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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?

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2 Answers

up vote 3 down vote accepted

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?

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This looks good. But then it seems to me, that this is in fact true for all $R$-modules $M$ and $N$. Or do we need the finiteness assumptions? Because i think, if $A\rightarrow B$ is the splitting extension, then $Hom_R(M,N)\otimes_A B \cong Hom_{R\otimes_A B}(M\otimes_A B,N\otimes_A B)$ isn't true for all $R$-modules. –  TonyS Oct 24 '10 at 17:18
    
I am pretty sure that being torsion-free is not required; but you are right, finiteness probably is needed in the argument. –  Angelo Oct 24 '10 at 17:56
    
Okay. Thanks a lot. –  TonyS Oct 24 '10 at 18:20
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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 $R\otimes_A \text{Hom}_R(M,N)$.

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Hold on, perhaps there is a problem: $Hom_A(R,N)$ as an $R$-module, might not be $N^r$?. I have to run now, so I will come back to this... –  Hailong Dao Oct 22 '10 at 19:57
    
Yes, i tried the Hom-Tensor adjunction. This should also give $Hom_R(R\otimes_A M,N)\cong Hom_A(M,N)$.But then i couldn't see why $R\otimes_A M$ is isomorphic to $M^r$ as an $R$-module. The same problem you encountered. –  TonyS Oct 23 '10 at 10:06
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