Cofiniteness of modules and local cohomology

Abstract

Let A be a commutative noetherian ring, let $\mathfrak a$ be an ideal of A and let n be a non-negative integer. In this paper, we study ${\mathcal {S}}_n(\mathfrak {a})$, a certain class of A-modules and we find some sufficient conditions so that a module belongs to ${\mathcal {S}}_n(\mathfrak {a})$. Moreover, we study the cofiniteness of local cohomology modules when $\dim A/\mathfrak {a}\ge 3$.

1 Introduction

Throughout this paper, assume that A is a commutative noetherian ring, $\mathfrak {a}$ is an ideal of A, M is an A-module and n is a non-negative integer. Following [5], the A-module M is said to be $\mathfrak {a}$-cofinite if ${{\,\textrm{Supp}\,}}_A(M)\subseteq V(\mathfrak {a})$ and ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a}, M)$ are finitely generated for all integers $i\ge 0$. The authors [7] studied a criterion for cofinitenss of modules. We denote by ${\mathcal {S}}_n(\mathfrak {a})$ the class of all A-modules M satisfying the following implication:

If ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a}, M)$ is finitely generated for all $i\le n$ and ${{\,\textrm{Supp}\,}}(M)\subseteq V(\mathfrak {a})$, then M is $\mathfrak {a}$-cofinite.

The abelianess of the category of $\mathfrak {a}$-cofinite modules is of interest to many mathematicians working in commutative algebra. This subject has been studied for small dimensions by various authors [1, 6, 8, 9]. As the Koszul complexes are effective tools for going down the dimensions, Sazeedeh [11] studied the cofiniteness of Koszul cohomology modules.

In Sect. 2, by means of the Koszul cohomologies of a module, we want to find out when this module belongs to ${\mathcal {S}}_n(\mathfrak {a})$. Let $\textbf{x}=x_1,\dots ,x_t$ be a sequence of elements of $\mathfrak {a}$. We show that if ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a},M)$ is finitely generated for all $i\le n+1$ and $H^i(\textbf{x},M)\in {\mathcal {S}}_1(\mathfrak {a})$ for all $i\le n$, then $H^i(\textbf{x},M)$ is $\mathfrak {a}$-cofinite for all $i\le n$ (see Proposition 2.2). Moreover, we prove the following theorem.

Theorem 1.1

If $H^i(\textbf{x},M)\in {\mathcal {S}}_1(\mathfrak {a})$ for all $i\ge 0$, then $M\in {\mathcal {S}}_{t+1}(\mathfrak {a})$.

As an application of this theorem, if $\dim H^i(\textbf{x},M)\le 1$ for all $i\ge 0$, then $M\in {\mathcal {S}}_{t+1}(\mathfrak {a})$ (see Corollary 2.4). Let ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a},M)$ be finitely generated for all $i\le n+2$ and let $H^i(\textbf{x},M)\in {\mathcal {S}}_2(\mathfrak {a})$ for all $i\le n$. Then we show that $H^i(\textbf{x},M)$ is $\mathfrak {a}$-cofinite for all $i\le n$ if and only if ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H^i(\textbf{x},M))$ is finitely generated for all $i\le n+1$ (see Proposition 2.5). Moreover, we have the following theorem.

Theorem 1.2

Let $H^i(\textbf{x},M)\in {\mathcal {S}}_2(\mathfrak {a})$ such that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H^i(\textbf{x},M))$ is finitely generated for all $i\ge 0$. Then $M\in {\mathcal {S}}_{t+2}(\mathfrak {a})$.

Let ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a},M)$ be finitely generated for all $i\le n+3$ and let $H^i(\textbf{x},M)\in {\mathcal {S}}_3(\mathfrak {a})$ for all $i\le n$. Then we prove that if $H^i(\textbf{x},M)$ is $\mathfrak {a}$-cofinite for all $i\le n$, then ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},H^i(\textbf{x},M))$ is finitely generated for all $i\le n+1$; $j=0,1$; furthermore if ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H^{n+2}(\textbf{x},M))$ is finitely generated, then the converse also holds (see Proposition 2.8). Moreover, we have the following theorem.

Theorem 1.3

Let $H^i(\textbf{x},M)\in {\mathcal {S}}_3(\mathfrak {a})$ such that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H^i(\textbf{x},M))$ and ${{\,\textrm{Ext}\,}}_A^1(A/\mathfrak {a},H^i(\textbf{x},M))$ are finitely generated for all $i\ge 0$. Then $M\in {\mathcal {S}}_{t+3}(\mathfrak {a})$.

As an application of this theorem, let A be a local ring such that $\dim A/((\textbf{x})+xA)\le 3$ for some $x\in \mathfrak {a}$. If ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H^i(\textbf{x},M)/xH^i(\textbf{x},M))$ and ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},H^i(\textbf{x},M))$ for all $i\ge 0$ and $j=0,1$ are finitely generated, then $M\in {\mathcal {S}}_{t+3}(\mathfrak {a})$ (see Corollary 2.10).

In Sect. 3, we study the cofiniteness of local cohomology modules. Let ${{\,\textrm{Ext}\,}}^i_A(A/\mathfrak {a},M)$ be finitely generated for all $i\ge 0$ and let $t<s$ be non-negative integers such that $H_{\mathfrak {a}}^i(M)$ is $\mathfrak {a}$-cofinite for all $i\ne t,s$. Then we show that $H_{\mathfrak {a}}^{t}(M)\in {\mathcal {S}}_{n+s-t+1}(\mathfrak {a})$ if and only if $H_{\mathfrak {a}}^s(M)\in {\mathcal {S}}_{n}(\mathfrak {a})$ (see Theorem 3.1). Moreover, we have the following theorem.

