aa r X i v : . [ m a t h . A C ] J a n A NOTE ON COHEN-MACAULAY DESCENT
MOHSEN ASGHARZADEH
Abstract.
We investigate the behavior of Cohen-Macaulay defect undertaking tensor product with aperfect module. Consequently, we study the perfect defect of a module. As an application, we connect toassociated prime ideals of tensor products. Introduction
We start with:
Conjecture . (See [13, Conjecture 2.11]) Let R be a Cohen-Macaulay local ring, M be a perfect R -moduleof positive dimension and let N be a module with dim( N ) = dim( R ) . If M ⊗ R N is a Cohen-Macaulay R -module, then N is a maximal Cohen-Macaulay R -module.Yoshida proved Conjecture 1.1 under the additional assumption that N is surjective Buchsbaum. Also,without loss of the generality, we may assume that d := dim R ≥ .In Example 2.4 we give a negative answer to the conjecture over any Cohen-Macaulay local ring ofdimension d > . The module M in Example 2.4 is of projective dimension at least two. In Section 3 weshow: Proposition 1.2.
Let R be a Cohen-Macaulay local ring with isolated singularity of dimension d , M beperfect of projective dimension at most one, and let N be torsion-free. If M ⊗ R N is Cohen-Macaulay, then N is maximal Cohen-Macaulay. Also, we present some partial positive answers in the case p . dim( M ) = 2 and give the correspondingobstruction. As an application, and in Section 4, we compute Cohen-Macaulay defect of tensor product,and compare it with the perfect defect. It may be interesting to find the situation for which these defectsare coincide to each other. We determine such a situation, see Corollary 4.17. Despite of simplicity of Ass(Hom( − , ∼ )) and Supp( −⊗ R ∼ ) , it seems Ass( −⊗ R ∼ ) rarely computed in literature, even in somespecial forms. For an achievement, see [9, Theorem 23.2] where one of the modules is relatively flat. As atypical example, see e.g. [1, Corollary 1.6]. Here, we show: Corollary 1.3.
Let R be local, M of finite projective dimension and N be maximal Cohen-Macaulay. Thefollowing holds:i) Ass( M ⊗ R N ) ⊆ Ass( M ) ∩ Supp( N ) .ii) If R is Cohen-Macaulay, then Ass( M ⊗ R N ) = Ass( M ) ∩ Supp( N ) .iii) If M becomes an ideal of positive height, then Ass( M ⊗ R N ) = Ass( N ) . Mathematics Subject Classification.
Primary 13C14; Secondary 13D07; 13D45.
Key words and phrases.
Associated primes; Cohen-Macaulay rings and modules; tensor products; perfect defect. iv) If M is Cohen-Macaulay then Ass( M ⊗ R N ) = Ass( M ) ∩ Supp( N ) . In particular, Ass( M ⊗ R N ) =Ass( M ) provided N is fully supported (e.g. R is a domain).v) Assume in addition to ii) that R is of prime characteristic and F -finite, then Ass( F n ( M )) =Ass( M ) . Part v) is well-known and presented here because of the motivation, it is true without Cohen-Macaulayand F -finite assumptions, see e.g. [8, Corollary 1.6] by Huneke-Sharp. Under the extra assumption M isprimary, i.e. Ass( M ) is singleton, Corollary 1.3(i) implies that Ass( M ⊗ R N ) = Ass( M ) . Also, we presenta connection to the support of maximal Cohen-Macaulay modules, see Corollary 4.12 and 4.7. Let M beof finite projective dimension over a local ring, and let Σ be the class of all maximal Cohen-Macaulaymodules. It follows by the above corollary that S L ∈ Σ Ass( M ⊗ R L ) is finite.Suppose the module M from Corollary 1.3(i) is primary and cyclic. The existence of such a module mayput some restrictions on the ring. In fact, Peskine and Szpiro [10, Corollary II.3.3] proved that the existenceof a prime ideal of finite projective dimension implies the ring is an integral domain. The correspondinginfinitesimal version may not be true. For example, let R := k [ X ] / ( X ) and p := ( x ) . Then p = p (2) = 0 ,and consequently p (2) is free. But, R is not an integral domain. Despite of this we have: Observation . Let R be a local ring, p ∈ Spec( R ) such that p . dim( R/ p ( n ) ) < ∞ for some n ∈ N . Adoptone of the following situations:i) p ( n ) = 0 , orii) R is reduced.Then R is an integral domain.It may be worth to mention that the dual version of this observation (working with injective dimension)imposes more restrictions on the ring. For the precise statement, see Corollary 5.5.2. Negative side of Conjecture 1.1
The notation ( R, m , k ) stands for a commutative, noetherian and local ring, with the maximal ideal m andthe residue field k . Also, modules are finitely generated, otherwise specialized. The notation p . dim R ( − ) stands for projective dimension of an R -module ( − ) . By grade R ( M ) we mean grade R (Ann M, R ) , i.e., grade R ( M ) = inf { i : Ext iR ( M, R ) = 0 } . Recall that a non-zero module M is called perfect if p . dim R ( M ) = grade R ( M ) . So, p . dim R ( M ) < ∞ . Forsimplicity of the reader, we bring the following well-known and easy facts: Fact . Let f : R → S be a homomorphism of local rings, and M an S -module which is finite as an R -module. Then depth R ( M ) = depth S ( M ) . Fact . Let R be a Cohen-Macaulay local ring, and M be of finite projective dimension. Then M isperfect if and only if M is Cohen-Macaulay.A module M is called quasi-perfect if inf { i : Ext iR ( M, R ) = 0 } = sup { i : Ext iR ( M, R ) = 0 } . Themonograph [3] is our reference for the concept of G -dimension.Observation . Let R be a local ring and M be Cohen-Macaulay module of finite G -dimension. Then dim( M ) + grade( M ) = depth( R ) and M is quasi-perfect. Proof.
