Let $I$ be an ideal of a commutative ring $R$ with unity such that for every non-zero finitely generated module $M$ over $R$ , $IM \ne M$ ; then how to show that $I$ is contained in the Jacobson radical of $R$ ?
I tried to go like this : Let $J(R)$ be the Jacobson radical . If $a \notin J(R)$ , then $(aR+J(R)) / J(R)$ is a non-zero , finitely generated $R$-module . Hence $I((aR+J(R)) / J(R))\neq (aR+J(R)) / J(R)$ ; and from this I somehow have to conclude $a\notin I$ ; but I am stuck here . I feel that I have to use some property of Jacobson radical , but I don't know what . From $a \notin J(R)$ , I can conclude $\exists y \in R$ such that $1+ay$ is not a unit ; but I don't know how to use that .
Please help . Thanks in advance
You're overthinking it. "Every nonzero finitely generated module $M$" is cleverly hiding the relevant subclass of simple modules.
$IM=\{0\}$ for every simple module, hence $I\subseteq J(R)$.