Nakayama's Lemma: If $M$ is a finitely generated $R$-module and there is an ideal $I\subset R$ with $IM=M$, then there exists $a\in I$ with $am=m$ for all $m\in M$.
I'm looking to prove: Let $M$ be a finitely generated module over a local ring $R$, then
- if there is a proper ideal $I\subset R$ with $IM=M$, then $M=0$;
Now: By Cayley-Hamilton, I know there is an element $a\in I$ such that $(1+a)M=0$. I thought of showing $1+a$ is a unit, but I don't really know how to use the fact that $R$ is local.
Since $a\in I$, we know in particular that $a$ belongs to the unique maximal ideal $m\subseteq R$. Clearly, $1+a$ cannot belong to $m$, as otherwise we would have $1=(1+a)-a\in m$, a contradiction. Thus, $1+a$ is invertible, and so $(1+a)M=0$ implies $M=0$.
More generally, if $R$ is any ring, $M$ a finitely generated module, and there is an ideal $I$ which is contained in the Jacobson radical of $R$ and satisfies $IM=M$, then $M=0$.