\begin{document}

\title[Remarks on subcategories of artinian modules]{\textbf{Remarks on subcategories of artinian modules}}

\author{Naoya Hiramatsu} 
\address{Department of general education, Kure National College of Technology, 2-2-11 Agaminami, Kure, Hiroshima, 737-8506 JAPAN}
\email{hiramatsu@kure-nct.ac.jp}

\subjclass[2000]{13C05, 16D90, 13J10}
\date{\today}
\keywords{artinian module, wide subcategory, Serre subcategory}

\begin{abstract}
We study two subcategories of the category of artinian modules, a wide subcategory and a Serre subcategory. 
We prove that all wide subcategories of artinian modules are Serre subcategories. 
We also provide the bijection between the set of Serre subcategories and the set of specialization closed subsets of the set of closed prime ideals of some completed ring. 
These results are artinian analogues of the theorems proved in \cite{T08}.    
\end{abstract}
\mathfrak maketitle

\section{Introduction}

Classification theory of subcategories has been studied by many authors in many areas \cite{G62, H85, N92, Th97, Ho01, T08, K08}. 
In 1990's, Hopkins \cite{H85} and Neeman \cite{N92} classify thick subcategories of the derived categories of perfect complexes in terms of the ring spectra. 
Thomason \cite{Th97} generalizes this result to quasi-compact and quasi-separated schemes.
Now the classification theorem by them is known as the Hopkins-Neeman-Thomason theorem.  

Let us recall the definitions of several subcategories of an abelian category. 
We say that a full subcategory is wide if it is closed under kernels, cokernels and extensions. 
A Serre subcategory is defined to be a wide subcategory which is closed under subobjects.  
Let $R$ be a commutative noetherian ring and $M$ be an $R$-module. 
We denote by $\mathfrak mathrm{Mod}(R)$ the category of $R$-modules and $R$-homomorphisms and by $\mathfrak mathrm{mod}(R)$ the full subcategory consisting of finitely generated $R$-modules. 
We also denote by $\mathfrak mathrm{Spec}R$ the set of prime ideals of $R$ and by $\mathfrak mathrm{Ass} _R M$ the set of associated prime ideals of $M$. 

Classifying subcategories of a module category also has been studied by many authors. 
Classically, Gabriel \cite{G62} gives a bijection between the set of Serre subcategories of $\mathfrak mathrm{mod}(R)$ and the set of specialization closed subsets of $\mathfrak mathrm{Spec}R$. 
Recently, the following result was proved by Takahashi \cite{T08} and Krause \cite{K08}.
Takahashi \cite{T08} first proved the theorem and Krause \cite{K08} generalized it to the category of arbitrary modules which is closed under submodules, extensions and direct unions. 
 

\begin{theorem}\cite[Theorem 4.1]{T08}\cite[Corollary 2.6]{K08}\mathfrak label{Theorem B}
Let $R$ be a noetherian ring. 
Then we have the following 1-1 correspondences;
{\small 
$$
\begin{array}{ccc}
\begin{CD}
&\mathfrak left\{ 
\begin{array}{c}
\text{subcategories of}\ \mathfrak mathrm{mod}(R) \ \text{closed under} \\ 
\text{submodules and extensions}\ 
\end{array}
\right\}
&\begin{matrix}
@>{\mathfrak Psi }>> \\ @<<{\mathfrak Phi }<
\end{matrix}
&\mathfrak left\{ 
\begin{array}{c}
\text{subsets of}\ \mathfrak mathrm{Spec}R 
\end{array}
\right\} \\
&\begin{matrix}
@AA{\subseteq}A 
\end{matrix}
&
&\begin{matrix}
@AA{\subseteq}A 
\end{matrix}
\\
&\mathfrak left\{ 
\begin{array}{c}
\text{Serre subcategories}\\ 
\text{of}\ \mathfrak mathrm{mod}(R) 
\end{array}
\right\}
&\begin{matrix}
@>{\mathfrak Psi }>> \\ @<<{\mathfrak Phi }<
\end{matrix}
&\mathfrak left\{ 
\begin{array}{c}
\text{specialization closed}\\
\text{subsets of}\ \mathfrak mathrm{Spec}R 
\end{array}
\right\}
\end{CD}
\end{array}
$$
}
where $\mathfrak Psi ({\mathfrak mathcal M}) = \cup _{M \in {\mathfrak mathcal M}} \mathfrak mathrm{Ass}_{R} M$ and $\mathfrak Phi ({\mathfrak mathcal S}) = \{ M \in \mathfrak mathrm{mod} | \mathfrak mathrm{Ass} _{R} M \subseteq {\mathfrak mathcal S} \}$. 
\end{theorem}

In addition, Takahashi \cite{T08} pointed out a property concerning wide subcategories of $\mathfrak mathrm{mod}(R)$. 
Actually he proved the following theorem. 

\begin{theorem}\cite[Theorem 3.1, Corollary 3.2]{T08}\mathfrak label{Theorem A}
Let $R$ be a noetherian ring. 
Then every wide subcategory of $\mathfrak mathrm{mod}(R)$ is a Serre subcategory of $\mathfrak mathrm{mod}(R)$.
\end{theorem}

\mathfrak noindent
It is worth nothing that Hovey \cite{Ho01} proved the theorem by using the Hopkins-Neeman-Thomason theorem, but in the case when $R$ is a quotient ring of a coherent regular ring by a finitely generated ideal. 

In the present paper we want to consider the artinian analogue of these results. 

In section \ref{Coh}, we consider wide subcategories of artinian modules. 
We shall show that the artinian analogue of Theorem \ref{Theorem A} also holds. 

\begin{theorem}[Theorem \ref{Theorem Coh}]
Let $R$ be a noetherian ring. 
Then every wide subcategory of $\mathfrak mathrm{Art}(R)$ is a Serre subcategory of $\mathfrak mathrm{Art}(R)$. 
\end{theorem} 

In section \ref{Class}, we propose to classify Serre subcategories of artinian modules. 
We consider some completion of a ring (see Proposition \ref{isom}), so that all of artinian modules can be regarded as modules over it. 
We classify Serre subcategories in terms of a specialization closed subset of the set consisting of closed prime ideals of the completed ring.

\begin{theorem}[Theorem \ref{classification}]
Let $R$ be a noetherian ring. 
Then one has an isomorphism
$$
\begin{array}{l}
\{ \ \text{subcategories of}\ \mathfrak mathrm{Art}(R) \ \text{closed under quotient modules and extensions}\ \} \\ 
\mathfrak qquad \cong \{ \ \text{subsets of the set consisting of closed prime ideals of }\ \hat{R} \ \}.  
\end{array}
$$
Moreover this induces the isomorphism
$$
\begin{array}{l}
\{ \ \text{Serre subcategories of}\ \mathfrak mathrm{Art}(R) \ \} \\ 
\mathfrak qquad \cong \mathfrak left\{ 
\begin{array}{c}
\text{specialization closed subsets of}\\
\text{the set consisting of closed prime ideals of }\ \hat{R} 
\end{array}
\right\}.
\end{array}
$$
\end{theorem}

In this paper, we always assume that $R$ is a commutative ring with identity, and by a subcategory we mean a nonempty full subcategory which is closed under isomorphism.  	

