It is easy to show that if $M$ is a Noetherian $R$-module then $R/\mbox{ann}(M)$ is a Noetherian ring (my thanks to all those who enlightened me on this). Is there a similar (or dual) result for Artinian modules?
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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 there exists non finitely generated Artinian $\mathbb{Z}$-module $\mathbb{Z}[1/p]/\mathbb{Z}$. And $\mathbb{Z}$ is not Artinian. Thus if $M$ is Artinian $R$-module, then $R$ is not necessary Artinian. (See the article of wiki about Artinian module.) |
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