Cohomological splitting, realization, and finiteness
aa r X i v : . [ m a t h . A C ] F e b COHOMOLOGICAL SPLITTING, REALIZATION, AND FINITENESS
MOHSEN ASGHARZADEHA
BSTRACT . We search some splitting (resp. finiteness) criteria by applying certain splitting (resp. finiteness)property of the cohomological functors. In particular, we deal with the cohomological splitting questionposted by Vasconcelos. We present a connection from our approach to the realization problem of Nunke.This equipped with several applications. For instance, we recover some results of Jensen (and others) byapplying simple methods. Additional applications include a computation of some numerical invariants suchas the projective dimension of some injective modules and et cetera.
1. I
NTRODUCTION
An exact sequence ζ : = → M → M → M → i )provided the local cohomology module H i m ( M ) decomposes into H i m ( M ) ⊕ H i m ( M ) . Here, we studythe cohomological splitting property and its connection with the classical splitting property of ζ . Theinitial motivation comes from the following beautiful theorem of Vasconcelos [27]: Theorem 1.1.
Let ( R , m ) be a -dimensional Gorenstein local ring and let ζ be the exact sequence → M → M → M → of finitely generated modules of projective dimension at most one. If tor ( M ) = tor ( M ) ⊕ tor ( M ) , then ζ splits. For each functor F , we assign the concept of F -splitting. In §2 we deal with the tensor and homfunctors, and study the corresponding splitting types. Then we investigate splitting with respect totheir derived functors, i.e., we study the Tor-splitting and Ext-splitting. These continue the work ofGuralnick [14], and may regard as a root of cohomological splitting property.Vasconcelos posted the following splitting question: Does Theorem 1.1 holds for any 1-dimensionalCohen-Macaulay rings? §3 equipped with a series of observations about the cohomological splittingquestion and presents a higher-dimensional analogue of Vasconcelos’ theorem. We do these, by apply-ing various aspects of cohomological splitting property. For a sample, see Proposition 3.4 and 3.13, andalso, Theorem 3.5 and its corollaries.The next goal is to find sub-functors or even direct summands of H i m ( − ) , see Proposition 4.1 andsubsequent examples. This is inspired from Auslander’s comments on the functor ext [3], where heinvestigated sub-functors of Ext. Using these, we select the following application: Proposition 1.2.
Let ( R , m ) be a complete Gorenstein local ring and M be finitely generated. Suppose H i m ( M ) is nonzero and injective. Then i = dim R. In particular, M is ( S ) if and only if M is free. We note that freeness of a reflexive module is a challenging problem. For instance, see [2] and refer-ences therein. Despite the importance of ( S ) , the following easy consequence of Proposition 1.2 holdseven the module is not assumed ( S ) : Let R : = k [[ x , . . . , x d ]] where k is a field of zero characteristic Mathematics Subject Classification.
Primary 13D45; 13D07.
Key words and phrases. cohomologically finite; Ext-modules; finiteness condtions; local cohomology; Gorenstein rings; splittingcriteria; torsion theory. and let M be a finitely generated R -module. We show M is holonomic if and only if M is free as an R -module. For the prime characteristic version, see Corollary 4.8.A module M is called cohomologically finite (length) if Ext iR ( M , R ) is finite (length) for all i . §5 dealswith cohomologically finite (length) modules, and connect them to finite (length) modules. For instance,see Proposition 5.2. This is inspired from a result of Bredon from his book [4].A module M is called cohomologically zero if Ext iR ( M , R ) = i . For a finitely generatedmodule, zero and cohomologically zero are equivalent notations. For more information on this see[1]. In §6 we deal with cohomologically zero injective modules. This is motivated from Auslander’spaper [3], where he proved the cohomologically zero property of the fraction field of a complete localdomain. We extend this result and present some partial converse. Then, we study the cohomologicalzero property of E R ( R / p ) where p ∈ Spec ( R ) . To state an application, recall that Osofsky [22] computedprojective dimension of injective envelop of regular rings. Her answer is very interesting: it dependsto your culture, i.e., accept generalized continuum hypothesis or not. In this regard, a natural questionarises: What can say about projective dimension of E R ( R / p ) ? We close §6 by answering this.We start §7 by a connection to the realization problem of Nunke [21]. This may consider as an ap-plication of §6. Our elementary approach reproves some technical (and important) results of Jensen[9]. Additional properties of (homologically) realizable modules are given. We do these by applyingsome basic properties of local cohomology modules, a tool introduced by Grothendieck some years af-ter than [21]. For more details, see Proposition 7.12. In particular, Matlis duality has an application tothe realization problem.In §8 we study cohomologically zero flat modules. Concerning this, there are series of interestingworks on a result of Jensen [9] (reproved by Buchweitz and Flenner [5, Corollary 1]). Both proofs are nottrivial. In Corollary 8.2, we use an easy natural transformation originally due to Auslander, to simplify[5, Corollary 1]. We apply the realization problem to present another homological property of Dedekinddomains. This was started by the seminal works [21] and [9].In the final section, we give a splitting criteria for short exact sequences of torsion-free modules.Again, some local cohomology arguments appear. For more details, see Proposition 9.1. Matlis in-troduced the concept of D -rings, and Kaplansky proved D-rings are complete in R -topology. As anapplication of the splitting criteria, we reprove this.2. R OOTS OF COHOMOLOGICAL SPLITTING
Let ( R , m , k ) be a commutative noetherian local ring. Let ζ : = → M → M → M → L be a module. By L ⊗ R ζ we mean the complex 0 → L ⊗ R M →L ⊗ R M → L ⊗ R M → R ( L , ζ ) stands for the complex0 → Hom R ( L , M ) → Hom R ( L , M ) → Hom R ( L , M ) → Fact . (See [27, Theorem 3.28]) If M ∼ = M ⊕ M and all modules under consideration are finitelygenerated, then ζ splits.In the case of maximal ideals the next result was proved by Striuli in her thesis [26]. Here, we extendit by a simple argument: Observation . Let ( R , m ) be local. Let a ✁ R and ζ : = → M → M → M → ζ splits if and only if R a n ⊗ R ζ splits for all n ≫ Proof.
Let a ✁ R . Suppose R a n ⊗ R ζ splits for all n ≫
0. By b − we mean m -adic completion. Thanks to Fact2.1 ζ splits if we can show that M ∼ = M ⊕ M . This is the case provided b M ∼ = c M ⊕ c M , since modulesare finitely generated. Then we may assume that R is m -adically complete. In particular, R is a -adicallycomplete. Recall that M a n M ∼ = M a n M ⊕ M a n M for all n ≫
0. Recall that completion commutes with finitedirect sums. Taking inverse limit, M b a ∼ = M b a ⊕ M b a . Since modules are finitely generated, M ∼ = M ⊕ M .This completes the proof. (cid:3) Let ( − ) v : = Hom R ( − , E R ( k )) be the Matlis functor, where E R ( k ) denotes the injective envelope of k . Observation . Let ( R , m ) be local. Let a ✁ R and ξ : = → M → M → M → ξ splits if and only if Hom R ( R a n , ξ ) splits for all n ≫ Proof.
Suppose Hom R ( R a n , ξ ) splits for all n ≫
0. It is easy to any artinian module has the structure of b R -module, compatible with the R -module structure. From this, and without loss of the generality, wemay assume that R is complete. Recall that Hom R ( R a n , ξ ) v splits. In view of Hom-Evaluation, R a n ⊗ R ξ v splits for all n ≫
0. According to Matlis theory, ζ : = ξ v is a short exact sequence of finitely generatedmodules. Due to Observation 2.2, ζ splits. Taking another Matlis duality, we see that ξ = ζ vv splits. (cid:3) Concerning Observation 2.3 (resp. 2.2), the artinian (resp. finitely generated) assumption of modulesin ξ (resp. ζ ) is important. To this end, by Syz n ( − ) we mean the n -th syzygy module of ( − ) . Example . Let ( R , m ) be a local integral domain of dimension d ≥
2. We look at the exact sequence ξ : = → Syz ( m ) → R β ( m ) → m → R ( R m n , ξ ) splits, in fact the complex Hom R ( R m n , ξ ) consists of zero modules, because ξ consists of modules ofpositive depth. If ξ splits, it follows by definition that m is projective. It turns out that m is principal, acontradiction with the generalized Krull’s principal ideal theorem. Definition 2.5.