Recall from [10, Lemma I.4.8] that dim M + grade( M ) ≥ depth( R ) . Also, in the case G -dimensionis finite we have Gdim R M = sup { i : Ext iR ( M, R ) = 0 } , see [3]. In particular, grade( M ) ≤ Gdim( M ) . Weapply these along with Auslander-Bridger formula to observe that dim M ≥ depth( R ) − grade( M ) ≥ depth( R ) − Gdim( M )= depth R ( M )= dim( M ) . So, dim( M ) + grade( M ) = depth( R ) and grade( M ) = Gdim( M ) , i.e., M is quasi-perfect. (cid:3) By H i m ( − ) we mean the i -th local cohomology module of ( − ) with respect to m .Example . Let ( R, m ) be a Cohen-Macaulay local ring of dimension d > and let x , . . . , x d be a systemof parameters and let N := I := ( x , . . . , x d − ) . The following holds:i) M := R/I is perfect,ii) dim( N ) = dim( R ) and M ⊗ R N is Cohen-Macaulay as an R -module,iii) if d > , then N is not maximal Cohen-Macaulay as an R -module,iv) if d = 2 , then N is maximal Cohen-Macaulay as an R -module. Proof. i): Clearly, M resolves by the Koszul complex with respect to x , . . . , x d − . In view of Fact 2.2 M is perfect, because it is Cohen-Macaulay.ii): Let p ⊂ I be a minimal prime ideal. It follows that N is of dimension equal to dim R , because p ∈ Ass( N ) and R is equidimensional. Since I is generated by a regular sequence, II is free as an R/I -module. It is easy to see
R/I is Cohen-Macaulay. From this, II is Cohen-Macaulay as an R/I -module.Note that dimension of II does not change as an R -module. We apply Fact 2.1 to deduce that II isCohen-Macaulay as an R -module. Note that M ⊗ R N = I/I .iii): Suppose d > . Here, we claim that N is not Cohen-Macaulay. Indeed, look at → I → R → R/I → . In view of the long exact sequence of local cohomology modules we have H m ( R ) −→ H m ( R/I ) −→ H m ( I ) ( ∗ ) Due to Grothendieck’s non-vanishing theorem, we know H m ( R/I ) = 0 . By local cohomology characteri-zation of depth, H m ( R ) = 0 . We put this into ( ∗ ) and we deduce that = H m ( R/I ) ⊂ H m ( I ) . By localcohomology characterization of depth, depth( I ) ≤ < d . Since d = dim( N ) we observe that N is notCohen-Macaulay, viewed as an R -module.iv): Suppose d = 2 . Note that M ⊗ R N = N/x N . Since x is regular over A , it is regular over itssubmodules. In particular, x is N -regular. Recall from item ii) that N/x N is Cohen-Macaulay as an R -module. From these, N is Cohen-Macaulay as an R -module. Since dim( N ) = d , and by definition, N ismaximal Cohen-Macaulay as an R -module. (cid:3) By Tor R + ( M, N ) we mean ⊕ i> Tor Ri ( M, N ) . The following inspired us to present §3.Remark . Adopt the notation of Example 2.4. The following holds:i) if d > , then Tor R + ( M, N ) = 0 ,ii) if d = 2 , then Tor R + ( M, N ) = 0 . Indeed, assume d > and apply − ⊗ R I to → I → R → R/I → , we obtain → Tor R ( M, N ) → I ⊗ I f −→ I → I/I → , where f ( x ⊗ y ) = xy . In particular, Tor R ( M, N ) = ker( f ) and = x ⊗ x − x ⊗ x ∈ ker( f ) . From this, Tor R + ( M, N ) = 0 . In the case d = 2 , we see N is free, and so claim is trivial.3. Positive side of Conjecture 1.1
Conjecture 1.1 is true over -dimensional rings by a trivial reason:Observation . Let R be a -dimensional local ring, M be perfect and of positive dimension, and let N be -dimensional. If M ⊗ R N is Cohen-Macaulay, then N is maximal Cohen-Macaulay. Proof.
