A note on Abelian categories of cofinite modules

TL;DR

This paper proves that under certain conditions, the category of I-cofinite modules over a Noetherian ring forms an Abelian subcategory, answering longstanding questions in the theory of cofinite modules.

Contribution

It establishes that the category of I-cofinite modules is Abelian when q(I,R) ≤ 1, resolving questions posed by Hartshorne and the author.

Findings

Confirmed the Abelian property of I-cofinite modules under specific conditions.

Extended understanding of the structure of cofinite modules in commutative algebra.

Provided affirmative answers to open questions in the theory of cofinite modules.

Abstract

Let $R$ be a commutative Noetherian ring and $I$ be an ideal of $R$. In this article we answer affirmatively a question raised by the present author in \cite{B2}. Also, as an immediate consequence of this result it is shown that the category of all $I$-cofinite $R$-modules $\mathscr{C}(R, I)_{cof}$ is an Abelian subcategory of the category of all $R$-modules, whenever $q(I,R)\leq 1$. These assertions answer affirmatively a question raised by R. Hartshorne in [{\it Affine duality and cofiniteness}, Invent. Math. {\bf9}(1970), 145-164], in some special cases.

Equations

$$H_I^i(M)=\underset{n\ge 1}{\underrightarrow{\lim }}{\mathrm{Ext}}_R^i(R/I^n,M).$$

$$\mathrm{cd}(I,M)=\sup \{i\in {\mathbb{N}}_0:H_I^i(M)\ne 0\}$$

$$q(I,M)=\sup \{i\in {\mathbb{N}}_0:H_I^i(M)\mathrm{is}\mathrm{not}\mathrm{Artinian}\},$$

$$\mathfrak{A}(I,M):=\{p\in {\mathrm{mAss}}_{R}M:I+p=R\mathrm{or}p\supseteq I\},$$

$$\mathfrak{B}(I,M):=\{p\in {\mathrm{mAss}}_{R}M:\mathrm{cd}(I,R/p)=1\},$$

$$\mathfrak{C}(I,M):=\{p\in {\mathrm{mAss}}_{R}M:q(I,R/p)=1\}\mathrm{and}$$

$$\mathfrak{D}(I,M):=\{p\in {\mathrm{mAss}}_{R}M:0\le \dim R/(I+p)\le 1\}.$$

$${\mathrm{mAss}}_{R}R=\mathfrak{A}(I,R)\cup \mathfrak{B}(I,R)\cup \mathfrak{C}(I,R)\cup \mathfrak{D}(I,R).$$

$${\mathrm{mAss}}_{R}M=\mathfrak{A}(I,M)\cup \mathfrak{B}(I,M)\cup \mathfrak{C}(I,M)\cup \mathfrak{D}(I,M)$$

$${\mathrm{mAss}}_{R}R=\mathfrak{A}(I,R)\cup \mathfrak{B}(I,R)\cup \mathfrak{C}(I,R)\cup \mathfrak{D}(I,R).$$

$${\mathrm{mAss}}_{\hat{R}}\hat{R}=\mathfrak{A}(I\hat{R},\hat{R})\cup \mathfrak{C}(I\hat{R},\hat{R})\cup \mathfrak{D}(I\hat{R},\hat{R}).$$

$${\mathrm{mAss}}_{R}R=\mathfrak{A}(I,R)\cup \mathfrak{B}(I,R)\cup \mathfrak{C}(I,R)\cup \mathfrak{D}(I,R),$$

$${\mathrm{mAss}}_{R}R=\mathfrak{A}(I,R)\cup \mathfrak{C}(I,R)\cup \mathfrak{D}(I,R).$$

$${\mathrm{mAss}}_{\hat{R}}\hat{R}\subseteq \mathfrak{A}(I\hat{R},\hat{R})\cup \mathfrak{C}(I\hat{R},\hat{R})\cup \mathfrak{D}(I\hat{R},\hat{R}).$$

$$\operatorname{\mathfrak{p}}\in\big{(}\mathfrak{A}(I,R)\cup\mathfrak{C}(I,R)\cup\mathfrak{D}(I,R)\big{)}.$$

$$\hat{R}=I\hat{R}+p\hat{R}\subseteq I\hat{R}+\mathfrak{P}\subseteq \hat{R}$$

$$q(I\hat{R},\hat{R}/p\hat{R})=1.$$

$$q(I\hat{R},\hat{R}/\mathfrak{P})\le q(I\hat{R},\hat{R}/p\hat{R})=1.$$

$$\dim \hat{R}/(I\hat{R}+\mathfrak{P})\le \dim \hat{R}/(I\hat{R}+p\hat{R})=\dim R/(I+p)\le 1,$$

