Is every Artinian module over an Artinian ring finitely generated?
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!