\section{wide subcategories of artinian modules}\mathfrak label{Coh}

In this section, we investigate wide subcategories of artinian modules. 
First we recall the definitions of the categories. 

\begin{definition}
A subcategory of an abelian category is said to be a wide subcategory if it is closed under kernels, cokernels and extensions. 
We also say that a subcategory is a Serre subcategory if it is a wide subcategory which is closed under subobjects.
\end{definition}

Let $M$ be an artinian $R$-module. 
We denote by $\mathfrak mathrm{Soc} (M)$ the sum of simple submodules of $M$. 
Since $\mathfrak mathrm{Soc} (M)$ is also artinian, there exist only finitely many maximal ideals $\mathfrak m$ of $R$ for which $\mathfrak mathrm{Soc} (M)$ has a submodule isomorphic to $R/\mathfrak m$. 
Let the distinct such maximal ideals be $\mathfrak m_1, \cdots ,  \mathfrak m_s$. 
Set $J_M = \bigcap _{i = 1}^{s}\mathfrak m $ and $\hat{R}M = \varprojlim R/J_{M}^n$. 

\begin{lemma}\cite[Lemma 2.2]{S92}\mathfrak label{Sharp}
Each non-zero element $m \in M$ is annihilated by some power of $J_M$. 
Hence $M$ has the natural structure of a module over $\hat{R}M$ in such a way that a subset of $M$ is an $R$-submodule if and only if it is an $\hat{R}M$-submodule. 
\end{lemma}

\begin{proof}
Although a proof of the lemma is given in \cite{S92}, we need in the present paper how the $\hat{R}M$-module structure is defined for an artinian module $M$. 
For this reason we briefly recall the proof of the lemma.

Since $\mathfrak mathrm{Soc} (M) = \oplus _{i=1}^{s}(R/\mathfrak m _{i})^{n_{i}}$, $M$ can be embedded in $\oplus _{i=1}^{s}(E_{R}(R/\mathfrak m _{i}))^{n_{i}}$ where $E_{R}(R/\mathfrak m )$ is an injective hull of $R/\mathfrak m$. 
Note that an element of $E_{R}(R/\mathfrak m)$ is annihilated by some power of $\mathfrak m$.  
Hence one can show that each element of $M$ is annihilated by some power of $ \mathfrak m _1 \cdots \mathfrak m _s = J_M$.  

Let $x \in M$ and $\hat{r} = (r_n + J_M ^n)_{n \in \mathfrak mathbb N} \in \hat{R}M$. 
Suppose that $J _M ^k x = 0$. 
It is straightforward to check that $M$ has the structure of an $\hat{R}M$-module such that $\hat{r}x = r_{k}x$. 

\end{proof}

\begin{remark}
As shown in the proof of Lemma \ref{Sharp},  $M$ can be embedded in $\oplus _{i=1}^{s}(E_{R}(R/\mathfrak m _{i}))^{n_{i}}$. 
Thus the maximal ideals $\mathfrak m_1, \cdots ,  \mathfrak m_s$ are just associative prime ideals of $M$ since $\mathfrak mathrm{Ass} _R \oplus _{i=1}^{s}(E_{R}(R/\mathfrak m _{i}))^{n_{i}} = \mathfrak mathrm{Ass} _R \mathfrak mathrm{Soc} (M) = \{ \mathfrak m_1, \cdots ,  \mathfrak m_s \}$. 

\end{remark}

By virtue of Lemma \ref{Sharp}, each artinian $R$-module can be regarded as a module over some complete semi-local ring. 
We note that Matlis duality theorem is allowed over a noetherian complete semi-local ring (cf. \cite[Theorem 1.6]{O76}). 
It is the strategy of the paper that we replace the categorical property on a subcategory of finitely generated (namely, noetherian) modules with that of artinian modules by using Matlis duality.    
We denote by $\mathfrak mathrm{Art}(R)$ the subcategory consisting of artinian $R$-modules.

\begin{lemma}\mathfrak label{duality}
Let $(R, \mathfrak m _{1}, \cdots , \mathfrak m _{s})$ be a noetherian complete semi-local ring and set $E = \oplus _{i = 1}^{s} E_{R}(R/\mathfrak m_{i})$. 
For each subcategory $\mathcal{X}$ of $\mathfrak mathrm{Mod}(R)$, we denote by $\mathcal{X} ^\vee = \{ M^{\vee}\ |\ M \in \mathcal{X} \}$ where $(-)^\vee = \mathfrak mathrm{Hom} _{R}(- , E)$. 
Then the following assertions hold.
\begin{itemize}
\item[(1)] If $\mathcal{X}$ is a subcategory of $\mathfrak mathrm{Art}(R)$ (resp. $\mathfrak mathrm{mod}(R)$) which is closed under quotient modules (resp. submodules) and extensions, then $\mathcal{X} ^\vee$ is a subcategory of $\mathfrak mathrm{mod}(R)$ (resp. $\mathfrak mathrm{Art}(R)$) which is closed under submodules (resp. quotient modules) and extensions.  

\item[(2)] If $\mathcal{X}$ is a wide subcategory of $\mathfrak mathrm{Art}(R)$ (resp. $\mathfrak mathrm{mod}(R)$), then $\mathcal{X} ^\vee$ is also a wide subcategory of $\mathfrak mathrm{mod}(R)$ (resp. $\mathfrak mathrm{Art}(R)$). 

\item[(3)] If $\mathcal{X}$ is a Serre subcategory of $\mathfrak mathrm{Art}(R)$ (resp. $\mathfrak mathrm{mod}(R)$), then $\mathcal{X} ^\vee$ is also a Serre subcategory of $\mathfrak mathrm{mod}(R)$ (resp. $\mathfrak mathrm{Art}(R)$). 
\end{itemize}
\end{lemma}

\begin{proof}
Since Matlis duality theorem is allowed over a noetherian complete semi-local ring, the assertions hold by Matlis duality. 

\end{proof}

\begin{definition}\mathfrak label{coh step}
Let $M$ be an $R$-module. 
For a nonnegative integer $n$, we inductively define a subcategory $\mathfrak mathrm{Wid} ^{n}_{R} (M)$ of $\mathfrak mathrm{Mod}(R)$ as follows:

\begin{itemize}
\item[(1)] Set $\mathfrak mathrm{Wid}_{R} ^{0}(M) = \{ M \}$.

\item[(2)] For $n \geq  1$, let $\mathfrak mathrm{Wid}_{R} ^{n} (M)$ be a subcategory of $\mathfrak mathrm{Mod}(R)$ consisting of all $R$-modules $X$ having an exact sequence of either of the following three forms: 
$$
\begin{array}{l}
A \to B \to X \to 0, \\
0 \to X \to A \to B, \\ 
0 \to A \to X \to B \to 0 
\end{array}
$$
where $A, B \in \mathfrak mathrm{Wid}_{R} ^{n-1} (M)$.
\end{itemize}
\end{definition}

\begin{remark}\mathfrak label{remark coh step}
Let $M$ be an $R$-module and $n$ be a nonnegative integer. 
Then the following hold. 
\begin{itemize}
\item[(1)] There is an ascending chain $\{ M \} = \mathfrak mathrm{Wid}_{R} ^{0}(M) \subseteq \mathfrak mathrm{Wid}_{R} ^{1} (M) \subseteq \cdots \subseteq \mathfrak mathrm{Wid}_{R} ^{n} (M) \subseteq \cdots \subseteq \mathfrak mathrm{Wid}_{R} (M)$ of subcategories $\mathfrak mathrm{Mod}(R)$, where $\mathfrak mathrm{Wid}_{R} (M)$ is the smallest wide subcategory of $\mathfrak mathrm{Mod}(R)$ which contains $M$.