Theorem 1.4

Let $\dim A/\mathfrak {a}=d\ge 3$ and let ${{\,\textrm{depth}\,}}({{\,\textrm{Ann}\,}}(M),A/\mathfrak {a})\ge d-2$. If ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a},M)$ is finitely generated for all $i\le n+1$, then ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^i(M))$ is finitely generated for all $i\le n$ if and only if $H_{\mathfrak {a}}^i(M)$ is $\mathfrak {a}$-cofinite for all $i<n.$

In the end of this paper we get a similar result for those rings A which $\dim A\ge 4$.

2 A criterion for cofiniteness of modules

Throughout this section, M is an A-module with ${{\,\textrm{Supp}\,}}_AM\subseteq V(\mathfrak {a})$ and n is a non-negative integer and $\textbf{x}=x_1,\dots , x_t$ is a sequence of elements of $\mathfrak {a}$.

An A-module M is said to be $\mathfrak {a}$-cofinite if ${{\,\textrm{Supp}\,}}_A(M)\subseteq V(\mathfrak {a})$ and ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a}, M)$ are finitely generated for all integers $i\ge 0$.

For a non-negative integer n, we denote by ${\mathcal {S}}_n(\mathfrak {a})$, the class of all A-modules M satisfying the following implication:

If ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a}, M)$ is finitely generated for all $i\le n$ and ${{\,\textrm{Supp}\,}}(M)\subseteq V(\mathfrak {a})$, then M is $\mathfrak {a}$-cofinite.

Examples 2.1

• (i) Assume that $\mathfrak {a}$ is an arbitrary ideal of A and M is an A-module of dimension d where $\dim M$ means the dimension of ${{\,\textrm{Supp}\,}}_A M$; which is the length of the longest chain of prime ideals in ${{\,\textrm{Supp}\,}}_AM$. Then $H_{\mathfrak {a}}^d(M)$ is in ${\mathcal {S}}_0(\mathfrak {a})$. To be more precise, if ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^d(M))$ is a finitely generated A-module, then it follows from [10, Theorem 3.11] that $H_{\mathfrak {a}}^d(M)$ is artinian and so, since ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^d(M))$ has finite length, according to [9, Proposition 4.1], the module $H_{\mathfrak {a}}^d(M)$ is $\mathfrak {a}$-cofinite.

• (ii) Given an arbitrary ideal $\mathfrak {a}$ of A, by virtue of [1, Proposition 2.6], $M\in {\mathcal {S}}_1(\mathfrak {a})$ for all modules M with $\dim M\le 1$. Especially, if $\dim A/\mathfrak {a}=1$, then it follows from [9, Theorem 2.3] that ${\mathcal {S}}_1(\mathfrak {a})=\textrm{Mod-}A$, the category of all A-modules. Furthermore, if $\dim A=2$, then it follows from [10, Corollary 2.4] that ${\mathcal {S}}_1(\mathfrak {a})=\textrm{Mod-}A$ for any ideal $\mathfrak {a}$ of A.

• (iii) Let $\mathfrak {a}$ be an ideal of a local ring A with $\dim A/\mathfrak {a}=2$. It follows from [2, Theorem 3.5] that ${\mathcal {S}}_2(\mathfrak {a})=\textrm{Mod-}A$. Furthermore, if A is a local ring with $\dim A=3$, then it follows from [10, Corollary 2.5] that ${\mathcal {S}}_2(\mathfrak {a})=\textrm{Mod-}A$ for any ideal $\mathfrak {a}$ of A.

The following result shows that the Koszul cohomology modules of M are $\mathfrak {a}$-cofinite when they belong to ${\mathcal {S}}_1(\mathfrak {a})$.

Proposition 2.2

Let ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a},M)$ be finitely generated for all $i\le n+1$. If $H^i(\textbf{x},M)\in {\mathcal {S}}_1(\mathfrak {a})$ for all $i\le n$, then $H^i(\textbf{x},M)$ is $\mathfrak {a}$-cofinite for all $i\le n$.

Proof

Consider the Koszul complex

$$\begin{aligned} K^*(\textbf{x},M):0\longrightarrow K^0{\mathop {\longrightarrow }\limits ^{d^0}}K^1{\mathop {\longrightarrow }\limits ^{d^1}}\dots {\mathop {\longrightarrow }\limits ^{d^{t-1}}}K^t\longrightarrow 0 \end{aligned}$$

and assume that $Z^i={{\,\textrm{Ker}\,}}d^i$, $B^i={{\,\textrm{Im}\,}}d^{i-1}$, $C^i={{\,\textrm{Coker}\,}}d^i$ and $H^i=H^i(\textbf{x},M)$ for each i. Then for each $j\ge 0$, we have an exact sequence of modules

$$\begin{aligned} 0\longrightarrow H^j\longrightarrow C^j\longrightarrow K^{j+1}\longrightarrow C^{j+1}\longrightarrow o\hspace{1cm}(\dag _j). \end{aligned}$$

We prove by induction on i that $H^i$ is $\mathfrak {a}$-cofinite and ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},C^{i+1})$ is finitely generated for all $i\le n$ and all $j\le n-i$. Assume that $i<n$ and so the induction hypothesis implies that ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},C^i)$ is finitely generated for all $j\le n+1-i$. In view of $(\dag _i)$, it is clear that ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},H^i)$ is finitely generated for $i=0,1$ and so the fact that $H^i\in {\mathcal {S}}_1(\mathfrak {a})$ forces $H^i$ is $\mathfrak {a}$-cofinite and ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},C^{i+1})$ is finitely generated for all $j\le n-i$. $\square$

The following theorem provides a sufficient condition so that a module belongs to ${\mathcal {S}}_{t+1}(\mathfrak {a})$.

