Symmetry in the vanishing of Ext over Gorenstein rings
aa r X i v : . [ m a t h . A C ] S e p SYMMETRY IN THE VANISHINGOF EXT OVER GORENSTEIN RINGS
Craig Huneke and David A. Jorgensen
September 17, 2001
Abstract.
We investigate symmetry in the vanishing of Ext for finitely generatedmodules over local Gorenstein rings. In particular, we define a class of local Goren-stein rings, which we call AB rings, and show that for finitely generated modules M and N over an AB ring R , Ext iR ( M, N ) = 0 for all i ≫ iR ( N, M ) = 0for all i ≫ Introduction
Let R be a local Gorenstein ring and let M and N denote finitely generated R -modules. This paper is concerned with the relation between the vanishing of allhigher Ext R ( M, N ) and the vanishing of all higher Ext R ( N, M ). As a means ofinvestigation we concentrate on the more natural duality between the vanishing ofall higher Ext R ( M, N ) and the vanishing of all higher Tor modules where either M or N is replaced by its dual M ∗ (:= Hom R ( M, R )) or N ∗ .Our interest in this topic came about in part from the following striking resultproved recently by Avramov and Buchweitz [AvBu, Thm. III]. Suppose M and N are finitely generated modules over a complete intersection R . Then the followingare equivalent:(1) Tor Ri ( M, N ) = 0 for all i ≫ iR ( M, N ) = 0 for all i ≫ iR ( N, M ) = 0 for all i ≫ . The first author was partially supported by the NSF and the second author was partiallysupported by the NSA. This work was done while the second author was visiting Kansas University.He thanks KU for their generous support. Typeset by
AMS -TEX CRAIG HUNEKE AND DAVID A. JORGENSEN
Their proof relies heavily on the use of certain affine algebraic sets associated to M and N , called support varieties. In their paper [AvBu], Avramov and Buchweitzraise the question of what class of rings satisfy these equivalences for all finitelygenerated modules M and N . They point out this class lies somewhere betweencomplete intersections and local Gorenstein rings, but mention that they do notknow whether this class is equal to either complete intersections or Gorenstein. Inthis paper we introduce a class of local Gorenstein rings, which we call AB rings ,and prove that AB rings satisfy the property that for all finitely generated modules M and N , the following are equivalent (Theorem 4.1):(1) Ext iR ( M, N ) = 0 for all i ≫ iR ( N, M ) = 0 for all i ≫ . Regular local rings are AB rings, and if R is an AB ring, then R/ ( x , ..., x c ) isalso an AB ring whenever x , ..., x c is a regular sequence (see Proposition 3.2). Thisimplies complete intersections are AB rings. Even when restricted to the case of acomplete intersection, our proof of the above equivalence avoids the use of supportvarieties, and in some ways is more direct than the methods of [AvBu]. We alsoprove that local Gorenstein rings of minimal possible multiplicity are AB rings, forthe strong reason that over such rings (except when the embedding dimension is2) all large Ext R ( M, N ) vanish if and only if either M or N has finite projectivedimension. See Theorem 3.5 for a precise statement. These rings are not completeintersections in general, so that in particular the class of AB rings is strictly largerthan that of complete intersections.An AB ring R is a local Gorenstein ring defined by the property that there is aconstant C , depending only on the ring, such that if Ext iR ( M, N ) = 0 for all i ≫ iR ( M, N ) = 0 for all i > C . As far as we know every
Gorenstein ring isan AB ring; we have been unable to find an example which is not. The name ‘AB’stands for both Auslander-Bridger and Avramov-Buchweitz.The paper is organized as follows. In Section 1 we give some preliminary andstraightforward results concerning the relationship of Ext and Tor. In Section 2 weprove a basic result concerning what holds over an arbitrary local Gorenstein ring.Specifically, if M and N are finitely generated maximal Cohen-Macaulay modulesover a local Gorenstein ring R , then the following are equivalent:(1) Tor Ri ( M, N ) = 0 for all i ≫ iR ( M, N ∗ ) = 0 for all i ≫
0, and(3) Ext iR ( N, M ∗ ) = 0 for all i ≫ d con- YMMETRY IN VANISHING EXT 3 secutive Ext modules ( d being the dimension of the ring) to the Cohen-Macaulayproperty of a related tensor product. We include some questions in a final section.
1. Preliminaries
In this section we set notation and discuss some basic facts which will be usedthroughout the paper.Unless otherwise stated, we will assume R to be a local Gorenstein ring. Also, M and N will denote finitely generated R -modules. For an R -module M we let M ∗ denote its dual Hom R ( M, R ). If M is maximal Cohen-Macaulay then it is alsoreflexive, meaning M ∗∗ ≃ M (assuming R is Gorenstein).By a complete intersection we mean a local ring whose completion with respectto the maximal ideal is the quotient of a regular local ring by a regular sequence.For a local ring R , we let embdim R denote its embedding dimension . Syzygies and Conversions for Ext and Tor.
Suppose M is an R -module. Then for i ≥ M i denote image f i , where f i is the i th differential in a minimal free resolution F : · · · → F f −→ F f −→ F f −→ M → M . These M i are the non-negative syzygies of M . They are unique up to isomor-phism, or if one considers a non-minimal resolution any two are stably isomorphic.Now suppose that M is a maximal Cohen-Macaulay R -module. Let G : · · · → G g −→ G g −→ G −→ M ∗ →
0a minimal free resolution of its dual M ∗ . Since M ∗ is maximal Cohen-Macaulay,the dual sequence G ∗ : 0 → M ∗∗ → G ∗ g ∗ −→ G ∗ g ∗ −→ G ∗ −→ · · · is exact. Using the fact that M is reflexive, we can splice F and G ∗ together, gettingthe doubly infinite long exact sequence C ( M ) : · · · → F f −→ F f −→ F → G − ∗ g ∗ −→ G − ∗ g ∗ −→ G − ∗ → · · · . (Note the degree convention.) For i ≤ − M i := image( g ∗− i ) These are the negative syzygies of M . They are unique up to isomorphism. Note that M i is againmaximal Cohen-Macaulay for all i when M is.We now list some properties of the long exact sequences C ( M ). CRAIG HUNEKE AND DAVID A. JORGENSEN
Let M be a finitely generated maximal Cohen-Macaulay R -module,and let N be a finitely generated R -module. (1) C ( M ) ∗ ≃ C ( M ∗ )(2) ( M i ) ∗ ≃ ( M ∗ ) − i for all i . (3) Tor Ri ( M, N ) ≃ H i ( C ( M ) ⊗ N ) for i ≥ . (4) Ext iR ( M, N ) ≃ H − i − ( C ( M ) ∗ ⊗ N ) for i ≥ . (5) for fixed t ≥ and for ≤ i ≤ t − we have ( i ) Ext iR ( M − t , N ) ≃ Tor Rt − i − ( M ∗ , N ) and ( ii ) Tor Ri ( M − t , N ) ≃ Ext t − i − R ( M ∗ , N ) . Proof.