We know that depth( M ) = dim( M ) > . By Auslander-Buchsbaum formula, p . dim( M ) = 0 . Recallthat M ⊗ R N = ⊕ N is Cohen-Macaulay. From this N is Cohen-Macaulay, and so the claim follows. (cid:3) The module M in Example 2.4(iii) is of projective dimension at least two. What can say if p . dim R ( M ) =1 ? First, we recall the following two results:Fact . (See [2, Theorem 1.2]) Let R be any local ring and p . dim R ( M ) < ∞ . Let q be the largest numbersuch that Tor Rq ( M, N ) = 0 . If depth R (Tor Rq ( M, N )) ≤ or q = 0 , then depth R ( N ) = depth R (Tor Rq ( M, N )) + p . dim R ( M ) − q. Fact . Let R be local, M of finite projective dimension and N be such that dim( N ) = dim( R ) . Thefollowing holds:i) depth( M ) ≤ dim( M ⊗ R N ) ≤ dim( M ) ,ii) if M is perfect, then dim( M ⊗ R N ) = dim( M ) . Proof. i): The second inequality always holds, because
Supp( M ⊗ R N ) = Supp( M ) ∩ Supp( N ) . Byintersection theorem [12], dim( M ⊗ R N ) ≥ dim( N ) − p . dim( M ) = dim( R ) − p . dim( M ) ≥ depth( R ) − p . dim( M ) . By Auslander-Buchsbaum, depth( M ) ≤ dim( M ⊗ R N ) .ii): This is in [13, Lemma 2.1]. (cid:3) Example . Concerning Fact 3.3, the first items shows that the assumption p . dim( M ) < ∞ is reallyneeded even if we assume M is quasi-perfect. The second item shows that the dimension restriction on N is important:i) Let R := k [[ x,y ]]( xy ) , M := R/ ( x ) and N := R/ ( y ) . Then dim( M ) = dim( N ) = dim( R ) = 1 , M ismaximal Cohen-Macaulay and so totally reflexive. But, dim( M ⊗ R N ) = 0 < dim( M ) .ii) Let ( R, m ) be a d -dimensional regular local ring with d > . There are d elements { x , . . . , x d } thatgenerates m . Let s ∈ [1 , d − be an integer, define p := ( x , . . . , x s ) and q := ( x s +1 , . . . , x d ) . Clearly, M := R/ p is perfect and let N := R/ q . Then dim( M ⊗ R N ) = 0 < depth( M ) = dim( M ) = s . In the case of integral domains, one can say a little more:
Remark . Let R be a local domain, M, N be finitely generated such that dim( N ) = dim( R ) . Then Supp( M ⊗ R N ) = Supp( M ) . In particular, dim( M ⊗ R N ) = dim M . Proof.
Since both modules are finitely generated,
Supp( M ⊗ R N ) = Supp( M ) ∩ Supp( N ) . Since R isdomain, Supp( N ) = Spec( R ) . From this, Supp( M ⊗ R N ) = Supp( M ) . (cid:3) Here, we present a partial positive answer to Conjecture 1.1:
Proposition 3.6.
Let R be a Cohen-Macaulay local ring with isolated singularity of dimension d , M beperfect of projective dimension at most one, and let N be torsion-free. If M ⊗ R N is Cohen-Macaulay, then N is maximal Cohen-Macaulay.Proof. Since N is torsion-free, it is of dimension d . We may assume that d > . The proof is by inductionon d . If d = 1 we get the claim because depth R ( N ) > . By applying the inductive step, it turns out that ℓ (Tor R ( M, N )) < ∞ , because R is regular over the punctured spectrum and that maximal Cohen-Macaulaymodules over regular rings are free. Let → R n → R m → M → be a free resolution of M . Apply − ⊗ R N to it we have the following exact sequence −→ Tor R ( M, N ) −→ N ⊗ R R n −→ N ⊗ R R m −→ M ⊗ R N −→ . Since N is of positive depth we deduce that Tor R ( M, N ) = 0 . Thus,
Tor R + ( M, N ) = 0 . This and finitenessof p . dim R ( M ) allow us to apply the depth formula of Auslander: depth R ( N ) = depth R ( M ⊗ R N ) + p . dim R ( M ) ( ∗ ) Also, in the light of Auslander-Buchsbaum formula, depth R ( M ) + p . dim R ( M ) = depth( R ) (+) Then, we have depth R ( N ) ( ∗ ) = depth R ( M ⊗ R N ) + p . dim R ( M )= dim( M ⊗ R N ) + p . dim R ( M )3 .
3= dim( M ) + p . dim R ( M )2 .
2= depth R ( M ) + p . dim R ( M ) (+) = depth( R )= dim( R ) . By definition, N is maximal Cohen-Macaulay. (cid:3) In the same vein we have
Proposition 3.7.
