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?



Why is multiplication with a scalar no global morphism?
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. 