Conditions (1)–(3) are straightforward. Condition (5) follows easily from(1)–(4), so only Condition (4) needs some explanation. The critical fact we need toshow is that for any complex of free R -modules F , Hom R ( F , N ) and Hom R ( F , R ) ⊗ R N are isomorphic as complexes: write F : · · · → F i +1 f i +1 −−−→ F i f i −→ F i − → · · · , where the F i are free R -modules. The natural maps h i : Hom R ( F i , R ) ⊗ R N → Hom R ( F i , N ) given by f ⊗ n
7→ { a f ( a ) n } are isomorphisms since F i is free. Itis easy to check that the diagramHom R ( F i , R ) ⊗ R N f ∗ i +1 ⊗ N −−−−−→ Hom R ( F i +1 , R ) ⊗ R N y h i y h i +1 Hom R ( F i , N ) Hom( f i +1 ,N ) −−−−−−−−→ Hom R ( F i +1 , N )is commutative, and this establishes our fact. (cid:3) Suppose M and N are R -modules with M maximal Cohen-Macaulay. ThenExt i ( M, R ) = 0 for all i ≥
1, and so by shifting along the short exact sequences0 → N n +1 → G n → N n → G n free) we obtain isomorphisms( ) Ext iR ( M, N ) ≃ Ext i + nR ( M, N n )for i ≥ n ≥ The Change of Rings Long Exact Sequences of Ext and Tor.
YMMETRY IN VANISHING EXT 5
Suppose that S is a commutative ring, x is a non-zerodivisor of S and R := S/ ( x ).Let M and N be R -modules. Then we have the change of rings long exact sequenceof Ext [Ro, 11.65](1.3) ... ... ...Ext R ( M, N ) ←− Ext S ( M, N ) ←− Ext R ( M, N ) ←− Ext R ( M, N ) ←− Ext S ( M, N ) ←− Ext R ( M, N ) ←− , and the change of rings long exact sequence of Tor [Ro. 11.64](1.4) ... ... ...Tor R ( M, N ) −→ Tor S ( M, N ) −→ Tor R ( M, N ) −→ Tor R ( M, N ) −→ Tor S ( M, N ) −→ Tor R ( M, N ) −→ .
2. Vanishing of Ext and Tor over Arbitrary Local Gorenstein Rings
In this section we prove what type of duality between the vanishing of Ext andTor holds over arbitrary local Gorenstein rings. It is possible an even stronger resultis true, as we discuss in Section 4, but the main result of this section is what is true‘on the surface’. In particular, Theorem 2.1 states that one can flip the argumentsin vanishing Ext modules ‘up to duals’.
Let R be a local Gorenstein ring, and let M and N be finitelygenerated maximal Cohen-Macaulay R -modules. Then the following are equivalent: (1) Tor Ri ( M, N ) = 0 for all i ≫ , (2) Ext iR ( M, N ∗ ) = 0 for all i ≫ , and (3) Ext iR ( N, M ∗ ) = 0 for all i ≫ .Proof. Suppose we have shown that (1) and (2) are equivalent. By replacing (1) bythe equivalent condition that Tor Ri ( N, M ) = 0 for all i ≫
0, we see then that (1) isequivalent to (3). Hence it suffices to prove (1) and (2) are equivalent, and for thiswe only need to assume that N is maximal Cohen-Macaulay.We induce upon the dimension of R , say d . If d = 0, then Ext iR ( M, N ∗ ) is theMatlis dual of Tor Ri ( M, N ), so the result is immediate in this case.Now assume that d >
0. Choose x ∈ R a non-zerodivisor on M d , N , N ∗ , and R .We have Tor Ri ( M, N ) = 0 for all i ≫ ⇐⇒ Tor Ri ( M d , N ) = 0 for all i ≫ CRAIG HUNEKE AND DAVID A. JORGENSEN and from the long exact sequence of Tor coming from the short exact sequence0 → N x −→ N → N/xN → ⇐⇒ Tor Ri ( M d , N/xN ) = 0 for all i ≫ Ri ( M d , N/xN ) ≃ Tor R/ ( x ) i ( M d /xM d , N/xN ), ⇐⇒ Tor R/ ( x ) i ( M d /xM d , N/xN ) = 0 for all i ≫ ⇐⇒ Ext iR/ ( x ) ( M d /xM d , ( N/xN ) ∗ ) = 0 for all i ≫ N is maximal Cohen-Macaulay and R is Gorenstein, N ∗ /xN ∗ ≃ ( N/xN ) ∗ (where the second module is Hom R/ ( x ) ( N/xN, R/ ( x ))), thus, ⇐⇒ Ext iR/ ( x ) ( M d /xM d , N ∗ /xN ∗ ) = 0 for all i ≫ iR ( M d , N ∗ /xN ∗ ) ≃ Ext iR/ ( x ) ( M d /xM d , N ∗ /xN ∗ ), ⇐⇒ Ext iR ( M d , N ∗ /xN ∗ ) = 0 for all i ≫ → N ∗ x −→ N ∗ → N ∗ /xN ∗ → ⇐⇒ Ext iR ( M d , N ∗ ) = 0 for all i ≫ ⇐⇒ Ext iR ( M, N ∗ ) = 0 for all i ≫ . (cid:3) Remark.
