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 $I\otimes J$. 



Duals and Tensor products
In general, we have a map $\mu:M^\vee\otimes N^\vee\to (M\otimes N)^\vee$ given by $\mu(\phi\otimes\psi)(\sum\_i m\_i\otimes n\_i)=\sum\_i\phi(m\_i)\psi(n\_i)$; this is presumably what mephisto is referring to, and it is an isomorphism in the case that he mentions. If the ring $A$ is a principal ideal domain and $M$ and $N$ are both finitely generated then the map is again an isomorphism, for rather uninteresting reasons, because the functor $M\mapsto M^\vee$ kills all torsion modules. 


Now let $M$ be the free module over $A$ with basis $\{e\_n\}\_{n\in\mathbb{N}}$ and take $N=M$ and define $\xi:M\otimes M\to A$ by $\xi(e\_i\otimes e\_i)=1$ and $\xi(e\_i\otimes e\_j)=0$ for $j\neq i$. I claim that this is not in the image of $\mu$. Indeed, if $\zeta=\mu(\sum\_{i=1}^r\phi\_i\otimes\psi\_i)$ then the matrix $(\zeta(e\_i\otimes e\_j))\_{i,j=0}^k$ has rank at most $r$ for all $k$, and it is clear that $\xi$ does not have this property. 


For another interesting example, take $A=k[x,y]/(xy)$ and $M=A/x$ and $N=A/y$. Multiplication by $y$ gives a map $\phi:M\to A$, so $\phi\in M^\vee$. Clearly $x\phi=0$ so multiplication by $\phi$ gives a well-defined map $M=A/x\to M^\vee$, which is an isomorphism. Similarly we have an isomorphism $N\to N^\vee$, so $M^\vee\otimes N^\vee\simeq M\otimes N=A/(x,y)=k$. On the other hand, $(M\otimes N)^\vee=\text{Hom}\_A(A/(x,y),A)=0$. Thus, we have a case where $\mu:M^\vee\otimes N^\vee\to(M\otimes N)^\vee$ is not injective. 


