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.

Since $M=IM$, we can write $m_1=a_1m_1+a_2m_2+\dots+a_nm_n$, for $a_i\in I$.  Rearranging terms, we find $(1-a_1)m_1\in IM'$.  Now use this to show that $(1-a_1)m_i\in IM'$ for $i=2,\dots,n$ as well.  Since $a_1\in I$, conclude that in fact $m_i\in IM'$ for $i=2,\dots,n$.
