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Let $R$ be a ring (commutative, with identity), $m$ a maximal ideal and $M$ an $R$-module. Suppose $m^nM=0$ for some $n>0$. Then

$M$ is Noetherian if and only if $M$ is Artinian

Do you have any idea how to solve this?

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2 Answers 2

up vote 5 down vote accepted

We will use the following standard commutative algebra fact:

Let $A$ be a commutative ring, and $E$ be an $A$-module. Suppose there is a filtration $$E = E_0\supseteq E_1\supseteq\cdots\supseteq E_n = 0$$ of $A$-submodules $E_i$. Then $E$ is Noetherian (resp. Artinian) $\iff$ each quotient $E_i/E_{i+1}$ is Noetherian (resp. Artinian).

Note that you have a filtration $$M \supseteq mM\supseteq m^2M\supseteq\cdots\supseteq m^nM = 0.$$ Thus $M$ is Noetherian (resp. Artinian) if and only if each of the successive quotients $m^kM/m^{k+1}M$ are Noetherian (resp. Artinian) $R$-modules. Note that each quotient $m^kM/m^{k+1}M$ is naturally an $R/m$-module, and moreover that $m^kM/m^{k+1}M$ will be a Noetherian (resp Artinian) $R$-module $\iff$ it is a Noetherian (resp Artinian) $R/m$-module. Since $R/m$ is a field, we know that $m^kM/m^{k+1}M$ is a Noetherian $R/m$-module $\iff$ it is an Artinian $R/m$-module $\iff$ it is finite dimensional over $R/m$. In particular, each of the quotients $m^kM/m^{k+1}M$ is Noetherian if and only if it is Artinian, and hence $M$ is Noetherian if and only if it is Artinian.

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+1 Nice answer. –  user38268 May 17 '12 at 23:30
    
Do we actually use the fact "$...\iff$ it is finite dimensional over $R/m$", or was this just included for completeness sake? –  Tom Apr 26 '13 at 15:01

(EDIT: Oof I thought this was the obvious answer, but I see no good reason the $\overline{R}$ to be Artinian, so the H-L Theorem may be too much. The other "piece" of the H-L theorem used in the other solution is better.)

Let $\overline{R}$ denote $R/ann(M)$, where $ann(M)=\{r\in R\mid Mr=\{0\}\}$.

Hint 1: the submodules of $M_R$ correspond to the submodules of $M_{\overline{R}}$.

Hint 2: The condition that the maximal ideal $m$ annihilates $M$ says that $rad(\overline{R})$ is nilpotent.

Hint 3: The Hopkins-Levitzki theorem says that, in this case, an $\overline{R}$ module is Noetherian iff Artinian.

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