\item[(2)] $\bigcup _{n \geq 0}\mathfrak mathrm{Wid}_{R} ^{n}(M)$ is wide and the equality $\mathfrak mathrm{Wid}_{R} (M) = \bigcup _{n \geq 0}\mathfrak mathrm{Wid}_{R} ^{n}(M)$ holds.
\end{itemize}
\end{remark}

\begin{definition}
Let $J$ be an ideal of $R$. 
For each $R$-module $M$, we denote by $\Gamma _{J} (M)$ the set of elements of $M$ which are annihilated by some power of $J$, namely $\Gamma _{J} (M) = \bigcup _{n \in \mathfrak mathbb N}( 0 :_M J^n )$.
An $R$-module $M$ is said to be $J$-torsion if $M = \Gamma _{J} (M)$. 
We denote by $\mathfrak mathrm{Mod} _{J}(R)$ the subcategory consisting of $J$-torsion $R$-modules. 
\end{definition}

\begin{lemma}\mathfrak label{lemma J-torsion}
For each object $M$ in $\mathfrak mathrm{Mod} _{J}(R)$, $M$ has the structure of an $\hat{R}^{(J)}$-module where $\hat{R}^{(J)}$ is a $J$-adic completion of $R$. 
\end{lemma}

\begin{remark}
By using an inductive argument on $n$, we can show that if $M$ is artinian (resp. $J$-torsion), then $\bigcup _{n \geq 0}\mathfrak mathrm{Wid}_{R} ^{n}(M)$, hence $\mathfrak mathrm{Wid}_{R} (M)$, is a subcategory of $ \mathfrak mathrm{Art}(R)$ (resp. $\mathfrak mathrm{Mod} _{J}(R)$) since $\mathfrak mathrm{Art}(R)$ (resp. $\mathfrak mathrm{Mod} _{J}(R)$) is a wide subcategory. 
\end{remark}

\begin{corollary}\mathfrak label{corollary coh}
Let $M$ be an artinian $R$-module. 
Then $\mathfrak mathrm{Wid}_{R} (M)$ and $\mathfrak mathrm{Wid} _{\hat{R}M} (M)$ are equivalent as subcategories of $\mathfrak mathrm{Art} (\hat{R}M )$. 
\end{corollary}

\begin{proof}
As remarked above, since $M$ is $J_M$-torsion, we can naturally identify $\mathfrak mathrm{Wid}_{R} (M)$ with a subcategory of $\mathfrak mathrm{Mod} _{J_M}(R)$. 
It is also a subcategory of $\mathfrak mathrm{Art} (\hat{R}M )$ by Lemma \ref{lemma J-torsion}. 

\end{proof}

\begin{theorem}\mathfrak label{Theorem Coh}
Let $R$ be a noetherian ring. 
Then every wide subcategory of $\mathfrak mathrm{Art} (R)$ is a Serre subcategory of $\mathfrak mathrm{Art} (R)$. 
\end{theorem}

\begin{proof}
Let $\mathcal{X}$ be a wide subcategory of $\mathfrak mathrm{Art}(R)$. 
It is sufficiently to show that $\mathcal{X}$ is closed under submodules. 
Assume that $\mathcal{X}$ is not closed under submodules. 
Then there exists an $R$-module $X$ in $\mathcal{X}$ and $R$-submodule $M$ of $X$ such that $M$ does not belong to $\mathcal{X}$. 
Applying Lemma \ref{Sharp} to $X$, $X$ is a module over the complete semi-local ring $\hat{R} : = \hat{R} ^{(J_X)}$ and $M$ is an $\hat{R}$-submodule of $X$. 
Now we consider the wide subcategory $\mathfrak mathrm{Wid}_{R} (X)$. 
By virtue of Corollary \ref{corollary coh}, $\mathfrak mathrm{Wid}_{R} (X) = \mathfrak mathrm{Wid}_{R}R (X)$ as a subcategory of $\mathfrak mathrm{Art} (\hat{R})$. 
Since $\hat{R}$ is a complete semi-local ring, by Matlis duality, we have the equivalence of the categories $\mathfrak mathrm{Wid}_{R}R (X) \cong \{ \mathfrak mathrm{Wid}_{R}R (X)^{\vee } \} ^{op} \cong \mathfrak mathrm{Wid}_{R}R (X^{\vee })^{op}$ where $( - ) ^{\vee} = \mathfrak mathrm{Hom} _{\hat{R}}(-, E_{\hat{R}}(\hat{R}/J_{X}\hat{R}))$. 
Since $\mathfrak mathrm{Wid}_{R}R (X^{\vee })$ is a wide subcategory of finitely generated $\hat{R}$-modules, it follows from Theorem \ref{Theorem A} that $\mathfrak mathrm{Wid}_{R}R (X^{\vee })$ is a Serre subcategory. 
Thus $M^{\vee}$ is contained in $\mathfrak mathrm{Wid}_{R}R (X^{\vee })$. 
Using Matlis duality again, we conclude that $M$ must be contained in $\mathfrak mathrm{Wid}_{R}R (X) = \mathfrak mathrm{Wid}_{R} (X)$, hence also in $\mathcal{X}$. 
This is a contradiction, so that $\mathcal{X}$ is closed under submodules. 

\end{proof}

\section{Classifying subcategories of artinian modules}\mathfrak label{Class}

In this section, we shall give the artinian analogue of the classification theorem of subcategories of finitely generated modules (Theorem \ref{classification}).  
First, we state the notion and the basic properties of attached prime ideals which are play a key role of our theorem. 
For the detail, we recommend the reader to refer to \cite{S76, S92} and \cite[\S 6 Appendix.]{M}.

\begin{definition}\mathfrak label{def of secondary}
Let $M$ be an $R$-module. 
We say that $M$ is secondary if for each $a \in R$ the endomorphism of $M$ defined by the multiplication map by $a$ is either surjective or nilpotent. 
\end{definition}

\begin{remark}
If $M$ is secondary then $\mathfrak p = \sqrt{\mathfrak mathrm{ann} _R (M)}$ is a prime ideal and $M$ is said to be $\mathfrak p$-secondary.
\end{remark}

\begin{definition}
$M = S_1 + \cdots + S_r$ is said to be a secondary representation if $S_i$ is a secondary submodule of $M$ for all $i$. 
And we also say that the representation is minimal if the prime ideals $\mathfrak p _i  = \sqrt{\mathfrak mathrm{ann} _R (S_i)}$ are all distinct, and none of the $S_i$ is redundant 
\end{definition}  

 

\begin{definition}\mathfrak label{def of attached prime}
A prime ideal $\mathfrak p$ is said to be an attached prime ideal of $M$ if $M$ has a $\mathfrak p$-secondary quotient. 
We denote  by $\mathfrak mathrm{Att}R M$ the set of the attached prime ideals of $M$. 
\end{definition}

\begin{remark}\mathfrak label{remark attach}
Let $M$ be an $R$-module. 
\begin{itemize}
\item[(1)] If $M = S_1 + \cdots + S_r$ is a minimal representation and $p _i = \sqrt{\mathfrak mathrm{ann} _R (S_i )}$. then $\mathfrak mathrm{Att}R M = \{ \mathfrak p _1 , \cdots , \mathfrak p _r \}$. 
See \cite[Theorem 6.9]{M}.

\item[(2)]Let $M$ be an $R$-module. 
Given a submodule $N \subseteq M$, we have
$$
\mathfrak mathrm{Att}R M/N \subseteq \mathfrak mathrm{Att}R M \subseteq \mathfrak mathrm{Att}R (N) \cup \mathfrak mathrm{Att}R M/N. 
$$
See \cite[Theorem 6.10]{M}.

\item[(3)]It is known that if $M$ is artinian then $M$ has a secondary representation. 
Thus it has a minimal one. 
See \cite[Theorem 6.11]{M}. 

\end{itemize}
\end{remark}

Let $(R, \mathfrak m)$ be a noetherian local ring and $\mathcal{X}$ a Serre subcategory of $\mathfrak mathrm{Art}(R)$. 
By virtue of Lemma \ref{Sharp}, $\mathfrak mathrm{Art}(R)$ is equivalent to $\mathfrak mathrm{Art} (\hat{R} )$ where $\hat{R}$ is an $\mathfrak m$-adic completion of $R$. 
Now we consider $\mathcal{X}$ as a subcategory of $\mathfrak mathrm{Art} (\hat{R})$. 
Since $\mathcal{X}^\vee$ is a Serre subcategory of $\mathfrak mathrm{mod} (\hat{R})$ (Lemma \ref{duality}), $\mathcal{X} ^{\vee}$, hence $\mathcal{X}$, corresponds to the specialization closed subset of $\mathfrak mathrm{Spec} \ \hat{R}$ by Theorem \ref{Theorem B}. 
Namely there is the bijection between the set of Serre subcategories of $\mathfrak mathrm{Art}(R)$ and the set of specialization closed subsets of $\mathfrak mathrm{Spec} \ \hat{R}$.  
This observation provides us that we should consider a larger set than $\mathfrak mathrm{Spec} \ R$ to classify subcategories of artinian modules. 

