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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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