Theorem 2.3

If $H^i(\textbf{x},M)\in {\mathcal {S}}_1(\mathfrak {a})$ for all $i\ge 0$, then $M\in {\mathcal {S}}_{t+1}(\mathfrak {a})$.

Proof

Assume that ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a},M)$ is finitely generated for all $i\le t+1$. Then, it follows from Proposition 2.2 that $H^i(\textbf{x},M)$ is $\mathfrak {a}$-cofinite for all $i\ge 0$. Consequently, [11, Theorem 2.4] implies that M is $\mathfrak {a}$-cofinite. $\square$

Corollary 2.4

If $\dim H^i(\textbf{x},M)\le 1$ for all $i\ge 0$, then $M\in {\mathcal {S}}_{t+1}(\mathfrak {a})$.

Proof

By virtue of Examples 2.1, $H^i(\textbf{x},M)\in {\mathcal {S}}_1(\mathfrak {a})$ for all $i\ge 0$ and so the result follows from Theorem 2.3. $\square$

When the Koszul cohomology modules of an A-module belong to ${\mathcal {S}}_2(\mathfrak {a})$, we have the following result about the $\mathfrak {a}$-cofiniteness of these modules.

Proposition 2.5

Let ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a}, M)$ be finitely generated for all $i\le n+2$ and let $H^i(\textbf{x},M)\in {\mathcal {S}}_2(\mathfrak {a})$ for all $i\le n$. Then the following conditions hold.

• (i) $H^i(\textbf{x},M)$ is $\mathfrak {a}$-cofinite for all $i\le n$.

• (ii) ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H^i(\textbf{x},M))$ is finitely generated for all $i\le n+1$.

Proof

Consider the same notation as in the proof of Proposition 2.2. (i)$\Rightarrow$(ii). By the assumption ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a}, H^i)$ is finitely generated for all $i\le n$. Thus, it is straightforward to see that ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a}, C^i)$ is finitely generated for all $i\le n+1$ and all $j\le n+2-i$. Since ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},C^{n+1})$ is finitely generated, $(\dag _{n+1})$ implies that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H^{n+1})$ is finitely generated. (ii)$\Rightarrow$(i). We prove by induction on i that $H^i$ is $\mathfrak {a}$-cofinite and ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},C^{i+1})$ is finitely generated for all $i\le n$ and all $j\le n+1-i$. The induction hypothesis implies that ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},C^{i})$ is finitely generated for all $j\le n+2-i$. Since by the assumption ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H^{i+1})$ is finitely generated, $(\dag _{i+1})$ implies that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},C^{i+1})$ is finitely generated; and hence $(\dag _i)$ implies that ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},H^i)$ is finitely generated for all $j\le 2$. Now, since $H^i\in {\mathcal {S}}_2(\mathfrak {a})$, it is $\mathfrak {a}$-cofinite; furthermore $(\dag _i)$ implies that ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},C^{i+1})$ is finitely generated for all $i\le n$ and all $j\le n+1-i$. $\square$

The following theorem provides a sufficient condition so that a module belongs to ${\mathcal {S}}_{t+2}(\mathfrak {a})$.

Theorem 2.6

Let $H^i(\textbf{x},M)\in {\mathcal {S}}_2(\mathfrak {a})$ such that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H^i(\textbf{x},M))$ is finitely generated for all $i\ge 0$. Then $M\in {\mathcal {S}}_{t+2}(\mathfrak {a})$.

Proof

Assume that ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a},M)$ is finitely generated for all $i\le t+2$. Then it follows from Proposition 2.5 that $H^i(\textbf{x},M)$ is $\mathfrak {a}$-cofinite for all $i\ge 0$. Consequently, [11, Theorem 2.4] implies that M is $\mathfrak {a}$-cofinite. $\square$

Corollary 2.7

Let A be a local ring such that $\dim A/(\textbf{x})\le 3$ and let ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H^i(\textbf{x},M))$ be finitely generated for all $i\ge 0$. Then $M\in {\mathcal {S}}_{t+2}(\mathfrak {a})$.

Proof

Put $B=A/(\textbf{x})$ and $\mathfrak {b}=\mathfrak {a}B$. We observe that $H^i(\textbf{x},M)$ is a B-module for each $i\ge 0$ and since $\dim B\le 3$, it follows from Examples 2.1 (iii) that $H^i(\textbf{x},M)\in {\mathcal {S}}_2(\mathfrak {a}B)$ for each $i\ge 0$. Now [7, Proposition 2.15] implies that $H^i(\textbf{x},M)\in {\mathcal {S}}_2(\mathfrak {a})$; and consequently, it follows from Theorem 2.6 that $M\in {\mathcal {S}}_{t+2}(\mathfrak {a})$. $\square$

When the Koszul cohomology modules of an A-module belong to ${\mathcal {S}}_3(\mathfrak {a})$, we have the following result about their $\mathfrak {a}$-cofiniteness

Proposition 2.8

Let ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a}, M)$ be finitely generated for all $i\le n+3$ and let $H^i(\textbf{x},M)\in {\mathcal {S}}_3(\mathfrak {a})$ for all $i\le n$. Consider the following statements.

