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?



Boundedness of $C.K$ on a surface with $-K$ pseudoeffective
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.