In the rest of this section, we always assume that $R$ is a noetherian ring.

As mentioned in Lemma \ref{Sharp}, we can determine some complete semi-local rings for each artinian module respectively, so that the artinian module has the module structure over such a completed ring. 
Now we attempt to treat all of artinian $R$-modules as modules over the same completed ring.  
For this, we consider the following set of ideals of $R$: 
$$
\mathcal T = \{ \ I \ | \  \text{a length of}\ R/I \ \text{is finite}\ \} . 
$$

The set $\mathcal T$ forms a directed set ordered by inclusion. 
Then we can consider the inverse system $\{ R/I, f_{I, I'} \}$ where $f_{I, I'}$ are natural surjections. 
That is, 
$$
I, I' \in \mathcal T \ \text{and}\  I' \subseteq I\Rightarrow  f_{I,I'}: R/I' \to R/I .
$$
We denote $\varprojlim _{I \in \mathcal T}R/I $ by $\hat{R}_{\mathcal T}$. 

\begin{lemma}\mathfrak label{extended sharp}
Every artinian $R$-module has the structure of an $\hat{R}TT$-module in such a way that a subset of an artinian $R$-module $M$ is an $R$-submodule if and only if it is an $\hat{R}TT$-submodule. 
Consequently, we have an equivalence of categories $\mathfrak mathrm{Art} (R) \cong \mathfrak mathrm{Art} (\hat{R}TT )$. 
\end{lemma}

\begin{proof}
The proof of the first part of the lemma will go through similarly to the proof of Lemma \ref{Sharp}. 
The last part of the lemma holds from the definition of the $\hat{R}TT$-module structure. 
\end{proof}

We set another family of ideals of $R$ as  
$$
\mathcal J = \{ \ \mathfrak m _{1}^{k_1} \cdots \mathfrak m_{s}^{k_s} \ | \ \mathfrak m _i  \ \text{is a maximal ideal of}\ R, \ k_i \in \mathfrak mathbb N \} . 
$$ 
It is also a directed set ordered by inclusion and we denote by $\hat{R}JJ$ its inverse limit on the system via natural surjections. 

\begin{proposition}\mathfrak label{1st isom}
There is an isomorphism of topological rings; 
$$
\hat{R}TT \cong \hat{R}JJ .
$$  
\end{proposition}

\begin{proof}
Let $I$ be an ideal in $\mathcal T$. 
Note that $\mathfrak mathrm{Ass} _R R/I = \{ \mathfrak m _1 , \cdots , \mathfrak m _s \}$ for some maximal ideals $\mathfrak m _i$ of $R$. 
Since $\mathfrak m _i$ are finitely generated, there exists a positive integer $k$ such that $(\mathfrak m _1 \cdots \mathfrak m_s )^k \subseteq I$. 
Thus, for each ideal $I$ in $\mathcal T$, we can take some ideal $J$ in $\mathcal J$ such that $J \subseteq I$. 
Hence $\mathcal T$ and $\mathcal J$ give the same topology on $R$, so that $\hat{R}TT \cong \hat{R}JJ$ as topological rings.  

\end{proof}

Now we consider a direct product of rings 
$$
\mathfrak prod _{\mathfrak n \in \mathfrak max (R)} \hat{R}n 
$$
where $\mathfrak max (R)$ is the set of maximal ideals of $R$ and $\hat{R}m$ is an $\mathfrak m$-adic completion of $R$. 
We regard the ring as a topological ring by a product topology, namely the linear topology defined by ideals which are of the form $\mathfrak m _{1}^{k_1}\hat{R} _{\mathfrak m _{1}} \times \cdots \times \mathfrak m _{s}^{k_s}\hat{R} _{\mathfrak m _{s}} \times \mathfrak prod _{\mathfrak n \mathfrak not= \mathfrak m_1, \cdots , \mathfrak m_s} \hat{R} _{\mathfrak n} $ for some $\mathfrak m _{i} \in \mathfrak max (R)$ and $k_i \in \mathfrak mathbb N$.   

\begin{proposition}\mathfrak label{2nd isom}\cite[\S 2.13. Proposition 17]{Bou}
There is an isomorphism of topological rings 
$$
\hat{R}JJ \cong \mathfrak prod _{\mathfrak n \in \mathfrak max (R)} \hat{R}n .
$$
\end{proposition}

\begin{proof}
Let $J$ be an ideal in $\mathcal J$ and suppose that $J = \mathfrak m _{1}^{k_1} \cdots \mathfrak m _{s}^{k_s}$. 
Note that $R/J$ is isomorphic to $\mathfrak prod _{i=1}^{s}R/\mathfrak m _{i}^{k_i}$ by Chinese remainder theorem. 
Let us set $A = \mathfrak prod _{\mathfrak n \in \mathfrak max (R)} \hat{R}n$. 
For all $J \in \mathcal J$, we define mappings $\varphi _{J} : A \to R/J$ by 
$$
\varphi _{J} : A \to R/J \cong \mathfrak prod _{i=1}^{s}R/\mathfrak m _{i}^k ; \mathfrak quad (\hat{a}_\mathfrak m )  \to (\bar{a_{\mathfrak m _1 ^{k_1}}},  \cdots , \bar{a_{\mathfrak m _s ^{k_s}}}). 
$$
Here we denote $(a _{\mathfrak m ^k} + \mathfrak m ^k) \in \hat{R}m$ by $\hat{a}_\mathfrak m $. 
It is easy to see that $\varphi = \{ \varphi _{J} \} _{J \in \mathcal J}$ is a morphism from $A$ to $\hat{R}JJ$. 
Write $p _J : \hat{R}JJ \to R/J$ for the projection.
We note that the topology of $\hat{R}JJ$ coincides with the linear topology defined by $\{ \mathfrak mathrm{Ker} \ p _J \} _{J \in \mathcal J}$ (cf. \cite[\S 8]{M}). 
Set $V_J = \ker \ p_J$.
For each $V_J$, we take the open set $W_J = \mathfrak m _{1}^{k_1}\hat{R} _{\mathfrak m _{1}} \times \cdots \times \mathfrak m _{s}^{k_s}\hat{R} _{\mathfrak m _{s}} \times \mathfrak prod _{\mathfrak n \mathfrak not= \mathfrak m_1, \cdots , \mathfrak m_s} \hat{R} _{\mathfrak n} $ in $A$.   
Then $p_J \circ \varphi (W_J ) = 0$. 
Thus $\varphi (W_J ) \subseteq V_J$, so that $\varphi$ is continuous. 

For each ideal $J \in \mathcal J$, we take an ideal $W_J$ of $A$ as above. 
As mentioned before, $A$ has a linear topology defined by $\{ W_J \} _{J \in \mathcal J}$, and $\varprojlim A/W_J = A$. 
We define mappings $\mathfrak psi _{J} : \hat{R}JJ \to A/W_J$ by 
$$
\mathfrak psi _{J} : \hat{R}JJ \to A/W_J \cong \mathfrak prod _{i=1}^{s}R/\mathfrak m _{i}^k \cong R/J ; \mathfrak quad (a_J + J)_{J \in \mathcal J} \to \bar{a_{J}}. 
$$
We also see that $\mathfrak psi _{J}$ induces the morphism $\mathfrak psi = \{ \mathfrak psi _{J} \} _{J \in \mathcal J}: \hat{R}JJ \to A$ which is a continuous mapping. 
In fact, $W_J$ is just a kernel of the natural projection $A \to A/W_J$ and $\mathfrak psi (V_J )$ goes to $0$ via the projections.  