Suppose M and N are maximal Cohen-Macaulay modules over thelocal Gorenstein ring R . ThenExt iR ( M, N ) ≃ Ext iR ( N ∗ , M ∗ ) . This isomorphism can be seen as follows: suppose that i = 1. As M and N aremaximal Cohen-Macaulay, they are reflexive, so short exact sequences 0 → N → YMMETRY IN VANISHING EXT 7 T → M → → M ∗ → T ∗ → N ∗ → then gives the isomorphism.For i > iR ( M, N ) ≃ Ext R ( M i − , N ) ≃ Ext R ( N ∗ , M ∗ i − ) by the i = 1 case ≃ Ext R ( N ∗ , ( M ∗ ) − i +1 ) by (2) of Lemma 1.1 ≃ Ext iR ( N ∗ , M ∗ ) by (1.2) . (cid:3) Presumably, one can directly prove this remark using the Yoneda definition ofExt and the fact that both M and N are maximal Cohen-Macaulay.Below is an example showing that the hypothesis that N is maximal Cohen-Macaulay in the equivalence of (1) and (2) in Theorem 2.1 cannot be dropped. Let R be the 3-dimensional hypersurface k [[ W, X, Y, Z ]] / ( W X − Y Z ), and set M := k and N := coker wxyz ! . Then pd R N = 1 (but pd R N ∗ = ∞ ),so we have Tor Ri ( M, N ) = 0 for all i ≫
0. However, Ext R ( M, N ∗ ) = 0. Bywhat is shown in the next section, R is an AB ring. If it were the case thatExt iR ( M, N ∗ ) is zero for all i ≫ R is AB, Proposition 3.1 shows then thatExt iR ( M, N ∗ ) = 0 for all i > dim R , which would be a contradiction. (cid:3)
3. AB Rings
Let R be a commutative ring. We define the Ext- index of R to besup { n | Ext iR ( M, N ) = 0 for all i > n and Ext nR ( M, N ) = 0 } , where the sup is taken over all pairs of finitely generated R -modules ( M, N ) withExt iR ( M, N ) = 0 for all i ≫ If R is a local Gorenstein ring of finite Ext-index, we say that R is an AB ring .We will prove that all complete intersections are AB rings. More generally, it isobvious that R is an AB ring if ˆ R is (where ˆ R is the completion of R ), and we show(3.2) that if R is an AB ring and x , ..., x c is a regular sequence, then R/ ( x , ..., x c )is also an AB ring. The class of AB rings also includes local Gorenstein rings of‘minimal’ multiplicity embdim( R ) − dim( R ) + 2 (see 3.5). CRAIG HUNEKE AND DAVID A. JORGENSEN
Suppose that R is an AB ring. Then the Ext -index of R equals dim R .Proof. Let n denote the Ext-index of R and d the dimension of R .Let x , . . . , x d be a maximal R -sequence. Set M := R/ ( x , . . . , x d ) and N = k ,the residue field of R . Then Ext iR ( M, N ) = 0 for i > d and Ext dR ( M, N ) ≃ k = 0.Hence n ≥ d .Suppose that n > d . There exists a pair of finitely generated R -modules ( M, N )such that Ext iR ( M, N ) = 0 for i > n and Ext nR ( M, N ) = 0. We have the isomor-phisms Ext i +1 R (( M d ) − d − , N ) ≃ Ext i − dR ( M d , N ) ≃ Ext iR ( M, N ) for i > d . HenceExt i (( M d ) − d − , N ) = 0 for i > n + 1 and Ext n +1 (( M d ) − d − , N ) = 0, which contra-dicts the definition of n . Therefore n = d . (cid:3) We of course have the dual notion of Tor- index . If R is local Gorenstein withfinite Tor-index, then it is also equal to dim R , by an argument analogous to thatof 3.1.Another related property of rings we are interested in is the following. We saythat Ext ∗ R ( M, N ) has a gap of length t if for some n ≥
0, Ext iR ( M, N ) = 0 for n + 1 ≤ i ≤ n + t , but Ext nR ( M, N ) and Ext n + t +1 R ( M, N ) are both nonzero. Wehave the analogous notion of gap for Tor R ∗ ( M, N ). (We allow gaps of length 0.) Weset Ext-gap( R ) := sup (cid:26) t ∈ N (cid:12)(cid:12)(cid:12)(cid:12) Ext ∗ R ( M, N ) has a gap of length t forfinite R -modules M and N (cid:27) , and Tor-gap( R ) := sup (cid:26) t ∈ N (cid:12)(cid:12)(cid:12)(cid:12) Tor R ∗ ( M, N ) has a gap of length t forfinite R -modules M and N (cid:27) . We say that R is Ext- bounded if it has finite Ext-gap. Similarly, we say R isTor- bounded if it has finite Tor-gap.We list some elementary properties involving finite Ext-index, Tor-index, Ext-boundedness and Tor-boundedness for local Gorenstein rings. Let x be a non-zerodivisor of the d -dimensional local Gorensteinring R . Then (1) R is an AB ring if and only if R/ ( x ) is an AB ring. (2) R has finite Tor-index if and only if R/ ( x ) does. (3) R is Ext-bounded if and only if R/ ( x ) is. (4) R is Tor-bounded if and only if R/ ( x ) is. YMMETRY IN VANISHING EXT 9
Proof. (1). Suppose that R is an AB ring. Let M and N be finitely generated R/ ( x )-modules such that Ext iR/ ( x ) ( M, N ) = 0 for all i ≫
0. By the change ofrings long exact sequence of Ext (1.3) we conclude that Ext iR ( M, N ) = 0 for all i ≫
0, and so Ext iR ( M, N ) = 0 for all i > d . Looking at (1.3) again, we see thatExt iR/ ( x ) ( M, N ) ≃ Ext i +2 R/ ( x ) ( M, N ) for i > d −
1. But as Ext iR/ ( x ) ( M, N ) = 0 for all i ≫
0, we have Ext iR/ ( x ) ( M, N ) = 0 for all i > d −
1. Hence R/ ( x ) is an AB ring.Now suppose that R/ ( x ) is an AB ring, and let M and N be finitely generated R -modules such that Ext iR ( M, N ) = 0 for all i ≫
0. We have the isomorphismsExt iR ( M, N ) ≃ Ext i − dR ( M d , N ) ≃ Ext iR ( M d , N d )which are valid for i > d , the second one being that of (1.2). The short exactsequence 0 → N d x −→ N d → N d /xN d → · · · → Ext iR ( M d , N d ) x −→ Ext iR ( M d , N d ) → Ext iR ( M d , N d /xN d ) → · · · . Since Ext iR ( M d , N d ) = 0 for all i ≫
0, we see that Ext iR ( M d , N d /xN d ) = 0 for all i ≫
0. We have the isomorphisms [R](3.2.2) Ext iR/ ( x ) ( M d /xM d , N d /xN d ) ∼ = Ext iR ( M d , N d /xN d )for all i ≥
0. Hence Ext iR/ ( x ) ( M d /xM d , N d /xN d ) = 0 for all i ≫
0, which meansthat Ext iR/ ( x ) ( M d /xM d , N d /xN d ) = 0 for all i > d −
1, since R/ ( x ) is an AB ring.Therefore Ext iR ( M d , N d /xN d ) = 0 for all i > d −
1. By (3.2.1) and Nakayama’sLemma, we conclude that Ext iR ( M d , N d ) = 0 for all i > d −
1, and so Ext iR ( M, N ) =0 for all i > d . Therefore R is an AB ring.The proof of (2) is exactly analogous to the proof of (1), using (1.4) and a longexact sequence of Tor this time.(3). Assume that e := Ext-gap( R ) < ∞ . Let M and N be finitely generated R/ ( x )-modules such that Ext iR/ ( x ) ( M, N ) = 0 for n ≤ i ≤ n + e + 1, some n ≥ iR ( M, N ) = 0for n + 1 ≤ i ≤ n + e + 1. Since Ext-gap( R ) = e we have Ext iR ( M, N ) = 0 forall i > n . Another look at (1.3) shows that Ext iR/ ( x ) ( M, N ) ≃ Ext i +2 R/ ( x ) ( M, N )for all i > n −
