# Infinitely many exceptional curves on ruled surfaces

Take an algebraically closed field $k$. Let $C$ be a smooth projective variety of dimension $1$ over $k$(a curve). Consider a geometrically ruled surface $S :=C\times\mathbb{P}^1$. Does there exists a blow-up, say $\widehat{S}$ of $S$ at distinct points $p_1,\ldots,p_n$ on $S$, which has infinitely many exceptional curves?

This is in spirit of the analogous result for $C=\mathbb P^1$, where the answer is affirmative. The pith of the proof relies on the fact that blowing-up $\mathbb P^2$ at $k$ points in the general position might introduce large number of exceptional curve even for very small $k$. But for curves of higher genus it seems not to be the case. Is this observation correct?

Also could we say something for higher Kodaira dimension surfaces?

• I believe you can prove this just by looking at the map $\widehat{S}\rightarrow C$ and observing that it's generally smooth. It follows that the exceptional curves must lie in finitely many fibers, but each of those fibers is dimension $1$ and finite type.
– dhy
Apr 19 '17 at 16:39
• The rational curves cannot dominate $C$, so they are contained in fibres. It follows that you must have finitely many of them. Apr 19 '17 at 16:41
• If $S$ has higher Kodaira dimension, there can only be finitely many exceptional curves (here I'm understanding this to mean "exceptional curves of the first kind", i.e. $-1$-curves). Pf: $|mK_S|$ has an effective representative $C$ for some $m$. A $-1$ curve has intersection $-1$ with $K_S$ by adjunction, so the only candidates are the (finitely many) components of $C$.
– user47305
Apr 19 '17 at 17:41

Let me write a short answer summarizing the comments above, so that the question will not appear unanswered anymore.

Proposition. The following holds.

(1) If $$C$$, $$D$$ are smooth curves and $$g(C) \geq 1$$, then any blow-up of $$C \times D$$ contains at most finitely many $$(-1)$$-curves.

(2) If $$S$$ is any smooth, complex, projective surface with non-negative Kodaira dimension, then $$S$$ contains at most finitely many $$(-1)$$-curves.

Thus, surfaces containing infinitely many $$(-1)$$-curves are necessarily rational.

Proof. (1) Let $$\pi \colon S \to C \times D$$ be a blow-up in a finite number of points, and $$p \colon S \to C$$ the composition of $$\pi$$ with the projection onto $$C$$. Since $$g(C) \geq 1$$, no rational curve in $$S$$ can dominate $$C$$. It follows in particular that the $$(-1)$$-curves of $$S$$ are contained in reducible fibres of $$p$$, so there are finitely many of them.

(2) If $$E \subset S$$ is a $$(-1)$$-curve, then by adjunction $$K_XE = -1$$.
If we take a positive integer $$m$$ such that $$mK_X$$ is effective, it follows that $$E$$ is necessarily one of the finitely many components of the base locus of the complete linear system $$|mK_X|$$.