TITLE: Duals and Tensor products
QUESTION [28 upvotes]: Let $A$ be a commutative ring with a unit element.  Let $M$ and $N$ be $A$-modules.  Let $M^v$ and $N^v$ be the dual modules.  In general, do we have $M^v \otimes N^v \cong (M\otimes N)^v$?  It is definitely true when M and N are free.  I believe (though haven't worked out the details) that it is true when M and N are projective.  Both Atiyah & Macdonald and Lang don't say anything on the matter.  
I came up with this question while studying $Pic(A)$: defined as the group of isomorphism classes of projective modules of rank 1.  If I and J are projective $A$-modules of rank 1, then the fact that $Pic(A)$ is a group immediately implies $I^v \otimes J^v \cong (I\otimes J)^v$, as both are the inverse of $I\otimes J$.

REPLY [4 votes]: I may as well add the following.
One can always consider the dual map:
$$(M^\vee\otimes N^\vee)^{\vee} \leftarrow ((M\otimes N)^\vee)^{\vee}.$$
I think that under moderate hypotheses (maybe just requiring the modules to be finitely generated is ok, but maybe one needs more: it's certainly ok if additionally the ambient ring is normal) this is an isomorphism.
Here's a brief discussion as to why.  Both sides of the equality are reflexive because they are both duals of something.  In the normal ring case, the statement reduces to codimension 1 (because reflexive modules in a normal ring are determined in codimension 1, let me know if you need references).  Then you are done because the ring becomes a DVR (and thus a PID) and Neil's original answer gives you an isomorphism even without the dual.