$${\mathrm{mAss}}_{R}R=\mathfrak{A}(I,R)\cup \mathfrak{B}(I,R)\cup \mathfrak{C}(I,R)\cup \mathfrak{D}(I,R).$$

$${\mathrm{mAss}}_{S^{-1}R}{S}^{-1}R=\mathfrak{A}(S^{-1}I,S^{-1}R)\cup \mathfrak{B}(S^{-1}I,S^{-1}R)\cup \mathfrak{C}(S^{-1}I,S^{-1}R)\cup \mathfrak{D}(S^{-1}I,S^{-1}R).$$

$${\mathrm{mAss}}_{R}R=\mathfrak{A}(I,R)\cup \mathfrak{B}(I,R)\cup \mathfrak{C}(I,R)\cup \mathfrak{D}(I,R).$$

$${\mathrm{mAss}}_{R}R=\mathfrak{A}(I,R)\cup \mathfrak{B}(I,R)\cup \mathfrak{C}(I,R)\cup \mathfrak{D}(I,R),$$

$${\mathrm{mAss}}_{\hat{R}}\hat{R}=\mathfrak{A}(I\hat{R},\hat{R})\cup \mathfrak{C}(I\hat{R},\hat{R})\cup \mathfrak{D}(I\hat{R},\hat{R}).$$

$$M{\otimes }_R\hat{R},N{\otimes }_R\hat{R}\in \mathcal{C}(\hat{R},I\hat{R}{)}_{cof}.$$

$${\mathrm{Ext}}_{\hat{R}}^j(\hat{R}/I\hat{R},K{\otimes }_R\hat{R})\simeq {\mathrm{Ext}}_R^j(R/I,K){\otimes }_R\hat{R}$$

$$0⟶\ker f⟶M⟶\mathrm{im}f⟶0,$$

$$0⟶\mathrm{im}f⟶N⟶\mathrm{coker}f⟶0.$$

Full text

2010 Mathematics Subject Classification: Primary 13D45; Secondary 14B15, 13E05.

This research of the author was supported by a grant from IPM (No. 96130018).

1. Introduction

Throughout this article, let $R$ denote a commutative Noetherian ring (with identity) and $I$ be an ideal of $R$. For an $R$-module $M$, the $i$th local cohomology module of $M$ with support in $V(I)$ is defined as:

[TABLE]

We refer the reader to [6] or [13] for more details about local cohomology.

For an $R$-module $M$, the notion $\operatorname{cd}(I,M)$, the cohomological dimension of $M$ with respect to $I$, is defined as:

[TABLE]

and the notion $q(I,M)$, which for first time was introduced by Hartshorne, is defined as:

[TABLE]

with the usual convention that the supremum of the empty set of integers is interpreted as $-\infty$. These two notions have been studied by several authors (see [3, 8, 10, 11, 12, 15, 16]).

Hartshorne in [14] defined an $R$-module $X$ to be $I$-cofinite, if $\operatorname{Supp}X\subseteq V(I)$ and ${\rm Ext}^{i}_{R}(R/I,X)$ is a finitely generated $R$-module for each integer $i\geq 0$. Then he posed the following question:

Question 1: Whether the category $\mathscr{C}(R,I)_{cof}$ of $I$-cofinite modules is an Abelian subcategory of the category of all $R$-modules? That is, if $f:M\longrightarrow N$ is an $R$-homomorphism of $I$-cofinite modules, are $\ker f$ and ${\rm coker}f$ $I$-cofinite?

With respect to the question (1), Hartshorne gave a counterexample to show that this question has not an affirmative answer in general, (see [14, Section 3]). On the positive side, Hartshorne proved that if $I$ is a prime ideal of dimension one in a complete regular local ring $R$, then the answer to his question is yes. Delfino and Marley extended this result to arbitrary complete local rings (see [7]). Kawasaki generalized the Delfino and Marley’s result for an arbitrary ideal $I$ of dimension one in a local ring $R$ (see [18]). Melkersson removed the local condition on the ring (see [20]). Finally, in [5] as a generalization of Melkersson’s result it is shown that for any ideal $I$ in any Noetherian ring $R$, the category $\mathscr{C}^{1}(R,I)_{cof}$ of all $I$-cofinite $R$-modules $M$ with $\dim M\leq 1$ is Abelian. For some other similar results, see also [4].