1. Since Ext iR/ ( x ) ( M, N ) = 0 for i = n, n + 1, we see then thatExt iR/ ( x ) ( M, N ) = 0 for i > n − R/ ( x )) ≤ e + 1.Now assume that e := Ext-gap( R/ ( x )) < ∞ . Suppose that M and N are finitelygenerated R -modules with Ext iR ( M, N ) = 0 for n ≤ i ≤ n + d + e + 1, some n ≥
1. We have Ext iR ( M d , N d ) ≃ Ext iR ( M, N ) = 0 for n + d ≤ i ≤ n + d + e + 1. Therefore, from (3.2.1), we get Ext iR ( M d , N d /xN d ) = 0 for n + d ≤ i ≤ n + d + e .Equivalently, Ext iR/ ( x ) ( M d /xM d , N d /xN d ) = 0 for n + d ≤ i ≤ n + d + e . SinceExt-gap( R/ ( x )) = e , Ext iR/ ( x ) ( M d /xM d , N d /xN d ) = 0 for all i ≥ n + d , whichimplies, by (3.2.1) and Nakayama’s lemma, Ext iR ( M d , N d ) = 0 for all i ≥ n + d ,which means Ext iR ( M, N ) = 0 for all i ≥ n . Therefore Ext-gap( R ) ≤ d + e + 1.The proof of (4) is similar to the proof of (3). (cid:3) Assume that R is a local Gorenstein ring. Then (1) R is an AB ring if and only if it has finite Tor-index; (2) R is Ext-bounded if and only if it is Tor-bounded; (3) if R is Ext-bounded, then it is an AB ring.Proof. We first prove (1). Choose a maximal regular sequence in R and let I bethe ideal generated by this sequence. Proposition 3.2 states that R is an AB ringif and only if R/I is an AB ring, and R has finite Tor-index if and only if R/I hasfinite Tor-index. Hence it suffices to prove (1) in case R is 0-dimensional. In thiscase Ext iR ( M, N ) is the Matlis dual of Tor Ri ( M, N ∗ ) so that the vanishing of oneimplies the vanishing of the other. This proves (1).Statement (2) is proved in a similar manner, using Proposition 3.2.We prove (3). Assume R is Ext-bounded. We prove that R has finite Tor-index.Let d denote the dimension of R and e := Ext-gap( R ), and suppose that for finite R -modules M and N , Tor Ri ( M, N ) = 0 for all i ≫
0. Let b = d − depth M so that M b is maximal Cohen-Macaulay. Choose n largest such that Tor Rn ( M b , N ) = 0.Using t = e + n + 3. in (5)( i ) of Lemma 1.1, we have Ext iR (( M ∗ b ) − e − n − , N ) ≃ Tor Re + n +2 − i ( M b , N ) = 0 for 1 ≤ i ≤ e + 1 . Hence we have a gap of zero Ext largerthan e . Therefore Ext iR (( M ∗ b ) − e − n − , N ) = 0 for all i ≥
1, which forces n = 0.Thus Tor Ri ( M, N ) = 0 for all i > d . (cid:3) The following Corollary is an almost immediate consequence of Proposition 3.3,as regular local rings are clearly Ext-bounded.
Let R be a local Gorenstein ring. If R is a complete intersection,then R is Ext-bounded. In particular, R is an AB ring.Proof. Since
R ֒ → ˆ R is a faithfully flat extension, Ext iR ( M, N ) = 0 if and only ifExt i ˆ R ( ˆ M , ˆ N ) = 0 and so R is an AB ring if ˆ R is. Therefore we may without lossof generality assume that R = S/ ( x , . . . , x c ) where S is a regular local ring and x , . . . , x c is an S -regular sequence. By Proposition 3.2 it suffices to prove that S is Ext-bounded. But this is trivial as every finitely generated module over S hasprojective dimension ≤ dim S , so that Ext iR ( M, N ) = 0 for i > dim S and Ext-gapscan occur of length no longer than dim S − (cid:3) YMMETRY IN VANISHING EXT 11
It could be that all local Gorenstein rings are AB rings; we have no counterexam-ple. The class of AB rings is strictly bigger than the class of complete intersectionsas the next Theorem proves, albeit for rather strong reasons.
Let ( R, m , k ) be a local Gorenstein ring with multiplicity equal to embdim R − dim R + 2 . Assume that embdim R > (so that R is not a completeintersection). Then for finitely generated R -modules M and N , Ext iR ( M, N ) = 0 for all i ≫ if and only if either M or N has finite projective dimension. Inparticular, R is an AB ring.Proof. We induce on d := dim R . d = 0. In this case by duality we have that Tor Ri ( M, N ∗ ) = 0 for all i ≫ M by a high enough syzygy such that Tor R ( M, N ∗ ) = Tor R ( M, N ∗ ) =Tor R ( M, N ∗ ) = 0. Now the following lemma, 3.6, says that either M or N is free. d >
0. By replacing M and N by syzygies we can assume they are both maximalCohen-Macaulay. Let x be a non-zerodivisor of R . Once again we use the fact thatExt iR ( M, N ) = 0 for all i ≫ iR/ ( x ) ( M/xM, N/xN ) = 0 for all i ≫
0. By induction either
M/xM or N/xN has finite projective dimension over R/ ( x ). But then either M or N has finite projective dimension over R . (cid:3) Let ( R, m , k ) be a -dimensional local Gorenstein ring with multiplic-ity embdim R +2 . Assume embdim R > (so that R is not a complete intersection).Let M and N be finitely generated R -modules. Then Tor R ( M, N ) = Tor R ( M, N ) =Tor R ( M, N ) = 0 if and only if either M or N is free.Proof. Let n denote the embedding dimension of R . Assume that M is not free. If k is a summand of M i for any 0 ≤ i ≤
4, then we get right away that N is free, sinceTor R ( M i , N ) = Tor Ri +1 ( M, N ) = 0 would then imply Tor R ( k, N ) = 0. Thereforeassume k is not a summand of M i for all 0 ≤ i ≤
4. Replace M by its first syzygy.Then as M ⊆ m F , for F a free module, we have m M = 0 (since m = 0). Let b i denote the i th Betti number of M and s := dim k m M . Then, as in Lescot’s paper[L, Lemma 3.3], M is 3-exceptional and b = nb − s , b = b ( n − − sn and b = b ( n − n ) − s ( n − N is also not free. Also replace N by its first syzygy, so that m N = 0. Write m N ≃ k d . We have a short exact sequence 0 → k d → N → k c →
0, where c is the minimal number of generators of N . Applying M ⊗ R tothis short exact sequence and using the fact that Tor R ( M, N ) = Tor R ( M, N ) =Tor R ( M, N ) = 0 we get cb = db and cb = db . Letting α := d/c we can writethese equations as b = αb and b = αb = α b . Substituting for the b i we get b ( n − − sn = α ( nb − s ) b ( n − n ) − s ( n −
1) = α ( nb − s ) . After rearranging we arrive at b ( n − αn −
1) = s ( n − α ) b ( n − α n − n ) = s ( n − α − . Now cross multiplying, cancelling off the b s terms, and simplifying we are left withthe condition α − nα + 1 = 0. This says that α ∈ Q is algebraic over Z . Hence α isan integer, and the only choice is α = 1. But then n = 2, which is a contradiction.Hence N must be free. (cid:3)
4. Vanishing of Ext and Tor over AB Rings
In this section we prove that AB rings are a class which gives the duality ofvanishing Ext discussed in the introduction. Our main theorem states:
Suppose that R is an AB ring, and let M and N be finitely generated R -modules. Then Ext iR ( M, N ) = 0 for all i ≫ if and only if Ext iR ( N, M ) = 0 for all i ≫ . Proof.
