---
abstract: 'Let $(R,{\mathfrak{m}})$ be a complete Noetherian local ring and let $M$ be a finite $R$–module of positive Krull dimension $n$. It is shown that any subset $T$ of ${\mbox{Assh}\,}_R(M)$ can be expressed as the set of attached primes of the top local cohomology module ${\mbox{H}\, }^n_{\mathfrak{a}}(M)$ for some ideal ${\mathfrak{a}}$ of $R$. Moreover if ${\mathfrak{a}}$ is an ideal of $R$ such that the set of attached primes of ${\mbox{H}\, }^n_{\mathfrak{a}}(M)$ is a non–empty proper subset of ${\mbox{Assh}\,}_R(M)$, then ${\mbox{H}\, }^n_{\mathfrak{a}}(M)\cong{\mbox{H}\, }^n_{\mathfrak{b}}(M)$ for some ideal ${\mathfrak{b}}$ of $R$ with ${\mbox{dim}\,}_R (R/{\mathfrak{b}})=1$.'
address:
- |
     Mohammad T. Dibaei\
    Faculty of Mathematical Sciences, Teacher Training University, Tehran, Iran, and Institute for Theoretical Physics and Mathematics (IPM), Tehran, Iran.
- |
    Raheleh Jafari\
    Faculty of Mathematical Sciences, Teacher Training University, Tehran, Iran
author:
- 'Mohammad T. Dibaei'
- Raheleh Jafari
title: |
    Top local cohomology modules\
    with specified attached primes
---

Introduction
============

Throughout $(R,{\mathfrak{m}})$ is a commutative Noetherian local ring with maximal ideal ${\mathfrak{m}}$, $M$ is a non-zero finite (i.e. finitely generated) $R$–module with positive Krull dimension $n:={\mbox{dim}\,}_R(M)$ and ${\mathfrak{a}}$ denotes an ideal of $R$. Recall that for an $R$–module $N$, a prime ideal ${\mathfrak{p}}$ of $R$ is said to be an [*attached prime*]{} of $N$, if ${\mathfrak{p}}={\mbox{Ann}\,}_R(N/K)$ for some submodule $K$ of $N$ (see [@MS]). The set of attached primes of $N$ is denoted by ${\mbox{Att}\,}_R(N)$. If $N$ is an Artinian $R$–module so that $N$ admits a reduced secondary representation $N=N_1+\cdots+N_r$ such that $N_i$ is ${\mathfrak{p}}_i$–secondary, $i=1,\ldots,r$, then ${\mbox{Att}\,}_R(N)=\{{\mathfrak{p}}_1,\ldots,{\mathfrak{p}}_r\}$ is a finite set.

Denote by ${\mbox{H}\, }^n_{\mathfrak{a}}(M)$ the $n$th right derived functor of $$\Gamma_{\mathfrak{a}}(M)=\{x\in M|\, {\mathfrak{a}}^rx=0 \ \mbox{for some positive
integer} \  r \}$$ applied to $M$. It is well-known that ${\mbox{H}\, }^n_{\mathfrak{a}}(M)$ is an Artinian module. Macdonald and Sharp, in [@MS], studied ${\mbox{H}\, }^n_{\mathfrak{m}}(M)$ and showed that ${\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{m}}(M))= {\mbox{Assh}\,}_R(M)$ where ${\mbox{Assh}\,}_R(M):=\{{\mathfrak{p}}\in
{\mbox{Ass}\,}_R(M)|\, {\mbox{dim}\,}_R(R/{\mathfrak{p}})=n\}$. It is shown in [@DY1 Theorem A], that for any arbitrary ideal ${\mathfrak{a}}$ of $R$, ${\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{a}}(M))=\{{\mathfrak{p}}\in{\mbox{Ass}\,}_R(M)|\, {\mbox{H}\, }^n_{\mathfrak{a}}(R/{\mathfrak{p}})\neq 0\}$ which is a subset of ${\mbox{Assh}\,}_R(M)$. In [@DY2], the structure of ${\mbox{H}\, }^n_{\mathfrak{a}}(M)$ is studied by the first author and Yassemi and they showed that, in case $R$ is complete, for any pair of ideals ${\mathfrak{a}}$ and ${\mathfrak{b}}$ of $R$, if ${\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{a}}(M))={\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{b}}(M))$, then ${\mbox{H}\, }^n_{\mathfrak{a}}(M) \cong {\mbox{H}\, }^n_{\mathfrak{b}}(M)$. They also raised the following question in [@DY3 Question 2.9] which is the main object of this paper.

