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Let $A$ be a commutative noetherian ring. Let $M, N$ be $A$-modules, and assume that $M$ is finite over $A$. Let $P$ be a flat $A$-module.

Is it true that there is an isomorphism $\operatorname{Hom}_A(M,N)\otimes_A P \cong \operatorname{Hom}_A(M,N\otimes_A P)$ ?

It appears that it holds at least locally (that is, when $P$ is free). Is it true in general? If so, is there a reference for this?

Thank you!

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If you assume $N$ is finite over $A$ rather than $M$, this will be ture. – Li Zhan Nov 12 '11 at 21:38
@LiZhan, this sounds suprising to me. Can you please elaborate? – the L Nov 12 '11 at 21:44
up vote 11 down vote accepted

Yes it is true for general flat modules $P$. We have a canonical map $$ \rho_M : \mathrm{Hom}_A(M, N)\otimes_A P\to \mathrm{Hom}_A(M, N\otimes_A P)$$ (it maps $\varphi \otimes p$ to the map $x\mapsto \varphi(x)\otimes p$). Write $M$ as the quotient of a free finite rank $A$-module $L$. Then we have an exact sequence \begin{equation} 0 \to \mathrm{Hom}_A(M, N)\to \mathrm{Hom}_A(L, N) \to \mathrm{Hom}_A(K, N) \end{equation} where $K$ is the kernel of $L\to M$. Tensoring by $P$, we get $$ 0 \to \mathrm{Hom}_A(M, N)\otimes_A P \to \mathrm{Hom}_A(L, N)\otimes_A P \to \mathrm{Hom}_A(K, N) \otimes_A P. $$ Similarly, replacing $N$ with $N\otimes_A P$, we have an exact sequence $$ 0 \to \mathrm{Hom}_A(M, N\otimes_A P) \to \mathrm{Hom}_A(L, N\otimes_A P) \to \mathrm{Hom}_A(K, N\otimes_A P) $$ and we have a commutative diagram with the above two lines and vertical maps $\rho_M$, $\rho_L$ and $\rho_K$. As $L\simeq A^d$, it is easy to see that $\rho_L$ is an isomorphism. Therefore $\rho_M$ is injective. As $A$ is noetherian, $K$ is finitely generated over $A$, the previous result applies to $K$ and $\rho_K$ is injective. Now coming back to our commutative diagram, it is easy to see that $\rho_M$ is then surjective.

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I'd like to add that you can get rid of noetherianity of A by insisting M to be of finite presentation. (which is automatic for finite modules over noetherian A) – Jacob Bell Nov 12 '11 at 23:46
@QiL: cf my question here. – Zhen Lin Nov 13 '11 at 0:49
@Zhen Lin: I see that in your question there you gave the right answer to the question here:). I think I can answer your question there. – user18119 Nov 13 '11 at 14:37

Here is sufficient condition for obtaining the isomorphism you are after.
Let $A$ be an arbitrary commutative ring (not supposed noetherian) and $M, N$ arbitrary modules . We have a canonical morphism
$$ M^*\otimes_A N\to Hom (M,N):\phi \otimes n \mapsto [m\mapsto\phi(m)n] \quad (\star)$$ Proposition ($\star$) is an isomorphism as soon as $M$ is finitely generated projective.
It is an isomorphism for $M=A$, then for finitely generated free modules $M=A^r$, and finally for summands of such i.e. finitely generated projectives.

Given three arbitrary modules $M,N,P$ over the commutative ring $A$ , with $M $ finitely generated projective, we have a natural isomorphism
$$Hom_A(M,N)\otimes_A P \cong Hom_A(M,N\otimes_A P)$$
Replace all $Hom$'s by $\otimes$'s and use associativity of tensor product.

a) The Corollary fails if $P$ is finitely generated but not projective.
Take for example $A=\mathbb Z, N=\mathbb Z, M=P=\mathbb Z/(2)$.
Then the left-hand side in the Corollary is $0$ and the right-hand side is $\mathbb Z/(2)$

b) As QiL pertinently comments, the Corollary also fails if $P$ is not assumed finitely generated.
Take for example $N=A$ and $M=P=$ an arbitrary non finitely generated module.
Then our canonical morphism $u: M^*\otimes M \to Hom(M,M)$ cannot be surjective.
Indeed any element $t\in M^*\otimes M$ gets sent to an endomorphism $u(t)=f:M\to M$ such that $u(M)$ is finitely generated, so that the identity $Id_M$ of $M$ will never be in the image of $u$.

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Nice answer. Compared with OP's hypotheses, $A, M$ are more general, and $P$ is more restrictive. I think it can not hold with $P$ not f.g. (even with $P, M$ free and $N=A$). – user18119 Nov 12 '11 at 22:49
Your sense of fair-play is heartwarming, @QiL. And I find that infinitely more valuable than any result on projective modules ( or anything else). Also, you are absolutely right about the necessity for $P$ to be finitely generated. I have added an Edit to confirm this. – Georges Elencwajg Nov 12 '11 at 23:46
@Georges: thanks for upvoting and for clearing my doubt: I was genuinely confused. – Jacob Bell Nov 12 '11 at 23:47
Dear @donkey kong: you are welcome. And I forgot to mention that you deserve the upvotes independently of that problem with comments! – Georges Elencwajg Nov 12 '11 at 23:49
an idle comment one could make is that this is the way to show that the identity remains true for arbitrary modules once you derive everything. (I think) – Jacob Bell Nov 12 '11 at 23:51

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