Recall that, for any proper ideal $I$ of $R$, the arithmetic rank of $I$, denoted by $\operatorname{ara}(I)$, is the least number of elements of $I$ required to generate an ideal which has the same radical as $I$.

Kawasaki proved that if $\operatorname{ara}(I)=1$ then the category $\mathscr{C}(R,I)_{cof}$ is Abelian (see [17]). Pirmohammadi et al. in [22] as a generalization of Kawasaki’s result proved that if $I$ is an ideal of a Noetherian local ring with $\operatorname{cd}(I,R)\leq 1$, then $\mathscr{C}(R,I)_{cof}$ is Abelian. Recently, Divaani-Aazar et al. in [9] have removed the local condition on the ring. Finally, the present author in [3] proved that if $I$ is an ideal of a Noetherian complete local ring $R$ with $q(I,R)\leq 1$, then $\mathscr{C}(R,I)_{cof}$ is Abelian.

We recall that the present author in [1], for any ideal $I$ of $R$ and any finitely generated $R$-module $M$, defined:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

He proved that if $\operatorname{mAss}_{R}R=\mathfrak{A}(I,R)\cup\mathfrak{B}(I,R)\cup\mathfrak{D}(I,R)$ then $\mathscr{C}(R,I)_{cof}$ is Abelian. Also, he asked the following question (see [1, Question D]):

Question 2: Let $R$ be a Noetherian ring and $I$ be an ideal of $R$ such that

[TABLE]

Whether $\mathscr{C}(R,I)_{cof}$ is Abelian?

In this article, we present an affirmative answer to Question 2. More precisely, we prove the following theorem:

Theorem 1. Let $R$ be a Noetherian ring and $I$ be an ideal of $R$ such that

[TABLE]

*Then $\mathscr{C}(R,I)_{cof}$ is Abelian.

Also, as an immediate consequence of Theorem 1, we deduce the following generalization of [3, Theorem 5.3] and [9, Theorem 2.2].

**Corollary. ** *Let $R$ be a Noetherian ring and $I$ be an ideal of $R$ with $q(I,R)\leq 1$. Then $\mathscr{C}(R,I)_{cof}$ is Abelian.

Throughout this paper, for any ideal $I$ of a Noetherian ring $R$, we denote the category of all $I$-cofinite $R$-modules by $\mathscr{C}(R,I)_{cof}$. Also, for each $R$-module $L$, we denote by $\operatorname{mAss}_{R}L$, the set of minimal elements of $\operatorname{Ass}_{R}L$ with respect to inclusion. For any ideal $\mathfrak{a}$ of $R$, we denote $\{\mathfrak{p}\in{\rm Spec}\,R:\,\mathfrak{p}\supseteq\mathfrak{a}\}$ by $V(\mathfrak{a})$. For any Noetherian local ring $(R,\operatorname{\mathfrak{m}})$, we denote by $\widehat{R}$ the $\operatorname{\mathfrak{m}}$-adic completion of $R$. Finally, we denote by $\operatorname{Max}(R)$ the set of all maximal ideals of $R$. For any unexplained notation and terminology we refer the reader to [6] and [19].

2. Results

The main purpose of this section is to prove Theorem 2.9, which presents an affirmative answer to a question raised by the present author in [1]. The following auxiliary lemmas are quite useful in the proof of Theorem 3.9.

Lemma 2.1**.**

$($See* [1, Corollary 3.6]$)$

Let $(R,\operatorname{\mathfrak{m}})$ be a Noetherian complete local ring and $I$ be an ideal of $R$ such that*

[TABLE]

Then $\mathscr{C}(R,I)_{cof}$ is Abelian.

∎

Lemma 2.2**.**

Let $(R,\mathfrak{m})$ be a Noetherian local ring and $T$ be an $R$-module such that the $\widehat{R}$-module $T\otimes_{R}\widehat{R}$ is finitely generated. Then the $R$-module $T$ is finitely generated.

Proof.