[**Question.**]{} For any subset $T$ of ${\mbox{Assh}\,}_R(M)$, is there an ideal ${\mathfrak{a}}$ of $R$ such that ${\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{a}}(M)) = T$?

This paper provides a positive answer for this question in the case $R$ is complete.

Main Result
===========

In this section we assume that $R$ is complete with respect to the ${\mathfrak{m}}$–adic topology. As mentioned above, ${\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{m}}(M)) =
{\mbox{Assh}\,}_R(M)$ and ${\mbox{Att}\,}_R({\mbox{H}\, }^n_R(M)) = \emptyset$ is the empty set. Also ${\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{a}}(M)) \subseteq {\mbox{Assh}\,}_R(M)$ for all ideals ${\mathfrak{a}}$ of $R$. Our aim is to show that as ${\mathfrak{a}}$ varies over ideals of $R$, the set ${\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{a}}(M))$ takes all possible subsets of ${\mbox{Assh}\,}_R(M)$ (see Theorem 2.8). In the following results we always assume that $T$ is a non–empty proper subset of ${\mbox{Assh}\,}_R(M)$

In our first result we find a characterization for a subset of ${\mbox{Assh}\,}_R (M)$ to be the set of attached primes of the top local cohomology of $M$ with respect to an ideal ${\mathfrak{a}}$.

Assume that $n:={\mbox{dim}\,}_R(M)\geq 1$ and that $T$ is a proper non-empty subset of ${\mbox{Assh}\,}_R(M)$. Set ${\mbox{Assh}\,}_R(M)\setminus
T=\{{\mathfrak{q}}_1,\ldots,{\mathfrak{q}}_r\}$. The following statements are equivalent.

1.  There exists an ideal ${\mathfrak{a}}$ of $R$ such that ${\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{a}}(M))=T$.

2.  For each $i,\,1\leq i\leq r$, there exists $Q_i\in {\mbox{Supp}\,}_R(M)$ with ${\mbox{dim}\,}_R(R/Q_i)=1$ such that $$\underset{{\mathfrak{p}}\in T}{\bigcap}{\mathfrak{p}}\nsubseteq Q_i \quad \mbox{and}
      \quad {\mathfrak{q}}_i\subseteq Q_i.$$

With $Q_i,\, 1\leq i\leq r$, as above, ${\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{a}}(M))=T$ where ${\mathfrak{a}}=\bigcap\limits_{i=1}^rQ_i$.

$(i)\Rightarrow (ii)$. By [@DY1 Theorem A], ${\mbox{H}\, }^n_{\mathfrak{a}}(R/{\mathfrak{p}})\neq 0$ for all ${\mathfrak{p}}\in T$, that is ${\mathfrak{a}}+{\mathfrak{p}}$ is ${\mathfrak{m}}$–primary for all ${\mathfrak{p}}\in T$ (by [Lichtenbaum-Hartshorne Theorem]{}). On the other hand, for $1\leq i\leq r, {\mathfrak{q}}_i\notin T$ which is equivalent to say that ${\mathfrak{a}}+{\mathfrak{q}}_i$ is not an ${\mathfrak{m}}$–primary ideal. Hence there exists a prime ideal $Q_i\in {\mbox{Supp}\,}_R(M)$ such that ${\mbox{dim}\,}_R(R/Q_i)=1$ and ${\mathfrak{a}}+{\mathfrak{q}}_i\subseteq Q_i$. It follows that $\underset{{\mathfrak{p}}\in
T}{\bigcap}{\mathfrak{p}}\nsubseteq Q_i$.\
$(ii)\Rightarrow (i)$. Set ${\mathfrak{a}}:=\bigcap\limits_{i=1}^rQ_i$. For each $i, 1\leq i\leq r$, ${\mathfrak{a}}+{\mathfrak{q}}_i\subseteq Q_i$ implies that ${\mathfrak{a}}+{\mathfrak{q}}_i$ is not ${\mathfrak{m}}$–primary and so ${\mbox{H}\, }^n_{\mathfrak{a}}(R/{\mathfrak{q}}_i)= 0$. Thus ${\mbox{Att}\,}_R{\mbox{H}\, }_{\mathfrak{a}}^n(M)\subseteq T$. Assume ${\mathfrak{p}}\in T$ and $Q\in
{\mbox{Supp}\,}(M)$ such that ${\mathfrak{a}}+{\mathfrak{p}}\subseteq Q$. Then $Q_i\subseteq Q$ for some $i, 1\leq i\leq r$. Since ${\mathfrak{p}}\nsubseteq Q_i$, we have $Q_i\neq
Q$, so $Q={\mathfrak{m}}$. Hence ${\mathfrak{a}}+ {\mathfrak{p}}$ is ${\mathfrak{m}}$–primary ideal. Now, by [Lichtenbaum-Hartshorne Theorem]{}, and by [@DY1 Theorem A], it follows that ${\mathfrak{p}}\in{\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{a}}(M))$.

