Divisor with positive Selfintersection Number Semi Ample

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.

• 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)? – Asal Beag Dubh Oct 4 '18 at 9:10
• @AsalBeagDubh: Good remark. I mean just a divisor. – KarlPeter Oct 4 '18 at 9:12

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 $$|mD|$$.
• 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? – KarlPeter Oct 4 '18 at 11:48