A short exact sequence ζ : = → M → M → M → i withrespect to a , if 0 → Ext iR ( R / a n , M ) → Ext iR ( R / a n , M ) → Ext iR ( R / a n , M ) → n ≫ a . Lemma 2.6.
Let ( R , m ) be local. Suppose a short exact sequence ζ : = → M → M → M → E-splits atlevel i with respect to a . Then ζ v T-splits at level i with respect to a . The converse holds if ζ consists of finitelygenerated modules.Proof. By definition, 0 → Ext iR ( R / a n , M ) → Ext iR ( R / a n , M ) → Ext iR ( R / a n , M ) → iR ( R / a n , L ) v ∼ = Tor Ri ( R / a n , L v ) . From this, ζ v is of T-splittingtype at level i . For the converse, without loss of the generality we assume that R is complete, and notethat ζ vv = ζ . (cid:3) Remark . One may extend the converse part of Lemma 2.6 by assuming ζ consists of Matlis reflexivemodules. Recall by a result of Enochs and Zoschinger that M is Matlis reflexive iff M contains a finitelygenerated submodule N such that M / N is artinian.Here, is a sample of E-splitting (resp. T-splitting): Proposition 2.8.
Let ( R , m ) be local. Let ζ : = → M → M → M → be the exact sequence of finitelygenerated modules and let a be an ideal generated by a regular element x. If ζ E-splits at level one with respect to a , then ζ splits. Proof.
The free resolution of R / a n is given by 0 → R x n −→ R → R / a n →
0. Then for any L , we haveExt R ( R / a n , L ) = H ( L x n −→ L → ) = L / x n L . The assumption says that 0 → Ext R ( R / a n , M ) → Ext R ( R / a n , M ) → Ext R ( R / a n , M ) → → M / x n M → M / x n M → M / x n M → R a n ⊗ R ζ splits for all n ≫
0. In view of Observation 2.2 ζ splits. (cid:3) Proposition 2.9.
Let ( R , m ) be local. Let ζ : = → M → M → M → be the exact sequence of artinianmodules and let a be an ideal generated by a regular element x. If ζ T-splits at level one with respect to a , then ζ splits.Proof. One may use Proposition 2.8 and Lemma 2.6. Here, is a direct proof. The free resolution of R / a n is given by 0 → R x n −→ R → R / a n →
0. Then for any L , we haveTor R ( R / a n , L ) = H (( → R x n −→ R ) ⊗ R L ) = ker ( L x n −→ L ) = ( L x n ) ∼ = Hom R ( R / x n R , L ) ( ∗ ) The assumption says that 0 → Tor R ( R / a n , M ) → Tor R ( R / a n , M ) → Tor R ( R / a n , M ) → ( ∗ ) , we observe that Hom R ( R a n , ζ ) splits for all n ≫
0. According to Observation2.3 we know that ζ splits. (cid:3) Definition 2.10.
Let
Z ⊂
Spec R and p ∈ Z . Then Z is called specialization-closed if V ( p ) ⊂ Z . Example . Here, are some examples of specialization-closed sets:a) Every closed subset of Spec ( R ) with respect to Zariski topology is specialization-closed.b) The set Spec ( R ) \ min ( R ) is specialization-closed.c) Suppose R is equidimensional. The set Spec ( R ) \ Ass ( R ) is specialization-closed.Let Z ⊂
Spec R be specialization-closed, and define Γ Z ( M ) : = { m ∈ M | Supp R ( Rm ) ⊆ Z } . For i ∈ N , the i -th right derived functor of Γ Z ( − ) , denoted by H i Z ( − ) . In the case Z : = V ( a ) this is H i a ( − ) .Here, we state two easy facts about them: Fact . Let ( R , m ) be equidimensional, and let Z : = Spec ( R ) \ Ass ( R ) . Then Γ Z ( L ) = tor ( L ) . Fact . Let ( R , m ) be 1-dimensional and equidimensional. Then H m ( L ) = Γ Z ( L ) = tor ( L ) . Definition 2.14.
Let ζ : = → M → M → M → ζ is called cohomologically splits with respect to a , if 0 → H i a ( M ) → H i a ( M ) → H i a ( M ) → i . Observation . E -splitting implies cohomologically splitting.
Proof.
This follows by H i a ( − ) = lim −→ n Ext iR ( R / a n , − ) . (cid:3) Here, is a sample to checking the cohomologically splitting property.
Observation . Let ( R , m ) be a 3-dimensional quasi-Gorenstein local ring and let ζ : = → M → M → M → H d − m ( M ) = H d − m ( M ) ⊕ H d − m ( M ) . If M is free, then ζ cohomologically splits with respect to m . Proof.
Quasi-Gorenstein rings satisfy Serre’s condition ( S ) . From this, H m ( M ) = H m ( M ) =
0. Inparticular, 0 → = H m ( M ) → H m ( M ) → H m ( M ) → H m ( M ) = = H m ( M ) → H m ( M ) → H m ( M ) → H m ( M ) f −→ H m ( M ) → H m ( M ) to deduce that H m ( M ) → H m ( M ) is surjective, because f is injective. By this, 0 → = H m ( M ) → H m ( M ) → H m ( M ) → H m ( M ) g → H m ( M ) → H m ( M ) → H m ( M ) → H m ( M ) → g is surjective, ξ : = → H m ( M ) → H m ( M ) → H m ( M ) → M is free, H m ( M ) is injective as an R -module. Thus ξ splits. In sum, ζ cohomologically splits with respect to m . (cid:3) In Corollary 3.6 (resp. 3.7) we remove the reflexivity (resp. Cohen-Macaulay) assumption of:
Corollary 2.17.
Let ( R , m ) be a d-dimensional Gorenstein local ring and let ζ : = → M → M → M → be an exact sequence of finitely generated reflexive modules such that H d − m ( M ) = H d − m ( M ) ⊕ H d − m ( M ) . IfM is free, then ζ splits. The notation ( − ) ∗ stands for Hom R ( − , R ) . Proof.
By using the proof of Observation 2.16, 0 → H d m ( M ) → H d m ( M ) → H d m ( M ) → M ∗ ∼ = M ∗ ⊕ M ∗ . Taking another ( − ) ∗ yields that M ∗∗ ∼ = M ∗∗ ⊕ M ∗∗ . By reflexivity, M ∼ = M ⊕ M . According to Fact 2.1 ζ splits. (cid:3) By pd R ( − ) we mean the projective dimension of ( − ) . Corollary 2.17 is not true if instead of M weassume M is free: Example . Let ( R , m , k ) be a Cohen-Macaulay local ring of dimension d > ζ : = → Syz d + ( k ) → R β d ( k ) → Syz d ( k ) →
0. This is a sequence of maximal Cohen-Macaulay modules, and reflexive modules, because they are second syzygy modules as d >
1. Thus,0 = H d − m ( M ) ∼ = H d − m ( M ) ⊕ H d − m ( M ) =
0. Suppose on the way of contradiction that ζ splits. Bydefinition, Syz d + ( k ) is projective. Thus pd R ( k ) < ∞ , and consequently, R is regular. This is excludedby the assumption. 3. C OHOMOLOGICAL SPLITTING
We start with
Question . (See [27, Page 78]) Does Theorem 1.1 holds for any 1-dimensional Cohen-Macaulay rings? Proposition 3.2.
