For vector spaces $V, W$ we have the following canonical isomorphisms, where $V^*$ denotes the dual space of $V$:
$$(V\otimes W)^*\cong V^*\otimes W^*$$
$$V^*\otimes W\cong \operatorname{Hom}(V, W)$$
Do these still hold true for modules $V, W$ over a commutative ring $A$?
I'm going to take a guess and say that the answer is probably yes, but apart from the case that $V, W$ are free, I was unable to prove this.
Best Answer
First of all, it does not always hold that $V^*\otimes W\cong \operatorname{Hom}(V, W)$, even for vector spaces! See this post for details. One general version of this that does hold in general is the tensor-hom adjunction $$ \mathrm{Hom}(V\otimes W,X)\;\cong\;\mathrm{Hom}(V,\mathrm{Hom}(W,X)). $$ This general fact gives us an interest result regarding $(V \otimes W)^*$. Letting $X$ be the base ring $R$ yields $$ (V \otimes W)^* = \mathrm{Hom}(V \otimes W,R) \cong \mathrm{Hom}(V,\mathrm{Hom}(W,R)) = \mathrm{Hom}(V,W^*). $$ From the first statement, we see that it does not necessarily hold that $\mathrm{Hom}(V,W^*) \cong V^* \otimes W^*$.