• (i) $H^i(\textbf{x},M)$ is $\mathfrak {a}$-cofinite for all $i\le n$.

• (ii) ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},H^i(\textbf{x},M))$ is finitely generated for $j=0, 1$ and all $i\le n+1.$

Then $\mathrm{(i)}\Rightarrow \mathrm{(ii)}$ holds. Moreover, if ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H^{n+2}(\textbf{x},M))$ is finitely generated, then $\mathrm{(ii)}\Rightarrow \mathrm{(i)}$ holds.

Proof

Consider the same notation as in the proof of Proposition 2.2. (i)$\Rightarrow$(ii). Clearly ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},H^i)$ is finitely generated for all $i\le n$ and $j=0,1$; furthermore ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a}, C^i)$ is finitely generated for all $i\le n+1$ and all $j\le n+3-i$. Since ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},C^{n+1})$ is finitely generated for $j=0,1$, the exact sequence $(\dag _{n+1})$ implies that ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},H^{n+1})$ is finitely generated for $j=0,1$. (ii)$\Rightarrow$(i). We prove by induction on i that $H^i$ is $\mathfrak {a}$-cofinite and ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},C^{i+1})$ is finitely generated for all $i\le n$ and all $j\le n+2-i$. The exact sequence $(\dag _{i+2})$ implies that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},C^{i+2})$ is finitely generated and so it follows from $(\dag _{i+1})$ that ${{\,\textrm{Ext}\,}}^j(A/\mathfrak {a}, C^{i+1})$ is finitely generated for $j=0,1$. Since by the induction hypothesis, ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},C^i)$ is finitely generated for all $j\le n+3-i$, the exact sequence $(\dag _i)$ implies that ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},H^i)$ is finitely generated for all $j\le n+3-i$; and hence since $H^i\in {\mathcal {S}}_3(\mathfrak {a})$, we deduce that $H^i$ is $\mathfrak {a}$-cofinite. $\square$

The following theorem provides a sufficient condition so that a module belongs to ${\mathcal {S}}_{t+3}(\mathfrak {a})$.

Theorem 2.9

Let $H^i(\textbf{x},M)\in {\mathcal {S}}_3(\mathfrak {a})$ such that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H^i(\textbf{x},M))$ and ${{\,\textrm{Ext}\,}}_A^1(A/\mathfrak {a},H^i(\textbf{x},M))$ are finitely generated for all $i\ge 0$. Then $M\in {\mathcal {S}}_{t+3}(\mathfrak {a})$.

Proof

Assume that ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a},M)$ is finitely generated for all $i\le t+3$. Then it follows from Proposition 2.8 that $H^i(\textbf{x},M)$ is $\mathfrak {a}$-cofinite for all $i\ge 0$. Consequently, [11, Theorem 2.4] implies that M is $\mathfrak {a}$-cofinite. $\square$

Corollary 2.10

Let A be a local ring such that $\dim A/((\textbf{x})+xA)\le 3$ for some $x\in \mathfrak {a}$. If ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H^i(\textbf{x},M)/xH^i(\textbf{x},M))$ and ${{\,\textrm{Ext}\,}}_A^j(A/\mathfrak {a},H^i(\textbf{x},M))$ for all $i\ge 0$ and $j=0,1$ are finitely generated, then $M\in {\mathcal {S}}_{t+3}(\mathfrak {a})$.

Proof

Set $B=A/xA$, $\mathfrak {b}=\mathfrak {a}/xA$. In view of the exact sequence

$$\begin{aligned} H^i(\textbf{x},x,M)\longrightarrow H^i(\textbf{x},M){\mathop {\longrightarrow }\limits ^{x.}}H^i(\textbf{x},M)\longrightarrow H^{i+1}(\textbf{x},x,M) \end{aligned}$$

for $i\ge 0$, there is an exact sequence

$$\begin{aligned} 0\longrightarrow (0:_{H^i(\textbf{x},M)}x)\longrightarrow H^i(\textbf{x},M){\mathop {\longrightarrow }\limits ^{x.}}H^i(\textbf{x},M)\longrightarrow H^i(\textbf{x},M)/xH^i(\textbf{x},M)\longrightarrow 0. \end{aligned}$$

As $(0:_{H^i(\textbf{x},M)}x)$ and $H^i(\textbf{x},M)/xH^i(\textbf{x},M)$ are B-modules, they belong to ${\mathcal {S}}_2(\mathfrak {b})$ by Examples 2.1 (iii) and so by virtue of [7, Proposition 2.15], they belong to ${\mathcal {S}}_2(\mathfrak {a})$. Thus, Theorem 2.6 implies that $H^i(\textbf{x},M)\in {\mathcal {S}}_3(\mathfrak {a})$ for each $i\ge 0$ and consequently, $M\in {\mathcal {S}}_{t+3}(\mathfrak {a})$ by using Theorem 2.9. $\square$

3 Cofiniteness of local cohomology modules

Throughout this section, M is an A-module, $\mathfrak {a}$ is an ideal of A and n is a positive integer. We study the cofiniteness of local cohomology.For more details about local cohomology, we refer the reader to the textbook of Brodmann and Sharp [3]. Throughout this section n is a non-negative integer.