If ${\mbox{H}\, }^n_{\mathfrak{a}}(M)\not=o$ then there is an ideal ${\mathfrak{b}}$ of $R$ such that ${\mbox{dim}\,}_R(R/{\mathfrak{b}})\leq 1$ and ${\mbox{H}\, }^n_{\mathfrak{a}}(M)\cong{\mbox{H}\, }^n_{\mathfrak{b}}(M)$.

If ${\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{a}}(M))= {\mbox{Assh}\,}_R(M)$, then ${\mbox{H}\, }^n_{\mathfrak{a}}(M)=
{\mbox{H}\, }^n_{\mathfrak{m}}(M)$. Otherwise $n\geq 1$ and ${\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{a}}(M))$ is a proper subset of ${\mbox{Assh}\,}_R(M)$. Set ${\mbox{Assh}\,}_R(M)\setminus
{\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{a}}(M)):=\{{\mathfrak{q}}_1, \cdots, {\mathfrak{q}}_r\}$. By Proposition 2.1, there are $Q_i\in {\mbox{Supp}\,}_R(M)$ with ${\mbox{dim}\,}_R(R/Q_i)= 1, \ i=1, \cdots,
r$, such that ${\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{a}}(M))= {\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{b}}(M))$ with ${\mathfrak{b}}= {\bigcap\limits_{i=1}^rQ_i}$. Now, by [@DY2 Theorem 1.6], we have ${\mbox{H}\, }^n_{\mathfrak{a}}(M)\cong {\mbox{H}\, }^n_{\mathfrak{b}}(M)$. As ${\mbox{dim}\,}(R/{\mathfrak{b}})= 1$, the proof is complete.

If ${\mbox{dim}\,}_R(M)=1$ then any subset $T$ of ${\mbox{Assh}\,}_R(M)$ is equal to the set ${\mbox{Att}\,}_R({\mbox{H}\, }^1_{\mathfrak{a}}(M))$ for some ideal ${\mathfrak{a}}$ of $R$.

With notations as in Proposition 2.1, we take $Q_i={\mathfrak{q}}_i$ for $i=1,\cdots, r$.

By a straightforward argument one may notice that the condition “complete" is superficial, for if $T$ is a non–empty proper subset of ${\mbox{Assh}\,}_R(M)$, then $T={\mbox{Att}\,}_R({\mbox{H}\, }^1_{\mathfrak{a}}(M))$, where ${\mathfrak{a}}=\underset{{\mathfrak{p}}\in{\mbox{Assh}\,}_R(M)\setminus T}{\cap}{\mathfrak{p}}$.

The following is an example to Proposition 2.1.

