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Let $R$ be a commutative Noetherian ring, $M$ is an Artinian $R$-module. Is the set $Supp_R(M)$ finite? Thanks.

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up vote 10 down vote accepted

The answer is yes.

You can find a proof in Leamer's thesis Homology of Artinian Modules Over Commutative Noetherian Rings, which contains the the following more precise result (see Lemma 3.1.11):

Let $R$ be a commutative noetherian ring and let $M$ be an artinian $R$-module. Then

1) the support of $M$ consists entirely of maximal ideals of $R$, that is, $\textrm{Supp}(M) \subseteq \textrm{m-Spec}(R)$;

2) we have $\textrm{Min}_R(M)=\textrm{Ass}_R(M)=\textrm{Supp}_R(M)$;

3) $\textrm{Supp}_R(M)$ is finite.

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A more general statement is true: Suppose $A$ is a Noetherian ring and $M$ is a finitely generated $A$-module, then there exists a finite increasing sequence $0=M_0\subseteq M_1\subseteq\dotsb\subseteq M_n=M$ such that $M_i/M_{i-1}\cong A/P_i$ for some prime ideal $P_i$. Moreover, $\operatorname{Ass}(M)\subseteq\{P_1,\dotsc,P_n\}\subseteq\operatorname{Supp}(M)‌​$, and three sets have the same minimal elements. (J-P.Serre, Local Algebra, section I.7) – Frank Science Dec 8 '15 at 18:30

It is well-known that for finitely generated Artinian $M$, the support is a finite set of maximal ideals. Since $\mathrm{Supp}(M)=\bigcup_{M'\subseteq M\ \mathrm{f.g.}}\mathrm{Supp}(M')$, we know that $\mathrm{Supp}(M)$ consists of maximal ideals. For any finite set $S\subseteq\mathrm{Supp}(M)$, it again follows easily from the finitely-generated case that $f_S\colon M\to\prod_{s\in S}M_s$ is surjective. In particular, for a strictly increasing chain $S_1\subset S_2\subset\dots$ of finite subsets of $\mathrm{Supp}(M)$, the chain of submodules $\ker(f_{S_i})$ is strictly decreasing, so $\mathrm{Supp}(M)$ must be finite.

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