The assertion easily follows from the fact that $\widehat{R}$ is a faithfully flat $R$-algebra.∎

Lemma 2.3**.**

$($See* [1, Lemma 3.1]$)$

Let $R$ be a Noetherian ring, $I$ be an ideal of $R$ and $M$ be a finitely generated $R$-module. Then, \mathfrak{B}(I,M)\subseteq\big{(}\mathfrak{C}(I,M)\cup\mathfrak{D}(I,M)\big{)}. In particular,*

[TABLE]

if and only if $\operatorname{mAss}_{R}M=\mathfrak{A}(I,M)\cup\mathfrak{C}(I,M)\cup\mathfrak{D}(I,M).$

∎

Lemma 2.4**.**

Let $(R,\operatorname{\mathfrak{m}})$ be a Noetherian local ring and $I$ be an ideal of $R$ such that

[TABLE]

Then,

[TABLE]

Proof.

Since, by the hypothesis we have

[TABLE]

it follows from Lemma 2.3 that

[TABLE]

Now, in order to prove the assertion, it is enough to prove that

[TABLE]

Assume that $\mathfrak{P}\in\operatorname{mAss}_{\widehat{R}}\widehat{R}$ and set $\operatorname{\mathfrak{q}}:=\mathfrak{P}\cap R$. Then, there exists $\operatorname{\mathfrak{p}}\in\operatorname{mAss}_{R}R$ such that $\operatorname{\mathfrak{p}}\subseteq\operatorname{\mathfrak{q}}$. From the hypothesis it follows that

[TABLE]

We consider the following three cases:

Case 1. Assume that $\operatorname{\mathfrak{p}}\in\mathfrak{A}(I,R)$. If $I+\operatorname{\mathfrak{p}}=R$ then,

[TABLE]

and hence $I\widehat{R}+\mathfrak{P}=\widehat{R}$. Therefore, $\mathfrak{P}\in\mathfrak{A}(I\widehat{R},\widehat{R})$. Also, if $\operatorname{\mathfrak{p}}\supseteq I$ then $I\widehat{R}\subseteq\operatorname{\mathfrak{p}}\widehat{R}\subseteq\mathfrak{P}$ and hence $\mathfrak{P}\in\mathfrak{A}(I\widehat{R},\widehat{R})$.

Case 2. Assume that $\operatorname{\mathfrak{p}}\in\mathfrak{C}(I,R)$. Then, as $q(I,R/\operatorname{\mathfrak{p}})=1$ it follows that

[TABLE]

So, as $\operatorname{Supp}\widehat{R}/\mathfrak{P}\subseteq\operatorname{Supp}\widehat{R}/\operatorname{\mathfrak{p}}\widehat{R},$ it follows from [8, Theorem 3.2] that

[TABLE]

If $q(I\widehat{R},\widehat{R}/\mathfrak{P})=1$, then $\mathfrak{P}\in\mathfrak{C}(I\widehat{R},\widehat{R})$. Also, if $q(I\widehat{R},\widehat{R}/\mathfrak{P})=0$ then it is clear that $\mathfrak{P}\in V(I\widehat{R})$ and hence $\mathfrak{P}\in\mathfrak{A}(I\widehat{R},\widehat{R})$. Finally, if $q(I\widehat{R},\widehat{R}/\mathfrak{P})=-\infty$ then in view of [3, Lemma 4.1] we have $\mathfrak{P}\in\mathfrak{D}(I\widehat{R},\widehat{R})\cup\mathfrak{A}(I\widehat{R},\widehat{R})$.

Case 3. Assume that $\operatorname{\mathfrak{p}}\in\mathfrak{D}(I,R)$. Then, by the definition we have

[TABLE]

which implies that $\mathfrak{P}\in\mathfrak{D}(I\widehat{R},\widehat{R})\cup\mathfrak{A}(I\widehat{R},\widehat{R})$. ∎

Lemma 2.5**.**

Let $R$ be a Noetherian ring and $I$ be an ideal of $R$ such that

[TABLE]

Then for each multiplicative subset $S$ of $R$ we have

[TABLE]

Proof.

The proof is straightforward and is left to the reader.∎

Lemma 2.6**.**

Let $(R,\mathfrak{m})$ be a Noetherian local ring and $I$ be an ideal of $R$ such that

[TABLE]

Then $\mathscr{C}(R,I)_{cof}$ is Abelian.

Proof.

