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Let $S$ be a projective surface with pseudoeffective anticanonical divisor $-K_S$. Is it true that if $C$ is an integral curve with $C^2<0$ and $C \cdot K_S >0$, then $\max_C (C \cdot K_S)$ is finte?

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Yes: any curve with positive intersection with the canonical divisor must be a component of any effective representative of a (multiple of) the anticanonical divisor. In particular, there are only finitely many integral curves $C$ with $C \cdot K > 0$. – M P Sep 16 '11 at 15:34
Dear MP: but here the assumption is only that -K is pseudoeffective. It's not obvious to me that that implies effective... – user5117 Sep 16 '11 at 15:36
Ah, I see. The issue seems to be with ruled non-rational surfaces (use Zariski decomposition and Riemann-Roch on the nef part to deduce effectivity of a multiple). I need to think more about this case... – M P Sep 16 '11 at 15:38
It seems that the Zariski decomposition of -K will work: the negative part is the only place where the intersection with -K is negative and this part is effective (up to a multiple). – M P Sep 16 '11 at 15:45
up vote 7 down vote accepted

Let $-K=N+E$ be the Zariski decomposition of $-K$, so that $N,E$ are $\mathbb{Q}$-divisors with $N$ nef, $E$ effective, such that $N \cdot E = 0$ and the restriction of the intersection pairing to the irreducible components of $E$ is negative definite. It follows that if $C$ is an integral curve such that $C \cdot (-K)<0$, then we must have $C \cdot E < 0$, so that $C$ is a component of $E$. Since $E$ has only finitely many components, we deduce that there are only a finite number of integral curves with negative intersection with $-K$, and the required boundedness follows.

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Note that the assumption $C^2 < 0$ is superfluous: the integral curves $C$ satisfying $K \cdot C > 0$ must have negative square by the argument above. – M P Sep 16 '11 at 16:07
Thank you very much for the answer. – fds Sep 17 '11 at 10:09

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