Throughout this paper, R is a commutative Noetherian ring with identity, a and b are ideals of R and M is a finitely generated R-module of dimension d. Recall that the i-th local cohomology module of M with respect to a is defined as:

The reader can refer to [3], for the basic properties of local cohomology.

Let a be an ideal of a local ring (R,m) and M a finitely generated R-module. For each i≥0; Fai(M):=lim←n∈ℕHmi(M/anM) is called the i-th formal local cohomology of M with respect to a. The basic properties of formal local cohomology modules are found in [1], [9] and [2].

In [8], we studied local cohomology module bHad(M) and formal local cohomology module bFad(M). In this paper, we obtain some new results about them.

Here, we obtain some relations between AnnR(bHad(M)), AttR(bHad(M))and SuppR(bHad(M)). Also, we get similar results for bFad(M).

We determine the support of the top local cohomology module bHadimM(M). More precisely, we will show that SuppR(bHad(M))=SuppR(Had(bM)). Also, we prove that for an arbitrary Noetherian local ring (R,m), SuppR(bFad(M))=SuppR(Fad(bM)).

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

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

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

Similar to the above method, we can show that

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

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

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

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

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

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

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

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

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

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

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.

The author would like to thank the referee for careful reading and many useful suggestions.