Let R be a Cohen-Macaulay local ring of dimension d , M be perfect of projective di-mension at most one, and let N be torsion-free and locally free over the punctured spectrum. If M ⊗ R N is Cohen-Macaulay, then N is maximal Cohen-Macaulay.Note that Conjecture 1.1 is not true when p . dim( M ) = 2 . Despite of this, we have:Remark . Let R be a Cohen-Macaulay local with isolated singularity of dimension d . Let M be a cyclicperfect module of projective dimension at most two, and let N be torsion-free. If M ⊗ R N is Cohen-Macaulayand (Ann R M ) ⊗ R N is torsion-free, then N is maximal Cohen-Macaulay. Proof.
We may assume that d > . The proof is by induction on d . If d = 1 we get the claim because depth R ( N ) > . By applying the inductive step, it turns out that ℓ (Tor R + ( M, N )) < ∞ . Let I := Ann R ( M ) .In the light of Hilbert-Burch, the free resolution of I is of the following from → R n → R n +1 → I → . Apply − ⊗ R N to it we have the following exact sequence −→ Tor R ( I, N ) −→ N ⊗ R R n −→ N ⊗ R R n +1 −→ M ⊗ R N −→ . Since N is of positive depth we deduce that Tor R ( I, N ) = 0 . From this,
Tor R ( M, N ) = 0 . We apply − ⊗ R N to → I → R → R/I → we have the following exact sequence −→ Tor R ( R/I, N ) −→ I ⊗ R N −→ N −→ M ⊗ R N −→ . By our assumption, depth( I ⊗ R N ) > . From this, Tor R ( R/I, N ) = 0 . Thus,
Tor R + ( M, N ) = 0 . Similar toProposition 3.6 we have depth R ( N ) = depth R ( M ⊗ R N ) + p . dim R ( M )= depth R ( M ) + p . dim R ( M )= depth( R ) . Since depth( R ) = dim R , N is maximal Cohen-Macaulay. (cid:3) Keep Remark 2.5 in mind and recall that p . dim( M ) = dim R − . This remark motivate us to restateConjecture 1.1 in the following obstruction format:Remark . Let R be a Cohen-Macaulay local ring, M be a perfect R -module of positive dimension andlet N be a module with dim( N ) = dim( R ) = d . If M ⊗ R N is a Cohen-Macaulay R -module. Suppose inaddition p . dim( M ) = d − . Then Conjecture 1.1 is true if and only if Tor R + ( M, N ) = 0 . Proof.
Suppose
Tor R + ( M, N ) = 0 . Then q := sup { i : Tor Ri ( M, N ) = 0 } ≥ . By Auslander-Buchsbaumformula, depth( M ) = 1 . Recall that depth R (Tor Rq ( M, N )) ≤ dim(Tor Rq ( M, N )) ≤ dim( M ) = 1 . In thelight of Fact 3.2 we see depth R ( N ) = depth R (Tor Rq ( M, N )) + p . dim R ( M ) − q ≤ d − − ≤ d − , i.e., N is not maximal Cohen-Macaulay. The reverse part is in the proof of Proposition 3.6. (cid:3) Perfect and Cohen-Macaulay defects
The following drops finiteness of N from [13, Lemma 2.2] . Lemma 4.1.
Suppose that p . dim R ( M ) < ∞ and N is big Cohen-Macaulay. Then Tor R + ( M, N ) = 0 .Proof.
Without loss of the generality we may assume that R is complete. This enable us to assume that M is balanced big Cohen-Macaulay. Now, the proof is similar to [13, Lemma 2.2], only note that insteadof the Eisenbud-Buchsbaum criterion we use a criterion of Northcott. (cid:3) The third item of the following result reproves [13, Proposition 2.4] , where Yoshida used the intersectiontheorem and a formula involved in Betti and Bass numbers.
Proposition 4.2.
Let R be local, M of finite projective dimension and N be maximal Cohen-Macaulay.The following holds: i) depth( M ) ≤ depth( M ⊗ R N ) ≤ dim( M ) ,ii) if R is Cohen-Macaulay, then depth( M ) = depth( M ⊗ R N ) ,iii) if M is perfect, then M ⊗ R N is Cohen-Macaulay.Proof. By Lemma 4.1,
Tor R + ( M, N ) = 0 . This allow us to apply Auslander’s depth formula. In view ofFact 3.2 we observe that depth R ( M ⊗ R N ) = depth R ( N ) − p . dim R ( M ) = dim( R ) − p . dim( M ) ( ∗ ) i): By Auslander-Buchsbaum formula we have: depth R ( M ⊗ R N ) ( ∗ ) = dim( R ) − p . dim( M ) ≥ depth( R ) − p . dim( M ) = depth( M ) . To see the second inequality, we note that depth( M ⊗ R N ) ≤ dim( M ⊗ R N ) ≤ dim( M ) (+) . ii): Since R is Cohen-Macaulay, we have dim( R ) = depth( R ) . Put this in ( ∗ ) and apply Auslander-Buchsbaum formula to see depth R ( M ⊗ R N ) = depth( R ) − p . dim( M ) = depth( M ) . iii): Recall that dim( R ) = dim( M ) + grade( M ) , because M is perfect. Then dim( M ) (+) ≥ depth R ( M ⊗ R N ) ( ∗ ) = dim( R ) − p . dim( M )= dim( R ) − grade( M )= dim( M ) . Therefore, dim( M ) = depth R ( M ⊗ R N ) . Since dim( M ⊗ R N ) ≤ dim( M ) , it follows that M ⊗ R N isCohen-Macaulay. (cid:3) Remark . Proposition 4.2(i) is not true even if we assume
Gdim( M ) < ∞ , see Example 3.4. Here, we present a series of corollaries and applications (also, see the next section):
Corollary 4.4.
