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> 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?
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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?



Infinitely many exceptional curves on ruled surfaces
Let me write a short answer summarizing the comments above, so that the question will not appear unanswered anymore.



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> **Proposition.** The following holds.
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> **(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.
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> **(2)** If $S$ is any smooth, complex, projective surface with non-negative Kodaira dimension, then $S$ contains at most *finitely many* $(-1)$-curves.
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> Thus, surfaces containing infinitely many $(-1)$-curves are necessarily rational.
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*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 $|mK\_X|$.


