#### Abstract

Let be a standard homogeneous Noetherian ring with local base ring and let be a finitely generated graded -module. Let be the th local cohomology module of with respect to . Let be a Serre subcategory of the category of -modules and let be a nonnegative integer. In this paper, if then we investigate some conditions under which the -modules and are in for all . Also, we prove that if , then the graded -module is in for all . Finally, we prove that if is the biggest integer such that , then for all .

#### 1. Introduction

Let be a Noetherian homogeneous ring with local base ring . So, is a Noetherian ring and there are finitely many elements such that . Let denote the irrelevant ideal of and let denote the graded maximal ideal of Finally, let be a finitely generated graded -module. Recall that a class of -modules is a Serre subcategory of the category of -modules when it is closed under taking submodules, quotients, and extensions. Always, stands for a Serre subcategory of the category of -modules. Hence, if is an exact sequence of the category of -modules and -homomorphisms such that both end terms belong to , then also belongs to . Note that the following subcategories are examples of Serre subcategory of the category of -modules: finite -modules; coatomic -modules [1]; minimax -modules [2]; -cofinite -modules; weakly Laskerian -modules; Matlis reflexive -modules; and trivially the zero -modules. Recently, some results have been proved concerning the local cohomology modules of a module 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 To give more details, let be an ideal of , let be an -module, and let The -module is said to satisfy condition on whenever the following condition holds:

Then is said to satisfy condition, whenever, every -module satisfies condition on . For example, the class of zero modules and Artinian -modules satisfy the condition .

In this paper, according to , we investigate some conditions under which the -modules and are in for all . Next, we prove that if , then is in for all Also if satisfies condition, then under some conditions, -modules and belong to . Finally, we prove that if is the biggest integer such that , then for all .

In Section 2, we give some notations and definition that we used in Section 3.

#### 2. Preliminaries

Let be a ring. The spectrum of is denoted by Spec , that is, the set of prime ideals of with the Zariski topology, which is the topology where the closed sets are for ideals .

The height of a prime ideal , denoted by height , is defined by the supremum of integers such that there exists a chain of prime ideals , where . The Krull dimension of is defined as .

Consider the complex

The differential on is denoted by , or sometimes simply . We say that is bounded below if for and bounded above if for ; if both conditions hold, then is bounded. On occasion, we consider a complex as a graded module with a graded endomorphism of degree 1.

Let be an -module. An injective resolution of is a complex of injective -modules , equipped with a quasi-isomorphism of complexes ; here one views as a complex concentrated in degree 0. Said otherwise, a complex of injective modules is a resolution of if there is an exact sequence .

Let and be -modules. Then are the right derived functors of , and are the left derived functors of .

For an exact sequence of -modules and each -module , one has exact sequences:

For an -module , the -torsion of is defined by . Observe that is a submodule of and is a left exact -torsion functor of -modules and -homomorphisms. Let and let be an -module. Choose an injective resolution of , so that we have an exact sequence

Then, apply the functor to the resolving cocomplex in order to obtain the cocomplex

Then, the th local cohomology module of with respect to is defined as the -module . If is an exact sequence of -modules and -homomorphisms, then we have the following long exact sequence:

As a general reference to homological and commutative algebra, we refer to [8].

#### 3. Main Results

In this paper, is a Noetherian homogeneous ring with local base ring . We start off this section with some technical lemmas to rich our main aims in this work.

Lemma 1. *Let be an -module and let be a finitely generated -module. Then for any the -module .*

*Proof. *Let be a minimal free resolution of . If for some integer , then is an -subquotient of . Thus .

Lemma 2. *If , then for all .*

*Proof. *Let be a finite free resolution of . If for some integer , then is a subquotient of . Since is a Serre class, . From definition, we have and so for all .

An immediate result of the above lemmas is the following result.

Corollary 3. *If and , then is in and for all .*

Lemma 4. *If , then as an -module for all .*

*Proof. *Since is a finitely generated -torsion module, there exists an such that . Then we obtain the -module . If , then from the Graded Independence Theorem (cf. ([8], 13.1.6)) there is an isomorphism of graded -modules . Since , Corollary 3 implies that .

Lemma 5. *Let and ; then .*

*Proof. *Let . Then we have the exact rows:such that is an epimorphism and is a monomorphism. Since is a submodule of , Lemma 4 leads to Also, from Lemma 1, we get . The short exact sequence (10) deduces the following exact sequence: According to our assumption, and then . From the exact sequence (9), we have the exact sequence , and hence .