Let ( R, m ) be Cohen-Macaulay, M of finite projective dimension and N be maximal Cohen-Macaulay. The following holds:i) Ass( M ⊗ R N ) = Ass( M ) ∩ Supp( N ) .ii) If R is a domain, then Ass( M ⊗ R N ) = Ass( M ) .Proof. i): Let p ∈ Ass( M ⊗ R N ) . First, we claim that N p is maximal Cohen-Macaulay. Indeed, p ∈ Supp( M ⊗ R N ) ⊆ Supp( N ) . Since R is Cohen-Macaulay, there is an R -sequence x , . . . , x h of length h := ht( p ) . Extend this to a full system of parameters of R and call it x . Recall that N is maximal Cohen-Macaulay. This implies that x is N -sequence. From this, x / , . . . , x h / is N p -sequence. Consequently, N p is maximal Cohen-Macaulay over R p . In view of Proposition 4.2(ii) we see that depth( M p ) = depth( M p ⊗ R p N p ) ( ∗ ) Since p ∈ Ass( M ⊗ R N ) , it turns out that depth( M ⊗ R N ) p = depth( M p ⊗ R p N p ) = 0 . We put this in ( ∗ ) and deduce that depth( M p ) = 0 . This is equivalent to saying that p ∈ Ass( M ) . The reverse inclusionholds by the same reasoning.ii): In the case R is a domain, we know Supp( N ) = Spec( R ) . So, the desired claim is an immediateapplication of i). (cid:3) Let us to recover:Fact . (Huneke-Sharp) Let ( R, m ) be Cohen-Macaulay of prime characteristic p and F -finite, M of finiteprojective dimension. Then Ass( F n ( M )) = Ass( M ) . Proof.
Each iteration of Frobenius defines a new R -module structure on R , and this R -module is denotedby n R . If x , . . . , x d is a full system of parameters, we know x p n , . . . , x p n d is a regular sequence. From this, n R is maximal Cohen-Macaulay and it is of full support. In view of Corollary 4.4(i) Ass( F n ( M )) = Ass( M ⊗ n R ) = Ass( M ) ∩ Supp( n R ) = Ass( M ) ∩ Spec( R ) = Ass( M ) , as claimed. (cid:3) Corollary 4.6.
Let R be a Cohen-Macaulay local ring equipped with a canonical module ω R and let M beof finite projective dimension. The following holds:i) Ass( M ⊗ R ω R ) = Ass( M ) .ii) M ⊗ R ω R has an ω R -resolution of length p . dim( M ) .Proof. i): Recall that Supp( ω R ) = Spec( R ) . This yields i), see Corollary 4.4.ii): Let p := p . dim( M ) . By definition, there is an exact sequence → R β p → . . . → R β → M → .Tensor it with − ⊗ R ω R we have → ω β p R → . . . → ω β R → M ⊗ R ω R → which is exact by Lemma 4.1. Bydefinition, M ⊗ R ω R has an ω R -resolution of length p . (cid:3) Concerning Corollary 4.4 a natural question arises: does maximal Cohen-Macaulay modules localize?Also, this was asked in MathOverflow, before than us. Here, we affirmatively answer it:
Corollary 4.7.
Let ( R, m ) be local, M of finite projective dimension and N be maximal Cohen-Macaulay.The following holds:i) N is locally maximal Cohen-Macaulay over its support.ii) Ass( M ⊗ R N ) ⊆ Ass( M ) ∩ Supp( N ) .iii) If M is Cohen-Macaulay then Ass( M ⊗ R N ) = Ass( M ) ∩ Supp( N ) . In particular, Ass( M ⊗ R N ) =Ass( M ) provided N is fully supported.Proof. i): Let p ∈ Supp( N ) . First, N p is Cohen-Macaulay and nonzero. It turns out that dim( R ) = dim( N ) = dim( N p ) + dim( N p N ) (1) Also,
Ann( N p N ) ⊇ p , and so(2) dim( N p ) ≤ dim( R p ) = ht( p ) ,(3) dim( N p N ) = dim( R Ann( N/ p N ) ) ≤ dim( R p ) ,(4) dim( R p ) + ht( p ) ≤ dim( R ) . Then we have dim( R ) (1) = dim( N p ) + dim( N p N ) (2) ≤ ht( p ) + dim( N p N ) (3) ≤ ht( p ) + dim( R p ) (4) ≤ dim( R ) . In particular, dim( N p ) = ht( p ) . Since N p is Cohen-Macaulay, it follows that N p is maximal Cohen-Macaulay.ii): Apply the first item along with the proof of Corollary 4.4 and use Proposition 4.2(i) to see Ass( M ⊗ R N ) ⊆ Ass( M ) ∩ Supp( N ) , as claimed.iii): Due to the second item, Ass( M ⊗ R N ) ⊆ Ass( M ) ∩ Supp( N ) . In view of Proposition 4.2(i) depth( M ) ≤ depth( M ⊗ R N ) ≤ dim( M ) . Since M is Cohen-Macaulay, depth( M ) = depth( M ⊗ R N ) . Now,the argument similar to Corollary 4.4 shows that Ass( M ⊗ R N ) = Ass( M ) ∩ Supp( N ) . (cid:3) Corollary 4.8.
