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I know that if $R$ is Artinian, then a f.g. $R$-module is Artinian. Is f.g. a necessary condition?

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  • $\begingroup$ Just to clarify (for myself, and for others): The question is if every artinian module over an artinian ring is finitely generated. $\endgroup$ – Martin Brandenburg Feb 12 '13 at 3:43
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Let $R$ be an artinian ring and $M$ an artinian $R$-module. Then $M$ is finitely generated.

If $M$ is not finitely generated one can assume that every proper submodule of $M$ is finitely generated. (In order to see this take the partial ordered set of submodules of $M$ which are not finitely generated and choose a minimal element.) Then $P=\operatorname{Ann}(M)$ is a prime ideal of $R$: pick $a,b\in R$ such that $ab\in P$. If $a\notin P$, then $(0:_Ma)\neq M$. This shows that $(0:_Ma)$ is finitely generated. Since $0\to(0:_Ma)\to M\to aM\to 0$ is a short exact sequence we get that $aM$ is not finitely generated, so $aM=M$. Then $0=abM=b(aM)=bM$, so $b\in P$.

Since $R$ is artinian and $P$ prime, $R/P$ is a field. But $M$ is an artinian $R/P$-module which is not finitely generated, false!

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    $\begingroup$ As one can see we can replace $R$ artinian by $\dim R=0$. $\endgroup$ – user26857 Feb 12 '13 at 2:01
  • $\begingroup$ Why does the set of non-finitely generated submodules contain a minimal element? $\endgroup$ – Tobias Kildetoft Feb 12 '13 at 13:42
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    $\begingroup$ @TobiasKildetoft Since it is nonempty and the module is artinian. $\endgroup$ – user26857 Feb 12 '13 at 21:09
  • $\begingroup$ Ahh yes, of course. $\endgroup$ – Tobias Kildetoft Feb 12 '13 at 23:45
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If $R$ is right Artinian, the Hopkins-Levitzki theorem holds, so that every $R$ module (left or right, it doesn't matter) is Artinian iff Noetherian iff it has finite composition length.

Of course, being Noetherian implies finitely generated, but as you see much more is true.

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