Suppose we have a ring $R$ and an $R$-module $M$. Suppose we have an ideal $I\subseteq R$ such that for all maximal ideals $\mathfrak{m} \supseteq I$ we have $M_\mathfrak{m} = 0$. I need to show that this implies $IM = M$.
I know that if $M$ was an $R/I$-module then the statement would imply $M=0$ since there is a correspondence between maximal ideals of $R$ with $I \subseteq \mathfrak{m}$ and maximal ideals of $R/I$, and being zero is a local property.
If I knew that $M$ was finitely generated then I could try to show $J + I = R$ with $J$ the Jacobson radical of $R$ and have
$J (M/IM) = (JM + IM/IM) = RM/IM = M/IM$ which gives me $M/IM = 0$ by Nakayama's Lemma. (Or more generally for $J$ an ideal with $J \subseteq \text{Jac}(R)$)
I'd be really happy if I could get some hints or if somebody could point me in the right direction. I don't know if my approaches are even useful or in the right direction. Thanks!
Best Answer
Note that $M = IM$ is equivalent to $R/I\otimes_R M = 0$, so what you want to show is that if $M_\mathfrak m=0$ for all maximal ideals $\mathfrak m$ over $I$, then $R/I\otimes_R M = 0$.
Naturally, $R/I\otimes_A M = N$ is an $R/I$-module, and checking that it is zero is equivalent to checking that $N_\mathfrak n = 0$ for all maximal ideals $\mathfrak n$ of $R/I$, which are in bijective correspondence with the maximal ideals above.
Can you conclude?