;;;!
G k G M M is injective, then ' k is injective too. This proves (a).
If ' For (b) one notes that its conclusion is equivalent to the injectivity of ' k. This is an easy consequence of Nakayama's lemma.
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Lemma 1.3.5. Let (R ) be a Noetherian local ring, and M a nite Rmodule. If x
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proj dimR M = proj dimR=(x) M=xM: Proof.
F.
Choose an augmented minimal free resolution F of M . Then
R=(x) is exact by 1.1.5, and therefore it is a minimal free resolution .
of M=xM over R=(x). Now apply 1.3.2. Proof of 1.3.3. Let depth R = 0 rst. By hypothesis M has a (minimal) free resolution F. : 0
' ;! Fn ;! Fn;1 ;! ;! F1 ;! F0 ;! M ;! 0 n
with n = proj dim M . Since depth R = 0, the maximal ideal is in Ass R . If n 1, i.e. if 'n is really present, then, as shown in 1.3.4, 'n maps Fn isomorphically onto a free direct summand of Fn;1, in contradiction to proj dim M = n. Therefore n = 0, and furthermore depth M = depth R = 0 since M is a free R -module. Let now depth R > 0. Suppose rst that depth M = 0. Then 1.2.9 yields depth M1 = 1 for a rst syzygy M1 of M . Since proj dim M1 = proj dim M ; 1, it is enough to prove the desired formula for M1 . Thus we may assume depth M > 0. Then 2= Ass R and 2= Ass M . So contains an element x which is both R -regular and M -regular. The formulas for the passage from M to M=xM in 1.2.10 and 1.3.5 yield m
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depthR=(x) R=(x) = depth R ; 1 depthR=(x) M=xM = depthR M ; 1 proj dimR=(x) M=xM = proj dim M:
Therefore induction completes the proof. Exercises 1.3.6. Let R be a Noetherian local ring, M a nite R -module, and x an M -sequence
of length n. Show proj dim(M=xM ) = proj dim M + n. 1.3.7. Let R be a Noetherian local ring, and N an n-th syzygy of a nite R -module in a nite free resolution. Prove that depth N min(n depth R ).
1.4 Some linear algebra In this section we collect several notions and results which may be classi ed as `linear algebra': torsion-free and reexive modules, the rank of a module, the acyclicity criterion of Buchsbaum and Eisenbud, and perfect modules.
19
1.4. Some linear algebra
Torsion-free and reexive modules. Let R be a ring, and M an R -module. If the natural map M ! M Q, where Q is the total ring of fractions of R , is injective, then M is torsion-free it is a torsion module if M Q = 0. The dual of M is the module HomR (M R ), which we usually denote by M the bidual then is M , and analogous conventions apply to homomorphisms. The bilinear map M M ! R , (x ') 7! '(x), induces a natural homomorphism h : M ! M . We say that M is torsionless if h is injective, and that M is reexive if h is bijective. Some relations between the notions just introduced are given in the exercises. Here we note a useful criterion: Proposition 1.4.1. Let R be a Noetherian ring, and M a nite R-module.
Then: (a) M is torsionless if and only if (i) M is torsionless for all Ass R, and (ii) depth M 1 for Spec R with depth R 1 (b) M is reexive if and only if (i) M is reexive for all with depth R 1, and (ii) depth M 2 for Spec R with depth R 2. p
p
p
p
p
2
2
p
2
p
p
p
Consider the natural map h : M ! M and set U = Ker h, C = Coker h. Note that the construction of h commutes with localization in the situation considered. Therefore the necessity of conditions (i) in (a) and (b) is obvious. Next Exercise 1.4.19 implies p
p
Proof.
depth M min(2 depth R ) p
p
for all 2 Spec R . That (b)(ii) is necessary for reexivity follows directly from this inequality. If M is torsionless, then M is isomorphic to a submodule of M , and we get depth M min(1 depth R ) for all 2 Spec R. So (a)(ii) is necessary for M to be torsionless. As to the suciency of (a)(i) and (ii), note that U = 0 for all 2 Ass R by (i), and, by (ii), depth U 1 if depth R 1. It follows that Ass U = , hence U = 0. For the suciency of (b)(i) and (ii) we may now use that (a) gives us an exact sequence 0 ! M ! M ! C ! 0. If depth R 1, then C = 0 by (i). If depth R 2, then depth M 2 by (ii), and depth M 2 by the inequality above. Therefore depth C 1, and it follows that Ass C = . p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
Rank. The dimension of a nite dimensional vector space over a eld
is given either by the minimal number of generators or by the maximal number of linearly independent elements. The second aspect of `dimension' is generalized in the notion of `rank':
20
1. Regular sequences and depth
De nition 1.4.2. Let R be a ring, M an R -module, and Q be the total ring of fractions of R . Then M has rank r if M Q is a free Q-module of rank r. If ' : M ! N is a homomorphism of R -modules, then ' has rank r if Im ' has rank r. Proposition 1.4.3. Let R be a Noetherian ring, and M an R-module with ' a nite free presentation F1 ;! F0 ;! M ;! 0. Then the following are
equivalent: (a) M has rank r (b) M has a free submodule N of rank r such that M=N is a torsion module (c) for all prime ideals Ass R the R -module M is free of rank r (d) rank ' = rank F0 r.
;
p
2
p
p
Proof. (a) ) (b): A free basis x1 . . . xr of M Q can be formed from elements xPi 2 M (multiply by a suitable common denominator). Now take N = Rxi. (b) ) (a): This is trivial. (a) ) (c): M is a localization of M Q. (c) ) (a): Q is a semi-local ring. Its localizations with respect to its maximal ideals are just the localizations of R with respect to the maximal elements of Ass R . By hypothesis M is therefore a projective module over Q, and moreover the localizations with respect to the maximal ideals of Q have the same rank r. Such a module is free see Lemma 1.4.4 below. (c) () (d): In view of the equivalence of (a) and (c) we can replace (d) by the condition that (Im ') is free and rank(Im ') = rank F0 ; r for all 2 Ass R . Now consider the exact sequence p
p
p
p
0 ;! (Im ') ;! (F0 ) ;! M ;! 0: If M is free, then (Im ') must be free. Since 2 Ass R , the converse is also true see 1.3.4. Lemma 1.4.4. Let R be a semi-local ring, and M a nite projective Rp
p
p
p
p
p
module. Then M is free if the localizations M have the same rank r for all maximal ideals of R. m
m
Proof. We use induction on r. The case r = 0 is trivial. Suppose that r > 0. Then 1.2.2 (with N = M and 1 . . . m denoting the maximal ideals of R ) yields an element x M such that x = M for all maximal ideals of M . Thus x is a member of a minimal system of generators of M . Since every such system is a basis of the free module M , one concludes that (M=Rx) is free of rank r 1. By the induction hypothesis M=Rx is free of rank r 1. Therefore M = Rx M=Rx. In particular Rx is a projective R -module. But Rx is also free: the natural epimorphism ' : R Rx yields an isomorphism ' : R (Rx) for every maximal ideal . Since (Ker ') = Ker ' it follows that Ker ' = 0.
2
p
m
m
!
;
m
m
m
m
p
; ! m
2
m
m
m
m
21
1.4. Some linear algebra
Rank is additive along exact sequences. Proposition 1.4.5. Let R be a Noetherian ring, and 0 ! U ! M ! N ! 0
an exact sequence of nite R-modules. If two of U, M, N have a rank, then so does the third, and rank M = rank U + rank N.
In view of 1.4.3 we may assume that R is local and of depth 0. Then two of U , M , N are free. If U and N are free, then so is M . Thus M is always free (after the reduction to depth 0), and the result follows from the equivalence of 1.4.3(a) and (d). Corollary 1.4.6. Let R be a Noetherian ring, and M an R-module with a nite freeP resolution F : 0 ! Fs ! Fs;1 ! ! F1 ! F0. Then rank M = sj =0(;1)j rank Fj . Proof. Observe 1.4.5 and use induction on s. Corollary 1.4.7. Let R be a Noetherian ring, and I 6= 0 an ideal with a Proof.
.
nite free resolution. Then I contains an R-regular element.
By 1.4.6 I has a rank, and that rank I + rank R=I = rank R = 1 follows immediately from 1.4.5. Since I is torsion-free and non-zero, the only possibility is rank I = 1, whence rank R=I = 0. Thus R=I is annihilated by an R -regular element. Proof.
Ideals of minors and Fitting invariants. Let U be an m n matrix over R where m n 0. For t = 1 . . . min(m n) we then denote by It(U ) the ideal generated by the t-minors of U (the determinants of t t submatrices). For systematic reasons one sets It(U ) = R for t 0 and It(U ) = 0 for t > min(m n). If ' : F G is a homomorphism of nite free R -modules, then ' is given by a matrix U with respect to bases of F and G. It is an elementary exercise to verify that the ideals It (U ) only depend on '. Therefore we may put It (') = It(U ). It is just as easy to show that It(') is already determined by the submodule Im ' of G. As proved by Fitting in 1936, these ideals are even invariants of Coker ' (when counted properly), and therefore called the Fitting invariants of Coker ': let
!
F1
' ;! F0 ;! M ;! 0
and G1 ;! G0 ;! M ;! 0
be nite free presentations of the R -module M , and n = rank F0 , p = rank G0 then In;u(') = Ip;u() for all u 0. (The proof is left as an exercise for the reader.) This justi es the term u-th Fitting invariant of M for In;u('). It is an important property of the ideals It (') that their formation commutes with ring extensions: if S is an R -algebra, then It (' S ) = It(')S . (Simply consider ' as given by a matrix.)
22
1. Regular sequences and depth
The ideals It (') determine the minimal number (M ) of generators of a localization in the same way that they control the vector space dimension of M if R is a eld. Lemma 1.4.8. Let R be a ring, M an R-module with a nite free presenta' tion F1 ;! F0 ;! M ;! 0, and a prime ideal. Then the following are p
p
equivalent: (a) It(') (b) (Im ') contains a (free) direct summand of (F0 ) of rank t (c) (M ) rank F0 t. Proof. It is no restriction to assume that R = R . Nakayama's lemma entails that (M ) = (M= M ). Similarly it implies that Im ' contains a (free) direct summand of F0 of rank t if and only if there are elements x1 . . . xt Im ' which are linearly independent modulo F0. (Note that
6 p
;
p
p
p
p
p
2
every direct summand of a nite free module over a local ring is free itself { again an application of Nakayama's lemma.) After these observations we may replace R by the eld R= . For vector spaces over elds the equivalence of (a), (b) and (c) is an elementary fact. Lemma 1.4.9. With the notation of 1.4.8, the following are equivalent: (a) It(') 6 and It+1(') = 0 (b) (Im ') is a free direct summand of (F0 ) of rank t (c) M is free and rank M = rank F0 ; t. Proof. We may assume that R = R . Then each of (b) and (c) is equivalent to the split exactness of the sequence 0 ! Im ' ! F0 ! M ! 0. If (a) holds, then, with respect to suitable bases of F1 and F0 , the matrix of ' has the form id 0 t 0 0 where idt is the t t identity matrix. This implies (b). The converse is seen similarly. Let M be a nite module over a Noetherian ring R . Then M is a projective module (of rank r) if and only if M is a free R -module (of rank r) for all 2 Spec R . Combining this fact with 1.4.9 we obtain the global version of 1.4.9: Proposition 1.4.10. Let R be a Noetherian ring, and M a nite R-module ' with a nite free presentation F1 ;! F0 ;! M ;! 0. Then the following p
p
p
p
p
p
p
p
p
p
p
p
are equivalent: (a) Ir (') = R and Ir+1(') = 0 (b) M is projective and rank M = rank F0 r. The rank of a homomorphism ' : F G is determined by the ideal It('), just as in elementary linear algebra:
; !
23
1.4. Some linear algebra
Proposition 1.4.11. Let R be a Noetherian ring, and let ' : F ! G be a homomorphism of nite free R-modules. Then rank ' = r if and only if grade Ir (') 1 and Ir+1(') = 0. The easy proof is left as an exercise for the reader. The Buchsbaum{Eisenbud acyclicity criterion. Let R be a ring. A complex G. :
;! Gm ;! Gm;1 ;! ;! G1 ;! G0 ;! 0 m
1
of R -modules is called acyclic if Hi (G ) = 0 for all i > 0, and split acyclic if it is acyclic and i+1(Gi+1) is a direct summand of Gi for i 0. Let R be a Noetherian ring, and .
F. : 0
' ' ;! Fs ;! Fs;1 ;! ;! F1 ;! F0 ;! 0 s
1
a complex of nite free R -modules. We want to develop a criterion for F. to be acyclic. This criterion will involve ideals generated by certain minors of the homomorphisms 'i . A rst relation between the ideals It(') and the acyclicity of complexes is given in the next proposition. Proposition 1.4.12. Let R be a ring, M an R-module,
' ' ;! Fs ;! Fs;1 ;! ;! F1 ;! F0 ;! 0 be a P complex of nite free R-modules, and R be a prime ideal. Set ri = sj =i(;1)j ;i rank Fj for i = 1 . . . s. Then the following are equivalent: (a) F R is split acyclic (b) Ir ('i) 6 for i = 1 . . . s.
F. : 0
s
1
p
.
p
i
p
Furthermore, It ('i) = 0 for all i = 1 . . . s and t > ri, if one of these conditions holds. If Ass M, then (a) and (b) are equivalent to (c) F. M is acyclic. Proof. We may suppose that R = R . (a) (b): If F. is split acyclic, then F. R= is a (split) acyclic complex of vector spaces over R= so we can refer to elementary linear p
2 ) p
p
p
p
algebra. (b) ) (a): We again use induction, and may assume that Coker '2 is a free R -module of rank r1 . According to 1.4.8, Im '1 contains a free direct summand U of F0 of rank r1 . So we get an induced epimorphism Coker '2 ! U of free R -modules, both of which have rank r1. Such a map must be an isomorphism. One easily concludes that Im '1 = U . Hence F is split acyclic. That It ('i) = 0 for t > ri , follows most easily from (a) in conjunction with 1.4.9. p
.
24
1. Regular sequences and depth
(c) ) (a): Let F 0 be the truncation 0 ! Fs ! ! F1 ! 0. Then M is acyclic arguing 0inductively, we may therefore suppose that is split acyclic. Then F1 = Coker '2 is free, and the induced map 0 1 M ! F0 M is injective by hypothesis. By virtue of 1.3.4, F1 is mapped isomorphically onto a free direct summand of F0 . (a) ) (c): This is evident. We have completed our preparations for the following important and extremely useful acyclicity criterion. Theorem 1.4.13 (Buchsbaum{Eisenbud). Let R be a Noetherian ring, and .
F.0 F.0 F0
' ' ;! Fs ;! Fs;1 ;! ;! F1 ;! F0 ;! 0 P nite free R-modules. Set ri = sj =i(;1)j ;i rank Fj . Then the
F. : 0
s
1
a complex of following are equivalent: (a) F. is acyclic (b) grade Iri ('i ) i for i = 1 . . . s.
Before we prove the theorem the reader should note that ri = rank 'i 0 when F is acyclic this is just a restatement of 1.4.6. Conversely, it is not necessary to require that ri 0 for the implication (b) ) (a) if ri < 0, then ri+1 > rank Fi , and Iri ('i+1) = 0 in contradiction with (b). In the situation of 1.4.13 we call ri the expected rank of 'i. Proof. (a) ) (b): By what has just been said and 1.4.11, we see that grade Iri ('i) 1 for i = 1 . . . s. In particular there is an R -regular element x contained in the product of the ideals Iri ('i ). If x is a unit, then Iri ('i ) = R for all i, and we are done. Otherwise we use induction. Let denote residue classes modulo x. It follows immediately from 1.1.5 that the induced complex 0 ! F s ! F s;1 ! ! F 2 ! F 1 ! 0 is acyclic. Furthermore Iri ('i ); = Iri ( 'i), and grade Iri ( 'i) i ; 1 by induction. Then grade Iri ('i) i for i = 2 . . . s. The reader may have noticed that this implication follows immediately from the Auslander{Buchsbaum formula 1.3.3. In view of the generalization 9.1.6 an independent proof is useful, however. (b) ) (a): Using induction again we may assume that F 0 : 0 ! Fs ! Fs;1 ! ! F2 ! F1 ! 0 is acyclic. We set Mi = Coker 'i+1 for i = 1 . . . s, and show by descending induction that depth(Mi) minfi depth R g for all 2 Spec R and i = 1 . . . s. As Ms = Fs , this is trivial for i = s. Let i < s and consider the exact sequence 0 ;! Mi+1 ;! Fi ;! Mi ;! 0: If depth R i + 1, then depth(Mi+1) i + 1, and we get depth(Mi ) i from 1.2.9. If depth R i, then Iri ('i+1) 6 by hypothesis on the .
+1
.
p
p
p
p
p
p
+1
p
p
25
1.4. Some linear algebra
other hand rank Mi+1 = rank 'i+1 = ri+1, and therefore It('i+1) = 0 for t > ri+1. So 1.4.9 yields that (Mi ) is free, hence depth(Mi) = depth R . We still have to show that the induced map '01 : M1 ! F0 is injective. Let N = Ker '01. In order to get N = 0, we derive that Ass N = . If depth R 1, then depth(M1) 1 as seen above therefore 2= Ass M1 Ass N . If depth R = 0, then Iri ('i) 6 for i = 1 . . . s, and F R is even split acyclic by 1.4.12. It follows that N = 0 since N = H1 (F R ). Often one only needs the following consequence of 1.4.13. Corollary 1.4.14. Let R be a Noetherian ring, and F be a complex as in 1.4.13. If F R is acyclic for all prime ideals with depth R < s, then p
p
p
p
p
p
p
p
.
p
p
.
p
p
.
.
p
F. is acyclic.
p
p
Proof. Let be a prime ideal with depth R < i s. The implication (a) ) (b) of the theorem applied to F R yields grade Iri ('i) i, which is only possible if Iri ('i) 6 . This shows grade Iri ('i) i, and the acyclicity of F follows from the implication (b) ) (a) of the theorem. p
p
.
p
p
p
.
Theorem 1.4.13 is the most important case of the acyclicity criterion of Buchsbaum and Eisenbud. Its general form will be discussed in Chapter 9. Perfect modules. Let R be a Noetherian ring, and M a nite R -module. Since one can compute ExtiR (M R ) from a projective resolution of M , one obviously has grade M proj dim M . Modules for which equality is
attained have especially good properties. De nition 1.4.15. Let R be a Noetherian ring. A non-zero nite R -module M is perfect if proj dim M = grade M . An ideal I is called perfect if R=I is a perfect module. Perfect modules are `grade unmixed': Proposition 1.4.16. Let R be a Noetherian ring, and M a perfect R-module. For a prime ideal 2 Supp M the following are equivalent: (a) 2 Ass M (b) depth R = grade M. Furthermore grade = grade M for all prime ideals 2 Ass M. Proof. For all nite R -modules M and 2 Supp M one has the inequalities grade M grade M proj dim M proj dim M and moreover proj dim M + depth M = depth R by the Auslander{ Buchsbaum formula 1.3.3. If M is perfect, then the inequalities become p
p
p
p
p
p
p
p
p
p
p
26
1. Regular sequences and depth
equations, and depth M = 0 if and only if depth R = grade M . This shows the equivalence of (a) and (b). If 2 Ass M , then Ann M , and so grade grade M . For perfect M the converse results from (b) and the inequality grade depth R . p
p
p
p
p
p
p
It follows easily from 1.3.6 that an ideal generated by a regular sequence in a Noetherian ring R is perfect. Some more examples are described in the following celebrated theorem: Theorem 1.4.17 (Hilbert{Burch). Let R be a Noetherian ring, and I an ideal with a free resolution F. : 0
' ;! Rn ;! R n+1 ;! I ;! 0:
Then there exists an R-regular element a such that I = aIn('). If I is projective, then I = (a), and if proj dim I = 1, then In(') is perfect of grade 2. Conversely, if ' : R n R n+1 is an R-linear map with grade In (') 2, then I = In(') has the free resolution F. . Proof. First we prove the converse part. Let ' : R n R n+1 be a map with grade In(') 2. Then ' is given by an (n + 1) n matrix U . Let i denote the i-minor of U with the i-th row deleted, and consider the homomorphism : R n+1 R which sends ei to ( 1)i i . Laplace
!
!
!
expansion shows that we have a complex
;
0 ;! R n ;! R n+1 ;! I ;! 0 which in fact is exact by 1.4.13. Suppose now that an ideal I with free resolution F is given. Then 1.4.13 yields grade In(') 2, and we can apply the rst part of the proof to obtain I = Coker ' = In(') equivalently, there exists an injective linear map : In (') ! R with I = Im . According to 1.2.24, is just multiplication by some a 2 R . Because of 1.4.7 (or 1.4.13) a cannot be a zero-divisor. If I is projective, then In(') = R by 1.4.10, and thus I = (a). If proj dim I = 1, then proj dim(R=In(')) = proj dim R=I = 2, and R=In(') is perfect of grade 2. '
.
Exercises Let R be a ring, and M a nite torsion-free module. Prove that if M has a rank, then M is isomorphic to a submodule of a nite free R -module of the same rank. 1.4.19. Let R be a Noetherian ring, I an ideal, and M , N nite modules. Prove grade(I HomR (M N )) min(2 grade(I N )).
1.4.18.
27
1.5. Graded rings and modules
Let R be a Noetherian ring, and M a nite R -module. Prove (a) if M is torsionless, then it is torsion-free, (b) M is torsionless if and only if it is a submodule of a nite free module, (c) if M is re exive, then it is a second syzygy, i.e. there is an exact sequence 0 ! M ! F1 ! F0 with Fi nite and free. 1.4.21. Let R be a Noetherian ring, and M a nite R -module. Suppose ' : G ! F is a homomorphism of nite free R -modules with M = Coker '. Then D(M ) = Coker ' is the transpose of M . (It is unique up to projective equivalence.) Show that Ker h = Ext1R (D(M ) R ) and Coker h = Ext2R (D(M ) R ) where h : M ! M is the natural homomorphism. 1.4.22. Let R be a Noetherian ring, and M a nite R -module such that M has nite projective dimension. Prove (a) if depth M min(1 depth R ) for all 2 Spec R , then M is torsionless, (b) if depth M min(2 depth R ) for all 2 Spec R , then M is re exive. Hint: proj dim M < 1 ) proj dim D(M ) < 1. 1.4.23. Let R be a Noetherian ring, and M a nite R -module. Show that M has a rank if and only if M has a rank (and both ranks coincide). Hint: It is enough to consider the case R = R , depth R = 0. Apply 1.4.22. 1.4.24. Let R be a Noetherian local ring, and 0 ! Ls ! Ls 1 ! ! L1 ! L0 ! 0 a complex of nite R -modules. Suppose that the following hold for i > 0: (i) depth Li i, and (ii) depth Hi (L. ) = 0 or Hi (L. ) = 0. Show that L. is acyclic. (This is Peskine and Szpiro's `lemme d'acyclicite' 297].) Hint: Set Ci = Coker(Li+1 ! Li), and show by descending induction that depth Ci i and Hi (L. ) = 0 for i > 0. 1.4.25. Let R be a Noetherian ring, I an ideal of nite projective dimension, and M a nite R=I -module. Prove the following inequality of Avramov and Foxby
29]: 1.4.20.
p
p
p
p
p
p
p
p
;
gradeR=I M + gradeR R=I gradeR M gradeR=I M + proj dimR R=I
if I is perfect, then equality is attained. (Use the Auslander{Buchsbaum formula.) 1.4.26. Let R be a Noetherian ring, and M a perfect R -module of grade n. Suppose P. is a projective resolution of M of length n and set M = ExtnR (M R ). Prove (a) P. is acyclic and resolves M , (b) M is perfect of grade n, and M = M , (c) Ass M = Ass M . 1.4.27. Let R be a Noetherian ring, x an R -sequence of length n, and I = (x). Show that R=I m is perfect of grade n for all m 1. (Theorem 1.1.8 is useful.) 0
0
0
00
0
1.5 Graded rings and modules In this section we investigate rings and modules which, like a polynomial ring, admit a decomposition of their elements into homogeneous components.
28
1. Regular sequences and depth
De nition L 1.5.1. A graded ring is a ring R together with a decomposition R = i2Z Ri (as a Z-module) such that RiRj Ri+j for all i j 2 Z. AL graded R-module is an R -module M together with a decomposition M = i2Z Mi (as a Z-module) such that RiMj Mi+j for all i j 2 Z. One calls Mi the i-th homogeneous (or graded) component of M . The elements x 2 Mi are called homogeneous (of degree i) those of Ri are also called i-forms. According to this de nition the zero element is homogeneous of arbitrary degree. The degree of x is denotedPby deg x. An arbitrary element x 2 M has a unique presentation x = i xi as a sum of homogeneous elements xi 2 Mi . The elements xi are called the homogeneous components of x. Note that R0 is a ring L with 1 2 R0, that all summands Mi are R0modules, and that M = i2Z Mi is a direct sum decomposition of M as an R0-module. De nition 1.5.2. Let R be a graded ring. The category of graded R-modules, denoted M0 (R ), has as objects the graded R -modules. A morphism ' : M ! N in M0 (R ) is an R -module homomorphism satisfying '(Mi) Ni for all i 2 Z. An R -module homomorphism which is a morphism in M0(R) will be called homogeneous. Let M be a graded R -module and N a submodule of M . N is called a graded submodule if it is a graded module such that the inclusion map is a morphism in M0 (R ). This is equivalent to the condition Ni = N \ Mi for all i 2 Z. In other words, N is a graded submodule of M if and only if N is generated by the homogeneous elements of M which belong to N . In particular, if x 2 N , then all homogeneous components of x belong to N . Furthermore, M=N is graded in a natural way. If ' is a morphism in M0(R), then Ker ' and Im ' are graded. A (not necessarily commutative) R -algebra A is graded if, in addition to the de nition, AiAj Ai+j . The graded submodules of R are called graded ideals. Let I be an arbitrary ideal of R . Then the graded ideal I is de ned to be the ideal generated by all homogeneous elements a 2 I . It is clear that I is the largest graded ideal contained in I , and that R=I inherits a natural structure as a graded ring. Examples 1.5.3. (a) Let S be a ring, and R = S X1 . . . Xn] a polynomial ring over S . Then for every choice of integers d1 . . . dn there exists a unique grading on R such that deg Xi = di and deg a = 0 for all a 2 S : the m-th graded component P is the S -module generated by all monomials X1e Xnen such that ei di = m. If one chooses di = 1 for all i, then one obtains the grading of the polynomial ring corresponding to the total degree of a monomial. Unless indicated otherwise we will always consider R to be graded in this way. 1
29
1.5. Graded rings and modules
(b) Every ring R has the trivial grading given by R0 = R and Ri = 0 for i 6= 0. A typical example of a graded module over R is a complex C. :
;!@ Cn ;!@ Cn;1 ;!@
of R -modules. L Such a complex may be equivalently described as a graded module C = 1i=;1 Ci together with an R -endomorphism @ such that @2 = 0 and @(Ci) Ci;1 for all i. (In the terminology to be introduced below, @ is a homogeneous endomorphism of degree ;1.) The most important graded rings arise in algebraic geometry as the coordinate rings of projective varieties. They have the form R = kX1 . . . Xn]=I where k is a eld and I is an ideal generated by homogeneous polynomials (in the usual sense). Then R is generated as a k-algebra by elements of degree 1, namely the residue classes of the indeterminates. Graded rings R which as R0-algebras are generated by 1-forms will be called homogeneous R0 -algebras. More generally, if R is a graded R0-algebra generated by elements of positive degree, then we say that R is a positively graded R0-algebra. We want to clarify which graded rings are Noetherian. Let us rst consider positively graded rings. Proposition 1.5.4. Let R be a positively graded R0-algebra, and x1 . . . xn .
homogeneous elements of positive degree. L Then the following are equivalent: (a) x1 . . . xn generate the ideal = 1i=1 Ri (b) x1 . . . xn generate R as an R0-algebra. In particular R is Noetherian if and only if R0 is Noetherian and R is a nitely generated R0-algebra. m
Proof. For the implication (a) ) (b) it is enough to write every homogeneous element y 2 R as a polynomial in x1 . . . xn with coecients in R0 , and this is very easy by induction on deg y. The rest is evident. The last assertion of 1.5.4 holds for graded rings in general. Theorem 1.5.5. Let R be a graded ring. Then the following are equivalent: (a) every graded ideal of R is nitely generated (b) R is a Noetherian ring (c) R0 is Noetherian, and R is a nitely L generated R0-algebra L (d) R0 is Noetherian, and both S1 = 1i=0 Ri and S2 = 1i=0 R;i are nitely
generated R0-algebras.
) ) )
)
Proof. The implications (d) (c) (b) (a) are obvious. For (a) (d) we rst note that R0 is a direct summand of R as an R0-module. It follows that IR R0 = I for every ideal I of R0, and thus (a) implies that R0 is Noetherian. (Extend an ascending chain of ideals of R0 to R , and
\
30
1. Regular sequences and depth
contract the extension back to R0 .) A similar argument shows that Ri is a nite R0-module L for every i 2 Z. Let = 1i=1 Ri . We claim that is a nitely generated ideal of S1. By hypothesis R has a nite system of generators x1 . . . xm , which may certainly be chosen to be homogeneous of positive degrees di. Let d be the maximum of d1 . . . dm . Then a homogeneous element y 2 with deg y d can be written as a linear combination of x1 . . . xm with coecients from S1. Thus x1 . . . xm together with a nite set of homogeneous elements spanning R1 . . . Rd ;1 over R0 generate as an ideal of S1 . According to 1.5.4, S1 is a nitely generated R0-algebra, and the claim for S2 follows by symmetry. m
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Very often we shall derive properties of a graded ring or module from its localizations with respect to graded prime ideals. The following lemma is basic for such arguments. Lemma 1.5.6. Let R be a graded ring. (a) For every prime ideal the ideal is a prime ideal. (b) Let M be a graded R-module. (i) If 2 Supp M, then 2 Supp M. (ii) If 2 Ass M, then is graded furthermore is the annihilator of p
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a homogeneous element.
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(a)PLet a b 2 R such that ab 2 . We write a = i ai, ai 2 Ri , and b = j bj , bj 2 Rj . Assume that a 62 and b 62 . Then there exist integers p q such that ap 62 , but ai 2 for i < p, and bq 62 , but j
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31
1.5. Graded rings and modules
is a prime ideal, we have ap 2 . Iterating this procedure we see that each homogeneous component of a belongs to . In order to prove the second assertion in (ii) one can now use the fact that is generated by homogeneous elements. It follows easily that annihilates all the homogeneous components T of x. Set i = Ann xi then, as just seen, i . On the other hand ni=m i . Since is a prime ideal, there exists j with j , and therefore j = . Let be a prime ideal of R , and let S be the set of homogeneous elements of R not belonging to . The set S is multiplicatively closed, and we put M( ) = MS for any graded R -module M . For x=a 2 M( ) , x homogeneous, we set deg x=a = deg x ; deg a. We further de ne a grading on M( ) by setting anp;m+1 annihilates x. As
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(M( ) )i = fx=a 2 M( ) : x homogeneous, deg x=a = ig: p
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It is easy to see that R( ) is a graded ring and that M( ) is a graded R( ) module M( ) is called the homogeneous localization of M . The extension ideal R( ) is a graded prime ideal in R( ), and the factor ring R( )= R( ) has the property that every non-zero homogeneous element is invertible. p
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Lemma 1.5.7. Let R be a graded ring. The following conditions are equiv-
alent: (a) every non-zero homogeneous element is invertible (b) R0 = k is a eld, and either R = k or R = kt t;1 ] for some homogeneous element t R of positive degree which is transcendental over k.
)
2
(a) (b): R0 = k is a eld. If R = R0, then R is a eld. Otherwise R = R0 , and there exist non-zero homogeneous elements of positive degree. Let t be an element of least positive degree, say deg t = d . As t is invertible there exists a homomorphism ' : kT T ;1] R of graded rings where ' maps k identically to R0 and where '(T ) = t. (The grading on kT T ;1] is of course de ned by setting deg T = d .) P We claim that ' is an isomorphism. Let f Ker ', f = i2Z ai T i, P ai k then 0 = '(f ) = i2Z ai ti, and so ai ti = 0 for all i. As t is invertible, we get ai = (ai ti) t;i = 0 for all i, which implies that f = 0. Hence ' is injective. In order to show that ' is surjective, we pick a non-zero homogeneous element a R of degree i. If i = 0, then a Im '. Thus we may assume that i = 0. Write i = jd + r with 0 r < d . The element at;j has degree r. As d was the least positive degree, we conclude that r = 0. Thus a = btj for some b R0, and hence a = '(bT j ) Im '.
Proof.
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(b) ) (a) is trivial.
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The following theorem contains the dimension theory of graded rings and modules:
32
1. Regular sequences and depth
Theorem 1.5.8. Let R be a Noetherian graded ring, M a nite graded R-module and 2 Supp M. (a) If is graded, then there exists a chain 0 d = , d = dim M , of graded prime ideals i 2 Supp M. (b) If is not graded, then dim M = dim M + 1. Proof. A very special case of (b) is the following: if is not graded, then height = = 1. In order to prove this equation we may replace R by R= and assume that = 0. Then does not contain a nonzero homogeneous element. Therefore it is harmless to invert all these elements. This yields the homogeneous localization R(0). Since R(0) is a non-zero prime ideal, R(0) has the form kt t;1 ] by 1.5.7, whence height = height R(0) = 1. Now let 2 Supp M be an arbitrary prime ideal, and d = dim M . Both claims will be proved once we show that there exists a chain 0 d = of prime ideals in Supp M such that 0 . . . d ;1 are graded. Note that in the case of (b) it follows that d ;1 , and therefore d ;1 = since there is no prime ideal properly between and d ;1 . Let 0 d = be a chain of prime ideals in Supp M . Then 0 is minimal in Supp M , and therefore graded by 1.5.6. In the case d = 1 we are already done. Arguing inductively we may therefore suppose that 0 . . . d ;2 are graded. If is not graded, we replace d ;1 by , which is properly contained in , and properly contains d ;2 because height = = 1, as was proved above. If is graded, then it contains a homogeneous element a 2= d ;2, and we replace d ;1 by a minimal prime of d ;2 + (a) contained in . Since height = = 1, it is impossible for to equal furthermore is a minimal element of Supp R=( d ;2 + (a)), and thus graded by 1.5.6. Our next goal is an equation similar to 1.5.8 for the depth of a graded module. We shall need the result that the ordinary Ext-groups ExtiR (M N ) of graded R -modules admit a natural grading, provided R is Noetherian and M is nite. If M is a graded R -module and i is an integer, then M (i) denotes the graded R -module with grading given by M (i)n = Mi+n. The category M0(R ) has enough projectives. In fact, each module M 2 0(R ) is a homomorphic image (in M0(R )) of a module of the form L R(M i). So every graded module has a graded free resolution. When we speak of a natural grading of modules appearing as the values of derived functors, then it is of course important that the standard argument of homological algebra (318], Theorem 6.11), which guarantees that derived functors are well de ned, can be made `graded'. That M0(R ) has enough injectives will be shown in 3.6.2. p
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1.5. Graded rings and modules
It is not hard to see that the tensor product M N of graded R modules is a graded R -module its homogeneous component (M N )n is generated (as a Z-module) by the products x y with x 2 M , y 2 N homogeneous such that deg x + deg y = n see 1.5.19. Together with the fact that each graded module has a graded free resolution this implies that the modules TorRi (M N ) admit a natural grading. Let M , N be graded R -modules. In general, the set of morphisms ' : M ! N in M0(R ) is not a submodule of HomR (M N ). Thus for the construction of a reasonable graded Ext functor one must consider a larger class of maps. An R -module homomorphism ' : M ! N is called homogeneous of degree i if '(Mn) Nn+i for all i. (A homogeneous homomorphism whose degree is not explicitly speci ed has degree 0.) Note that ' may be considered as a morphism ' : M (;i) ! N in M0(R ). Denote by Homi (M N ) the group of homogeneous homomorphisms of degree i. The Z-submodules Homi (M N ) ofLHomR (M N ) form a direct sum, and it is obvious that HomR (M N ) = i2Z Homi (M N ) is a graded R -submodule of HomR (M N ). In general HomR (M N ) 6= HomR (M N ), but equality holds if M is nite: see Exercise 1.5.19. For any N 2 M0(R ) we de ne ExtiR (M N ) as the i-th right derived functor of HomR ( N ) in M0(R ). Thus, if P is a projective resolution of M in M0 (R ), then .
Exti (M N ) = H i (Hom (P. N )) R R
for all i 0. It is immediate from this de nition and the above remarks that ExtiR (M N ) = ExtiR (M N ) for Noetherian R and nite M . Nevertheless we shall use the notation ExtiR (M N ) to emphasize that these modules are graded. Theorem 1.5.9. Let R be a Noetherian graded ring, M a nite graded R-module, and 2 Supp M a non-graded prime ideal. Then depth M = depth M + 1 and r(M ) = r(M ): p
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In order to compute the depths and types of M and M we may consider both modules as modules over the homogeneous localization R( ) of R with respect to . Thus we may assume that R= = kt t;1 ] where k is a eld and t is an element of positive degree which is transcendental over k. It follows that = aR + for some a 2 R n . Hence we have an exact sequence
Proof.
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a 0 ;! R= ;! R= ;! R= which yields the long exact sequence p
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;! ExtiR (R= M) ;!a ExtiR (R= M) ;! ExtiR+1(R= M) ;! p
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34
1. Regular sequences and depth
The ExtiR (R= M ) are graded R= (= kt t;1 ])-modules. Since every graded kt t;1 ]-module is free (Exercise 1.5.20) and a 62 , the map p
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Exti (R= M ) R p
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is injective. Therefore ExtiR+1(R= M ) = ExtiR (R= M ) a ExtiR (R= M ): p
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The equation = aR + implies that ExtiR+1(R= M ) is a free (R= )module of the same rank as the free (R= )-module ExtiR (R= M ). Hence dimk( ) ExtiR+1(k( ) M ) = rankR= ExtiR+1(R= M )) = rankR= ExtiR (R= M ) = dimk( ) ExtiR (k( ) M ): p
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This equation in particular entails the assertion of the theorem. What makes the proof of 1.5.9 more dicult than one might expect at rst sight is illustrated by the following example. Let k be a eld and S = kX Y ] be graded such that deg X = 0 and deg Y = 1. The residue class ring R = S=(XY ) is graded, and (x y) is a graded maximal ideal of grade 1. Nevertheless every homogeneous element of (x y) is a zero-divisor, in fact contained in a minimal prime ideal. However, as we shall see in 1.5.11, under suitable hypotheses there exist homogeneous regular sequences. First we prove a graded version of prime avoidance: Lemma 1.5.10. Let R be a graded ring and I an ideal generated by elements of positive degree. Let 1 . . . n be prime ideals such that I 6 i for i = 1 . . . n. Then there exists a homogeneous element x 2 I, x 2= 1 n . L1 Rj . Since I is generated by elements of positive Proof. Let S = j =0 degree, one has I \ S 6 i \ S for i = 1 . . . n. Therefore we may assume that R is positively graded. Furthermore it is harmless to replace i by for all i. i Using induction on n, we may suppose that is a minimal element of f 1 .0 . . ng and that there is a homogeneous x0 2 nI with x0 2= 1 n;1. If x T2= n, then we are done. Otherwise there exists a homogeneous ;1 ) n . We choose a homogeneous y 2 I n . Then deg x0 > 0 r 2 ( ni=1 i n n and deg ry > 0 so that (x0 )u +(ry)v is homogeneous for suitable exponents u v. Furthermore, (x0 )u + (ry)v 2= 1 n . Proposition 1.5.11. Let R be a Noetherian graded ring, and let I be p
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an ideal in R generated by homogeneous elements of positive degree. Set h = height I and g = grade(I M ) where M is a nite R-module. Then there exist sequences x = x1 . . . xh and y = y1 . . . yg of homogeneous elements of I such that height(x1 . . . xi ) = i for i = 1 . . . h and y is an M-sequence.
35
1.5. Graded rings and modules
It is enough to nd x1 and y1 because we may use induction on n after having replaced all objects by their reductions modulo x1 or y1 . But the choice of x1 or y1 only requires the avoidance of nitely many prime ideals none of which contains I .
Proof.
Often one needs a stronger version of 1.5.11. Proposition 1.5.12. In addition to the hypotheses of 1.5.11 assume that R0
is a local ring with an in nite residue class eld and that I is generated by elements of degree 1. Then the sequences x = x1 . . . xh and y = y1 . . . yg can be composed of elements of degree 1. Proof. We choose a system z1 . . . zn
of degree 1 elements generating I . If height I > 0 and is a minimal prime ideal of R , then I1 6 . Therefore I1 \ is a proper R0-submodule of I1. As k0 is in nite, it is impossible for I1 to be the union of the nitely many proper submodules obtained in this manner. (Modulo the maximal ideal 0 of R0 this turns into an elementary fact of linear algebra.) So I1 has an element x1 which is not in any minimal prime ideal of R . In order to construct x2 . . . xh one proceeds by induction. The construction of y is similar. p
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Local rings.
In the following de nition we introduce the graded counterparts of local rings. De nition 1.5.13. Let R be a graded ring. A graded ideal of R is called maximal, if every graded ideal that properly contains equals R . The ring R is called local, if it has a unique maximal ideal . A local ring with maximal ideal will be denoted by (R ). Let (R ) be a local ring. All non-zero homogeneous elements of the graded ring R= are invertible, and so R= is either a eld, or else R= = kt t;1 ], where k is a eld and t is a homogeneous element of positive degree which is transcendental over k see 1.5.7. In the rst case is a maximal ideal, and in the second is a prime ideal with dim R= = 1. Note that R0 is a local ring with maximal ideal 0 = \ R0, and that all homogeneous elements a 2 R n are units. We de ne the dimension of R as the height of and denote it by dim R . According to 1.5.8, dim R equals the supremum of all numbers h for which there exists a chain of graded prime ideals 0 1 h in R . If x1 . . . xn , n = dim R , are homogeneous elements such that (x1 . . . xn) is -primary, then x1 . . . xn is called a homogeneous system of parameters. Examples 1.5.14. (a) Let be a graded prime ideal. Then R( ) is a local ring. (b) Let R be a positively graded ring for which R0 is a local ring with maximal ideal 0. Then R is a local ring with maximal ideal m
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1. Regular sequences and depth
L
= 0 n>0 Rn. In particular a positively graded algebra over a eld is local. (c) Let (S ) be a local ring and t an indeterminate over S . Then R = S t t;1] is in a natural way a local ring with maximal ideal S t t;1 ], and one has dim S = dim R = dim R ; 1. With respect to its nite graded modules M , a local ring (R ) behaves like a local ring, as we shall now see. Let g1 . . . L gn be a homogeneous minimal system of generators of M . Let F0 = ni=1 R (; deg gi ), the i-th summand being generated by an element ei satisfying deg ei = deg gi. The R -module F0 is free of rank n, and the assignment ei 7! gi induces a surjective morphism '0 of graded modules. Of course Ker '0 is a graded submodule of F0 . Suppose that Ker '0 6 F0 . Then there exists a homogeneous P element u 2 Ker '0, u 2= F0 , and one of the coecients ai in u = ai ei is not in call it aj . But all the ai are homogeneous, and so aj is a unit by hypothesis on (R ). It follows that the given system of generators is not minimal, which is a contradiction. Localizing with respect to we see that n = (M ). In particular all homogeneous minimal systems of generators have the same number of elements. Furthermore, iterating the construction of F0 and '0, one obtains an (augmented) free resolution of M which for the reasons given is called a minimal graded free resolution of M . It is easy to show that such a resolution is unique up to an isomorphism in M0. Proposition 1.5.15. Let (R ) be a Noetherian local ring, M a nite graded m
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R-module, and I a graded ideal. Then (a) every minimal homogeneous system of generators of M has exactly (M ) elements, (b) if F. is a minimal graded free resolution of M, then F. R is a minimal free resolution of M , R is faithfully exact on the category 0 , (c) the functor (d) M is projective if and only if it is free, (e) one has
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proj dim M = proj dim M grade M = grade M m
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grade( M ) = depth M grade(I M ) = grade(I M ): m
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Proof. (a) and (b) were shown above, and (a) implies that R is faithful, which proves (c) because localization is always exact. Part (d) follows from (b) since the rst Betti number 1 (M ) = 0 if M is projective, and therefore F1 = 0 in a minimal graded free resolution of M . The rst equation in (e) is also a consequence of (b), whereas the remaining ones result from (c) and the fact that the modules ExtiR (R=I M ) and ExtiR (M R ) are graded. (One must of course use the description of grade by 1.2.5). m
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1.5. Graded rings and modules
It is customary to collect the terms with the same `shift' in each free module of a graded free resolution and to write it in the form ;! M R(;j)ij ;! ;! M R(;j) j ;! M ;! 0: 0
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Though a minimal graded free resolution is uniquely determined, this is not true for the numbers ij if one only requires that (R ) is local. We need a slightly stronger hypothesis which is satis ed for example by all positively graded algebras over local rings: Proposition 1.5.16. Let (R ) be a Noetherian local ring such that is a maximal ideal of R (in the ordinary sense). Then for every nite graded m
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R-module M the numbers ij in a minimal graded free resolution of M are uniquely determined by M. L Proof. Let F = j R ( j )j . Then j = dimR= (F R= )j since one has R ( j ) R= = R= ( j ).
