Converse of Nakayama lemma

Question

Let $A$ be a commutative ring with $1$ and $J(A)$ denotes the jacobson radical of $A.$ Let $I$ be an ideal of $A$ such that for every finitely generated $A$ module $M$ whenever $IM=M,$ $M=0.$ Then how do I show that $I \subset J(A).$

Give me some idea. Thanks.

Answer 1

The elementary version of rschwieb's answer : for $m$ a maximal ideal, what's $I(R/m)$ ? ($R/m$ is seen as an $R$-module)

Answer 2

In particular for simple modules, $IS=\{0\}$, so $I$ annihilates all simple modules. The Jacobson radical is the intersection of annihilators of all simple modules, so $I$ is contained in it.