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 $N$?
$\textbf{Edit}:$ Would it be easier if \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 assume have $M$ to be R\otimes_A Hom_R(M,N) \cong R\otimes_A N$. Now because $R$-projective? For my purposes it would be enough to show it it for "rank 1" R$ is Azumaya we have $R$-modules, i.eR\cong Hom_A(R,A)$. they So we really have the same $A$-rank as R\otimes_A Hom_R(M,N) \cong Hom_A(M,N)$.
So we have an isomorphism for $R$.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?