;
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In the situation of 1.5.16 not only is the cardinality of a minimal homogeneous system of generators unique, but also their degrees are xed (up to a permutation).
Graded Noether normalization. The existence of Noether normalizations
of ane algebras is stated in A.14. Here we want to prove its graded variant. Theorem 1.5.17. Let k be a eld and R a positively graded ane k-algebra. Set n = dim R. (a) The following are equivalent for homogeneous elements x1 . . . xn : (i) x1 . . . xn is a homogeneous system of parameters (ii) R is an integral extension of kx1 . . . xn] (iii) R is a nite kx1 . . . xn ]-module. (b) There exist homogeneous elements x1 . . . xn satisfying one, and therefore all, of the conditions in (a). Moreover, such elements are algebraically
independent over k. (c) If R is a homogeneous k-algebra and k is in nite, then such x1 . . . xn can be chosen to be of degree 1. Proof. We set S = kx1 . . . xn ] and I = (x1 . . . xn ). The existence of x1 . . . xn as claimed in (b) or (c) follows immediately from 1.5.11 and 1.5.12 if one observes that the maximal ideal of R has height n. The algebraic independence of x1 . . . xn results from (a)(ii) in conjunction with A.8 because dim S = n if and only if x1 . . . xn are m
algebraically independent. The equivalence of (a)(ii) and (iii) is a general fact: R is a nitely generated S -algebra. That (a)(ii) ) (i) follows from A.8 which entails that dim R=I = dim(S=I \ S ) = 0. Thus I is -primary. m
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1. Regular sequences and depth
There remains the proof of (a)(i) ) (iii). We choose a system y1 . . . ym of homogeneous elements of positive degree generating R over k. If (i) holds, then I is -primary, and there exists an e such that z = y1e ymem 2 I whenever deg z e (deg is the degree in R ). Let M be the S -submodule of R generated by those monomials z with deg z < e. We claim that R = M . In fact, every r 2 R is a k-linear combination of monomials y1e ymem , and thus it is enough that s = y1e ymem 2 M for all ei 2 N. If deg s < e, thenPs 2 M for trivial reasons. So assume deg s e. Then s 2 I , and s = ni=1 fixi with elements fi 2 R . Since s and the xi are homogeneous of positive degree, the fi can be chosen homogeneous of degree < deg s. Now write fi as a k-linear combination of monomials in y1 . . . ym , and apply an inductive argument. 1
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Dehomogenization. In concluding we want to study the relation between a graded ring R and a residue class ring A = R=(x 1) where x is a non-nilpotent homogeneous element of degree 1. One calls A the dehomogenization of R with respect to x. The relationship between R and A is much closer than between a ring and a residue class ring in general. A typical example for R and A arises in algebraic geometry: R is the homogeneous coordinate ring of a projective variety, and A is
;
the coordinate ring of the ane open subvariety complementary to the hyperplane given by the vanishing of x. Let : R ! A be the natural homomorphism, and S = Rx . Then factors in a natural way through a homomorphism : S ! A. Since x is homogeneous, the grading of R induces a grading on S . Proposition 1.5.18. (a) The homomorphism S0X X ;1] ! S which is the
identity on S0 and sends X to x is an isomorphism. (b) The restriction of to S0 is an isomorphism S0 = A.
(a) This is a general fact: if T is a graded ring which has a unit x of degree 1, then T = T0X X ;1]. (b) The kernel of is the ideal (x 1)S , and therefore induces an isomorphism A = S=(x 1)S = S0.
Proof.
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It follows easily that several properties transfer from R to A. For example, it is immediate that if R is reduced or an integral domain, then so is A. Also see Exercises 1.5.26, 2.2.34, and 2.2.35. Exercises Let R be a graded ring. All the modules in this problem are supposed to be graded. L (a) Prove Mi has a unique gradingLfor which the natural embeddings L that Mj ! Mi are morphisms in M0 , that is, Mi is the direct sum in M0 .
1.5.19.
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1.5. Graded rings and modules Q
(b) The direct product Mi lacks this propertyQof the direct sum nevertheless, Q prove there exists a direct product in M0 : let Mi be the submodule of Mi generated by the elements (xi ) such that all the xi are homogeneous of degree n, n 2 Z. (c) What can be said about direct and inverse limits in M0 ? (d) Prove that the tensor product M N is a graded R -module with (M N )u generated over Z by the tensor products x y of homogeneous elements with deg xL+ deg y = u. (Choose a presentation G ! F ! M ! 0 in M0 with L F = R (i) and G = R (j ).) L (e)Q Show the functor HomR ( N ) is left exact, and one has HomR ( Mi N )
= HomR (Mi N ). (f) Verify that HomR (M N ) = HomR (M N ) if M is nite. In general, however, HomR (M N ) is a proper submodule of HomR (M N ). (g) Prove HomR (M (;i) N (;j ))
= HomR (M N )(i ; j ). 1.5.20. Let R = k t t 1 ] be a graded ring where R0 = k is a eld, and t 2 R is a homogeneous element of positive degree which is transcendental over k. Show that every graded R -module is free. 1.5.21. Let k be a eld, S = k X1 X2 Y1 Y2 ] the polynomial ring with the grading determined by deg Xi = 0 and deg Yi = 1, and R = S=(X1 Y1 ; X2 Y2 ). Prove that the grade of the ideal I = (x1 x2 y1 y2 ) in R is 3, but I does not even contain a homogeneous R -sequence of length 2. 1.5.22. Let k be a eld. We consider the polynomial ring R = k X1 . . . Xn ] as a graded k-algebra with deg Xi = ai for i = 1 . . . n. Show that R is local if and only if all ai are positive or all ai are negative. 1.5.23. Let (R ) be a Noetherian local ring, and M a nite graded R -module. Show that every permutation of a homogeneous M -sequence is an M -sequence. 1.5.24. Prove the following variants of Nakayama's lemma. (a) Let (R ) be a local ring, M a nite graded R -module, and N a graded submodule. If M = N + M , then M = N . (b) Let R be a graded ring for which (R0 0 ) is local. Suppose that M is a graded R -module such that Mi is nite over R0 for all i. If one has M = N + 0 M for a graded submodule N of M , then M = N . 1.5.25. Let R be a Noetherian positively graded ring, and M a nite graded R -module. Prove dim M = supfdim M : 2 Supp M gradedg. Hint: consider an ideal which is maximal among the graded members of Supp M and use 1.5.7. 1.5.26. Let R be a graded ring, x a non-nilpotent element of degree 1, A = R=(x ; 1), and : R ! A the natural homomorphism. As in 1.5.18 we set S = Rx , and identify A and S0 . (a) One has (I ) = IS \ A for every homogeneous ideal I R and J = (JS \ R ) for every ideal J of A. (One calls (I ) the dehomogenization of I , and JS \ R the homogenization of J .) (b) The homomorphism induces a bijective correspondence between the set of homogeneous ideals of R modulo which x is regular and the set of all ideals of A. (c) This correspondence preserves inclusions and intersections, and the properties of being a prime, primary, or radical ideal.
;
m
m
m
m
m
p
p
m
40
1. Regular sequences and depth
1.6 The Koszul complex We introduce the Koszul complex K (x) of a sequence x = x1 . . . xn of elements of a ring R . Under suitable hypotheses one can determine grade(I M ) from the homology of K (x)M where I is the ideal generated by x. This fact and its `universal' properties make the Koszul complex an indispensable tool. Moreover, the Koszul complex is the paradigm of a complex with an algebra structure. In order to emphasize this fact we introduce more generally the Koszul complex of a linear form. A review of exterior algebra has been included for the reader's convenience. Review of exterior algebra. The following is an excerpt from Bourbaki 48], Ch. III, which we recommend as a source for exterior algebra. We hope that the details included will enable the reader to carry out the calculations on which the theory is based. When one has to check whether a map is well de ned, it is usually the best strategy to exploit the universal properties of the objects under consideration. Let R be a ring, and M an R -module. We consider R as a graded ring by giving it the trivial grading. Let M i denote the i-th tensor power of M , i.e. the tensor product M M of i factors M for i > 0, and R for i = 0. The tensor powers form a graded R -module .
.
1 NM = M M i : i=0
The assignment ((x1 . . . xm ) (y1 . . . yn)) 7;! x1 xm y1 yn m M n ! M (m+n), and its additive induces an RN-bilinear map MN N extension to M M gives M the structure of a graded associative R -algebra. Henceforth `R -algebra' always means `associative R -algebra'. N (Obviously M is not commutative in general.) The tensor algebra is characterized by a universal property: given an R -linear map ' : M ! A where N A is an R-algebra, there exists a unique R-algebra homomorphism : M ! A extendingV' here we identify M and M 1 . The exterior algebra M is the residue class algebra ^ N M = ( M )= where is the two-sided ideal generated by theV elements x x, x 2 M . Since is generated by homogeneous elements, M inherits the structure V M is denoted of a graded R -algebra. The product in V M is not commutative it is however alternating : one xhas^ y. In general ^ for homogeneous x y 2 M and x ^ y = (;1)(deg x)(deg y) y ^ x x^x= 0 for homogeneous x deg x odd: J
J
J
41
1.6. The Koszul complex
Let x1 . . . xn be elements of M , and a permutation of f1 . . . ng. Then x(1) ^ ^ x(n) = ( )x1 ^ ^ xn here ( ) is the sign of . Furthermore x1 ^ ^ xn = 0 if xi = xj for some indices i 6= j . For a subset I of f1 . . . ng we set xI = xi ^ ^ xim when I = fi1 . . . im g with i1 < < im : For subsets J , K f1 . . . ng with J \ K = let (J K ) = (;1)i where i is the number of elements (j k) 2 J K with j > k if J \ K 6= , let (J K ) = 0. Then xJ ^ xK = (J K )xJ K : Useful identities satis ed by are given in Exercise 1.6.23. It is clear that the notation xI can be extended to the more general case in which (xg )g2G is a family of elements of M indexed by a linearly ordered set G and I is a nite subset of G. V V The i-th graded component of M is denoted byV i M and is called the i-th exterior power of M . From the de nitionVof M it follows easily V 0 that one has natural isomorphisms M = R, 1 M = M so we may V V 0 1 identify R and M , M and M . V Let (xg )g2G be a system of generators of M . Then j M is generated by the exterior products xI with V I G and jI j = j. In particular, if M is generated by x1 . . . xn, then i M = 0 for all i > n. The exterior algebra is characterized by a universal property which it inherits from that of the tensor algebra: given an R -linear map ' : M ! E from M to an R -algebra E such that '(xV)2 = 0 for all x 2 M , there exists a unique R -algebra homomorphism : M ! E extending '. It follows immediately that for every RV-linear map ' : M ! N there exists a unique R -algebra homomorphism ' for which the diagram 1
' ;;;;! N ?? ??
M
y V ynat ' V M ;;;;! VN
nat
V
commutes ' is homogeneous of degree 0, and one has
^
'(x1
^ ^ xn) = '(x1) ^ ^ '(xn) V
for all and V x1 . . . xn 2 M. If ' is surjective, then ' is also surjective, Ker ' is the ideal generated by Ker '. (This is neither obvious V ' need notnor indeed true in general for example, if ' is injective, V V V V be injective.) The map i M ! i N Vinduced by ' is denoted by i '. Suppose that ' is surjective then i ' is also surjective, and from the
42
1. Regular sequences and depth
V
description of Ker ' just mentioned (and the alternating property of V M) it follows easily that the sequence i;1 ^ ^i Vi ' ^i M Ker ' ;! M ;;! N ;! 0 is exact where Vthe map on the leftVhand side is induced by the exterior multiplication i;1 M VKer ' ! i M . The exterior powers i M are also characterized by a universal property: for every alternating i-linear map : M i ! N , N an R -module, V i there exists a unique R -linear map : M ! N such that (x1 . . . xi ) = (x1 ^ ^ xi ) for all x1 . . . xi 2 M . An important property of the exterior algebra is that it commutes with base extensions: if R ! S is a homomorphism of commutative rings, then one has a natural isomorphism
^
( M ) R S =
^
(M R S )
of graded S -algebras. V V Let M1, M2 be R -modules. On ( M1) ( M2) one de nes a multiplication by setting (x y)(x0 y0 ) = (;1)(deg y)(deg x0 )(x ^ x0 ) (y ^ y0 ) for all homogeneous elements x, x0 2 M1, y, y0 2 M2. It is straightforward V V to verify that ( M1 ) ( M2) is an alternating graded R -algebra under this multiplication. Its degree 1 component is (M1 R ) (R M2 ) = M1M2 . By the universal property of the exterior algebra the natural map V (V M2) extends to an R-algebra homomorphism M1 M2 ! ( M1 ) V V V : (M1 M2 ) ! ( M1 ) V ( M2). V V One gets an inverse : ( M1) ( M2) ! (M1 M2) to by setting (x y) = 1 (x) ^ 2(y) V V where i : Mi ! (M1 M2 ) is the extension of the natural embedding M and are the identities on Vi ! M1 VM2. The compositions V ( M1) ( M2 ) and (M1 M2). Therefore we have an isomorphism
^
^
( M1) ( M2) =
^
(M1 M2 )
of alternating graded R -algebras. In what follows, the most important case for M is that of a nite free R -module F . Suppose e1 . . . en is a basis of F . The elements eI I f1 . . . ng jI j = i
43
1.6. The Koszul complex
V
form a basis of i F this non-trivial existence of ;nto. Athemultiplication Vi F is freefactofamounts determinants. In particular rank i V table of F with respect to this basis in given by
^ eJ = (I J )eIJ :
eI
L
Suppose R is a graded V ring, and M = i2Z Mi is a graded R-module. V Then one can endow M Vwith a unique grading such that M M has the given grading, and MLis a graded algebra over R . We restrict ourselves to the case M = F = ni=1 R (;ai). Let e1 . . . en be the basis of F corresponding to this decomposition. Then one assigns eI the degree P V toF makes VF a a , and veri es easily that the induced grading on i2I i graded (in fact, a bigraded) R -algebra. Basic properties of the Koszul complex. Let R be a ring, L an R -module, and f : L R an R -linear map. The assignment
!
(x1 . . . xn) 7!
Xn i=1
(;1)i+1f (xi )x1 ^ ^ bxi ^ ^ xn
V
de nes an alternating n-linear map Ln ! n;1 L. (By bxi we indicate that xi is to be omitted from the exterior product.) By the universal V property V of the n-th exterior power there exists an R -linear map df(n) : n L ! n;1 L with df(n)(x1
^ ^ xn) = X(;1)i+1f(xi)x1 ^ ^ bxi ^ ^ xn n
i=1
for all x1 . . . xn 2 L. The collection of the maps df(n) de nes a graded R -homomorphism ^ ^ df : L ! L
of degree ;1. By a straightforward calculation one veri es the following identities:
^
^
;
^
df df = 0 and df (x y) = df (x) y + ( 1)deg x x df (y)
V for all homogeneous x 2 L. To say that df df = 0 is to say that ^ d f d ^ ;! ^ L ;! L ;! ;! L ;! L ;! R ;! 0 n
f
n;1
2
f
is a complex. The second equation expresses that df is an antiderivation (of degree ;1).
44
1. Regular sequences and depth
De nition 1.6.1. The complex above is the Koszul complex of f , denoted by K (f ). More generally, if M is an R -module, then K (f M ) is the complex K (f ) R M , called the Koszul complex of f with coecients in M its di erential is denoted by dfM . Proposition 1.6.2. Let R be a ring, L an R-module, and f : L ! R an .
.
.
R-linear map. (a) The Koszul complex K. (f ) carriesVthe structure of an associative graded alternating algebra, namely that of L. (b) Its dierential df is an antiderivation of degree 1. (c) For every R-module M the complex K. (f M ) is a K. (f )-module in a natural way. (d) One has dfM (x : y) = df (x) : y + ( 1)deg x x : dfM (y) for all homogeneous elements x of K. (f ) and all elements y K. (f M ).
;
;
2
(a) and (b) are part of the discussion preceding the proposition. (c) is obvious: if A is an R -algebra, then A R M is an A-module for every R -module M . (d) It is enough to verify the equation for elements y = w z with w 2 K (f ), z 2 M . Then dfM (x : w z ) = dfM ((x ^ w) z ) = df (x ^ w) z , and the rest follows from the fact that df is an antiderivation. For a subset S of K (f ) and a subset U of K (f M ) let S : U denote the R -submodule of K (f M ) generated by the products s : u, s 2 S , u 2 U . Set Z (f ) = Ker df Z (f M ) = Ker dfM B (f ) = Im df B (f M ) = Im dfM : De nition 1.6.3. The homology H (f ) = Z (f )=B (f ) is the Koszul homology of f . For every R -module M the homology Z (f M )=B (f M ) is denoted by H (f M ) and called the Koszul homology of f with coecients in M . From 1.6.2(d) one easily derives the following relations: Z (f ) : Z (f M ) Z (f M ) Z (f ) : B (f M ) B (f M ) B (f ) : Z (f M ) B (f M ): We have a natural isomorphism K (f ) = K (f R ). So the rst relation entails that Z (f ) is a graded R -subalgebra of K (f ), and the second and third show that B (f ) is a two-sided ideal in Z (f ). Proposition 1.6.4. Let R be a ring, L an R-module, and f : L ! R an Proof.
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R-linear map. (a) The Koszul homology H. (f ) carries the structure of an associative graded alternating R-algebra. (b) For every R-module M the homology H. (f M ) is an H. (f )-module in a natural way.
45
1.6. The Koszul complex
(a) That H (f ) is an R -algebra follows from the discussion preceding the proposition. The asserted properties are inherited by quotients of graded R -subalgebras of K (f ) modulo graded ideals. (b) The rst of the relations above shows that Z (f M ) is a Z (f )module the second says that B (f M ) is a Z (f )-submodule, and the third implies that Z (f M )=B (f M ) is annihilated by B (f ). It results immediately from 1.6.4 that H (f M ) is an R=I -module where I = Im f . This will be stated in 1.6.5 where it follows from a somewhat stronger statement. It is useful also to introduce the Koszul cohomology (with coecients in M ): we set K (f ) = HomR (K (f ) R ) K (f M ) = HomR (K (f ) M ) H (f ) = H (K (f )) H (f M ) = H (K (f M )): Proof.
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Let I = Im f R then, by construction, H0 (f ) = R=I and H0 (f M ) = M=IM . Proposition 1.6.5. Let R be a ring, L an R-module, and f : L ! R an R-linear map. Set I = Im f. (a) For every a 2 I multiplication by a on K (f ), K (f M ), K (f ), K (f M ) .
.
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.
is null-homotopic. (b) In particular I annihilates H. (f ), H. (f M ), H . (f ), H . (f M ). (c) If I = R, then the complexes K. (f ), K. (f M ), K . (f ), K . (f M ) are null-homotopic. In particular their (co)homology vanishes.
We choose x 2 L with a = f (x). Let #a denote the multiplication by a on K (f ), and x the left multiplication by x on K (f ). Then #a = df x + x df as is easily veri ed. Thus multiplication by a is null-homotopic on K (f ). Of course #a M and HomR (#a M ) are the multiplications by a on K (f M ) and K (f M ), and the rest of (a) follows immediately. Part (b) is a general fact: if ' is a null-homotopic complex homomorphism, then the map induced by ' on homology is zero. For (c) we choose a = 1, and apply (a) and (b). Let L1 and L2 be R -modules, and f1 : L1 ! R , f2 : L2 ! R be R linear maps. Then f1 and f2 induce a linear form f : L1 L2 ! R by f (x1 x2) = f1 (x1) + f2(x2 ). Proposition 1.6.6. With the notation just introduced, one has an isomorphism of complexes K (f1) R K (f2) = K (f ). Proof. The f ), V gradedV R-algebras V underlying K (f1) K (f2) and K (We namely ( L1) ( L2 ) and L, are isomorphic, as noted above. V may identify them. The di erential df is an antiderivation on L which on the degree 1 graded component L = L1 L2 coincides with df df . Proof.
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1
2
46
1. Regular sequences and depth
V
An antiderivation on the exterior algebra L is uniquely determined by its values on L. Therefore it is enough to check that df df is an antiderivation too. The straightforward veri cation of this fact is left as an exercise for the reader. One has of course to remember the de nition of tensorLproduct the n-th graded component of V of complexes: V K (f1) K (f2) is ni=0 i L1 n;i L2 , and 1
.
2
.
df (x y) = df (x) y + (;1)ix df (y) V V for x y 2 i L1 n;i L2. df
1
2
1
2
The Koszul complex `commutes' with ring extensions, and so does Koszul homology if the extension is at: Proposition 1.6.7. Let R be a ring, L an R-module, and f : L ! R an R-linear map. Suppose ' : R ! S is a ring homomorphism. (a) Then one has a natural isomorphism K (f ) R S = K (f S ). (b) Moreover, if ' is at, then H (f M ) S = H (f S M S ) for every .
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R-module M.
.
V
V
There is a natural isomorphism ( L) S = (L S ), and df S and dfS are antiderivations which coincide in degree 1. So we can use the same argument as in the previous demonstration. This proves (a), and (b) follows immediately since H (C S ) = H (C ) S for an arbitrary complex C over R if S is R -at. Suppose L and L0 are R -modules with linear forms f : L ! R and 0 f : L0 ! R . Every RV -homomorphism ' : L ! L0 extends to a homomorV V phismV ' : L ! L0 of R -algebras, as discussed above. If f = f 0 ', then ' is a homomorphism of Koszul complexes: Proposition With the notation just introduced, if f = f 0 ', then V ' : K (f) !1.6.8. K (f 0 ) is a complex homomorphism. Proof.
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The Koszul complex of a sequence. Let L be a nite free R -module with basis e1 . . . en . Then a linear form f on L is uniquely determined by the values xi = f (ei ), i = 1 . . . n. Conversely, given a sequence x = x1 . . . xn, there exists a linear form f on L with f (ei) = xi . We set K. (x) = K. (f )
and the rest of the notation is to be modi ed accordingly. Henceforth we shall only consider Koszul complexes K (x). Since f is just the `direct sum' of the linear forms fi : R ! R , fi (1) = xi , 1.6.6 specializes to the isomorphism .
K. (x) = K. (x0)
K (xn) = K (x1) K (xn) .
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47
1.6. The Koszul complex
where x0 = x1 . . . xn;1. Furthermore one should note that, by 1.6.8, K (x) is essentially invariant under a permutation of x. We set I = (x). Let F be a free resolution of R=I . As H0 (x) = R=I , there exists a complex homomorphism ' : K (x) ! F lifting the identity on R=I note that ' is unique up to homotopy. .
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Proposition 1.6.9. Let R be a ring, x = x1 . . . xn a sequence in R, and I = (x). For all i there exist natural homomorphisms Hi (x M ) Proof.
! ToriR (R=I M)
and ExtiR (R=I M )
! H i(x M):
The map ' introduced above yields complex homomorphisms ! F M and HomR (' M): HomR (F M) ! K (f M).
' M : K. (f M )
.
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.
Let L be V a nite free R-module with basis e1 . . . en. Then! e:1 ^^ Vn L !en Ris a basis of n L, and there exists a unique R -isomorphism n Vn L = R is usually called with !n (e1 ^ ^ en ) = 1. (An isomorphism V V an orientation on L.) We de ne !i : i L ! ( n;i L) by setting (!i (x))(y) = !n(x ^ y)
for x 2
^i
L y
2 ^ L: n;i
(This causes no ambiguity for i = n if we identify R and R under the natural isomorphism.) It follows immediately that
0
I \ J 6= , (!i (eI ))(eJ ) = (I J ) for for I \ J = .
In this formula I and J are multi-indices as introduced above. It shows that !i is an isomorphism. If we denote the dual basis of (eI ) by (eI ), the formula says that !i (eI ) = (I I )eI where I = f1 . . . ng n I . Thus !i is an isomorphism. We consider the diagram K. (x): 0 K . (x): 0
;! ^?L ;!d ^?L ;!d ;!d L? ;!d R? ;! 0 ?! ?! ?! ?! n;1
n
y
;!
R
y
n
d ;!
L
with d = dx and d = (dx ).
n;1
d d ;! ;! (
y ^
n;1
y ^
1
L)
d ;! (
n
0
L)
;! 0
48
1. Regular sequences and depth
Proposition 1.6.10. Let x = x1 . . . xn be a sequence in a ring R. (a) With the notation just introduced, one has !i;1 di = (;1)i;1dn;i+1 !i
for all i. (b) The complexes K. (x) and K . (x) = (K. (x)) are isomorphic (we say that K. (x) is self-dual). (c) More generally, for every R-module M the complexes K. (x M ) and K . (x M ) are isomorphic, and (d) Hi (x M ) = H n;i (x M ) for i = 0 . . . n.
The veri cation of (a) is left as an exercise for the reader (1.6.23 is helpful). We observed above that !i is an isomorphism so that the maps i = (;1)i(i;1)=2!i de ne an isomorphism K (x) = K (x) = (K (x)). For (c) we note that there is a natural homomorphism N M ! HomR (N M ) for all R -modules N , M . If N is nite and free, this homomorphism is an isomorphism, and it induces an isomorphism K (x) M = HomR (K (x) M ). Now one uses (b). Part (d) is a trivial consequence of (c). The reader may have noticed that for a formally correct formulation of 1.6.10(b) one would rst have to convert the cochain complex K (x) into a chain complex C (by setting Ci = K ;i (x)) and then state that K (x) = C (;n). A similar manipulation would be necessary for (c). The Koszul complex is an exact functor: Proposition 1.6.11. Let R be a ring, x = x1 . . . xn a sequence in R, and 0 ! U ! M ! N ! 0 an exact sequence of R-modules. Then the induced Proof.
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sequence
0 ;! K (x U ) ;! K (x M ) ;! K (x N ) ;! 0 .
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is an exact sequence of complexes. In particular one has a long exact sequence
;! Hi (x U) ;! Hi(x M) ;! Hi (x N) ;! Hi;1(x U) ;!
of homology modules.
The components of K (x) are free, hence at R -modules. In place of an R -module M one can more generally consider a complex C , and then de ne the Koszul homology of C to be the homology of K (x) C etc. We consider this construction only for the special case in which x = x: Proposition 1.6.12. Let R be a ring, and x 2 R. (a) For every complex C of R-modules one has an exact sequence 0 ;! C ;! C K (x) ;! C (;1) ;! 0:
Proof.
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49
1.6. The Koszul complex
(b) The induced long exact sequence of homology is
x ;! Hi (C ) ;! Hi(C K (x)) ;! Hi;1(C ) ;! Hi;1 (C ) ;! (c) Moreover, if x is C -regular, then there is an isomorphism H (C K (x)) = H (C =xC ): (According to our convention for graded modules C (;1) is just the .
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complex C with all degrees increased by 1.) x Proof. The complex K (x) is simply 0 ;! R ;! R ;! 0. The i-th graded component of K (x) C is therefore (R Ci ) (R Ci;1) = Ci Ci;1. So we have in each degree an exact sequence 0 ;! Ci ;! Ci Ci;1 ;! Ci;1 ;! 0 where and are the natural embedding and projection. If @ is the di erential of C , then the di erential d : Ci Ci;1 ! Ci;1 Ci;2 is given by the matrix @ (;1)i;1x 0 @ according to the de nition of tensor products of complexes. Now (a) is obvious. For (b) one looks up the de nition of connecting homomorphism. It is de ned by the following chain of assignments starting from z 2 Ci;1 with @(z ) = 0: .
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;7 ! (0 z) 7;d! ((;1)i;1xz 0) 7;! (;1)i;1xz: So the connecting homomorphism Hi (C (;1)) = Hi;1(C ) ! Hi;1 (C ) is multiplication by (;1)i;1x. (c) The natural maps Ci Ci;1 ! Ci ! Ci=xCi constitute a complex homomorphism C K (x) ! C =xC . We claim that the associated map of homology is an isomorphism. In fact, let z 2 Ci such that @(z ) 2 xCi;1. Then there exists z 0 2 Ci;1 with @(z ) = xz 0 , and d (z (;1)i z 0) = (0 (;1)i@(z 0 )). Next one has x@(z 0 ) = @(@(z )) = 0 so @(z 0 ) = 0 since multiplication by x is injective on C : (z (;1)i z 0 ) is a cycle mapped to the cycle z 2 Ci=xCi. That the map of homology is injective can be veri ed z
;1
;1
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similarly. Corollary 1.6.13. Let R be a ring, x = x1 . . . xn a sequence in R, and M an R-module. (a) Set x0 = x1 . . . xn;1. Then one has an exact sequence
x x ;;! Hi (x0 M ) ! Hi (x M ) ! Hi;1 (x0 M ) ;;! Hi;1 (x0 M ) ! (b) Let p n, x0 = x1 . . . xp, and x00 = xp+1 . . . xn . If x0 is weakly M-regular, then one has an isomorphism H (x M ) = H (x00 M=x0 M ). n
n
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50
1. Regular sequences and depth
Part (a) is a special case of 1.6.12(b) when we take C = K (x0 M ) and use the isomorphisms K (x0 M ) K (xn ) = K (x0 ) M K (xn ) = K (x M ): For part (b) it is enough to do the case p = 1 from which the general case follows by induction. Next we may permute x to the sequence x2 . . . xn x1, and then the assertion follows from 1.6.12(c). It is an immediate consequence of 1.6.13 that Hi (x M ) = 0 for i = n ; p + 1 . . . n if (the rst) p elements of x1 . . . xn form an M -sequence. As we shall see in 1.6.16, there is a somewhat stronger vanishing theorem. Corollary 1.6.14. Let R be a ring, x a sequence in R, and M an R-module. (a) If x is an M-sequence, then K (x M ) is acyclic. (b) If x is an R-sequence, then K (x) is a free resolution of R=(x). Remark 1.6.15. Let R be a graded ring and x = x1 . . . xn a sequence of homogeneous elements. Then x induces a linear form of degree 0 on L F = ni=1 R (; deg xi ). The Koszul complex (x) is a graded complex V F theK grading with a di erential of degree 0 if we give discussed above. V = R(; Pn deg x ). In particular one has n F i i=1 Proof.
.
.
.
.
.
.
.
.
.
.
The main importance of the Koszul complex stems from the fact that H (x M ) measures grade(I M ) if M is a nite module over a Noetherian ring R and I = (x). This will be made precise in 1.6.17. The niteness assumption just stated will be necessary to establish the existence of an M -sequence in I from the vanishing of certain homology modules Hi (x M ). The converse holds without such an assumption: Theorem 1.6.16. Let R be a ring, x = x1 . . . xn a sequence in R, and M an R-module. If I = (x) contains a weak M-sequence y = y1 . . . ym , then Hn+1;i(x M ) = 0 for i = 1 . . . m and Hn;m (x M ) = HomR (R=I M=y M ) = ExtmR (R=I M ): The Koszul complex and grade.
.
The last isomorphism is given by Lemma 1.2.4. The remaining claims are proved by induction on m. For m = 0 we must show that Hn (x M ) = HomR (R=I M ): In fact, by 1.6.10 one has Hn (x M ) = H 0 (x M ), and the latter is naturally isomorphic with HomR (R=I M ), as follows from the exactness of R n ! R ! R=IV ! 0 and the left exactness of HomR ( M ). Explicitly, if we identify n R n M and R M = M via an orientation !n of R n, then Hn (x M ) is just the submodule fy 2 M : Iy = 0g = HomR (R=I M ) of M . Proof.
51
1.6. The Koszul complex
= M=y1M . The exact sequence Let m 1. Then we set M
y ;! 0 0 ;! M ;! M ;! M induces an exact sequence y y ) ;! Hi;1 (x M ) ;! ;! Hi(x M) ;! Hi (x M ) ;! Hi (x M 1
1
1
see 1.6.11. Since, by 1.6.5, y1 annihilates Hi (x M ) for all i, this exact sequence breaks up into exact sequences ) ;! Hi;1(x M ) ;! 0: 0 ;! Hi (x M ) ;! Hi (x M It only remains to apply the induction hypothesis. Theorem 1.6.17. Let R be a Noetherian ring, and M a nite R-module. Suppose I is an ideal in R generated by x = x1 . . . xn . (a) All the modules Hi (x M ), i = 0 . . . n, vanish if and only if M = IM. (b) Suppose that Hi (x M ) 6= 0 for some i, and let h = maxfi : Hi (x M ) 6= 0g: Then every maximal M-sequence in I has length g = n ; h in other words, grade(I M ) = n ; h. Proof. (a) The implication `)' is trivial: M = IM () H0 (x M ) = M=IM = 0. For the converse choose a prime ideal . By 1.6.7 and the atness of localization one has (Hi (x M )) = Hi (x M ) where x is considered a sequence in R on the right hand side. If I 6 , then Hi (x M ) = 0 by 1.6.5. If I , then M = 0 by Nakayama's lemma, and again we have Hi (x M ) = 0. (b) We give two proofs. (A third proof for the case M = R is indicated in Exercise 1.6.30.) (i) By part (a) we have M 6= IM . Let y be a maximal M -sequence in I then y has length g = grade(I M ). It follows immediately from 1.6.16 and 1.2.5 that Hi (x M ) = 0 for i = n ; g + 1 . . . n and Hn;g (x M ) = ExtgR (R=I M ) 6= 0. (ii) Let y be a maximal M -sequence in I , and suppose that y has length g. Then Hi (x M ) = 0 for i = n ; g + 1 . . . n by 1.6.16, and furthermore Hn;g (x M ) = HomR (R=I M=y M ). Since I consists of zerodivisors of M=y M , this module is non-zero see 1.2.3. The second proof just given is independent of the `homological' Lemma 1.2.4, and shows again that all maximal M -sequences in I have the same length. Therefore one could build the theory of grade upon 1.6.17. Corollary 1.6.14 can be reversed for local rings. We need the following lemma: p
p
p
p
p
p
p
p
p
52
1. Regular sequences and depth
Lemma 1.6.18. Let (R ) be a Noetherian local ring, M a nite R-module, and x = x1 . . . xn a sequence in . Set x0 = x1 . . . xn;1. If Hi (x M ) = 0, then Hi (x0 M ) = 0. Proof. By 1.6.6 we have K (x) = K (x0) K (xn ). So 1.6.13 gives us an exact sequence m
m
.
Hi (x0 M )
.
.
x ;;! Hi (x0 M ) ;! Hi (x M ): n
These modules are nite. If Hi (x M ) = 0, then multiplication by xn on Hi (x0 M ) is surjective, whence Hi (x0 M ) = 0 by Nakayama's lemma. Corollary 1.6.19. Let (R ) be a Noetherian local ring, M 6= 0 a nite Rmodule, and I an ideal generated by x = x1 . . . xn. Then the following m
m
are equivalent: (a) grade(I M ) = n (b) Hi (x M ) = 0 for i > 0 (c) H1 (x M ) = 0 (d) x is an M-sequence.
The equivalence of (a) and (b) follows from 1.6.17, and (b) ) (c) and (d) ) (a) are trivial. The proof of (c) ) (d) is an easy induction based on 1.6.18 and 1.6.13. We saw in 1.1.6 that under the hypotheses of 1.6.19 every permutation of an M -sequence is again an M -sequence. Since, by 1.6.8, the Koszul complexes of x and every permutation of x are isomorphic, 1.6.19 yields another proof of 1.1.6. Remark 1.6.20. For an arbitrary ring R and an arbitrary module M it follows from H1 (x M ) = 0 that x is M -quasi-regular, provided xM 6= M see 48], Ch. X, x9, Theor!eme 1. Proof.
The Koszul complex as an invariant. Let R be a Noetherian local ring, I an ideal, and x = x1 . . . xn and y = y1 . . . yn minimal systems of generators of I . Then any n n matrix A = (apq) such that
xi =
Xn j =1
ajiyj
i = 1 . . . n
is invertible since the residue classes of x and y are bases of I= I over R= . If f and f 0 are the linear forms on R n de ned by x and y respectively, then there exists an R -automorphism ' of R n (de ned by A) such that f = f 0 ', and it follows from 1.6.8 that the Koszul complexes K (x) and K (y ) are isomorphic. This obviously fails if x and y have m
m
.
.
53
1.6. The Koszul complex
di erent lengths. Nevertheless the Koszul complexes K (x) and K (y ) are closely related. The following proposition shows how to compare each of them to K (x y ). Proposition 1.6.21. Let R be a ring, x = x1 . . . xn a sequence in R, and x0 = x1 . . . xn xn+1 . . . xm with xn+1 . . . xm 2 (x). Then .
.
.
K. (x0 ) = K. (x)
V
^ Rm;n
as graded R-algebras here R m;n is considered a complex with zero differential. In particular, for every R-module M one has
H. (x0 M ) = H. (x M )
V
V
^ Rm;n:
V
Since PR k+1 = R k R it suces to treat the case m = n + 1. n Write xn+1 = j =1 aj xj . Let f be the linear form on R n+1 de ned by x0 and f 0 the linear form de ned by x00 = x 0. The assignment ei 7! ei for P n i = 1 . . . n and en+1 7! j =1 aj ej + en+1 induces an automorphism ' of R n+1 such that f = f 0 '. As above one concludes that K (x0 ) = K (x00) in other words, there is no restriction in assuming that xn+1 = 0. In the special situation we have reached, the rst claim is a trivial consequence of 1.6.6. The second claim is easily veri ed. Corollary 1.6.22. Let R be a ring, I a nitely generated ideal, and M an Rmodule. Suppose x = x1 . . . xm and y = y1 . . . yn are systems of generators of I, and let g 2 N. Then Hi (x M ) = 0 for i = m ; g + 1 . . . m if and only if Hj (y M ) = 0 for j = n ; g + 1 . . . n. The corollary follows easily from 1.6.21. Note that for a nite module M over a Noetherian ring R it just restates part of 1.6.17. However, when we de ne the grade of a nitely generated ideal with respect to an arbitrary module in Chapter 9, 1.6.22 will be an essential result. Proof.
.
Exercises
.
f
g
Let I , J , I1 , I2 , I3 be nite subsets of N. Suppose I = i1 . . . ip , J = ip+1 . . . ip+q , the elements given in ascending order. (a) Suppose I J = , and let be the permutation of I J given by (jk ) = ik where I J = j1 . . . jp+q is given in ascending order. Prove (I J ) = () = ( 1)pq (J I ). (b) Deduce that (I1 I2 ) (I1 I2 I3 ) = (I1 I2 I3 ) (I2 I3 ). 1.6.24. Let R be a local ring, and M a nite R -module. ; V (a) Show ( i M ) = (Mi ) for all i N. V (b) Let 1 i (M ). Prove that M is free if and only if i M is free. 1.6.23.
f
;
g
\ f g
2
54
1. Regular sequences and depth ;
V
(a) Let R be a ring, and M an R -module of rank r. Prove rank i M = ri for all i 2 N. (b) Show the analogue for a homomorphism ' : F ! G of nite free modules over a Noetherian ring. Hint for (b): One may assume that R is local and of depth 0. Then Im ' is a free direct summand of G. 1.6.26. Let R be a Noetherian local ring, F a nite free R -module, U F a V submodule ofV rank r and the natural embedding. Show that j is V injective if V and only if j U is torsion-free. In particular j U is non-zero, but j is not injective for rank U < j (U ). 1.6.27. Let R be a ring, and M an RV -module. For f1 . . . fp 2 M let '(f1 . . . fp ) p be the restriction of d . . . d to M . Show that ' induces an R -linear map f f p Vp Vp : (M ) ! ( M ) . Prove that is an isomorphism, if M is nite and free. 1.6.28. Let L be an R -module, x 2 L , and the right multiplication by x on V L . Prove 1.6.25.
1
0
0
0
e . (x): 0 K
2
^ ;;! R ;;! L ;;! L ;;! 0
0
is a complex. Suppose that L = (R n ) and f 2 L . Then the complexes Ke . (f ) and K . (f ) are isomorphic. (Since K . (f )
= K. (f ) by 1.6.10, one can introduce the Koszul complex via Ke . (f ) if one is satis ed with having it only for linear forms on nite free modules.) 1.6.29. Let R be a Noetherian ring, x = (x1 . . . xn ) an element of R n , M = R n =Rx, and I the ideal generated by x1 . . . xn . Prove that grade I k if and only if 0
0
0 ;;! R ;;! R n ;;!
2 ^
Rn
V
k
^ n ;;! ;;! R ;;! 0
is a free resolution of k M . (The map
is right multiplication by x as in 1.6.28 Vk V V one always has M
= ( k R n )= ( k 1 R n ).) 1.6.30. Let x = x1 . . . xn be a sequence in R , and denote by 'i the di erential ; Vi n V R ! i 1 R n in the Koszul complex of x. Let ri = ni 11 be the expected rank of 'i . (a) Show that Rad Iri ('i ) = Rad(x). (b) Derive 1.6.17 for M = R from the Buchsbaum{Eisenbud acyclicity criterion. 1.6.31. Let R be a Noetherian ring, and M 6= 0 a nite R -module. Let I be an ideal, x = x1 . . . xn a system of generators of I , and g = grade(I M ). Show Hi (x M ) = 0 for i = n ; g + 1 . . . n, and Hi (x M ) 6= 0 for i = 0 . . . n ; g. (This property is called the rigidity of the Koszul complex.) Hint: Reduce to the local case and use 1.6.18 for an inductive argument. 1.6.32. Let (R ) be a Noetherian local ring, I an ideal, x 2 , and M a nite R -module. Prove grade(I + (x) M ) grade(I M ) + 1. 1.6.33. Let (R ) be a local ring, and 6= be a prime ideal such that = . Choose a 2 with = + (a) (see the proof of 1.5.9). Then a is R -regular (why?). If F. is a graded minimal free resolution of R= , show that (F. K. (a)) is a minimal free resolution of R = R . ;
;
;
;
m
m
m
m
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p
Notes
55
Notes After the foundations of homological algebra had been laid by Cartan and Eilenberg 67], it invaded commutative ring theory through the epochal work of Auslander and Buchsbaum 17], 18], 19], Rees 303], and Serre 332]. These works cover the contents of Sections 1.1{1.3, and much more, to be developed in Chapters 2{4. Previously commutative algebra had been ideal theory (under which title Krull (in German) and Northcott published inuential monographs) now modules were considered the objects that give structure to a ring. An intermediate position was taken by Grobner's rather `modern' treatise 141], but it introduced modules only as `Vektormoduln', i.e. submodules of free modules over polynomial rings. Proposition 1.2.16 and several theorems in Chapters 2 and 3 resemble a very successful method in topology, namely to relate the properties of the total space of a bration to those of the base and the bre. The algebraic analogue of this principle was studied systematically by Grothendieck 142] (which, by the way, contains various results on regular sequences not reproduced by us). Torsion-freeness, reexivity, and their `higher' analogues are treated in the monograph 16] of Auslander and Bridger see Bruns and Vetter 61] for a compact presentation. The de nition of rank is taken from Scheja and Storch 323]. The very useful acyclicity criterion of Buchsbaum and Eisenbud appeared in 63]. It is closely related to Peskine and Szpiro's equally important `lemme d'acyclicite' 297] which we reproduced in Exercise 1.4.24. The notion of perfect ideal or module appeared in Rees 303]. It is an abstract version of Grobner's 141] which in turn goes back to Macaulay 262]. A special form of the Hilbert{Burch theorem was proved by Hilbert 171] (and had been previously conjectured by Meyer 274]) whereas Burch 66] provided the rst `modern' version. The theorem has been re-proved several times we have essentially reproduced the version of Buchsbaum and Eisenbud 64] who generalized the theorem to a factorization theorem for the ideals Iri ('i) appearing in their acyclicity criterion. Because of their importance for algebraic geometry, graded rings have been a standard topic in commutative algebra. Their enumerative theory will be developed in Chapter 4. Rees 305] ascribes Theorem 1.5.5 to Samuel. Theorem 1.5.8 is due to Matijevic 267], and 1.5.9 was given by Matijevic and Roberts 268]. (The proof of 1.5.9 has been drawn from Fossum and Foxby 109] and Goto and Watanabe 134].) These theorems are part of a programme aiming at characterizations of graded rings which only use localizations with respect to graded prime ideals.
