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Let $X$ be a surface (therefore $2$-dimensional, proper $k$-scheme) and $D$ a divisor with positive self intersection number $(D \cdot D) >0$. Futhermore it is nef therefore for each irreducible curve $C$ we have $(D \cdot C) \ge 0$.

Is then $D$ semi ample? By semi ampleness I mean here 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 $s(x) \neq 0$)

such that $\mathcal{L} = \mathcal{O}_X(D)$ holds.

Remark: Since we demand only $(D \cdot C) \ge 0$ and not $(D \cdot C) > 0$ we can't apply Nakai-Moishezon.

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  • $\begingroup$ The phrasing is a little confusing. You say "$C$ a divisor (so a curve)". Is this divisor supposed to be effective? Is it supposed to be irreducible (as "curve" might suggest)? $\endgroup$ – Asal Beag Dubh Oct 4 '18 at 9:10
  • $\begingroup$ @AsalBeagDubh: Good remark. I mean just a divisor. $\endgroup$ – KarlPeter Oct 4 '18 at 9:12
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No, this is not true. Note that for $D$ nef, the condition that $D^2>0$ is equivalent to $D$ being big.

There is a famous example due to Zariski of a surface $S$ and a divisor $D$ on $S$ such that $D$ is nef and big, but $D$ is not semi-ample. Roughly, we construct $S$ by blowing up $\mathbf P^2$ in a certain set of 12 points on an elliptic curve $C$. The fact that there are non-torsion line bundles of degree 0 on $C$ allows us to choose the points in such a way that the line bundle $D=4H-\sum_i E_i$ on $S$ is nef and big, but for every $m>0$ the proper transform of $C$ is in the base locus of $|mD|$.

For details, see Section 2.3.A of Positivity in Algebraic Geometry by Lazarsfeld.

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  • $\begingroup$ Hi. Thank you for your answer. So indeed the statement do not holds generally. The background of my question was the case if $S = \mathbb{P}(\mathcal{O}_{\mathbb{P}^2} \oplus \mathcal{O}_{\mathbb{P}^2}(2))$ is the Hirzebruch surface with divisor $C:=\mathbb{P}(\mathcal{O}_{\mathbb{P}^2}(2))$. Here I intended to treat this special case: math.stackexchange.com/questions/2942014/…. Do you see if concretely in this case $C$ is semi ample and if yes, why? $\endgroup$ – KarlPeter Oct 4 '18 at 11:48

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