# Is every nontrivial morphism already injective in this case?

I'm a little bit suprised at the moment, so i'll ask here if I see this wrong:

Given a sheaf of algebras $R$ ( e.g. maximal order or Azumaya) on a smooth projective scheme $X$ with generic point $p$. Asumme $M$ and $N$ are two left $R$-modules, coherent and torsion free over $O_X$, such that $M_p$ and $N_p$ are simple $R_p$-modules. Now given a nontrivial R-morphism $f: M \rightarrow N$. I think that f is automatically injective in this case.

Injectivity can be check on the open subsets $U \subset X$, there we have $f_U : M(U) \rightarrow N(U)$. We also have the injections $i_{U,p} :M(U)\rightarrow M_p$ and $j_{U,p}: N(U)\rightarrow N_p$ and the induced map $f_p: M_p \rightarrow N_p$, with the property $j_{U,p}\circ f_U=f_p\circ i_{U,p} (1)$.

Assume $f_p=0$, then we have $j_{U,p}\circ f_U=0$ by $(1)$ for every open subset $U$. But since $j_{U,p}$ is injective, this would imply that $f_U=0$ for every $U$ so $f=0$, which contradicts $f\neq 0$. So $f_p: M_p \rightarrow N_p$ is a nontrivial map between simple modules, so it is an isomorphism, especially injective. This implies that $f_U$ must be injective for every open subset $U \subset X$ because of $(1)$. So f is injective.

What do you think? Am I missing something?

I think you are right. Another way to prove this is the following: Let $K=\ker [f:M\to N]$ and $I={\rm im}[f:M\to N]$. Since $f$ is non-trivial, $I\neq 0$ and since it is torsion-free (as a subsheaf of $N$), $I_p\neq 0$. Then $K_p\subsetneq M_p$, so $K_p=0$, but then $K$ is a torsion-sheaf and hence $0$ since it is a subsheaf of the torsion-free sheaf $M$.