56
1. Regular sequences and depth
We shall reproduce the pertinent results in the exercises of Chapters 2 and 3. The Koszul complex 240] appeared for the rst time in Hilbert 171]: after having proved his syzygy theorem (see 2.2.14) Hilbert determined the free resolution of the kX1 . . . Xn]-module k. That the Koszul complex is an utterly useful construction even when it is not acyclic seems to have been recognized by Auslander and Buchsbaum 19] and Serre 334]. Auslander and Buchsbaum established the main results of Section 1.6 whereas Serre found the connection with multiplicity theory see Chapter 4.
2 Cohen{Macaulay rings
In this chapter we introduce the class of Cohen{Macaulay rings and two subclasses, the regular rings and the complete intersections. The de nition of Cohen{Macaulay ring is suciently general to allow a wealth of examples in algebraic geometry, invariant theory, and combinatorics. On the other hand it is suciently strict to admit a rich theory: in the words of Hochster, `life is really worth living' in a Cohen{Macaulay ring (183], p. 887). The notion of Cohen{Macaulay ring is a workhorse of commutative algebra. Regular local rings are abstract versions of polynomial or power series rings over a eld. The fascination of their theory stems from a unique interplay of homological algebra and arithmetic. Complete intersections arise as residue class rings of regular rings modulo regular sequences, and, in a sense, are the best singular rings. Their exploration is dominated by methods related to the Koszul complex. 2.1 Cohen{Macaulay rings and modules Let R be a Noetherian local ring, and M a nite module. If the `algebraic' invariant depth M equals the `geometric' invariant dim M , then M is called a Cohen{Macaulay module: De nition 2.1.1. Let R be a Noetherian local ring. A nite R -module M 6= 0 is a Cohen{Macaulay module if depth M = dim M . If R itself is a Cohen{Macaulay module, then it is called a Cohen{Macaulay ring. A maximal Cohen{Macaulay module is a Cohen{Macaulay module M such that dim M = dim R . In general, if R is an arbitrary Noetherian ring, then M is a Cohen{ Macaulay module if M is a Cohen{Macaulay module for all maximal ideals 2 Supp M . (So we consider the zero module to be Cohen{ Macaulay.) However, for M to be a maximal Cohen{Macaulay module we require that M is such an R -module for each maximal ideal of R . As in the local case, R is a Cohen{Macaulay ring if it is a Cohen{ Macaulay module. If I is an ideal contained in Ann M , then it is irrelevant for the Cohen{Macaulay property whether we consider M as an R -module or m
m
m
m
m
57
58
2. Cohen{Macaulay rings
an R=I -module. In particular, if R is local and M a Cohen{Macaulay module, then M is a maximal Cohen{Macaulay module over R= Ann M . The next theorem exhibits the fact that for a Cohen{Macaulay module the grade of an arbitrary ideal is given by its `codimension'. Theorem 2.1.2. Let (R ) be a Noetherian local ring, and M 6= 0 a m
Cohen{Macaulay R-module. Then (a) dim R= = depth M for all Ass M, (b) grade(I M ) = dim M dim M=IM for all ideals I , (c) x = x1 . . . xr is an M-sequence if and only if dim M=xM = dim M r, (d) x is an M-sequence if and only if it is part of a system of parameters of M. Proof. (a) We saw depth M dim R= in 1.2.13, and dim R= dim M holds since Ass M Supp M . (b) If grade(I M ) = 0, then there exists Ass M with I therefore dim M=IM = dim M follows from (a). If grade(I M ) > 0, then we choose x I regular on M . One has grade(I M=xM ) = grade(I M ) 1, depth M=xM = depth M 1, and dim M=xM = dim M 1
;
p
p
2
p
p
2
p
2
;
;
m
;
p
;
so that induction completes the argument. (c) It suces now to quote 1.6.19. (d) This is just a reformulation of (c). The Cohen{Macaulay property is stable under specialization and localization: Theorem 2.1.3. Let R be a Noetherian ring, and M a nite R-module. (a) Suppose x is an M-sequence. If M is a Cohen{Macaulay module, then M=xM is Cohen{Macaulay (over R or R=(x)). The converse holds if R is
local. (b) Suppose that M is Cohen{Macaulay. Then for every multiplicatively closed set S in R the localized module MS is also Cohen{Macaulay. In particular, M is Cohen{Macaulay for every Spec R. If M = 0, then depth M = grade( M ) if in addition R is local, then dim M = dim M + dim M= M.
2
p
p
p
6
p
p
p
p
(a) By the de nition of Cohen{Macaulay module one may evidently assume that R is local. Let n be the length of x. Then dim M=xM = dim M ; n by 1.2.12 and depth M=xM = depth M ; n by 1.2.10. (b) Let be a maximal ideal of RS . The ideal is the extension of a prime ideal in R , and so (RS ) = R . Let be a maximal ideal of R containing . Then R is a localization of the Cohen{Macaulay local ring R . So we may again assume that R is local. There is nothing to prove if M = 0. When M 6= 0, we use induction on depth M . If depth M = 0, then 2 Ass M , and is a minimal Proof.
q
q
p
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p
59
2.1. Cohen{Macaulay rings and modules
prime of Supp M by 2.1.2 therefore dim M = 0. The same argument shows that cannot be contained in any 2 Ass M if depth M > 0. So contains an M -regular element x, and the induction hypothesis applies to M=xM . It follows easily that M is Cohen{Macaulay and that depth M = grade( M ). The second equation results from that and 2.1.2. Corollary 2.1.4. Let R be a Cohen{Macaulay ring, and I 6= R an ideal. Then grade I = height I, and if R is local, height I + dim R=I = dim R. Proof. One has height I = minfdim R : 2 V (I )g and furthermore grade I = minfdepth R : 2 V (I )g. Theorem 2.1.3 yields dim R = depth R for all 2 Spec R . This proves the rst equation, and the second follows from that and 2.1.2. Let k be a eld. We shall see in the next section that every nite module over a polynomial ring kX1 . . . Xn] or a power series ring kX1 . . . Xn]] has nite projective dimension. Furthermore these rings are Cohen{Macaulay as will be shown below. This explains why the following theorem is a very e ective Cohen{Macaulay criterion. Theorem 2.1.5. Let R be a Cohen{Macaulay ring, and M a nite R-module p
p
q
p
p
p
p
p
p
p
p
p
p
p
p
of nite projective dimension. (a) If M is perfect, then it is a Cohen{Macaulay module. (b) The converse holds when R is local. Proof. Let M be perfect and Supp M . Then M is a perfect module as shown in the proof of 1.4.16. So we may assume that R is local. The Auslander{Buchsbaum formula gives proj dim M = dim R depth M , and 2.1.4 yields grade M = dim R dim M . Thus depth M = dim M if and only if proj dim M = grade M . One says that an ideal I is unmixed if I has no embedded prime divisors or, in modern language, if the associated prime ideals of R=I are the minimal prime ideals of I . Macaulay showed in 1916 that an ideal I = (x1 . . . xn ) of height n in a polynomial ring over a eld is unmixed, and for regular local rings this was proved by Cohen in 1946. (An ngenerated ideal of height n is said to be of the principal class.) These facts
2
p
p
;
;
and the following theorem explain the nomenclature `Cohen{Macaulay'. Theorem 2.1.6. A Noetherian ring R is Cohen{Macaulay if and only if every ideal I generated by height I elements is unmixed. Proof. `)': Suppose I = (x), x = x1 . . . xn , and let 2 Ass R=I , . Then there is a maximal ideal with , and R , R 2 Ass(R =I ). If height I = n, then dim(R =I ) = dim R ; n, and x is an R -sequence by 2.1.2. Therefore R =I is Cohen{Macaulay, whence R = R (again by 2.1.2), and so = . p
m
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60
2. Cohen{Macaulay rings
`(': Let J R be an arbitrary ideal, say height J = n. Then there exist x1 . . . xn 2 J with height(x1 . . . xi ) = i for all i = 0 . . . n (see A.2). It is impossible for xi+1 to be contained in a minimal prime ideal of (x1 . . . xi ). By hypothesis it therefore is an (R=(x1 . . . xi ))regular element. So x1 . . . xn is an R -sequence. We have shown that grade J = height J for every proper ideal J of R . Then R is certainly Cohen{Macaulay. Flat extensions of Cohen{Macaulay rings and modules. The behaviour of
depth under at local extensions was studied in Section 1.2. That makes it easy to prove an analogous theorem for the Cohen{Macaulay property. Theorem 2.1.7. Let ' : (R ) ! (S ) be a homomorphism of Noetherian m
n
local rings. Suppose M is a nite R-module and N is an R-at nite Smodule. Then M R N is a Cohen{Macaulay S-module if and only if M is Cohen{Macaulay over R and N= N is Cohen{Macaulay over S.
m
In fact, according to 1.2.16 we have depthS M N = depthR M + depthS N= N . Since depth is bounded above by dimension, the theorem follows from the analogous equation for dimension see A.11. Corollary 2.1.8. Let (R ) be a Noetherian local ring, M a nite R^ its -adic completion. module, and M ^ (a) Then dimR M = dimR^ M^ and depthR M = depthR^ M. (b) M is Cohen{Macaulay if and only if M^ is Cohen{Macaulay. ^ is local and at, and M^ = M R R^ since M Proof. The extension R ! R is nite. One can of course use more direct arguments in order to prove the previous corollary. Similarly there is a more `elementary' approach to the following theorem see for example 231]. Theorem 2.1.9. Let R be a Noetherian ring, M a nite R-module, and S = R X1 . . . Xn ] or S = R X1 . . . Xn]]. Then M S is a Cohen{Macaulay m
m
m
S-module if and only if M is a Cohen{Macaulay module.
Since the indeterminates can be adjoined successively, we may assume n = 1, X = X1. The `only if' part is easy: in both cases X is (M S )-regular, and R = S=(X ), M = (M S )=X (M S ). (That X is (M S )-regular is evident for S = R X ] the reader should nd a justi cation for S = R X ]].) Conversely, let be a maximal ideal of S , and set = \ R . As outlined below A.12 the bre S = S is a discrete valuation ring, and thus Cohen{Macaulay. Now we invoke 2.1.7 and complete the proof. Proof.
m
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m
61
2.1. Cohen{Macaulay rings and modules
For polynomial extensions the proof of 2.1.9 shows that a stronger local version of 2.1.9 is valid: for 2 Spec R X1 . . . Xn ] the localization R X1 . . . Xn ] is Cohen{Macaulay if and only if R is Cohen{Macaulay for = \ R . Similarly, there is a local version of the following theorem: Theorem 2.1.10. Let k be a eld, R a Noetherian k-algebra, and K an q
q
p
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extension eld of k. Suppose that R is a nitely generated k-algebra, or that K is nitely generated as an extension eld of k. Then R is a Cohen{ Macaulay ring if and only if R k K is. Proof. If R is a nitely generated k-algebra, then R k K is a nitely generated K -algebra, and therefore Noetherian. Suppose that K is a nitely generated extension eld. Then K is a nite algebraic extension of a nite purely transcendental extension K 0 of k. Since K 0 is the eld of fractions of a polynomial ring kT1 . . . Tn], we nd again that R k K 0 is Noetherian, whence R k K = (R k K 0 ) K 0 K is also Noetherian. Evidently R K is a faithfully at R -algebra. Therefore, given a prime ideal of R , there exists Spec R K such that = R . The bre of the extension R (R K ) is a localization of k( ) K . In conjunction with 2.1.7 this argument reduces the theorem to the assertion that L k K is Cohen{Macaulay for extension elds L and K of k, provided one of
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\
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them is nitely generated. This follows from the next proposition. Proposition 2.1.11. Let k be a eld, R a k-algebra, and K a nitely generated extension eld of k. Then R
k K is isomorphic to a ring
R X1 . . . Xn ]S =(f1 . . . fm ) where S is a multiplicatively closed subset of R X1 . . . Xn], and f1 . . . fm is a R X1 . . . Xn]S -sequence. Proof. The extension k K decomposes into a series of cyclic extensions k = K0 Kt = K . We use induction on t. Suppose that T = R k Ki = R X1 . . . Xn]S =(f1 . . . fm ). If Ki+1 = Ki Y ]=(g) with a monic irreducible polynomial g, then
R
k Ki+1 = T K Ki+1 = T Y ]=(g): i
Since T is a at Ki -algebra, g is not a zero-divisor of T Y ] = R X1 . . . Xm Y ]S =(f1 . . . fm ): If Ki+1 = Ki (Y ), then R k Ki+1 = R X1 . . . Xn Y ]S 0 =(f1 . . . fm ) where S 0 is generated by the image of S and the image of Ki Y ] n f0g. Chain conditions in Cohen{Macaulay rings. Cohen{Macaulay rings were
introduced as those rings for which depth equals dimension. Corollary 2.1.4 and the next theorem show that dimension theory itself is simpler
62
2. Cohen{Macaulay rings
for Cohen{Macaulay rings than for general Noetherian rings. One says a Noetherian ring R is catenary if every saturated chain joining prime ideals and , , has (maximal) length height = R is universally catenary if all the polynomial rings R X1 . . . Xn ] are catenary. It is easy to see that R is universally catenary if and only if every nitely generated R -algebra is (universally) catenary. Theorem 2.1.12. A Cohen{Macaulay ring R is universally catenary. Proof. `Universally' may be dropped because of 2.1.9. So let be prime ideals of R . The localization R is Cohen{Macaulay, and 2.1.4 applied to R yields height = dim R = height R + dim(R = R ) = height + height = : It is an easy exercise to show that R is catenary if this equation holds for all prime ideals . Corollary 2.1.13. A Noetherian complete local ring R is universally catep
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nary.
Cohen's structure theorem (see A.21) tells us that R is a residue class ring of a formal power series ring A = kX1 . . . Xn]] where k is a eld or a discrete valuation ring. By 2.1.9, A is Cohen{Macaulay and therefore universally catenary R inherits this property as a residue class ring of A. Remark 2.1.14. For the sake of clarity 2.1.4 and 2.1.12 were kept more special than necessary. If R has a Cohen{Macaulay module M with Supp M = Spec R , then we need only replace grade I by grade(I M ) in 2.1.4 to obtain an equally valid result. It follows that a local Noetherian domain which has a maximal Cohen{Macaulay module is universally catenary. One of Nagata's famous counter-examples is a non-catenary such domain (284], Example 2, p. 203). However, to be universally catenary is not the only necessary condition for R to have a Cohen{Macaulay module M with Supp M = Spec R it must also satisfy Grothendieck's condition (CMU). This condition requires that for every prime ideal of R the spectrum of R= contains a non-empty open subset U such that (R= ) is Cohen{Macaulay for all 2 U see 142], IV, 6.11. A local ring violating (CMU) was constructed by Ferrand and Raynaud 106]. A Noetherian complete local ring is universally catenary since it is a residue class ring of a Cohen{Macaulay ring, and for the same reason it satis es (CMU). It is an open question whether every Noetherian complete local ring has a maximal Cohen{Macaulay module. We shall see in Chapter 9 that the existence of maximal Cohen{ Macaulay modules implies a wealth of homological theorems. Fortunately, it will not be essential that these Cohen{Macaulay modules M Proof.
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2.1. Cohen{Macaulay rings and modules
are really nite we `only' need every system of parameters of the ring to be an M -sequence. In Chapter 8 such modules will be shown to exist for local rings containing a eld. For a prime ideal in a Cohen{Macaulay local ring R the residue class ring R= is not Cohen{Macaulay in general it is however unmixed in the sense of Nagata 284]: Theorem 2.1.15. Let R be a Cohen{Macaulay local ring, and a prime ^ = dim R= for all 2 Ass(R= ^ R^ ). In particular R^ ideal. Then dim R= p
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p
is an unmixed ideal. ^ R^ )-regular element. Proof. If R = , then would contain an (R= Therefore R = , and we have a at local ring extension R R^ . Applying 1.2.16 and since R^ and R are Cohen{Macaulay we get
\ 6 \
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dim R^ = depth R^ = depth R + depth(R^ = R^ ) = dim R : In view of 2.1.4 this equation is equivalent to the theorem. p
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Serre's condition (Sn).
Sometimes one only needs a ring or a module to be Cohen{Macaulay in low `codimension'. A nite module over a Noetherian ring R satis es Serre's condition (Sn) if depth M min(n dim M ) for all 2 Spec R . The theorems of this section need some modi cation when Cohen{Macaulay is replaced by (Sn ). As an example we treat the (Sn) analogue of 2.1.7. Proposition 2.1.16. Let ' : R ! S be a at homomorphism of Noetherian p
p
p
rings. (a) Let Spec S and = R. If S satis es (Sn), then so does R . (b) Suppose R and all the bres k( ) S with Spec R satisfy (Sn). Then S satis es (Sn ). Proof. (a) Replacing R by R and S by S we may assume that ' is a at homomorphism of local rings. For Spec R we now choose a minimal prime of S . Then dim(S = S ) = 0, and according to 1.2.16 we have q
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depth R = depth S min(n dim S ) = min(n dim R ): (b) For 2 Spec S and = R \ one similarly deduces depth S = depth R + depth(S = S ) min(n dim R ) + min(n dim(S = S )) min(n dim R + dim(S = S )) = min(n dim S ): q
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64
2. Cohen{Macaulay rings
Exercises Let k be a eld, and R = k X1 . . . Xn ]. If is a prime ideal in R with 2 f0 1 n ; 1 ng, show that R= is Cohen{Macaulay. 2.1.18. Let k be a eld. Show (a) the subalgebra S = k U 4 U 3 V UV 3 V 4 ] of k U V ] is not Cohen{Macaulay, (b) for each m with 2 m n ; 2 there exists a prime ideal of height m in R = k X1 . . . Xn ] for which R= is not Cohen{Macaulay. 2.1.19. Let k be a eld, and S the subalgebra of k X1 . . . Xn ] generated by the monomials of degrees 2 and 3. Show S is an n-dimensional domain the maximal ideal (X1 . . . Xn ) \ S has height n and grade 1. 2.1.20. Prove (a) a one dimensional reduced Noetherian ring is Cohen{Macaulay, (b) a one dimensional Noetherian local ring has a maximal Cohen{Macaulay module. 2.1.21. Characterize (Sn ) by an unmixedness property. 2.1.22. Prove that a module M satis es (Sn ) if and only if M is Cohen{Macaulay for all prime ideals with depth M < n. 2.1.23. Let R ! S be a faithfully at homomorphism of Noetherian rings. Show the following are equivalent: (a) S is Cohen{Macaulay (b) R and all the bres S = S are Cohen{Macaulay where 2 Spec S and = \ R. Hint: use A.10. 2.1.24. Prove the analogues of 2.1.8(b), 2.1.9, and 2.1.10 for (Sn ). For the passages from R to R^ and to R
X1 . . . Xn ]] assume that R is a residue class ring of a Cohen{Macaulay ring. 2.1.25. Prove the converse of 2.1.5(a) under the hypothesis that Supp M is connected. (The crucial point is to show that the function 7! proj dim M is locally constant on Supp M if M is locally perfect.) 2.1.26. Let R be a Cohen{Macaulay local ring of dimension d and M a nite R -module. Deduce that the d -th syzygy of M in an arbitrary nite free resolution is either 0 or a maximal Cohen{Macaulay module. 2.1.27. Let R be a Noetherian graded ring, and M a nite graded R -module. Show: (a) For 2 Spec R the localization M is Cohen{Macaulay if and only if M is. (This follows easily from the results of Section 1.5.) (b) The following are equivalent: (i) M is Cohen{Macaulay (ii) M is Cohen{Macaulay for all graded prime ideals (iii) M( ) is Cohen{Macaulay for all graded prime ideals . (c) Suppose in addition that (R ) is local. Then M is Cohen{Macaulay if and only if M is. 2.1.28. Let (R ) be a Noetherian local ring, and x 2 a homogeneous R -regular element. Then R is Cohen{Macaulay if and only if so is R=(x). 2.1.17.
height
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65
2.2. Regular rings and normal rings
Let the Noetherian ring R be a free Z-module such that R K is Cohen{Macaulay for some eld K of characteristic p > 0. Show that R L is Cohen{Macaulay for every eld L of characteristic 0. Hint: reduce the problem to the case in which K = Z=(p), L = Q and use R Z , = (p), as a `bridge'. This is the rst and easiest example of reduction to characteristic p.
2.1.29.
p
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2.2 Regular rings and normal rings The most distinguished of all Noetherian local rings are those whose maximal ideal can be generated by a system of parameters: De nition 2.2.1. A Noetherian local ring (R ) is regular if it has a system of parameters generating such a system of parameters is called a regular system of parameters. Evidently, when dim R = 0, then R is regular if and only if it is a eld, and when dim R = 1, R is regular if and only if it is a discrete valuation ring. Other examples of regular local rings are kX1 . . . Xn ]] where k is a eld, and kX1 . . . Xn] , = (X1 . . . Xn ). We may rephrase the de nition above as follows: R is regular if and only if ( ) = dim R . In fact, ( ) dim R by Krull's principal ideal theorem, and a system of generators of has dim R elements exactly when it is a system of parameters. Proposition 2.2.2. A Noetherian local ring (R ) is regular if and only if its -adic completion R^ is regular. Proof. The maximal ideal of R^ is R^ , and we have natural isomor^ R^ , = 2 phisms R= = R= = ( R^ )=( R^ )2. Therefore ( ) = ( R^ ). ^ Furthermore dim R = dim R , and by de nition R is regular if and only if dim R = ( ). It is easily proved that regular local rings are integral domains. Proposition 2.2.3. Let (R ) be a regular local ring. Then R is an integral m
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domain.
We use induction on dim R . When dim R = 0, R is a eld. So suppose dim R > 0, and let 1 . . . m be the minimal prime ideals of R . There exists an element x 2 which is not contained in any of the ideals 2 . . . . (This follows easily from 1.2.2 with M = N = .) Since x is 1 m part of a minimal system of generators of , it is part of a regular system of parameters, and thus R=(x) is regular (use that dim R=(x) = dim R ; 1). As dim R=(x) < dim R we may assume that R=(x) is a domain. Thus (x) is a prime ideal, and therefore contains a minimal prime ideal of R , say = 1 , z is an element 1 . Every y 2 1 has the form y = xz , and since x 2 of 1. It follows that 1 = x 1, which, by Nakayama's lemma, implies 1 = 0, as required.
Proof.
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2. Cohen{Macaulay rings
Using the previous proposition, one can say precisely which residue class rings of a regular local ring are also regular: Proposition 2.2.4. Let R be a regular local ring, and I R an ideal. Then
R=I is regular if and only if I is generated by a subset of a regular system of parameters. Proof. The `if' part is trivial. So suppose that R=I is regular. Then ( =I ) = dim R=I set m = dim R dim R=I . By Nakayama's lemma I contains elements x1 . . . xm which are part of a minimal system of generators of . Then R=(x1 . . . xm ) is regular of dimension dim R m = dim R=I . Since I and (x1 . . . xm ) are prime ideals, one must have I = (x1 . . . xm).
;
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The next proposition gives useful characterizations of regularity. Proposition 2.2.5. Let (R k) be a Noetherian local ring, and x1 . . . xn a m
minimal system of generators of . Then the following are equivalent: (a) R is regular (b) x1 . . . xn is an R-sequence (c) the substitution Xi x i = 2 yields an isomorphism kX1 . . . Xn] = gr (R ). Proof. (a) (b): Since x1 . . . xn is a minimal system of generators of , it is a regular system of parameters, and R=(x1 . . . xi) is also regular for each i. Therefore R=(x1 . . . xi ) is a domain, and xi+1 is regular on R=(x1 . . . xi ). (b) (a): An R -sequence is part of a system of parameters by 1.2.12. m
7! 2
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)
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(b) () (c): This follows from 1.1.8 and its converse 1.1.15. Corollary 2.2.6. A regular local ring is Cohen{Macaulay.
The Auslander{Buchsbaum{Serre theorem. Whereas the characterizations
of regular local rings in 2.2.5 are rather close to the de nition, this can hardly be said of the following theorem. Together with 2.2.19 below, it is considered to be the most important achievement of the use of homological algebra in the theory of commutative rings. Theorem 2.2.7 (Auslander{Buchsbaum{Serre). Let (R k) be a Noem
therian local ring. Then the following are equivalent: (a) R is regular (b) proj dim M < for every nite R-module M (c) proj dim k < . Proof. (a) (b): Let d = dim R , and N a d -th syzygy module of M . Since R is Cohen{Macaulay, N is a maximal Cohen{Macaulay module or 0 by Exercise 2.1.26. If N = 0, we are done otherwise every regular system of parameters x is a (maximal) N -sequence. Lemma 1.3.5 gives
)
1 1
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2.2. Regular rings and normal rings
proj dimR N = proj dimR=(x) (N=xN ) = proj dimk (N= N ) = 0. So N is free, and proj dim M d . (b) ) (c): This is trivial. (c) ) (a): This is a special case of the following theorem. Theorem 2.2.8 (Ferrand, Vasconcelos). Let (R ) be a Noetherian local ring, and I 6= 0 a proper ideal with proj dim I < 1. If I=I 2 is a free m
m
R=I-module, then I is generated by a regular sequence. Proof. Since I has a nite free resolution, it contains an R -regular element x by 1.4.7. It is no restriction to assume x = I : if x I , then we choose some y I I Ry + Rx is not contained in any Ass R , and by 1.2.2 there is a R for which y + ax has the same property. This proves the theorem when (I ) = 1, and when (I ) > 1, we use induction, passing from R to R=(x) and from I to I=(x). Of course, we must rst verify that proj dimR=(x) I=(x) < . Since x = I , the residue class of x in I=I 2 is part of a basis x x 2 . . . x m of this free module. Set J = (x2 . . . xm ) we claim: J (x) xI . In fact, if z = ax = a2 x2 + + amxm , then a I because x x 2 . . . x m are linearly independent modulo I . Therefore we get a composition of maps
2 n
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\ 2 I=(x) = (J + (x))=(x) = J=J \ (x) ;! I=xI ;! I=(x)
m
in which the residue class of xi is sent to itself, and which therefore is the identity on I=(x). So I=(x) is a direct summand of I=xI as x is I -regular, the latter has nite projective dimension over R=(x) by 1.3.5. Finally we need that I =I 2 is a free R=I -module where I = I=(x). But this is a very easy consequence of the linear independence of x x2 . . . xm modulo I . The proof of 2.2.7 can be varied: the Koszul complex of a regular system of parameters resolves k by 2.2.5 and 1.6.14, whence (a) ) (c). Moreover, the implication (c) ) (b) follows from 1.3.2: proj dim k < 1 ) TorRi (M k) = 0 for i 0, and this in turn gives proj dim M < 1. While this reasoning uses a truly homological argument, namely the fact that Tor can be computed from a free resolution of either module, the proof above merely exploits the existence of minimal free resolutions. Serre's original argument for (c) ) (a) will be indicated in Exercise 2.3.22. Corollary 2.2.9. Let R be a regular local ring, and a prime ideal in R. p
Then R is regular.
By 2.2.7(a) ) (b) we have proj dim R= < 1. It follows that proj dimR (R= ) = proj dimR (R = R ) < 1, whence R is regular by 2.2.7(c) ) (a). Over a regular local ring the Cohen{Macaulay property is equivalent to perfection (see Section 1.4 for this notion): p
Proof.
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68
2. Cohen{Macaulay rings
Corollary 2.2.10. A nite module M over a regular local ring is Cohen{ Macaulay if and only if it is perfect.
The corollary is an immediate consequence of 2.1.5 and 2.2.7. Let (R ) be a Noetherian local ring. By 2.1.8, R is Cohen{Macaulay if and only if its -adic completion R^ is Cohen{Macaulay. Furthermore, if R contains a eld or is a domain, then it is a nite module over a regular local subring (see A.22). Thus the following proposition may almost be considered a new description of the Cohen{Macaulay property for rings. Proposition 2.2.11. Let R be a Noetherian local ring and S a regular local m
m
subring such that R is a nite S-module. Then R is Cohen{Macaulay if and only if it is a free S-module.
By 2.2.7 one has proj dimS R < 1 therefore R is S -free if and only if depthS R = dim S . Choose a (regular) system of parameters x in S . Then x is also a system of parameters of R , and therefore R is Cohen{Macaulay () x is an R -sequence () depthS R = dim S . (One could also use Exercise 1.2.26.) Proof.
Flat extensions of regular rings. The behaviour of regularity under at
local extensions is described by the following theorem. Theorem 2.2.12. Let ' : (R k) ! (S l ) be a at homomorphism of m
n
Noetherian local rings. (a) If S is regular, then so is R. (b) If R and S= S are regular, then so is S. Proof. (a) Let F. be a minimal free resolution of the R -module k. Then F. S is a free resolution of k S = S= S because of atness, and even a minimal one since '( ) . Thus proj dimR k = proj dimS S= S < , and R is regular by 2.2.7. (b) Let m = dim R , n = dim S= S , and choose minimal systems of generators x1 . . . xm of and y1 . . . yn of = S . Then '(x1) . . . '(xm ), y1 . . . yn generate , and S must be regular because dim S = dim R + dim S= S see A.11. An easy example shows that S= S need not be regular in the situation of 2.2.12(a): Let k be a eld, and choose S = kX Y ]]=(Y X 2) and R = ky]] S . Then R and S are regular, and S is a free R -module generated by 1 and x, but S=yS = kX ]]=(X 2) is not regular. (The reader m
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should imagine the geometry of this example.) In order to formulate a theorem relating the regularity of R and that of R X1 . . . Xn] we must rst agree on calling a Noetherian ring R regular if its localizations R with respect to maximal ideals are regular. m
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2.2. Regular rings and normal rings
Theorem 2.2.13. A Noetherian ring R is regular if and only if R X1 . . . Xn] is regular. The same holds for R and R X1 . . . Xn]]. Proof. We may assume that n = 1 and set X = X1. Suppose that R X ] is regular, and, given a maximal ideal of R , choose = ( X ) R X ]. Then X 2= 2 , equivalently X 2= ( R X ] )2 , and it follows immediately from the de nition of regularity that R is regular. The same argument shows that regularity descends from R X ]] to R . Of course, what has just been shown can also be derived from 2.2.12(a), and the reader is invited to use 2.2.12(b) in proving that regularity ascends from R to R X ] and R X ]]. (Compare the proof of 2.1.9.) In particular, a polynomial ring kX1 . . . Xn] over a eld k is regular. Corollary 2.2.14. Let k be a eld, and R = kX1 . . . Xn]. (a) (Hilbert's syzygy theorem) Every nite graded R-module M has a nite m
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graded free resolution of length n. (b) Moreover, proj dim M n for every nite R-module M. (c) In fact, every nite R-module has a nite free resolution of length n. Proof. (a) Set = (X1 . . . Xn) and consider a minimal graded free resolution F. of M . Such a resolution exists, and furthermore F. R is a minimal free resolution of M see 1.5.15. Now R is a regular local ring. Therefore F. R has length at most n, and the same holds true for F. . (b) Consider an arbitrary maximal ideal of R . Then dim R dim R = n, and R is a regular local ring. Hence proj dimR M n. Taking the supremum over all maximal ideals, we get proj dim M n. (In fact, let N be the n-th syzygy of M in a resolution by nite projective R -modules. Then N is a nite free R -module for all , and therefore N
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is projective.) (c) By the theorem of Quillen and Suslin (318], Theorem 4.59) every nite projective R -module is free. In 2.2.14(a) and 2.2.15 below it is not essential that deg Xi = 1 for all i. One may replace the standard grading of R by any grading which makes R a local ring, for example by a grading such that deg Xi > 0 for all i. Corollary 2.2.15. Let k be a eld, R = kX1 . . . Xn ], = (X1 . . . Xn), n
n
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and M a nite graded R-module. Then the following are equivalent: (a) M is Cohen{Macaulay (a0) M is perfect (b) M is Cohen{Macaulay (b0) M is perfect. m
The implications (a0 ) ) (a) ) (b) ) (b0) follow from 2.1.5 and 2.2.10. The remaining implication (b0 ) ) (a0 ) is an immediate consequence m
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of the equations proj dim M = proj dim M and grade M = grade M proved in 1.5.15. Remark 2.2.16. Let R be a Noetherian k-algebra where k is a eld, and K an extension eld of k. If R is nitely generated as a k-algebra or K is a nitely generated extension eld, then R k K is a Noetherian ring as shown in the proof of 2.1.10. Since R k K is a at R -algebra, it follows readily from 2.2.12 that R is regular if R k K is regular. We saw in the proof of 2.1.10 that the bres of the extension R ! R k K are of the form (L k K ) where L is an extension eld of k and 2 Spec(L k K ). If L k K is regular for every extension eld L of k (provided one of K L is nitely generated), then one obtains from 2.2.12 that R k K is regular if R is regular. The elds K satisfying the condition just formulated are the separable extensions of k. We refer the reader to 270], x26 for a discussion of separability, and to 142], IV, 6.7.4.1 for the theorem concerning the regularity of L k K . m
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Factoriality of regular local rings. Our next goal is to show that a regular local ring is a factorial domain (a UFD in other terminology). We need two elementary lemmas whose proofs are left as an exercise for the reader. Lemma 2.2.17. A Noetherian domain R is factorial if and only if every prime ideal of height 1 is principal. Lemma 2.2.18. Let R be a Noetherian domain and a prime element in R. p
Then R is factorial if and only if R is factorial.
Theorem 2.2.19 (Auslander{Buchsbaum{Nagata). A regular local ring R is factorial.
We use induction on dim R . If dim R = 0, then R is a eld, and there is nothing to prove. So suppose dim R > 0, and choose 2 n 2 . Since R=( ) is again a regular local ring, is a prime element. According to the previous lemma, it is enough to show that S = R is factorial. Let be a prime ideal of S with height = 1. Every localization S is a localization of R with respect to a prime ideal 6= , and therefore a regular local ring by 2.2.9. By induction S is factorial. If 6 , then S = S for trivial reasons, and if , then also S = S as follows from 2.2.17 in conjunction with the factoriality of S . This implies that is a projective S -module of rank 1. Of course is of the form S with a prime ideal of R . The R -module has a nite free resolution F , whence = has an augmented resolution Proof.
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G. : 0
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' ' Gs;1 ;! ;! G1 ;! G0 ;! ;! 0 ;! Gs ;! s
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2.2. Regular rings and normal rings
by nite free S -modules. However, is a projective S -module, and its syzygy modules with respect to G are likewise projective. In particular Im 's;1 Gs = Gs;1. If s > 1, we can modify the tail of G to obtain the free resolution ' G : 0 ;! Im 's;1 Gs ;! Gs;2 Gs ;! ;! G1 ;! G0 ;! ;! 0: Therefore, by induction on the length of G , in fact has a free resolution p
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0 ;! S n ;! S n+1 ;! ;! 0: The Hilbert{Burch theorem 1.4.17 yields that = aIn(') with some a 2 S , and furthermore that = aS since is projective. So is a principal ideal. A ring is normal if all its localizations are integrally closed domains a Noetherian ring is normal if and only if it is the direct product of nitely many integrally closed domains (see 270] for a detailed discussion of normality). Corollary 2.2.20. A regular local ring is a normal domain. A regular ring '
p
p
p
p
p
is the direct product of regular domains. In fact, every factorial ring R is a normal domain. (One proves this just as for the special case R = Z.) The `classical' proof of the corollary
uses 2.2.5 and the fact that a Noetherian local ring is a normal domain if gr (R ) is a normal domain see 4.5.9. There is even a third proof, as we shall see now. m
Serre's normality criterion. A Noetherian ring R satis es Serre's condition (Rn) if R is a regular local ring for all prime ideals in R with dim R n. (Note the similarity with (Sn ) contrary to (Sn) however, (Rn) says nothing about localizations R with dim R > n.) We leave it as an exercise for the reader to prove that the behaviour of (Rn) under at local extensions is the same as that of (Sn): Proposition 2.2.21. Let ' : R ! S be a at homomorphism of Noetherian p
p
p
p
p
rings. (a) Let Spec S and = R. If S satis es (Rn), then so does R . (b) If R and all bres k( ) S, Spec R, satisfy (Rn), then so does S. It is easy to see that a Noetherian ring R is reduced if and only if it satis es (R0) and (S1). Serre characterized normality in a similar way: Theorem 2.2.22 (Serre). A Noetherian ring R is normal if and only if it satis es (R1) and (S2). q
2
p
q
p
\
p
2
q
p
We refer the reader to Serre 334], IV.4, or 270], x23, for a proof of 2.2.22. The following corollary is an evident consequence of 2.1.16, 2.2.21, and 2.2.22:
72
2. Cohen{Macaulay rings
Corollary 2.2.23. Let ' : R
! \
S be a at homomorphism of Noetherian rings. (a) Let Spec S and = R. If S is normal, then so is R . (b) If R and all the bres k( ) S, Spec R, are normal, then so is S.
2
q
p
q
p
2 q
p
p
Suppose that (R ) and (S ) are local, and that ' is at and local. Then, for S to be normal, it is not sucient to have R and S= S normal: there are normal local domains whose completions are not even domains see 284], p. 209, Example 7. m
n
m
Exercises Let R be a Noetherian graded ring. Show: (a) For 2 Spec R the localization R is regular if and only if R is. (b) The following are equivalent: (i) R is regular (ii) R is regular for all graded prime ideals (iii) R( ) is regular for all graded prime ideals . (c) Suppose moreover that (R ) is local. Then R is regular if and only if R is. Hint: Use 1.6.33. 2.2.25. Let R be a positively graded k-algebra over a eld k. Prove the following are equivalent: (a) R is regular (b) R is regular where is the maximal ideal (c) there exist homogeneous elements x1 . . . xn of positive degree for which the assignment Xi 7! xi induces an isomorphism k X1 . . . Xn ]
= R. Hint: For the non-trivial implication (b) ) (c) choose a minimal homogeneous system of generators x1 . . . xn of then apply 1.5.15 and 1.5.4. The rest is a simple dimension argument. 2.2.26. In the situation of 2.2.11 characterize the Cohen{Macaulay R -modules by a property they have as S -modules. 2.2.27. Let R be a Noetherian ring over which every nite module has a nite free resolution. Show R is a factorial domain. 2.2.28. Let R be a regular local ring, and I an ideal of height 1. Prove that the following are equivalent: (a) R=I is Cohen{Macaulay (b) height = 1 for all prime ideals 2 Ass R=I (c) I is a principal ideal. Hint: For (b) ) (c) one uses primary decomposition and the factoriality of R . 2.2.29. Prove that a Noetherian ring R satis es (Ri ) and (Si+1 ) if and only if R is regular for every prime ideal such that depth R i. 2.2.30. (a) Show a Noetherian normal ring of dimension 2 is Cohen{Macaulay. (b) A Cohen{Macaulay ring is normal if and only if it satis es (R1 ). 2.2.31. (a) Let R be a Noetherian complete local domain. Then R is a nite module over a regular local ring S contained in R see A.22. Set M = HomS (R S ) then 2.2.24.
p
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p
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p
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m
m
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p
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73
2.3. Complete intersections M is an R -module in a natural way. Show that depthR M = depthS M min(dim R 2).
(b) Prove that every Noetherian complete local ring of dimension 2 has a maximal Cohen{Macaulay module. 2.2.32. Let R be a Noetherian complete local domain. It is known that the integral closure of R in its eld of fractions is a nite R -module ( 284], (32.1) or 47], Ch. IX, x4). Use this to give a fresh proof of the fact that a Noetherian complete local ring of dimension 2 has a maximal Cohen{Macaulay module. 2.2.33. Let R be a Noetherian ring, and x 2 R an R -regular element. (a) Assume that Rx ful lls (Rn ) and (Sn+1 ), and that (Rn 1 ) and (Sn ) hold for R=(x). Show that R satis es (Rn ) and (Sn+1 ). (b) Assume Rx is a normal domain, and R=(x) is reduced. Show that R is a normal domain. 2.2.34. Let R be a Noetherian graded ring, and A its dehomogenization with respect to an element of degree 1 (see 1.5.18). Show that if R is a normal domain, then so is A. 2.2.35. We keep the notation of 2.2.34. Let be a graded prime ideal of R with x 2= , and 2 Spec A its dehomogenization (see 1.5.26). Show that R is a at local extension of A , and determine its bre. Compare R and S with respect to the following quantities and properties: dimension, depth, type being reduced, an integral domain, Cohen{Macaulay, normal, regular. ;
p
p
q
p
q
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q
2.3 Complete intersections We observed that the homological relationship between a local ring S and a residue class ring R = S=I is particularly strong if I is generated by an S -sequence. In this section we investigate such residue class rings of regular local rings. Slightly more generally we de ne: De nition 2.3.1. A Noetherian local ring R is a complete intersection (ring) if its completion R^ is a residue class ring of a regular local ring S with respect to an ideal generated by an S -sequence. Note that R^ is always a residue class ring of a regular local ring (see A.21). It follows immediately from 2.1.3, 2.1.8, and 2.2.6 that a complete intersection is Cohen{Macaulay. The nomenclature `complete intersection' comes from algebraic geometry. Suppose R is the coordinate ring of an ane variety over an algebraically closed eld k. Then R has the form R = S=I where S is a polynomial ring over k, and R is called a complete intersection if I is generated by the least possible number of elements, namely codim V = height I . Then V is the intersection of codim V hypersurfaces, and I is generated by an S -sequence. Let (S ) be a regular local ring, and (R ) a residue class ring, R = S=I . Suppose that I 6 2 . Then there exists x 2 I , x 2= 2, and we n
m
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n
74
2. Cohen{Macaulay rings
obtain a representation R = S 0 =I 0 with S 0 = S=(x), I 0 = I=(x). The ring S 0 is regular again, and I is generated by a regular sequence if and only if I 0 is: the element x is part of a minimal system of generators of I , and I can be generated by an S -sequence if and only if every minimal system of generators is an S -sequence see 1.6.19. Iterating this procedure we eventually obtain a minimal presentation R = S 00 =I 00 in which S 00 is regular and I 00 ( 00 )2 . It follows that ( ) = ( 00) = dim S 00 . For an arbitrary local ring (R ) the number ( ) is called the embedding dimension of R, emb dim R = ( ): (This terminology is again to be illustrated by the geometric analogue.) The discussion above shows that we may freely assume that I 2 when it is only to be veri ed whether I is generated by an S -sequence or otherwise. Nevertheless our de nition has two aws: rst, it does not use intrinsic characteristics of R second, it is not clear whether for an arbitrary presentation R^ = S=I with S regular local, the ideal I is generated by an S -sequence if R is a complete intersection. The intrinsic characteristics we are seeking are hidden in a Koszul complex. Let us x some standard notation which we shall use frequently throughout this section: (R k) is a Noetherian local ring, and x = x1 . . . xn is a minimal system of generators of . If present, (S k) is a regular local ring such that R = S=I with I 2 the ideal I is minimally generated by a = a1 . . . am , and y = y1 . . . yn is a regular system of parametersP such that xi is the residue class of yi . Furthermore we write ai = ajiyj with aji 2 (necessarily). Let ' : S m ! S n be given by the matrix (aji), and g : S n ! S and h : S m ! S be the linear forms de ned by y and a respectively. Then h = g ', and ^ ' : K (a) ! K (y ) is a complex homomorphism see 1.6.8. By 1.6.7, K (y ) R is just V K (x), and K (a) R has zero di erential forgetting it, we write R m for K (a) R . So we have a complex homomorphism ^ ^ ' R : R m ! K (x): V V Since R m has zero di erential, this yields a map R m ! H (x), and nally, as H (x) = 0, a map ^ : km ! H (x): V As is induced by ', it is a homomorphism of graded k-algebras. Sometimes it will be necessary to use the canonical bases f1 . . . fn of S n, e1 . . . em of S m , and the elements ui = '(ei ) 2 S n . By the choice of ', da(ei ) = ai = dy (ui ). Finally, denotes residue classes mod I . n
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2.3. Complete intersections
Theorem 2.3.2. With the notation just introduced, (a) 1 : km ! H1 (x) is an isomorphism of k-vector spaces, (b) (I ) = dim ; k H1(Rx),+ dim H (x). (c) 2 (k) = emb dim k 1 2 (Here 2 (k) is the second Betti number of k as an R -module see Section 1.3.) Proof. We constructed such that it maps the canonical basis of km to the homology classes of u1 . . . um. A minimal free resolution of k starts as x R n ;! R ;! 0 and therefore 2 (k) = (Z1(x)). So it is enough for (a), (b), and (c) to prove that dx (f p ^ f q ) 1 p < q n and u i i = 1 . . . m form a minimal system of generators of Z1 (x). Suppose that b 2 Z1(x), b 2 S n . Then dy (b) 2 I , dy (b) = c1 a1 + + cm am = c1dy (u1 ) + + cm dy (um): P Since K (y ) is acyclic, b ; ci ui is a linear combination of the elements dy (fp ^ fq ). That is, the P elements considered P generate Z1(x). Now assume that pq dx(f p ^ f q ) + i u i = 0. We have to show that all the coecients pq , i are in . Lifting the equation to S n gives .
