Let $\mathcal{F}$ and $\mathcal{G}$ be semistable sheaves with the same Hilbert polynomial, i.e. $P(\mathcal{F})=P(\mathcal{G})$ (pay attention that these are Hilbert polynomial and not reduced Hilbert polynomial). If $\mathcal{F}$ or $\mathcal{G}$ is stable, than why is any non-zero morphism $f:\,\mathcal{F}\to\mathcal{G}$ an isomorphism? Alternatively, why every non-zero morphism, that is injective or suriective, an isomorphism?

  • $\begingroup$ Why do you know that your assertions are true? $\endgroup$ – Stefan Kohl Jul 5 '14 at 17:23
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    $\begingroup$ Dear Stephan, the first assertion is proposition 1.2.7 of the book "The Geometry of moduli spaces of Sheaves" by Huybrechts-Lehn and the second assertion is their answer to the first one. I don't understand their reason and I am not able to prove the first assertion by myself. $\endgroup$ – User3773 Jul 5 '14 at 17:35

Suppose $f:F \to G$ is injective, and let $C$ be the cokerel, so we have an exact sequence $0 \to F \to G \to C \to 0$. Since the Hilbert polynomial is additive on short exact sequences, and $P(F)=P(G)$, we get that $P(C)$=0. So for all $m>>0$, $h^0(X,C(m))=P(C)(m)=0$. Serre has a theorem that says that for all $m>>0$ $C(m)$ is globally generated. But we just saw that for $m>>0$, $C(m)$ has no global sections besides $0$, and so $C(m)=0$, so $C=0$.

The case when $f$ is surjective is similar; look at the kernel of $f$.


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