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It is easy to show that if $M$ is a Noetherian $R$-module then $R/\mbox{ann}(M)$ is a Noetherian ring. Is there a similar (or dual) result for Artinian modules?

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    simple modules, and finite length modules are intensively studied for all rings, including non-artinian rings. That's what representation theory mostly does!2011-07-08

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If $M$ is an Artinian $R$-module, then so is any submodule and any quotient of $M$. Thus if $M$ is finitely generated, then $R/\mathrm{Ann}(M)$ is Artinian.
But $\mathbb{Z}[1/p]/\mathbb{Z}$ is a non finitely generated Artinian $\mathbb{Z}$-module and $\mathbb{Z}$ is not Artinian. Thus if $M$ is an Artinian $R$-module, then $R$ is not necessary Artinian. (See the article of wiki about Artinian module.)

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    Why is $R/\mathrm{Ann}(M)$ Artinian exactly?2015-09-22