Is every Artinian module over an Artinian ring finitely generated?

Question

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

Answer 1

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!

Answer 2

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.

Answer 3

This is more elementary proof I think.

Let $R$ be a left Artinian ring, then every left Artinian module is finite generated module.

Let $M$ be a left Artinian module, denote $J$ the Jacobson radical of $R$. Then consider $M\supseteq JM\supseteq J^2M\supseteq...$. Since $M$ is Artinian, we know the chain is stable, say $J^NM=J^{N+1}M=...$. Since $R$ is left Artinian, so we know $J$ is nilpotent. So $J^NM=J^{N+1}M...=0$. Remark $\dfrac{J^iM}{J^{i+1}M}$ is sub-quotient of $M$, hence is also Artinian. So $\dfrac{J^iM}{J^{i+1}M}$ is finite length module since it is also semisimple module. So we know $M$ is finite length module. So when $R$ is left Artinian, there is no difference between left Noetherian module and left Artinian module.