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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1$\begingroup$ The title you picked for your question is almost completely unrelated to the question itself! $\endgroup$– Mariano Suárez-ÁlvarezJul 8, 2011 at 3:37
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$\begingroup$ I hope there is no mathematical term for "relevant", but what I meant was that if any Artinian module can be reduced to an Artinian module over an Artinian ring (as is the case for Notherian modules), then there is no point considering Artinian modules over non-Artinian rings. $\endgroup$– ashpoolJul 8, 2011 at 19:17
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$\begingroup$ simple modules, and finite length modules are intensively studied for all rings, including non-artinian rings. That's what representation theory mostly does! $\endgroup$– Mariano Suárez-ÁlvarezJul 8, 2011 at 22:24
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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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1$\begingroup$ In fact, every ring has artinian modules, becaue it has maximal left ideals. $\endgroup$ Jul 9, 2011 at 20:59
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$\begingroup$ right. if we donot require the ring is commutative. then,the maximal left ideal is not necessary a two side ideal. $\endgroup$– wxuJul 10, 2011 at 6:04
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$\begingroup$ I don't see how that is related... $\endgroup$ Jul 10, 2011 at 6:07
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$\begingroup$ what i mean is that if ann(M) is an two side ideal, we may view M as a R/ann(M) module. anyway, since the op taged the post as commutative algebra, i think he prefer to thinking commutative rings. $\endgroup$– wxuJul 10, 2011 at 6:21
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