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?



Comparing homomorphisms over different base rings
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?