Question 3.1 is true if M is torsion-free.Proof.
In view of tor ( M ) = tor ( M ) ⊕ tor ( M ) we deduce that tor ( M ) =
0. Without loss of the general-ity we assume that M =
0. This implies that M is of positive depth. Thanks to Auslander-Buchsbaumformula, we know M is projective. By definition, the sequence 0 → M → M → M → (cid:3) The above result shows the assumption d > Observation . Question 3.1 is true if in addition M is torsion-free.By id R ( − ) we mean the injective dimension of ( − ) . Proposition 3.4.
Let R be a local ring of dimension one, M be torsion-free and M be of finite injective dimension.Let ζ : = → M → M → M → be an exact sequence of finitely generated modules. If tor ( M ) = tor ( M ) ⊕ tor ( M ) , then ζ splits. Proof.
According to a theorem of Bass, R is Cohen-Macaulay. Also, id R ( M ) = depth R ( R ) =
1. Fromtor ( M ) = tor ( M ) ⊕ tor ( M ) we deduce that tor ( M i ) =
0. In particular, if y is a regular element of R , itis regular over any of { M , M } . Such a thing exists, as R is Cohen-Macaulay. This true for any powersof y n . Set R : = Ry n R . We have Tor R ( R , M ) = ker ( M y n −→ M ) =
0. We apply − ⊗ R R to ζ and obtain0 = Tor R ( R / y n , M ) −→ M ⊗ R R −→ M ⊗ R R −→ M ⊗ R R −→ R ( M ⊗ R R ) = id R ( M ) − =
0, the above sequence splits as an R -module. Recall thatHom R ( M ⊗ R R , M ⊗ R R ) = Hom R ( M ⊗ R R , M ⊗ R R ) , e.g., the above sequence splits as an R -module.Now, recall from [25, Proposition 2.8] that:Fact A): Let A , B be finitely generated, let x ∈ R be a non-zerodivisor on R , A , B and let α : = → B → X α → A → α ⊗ R / xR splits. Then α ∈ x Ext R ( A , B ) .In the light of Fact A) we see ζ ∈ \ n ∈ N y n Ext R ( M , M ) = ζ splits. (cid:3) In order to present a higher dimensional version of Theorem 1.1, we borrow some lines from [27].
Theorem 3.5.
Let ( R , m ) be a d-dimensional Gorenstein local ring with d > and let ζ be the exact sequence → M → M → M → of finitely generated modules of projective dimension at most . If H d − m ( M ) ∼ = H d − m ( M ) ⊕ H d − m ( M ) , then ζ splits.Proof. We apply the local duality theorem to see Ext R ( M , R ) ∼ = ⊕ i Ext R ( M i , R ) . This yields Ext R ( M , R ) ⊗ R M ∼ = ⊕ i Ext R ( M i , R ) ⊗ R M . Let C be the family of finitely generated modules. Let F : C → Ab be thefunctor defined by F ( L ) : = Ext R ( M , L ) . Since pd ( M ) is at most 1, this functor is right exact. Also,it preserves direct sums. By Watts’ theorem, Ext R ( M , L ) = F ( L ) ∼ = F ( R ) ⊗ R L = Ext R ( M , R ) ⊗ R L .We apply this to deduce that Ext R ( M , M ) ⊕ Ext R ( M , M ) ∼ = Ext R ( M , M ) . Thanks to Fact 2.1 thefollowing natural sequence0 −→ Ext R ( M , M ) −→ Ext R ( M , M ) −→ Ext R ( M , M ) −→ R ( M , M ) → Hom R ( M , M ) is surjective. By looking at the preimage of 1 ∈ Hom R ( M , M ) , ζ splits. (cid:3) Corollary 3.6.
Let ( R , m ) be a d-dimensional Gorenstein local ring and let ζ : = → M → M → M → bean exact sequence of finitely generated modules such that H d − m ( M ) ∼ = H d − m ( M ) ⊕ H d − m ( M ) . If M is free,then ζ splits.Proof. This is in the proof of Theorem 3.5. (cid:3)
Corollary 3.7.
Let ( R , m , k ) be a d-dimensional quasi-Gorenstein local ring and let ζ : = → M → M → M → be an exact sequence of finitely generated reflexive modules such that H d − m ( M ) = H d − m ( M ) ⊕ H d − m ( M ) . If M is free, then ζ splits.Proof. A ring is quasi-Gorenstein if and only if its completion is quasi-Gorenstein. Let L : = M and L : = M ⊕ M be finitely generated modules. If L ⊗ R b R ∼ = L ⊗ R b R , then L ∼ = L . According toFact 2.1, we may and do assume that R is complete. It follows from the proof of Observation 2.16 thatthe sequence 0 → H d m ( M ) → H d m ( M ) → H d m ( M ) → H > d m ( − ) =
0. So, the functor H d m ( − ) is right exact. Also, it preserves direct sums. Due to the Watts’theorem we know that H d m ( − ) ∼ = ( − ) ⊗ R H d m ( R ) ∼ = ( − ) ⊗ R E R ( k ) . We combine this with the previousobservation and conclude that L i M i ⊗ R E R ( k ) ∼ = M ⊗ R E R ( k ) . We apply the Matlis functor to see M i = Hom R ( M i ⊗ R E R ( k ) , E R ( k )) ∼ = Hom R ( M ⊗ R E R ( k ) , E R ( k )) .Recall that Hom R ( E R ( k ) , E R ( k )) ∼ = b R = R . In the light of tensor-hom adjunction we observe ⊕ i M ∗ i ∼ = ⊕ i Hom R ( M i , Hom R ( E R ( k ) , E R ( k )) ∼ = Hom R ( M , Hom R ( E R ( k ) , E R ( k )) ∼ = M ∗ .Taking another ( − ) ∗ yields that M ∗∗ ∼ = M ∗∗ ⊕ M ∗∗ . Applying reflexivity yields that M ∼ = M ⊕ M . Inview of Fact 2.1 ζ splits. (cid:3) Discussion . i) Let P ∂ → P ∂ → M → M . The transpose of M isTr M : = coker ( ∂ ∗ ) . There is a useful exact sequence:Tor R ( Tr Syz n M , N ) → Ext nR ( M , R ) ⊗ R N f n −→ Ext nR ( M , N ) → Tor R ( Tr Syz n M , N ) → ( ∗ ) ii) Concerning Theorem 3.5, we present a replacement for Watts’ theorem. Since pd ( M ) ≤
1, Syz ( M ) is free. In particular, its presentation is given by 0 ∂ → P ∂ → Syz M →
0. By definition Tr Syz ( M ) =
0, and so Tor R ( Tr Syz ( M ) , M ) = Tor R ( Tr Syz ( M ) , M ) =
0. In view of ( ∗ ) , Ext R ( M , R ) ⊗ R M ∼ = Ext R ( M , M ) .iii) One may control the error term coker ( f ) in a nontrivial case. Suppose pd ( M ) =
2. ThenTr Syz ( M ) ∼ = Ext R ( Syz ( M ) , R ) ∼ = Ext R ( M , R ) . So, coker ( f ) = Tor R ( Ext R ( M , R ) , M ) .One may like to conclude the splitting of a short exact sequence of modules of projective dimensionat most one, from splitting of the corresponding sequence of torsion modules. This is not the case: Example . Let ( R , m ) be a 2-dimensional regular local ring and note that ζ : = → R → R → m → ζ is not of splitting type.Concerning Theorem 3.5, the presented bound on projective dimension is optimal: Example . Let ( R , m , k ) be a 3-dimensional regular local ring and note that ζ : = → R → R → m → = H d − m ( M ) ∼ = H d − m ( M ) ⊕ H d − m ( M ) =
0, but ζ is not of splitting type. Indeed, we use 0 → m → R → k → = H m ( k ) → H m ( m ) → H m ( R ) = H d − m ( M ) = ζ splits. Then m is principal. This implies that R is 1-dimensional,which is excluded by the assumption.Concerning Theorem 3.5, one can not replace projective dimension with G -dimension: Example . Let ( R , m ) be a 1-dimensional Gorenstein ring which is not regular. We look at the exactsequence ζ : = → Syz ( m ) → R β ( m ) → m → G -dimension zerosuch that the corresponding sequence of torsions splits. Suppose on the way of contradiction ζ splits.Then m is free. This implies that R is regular, which is excluded by the assumption. Fact . (See [27, Proposition 2.22]) Let R be a commutative ring and f : M m → M n be a homomor-phism of R –modules given by an n × m matrix ( a ij ) . Then f is injective iff the ideal generated by theminors of order m does not annihilated a nonzero element of M . Proposition 3.13.