Set $R=k[[X,Y,Z,W]]$, where $k$ is a field and $X,Y,Z,W$ are independent indeterminates. Then $R$ is a complete Noetherian local ring with maximal ideal ${\mathfrak{m}}=(X,Y,Z,W)$. Consider prime ideals $${\mathfrak{p}}_1=(X,Y) \quad  ,  \quad {\mathfrak{p}}_2=(Z,W)\quad , \quad  {\mathfrak{p}}_3=(Y,Z)
\quad , \quad  {\mathfrak{p}}_4=(X,W)$$ and set $\displaystyle
M=\frac{R}{{\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4}$ as an $R$–module, so that we have ${\mbox{Assh}\,}_R(M)=\{{\mathfrak{p}}_1,{\mathfrak{p}}_2,{\mathfrak{p}}_3,{\mathfrak{p}}_4\}$ and ${\mbox{dim}\,}_R(M)=2$. We get $\{{\mathfrak{p}}_i\}={\mbox{Att}\,}_R({\mbox{H}\, }^2_{{\mathfrak{a}}_i}(M))$, where ${\mathfrak{a}}_1={\mathfrak{p}}_2,
{\mathfrak{a}}_2={\mathfrak{p}}_1, {\mathfrak{a}}_3={\mathfrak{p}}_4, {\mathfrak{a}}_4={\mathfrak{p}}_3$, and $\{{\mathfrak{p}}_i,{\mathfrak{p}}_j\}={\mbox{Att}\,}_R({\mbox{H}\, }^2_{{\mathfrak{a}}_{ij}}(M))$, where $$\begin{array}{l}
{\mathfrak{a}}_{12}=(Y^2+YZ,Z^2+YZ,X^2+XW,W^2+WX),\\
{\mathfrak{a}}_{34}=(Z^2+ZW,X^2+YX,Y^2+YX,W^2+WZ),\\
{\mathfrak{a}}_{13}=(Z^2+XZ,W^2+WY,X^2+XZ),\\
{\mathfrak{a}}_{14}=(W^2+WY,Z^2+ZY,Y^2+YW),\\
{\mathfrak{a}}_{23}=(X^2+XZ,Y^2+WY,W^2+ZW),\\
{\mathfrak{a}}_{24}=(X^2+XZ,Y^2+WY,Z^2+ZW).\\
\end{array}$$ Finally, we have $\{{\mathfrak{p}}_i,{\mathfrak{p}}_j,{\mathfrak{p}}_k\}={\mbox{Att}\,}_R({\mbox{H}\, }^2_{{\mathfrak{a}}_{ijk}}(M))$, where ${\mathfrak{a}}_{123}=(X,W,Y+Z)$, ${\mathfrak{a}}_{234}=(X,Y,W+Z)$, ${\mathfrak{a}}_{134}=(Z,W,Y+X)$.\

Assume that $n:={\mbox{dim}\,}_R(M)\geq 2$, and that $T$ is a non-empty subset of ${\mbox{Assh}\,}_R(M)$ such that $\underset{{\mathfrak{p}}\in T}{\bigcap} {\mathfrak{p}}\nsubseteq \underset
   {{\mathfrak{q}}\in {\mbox{Assh}\,}_R(R/\sum\limits_{{\mathfrak{p}}\in T'}{\mathfrak{p}})}{\bigcap} {\mathfrak{q}}$, where $T'={\mbox{Assh}\,}_R(M)\setminus T$. Then there exists a prime ideal $Q\in {\mbox{Supp}\,}_R(M)$ with ${\mbox{dim}\,}_R(R/Q)=1$ and ${\mbox{Att}\,}_R({\mbox{H}\, }^n_Q(M))=T.$

