Let R=⊕n≥0Rn be a standard homogeneous Noetherian ring with local base ring (R0,m0) and let M be a finitely generated graded R-module. Let HR+i(M) be the ith local cohomology module of M with respect to R+=⊕n>0Rn. Let S be a Serre subcategory of the category of R-modules and let i be a nonnegative integer. In this paper, if dimR0≤1, then we investigate some conditions under which the R-modules R0/m0⊗R0HR+i(M),Γm0R(HR+i(M)) and Hm0R1(HR+i(M)) are in S for all i≥0. Also, we prove that if dimR0≤2, then the graded R-module Hm01(HR+i(M)) is in S for all i≥0. Finally, we prove that if n is the biggest integer such that Hai(M)∉S, then HR+i(M)/m0HR+i(M)∈S for all i≥n.
1. Introduction
Let R=⊕n≥0Rn be a Noetherian homogeneous ring with local base ring R0,m0. So, R0 is a Noetherian ring and there are finitely many elements l1,l2,…,lr∈R1 such that R=R0l1,l2,…,lr. Let R+=⊕n>0Rn denote the irrelevant ideal of R and let m=m0⊕R+ denote the graded maximal ideal of R. Finally, let M=⊕n∈ZMn be a finitely generated graded R-module. Recall that a class of R-modules is a Serre subcategory of the category of R-modules when it is closed under taking submodules, quotients, and extensions. Always, S stands for a Serre subcategory of the category of R-modules. Hence, if L→M→N is an exact sequence of the category of R-modules and R-homomorphisms such that both end terms belong to S, then M also belongs to S. Note that the following subcategories are examples of Serre subcategory of the category of R-modules: finite R-modules; coatomic R-modules [1]; minimax R-modules [2]; a-cofinite R-modules; weakly Laskerian R-modules; Matlis reflexive R-modules; and trivially the zero R-modules. Recently, some results have been proved concerning the local cohomology modules HaiM of a module M in some certain Serre subcategory of the category of modules (cf. [3–7]). Aghapournahr and Melkersson (cf. [4]) gave a condition on a Serre subcategory S. To give more details, let a be an ideal of R, let M be an R-module, and let 0:Ma=x∈M∣ax=0. The R-module M is said to satisfy Ca condition on S whenever the following condition holds: (1)If M=ΓaM and if 0:Ma is in S then M is in S.
Then S is said to satisfy Ca condition, whenever, every R-module satisfies Ca condition on S. For example, the class of zero modules and Artinian R-modules satisfy the condition Ca.
In this paper, according to dimR0≤1, we investigate some conditions under which the R-modules R0/m0⊗R0HR+iM,Γm0RHR+iM and Hm0R1(HR+i(M)) are in S for all i≥0. Next, we prove that if dimR0≤2, then Hm01(HR+i(M)) is in S for all i≥0. Also if S satisfies Ca condition, then under some conditions, R-modules HR+i(Hm0R1(M)) and ΓR+Hm0R1M belong to S. Finally, we prove that if n is the biggest integer such that HaiM∉S, then HR+iM/m0HR+iM∈S for all i≥n.
In Section 2, we give some notations and definition that we used in Section 3.
2. Preliminaries
Let R be a ring. The spectrum of R is denoted by Spec R, that is, the set of prime ideals of R with the Zariski topology, which is the topology where the closed sets are Va=p∈SpecR∣a⊆p for ideals a⊆p.
The height of a prime ideal p, denoted by height p, is defined by the supremum of integers t such that there exists a chain of prime ideals p=p0⊋p1⊋p2⊋⋯⊋pt, where pi∈SpecR. The Krull dimension of R is defined as dimR=supheightp∣p∈SpecR.
Consider the complex K∙(2)⋯⟶Ki-1⟶δKi-1Ki⟶δKiKi+1⟶⋯.
The differential on K∙ is denoted by δK, or sometimes simply δ. We say that K∙ is bounded below if Ki=0 for i≪0 and bounded above if Ki=0 for i≫0; if both conditions hold, then K∙ is bounded. On occasion, we consider a complex as a graded module with a graded endomorphism of degree 1.
Let M be an R-module. An injective resolution of M is a complex E∙ of injective R-modules 0→E0→E1→E2→⋯, equipped with a quasi-isomorphism of complexes l:M→E∙; here one views M as a complex concentrated in degree 0. Said otherwise, a complex E∙ of injective modules is a resolution of M if there is an exact sequence 0→M→lE0→E1→E2→⋯.