X
^ X iui 2 IS n P iai 2 I . So i 2 by the choice of a. As and applying d yields P I 2 one obtains pq d (fp ^ fq ) 2 2 S n . Looking at the components of the elements d (fp ^ fq ) 2 S n and since y is a minimal system of generators of , one sees that pq 2 for all p, q. m
pq dy (fp fq ) +
y
n
n
y
n
n
y
n
n
The Koszul complexes with respect to di erent minimal systems of generators of are isomorphic R -algebras see the discussion before 1.6.21. In particular H (x) is essentially independent of x this justi es the notation H (R ) = H (x) and we call H (R ) the Koszul algebra of R . The number "1(R ) = dimk H1 (R ) is the rst deviation of R . It follows immediately from 2.2.5 and 1.6.19 that R is regular if and only if "1 (R ) = 0. So "1(R ) may be considered a measure of how far R deviates from regularity. The following theorem contains the desired intrinsic characterization of complete intersections. m
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2. Cohen{Macaulay rings
Theorem 2.3.3. Let (R k) be a Noetherian local ring. (a) One has "1(R ) = "1 (R^ ). (b) The following are equivalent: (i) R is a complete intersection (ii) "1(R ) = emb ; dim R ; dim R (iii) 2 (k) = 2(k) + 1(k) ; dim R. (c) Suppose that R = S=I with S regular and local. Then R is a complete m
1
intersection if and only if I is generated by an S-sequence.
(a) Choose a minimal system x of generators of . We write x^ for x considered as a sequence in R^ . Then H (R^ ) = H (^x) = H (x) R^ = ^ H (R ) R by 1.6.7. Since H (R ) has nite length, one has H (R ) R^ = H (R ). (b) Because of (a) and the de nition of complete intersection we may assume that R is complete and has a minimal presentation R = S=I . If R is a complete intersection, then there is such a presentation with I generated by an S -sequence, hence "1(R ) = (I ) = dim S ; dim R . Conversely, if "1(R ) = emb dim R ; dim R , then (I ) = dim S ; dim R in an arbitrary minimal presentation, and so I is generated by an S -sequence see 1.6.19. The equivalence of (ii) and (iii) follows immediately from 2.3.2. (c) is proved along the same lines as (b). Proof.
m
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Permanence properties of complete intersections. As we did for the Cohen{
Macaulay property and regularity we want to discuss how complete intersections behave under certain standard ring extensions. Theorem 2.3.4. Let (R k) be a Noetherian local ring. (a) Suppose x is an R-sequence. Then "1(R=(x)) ; (emb dim R=(x) ; dim R=(x)) = "1 (R ) ; (emb dim R ; dim R ) in particular R is a complete intersection if and only if R=(x) is a complete m
intersection. (b) Suppose R is a residue class ring of a regular local ring. Then if R is a complete intersection, so is R for every Spec R. p
p
2
Proof. Using induction we only need to prove (a) in the case in which x = x R . Suppose rst that x = 2 . Then emb dim R=(x) = emb dim R 1
2
2
;
and dim R=(x) = dim R ; 1 furthermore H1 (R ) = H1 (R=(x)) as k-vector spaces by 1.6.13(b). So "1(R ) = "1 (R=(x)). Now suppose that x 2 2 . Then emb dim R=(x) = emb dim R and dim R=(x) = dim R ; 1 moreover we have an exact sequence 0 ;! H1 (R ) ;! H1 (R=(x)) ;! H0 (R ) = k ;! 0 as in the proof of 1.6.16. Thus "1(R=(x)) = "1 (R ) + 1. The proof of (b) is very easy it uses 2.3.3 and basic properties of regular sequences. m
m
77
2.3. Complete intersections
Remark 2.3.5. Having studied an `abstract' characterization of complete intersections, the reader may expect an `abstract' version of 2.3.4(b) without any restrictions. In fact, such an assertion holds for arbitrary complete intersections as was proved by Avramov 22]. Actually Avramov proved a stronger result, namely the analogue of 2.1.7: suppose (R ) ! (S ) is a at homomorphism of Noetherian local rings then S is a complete intersection if and only if R and S= S are complete intersections. It is not dicult to deduce the localization property from the theorem on at extensions: there is (by faithful atness) a prime ideal R^ such that = \ R the extension R ! R^ is local and at, and R^ is a complete intersection by 2.3.4(b). In 23], Avramov gave quantitatively precise results concerning at extensions (and localizations): let (R ) = "1 (R ) ; (emb dim R ; dim R ) be the complete intersection defect of R then, in the situation of a at extension, (S ) = (R ) + (S= S ). (Limitation of space prevents us including a proof.) Using Avramov's theorem one can also remove the undesirable restrictions in 2.3.6 and 2.3.7 below. In the following we say that a Noetherian ring is a locally complete intersection if all its localizations are complete intersections. Theorem 2.3.6. Let R be a Noetherian ring which is a residue class ring m
n
m
q
p
q
p
q
q
m
of a regular ring S. Then R is a locally complete intersection if and only if R X1 . . . Xn ] is a locally complete intersection. The same holds for R and R X1 . . . Xn ]].
The proof follows the pattern of that of 2.1.9 one notes that
S X1 . . . Xn] and S X1 . . . Xn]] are regular rings by 2.2.13, and replaces
2.1.7 by Exercise 2.3.20. As with the Cohen{Macaulay property, the argument outlined really proves the stronger local version of 2.3.6: R is a complete intersection if and only if R X1 . . . Xn] is a complete intersection for 2 Spec R , 2 Spec RX1 . . . Xn] with R \ = . A similar remark applies to the following theorem and its proof. Theorem 2.3.7. Let k be a eld, R a Noetherian k-algebra, and K an extension eld of k. Suppose that R is a (ring of fractions of a) nitely p
p
q
q
q
p
generated k-algebra or K is nitely generated as an extension eld. Moreover, suppose that R is a residue class ring of a regular ring. Then R is a locally complete intersection if and only if R k K is a locally complete intersection.
Proof. We saw in 2.1.10 that R k K is a Noetherian ring. Given a prime ideal in R k K , we set = R \ conversely, by faithful atness, for every 2 Spec R there exists 2 Spec R K such that = R \ . Furthermore the extension R ! (R K ) factors through R K so q
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2. Cohen{Macaulay rings
we may replace R by R . By hypothesis, R = S=I with a regular local k-algebra S . Let R be a complete intersection. Then I is generated by a regular sequence g1 . . . gr . Because of faithful atness g1 . . . gr is also a regular sequence in S K . So it suces that S K is a locally complete intersection, and this is immediate from 2.1.11 (in conjunction with 2.2.13). As to the converse, we only do the more dicult case in which K is a nitely generated eld extension. By 2.1.11 again, one has K = (kX1 . . . Xn])T =(h1 . . . hm ) where h1 . . . hm is a regular sequence in (kX1 . . . Xn])T and T is a multiplicatively closed set. Therefore h1 . . . hm is a regular sequence in the faithfully at extension R (kX1 . . . Xn])T = (R X1 . . . Xn])T 0 here T 0 is the image of the natural map kX1 . . . Xn] ! R X1 . . . Xn]. Moreover, (R K ) has the form R X1 . . . Xn] =(h1 . . . hm) with 2 Spec R X1 . . . Xn] such that \ R = and T 0 \ = . By 2.3.4, R X1 . . . Xn ] is a complete intersection. So we can apply the local version of 2.3.6. p
Q
q
Q
Q
Q
p
Q
Q
The Koszul algebra of a complete V intersection. Above we constructed an algebra homomorphism : km H. (R ), m = "1 (R ), starting from a minimal presentation R = S=I , and we saw that 1 : km H1 (R ) is an isomorphism. Such a homomorphism is always present. In fact, H. (R ) is an alternating graded k-algebra therefore, by the universal property
!
!
of the a unique algebra homomorphism V exterior algebra, there exists 0 : H1 (R ) ! H (R ) extending the identity on H1 (R ). Moreover, algeV bra homomorphisms 0 : H1 (R ) ! H (R ) such that 1 andV 01 are V isomorphisms, only di er by the automorphism (01;1 1) of H1 (R ). So we may replace the `abstract' homomorphism 0 by the `concrete' whenever we have a minimal presentation. The situation under consideration can be generalized as follows: S is a ring, I and are ideals generated by sequences a = a1 . . . am and y = y1 . . . yn, and we have I . Then, as above 2.3.2, there is a homomorphism .
.
n
n
: H. (a S= ) =
^
n
(S= )m ! H (y S=I ): .
n
Choose S -free resolutions F of S=I and G of S= . Then there exist complex homomorphisms K (a) ! F and K (y ) ! G . These in turn induce maps : H (a S= ) ;! H (F S= ) = TorS (S=I S= ) : H (y S=I ) ;! H (S=I G ) = TorS (S=I S= ): .
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2.3. Complete intersections
79
V
Hence there exist two maps from H (a S= ) = (S= )m to TorS (S=I S= ), namely and . It is crucial that these maps are essentially equal { of course, we must use the proper identi cation of H (F S= ) and H (S=I G ). To this end one forms the double complex F G and considers S=I and S= as complexes concentrated in degree 0. Then one has complex homomorphisms K (a) ! F ! S=I and K (y ) ! G ! S= . Taking tensor products yields a commutative diagram .
n
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n
? S= ;;;; K (a) K (y ) ;;;;! S=I K (y ) ? ?
K. (a)
?y
F.
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n
S= ;;;;
?y
F.
n
G
.
?y
;;;;!
.
S=I
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G
.
By a fundamental theorem of homological algebra (318], Theorem 11.21) the bottom row induces an isomorphism H. (F.
= = S= ) ; H (F G ) ;! H (S=I G ): .
n
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.
It is this identi cation we need: Lemma 2.3.8. With the notation introduced, s = (;1)s s s . Proof. Let e1 . . . em be a basis of S m and choose elements ui 2 S n with dy (ui) = da (ei ). One has s ( ei ^^ eis ) = u i ^^ u is . Thus it is enough to show that s ( ei ^ ^ eis ) = (;1)s s ( ui ^ ^ u is ). Let z 2 K (a) K (y ) be a cycle. Then the commutativity of the diagram above implies that z , (z ), and (z ) are all mapped to the same homology class. So it suces to exhibit a cycle z with (z ) = ei ^^ eis and (z ) = (;1)s ( ui ^ ^ u is ). Simply take z = (ei 1 ; 1 ui ) (eis 1 ; 1 uis ). In order to see that it is a cycle, one uses the de nition of the di erential of K (a) K (y ) and the fact that a product of cycles is again a cycle. Theorem 2.3.9. Let S be a ring, and a = a1 . . . am and y = y1 . . . yn be Ssequences such that I = (a) = (y ). Then H (y S=I ) = TorS (S=I S= ) V m is (isomorphic with) the exterior algebra (S= ) . Proof. The isomorphism H (y S=I ) = TorS (S=I S= ) results from the fact that K (y ) is a free resolution of S= see 1.6.14. So, with the notation above, is an isomorphism. Similarly is an isomorphism hence , being an algebra homomorphism, is an isomorphism of graded algebras. V (In order to remove the sign in 2.3.8 one would have to replace by (;1).) Corollary 2.3.10. P (a) With the hypotheses of 2.3.9 suppose that m = n. Write ai = ajiyj , i = 1 . . . n. Then I : = I + S = det(aji): (b) In particular, suppose that y is a regular system of parameters in a regular local ring S. Then Soc(S=I ) = (S=I ). 1
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(a) That (I : )=I = HomS (S= S=I ) and Hn (y S=I ) can be identi ed was shown in the proof of 1.6.16. P Now let f1 . . . fn be a basis of S n with dy (fi ) = yi, and set ui = nj=1 aji fj . Then if e1 . . . en is a basis of S n with da(ei ) = ai , one has dy (ui) = ai = da (ei). The theorem implies that Hn (y S=I ) is generated by u 1 ^ ^ un = f 1 ^ ^ f n. So u 1 ^ ^ u n is mapped to the residue class of in S=I by the homomorphism which sends f1 ^^ fn to 1 (and thus gives the identi cation (I : )=I = Hn (y S=I )). (b) By de nition, Soc(S=I ) = (I : )=I . We have completed our preparations for the following beautiful characterization of complete intersections: Theorem 2.3.11 (Tate, Assmus). Let (R k) be a Noetherian local ring. Proof.
n
n
n
n
m
Then the following are equivalent: (a) R is a complete intersection (b) H. (R ) is (isomorphic to) the exterior algebra of H1 (R ) (c) H. (R ) is generated by H1 (R ) (d) H2 (R ) = H1 (R )2.
Here H1 (R )2 is the k-vector space generated by the products w ^ z with w, z 2 H1 (R ). Proof. It was observed in the proof of 2.3.3 that H (R ) is invariant under completion. So we may assume that R is complete and has a minimal presentation R = S=I . The implication (a) ) (b) is a special case of 2.3.9, and (b) ) (c) ) (d) is trivial. For (d) ) (a) we note rst that the map above is an isomorphism. Next, (d) says that 2 : K2 (a) S= ! H2 (y S=I ) = TorS2 (R S= ) is surjective. So 2 is surjective. Choose F as a minimal free resolution of S=I . Then we have a commutative diagram .
n
n
.
V2 S m ;;;;! S m ;;;;! S ;;;;! 0 ?? ?? y y=
;;;;! F1 ;;;;! S ;;;;! 0 The map 2 is just k, and k being surjective, is surjective itself. F2
It follows immediately that H1 (a) = 0, whence a is an S -sequence by 1.6.19. Theorem 2.3.3 contains a characterization of complete intersections in terms of the numerical invariants dim R , emb dim R = 1(k), and 2(k). It is possible to remove the `non-homological' Krull dimension, and to give a description of complete intersections using 1 (k), 2 (k), and 3(k).
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In order to construct the rst steps in a free resolution of k, we start with the Koszul complex
^2
Rn
;! Rn ;! R ;! 0
unless R is regular, H1 (R ) is non-zero. So we add a free direct summand R m with m = dimk H1 (R ) = "1 (R ), and send its generators e1 . . . em to cycles u1 . . . um whose homology classes generate H1 (R ):
^2
d d ;! R n ;! R ;! 0: The kernel of d2 contains the Koszul cycles d (fi ^ fj ^ fl ) as well as the elements xi ep ; fi ^ up again, f1 . . . fn denotes a basis of R n. In order to
Rm
Rn
2
1
x
`kill' at least these cycles we form the complex T. : (R
with
n
R ) m
^3
R
;! R
n d3
^ ^
m
^2
Rn
d d ;! R n ;! R ;! 0 2
^ ^ ; ^
d3 (fi fj fl ) = dx (fi fj fl ) d3(fi ep ) = xi ep fi up = dx (fi )ep
1
; fi ^ d2(ep):
Part (a) of the following theorem shows that (H2 (T )) is an invariant of R . One writes "2 (R ) = (H2 (T )) and calls this number the second deviation of R . Theorem 2.3.12. Let (R k) be a Noetherian local ring. Then (a) H2 (T ) = H2 (R )=H1(R )2, (b) R is a complete intersection if and only if "2 (R ) = 0, ; ;
(c) "2 (R ) = 3(k) ; 3(k) ; 1 (k) 2(k) ; 2(k) . .
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m
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1
Proof.
1
(a) Let K be the complex .
K. :
^3
R
n
;!
^2
Rn
;! Rn ;! R ;! 0
obtained by truncating the Koszul complex K is a subcomplex of T . The quotient T =K is isomorphic to .
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L. : R n
d id Rm ;;;! R m ;! 0 1
with R n R m in degree 3. Consider the exact sequence of homology H3 (L ) ! H2 (K ) ! H2 (T ) ! H2 (L ) ! H1 (K ) ! H1 (T ) = 0: .
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2. Cohen{Macaulay rings
The map H2 (L ) ! H1 (K ) is an isomorphism since both vector spaces have dimension m. Hence we have an exact sequence H3 (L ) ! H2 (K ) ! H2 (T ) ! 0. The kernel of R n R m ! R m is obviously generated by the elements dx (fi ^ fj ) eq and up eq . An analysis of the connecting homomorphism shows that the class of dx (fi ^ fj ) eq goes to that of dx(fi ^ fj ) ^ uq which is a boundary in the Koszul complex (see the formulas above 1.6.4) the class of up eq goes to that of up ^ uq whence the image of H3 (L ) ! H2 (K ) is H1 (K )2. (b) follows immediately from (a) and 2.3.11. (c) The veri cation is similar to that of 2.3.2(c), and therefore left to the reader. Remark 2.3.13. The equation d3(fi ep) = dx (fi )ep ; fi ^ d2(ep ) suggests that d3 is a component of an antiderivation of an alternating algebra. In fact, the choice of d3 is part of Tate's construction of resolutions with algebra structures 369]. Suppose that A is an alternating graded R -algebra equipped with an antiderivation @ of degree ;1 such that @2 = 0, and consider the homology H (A ) = Ker @= Im @. Let z 2 Hp (A ) be a non-zero homology element. Then one may adjoin a variable to `kill' the cycle z representing z : (i) If p is even, let B be the exterior algebra in a variable of degree p + 1, i.e. B = R Re with R in degree 0 and Re = R in degree p + 1, the multiplication being de ned by e2 = 0. (ii) For p odd let B be theL`divided power algebra' over R in a variable of degree p + 1, i.e. B = 1j =0 Rej with Rej = R in degree j (p + 1), the multiplication being de ned by ej el = ((j + l )!=j ! l !)ej +l. In both cases A B is again an alternating algebra, and there is a unique antiderivation d on A B such that d jA 1 = @, d (e) = z in case (i), and d (ej ) = zej ;1 for all j in case (ii) moreover, one has d 2 = 0. It follows easily that Hq (A B ) = Hq (A ) for q < p and Hp (A B ) = Hp (A )=R z. In order to resolve the residue class eld k of a local ring (R k) one starts with the R -algebra T (0) = R . Let "0(R ) = emb dim R = ( ), and successively adjoin "0 (R ) variables of degree 1 `to kill the zero-cycles'. The resulting algebra T (1) is the Koszul complex of a minimal system of generators of . Next one adjoins "1 (R ) variables of degree 2 to kill "1(R ) cycles generating H1 (T (1)). The algebra T (2) thus constructed has H1 (T (2)) = 0. It is a theorem of Tate 369] that in the case of a complete intersection the complex T (2) is a minimal free resolution of k. However, if R is not a complete intersection, then one has to adjoin "2 (R ) variables of degree 3 etc. A famous theorem of Gulliksen 146] and Schoeller 330] says that the resolution of k obtained in this way is always minimal. For a comprehensive study of resolutions with an algebra structure we refer the reader to Gulliksen and Levin 147]. .
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2.3. Complete intersections
As above, let (R k) be a Noetherian local ring. V We saw in 2.3.11 that surjectivity of the natural homomorphism : H1 (R ) ! H (R ) is already sucient for R to be a complete intersection. This is also true for injectivity, at least when R contains a eld. Theorem 2.3.14. Let (R k) be a Noetherian local ring containing a eld. m
.
m
Then (a) H1 (R )j = 0 for j > emb dim R dim R (b) in particular, R is a complete intersection if (and only if) the natural V map : H1 (R ) H. (R ) is injective.
;
!
Proof. It is harmless to complete R so that we may assume that R has a minimal presentation R = S=I as above. In order to prove (a) we must anticipate Corollary 9.5.3: it says that, with the notation of 2.3.8, j = 0 for j > emb dim R ; dim R . Since, in the present circumstances, j is an isomorphism, one has j = 0 for j > emb dim R ; dim R . As V H1 (R )j = j ( j H1 (R )), one has H1 (R )j = 0. This proves (a), and (b) is an obvious consequence of (a).
The restriction to local rings containing a eld is forced upon us since there does not yet exist a proof of 9.5.3 without this restriction. However, one always has H1 (R )j = 0 for j > emb dim R ; dim R + 1 so that the gap in 2.3.14 is as small as it could be see 9.5.7. The reader may have noticed that 2.3.14 is trivial for Cohen{Macaulay rings R : if j > emb dim R ; depth R , then even Hj (R ) = 0 by 1.6.17. On the other hand, for Cohen{Macaulay R the non-vanishing of H1 (R )p for p = emb dim R ; dim R conveys the strongest possible information: R is a complete intersection. More generally, we have the following theorem. Theorem 2.3.15. Let (R k) be a Noetherian local ring. Then R is a complete intersection if (and only if) H1 (R )p 6= 0 for p = emb dim R ; depth R. Proof. By virtue of 2.3.14 the hypothesis H1 (R )p 6= 0 for p = emb dim R ; depth R forces R to be Cohen{Macaulay if it contains a eld. Because of this restriction we give a proof not using 2.3.14. First we reduce to the case depth R = 0. So suppose that depth R > 0. Then there exists an x 2 n 2 which is not a zero-divisor, and 1.6.13 furnishes us with an isomorphism : H (R ) = H (R 0 ), R 0 = R=(x). It is not dicult to verify that is a k -algebra isomorphism after all, is induced V by , being the composition R n ! R n;1 ! (R 0)n;1 (see the proof of 1.6.12). Furthermore emb dim R 0 ; depth R 0 = emb dim R ; depth R , and R is a complete intersection if and only if this holds for R 0 . It remains to show that R is a zero dimensional complete intersection if H1 (R )n 6= 0 for n = emb dim R . The complex K (R ) has length n so m
m
m
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2. Cohen{Macaulay rings
Bn+1(R ) = 0, and therefore
H1 (R )n = (Z1 (R )n + Bn+1(R )) Bn+1(R ) = Z1 (R )n:
V
Consider an exact sequence R m 2 R n ;! R nP;! ;! 0 as above. Choose elements v1 . . . vn 2 Im d2 = Z1(R ), vi = vji fj where f1 . . . fn is a basis of R n . Then v1 ^^ vn = det(vji )f1 ^^ fn, whence H1 (R )n 6= 0 is equivalent to In (d2) 6= 0. It remains to apply the next theorem. Theorem 2.3.16 (Wiebe). Let (R k) be a Noetherian local ring, and ' R r ;! R n ! ! 0 a presentation of its maximal ideal. If In (') 6= 0, then R is a complete intersection of dimension zero (and conversely). Proof. The ideal In(') is the zeroth Fitting ideal of , which is an invariant of . Therefore it is enough to consider a special presentation. Moreover, we may assume that R is complete. Then R = S=I where (S ) is a regular local ring and I 2 . Let y = y1 . . . yn be a regular system of parametersPof S , and a = a1 . . . am a minimal system of generators of I . Write ai = ajiyj . The converse of the theorem is part of Corollary 2.3.10 it implies the following claim which is crucial in what follows. Let b = b1 . . . bn be a maximal S-sequence, and J 0 an ideal properly containing P J = (b) then det(bji ) 2 J 0 where the bji are chosen such that bi = bjiyj . In fact, det(bji )S=J is the socle of S=J . Since it has dimension 1 over k, it is contained in every non-zero ideal of S=J . Let f1V . . . fn be a basis of S n , and e1V . . . em a basis of S m . De ne 2 m n n : S SP ! S by the Koszul map 2 S n ! S n with respect to y and (ei ) = ajifj . We saw in 2.3.2 that Coker S=I = . So the theorem claims that In() I unless S=I is a zero dimensional complete intersection. Choose an n n submatrix U of (a matrix of) . If U involves a column corresponding to one of the elements fi ^ fj , then det U 2 I , since, on the level of R , we are taking the exterior product of n cycles at least one of which is a boundary: (det U )f 1 ^ ^ f n 2 Bn+1(R ) = 0. Therefore it is enough to consider submatrices U of (aji). For simplicity of notation one may assume U consists of the rst n columns. If a1 . . . an is a regular sequence, then I contains J = (a1 . . . an) properly since S=I is not a complete intersection. So det U I by the claim above. If a1 . . . an is not a regular sequence, then dim S=J > 0, and it is certainly enough to show that det U 2 J , that is, we may assume that I = (a1 . . . an ). We will show that det U 2 I + p for all p 2 N. Then det U 2 I follows from Krull's intersection theorem. Fix p 2 N. According to Exercise 2.3.17 one nds elements a00i 2 p+1 such that a0i = ai + a00i , i = 1 . . . n, is a d2
m
m
m
m
m
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m
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n
Notes
85
P
regular sequence. Write a0i = a0jiyj then aji ; a0ji 2 p follows from the quasi-regularity of the regular sequence y , in other words, from the fact that the associated graded ring gr (S ) is a polynomial ring (see 1.1.8). Therefore det U ; det(a0ji) 2 p : n
n
n
The ideal I + p properly contains (a01 . . . a0n) I + p+1. Once more the auxiliary claim above is applied, and it yields det(a0ji) 2 I + p . n
n
n
Exercises
Let (R ) be a Noetherian local ring of depth t, and a1 . . . at 2 . Then, given p 2 N, show there exist a1 . . . at 2 p such that a1 + a1 . . . at + at is an R -sequence. 2.3.18. Let S be a regular local ring of dimension 4, and y1 . . . y4 a regular system of parameters. Let I = (y1 y2 y3 y4 y1 y3 + y2 y4 ) and R = S=I . (a) Construct a minimal free resolution of R . (b) Prove depth R = 0 and dim R = 2. (c) Show that the vector space H1 (R )2 has dimension 3, the maximal value for an ideal generated by 3 elements, but R is not a complete intersection. 2.3.19. Prove all the claims in the second paragraph of 2.3.13. 2.3.20. Let ' : (S1 1 ) ! (S2 2 ) be a at local homomorphism of regular rings, and I S1 an ideal. Verify S1 =I is a complete intersection if and only if S2 =IS2 is a complete intersection. 2.3.21. Let R be a Noetherian graded ring. Show: (a) For 2 Spec R the localization R is a complete intersection if and only if R is. (b) The following are equivalent: (i) R is locally a complete intersection (ii) R is a complete intersection for all graded prime ideals (iii) R( ) is locally a complete intersection for all graded prime ideals . (c) Suppose in addition that (R ) is local. Then R is locally a complete intersection if and only if R is a complete intersection. Hint: 1.6.33. 2.3.22. Extend 2.3.2 and 2.3.12 to the following theorem which Serre 332] used to prove 2.2.7(c) ) (a): let (R k) be a Noetherian local ring, and x a minimal system of generators of then the natural map K. (x k) ! TorR. (k k) (see 1.6.9) is injective. 2.3.17.
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Notes The origins of the theory of Cohen{Macaulay rings are the unmixedness theorems of Macaulay 263] and Cohen 71] and the notion of perfect ideals, which also goes back to Macaulay and was clari ed by
86
2. Cohen{Macaulay rings
Grobner 141]. The present shape of the theory was formed by Auslander and Buchsbaum 18], Nagata 283], and Rees 303]. It seems that Cohen{Macaulay modules made their rst appearance in Auslander and Buchsbaum 19]. The characterization 2.1.27 of graded Cohen{Macaulay rings is essentially due to Hochster and Ratli 200] and Matijevic and Roberts 268]. By analogy to the desingularization, one can try to `Macaulayfy' a Noetherian scheme. Special results in this direction were obtained by Brodmann 51] and Faltings 100]. Recently Kawasaki 234] has proved a general theorem on the existence of `Macaulay cations'. Of all the notions generalizing Cohen{Macaulay rings and modules, the concept of Buchsbaum ring or module is the most important see Stuckrad and Vogel 365] and Schenzel 329]. The `classical' theory of regular local rings, to be found in Zariski and Samuel 397], Vol. II, was developed by Krull 242], Chevalley 68], Cohen 71], and Zariski 396]. It depends in an essential way on power series methods, and is therefore mainly restricted to local rings containing a eld. The problems it could not solve were (i) the regularity of a localization of a regular local ring R (even if R contains a eld), and (ii) the factoriality of such rings (because of the Cohen structure theorem this is easy if R contains a eld). The breakthrough was the theorem 2.2.7 of Auslander and Buchsbaum 17], 18] and Serre 332] which not only solved the localization problem: `this resounding triumph of the new homological method marked a turning point of the subject of commutative Noetherian rings' (Kaplansky 230], p. 159). Theorem 2.2.8 was independently given by Ferrand 105] and Vasconcelos 379] it generalizes Kaplansky's proof of 2.2.14(c) ) (a) (see 270], x19). The problem of factoriality was solved by Auslander and Buchsbaum 20], using results of Zariski and Nagata who reduced the theorem to the case of Krull dimension 3. See Nagata 284], p. 217 for a minute history. The proof we have reproduced is due to Kaplansky (except for the application of the Hilbert{Burch theorem). That regular local rings are factorial can be expressed by saying that every ideal I has a greatest common divisor: there is a regular element a and an ideal J of grade 2 such that I = aJ . MacRae 265] proved this fact for every ideal with a nite free resolution. Another (related) generalization is that every module with a nite free resolution over a normal domain has divisor class zero see 47], Ch. VII, x4. The most concrete and computationally e ective result is the factorization theorem of Buchsbaum and Eisenbud 64]. The notion of complete intersection is classical in algebraic geometry. An abstract de nition in terms of local algebra was given by Scheja
87
Notes
322], together with 2.3.3. Our de nition is that of Grothendieck 142]. Avramov's contributions 22], 23] have been described in 2.3.5 they ultimately justi ed the abstract notion of complete intersection. The program of Tate's seminal paper 369] has been outlined in Remark 2.3.13. Assmus 14] used Tate's method to give the description 2.3.11 of complete intersections in terms of their Koszul algebras. There are several papers devoted to the characterization of complete intersections by the vanishing of a deviation "i (which we de ned only for i = 1 2) the question was nally settled by Halperin 148] who showed that "i > 0 for all i if R is not a complete intersection. Wiebe's theorem 2.3.16 appeared in 395] see Kunz 244], Hilfssatz 1, for a related result. A driving force in this area of researchP was the problem (posed by Serre 334]) of whether the Poincare series i (k)ti of a Noetherian local ring (R k) is a rational function of t. After several special cases had been solved positively, the general question was answered negatively by Anick 9]. In several theorems we studied the behaviour of ring-theoretic properties under at extensions R ! S . Avramov, Foxby, and Halperin 31] investigated the more general situation in which S is supposed of nite at dimension over R . As we have seen, another `homologically nice' type of extension is that of passing to a residue class modulo a regular sequence. Avramov and Foxby have essentially completed a program which aims at the uni cation of these types of extensions by introducing a suitable notion of bre see 27], 28], 29], 30]. m
3 The canonical module. Gorenstein rings
The concept of a canonical module is of fundamental importance in the study of Cohen{Macaulay local rings. The purpose of this chapter is to introduce the canonical module and derive its basic properties. By de nition it is a maximal Cohen{Macaulay module of type 1 and of nite injective dimension. In the rst two sections we investigate the injective dimension of a module, and prove Matlis duality which plays a central role in Grothendieck's local duality theorem. Actually the canonical module has its origin in this theory. Here the canonical module is introduced independently of local cohomology which is an important notion in itself and will be treated later in this chapter. A ring which is its own canonical module is called a Gorenstein ring. Next to regular rings and complete intersections, Gorenstein rings are in many ways the `nicest' rings. Distinguished by the fact that they are of nite injective dimension, they have various symmetry properties, as reected in their free resolution, their Koszul homology, and their Hilbert function. The last aspect will be discussed in the next chapter. Gorenstein rings of embedding dimension at most two are complete intersections. The rst non-trivial Gorenstein rings occur in embedding dimension three, and they are classi ed by the Buchsbaum{Eisenbud structure theorem. In the nal section the canonical module of a graded ring is introduced. 3.1 Finite modules of nite injective dimension In this section we study injective resolutions of nite modules. We shall see that the injective dimension of a nite module M over a Noetherian local ring R either is in nite or equals the depth of R , and is bounded below by the dimension of M . Thus, quite contrary to the behaviour of projective dimension, the injective dimension, if it is nite, does not depend on the module. We introduce Gorenstein rings and show that Gorenstein rings are Cohen{Macaulay rings. De nition 3.1.1. Let R be a ring. An R -module I is injective if the functor HomR ( I ) is exact. 88
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3.1. Finite modules of nite injective dimension
Notice that HomR ( I ) is always left exact. Thus the R -module I is injective if and only if HomR ( I ) is right exact as well. We now list some useful characterizations of injective modules. Proposition 3.1.2. Let R be a ring and I an R-module. The following
conditions are equivalent: (a) I is injective (b) given a monomorphism ' : N M of R-modules, and a homomorphism : N I, there exists a homomorphism : M I such that = ' (c) given R-modules N M, and a homomorphism : N I, there exists a homomorphism : M I such that N = in other words, : N I can be extended to a homomorphism : M I (d) for all ideals J R, every homomorphism J I can be extended to R, that is, Ext1R (R=J I ) = 0 (e) let M be an R-module with I M then I is a direct summand of M (f) Ext1R (M I ) = 0 for all R-modules M (g) ExtiR (M I ) = 0 for all R-modules M and all i > 0.
!
!
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j
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The ExtiR ( I ) are the right derived functors of HomR ( I ). The equivalence of (a), (f) and (g) follows therefore from the general properties of right derived functors. For details we refer to 318], Section 6. (a) ) (b): The monomorphism ' : N ! M induces the homomorphism Hom(' I ): HomR (M I ) ;! HomR (N I ) where Hom(' I )( ) = ' for all 2 HomR (M I ). By assumption, Hom(' I ) is an epimorphism, and so 2 HomR (N I ) is of the form ' for some 2 HomR (M I ). Similarly one proves (b) ) (a). The implications (b) () (c) and (c) ) (d) are clear. (d) ) (c): We consider the set of all pairs (U '), where U is a submodule of M with N U , and where ' extends . We order this set partially: (U1 '1) (U2 '2) if and only if U1 U2 and '1 = '2jU . By Zorn's lemma there exists a maximal element (U 0 '0) in this set. Suppose U 0 6= M then we may choose x 2 M n U 0 . Set W = U 0 + Rx then W =U 0 = R=J for some ideal J in R . Applying the functor HomR ( I ) to the exact sequence 0 ;! U 0 ;! W ;! R=J ;! 0 we obtain the exact sequence HomR (W I ) ;! HomR (U 0 I ) ;! Ext1R (R=J I ): Since by assumption Ext1R (R=J I ) = 0, it follows from the exact sequence that any homomorphism from U 0 to I can be extended to a homomorphism W ! I , contradicting the maximality of (U 0 '0 ). Thus we have shown that U 0 = M . Proof.
1
90
3. The canonical module. Gorenstein rings
(c) ) (e): There exists a homomorphism : M ! I with jI = idI . Therefore, M = I Ker : (e) ) (b): Given a monomorphism ' : N ! M and a homomorphism : N ! I , we want to nd : M ! I such that = '. In order to do this, we construct a commutative diagram ' ;;;;! M ? ?
N
?y
I
?y
;;;;! W
where is injective. In fact, we may choose W = (M I )=C with C = f('(x) ; (x)): x 2 N g and are the natural homomorphisms arising from this situation. (This diagram is called the pushout of and '.) Since is injective by construction, it is split injective by our assumption (e). This means that there exists a homomorphism : W ! I with = idI . The homomorphism : M ! I , = , is the desired extension of . Corollary 3.1.3. Let R be a Noetherian ring. (a) If I is an injective R-module and S is a multiplicatively closed set of R, then IS is an injective RS -module. (b) If (I ) 2 is a family of injective R-modules, then the direct sum I=
M 2
I
is an injective R-module. Proof. (a) Let J be an ideal of R . Since R is Noetherian one has Ext1RS (RS =JRS IS ) = Ext1R (R=J I )S = 0: Since every ideal of RS is extended from R , 3.1.2 yields that IS is an injective RS -module. (b) By 3.1.2 it is L enough to show that for an ideal J of R , any homomorphism ' : J 2 I extends to R . Since J is nitely generated Ln I . there exists a nite subset 1 . . . n of such that Im ' i=1 i We denote by 'j the j -th component of '. Since I i is injective we can extend 'L I i to a homomorphism I i . It is clear that i : J Pn i(a), ai : RR, extends : R ' to all of R . 2 I with (a) = i=1
!
!
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g
2
!
Remark 3.1.4. It is a simple exercise to see that for an arbitrary ring R any direct product of injective modules is injective. It is however essential to require that R is Noetherian (as we have done in 3.1.3) to obtain a similar result for direct sums. In fact, this property characterizes Noetherian rings see 318], Theorem 4.10.
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3.1. Finite modules of nite injective dimension
An R -module M is divisible if for every regular element r 2 R , and every element m 2 M , there exists an element m0 2 M such that m = rm0 . Condition 3.1.2(d) has the following consequence. Corollary 3.1.5. Let R be a ring and I an R-module. (a) If I is injective, then I is divisible. (b) If R is a principal domain and I is divisible, then I is injective. Proof. The property that I is divisible is equivalent to the property that every homomorphism : (r) ! I , r regular, can be extended to R . Therefore (a) and (b) follow from 3.1.2(d). For later applications we note the following result about change of rings. Lemma 3.1.6. Let ' : R ! S be a ring homomorphism, and let I be an injective R-module. Then HomR (S I ) (equipped with the natural S-module structure) is an injective S-module. Proof. Let M be an S -module. There is a natural isomorphism HomS (M HomR (S I )) = HomR (M I ) of S -modules. Indeed, to 2 HomS (M HomR (S I )) one assigns 0 2 HomR (M I ) where 0 (x) = (x)(1) for all x 2 M . Thus the exactness of the functor HomR ( I ) on the category of S -modules (considered as R modules via ') implies the exactness of the functor HomS ( HomR (S I )). This means that HomR (S I ) is an injective S -module. De nition 3.1.7. Let R be a ring and M an R -module. A complex I : 0 ;! I 0 ;! I 1 ;! I 2 ;! with injective modules I i is an injective resolution of M if H 0 (I ) =M and H i (I ) = 0 for i > 0. While it is obvious that every module has a projective resolution, it is less obvious that it has an injective resolution. It is however clear that an injective resolution can be constructed by resorting to the following result. Theorem 3.1.8. Let R be a ring. Every R-module can be embedded into an .
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injective R-module.
The Z-module Q is divisible, and hence injective. Therefore any free Z-module F can be embedded into an injective Z-module I . We just take suciently many copies of Q. If G is an arbitrary Z-module, then G = F=U , and we can embed G into I=U . It is immediate that I=U is again divisible, and hence injective. Thus the theorem is proved for Z-modules. Proof.
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Now let R be an arbitrary ring, and let M be an R -module. The map : M HomZ (R M ) with (x)(a) = ax for all x M and all a R is an R -module monomorphism. By our considerations above, the R -module M can be embedded as a Z-module into an injective Z-module I . This
!
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inclusion induces a monomorphism : HomZ (R M ) ;! HomZ (R I ): By 3.1.6, the R -module J = HomZ (R I ) is injective, and thus : M ! J is the desired embedding.
Injective dimension. Let R be a ring and M an R -module. The injective dimension of M (denoted inj dim M or inj dimR M ) is the smallest integer n for which there exists an injective resolution I . of M with I m = 0 for m > n. If there is no such n, the injective dimension of M is in nite.
The following observation is an immediate consequence of 3.1.3 and the exactness of localization. Proposition 3.1.9. Let R be a Noetherian ring, M an R-module and S a multiplicatively closed set. Then inj dimRS MS inj dimR M. In the next proposition we characterize the injective dimension of a module homologically. Proposition 3.1.10. Let R be a ring and M an R-module. The following conditions are equivalent: (a) inj dim M n (b) ExtnR+1(N M ) = 0 for all R-modules N (c) ExtRn+1(R=J M ) = 0 for all ideals J of R.
Proof. (a)
) (b) follows from the fact that ExtnR+1(N M) can be computed
from an injective resolution of M . (b) ) (c) is trivial. (c) ) (a): Let 0 ;! M ;! I 0 ;! I 1 ;! ;! I n;1 ;! C ;! 0 be an exact sequence, where the modules I j are injective. From the fact that ExtiR (R=J I ) = 0 for i > 0 if I is an injective R -module, the above exact sequence yields the isomorphism Ext1R (R=J C ) = ExtnR+1(R=J M ) and so Ext1R (R=J C ) = 0 for all ideals J of R . This is condition (d) of 3.1.2, and so C is injective. Proposition 3.1.10 can be sharpened if R is Noetherian. We rst observe:
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3.1. Finite modules of nite injective dimension
Lemma 3.1.11. Let R be a Noetherian ring, M an R-module, N a nite R-module and n > 0 an integer. Suppose that ExtnR (R= M ) = 0 for all 2 Supp N. Then ExtnR (N M) = 0. Proof. N has a nite ltration whose factors are isomorphic to R= for certain 2 Supp N . Hence the lemma follows from the `additivity' of the vanishing of ExtnR ( M ). Corollary 3.1.12. Let R be Noetherian and M an R-module. The following p
p
p
p
conditions are equivalent: (a) inj dim M n (b) ExtnR+1(R= M ) = 0 for all
p
p
2 Spec R.
Lemma 3.1.11 has another remarkable consequence. Proposition 3.1.13. Let (R k) be a Noetherian local ring, a prime ideal dierent from , and M a nite R-module. If ExtnR+1(R= M ) = 0 for all prime ideals 2 V ( ), 6= , then ExtnR (R= M ) = 0. Proof. We choose an element x 2 n . The element is R= -regular, and therefore we get the exact sequence m
p
m
q
q
p
q
p
p
m
p
0 ;! R= ;! R= which induces the exact sequence x
p
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p
;! R=(x ) ;! 0 p
ExtnR (R= M ) ;! ExtnR (R= M ) ;! ExtnR+1(R=(x ) M ): Since V (x ) f 2 V ( ): 6= g, Lemma 3.1.11 and our assumption imply ExtnR+1(R=(x ) M ) = 0 so that multiplication by x on the nite R -module ExtnR (R= M ) is a surjective homomorphism. The desired result follows from Nakayama's lemma. It is now easy to derive the following useful formula for the injective dimension of a nite module. Proposition 3.1.14. Let (R k) be a Noetherian local ring, and M a nite x
p
p
p
q
p
q
p
p
p
p
R-module. Then
m
inj dim M = supfi : ExtiR (k M ) 6= 0g:
We set t = supfi : ExtiR (k M ) 6= 0g. It is clear that inj dim M t. To prove the converse inequality, note that the repeated application of 3.1.13 yields ExtiR (R= M ) = 0 for all 2 Spec R and all i > t. According to 3.1.12 this implies inj dim M t.
Proof.
p
p
94
3. The canonical module. Gorenstein rings
Corollary 3.1.15. Let (R k) be a Noetherian local ring and M a nite
R-module. If x
2
m
m
is an element which is R- and M-regular, then
inj dimR=(x) M=xM = inj dimR M ; 1:
The proof is an immediate consequence of 3.1.14 and the following result of Rees 302], Theorem 2.1. Lemma 3.1.16. Let R be a ring, and let M and N be R-modules. If x is an R- and M-regular element with x N = 0, then for all i
0.
ExtiR+1(N M ) = ExtiR=(x)(N M=xM )
Proof. We set R i +1 ExtR ( M ), i
= M=xM , and show that the functors = R=(x) and M 0, from the category of R -modules into itself are the ). To see this, we have to verify right derived functors of HomR ( M i +1 (1) the functors ExtR ( M ), i 0, are strongly connected, ) are equivalent, (2) the functors Ext1R ( M ) and HomR ( M i +1 (3) ExtR (F M ) = 0 for all i > 0 and every free R -module F . (An axiomatic description of the Ext groups as functors in the second variable is given in 318], Theorem 7.22. Similarly the Ext groups can be described axiomatically as functors in the rst variable see 318], Exercise 7.27.) x ;! 0 (1) is obvious. The exact sequence 0 ;! M ;! M ;! M yields the exact sequence x HomR (N M ) ;! HomR (N M ) ;! Ext1R (N M ) ;!
