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Let $A$ be a commutative ring, and $L, M, N$ be $A$-modules. Then is it true that $$\text{Hom}_A (L, M)\otimes_A N \cong \text{Hom}_A (L, M\otimes_A N)$$ as $A$-modules?

(Note that there is a natural morphism from the left to the right, I think it's not easy to check it is injective or surjective, but I didn't really do it; also, I think elements of the RHS are hard to decompose, so I don't hope for a (natural) arrow in the opposite direction.)

If this is not true, how about we assume that $A$ is a local ring and $N$ is a flat $A$-module or even a flat local $A$-algebra?

Could anyone give some hint or a proof, or a counterexample?

Other appropriate conditions that guarantee the isomorphism are appreciated.

$\textbf{Edit:}$ My main concern is the case when $A=\mathscr{O}_{\mathbb{C}^n,0}=M, N=\mathscr{E}_{\mathbb{C}^n,0}$, and $L$ is the stalk at $0\in \mathbb{C}^n$ of some coherent $\mathscr{O}_{\mathbb{C}^n}$-module, where $\mathscr{O}_{\mathbb{C}^n}$ and $\mathscr{E}_{\mathbb{C}^n}$ mean the sheaves of holomorphic functions and complex-valued smooth functions on $\mathbb{C}^n$ respectively. The flatness of $\mathscr{E}_{\mathbb{C}^n}$ over $\mathscr{O}_{\mathbb{C}^n}$ can be found here (Theorem 7.2.1), which cites Bernard Malgrange's book 'Ideals of differentiable functions' (page 88, Coro 1.12), and here is another discussion on MO.

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    $\begingroup$ $A=M=\mathbb{Z}$, $L=N=\mathbb{Q}$ $\endgroup$ – Neil Strickland Dec 17 '15 at 12:31
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    $\begingroup$ @Lau-tzu $Hom_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})=0$ $\endgroup$ – Aurora Dec 17 '15 at 13:22
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    $\begingroup$ @Neil Strickland Yours is indeed a counterexample, thanks! $\endgroup$ – Lao-tzu Dec 17 '15 at 13:30
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    $\begingroup$ I think you should be looking on conditions on L (rather than N) for that iso to be true. If L is finitely generated projective, then $Hom(L,N) = L^\vee \otimes N$ and so your iso follows. $\endgroup$ – pro Dec 17 '15 at 13:51
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    $\begingroup$ Perhaps the best strategy is for @NeilStrickland to post his comment as an answer so that OP can accept it; this is better than closing because the only flaw with the question is that it happens to have a trivial answer. Then the OP can ask a new question which focuses on the more interesting special case identified in the edit. $\endgroup$ – Paul Siegel Dec 17 '15 at 15:25
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The example $A=M=\mathbb{Z}$, $L=N=\mathbb{Q}$ shows that the answer is negative: we have $\text{Hom}(L,M)=0$ so $\text{Hom}_A(L,M)\otimes_AN=0$, but $\text{Hom}_A(L,M\otimes_AN)=\text{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q})=\mathbb{Q}\neq 0$

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You can think about tensor products as a kind of colimit; you're asking the hom functor $\text{Hom}_A(L, -)$ to commute with this colimit in the second variable, but usually the hom functor only commutes with limits in the second variable. Dually, you can think about homs as a kind of limit (in the second variable); you're asking the tensor product functor $(-) \otimes_A N$ to commute with this limit, but usually tensor products only commute with colimits.

This sort of reasoning not only suggests that your statement should be false but suggests what extra hypotheses might make it true: namely, some kind of projectivity hypothesis on $L$, or some kind of flatness hypothesis on $N$. In fact the statement is true if either $L$ or $N$ is finitely presented projective; these conditions are equivalent to requiring that $\text{Hom}_A(L, -)$ commutes with all colimits or that $(-) \otimes_A N$ commutes with all limits respectively.

But it's also true if $L$ is finitely presented and $N$ is flat! In this case $\text{Hom}_A(L, M)$ is a finite limit (really an iterated finite limit, but this isn't an issue) in $M$, and $(-) \otimes_A N$ preserves it. Dually, it's also true if $N$ is finitely presented and $L$ is projective: in this case $M \otimes_A N$ is a finite colimit in $M$, and $\text{Hom}_A(L, -)$ preserves it. Note that in Neil Strickland's example neither $L$ nor $N$ is finitely presented.

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    $\begingroup$ Could you please give some reference for these materials concerning (co)limits? I didn't quite understand your argument, but I believe you had solve all of my problem. $\endgroup$ – Lao-tzu Dec 18 '15 at 0:39
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    $\begingroup$ @Lao-tzu: I don't have any references, but this isn't a difficult exercise. Giving a presentation of a module $N$ means writing it as the cokernel of a map between direct sums of copies of a free module. This lets you write $(-) \otimes_A N$, as a functor, as the cokernel of a map between direct sums of copies of the identity functor. Any functor that commutes with direct sums and cokernels (equivalently, any cocontinuous functor) therefore commutes with this functor. If $N$ is finitely presented then you can replace "direct sums" with "finite direct sums" above, and since... $\endgroup$ – Qiaochu Yuan Dec 18 '15 at 1:37
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    $\begingroup$ Bourbaki Commutative Algebra Section 9 (Extension of scalars in homomorphisms of modules) Proposition 10. $\endgroup$ – pro Dec 18 '15 at 1:38
  • $\begingroup$ ...any $\text{Ab}$-enriched functor between $\text{Ab}$-enriched categories commutes with finite direct sums, it now suffices for a functor to commute with cokernels in order for it to also commute with $(-) \otimes_A N$. Meditate, for example, on the special case that $N = A/a$ for some $a \in A$, which gives $M \otimes_A N \cong M/aM$, the cokernel of the diagram $M \xrightarrow{a} M$. There's a dual story for $\text{Hom}_A(L, -)$. All of this fits into a broader context of tensoring and powering in enriched category theory. $\endgroup$ – Qiaochu Yuan Dec 18 '15 at 1:38
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    $\begingroup$ @pro I checked Bourbaki Commutative Algebra (on page 22), it's really cool, Bourbaki did everything in the most general form! Thanks again! $\endgroup$ – Lao-tzu Dec 18 '15 at 5:59

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