A non-zero R-module M is called secondary if its multiplication map by any element a of R is either surjective or nilpotent. A secondary representation for an R-module M is an expression for M as a finite sum of secondary submodules. If such a representation exists, we will say that M is representable. A prime ideal p of R is said to be an attached prime of M if p=(N:RM) for some submodule N of M. If M admits a reduced secondary representation, M=S1+S2+…+Sn, then the set of attached primes AttR(M) of M is equal to {0:RSi:i=1,…,n}.
Recall that for any R-module M, the cohomological dimension of M with respect to an ideal a is defined as cd(a,M)=max{i:Hai(M)≠0} (see [5]). Also, AsshRM is defined as {p∈ AssRM:dimR/p=dimM}.
We need the following lemmas in the proof of main results.
Lemma 2.1. Let a and b be two ideals of R and M be a non-zero finitely generated R-module with finite dimension d>0. Then bHad(M)=0 if and only if Had(bM)=0.
Proof. See [8, Theorem 2.3].
Lemma 2.2. Let a and b be two ideals of R. Let M be a non-zero finitely generated R-module with finite dimension d>0 such that bHad(M)≠0. Then cd(a,M)=cd(a,bM)=dim(bM)=dimM=d.
Proof. See [8, Corollary 2.4].
The following theorem is a main result of [8] and has a key role in our proofs.
Theorem 2.3. Let a and b be two ideals of R. Let M be a finitely generated R-module with dimension d>0. Then
i) Ann(bHad(M))=Ann(Had(bM)).
ii) AttR(bHad(M))=AttR(Had(bM)).
Proof. i) See [8, Theorem 2.5].
ii) See [8, Theorem 2.15].
In the following, we obtain a generalization of [6, Theorem 2.6].
Theorem 2.4. Let M be a non-zero finitely generated R-module of dimension d, and let b be an ideal of R such that dimbM=d>0. If T⊆AsshR(bM), then there exists an ideal a of R such that AttR(bHad(M))=T.
Proof. Let T be a subset of AsshR(bM). By [6, Theorem 2.6], there exists an ideal a of R such that AttR(Had(bM))=T. Therefore Theorem 2.3 (ii) implies that AttR(bHad(M))=T, as required.
Theorem 2.5. Let a,b and c be three ideals of a complete local ring (R,m) and M a finitely generated R-module of dimension d>0. Then
AttR(bHad(M))=AttR(bHcd(M))⇔ AnnR(bHad(M))=AnnR(bHcd(M)).
Proof. Clearly, we can assume that bHad(M)≠0 and bHcd(M)≠0 and so by Lemma 2.2 we have dimbM=d. Assume that AttR(bHad(M))=AttR(bHcd(M)). Theorem 2.3 (ii) implies that AttR(Had(bM))=AttR(Hcd(bM)). By [4, Corollary 3.4], we conclude that Had(bM)≅Hcd(bM) and so AnnR(Had(bM))=AnnR(Hcd(bM)). Thus by Theorem 2.3 (i) it follows that AnnR(bHad(M))=AnnR(bHcd(M)). Conversely, let AnnR(bHad(M))=AnnR(bHcd(M)). By Theorem 2.3 (i), we have AnnRHad(bM)=AnnRHcd(bM). Then by Theorem [7, 2.9] we conclude that Had(bM)≅Hcd(bM) and so AttR(Had(bM))=AttR(Hcd(bM)). Now the result follows from Theorem 2.3,(ii).
Theorem 2.6. Let a and b be two ideals of R and M a finitely generated R-module of dimension d>0. Then
SuppR(bHad(M))=SuppR(Had(bM)).
Proof. By Lemma 2.1 we can assume that bHad(M)≠0 and Had(bM)≠0. Thus, by Lemma 2.2 we have dimR(bM)=d. By [3, 7.1.7], Had(M) is artinian and so bHad(M) is artinian. Thus SuppR(bHad(M))⊆ Max(R). Assume that m∈Max(R) such that m∈SuppR(bHad(M)). Thus bmHaRmd(Mm)≠0 and so by [3, Theorem 6.4.1], it follows that dimRm(Mm)=d. By Lemma 2.1 we conclude that HaRmd(bmMm)≠0. Thus m∈SuppR(Had(bM)). Therefore
SuppR(bHad(M))⊆SuppR(Had(bM)).
Similar to the above method, we can show that
SuppR(Had(bM))⊆SuppR(bHad(M))
and the proof is complete.
Corollary 2.7. Let a and b be two ideals of R and M a finitely generated R-module of dimension d>0. Let bHad(M)≠0. Then
SuppR(bHad(M))={m∈Max R:∃p∈Assh (bM)s.tp⊆m,cdRm(aRm,RmpRm)=d}.
Proof. The assertion follows immediately from Theorem 2.6 and [4, 2.7].
Corollary 2.8. Let a,b and c be three ideals of R and M,N two finitely generated R-modules of dimension d>0. If SuppR(bM)=SuppR(cN) then
i) SuppR(bHad(M))=SuppR(cHad(N)).
ii) AttR(bHad(M))=AttR(cHad(N)).
