Let $M$ be a finitely generated $R$-module and $I \subset R$ an ideal such that $IM = M$. If $M'$ is a particular submodule, does $IM '= M'$? 
Suppose $R$ is a commutative ring, $I$ an ideal, and $M$ a finitely generated $R$-module with generators $\{m_1, \ldots, m_n\}$; suppose further that $IM = M$. Let $M'$ be the submodule of $M$ generated by $\{m_2, \ldots, m_n\}$. I would like to prove that $IM' = M'$. 

$IM' \subseteq M'$ is of course trivial, following from the closure of submodules under scalar multiplication by elements of $R$; I am having some difficulty proving containment in the opposite direction.

This is essentially a corollary of Nakayama's Lemma. This states that $IM=M$ is equivalent to $(1-a)M=0$ for some $a \in I$. Now, if $M'$ is any submodule of $M$, it follows $(1-a)M'=0$ and hence $IM'=M'$.