Theorem 3.1

Let ${{\,\textrm{Ext}\,}}^i_A(A/\mathfrak {a},M)$ be finitely generated for all $i\ge 0$ and let $t<s$ be non-negative integers such that $H_{\mathfrak {a}}^i(M)$ is $\mathfrak {a}$-cofinite for all $i\ne t,s$. Then $H_{\mathfrak {a}}^{t}(M)\in {\mathcal {S}}_{n+s-t+1}(\mathfrak {a})$ if and only if $H_{\mathfrak {a}}^s(M)\in {\mathcal {S}}_{n}(\mathfrak {a})$.

Proof

We first assume $H_{\mathfrak {a}}^{t}(M)\in {\mathcal {S}}_{n+s-t+1}(\mathfrak {a})$ and that ${{\,\textrm{Ext}\,}}_A^p(A/\mathfrak {a},H_{\mathfrak {a}}^s(M))\in {\mathcal {S}}_{n}(\mathfrak {a})$ is finitely generated for all $p\le n$. There is the Grothendieck spectral sequence

$$\begin{aligned} E_2^{p,q}:={{\,\textrm{Ext}\,}}_A^p(A/\mathfrak {a}, H_{\mathfrak {a}}^q(M))\Rightarrow {{\,\textrm{Ext}\,}}_A^{p+q}(A/\mathfrak {a},M). \end{aligned}$$

For each $r\ge 3$, consider the sequence $E_{r-1}^{p-r+1,t+r-2}{\mathop {\longrightarrow }\limits ^{d_{r-1}^{p-r+1,t+r+2}}}E_{r-1}^{p,t}{\mathop {\longrightarrow }\limits ^{d_{r-1}^{p,t}}}E_{r-1}^{p+r-1,t-r+2}$ and so $E_r^{p,t}={{\,\textrm{Ker}\,}}d_{r-1}^{p,t}/{{\,\textrm{Im}\,}}d_{r-1}^{p-r+1,t+r-2}$. Considering $p\le n+s-t+1$, we show that $E_2^{p,t}$ is finitely generated. If $r=s-t+2$, then $p-r+1\le n$ and so $E_{r-1}^{p-r+1,t+r-2}$ is finitely generated by the argument in the beginning of proof. If $r\ne s-t+2$, then the assumption implies that $E_{r-1}^{p-r+1,t+r-2}$ is finitely generated for all $p\ge 0$ (we observe that $t+r-2\ne t$). Consequently, ${{\,\textrm{Im}\,}}d_{r-1}^{p-r+1,t+r-2}$ is finitely generated for all $r\ge 3$ and $p\le n+s-t+1$. But there is a finite filtration

$$\begin{aligned} 0=\Phi ^{p+t+1}H^{p+t}\subset \Phi ^{p+t}H^{p+t}\subset \dots \subset \Phi ^1H^{p+t}\subset \Phi ^0H^{p+t}\subset H^{p+t} \end{aligned}$$

such that $\Phi ^pH^{p+t}/\Phi ^{p+1}H^{p+t}\cong E_{\infty }^{p,t}$ for all $p\ge 0$. In view of the assumption, $H^{p+t}={{\,\textrm{Ext}\,}}_A^{p+t}(A/\mathfrak {a},M)$ is finitely generated and so $E_{\infty }^{p,t}$ is finitely generated for all $p\ge 0$. On the other hand, $E_r^{p,t}=E_{\infty }^{p,t}$ for sufficiently large r and so $E_r^{p,t}$ is finitely generated for all $p\ge 0$. The previous argument implies that ${{\,\textrm{Ker}\,}}d_{r-1}^{p,t}$ is finitely generated for all $p\le n+s-t+1$; moreover since $r\ge 3$, we have $t-r+2\le t-1$ and so the assumption implies that $E_{r-1}^{p+r-1,t-r+2}$ is finitely generated, and hence $E_{r-1}^{p,t}$ is finitely generated for all $p\le n+s-t+1$. Continuing this way, we deduce that $E_2^{p,t}$ is finitely generated for all $p\le n+s-t+1$, and since $H_{\mathfrak {a}}^t(M)\in {\mathcal {S}}_{n+s-t+1}(\mathfrak {a})$, we deduce that $H_{\mathfrak {a}}^t(M)$ is $\mathfrak {a}$-cofinite. Therefore, $E_2^{p,q}$ is finitely generated for all $q\ne s$ and all $p\ge 0$. By a similar argument, we have $E_{\infty }^{p,s}=E_r^{p,s}$ for sufficiently large r and all $p\ge 0$ and since $E_{\infty }^{p,s}$ is a subquotient of ${{\,\textrm{Ext}\,}}_A^{p+s}(A/\mathfrak {a},M)$, it is finitely generated so that $E_r^{p,s}$ is finitely generated. In view of the sequence $E_{r-1}^{p-r+1,s+r-2}{\mathop {\longrightarrow }\limits ^{d_{r-1}^{p-r+1,s+r+2}}}E_{r-1}^{p,s}{\mathop {\longrightarrow }\limits ^{d_{r-1}^{p,s}}}E_{r-1}^{p+r-1,s-r+2}$, since $E_{r-1}^{p-r+1,s+r-2}$ is finitely generated and $E_r^{p,s}={{\,\textrm{Ker}\,}}d_{r-1}^{p,s}/{{\,\textrm{Im}\,}}d_{r-1}^{p-r+1,s+r-2}$, we conclude that ${{\,\textrm{Ker}\,}}d_{r-1}^{p,s}$ is finitely generated; and hence $E_{r-1}^{p,s}$ is finitely generated. Continuing this way, we deduce that $E_3^{p,s}$ is finitely generated and so is ${{\,\textrm{Ker}\,}}d_2^{p,s}$ for all $p\ge 0$. Now the exact sequence $0\longrightarrow {{\,\textrm{Ker}\,}}d_2^{p,s}\longrightarrow E_2^{p,s}\longrightarrow E_2^{p+2,s+1}$ implies that $E_2^{p,s}$ is finitely generated for all $p\ge 0$; and consequently $H_{\mathfrak {a}}^s(M)$ is $\mathfrak {a}$-cofinite. A similar proof gets the converse. $\square$