Let ( R, m ) be local, M of finite projective dimension. Let Σ be the class of all maximalCohen-Macaulay modules. Then S L ∈ Σ Ass( M ⊗ R L ) is finite.Proof. In the light of Corollary 4.7(ii) we observe that S L ∈ Σ Ass( M ⊗ R L ) ⊆ Ass( M ) . Since M is finitelygenerated, the set of its associated prime ideals is finite. So, S L ∈ Σ Ass( M ⊗ R L ) is finite. (cid:3) We say a module M is primary if its zero submodule is primary in M , i.e., Ass( M ) = { p } . Corollary 4.9.
Let ( R, m ) be local, M be primary of finite projective dimension and N be maximal Cohen-Macaulay. Then M ⊗ R N is primary. In fact, Ass( M ⊗ R N ) = Ass( M ) .Proof. Let p ∈ Spec( R ) be such that Ass( M ) = { p } . Recall that Ass( L ) = ∅ if and only if L = 0 . Inview of Corollary 4.7(ii) we see ∅ 6 = Ass( M ⊗ R N ) ⊆ Ass( M ) ∩ Supp( N ) = { p } ∩ Supp( N ) ⊂ { p } . So,
Ass( M ⊗ R N ) = { p } . By definition, M ⊗ R N is primary. (cid:3) The following completes the proof of Corollary 1.3 from the introduction.
Corollary 4.10.
Let R be a local ring, I ⊳ R be of finite projective dimension and N be maximal Cohen-Macaulay. If ht( I ) > , then Ass( I ⊗ R N ) = Ass( N ) .Proof. We look at → I → R → R/I → and recall from Lemma 4.1 that Tor R ( R/I, N ) = 0 . We tensorthe short exact sequence with − ⊗ R N and obtain the following exact sequence R ( R/I, N ) −→ I ⊗ R N −→ N −→ R/I ⊗ R N −→ . This yields
Ass( I ⊗ R N ) ⊆ Ass( N ) ⊆ Ass(
N/IN ) ∪ Ass( I ⊗ R N ) (+) Recall from Corollary 4.7(ii) that
Ass(
R/I ⊗ R N ) ⊆ Ass(
R/I ) ∩ Supp( N ) . We plug this (+) and observethat Ass( I ⊗ R N ) ⊆ Ass( N ) ⊆ (Ass( R/I ) ∩ Supp( N )) ∪ Ass( I ⊗ R N ) ( ∗ ) Recall that N is equi-dimensional and of dimension equal to dim( R ) . Now, let p ∈ Ass( N ) . Since ht( I ) > , p / ∈ Ass(
R/I ) . By ( ∗ ) , p ∈ Ass( I ⊗ R N ) , i.e., Ass( N ) ⊆ Ass( I ⊗ R N ) . In sum, Ass( I ⊗ R N ) ⊆ Ass( N ) ⊆ Ass( I ⊗ R N ) , and the desired claim follows. (cid:3) Example . Concerning Corollary 4.10, the first item shows that the assumption p . dim( I ) < ∞ is neededeven if we assume Gdim( I ) < ∞ . The second item shows that the maximal Cohen-Macaulay assumptionis needed.i) Let ( R, m , k ) be any one-dimensional local integral domain which is not regular. Let I := m and lookat the maximal Cohen-Macaulay module N := m . We claim that Ass( I ⊗ R N ) = Ass( N ) . Indeed,we look at → I → R → k → and we drive the following exact sequence → Tor R ( k, N ) → I ⊗ R N → N → N ⊗ R k → . Recall that and Tor R ( k, N ) ≃ Tor R ( k, R/I ) ≃ k β ( R/I ) = 0 . Sincethis nonzero module is annihilated by m , we deduce that H m ( I ⊗ R N ) = 0 , i.e., m ∈ Ass( I ⊗ R N ) .This yields the first claim. If we assume in addition that R is Gorenstein, then Gdim( I ) = 0 < ∞ .ii) Let ( R, m ) be any regular local ring of dimension greater than one. Let I := m which is of finiteprojective dimension and look at N := m . Similar to the first item, we have Ass( I ⊗ R N ) = Ass( N ) . It may be nice to know when a prime ideal is in the support of a maximal Cohen-Macaulay module:
Corollary 4.12.
