Given a smooth and projective surface $S$ over an algebraically closed field $k$ and a sheaf of Azumaya algebras $R$, i.e. $R$ is a locally free $O_S$-module of finite rank. Let $M$ be a coherent and torsion free $O_S$-module, which is also a left $R$-module, such that generically $M_\eta$ is a simple $R_\eta$-module. Then we have $Hom_R(M,M)=k$.
Now $M^\*:=Hom_{O_S}(M,O_S)$$M^*:=Hom_{O_S}(M,O_S)$ is a right $R$-module and $M^{\*\*}$$M^{**}$ is a left $R$-module. We have the canonical map $\iota: M \rightarrow M^{\*\*}$$\iota: M \rightarrow M^{**}$.
Is it true that $Hom_R(M,M^{\*\*})$$Hom_R(M,M^{**})$ just consists of the muliples of $\iota$, i.e. is it a one dimensional $k$-vector space?
I tried to use the sequence $0\rightarrow M\rightarrow M^{\*\*} \rightarrow Q\rightarrow 0$$0\rightarrow M\rightarrow M^{**} \rightarrow Q\rightarrow 0$. Since $M$ is torsion free $Q$ has support in codimension 2. Then apply $Hom_R(M, - )$, which is left exact, so we get, with $Hom_R(M,M)=k$: $0\rightarrow k\rightarrow Hom_R(M,M^{\*\*}) \rightarrow Hom_R(M,Q)$$0\rightarrow k\rightarrow Hom_R(M,M^{**}) \rightarrow Hom_R(M,Q)$. But here i am stuck.
Or is this assertion wrong, i.e. are there more morphisms? If it is right, can it be generalized to a bigger class of algebras $R$?
