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We consider the Hirzebruch surface $S = \mathbb{P}(\mathcal{E})$ with locally free sheaf $\mathcal{E} = \mathcal{O}_{\mathbb{P}^1} \oplus \mathcal{O}_{\mathbb{P}^1}(2)$. By definition of projectivation every section $i:C \subset S$ coresponds bijectively to a surjective bundle morphism $\mathcal{E} \to \mathcal{L}$ by taking into mind that $\mathbb{P}(\mathcal{E})$ represents a functor. Under this condition we have for this section $C= \mathbb{P}(\mathcal{L})$.

Futhermore it is known that $C^2 = deg(\mathcal{L}) - deg(K_{\mathcal{L}})$ where $K_{\mathcal{L}} = Ker(\mathcal{E} \to \mathcal{L})$.

Now back to our situation:

The considerations above provides us two obviously sections:

$C:= \mathbb{P}(\mathcal{O}_{\mathbb{P}^1}(2))$ coming from the sequence

$0 \to \mathcal{O}_{\mathbb{P}^1} \to \mathcal{E} \to \mathcal{O}_{\mathbb{P}^1}(2) \to 0$

and $D:= \mathbb{P}(\mathcal{O}_{\mathbb{P}^1})$ coming from the sequence

$0 \to \mathcal{O}_{\mathbb{P}^1}(2) \to \mathcal{E} \to \mathcal{O}_{\mathbb{P}^1} \to 0$

One can show that $(C \cdot D)=0$ and $C^2 = 2$.

My question is why and how to show that $C$ is a semi ample divisor?

Two remarks:

  1. by semi ample divisor I mean that there exist a semi ample invertible sheaf $\mathcal{L}$ (so for exery $x \in X$ there exist $n \in \mathbb{N}$ and a global section $s \in \Gamma(X, \mathcal{L}^{\otimes n}$) with property $s(x) \neq 0$)

  2. Here: Divisor with positive Selfintersection Number Semi Ample I asked about a more general case: If we consider a general surface with a nef divisor $N$ with $N^2 >0$ then - by considering @Asal Beag Dubh's answer - $D$ is in general not semi ample.

So my question here is if concretely in the setting above with a Hirzebruch surface the section $C$ is semi ample? If yes, why?

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  • $\begingroup$ Are you sure you mean $\mathcal{O}_{\mathbb{P}^2}$ etc. and not $\mathcal{O}_{\mathbb{P}^1}$? In the previous case you get a 3-fold and not a surface. $\endgroup$ – Mohan Oct 4 '18 at 12:05
  • $\begingroup$ @Mohan: Yes, you are right. Thanks. $\endgroup$ – KarlPeter Oct 4 '18 at 12:18
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Yes, in this more specific case $C$ is semi-ample, in fact globally generated. To show it:

  1. Note that it suffices to show that $O_S(C)$ has no base points along $C$.

  2. Use the twisted ideal sheaf seqeuence $$ \begin{align*} 0 \rightarrow O_S \rightarrow O_S(C) \rightarrow O_C(C) \rightarrow 0 \end{align*}$$ to show that every global section of $O_C(C)$ comes from a global section of $O_S(C)$.

  3. Note that $C$ is isomorphic to $\mathbf P^1$, and hence $O_C(C)$ is just $O(2)$, which is clearly globally generated.
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