1
$\begingroup$

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?

$\endgroup$

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
    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... $\endgroup$ 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
    Oct 9, 2010 at 11:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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