# In what generality is the natural map $\textrm{Hom}_R(L,M)\otimes S \to \operatorname{Hom}_{R \otimes S}(L\otimes S, M \otimes S)$ an isomorphism?

Note: A version of this question has been reposted to MathOverflow.

Let $k$ be a commutative ring, $R$ and $S$ commutative $k$-algebras. Let $L$ and $M$ be $R$-modules.

Unless I am much mistaken, there is a natural map $$\operatorname{Hom}_R(L,M)\otimes_k S \to \operatorname{Hom}_{R \otimes_k S}(L\otimes_k S, M \otimes_k S)$$ where $\phi \otimes s$ defines the $(R \otimes_k S)$-morphism $$l \otimes s_1 \mapsto \phi(l) \otimes s s_1.$$ Based on a dimension count, I believe that this map is an isomorphism under sufficiently nice hypotheses.

Under what hypotheses is this natural map an isomorphism? In particular, do we need to assume that some or all of the modules are (locally) free and/or finitely generated?

I'm also going to take the liberty of throwing in an additional, more general question. When I am trying to follow an argument that involves an unexplained "equality" of two expressions involving tensor product and/or Hom, I often find myself wanting to write down intermediate steps such as the one above, or the statement $$(M \otimes_k S) \otimes_{R \otimes_k S} (N \otimes_k R) = M \otimes_k N$$ for $M$ an $R$-module and $N$ an $S$-module. In this situation, I tend to find myself thinking, "This looks familiar and plausible, and I can write down something on simple tensors that probably defines a natural map in at least one direction, but I can't remember for sure if this is actually an isomorphism under my hypotheses." Is there a good "cheat sheet" (ideally a reference that is legally available online) for these sorts of things?

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Aaron and I were talking today and came up with a partial answer, as follows.

For simplicity, let $T$ denote $R \otimes_k S$. The tensor-Hom adjunction (in a version that allows for change of rings) tells us that \begin{align*} \operatorname{Hom}_T(L \otimes_k S, M \otimes_k S) & = \operatorname{Hom}_T(L \otimes_R T, M \otimes_k S) \\ & = \operatorname{Hom}_R(L, \operatorname{Hom}_T(T, M \otimes_k S)) \\ & = \operatorname{Hom}_R(L, M \otimes_k S). \end{align*} Now, we come to a hypothesis required for all the remaining reasoning: Namely, we assume that $S$ is free as a $k$-module. (This is automatically true if $k$ is a field.) Thus, considering $S$ as a $k$-module, we write $$S = \bigoplus_i k,$$ where $i$ ranges over some indexing set $I$. Then, we have $$\operatorname{Hom}_R(L, M) \otimes_k S = \bigoplus_i \operatorname{Hom}_R(L, M)$$ while $$\operatorname{Hom}_R(L, M \otimes_k S) = \operatorname{Hom}_R(L, \bigoplus_i M).$$ If $L$ is finitely generated as an $R$-module, or if the direct sum is finite, then these two are naturally isomorphic. Thus, sufficient hypotheses to ensure an isomorphism are that $S$ is free over $k$ and either ($L$ is finitely generated over $R$) or ($S$ is finite over $k$).

On the other hand, if $L$ is not finitely generated, then the natural map need not be surjective. For instance, if $L = \bigoplus_{j=1}^{\infty} R e_j$, $M = R$, and $i$ ranges over the integers, then we have \begin{align*} L & \to \bigoplus_i M e_i \\ e_j & \mapsto e_1 + \dotsb + e_j, \end{align*} a morphism that does not come from $\bigoplus_i \operatorname{Hom}_R(L, M)$. So in this case, the map is not an isomorphism.

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The counterexample appears to violate the accepted wisdom that "the map is an isomorphism if all modules in sight are free." Thus, I would appreciate it if people would point out any errors here. –  Charles Staats Aug 2 '11 at 18:45
The accepted wisdom was wrong. If it wasn't, then we would have a cannonical isomorphism $\operatorname{End}(V)\cong V^{\star}\otimes V$ for all vector spaces $V$, (take $R=M=k, L=S=V$, where you have put an algebra structure on $S$). The issue is that tensor products do not commute with direct products, even though they commute with direct sums. –  Aaron Aug 4 '11 at 5:23

It looks (to me) like this map is an isomorphism if $L$ is a free $R$-module.

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Of course, I assume all modules finitely generated in the answer above. –  curious Jul 11 '11 at 15:09
ah right, my bad. I will delete the remark, the first sentence is still alright I think. –  curious Jul 11 '11 at 22:08