I have proven the following form of Nakayama's:
Suppose $M$ is a finitely-generated module over $R, J(R)$ is the Jacobson radical of $R$ and $I\subseteq J(R)$ where $I$ is an ideal. If $I M=M$, then $M=0$.
However, I want to show that if $I$ is an ideal of $R$ such that $IM=M$ then there exist $i\in I$ such that $im=m$ for all $m\in M$.
Any ideas? Im stuck at this point.