Projective surface has an ample divisor

Question

Let $S$ be a projective surface. In https://mathoverflow.net/questions/63999/nef-divisor-on-surface, there is an argument proving that any nef divisor $D$ on $S$ has $D^2\geq 0$. But the argument implicitly assumes that $S$ has an ample divisor. Is it true that any projective surface has an ample divisor?

Answer 1

Yes: saying $S$ is projective means it admits a closed embedding in to $\Bbb P^n$. Then $S$ has a very ample divisor (pull back $\mathcal{O}_{\Bbb P^n}(1)$ along the closed embedding) and thus an ample divisor. See for instance 01VT, Hartshorne theorem II.7.6, or Vakil 17.3.9 on the interaction between very ample and ample.