Let $S$ the blow up of $P^2$ in nine points. Why is the anticanonical divisor $K_S$ not semiample?

1$\begingroup$ I would advise you to have a look (if you have not seen it yet) at the following beautiful article: arxiv.org/abs/0808.0695 , Hilbert's fourteenth problem over finite fields, and a conjecture on the cone of curves $\endgroup$– Dmitri PanovSep 26, 2011 at 10:21
2 Answers
Your nine points must be [EDIT: very] general [see MP's answer], otherwise it can be semiample.
The only effective anticanonical divisor is then (the strict transform of) the cubic C through the nine points. Since there is no other cubic curve cutting C in your nine points, the restriction of K_S to C is a noneffective divisor of degree 0 (C has genus 1). So the restriction of mK_S is also noneffective for all m (the points are [very] general in C! [I guess] Torsion points can make a difference) which means the only effective divisor in mK_S is mC. Thus mK_S is never base point free.
Just to complement quim's answer, note that if $S$ is defined over the algebraic closure of a finite field, then the anticanonical divisor of the blow up of 9 general points in $\mathbb{P}^2$ is actually semiample!
The reason this is not in contradiction with quim's answer lies in the subtlety of the word "general". Here the correct genericity assumption is what is sometimes called 'very general": the points need to lie in the complement of a countable union of closed subsets. Indeed, if you read quim's answer you see that there is a degree zero line bundle that needs to be nontorsion, and this condition translates to not being of order at most $n$ for every positive integer $n$: clearly a countable union of closed conditions. On the other hand, every degree zero line bundle on smooth curve defined over a finite field s torsion!!