Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

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 $M$.

share|improve this answer
    
Thanks, that's very nice & short. I like it. –  TonyS Oct 27 '10 at 19:54
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.