Set $s:={\mbox{ht}\,}_M(\sum\limits_{{\mathfrak{p}}\in T'}{\mathfrak{p}})$. We have $s\leq n-1$, otherwise ${\mbox{Assh}\,}_R(R/\sum\limits_{{\mathfrak{p}}\in T'}{\mathfrak{p}})= \{{\mathfrak{m}}\}$ which contradicts the condition $\underset{{\mathfrak{p}}\in T}{\bigcap} {\mathfrak{p}}\nsubseteq \underset
   {{\mathfrak{q}}\in {\mbox{Assh}\,}_R(R/\sum\limits_{{\mathfrak{p}}\in T'}{\mathfrak{p}})}{\bigcap} {\mathfrak{q}}$. As $R$ is catenary, we have ${\mbox{dim}\,}_R(R/\sum\limits_{{\mathfrak{p}}\in
T'}{\mathfrak{p}})=n-s$. We first prove, by induction on $j$, $0\leq j\leq
n-s-1$, that there exists a chain of prime ideals $Q_0 \subset Q_1
\subset \cdots \subset Q_j \subset {\mathfrak{m}}$ such that $Q_0\in{\mbox{Assh}\,}_R(R/\sum\limits_{{\mathfrak{p}}\in T'}{\mathfrak{p}})$, ${\mbox{dim}\,}_R(R/Q_j)=n-s-j$ and $\underset{{\mathfrak{p}}\in T}{\bigcap}{\mathfrak{p}}\nsubseteq Q_j$. There is $Q_0\in{\mbox{Assh}\,}_R(R/\sum\limits_{{\mathfrak{p}}\in T'}{\mathfrak{p}})$ such that $\underset{{\mathfrak{p}}\in T}{\bigcap}{\mathfrak{p}}\nsubseteq Q_0$. Note that ${\mbox{dim}\,}_R(R/Q_0)={\mbox{dim}\,}_R(R/\sum\limits_{{\mathfrak{p}}\in T'}{\mathfrak{p}})=n-s$. Now, assume that $0<j\leq n-s-1$ and that we have proved the existence of a chain $Q_0 \subset Q_1 \subset \cdots \subset Q_{j-1}$ of prime ideals such that $Q_0\in{\mbox{Assh}\,}_R(R/\sum\limits_{{\mathfrak{p}}\in T'}{\mathfrak{p}})$, ${\mbox{dim}\,}_R(R/Q_j)=n-s-(j-1)$ and that $\underset{{\mathfrak{p}}\in
T}{\bigcap}{\mathfrak{p}}\nsubseteq Q_{j-1}$. Note that we have $n-s-(j-1)=n-s+1-j\geq 2$. Therefore the set $V$ defined as\
$$\begin{array}{ll}
V= \{{\mathfrak{q}}\in {\mbox{Supp}\,}_R(M) |&  Q_{j-1}\subset {\mathfrak{q}}\subset {\mathfrak{q}}'\subseteq
 {\mathfrak{m}}, {\mbox{dim}\,}_R(R/{\mathfrak{q}})=n-s-j,\\ &
{\mathfrak{q}}'\in{\mbox{Spec}\,}(R)\, \mbox{and}\,  {\mbox{dim}\,}_R(R/{\mathfrak{q}}')=n-s-j-1\}
\end{array}$$\
is non-empty and so, by Ratliff’s weak existence theorem [@M Theorem 31.2], is not finite. As $\underset{{\mathfrak{p}}\in
T}{\bigcap}{\mathfrak{p}}\nsubseteq Q_{j-1}$, we have $Q_{j-1}\subset
Q_{j-1}+\underset{{\mathfrak{p}}\in T}{\bigcap}{\mathfrak{p}}$. If, for ${\mathfrak{q}}\in V$, $\underset{{\mathfrak{p}}\in T}{\bigcap}{\mathfrak{p}}\subseteq {\mathfrak{q}}$, then ${\mathfrak{q}}$ is a minimal prime of $Q_{j-1}+\underset{{\mathfrak{p}}\in T}{\bigcap}{\mathfrak{p}}$. As $V$ is an infinite set, there is $Q_j\in V$ such that $\underset{{\mathfrak{p}}\in
T}{\bigcap}{\mathfrak{p}}\nsubseteq Q_j$. Thus the induction is complete. Now by taking $Q:=Q_{n-s-1}$ and by Proposition 2.1, the claim follows.

Assume that $n:={\mbox{dim}\,}_R(M)\geq 2$ and $T$ is a non-empty subset of ${\mbox{Assh}\,}_R(M)$ with $|T|=|{\mbox{Assh}\,}_R(M)|-1$. Then there is an ideal ${\mathfrak{a}}$ of $R$ such that ${\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{a}}(M))=T$.

Note that ${\mbox{Assh}\,}_R(M)\setminus T$ is a singleton set $\{{\mathfrak{q}}\}$, say, and so ${\mbox{ht}\,}_M({\mathfrak{q}})=0$ and $\underset{{\mathfrak{p}}\in T}{\bigcap}{\mathfrak{p}}\nsubseteq
{\mathfrak{q}}$. Therefore, by Lemma 2.5, the result follows.

