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!

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    $\begingroup$ It seems that if $IM$ is idempotent if and only if $IM = I^2M$. But I doubt that this implies $I = I^2$. $\endgroup$ – Claudius Jan 6 '17 at 9:51
  • $\begingroup$ @user218931 Why $IM=I^2M$ gives idempotent-ness of $IM$? $\endgroup$ – karparvar Jan 7 '17 at 5:52
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    $\begingroup$ $\hom(M,IM)IM \subseteq I^2M$ 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, $IM = I^2M = \hom(M,IM)IM$. $\endgroup$ – 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 $(2)\neq (4)$.

  • $\begingroup$ @karparvar I'm not trying to characterize idempotentness, just that it is a necessary condition, and it suggests the counterexample I gave. $\endgroup$ – rschwieb Jan 7 '17 at 5:56

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