For $s\leq 8$, if we blowup $\mathbb{P}^2$ at $s$ general points, we get a Del Pezzo surface. I am wondering what happens if $s\geq 9$? How does this 8 being calculated?
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8Don't do it! P^2 can take only so many blowups! If you overdo it... it'll blow up! – Mariano Suárez-Álvarez Jan 17 '13 at 06:50
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@MarianoSuárez-Alvarez: Interesting! Mind elaborate more on this? – minimax Jan 17 '13 at 07:07
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I was joking ;-) – Mariano Suárez-Álvarez Jan 17 '13 at 07:24
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@MarianoSuárez-Alvarez: Ah great :) How was the number 8 correspond to anti-canonical being ample? – minimax Jan 17 '13 at 17:14
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The canonical class of the blow-up in question is $K_X = -3h + \sum_{i=1} e_i$, where $e_i$ are the exceptional curves. In particular, this gives $(-K_X)^2 = 9 -k$. If $k \geq 9$, this is less than or equal $0$. But an ample divisor must have positive self-intersection! So there is no way it's del Pezzo.
Note that positive self-intersection doesn't imply ampleness; you have to actually check that it works for the smaller cases. For $1 \leq s \leq 6$ in fact $-K_X$ is very ample, while for $s = 7,8$ if memory serves you get $-K_X$ not very ample, but $-2K_X$ is. At least the first of these is proved in Hartshorne, in the chapter about the cubic surface (V.4 maybe? I'm afraid I don't have the book in front of me)
The case of $k \geq 9$ points has very interesting geometry, and some quite basic looking conjectures (most famously Nagata's conjecture) are still open. One place to read a bit about this is here: http://www.uni-due.de/~mat903/preprints/nagata1.pdf
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