Finally, we shall show $\varphi \circ \mathfrak psi = 1_{\hat{R}JJ}$ and $\mathfrak psi \circ \varphi = 1_{A}$, but this is clear from the definition of $\varphi$ and $\mathfrak psi$. 

\end{proof}

Combining Proposition \ref{1st isom} with Proposition \ref{2nd isom}, we can show the following. 

\begin{corollary}\mathfrak label{isom}
There are isomorphisms of topological rings 
$$
\hat{R}TT \cong \hat{R}JJ \cong \mathfrak prod _{\mathfrak n \in \mathfrak max (R)} \hat{R}n .
$$  
\end{corollary}

For closed prime ideals of $\mathfrak prod _{\mathfrak n \in \mathfrak max (R)} \hat{R}n$, we have the following result. 

\begin{proposition}\mathfrak label{closed prime}
Every proper closed prime ideal of $\mathfrak prod _{\mathfrak n \in \mathfrak max (R)} \hat{R}n$ is of the form $\mathfrak p \times \mathfrak prod _{\mathfrak n \in \mathfrak max (R), \mathfrak m \mathfrak not= \mathfrak n  } \hat{R} _{\mathfrak n}$ for some prime ideal $\mathfrak p \in \mathfrak mathrm{Spec} \ \hat{R}m$. 
Hence we can identify the set of closed prime ideals of $\mathfrak prod _{\mathfrak n \in \mathfrak max (R)} \hat{R}n$ with the disjoint union of $\mathfrak mathrm{Spec} \ \hat{R}m$, i.e. $\coprod _{\mathfrak n \in \mathfrak max (R)} \mathfrak mathrm{Spec} \ \hat{R}n$.
\end{proposition}

\begin{proof}
Let us set $A = \mathfrak prod _{\mathfrak n \in \mathfrak max (R)} \hat{R}n$. 
We take an element $\hat{e}_\mathfrak m = (\hat{e}_{\mathfrak m, \mathfrak n})$ of $A$ defined by 
$$
\hat{e}_{\mathfrak m, \mathfrak n}  =
\mathfrak left\{
\begin{array}{ll}
\hat{1},&\text{if} \ \mathfrak m = \mathfrak n , \\
0,&\text{otherwise}.
\end{array}
\right.
$$
Let $\mathfrak P$ be an arbitrary closed prime ideal of $A$.
Since $\hat{e}_\mathfrak l \cdot \hat{e}_\mathfrak m = 0$, we have $\hat{e}_\mathfrak l \cdot \hat{e}_\mathfrak m \in \mathfrak P$. 
Thus if there is a maximal ideal $\mathfrak m$ such that $\hat{e}_\mathfrak m$ is not contained in $\mathfrak P$, $\hat{e}_\mathfrak l$ is in $\mathfrak P$ whenever $\mathfrak l$ is not equal to $\mathfrak m$.   

Suppose that $\hat{e}_\mathfrak m \mathfrak not \in \mathfrak P$. 
Then $\mathfrak P$ contains $\hat{e}_\mathfrak l$ for all maximal ideals $\mathfrak l \mathfrak not = \mathfrak m$. 
First we shall show the family $(\hat{e}_{\mathfrak l} )$ where $\mathfrak l$ runs through all maximal ideals of $R$ except $\mathfrak m$ is summable in $A$. 
Namely the sum $\varepsilon = \Sigma \hat{e}_\mathfrak l$ is an element of $A$. 
For each neighborhood $W_J $ (of $0$) in $A$, we take a finite set of maximal ideals $H_{J} = \{ \mathfrak m_1 , \cdots , \mathfrak m_s , \mathfrak m \}$. 
Then one can show that 
$$
\Sigma _{\mathfrak l \in H} \hat{e}_\mathfrak l \in W _J
$$
for every finite set of maximal ideals $H$ which does not intersect with $H_{J}$. 
Thus our claim follows from Cauchy's criterion (\cite[Chapter 3, \S 5, no. 2, Theorem 1]{Bou98}).   
Note that $\varepsilon$ is contained in $\mathfrak P$ since $\mathfrak P$ is a closed ideal. 
Thus $\varepsilon A = 0 \times \mathfrak prod _{\mathfrak n \mathfrak not = \mathfrak m } \hat{R} _{\mathfrak n}$ is an $A$-submodule of $\mathfrak P$. 
Then we have the sequence:
$$
0 \to \mathfrak P / \varepsilon A \to A/ \varepsilon A \cong \hat{R} _{\mathfrak m} \to A/\mathfrak P \to 0.
$$   
Since $\mathfrak P / \varepsilon A$ is a prime ideal of $\hat{R}m$, we conclude that $\mathfrak P$ is of the form $\mathfrak p \times \mathfrak prod _{\mathfrak n \in \mathfrak max (R), \mathfrak m \mathfrak not= \mathfrak n  } \hat{R} _{\mathfrak n}$ for some prime ideal $\mathfrak p \in \mathfrak mathrm{Spec} \ \hat{R}m$. 

Suppose that all of elements $\hat{e}_\mathfrak m$ are contained in $\mathfrak P$. 
Then we can easily show that $\mathfrak P = A$ and this is a contradiction. 

\end{proof}

By virtue of Corollary \ref{isom}, we can equate the rings $\hat{R}TT$, $\hat{R}JJ$ and $\mathfrak prod _{\mathfrak n \in \mathfrak max (R)} \hat{R}n$. 
In the rest of this paper we always denote them by $\hat{R}$ and identify the set of closed prime ideals of $\hat{R}$ with $\coprod _{\mathfrak n \in \mathfrak max (R)} \mathfrak mathrm{Spec} \ \hat{R}n$.

Let $M$ be an artinian $R$-module. 
It follows from Lemma \ref{extended sharp} that $M$ is also an artinian $\hat{R}$-module. 

\begin{proposition}\mathfrak label{closedness} 
Let $M$ be an artinian $R$-module. 
Then $\mathfrak mathrm{ann} _{\hat{R}} (M)$ is a closed ideal of $\hat{R}$. 
\end{proposition}

\begin{proof}
We denote by $U_I$ a kernel of the natural projection $\hat{R}TT \to R/I$ for each ideal $I \in \mathcal T$. 
It suffices to prove that the inclusion $\mathfrak mathrm{ann} _{\hat{R}} (M) \supseteq \cap _{I \in \mathcal T} (\mathfrak mathrm{ann} _{\hat{R}} (M) + U_I)$ holds. 
Take an arbitrary element $\hat{a} \in \cap _{I \in \mathcal T} (\mathfrak mathrm{ann} _{\hat{R}} (M) + U_I )$. 
Then there exist some elements $\hat{b}_{I} \in \mathfrak mathrm{ann} _{\hat{R}} (M)$ and $\hat{c}_{I} \in U_I$ such that $\hat{a} = \hat{b}_{I} + \hat{c}_{I}$ for all $I$. 
Let $x$ be an element of $M$. 
Then there exists some ideal $I \in \mathcal T$ such that $Ix = 0$. 
Thus
$$
\hat{a}x = (\hat{b}_{I} + \hat{c}_{I})x = \hat{c}_{I}x = c_{I, I} x \subseteq Ix = 0. 
$$
Hence $\hat{a}$ is an annihilator of $M$. 

