Comparing homomorphisms over different base rings I am trying to compare some homomorphism groups over different base rings, so given a commutative local ring $(A,\mathfrak{m})$ and a finite dimensional Azumaya algebra $R$ over $A$. 
If $M$ and $N$ are two left $R$-modules which are finitely generated and torsion free over $A$, is there an $A$-isomorphism $R \otimes_A Hom_R(M,N) \rightarrow Hom_A(M,N)$?
So for example if $M=N=A^n$ and $R=M_n(A)$, then we have $Hom_R(M,N)\cong A$ and $Hom_A(M,N)\cong M_n(A)$ so we have an isomorphism in this case.
What about "non trivial" examples? Or do we need to have put stronger conditions on $M$ and $N$?
$\textbf{New idea}$: I think the Hom-Tensor adjunction won't help here. So next try:
Put $M=R$, this gives $Hom_R(M,N)=Hom_R(R,N)\cong N$ as $A$-modules. So we have $R\otimes_A Hom_R(M,N) \cong R\otimes_A N$. Now because $R$ is Azumaya we have $R\cong Hom_A(R,A)$. So we really have $R\otimes_A Hom_R(M,N) \cong Hom_A(M,N)$.
So we have an isomorphism for $M=R^k$. Now what about if $M$ is a projective $R$-module? Then there is some $R$-module $P$ such that $M\oplus P\cong R^k$. Can we somehow conclude that we have an isomorphism for M in this case?
 A: I think this is true. There is a homomorphism $R \otimes_A Hom_R(M,N) \rightarrow Hom_A(M,N)$, given from the obvious $R$-module structure of $Hom_A(M,N)$. To check that this is an isomorphism we can make a faithfully flat extension and split $R$; so we may assume that $R$ is a matrix algebra $M_n(A)$.
Now we use Morita equivalence: tensoring with the $(R-A)$-bimodule $A^n$ gives an equivalence of categories between $A$-modules and $R$-modules. If $M = A^n \otimes_A M'$ and $N = A^n \otimes_A N'$, we have
$$
Hom_A(M,N) = Hom_A(A^n \otimes_A M', A^n \otimes_A N') = M_n(A) \otimes_A Hom_A(M',N') = R \otimes_A Hom_R(M,N)
$$
and it should be easy to see that this is the desired isomorphism.
Does this work?
A: May be I am missing something, but the adjoint isomorphism gives:
$$ \text{Hom}_R(M, \text{Hom}_A(R,N)) \cong \text{Hom}_A(R\otimes_R M, N) = \text{Hom}_A( M, N)$$
Since $R$ is Azumaya, $R$ is a projective, hence free $A$-module of finite rank $r>0$. So the left hand side is isomorphic to $  \text{Hom}_R(M,N)^{\oplus r}$, which is also isomorphic (as $A$-modules) to $R\otimes_A \text{Hom}_R(M,N)$. 