Let R be local and N be maximal Cohen-Macaulay. Then { p ∈ Spec( R ) : ∃ = I ⊳ R s.t. p . dim( I ) < ∞ and Ass( RI ) = { p }} ⊆ Supp( N ) \ Ass( R ) . Proof.
Let p ∈ Spec( R ) be such that there is a p -primary ideal I of finite projective dimension. Set M := R/I and recall that ∅ 6 = Ass( M ⊗ R N ) ⊆ { p } ∩ Supp( N ) . So, p ∈ Supp( N ) . In order to prove p Ass( R ) we remark that ∪ q ∈ Ass( R ) q consists of zero-divisors of R and that any non-trivial ideal of finiteprojective dimension equipped with a regular element. (cid:3) To see maximal Cohen-Macaulay modules are not necessarily full supported, we look at R := k [[ x, y ]] / ( xy ) and the maximal Cohen-Macaulay module N := R/ ( x ) . Then Supp( N ) = { ( x ) , m } $ { ( x ) , ( y ) , m } =Spec( R ) . In this example, { p ∈ Spec( R ) : ∃ I ⊳ R s.t. p . dim( I ) < ∞ and I is p − primary } = Supp( N ) \ Ass( R ) . Definition 4.13.
By Cohen-Macaulay defect we mean dim( M ) − depth( M ) and we denote it by CMdef( M ) . Corollary 4.14.
Let R be Cohen-Macaulay, M of finite projective dimension and N be maximal Cohen-Macaulay. Then CMdef( M ⊗ R N ) = CMdef( M ) .Proof. Recall from Proposition 4.2 ( ∗ ) that depth R ( M ⊗ R N ) = dim( R ) − p . dim R ( M ) . Since R is Cohen-Macaulay, and in view of Auslander-Buchsbaum formula, depth R ( M ⊗ R N ) = depth( M ) . We combine thiswith Fact 3.3 to observe that dim( M ⊗ R N ) − depth( M ⊗ R N ) = dim( M ) − depth( M ) . By definition,
CMdef( M ⊗ R N ) = CMdef( M ) . (cid:3) Definition 4.15.
Suppose p . dim( M ) < ∞ . By perfect defect of M , we mean pdef( M ) := p . dim( M ) − grade( M ) . Lemma 4.16.
Let R be formally equidimensional, M of finite projective dimension and N maximal Cohen-Macaulay. Then depth( M ⊗ R N ) = dim( M ) − pdef( M ) .Proof. The assumptions guarantee that dim( R ) = dim( M ) + grade( M ) (see [11]) and Tor R + ( M, N ) = 0 .Recall from Fact 3.2 that depth R ( M ⊗ R N ) = depth R ( N ) − p . dim R ( M ) . Combining these, we get thedesired claim. (cid:3) The following is a generalization of Fact 2.2:
Corollary 4.17.
A local ring R is Cohen-Macaulay iff pdef( M ) = CMdef( M ) for all p . dim( M ) < ∞ .Proof. Suppose R is Cohen-Macaulay and p . dim( M ) < ∞ . Apply N := R in Lemma 4.16 to see depth( M ) = dim( M ) − pdef( M ) . By definition, depth( M ) = dim( M ) − CMdef( M ) . Combining these, pdef( M ) = CMdef( M ) . Conversely, assume pdef( M ) = CMdef( M ) for all p . dim( M ) < ∞ . Apply this for M := R , we see R ) = CMdef( R ) , i.e., R is Cohen-Macaulay. (cid:3) Being integral domain
In Corollary 4.9 we were interested on primary modules of finite projective dimension. Symbolic powersof primes is the natural source to produce primary modules. Recall that the n -th symbolic power of I is denoted by I ( n ) . For example, R p ( n ) is primary. If it has finite projective dimension it imposes someconditions on the ring. Concerning the following observation, the case n = 1 is known, see [10, CorollaryII.3.3] . For more partial results with n = 1 , see [5, Lemma 4.9] and [6, Abstract] .Observation . Let R be a local ring and p ∈ Spec( R ) . If p . dim( R/ p ( n ) ) < ∞ for some n ∈ N and p ( n ) = 0 , then R is an integral domain. Proof.