Since HomR (N M ) = 0, and since x annihilates Ext1R (N M ), we obtain the natural isomorphism ) HomR (N M = Ext1R (N M ): This proves (2). Finally, (3) is clear since proj dimR F 1 for every free R -module F . We now present the main result of this section. Theorem 3.1.17. Let (R k) be a Noetherian local ring, and let M be a m
nite R-module of nite injective dimension. Then
dim M inj dim M = depth R:
95
3.1. Finite modules of nite injective dimension
Let 0 1 d = be a maximal chain of prime ideals in Supp M . We show by induction on i that ExtiR i (k( i ) M i ) 6= 0. In particular, it will follow that ExtdR (k M ) 6= 0 for d = dim M , so that dim M inj dim M by 3.1.14. If i = 0, then 0R 2 Ass M , and therefore HomR (k( 0 ) M ) 6= 0. Now suppose i > 0. We set B = R i then Proof.
p
p
p
m
p
p
p
p
p
0
p
0
p
0
p
p
0
ExtiB;1 (B= i;1B M i ) i; = ExtiR;1i; (k( i;1) M i; ) 6= 0 p
p
p
p
1
1
p
p
p
1
by the induction hypothesis, and so ExtiB;1(B= i;1B M i ) 6= 0. It follows from 3.1.13 that ExtiB (k( i ) M i ) 6= 0: To prove the equality inj dim M = depth R , we set r = inj dim M and t = depth R . Let x = x1 . . . xt be a maximal R -sequence. Then the Koszul complex K (x) is a minimal free resolution of R=(x) by 1.6.19 so that proj dim R=(x) = t and furthermore ExttR (R=(x) M ) is isomorphic to the t-th Koszul cohomology H t (x M ). It follows from 1.6.10 that H t (x M ) = H0 (x M ) = M=xM 6= 0. This implies r t. On the other hand, since depth R=(x) = 0, there is an embedding k ! R=(x) which induces an epimorphism ExtrR (R=(x) M ) ;! ExtrR (k M ) p
p
p
p
.
since ExtrR+1(N M ) = 0 for all R -modules N . But ExtrR (k M ) 6= 0 by 3.1.14, and so ExtrR (R=(x) M ) 6= 0. It follows that t = proj dimR R=(x) r. Gorenstein rings. We are now going to introduce an important class of
local rings. As for regular rings, this class can be characterized in terms of homological algebra. De nition 3.1.18. A Noetherian local ring R is a Gorenstein ring if inj dimR R < 1. A Noetherian ring is a Gorenstein ring if its localization at every maximal ideal is a Gorenstein local ring. The Gorenstein property is stable under standard ring operations. To begin with we show Proposition 3.1.19. Let R be a Noetherian ring. (a) Suppose R is Gorenstein. Then for every multiplicatively closed set S in R the localized ring RS is also Gorenstein. In particular, R is Gorenstein for every Spec R. (b) Suppose x is an R-regular sequence. If R is Gorenstein, then so is R=(x). The converse holds when R is local. (c) Suppose R is local. Then R is Gorenstein if and only if its completion R^ is Gorenstein. p
2
p
96
3. The canonical module. Gorenstein rings
(a) Let be a maximal ideal of RS . The ideal is the extension of a prime ideal in R , and so (RS ) = R . Let be a maximal ideal of R containing . Then R is a localization of the Gorenstein local ring R . From 3.1.9 the conclusion follows. (b) Without restriction we may assume that R is local. Thus (b) is an immediate consequence of 3.1.15. (c) Let k be the residue eld of R . Use that ExtiR (k R )b = ExtiR^ (k R^ ). Proof.
q
q
p
m
q
p
p
p
m
In concluding this section we clarify the position of the Gorenstein rings in the hierarchy of Noetherian local rings. Proposition 3.1.20. Let (R k) be a Noetherian local ring. Then we have m
the following implications: R is regular R is a complete intersection R is Cohen{Macaulay:
)
)
) R is Gorenstein
The rst implication is trivial. If R is regular, then its global homological dimension is nite (see 2.2.7), and hence ExtiR (k R ) = 0 for i 0. It follows from 3.1.14 that R is Gorenstein. In view of 3.1.19(c) we may as well assume that R is complete. Now 3.1.19(b) implies that a complete intersection is Gorenstein. The last implication follows from 3.1.17. All the implications of 3.1.20 are strict. This is clear for the rst, and will be shown for the other implications in the next section (see 3.2.11), where we derive a di erent, more easily veri able, characterization of Gorenstein rings. Proof.
Exercises 3.1.21. Let R be a principal ideal domain with eld of fractions K . Prove that 0 ! K ! K=R ! 0 is an injective resolution of R . 3.1.22. Let k be a eld, and let R be a local k-algebra of nite k-dimension. Show that the R -module Homk (R k) is an indecomposable (see the de nition before 3.2.6) injective R -module. 3.1.23. Let R be a Noetherian local ring. If there exists a non-zero nite injective R -module, then deduce R is Artinian. 3.1.24. Let (R k) be a Noetherian local ring, M 6= 0 and N 6= 0 nite R -modules. If inj dim N < 1, then deduce the following result of Ischebeck 226]: depth R ; depth M = supfi : ExtiR (M N ) 6= 0g: In particular, if R admits a nite module of nite injective dimension, then show that the depth of any nite R -module does not exceed the depth of R . (Bass' conjecture claims more: in the above situation R is Cohen{Macaulay. In 9.6.2 a proof of this conjecture will be given, provided R contains a eld.) m
97
3.2. Injective hulls. Matlis duality
Let R be a Gorenstein local ring, and M a nite R -module. Show proj dim M < 1 if and only if inj dim M < 1. (Foxby 114] proved the following remarkable characterization of Gorenstein rings: if a Noetherian local ring possesses a nite module M for which inj dim M < 1 and proj dim M < 1, then it is Gorenstein.) 3.1.26. Let (R k) be a Noetherian local ring. If inj dim k < 1, show R is regular. 3.1.25.
m
3.2 Injective hulls. Matlis duality We saw in Section 3.1 that any module M can be embedded into an injective module. Here we will show that such an embedding can be chosen minimal. In this case the corresponding injective module is unique up to isomorphism, and is called the injective hull of M . We will see that for a Noetherian ring R an injective module can be uniquely written as a direct sum of indecomposable injective modules, and the indecomposable injective R -modules are just the injective hulls of the cyclic R -modules R= , where 2 Spec R . If (R k) is a complete Noetherian local ring, and E is the injective hull of k, then the functor HomR ( E ) establishes an anti-equivalence between the category of Artinian R -modules and the category of nite R -modules. This result is known as the main theorem of Matlis duality. De nition 3.2.1. Let R be a ring and let N M be R -modules. M is an essential extension of N if for any non-zero R -submodule U of M one has U \ N 6= 0. An essential extension M of N is called proper if N 6= M . The following proposition gives a new characterization of injective modules. Proposition 3.2.2. Let R be a ring. An R-module N is injective if and only p
p
m
if it has no proper essential extension.
Let N M be an extension. If N is injective, then N is a direct summand of M . Let W be a complement of N in M . Then N \ W = 0, and so, if the extension is essential, W = 0. It follows that N = M . Conversely, suppose that N has no proper essential extension. Given a monomorphism ' : U ! V and a homomorphism : U ! N , we want to construct : V ! N such that = '. As in the proof of 3.1.2 we consider the pushout diagram Proof.
' ;;;;! V ?? ??
U
y
N
y
;;;;! W
98
3. The canonical module. Gorenstein rings
Here is a monomorphism, since ' is a monomorphism. Thus we may consider N as a submodule of W . Employing Zorn's lemma one shows that there exists a maximal submodule D W such that N \ D = 0, and so N may even be considered as a submodule of W =D obviously, W =D is an essential extension of N . It follows that N = W =D, since N has no proper essential extension, and so W = N D. Let : W ! N be the natural projection of W onto the rst summand. The composition : V ! N is an extension of . De nition 3.2.3. Let R be a ring and M an R -module. An injective module E such that M E is an essential extension is called an injective hull of M . Our notation will be E (M ) or ER (M ). The next proposition justi es this name. Proposition 3.2.4. Let R be a ring and M an R-module. (a) M admits an injective hull. Moreover, if M I and I is injective, then a maximal essential extension of M in I is an injective hull of M. (b) Let E be an injective hull of M, let I be an injective R-module, and : M I be a monomorphism. Then there exists a monomorphism ' : E I such that the diagram M E
!
?? ;;;;!
y = '
!
I is commutative, where M E is the inclusion map. In other words, the injective hulls of M are the `minimal' injective modules in which M can be embedded. (c) If E and E 0 are injective hulls of M, then there exists an isomorphism ' : E E 0 such that the diagram
!
!
M
' JJ^ E ;;;! E 0 commutes. Here M ! E and M ! E 0 are the inclusion maps. Proof. (a) We embed M into an injective R -module I . Consider the set S of all essential extensions M N with N I . Zorn's lemma applied to S yields the existence of a maximal essential extension M E with E I . We claim that E has no proper essential extension, and this
together with 3.2.2 implies then that E is an injective hull of M . Indeed, assume that E has a proper essential extension E 0 . Since I is injective there exists : E 0 ! I extending the inclusion E I . Suppose Ker = 0 then Im I is an essential extension of M (in I ) properly containing
99
3.2. Injective hulls. Matlis duality
E , a contradiction. On the other hand, since extends the inclusion E I we have E Ker = 0. But this contradicts the essentiality of the extension E E 0 . (b) Since I is injective, can be extended to a homomorphism ' : E I . We have ' M = , and so M Ker ' = Ker = 0. Thus, since the extension M E is essential, we even have Ker ' = 0. (c) By (b) there is a monomorphism ' : E E 0 such that ' M equals the inclusion M E 0 . Im ' is injective and hence a direct summand of E 0 . However, since the extension M E 0 is essential, ' is surjective, and
!
\
j
\
!
j
therefore an isomorphism. We may apply 3.2.4 to construct an injective resolution E (M ) of a module M which for obvious reasons is called the minimal injective resolution of M : we let E 0 (M ) = E (M ), and denote by @;1 the embedding M ! E 0(M ). Suppose the injective resolution has already been constructed up to the i-th step: .
0 ;! E 0(M ) ;! E 1 (M ) ;! ;! E i;1(M ) ;;! E i (M ): We then de ne E i+1(M ) = E (Coker @i;1), and @i is de ned in the obvious way. It is clear that any two minimal injective resolutions of M are isomorphic. Moreover, if I is an arbitrary injective resolution of M , then, as is readily seen, E (M ) is isomorphic to a direct summand of I . We note a technical result about injective hulls which will be needed later in this section. Lemma 3.2.5. Let R be a Noetherian ring, S R a multiplicatively closed set and M an R-module. Then ER (M )S = ERS (MS ): Proof. We show that ER (M )S is an injective hull of the RS -module MS . We know from 3.1.3 that ER (M )S is an injective RS -module. It remains to be shown that ER (M )S is an essential extension of MS . To simplify notation we set N = ER (M ), and pick x 2 NS , x 6= 0. We want to prove that RS x \ MS 6= 0. There exists y 2 N such that RS y = RS x. Thus we may as well assume that x 2 N . We consider the set of ideals S = fAnn(tx): t 2 S g. Since R is Noetherian this set has a maximal element, say Ann(sx), and since RS x = RS (sx), we may replace x by sx, and thus may assume that Ann(x) is maximal in the set S. Since N is an essential extension of M , we have Rx \ M = Ix 6= 0 where I is an ideal in R . Let I = (a1 . . . an ), and assume that ai x = 0 in NS for i = 1 . . . n. Then there exists t 2 S such that t(aix) = 0 in N for i = 1 . . . n. But Ann(tx) = Ann(x), by the choice of x, and so Ix = 0. This is a contradiction. Hence aix 6= 0 in NS for some i, and it follows that RS x \ MS 6= 0. @i;1
@0
.
.
.
100
3. The canonical module. Gorenstein rings
In the next theorem we determine the indecomposable injective R modules of a Noetherian ring R . Recall that an R -module M is decomposable if there exist non-zero submodules M1 M2 of M such that M = M1 M2 otherwise it is indecomposable. Theorem 3.2.6. Let R be a Noetherian ring. (a) For all 2 Spec R the module E (R= ) is indecomposable. (b) Let I 6= 0 be an injective R-module and let 2 Ass I. Then E (R= ) is p
p
p
p
a direct summand of I. In particular, if I is indecomposable, then
I = E (R= ):
(c) Let 2 Spec R. Then E (R= ) = E (R= ) () = . Proof. (a) Suppose E (R= ) is decomposable. Then there exist non-zero submodules N1 N2 of E (R= ) such that N1 \ N2 = 0. It follows that (N1 \ R= ) \ (N2 \ R= ) = (N1 \ N2 ) \ R= = 0. On the other hand, since R= E (R= ) is an essential extension, we have N1 \ R= 6= 0 6= N2 \ R= . This contradicts the fact that R= is a domain. (b) R= may be considered as a submodule of I since 2 Ass I . It follows from 3.2.4 that there exists an injective hull E (R= ) of R= such that E (R= ) I . As E (R= ) is injective, it is a direct summand of I . Statement (c) follows from the next lemma. Lemma 3.2.7. Let R be a Noetherian ring, 2 Spec R, and M a nite p
q
p
p
q
p
q
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
R-module. Then (a) Ass M = Ass E (M ) in particular one has = Ass E (R= ) (b) k( ) = HomR (k( ) E (R= ) ). Proof. (a) It is clear that Ass M Ass E (M ). Conversely, suppose Ass E (M ). Then there exists a submodule U E (M ) which is isomorphic to R= . We have U M = 0 since the extension M E (M ) is essential, and so Ass(U M ) Ass M . (b) Since E (R= ) = ER (k( )), we assume that (R k) is local and = is the maximal ideal. The k-vector space HomR (k E (k)) may be identi ed with V = x E (k): x = 0 it contains k. If V = k, then there exists a non-zero vector subspace W of V with k W = 0. This, however, contradicts the essentiality of the extension k E (k). The importance of the indecomposable injective R -modules results p
q
fg p
p
p
2
q
q
p
p
p
m
\ 6 \
p
p
p
2 f 2
p
m
p
g
m
\
6
from the following: Theorem 3.2.8. Every injective module I over a Noetherian ring R is a
direct sum of indecomposable injective R-modules, and this decomposition is unique in the following sense: for any Spec R the number of indecomposable summands in the decomposition of I which are isomorphic to E (R= ) depends only on I and (and not on the particular decomposition). In fact, this number equals dimk( ) HomR (k( ) I ). p
p
2
p
p
p
p
p
101
3.2. Injective hulls. Matlis duality
Consider the set S of all subsets of the set of indecomposable injective submodules of I with the property: if F 2 S, then the sum of all modules belonging to F is direct. The set S is partially ordered by inclusion. By Zorn's lemma it has a maximal element F0 . Let E be the sum of all the modules in F0. The module E is a direct sum of injective modules, and hence by 3.1.3 is itself injective. Therefore E is a direct summand of I , and we can write I = E H , where H is injective since it is a direct summand of I . Suppose H 6= 0 then there exists 2 Ass H , and so E (R= ) is a direct summand of H see 3.2.6(b). Thus we may enlarge F0 by E (R= ), contradicting the maximality of F0. We conclude that H = 0 and I = EL . Suppose that I = 2 I is the given decomposition. Then Proof.
p
p
p
HomR (k( ) I ) = HomR (k( ) p
M
p
p
p
p
2
By 3.2.7 we have
M
HomR (k( ) (I ) ) =
(I ) ) = p
M
p
2
p
p
M
HomR (k( ) (I ) ): p
2
p
p
HomR (k( ) (I ) ) p
2 0
p
p
where 0 = f 2 : I = E (R= )g. If we again use 3.2.7, we nally get
M p
HomR (k( ) I ) = p
p
p
2 0
HomR (k( ) (I ) ) = k( )( ): 0
p
p
p
p
Bass numbers. Let R be a Noetherian ring, M a nite R -module and Spec R . The ( nite) number i ( M ) = dimk( ) ExtiR (k( ) M ) is called the i-th Bass number of M with respect to .
2
p
p
p
p
p
p
These numbers have an interpretation in terms of the minimal injective resolution of M . Proposition 3.2.9. Let R be a Noetherian ring, M a nite R-module, and E (M ) the minimal injective resolution of M. Then p
.
M
E i (M ) =
p
Let 0 ;! M ;! E 0(M ) injective resolution of M , and let it follows from 3.2.5 that p
0 ;! M
p
p
2Spec R
Proof.
E (R= )i (
M ):
@ @ ;! E 1(M ) ;! be the minimal 2 Spec R. Since localization is exact, 0
1
d d ;! E0(M) ;! E 1(M ) ;! 0
p
p
1
p
is the minimal injective resolution of M here d i is the localization of @i . p
102
3. The canonical module. Gorenstein rings
The complex HomR (k( ) E (M ) ) is isomorphic to the subcomplex C of E (M ) , where .
p
p
p
.
.
p
f 2 Ei(M) :
Ci = x
p
p
g
R x=0 : p
Let x be a non-zero element of C i . Since the extension Im d i;1 E i (M ) is essential, there exists a 2 R with ax 2 Im d i;1 and ax 6= 0. Since R annihilates x, we see that a 2= R . Hence a is a unit in R , and x 2 Im d i;1. It follows that d i (x) = 0, and hence d i jC i = 0 for all i. Consequently we get ExtiR (k( ) M ) = HomR (k( ) E i (M ) ), which by 3.2.8 implies the isomorphism asserted. Among the Bass numbers the type of a module or a local ring is of particular importance. Let (R k) be a Noetherian local ring and M a nite module of depth t. In Chapter 1 we have already considered the Bass number r(M ) = t ( M ), and called it the type of M . In the next theorem we give a new, extremely useful characterization of Gorenstein rings. Theorem 3.2.10. Let (R k) be a Noetherian local ring. The following p
p
p
p
p
p
p
p
p
p
p
p
p
m
m
m
conditions are equivalent: (a) R is a Gorenstein ring (b) R is a Cohen{Macaulay ring of type 1.
Let x be a maximal R -sequence. By 3.1.19, R is Gorenstein if and only if R=(x) is. Similarly the properties in (b) are stable under specialization modulo x see 1.2.19 and 2.1.3. Thus we may assume dim R = 0. (a) ) (b): By 3.1.17, R is an injective R -module. Since R is local, it is indecomposable as an R -module, and so, since Ass R = f g, we have that R = ER (k) see 3.2.6. It follows from 3.2.7 that R is of type 1. (b) ) (a) follows from statement (e) in 3.2.12 below. We use this new characterization of Gorenstein rings to give examples of Cohen{Macaulay rings which are not Gorenstein, and of Gorenstein rings which are not complete intersections. Examples 3.2.11. (a) Let (R k) be an Artinian local ring for which 2 = 0. For instance, R = kX . . . X ]=(X . . . X )2 is such a ring. It 1 n 1 n is easily seen that = Soc R . Hence we have r(R ) = emb dim R , and conclude that R is Gorenstein if and only if emb dim R = 1. When R is Gorenstein, it is even a complete intersection. (b) In the following we present a method to produce a large class of Artinian Gorenstein rings: let k be a eld, S = kX1 . . . Xn] the polynomial ring in n variables over k, m an integer, Sm the m-th homogeneous part of S , and ' : Sm ! k a non-trivial k-linear map. Proof.
m
m
m
m
103
3.2. Injective hulls. Matlis duality
For every jL2 N, we de ne Ij = fa 2 Sj : '(a Sm;j ) = 0g. It is readily seen that I = j 0 Ij is a graded ideal with Ij = Sj for j > m. Thus we conclude that R = S=I is an Artinian (graded) local ring. We claim that R is a Gorenstein ring. To see this, we determine the socle of R . For any element a 2 S , we denote by a its residue class modulo I . Let j 2 N with 0 < j < m, and let a 2 Rj a 6= 0. Then by the de nition of I , there exists b 2 Sm;j such that '(a b) 6= 0, and so a b 6= 0. But since b belongs to the maximal ideal of R , it follows that a 2= Soc R . Therefore, Soc R = Rm . As dimk Rm = 1, it follows that R is Gorenstein. We give an explicit example for this construction: let ' : S2 ! k be the k-linear map with '(XiXj ) = 0 1
i < j n
'(Xi2) = 1 i = 1 . . . n:
For this linear form ' we get I = (X12
; X22 . . . X12 ; Xn2 X1X2 X1X3 . . . Xn;1Xn):
Therefore, R = S=I is Gorenstein, and is a complete intersection if and only if n 2. Matlis duality. Let (R k) be a Noetherian local ring. We are going to study the functor which takes the dual M 0 of an R -module M with respect to the injective hull E of k. If M is a nite module, the dual M 0 need not be nite. Indeed, we know from Exercise 3.1.23 that R 0 = E is nite only if R is Artinian. However, the E -dual of a module of nite m
length also has nite length, as we shall see now. Proposition 3.2.12. Let (R k) be a Noetherian local ring, E the injective m
hull of k, and N an R-module of nite length. For any R-module M we set M 0 = HomR (M E ). Then: (a) one has i = 0, ExtiR (k E ) = k0 for for i > 0
(b) `(N ) = `(N 0 ) (c) the canonical homomorphism N ! N 00 is an isomorphism (d) (N ) = r(N 0 ) and r(N ) = (N 0 ) (e) if R is Artinian, then E is a nite faithful R-module satisfying (i) `(E ) = `(R ), (ii) the canonical homomorphism R ! EndR (E ), a 7! 'a, where 'a(x) =
2
ax for all x E, is an isomorphism, (iii) r(E ) = 1 and (E ) = r(R ) conversely, any nite faithful R-module of type 1 is isomorphic to E.
104
3. The canonical module. Gorenstein rings
(a) ExtiR (k E ) = 0 for i > 0, as E is injective furthermore HomR (k E ) = k see 3.2.7. (b) We prove the equality asserted by induction on the length of N . If `(N ) = 1, then N = k, and the equality follows from (a). Now suppose that `(N ) > 1. Then there exists a proper submodule U N , and we obtain an exact sequence 0 ;! U ;! N ;! W ;! 0 with `(U ) < `(N ) and `(W ) < `(N ). Since E is injective this sequence yields the dual exact sequence 0 ;! W 0 ;! N 0 ;! U 0 ;! 0: The induction hypothesis applies to U and W , and the additivity of length gives the result. (c) Again we use induction on `(N ). If `(N ) = 1, then N = k, and N 00 = k by (a). Therefore it suces to show that the canonical homomorphism : k ! HomR (HomR (k E ) E ) is not the zero map. Let x 2 E , x 6= 0, be a socle element of E . There exists ' 2 HomR (k E ) with '(1) = x. Then (1)(') = x 6= 0, and so 6= 0. If `(N ) > 1, we choose as before an exact sequence 0 ;! U ;! N ;! W ;! 0 with `(U ) < `(N ) and `(W ) < `(N ). The natural homomorphisms into the bidual modules induce a commutative diagram 0 ;;;;! U ;;;;! N ;;;;! W ;;;;! 0 Proof.
?? y
?? y
?? y
0 ;;;;! U 00 ;;;;! N 00 ;;;;! W 00 ;;;;! 0 where the outer vertical arrows are isomorphisms by our induction hypothesis. The snake lemma (318], Theorem 6.5) applied to this diagram implies N ! N 00 is an isomorphism. (d) The module (N= N )0 is the kernel of the linear map N 0 ! ( N )0 which assigns to every ' 2 N 0 its restriction to N . Hence ' 2 (N= N )0 if and only if '(N ) = '( N ) = 0. In other words, (N= N )0 = f' 2 N 0 : ' = 0g = Soc N 0 : Thus we get (N ) = dimk N= N = dimk (N= N )0 = dimk Soc N 0 = r(N 0 ). The second equality follows from the rst by (c). (e) By (b) we have `(E ) = `(R 0) = `(R ) < 1. In particular, E is a nite R -module. Next it follows from (c) that the canonical homomorphism : R ! HomR (HomR (R E ) E ) is an isomorphism. If we identify m
m
m
m
m
m
m
m
m
m
105
3.2. Injective hulls. Matlis duality
HomR (R E ) with E , then identi es with the canonical homomorphism R ! EndR (E ). A module whose endomorphism ring is R is necessarily faithful. Statement (e)(iii) follows from (d). Finally, let N be a faithful R -module of type 1. Then N 0 is cyclic, and so N = HomR (R=I E ) for some ideal I . Here we have used (c) and (d). But since N is faithful, I = 0 and so N = E. Proposition 3.2.12 may be viewed as the Matlis duality theorem for nite Artinian modules. Now we prove its general form. It will be of crucial importance for the local duality theorem of Grothendieck, which we will discuss in Section 3.5. Let (R k) be a complete local ring. We denote by M(R ) the category of R -modules, by A(R ) the full subcategory of Artinian R modules and by F(R ) the full subcategory of nite R -modules. Let E be an injective hull of k. We set T ( ) = HomR ( E ). The contravariant functor T : M(R ) ! M(R ) is exact. Its restriction to A(R ) or F(R ) will again be denoted by T . m
Theorem 3.2.13 (Matlis). Let (R k) be a Noetherian complete local ring, N 2 A(R ) and M 2 F(R ). Then (a) T (R ) = E and T (E ) = R, (b) T (M ) 2 A(R ) and T (N ) 2 F(R ), (c) there are natural isomorphisms T (T (N )) = M, = N and T (T (M )) (d) the functor T establishes an anti-equivalence between the categories A(R) and F(R). m
We proceed in several stages. (1) For all n 2 N we set En = n x = 0g. Let x 2 E , x 6= 0 then Ass(Rx) Ass E = f g, see 3.2.7.SHence there exists an integer n such that nx = 0. This proves that E = n 0 En = lim ;! En. (2) The natural homomorphism R ! EndR (E ) = T (E ) is an isomorphism: by 3.2.12(e)(ii), the natural homomorphisms n : R= n ! T (En) are isomorphisms, and we obtain commutative diagrams such as Proof.
fx 2 E :
m
m
m
m
?? ;;;;! T ??(E) y y
R R=
m
n
;;;;! T (En) n
in which the only homomorphisms are the natural ones. As R is complete, the map R ! lim R= n is an isomorphism. Likewise T (E ) ! lim ; ; T (En) is an isomorphism since by 318], Theorem 2.27, and (1) we have lim = T (E ). It follows that the natural homomorphism ; T (ERn)!=TT(E(lim ;)!isEann) isomorphism as well. This proves (a). m
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3. The canonical module. Gorenstein rings
(3) E is Artinian: let E = U0 U1 U2 be a descending chain of submodules of E . This chain induces a sequence of epimorphisms
;! T (U1) ;! T (U2) ;! Thus we can write T (Ui) = R=Ii, where (0) = I0 I1 I2 is an ascending chain of ideals. Since R is Noetherian this chain stabilizes, and so there exists an integer i0 such that T (Ui) = T (Ui+1) for i i0 . We will show that Ui = Ui+1 for i i0 . Suppose that Ui = 6 Ui+1, but T (Ui) = T (Ui+1). Let V = Ui =Ui+1 then V = 6 0, but T (V ) = 0. However, V is a subquotient of E , and so Ass V = f g see 3.2.7. In other words, there exists a monomorphism k ! V . Applying T , we obtain an epimorphism 0 = T (V ) ! T (k) = k, a contradiction. (4) If N is Artinian, then there exists an embedding N ! E n for some integer n: Soc N is a nite dimensional k-vector space since N is Artinian. Moreover, the extension Soc N N is essential. In fact, if x 2 N , then Rx is a nite Artinian module, and therefore Rx \ Soc N = Soc Rx = 6 0. Let N I be an embedding of N into an injective R -module. By 3.2.4, an injective hull E (N ) can be chosen as a maximal essential extension of N in I . Since the extension Soc N N is essential, E (N ) is likewise an injective hull of Soc N . Suppose Soc N = kn then it follows that n N E (Soc N ) = E . The remaining assertions of the theorem now follow easily. (b) Let N 2 A(R ) then by (4) there exists an embedding N E n which by (2) induces an epimorphism R n ! T (N ) therefore T (N ) 2 F(R). Conversely, suppose M 2 F(R). We choose an epimorphism R n ! M . This epimorphism yields an embedding T (M ) ! E n . The module E is Artinian by (3), and so any submodule of E n is Artinian. It follows that T (M ) 2 A(R ). (c) By (4) there exists an integer n and an exact sequence 0 ! N ! E n ! W ! 0 which we may complete to a commutative diagram n 0 ;;;;! N ? ;;;;! E? ;;;;! W? ;;;;! 0 R = T (E )
m
?y
?y
?y
0 ;;;;! T (T (N )) ;;;;! T (T (E n)) ;;;;! T (T (W )) ;;;;! 0 whose vertical maps are just the canonical homomorphisms. It follows from (2) that is an isomorphism. Therefore, by the snake lemma, is an isomorphism if and only if is a monomorphism. Let x 2 Ker then '(x) = 0 for all ' 2 HomR (W E ). Suppose x 6= 0, and let : Rx ! E be the homomorphism which maps x to a non-zero socle element of E . Then (x) 6= 0, and since E is injective, can be extended to a homomorphism ' : W ! E . We then have '(x) 6= 0, a contradiction.
3.3. The canonical module
107
Similarly one proves that the natural linear map M ! T (T (M )) is an isomorphism, starting with the exact sequence 0 ! U ! R n ! M ! 0 and using the fact that the natural homomorphism R ! T (T (R )) is an isomorphism (which is an immediate consequence of (2)). Exercises Let (R k) be a Noetherian local ring, E an injective hull of k. Prove: (a) The natural homomorphism E ! E R R^ is an isomorphism. In particular, E is an R^ -module. (b) As an R^ -module, E
= ER^ (k). (c) For all nite R -modules N there exists a natural isomorphism HomR (N E )
= HomR^ (N^ E ): (If this problem seems to be too di cult, the reader may consult 270], Theorem 18.6.) 3.2.15. Let (R ) be an Artinian local ring. Show the following conditions are equivalent: (a) R is a Gorenstein ring (b) all nite R -modules are re exive (c) I = Ann Ann I for all ideals I of R (d) for all non-zero ideals I and J one has I \ J 6= 0. 3.2.14.
m
m
3.3 The canonical module So far we have studied nite modules of nite injective dimension over Noetherian local rings, but we have ignored the question as to under what circumstances such modules actually exist. A Gorenstein ring R admits plenty of nite modules of nite injective dimension: any module of nite projective dimension has nite injective dimension as well, simply because R itself has nite injective dimension by de nition. Also any Artinian local ring (R k), Gorenstein or not, admits a nite injective module { the injective hull of k. The question becomes more delicate for non-Gorenstein local rings of positive dimension. One of the main results of this section will be that any Cohen{Macaulay ring which is a homomorphic image of a Gorenstein ring has a nite module of nite injective dimension. Moreover, this module can be chosen to be a maximal Cohen{Macaulay module of type 1. It will be shown that such a module is unique up to isomorphism. It is called the canonical module of R . For a Gorenstein ring the canonical module is just the ring itself. We shall study the behaviour of the canonical module under at extensions, localizations, and specializations. De nition 3.3.1. Let (R k) be a Cohen{Macaulay local ring. A maximal Cohen{Macaulay module C of type 1 and of nite injective dimension is called a canonical module of R . m
m
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It is immediate (see 1.2.8 and 3.1.14) that C is a canonical module of
R if and only if
dimk ExtiR (k C ) = id d = dim R: Two questions arise: when does a canonical module exist, and is it uniquely determined up to isomorphism? This question has a simple answer in the case dim R = 0: by 3.2.8, ER (k) is the uniquely determined canonical module. To prove uniqueness in general, we will need the following two results. Lemma 3.3.2. Let (R k) be a Noetherian local ring, ' : M ! N a homomorphism of nite R-modules, and x an N-sequence. If ' R=(x) is m
an isomorphism, then ' is an isomorphism. Proof. The surjectivity of ' follows from Nakayama's lemma. In order to prove that ' is injective, we may assume without loss of generality that the sequence x consists of one element, say x. Let K = Ker ' since x is N -regular, the exact sequence 0 K M N 0
;! ;! ;! ;!
induces the exact sequence 0 ;! K=xK ;! M=xM ;! N=xN ;! 0: By assumption, K=xK = 0, and hence K = 0, by Nakayama's lemma. Proposition 3.3.3. Let (R k) be a Cohen{Macaulay local ring of dimenm
sion d, and C a maximal Cohen{Macaulay R-module. (a) Suppose M is a maximal Cohen{Macaulay R-module with ExtjR (M C ) = 0 for all j > 0. Then HomR (M C ) is a maximal Cohen{Macaulay module, and for any R-sequence x we have HomR (M C ) R=xR = HomR=xR (M=xM C=xC ): (b) Assume in addition that C has nite injective dimension, and M is a Cohen{Macaulay R-module of dimension t. Then (i) ExtjR (M C ) = 0 for j = d t, (ii) ExtdR;t(M C ) is a Cohen{Macaulay module of dimension t. Proof. (a) Let x be an R -regular element. Since C is a maximal Cohen{Macaulay module, the element x is C -regular as well, and one
6 ;
2
m
has the exact sequence
0 ;! C ;! C ;! C=xC ;! 0 which by our assumtion induces the exact sequence x
0 ;! HomR (M C ) ;! HomR (M C ) ;! HomR (M C=xC ) ;! 0: x
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Therefore,
HomR=xR (M=xM C=xC ) = HomR (M C=xC ) = HomR (M C )=x HomR (M C ) = HomR (M C ) R=xR:
For an arbitrary R -sequence x one proceeds by induction on the length of the sequence. (b)(i) It follows from 1.2.10(e) that ExtjR (M C ) = 0 for j < d ; t. Next we show by induction on t that ExtjR (M C ) = 0 for any t-dimensional Cohen{Macaulay module M and all j > d ; t. If t = 0, then the claim follows from 3.1.17. Now suppose that t > 0, and let x 2 be an M -regular element. The exact sequence m
0 ;! M ;! M ;! M=xM ;! 0 x
induces the exact sequence
ExtjR (M C ) ;! ExtjR (M C ) ;! ExtjR+1(M=xM C ): x
; 1)-dimensional Cohen{Macaulay module. Hence by the induction hypothesis we have ExtjR+1(M=xM C ) = 0 for j > d ; t, and so Nakayama's lemma implies that ExtjR (M C ) = 0 for j > d ; t. (ii) We proceed by induction on t. The assertion is trivial if t = 0. Assume now dim M = t > 0, and let x 2 be an M -regular element. By (i), the exact sequence
M=xM is a (t
m
0 ;! M ;! M ;! M=xM ;! 0 x
yields the exact sequence
0 ;! ExtdR;t(M C ) ;! ExtdR;t(M C ) ;! ExtdR;(t;1)(M=xM C ) ;! 0: x
Thus x is regular on ExtdR;t(M C ), and so it follows from our induction hypothesis that ExtdR;t(M C ) is Cohen{Macaulay. We are now ready to prove the uniqueness of the canonical module. Theorem 3.3.4. Let (R k) be a Cohen{Macaulay local ring, and let C and C 0 be canonical modules of R. Then (a) C=xC = ER=(x)(k) for any maximal R-sequence x, (b) the canonical modules C and C 0 are isomorphic, (c) HomR (C C 0) = R, and any generator ' of Hom(C C 0) is an isomorphism, (d) the canonical homomorphism R EndR (C ) is an isomorphism. m
!
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(a) By 3.1.15, C=xC is an injective R=(x)-module of type 1. Since Spec(R=(x)) = f =(x)g, 3.2.8 yields the assertion. (b) and (c): It follows from (a) that C=xC = ER=(x) (k) = C 0=xC 0: Now 3.3.3 and 3.2.12 imply that HomR (C C 0) R R=(x) = HomR=(x) (C=xC C 0=xC 0) = R=(x) 0 and so HomR (C C ) is cyclic by Nakayama's lemma. Let ' be a generator of this module. Then the natural inclusion R' ! HomR (C C 0) induces the above isomorphism modulo x. By 3.3.3, HomR (C C 0) is a maximal Cohen{Macaulay module. Thus 3.3.2 implies that R' ! HomR (C C 0) is an isomorphism. In particular it follows that R' is a maximal Cohen{ Macaulay module. We may therefore apply 3.3.2 once again to conclude that R ! R' is an isomorphism, too. Next we show that ' : C ! C 0 is an isomorphism. Indeed, ' R=(x) can be identi ed with a generator of End(ER=(x)(k)). It follows therefore from 3.2.12(e)(ii) that ' R=(x) is an isomorphism. Since C 0 is a maximal Cohen{Macaulay module, 3.3.2 implies that ' is an isomorphism. It is clear that any other isomorphism C ! C 0 is a generator of HomR (C C 0), too. (d) is proved similarly. In view of this result we may talk of the canonical module of R provided it exists. From now on we will denote the canonical module of R by !R . The next theorem lists some useful and often applied change of ring formulas for the canonical module. Theorem 3.3.5. Let (R k) be a Cohen{Macaulay local ring with canonProof.
m
m
ical module !R . Then (a) !R =x!R = !R=xR for all R-sequences x, that is, the canonical module specializes, (b) (!R ) = !R for all Spec R, that is, the canonical module localizes, (c) (!R )b = !R^ . Proof. (a) First notice that x is an !R -sequence, too. The (R=xR )-module !R =x!R has nite injective dimension see 3.1.15. Since r(!R =x!R ) = r(!R ) = 1, the module !R =x!R is the canonical module of R=xR , by p
p
p
2
de nition. (b) The R -module (!R ) has nite injective dimension (see 3.1.9), and is again a maximal Cohen{Macaulay module. It remains to be shown that r((!R ) ) = 1. Let x be a sequence of elements of R whose image in R is a maximal R -sequence. Then by 3.1.15, M = (!R ) =x(!R ) p
p
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p
p
p
p
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3.3. The canonical module
is an injective module over the Artinian local ring A = R =xR . It follows from 3.2.7 and 3.2.8 that M r = r (M ): = EA (k( ))r From 3.2.12 we get p
p
p
(1) HomA (M M ) = Ar : On the other hand, from 3.3.5(a) we obtain (2) HomA (M M ) = HomR=xR (!R =x!R !R =x!R ) = HomR=xR (!R=xR !R=xR ) = (R=xR) = A: For the last isomorphism we used the fact that the endomorphism ring of the canonical module of S = R=xR is isomorphic to S see 3.3.4. A comparison of (1) and (2) yields r = 1, as desired. (c) The bre of R ! R^ is k, so that by atness, ExtiR (k !R )b = ExtiR^ (k (!R )b) for all i. This implies the assertion. 2
p
p
p
Existence of the canonical module. Our next goal is to clarify for which
Cohen{Macaulay local rings the canonical module exists. Theorem 3.3.6. Let (R k) be a Cohen{Macaulay local ring. The folm
lowing conditions are equivalent: (a) R admits a canonical module (b) R is the homomorphic image of a Gorenstein local ring.
One direction of the proof resorts to the principle of idealization due to Nagata: let R be a ring and M an R -module. We construct a ring extension R R M of R , called the trivial extension of R by M . As an R -module, R M is just the direct sum of R and M . The multiplication is de ned by (a x)(b y) = (ab ay + bx) for all a b 2 R and x y 2 M . Some basic facts on trivial extensions are the subject of Exercise 3.3.22. Here we will only use that R M is a ring, and if M is nite and R is a Noetherian (or Artinian) local ring with maximal ideal , then so is R M with maximal ideal M = f(a x) 2 R M : a 2 g. Proof of 3.3.6. (a) ) (b): The ring R is a homomorphic image of the trivial extension R !R . We will show that R !R is a Gorenstein ring. Let x be an R -regular sequence of maximal length. It is easy to see that x is a maximal (R !R )-sequence as well, and that (R !R )=x(R !R ) = (R=xR ) (!R =x!R ): m
m
m
112
3. The canonical module. Gorenstein rings
By 3.3.4, !R =x!R = ER=xR (k). Bearing in mind the characterization 3.2.10 of Gorenstein rings, we may assume that R is Artinian, and it remains to be shown that the type of the Artinian local ring R 0 = R ER (k) is 1. Let (a x) 2 Soc R 0 then (b 0)(a x) = (ba bx) = (0 0) for all b 2 . This implies that a 2 Soc R and x 2 Soc ER (k). Assume that a 6= 0. The exact sequence a R ;! R ;! R=(a) ;! 0 a induces the exact sequence 0 ;! ER=(a)(k) ;! ER (k) ;! ER (k) (see 3.1.6). As `(ER=(a)(k)) = `(R=(a)) < `(R ) = `(ER (k)) (see 3.2.12), multiplication by a on ER (k) cannot be the zero map. Therefore there exists y 2 ER (k) with ay 6= 0, and so (0 y)(a x) = (0 ay) 6= (0 0), a contradiction. Our conclusion is that Soc(R ER (k)) = Soc ER (k), and therefore by 3.2.12, r(R E (k)) = 1. For the proof of 3.3.6(b) ) (a) we note the following more general result. Theorem 3.3.7. Let (R ) be a Cohen{Macaulay local ring. (a) The following conditions are equivalent: (i) R is Gorenstein (ii) !R exists and is isomorphic to R. (b) Let ' : (R ) ! (S ) be a local homomorphism of Cohen{Macaulay m
m
m
n
local rings such that S is a nite R-module. If !R exists, then !S exists and !S = ExttR (S !R ) t = dim R dim S:
;
(a)(i) () (ii) follows from 3.1.18 and 3.2.10. (b) By virtue of 2.1.2(b), and since dim S = dim(R= Ker '), there exists an R -sequence x = x1 . . . xt with xi 2 Ker ', t = dim R ; dim S . Set R = R=(x)R as !R =(x)!R = !R (see 3.3.5), we have ExttR (S !R ) = HomR (S !R ), by 3.1.16. Thus we may assume from the beginning that dim R = dim S . Let d = dim R , and y = y1 . . . yd an R -sequence. Then y is !R regular and HomR (S !R )-regular as well, since both modules are Cohen{ Macaulay modules of dimension d see 3.3.3. It follows from 3.3.3(a) that HomR (S !R ) R R 0 = HomR0 (S 0 !R0 ) where R 0 = R=(y )R , and S 0 = S=(y )S . In view of Exercise 3.3.23 it suces to show that HomR0 (S 0 !R0 ) is the canonical module of S 0 . Since !R0 = ER0 (k), 3.1.6 implies that HomR0 (S 0 ER0 (k)) is an injective S 0 -module, and so HomR0 (S 0 ER0 (k)) = ES 0 (k)r for some r > 0. By 3.2.12(b) and (e)(i) Proof.
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3.3. The canonical module
we get `(ES 0 (k)) = `(S 0 ) = `(HomR0 (S 0 ER0 (k))) = r `(ES 0 (k)) therefore r = 1. A noteworthy case of 3.3.7 is the following: let k be a eld, and R an Artinian local k-algebra. Then Homk (R k) is the canonical module of R . A Noetherian complete local ring is a homomorphic image of a regular local ring see A.21. Regular local rings are Gorenstein (see 3.1.20), and so 3.3.6 implies Corollary 3.3.8. A complete Cohen{Macaulay local ring admits a canonical module.
Corollary 3.3.9. Let (R k) be a regular local ring and I an ideal of height g such that S = R=I is Cohen{Macaulay. Let F : 0 ;! Fg ;! Fg;1 ;! ;! F0 ;! 0 be the minimal free R-resolution of S, and let G = HomR (F R ) be the m
m
.
dual complex
.
;! ;!
.
;! ;! ;!
G. : 0 Gg Gg;1 G0 0 where Gi = Fg;i for i = 0 . . . g. Then G. is a minimal free R-resolution of !S .
Note that g is indeed the length of the minimal free resolution of S see 2.2.10. One has ExtiR (S R ) = H i (F. ) for all i 0. The corollary Proof.
follows therefore from 3.3.3(b) and 3.3.7.
Further properties of the canonical module. In the next theorem some
useful characterizations of the canonical module will be given. Theorem 3.3.10. Let (R k) be a Cohen{Macaulay local ring of dimenm
sion d, and let C be a nite R-module. Then the following conditions are equivalent: (a) C is the canonical module of R (b) i ( C ) = ih for all i 0 and all Spec R, where h = height (c) for all integers t = 0 1 . . . d, and all Cohen{Macaulay R-modules M of dimension t one has (i) ExtRd ;t(M C ) is a Cohen{Macaulay R-module of dimension t, (ii) ExtiR (M C ) = 0 for all i = d t, (iii) there exists an isomorphism M ExtdR;t (ExtdR;t(M C ) C ) which in the case d = t is just the natural homomorphism from M into the bidual of M with respect to C (d) for all maximal Cohen{Macaulay R-modules M one has (i) HomR (M C ) is a maximal Cohen{Macaulay R-module, (ii) ExtiR (M C ) = 0 for i > 0, (iii) the natural homomorphism M HomR (HomR (M C ) C ) is an isomorphism. p
p
6 ;
2
!
