hello everybody. someone can suggest me some reference or an example of a divisor nef $D$ on a surface such that $D^{2}<0$ if it exists?
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3$\begingroup$ As Francesco's answer shows, this is impossible. More generally, a nef divisor $D$ on a projective variety $X$ of dimension $n$ must satisfy $D^n \geq 0$. This is an implication of the so-called Nakai--Moishezon criterion, which is very useful to know. $\endgroup$– user5117May 5, 2011 at 18:35
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This example cannot exist, indeed every nef divisor $D$ on a projective surface satisfies $D^2 \geq 0$.
The following easy argument is borrowed from Miles Reid's "Chapters on algebraic surfaces".
Fix an ample divisor $H$ and consider the quadratic function
$p(t)=(D+tH)^2$.
Then $p(t)$ is continuous and increasing for $t \in \mathbb{Q}$, $t >0$ and $p(t)>0$ for sufficiently large $t$.
CLAIM. Let $t \in \mathbb{Q}$, $t >0$. Then $p(t)>0$ implies that also $p(t/2)>0$.
In fact, assume $p(t)=(D+tH)^2 >0$. Then since $H(D+tH)>0$ it follows that $n(D+tH)$ is effective for $n$ big enough. By the assumption that $D$ is nef, we have $D(D+tH) \geq 0$, hence
$p(t/2)=(D+(t/2)H)^2=D(D+tH)+(t/2)^2H^2 >0$.
This proves our claim. Therefore, taking the limit for $t \to 0$, it follows $D^2 \geq 0$.
answered May 5, 2011 at 14:24
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$\begingroup$ In a similar spirit, one could prove the claim by writing $D$ as a limit of ample divisors, all with positive self-intersection, and argue by continuity. $\endgroup$ May 5, 2011 at 18:35
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$\begingroup$ J.C.: This argument is the most satisfactory, I think, but looks slightly circular. Indeed, one needs the Nakai-Moishezon theorem to see that nef line bundles are limit of ample ones. And unless I'm mistaken, Francesco's argument is used in its proof. $\endgroup$– ACLMay 14, 2011 at 10:48