TITLE: Is it true that $\operatorname{Hom}_R(M,N) \otimes_R P=\operatorname{Hom}_R(M ,N \otimes P)$?
QUESTION [5 upvotes]: Let $M, N, P$ be modules over a commutative ring $R$. The above identity is true for $R$ a field:
Since the RHS $\cong$ (by passing to double dual)  $$\operatorname{Hom}(N^* \otimes P^*, M^*)\cong\operatorname{Hom}(P^*,\operatorname{Hom}(N^*,M^*)) \cong P^{**}  \otimes\operatorname{Hom}(M,N)\cong P \otimes\operatorname{Hom}(M,N).$$
In a couple examples that I worked out, it held true but the steps do not hold true if $N$ and $M$ have torsion.  Is the identity true?    

edit:  I had originally made a mistake when writing my question.  I have changed the question title to make it correct.  Originally I had written $\operatorname{Hom}_R(M,N) \otimes_R P=\operatorname{Hom}_R(M\otimes P,N)$ which as Qiaochu Yuan remarks is ludicrous (for the vector space case $RHS= P^*\otimes\operatorname{Hom}(M,N)$). Other than cosmetic changes the body of the text remain unchanged.

REPLY [2 votes]: As explained here, this does not hold in general, but it does hold if any of the following sets of hypotheses holds:

$P$ is finitely presented projective,
$M$ is finitely presented and $P$ is flat,
$P$ is finitely presented and $M$ is projective.

As an example of the second case, $P$ can be a localization of the base ring, and you get that if $M$ is finitely presented then hom commutes with localization. 
Your argument can be made to work under the additional assumption of dualizability, which for modules is equivalent to being finitely presented projective; see for example this blog post for details. But you assume in your argument that $M, N, P$ are all dualizable, and as mentioned above you actually only need to assume that $P$ is dualizable, although this doesn't cover all cases of interest.