!
p
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3. The canonical module. Gorenstein rings
(a) () (b): The canonical module localizes see 3.3.5 therefore (a) implies (b). Choosing = , we obtain (a) from (b). (a) ) (c): (i) and (ii) have already been shown in 3.3.3. From the Rees lemma 3.1.16 and (i) one deduces that Proof.
p
m
ExtdR;t (ExtdR;t(M C ) C ) = HomR=xR (HomR=xR (M C=xC ) C=xC )
for an R -sequence x of length d ; t which is contained in AnnR M . Replacing R by R=xR , we may as well assume that t = d . Since by 3.3.3, HomR (M C ) R=y R = HomR=yR (M=y M C=y C ) for any R -sequence y , we may nally assume that dim R = 0. In this case however C = ER (k), and the assertion follows from 3.2.12. (d) is a special case of (c). (d) ) (a): If we choose M = R , then it follows from (i) that C is a maximal Cohen{Macaulay module. According to Exercise 2.1.26, for all i d the i-th syzygy module of the residue class eld k of R is 0 or a maximal Cohen{Macaulay module. Therefore (ii) implies that ExtiR (k C ) = 0 for i > d , and hence we have inj dim C < 1 see 3.1.14. It remains to be shown that r(C ) = 1. By 3.3.3(a), the conditions in (d) are stable under reduction modulo R -sequences. Thus, since the type of C is also stable under reduction modulo R -sequences, we may restrict ourselves to the case where R is Artinian. Then the module C is necessarily injective, and so it must be isomorphic to ER (k)r , r = r(C ). Now it follows from 3.2.12 that R r = HomR (HomR (R C ) C ). Consequently condition (iii) implies r(C ) = 1. We complement the previous theorem with some extra information about the Exti ( !R ). Observe the analogy of the statements with 3.2.12. The canonical module takes the position of the injective hull when one deals with arbitrary Cohen{Macaulay local rings rather than Artinian local rings. Proposition 3.3.11. Let R be a Cohen{Macaulay local ring of dimension 2
d with canonical module !R , and M a Cohen{Macaulay R-module of dimension t. Then (a) (M ) = r(ExtdR;t (M !R )), (b) r(M ) = (ExtdR;t(M !R )), (c) !R is a faithful R-module, and (i) r(!R ) = 1, (!R ) = r(R ), (ii) End(!R ) = R. Proof.
get
There exists an !R -sequence x of length d ; t in Ann M , and we ExtRd ;t (M !R ) = HomR=xR (M !R=xR ):
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3.3. The canonical module
So we may assume that dim R = dim M . By 3.3.3, we may further assume that dim R = 0. Since the canonical module of an Artinian local ring is the injective hull of the residue class eld, all assertions follow from 3.2.12. The previous proposition has an interesting application. Corollary 3.3.12. Let R be a Cohen{Macaulay local ring, M a Cohen{ Macaulay R-module and 2 Supp M. Then r(M ) r(M ). ^ R^ ) then dim R= ^ = dim R= see 2.1.15. ThereProof. Pick 2 Ass(R= fore we obtain a at local homomorphism R ! R^ whose bre is of dimension zero. From 1.2.16 it follows that r(M ) r(M^ ). Since R ! R^ is at with bre k, 1.2.16 once again applied gives r(M ) = r(M^ ). We may therefore assume that R is complete. By A.11, R is the epimorphic image of complete regular local S , and by Exercise 1.2.26(c) we have rS (M ) = rR (M ). Thus we may assume that R is regular. In particular R is Gorenstein. Hence, by 3.3.7, R has a canonical module and is isomorphic to R , and so 3.3.11 yields r(M ) = (ExtdR;t(M R )) = (ExtdR;t(M R ) ) (ExtdR;t(M R)) = r(M) where d = dim R and t = dim M . Here we have used that, by 2.1.3, d ; t = dim R ; dim M = dim R ; dim R= ; (dim M ; dim M= M ) = dim R ; dim M : p
p
q
p
q
p
p
q
p
p
p
p
q
p
p
p
p
p
p
The canonical module and at extensions. We will show that the canonical module behaves well under at ring extensions. For the proof we need Proposition 3.3.13. Let (R ) be a Cohen{Macaulay local ring, and C a m
nite R-module. The following conditions are equivalent: (a) C is the canonical module of R (b) C is a faithful maximal Cohen{Macaulay R-module of type 1.
Proof. (a) ) (b): The canonical module is a maximal Cohen{Macaulay module of type 1, by de nition, and faithful by 3.3.11. (b) ) (a): Note rst that C has one of the properties in (a) or (b) if and only if the completion C^ has this property. For instance, the property of being faithful means that the canonical homomorphism ' : R ! HomR (C C ) is injective. Since R ! R^ is faithfully at, ' is injective if and only if its completion is injective. The reader should check the other properties. We may now assume that R is complete. Then, by 3.3.8, R has a canonical module !R . By 3.3.11(b), HomR (C !R ) is a cyclic module, say
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3. The canonical module. Gorenstein rings
R=I , so that by 3.3.10(d)(iii) we have C = HomR (R=I !R ). It follows that I annihilates C . Since we assume that C is faithful, we get I = 0, and hence C = HomR (R !R ) = !R .
Theorem 3.3.14. Let (R ) be a Cohen{Macaulay local ring, and (R ) ! (S ) a at homomorphism of local rings. (a) If !R exists and S= S is Gorenstein, then !S = !R S. (b) If C is a nite R-module, and S a Cohen{Macaulay ring with canonical module !S = C S, then S= S is Gorenstein and C = !R . Proof. A at local homomorphism is faithfully at. Thus we see as in the proof of the previous proposition that a nite R -module M is faithful if and only if M R S is a faithful S -module. By 1.2.16, we have r(!R S ) = r(S= S )r(!R ) = 1 m
m
n
m
m
m
hence (a) follows from 3.3.13. Part (b) is proved similarly: by 1.2.16, 1 = r(S= S )r(C ) and C is a maximal Cohen{Macaulay module. It follows that r(S= S ) = 1 and r(C ) = 1. Therefore S= S is Gorenstein (see 3.2.10), and in view of 3.3.13, C is the canonical module of R . Corollary 3.3.15. Let ' : (R ) ! (S ) be a at homomorphism of m
m
m
m
n
Noetherian local rings. Then S is Gorenstein if and only if R and S= S are Gorenstein. m
The canonical module for non-local rings. We saw in 3.3.5 that the canon-
ical module localizes. This suggests the following De nition 3.3.16. Let R be a Cohen{Macaulay ring. A nite R -module !R is a canonical module of R if (!R ) is a canonical module of R for all maximal ideals of R . Remark 3.3.17. In contrast to the local case, a canonical module is in general not unique (up to isomorphism). Indeed, let R be a Cohen{ Macaulay ring (not necessarily local), and let !R and !R0 be canonical modules of R . We set I = HomR (!R !R0 ). Localizing at a prime ideal and using 3.3.4 and 3.3.5 we see that I = R for all 2 Spec R . We de ne an R -module homomorphism : I !R ! !R0 by (' x) = '(x) for all ' 2 I and all x 2 !R . Then is an isomorphism since it is locally an isomorphism. Conversely, suppose !R is a canonical module of R and I is a locally free R -module of rank 1. Then I !R is locally isomorphic to !R , and so is a canonical module of R . Thus a canonical module of a Cohen{ Macaulay ring is only unique up to a tensor product with a locally free module of rank 1. m
m
m
p
p
p
117
3.3. The canonical module
Proposition 3.3.18. Let R be a Cohen{Macaulay ring and !R a canonical module of R. (a) The following conditions are equivalent: (i) !R has a rank (ii) rank !R = 1 (iii) R is generically Gorenstein, i.e. R is Gorenstein for all minimal prime ideals of R. (b) If the equivalent conditions of (a) hold, then !R can be identi ed with an ideal in R. For any such identi cation, !R is an ideal of height 1 or equals R. In the rst case, the ring R=!R is Gorenstein. Proof. (a)(i) (ii): Let Spec R be a minimal prime ideal. Then !R = (!R ) is a free R -module, and R is Artinian. The canonical module of an Artinian local ring is the injective hull E of the residue class eld, and so the free R -module (!R ) has rank 1 since E is indecomposable see p
p
)
p
p
2
p
p
p
3.2.6. The implications (ii) ) (iii) and (iii) ) (i) are clear in view of 3.3.7. (b) The canonical module !R is torsion-free since all R -regular elements are !R -regular as well. According to Exercise 1.4.18, !R is isomorphic to a submodule of R . Therefore it may be identi ed with an ideal in R which we again denote by !R . If dim R = 0, then necessarily !R = R . We may therefore assume that dim R > 0, and that !R is a proper ideal of R . Then !R must contain an R -regular element since rank !R = 1. Let be a prime ideal containing !R . Using the fact that !R R is a maximal Cohen{Macaulay R -module we then get dim R ; 1 dim(R =!R R ) depth(R =!R R ) depth R ; 1 = dim R ; 1. This shows that height !R = 1, and that R=!R is Cohen{Macaulay. Finally we prove that R=!R is Gorenstein. To show this, we may assume that R is local. Applying the functor HomR ( !R ) to the exact sequence 0 ;! !R ;! R ;! R=!R ;! 0 and using 3.3.4(d) we obtain the exact sequence 0 ;! !R ;! R ;! Ext1R (R=!R !R ) ;! Ext1R (R !R ) = 0: This implies R=!R = Ext1R (R=!R !R ). Thus the conclusion follows from 3.3.7. Corollary 3.3.19. Let R be a Cohen{Macaulay normal domain with canonp
p
p
p
p
p
p
p
p
p
p
p
ical module !R . Then !R is isomorphic to a divisorial ideal. In particular, if R is factorial, then R is Gorenstein. Proof. By 3.3.18, !R is an ideal. It satis es the Serre condition (S2 ), and moreover, (!R ) = !R = R for all prime ideals of height 1. This follows from 3.3.7 since, by normality, R is regular for all prime ideals of height 1. Thus we have shown that !R is a reexive ideal see 1.4.1. p
p
p
p
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3. The canonical module. Gorenstein rings
A reexive ideal is divisorial. (We refer to Fossum 108] for the theory of divisorial ideals.) In a factorial ring all divisorial ideals are principal, and so !R is principal and R is Gorenstein see 3.3.7. In concluding these considerations we show that formula 3.3.14(a) for the canonical module under at extensions has a non-local counterpart. Proposition 3.3.20. Let ' : R ! S be a at homomorphism of Noetherian rings whose bres S R k( ) are Gorenstein for all 2 Spec R for which there exists a maximal ideal in S with = \ R. If !R is a canonical p
p
R S is a canonical module of S. Proof. Let be a maximal ideal of S , = \ R : then R ! S is a at local homomorphism whose bre is a localization of S R k( ), and thus is Gorenstein. It follows from 3.3.14(a) that !R R S is a canonical module of S . Since (!R R S ) = !R R S , the proposition q
module of R, then !R
p
q
q
p
q
p
q
p
p
q
p
is proved. Corollary 3.3.21. Let R be a Cohen{Macaulay ring with canonical module !R , and let S be either the polynomial ring R X1 . . . Xn] or the formal power series ring R X1 . . . Xn]]. Then !R R S is a canonical module of q
q
p
p
q
S. In particular, if R is Gorenstein, then so is S.
We may assume that n = 1. The result then follows from 3.3.20 since in both cases the bres considered there are regular rings see the proof of A.12. Proof.
Exercises
Let R be a ring, M an R -module, and R M the trivial extension of R by M . (The de nition of R M is given after Theorem 3.3.6.) Prove: (a) R M is a ring. (b) R can be identi ed with the subring R 0 = (a x) R M : x = 0 . (c) 0 M = (a x) R M : a = 0 is an ideal in R M with (0 M )2 = 0. As R -modules, M and 0 M are isomorphic. (d) If (R ) is local, then R M is local with maximal ideal M = (a x) R M: a . (e) The natural inclusion R R M composed with the natural epimorphism R M (R M )=(0 M ) is an isomorphism. (e) If R is Noetherian and M is a nite R -module, then R M is Noetherian and dim R = dim R M . 3.3.23. Let R be a Cohen{Macaulay local ring, C a maximal Cohen{Macaulay R -module, and x an R -sequence. If C=xC is the canonical module of R=(x), show that C is the canonical module of R . 3.3.24. Let (R ) be a Gorenstein local ring and I R an ideal of grade g such that S = R=I is a Cohen{Macaulay ring. Let x = x1 . . . xn be a system of generators of I . Show that !S = Hn g (x): 3.3.22.
f
2 g !
2
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!
g
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;
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f
2
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3.3. The canonical module 3.3.25.
and let
Let (R ) be a Gorenstein local ring, I R a perfect ideal of grade g, m
0 ;;! Fg ;;! ;;! F0 ;;! 0 be a minimal free R -resolution of S = R=I . Prove: @ @g (a) The dual complex 0 ;;! F0 ;;! ;;! Fg ;;! 0 is acyclic, and Coker @g
= !S . (b) HomS (!S S )
= Ker(Fg S ! Fg 1 S )
= TorRg (S S ). R (c) S is Gorenstein if and only if Torg (S S )
= S . (If you nd this problem too di cult, consult 1.4.22 or 159], Section 7.5.) (d) Suppose g = 2 then (I ) = r(S ) + 1. In particular, if R is regular and S is Gorenstein, then S is a complete intersection 3.3.26. Let (R k) be a Gorenstein local ring of dimension d , and M a nite module of nite projective dimension. Show that @g
@1
1
;
m
TorRi (k M )
= ExtdR i (k M )
for all i.
;
Let (R ) be a Cohen{Macaulay local ring with canonical module !R . Suppose for all nite R -modules M there exist an integer n and an epimorphism !Rn ! M . Prove R is a Gorenstein ring. 3.3.28. Let (R ) be a Cohen{Macaulay local ring with canonical module !R . (a) Suppose M is a maximal Cohen{Macaulay R -module of nite injective dimension. Show M is isomorphic to a direct sum of nitely many copies of !R . (b) Let M be a nite R -module. Show inj dim M < 1 if and only if M has a nite !R -resolution, that is, there exists an exact sequence 3.3.27.
m
m
0 ;;! !Rrp ;;! ;;! !Rr 'p
'1
0
;;! M ;;! 0:
Hint: For all nite R -modules M there exists an exact sequence 0 ! Y ! X ! M ! 0 where X is a maximal Cohen{Macaulay R -module, and Y a module of nite injective dimension see 21]. Such an exact sequence is called a Cohen{Macaulay approximation. (c) The !R -resolution is minimal if Im 'i !Rri; for i = 1 . . . p. Show that a module M of nite injective dimension even has a minimal !R -resolution, and that ri = d i ( M ) for all i when the resolution is minimal. 3.3.29. Let (R ) be a Cohen{Macaulay local ring of dimension 1. A subset I of the total ring Q of fractions of R is called a fractionary ideal if there exist R -regular elements x, y such that y 2 xI R . The inverse of a fractionary ideal I is the set I 1 = fa 2 Q : aI R g. We denote by F the set of fractionary ideals of R . Show: (a) If I 2 F , then I 1 2 F and I (I 1 ) 1 . (b) If I R is a fractionary ideal, then `(R=I ) < 1, R I 1 and `(I 1 =R ) < 1. (c) The following conditions are equivalent: (i) R is a Gorenstein ring (ii) I = (I 1 ) 1 for all I 2 F, (iii) `(R=I ) = `(I 1 =R ) for all I 2 F, I R . 1
m
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3. The canonical module. Gorenstein rings
!
Let R S be a faithfully at homomorphism of Noetherian rings, and C a nite R -module. Show the following are equivalent: (a) C S is a canonical module of S (b) C is a canonical module of R , and for every prime ideal Spec S the bre S= S, = R , is Gorenstein.
3.3.30.
p
q
p
q
q
\
q
2
Let k be a eld, and R a k-algebra which is Cohen{Macaulay and admits a canonical module. Let K be a eld, and suppose that either R is a nitely generated k-algebra or K is a nitely generated extension eld of k. Show that !R k K is a canonical module of R k K . Hint: apply 2.1.11. 3.3.31.
3.4 Gorenstein ideals of grade 3. Poincare duality The Hilbert{Burch theorem 1.4.17 identi es perfect ideals of grade 2 as the ideals of maximal minors of certain matrices. For Gorenstein ideals of grade 3 there exists a similar `structure theorem' due to Buchsbaum and Eisenbud 65]. Let R be a Noetherian local ring. An ideal I R is a Gorenstein ideal (of grade g) if I is perfect and ExtgR (R=I R ) = R=I . Note that if R is Gorenstein and I is perfect, then I is Gorenstein if and only if R=I is Gorenstein. This follows from 3.3.7(b). To describe the structure theorem we recall a few facts from linear algebra: let R be a commutative ring, and F a nite free R -module. An R -module homomorphism ' : F ! F is said to be alternating if with respect to some (and therefore with respect to any) basis of F and the corresponding dual basis F , the matrix of ' is skew-symmetric and all its diagonal elements are 0. Suppose now that ' is alternating, choose a basis of F and the basis dual to this, and identify ' with the corresponding matrix (aij ). If rank F is odd, then det ' = 0, and if rank F is even, there exists an element pf(') 2 R , called the Pfaan of ', which is a polynomial function of the entries of ', such that det(') = pf(')2. For more details about Pfaans we refer the reader to 48], Ch. IX, x5, no. 2. We set pf(') = 0 if rank F is odd. Just like determinants, Pfaans can be developed along a row. Denote by 'ij the matrix obtained from ' by deleting the i-th and j -th rows and columns of ' then for all i, pf(') =
Xn
(;1)i+j ;1(i j )aij pf('ij )
j =1
((i j ) is the sign of j ; i). From now on we assume that the rank of F is odd, and consider the matrix derived from ' by repeating the i-th row
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3.4. Gorenstein ideals of grade 3. Poincare duality
and column as indicated in the following picture 0 0 ai1 ain
B ;ai1 @ ...
=B B
'
;ain
1 CC CA :
Expansion with respect to the rst row of yields the equations 0 = ; pf() =
Xn j =1
(;1)j aij pf('j )
for i = 1 . . . n, where 'j is the matrix obtained from ' by deleting the j -th row and column. In other words, if we let : R ! F be the linear map de ned by (p1 . . . pn) (with respect to the given basis) where pj = (;1)j pf('j ), j = 1 . . . n, are the submaximal Pfaans of ', then we obtain the complex F. ('): 0
' ;! R ;! F ;! F ;! R ;! 0:
Theorem 3.4.1 (Buchsbaum{Eisenbud). Let (R ) be a Noetherian local m
ring. (a) Suppose n 3 is an integer, F a free R-module of rank n, and ' : F F an alternating map of rank n 1 whose image is contained in F . Then n is odd. Moreover, if Pf(') denotes the ideal generated by the submaximal Pfaans of ', then grade Pf(') 3. If grade Pf(') = 3, then F. (') is acyclic and Pf(') is a Gorenstein ideal. (b) Conversely, let I be a Gorenstein ideal of grade 3. Then there exist a free module F of odd rank and an alternating homomorphism ' : F F such that F. (') is a minimal free R-resolution of R=I. In particular, any Gorenstein ideal of grade 3 is minimally generated by an odd number of Pfaans.
;
!
m
!
Part (a) of the theorem is a consequence of the Buchsbaum{Eisenbud acyclicity criterion 1.4.13 and the following simple observation relating the ideal of (n ; 1)-minors of ' to the ideal of submaximal Pfaans. Lemma 3.4.2. Let (R ) be a Noetherian ring, F a free R-module of rank n, and ' : F ! F an alternating map of rank n ; 1. Then Rad Pf(') = Rad In;1('), and n is odd. Proof. For all i j = 1 . . . n, we denote by ij the matrix which is obtained from ' by deleting the i-th row and j -th column. Then In;1(') is generated by the elements det( ij ), and since pf('i)2 = det( ii) it follows right away that a power of Pf( ') is contained in In;1('). V Conversely, we consider the matrix of n;1 ' with entries det( ij ), m
122
3. The canonical module. Gorenstein rings
V
i j = 1 . . . n. It follows from Exercise 1.6.25 that rank n;1 ' = 1 since, by assumption, rank ' = n 1. Now 1.4.11 implies that all 2V minors of n;1 ' are zero. Therefore, since ' is skew-symmetric, we have det( ij )2 = det( ij ) det( ji) = det( ii) det( jj ) = pf('i)2 pf('j )2 for all i j = 1 . . . n. This implies that a power of In;1(') is contained in Pf('). Finally, since In;1(') = 0, we conclude that Pf(') = 0. This is only possible if n is odd.
;
;
6
6
For the proof of part (b) of 3.4.1 a little excursion to resolutions with algebra structures is needed. Let R be a commutative ring, and let P. :
;! P2 ;!d P1 ;!d P0 = R ;! 0
be an acyclic complex of projective modules. We may consider P as a graded module equipped with an endomorphism d : P ! P of degree ;1 satisfying d d = 0. The question is whether there can be de ned an associative multiplication on P satisfying the following rules: (a) PpPq Pp+q for all p q 0 (b) 1 2 P0 acts as the unit element, i.e. 1a = a1 = a for all a 2 P (c) ab = (;1)(deg a)(deg b)ba for all homogeneous elements a b 2 P (d) aa = 0 for all homogeneous elements a 2 P of odd degree (e) d (ab) = (da)b + (;1)deg aa(db) for all homogeneous elements a b 2 P . An example of a complex admitting such a multiplication is the Koszul complex. Unfortunately not all nite projective resolutions can be given an algebra structure with these properties. Avramov 25] found obstructions for this, and gave explicit examples of nite projective resolutions which fail to have such a structure. Nevertheless, if we do not insist on the associativity of the multiplication, we surprisingly have Theorem 3.4.3 (Buchsbaum{Eisenbud). Any projective resolution P with P0 = R admits a (possibly non-associative) multiplication satisfying the conditions (a){(e). Proof. We form the tensor product P P of complexes, and de ne the second symmetric power S2 (P ) of P to be .
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.
P )=U where U is the graded submodule of P P which is generated by the elements a b ; (;1)(deg a)(deg b)b a with homogeneous a b 2 P , and the elements aa with homogeneous a 2 P of odd degree. Let d again denote the di erential of P P then d (U ) U . This implies that d induces S2 (P.) = (P.
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a di erential on S2 (P ), so that S2 (P ) inherits a complex structure. We .
.
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3.4. Gorenstein ideals of grade 3. Poincare duality
claim (and this is crucial for the proof) that the homogeneous components S2(P )k of this complex are all projective modules. Indeed, we have
M Pi Pj ) Tk
.
S2 (P.)k = (
8< 0 V Tk = : 2 Pk=2 S2 (Pk=2)
where
i+j =k i<j
if k is odd, if k is of the form 4n + 2, if k is of the form 4n. Thus S2 (P ) is a complex of projective R -modules which coincides with P in degrees 0 and 1. Therefore there exists a complex homomorphism : S2 (P ) ! P extending the identity in degrees 0 and 1, and we may assume that is chosen such that its restriction to R Pk is just the natural homomorphism to Pk . For all homogeneous elements a b 2 P we denote by ab the image of a b under the composition of the maps P P ;! S2 (P ) ;! P , and extend this multiplication by linearity to all other elements of P . It is clear that it has all the desired properties. Suppose now we are given a Noetherian local ring R and a Gorenstein ideal I R of grade g. Let F : 0 ;! Fg ;! ;! F1 ;! F0 ;! 0 be the minimal free resolution of R=I . The dual complex F is a minimal free resolution of ExtgR (R=I R ) = R=I , and hence must be isomorphic to F . Such an isomorphism is unique up to homotopy, and we are now choosing one which is derived from the multiplicative structure on F as given by 3.4.3. Observe that the multiplication de nes maps Fi Fg;i ! Fg = R which in turn induce R -module homomorphisms si : Fi ! Fg;i . For i = 0 . . . g we let s if i 0 1 mod 4, ti = ;i s if i 2 3 mod 4. i Proposition 3.4.4. t : F ! F is an isomorphism of complexes. In particular, si : Fi ! Fg;i is an isomorphism for i = 0 . . . g. Proof. We denote by d the di erential of F . Let a 2 Fi and b 2 Fg+1;i. Then ab = 0, and therefore 0 = d (ab) = d (a)b + (;1)i ad (b), or d (a)b = (;1)i+1ad (b). It follows that si;1(d (a))(b) = d (a)b = (;1)i+1ad (b) = (;1)i+1si (a)(d (b)) = (;1)i+1d (si(a))(b): .
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Thus si;1 d = (;1)i+1d si which implies that t is a homomorphism of complexes. The induced homomorphism H0 (t ): R=I ! R=I must be an isomorphism since t0 = s0 is an isomorphism. Since t extends H0 (t ) it must be an isomorphism as well. We are now ready to prove 3.4.1(b): let .
.
.
F. : 0
.
' ;! F3 ;! F2 ;! F1 ;! R ;! 0
be the minimal free resolution of R=I equipped with a multiplication as in 3.4.3. Let e1 . . . en be a basis of F2 . Then, as we have just seen, s2(e1 ) . . . s2(en ) is a basis of F1 , and we may choose basis elements f1 . . . fn of F1 such that fi = s2 (ei ) for i = 1 . . . n. Then ei fj = ij g for all i j = 1 . . . n where g is a basis P element of F3 and ij denotes the Kronecker symbol. Let '(ei ) = nj=1 aij fj we claim that (aij ) is skew-symmetric, and all its diagonal elements are 0. To see this, notice that ej '(ei) = aij g. Therefore, aij (g) = (ej '(ei )) = '(ej )'(ei): The claim follows since is injective, and since '(ei)'(ej ) = ;'(ej )'(ei) and '(ei )'(ei) = 0 according rules. P to the multiplication Now let (g) = ni=1 ai ei since F = F , is isomorphic to the transpose of , and we conclude that I = (a1 . . . an). On the other hand, rank ' = n ; 1 and grade Pf(') = grade In;1('), by 3.4.2. Now grade In;1(') 3 since F becomes split exact after localizations at prime ideals of height 2. Thus part (a) of Theorem 3.4.1 implies that P F (') is acyclic. In particular it follows that Ker ' is generated by ni=1 pi ei where, up to signs, the pi are the submaximal Pfaans of '. Thus we have Pf(') = I , as desired. .
.
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.
Poincare duality. Buchsbaum and Eisenbud 65] remark that the multiplication de ned on F. induces a multiplication on Tor. (k R=I ) giving
it the structure of an associative graded alternating algebra. They further point out that in view of 3.4.4, Tor (k R=I ) is a Poincare algebra if I is a Gorenstein L ideal. Recall that an associative graded alternating algebra A = gi=0 Ai is a Poincare algebra if for all i = 0 . . . g the A0homomorphisms Ai ! HomA (Ag;i Ag ), a 7! 'a with 'a(b) = ab, are isomorphisms. Notice that if R is regular, then there is a natural isomorphism between Tor (k R=I ) and the Koszul homology H (R=I ) = H (x R=I ) where x is a minimal set of generators of the maximal ideal of R see 1.6.9. It can be shown that this is an isomorphism of algebras. In particular, the Koszul homology H (R ) of a Gorenstein ring is a Poincare algebra. This is one direction of the theorem of Avramov and Golod 32] which asserts that a Gorenstein ring is characterized by its Koszul homology. Their theorem .
0
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3.4. Gorenstein ideals of grade 3. Poincare duality
complements the result 2.3.11 of Tate and Assmus according to which the Koszul algebra of a complete intersection is an exterior algebra. We will present their proof which is independent of the above considerations. Theorem 3.4.5 (Avramov{Golod). Let (R k) be a Noetherian local ring, and let n = emb dim R ; depth R. The following conditions are equivalent: (a) R is a Gorenstein ring (b) H (R ) is a Poincare algebra (c) the k-linear map Hn;1 (R ) ! Homk (H1 (R ) Hn(R )) induced by the multiplication on H (R ) is a monomorphism. We begin with a few preliminary remarks. Suppose t = depth R > 0. By 1.2.2 (choose M = R and N = ), there exists an R -regular element y1 2 n 2 . Hence by induction on t we may construct an R -sequence y = y1 . . . yt such that y is part of a minimal system of generators of . By 1.6.13, one has H (R ) = H (R=y R ) as graded k-vector spaces. Inspecting this isomorphism we see that it is actually a k-algebra isomorphism. On the other hand, R is Gorenstein if and only if R=y R is too. Thus we may assume that depth R = 0. Let x = x1 . . . xn be a minimal system of generators of , and K = K (x) the Koszul complex of this sequence then, by de nition, H (R ) = H (K ). Note that K is a Poincare algebra: let e1 . . . en be an R basis of K1 then, in the terminology of Section 1.6, the eI , I f1 . . . ng, jI j = i, form an R-basis of Ki , and we have eI ^ eJ = (I J )e1 ^^ en for J f1 . . . ng, jJ j = n ; i. Here (I J ) = 1 if I \ J = , and 0 otherwise. This clearly proves that the maps !i : Ki ! HomR (Kn;i Kn ), !i (a) = 'a with 'a (b) = a ^ b, are isomorphisms as asserted. We denote by d the di erential of K . Then d and its dual anticommute with ! . In other words, we have !i;1 di = (;1)i;1 Hom(dn;i+1 Kn ) !i for all i = 0 . . . n. This equation is stated in 1.6.10, the only di erence being that there Kn is identi ed with R . It follows that the isomorphisms !i induce isomorphisms ! e i : Hi(R) ! H n;i(R) where we identify H n;i(R) n ; i with H ((K ) ), and where (K ) = HomR (K Kn ). Consider the diagram m
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m
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m
m
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m
. .
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;;;;! Homk (Hn;i (R ) Hn(R )) ?? ??
Hi (R )
i
~y yi i H n;i (R ) ;;;;! HomR (Hn;i(R ) Kn ) Here the upper map i is induced by the multiplication on H (R ). We have just seen that !e i is an isomorphism. The lower map i is the natural homomorphism which assigns to a homology class , an (n ; i)-cycle in !i
.
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(K ), the (well de ned) homomorphism i ( ) 2 HomR (Hn;i(R ) Kn) with i( )( a) = (a), a 2 Kn;i a cycle. Next note that Hn (R ) = Soc Kn Kn . We de ne i to be Hom(Hn;i(R ) ) where : Soc Kn ! Kn is the natural inclusion. It is clear that i is an isomorphism. In fact, since Hn;i(R ) is annihilated by , any homomorphism Hn;i (R ) ! Kn necessarily maps Hn;i(R ) into Soc Kn . We leave it to the reader to check the commutativity of the diagram. In conclusion we have that i is a mono-, epi-, or isomorphism if and only if i is too. We will determine the kernel and cokernel of i. Lemma 3.4.6. Let B denote the boundaries of K . Then for any i we have .
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a long exact sequence
n;i 0 ;! Ext1R (Ki;1 =Bi;1 Kn ) ;! H i (R ) ;;! HomR (Hi (R ) Kn) 1 ;! ExtR (Bi;1 Kn) ;!
The short exact sequence 0 ! Hi (R ) ! Ki =Bi ! Bi;1 ! 0 gives rise to the long exact sequence Proof.
i 0 ;! HomR (Bi;1 Kn ) ;! HomR (Ki =Bi Kn ) ;! HomR (Hi (R ) Kn) 1 ;! ExtR (Bi;1 Kn) ;! It is immediate to see that Im n;i = Im i, so that the sequence
H i (R )
;;! HomR (Hi (R ) Kn) ;! Ext1R (Bi;1 Kn ) ;! n;i
is exact. The module U of i-cycles of (K ) whose homology classes belong to Ker n;i is the module of homomorphisms 2 (Ki ) for which jZi = 0 (Zi cycles in Ki ). Therefore U is isomorphic to HomR (Bi;1 Kn ). Under this identi cation Ker n;i equals U=V where V is the module of homomorphisms : Bi;1 ! Kn which can be extended to Ki;1 . This means that Ker n;i = Ext1R (Ki;1=Bi;1 Kn ). In order to complete the proof of 3.4.5 we need the following Lemma 3.4.7. Let (R k) be a Noetherian local ring of depth 0. If Ext1R (k R ) = 0, then R is Gorenstein. Proof. The hypothesis implies that the functor HomR ( R ) is exact on the category of R -modules of nite length. This yields `(HomR (M R )) = `(M ) `(HomR (k R )) for any R -module of nite length M . Now assume dim R > 0. Then `(R= n ), and so `(HomR (R= n R )), tends to in nity with n. On the other hand, HomR (R= n R ) = 0 : n. 2 Since 0 : 0 : is an ascending chain of ideals, and since R .
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is Noetherian, this chain stabilizes. Consequently, `(HomR (R= n R )) is bounded, a contradiction. Thus R is a zero dimensional ring for which HomR ( R ) is an exact functor. Hence R is an injective R -module, and so R is Gorenstein by de nition. End of the proof of 3.4.5: We have already accomplished the reduction to the case depth R = 0. (a) ) (b): Since R is Gorenstein and Kn = R , all Ext groups in the exact sequence 3.4.6 vanish, and so i is an isomorphism for all i. (b) ) (c) is trivial. (c) ) (a): By assumption n;1 is injective, and this implies that n;1 is injective. Thus it follows from 3.4.6 that Ext1R (K0=B0 R ) = 0. Now 3.4.7 completes the proof since K0 =B0 = k. Corollary 3.4.8. Let R be a Gorenstein local ring which is not a complete intersection. Then H1 (R )n;1 = 0 for n = emb dim R ; dim R. Proof. Suppose the vector subspace H1 (R )n;1 of Hn;1(R ) is not zero. Then H1 (R )n 6= 0 since 1 : H1 (R ) ! Homk (Hn;1 (R ) Hn(R )) is an isomorphism. This contradicts 2.3.15. m
Exercises 3.4.9. Let k be a eld and I
k
X1 X2 X3 ]] the ideal generated by the polynomials X12 X22 , X12 X32 , X1 X2 , X1 X3 , X2 X3 . By 3.2.11 it is a Gorenstein ideal of grade 3. Compute its free resolution (as a k
X1 X2 X3 ]]-module). 3.4.10. Let (R k) be a Cohen{Macaulay local ring with canonical module !R , and x a minimal set of generators of . We denote by H. (M ) the Koszul homology of an R -module M with respect to x. Recall that H. (M ) is an H. (R )module. Let n = embdim R dim R . Show that for all i, 0 i n, the k-linear map Hi (R ) Homk (Hn i (!R ) Hn (!R )) which is induced by the scalar multiplication of H. (R ) on H. (!R ) is an isomorphism.
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3.5 Local cohomology. The local duality theorem The canonical module was introduced by Grothendieck in connection with the local duality theorem which relates local cohomology with certain Ext functors. We will describe this approach to the canonical module in this section. First local cohomology functors will be introduced, and it will be shown that the depth and the dimension of a module can be expressed in terms of their vanishing and non-vanishing. We end with the local duality theorem. Let (R k) be a Noetherian local ring and M an R -module. Denote by ; (M ) the submodule of M consisting of all elements of M with m
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support in f g. That is, m
f 2 M:
= 0 for some k 0g: Let F = (Ik )k 0 be a family of ideals of R such that Ij Ik for all j > k . Then F de nes a topology on R see 270], Section 8. F gives the -adic topology on R if and only if for each Ik there is a j 2 N such that j Ik , and for each i there is an l 2 N such that Il i . It is clear that for any such family one has ; (M ) = fx 2 M : Ik x = 0 for some k 0g: Let x = x1 . . . xn be a sequence of elements in R generating an -primary ideal. We set xk = xk1 . . . xkn for all k 0: The family (xk ) gives the -adic topology on R , and so ; (M ) = x m
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f 2 M : (xk)y = 0 for some k 0g: Noting that HomR (R=I M ) = fx 2 M : Ix = 0g for any ideal I of R , we obtain natural isomorphisms ; (M ) = lim ;! HomR (R= k M) = lim ;! HomR (R=(xk) M): ; (M ) = y m
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Proposition 3.5.1. ; ( ) is a left exact additive functor. Proof. The additivity of ; ( ) is trivial. We show that ; ( ) is left exact. If
0 ;! M1 ;! M2 ;! M3 m
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is exact, then we have a sequence 0 ;! ; (M1 ) ;! ; (M2) ;! ; (M3), where 0 = ; ( ) = j; (M ) and 0 = ; ( ) = j; (M ) . It is obvious that 0 is injective. Let x 2 Ker 0 then (x) = 0, and so there exists y 2 M1 such that x = (y). Since x 2 ; (M2), there exists an integer k 0 such that k x = 0. It follows that k (y) = ( k y) = 0. But is injective, and so y 2 ; (M1) and 0 (y) = x. De nition 3.5.2. The local cohomology functors, denoted by H i ( ), are the right derived functors of ; ( ). In other words, if I is an injective resolution of the R -module M , then H i (M ) = H i (; (I )) for all i 0. Remarks 3.5.3. (a) Let M be an R -module then H 0 (M ) = ; (M ) and H i (M ) = 0 for i < 0. (b) If I is an injective R -module, then H i (I ) = 0 for all i > 0. (c) For any R -module M and all i 0 one has 0
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H i (M ) = lim ExtiR (R= k M ) = lim ExtiR (R=(xk) M ) m
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where x is a sequence in R generating an -primary ideal. (d) A short exact sequence of R -modules 0 ;! M1 ;! M2 ;! M3 ;! 0 gives rise to a long exact sequence 0 ;! ; (M1 ) ;! ; (M2) ;! ; (M3) ;! H 1 (M1) ;! ;! H i;1(M3) ;! H i (M1) ;! H i (M2) ;! Only (c) needs some explanation: ; lim is an exact functor see 318], ! Theorem 2.18. Therefore if I is an injective resolution of M , then k H i (M ) H i (HomR (R= k I )) = H i (lim = lim R (R= I )) ;! Hom ; ! = lim ;! ExtiR (R= k M): Note that E(k) if = , ; (E (R= )) = 0 otherwise m
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see 3.2.7 and part (1) of the proof of 3.2.13. Using the structure of the minimal injective resolution E (M ) of M given in 3.2.9, we conclude that ; (E (M )) is a complex of the form 0 ;! E (k) ( M ) ;! E (k) ( M ) ;! ;! E (k)i ( M ) ;! This entails Proposition 3.5.4. Let (R k) be a Noetherian local ring and M a nite .
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R-module. (a) The modules H i (M ) are Artinian. (b) One has H i (M ) = 0 if and only if i < depth M. (c) If R is Gorenstein, then n H i (R ) = E (k) for i = dim R, 0 otherwise. ^ (d) Let N denote the -adic completion of an R-module N. Then ^) H i (M ) = H i (M ) R R^ = H i^ (M for all i 0: m
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(a), (b) and (c) follow from the structure of ; (E (M )) and the fact that depth M = inf fi : i( M ) 6= 0g. (d) As H i (M ) is Artinian, it is the direct limit of submodules Uj of nite length. For each Uj one has Uj R R^ = Uj , and so H i (M ) = lim ;!(Uj R R^ ) = (lim ;! Uj ) R R^ = H i (M) R R:^ Using the R -atness of R^ , we get ^ ^ j M^ ) H i (M ) R R^ = lim ;!i ExtiR (R= j M) R R^ = lim ;! ExtiR^ (R= = H ^ (M^ ): Proof.
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Local cohomology and the Koszul complex. Our next goal is to construct a more explicit complex whose cohomology gives us H . (M ). Let x = x1 . . . xn be a system of parameters of R . For all l 0 we get a m
commutative diagram
' ;;;;! K1 (xl ) ?? ?? (l ) 1
K1 (xl+1)
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with '(1l) (ei ) = xi ei for i = 1 . . . n. (In both Koszul complexes we denote the natural basis of K1 = R n by e1 . . . en ). V (l) i (l ) Let 'i = '1 then '(l) : K (xl+1) ;! K (xl ) is a complex homomorphism see 1.6.8. We denote by 'l : K (xl ) ;! K (xl+1) the dual complex homomorphism. This can be done for each l , and so we obtain a direct system of complexes. Thus we may form the complex lim ;! K (xl ): .
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On the other hand, one de nes a complex C : 0 ;! C 0 ;! C 1 ;! ;! C n ;! 0 M Ct = Rxi xi xit .
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;! Rx x to be the homomorphism (;1)s;1 nat: Rx x ! (Rx x )x if fi1 . . . it g = fj1 . . . jbs . . . jt+1g and 0 otherwise. Rxi xit
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The complex C is called the modi ed Cech complex. In the usual 0 % complex, C is replaced by 0 and the homological degree is shifted Cech by 1. Proposition 3.5.5. lim ;! K (xl ) = C . Proof. For all l 0 we de ne a complex homomorphism 1 l : K (xl ) ;! C by lt ((ej ^ ^ ejt )) = (xj xj xjt )l V here (ej ^ ^ ejt ) is anVelement of the basis of ( t R n) which is dual to the standard basis of t R n. A straightforward calculation shows .
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(i) l is indeed a complex homomorphism, (ii) l = l+1 'l for all l 0, and therefore the family (l ) induces a complex homomorphism : ; lim K (xl ) ;! C , ! (iii) is an isomorphism. Note that for (iii) one essentially has to verify that Rxi xit is the limit of the direct system (Fi )i 0 in which Fi = R for all i and the map Fi ! Fi+1 is just multiplication by xi xit . The importance of these complexes results from Theorem 3.5.6. Let M be an R-module. Then H i (M ) = H i (M R C ) = lim ;! H i (xl M) for all i 0: .
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functor, and hence commutes with cohomology. In order to prove the rst isomorphism, we show that the functors H i ( C ) are the right derived functors of ; ( ). Identifying M Rxi with Mxi we have .
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R C ) = Ker(M ;! M Mx ): j =1 The kernel consists of all m 2 M for which there exist integers lj , l H 0 (M
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j = 1 . . . n, such that xjj m = 0, and this set is obviously equal to ; (M ). Since C . is a complex of at R -modules, the exact sequence of R m
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0 ;! M1 ;! M2 ;! M3 ;! 0 yields the exact sequence 0 ;! M1 R C ;! M2 R C ;! M3 R C ;! 0 from which we obtain the long exact sequence 0 ;! H 0 (M1 R C ) ;! H 0 (M2 R C ) ;! H 0 (M3 R C ) ;! H 1(M1 R C ) ;! H 1 (M2 R C ) ;! It remains to show that H i (I R C ) = 0 for i > 0 and any injective R -module I . Of course we may assume that I is indecomposable. Let I = E (k) and a 2 E (k). Then for j = 1 . . . n there exist integers lj > 0 such that xljj a = 0, and so E (k) C i = 0 for i > 0. Next assume I = E (R= ), 6= then there exists j 2 f1 . . . ng such that xj 2= . We claim that multiplication by xj on E (R= ) is an isomorphism. It is certainly a monomorphism since Ass E (R= ) = f g (see 3.2.7) and since xj 2= . The submodule xj E (R= ) of E (R= ) is an injective .
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module since xj E (R= ) = E (R= ), and hence is a direct summand of E (R= ). But E (R= ) is indecomposable, and so xj E (R= ) = E (R= ). We can then de ne a homotopy of the complex E (R= ) R C : l : E (R= ) R C l ;! E (R= ) R C l;1 is de ned on the component E (R= )xi xil ;! E (R= )xj xjl; to be (;1)s;1 nat, if fj1 . . . jl;1g = fi1 . . . ibs . . . il g and is = j , and 0 otherwise. It is easily veri ed that is a contracting homotopy, that is, the identity and the zero-map of the complex are homotopic via . This implies that E (R= ) R C is exact. p
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Grothendieck's theorems. We are now in the position to prove the follow-
ing important vanishing theorem. Theorem 3.5.7 (Grothendieck). Let (R k) be a Noetherian local ring m
and M a nite R-module of depth t and dimension d. Then (a) H i (M ) = 0 for i < t and i > d, (b) H t (M ) = 0 and H d (M ) = 0.
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We rst note the following rule which will be used several times in the proof of (a) and (b): Let ' : (R k) ! (R 0 0 k0 ) be a local ring homomorphism such that R 0 is an 0 -primary ideal. Then for any R 0-module M one has (3) H i (M ) = H i 0 (M ) for all i 0: Of course, on the left hand side of this formula M is considered as an R -module. In fact, if = (x) with x = x1 . . . xn and if x0 = '(x1) . . . '(xn), then C R M = C 0 R0 M , where C and C 0 are the complexes of 3.5.6 de ned with respect to x and x0 . The isomorphism (3) follows from 3.5.6. (a) We only need to prove that H i (M ) = 0 for i > d . The other part of statement (a) has already been shown in 3.5.4. Let R ! R 0 = R= Ann M be the canonical epimorphism. Then M is an R 0-module with dim M = dim R 0. Using (3) we may therefore assume that dim R = dim M = d . Let x = x1 . . . xd be a system of parameters of R , and let C be the complex 3.5.6 de ned with respect to x. Then C i = 0 for i > d , and so H i (M ) = H i (M R C ) = 0 for i > d see 3.5.6. (b) We proceed by induction on t in order to show that H t (M ) 6= 0. If t = 0, then 0 6= Soc M H 0 (M ). Now suppose t > 0 then there exists an M -regular element x 2 . The exact sequence x 0 ;! M ;! M ;! M=xM ;! 0 yields the exact sequence 0 = H t;1 (M ) ;! H t;1 (M=xM ) ;! H t (M ). By our induction hypothesis we have H t;1 (M=xM ) 6= 0 this implies H t (M ) 6= 0. Proof.