\end{proof}

\begin{remark}\mathfrak label{closedness radiacl}
Under the same assumption in Proposition \ref{closedness}, a radical of $\mathfrak mathrm{ann} _{\hat{R}}(M)$ is also a closed ideal. 
In fact, let $\hat{a}$ be an element of a closure of $\sqrt{\mathfrak mathrm{ann} _{\hat{R}}(M)}$. 
Take $x \in M$ and suppose that $Ix = 0$ for some $I \in \mathcal T$. 
Since $\hat{b}_{I} \in \sqrt{\mathfrak mathrm{ann} _{\hat{R}} (M)}$, $\hat{b}_{I}^k \in \mathfrak mathrm{ann} _{\hat{R}} (M)$. 
Hence we see that $\hat{a}^k x = (\hat{b}_{I} + \hat{c}_{I})^k x = 0$ holds, so that $\hat{a} \in \sqrt{\mathfrak mathrm{ann} _{\hat{R}} (M)}$.
Consequently, $\mathfrak mathrm{Att} _{\hat{R}} M$ is a subset of the set of closed prime ideals of $\hat{R}$. 
\end{remark}

\begin{lemma}\cite[Exercise 8.49]{S00}\mathfrak label{torsion decomp}
Let $M$ be an artinian $R$-module. 
Set $\mathfrak mathrm{Ass}R M = \{ \mathfrak m _{1}, \dots , \mathfrak m _{s} \}$.
Then $M$ is direct sums of $\Gamma _{\mathfrak m _i}(M)$, namely $M = \oplus _{i=1}^{s}\Gamma _{\mathfrak m _i}(M)$. 
\end{lemma}

\begin{proof}
It is clear that $M$ contains $\Sigma _{i=1}^{s} \Gamma _{\mathfrak m _i}(M)$. 

For each element $x \in M$, there is some positive integer $k$ such that $J^k x = 0$ where $J = \mathfrak m _{1} \dots \mathfrak m _{s}$. 
Since $\mathfrak m _{1}, \dots \mathfrak m _{s}$ are all distinct maximal ideals, we have 
$$
\mathfrak m _{1}^{k} \mathfrak m _{2}^{k} \cdots \mathfrak m _{s-1}^{k} + \mathfrak m _{1}^{k} \mathfrak m _{2}^{k} \cdots \mathfrak m _{s-2}^{k}\mathfrak m _{s}^{k} + \cdots + \mathfrak m _{2}^{k} \mathfrak m _{3}^{k} \cdots \mathfrak m _{s}^{k} = R.
$$
Thus there are elements $r_i \in \mathfrak m _{1}^{k}\cdots \mathfrak m _{i-1}^{k} \mathfrak m _{i+1}^{k} \cdots \mathfrak m _{s}^{k}$ such that $\Sigma _{i=1}^{s}r_i = 1$, and we get the equality $x = \Sigma _{i=1}^{s}r_{i}x$. 
Then we can show that each $r_i x$ is an element of $\Gamma _{\mathfrak m _i}(M)$. 
In fact,  
$$
\mathfrak m _{i}^{k}r_{i}x \subseteq \mathfrak m _{1}^{k} \mathfrak m _{2}^{k} \cdots \mathfrak m _{s}^{k}x = 0. 
$$
Therefore we obtain $M = \Sigma _{i=1}^{s} \Gamma _{\mathfrak m _i}(M)$. 
 

It remains to show the sum above is a direct sum. 
This follows from the facts that $\mathfrak mathrm{Ass}R (\Gamma _{\mathfrak m _{i}}(M) ) = \{ \mathfrak m _i \}$ and all $\mathfrak m _i$ are distinct. 

\end{proof}

Let $M$ be an $\mathfrak m$-torsion $R$-module. 
Then $M$ has the structure of an $\hat{R}$-module and an $\hat{R}m$-module. 
Note that the $\hat{R}m$-module action on $M$ is identical with the action by means of the natural inclusion $\hat{R}m \to \mathfrak prod _{\mathfrak n \in \mathfrak max (R)} \hat{R}n \cong \hat{R}$. 
We also note from Lemma \ref{extended sharp} or Lemma \ref{Sharp} that $N$ is an $\hat{R}$-submodule (resp. a quotient $\hat{R}$-module) of $M$ if and only if it is an $\hat{R}m$-submodule (resp. a quotient $\hat{R}m$-module) of $M$.   

\begin{proposition}\mathfrak label{prop A}
Let $M$ be an $\mathfrak m$-torsion $R$-module. 
Then 
$$\mathfrak mathrm{Att} _{\hat{R}} M = \mathfrak mathrm{Att} _{\hat{R}m} M$$ 
as a subset of $\coprod _{\mathfrak n \in \mathfrak max (R)} \mathfrak mathrm{Spec} \ \hat{R_{\mathfrak n }}$. 
\end{proposition}

\begin{proof}
Let $\mathfrak P \in \mathfrak mathrm{Att} _{\hat{R}} M$ and $W$ be a $\mathfrak P$-secondary quotient $\hat{R}$-module of $M$. 
Note that $W$ is also a quotient $\hat{R}m$-module of $M$. 
As noted in Remark \ref{closedness radiacl}, $\mathfrak P = \sqrt{\mathfrak mathrm{ann} _{\hat{R}} (W)}$ is a closed prime ideal. 
Thus $\mathfrak P$ is of the form $\mathfrak p \times \mathfrak prod _{\mathfrak l \mathfrak not= \mathfrak n} \hat{R} _\mathfrak n$ where $\mathfrak p$ is a prime ideal of $\hat{R} _\mathfrak l$ for some maximal ideal $\mathfrak l$. 
First we shall show $\mathfrak l = \mathfrak m$. 
For this, we show $\hat{e}_\mathfrak n \in \mathfrak P$ if $\mathfrak n \mathfrak not= \mathfrak m$ (see Proposition \ref{closedness} for the definition $\hat{e}_\mathfrak n$). 
Let $x$ be an element of $W$ and suppose that $\mathfrak m ^k x = 0$. 
The $\hat{R}$-module action $\hat{r}x$ is defined by $r_{\mathfrak m^k}x$ for each $\hat{r} \in \hat{R}$. 
Then we see that  
$$
\hat{e}_\mathfrak n x = \hat{e}_{\mathfrak n , \mathfrak m}x = 0_{\mathfrak m^k} x = 0.
$$ 
Thus, if $\mathfrak n \mathfrak not= \mathfrak m$, $e _\mathfrak n$ is contained in $\mathfrak mathrm{ann} _{\hat{R}} (W)$, so that in $\sqrt{\mathfrak mathrm{ann} _{\hat{R}} (W)} = \mathfrak P$. 
Hence $\hat{e}_\mathfrak n \hat{R} = \hat{R} _{\mathfrak n} \subseteq \mathfrak P $ whenever $\mathfrak n \mathfrak not= \mathfrak m$, so that $\mathfrak l$ must be $\mathfrak m$. 
Consequently, $\mathfrak p$ is a prime ideal of $\hat{R}m$. 

Since the $\hat{R}m$-action on $W$ is the same as the action via the natural inclusion $\hat{R}m \to \hat{R}$, we have $\sqrt{\mathfrak mathrm{ann} _{\hat{R}} (W)} \cap \hat{R}m = \sqrt{\mathfrak mathrm{ann} _{\hat{R}m}(W)}$.  
Therefore $\mathfrak p \in \mathfrak mathrm{Att} _{\hat{R}m} M$. 

Conversely, let $\mathfrak q$ be an attached prime ideal of $M$ as $\hat{R}m$ modules and $V$ be a $\mathfrak q$-secondary quotient $\hat{R}m$-module of $M$. 
Then $V$ is also an $\hat{R}$-quotient module of $M$, and $\mathfrak Q = \mathfrak q \times \mathfrak prod _{\mathfrak n \mathfrak not= \mathfrak m}\hat{R} _{\mathfrak n}$ is equal to $\sqrt{\mathfrak mathrm{ann} _{\hat{R}} (V)}$. 
Hence $\mathfrak Q \in \mathfrak mathrm{Att} _{\hat{R}} M$. 