First assume the theorem is true if both M and N are maximal Cohen-Macaulay. For the general case, take syzygies M m and N n ( m, n ≥
0) of M and N ,respectively, which are maximal Cohen-Macaulay. We have Ext iR ( M, N ) = 0 for all i ≫ iR ( M m , N ) = 0 for all i ≫ iR ( M m , N n ) = 0 for all i ≫
0. Thus Ext iR ( M, N ) = 0 for all i ≫ iR ( M m , N n ) = 0 for all i ≫
0, and so the theorem holds generally.Now suppose that M and N are maximal Cohen-Macaulay and Ext iR ( M, N ) =0 for all i ≫
0. Then for all t ≥
1, Ext iR ( M − t , N ) = 0 for all i ≫
0. Since R is an AB ring, it follows from Proposition 3.1 that for all t ≥ i >d := dim( R ), Ext iR ( M − t , N ) = 0. However, Ext iR ( M − t , N ) ≃ Ext R ( M i − t − , N ) . Hence for all t ≥ i > d , Ext R ( M i − t +1 , N ) = 0 . By varying i and t , weobtain that Ext R ( M − t , N ) = 0 for all t ≥
1. Therefore by (5)( i ) of Lemma 1.1,Tor Rt − ( M ∗ , N ) = Tor Rt − ( N, M ∗ ) = 0 for all t ≥
3. Applying Theorem 2.1 thenshows that Ext iR ( N, M ) = 0 for all i ≫ (cid:3) As an immediate corollary, we have an analogue of Theorem 2.1
YMMETRY IN VANISHING EXT 13
Suppose that R is an AB ring, and let M and N be finitely gener-ated maximal Cohen-Macaulay R -modules. Then the following are equivalent: (1) Tor Ri ( M, N ) = 0 for all i ≫ , (2) Ext iR ( M ∗ , N ) = 0 for all i ≫ , and (3) Ext iR ( N ∗ , M ) = 0 for all i ≫ .Proof. Due to the natural symmetry in Tor, it suffices to prove just the equivalencebetween (1) and (2), and for this we only need to assume that M is maximal Cohen-Macaulay.We haveTor Ri ( M, N ) = 0 for all i ≫ ⇐⇒ Tor Ri ( N, M ) = 0 for all i ≫ ⇐⇒ Ext iR ( N, M ∗ ) = 0 for all i ≫ ⇐⇒ Ext iR ( M ∗ , N ) = 0 for all i ≫ (cid:3) Our final proposition in this section is an observation that there are circumstancesother than where one module has finite projective dimension or where the ring is acomplete intersection in which all large Ext modules vanish.
Let ( R, m R , k ) and ( S, m S , k ) be two local Gorensteinrings essentially of finite type over the same field k , and let M R be a finitely gen-erated R -module and N S a finitely generated S -module. Set A := ( R ⊗ k S ) P where P := ( m R ⊗ k S + R ⊗ k m S ) , M := ( M R ⊗ k S ) P and N := ( R ⊗ k N S ) P . Then A islocal Gorenstein and Ext iA ( M, N ) = 0 for all i > dim A .Proof. Of course A is Noetherian, being a localization of a finitely generated k -algebra. A is also Gorenstein by applying [WITO].For the last statement, we induce on d := dim A = dim R + dim S . Suppose that A has dimension 0. In this case duality yields Ext iA ( M, N ) ∗ ≃ Tor Ai ( M, N ∗ ). Thusit suffices to prove Tor Ai ( M, N ∗ ) = 0 for all i > R ⊗ k S ( M R ⊗ k S, R ⊗ k S ) ≃ Hom R ( M R , R ) ⊗ k S . To seethis, note that these modules are naturally isomorphic if M R is free R -module. Ingeneral, let R m ρ −→ R n → M R → M R over R . Let A ′ := R ⊗ k S and M ′ := M R ⊗ k S . This yields a presentation ( A ′ ) m −→ ( A ′ ) n → M ′ → M ′ over A ′ . We obtain a commutative diagram0 −−→ Hom A ′ ( M ′ , A ′ ) −−→ Hom A ′ (( A ′ ) n , A ′ ) −−→ Hom A ′ (( A ′ ) m , A ′ ) y y −−→ Hom R ( M R , R ) ⊗ k S −−→ Hom R ( R n , R ) ⊗ k S −−→ Hom R ( R m , R ) ⊗ k S, where the first row is exact and the vertical arrows are isomorphisms. To establishthe claim we only need to know that the bottom row is exact, but this follows fromthe fact that 0 → Hom R ( M R , R ) → Hom R ( R n , R ) → Hom R ( R m , R ) is an exactsequence of k -modules and S is flat as a k -module.Localizing the isomorphism in the claim above at P , we see that the A -module M ∗ := Hom A ( M, A ) comes from the R -module Hom R ( M R , R ). Similarly N ∗ :=Hom A ( N, A ) comes from the S -module Hom S ( N S , S ). Hence there is no distinctionbetween proving Tor Ai ( M, N ∗ ) = 0 for all i > Ai ( M, N ) = 0 forall i >
0. We will prove the latter.Let ( F , f ) be an R -free resolution of M R . Then F is an exact sequence of k modules, and since S is flat as a k -module, F ⊗ k S is an exact sequence, of R ⊗ k S -modules. Thus ( F ⊗ k S ) P is an A -free resolution of M . To show that Tor Ai ( M, N ) =0 for all i > F ⊗ k S ) P ⊗ A N is acyclic(meaning the homology is zero except in degree zero).For all i we have a commutative diagram( F i ⊗ k S ) ⊗ R ⊗ k S ( R ⊗ k N S ) ( f i ⊗ S ) ⊗ ( R ⊗ N S ) −−−−−−−−−−−→ ( F i − ⊗ k S ) ⊗ R ⊗ k S ( R ⊗ k N S ) y ≃ y ≃ F i ⊗ k N S f i ⊗ N S −−−−→ F i − ⊗ k N S , where the vertical arrows are the natural isomorphisms. Hence ( F ⊗ k S ) ⊗ R ⊗ k S ( R ⊗ k N S ) and F ⊗ k N S are isomorphic complexes of R ⊗ k S -modules. Since N S is flat asa k -module, the latter is acyclic, and therefore so is ( F ⊗ k R S ) ⊗ R ⊗ k S ( R ⊗ k N S ).Finally, localizing at P we get that ( F ⊗ k S ) P ⊗ A N is acyclic, and this finishes theproof in the d = 0 case.Now without loss of generality assume that dim R >