Let ( R , m ) be a -dimensional Cohen-Macaulay local ring with a canonical module ω R and let α : = → M → M → M → be a short exact sequence of finitely generated modules of projective dimensionat most one. The following holds:a) The sequence ζ : = → M ⊗ ω R → M ⊗ ω R → M ⊗ ω R → is exact.b) If tor ( M ⊗ ω R ) = tor ( M ⊗ ω R ) ⊕ tor ( M ⊗ ω R ) , then ζ splits.Proof. a ) : Let ξ : = → R m f −→ R n → M → M . Note that f is represents byan n × m matrix ( a ij ) . Let I be the ideal generated by the minors of order m . In the light of Fact 3.12, I does not annihilated a nonzero element of R . Tensor ξ with ω R we have the following exact sequence0 −→ Tor R ( M ⊗ , ω R ) −→ R m ⊗ ω R f ⊗ −→ R n ⊗ ω R −→ M ⊗ ω R −→ (+) Note that f ⊗ ( a ij ) . If I is annihilated a nonzero element of ω R , then I is annihilated anonzero element of Hom R ( ω R , ω R ) . Since Hom R ( ω R , ω R ) = R we get to a contradiction. Thus, I isnot annihilated a nonzero element of ω R . Thanks to Fact 3.12, f ⊗ (+) we seeTor R ( M , ω R ) =
0. Tensor α with ω R we have the following exact sequence0 = Tor R ( M , ω R ) −→ M ⊗ ω R −→ M ⊗ ω R −→ M ⊗ ω R −→ a ) .b): Since R is one-dimensional, and in view of Fact 2.13, H m ( − ) = tor ( − ) . We use this along withour assumption to see H m ( M ⊗ ω R ) ∼ = ⊕ i H m ( M i ⊗ ω R ) . We apply the local duality theorem to seeExt R ( M ⊗ ω R , ω R ) ∼ = ⊕ i Ext R ( M i ⊗ ω R , ω R ) . Tensoring this with M , we obtain Ext R ( M ⊗ ω R , ω R ) ⊗ R M ∼ = ⊕ i Ext R ( M i ⊗ ω R , ω R ) ⊗ R M . Let C be the family of modules of projective dimension at most one.Let F : C → Ab be the functor defined by F ( L ) : = Ext R ( M ⊗ ω R , L ⊗ ω R ) . Let 0 → R m → R n → L → L . We observed in part a ) that 0 → R m ⊗ ω R → R n ⊗ ω R → L ⊗ ω R → R ( ⊕ ω R ) < ∞ . From this, id R ( L ⊗ ω R ) < ∞ , and so id R ( L ⊗ ω R ) = depth R ( R ) =
1. Thisimplies that the functor F is right exact. Also, it preserves direct sums. In the light of Watts’ theorem wesee Ext R ( M ⊗ ω R , L ⊗ ω R ) = F ( L ) ∼ = F ( R ) ⊗ R L = Ext R ( M ⊗ ω R , ω R ) ⊗ R L .We apply this along with Fact 2.1 to deduce that the following natural sequence0 −→ Ext R ( M ⊗ ω R , M ⊗ ω R ) −→ Ext R ( M ⊗ ω R , M ⊗ ω R ) −→ Ext R ( M ⊗ ω R , M ⊗ ω R ) −→ R ( M ⊗ ω R , M ⊗ ω R ) → Hom R ( M ⊗ ω R , M ⊗ ω R ) is surjective. Bylooking at the pre-image of 1 ∈ Hom R ( M ⊗ ω R , M ⊗ ω R ) we deduce M ⊗ ω R ∼ = L i = ( M i ⊗ ω R ) , andwe get the desired claim. (cid:3)
4. S UB - FUNCTORS OF COHOMOLOGY
Proposition 4.1.
Let ( R , m ) be a d-dimensional Cohen-Macaulay complete local ring and let F be an additivefunctor which is a direct summand of H d m ( − ) and preserves direct sums. Then F is trivial. Proof.
By Grothendieck’s vanishing theorem, F is right exact. According to Watts’ theorem, F ( L ) = F ( R ) ⊗ R L . In the light of Matlis theory, H d m ( R ) is indecomposable if and only if H d m ( R ) v is inde-composable. We apply the local duality theorem to deduce that H d m ( R ) v = ω R which is indecom-posable. Note that F ( R ) is a direct summand of H d m ( R ) . Since H d m ( R ) is indecomposable, either F ( R ) = F ( R ) = H d m ( R ) . We plug this in the previous observation to see either F ( L ) = F ( L ) = F ( R ) ⊗ R L = H d m ( R ) ⊗ R L ∼ = H d m ( L ) , i.e., F = F ( − ) = H d m ( − ) . (cid:3) The Cohen-Macaulay assumption is important:
Example . Let R : = Q [[ x ,..., x ]]( x , x ) ∩ ( x , x ) . Then H ( x , x ) ( − ) ⊕ H ( x , x ) ( − ) ∼ = H m ( − ) is a nontrivial decompo-sition of functors. Proof.
Recall that H i ( − ) = i >
0. By Mayer–Vietoris sequence, we have0 = H ( x , x ) ∩ ( x , x ) ( − ) → H ( x , x ) ( − ) ⊕ H ( x , x ) ( − ) → H m ( − ) → H ( x , x ) ∩ ( x , x ) ( − ) = H m ( − ) ∼ = H ( x , x ) ( − ) ⊕ H ( x , x ) ( − ) . Due to Lichtenbaum–Hartshorne vanishing theorem, bothof H ( x , x ) ( − ) and H ( x , x ) ( − ) are nonzero. (cid:3) Concerning Proposition 4.1, the direct summand assumption is needed:
Example . Let ( R , m ) be a d -dimensional regular ring and let F ( − ) : = Ext dR ( R / m , − ) . We left to thereader to check that F ( − ) ֒ → H d m ( − ) and that F is nontrivial.Finding direct summand of H ∗ m ( − ) inspired from [3], and has the following applications: Fact . (See [1, Claim 4.1.A]) Let M be a finitely generated module over any commutative ring and let L be any module. Then Ext iR ( M , L ) has no nonzero projective submodule for all i > Proposition 4.5.
