Let $R$ be a commutative ring with unit element, and let $M$ be a multiplication $R$-module, i.e., each submodule of $M$ is of the form $IM$ for some ideal $I$ of $R$. I call a submodule $N$ of $M$ idempotent if $N=\operatorname{Hom}(M,N)N$.
My question:"If a submodule $IM$ of $M $ is idempotent, is it necessarily true that $I$ is an idempotent ideal of $R$?"
By the definition, for any $x\in I$ and $m\in M$ we could write $$xm=f_1(m_1)+...+f_k(m_k),$$ where $k$ is an integer and each $f_i\in \operatorname{Hom}(M,IM)$. And, somehow, we should write $x$ as a finite sum in $I^2$ if the assertion holds! Thanks for any suggestion!
I think the comment of user218931 above is relevant: if $IM=\hom(M, IM)IM$, then (for obvious conventions that $i$'s lie in $I$ and $m$'s lie in $M$) $$im=\sum f_j(i_jm_j)=\sum i_jf_j(m_j)\in I^2M\,.$$ It follows that $IM\subseteq I^2M$. Since you always have $I^2M\subseteq IM$, we have that $I^2M=IM$.
If you take, say, $M=\mathbb Z/7\mathbb Z$ as a $\mathbb Z$ module, then $(2)M=(4)M$, but $(2)\neq (4)$.