0. ¿From the discussionabove we know that M ≃ (( M R ) ⊗ k S ) P . Let x be a non-zerodivisor on both( M R ) and R . Then x ⊗ M , A and N , and we have A/ ( x ⊗ ≃ ( R/ ( x ) ⊗ k S ) P ,M / ( x ⊗ M ≃ (( M R ) /x ( M R ) ⊗ k S ) P and N/ ( x ⊗ N ≃ ( R/ ( x ) ⊗ k N S ) P . Hence by induction we have thatExt iA/ ( x ⊗ ( M / ( x ⊗ M , N/ ( x ⊗ N ) = 0for all i > d −
1. Now (3.2.1) and (3.2.2) show that Ext iA ( M , N ) = 0 for all i > d − iA ( M, N ) = 0 for all i > d . (cid:3) YMMETRY IN VANISHING EXT 15
5. What does the vanishing of Ext mean?
Many of the results in this section are closely related to the work of Auslanderand Bridger. See [AB], and the writeup [M] of the contents of [AB]. However, noneof the results below is explicitly in these works, and we found they gave us a betterunderstanding of what the vanishing of Ext means.
The natural maps M ∗ ⊗ R N → Hom R ( M, N ) and M ⊗ R N ∗ → Hom R ( M, N ) ∗ . Assume that M is maximal Cohen-Macaulay. From the short exact sequence0 → M → F → M → → M ∗ → F ∗ → M ∗ →
0, and these yield a commutative diagram M ∗ ⊗ R N α −−→ F ∗ ⊗ R N −−→ M ∗ ⊗ R N −−→ y f y g y f −−→ Hom R ( M, N ) −−→ Hom R ( F, N ) β −−→ Hom R ( M , N ) , where the vertical arrows are the natural maps M ∗ ⊗ R N → Hom R ( M, N ) givenby φ ⊗ n
7→ { m φ ( m ) n } . Note that g is an isomorphism (since F is free),ker α ≃ Tor R ( M ∗ , N ) and coker β ≃ Ext R ( M, N ). ¿From this diagram one easilydeduces the following three facts.(1) ker f ≃ coker f ,(2) Tor R ( M ∗ , N ) ≃ ker f , and(3) Ext R ( M, N ) ≃ coker f .Building a diagram as above for each of the exact sequences 0 → M i +1 → F i → M i → → Ext R ( M i − , N ) → M ∗ i ⊗ R N → Hom R ( M i , N ) → Ext R ( M i − , N ) → , and0 → Tor R ( M ∗ i +1 , N ) → M ∗ i ⊗ R N → Hom R ( M i , N ) → Tor R ( M ∗ i +2 , N ) → . For i ≥ ) 0 → Ext i − R ( M, N ) → M ∗ i ⊗ R N → Hom R ( M i , N ) → Ext iR ( M, N ) → . An immediate observation is
Let R be a local Gorenstein ring, and let M and N be finitelygenerated R -modules with M maximal Cohen-Macaulay. Then Ext iR ( M, N ) = 0 for all i ≫ if and only if the natural maps M ∗ i ⊗ R N → Hom R ( M i , N ) areisomorphisms for all i ≫ . (cid:3) Note also that building exact sequences (5.1) for arbitrarily large negative syzy-gies of M , and then splicing the resulting exact sequences together, we obtain adoubly infinite long exact sequence( ) · · · → M ∗ i ⊗ R N → Hom R ( M i , N ) → M ∗ i + i ⊗ R N → Hom R ( M i +1 , N ) → · · · . Now suppose that N is maximal Cohen-Macaulay and that Ext R ( M, N ) = 0.¿From the short exact sequence 0 → M → F → M → → Hom R ( M, N ) → Hom R ( F, N ) → Hom R ( M , N ) →
0, and acommutative diagram M ⊗ R N ∗ γ −−→ F ⊗ R N ∗ −−→ M ⊗ R N ∗ −−→ y h y g ′ y h −−→ Hom R ( M , N ) ∗ −−→ Hom R ( F, N ) ∗ δ −−→ Hom R ( M, N ) ∗ , where the vertical arrows are the natural maps M ⊗ R N ∗ → Hom R ( M, N ) ∗ given by m ⊗ φ
7→ { ψ φ ( ψ ( m )) } . Note that g ′ is an isomorphism (since F is free), ker γ ≃ Tor R ( M, N ∗ ) and coker δ ≃ Ext R (Hom R ( M , N ) , R ). Regarding this diagram, wehave the following three facts.(1) ker h ≃ coker h ,(2) Tor R ( M, N ∗ ) ≃ ker h , and(3) Ext R (Hom R ( M , N ) , R ) ≃ coker h .Now assume that Ext iR ( M, N ) = 0 for all i >
0, equivalently Ext R ( M i , N ) = 0for all i ≥
0. Constructing such a diagram as above for each of the short exactsequences 0 → M i +1 → F i → M i → ) 0 → Tor Ri ( M, N ∗ ) → M i ⊗ R N ∗ h i −→ Hom R ( M i , N ) ∗ → Tor Ri − ( M, N ∗ ) → i ≥
2. From Theorem 2.1 we know that Ext iR ( M, N ) = 0 for all i ≫ Ri ( M, N ∗ ) = 0 for all i ≫
0. Hence
YMMETRY IN VANISHING EXT 17
Let R be a local Gorenstein ring, and let M and N be finitelygenerated R -modules with N maximal Cohen-Macaulay. Then Ext i ( M, N ) = 0 forall i ≫ implies the natural maps M i ⊗ R N ∗ → Hom R ( M i , N ) ∗ are isomorphismsfor all i ≫ . (cid:3) Theorem 5.9 below contains a similar result.