Theorem 6. *Let and ; then the following assertions hold: *(1)*If , then .*(2)*If , such that , and , then the -modules and are in .*

*Proof. * If , then from Corollary 3 we have and then from Lemma 1 we get . Let and . Then there exists an -regular element which avoids all minimal primes of . Now, the exact sequence induces an exact sequenceSince avoids all minimal primes of , . Then by the Independence Theorem, and then . The exact sequence (13) gives the exact sequence As , the multiplication map is zero. Then there is an isomorphism Therefore, .

Let such that . Therefore, for all , there is an exact sequence of graded -modules such that and are the natural homomorphisms (cf. [8], 13.1.12). Now from we obtain an exact sequence Therefore, and are in .

Corollary 7. *Let and . Moreover, let be an -primary ideal. Then *(1)*the -module is in ;*(2)*the -module is in .*

*Proof. * Since is an -primary ideal, there is some such that . Then it suffices to prove that . We proceed the assertion by induction on . For , use part of Theorem 6. Now suppose, inductively, that and the result has been proved for all values smaller than . We have the exact sequence and the natural isomorphism Now, use part of Theorem 6 and Lemma 1.

Since , part (2) of Theorem 6 implies that .

The following result is an extension of Theorem 6 to the case .

Theorem 8. *If , then for all .*

*Proof. *If , this statement is the same as Theorem 6. Let , then there is a system of parameters of such that . From the exact sequence which obtains an epimorphism , also, there is a monomorphism Since , . Application of the functor to the monomorphism yields a monomorphism . Also, there is an epimorphism Because , . The exact sequence implies that .

Proposition 9. *Let and satisfies condition on . If is a local ring of dimension one, then for all .*

*Proof. *Since , one can assume that . Then, there exists an element such that it is an -sequence. By the exact sequence and by application of the functor to this exact sequence, we have the following exact sequence Lemma 4 leads to and then . Furthermore, by the basic properties of local cohomology, we have . Since satisfies condition on , for all .

Proposition 10. *Suppose that satisfies condition on and is the least integer such that . Then for all .*

*Proof. *If , then and therefore . Let and from the isomorphism for all , one can suppose that Therefore, there exists an element such that it is an -sequence. In view of the exact sequence we have the exact sequence By the above sequence, we obtain the exact sequence and isomorphism . Since , and . Thus . Then , because is -torsion.

*Definition 11. *Let be a graded ideal of and . Define .

Lemma 12. *Let be a homogeneous nonzero divisor of Then *

*Proof. *Let . Since is a nonzero divisor of , there is an exact sequence Application of the functor to the above exact sequence induces the exact sequence If , then for all we have . This means that . Thus, .

*Remark 13. * Any local flat morphism of local Noetherian rings is faithfully flat. So, if is flat over and , then is faithfully flat over . Moreover, if is a faithfully flat local -algebra, then from ([9], Theorem ) we obtain that of the graded -module if and only if of the graded module over .

If is a faithfully flat local -algebra, then .

Theorem 14. *Let ; then for all .*

*Proof. *We proceed by induction on . If , then and then . Let . If , then the result follows Theorem 6. Assume inductively that the result has been proved for all values smaller than , so we prove it for . Let be an indeterminate and let and . Then by the flat base change property of local cohomology, we have Since is a faithfully flat local -algebra, in view of Remark 13 it is enough to show that . Then, we may replace and by and , respectively. Hence we assume that is infinite residue field. From application of the functor over the exact sequence we have the following exact sequence: From Lemma 4, we have that for all . Now, let , , and . It follows from Lemma 1 that and are in for all . If , then . This implies that . Since , we may assume that . If is a nonzero divisor of and a part of a system of parameters of , then we have the exact sequence Application of the functor to the above obtained exact sequence induces the exact sequence Let and . Since , the graded -module is in and therefore from Lemma 1 we get if and only if . On the other hand, application of the functor induces the exact sequence such that Since , the map is zero and . Now, similar to the above arguments, one can conclude that if and only if . Let and . Using the Independence Theorem for graded local cohomology, we have .

Note that the local base ring of the graded ring is and . By the following isomorphism together with Lemma 12 and using induction hypothesis, this module is in .

#### Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.