Let ( R , m ) be a complete Gorenstein local ring and M be finitely generated. Suppose H i m ( M ) is nonzero and injective. Then i = dim R. In particular, the following are equivalent:a) M is ( S ) ,b) M is free,c) M is Cohen-Macaulay.Proof. Let d : = dim R . First, assume that d − i >
0. By local duality, Ext d − iR ( M , R ) is free. According toFact 4.4, this is possible only if Ext d − iR ( M , R ) =
0. In dual words, H i m ( M ) =
0. But, this case excludedfrom the assumptions. Hence, i = d . Now, we prove the particular case. The only nontrivial implicationis a ) ⇒ b ) . Thus, we assume that M is ( S ) . Recall that H d m ( M ) is injective. In view of Matlis duality, M ∗ is free. So, M ∗∗ is as well. Since M is ( S ) it follows from [7, Theorem 3.6] that M is reflexive. Therefore, M ∼ = M ∗∗ is free. (cid:3) The ( S ) -condition is not a consequence of the injectivity of the nonzero module H i m ( M ) . Example . Let ( R , m , k ) be a quasi-Gorenstein ring of dimension d ≥
2. In view of 0 → m → R → k → H d m ( m ) = H d m ( R ) = E R ( k ) is injective, but m is not free. As another example, look at the toplocal cohomology of M : = R ⊕ k .We conclude the following easy facts from Proposition 4.5. Corollary 4.7.
Let R : = k [[ x , . . . , x d ]] where k is a field of zero characteristic and let M be a finitely generatedR-module. Then M is holonomic (or, more generally a Der ( R , k ) -module) if and only if M is free as an R-module. Proof.
Without loss of the generality we may assume that M is nonzero. Suppose M is holonomic. Let r : = depth R ( M ) . Recall that H r m ( M ) =
0. Due to a result of Lyubeznik [11] we know id ( H r m ( M )) ≤ dim ( H r m ( M )) =
0. In the light of Proposition 4.5 we see r = dim R . By definition, M is maximal Cohen-Macaulay. We conclude from Auslander-Buchsbaum formula that M is free. The reverse implication isalways true. (cid:3) Corollary 4.8.
Let R be a regular ring of prime characteristic, and let M be a finitely generated R-module. ThenM is F -module if and only if M is projective as an R-module.Proof. Without loss of the generality we may assume that R is local. Along the same lines as Corollary4.7 we get the desired claim. (cid:3) Remark . We left to the reader to formulate Proposition 4.5 in the setting of Cohen-Macaulay rings.5. C
OHOMOLOGICAL FINITENESS
Recall that a module M is called cohomologically finite if Ext iR ( M , R ) is finitely generated as an R -module for all i ≥
0. Here, we use some ideas of Bredon:
Lemma 5.1.
Let ( R , m ) be a regular local ring of dimension d > and M be torsion-free and cohomologicallyfinite. Then Ext dR ( M , N ) = for any finitely generated R-module N.Proof. First, we show that Ext dR ( M , R ) =
0. Let T : = ker ( M f −→ M ∗∗ ) and F : = im ( f ) . There is anexact sequence 0 → T → M → F →
0. Submodules of a torsion-free is torsion-free, e.g., T is torsion-free. Let r ∈ R . Since T is torsion-free, there is an exact sequence 0 → T r −→ T → T / r T →
0. Thisinduces Ext dR ( T , R ) r −→ Ext dR ( T , R ) → Ext d + R ( T / r T , R ) =
0, i.e., Ext dR ( T , R ) is divisible. Since R is ofglobal dimension d , any submodule of a free module is of projective dimension at most d −
1. Recall thatany module of the form ( − ) ∗ is a submodule of a free module. Since F ⊂ M ∗∗ we deduce the following.Fact A): One has pd R ( F ) < d .In view of 0 = Ext dR ( F , R ) → Ext dR ( M , R ) −→ Ext dR ( T , R ) −→ Ext d + R ( F , R ) = ( ∗ ) we see Ext dR ( T , R ) is finitely generated, and recall that it is divisible. These yield that Ext dR ( T , R ) = ( ∗ ) , we get Ext dR ( M , R ) = N be a finitely generated R -module. We look at 0 → Syz ( N ) → R n → N →
0. This induces0 = Ext dR ( M , R n ) → Ext dR ( M , N ) → Ext d + R ( M , Syz ( N )) =
0, and consequently, Ext dR ( M , N ) = (cid:3) Proposition 5.2.
Let ( R , m ) be a PID which is not complete and M be torsion-free and cohomologically finite.Then M is finitely generated.Proof. Let T : = ker ( M f −→ M ∗∗ ) and F : = im ( f ) . By definition, T ∗ =
0. Due to Fact 5.1.A), F is free and of finite rank (recall that M ∗ is finitely generated). According to the proof of Lemma 5.1Ext R ( T , R ) = Ext R ( M , R ) =
0. By a result of Nunke [21, Theorem 8.5], T =
0. In view of the exactsequence 0 → T → M → F →
0, we deduce M is free and of finite rank. (cid:3) We recall that a ring is slender if for each countable family {M n } the natural map L n Hom R ( M n , R ) → Hom R ( ∏ ∞ n = M n , R ) is an isomorphism. Here, we present a connection to set theory of rings. Corollary 5.3.
Let R be a discrete valuation domain of positive dimension. The following are equivalent: a) cohomologically finite and finite are equivalent notions over torsion-free modules;b) R is slender.Proof. a ) ⇒ b ) : Suppose on the way of contradiction that R is not slender. By [6, Theorem 2.9] R iscomplete. In view of Fact 6.3 (see below), the fraction field of R is cohomologically finite. It is well-known and easy to see that the fraction field of R is not finitely generated. Since it is torsion-free, we getto a contradiction. b ) ⇒ a ) : In view of [6, Theorem 2.9], R is not complete. In particular, we are in the situation ofProposition 5.2 to deduce a). (cid:3) Definition 5.4.
A module M is called cohomologically of finite length if Ext iR ( M , R ) is of finite lengthas an R -module for all i ≥ Proposition 5.5.
Let G be an abelian group. Then G is cohomologically of finite length iff ℓ Z ( G ) < ∞ .Proof. Thanks to [4, Proposition V.14.7] we know that G is finitely generated. By fundamental theoremof finitely generated abelian groups, G ∼ = Z n ⊕ L i Z ri n i Z ri for some n , n i , r i ∈ N . Since Hom ( G , Z ) is offinite length, it follows that n =
0. In other words, G is of finite length. (cid:3) Definition 5.6.
Let ( R , m ) be local. A module M is called H-finite if H i m ( M ) is artinian as an R -modulefor all i ≥ Q ( R ) we mean the fraction field of a local domain R . Remark . For example, any finite module is H-finite, but not the converse. Indeed, let R be a localdomain R of positive dimension. Since H i m ( Q ( R )) = i ≥
0, it is non-finite and H-finite.
Proposition 5.8.
Let ( R , m ) be a complete quasi-Gorenstein ring and M be reflexive and H-finite. Then M isfinite.Proof. Let d : = dim R . Let L be an R -module. According to Watts’ theorem, H d m ( L ) = H d m ( R ) ⊗ R L = E R ( k ) ⊗ R L . In the light of Matlis theory, H d m ( M ) v is finitely generated. By tensor-hom adjunction weobserve H d m ( M ) v ∼ = Hom R ( M , Hom R ( E R ( k ) , E R ( k )) ∼ = M ∗ is finitely generated. We apply another ( − ) ∗ and use the reflexivity assumption, to see M is finitely generated. (cid:3) Example . The first (resp. second) item shows that the reflexivity (resp. H-finite) assumption is im-portant.a) Let ( R , m , k ) be a local ring of positive dimension. Then E R ( k ) is H-finite, and is not finite.b) Let ( R , m ) be a discrete valuation domain which is not complete. Clearly, M : = ⊕ N R is notfinitely generated. It is easy to see M ∗ = ∏ N R . Since R is Slender, Hom R ( ∏ N R , R ) ∼ = L n Hom R ( R , R ) = M . So, M is reflexive. Proposition 5.10.