Assume that $n:={\mbox{dim}\,}_R(M)\geq 2$ and ${\mathfrak{a}}_1$ and ${\mathfrak{a}}_2$ are ideals of $R$. Then there exists an ideal ${\mathfrak{b}}$ of $R$ such that ${\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{b}}(M))={\mbox{Att}\,}_R({\mbox{H}\, }^n_{{\mathfrak{a}}_{1}}(M))\cap{\mbox{Att}\,}_R({\mbox{H}\, }^n_{{\mathfrak{a}}_{2}}(M))$.

Set $T_{1}={\mbox{Att}\,}_R({\mbox{H}\, }^n_{{\mathfrak{a}}_{1}}(M))$ and $T_{2}={\mbox{Att}\,}_R({\mbox{H}\, }^n_{{\mathfrak{a}}_{2}}(M))$. We may assume that $T_1\bigcap
T_2$ is a non–empty proper subset of ${\mbox{Assh}\,}_R(M)$. Assume that ${\mathfrak{q}}\in {\mbox{Assh}\,}_R(M)\setminus (T_1\bigcap T_2)=({\mbox{Assh}\,}_R(M)\setminus
T_1)\bigcup({\mbox{Assh}\,}_R(M)\setminus T_2) $. By Proposition 2.1, there exists $Q\in {\mbox{Supp}\,}_R(M)$ with ${\mbox{dim}\,}_R(R/Q)=1$ such that ${\mathfrak{q}}\subseteq Q$ and $\bigcap_{{\mathfrak{p}}\in T_1\bigcap T_2}{\mathfrak{p}}\nsubseteq
Q$. Now, by Proposition 2.1, again there exists an ideal ${\mathfrak{b}}$ of $R$ such that ${\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{b}}(M))=T_1\bigcap T_2$.

Now we are ready to present our main result.

Assume that $T\subseteq {\mbox{Assh}\,}_R(M)$, then there exists an ideal ${\mathfrak{a}}$ of $R$ such that $T={\mbox{Att}\,}_R({\mbox{H}\, }^n_{\mathfrak{a}}(M))$.

By Corollary 2.3, we may assume that ${\mbox{dim}\,}_R(M)\geq 2$ and that $T$ is a non-empty proper subset of ${\mbox{Assh}\,}_R(M)$. Set $T=\{{\mathfrak{p}}_1,\ldots,{\mathfrak{p}}_t\}$ and ${\mbox{Assh}\,}_R(M)\setminus
T=\{{\mathfrak{p}}_{t+1},\ldots,{\mathfrak{p}}_{t+r}\}$. We use induction on $r$. For $r=1$, Corollary 2.6 proves the first step of induction. Assume that $r>1$ and that the case $r-1$ is proved. Set $T_1=\{{\mathfrak{p}}_1,\ldots,{\mathfrak{p}}_t,{\mathfrak{p}}_{t+1}\}$ and $T_2=\{{\mathfrak{p}}_1,\ldots,{\mathfrak{p}}_t,{\mathfrak{p}}_{t+2}\}$. By induction assumption there exist ideals ${\mathfrak{a}}_1$ and ${\mathfrak{a}}_2$ of $R$ such that $T_1={\mbox{Att}\,}_R({\mbox{H}\, }^n_{{\mathfrak{a}}_1}(M))$ and $T_2={\mbox{Att}\,}_R({\mbox{H}\, }^n_{{\mathfrak{a}}_2}(M))$. Now by the Lemma 2.7 there exists an ideal ${\mathfrak{a}}$ of $R$ such that $T=T_1\bigcap T_2={\mbox{Att}\,}_R({\mbox{H}\, }^n_{{\mathfrak{a}}}(M))$.

(See [@C Corollary 1.7]) With the notations as in Theorem 2.8, the number of non-isomorphic top local cohomology modules of $M$ with respect to all ideals of $R$ is equal to $2^{|{\mbox{Assh}\,}_R (M)|}$.

It follows from Theorem 2.8 and [@DY2 Theorem 1.6].

[**Acknowledgment.**]{}The authors would like to thank the referee for her/his comments.

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