Let $M,\,N\in\mathscr{C}(R,I)_{cof}$ and let $f:M\longrightarrow N$ be an $R$-homomorphism. It is enough to prove that the $R$-modules ${\rm ker}f$ and ${\rm coker}f$ are $I$-cofinite. Set $K:={\rm ker}f$. Since,

[TABLE]

from Lemma 2.4 it follows that

[TABLE]

Moreover, from the hypothesis $M,\,N\in\mathscr{C}(R,I)_{cof}$ it follows that

[TABLE]

Since, $\widehat{R}$ is a flat $R$-algebra it follows that $K\otimes_{R}\widehat{R}={\rm ker}(f\otimes_{R}\widehat{R})$. Therefore, Lemma 2.1 yields that the $\widehat{R}$-module $K\otimes_{R}\widehat{R}$ is $I\widehat{R}$-cofinite. Hence, for each integer $j\geq 0$ the $\widehat{R}$-module

[TABLE]

is finitely generated. Now, Lemma 2.2 implies that for each integer $j\geq 0$ the $R$-module $\operatorname{Ext}^{j}_{R}(R/I,K)$ is finitely generated, which means that $K$ is $I$-cofinit. Now, the assertion follows from the exact sequences

[TABLE]

and

[TABLE]

∎

Lemma 2.7**.**

$($See* [1, Theorem 2.5]$)$

Let $R$ be a Noetherian ring and $I$ be an ideal of $R$ such that*

[TABLE]

Then $\mathscr{C}(R,I)_{cof}$ is Abelian.

∎

Lemma 2.8**.**

Let $R$ be a Noetherian ring and $J=(x_{1},\dots,x_{k})$ be an ideal of $R$. Let $\operatorname{\mathfrak{p}}_{1},\operatorname{\mathfrak{p}}_{2},...,\operatorname{\mathfrak{p}}_{n}$ be prime ideals of $R$ such that $J\not\subseteq\bigcup_{i=1}^{n}\operatorname{\mathfrak{p}}_{i}$. Then, there are elements $y_{1},\dots,y_{k}\in J$ such that $J=(y_{1},\dots,y_{k})$ and $\left(\bigcup_{i=1}^{n}\operatorname{\mathfrak{p}}_{i}\right)\bigcap\{y_{1},\dots,y_{k}\}=\emptyset$.

Proof.

Since, $J=(x_{1},\dots,x_{k})\nsubseteq\bigcup_{i=1}^{n}\operatorname{\mathfrak{p}}_{i}$ it follows that there is $a_{1}\in(x_{2},\dots,x_{k})$ such that

[TABLE]

Set $y_{1}:=x_{1}+a_{1}$. Then, $J=(y_{1},x_{2},\dots,x_{k})$. We shall construct the sequence $y_{1},\dots,y_{k}$ which are not belong to $\bigcup_{i=1}^{n}\operatorname{\mathfrak{p}}_{i}$ and $J=(y_{1},\dots,y_{k})$ by an inductive process. To do this end, assume that $1\leq i<k$, and that we have already constructed elements $y_{1},\dots,y_{i}$ such that $J=(y_{1},\dots,y_{i},x_{i+1},\dots,x_{k}).$ We show how to construct $y_{i+1}$.

To do this, as

[TABLE]

it follows that there is $a_{i+1}\in(y_{1},\dots,y_{i},x_{i+2},\dots,x_{k})$ such that

[TABLE]

Set $y_{i+1}:=x_{i+1}+a_{i+1}$. Then, $J=(y_{1},\dots,y_{i},y_{i+1},x_{i+2}\dots,x_{k}).$ This completes the inductive step in the construction.∎

Now, we are ready to state and prove the main result of this article, which answers affirmatively [1, Question D].

Theorem 2.9**.**

Let $R$ be a Noetherian ring and $I$ be an ideal of $R$ such that

[TABLE]

Then $\mathscr{C}(R,I)_{cof}$ is Abelian.

Proof.

Let $M,\,N\in\mathscr{C}(R,I)_{cof}$ and let $f:M\longrightarrow N$ be an $R$-homomorphism. By the proof of Lemma 2.6 it is enough to prove that the $R$-module ${\rm ker}f$ is $I$-cofinite. Set $K:={\rm ker}f$.

If $\operatorname{mAss}_{R}R=\mathfrak{A}(I,R)\cup\mathfrak{B}(I,R)\cup\mathfrak{D}(I,R),$ then the assertion holds by Lemma 2.7. So, we may assume that $\operatorname{mAss}_{R}R\neq\mathfrak{A}(I,R)\cup\mathfrak{B}(I,R)\cup\mathfrak{D}(I,R).$

