# Some Properties of Serre Subcategories in the Graded Local Cohomology Modules

Authors: Feysal Hassani; Rasul Rasuli
Journal: Journal of Mathematics
 (2017)

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## Abstract

LetR=⊕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 dim⁡R0≤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 dim⁡R0≤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.


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## Body

## 1. Introduction

LetR=⊕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)IfM=ΓaMand  if0:Mais  inSthenMis  inS.ThenS 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 todim⁡R0≤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 dim⁡R0≤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 Section2, we give some notations and definition that we used in Section 3.


## 2. Preliminaries

LetR 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 idealp, 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 dim⁡R=sup⁡heightp∣p∈SpecR.Consider the complexK∙(2)⋯⟶Ki-1⟶δKi-1Ki⟶δKiKi+1⟶⋯.The differential onK∙ 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.LetM 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→⋯.LetM 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 ofR-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 anR-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 functorΓa(-) 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, thenth local cohomology module of M with respect to a is defined as the R-module HanM=ker⁡Γadn/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.
LetH∈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.
IfM,R/a∈S, then HaiM∈S for all i≥0.Proof.
LetF∙:⋯→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∈N⁡ExtRiR/an,M and so HaiM∈S for all i≥0.An immediate result of the above lemmas is the following result.Corollary 3.
Ifdim⁡R0=0 and M∈S, then R0=R/R+ is in S and HR+iM∈S for all i≥0.Lemma 4.
IfM∈S, then HR+iΓm0RM∈S as an R-module for all i≥0.Proof.
SinceΓm0RM 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 dim⁡R/m0nR0=dim⁡R0/m0n=0, Corollary 3 implies that HR+iΓm0RM∈S.Lemma 5.
Leti≥0 and R0/m0⊗R0HR+iM/Γm0RM∈S; then R0/m0⊗R0HR+iM∈S.Proof.
LetL≔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.
Letdim⁡R0≤1 and i≥0; then the following assertions hold: (1)
IfM∈S, then R0/m0⊗R0HR+i(M)∈S.(2)
Ifx∈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 dim⁡R0=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 dim⁡R0=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.
Sincex avoids all minimal primes of R0, dim⁡R0/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.
Asx∈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)coker⁡fxi-1≅HxR1HR+iM≅Hm0R1HR+iM,ker⁡fxi≅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.
Letdim⁡R0≤1 and i≥0. Moreover, let q0⊆R0 be an m0-primary ideal. Then (1)
theR-module R0/m0/q0HR+iM is in S;(2)
theR-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 Theorem6 to the case dim⁡R0=2.Theorem 8.
Ifdim⁡R0≤2, then Hm01HR+iM∈S for all i≥0.Proof.
Ifdim⁡R0≤1, this statement is the same as Theorem 6. Let dim⁡R0=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.
SinceHmi+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.
BecauseHmi+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.
LetM∈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.
SinceHm0R1(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.
Lemma4 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 thatM 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.
Ifi<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⟶ker⁡f⟶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.
Leta be a graded ideal of R and M∈S. Define taS(M)=sup⁡i∣Hai(M)∉S.Lemma 12.
Letx∈m be a homogeneous nonzero divisor of M. Then (30)tR+SMxM≤tR+SM.Proof.
Letdeg⁡x=d. Since x is a nonzero divisor of M, there is an exact sequence (31)0⟶M-d⟶xM⟶MxM⟶0.
Application of the functorHR+i- to the above exact sequence induces the exact sequence (32)HR+iM⟶HR+iMxM⟶xHR+i+1M-d⟶HR+i+1M.
IftR+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.
Letn=tR+S(M); then HR+i(M)/m0HR+i(M)∈S for all i≥n.Proof.
We proceed by induction ond=dim⁡R0. 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´.
SinceR0´ 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 Lemma4, 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 functorHR+n(-) to the above obtained exact sequence induces the exact sequence (37)HR+nM⟶x0HR+nM⟶HR+nMx0M⟶kHR+n+1M.
Letker⁡k=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 ringR0¯⊗R0R is R0¯,m0¯ and dim⁡R0¯=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.



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