Is every Artinian module over an Artinian ring finitely generated?

Solution 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!

Solution 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.