Ext and Stable Hom.
Recall that the stable Hom , Hom R ( M, N ), is the cokernel of the natural map M ∗ ⊗ R N → Hom R ( M, N ). Equivalently, it is the quotient of Hom R ( M, N ) by maps f : M → N which factor through a free module. Stable Homs offer a convenientway of interpreting the vanishing of all higher Ext iR ( M, N ): from (5.1) (and theexact sequence involving Ext preceding it) we see that( ) Ext iR ( M, N ) ≃ Hom R ( M i , N )for i ≥
1. Hence we may record the following as a corollary of 5.2.
Let R be a local Gorenstein ring, and let M and N be finitelygenerated R -modules with M maximal Cohen-Macaulay. Then Ext iR ( M, N ) = 0 forall i ≫ if and only if for all i ≫ every map M i → N factors through a freemodule. The next Proposition allows us to shift among the stable Homs with ease, whichoften can clarify basic vanishing results concerning Ext. ( cf.
Remark 2.2 and (1.2).)
Let M and N be finitely generated maximal Cohen-Macaulaymodules over the local Gorenstein ring R . Then we have (1) Hom R ( M, N ) ≃ Hom R ( N ∗ , M ∗ ) . (2) Hom R ( M, N ) ≃ Hom R ( M t , N t ) for all t ∈ Z .Proof. (1). The isomorphism is induced by the obvious mapping Hom R ( , R ) :Hom R ( M, N ) → Hom R ( N ∗ , M ∗ ). The fact that the induced map is an isomorphismis straightforward (since M and N are reflexive) provided it is well-defined. But thisis clear since if f ∈ Hom R ( M, N ) factors through a free module F , then Hom R ( f, R )factors through F ∗ .(2). It is enough to prove (2) in the case t = 1. Given a map f ∈ Hom R ( M, N )we get a map f ∈ Hom R ( M , N ) by completing the diagram(5.8.1) 0 −−−−→ M −−−−→ F ǫ −−−−→ M −−−−→ y f y f y f −−−−→ N −−−−→ G ǫ ′ −−−−→ N −−−−→ . Define Φ : Hom R ( M, N ) → Hom R ( M , N ) by Φ( ¯ f ) = ¯ f .We first show that Φ( ¯ f ) is determined independent of the choice of chain map { f , f } . Suppose that g : F → G and g : M → N are two other maps makingthe diagram (5.8.1) commute and such that ǫ ′ g = f ǫ . Then we have the standardhomotopy h : F → N such that f − g = h ( M ֒ → F ). That is, f − g factorsthrough a free module, so that ¯ f = ¯ g in Hom R ( M , N ).Next we show Φ is well-defined. Suppose that f ∈ Hom R ( M, N ) factors througha free module H . Then f = 0 completes the diagram0 −−−−→ M −−−−→ F ǫ −−−−→ M −−−−→ y y y −−−−→ H −−−−→ H −−−−→ y y y −−−−→ N −−−−→ G ǫ ′ −−−−→ N −−−−→ . Hence, Φ( ¯ f ) = 0, as desired.In order to show that Φ is an isomorphism we exhibit its inverse. Let g be inHom R ( M , N ). We dualize and complete the diagram0 −−−−→ M ∗ −−−−→ F ∗ −−−−→ M ∗ −−−−→ x ( g ∗ ) x ( g ∗ ) x g ∗ −−−−→ N ∗ −−−−→ G ∗ −−−−→ N ∗ −−−−→ . Define Ψ : Hom R ( M , N ) → Hom R ( M, N ) by Ψ(¯ g ) = ( g ∗ ) ∗ . It’s not hard to seethat Φ and Ψ are inverses of one another (since M and N are reflexive). (cid:3) Vanishing Ext and Cohen-Macaulayness.
We end with a theorem which shows the relationship between the vanishing of d consecutive Ext modules and the Cohen-Macaulayness of a certain tensor product. Let R be a d -dimensional local Gorenstein ring, and let M and N be maximal Cohen-Macaulay modules. Consider the following two conditions. (1) M ∗ ⊗ R N is maximal Cohen-Macaulay, (2) Ext R ( N, M ) = · · · = Ext dR ( N, M ) = 0 .Then (2) implies (1). If we assume that
Ext R ( N, M ) , . . . , Ext dR ( N, M ) have finitelength, then (1) implies (2). Furthermore, if (1) holds then Hom R ( N, M ) is max-imal Cohen-Macaulay and M ∗ ⊗ R N ≃ Hom R ( N, M ) ∗ . If (1) holds and R is alsointegrally closed, then M ∗ ⊗ R N ≃ Hom R ( M, N ) . YMMETRY IN VANISHING EXT 19
Proof.
