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Let $A$ be a $\mathbb Z$-algebra that is finitely-generated and free as a $\mathbb Z$-module and let $\pi: A \rightarrow \mathbb Z$ a nontrivial $\mathbb Z$-module homomorphism. For a positive integer, $m$, I am interested in ${\rm Hom}_{\,\mathbb Z}(A/mA, \mathbb Z/m\mathbb Z)$, the $\mathbb Z$-module homomorphisms, considered as an $A$-module.

Under suitable conditions (e.g., $m$ relatively prime to some integer depending on $A$ and $\pi$, probably), it seems that ${\rm Hom}_{\,\mathbb Z}(A/mA, \mathbb Z/m\mathbb Z)$ and $A/mA$ are isomorphic as $A$-modules.

The (conjectured) isomorphism I have been looking at maps $\overline{a}$ (the coset of $a \in A$ in $A/mA$) to the homomorphism defined by taking $\overline{b}$ to $\pi(ab) \bmod m$ (where $ab$ is the product of $a$ and $b$ in $A$).

However, I am stuck trying to prove this is actually an isomorphism. Or disproving it.

Any suggestions?

Duncan
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    Wouldn't $A\cong{\Bbb Z}^n$ imply $A/mA\cong({\Bbb Z}/m{\Bbb Z})^n$ and hence $$\begin{array}{ll} \hom(A/mA,{\Bbb Z}/m{\Bbb Z}) & \cong\hom(\bigoplus^n{\Bbb Z}/m{\Bbb Z},{\Bbb Z}/m{\Bbb Z}) \ & \cong\bigoplus^n\hom({\Bbb Z}/m{\Bbb Z},{\Bbb Z}/m{\Bbb Z}) \ & \cong\bigoplus^n{\Bbb Z}/m{\Bbb Z} \ & \cong A/mA~? \end{array}$$ (Notice multiplication in $A$ doesn't factor in.) – anon Aug 15 '13 at 14:01
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    What do you mean by $\mathrm{Hom}$: module homomorphisms or algebra homomorphisms? – Zhen Lin Aug 15 '13 at 14:14
  • @ZhenLin hi and thanks for your comment. I have specified that now. Notice that all we know about $\pi$ is that it is a $\mathbb Z$-module homomorphism, so my conjectured isomorphism would also map to $\mathbb Z$-module homomorphisms. – Duncan Aug 15 '13 at 15:26
  • In that case, anon's comments are right: the algebra structure on $A$ is irrelevant, as is the homomorphism $\pi$. – Zhen Lin Aug 15 '13 at 15:30
  • @anon: Dear anon, Note that $\mathrm{Hom}(A/m, \mathbb Z/m)$ has an $A$-module structure, and the question of whether this is isomorphisc as an $A$-module to $A/mA$ is substantially more involved than the analogous question as $\mathbb Z$-modules (which, as you note, is pretty obvious). It is related to the question of whether $A$ is Gorenstein as a $\mathbb Z$-module. Regards, – Matt E Aug 15 '13 at 15:46
  • @ZhenLin: Dear Zhen, See my comment addressed to anon above. It matters a lot whether the question is about an isomorphism of $A$-modules or of $\mathbb Z$-modules. Regards, – Matt E Aug 15 '13 at 15:47
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    Dear Duncan, You should clarify whether you are asking about an isomorphism of abelian groups (in which case the answer is trivially yes, as already noted in the comments and in the answer below), or whether you are asking about an isomorphism of $A$-modules (in which case the question is much more subtle). Regards, – Matt E Aug 15 '13 at 15:48
  • @MattE (and others) apologies for the confusion by me not being clear here. And thanks too for your time and comments. – Duncan Aug 15 '13 at 16:10

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