The following lemma extends [7, Proposition 2.6].

Lemma 3.2

Let $\dim A/\mathfrak {a}=3$ and ${{\,\textrm{depth}\,}}({{\,\textrm{Ann}\,}}(M),A/\mathfrak {a})>0$. If ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a},M)$ is finitely generated for all $i\le n+1$, then the following conditions are equivalent.

• (i) ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^i(M))$ is finitely generated for all $i\le n$.

• (ii) $H_{\mathfrak {a}}^i(M)$ is $\mathfrak {a}$-cofinite for all $i<n$.

Proof

Since ${{\,\textrm{depth}\,}}({{\,\textrm{Ann}\,}}(M),A/\mathfrak {a})>0$, there exists an element $x\in {{\,\textrm{Ann}\,}}(M)$ such that x is a non-zerodivisor on $A/\mathfrak {a}$. Taking $\mathfrak {b}=\mathfrak {a}+xA$, we have $\dim A/b=2$ and it follows from [4, Proposition 1] that ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {b},M)$ is finitely generated for all $i\le n+1$. Moreover, we have $\Gamma _{xA}(M)=\Gamma _{x\mathfrak {a}}(M)=M$ and $\Gamma _{\mathfrak {a}}(M)=\Gamma _{\mathfrak {b}}(M)$. Thus the ideals $\mathfrak {a}$ and xA of A provides the following Mayer-Vietoris exact sequence

$$\begin{aligned} 0\longrightarrow \Gamma _{\mathfrak {b}}(M)\longrightarrow \Gamma _{\mathfrak {a}}(M)\oplus M\longrightarrow M\longrightarrow H_{\mathfrak {b}}^1(M)\longrightarrow H_{\mathfrak {a}}^1(M)\longrightarrow 0 \end{aligned}$$

and the isomorphism $H_{\mathfrak {a}}^i(M)\cong H_{\mathfrak {b}}^i(M)$ for each $i\ge 2$. (i)$\Rightarrow$(ii). The case $n=1$ follows from [7, Proposition 2.6]. For $n\ge 2$, $\Gamma _{\mathfrak {a}}(M)$ is $\mathfrak {a}$-cofinite. Then $\Gamma _{\mathfrak {b}}(M)=\Gamma _{\mathfrak {a}}(M)$ is $\mathfrak {b}$-cofinite too. Moreover, it is clear that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {b},H_{\mathfrak {a}}^i(M))$ is finitely generated for all $i\le n$. Applying the functor ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {b},-)$ to the above exact sequence, we deduce that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {b},\Gamma _{\mathfrak {b}}(M))$ and ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {b},H_{\mathfrak {b}}^1(M))$ are finitely generated. Furthermore ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {b},H_{\mathfrak {b}}^i(M))\cong {{\,\textrm{Hom}\,}}_A(A/\mathfrak {b},H_{\mathfrak {a}}^i(M))$ is finitely generated for each $2\le i\le n$. Now, [10, Theorem 3.7] implies that $H_{\mathfrak {b}}^i(M)$ is $\mathfrak {b}$-cofinite and consequently $H_{\mathfrak {a}}^i(M)$ is $\mathfrak {a}$-cofinite for all $i<n$ using [4, Proposition 2]. (ii)$\Rightarrow$(i) Since $H_{\mathfrak {a}}^i(M)$ is $\mathfrak {a}$-cofinite for all $i<n$, by the previous argument, $H_{\mathfrak {b}}^i(M)$ is $\mathfrak {b}$-cofinite for all $i<n$; and hence it follows from [10, Theorem 3.7] that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {b},H_{\mathfrak {b}}^i(M))$ is finitely generated for all $i\le n$. We observe that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},\Gamma _{\mathfrak {b}}(M))={{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},\Gamma _{\mathfrak {a}}(M))={{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},M)$ is finitely generated. Furthermore, ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {b}}^1(M))\cong {{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},{{\,\textrm{Hom}\,}}_A(A/xA,H_{\mathfrak {b}}^1(M)))\cong {{\,\textrm{Hom}\,}}_A(A/\mathfrak {b},H_{\mathfrak {b}}^1(M))$ is finitely generated. Now, since $\Gamma _{\mathfrak {a}}(M)$ is $\mathfrak {a}$-cofinite, applying the functor ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},-)$ to the above exact sequence, we conclude that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H^1_{\mathfrak {a}}(M))$ is finitely generated. By the argument mentioned in the beginning of the proof ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^i(M))\cong {{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {b}}^i(M))\cong {{\,\textrm{Hom}\,}}_A(A/\mathfrak {b},H_{\mathfrak {b}}^i(M))$ is finitely generated for all $2\le i\ge n$. $\square$

It was proved in [10, Theorems 3.3, 3.7] that if $\dim A/\mathfrak {a}\le 2$, then the conditions in Lemma 3.2 are equivalent. In the following theorem, we extend this result for $\dim A/\mathfrak {a}\ge 3$, but by an additional assumption.