Proof. Since SuppR(bM)=SuppR(cN) we conclude that cd(a,bM)=cd(a,cN) by [5, Theorem 2.2]. If bHad(M)=0 then Had(bM)=0 and so cd(a,bM)<d by Lemma 2.1. Thus cd(a,cN)<d and so Had(cN)=0. By Lemma 2.1, it follows that cHad(N)=0 and the result follows in this case.
Now, assume that bHad(M)≠0. By Lemma 2.2 we have dimR(bM)=cd(a,bM)=d. But SuppR(bM)=SuppR(cN) implies that dimR(cN)=cd(a,cN)=d and AsshRbM=AsshRcN. Now Corollary 2.7 shows that, SuppRbHad(M)=SuppRcHad(N). On the other hand, since AsshR(bM)=AsshR(cN) we have
{p∈AsshR(bM):cd(a,R/p)=d}={p∈AsshR(cN): cd(a,R/p)=d}.
Now, the result (ii) follows immediately from [8, Corollary 2.16], as required.
In the above results, we saw that some basic properties of two R-modules bHad(M) and Had(bM) are the same. The following result shows that these two R-modules are isomorphic under certain conditions.
Theorem 2.9. Let a and b be two ideals of R and M a finitely generated R-module of dimension d>0. In each of the following cases, bHad(M)≅Had(bM).
i) cd(a,M/bM)<d−1.
ii) bM=M.
iii) b=Rx where x is a non-zerodivisor on M.
Proof. By Lemma 2.1 we can assume that bHad(M)≠0 and Had(bM)≠0. Thus, by Lemma 2.2, we have dimR(bM)=cd(a,M)=d. Assume that c:=AnnRM. Since SuppR(R/c)=SuppR(M), by [5, Theorem 2.2] cd(a,R/c)=cd(a,M)=d. Thus by the Independence Theorem ([3, Theorem 4.2.1]) cd(aR/c,R/c)=d and so HaR/ci(R/c)=0 for all i>d. It follows that HaR/cd(−) is a right exact functor on the category of R/c-modules and R/c-homomorphisms. Thus
Had(M)/b Had(M)≃(HaR/cd(R/c)⊗R/cM)⊗RR/b
≃HaR/cd(R/c)⊗R/cM/bM
≃Had(M/bM).
If cd(a,M/bM)<d−1 then Had(M/bM)=0 and so by the above isomorphism we conclude that Had(M)=bHad(M). On the other hand, the exact sequence 0→bM→M→M/bM→0 induces an exact sequence
⋯→Had−1(M/bM)→Had(bM)→Had(M)→Had(M/bM)→0.
Since cd(a,M/bM)<d−1 we have Had−1(M/bM)=Had(M/bM)=0 and so the above exact sequence implies that Had(bM)≅Had(M). But, we showed that Had(M)=bHad(M) and so the proof is complete in this case.
If bM=M then Had(M)/b Had(M)≃Had(M/bM)=0. Thus bHad(M)=Had(M). Since bM=M we have Had(bM)=Had(M) and the result follows in this case.
Now assume that b=Rx and x is a non-zerodivisor on M. The exact sequence 0→M→xM→M/xM→0 induces an exact sequence
⋯→Had(M)→xHad(M)→Had(M/xM)→0.
Since dimM/xM=d−1, by [3][6.1.2] we have Had(M/xM)=0. Thus, the above exact sequence implies that xHad(M)=Had(M). On the other hand, it is easy to see that the R-module homomorphism φ:M→xxM is an isomorphism and it follows that Had(M)≅Had(xM). Therefore xHad(M)≅Had(xM), as required.
In the remainder, we prove some results about Fad(bM). For our proofs, we need the following theorems.
Theorem 2.10. Let a be an ideal of a local ring (R,m) and M a finitely generated R-module. Then
dimRM/aM=sup{i∈ℕ0:Fai(M)≠0}.
Proof. See [9, Theorem 4.5].
Theorem 2.11. Let a and b be two ideals of a local ring (R,m) and M a finitely generated R-module of dimension d>0. Then
bFad(M)=0⇔Fad(bM)=0.
Proof. See [8, Theorem 3.3].
Corollary 2.12. Let a and b be two ideals of a local ring (R,m) and M a finitely generated R-module of dimension d>0 such that bFad(M)≠0. Then Fad(bM)≠0 and dim(M/aM)=dim(bM/abM)=dimbM=d.
Proof. See [8, Corollary 3.4].
Theorem 2.13. Let a and b be two ideals of a local ring (R,m) and M a finitely generated R-module of dimension d>0. Then
i) Ann(bFad(M))=Ann(Fad(bM)).
ii) AttR(bFad(M))=AttR(Fad(bM)).
Proof. i) See [8, Theorem 3.6].
ii) See [8, Theorem 3.13].