Let ( R , m ) be a complete Gorenstein ring. Then M is H-finite iff M is cohomologically finite.Proof. Let d : = dim R . The point is to establish the local duality theorem for H-finite modules overcomplete Gorenstein rings. Note that f M : H d m ( M ) v ∼ = −→ M ∗ , as we observed in the previous proposi-tion. Since H d − i m ( − ) and Ext iR ( − , R ) are zero over projective modules and all i >
0, it turns out from aninductive argument that H d − i m ( M ) v ∼ = Ext iR ( M , R ) . This completes the proof. (cid:3) In general, H-finite and cohomologically finite are not the same. For example, let R be a local domainof positive dimension. Then, H i m ( Q ( R )) =
0, i.e., Q ( R ) is H-finite. By a result of Jensen [8] there arecases for which Ext R ( Q ( R ) , R ) is a non-empty direct sum of Q ( R ) , i.e., Q ( R ) is not cohomologicallyfinite. The next section talks more on this topic.6. C OHOMOLOGICALLY ZERO INJECTIVE MODULES
By hSupp ( − ) , we mean S i Supp ( Ext iR ( − , R )) . Definition 6.1.
A module ( − ) is called cohomologically zero if hSupp ( − ) = ∅ .Recall that a finitely generated module is cohomologically zero iff it is zero. Let us connect to theprevious section: Corollary 6.2.
Let ( R , m ) be a PID. Then Q ( R ) is cohomologically finite iff R is complete. To see the corollary, we may assume that it is of positive dimension, then it follows by Proposition 5.2and the following result of Auslander [3, Page 166]:
Fact . Let ( R , m ) be a complete local integral domain of positive dimension. Then Q ( R ) is cohomo-logically zero.This result of Auslander can be extended in 3 different directions: Observation . Let R be any integral domain of positive dimension which is not necessarily noetherian.Suppose R is complete with respect to a nontrivial ideal I . Then Ext iR ( Q ( R ) , R ) = i . Proof.
Let F be a flat module with the property that F ⊗ R R / I =
0. The desired claim follows immedi-ately from [24, Theorem 4.4], where it were shown that Ext iR ( F , M b I ) = M . It remains to notethat Q ( R ) is flat and that Q ( R ) ⊗ R R / I = (cid:3) Remark . Recall from [18] that a module M is called strongly cotorsion if Ext iR ( Q ( R ) , M ) = i . By this terminology, the previous observation says that R is strongly cotorsion.Let ( R , m ) be a complete-local integral domain. In [20, Corollary 2], Matlis proved that Ext R ( E R ( k ) , R ) = R =
1. This is true for another class of rings as item b) of the next result indicates:
Proposition 6.6.
Let ( R , m , k ) be a local Gorenstein integral domain of dimension d > . Then Ext iR ( E R ( k ) , R ) ≃ ( b R if i = d otherwiseIn particular, the following properties are true for the non-finitely generated module E R ( k ) :a) E R ( k ) is cohomologically finite iff R is complete,b) Ext R ( E R ( k ) , R ) = iff dim R = ,c) E R ( k ) is not cohomologically zero,d) hSupp ( E R ( k )) = Spec ( R ) = { m } = Supp ( E R ( k )) .Proof. The injective resolution of R is given by0 → R → ⊕ ht ( q )= E R ( R / q ) → . . . → ⊕ ht ( q )= d − E R ( R / q ) → E R ( k ) → Recall that Hom R ( E R ( k ) , R ) = R ( E R ( k ) , E R ( R / p )) ≃ ( b R if p = m iR ( E R ( R / p ) , R ) = H i ( → → . . . → → b R → ) . This proves the first part.a): Note that b R is finitely generated as an R -module iff R is complete. b): The existence of a finitelygenerated injective module implies that the ring is artinian. Items c) and d ) are clear, because d > (cid:3) Here, we deal with the corresponding property of E R ( R / p ) for a middle p . Lemma 6.7.
Let ( R , m , k ) be a regular local ring of dimension d > and p be a prime ideal of height d − . Thereis a number µ (finite, or infinite) and a module H isomorphic to c R p such that Ext iR ( E R ( R / p ) , R ) ≃ c R µ p H if i = d otherwiseProof. Let X ( i ) = { q ∈ Spec ( R ) : ht ( q ) = i } . The minimal injective resolution of R is given by ζ : = → R → ⊕ q ∈ X ( ) E R ( R / q ) → . . . → ⊕ q ∈ X ( d − ) E R ( R / q ) f −→ ⊕ q ∈ X ( d − ) E R ( R / q ) g −→ E R ( k ) → ( E R ( k ) , R ) = i >
0. Recall Hom R ( E R ( R / p ) , E R ( R / p )) = Hom R p ( E R ( R / p ) p , E R ( R / p ) p ) = c R p . Also, there is an µ such that E R ( R / p ) v = c R µ p (see [23, Page 2392]). Now, use the fact:Hom R ( E R ( R / p ) , E R ( R / q )) ≃ c R p if p = q c R µ p if q = m iR ( E R ( R / p ) , R ) = H i (cid:16) → → . . . → → c R p h −→ c R µ p → (cid:17) (+) Let g : = ( g q ) where g q : E R ( R / q ) → E R ( R / m ) . Suppose on the way of contradiction that g q = E R ( R / q ) ⊂ ker ( g ) = im ( f ) . Let f be the composition map ⊕ p ∈ X ( d − ) E R ( R / p ) f −→ ⊕ Q ∈ X ( d − ) E R ( R / Q ) ։ ⊕ Q ∈ X ( d − ) \ q E R ( R / Q ) .We look at ⊕ q ∈ X ( d − ) E R ( R / q ) f −→ ⊕ q ∈ X ( d − ) \ q E R ( R / q ) g −→ E R ( k ) → ( f ) ⊕ E R ( R / q ) = im ( f ) and ker ( g ) = ker ( g ) ⊕ E R ( R / q ) . Since ker ( g ) = im ( f ) we deduce that ker ( g ) = im ( f ) . This is in contradiction with the minimality of ζ . In sum, g q = g p : E R ( R / p ) → E R ( R / m ) is nonzero, we know h : = Hom ( E R ( R / p ) , g p ) is nonzero. Let A be aring and F : A → ⊕ i ∈ I A be nonzero. Then F induces from f i : A → A . In particular, f i is a multiplicationby r i ∈ R . By assumption, c R p is an integral domain. From these, we deduce that h is injective.Let H : = im ( h ) . Since h is injective, it is isomorphic to c R p , and in view of (+) we deduce thatExt iR ( E R ( R / p ) , R ) ≃ c R µ p H if i = d (cid:3) Corollary 6.8.
Adopt the above notation and suppose in addition R / p is not complete. Then E R ( R / p ) is notcohomologically zero.Proof. By a result of Schenzel [23, Proposition 4.3] we know µ >
1. Since Ext dR ( E R ( R / p ) , R ) = c R µ p H and H ∼ = c R p , we deduce that Ext dR ( E R ( R / p ) , R ) is not zero. (cid:3) Suppose R is a local ring of a polynomial ring over a field in m variables with m ≥ n +
3. ThenOsofsky [22] proved that pd R ( Q ( R )) = n + ℵ n = ℵ n + . Now, we state two corollaries on projectivedimension of certain injective modules. Corollary 6.9.
Let ( R , m ) be a regular local ring of dimension d > and p be a prime ideal of height d − suchthat R / p is not complete. Then pd R ( E R ( R / p )) = d.Proof. We observed in the previous corollary that Ext dR ( E R ( R / p ) , R ) is not zero. Since global-dimensionof R is d we get the desired claim. (cid:3) Corollary 6.10.
Let ( R , m , k ) be a regular local ring of dimension d > . Then pd R ( E R ( k )) = d.Proof. Recall from Proposition 6.6 that Ext dR ( E R ( k ) , R ) =
0. Since global-dimension of R is d we get thedesired claim. (cid:3)
7. R
EALIZATION
Definition 7.1. (Jensen-Nunke) Let i >
0. Recall from [9] that a module M is called i -realizable if thereare modules N , L such that M ∼ = Ext iR ( N , L ) . When i : = dim R < ∞ we say M is realizable.The following result is not new. It was proved by Jensen [9, Theorem 2] via some spectral sequencesand derived functors lim ←− ( i ) . The following proof is elementary. Corollary 7.2.
