Given a commutative ring $A$ and an $A$-module $M$, does there exist $A$-modules $M'$ and $M''$ so that $M\cong\operatorname{Hom}_A(M',M'')$?

So far, I have that if $M$ is free of finite rank then there does exists $A$-modules $M'$ and $M''$ so that $M=\operatorname{Hom}(M',M'')$. To see this, let $M$ be free of finite rank, so

$$M\cong M^{\vee \vee}=\operatorname{Hom}_A(M^{\vee},A)$$

(where $M^{\vee}=\operatorname{Hom}_A(M,A)$). Since $A$ is commutative, $M^{\vee}$ is a module over $A$, so $M$ is of the desired form.

I cannot make progress in the more general question, so any help will be appreciated. Thanks!


How about $M\cong\mathrm{Hom}_A(A,M)$ with the natural identification $m\mapsto(a\mapsto am)$?

  • $\begingroup$ I was making that overly complicated! Thank you! I'll accept the answer when the wait time is over. $\endgroup$ – user304718 Jan 30 '17 at 19:42

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