Theorem 3.3

Let $\dim A/\mathfrak {a}=d\ge 3$ and let ${{\,\textrm{depth}\,}}({{\,\textrm{Ann}\,}}(M),A/\mathfrak {a})\ge d-2$. If ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a},M)$ is finitely generated for all $i\le n+1$, then the conditions in Lemma 3.2 are equivalent.

Proof

We proceed by induction on d. The case $d=3$ follows from Lemma 3.2 and so we may assume that $d\ge 4$. Since ${{\,\textrm{depth}\,}}({{\,\textrm{Ann}\,}}(M),A/\mathfrak {a})>0$, there exists an element $x\in {{\,\textrm{Ann}\,}}_RM$ which is a non-zerodivisor on $A/\mathfrak {a}$. Taking $\mathfrak {b}=\mathfrak {a}+xA$, we have $\dim A/\mathfrak {b}=d-1$. Since ${{\,\textrm{Supp}\,}}_AA/\mathfrak {b}\subseteq {{\,\textrm{Supp}\,}}_AA/\mathfrak {a}$, it follows ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {b},M)$ is finitely generated for all $i\le n+1$. If $H_{\mathfrak {a}}^i(M)$ is $\mathfrak {a}$-cofinite for each $i<n$, then $H_{\mathfrak {b}}^i(M)$ is $\mathfrak {b}$-cofinite for each $i<n$. The induction hypothesis implies that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {b},H_{\mathfrak {b}}^i(M))$ is finitely generated for all $i\le n$. By the same reasoning in the proof of Lemma 3.2, we have ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^i(M))$ is finitely generated for all $i\le n$. Conversely, assume that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^i(M))$ is finitely generated for all $i\le n$ and so ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {b},H_{\mathfrak {b}}^i(M))$ is finitely generated for all $i\le n$. The induction hypothesis implies that $H_{\mathfrak {b}}^i(M)$ is $\mathfrak {b}$-cofinite for each $i<n$ and so $H_{\mathfrak {a}}^i(M)$ is $\mathfrak {a}$-cofinite for each $i<n$. $\square$

Corollary 3.4

Let $\mathfrak {b}$ an ideal of A and $\mathfrak {a}=\Gamma _{\mathfrak {b}}(A)$ such that $\dim A/\mathfrak {a}=3$. Then ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^i(A/\mathfrak {b}))$ is finitely generated for all $i\le n$ if and only if $H^i_{\mathfrak {a}}(A/\mathfrak {b})$ is $\mathfrak {a}$-cofinite for all $i<n$.

Proof

As $\Gamma _{\mathfrak {b}}(A/\mathfrak {a})=0$, we have ${{\,\textrm{depth}\,}}(\mathfrak {b},A/\mathfrak {a})>0$. Now the assertion is obtained by using Lemma 3.2. $\square$

Corollary 3.5

Let $\mathfrak {b}$ an ideal of A and $\mathfrak {a}=\Gamma _{\mathfrak {b}}(A)$ such that $\dim A/\mathfrak {a}=3$. Then ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^i(A))$ is finitely generated for all $i\le n$ if and only if $H^i_{\mathfrak {a}}(A)$ is $\mathfrak {a}$-cofinite for all $i<n$. In particular, ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^1(A))$ is finitely generated.

Proof

There exists a positive integer t such that $\mathfrak {b}^t\mathfrak {a}=0$ and so $\Gamma _{\mathfrak {a}}(\mathfrak {b}^t)=\mathfrak {b}^t$. Thus applying the functor $\Gamma _{\mathfrak {a}}(-)$ to the exact sequence $0\longrightarrow \mathfrak {b}^t\longrightarrow A\longrightarrow A/\mathfrak {b}^t\longrightarrow 0$, we deduce that $H_{\mathfrak {a}}^i(A)\cong H_{\mathfrak {a}}^i(A/\mathfrak {b}^t)$ for each $i> 0$. Since $\Gamma _{\mathfrak {b}}(A/\mathfrak {a})=0$, we have ${{\,\textrm{depth}\,}}(\mathfrak {b}^t,A/\mathfrak {a})={{\,\textrm{depth}\,}}(\mathfrak {b},A/\mathfrak {a})>0$. If ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^i(A))$ is finitely generated for all $i\le n$, then ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^i(A/\mathfrak {b}^t))$ is finitely generated for all $i\le n$. Now Lemma 3.2 implies that $H_{\mathfrak {a}}^i(A/\mathfrak {b}^t)$ is $\mathfrak {a}$-cofinite for all $i<n$; and hence $H_{\mathfrak {a}}^i(A)$ is $\mathfrak {a}$-cofinite for all $i<n$. Conversely, if $H_{\mathfrak {a}}^i(A)$ is $\mathfrak {a}$-cofinite for all $i<n$, then $H_{\mathfrak {a}}^i(A/\mathfrak {b}^t)$ is $\mathfrak {a}$-cofinite for all $i<n$ and so using again Lemma 3.2, ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^i(A/\mathfrak {b}^t))$ is finitely generated for all $i\le n$ so that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^i(A))$ is finitely generated for all $i\le n$ $\square$

Corollary 3.6

Let $\mathfrak {p}$ be a prime ideal of A with $\dim A/\mathfrak {p}=3$ and let $\mathfrak {b}$ be an ideal of A such that $\mathfrak {b}\nsubseteq \mathfrak {p}$. Then ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {p},H_{\mathfrak {p}}^i(A/\mathfrak {b}))$ is finitely generated for all $i\le n$ if and only if $H_{\mathfrak {p}}^i(A/\mathfrak {b})$ is $\mathfrak {p}$-cofinite for all $i<n$.