\end{proof}

Combing Proposition \ref{prop A} with Lemma \ref{torsion decomp}, we have the following corollary. 

\begin{corollary}\mathfrak label{cor A}
Let $M$ be an artinian $R$-module. 
Then 
$$
\mathfrak mathrm{Att} _{\hat{R}} M = \coprod _{\mathfrak m \in \mathfrak mathrm{Ass}R M}\mathfrak mathrm{Att} _{\hat{R}m} \Gamma _{\mathfrak m} (M ) 
$$ 
as a subset of $\coprod _{\mathfrak n \in \mathfrak max (R)} \mathfrak mathrm{Spec} \ \hat{R_\mathfrak n}$.
\end{corollary}

Let us state the result which is a key to classify the subcategory of the category of noetherian modules. 

\begin{theorem}\cite[Corollary 4.4]{T08}\cite[Corollary 2.6]{K08}\mathfrak label{Krause}
Let $M$ and $N$ be finitely generated $R$-modules. 
Then $M$ can be generated from $N$ via taking submodules and extension if and only if $\mathfrak mathrm{Ass} _{R} M \subseteq \mathfrak mathrm{Ass} _{R} N$. 
\end{theorem}

The following lemma is due to Sharp \cite{S76}. 

\begin{lemma}\cite[3.5.]{S76}\mathfrak label{att ass}
Let $(R, \mathfrak m _{1}, \cdots , \mathfrak m _{s})$ be a commutative noetherian complete semi-local ring and set $E = \oplus _{i = 1}^{s} E_{R}(R/\mathfrak m_{i})$.
For an artinian $R$-module $M$, we have  
$$
\mathfrak mathrm{Att} _{R} M = \mathfrak mathrm{Ass} _{R} \mathfrak mathrm{Hom} _R (M, E). 
$$ 
\end{lemma}

The next claim is reasonable as the artinian analogue of Theorem \ref{Krause}.

 

\begin{theorem}\mathfrak label{theorem 1}
Let $M$ and $N$ be artinian $R$-modules. 
Then $M$ can be generated from $N$ via taking quotient modules and extensions as $R$-modules if and only if $\mathfrak mathrm{Att} _{\hat{R}} M \subseteq \mathfrak mathrm{Att} _{\hat{R}} N$. 
\end{theorem}

\begin{proof}
Suppose that $M$ is contained in $\mathfrak mathrm{quotext} _R (N)$. 
It is clear from the property of attached prime ideals (Remark \ref{remark attach}) that $\mathfrak mathrm{Att} _{\hat{R}} M \subseteq \mathfrak mathrm{Att} _{\hat{R}} N$ holds. 

Conversely, suppose that $\mathfrak mathrm{Att} _{\hat{R}} M \subseteq \mathfrak mathrm{Att} _{\hat{R}} N$.
First, we shall show that we may assume that $M$ and $N$ are $\mathfrak m$-torsion $R$-modules for some maximal ideal $\mathfrak m$. 
In fact, $M$ (resp. $N$) can be decomposed as $M = \oplus _{\mathfrak m \in \mathfrak mathrm{Ass}R M}\Gamma _{\mathfrak m }(M)$ (resp. $N = \oplus _{\mathfrak n \in \mathfrak mathrm{Ass}R N}\Gamma _{\mathfrak n }(N)$) and the assumption implies that $\mathfrak mathrm{Att} _{\hat{R}m} \Gamma _{\mathfrak m} (M ) \subseteq \mathfrak mathrm{Att} _{\hat{R}m} \Gamma _{\mathfrak m} (N )$ for all $\mathfrak m \in \mathfrak mathrm{Ass}R M$ by Corollary \ref{cor A}.   
If we show that $\Gamma _{\mathfrak m} (M )$ is contained in $\mathfrak mathrm{quotext} _R (\Gamma _{\mathfrak m} (N ))$, we can get the assertion since $\mathfrak mathrm{quotext} _R (N)$ is closed under direct sums and direct summands. 

Let $M$ and $N$ be $\mathfrak m$-torsion $R$-modules and $E$ be an injective hull of $\hat{R}m /\mathfrak m \hat{R}m $ as an $\hat{R}m$-module. 
Since $M$ and $N$ are also artinian $\hat{R}m$-modules, $M^{\vee}$ and $N^{\vee}$ are finitely generated $\hat{R}m$-modules by Matlis duality, where $(-)^\vee = \mathfrak mathrm{Hom} _{\hat{R}m}(- , E)$. 
Since $\mathfrak mathrm{Att} _{\hat{R}m} M$ (resp. $\mathfrak mathrm{Att} _{\hat{R}m} N$) is equal to $\mathfrak mathrm{Ass} _{\hat{R}m} M^{\vee}$ (resp. $\mathfrak mathrm{Ass} _{\hat{R}m} N^{\vee}$) (Lemma \ref{att ass}), the inclusion 
$$
\mathfrak mathrm{Ass} _{\hat{R}m} M^{\vee} \subseteq \mathfrak mathrm{Ass} _{\hat{R}m} N^{\vee}
$$
holds.   
By virtue of Theorem \ref{Krause}, we conclude that $M^{\vee}$ can be generated from $N^{\vee}$ via taking submodules and extensions, i.e. $M^{\vee} \in \mathfrak mathrm{subext} _{\hat{R}m}(N^{\vee})$. 
Hence it follows from Matlis duality and Lemma \ref{duality} that  
$$
M^{\vee \vee} \cong M \in \mathfrak mathrm{subext} _{\hat{R}m}(N^{\vee})^{\vee} = \mathfrak mathrm{quotext} _{\hat{R}m}(N). 
$$
Since artinian $\hat{R}m$-modules are also artinian $R$-modules (cf. Lemma \ref{Sharp}), we conclude that $M \in \mathfrak mathrm{quotext} _{R}(N)$.

\end{proof}

We define by $\mathfrak Psi $ the map sending a subcategory $\mathcal{X}$ of $\mathfrak mathrm{Art}(R)$ to 
$$
\mathfrak mathrm{Att} \mathcal{X} = \cup _{M \in \mathcal{X}}\mathfrak mathrm{Att} _{\hat{R}} M
$$ 
and by $\mathfrak Phi$ the map sending a subset $S$ of $\coprod _{\mathfrak n \in \mathfrak max (R)} \mathfrak mathrm{Spec} \ \hat{R_\mathfrak n}$ to 
$$
\{ M \in \mathfrak mathrm{Art}(R) \ | \mathfrak mathrm{Att} _{\hat{R}} M \subseteq S \}.
$$ 

Note from Corollary \ref{cor A} that $\mathfrak Psi (\mathcal{X} )$ determine the subset of $\coprod _{\mathfrak n \in \mathfrak max (R)} \mathfrak mathrm{Spec} \ \hat{R_\mathfrak n}$. 
On the other hand, it follows from Remark \ref{remark attach} (2) that $\mathfrak Phi (S)$ is closed under quotient modules and extensions. 

Now we state a main theorem of this section. 

\begin{theorem}\mathfrak label{classification}
Let $R$ be a commutative noetherian ring. 
Then $\mathfrak Psi$ and $\mathfrak Phi$ induce a bijection between the set of subcategories of $\mathfrak mathrm{Art}(R)$ which are closed under quotient modules and extensions and the set of subsets of $\coprod _{\mathfrak n \in \mathfrak max (R)} \mathfrak mathrm{Spec} \ \hat{R_\mathfrak n}$. 

