Let $R$ be a commutative ring, and let $M,N,P$ be $R$-modules. In what generality can we say that $Hom_R(M,N\otimes_R P)\cong Hom_R(M,N)\otimes_R Hom_R(M,P)$. This is true in a cartesian monoidal category, and from that, it seems like we might be able to force it if all of the modules are free (look at basis-to-basis mappings and distribute the Hom first), but do we need any finiteness conditions? Do we need freeness at all?
In what generality does the second argument of Hom distribute over tensor?
1 Answers
This is not true even if all the modules are free; let $R$ be a field, let $M, N, P$ be finite-dimensional, and compare dimensions. (It is true if $\otimes$ is replaced with $\oplus$; this is because $\oplus$ is both a finite product and a finite coproduct.)
The tensor product is not a Cartesian product, and in this sense it does not even behave like a Cartesian product. The sense in which it does behave like a Cartesian product is the adjunction
$\text{Hom}_R(M \otimes_R N, P) \simeq \text{Hom}_R(M, \text{Hom}_R(N, P))$
where $\text{Hom}_R(N, P)$ is the $R$-module of all $R$-module homomorphisms. This is analogous to the adjunction
$\text{Hom}(M \times N, P) \simeq \text{Hom}(M, P^N)$
in $\text{Set}$, where $P^N$ is the set of all functions $N \to P$. This and the hom above are examples of "internal homs," and if internal hom has a left adjoint it defines a notion of tensor product which does not necessarily have to behave like a Cartesian product; when it does, you might be in a topos, and the category of $R$-modules is not a topos but an abelian category.