Let ( R , m ) be a regular of positive dimension d and M be finitely generated. Then b M ∼ = Ext dR ( E R ( k ) , M ) . In particular, if M is complete then M is realizable.Proof. Let ( − ) be a finitely generated module and set F ( − ) : = Ext dR ( E R ( k ) , − ) . Since injective dimensionof any module is at most d , F ( − ) is right exact. Also, it preserves finite direct sum. By Watts’ theorem, F ( − ) ∼ = F ( R ) ⊗ R ( − ) for any finitely generated module. Recall from Proposition 6.6 that F ( R ) = b R . Byplugging this in Watts’ isomorphism, we get that b M ∼ = M ⊗ R b R ∼ = Ext dR ( E R ( k ) , M ) . (cid:3) Remark . If we allow the module N in the above definition to be finitely generated, the story ofCorollary 7.2 will changes, see Fact 4.4. Example . The finitely generated assumption of M in Corollary 7.2 is important: Let R be a dis-crete valuation domain which is not a field. Then pd R ( Q ( R )) =
1. There is a module M such thatExt R ( Q ( R ) , M ) =
0. Let d M be the largest divisible submodule of M . In view of [21, Theorem 7.1], M d M is not realizable.Recall from [18] that a module is called corosion if Hom R ( Q ( R ) , M ) = Ext R ( Q ( R ) , M ) = Example . (Matlis) Any cotorsion module M over an integral domain is 1-realizable. Proof.
We apply Hom R ( − , M ) to 0 → R → Q ( R ) → Q ( R ) R → = Hom R ( Q ( R ) , M ) → Hom R ( R , M ) −→ Ext R ( Q ( R ) / R , M ) −→ Ext R ( Q ( R ) , M ) = M ∼ = Hom R ( R , M ) ∼ = Ext R ( Q ( R ) / R , M ) . (cid:3) In order to extend Corollary 7.2, we need to state the following:
Definition 7.6.
Let i >
0. We say a module M is called homologically realizable at level i if there aremodules N , L such that M ∼ = Tor Ri ( N , L ) . When i : = dim R < ∞ we say M is homologically realizable. Fact . (Matlis) Let R be a domain. Any torsion module M is homologically realizable at level one. Inparticular, over 1-dimensional rings a module is homologically realizable iff it is torsion. Proof.
Since M is torsion, M ⊗ R Q ( R ) =
0. Apply − ⊗ R M to 0 → R → Q ( R ) → Q ( R ) R → = Tor R ( Q ( R ) , M ) → Tor R ( Q ( R ) / R , M ) → R ⊗ R M → M ⊗ R Q ( R ) = Q ( R ) is flat. So, M ∼ = Tor R ( Q ( R ) / R , M ) . (cid:3) Proposition 7.8.
Let ( R , m ) be a local ring , x : = x , . . . , x n be a regular sequence and let A be artinian. Then A is homologically realizable at level i for each < i ≤ n. In fact, A ∼ = Tor Ri ( H ix i ( R ) , A ) where x i : = x , . . . , x i . The following proof works for m -torsion modules. Proof.
Let a i : = ( x , . . . , x i ) . Since A is an m -torsion module, H a i ( A ) = A . Denote the ˇ Check complex of R with respect to a i by ˇ C ( x , . . . , x i ; R ) . Since x is a regular sequence, ˇ C ( x , . . . , x i ; R ) is a flat resolutionof H i a i ( R ) . Then, Tor Rj ( H i a i ( R ) , A ) ∼ = H j ( ˇ C ( x , . . . , x i ; R ) ⊗ R A ) ∼ = H j − i a i ( A ) .Thus, Tor Ri ( H i a i ( R ) , A ) ∼ = H a i ( A ) ∼ = A ,as claimed. (cid:3) Corollary 7.9.
Let ( R , m ) be a Cohen-Macaulay local ring of positive dimension and let {A j } be a directed familyof artinian modules. Then A : = lim −→ j A j is homologically realizable.Proof. Let d : = dim R . In view of Proposition 7.8, we know Tor Rd ( H d m ( R ) , A j ) ∼ = A j . Since Tor-functorscommutes with directed limits, we have Tor Rd ( H d m ( R ) , A ) ∼ = lim −→ Tor Rd ( H d m ( R ) , A j ) ∼ = lim −→ A j = A . (cid:3) Let us give examples that can’t be written as a direct limit of artinian:
Example . Let I be any ideal. Then the conormal module II is homologically realizable at level one.There is a choice for I such that the conormal module is not of finite length. Indeed, use the naturalisomorphism Tor R ( RI , RJ ) ∼ = I ∩ JIJ .Let M , N be finitely generated modules. The notation P ( M , N ) stands for the set of all M → N whichfactor through free modules. By stable-hom, we mean Hom R ( M , N ) : = Hom R ( M , N ) P ( M , N ) . Example . (Auslander) The stable-hom is homologically realizable at level one. Indeed, Hom R ( M , N ) = coker ( f : Hom R ( M , R ) ⊗ R N → Hom R ( M , N )) . Recall from Discussion 3.8 that coker ( f ) = Tor R ( Tr M , N ) .Now, we present a simple proof of [9, Proposition 3] by Jensen: Proposition 7.12.
Let ( R , m ) be a complete local ring, x : = x , . . . , x n be a regular sequence and let M be finitelygenerated. Then M ∼ = Ext iR ( H ix i ( R ) , M ) for each < i ≤ n. In particular, M is realizable at level i. Proof.
Let a i : = ( x , . . . , x i ) and A : = M v . Then A is artinian. Thanks to Proposition 7.8 we know A ∼ = Tor Ri ( H i a i ( R ) , A ) . By taking another Matlis duality and using Matlis theory we observe that M ∼ = M vv = A v ∼ = Tor Ri ( H i a i ( R ) , M v ) v ∼ = Ext iR ( H i a i ( R ) , M vv ) ∼ = Ext iR ( H i a i ( R ) , M ) ,as claimed. (cid:3) Corollary 7.13.
Let ( R , m ) be a complete Cohen-Macaulay local ring of dimension d > and let M be finitelygenerated. Then M ∼ = Ext dR ( H d m ( R ) , M ) . In particular, M is realizable.Proof. It is enough to note that H d m ( R ) ∼ = H dx ( R ) , where x is a full parameter sequence. (cid:3) Example . Here, we collect some example of realizable flat module over non-artinian rings that arenot finitely generated:i) Let ( R , m ) be a Gorenstein local ring. Then b R is realizable.ii) Let ( R , m ) be a complete Cohen-Macaulay local ring and let { M j } be a family of finitely gener-ated modules. Then ∏ j M j is i -realizable for any 0 < i ≤ dim R .iii) Let R be as item ii). Then ∏ N R is realizable. Proof.
Let d : = dim R .i) In view of Proposition 6.6 we know that b R ∼ = Ext dR ( H d m ( R ) , R ) . By definition, b R is realizable.ii) For simplicity, we assume i = d . By the above corollary we know that M j = Ext dR ( H d m ( R ) , M j ) .Thus, ∏ j M j ∼ = Ext dR ( H d m ( R ) , ∏ j M j ) , and so ∏ j M j is realizable.iii) This is a special case of ii) (cid:3) The following result was proved by Jensen [8, Proposition 5]. Here, we present an elementary proof:
Fact . Let R be a complete local ring and { M i } be an inverse system of finitely generated modules.Suppose id R ( M i ) ≤ n for all i . Then id R ( lim ←− M i ) ≤ n . Proof.