Moreover, they also induce a bijection between the set of Serre subcategories of $\mathfrak mathrm{Art}(R)$ and the set of specialization closed subsets of $\coprod _{\mathfrak n \in \mathfrak max (R)} \mathfrak mathrm{Spec} \ \hat{R_\mathfrak n}$. 
\end{theorem}

\begin{proof}
Let $\mathcal{X}$ be a subcategory of $\mathfrak mathrm{Art}(R)$ which is closed under quotient modules and extensions. 
The subcategory $\mathfrak Phi \mathfrak Psi (\mathcal{X} )$ consists of all artinian $R$-modules $M$ with $\mathfrak mathrm{Att} _{\hat{R}} M \subseteq \cup _{X \in \mathcal{X}}\mathfrak mathrm{Att} _{\hat{R}} X$. 
It is clear that $\mathcal{X}$ is a subcategory of $\mathfrak Phi \mathfrak Psi (\mathcal{X} )$. 
Let $M$ be an artinian $R$-module with $\mathfrak mathrm{Att} _{\hat{R}} M \subseteq \cup _{X \in \mathcal{X}}\mathfrak mathrm{Att} _{\hat{R}} X$. 
For each ideal $\mathfrak P \in \mathfrak mathrm{Att} _{\hat{R}} M$, there exists $X^{(\mathfrak P)} \in \mathcal{X}$ such that $\mathfrak P \in \mathfrak mathrm{Att} _{\hat{R}} X^{(\mathfrak P)} $. 
Take the direct sums of such objects, say $X = \oplus _{\mathfrak P \in \mathfrak mathrm{Att} _{\hat{R}} M} X^{(\mathfrak P)}$, $X$ is also an object of $\mathcal{X}$. 
In fact, $\mathfrak mathrm{Att} _{\hat{R}} M$ is a finite set and $\mathcal{X}$ is closed under finite direct sums.  
It follows from the definition of $X$ that $\mathfrak mathrm{Att} _{\hat{R}} M \subseteq \mathfrak mathrm{Att} _{\hat{R}} X$. 
By virtue of Theorem \ref{theorem 1}, $M$ is contained in $\mathfrak mathrm{quotext} _R (X)$, so that  in $\mathcal{X}$. 
Hence we have the equality $\mathcal{X} = \mathfrak Phi \mathfrak Psi (\mathcal{X} )$. 
 

Let $S$ be a subset of $\coprod _{\mathfrak n \in \mathfrak max (R)} \mathfrak mathrm{Spec} \ \hat{R_\mathfrak n}$. 
It is trivial that the set $\mathfrak Psi \mathfrak Phi (S)$ is contained in $S$. 
Let $\mathfrak p$ be a prime ideal in $S$. 
Take a maximal ideal $\mathfrak m$ so that $\mathfrak p$ is a prime ideal of $\hat{R}m$. 
We consider an $\hat{R}m$-module $E_{\hat{R}m /\mathfrak p \hat{R}m }(\hat{R}m / \mathfrak m \hat{R}m)$. 
Then we have the equality: 
$$
\mathfrak mathrm{Att} _{\hat{R}m } E_{\hat{R}m /\mathfrak p \hat{R}m}(\hat{R}m / \mathfrak m \hat{R}m) = \mathfrak mathrm{Ass} _{\hat{R}m } \hat{R}m /\mathfrak p \hat{R}m = \{ \mathfrak p \}.

$$
Note that $E_{\hat{R}m}(\hat{R}m / \mathfrak m \hat{R}m)$ is artinian as an $R$-module. 
Indeed, we have the equality $E_{\hat{R}m}(\hat{R}m / \mathfrak m \hat{R}m) = E_{R}(R / \mathfrak m R)$ as $R$-modules (\cite[Theorem 18.6 (iii)]{M}). 
Since $E_{\hat{R}m /\mathfrak p \hat{R}m }(\hat{R}m / \mathfrak m \hat{R}m)$ is an $\hat{R}m$-submodule (thus an $R$-submodule) of $E_{\hat{R}m}(\hat{R}m / \mathfrak m \hat{R}m)$, it is an artinian $R$-module. 
Hence $E_{\hat{R}m /\mathfrak p \hat{R}m}(\hat{R}m / \mathfrak m \hat{R}m)$ is an artinian $R$-module which is a $\mathfrak p$-secondary $\hat{R}m$-module. 
Consequently, $E_{\hat{R}m /\mathfrak p \hat{R}m }(\hat{R}m / \mathfrak m \hat{R}m)$ belongs to $\mathfrak Phi (S)$, so that $\mathfrak p \in \mathfrak Psi \mathfrak Phi (S)$. 

Suppose that $\mathcal{X}$ is a Serre subcategory of $\mathfrak mathrm{Art}(R)$. 
Let $\mathfrak p$ be a prime ideal of $\hat{R}m$ which is contained in $\mathfrak Psi (\mathcal{X} )$.    
Chose $\mathfrak q \in \mathfrak mathrm{Spec} \ \hat{R}m$ such that $\mathfrak p \subseteq \mathfrak q$. 
Then we have the inclusion of $\hat{R}m$-modules (hence, of $R$-modules):
$$
0 \to E_{\hat{R}m /\mathfrak q \hat{R}m}(\hat{R}m/ \mathfrak m \hat{R}m) \to E_{\hat{R}m /\mathfrak p \hat{R}m}(\hat{R}m/ \mathfrak m \hat{R}m). 
$$
Since $E_{\hat{R}m /\mathfrak p \hat{R}m}(\hat{R}m/ \mathfrak m \hat{R}m)$ is an artinian $R$-module which is a $\mathfrak p$-secondary $\hat{R}m$-module, $E_{\hat{R}m /\mathfrak p \hat{R}m}(\hat{R}m/ \mathfrak m \hat{R}m)$ is contained in $\mathcal{X}$. 
Thus $E_{\hat{R}m /\mathfrak q \hat{R}m}(\hat{R}m/ \mathfrak m \hat{R}m)$ is also in $\mathcal{X}$ since $\mathcal{X}$ is closed under submodules. 
Hence we have that $\mathfrak q \in \mathfrak Psi (\mathcal{X} )$, so that $\mathfrak Psi (\mathcal{X} )$ is closed under specialization. 

On the other hand, let $S$ be a specialization closed subset of $\coprod _{\mathfrak n \in \mathfrak max (R)} \mathfrak mathrm{Spec} \ \hat{R_\mathfrak n}$. 
We shall show $\mathfrak Phi (S)$ is a Serre subcategory. 
Since $\mathfrak Phi (S)$ is closed under quotient modules and extensions, we sufficiently show that it is closed under submodules. 
Let $M$ be in $\mathfrak Phi (S)$ and $N$ be an $R$-submodule of $M$. 
Set $J_M = \cap _{\mathfrak m \in \mathfrak mathrm{Ass} _{R} M}\mathfrak m$ and $\hat{R}M = \varprojlim R/J_{M}^n$. 
Then $M$ is an artinian $\hat{R}M$-module and $N$ is also an artinian $\hat{R}M$-submodule of $M$ (Lemma \ref{Sharp}). 
Since $\hat{R}M$ is a complete semi-local ring, Matlis duality is allowed. 
By using Matlis duality, we can show that $N^{\vee}$ is contained in the Serre subcategory generated by $M^{\vee}$, where $(-)^\vee = \mathfrak mathrm{Hom} _{\hat{R}M}(- , E_{\hat{R}M}(\hat{R}M/J_{M}\hat{R}M))$. 
Thus, by Theorem \ref{Theorem B}, $\mathfrak mathrm{Ass} _{\hat{R}M} N^{\vee}$ is in $\cup _{\mathfrak p \in \mathfrak mathrm{Ass} _{\hat{R}M} M^{\vee}} V(\mathfrak p)$. 
Since we have the equalities $\mathfrak mathrm{Att} _{\hat{R}} N = \mathfrak mathrm{Att} _{\hat{R}M} N = \mathfrak mathrm{Ass} _{\hat{R}M} N^{\vee}$, and $\cup _{\mathfrak p \in \mathfrak mathrm{Ass} _{\hat{R}M} M^{\vee}} V(\mathfrak p) = \cup _{\mathfrak p \in \mathfrak mathrm{Att} _{\hat{R}M} M} V(\mathfrak p) \subseteq S$, $\mathfrak mathrm{Att} _{\hat{R}} N$ is contained in $S$. 
Therefore $N$ is in $\mathfrak Phi (S)$. 

\end{proof}

\subsection*{Acknowledgments}
The author express his deepest gratitude to Ryo Takahashi and Yuji Yoshino for valuable discussions and helpful comments. 

\end{document}