Set \Phi:=\operatorname{mAss}_{R}R\backslash\big{(}\mathfrak{A}(I,R)\cup\mathfrak{B}(I,R)\cup\mathfrak{D}(I,R)\big{)}. Then, it is clear that $\Phi\subseteq\mathfrak{C}(I,R)$. Now, set $B:=\bigoplus_{\operatorname{\mathfrak{p}}\in\Phi}R/\operatorname{\mathfrak{p}}$ and assume that the ideal $I$ is generated by $n$ elements. Then, in view of [6, Theorem 3.3.1] we have ${\rm cd}(I,B)\leq n$. Since, the $R$-module $H^{i}_{I}(B)$ is Artinian for each integer $i\geq 2$ it follows that

[TABLE]

is a finite subset of $\operatorname{Max}(R)$. Assume that $W=\{\operatorname{\mathfrak{m}}_{1},\operatorname{\mathfrak{m}}_{2},...,\operatorname{\mathfrak{m}}_{t}\}.$ Then, using the Grothendieck’s Vanishing Theorem we can deduce that $0pt\operatorname{\mathfrak{m}}_{c}\geq 2$, for each $1\leq c\leq t$. Therefore, we have 0pt\big{(}\bigcap_{c=1}^{t}\operatorname{\mathfrak{m}}_{i}\big{)}\geq 2 and hence

[TABLE]

Assume that

[TABLE]

Then, by Lemma 2.8 there are elements $y_{1},\dots,y_{k}\in\bigcap_{c=1}^{t}\operatorname{\mathfrak{m}}_{c}$ such that

[TABLE]

In particular, $y_{\ell}$ is not a nilpotent element of $R$ and hence $S_{\ell}=\{1_{{}_{R}},y_{\ell},y_{\ell}^{2},y_{\ell}^{3},...\}$ is a multiplicative subset of $R$, for each $1\leq\ell\leq k$.

Now, for each integer $i\geq 0$, set $U_{i}:=\operatorname{Ext}^{i}_{R}(R/I,K)$. Then, for each $1\leq c\leq t$, since by the hypothesis we have

[TABLE]

from Lemma 2.5 it follows that

[TABLE]

Moreover, since $M,\,N\in\mathscr{C}(R,I)_{cof}$ and $f:M\longrightarrow N$ is an $R$-homomorphism it follows that, for each $1\leq c\leq t$, we have $M_{\operatorname{\mathfrak{m}}_{c}},\,N_{\operatorname{\mathfrak{m}}_{c}}\in\mathscr{C}(R_{\operatorname{\mathfrak{m}}_{c}},IR_{\operatorname{\mathfrak{m}}_{c}})_{cof}$ and $f_{\operatorname{\mathfrak{m}}_{c}}:M_{\operatorname{\mathfrak{m}}_{c}}\longrightarrow N_{\operatorname{\mathfrak{m}}_{c}}$ is an $R_{\operatorname{\mathfrak{m}}_{c}}$-homomorphism with the kernel $K_{\operatorname{\mathfrak{m}}_{c}}$. So, in view of Lemma 2.6 for each $1\leq c\leq t$ and each $i\geq 0$, the $R_{\operatorname{\mathfrak{m}}_{c}}$-module, $(U_{i})_{\operatorname{\mathfrak{m}}_{c}}\simeq\operatorname{Ext}^{i}_{R_{\operatorname{\mathfrak{m}}_{c}}}(R_{\operatorname{\mathfrak{m}}_{c}}/IR_{\operatorname{\mathfrak{m}}_{c}},K_{\operatorname{\mathfrak{m}}_{c}})$ is finitely generated.

On the other hand, for each $1\leq\ell\leq k$ and each $j\geq 2$, since

[TABLE]

it follows that

[TABLE]

which means that

[TABLE]

Furthermore, since $M,\,N\in\mathscr{C}(R,I)_{cof}$ and $f:M\longrightarrow N$ is an $R$-homomorphism it follows that $M_{y_{\ell}},\,N_{y_{\ell}}\in\mathscr{C}(R_{y_{\ell}},IR_{y_{\ell}})_{cof}$ and $(S_{\ell})^{-1}f:M_{y_{\ell}}\longrightarrow N_{y_{\ell}}$ is an $R_{y_{\ell}}$-homomorphism with the kernel $K_{y_{\ell}}$. Therefore, by Lemma 2.7, the $R_{y_{\ell}}$-module

[TABLE]

is finitely generated, for each $1\leq\ell\leq k$ and each $i\geq 0$. Thus, for each $1\leq\ell\leq k$ and each $i\geq 0$, there is a finitely generated submodule $G_{i,\ell}$ of the $R$-module $U_{i}$ such that $(U_{i})_{y_{\ell}}=(G_{i,\ell})_{y_{\ell}}$ and so,