Let M and N be R-modules. Then ExtRiM,-i≥0 are the right derived functors of HomRM,-, and ToriR-,Ni≥0 are the left derived functors of -⊗RN.
For an exact sequence of R-modules 0→J→M→L→0 and each R-module N, one has exact sequences:(3)⋯⟶ExtRiN,M⟶ExtRiN,L⟶ExtRi+1N,J⟶ExtRi+1N,M⟶⋯,⋯⟶ExtRiM,N⟶ExtRiJ,N⟶ExtRi+1L,N⟶ExtRi+1M,N⟶⋯,⋯⟶Tori+1RM,N⟶Tori+1RL,N⟶ToriRJ,N⟶ToriRM,N⟶⋯.
For an R-module M, the a-torsion of M is defined by Γa(M)≔⋃n∈N0:Man. Observe that Γa(M) is a submodule of M and Γa- is a left exact a-torsion functor of R-modules and R-homomorphisms. Let n∈N0 and let M be an R-module. Choose an injective resolution of M, so that we have an exact sequence (4)0⟶M⟶lE0⟶d0E1⟶d1E2⟶d2E3⟶⋯.
Then, apply the functor Γa(-) to the resolving cocomplex (5)⋯⟶0⟶d-1E0⟶d0E1⟶d1E2⟶d2E3⟶⋯in order to obtain the cocomplex (6)⋯⟶0⟶Γad-1ΓaE0⟶Γad0ΓaE1⟶Γad1ΓaE2⟶Γad2⋯.
Then, the nth local cohomology module of M with respect to a is defined as the R-module HanM=kerΓadn/ImΓadn-1. If 0→L→fM→gN→0 is an exact sequence of R-modules and R-homomorphisms, then we have the following long exact sequence: (7)0⟶Ha0L⟶Ha0fHa0M⟶Ha0gHa0N⟶Ha1L⟶Ha1fHa1M⟶Ha1gHa1N⟶⋯.
As a general reference to homological and commutative algebra, we refer to [8].
3. Main Results
In this paper, R=⨁n≥0Rn is a Noetherian homogeneous ring with local base ring R0,m0. We start off this section with some technical lemmas to rich our main aims in this work.
Lemma 1.
Let H∈S be an R-module and let M0 be a finitely generated R0-module. Then for any i≥0 the R-module ToriR0M0,H∈S.
Proof.
Let (8)⋯⟶Fi+1⟶Fi⟶Fi-1⟶⋯⟶F1⟶F0⟶M⟶0be a minimal free resolution of M0. If Fi=Rni for some integer ni, then ToriR0(M0,H) is an R0-subquotient of Fi⊗R0H≅Hni. Thus ToriR0M0,H∈S.
Lemma 2.
If M,R/a∈S, then HaiM∈S for all i≥0.
Proof.
Let F∙:⋯→F1→F0→0 be a finite free resolution of R/an. If Fi=Rni for some integer ni, then ExtRiR/an,M=HiHomRF∙,M is a subquotient of Mni. Since S is a Serre class, ExtRiR/an,M∈S. From definition, we have Hai(M)=limn∈NExtRiR/an,M and so HaiM∈S for all i≥0.
An immediate result of the above lemmas is the following result.
Corollary 3.
If dimR0=0 and M∈S, then R0=R/R+ is in S and HR+iM∈S for all i≥0.
Lemma 4.
If M∈S, then HR+iΓm0RM∈S as an R-module for all i≥0.
Proof.
Since Γm0RM is a finitely generated m0-torsion module, there exists an n≥0 such that m0nΓm0RM=0. Then we obtain the R/m0n-module Γm0RM. If i≥0, then from the Graded Independence Theorem (cf. ([8], 13.1.6)) there is an isomorphism of graded R-modules HR+iΓm0RM≅HR/m0nR+iΓm0RM. Since dimR/m0nR0=dimR0/m0n=0, Corollary 3 implies that HR+iΓm0RM∈S.
Lemma 5.
Let i≥0 and R0/m0⊗R0HR+iM/Γm0RM∈S; then R0/m0⊗R0HR+iM∈S.
Proof.