We look at the localization map f : R → R p . Let r ∈ ker( f ) . Then there is x ∈ R \ p such that rx = 0 . Recall that p ( n ) is p -primary. Then R p ( n ) x −→ R p ( n ) is injective as x / ∈ rad( p ( n ) ) = p . In other words, x is R p ( n ) -sequence. Now, we apply the assumption p . dim( R p ( n ) ) < ∞ along with Auslander’s zero-divisorconjecture (follows from Roberts’ intersection theorem) to observe that x is R -sequence. From this r = 0 ,i.e., f is injective. Now recall that p ( n ) = p n R p ∩ R . Thus, p ( n ) R p = ( p n R p ∩ R ) R p = p n R p . Since p ( n ) = 0 we see p n R p = 0 . In particular, some nonzero powers of the maximal ideal of R p is of finiteprojective dimension. By a result of Levin and Vasconcelos [7, Theorem 1.1] we know R p is regular andlocal. Regular local rings are domain. From this R is an integral domain. (cid:3) The nonvanishing of p ( n ) is important. For example, let R := k [ X ] / ( X ) and p := ( x ) . Then p = 0 and consequently, p (2) = 0 . So, p (2) is free and R is not an integral domain. In this regard, we have: Corollary 5.2.
Let R be a reduced local ring and p ∈ Spec( R ) . If p . dim( R/ p ( n ) ) < ∞ for some n ∈ N ,then R is an integral domain.Proof. We may assume that p = 0 . Since = p n ⊂ p ( n ) , we deduce that p ( n ) is nonzero and of finiteprojective dimension, in particular it has a regular element. Since rad( p ( n ) ) = p it follows that p has aregular element. From this, dim( R p ) ≥ depth( R p ) ≥ . In other words, p n R p = 0 . Put this in the previousresult to see that R is an integral domain. (cid:3) Remark . Observation 5.1 is not true if we replace p . dim( − ) with Gdim( − ) , e.g. we look at R := k [ X ] / ( X ) , p := ( x ) and n := 1 . By id R ( − ) we mean the injective dimension. Here, we present a dual version of Observation 5.1: Corollary 5.4.
Let R be a local ring and p ∈ Spec( R ) . If id R ( R/ p ( n ) ) < ∞ for some n ∈ N and p ( n ) = 0 ,then R is a (Gorenstein) integral domain.Proof. By an application of Bass’ conjecture (follows by Rorbert’s intersection theorem), R is Gorenstein.For more details, see [10, Corollary II.5.3]. Over Gorenstein rings, a module has finite injective dimensionif and only if it has finite projective dimension. From this, p . dim( R/ p ( n ) ) < ∞ . In view of Observation5.1, R is an integral domain. (cid:3) Corollary 5.5.
Let R be a reduced local ring and p ∈ Spec( R ) . If id R ( R/ p ( n ) ) < ∞ for some n ∈ N , then R is a (Gorenstein) integral domain.Example . There is a ring with a prime ideal p such that the following assertions holds:i) It may be id( p ) < ∞ and id( p (2) ) = ∞ .ii) It may be p . dim( p ) = ∞ and p . dim( p (2) ) < ∞ . Proof.
Let X := ( X ij ) be a symmetric n × n matrix of indeterminates and let I r +1 ( X ) be the ideal generatedby the ( r + 1) -minors of X . The rings R = k [ X ] /I r +1 ( X ) is Cohen-Macaulay. Let p be the ideal generatedby the r -minors of the first r rows of X . It is well-known that p (2) is principal. Suppose r + 1 ≡ n is nottrue. Then p is the canonical module of R , see [4, Exercise 7.3.10].i) Since p is the canonical module we have id( p ) < ∞ . Since p (2) is principal ([4, Exercise 7.3.10]) and S r +1 is an integral domain, id( p (2) ) = id( R ) = ∞ , because R is not Gorenstein.ii) If p . dim( ω R ) < ∞ it follows from Auslander-Bridger formula that ω R is free, and so R is Goren-stein. This contraction shows that p . dim( ω R ) = ∞ . Since p (2) is principal and R is an integral domain, p . dim( p (2) ) = p . dim( R ) = 0 < ∞ (cid:3) Example . For each t > there is a ring with a prime ideal p such that id( p ) = ∞ and id( p ( t ) ) = id( p t ) < ∞ . Proof.
For simplicity we assume t = 2 . Let X := ( X ij ) be an n × ( n + 2) matrix of indeterminates andlet I r +1 ( X ) be the ideal generated by the ( r + 1) -minors of X . The ring R := k [ X ] /I r +1 ( X ) is Cohen-Macaulay. Let p be the ideal generated by the r -minors of the first r rows of X . It is well-known that p (2) is the canonical module of R , see [4, Theorem 7.3.6].Since p (2) is the canonical module we have id( p (2) ) < ∞ . Suppose on the way of contradiction that id( p ) < ∞ . Since p is a height one prime ideal and R/ p is a Cohen-Macaulay ring, it follows from −→ p −→ R −→ R/ p −→ that p is maximal Cohen-Macaulay. We apply these to find an integer ℓ such that p ∼ = ⊕ ℓ ω R = ⊕ ℓ p (2) .Taking the rank, it follows that ℓ = 1 . Consequently, p ∼ = p (2) . Now, we look at them in the divisor classgroup. We know that [ p (2) ] = 2[ p ] . In particular, p and p (2) are not isomorphic. This contradiction saysthat id( p ) = ∞ . (cid:3) References [1] M. Asgharzadeh,
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