[TABLE]

Now, for each $i\geq 0$, set $X_{i}:=G_{i,1}+G_{i,2}+\cdots+G_{i,k}$. Then, for each $i\geq 0$, the $R$-module $X_{i}$ is a finitely generated submodule of $U_{i}$ such that \big{(}U_{i}/X_{i}\big{)}_{y_{\ell}}=0, for each $1\leq\ell\leq k$. In particular, the $R$-module $U_{i}/X_{i}$ is $Ry_{\ell}$-torsion, for each $1\leq\ell\leq k$ and each $i\geq 0$. Consequently, the $R$-module $U_{i}/X_{i}$ is $(y_{1},\dots,y_{k})$-torsion, for each $i\geq 0$. So, we have

[TABLE]

Since, for each $1\leq c\leq t$ and each $i\geq 0$, the $R_{\operatorname{\mathfrak{m}}_{c}}$-module \big{(}U_{i}/X_{i}\big{)}_{\operatorname{\mathfrak{m}}_{c}} is finitely generated, it follows that $U_{i}/X_{i}$ is a finitely generated $R$-module and hence $U_{i}$ is finitely generated too. This means that $K$ is an $I$-cofinite $R$-module, as required. ∎

The following consequence of Theorem 2.9 is a generalization of [3, Theorem 5.3] and [9, Theorem 2.2].

Corollary 2.10**.**

Let $R$ be a Noetherian ring and $I$ be an ideal of $R$ with $q(I,R)\leq 1$. Then $\mathscr{C}(R,I)_{cof}$ is Abelian.

Proof.

Since, by the hypothesis we have $q(I,R)\leq 1$, from [8, Theorem 3.2] it follows that $q(I,R/\operatorname{\mathfrak{p}})\leq 1$, for each $\operatorname{\mathfrak{p}}\in\operatorname{mAss}_{R}R$. Now, let $\operatorname{\mathfrak{p}}\in\operatorname{mAss}_{R}R$. If $q(I,R/\operatorname{\mathfrak{p}})=1$, then $\operatorname{\mathfrak{p}}\in\mathfrak{C}(I,R)$. Also, if $q(I,R/\operatorname{\mathfrak{p}})=0$ then it is clear that $\operatorname{\mathfrak{p}}\in V(I)$ and hence $\operatorname{\mathfrak{p}}\in\mathfrak{A}(I,R)$. Also, if $q(I,R/\operatorname{\mathfrak{p}})=-\infty$ then in view of [3, Lemma 4.1] we have $\operatorname{\mathfrak{p}}\in\mathfrak{D}(I,R)\cup\mathfrak{A}(I,R)$. Therefore, we have

[TABLE]

and hence the assertion holds by Theorem 2.9.∎

Corollary 2.11**.**

Let $R$ be a Noetherian ring of dimension at most $2$ and $I$ be an arbitrary ideal of $R$. Then $\mathscr{C}(R,I)_{cof}$ is Abelian.

Proof.

Let $d:=\dim R$. Then, by [21, Proposition 5.1] the $R$-module $H^{d}_{I}(R)$ is Artinian and by Grothendieck’s Vanishing Theorem, for each $i>d$ we have $H^{i}_{I}(R)=0$. Since, by the hypothesis we have $d\leq 2$ it is clear that $q(I,R)\leq 1$ and hence the assertion follows from Corollary 2.10.∎

Corollary 2.12**.**

Let $R$ be a Noetherian ring and $I$ be an ideal of $R$ such that

[TABLE]

Let

[TABLE]

be a complex such that $X^{i}\in\mathscr{C}(R,I)_{cof}$ for all $i\in\mathbb{Z}$. Then for each $i\in\mathbb{Z}$ the $i$-th cohomology module $H^{i}(X^{\bullet})$ is in $\mathscr{C}(R,I)_{cof}$.

Proof.

The assertion follows from Theorem 2.9.∎

Corollary 2.13**.**

Let $R$ be a Noetherian ring and $I$ be an ideal of $R$ such that $q(I,R)\leq 1$. Let

[TABLE]

be a complex such that $X^{i}\in\mathscr{C}(R,I)_{cof}$ for all $i\in\mathbb{Z}$. Then for each $i\in\mathbb{Z}$ the $i$-th cohomology module $H^{i}(X^{\bullet})$ is in $\mathscr{C}(R,I)_{cof}$.

Proof.