Theorem 2.14. Let a,b and c be three ideals of a local ring (R,m) and M,N two finitely generated R-modules of dimension d>0 such that SuppR(bM)=SuppR(cN). Then AttR(bFad(M))=AttR(cFad(N)).
Proof. If bFad(M)=0 then by Theorem 2.11, Fad(bM)=0 and so by Theorem 2.10, dim(bM/abM)<d. Since SuppR(bM)=SuppR(cN) we conclude that dim(cN/acN)<d and by Theorem 2.10, Fad(cN)=0. Therefore by Theorem 2.11, cFad(N)=0. Thus AttR(bFad(M))=AttR(cFad(N))=∅ and so the result follows in this case.
Now, assume that bFad(M)≠0. If cFad(N)=0 then similar to the above argument it follows that bFad(M)=0. Thus cFad(N)≠0 and so by Corollary 2.12 we have dimbM=dimcN=d. Since SuppR(bM)=SuppR(cN) it is easy to see that AsshR(bM)=AsshR(cN) and so AsshR(bM)∩Var(a)=AsshR(cN)∩Var(a). Therefore, by [8, Corollary 3.14], we have AttR(bFad(M))=AttR(cFad(N)), as required.
Theorem 2.15. Let b be an ideal of a local ring (R,m), and let M be a finitely generated R-module of dimension d>0 such that dimR(bM)=d. If T⊆AsshR(bM), then AttR(bFad(M))=T where a:=∩ pi∈Tpi is an ideal of R.
Proof. By [7, Theorem 2.2], AttR(Fad(bM))=T where a:=∩ pi∈Tpi is an ideal of R. Now Theorem 2.13 (ii) implies that AttR(bFad(M))=T.
Theorem 2.16. Let a,b and c be three ideals of a local ring (R,m) and M a finitely generated R-module of dimension d>0. Then
AttR(bFad(M))=AttR(bFcd(M))⇔AnnR(bFad(M))=AnnR(bFcd(M)).
Proof. ⇒): If bFad(M)=0 then AttR(bFad(M))=∅. By assumption we conclude that AttR(bFcd(M))=∅ and so bFcd(M)=0.
Now we assume that bFad(M)≠0 and bFcd(M)≠0. Thus, by Corollary 2.12 dimbM=d and by Theorem 2.13 (ii) and our assumption, we have AttR(Fad(bM))=AttR(Fcd(bM)). Therefore, by [2, 3.4], we have Fad(bM)≅Fcd(bM). Thus AnnR(Fad(bM))=AnnR(Fcd(bM)) and so the assertion follows from Theorem 2.13 (i).
⇐): We can (and do) assume that bFad(M)≠0 and bFcd(M)≠0. Then, by Theorem 2.13 (i) and our assumption, we have AnnR(Fad(bM))=AnnR(Fcd(bM)). Therefore, by [6, Corollary 3.9 (i)], we conclude that Fad(bM)≅Fcd(bM). Thus AttR(Fad(bM))=AttR(Fcd(bM)). Therefore, the assertion follows from Theorem 2.13 (ii).
Theorem 2.17. Let a,b and c be three ideals of a complete local ring (R,m) and M a finitely generated R-module of dimension d>0. Then
AttR(bHad(M))=AttR(bFcd(M))⇔AnnR(bHad(M))=AnnR(bFcd(M)).
Proof. ⇒): We can assume that bHad(M)≠0 and bFcd(M)≠0 and so dimbM=d. If Att(bHad(M))=AttR(bFcd(M)) then by Theorem 2.3 (ii) and Theorem 2.13 (ii) we conclude that AttR(Had(bM))=AttR(Fcd(bM)). By [7][Theorem 2.5], it follows that Had(bM)≅Fcd(bM) and so AnnR(Had(bM))=AnnR(Fcd(bM)). Thus by Theorem 2.3 (i) and Theorem 2.13 (i) we have AnnR(bHad(M))=AnnR(bFcd(M)), as required.
⇐): Assume that AnnR(bHad(M))=AnnR(bFcd(M)). By Theorem 2.3 (i) and Theorem 2.13 (i), we have AnnRHad(bM)=AnnRFcd(bM). Then by [7][Corollary 2.9] we conclude that Had(bM)≅Fcd(bM) and so AttR(Had(bM))=AttR(Fcd(bM)). Now, the result follows from Theorem 2.3 (ii) and Theorem 2.13 (ii).
Theorem 2.18 Let a and b be two ideals of a local ring (R,m) and M a finitely generated R-module of dimension d>0. Then
SuppR(bFad(M))=SuppR(Fad(bM))
Proof. By Theorem 2.11, we can (and do) assume that bFad(M)≠0 and so by Corollary 2.12 we have Fad(bM)≠0 and dimbM=d. But, by [2, Lemma 2.2] Fad(bM) is an artinian R-module and so Supp(Fad(bM))={m}. On the other hand, Fad(M) is artinian and so bFad(M)≠0 is artinian. Thus SuppR(bFad(M))={m}. Since SuppR(Fad(bM))={m} we conclude that SuppR(Fad(bM))=SuppR(bFad(M)), as required.