Let L≔M/Γm0RM. Then we have the exact rows:(9)HR+iΓm0RM⟶fHR+iM⟶gHR+iL⟶HR+i+1Γm0RM,(10)0⟶HR+iMImf⟶HR+iL⟶HR+iLImg⟶0,such that h:HR+iM→HR+i(M)/Imf is an epimorphism and k:HR+iL/Img→HR+i+1Γm0RM is a monomorphism. Since T≔HR+iL/Img is a submodule of HR+i+1Γm0RM, Lemma 4 leads to T∈S. Also, from Lemma 1, we get Tor1R0R0/m0,T∈S. The short exact sequence (10) deduces the following exact sequence: (11)Tor1R0R0m0,T⟶R0m0⊗R0HR+iMImf⟶R0m0⊗R0HR+iL.
According to our assumption, R0/m0⊗R0HR+i(L)∈S and then R0/m0⊗R0HR+i(M)/Imf∈S. From the exact sequence (9), we have the exact sequence R0/m0⊗R0HR+i(M/Γm0R(M))→R0/m0⊗R0HR+i(M)→R0/m0⊗R0HR+i(M)/Imf, and hence R0/m0⊗R0HR+i(M)∈S.
Theorem 6.
Let dimR0≤1 and i≥0; then the following assertions hold:
If M∈S, then R0/m0⊗R0HR+i(M)∈S.
If x∈m0, such that xR0=m0, and H(R+,x)i(M)∈S, then the R-modules Γm0R(HR+i(M)) and Hm0R1(HR+i(M)) are in S.
Proof.
(1) If dimR0=0, then from Corollary 3 we have HR+i(M)∈S and then from Lemma 1 we get R0/m0⊗R0HR+i(M)∈S. Let dimR0=1 and Γm0R(M)=0. Then there exists an M-regular element x∈m0 which avoids all minimal primes of R0. Now, the exact sequence (12)0⟶M⟶xM⟶MxM⟶0induces an exact sequence(13)HR+i-1MxM⟶HR+iM⟶xHR+iM⟶fHR+iMxM.
Since x avoids all minimal primes of R0, dimR0/xR0=0. Then by the Independence Theorem, HR+i(M/xM)∈S and then Imf∈S. The exact sequence (13) gives the exact sequence (14)R0m0⊗R0HR+iM⟶xR0m0⊗R0HR+iM⟶R0m0⊗R0Imf⟶0.
As x∈m0, the multiplication map x is zero. Then there is an isomorphism (15)R0m0⊗R0HR+iM≅R0m0⊗R0Imf.
Therefore, R0/m0⊗R0HR+i(M)∈S.
(2) Let x∈m0 such that xR0=m0. Therefore, for all i≥0, there is an exact sequence of graded R-modules (16)HR+i-1M⟶fxi-1HR+i-1Mx⟶HR+,xiM⟶HR+iM⟶fxiHR+iMxsuch that fxi-1 and fxi are the natural homomorphisms (cf. [8], 13.1.12). Now from (17)cokerfxi-1≅HxR1HR+iM≅Hm0R1HR+iM,kerfxi≅ΓxRHR+iM=Γm0RHR+iM,we obtain an exact sequence (18)0⟶Hm0R1HR+i-1M⟶HR+,xiM⟶Γm0RHR+iM⟶0.
Therefore, Γm0R(HR+i(M)) and Hm0R1(HR+i(M)) are in S.
Corollary 7.
Let dimR0≤1 and i≥0. Moreover, let q0⊆R0 be an m0-primary ideal. Then
the R-module R0/m0/q0HR+iM is in S;
the R-module 0:HR+iMq0 is in S.
Proof.
(1) Since q0 is an m0-primary ideal, there is some n≥0 such that m0n⊆q0. Then it suffices to prove that R0/m0n⊗R0HR+iM∈S. We proceed the assertion by induction on n. For n=1, use part (1) of Theorem 6. Now suppose, inductively, that n>1 and the result has been proved for all values smaller than n. We have the exact sequence (19)m0n-1m0n⊗R0HR+iM⟶R0m0n⊗R0HR+iM⟶R0m0n-1⊗R0HR+iM,and the natural isomorphism (20)m0n-1m0n⊗R0HR+iM≅m0n-1⊗R0R0m0⊗R0HR+iM.
Now, use part (1) of Theorem 6 and Lemma 1.
(2) Since 0:HR+iMq0⊆Γm0RHR+iM, part (2) of Theorem 6 implies that 0:HR+iMq0∈S.