Using Corollary 2.12, the assertion follows from the proof of Corollary 2.10.∎

Corollary 2.14**.**

Let $R$ be a Noetherian ring of dimension at most $2$ and $I$ be an ideal of $R$. Let

[TABLE]

be a complex such that $X^{i}\in\mathscr{C}(R,I)_{cof}$ for all $i\in\mathbb{Z}$. Then for each $i\in\mathbb{Z}$ the $i$-th cohomology module $H^{i}(X^{\bullet})$ is in $\mathscr{C}(R,I)_{cof}$.

Proof.

Using Corollary 2.13, the assertion follows from the proof of Corollary 2.11.∎

Corollary 2.15**.**

Let $R$ be a Noetherian ring and $I$ be an ideal of $R$ such that

[TABLE]

Let $M$ be an $I$-cofinite $R$-module and $N$ be a finitely generated $R$-module. Then the $R$-modules $\operatorname{Tor}_{i}^{R}(N,M)$ and $\operatorname{Ext}^{i}_{R}(N,M)$ are $I$-cofinite, for all integers $i\geq 0$.

Proof.

Since $N$ is finitely generated it follows that, $N$ has a free resolution with finitely generated free $R$-modules. Now the assertion follows using Corollary 2.12 and computing the $R$-modules $\operatorname{Tor}_{i}^{R}(N,M)$ and $\operatorname{Ext}^{i}_{R}(N,M)$, by this free resolution. ∎

Corollary 2.16**.**

Let $R$ be a Noetherian ring and $I$ be an ideal of $R$ such that $q(I,R)\leq 1$. Let $M$ be an $I$-cofinite $R$-module and $N$ be a finitely generated $R$-module. Then the $R$-modules $\operatorname{Tor}_{i}^{R}(N,M)$ and $\operatorname{Ext}^{i}_{R}(N,M)$ are $I$-cofinite, for all integers $i\geq 0$.

Proof.

Using Corollary 2.15, the assertion follows from the proof of Corollary 2.10.∎

Corollary 2.17**.**

Let $R$ be a Noetherian ring of dimension at most $2$ and $I$ be an ideal of $R$. Let $M$ be an $I$-cofinite $R$-module and $N$ be a finitely generated $R$-module. Then the $R$-modules $\operatorname{Tor}_{i}^{R}(N,M)$ and $\operatorname{Ext}^{i}_{R}(N,M)$ are $I$-cofinite, for all integers $i\geq 0$.

Proof.

Using Corollary 2.16, the assertion follows from the proof of Corollary 2.11.∎

Corollary 2.18**.**

Let $R$ be a Noetherian ring and $I$ be an ideal of $R$ such that

[TABLE]

Let $M,\,N$ be two finitely generated $R$-modules. Then the $R$-modules $\operatorname{Tor}_{i}^{R}(N,H^{j}_{I}(M))$ and $\operatorname{Ext}^{i}_{R}(N,H^{j}_{I}(M))$ are $I$-cofinite, for all integers $i\geq 0$ and $j\geq 0$.

Proof.

The assertion follows from [1, Theorem 3.8] and Corollary 2.15.∎

Corollary 2.19**.**

Let $R$ be a Noetherian ring and $I$ be an ideal of $R$ such that $q(I,R)\leq 1$. Let $M,\,N$ be two finitely generated $R$-modules. Then the $R$-modules $\operatorname{Tor}_{i}^{R}(N,H^{j}_{I}(M))$ and $\operatorname{Ext}^{i}_{R}(N,H^{j}_{I}(M))$ are $I$-cofinite, for all integers $i\geq 0$ and $j\geq 0$.

Proof.

The assertion follows from [1, Theorem 4.10] and Corollary 2.16.∎

Corollary 2.20**.**

Let $R$ be a Noetherian ring of dimension at most $2$ and $I$ be an ideal of $R$. Let $M,\,N$ be two finitely generated $R$-modules. Then the $R$-modules $\operatorname{Tor}_{i}^{R}(N,H^{j}_{I}(M))$ and $\operatorname{Ext}^{i}_{R}(N,H^{j}_{I}(M))$ are $I$-cofinite, for all integers $i\geq 0$ and $j\geq 0$.

Proof.

Using the proof of Corollary 2.11 we have $q(I,R)\leq 1$. So, the assertion follows from Corollary 2.19.∎

Acknowledgements

The author would like to thank to School of Mathematics, Institute for Research in Fundamental Sciences (IPM) for its financial support.