# Idempotent Modules and Ideals

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!

• It seems that if $IM$ is idempotent if and only if $IM = I^2M$. But I doubt that this implies $I = I^2$. – Claudius Jan 6 '17 at 9:51
• @user218931 Why $IM=I^2M$ gives idempotent-ness of $IM$? – karparvar Jan 7 '17 at 5:52
• $\hom(M,IM)IM \subseteq I^2M$ is clear. Conversely, let $m = x_1m_1 + \dotsb + x_nm_n \in I^2M$, where $x_i\in I$ and $m_i\in IM$. Let $f_i\in \hom(M, IM)$ be multiplication by $x_i$. Then $m = f_1(m_1) + \dotsb + f_n(m_n)\in \hom(M, IM)IM$. This shows $I^2M\subseteq \hom(M,IM)IM$. Therefore, $IM = I^2M = \hom(M,IM)IM$. – Claudius Jan 7 '17 at 15:09

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 $(2)\neq (4)$.