We first prove that if (1) holds then the R -module Hom R ( N, M ) is maximalCohen-Macaulay and M ∗ ⊗ R N ≃ Hom R ( N, M ) ∗ . Note that as both M and N are reflexive, the natural map Hom R ( N, M ) → Hom R ( M ∗ , N ∗ ) is an isomorphism.Since M ∗ ⊗ R N is maximal Cohen-Macaulay so is ( M ∗ ⊗ R N ) ∗ , and we have( M ∗ ⊗ R N ) ∗ = Hom R ( M ∗ ⊗ R N, R ) ≃ Hom R ( M ∗ , N ∗ ) by Hom-tensor adjointness ≃ Hom R ( N, M )(5.9.1)Hence Hom R ( N, M ) is maximal Cohen-Macaulay and M ∗ ⊗ R N ≃ ( M ∗ ⊗ R N ) ∗∗ ≃ Hom R ( N, M ) ∗ .We prove (1)= ⇒ (2) under the assumption that Ext R ( N, M ) , . . . , Ext dR ( N, M )have finite length. We induce on d . The case in which d = 0 is vacuous. d = 1. Since Ext R ( N, M ) has finite length and M and N are maximal Cohen-Macaulay, we can choose a non-zerodivisor x ∈ R such that x is a non-zerodivisoron both M and N and x Ext R ( N, M ) = 0. For any R -module X we let X denote X/xX , and we let ∗ indicate Hom into either R or R depending on the module inquestion. We have the short exact sequence 0 → M x −→ M → M →
0, which yieldsthe exact sequence0 → Hom R ( N, M ) x −→ Hom R ( N, M ) → Hom R ( N , M ) → Ext R ( N, M ) → . Hence length(Hom R ( N , M )) = length(Hom R ( N, M )) + length(Ext R ( N, M )). Notethat for any maximal Cohen-Macaulay R -module X , X ∗ ≃ X ∗ . We havelength(Hom R ( N , M )) = length(( M ∗ ⊗ R N ) ∗ ) by (5.9.1)= length( M ∗ ⊗ R N ) since R is 0-dimensional= length( M ∗ ⊗ R N ) since M ∗ ≃ M ∗ = length( M ∗ ⊗ R N )= length(Hom R ( N, M ) ∗ ) by (5.9.1)= length(Hom R ( N, M ) ∗ )= length(Hom R ( N, M )) since R is 0-dimensional . Therefore length(Hom R ( N, M )) = length(Hom R ( N, M )) + length(Ext R ( N, M )),and so Ext R ( N, M ) = 0. d >
1. Choose a parameter x ∈ ∩ di =1 ann R Ext iR ( N, M ). We have M ∗ ⊗ R N ≃ M ∗ ⊗ R N is maximal Cohen-Macaulay. The short exact sequence 0 → M x −→ M → M → → Ext iR ( N, M ) → Ext iR ( N, M ) → Ext i +1 R ( N, M ) → i = 1 , . . . , d −
1. Hence Ext iR ( N, M ) have finite length for 1 ≤ i ≤ d −
1. SinceExt iR ( N, M ) ≃ Ext iR ( N , M ) for all i [R], it follows by induction that Ext R ( N , M ) = · · · = Ext d − R ( N , M ) = 0. Now the exact sequences (5.9.2), for i = 1 , . . . , d −
1, andthe fact that Ext iR ( N, M ) ≃ Ext iR ( N , M ) for all i ≥ iR ( N, M ) = 0 for i = 1 , . . . , d .Assume (2). Note then that Hom R ( N, M ) is maximal Cohen-Macaulay: let F : · · · → F → F → N → R -free resolution of N . Applying Hom R ( , M )and using our hypothesis we get the exact sequence0 → Hom R ( N, M ) → Hom R ( F , M ) → · · · → Hom R ( F d +1 , M ) . Each Hom R ( F i , M ) is maximal Cohen-Macaulay since M is. By counting depthsalong this exact sequence we get the desired conclusion.For (2) = ⇒ (1) we again induce on d . The case d = 0 is trivial. d = 1. Let x be a non-zerodivisor on both M and N . The short exact sequence0 → M x −→ M → M → → Hom R ( N, M ) x −→ Hom R ( N, M ) → Hom R ( N , M ) → . Therefore(5.9.3) Hom R ( N, M ) ≃ Hom R ( N , M )Consider the natural map M ∗ ⊗ R N h −→ Hom R ( N, M ) ∗ . We have a commutativediagram M ∗ ⊗ R N h −−−−→ Hom R ( N, M ) ∗≃ y ≃ y M ∗ ⊗ R N ≃ −−−−→ Hom R ( N , M ) ∗ where the right vertical arrow comes from (5.9.3) and the bottom arrow is theisomorphism of (5.9.1). Thus h is an isomorphism modulo x . By Nakayama’slemma, h must be onto. Now we have a short exact sequence0 → K → M ∗ ⊗ R N h −→ Hom R ( N, M ) ∗ → . YMMETRY IN VANISHING EXT 21
The fact that Hom R ( N, M ) is maximal Cohen-Macaulay implies that K = 0, andtherefore K = 0. Thus h is an isomorphism and M ∗ ⊗ R N is maximal Cohen-Macaulay. d >
1. For x a non-zerodivisor on M and N , the hypothesis yields Ext R ( N , M ) = · · · = Ext d − R ( N , M ) = 0, so by induction M ∗ ⊗ R N is maximal Cohen-Macaulay,which means so is M ∗ ⊗ R N .Finally suppose that R is integrally closed. There is always a natural map fromHom R ( M, N ) → Hom R ( N, M ) ∗ obtained by composition, and this map is an iso-morphism if either N or M is free. Since M and N are maximal Cohen-Macaulaymodules and R P is regular if the height of P is at most one, it follows that thisnatural map is an isomorphism in codimension one. It is a standard result that amap between reflexive modules which is an isomorphism in codimension one mustitself be an isomorphism. Hence, as both Hom R ( M, N ) and Hom R ( N, M ) ∗ are re-flexive, the natural map map Hom R ( M, N ) → Hom R ( N, M ) ∗ is an isomorphism.The stated isomorphism of M ∗ ⊗ R N with Hom R ( M, N ) follows from the aboveparagraph. (cid:3)
6. Questions
This work leaves quite a few questions unresolved. We list a few for further study.Perhaps the most intriguing is1. Are all local Gorenstein rings AB rings?Some other interesting questions are:2. Let R and S be AB rings which are essentially of finite type over a field k . Is R ⊗ k S locally an AB ring?3. Are localizations of AB rings AB rings?4. Are AB rings Ext-bounded? References [AB] M. Auslander and M. Bridger,
Stable Module Theory , Memoirs of the A.M.S. (1969),American Math. Society, Providence, R.I..[AvBu] L. L. Avramov and R-.O. Buchweitz, Support varieties and cohomology over completeintersections , Invent. Math. (2000), 285-318.[L] J. Lescot,
Asymptotic properties of Betti numbers of modules over certain rings , J. PureAppl. Algebra (1985), 287-298.[M] V. Ma¸sek, Gorenstein dimension and torsion of modules over commutative Noetherianrings , preprint.[R] D. Rees,
A theorem of homological algebra , Proc. Camb. Phil. Soc. (1956), 605-610.2 CRAIG HUNEKE AND DAVID A. JORGENSEN[Ro] J. Rotman, An Introduction to Homological Algebra , Academic Press, New York, 1979.[WITO] K.I. Watanabe, T. Ishikawa, S. Tachibana, and K. Otsuka,
On tensor products of Goren-stein rings , J. Math. Kyoto Univ. (1969), 413–423. Department of Mathematics, University of Kansas, Lawrence, KS 66045
E-mail address : [email protected] Department of Mathematics, University of Texas at Arlington, Arlington, TX76019
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