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Finally we show that H d (M ) 6= 0. Using 3.5.4 and the fact that dim M^ = dim M for the -adic completion M^ of M , we may assume that R is complete. Let 2 Supp M with dim M = dim R= . Then dim M= M = dim M = d , and we get an exact sequence of R -modules 0 ;! U ;! M ;! M= M ;! 0 m
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inducing the exact sequence H d (M ) ;! H d (M= M ) ;! H d +1(U ). According to (a) we have H d +1(U ) = 0, and so, if H d (M= M ) 6= 0, then H d (M ) 6= 0. As M= M is an R= -module we may as well assume, by (3), that R is a domain and dim R = dim M . Any complete Noetherian domain has a Noether normalization: there exists a regular local subring (S ) such that R is a nite S -module see A.22. In particular, the extension ideal R is -primary. Again using (3) we may replace R by S , and so may assume that R itself is regular. Let K be the fraction eld of R , and let : M ! K R M be the canonical homomorphism. We set U = Ker and N = Im . Then we obtain the exact sequence (4) 0 ;! U ;! M ;! N ;! 0 and, as a consequence of Exercise 1.4.18, an exact sequence (5) 0 ;! N ;! R s ;! W ;! 0 where s = rank M = rank N and consequently dim W < dim R = d . As dim W < d , (5) yields the exact sequence H d (N ) ;! H d (R s ) ;! H d (W ) = 0: p
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We have H d (R s ) = H d (R )s = E (k)s (see 3.5.4) and so H d (N ) 6= 0. Finally, from the exact sequence (4) it follows that H d (M ) 6= 0. The next theorem is known as the local duality theorem. Theorem 3.5.8 (Grothendieck). Let (R k) be a Cohen{Macaulay comm
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plete local ring of dimension d. Then for all nite R-modules M and all integers i there exist natural isomorphisms
H i (M ) = HomR (ExtdR;i(M !R ) E (k)) and ExtiR (M !R ) = HomR (H d ;i (M ) E (k)): m
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The rst isomorphisms result from the second by Matlis duality 3.2.13. For the proof of the second isomorphisms note that both sides vanish for i < 0 see 3.5.7. For i 0 we set T i( ) = HomR (H d ;i ( ) E (k)). It is clear that T 0( ) is a contravariant left exact functor which maps
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direct sums to direct products. Hence there exists an R -module C such that T 0( ) = HomR ( C ) see 318], Theorem 3.36. It follows that C = T 0(R ). As H d (R ) is an Artinian module, Matlis duality 3.2.13 implies that C is a nite R -module. In order to conclude the proof we will show that the functors T i( ) are the right derived functors of T 0( ), and that C = !R . Remark 3.5.3 implies immediately that the functors T i ( ) are strongly connected (see 318], p. 212). Thus the T i( ) are the right derived functors of T 0( ), once we have shown that T i(F ) = 0 for every free R -module F and all i 1. The functors T i ( ) map direct sums to direct products, and so it suces to show that T i(R ) = 0 for i 1, or equivalently that H i (R ) = 0 for i < d . This however follows from 3.5.7 since R is Cohen{Macaulay. Summing up we have (6) HomR (H i (M ) E (k)) = ExtdR;i(M C ) for all i and all R -modules M . Now 3.5.7 yields i = 0, and therefore Exti (k C ) k for i = d , H i (k) = k0 for = R for i > 0, 0 for i 6= d , by (6). Thus it follows from the remark after 3.3.1 that C = !R . Grothendieck's duality theorem has the following often applied variant: Corollary 3.5.9. Let (R k) be a Cohen{Macaulay local ring of dimension m
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d which is the homomorphic image of a Gorenstein local ring. Then R has a canonical module, and for all nite R-modules M and all integers i there exist natural isomorphisms
H i (M ) = HomR (ExtdR;i(M !R ) E (k)): m
For the proof we apply 3.5.4, 3.3.5, and Exercise 3.2.14. Then ^ !R^ ) E (k)) H i (M ) = H i^ (M^ ) = HomR^ (ExtdR^ ;i(M = HomR (ExtdR;i(M !R ) E(k)): Remark 3.5.10. Let (R k) be a complete local ring. The proof of 3.5.8 shows that the functor HomR (H d ( ) E (k)) is representable, even if R is not a Cohen{Macaulay ring. In other words, there exist a unique R -module KR (in the proof of 3.5.8 this module was denoted by C ) and a canonical isomorphism HomR (H d (M ) E (k)) = HomR (M KR )
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for all R -modules M . Of course, KR = !R if R is Cohen{Macaulay. Even in the more general situation when the ring is not Cohen{Macaulay, the module KR is often called the canonical module of R . Its properties have been investigated by Aoyama 10]. Schenzel 329] has introduced the canonical module KM of an R -module M . The local duality theorem combined with 3.5.7 allows us to generalize 3.3.10(d). Corollary 3.5.11. Let (R k) be a Cohen{Macaulay local ring of dimenm
sion n with canonical module !R , and M a nite R-module of depth t and dimension d. Then (a) ExtiR (M !R ) = 0 for i < n d and i > n t, (b) ExtiR (M !R ) = 0 for i = n d and i = n t, (c) dim ExtiR (M !R ) n i for all i 0. ^ !R^ ) for all i 0, since (!R )b = Proof. We have ExtiR (M !R )b = ExtiR^ (M !R^ (see 3.3.5). Under completion depth and dimension of a module are preserved. We may therefore assume that R is complete, and so (a) and
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Exercises Let (R ) be a Noetherian local ring, and M a nite R -module. Prove: (a) If i 1 ( M ) = 0 and i( M ) 6= 0, then H i (M ) 6= 0. (b) Suppose inj dim M = 1 then i( M ) 6= 0 for all i > dim M . 3.5.13. Find a Noetherian local ring (R ) of dimension d and depth t with (a) H i (R ) 6= 0 for i = t . . . d , (b) H i (R ) = 0 for i 6= t and i 6= d . 3.5.14. Let (S k) be a complete Cohen{Macaulay local ring, (R k) a residue class ring of S , and M a nite R -module. Show that HomR (H i (M ) ER (k))
= HomS (H i (M ) ES (k)) and derive the following version of the local duality theorem: for all integers i there exist natural isomorphisms HomR (H i (M ) ER (k))
d = dim S: = ExtdS i (M !S ) Hint: See the rst step in the proof of 3.5.7, and use 3.1.6. 3.5.12.
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Let (R k) be a regular local ring of dimension d 2. Let E be a nite R -module which is locally free on the punctured spectrum of R . That is, E is free for all Spec R , = . Show (a) `(H i (E )) < for all i < d , (b) the R -dual E of E is again locally free on the punctured spectrum of R , and H i (E ) = 0 for i = 0 1, (c) H i+1 (E ) = HomR (H d i (E ) E (k)) for i = 1 . . . d 2. 3.5.15.
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3.6 The canonical module of a graded ring For a graded ring R we de ne the canonical module in the category of graded R -modules and establish the graded version of the local duality theorem. Under certain restrictive assumptions on R the degrees of the generators in a minimal set of generators of the canonical module are uniquely determined, and one de nes the a-invariant of R to be the smallest of these degrees, multiplied by ;1. We adopt the assumptions and notation of Section 1.5. Thus R will be a Noetherian graded ring, and M0(R ) the category of graded R -modules. M0(R) is an Abelian category which has direct sums and direct products see 1.5.19. Likewise limits and colimits exist in M0(R ). As we have already mentioned in Section 1.5, M0(R ) has enough projectives. Our next concern will be to show that M0(R ) has enough injectives as well. Injective modules.
A graded R -module M is called injective if it is an injective object in M0 (R ). One sees easily that this is the case if and only if the functor HomR ( M ): M (R ) ;! M (R ) 0 0 is exact. A injective module M need not be injective (in the category M(R ) see 3.6.5). Just as in the category of all R -modules, one calls an extension N M of graded R -modules essential if for any graded submodule 0 6= U M one has U \ N 6= 0. If, in addition, N 6= M , the extension is called a proper essential extension. Similarly as in the non-graded case (see 3.2.2) one shows Proposition 3.6.1. A graded module is injective if and only if it has no proper essential extension.
We now prove that any graded R -module has a injective hull. In analogy to the de nition in the non-graded case, E is called a injective hull of M if it is injective and a essential extension of M . Theorem 3.6.2. Any graded R-module M admits a injective hull, and any
two injective hulls of M are isomorphic.
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3.6. The canonical module of a graded ring
We embed M into a (not necessarily graded) injective R -module I . According to 3.1.8 this is possible. Similarly as in the proof of 3.2.4 we consider the set = N : M N I M N is essential . We de ne a partial order on by setting N1 N2 if N1 is a graded submodule of N2 . Zorn's lemma applied to this set yields a maximal essential extension M E with E I . Suppose E is not injective then E has a proper essential extension E E 0 by 3.6.1. As I is injective, there exists an R -module homomorphism ' : E 0 I (not necessarily homogeneous), extending the inclusion E I . We claim that ' is injective. In fact, assume that there is a non-zero element x Ker ', say x = xr + + xs , with xi homogeneous of degree i, r s, and xr = 0. We show by induction on s r that there exists a homogeneous element a R such that ax E and ax = 0. Since ' E is injective, this gives a contradiction. If s r = 0, x is homogeneous, and the assertion follows since the extension E E 0 is essential. Now suppose that s r > 0. We choose a homogeneous element a R such that axr E 0 . Let Proof.
S f S
;
6
;
!
2
j
g
6
2
2
; 2 nf g x0 = x ; xr = xr+1 + + xs : If ax0 = 0, then ax = axr 2 E n f0g and we are done. Otherwise ax0 6= 0, and by our induction hypothesis we may choose a homogeneous element b 2 R such that bax0 2 E and bax0 = 6 0. Then bax = bax0 + baxr 2 E, and bax = 6 0. Next let Ee = Im '. As ' is injective we may give Ee a natural graded structure (Ee i = '(Ei) for all i 2 Z). Then E Ee is a proper essential extension with Ee I , contradicting the maximality of E . 2
The uniqueness of the injective hull is proved as in the non-graded case. We denote the injective hull of a graded R -module M by E (M ) or ER (M ) . The preceding theorem implies in particular that any graded R -module N has a injective resolution. That is, there exists a complex
;! I 0 ;! I 1 ;! I 2 ;! with injective modules I i such that H 0 (I ) = N and H i (I ) = 0 for i > 0. I. : 0
.
.
Given such a injective resolution I of N we have .
Exti (M N ) = H i (HomR (M I . )) R
for all i 0 and all graded R -modules M . We omit the proofs of the following two results which have their analogues in 3.2.6, 3.2.7, 3.2.8, and 3.2.9, and which can be proved along the same lines as in the corresponding local case.
138
3. The canonical module. Gorenstein rings
Theorem 3.6.3. Let R be a Noetherian graded ring. Then (a) Ass E (M ) = Ass M for all M 2 M0(R ), (b) E 2 M0(R ) is a indecomposable injective module if and only if E = E (R= )(n)
2
2
p
for some graded prime ideal R and some integer n Z, (c) every injective module can be decomposed into a direct sum of indecomposable injective modules, and this decomposition is unique up to homogeneous isomorphism. Proposition 3.6.4. Let R be a Noetherian graded ring, and M a graded R-module. Consider the minimal injective resolution p
0 ;! M ;! E 0 (M ) ;! E 1(M ) ;! of M (which is obtained recursively by setting E i (M ) = E (Im d i;1 )). Then, for every graded prime ideal of R and for every integer i 0, the Bass number i ( M ) equals the number of graded R-modules of the form E (R= )(n), n 2 Z, that appear in E i (M ) as direct summands. For any M 2 M0 (R ) we denote by inj dim M the injective dimension. Theorem 3.6.5. Let R be a Noetherian graded ring and M 2 M0 (R ). Then (a) inj dim M inj dim M + 1, (b) if M is injective, then inj dim M = 1 if and only if 2 Ass M for d0
d1
p
p
p
some non-graded prime ideal
p
p
of R.
For the proof of 3.6.5 we will use Proposition 3.6.6. Let M 2 M0(R ) and a non-graded prime ideal in R. Then 0( M ) = 0, and i+1( M ) = i ( M ) for every integer i 0. The proof of this proposition is already given in 1.5.9, where we actually prove more than is stated in that theorem itself. Proof of 3.6.5. (a) We may assume that inj dim M = t < 1. Let 2 Spec R we want to prove that i( M) = 0 for i t + 2. This is certainly true when is a graded prime ideal see 3.6.4. Now suppose that is not a graded prime ideal. Then i+1( M ) = i ( M ) by 3.6.6 and i ( M ) = 0 for i t + 1, hence the assertion follows. (b) inj dim M = 1 happens if and only if 1( M ) 6= 0 for some nongraded prime ideal of R . But 1 ( M ) = 0 ( M ), and so 1 ( M ) 6= 0 if and only if 2 Ass M . Corollary 3.6.7. Let R be a Noetherian graded ring and a graded maximal p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
ideal of R. Then
m
) = E (R= ): Proof. By 3.6.3 we have Ass E (R= ) = f g, and so 3.6.5 implies that E (R= ) is injective as an object in M(R ). Since 0 ( E (R= )) = 1 (see E (R=
m
m
m
m
m
m
m
139
3.6. The canonical module of a graded ring
3.6.4), we conclude that E (R= ) is indecomposable in M(R ), and hence by 3.2.6 it must be isomorphic to E (R= ). m
m
The canonical module of a graded ring. Recall from 1.5 that a graded
ring is a local ring if it has a unique maximal ideal, that is, a graded ideal which is not properly contained in any graded ideal 6= R . De nition 3.6.8. Let (R ) be a Cohen{Macaulay local ring of dimension d . A nite graded R -module C is a canonical module of R if there exist homogeneous isomorphisms 0 for i 6= d , Exti (R= C ) = R= for i = d . R m
m
m
m
For a nite graded R -module M it may happen that there exists a homogeneous isomorphism M = M (i) with i 6= 0. To avoid this phenomenon, one has to require that R has no homogeneous units of positive degree. For a local ring (R ) this is the case if and only if is maximal (in the usual sense) see 1.5.16. Proposition 3.6.9. Let (R ) be a Cohen{Macaulay local ring, and C be m
m
a canonical module of R. Then (a) C is a canonical module of R, (b) C is uniquely determined up to homogeneous isomorphism, provided is maximal. Proof. (a) We need to show that C is a canonical module of R for all Spec R . First, let Spec R be a graded prime ideal then . The de nition of the canonical module implies that C is a canonical module of R , and so C is a canonical module of R see 3.3.5. Now let Spec R be a non-graded prime ideal. Then i+1( C ) = i ( C ), by m
m
p
2
p
2
p
p
p
m
m
2
m
p
p
3.6.6, and the assertion follows again. (b) Let C 0 be another canonical module. Remark 3.3.17 implies that HomR (C C 0) is a projective module of rank 1, and hence, as a graded module, is free (see 1.5.15). Therefore, HomR (C C 0) = R (i) for some i 2 Z. This implies HomR (C C 0(;i)) = R . Let ' 2 HomR (C C 0(;i)) be an element corresponding to 1 under this identi cation. Then, since HomR (C C 0(;i)) = HomR (C C 0) by Exercise 1.5.19(f), it follows from (a) and 3.3.4(c) that ' is locally an isomorphism. But then ' is a homogeneous isomorphism, and we have R= = ExtdR (R= C 0 (;i)) = ExtdR (R= C ) = ExtdR (R= C 0)(;i) = (R= )(;i): Therefore i = 0, and C = C 0. Example 3.6.10. Let R = kX1 . . . Xn] be a polynomial ring over a eld, and assign to the indeterminates the degree deg Xi = ai > 0 for i = 1 . . . n. p
p
m
m
m
m
m
p
140
3. The canonical module. Gorenstein rings
The maximal ideal of R is = (X1 . . . Xn), and the Koszul complex of X1 . . . P Xn yields a homogeneous free resolution of R= whose last term that Exti (R= R ) = 0 for i 6= n, is R (; ni=1 ai ). From this one concludes P n n and Ext (R= R ) = (R= )( i=1 ai ). In other words, the canonical P n module of R is R (; i=1 ai ). Proposition 3.6.11. Let (R ) be a Cohen{Macaulay local ring with m
m
m
m
m
canonical module !R . The following conditions are equivalent: m
(a) R is a Gorenstein ring (b) !R = R (a) for some integer a 2 Z. Proof. R is Gorenstein if and only if !R is locally free. By 1.5.15(d) this is the case if and only if !R = R (a) for some a 2 Z. The number a occurring in 3.6.11 is a numerical invariant of the Gorenstein local ring (R ), provided is maximal. In the case of a positively graded algebra over a eld it will be given a special name see 3.6.13. Let (R ) be a Cohen{Macaulay local ring with canonical module !R . The canonical module is a graded module, and by 1.5.15, every minimal system of homogeneous generators of !R has exactly ((!R ) ) elements. In analogy to the local case we de ne this number to be the type of R , and denote it by r(R ). In view of 3.6.11 it is clear that R is Gorenstein if and only if r(R ) = 1. For the sake of completeness we list a few change of rings properties of the canonical module. While part (a) of the next proposition follows easily from the results proved so far, it is best to use the change of rings spectral sequence Extp (k Extq (S !R )) =) Extn (k !R ) S R R p m
m
m
m
for (b) (see 318], 11.66) the reduction we used for the corresponding local result 3.3.7 is only possible if homogeneous systems of parameters are available. Proposition 3.6.12. Let (R ) be a Cohen{Macaulay local ring with canonical module !R .
m
(a) If is a graded prime ideal of R, then !R = (!R )( ) up to a shift. (b) Let ' : (R ) ! (S ) be a ring homomorphism of Cohen{Macaulay ( )
p
p
local rings satisfying m
p
n
(i) '(Ri) Si for all i 2 Z, (ii) '( ) , (iii) S is a nite graded R-module. Then !S = ExttR (S !R ), where t = dim R ; dim S. Example 3.6.10 and 3.6.12 imply that any Cohen{Macaulay positively graded algebra over a eld admits a canonical module. Following Goto and Watanabe 134] we de ne: m
n
141
3.6. The canonical module of a graded ring
De nition 3.6.13. Let k be a eld, and R a Cohen{Macaulay positively graded k-algebra. Then a(R ) = ; minfi : (!R )i 6= 0g is called the a-invariant of R . As a consequence of 3.6.12 we have Corollary 3.6.14. Let R be a Cohen{Macaulay local ring with canonical module !R , and let x = x1 . . . xn be an R-sequence of homogeneous elements with deg xi = ai for i = 1 . . . n. Then
!R=xR = (!R =x!R )(
Xn i=1
ai ):
In particular, if k be a eld, and RP a Cohen{Macaulay positively graded k-algebra, then a(R=xR ) = a(R ) + ni=1 ai . Proof. The Koszul complex ) is a graded free R -resolution of PnK.(ax). RFrom R=xR , and Kn (x R ) = R ( 3.6.12 we obtain i=1 i
;
Xn
!R=xR = ExtnR (R=xR !R ) = H n (x !R ) = (!R =x!R )(
i=1
ai):
Examples 3.6.15. (a) A graded polynomial ring R = kX1P . . . Xn] over a eld k with deg Xi = ai > 0 has the a-invariant a(R ) = ; ni=1 deg ai . (b) Let k be a eld, and R ! S a homomorphism of Cohen{Macaulay positively graded rings with maximal ideals and , respectively. Suppose the homomorphism satis es the conditions of 3.6.12(b), and suppose further that S has a nite free homogeneous R -resolution 0 ;! Ft ;! Ft;1 ;! ;! F0 ;! S ;! 0 L where t = dim R ; dim S . Write Ft = i2Z R (;ai ) then a(S ) = a(R ) + maxfai : i 2 Zg. This is proved exactly as in the special case 3.6.14. m
n
Local Duality. Our nal objective is to derive the graded version of the
local duality theorem. We begin with Matlis duality. Let (R ) be a Noetherian local ring then R0 is local with maximal ideal 0. We consider R0 as a graded ring by de ning (R0)i = 0 for i 6= 0. Similarly any R0-module may be considered a graded R0 -module concentrated in degree 0. Moreover, if M is a graded R -module, it may be viewed as a graded R0-module as well. Thus we can de ne M _ = HomR (M ER (R0= 0)): A priori, M _ is a graded R0 -module whose grading is given by (M _ )i = HomR (M;i ER (R0= 0)) m
m
0
0
0
m
0
m
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3. The canonical module. Gorenstein rings
for all i 2 Z. But it is obvious that M _ has a natural structure as a graded R -module. The Noetherian local ring (R ) is said to be complete if (R0 0 ) is complete. If (R ) is complete and M is a nite graded R -module, then all homogeneous components Mi of M are complete R0-modules (since they are nite R0-modules). Proposition 3.6.16. Let (R ) be a Noetherian complete local ring. Then (a) the additive contravariant functor ( )_ : M0 (R ) ;! M0(R ) is exact (b) M _ = HomR (M R _ ) for all graded R-modules M (c) one has R _ = ER (R= ). Proof. (a) is obvious. (b) We de ne ' : HomR (M HomR (R E )) ! HomR (M E ) by setting '( )(x) = (x)(1) for all 2 Hom(M HomR (R E )) and all x 2 M . It is readily seen that ' is an isomorphism. (c) It follows from (a) and (b) that HomR ( R _ ) is an exact functor, and so R _ is injective. R _ is indecomposable since R __ = R . Note further that (R= )_ = R= . This is clear in the case where R= =k ; 1 is a eld, and it is easy to see in the case R= k t t ], since then = all homogeneous components of kt t;1 ] are isomorphic to k. Therefore the canonical epimorphism R ! R= yields a monomorphism R= = (R= )_ ;! R _ , and the assertion follows from 3.6.3. M 2 M0(R ) is called Artinian if every descending chain of graded submodules terminates. The homogeneous socle of a graded R -module M is de ned to be Soc M = HomR (R= M ). It is an R= -module and can be viewed as a graded submodule ofLM . As an R= -module it is free (see Exercise 1.5.20), and so Soc M = i2I (R= )(ai). If M is Artinian, then Soc M can have only nitely many summands (R= )(ai). Hence we may write n M Soc M = (R= )(ai ): m
m
m
m
m
0
0
0
m
m
m
m
m
m
m
m
m
m
m
m
m
i=1
As in the proof of 3.2.13 we conclude that M is Artinian if and only if there exist an integer n 0 and integers a1 . . . an such that (7)
M
M R_(ai): n
i=1
Let A0(R ) denote the full subcategory of M0 (R ) consisting of all Artinian R -modules and F0(R ) the full subcategory of all nite graded R -modules. Theorem 3.6.17 (Matlis duality for graded modules). Let (R ) be a Noetherian complete local ring, and let M 2 F0 (R ) and N 2 A0(R ). m
Then
143
3.6. The canonical module of a graded ring
(a) M _ 2 A0(R ) and N _ 2 F0(R ), (b) M __ = M and N __ = N, (c) the functor ( )_ : F0(R ) ;! A0(R ) establishes an anti-equivalence of categories.
Using (7) one proves the theorem in the same way as 3.2.13. For example, in order to show (b) we set E = ER (R0= 0 ). Then we have Proof.
0
(M __ )i = HomR (HomR (Mi E ) E ) = Mi 0
m
0
by Matlis duality see 3.2.13. Now let (R ) be a Noetherian local ring. For M 2 M0(R ) we de ne m
H i (M ) = lim Exti (R= R
;!
m
k M
m
)
it is called the i-th local cohomology functor. H 0 ( ) is left exact and the functors H i ( ), i 0, are the right derived functors of H 0 ( ). Remark 3.6.18. Assume in addition that the maximal ideal of R is maximal. Since for all i and j , and all M 2 M0 (R ), we have Exti (R= j M ) = ExtiR (R= j M ) = ExtiR (R = j R M ), we see that R in this case H i (M ) = HRi (M ). Theorem 3.6.19 (The local duality theorem for graded modules). Let (R ) be a Cohen{Macaulay complete local ring of dimension d. Then (a) R has a canonical module !R , and !R = (H d (R ))_ , (b) for all nite graded R-modules M and all integers i there exist natural m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
homogeneous isomorphisms
(H i (M ))_ = ExtdR;i (M !R ): m
The proof follows as in the non-graded case see 3.5.8. Let R be a positively graded k-algebra. Then it follows from 3.6.19(a) that a(R ) = maxfi : H d (R )i 6= 0g: If in addition dim R = 0, then a(R ) = maxfi : Ri 6= 0g. m
Exercises Let R be a Noetherian graded ring. (a) For 2 Spec R show that R is Gorenstein if and only if R is Gorenstein. (b) Show the following conditions are equivalent: (i) R is a Gorenstein ring (ii) R is a Gorenstein ring for all graded prime ideals 2 Spec R (iii) R( ) is a Gorenstein ring for all graded prime ideals 2 Spec R . 3.6.20.
p
p
p
p
p
p
p
144
3. The canonical module. Gorenstein rings
(c) Let (R ) be local ring. Deduce that R is Gorenstein if and only if R is Gorenstein. 3.6.21. The purpose of this exercise is to re-prove a few results of Goto and Watanabe. L Let R be a graded ring, d a positive integer. The ring R (d ) = i Z Rid is called the d-th Veronese subring of R . It is a graded subring of R with grading (d ) ( R (d ) ) i = LRid for all i 2 Z. For j = 0 . . . d ; 1 we consider the graded R -modules Mj = i Z Rid +j with grading (Mj )i = Rid +j for all i 2 Z. We assume that R is Noetherian, and for (d) and (e) that it is a positively graded algebra over a eld. Show: L (a) R = dj=01 Mj (as R (d ) -module). In particular, R (d ) is a direct summand of R R (d ) is Noetherian, and the Mj are nite R (d ) -modules. (Hint: Compare the proof of 1.5.5.) (b) R is Cohen{Macaulay if and only if all Mj are maximal Cohen{Macaulay R (d ) -modules. L (c) If R is Cohen{Macaulay, then !R d
= i Z(!R )id . (d) If R is Gorenstein and a(R ) = bd + j , 0 j d ; 1, then !R d
= Mj (b). (e) If R is Gorenstein and a(R ) 0 mod d , then R (d ) is Gorenstein. Is a(R ) 0 mod d if R and R (d ) are Gorenstein? 3.6.22. Let k be a eld. We consider R = k X1 . . . Xn ] as a graded k-algebra with deg Xi = ai > 0 for i = 1 . . . n. Determine all Veronese subrings of R which are Gorenstein. 3.6.23. Let R be a homogeneous k-algebra, k a eld. Express a(R (d ) ) in terms of a(R ) and d .
m
m
2
2
;
( )
2
( )
Notes Grothendieck introduced the canonical module (often called dualizing module) and proved the local duality theorem. A comprehensive presentation of this theory including local cohomology is given in 143]. Equally fundamental is the famous paper of Bass 37]. The interested reader can nd some more historical background there. We were guided by the books of Kaplansky 231] and Matsumura 270] in Sections 3.1 and 3.2. In Sections 3.3 and 3.5 we follow partly the lecture notes of Herzog and Kunz 159]. Part of Section 3.5 has been inuenced by the notes of P. Roberts 311]. In particular the description % complex has been taken from this source. In Section of the modi ed Cech 3.6 we follow to a large extent the papers 134] by Goto and Watanabe and 109] by Fossum and Foxby. The main result 3.1.17 of Section 3.1 is due to Bass 37]. The characterization 3.2.10 of Gorenstein rings in terms of the type was rst proved by Bass 36]. Bass 37] gives a list of other equivalent conditions for the Gorenstein property. In 115] Foxby proves the following conjecture of Vasconcelos, which is a remarkable characterization of Gorenstein rings. Suppose (R k) is a Noetherian local ring of dimension d containing a m
145
Notes
eld then R is a Gorenstein ring if d ( R ) = 1. (We will give a proof of this result in 9.6.3.) The main theorem of Matlis duality and the structure theorem for injective modules are proved by Matlis in 269], and can also be found in the more general framework of Abelian categories in Gabriel 122]. Theorem 3.3.10 which characterizes the canonical module is taken from 159]. However in the proof given here we do not use local cohomology. Theorem 3.3.6 on the existence of the canonical module is independently due to Foxby 113] and Reiten 307], and 3.3.19 is a theorem of Murthy 281]. It says that every factorial Cohen{Macaulay ring with a canonical module is Gorenstein. In 372] Ulrich proves a certain converse of Murthy's theorem: any Gorenstein ring which is a factor ring of a regular local ring and which is locally a complete intersection in codimension one can be realized as a specialization of a Cohen{Macaulay factorial domain. For a while it had been open whether or not there exist non{Cohen{ Macaulay factorial local rings. Such examples were found by Bertin 41] (see 6.4.7) and also by Fossum and Grith 111]) in characteristic p and by Freitag and Kiehl 118] in characteristic 0. There are two remarkable extensions of the theory of the canonical module in its basic form as presented here: Sharp introduced Gorenstein modules in 337] as those nite modules G whose Cousin complex provides a (minimal) injective resolution for G. A Gorenstein module shares many properties with the canonical module. It is a Cohen{Macaulay module of nite injective dimension whose type and rank, however, may be bigger than one. We refer the reader to the papers on Cousin complexes and Gorenstein modules 336], 337], 338], and 341] by Sharp, and the article 113] by Foxby. It is shown in 339] that for a Noetherian local ring admitting a canonical module !R , any Gorenstein module is a direct sum of copies of !R . However there exist Cohen{Macaulay local rings not admitting a canonical module. A rst example of a one dimensional ring with this property was given by Ferrand and Raynaud 106], and an example of a factorial Cohen{Macaulay ring without a canonical module is due to Ogoma 294]. In 392] Weston gives an example of a ring with a Gorenstein module of rank 2, admitting no canonical module. The second extension of the basic concept gives a duality theory even for non{Cohen{Macaulay rings. In this theory the canonical module has to be replaced by the so-called dualizing complex, and duality is obtained in the derived category. We refer the reader to the book of Hartshorne 151]. A more elementary account of the theory can be found in Sharp 340]. As a consequence of the structure theorem 3.4.1 for Gorenstein ideals of grade three, these ideals have an odd number of generators. This had been observed before by J. Watanabe 386]. He uses linkage arguments m
146
3. The canonical module. Gorenstein rings
in his proof. Linkage had already been considered in 1945 by Apery 11] and by Gaeta 123] in 1952. It has become popular as a result of the paper of Peskine and Szpiro 298]. Linkage provides a technique to construct large and interesting classes of perfect ideals or of Gorenstein ideals whose structure is well understood. Of particular interest are the ideals in the linkage class of a complete intersection, called licci ideals. The simplest examples are the so-called Northcott ideals 288] and the Gorenstein ideals de ned in 156]. More important is the fact that perfect ideals of grade two 298] and Gorenstein ideals of grade three 386] are in the linkage class of a complete intersection. They are in a sense the archetypes of licci ideals as shown by Huneke and Ulrich 220]. For further study of linkage theory we refer the reader to the papers of Huneke 210], 213], Huneke and Ulrich 219], 221], 222], Kustin and Miller 251], 252], 254], and Ulrich 374], 373]. The height 3 monomial Gorenstein ideals have been completely classi ed by Bruns and Herzog 58]. There have been attempts to obtain structure theorems for nonGorenstein ideals of grade 3 or even for ideals of grade higher than 3. The next case of interest is Gorenstein ideals of grade 4. As a rst approach to the problem one may try to classify the Tor-algebras of these ideals I , i.e. TorR (k R=I ) when I is an ideal in the local ring R with residue class eld k. For ideals I such that proj dim R=I = 3 this has been done by Weyman 393] in characteristic 0 and by Avramov, Kustin, and Miller 33] in all characteristics. The next case of interest is Gorenstein ideals of grade 4. At the moment a general structure theorem for these ideals seems to be out of range. Kustin and Miller 253] succeeded in classifying their Tor-algebras. A remarkable result, valid for ideals of arbitrary grade, is due to Kunz 246]: if I is a Gorenstein ideal, then (I ) 6= grade I + 1. Duality theory is a classical and fundamental topic in algebraic geometry, and has also several algebraic aspects we have not even touched upon. We must content ourselves with a list of keywords and references: Riemann{Roch theorem, Serre duality, modules of regular di erentials, residue symbols, trace maps see Hartshorne 151], 152], Kunz 247], Kunz and Waldi 250], Lipman 259], Scheja and Storch 324], 325], and Serre 333].
4 Hilbert functions and multiplicities
The Hilbert function H (M n) measures the dimension of the n-th homogeneous piece of a graded module M . In the rst section of this chapter we study the Hilbert function of modules over homogeneous rings, prove that it is a polynomial for large values of n, and introduce the Hilbert series and multiplicity of a graded module. The next section is devoted to the proof of Macaulay's theorem which describes the possible Hilbert functions. The third section complements these results by Gotzmann's regularity and persistence theorem. The Hilbert function behaves quite regular, even for graded, nonhomogeneous rings. Such rings will be considered in the fourth section, where we will also investigate the Hilbert function of the canonical module. The passage to the associated graded ring with respect to a ltration allows us to extend some concepts for graded rings like `Hilbert function' or `multiplicity' to non-graded rings, and leads to the Hilbert{Samuel function and the multiplicity of a nite module with respect to an ideal of de nition. We shall study basic properties of ltrations and their associated Rees rings and modules, and sketch the theory of reduction ideals. Finally we prove Serre's theorem which interprets multiplicity as the Euler characteristic of a certain Koszul homology. 4.1 Hilbert functions over homogeneous rings We begin by studying numerical properties of nite graded modules over a graded ring R . Our standard assumption in this section will be that R0 is an Artinian local ring, and that R is nitely generated over R0 . Notice that for each nite graded R -module M , the homogeneous components Mn of M are nite R0-modules, and hence have nite length. De nition 4.1.1. Let M be a graded R -module whose graded components Mn have nite length for all n. The numerical function H (M ): Z ! Z with n) = `(Mn) for all n 2 Z is the Hilbert function, and HM (t) = P HH((M M n)tn is the Hilbert series of M . n2Z For the rest of this section we will assume that R is generated over R0 by elements of degree 1, that is, R = R0 R1]. Recall that such a ring is
said to be homogeneous.
147
148
4. Hilbert functions and multiplicities
! 2
We say that a numerical function F : Z Z is of polynomial type (of degree d ) if there exists a polynomial P (X ) QX ] (of degree d ) such that F (n) = P (n) for all n 0. By convention the zero polynomial has
degree ;1. We de ne the dierence operator on the set of numerical functions by setting (F )(n) = F (n + 1) ; F (n) for all n 2 Z. Notice that maps polynomial functions to polynomial functions, lowering the degree of non-zero polynomials by 1. The d times iterated -operator will be denoted by d . We further set 0F = F . Lemma 4.1.2. Let F : Z ! Z be a numerical function, and d 0 an integer. The following conditions are equivalent: (a) d F (n) = c, c = 0, for all n 0 (b) F is of polynomial type of degree d.
6
(b) ) (a) is easy. We prove the other implication by induction on d . The assertion is trivial for d = 0. Now assume that d > 0, and d F (n) = d ;1(F (n + 1) ; F (n)) = c, c 6= 0, for all n 0. By the induction hypothesis it then follows that there exist an integer n0 and a polynomial P (X ) 2 QX ] of degree d ; 1 such that P F (n + 1) ; F (n) = P (n) for all n n0 . Then F (n + 1) = F (n0) + nk=n P (k), and this last sum is a polynomial function in n of degree d . After these preparations we can state the main result of this section as follows: Theorem 4.1.3 (Hilbert). Let M be a nite graded R-module of dimension d. Then H (M n) is of polynomial type of degree d ; 1. Proof. We prove the theorem by induction on the dimension d of M . First note that there is a chain 0 = N0 N1 Nn = M of graded submodules of M such that for each i we have Ni+1=Ni = (R= i )(ai) where i is a graded prime ideal. Indeed, we may assume that M 6= 0. Choose 1 2 Ass(M ). The prime ideal 1 is graded see 1.5.6. There exists a graded submodule N1 M with N1 = (R= 1 )(a1). If N1 6= M we choose 2 2 Ass(M=N1 ). Then there exists a graded submodule N2 M with N2 =N1 = (R= 2 )(a2). If N2 6= M , we may proceed in the same way. But M is Noetherian, and so this process terminates eventually. Now, since the Hilbert P function is additive on short exact sequences, it follows that H (M n) = i H ((R= i)(ai ) n). Notice that d is the supremum of the numbers dim R= i . Hence the theorem will follow once we have shown it for M = R= , a graded prime ideal. (Here, of course, one has to observe that the polynomials describing Hilbert functions are zero or have positive leading coecients since their values are non-negative for n 0. As a consequence, the degree of the sum of such polynomials is the maximum of their degrees.) Proof.
0
p
p
p
p
p
p
p
p
p
p
p
149
4.1. Hilbert functions over homogeneous rings
R= = 0, then is the unique graded maximal ideal 0 L If Rdim n of R , where 0 is the maximal ideal of R0. It follows that p
p
n>0
m
H (R= n) = 0 for n > 0. If dim R= > 0, we may choose a homogeneous element x R= , x = 0, of degree 1. Here we use the fact that R is homogeneous. The m
2
p
6
p
p
exact sequence
0 ;! (R= )(;1) ;! R= ;! R=(x ) ;! 0 gives the equation H (R= n) = H (R= n + 1) ; H (R= n) = H (R=(x ) n + 1): As dim R=(x ) = d ; 1, our induction hypothesis implies that H (R= n) is of polynomial type of degree d ; 2. Hence if d > 1, then 4.1.2 implies that d ;1H (R= n) = d ;2(H (R= n)) is a non-zero constant function for n, and if d = 1, then d ;1H (R= n) = H (R= n) = H (R= 0) + Pn large i=1 H (R=( x) i) is constant for large n since H (R=( x) i) = 0 for i 0. Again this constant is not zero since H (R= 0) 6= 0. Now 4.1.2, (a) ) (b) yields the assertion. Hilbert's original proof (see 171]) makes use of his syzygy theorem 2.2.14. This approach will be described in 4.1.13. The next lemma clari es which polynomials in QX ] have integer values. Lemma 4.1.4. Let P (X ) 2 QX ] be a polynomial of degree d ; 1. Then x
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
the following conditions are equivalent: (a) P (n) Z for all n Z (b) there exist integers a0 . . . ad ;1 such that
2
2
P (X ) =
X X + i d ;1 i=0
ai
i
:
is trivial. For the converse observe that the P polynomials ; 2 N), (a) d ;1 ;X +i form a Q -basis of Q X ]. Therefore P ( X ) = i=0 ai i i ; ; ;
Proof. (b) X +i ,i
with ai 2 Q. The identity X +ii+1 ; Xi+i = Xi;+1i immediately implies that ai = i P (;i ; 1) 2 Z for i = 0 . . . d ; 1. De nition 4.1.5. Let M be a nite graded R -module of dimension d . The unique polynomial PM (X ) 2 QX ] for which H (M n) = PM (n) for all n 0 is called the Hilbert polynomial of M . We write PM (X ) =
X
d ;1 i=0
(;1)d ;1;ied ;1;i
X + i i
:
150
4. Hilbert functions and multiplicities
Then the multiplicity of M is de ned to be e d > 0, e(M ) = `0(M ) if if d = 0.
Remark 4.1.6. The higher iterated Hilbert functions Hi (M n), i 2 N, of a nite graded R -module M are de ned recursively as follows: X H0 (M n) = H (M n) and Hi (M n) = Hi;1 (M j ) j n
for i > 0. Occasionally the functions Hi (M ) are called the sum transforms of H (M ). It follows from 4.1.2 and 4.1.3 that Hi (M n) is of polynomial type of degree d + i ; 1, where d = dim P M. ;In particular, for all n 0 there is a representation H1 (M n) = di=0 ai n+i i with ai 2 Z, and it is easy to see that ad = e(M ). Another formula for the multiplicity will be given in 4.1.9. Theorem 4.1.3 together with the next lemma yields a structural result about Hilbert series. P Lemma 4.1.7. Let H (t) = antn be a formal Laurent series with integer coecients, and ai = 0 for i 0. Further, let d > 0 be an integer. Then the following conditions are equivalent: (a) there exists a polynomial P (X ) QX ] of degree d 1 such that P (n) = an for large n (b) H (t) = Q(t)=(1 t)d where Q(t) Zt t;1] and Q(1) = 0. Proof. Assume (a), and set F (n) = an for all n Z. Then
2 2
;
6
;
2 (1 ; t)d H (t) = d F (n ; d )tn n and it follows from 4.1.2 that (1 ; t)d H (t) 2 Zt t;1] . We set Q(t) = P d n n F (n ; d )t . Suppose Q(1) = 0 then X X 0 = d F (n ; d ) = (d ;1F (n + 1 ; d ) ; d ;1F (n ; d )) = d ;1F (m) X
n
n
for large m. This contradicts 4.1.2, and thus proves the implication (a) ) (b). The converse is proved similarly. Corollary 4.1.8. Let M 6= 0 be a nite graded R-module of dimension d. Then there exists a unique QM (t) 2 Zt t;1] with QM (1) 6= 0 such that Q (t) HM (t) = M d : (1 ; t) P Moreover, if QM (t) = i hi ti , then minfi : hi 6= 0g is the least number such that Mi 6= 0.
151
4.1. Hilbert functions over homogeneous rings
Proof. The rst part of the assertion is clear for d = 0, and for d > 0 it follows from 4.1.3 and 4.1.7. In order to prove the second part multiply both sides of HM (t) = QM (t)=(1 ; t)d by (1 ; t)d and compare coecients.
In the next proposition we show how one can recover the coecients
ei of the Hilbert polynomial of a module M from QM . We will denote by P (i) the i-th formal derivative of an element P Zt t;1]. Proposition 4.1.9. Under the assumptions of 4.1.8 the following formulas hold: Q(i) (1) ei = M i! for i = 0 . . . d 1. Moreover, e(M ) = QM (1).
2
;
Proof.
We write
HM (t)
; X (;i!1) (1Q;Mt(1))d;i = (1D;(tt))d d ;1
i
(i)
i=0
P where D(t) = QM (t); d ;1 (;1) Q(i) (1)(1;t)i is the remainder of the Taylor i=0
i
expansion of QM (t) up to degree d ; 1. The element D(t) 2 Zt t;1] is (j )(1) = 0 for j = 0 . . . d ; 1. It follows that the divisible by (1 ; t)d since DP ;1( (;1)i Q(i) (1)=(1 ; t)d ;i) coincide for large coecients of HM (t) and di=0 M i! n. Hence d ;1 X (;1)i Q(Mi) (1) = X P (n)tn M i! (1 ; t)d ;i i=0
i!
M
n 0
since the coecients of both series are polynomial functions in n which are equal for large n (and hence must be equal for all n). Expanding the left hand side of the equation as a power series, and comparing coecients we get ei = Q(Mi) (1)=i!. Finally, by what we have just proved, P we have e(M) = e0 = QM (1) if d > 0, and, if d = 0, e(M ) = `(M ) = n H (M n) = HM (1) = QM (1), since in this case HM (t) = QM (t). Corollary 4.1.10. Assume that in addition to P the assumptions of 4.1.8 the module M is Cohen{Macaulay. Let QM (t) = hi ti . Then hi 0 for all i. Moreover, ei 0 for all i if Mj = 0 for all j < 0. Proof. Without loss of generality we may assume that the residue class eld of R0 is in nite. Otherwise we resort to a standard trick: we replace R by R 0 = R R R0(Y ) and M by M 0 = M R R0(Y ) where Y is an indeterminate over R0 , and where R0(Y ) is the local ring R0Y ]S , S being the multiplicatively closed set of polynomials P (Y ) 2 R0Y ] which have at least one unit among their coecients. The natural ring 0
0
152
4. Hilbert functions and multiplicities
homomorphism R0 ! R0(Y ) is local and at, and its bre is the residue class eld of R0(Y ), namely the eld k(Y ) of rational functions over k = R0= 0 . If we assign the degree 0 to the elements of R0 (Y ), then both R 0 and M 0 are naturally graded, and because of atness, M 0 is a Cohen{Macaulay R 0-module of dimension d with HM 0 (t) = HM (t) see 2.1.7 and 1.2.25. Moreover, R00 = R0(Y ), and hence has an in nite residue class eld. In view of 4.1.9 it suces to show that hi 0 for all i. We prove the assertion by induction on d . If d = 0, then QM (t) = HM (t) (see 4.1.8), and so all coecients of QM (t) are non-negative. Suppose L now that d > 0. The unique homogeneous maximal ideal =L 0 n>0 Rn of R does not belong to Ass M , and the ideal I = n>0 Rn, generated by the elements of R1 , is -primary. Thus, since R= is in nite there exists an element a 2 R1 which is M -regular see 1.5.12. Let N = M=aM then N is a Cohen{Macaulay graded R -module of dimension d ; 1, and the exact sequence m
M
m
M
M
0 ;! M (;1) ;! M ;! N ;! 0 gives the equation (1 ; t)HM (t) = HN (t). It follows that QM (t) = QN (t), which by our induction hypothesis yields the conclusion. Remark 4.1.11. The arguments in the previous proof show the following notable result: suppose M is a nite graded R -module, and x is an M -sequence of elements of degree 1 then QM (t) = QM=(x)M (t). Hilbert's theorem tells us that the Hilbert function of a nite graded module is a polynomial function for large n. We will determine from which integer n onwards this happens. Proposition 4.1.12. P Let M 6= 0 be a nite graded R-module of dimension d, and QM (t) = bi=a hiti with hb 6= 0. Then H (M b ; d ) 6= PM (b ; d ) and H (M i) = PM (i) for all i b ; d + 1. ; Proof. For i = a . . . b we set Hi (t) = hi ti =(1 ; t)d and Pi(n) = hi n;di+;d1;1 . P Then Hi (t) = 1n=i Pi(nP )tn, but since Pi (n) = 0 for n = i ; (d ; 1)P . . . i ; 1 we even have Hi (t) = 1n=i;(d ;1) Pi(n)tn. Furthermore PM (n) = bi=a Pi (n) P for all n 2 Z. For n b ; d + 1 one has H (M n) = bi=a Pi (n), whereas P H (M b ; d ) = bi=;a1 Pi(b ; d ) and Pb(b ; d ) 6= 0. Thus H (M b ; d ) 6= PM (b ; d ). In Section 4.4 we will give a homological interpretation of the di erence between the Hilbert function and the Hilbert polynomial. a
Hilbert series and free resolutions. The Hilbert series of a graded module
can be expressed in terms of its graded resolution.