If 0 → M i → E → · · · → E n → → E vn → · · · → E v → M vi → R ( lim −→ M vi ) ≤ n . Let0 → F n → · · · → F → lim −→ ( M vi ) → → ( lim −→ M vi ) v → F v → · · · → F vn → ( lim −→ M vi ) v = lim ←− M vvi = lim ←− M i . This completes the proof. (cid:3) In fact:
Fact . Let R be a complete local ring and {A i } be an inverse system of Matlis reflexive modules (e.g.artinian modules). Suppose id R ( A i ) ≤ n for all i . Then id R ( lim ←− A i ) ≤ n .8. H OMOLOGICALLY ZERO FLAT MODULES
We recall the following elementary result of Auslander:
Fact . (See [3, Proposition 4.3]) Let ( R , m ) be a complete local, M be finitely generated. ThenExt pR ( lim −→ A i , M ) ∼ = lim ←− Ext pR ( A i , M ) for any directed family {A i } .We simplify the following result of Jensen and Buchweitz-Flenner: Corollary 8.2. (See [5, Corollary 1] ) If ( R , m ) is a complete local noetherian ring and F is a flat R-module then Ext pR ( F , M ) = for all p ≥ and all finite R-modules M. Proof.
Due to Lazard’s theorem we know that any flat module F is a direct limit of free modules { F i } .By the above fact, Ext pR ( F , M ) = Ext pR ( lim −→ F i , M ) ∼ = lim ←− Ext pR ( F i , M ) =
0, as claimed. (cid:3)
Corollary 8.3. (Compare with [21, Theorem 7.1] ) Let ( R , m ) be a PID. The following are equivalent:a) Ext R ( F , N ) = for any torsion-free module F and any finitely generated R-module N,b) Ext R ( F , N ) is finitely generated for any torsion-free module F and any finitely generated R-module N,c) Ext R ( F , R ) is finitely generated for any torsion-free module F ,d) Ext R ( F , R ) = for any countably generated torsion-free module F ,e) Ext R ( F , R ) = for any torsion-free module F ,f) R is complete,g) R is realizable,h) any finitely generated module is realizable.Proof. a ) ⇒ b ) ⇒ c ) : These are trivial. c ) ⇒ d ) : This is in Lemma 5.1. d ) ⇒ e ) : This is trivial. e ) ⇒ f ) : Recall that flat and torsion-free are the same notions. Now, use [9, Theorem 1]. f ) ⇒ g ) : SeeCorollary 7.2. g ) ⇒ h ) : It follows that any finitely generated module is complete. Now, see Corollary7.2. h ) ⇒ a ) : This implies that R is complete, and recall that torsion-free are the same notions. Now,use Corollary 8.2. (cid:3) Here, we collect some examples of cohomogically (non-) zero flat modules from literature.
Example . (Gruson) Let k be an uncountable field, and let R : = k [ X , Y ] ( X , Y ) . Then Ext iR ( Q ( R ) , R ) = i =
2. In particular, Ext R ( Q ( R ) , R ) is not finitely generated. Proof.
Suppose on the way of contradiction that Ext R ( Q ( R ) , R ) is finitely generated. By a result ofGruson [15, Proposition 3.2] we know that R is complete with respect to adic topology of the collection S : = { rR : r ∈ R } , i.e., the map f in the following exact sequence0 −→ R f −→ lim ←− r ∈ S RrR −→ Ext R ( Q ( R ) , R ) −→ R ( Q ( R ) , R ) =
0. Also, Hom R ( Q ( R ) , R ) =
0. So, Q ( R ) iscohomologically finite. In view of Lemma 5.1 we see Ext R ( Q ( R ) , R ) =
0. Since id R ( R ) =
2, we haveExt iR ( Q ( R ) , R ) = i . According to [10, Theorem 9.19], Q ( R ) =
0. This is a contradiction. (cid:3)
The uncountable assumption on k is really needed, as the next example says: Example . Let ( R , m ) be a countable Gorenstein integral domain. Then Ext iR ( Q ( R ) , R ) = i = Proof.
Recall that pd R ( Q ( R )) = R ( Q ( R ) , R ) =
0. It remains to recall from [10, Theorem 9.18]that over a countable local Gorenstein ring, a flat module is zero iff it is cohomologically zero. (cid:3)
Example . Let ( R , m ) be complete and f ∈ m be regular. Then R f is homologically zero flat module. Proof.
It is easy to see Hom R ( R f , R ) =
0. By Corollary 8.2 Ext + R ( R f , R ) = (cid:3) We need the reverse part of Example 8.6:
Fact . (See [23, Theorem 1.1]) Let x , . . . , x d be a full system of parameters for ( R , m ) . ThenExt R ( ⊕ R x i , R ) = R is complete in m -adic topology.
9. S
PLITTING ON A THEME OF K APLANSKY
We start with the following splitting criteria:
Proposition 9.1.
Let ( R , m ) be a local domain and f ∈ R be nonzero. Let ζ : = → R → A g −→ R f → besuch that A decomposes into nonzero modules. Then ζ splits.Proof. We may assume that f is not invertible. The flat resolution of R f R is given by 0 → R → R f → R f R →
0. Then Tor R ( R f R , A ) = H ( ˇ C ( f , R )) ⊗ R A = H ( f ) ( A ) = A ∼ = A ⊕ A be a nontrivial decomposition. Recall that R f R ⊗ R R f = R f R f =
0. We apply R f R ⊗ R − to ζ to deduce that 0 = Tor R ( R f R , A ) −→ R f R −→ R f R ⊗ R A −→ R f R ⊗ R R f = R f R ∼ = R f R ⊗ R A .Claim A): R f R is indecomposable.Indeed, thanks to ˇ Check -complex, we know that R f R = H ( ˇ C ( f , R )) = H ( f ) ( R ) . It is easy to seeEnd R ( H ( f ) ( R )) ∼ = R b ( f ) which is a commutative integral domain. From this, R f R is indecompos-able.Now, recall that A ∼ = A ⊕ A . Apply R f R ⊗ R − to it and use R f R ∼ = R f R ⊗ R A to see R f R ∼ = R f R ⊗ R A ∼ = R f R ⊗ R A ⊕ R f R ⊗ R A .We combine Claim A) with this, to assume without loss of the generality that R f R ⊗ R A =
0. We apply − ⊗ R A to 0 → R → R f → R f R → = H ( f ) ( A ) ∼ = Tor R ( R f R , A ) −→ A −→ R f ⊗ R A −→ R f R ⊗ R A = A ∼ = R f ⊗ R A . Then R f = g ( A ) = g ( A ) f = g ( A ⊕ A ) f = g ( A ) f ⊕ g ( A ) f = g (( A ) f ) ⊕ g ( A ) f = g ( A ) ⊕ g ( A ) f .Since R f is indecomposable, either g ( A ) = g ( A ) f =
0. Suppose on the way of contradiction that g ( A ) =
0. Then A ⊂ ker ( g ) = R , but A is not finitely generated. This contradiction implies that g ( A ) f =
0. Thus, g ( A ) = R f . From this, R + A = A . Since R ∩ A = R ⊕ A = A . Thedesired claim follows by this. (cid:3) Question . (Kaplansky, [16]) For what integral domains is it true that any torsionfree module of ranktwo is a direct sum of modules of rank one?Here, we reprove [19, Theorem 61] and an essential part of [16]. Corollary 9.3.
Let ( R , m ) be a local domain such that every torsion-free R-module of rank that is not finitelygenerated is a direct sum of modules of rank . Then R is complete in the m -adic topology.Proof. Let x , . . . , x d be a system of parameters of m . Apply the assumption along with Proposition 9.1to see that Ext R ( R x i , R ) = i . By Fact 8.7 R is complete in m -adic topology. (cid:3) Remark . Among other things, [2] talks about rings for which any finitely generated reflexive moduleis a direct sum of modules of rank 1. R
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