The following result is an extension of Theorem 6 to the case dimR0=2.
Theorem 8.
If dimR0≤2, then Hm01HR+iM∈S for all i≥0.
Proof.
If dimR0≤1, this statement is the same as Theorem 6. Let dimR0=2, then there is a system of parameters x,y of R0 such that Hm0R1HR+iM=H(x,y)R1(HR+i(M)). From the exact sequence (21)0⟶HyR1HR+i-1M⟶Hy,R+iM⟶ΓyRHR+iM⟶0,which obtains an epimorphism HxR1Hy,R+iM→HxR1ΓyRHR+iM, also, there is a monomorphism (22)HxR1Hy,R+iM⟶Hx,y,R+i+1M=Hmi+1M.
Since Hmi+1(M)∈S, HxR1ΓyRHR+iM∈S. Application of the functor ΓxR- to the monomorphism HyR1HR+iM→Hy,R+i+1M yields a monomorphism ΓxRHyR1HR+iM→ΓxRHy,R+i+1M. Also, there is an epimorphism (23)Hmi+1M=Hx,y,R+i+1M⟶ΓxRHy,R+i+1M.
Because Hmi+1(M)∈S, ΓxRHyR1HR+iM∈S. The exact sequence (24)0⟶HxR1ΓyRHR+iM⟶Hx,yR1HR+iM⟶ΓxRHyR1HR+iM⟶0implies that Hm01HR+iM∈S.
Proposition 9.
Let M∈S and M satisfies Ca condition on S. If R0 is a local ring of dimension one, then HR+iHm0R1M∈S for all i≥0.
Proof.
Since Hm0R1(M)≅Hm0R1M/Γm0RM, one can assume that Γm0R(M)=0. Then, there exists an element x0∈m0 such that it is an M-sequence. By the exact sequence (25)0⟶M⟶x0M⟶Mx0M⟶0and by application of the functor HR+i- to this exact sequence, we have the following exact sequence (26)HR+iΓm0RMx0M⟶HR+iHm0RiM⟶x0HR+iHm0RiM⟶HR+i+1Γm0RMx0M.
Lemma 4 leads to HR+iΓm0RM/x0M∈S and then 0:HR+i(Hm0R1M)x0∈S. Furthermore, by the basic properties of local cohomology, we have Γa(HR+iHm0R1M)=HR+i(Hm0R1(M)). Since HR+iHm0R1M satisfies Ca condition on S, HR+i(Hm0R1(M))∈S for all i≥0.
Proposition 10.
Suppose that M satisfies Ca condition on S and n is the least integer such that Hm0Rn(M)∉S. Then ΓR+Hm0R1M∈S for all i≤n.
Proof.
If i<n, then Hm0RiM∈S and therefore ΓR+(Hm0Ri(M))∈S. Let i=n and from the isomorphism Hm0RiM≅Hm0RiM/Γm0RM for all i, one can suppose that Γm0RM=0. Therefore, there exists an element x∈m0 such that it is an M-sequence. In view of the exact sequence (27)0⟶M⟶xM⟶MxM⟶0,we have the exact sequence (28)Hm0Rn-1M⟶fHm0Rn-1MxM⟶Hm0RnM⟶xHm0RnM.
By the above sequence, we obtain the exact sequence (29)0⟶kerf⟶Hm0Rn-1M⟶Imf⟶0,and isomorphism Imf≅0:Hm0RnMx. Since Hm0Rn-1(M)∈S, Imf∈S and (0:Hm0Rn(M)x)∈S. Thus ΓR+0:Hm0RnMx=(0:ΓR+(Hm0Rn(M))x)∈S. Then ΓR+Hm0RnM∈S, because ΓR+Hm0RnM is x-torsion.
Definition 11.
Let a be a graded ideal of R and M∈S. Define taS(M)=supi∣Hai(M)∉S.
Lemma 12.
Let x∈m be a homogeneous nonzero divisor of M. Then (30)tR+SMxM≤tR+SM.
Proof.
Let degx=d. Since x is a nonzero divisor of M, there is an exact sequence (31)0⟶M-d⟶xM⟶MxM⟶0.
Application of the functor HR+i- to the above exact sequence induces the exact sequence (32)HR+iM⟶HR+iMxM⟶xHR+i+1M-d⟶HR+i+1M.