153
4.1. Hilbert functions over homogeneous rings
Lemma 4.1.13. Let M be a nite graded R-module of nite projective dimension, and let
0 ;!
M j
; ;! ;! M R(;j) ;! M ;! 0
R ( j )pj
0j
j
be a graded free resolution of M. Then HM (t) = SM (t)HR (t)
P where SM (t) = ij (;1)i ij tj .
In particular, if R = kX1 . . . Xn] is the polynomial ring over the eld k, then S (t) HM (t) = M n : (1 t)
;
For the proof we simply note that thePHilbert function is additive on short exact sequences, so that HM (t) = i (;1)iij HR(;j ) (t). Taking into account that HR(;j ) (t) = tj HR (t), we obtain the required formula. If R = kX1 . . . Xn] is the polynomial ring, then H (R i) equals the number of monomials; in degree i. One easily Pproves ; by induction on n that this number is nn+;i;1 1 , whence HR (t) = i nn+;i;1 1 ti = 1=(1 ; t)n. Proof.
Corollary 4.1.14. Let R = kX1 . . . Xn] be a polynomial ring over a eld k, and let M be a nite graded R-module of dimension d. Then (a) SM (t) = (1 t)n;d QM (t), (b) n d = inf i : SM(i)(1) = 0 , ; (c) SM(n;d +i)(1) = ( 1)n;d n;di +i ei .
;
Proof.
; f
;
6 g
(a) and (b) are immediate while (c) follows from (a) and 4.1.9.
We conclude this section with an application to a special class of graded rings. Let R = kX1 . . . Xn] be a polynomial ring over a eld k, I R a graded ideal. We say that R=I has a pure resolution of type (d1 . . . dp) if its minimal resolution has the form 0 ;! R (;dp )p ;! ;! R (;d1 ) ;! R ;! R=I ;! 0: Note that d1 < d2 < < dp. Theorem 4.1.15. Suppose R=I is Cohen{Macaulay and has a pure resolution of type (d1 . . . dp ). Then 1
(a) i = (;1)i+1
Y j 6=i
dj
(dj ; di )
Yp (b) e(R=I ) = p1! di : i=1
154
4. Hilbert functions and multiplicities
As R=I is Cohen{Macaulay, the Auslander{Buchsbaum formula 1.3.3 (in conjunction with 1.5.15) implies that p = dim R ; dim R=I therefore, with 0 = 1, d0 = 0, we have Proof.
SR=I (t) =
Xp i=0
(j ) (1) = 0 (;1)i i tdi and SR=I
for j = 0 . . . p ; 1 see 4.1.14. We obtain the following system of linear equations:
Xp i=1
Xp
(;1)ii = ;1
i=1
(;1)ii di (di ; 1) (di ; j + 1) = 0
for j = 1 . . . p ; 1:
Upon applying elementary row operations, which do not a ect the solution of this system of linear equations with coecient matrix (di!=(di ; j )!) j =0...p;1 i=1...p
we are led to the Vandermonde matrix whose determinant is Now Cramer's rule gives the stated solutions for the i. (b) According to 4.1.14 we have
;
e(R=I ) = ( 1)p
Q
i>j
(di ; dj ).
(p) (1) X p SR=I p+i di : = ( 1) i p! p i=0
;
Thus (a) implies that
Q
p Xp p;1(di ; j) 1Y e(R=I ) = di Q j =1 : p! i=1 i=1 j 6=i(di ; dj )
It remains to show that the sum in this expression equals 1. We introduce the rational complex function
Qp;1(z ; j) f (z ) = Qpj =1 (z ; d ) : j =1
j
This function has simple poles at worst in the points d1 . . . dp , and the residues in these points are
Qp;1(d ; j) Resd f (z ) = Q j =1 i : (di ; dj ) i
j 6=i
155
4.1. Hilbert functions over homogeneous rings
The sum of all residues of a rational function at all points including 1 is zero, and Res1 f (z ) = ; Res0 f (1=z )=z 2. Therefore p;1 Xp Y
Y
i=1
j 6=i
(di ; j )
j =1
(di ;
dj );1
! Xp =
i=1
Y = Res0 f (1z 2=z ) = Res0 z1 (1 j =1 p;1
Resdi f (z )
; jz)
Yp
(1 ;
j =1
dj z );1
!
= 1:
Exercises Let k be a eld, and M a nite graded module over the polynomial ring R = k X1 . . . Xn ] with minimal graded resolution 4.1.16.
0 ;;!
M j
;
R ( j )pj
;;! ;;! M R(;j) ;;! M ;;! 0: 0j
j
We say that two modules have numerically the same resolution if their graded Betti numbers ij are the same. Show: (a) The homogeneous rings k X Y ]=(X 2 Y 2 ) and k X Y ]=(X 2 XY Y 3 ) have the same Hilbert series, but their minimal graded free k X Y ]-resolutions are numerically di erent. (b) The homogeneous rings k X Y ]=(X 2 Y 2 ) and k X Y ]=(XY X 2 ; Y 2 ) have numerically the same graded k X Y ]-resolution, but are not isomorphic when k = R. 4.1.17. Let k be a eld, and let R = k X1 . . . Xn ]=I be a homogeneous Cohen{ Macaulay ring. The ring R has an m-linear resolution if it has a pure resolution of type (m m + 1 . . . m + p ; 1), p = n ; dim R . (a) Suppose R has an m-linear resolution. What are the ranks of the free modules in the free resolution of R , and what is the multiplicity of R ? (b) Suppose dim R = 0 prove R has an m-linear resolution if and only if I = (X1 . . . Xn )m . Hint: relate the last shifts in the resolution of R with the degrees of the socle elements of R . (c) Prove the homogeneous Cohen{Macaulay ring R = k X1 . . . ;Xn ]=I has an m-linear resolution if and only if Ij = 0 for j < m, and dimk Im = m+mg 1 , where g = height I . Hint: reduce to dimension zero. 4.1.18. Let k be a eld, and let R = k X1 . . . Xn ]=I be a homogeneous Gorenstein ring of dimension 0. Assume that all generators of I have the same degree c. (a) Show a(R ) 2c ; 2. (b) Show a(R ) = 2c ; 2 if and only if R has a pure resolution of type (c c + 1 . . . c + n ; 2 2c + n ; 2). In this case R is called an extremal Gorenstein ring. This class of Gorenstein rings was rst considered by Schenzel 327]. (c) Compute the Betti numbers i (R ) of an extremal Gorenstein ring R in a minimal graded free k X1 . . . Xn ]-resolution of R . ;
156
4. Hilbert functions and multiplicities
4.2 Macaulay's theorem on Hilbert functions This section is devoted to a theorem of Macaulay describing exactly those numerical functions which occur as the Hilbert function H (R n) of a homogeneous k-algebra R , k a eld. Macaulay's theorem says that for each n there is an upper bound for H (R n + 1) in terms of H (R n), and this bound is sharp in the sense that any numerical function satisfying it can indeed be realized as the Hilbert function of a suitable homogeneous k-algebra. One part of the proof of Macaulay's theorem will be based on a theorem of Green which relates the Hilbert function of a homogeneous ring R with the Hilbert function of the factor ring R=hR by a general linear form h.L Let R = n 0 Rn be a homogeneous k-algebra, where R0 = k is a eld. We will show that R has a k-basis consisting of monomials in a basis x1 . . . xm of R1 . We are going to de ne this basis of monomials on the level of the polynomial ring. So let : kX1 . . . Xm ]
;! R
be the surjective k-algebra homomorphism with (Xi) = xi . De nition 4.2.1. A non-empty set of monomials in the indeterminates X1 . . . Xm is called an order ideal of monomials if the following holds: whenever m 2 and a monomial m0 divides m, then m0 2 . Equivalently, if X1a Xmam 2 and 0 bi ai for i = 1 . . . m, then X1b Xmbm 2 . Remarks 4.2.2. (a) In Chapter 9 we introduce the order ideal of an element in a module. This notion has nothing to do with the order ideal of monomials, and they should not be confused. (b) Of course an order ideal of monomials is not a k-basis of an ideal, let alone an ideal. Quite the contrary, if we let C be the complement of in the set of all monomials, then C is a k-basis of the ideal generated by the monomials m 2 C . Theorem 4.2.3 (Macaulay). Let R be a homogeneous k-algebra, k a eld. Further let x1 . . . xm be a k-basis of R1, and : kX1 . . . Xm ] ! R the k-algebra homomorphism with (Xi) = xi for i = 1 . . . m. Then there exists an order ideal of monomials such that ( ) is a k-basis of R. Proof. Let S denote the set of all monomials in the indeterminates X1 . . . Xm. We de ne a total order, the so-called reverse degree-lexicographical order, on S: if u = X1a Xmam and v = X1b XPmbm , then Pu < v if the last non-zero component of (b1 ; a1 . . . bm ; am , bi ; ai ) is positive. (The usage of the term `reverse degree-lexicographical' is not coherent in the literature.) M
M
M
1
M
1
M
M
M
M
M
M
M
M
1
1
157
4.2. Macaulay's theorem on Hilbert functions
It is clear that (S <) is an ordered semigroup, i.e. u < v ) mu < mv for all u v m 2 S. Moreover, any descending chain v1 > v2 > of elements of S must stop after a nite number of steps. Equivalently, every non-empty set of elements in S has a minimal element { a fact that will be used later. Now we de ne recursively a sequence of monomials u1 u2 . . . according to the following rule. We set u1 = 1 assume u1 . . . ui have been de ned then we let ui+1 be the least element in the reverse degreelexicographical order such that (u1 ) . . . (ui ), (ui+1) are linearly independent over k. If such ui+1 does not exist, the sequence terminates with ui . We claim that = fu1 u2 . . .g is the required order ideal of monomials. By construction ( ) is a k-basis of R . Assume is not an order ideal of monomials. Then there exist ui 2 and u 2 S nP such that ui = u Xj for some Xj . As u 2= , we can P write (u) = i (ui) with ui 2 , ui < u, and i 2 k. Then (ui ) = i (ui Xj ), and ui Xj < ui , for all i in the sum, a contradiction. We saw in the proof that our k-basis (u1) (uP2 ) . . . of R has a remarkable property: let u 2 S, and write (u) = i6=0 i (ui ). Then ui u for all i, and if u 2= , these inequalities are strict. The previous theorem and 4.2.2(b) immediately imply Corollary 4.2.4. Let J be the ideal which is generated by the monomials in C . Then the homogeneous k-algebra R and kX1 . . . Xm ]=J have the M
M
M
0
0
M
M
M
0
M
0
M
M
same Hilbert function. In particular, all Hilbert functions of homogeneous rings arise as Hilbert functions of homogeneous rings whose de ning ideal is generated by monomials. The set of monomials C associated with R can be described di erently. Let I = Ker , and set L(I ) = L(f ): f I and I ? = L(I )R where L(f ) denotes the leading monomial of f , that is, the monomial occurring in f which is maximal in the reverse degree-lexicographical order.P Then L(I ) = C . Indeed, let v L(I ), and choose f I , f = ni=1 i vi with monomials vi such that v = L(f ) = vn . Assume vn = C then vn , and so M
f
2g
2
M
2
M
2
2
M
0 6= (vn ) = ;
n;1 X
;n 1i (vi ):
i=1 P Each (vi ) is a linear combination ij (uj ), ij 2 k, uj 2
, uj vi < vn . Replacing the (vi ) in the above equation by their linear combinations gives a representation as a non-trivial linear combination of elements in ( ). This contradicts 4.2.3. M
M
158
4. Hilbert functions and multiplicities
P
Conversely, suppose v 2 C P. Then (v) = i (ui ) with ui 2 , ui < v. Hence, if we set f = v ; iui , then (f ) = 0 and L(f ) = v. The ideal I ? is nitely generated. Therefore there exist polynomials f1 . . . fn 2 I such that I ? = (L(f1) . . . L(fn)). Any such subset of I is called a Grobner or standard basis of I . NotePthat any Grobner basis of I generates I : let f 2 I then L(f ) =P giL(fi) for some P gi 2 kX1 . . . Xm], and it follows that either f = gi fi, or L(f ; gi fi ) < L(f ) for a suitable 2 k. In the rst case, f is an element of (fP1 . . . fn ). In the second case we apply the same procedure to f 0 = f ; gifi to obtain an element f 00 which is either zero, in which case f 2 (f1 . . . fn ), or which has L(f 00) < L(f 0 ). Since any descending sequence of elements in S terminates, we eventually arrive at the required conclusion. We should warn the reader that the converse is not true. Consider for example the ideal I = (f1 f2) with f1 = X1X2 + X32, f2 = X2X3. Then X1X22 = X2f1 ; X3f2 is an element of I ? , but not of (L(f1) L(f2)) = (X32 X2X3). Even though a Grobner basis of an ideal I is not simply given by the leading forms of a system of generators of I , there does exist an algorithm to compute a Grobner basis { the so-called Buchberger algorithm. This, and the fact that most explicit calculations in commutative algebra are performed using Grobner bases, explain their importance. Buchberger's algorithm has been implemented in various computer algebra programs. M
M
Macaulay representations and lexsegment ideals. Let S = kX1 . . . Xm] denote the polynomial ring over a eld k. The problem of determining the Hilbert function of a homogeneous factor ring R of S boils down to the following question: given a subspace V Sd , what can be said about the k-dimension of the subspace S1 V in Sd +1? We will give the answer in a special but important case. Let u Sd
2
be a monomial. We de ne the sets Lu = fv 2 Sd : v < ug and Ru = fv 2 Sd : v ug of the monomials of degree d which are `left' and `right' of u. Note that RX = fX1 . . . Xng. Monomial sets of the form Ru Sd are called lexsegments (of degree d ). The next lemma says that a lexsegment of degree d spans a lexsegment of degree d + 1. Lemma 4.2.5. RX Ru = RX u. Proof. Let v 2 Ru then Xi v X1v X1u. Conversely, let v 2 RX u. We may assume that X1 does not divide v then v > X1u. Let u = X1a Xmam , v = X2b Xmbm , and i be the largest integer such that bi > ai . If there exists j < i with bj > 0, then Xj;1v 2 Ru otherwise, Xi;1v 2 Ru . In both cases it follows that v 2 RX Ru. 1
1
1
1
2
1
1
159
4.2. Macaulay's theorem on Hilbert functions
The sets Lu admit a natural decomposition: let i be the largest integer such that Xi divides u. Then we can write Lu = L0u L00u Xi where Xi does not divide any element in L0u. It is clear that this union is disjoint, that L0u consists of all monomials of degree d in the variables X1 . . . Xi;1, and that L00u = LXi; u . An example illustrates this decomposition. Let S = kX1 . . . X4] and u = X22X3. Then Lu = fX13 X12X2 X1X22 X23 X12X3 X1X2X3g L0u = fX13 X12X2 X1X22 X23g L00u = fX12 X1X2g = LX : It is convenient to denote the set of all monomials of degree d in the variables X1 . . . Xi by X1 . . . Xi]d . We may again decompose L00u , etc. Thus if we write u = Xj (1)Xj (2) Xj (d ) with 1 j (1) j (2) j (d ), then Lu = X1 . . . Xj(d);1]d LXj;d uXj(d) = X1 . . . xj (d );1]d X1 . . . Xj (d ;1);1]d ;1Xj (d ) 1
2 2
1 ( )
= and we end up with the disjoint union
LX
;1
;1
j (d ;1) Xj (d ) u
Xj (d ;1)Xj (d )
Lu = X1 . . . Xj(i);1]iXj(i+1) Xj(d) i=1 called the natural decomposition of Lu. d
It follows that
jLuj =
Xd k(i) i=1
i
with k(i) = j (i) + i ; 2. Note that ; k(d ) > k(d ; 1) > > k(1) 0. Here and in the sequel we use that kl = 0 for 0 k < l . The above considerations show that any non-negative integer has such a binomial sum expansion. We prove this directly. Lemma 4.2.6. Let d be a positive integer. Any a 2 N can be written
uniquely in the form
k(d ) k(d ; 1) k(1) a= + + +
where k(d ) > k(d
;
d 1) >
;1 > k(1) 0. d
1
160
4. Hilbert functions and multiplicities
Proof. ; In order to prove; the existence, Pwe choose ; k(d ) maximal such that k(dd ) a. If a = k(dd ) , then a = di=1; k(ii) with k(i) = i ; 1 for i = 1 . . . d ; 1. Now assume that a0 = a ; k(dd ) > 0. By the induction P ;1 ;k(i) with k(d ; 1) > k(d ; 2) > hypothesis we may assume that a0 = di=1 i > k(1) 0. It remains to show that k(d ) > k(d ; 1): since ;k(dd)+1 > a, it follows that k(d ) k(d ) + 1 k(d ) k(d ; 1) = ; > a0 : d ;1 d d d;1 Hence k(d ) > k(d ; 1). The uniquenessPfollows by induction on a, once we have shown the d ;k(i) following: if a = i=1 i with; k(d ) > k(d ; 1) > > k(1) 0, then k(d ) is the largest integer with k(dd ) a. Again we prove this statement by induction ; on a. For a = 1 the assertion is trivial. Now assume that a > 1, and k(dd)+1 a. Then
X k(i) k(d ) + 1 k(d ) k(d ) k(d ; 1) + 1 ; = d ;1 i=1
i
d
d
d
;1
d
;1
and this contradicts the induction hypothesis. Following Green 138] we refer to the sum 4.2.6 as the d-th Macaulay representation of a, and call k(d ) . . . k(1) the d-th Macaulay coecients of a. Note that for all i d the coecient k(i) is determined by the property of being the maximal integer j such that j k(d ) k(i + 1) a; ; ; : i d i+1 The d -th Macaulay coecients have the following nice property. Lemma 4.2.7. Let k(d ) . . . k(1), respectively k0 (d ) . . . k0 (1), be the d-th Macaulay coecients of a, respectively a0. Then a > a0 if and only if
(k(d ) . . . k(1)) > (k0 (d ) . . . k0 (1)) in the lexicographical order.
We prove both implications by induction on d . For d = 1 the assertion is trivial. We now assume that d > 1. If k(d ) = k0 (d ), then k(d ; 1) . . . k(1) (respectively k0 (d ; 1) . . . k0 (1)); are the (d ; 1)-th ; 0 k(d ) Macaulay coecients of a ; d (respectively a0 ; k d(d ) ), and we may apply the induction hypothesis. If k(d ) 6= k0 (d ), then k(d ) > k0 (d ) if
Proof.
161
4.2. Macaulay's theorem on Hilbert functions
and only if a > a0 . This follows from the characterization of the d -th Macaulay coecients preceding this lemma. Skipping the summands which are zero in the d -th Macaulay representation of a we get the following unique sum expansion: k(d ) k(d ; 1) k(j) a= + d ; 1 + + j d where k(d ) > k(d ; 1) > > k(j ) j 1. We de ne k(d ) + 1 k(d ; 1) + 1 ahd i = + + + k(1) + 1 k(dd +) +1 1 k(d ;d1) + 1 k(j)2+ 1 = + + + d+1 d j+1 and set 0hd i = 0. Proposition 4.2.8. Let u be a monomial of degree d in the polynomial ring S. Then jLX uj = jLujhd i . 1
S
Let Lu = di=1X1 . . . Xj (i);1]i Xj (i+1) Xj (d ) be the canonical decomposition of Lu. We claim that Proof.
d
X1 . . . Xj (i);1]i+1Xj (i+1) Xj (d )
i=1
is the canonical decomposition of LX u. Indeed, the canonical decomposition of Lu is completely determined by the sequence j (1) . . . j (d ) attached to u. Let l (1) . . . l (d +1) be the corresponding sequence for X1u. Then l (1) = 1, and l (i) = j (i ; 1) for i = 2 . . . d + 1. This proves the claim and the proposition. Let S = kX1 . . . Xm ] be a polynomial ring over a eld, and u 2 S a monomial. An ideal which in each degree is spanned by a lexsegment will be called a lexsegment ideal. In view of 4.2.5 and 4.2.8 we obtain Corollary 4.2.9. Let I S be a lexsegment ideal, and set R = S=I. Then 1
H (R n + 1)
H (R n)hni
for all
n:
Equality holds for a given n if and only if In+1 = (X1 . . . Xm)In . Macaulay's theorem. We now come to the main result of this section. It
will follow that the growth of the Hilbert function of a homogeneous ring de ned by a lexsegment ideal is, in a sense, the maximum possible.
162
4. Hilbert functions and multiplicities
Theorem 4.2.10 (Macaulay). Let k be a eld, and let h : N ! N be a
numerical function. The following conditions are equivalent: (a) there exists a homogeneous k-algebra R with Hilbert function H (R n) = h(n) for all n 0 (b) there exists a homogeneous k-algebra R with monomial relations and with Hilbert function H (R n) = h(n) for all n 0 (c) one has h(0) = 1, and h(n + 1) h(n)hni for all n 1 (d) let m = h(1), and for each n 0 let n be the rst h(n) monomials in the variables X1S . . . Xm of degree n in the reverse degree-lexicographical order set = n 0 n then is an order ideal of monomials.
M
M M
M
The following example demonstrates the e ectiveness of Macaulay's theorem: let us check that 1 + 3t + 5t2 + 8t3 is not the Hilbert series ;3 + of;2a, homogeneous ring. In fact, condition (c) is violated since 5 = 2 1 ; ; and 5h2i = 43 + 32 = 7 < 8. Instead we also could apply (d), and get M1 = fX1 X2 X3g, M2 = fX12 X1X2 X22 X1X3 X2X3g, M3 = fX13 2X12X2 X1X22 ;X1 23 X122X3 X1X2X3 X22X3 X1X32g. Thus we see that X1X3 2 M3 , but X1 (X1X3 ) 2= M2. Therefore M is not an order ideal of monomials. Most parts of the theorem have already been shown: the equivalence of (a) and (b) is the content of 4.2.4, and the implication (d) ) (b) is trivial. For the proof of (c) ); (d) we assume that h(1) = m. ;Then condition (c);implies that h(n) n+mn ;1 . Suppose that h(n + 1) = nn++1m , then h(n) = n+mn ;1 , and so Mi = X1 . . . Xm]i for i = n and i = n + 1. Therefore, if u 2 Mn+1, and Xi divides u, then trivially Xi;1u 2 Mn . ; n+m Now we suppose that h(n + 1) < n+1 , then there exist a monomial un+1 such that Mn+1 = Lun . If, as before, Mn = X1 . . . Xm ]n, there is nothing to show. Otherwise, there exists a monomial un such that Mn = Lun . Condition (c) and 4.2.8 imply that RX Run Run . Therefore, if u 2S Mn+1, and Xi divides u, then Xi;1u 2 Mn . In other words, M = n 0 Mn is an order ideal of monomials. For the most dicult implication (a) ) (c) we present the elegant proof of Green 138]. This needs some preparations. If a positive integer a has d -th Macaulay coecients k(d ) . . . k(1), then let k(d ) ; 1 k(d ; 1) ; 1 k(1) ; 1 ahd i = + + + d;1 1 k(d )d; 1 k(j) ; 1 = + + +1
1
+1
d j where j = min i : k(i) i . Note that ahd i has d -th Macaulay coecients k(d ) 1 . . . k(j ) 1 j 2 j 3 . . . 0.
;
f
g
;
; ;
163
4.2. Macaulay's theorem on Hilbert functions
Lemma 4.2.11. (a) If a a0 , then ahd i a0hd i . (b) If k(j ) 6= j for j = minfi : k(i) ig, then (a ; 1)hd i < ahd i . Proof. (a) follows from the observation preceding this lemma and 4.2.7. For (b) let k(d ) k(d ; 1) . . . k(1) be the d -th Macaulay coecients of a, and k0 (d ) k0 (d ; 1) . . . k0 (1) the d -th Macaulay coecients ; ofa ; 1 then k0 (d ) k(d ) by 4.2.7. If k0 (d ) = k(d ), we set a0 = a ; k(dd ) . Convince yourself that a0 ; 1 > 0. Then it follows that a0 (respectively a0 ; 1) has (d ; 1)-th Macaulay coecients k(d ; 1) k(d ; 2) . . . k(1) (respectively k0 (d ; 1) k0 (d ; 2) . . . k0 (1)). Moreover, a0 satis es the hypothesis of (b). Therefore, if we argue by induction on d , we may assume that (a0 ; 1)hd ;1i < a0hd ;1i . Hence the required inequality in the case in ; which k0 (d ) = k(d ) follows from the equalities ahd i = a0hd ;1i + k(dd);1 and ; (a ; 1)hd i = (a0 ; 1)hd ;1i + k(dd);1 . Now suppose that k0 (d ) < k(d ). Our assumption implies that the d -th Macaulay coecient of ahd i is k(d ) ; 1, and that the d -th Macaulay coecient of (a ; 1)hd i is less than or equal to k0 (d ) ; 1. Therefore the conclusion follows from 4.2.7. Theorem 4.2.12, interesting in its own right, is the key to the still unproved implication (a) ) (c) of 4.2.10. Let R be a homogeneous k-algebra, k an in nite eld. The ane k-space R1 is irreducible, and so any non-empty (Zariski-) open subset is dense in R1. This suggests the following terminology: a property P holds for a general linear form of R1 if there exists a non-empty open subset U of R1 such that P holds for all h 2 U . Theorem 4.2.12 (Green). Let R be a homogeneous k-algebra, k an in nite eld, and let n 1 be an integer. Then H (R=hR n) H (R n)hni for a general linear form h. Proof. Let s = sup dimk hRn;1 : h R1 then dimk hRn;1 = s for a general linear form. Indeed, let U R1 be the subset of elements h R1 such that dimk hRn;1 = s. It is obvious that U = . In order to see that U is open, we choose a basis a1 . . . am of R1 and bases P of Rn;1 and Rn. Then the multiplication map h : Rn;1 Rn, h = mi=1 xi ai, can be described by a matrix of linear forms in x1 . . . xm . Replacing the xi by indeterminates yields a matrix A of linear polynomials with coecients in k, and it is clear that U is the complement of V (Is (A)) in R1. Let h U , and set S = R=hR . We claim that H (S n) H (R n)hni , and prove it by induction on min n dimk R1 . If either n = 1 or dimk R1 = 1, then the assertion is trivial. Now assume that n > 1 and dimk R1 > 1. Let V S1 be the subset of linear forms g for which dimk gSn;1 is maximal,
f
2 g
!
2
f
g
2
6
164
4. Hilbert functions and multiplicities
and denote by ' the canonical epimorphism R ! S . We consider the open subset W = (U n kh) \ ';1(V ) of R1. The set W is non-empty since R1 is irreducible and both ';1(V ) and U n kh are non-empty. In fact, assume that U kh. Then, since U is a dense and kh is a closed subset in R1 it follows that R1 = kh, contradicting the assumption dimk R1 > 1. Now we choose h 2 W , and get H (S n) = dimk (Sn =hSn;1) + dimk h Sn;1: By our choice of h, the induction hypothesis yields the inequality dimk (Sn =hSn;1) H (S n)hni :
To obtain an upper bound for the second summand note rst that
dimk hSn;1 dimk (h Rn;1=h(hRn;2)) = dimk (hRn;1=h (hRn;2)): The last equality holds true since the di erence of both sides equals dimk (Rn =hRn;1) ; dimk (Rn=hRn;1), and this di erence is zero since both h and h belong to U . Let W R1 be the (non-empty) open set of linear forms l for which l (hRn;2) has maximal dimension. Then, if we actually choose h 2 W \ W , noting that hRn;1 may be viewed as the (n ; 1)-th homogeneous component of P = R=(Ann h), we may apply our induction hypothesis to conclude that dimk hSn;1 (H (R n) ; H (S n))hn;1i . The rest of the proof is a purely numerical argument. What we need is this: given integers 0 < b < a such that b
bhni + (a ; b)hn;1i
then b ahni . ; ; Assume this fails, and write b = k(nn) + + k(jj ) with k(n) > ; ; j ) j > 0. Then a < k(nn)+1 + + k(jj)+1 , and so a ; b < ;k(n)> +k( ; + jk;(j1) . n;1 ;k(n) + + ;k(2), and We distinguish two cases. If j = 1, then a ; b n;;1 1; ; ; k(n);1 k(1);1 hence (a ; b)hn;1i k(nn;);1 1 + + k(2);1;1 , and bhni = + + n ; 1 . ; k(1);1 Thus our hypothesis implies b k(nn) + + k(2) + < b, a 2 1 contradiction. ; ; If j > 1, then (a ; b)hn;1i < k(nn;);1 1 + + k(jj;);1 1 , and this together with our assumption again yields a contradiction. In order to complete the proof of Macaulay's theorem another numerical result is needed.
165
4.2. Macaulay's theorem on Hilbert functions
Lemma 4.2.13. Let a, a0, and d be positive integers. (a) If a a0 , then ahd i a0hd i . (b) Let k(d ) . . . k(1) be the d-th Macaulay coecients of a, and j = minfi : k(i) ig. Then (a + 1)hd i
=
ahdi + k(1) + 1 ahd i + 1
if j = 1, if j > 1.
Claim (a) follows from 4.2.7, and (b) is immediate for j > 1. Now assume that j = 1, and let i be the maximal integer such that k(i) = k(1) + i ; 1. Then Proof.
a=
=
k(d ) d
+ +
d
+ +
k(d )
and hence a+1 =
k(d ) d
k(i + 1) Xi k(1) + r ; 1 i+1
+
r
k(i + 1) r=1k(1) + i i+1
+ +
+
i
; 1
k(i + 1) k(1) + i i+1
+
i
is the d -th Macaulay expansion of a + 1 since k(i + 1) > k(1) + i. Now we get ahd i =
k(d ) + 1
d+1
Xi k(1) + r + + k(i +i +1)2+ 1 + r+1 r=1
i+1 X k(1) + r ; 1 = k(d ) + 1 + + k(i + 1) + 1 + d+1 i+2 r k(d ) + 1 k(i + 1) + 1 r=2k(1) + i + 1 = d + 1 + + + ; k(1) ; 1 i+2 i+1 and so k(d ) + 1 k(i + 1) + 1 k(1) + i + 1 h d i (a + 1) = d + 1 + + + i+2 i+1 h d i = a + k(1) + 1 as asserted. Proof of 4.2.10, (a) ) (c). We may assume that k is in nite: if necessary replace R by l k R where l is an in nite extension eld of k. Let g be a linear form, and set S = R=gR . The exact sequence 0 ;! gRn ;! Rn+1 ;! Sn+1 ;! 0
166
4. Hilbert functions and multiplicities
yields the inequality H (R n + 1) H (R n) + H (S n + 1). Set a = H (R n) and b = H (R n + 1). For a general linear form g the inequality and 4.2.12 give b a + bhn+1i. Let k(n + 1) . . . k(1) be the (n + 1)-th Macaulay coecients of b. Then k(n + 1) ; 1 k(1) ; 1 bhn+1i = + + n+1 1 and so k(n + 1) ; 1 k(2) ; 1 k(1) ; 1 a + + + : n 1 0 Let, as before, j = minfi : k(i) ig. If j > 1, then k(1) = 0, and k(n + 1) ahni + + k(2) = b: n+1 2 If j = 1, then k(n + 1) k(3) k(2) h n i a + + + + k(2) n+1 3 2 by 4.2.13. But k(2) > k(1), and hence ahni > b. Corollary 4.2.14. Let R be a homogeneous k-algebra, k a eld. Then H (R n + 1) = H (R n)hni for n 0. Proof. Write R = S=I , S = kX1 . . . XS m]. According to 4.2.10(d) there exists an order ideal of monomials M = n 0 Mn in S such that H (R n) = H (S=J n) for all n, where J is the ideal generated by all the monomials not in M. Moreover, the choice of M was such that Mn consists of all monomials of degree n if In = 0 and Mn = Lun for a suitable monomial un otherwise. Since J is nitely generated, there exists an integer r such that Run = RX Run for all n r. Thus the assertion follows from 4.2.8. If we combine 1.5.12 with Macaulay's theorem, we obtain the following characterization of the Hilbert series of Cohen{Macaulay homogeneous algebras. Proposition 4.2.15. Let k be a eld, and h0 . . . hs a nite sequence of +1
1
positive integers. The following conditions are equivalent: (a) there exist an integer d, and a Cohen{Macaulay (reduced) homogeneous k-algebra R of dimension d (whose de ning ideal is generated by squarefree monomials) such that P s h ti HR (t) = i=0 id (1 t)
(b) h0 = 1, and 0 hi+1
;
hhi ii for all i = 1 . . . s
; 1.
167
4.2. Macaulay's theorem on Hilbert functions
(a) ) (b): By 1.5.11 there exists an R -sequence x = x1 . . . xd of degree 1 elements. According to 4.1.8 we have
Proof.
HR (t) =
QR (t) (1 t)d
;
Xs
QR (t) =
i=0
hi ti:
Let R = R=xR then HR (t) = (1 ; t)d HR (t) = QR (t). It follows that H (R n) = hn for all n 0. Therefore 4.2.10 yields the assertion. (b) ) (a): By 4.2.10 there exists a homogeneous k-algebra R = kPX1 . . . Xm]=I , where I is generated by monomials, such that HR (t) = s i i=0 hi t . The k-algebra R is Cohen{Macaulay, simply because R is of dimension zero. In order to get a reduced such k-algebra with the required Hilbert series we consider a certain `deformation' of R as described in the next lemma. Lemma 4.2.16. Let R = kX1 . . . Xm ]=I be a homogeneous k-algebra,
where k is a eld and I is generated by monomials. Then there exist a reduced homogeneous k-algebra S whose de ning ideal is generated by squarefree monomials, and an S-sequence y of elements of degree 1 such that R = S=y S. Proof. Assume I = (u1 . . . un), ui = X1ai Xmaim for i = 1 . . . n. If all exponents aij are at most 1, then I is a radical ideal see Exercise 4.4.17. Suppose now that at least one aij > 1, say ai1 > 1 for some i. We introduce a new indeterminate Y , and set
1
vk = Y ak ;1X1 X2ak 1
2
Xma
km
if ak1 > 1, and vk = uk otherwise. The vi satisfy the following conditions: (i) if Y divides vi , then X1 divides vi (ii) the indeterminate X1 occurs in each vi with multiplicity at most 1. We claim that Y ; X1 is regular modulo the ideal J = (v1 . . . vn ). Indeed, assume the contrary is true. Then there exists an associated prime ideal of kX1 . . . Xm Y ]=J with Y ; X1 2 . By Exercise 4.4.15, is generated by a set of variables, and so X1 Y 2 . It follows that there exists w 2 kX1 . . . Xm Y ], w 2= J , with X1w 2 J and Y w 2 J . As J is generated by monomials, we may assume that w is a monomial. Then there exist integers i, j and monomials u1 , u2 such that X1w = vi u1 and Y w = vj u2. As Y divides vj , it follows from (i) that X1 does also, and so X1 divides w. But then the multiplicity of X1 in vi is at least 2, a contradiction. If all variables in the vi occur with multiplicity one, then J is a radical ideal. Otherwise we repeat this construction, and eventually reach the goal, since at each step we lower the multiplicities of the variables in the generators. p
p
p
p
168
4. Hilbert functions and multiplicities
Exercises Let R be a homogeneous k-algebra, k a eld. (a) Establish from 4.2.14 that there exist integers a1 a2 aj 0 such that n + a1 n + a2 ; 1 n + aj ; (j ; 1) PR (n) = + + + : a a a 4.2.17.
1
2
j
(b) Determine the dimension and the multiplicity of R in terms of the integers a1 . . . aj . 4.2.18. Let k be a eld, S = k X1 . . . Xm ], and u = X1a1 X2a2
Xma
m
= Xj(1) Xj(2) Xj(d )
j (1)
j(2) j(d )
a monomial of degree d . Set R = S=I , where I is the lexsegment ideal generated by Ru . Then deduce (a) dim R = j (d ) ; 1, and (b) e(R ) = ai where i = maxfj : aj 6= 0g, provided dim R > 0.
4.3 Gotzmann's regularity and persistence theorem Gotzmann's 136] regularity and persistence theorems give some deeper insight into the nature of the Hilbert polynomial and the Hilbert function. As before let S = kX1 . . . Xm ] be the polynomial ring in m variables de ned over a eld k, I S a graded ideal, and R = S=I . The regularity theorem is a statement about the regularity of the ideal sheaf I associated with I in projective space. Note that di erent ideals may yield the same ideal sheaf. The ideal I = Ker(S ! R=H 0 (R )) is called the saturation of I the sheafs associated with ideals I and J coincide if and only if I = J . We will formulate Gotzmann's theorems in the language of commutative algebra. So we de ne the (Castelnuovo{Mumford) regularity of a nite graded S -module M , rather than that of a sheaf. It is the number m
reg M = maxfi + j : H i (M )j 6= 0g:
Let q be an integer. Then M is called q-regular if q reg(M ), equivalently, if H i (M )j ;i = 0 for all i and all j > q. Before we set out for Gotzmann's theorems, we include an interesting description of regularity in terms of graded Betti numbers. It shows that reg(M ) measures the `complexity' of the minimal graded free resolution of M . Therefore regularity plays an important r^ole in algorithmic commutative Denoting by M q the truncated graded R -module L Mj , onealgebra. has j q Theorem 4.3.1 (Eisenbud{Goto). The following conditions are equivalent: (a) M is q-regular (b) TorSi (M k)j +i = 0 for all i and all j > q m
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4.3. Gotzmann's regularity and persistence theorem
(c) M q admits a linear S-resolution, i.e., a graded resolution of the form 0 ;! S (;q ; l )cl
Proof. (b)
;! ;! S (;q ; 1)c ;! S (;q)c ;! M q ;! 0: 1
0
() (c): By de nition, the module M q has a linear resolution
if and only if
TorS (M q k)r i
= Hi (x M q )r = 0
for all i r, r 6= i + q:
Here H (x M ) is the Koszul homology of M with respect to the sequence x = X1 . . . Xm . Since (M q )j = 0 for j < q, we always have Hi (x M q )r = 0 for .
r < i + q, while for r > i + q
Hi (x M q )r = Hi (x M )r = TorSi (M k)r :
Thus the desired result follows. (a) ) (c): We may assume q = 0 and M = M 0 . Then it is immediate that H 0 (M ) is concentrated in degree 0. This implies M = H 0 (M ) M=H 0 (M ). The rst summand is a direct summand of copies of k. Hence M is 0-regular if and only if M=H 0 (M ) is 0-regular. In other words, we may assume that depth M > 0. We may further assume that k is in nite. Then there exists an element y 2 S of degree 1 which is M -regular. From the cohomology exact sequence associated with m
m
m
m
0 ;! M (;1) ;! M ;! M=yM ;! 0 y
it follows that M=yM is 0-regular. By induction on the dimension on M , we may suppose that M=yM has a linear S=yS -resolution. But if F is a minimal graded free S -resolution, then F =yF is a minimal graded S=yS -resolution of M=yM . This implies that F is a linear S -resolution of M . (c) ) (a): Again one may assume q = 0 and M = M 0 . Then M has a linear resolution .
.
.
.
;! S (;2)c ;! S (;1)c ;! S c ;! M ;! 0: 2
1
0
Computing ExtiS (M S ) from this resolution we see that ExtiS (M S )j = 0 for j < ;i. By duality (see 3.6.19) there exists an isomorphism of graded R -modules ; H i (M ) = Homk ExtmS ;i (M S (;m)) k : Therefore H i (M )j ;i = 0 for all j > 0, as desired. The regularity theorem says that the regularity of the saturation of an ideal I can be read o the Hilbert polynomial of S=I . m
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4. Hilbert functions and multiplicities
Theorem 4.3.2 (Gotzmann). Write the Hilbert polynomial PR (n) of R = S=I
in the unique form PR (n) =
n + a1 n + a2 ; 1 +
a1
a2
+ +
n + as ; (s ; 1) as
as 0, as described in 4.2.17. Then the saturation I
with a1 a2 of I is s-regular.
We prove the theorem by induction on the dimension of S . For m = 1 the assertion is trivial. Now let m > 1, and choose be a general linear form h. Since PR (n) = PS=I (n) we may assume that I = I . We may further assume that I = S . Then depth R > 0, and h is R -regular. Hence Proof.
6
we get an exact sequence
0 ;! R (;1) ;! R ;! R=hR ;! 0 h
yielding the equation (1)
PR=hR (n) = PR (n)
; PR (n ; 1):
Set Re = R=hR , Se = S=hS then Re = Se=J for some ideal J Se. Furthermore P~R (n) = P~S =J (n). Suppose that (2) P~S =J (n) =
n + b1 n + b2 ; 1 +
b1
b2
+ +
n + br ; (r ; 1)
then J is r-regular by the induction hypothesis. (1) and (2) imply PR (n) =
n + a1 n + a2 ; 1 +
a1
a2
br
+ + n + ar ; (r ; 1) + c ar
where c is a constant and ai = bi + 1 for all i. We claim that c 0, and that I is s-regular for s = r + c. These two claims complete the proof. Indeed, we may set ar+1 = = ar+c = 0. In order to derive the rst claim, assume that c < 0. For n 0 we then have n + a1 n + a2 ; 1 n + ar ; (r ; 1) (3) H (R n) < + + + : a1
a2
ar
The right hand side b of this inequality satis es the equation b=
n + a n + a ; 1 1 2 + + + n + ar ; (r ; 1) n
n
;1
n
; (r ; 1)
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4.3. Gotzmann's regularity and persistence theorem
so that bhni =
=
n + b1 n + b2 ; 1 +
n +n b1 n +n b;2 1; 1 b1
+
b2
+ + + +
n + br ; (r ; 1) n +n ;br (;r ;(r 1); 1) br
:
Observing that n + ar ; (r ; 1) > n ; (r ; 1), one deduces from (3) (see 4.2.11) that H (R n)hni < bhni . Therefore by Green's theorem, n + b1 n + b2 ; 1 n + br ; (r ; 1) e H (R n) < + + + : b1
b2
br
This contradicts (2). For the proof of the second claim note rst that H i (J ) = H i (J ) for i > 1. Therefore and since J is r-regular we deduce from the local cohomology sequence associated with m
m
0 ;! I (;1) ;! I ;! J ;! 0 that H i (I )j ;i = 0 for all i > 2 and j > r (and thus for j > s). It remains to be shown that H 2 (I )j ;2 = 0 for j > s. Suppose this is not the case, and let j be the largest number with H 2 (I )j ;2 6= 0. Then by 4.4.3(b) below, H (R j ; 2) ; PR (j ; 2) = ;H 1 (R )j ;2 < 0 since H 0 (R ) = 0 and H i;1 (R )j ;2 = H i (I )j ;2 = 0 for i > 2, as we have already seen. By our choice of j we have H 1 (R )j ;1 = 0, so that H (R j ; 2) < PR (j ; 2) but H (R j ; 1) = PR (j ; 1): If j = s + 1, then j ; 2 = s ; 1, and s ; 1 + a1 1 + as;1 + + + 1 PR (s ; 1) = 1 s;1 whence s ; 1 + a1 1 + as;1 H (R j ; 2) = H (R s ; 1) + + : s;1 1 Thus Macaulay's theorem implies s + a1 2 + as;1 h j ;2i H (R j ; 1) H (R j ; 2) s + + 2 = PR (s) ; (as + 1) < PR (s) = PR (j ; 1) which is a contradiction. h
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m
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4. Hilbert functions
