I'd like to see that $K_S$ is anti-ample for S a rational elliptic surface. I don't want to use the fact that $\chi(S) = 1$, where $\chi$ denotes the arithmetic genus (or equivalently by the canonical bundle formula that $K_S = -F$ for a fiber $F$). This is mentioned in the following paper on page 39. http://arxiv.org/pdf/0907.0298v3.pdf
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1$\begingroup$ Yes, I believe that sentence in the paper is mistaken. By definition, a surface $S$ is called Fano if $-K_{S}$ is an ample divisor. In particular, as Rhys points out, this requires $K_{S}^{2} >0$. The confusion is that Fano surfaces are rational, but not the other way around. So a rational elliptic surface is a nice example of a rational surface which is not Fano. $\endgroup$– BenightedCommented May 17, 2019 at 18:41
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The statement is false (assuming that $K_S$ 'anti-ample' means that $-K_S$ is ample). For $S$ a rational elliptic surface, we have $K_S^2 = 0$, and the Nakai-Moishezon criterion states (in part) that a divisor $D$ on a surface is ample only if $D^2 > 0$.