If tR+S(M)=n, then for all i>n we have HR+i(M)∈S. This means that HR+i(M/xM)∈S. Thus, tR+S(M/xM)≤n.
Remark 13.
(1) Any local flat morphism of local Noetherian rings is faithfully flat. So, if R0´ is flat over R0 and m0´=m0R0´, then R0´ is faithfully flat over R0. Moreover, if R0´,m0´ is a faithfully flat local R0-algebra, then from ([9], Theorem 1) we obtain that A∈S of the graded R-module if and only if A´≔R0´⊗R0A∈S of the graded module over R´≔R0´⊗R0R.
(2) If R0´,m0´ is a faithfully flat local R0-algebra, then tR+S(M)=tR0´⊗R0R+SR0´⊗R0M.
Theorem 14.
Let n=tR+S(M); then HR+i(M)/m0HR+i(M)∈S for all i≥n.
Proof.
We proceed by induction on d=dimR0. If i>n, then HR+iM∈S and then HR+i(M)/m0HR+iM∈S. Let i=n. If d=1, then the result follows Theorem 6. Assume inductively that the result has been proved for all values smaller than d, so we prove it for d. Let x be an indeterminate and let R0´≔R0xm0R0,m´=m0R0´,R´=R0´⊗R0R and M´≔R0´⊗R0M. Then by the flat base change property of local cohomology, we have (33)R0´⊗R0HR+nMm0HR+nM≅HR´+nMm0´HR´+nM´.
Since R0´ is a faithfully flat local R0-algebra, in view of Remark 13(1) it is enough to show that HR´+nM/m0´HR´+nM´∈S. Then, we may replace R and M by R´ and M´, respectively. Hence we assume that R0/m0 is infinite residue field. From application of the functor HR+i(-) over the exact sequence (34)0⟶Γm0M⟶M⟶MΓm0M⟶0,we have the following exact sequence: (35)HR+iΓm0M⟶fHR+iM⟶gHR+iMΓm0M⟶hHR+i+1Γm0M.
From Lemma 4, we have that HR+iΓm0M∈S for all i. Now, let U=Imf, V=Img, and W=Imh. It follows from Lemma 1 that ToriR0R0/m0,U and ToriR0R0/m0,W are in S for all i. If HR+iM/Γm0M/m0HR+iM/Γm0M∈S, then V/m0V∈S. This implies that HR+i(M)/m0HR+i(M)∈S. Since tSR+(M)=tSR+(M/Γm0(M)), we may assume that Γm0(M)=Γm0R(M)=0. If x0 is a nonzero divisor of M and a part of a system of parameters of m0, then we have the exact sequence (36)0⟶M⟶x0M⟶Mx0M⟶0.
Application of the functor HR+n(-) to the above obtained exact sequence induces the exact sequence (37)HR+nM⟶x0HR+nM⟶HR+nMx0M⟶kHR+n+1M.
Let kerk=X and Imk=Y. Since HR+n+1(M)∈S, the graded R-module Y is in S and therefore from Lemma 1 we get HR+n(M/x0M)/m0HR+n(M/x0M)∈S if and only if X/m0X∈S. On the other hand, application of the functor R0/m0⊗R0- induces the exact sequence (38)R0m0⊗R0HR+nM⟶pR0m0⊗R0HR+nM⟶Xm0X⟶0,such that p≔idR0/m0⊗R0x0. Since x0∈m0, the map p is zero and HR+n(M)/m0HR+n(M)≅X/m0X. Now, similar to the above arguments, one can conclude that HR+n(M)/m0HR+n(M)∈S if and only if HR+n(M/x0M)/m0HR+n(M/x0M)∈S. Let R0¯=R0/x0R0 and m0¯=m0/x0m0. Using the Independence Theorem for graded local cohomology, we have HR+nM/x0M≅H(R0¯⊗R0R)+n(M/x0M).
Note that the local base ring of the graded ring R0¯⊗R0R is R0¯,m0¯ and dimR0¯=d-1. By the following isomorphism (39)HR+nM/x0Mm0HR+nM/x0M≅HR0¯⊗R0R+nM/x0Mm0¯HR0¯⊗R0R+nM/x0Mtogether with Lemma 12 and using induction hypothesis, this module is in S.
Conflicts of Interest
The authors declare that there are no conflicts of interest regarding the publication of this paper.
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