Proof

Since $\mathfrak {b}\nsubseteq \mathfrak {p}$, we have ${{\,\textrm{depth}\,}}(\mathfrak {b},A/\mathfrak {p})>0$ and so the the result follows from Theorem 3.3. $\square$

If $\dim A\ge 4$, then we have the following result.

Proposition 3.7

Let $\dim A=d\ge 4$ with ${{\,\textrm{depth}\,}}({{\,\textrm{Ann}\,}}(M),A/(\mathfrak {a}+\Gamma _{\mathfrak {a}}(A)))\ge d-3$ and let ${{\,\textrm{Ext}\,}}_A^i(A/\mathfrak {a},M)$ be finitely generated for all $i\le n+1$. Then the conditions in Lemma 3.2 are equivalent.

Proof

We can choose an integer t such that $(0:_A \mathfrak {a}^t)= \Gamma _{\mathfrak {a}}(A)$. Put ${\overline{A}} = A/{\Gamma _{\mathfrak {a}}(A)}$ and ${\overline{M}} = M/{(0:_M \mathfrak {a}^t)}$ which is an ${\overline{A}}$-module. Taking $\overline{\mathfrak {a}}$ as the image of $\mathfrak {a}$ in ${\overline{A}}$, we have $\Gamma _{\overline{\mathfrak {a}}}({{\overline{A}}})=0$. Thus $\overline{\mathfrak {a}}$ contains an ${\overline{A}}$-regular element so that $\dim A/(\mathfrak {a}+\Gamma _{\mathfrak {a}}(A))=\textrm{dim}{\overline{A}}/\overline{\mathfrak {a}}\le d-1$. The assumption on M together with the fact that ${{{\,\textrm{Supp}\,}}} _{A}(A/(\mathfrak {a}+\Gamma _{\mathfrak {a}}(A)))\subset {{{\,\textrm{Supp}\,}}} _{A}(A/\mathfrak {a})$ and [4, Proposition 1] imply that $\textrm{Ext}^i_A(A/(\mathfrak {a}+\Gamma _{\mathfrak {a}}(A)),M)$ is finitely generated for all $i\le n$. Since by the assumption $(0:_M\mathfrak {a})$ is finitely generated, it is clear that $(0:_M\mathfrak {a}^t)$ is finitely generated and we have an exact sequence

$$\begin{aligned} 0\longrightarrow (0:_M\mathfrak {a}^t)\longrightarrow \Gamma _{\mathfrak {a}}(M)\longrightarrow \Gamma _{\mathfrak {a}}({\overline{M}})\longrightarrow 0 \end{aligned}$$

and the isomorphism $H_{\mathfrak {a}}^i(M)\cong H_{\mathfrak {a}}^i({\overline{M}})$ for all $i>0$. In order to prove (i)$\Rightarrow$(ii), assume that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^i(M))$ is finitely generated for all $i\le n$. Then in view of the previous argument and the independence theorem for local cohomology ${{\,\textrm{Hom}\,}}_A(A/(\mathfrak {a}+\Gamma _{\mathfrak {a}}(A)),H_{\mathfrak {a}+\Gamma _{\mathfrak {a}}(A)}^i({\overline{M}}))$ is finitely generated for all $i\le n$. It now follows from Theorem 3.3 that $H_{\mathfrak {a}+\Gamma _{\mathfrak {a}}(A)}({\overline{M}})$ is $\mathfrak {a}+\Gamma _{\mathfrak {a}}(A)$-cofinite for all $i<n$; and hence using the change of ring principle [4, Proposition 2], $H_{\mathfrak {a}}^i({\overline{M}})$ is $\mathfrak {a}$-cofinite for all $i<n$. Consequently, the previous argument implies that $H_{\mathfrak {a}}^i(M)$ is $\mathfrak {a}$-cofinite for all $i<n$. (ii)$\Rightarrow$(i). Assume that $H_{\mathfrak {a}}^i(M)$ is $\mathfrak {a}$-cofinite for all $i<n$. By the same reasoning as mentioned before, we deduce that $H_{\mathfrak {a}+\Gamma _{\mathfrak {a}}(A)}^i({\overline{M}})$ is $\mathfrak {a}+\Gamma _{\mathfrak {a}}(A)$-cofinite for all $i<n$. Now, using again Theorem 3.3, we deduce that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^i({\overline{M}}))\cong {{\,\textrm{Hom}\,}}_A(A/(\mathfrak {a}+\Gamma _{\mathfrak {a}}(A)),H_{\mathfrak {a}+\Gamma _{\mathfrak {a}}(A)}^i({\overline{M}}))$ is finitely generated for all $i\le n$ and consequently the previous argument yields that ${{\,\textrm{Hom}\,}}_A(A/\mathfrak {a},H_{\mathfrak {a}}^i(M))$ is finitely generated for all $i\le n$. $\square$