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Given a smooth projective surface $S$ over an algebraically closed field, a sheaf rings or algebras $R$ on $S$ and a simple left $R$-module $M$, i.e. $Hom_R(M,M)=k$.Then we have $Hom_R(M,M(-i))=H^{0}(S,\mathcal{H}om_R(M,M)\otimes O(-i))=0$ for $i>0$.

Now given some $a\in k, a\neq 0$. Then $Hom_R(M,M(-i))=0$ implies multiplication with $a$ doesn't give a global morphism $M\rightarrow M(-i)$. But what is the reason for this? Is this because as a constant $a$ doesn't have any zeroes or poles?

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The element $a \in k$ has degree zero so it gives a global morphism $M \to M$. These are the only global morphisms because of your simplicity assumption.

If you take instead an element of degree $i > 0$, multiplication with it gives a global morphism $M(-i) \to M$.

Think about $S=\mathbb{P}^2$ and $M=\mathcal{O}$ if you want to convince yourself with a basic example.

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  • $\begingroup$ Okay, can i rephrase this like: Assume multiplication with $a \in k$ is a global morphism $M\rightarrow M(-i)$, multiplication with an element $b$ of degree $i>0$ gives a global morphism $M(-i)\rightarrow M$, so we have a global morphism $M\rightarrow M$, which must be multiplication with an element $c$ of degree zero. But then $a$ cannot have degree zero because then we would have $0=deg(c)=deg(ba)=deg(b)+deg(a)=i$ ? $\endgroup$
    – TonyS
    Commented Oct 9, 2010 at 11:06
  • $\begingroup$ Yes, if you like you can see it in this way... The point is that the composition will be the zero map, since $M \to M(−i)$, as you observed, is necessarily the zero map. Think about the easiest situation, when $M$ is an invertible sheaf. Then a global morphism $M \to M(−i)$ corresponds to a section of $\mathcal{O}(−i)$, so it must be zero... $\endgroup$
    – Francesco Polizzi
    Commented Oct 9, 2010 at 11:24
  • $\begingroup$ Okay. Thanks for your help. At first i thought it is unsual that "just" multiplication with a scalar doesn't give a global morphism. But now it makes sense. $\endgroup$
    – TonyS
    Commented Oct 9, 2010 at 11:39

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