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Let $X$ be a projective complex surface ($\dim_\mathbb C X = 2$). If $(K_X, K_X) > 0$, is it true that $K_X$ is an ample line bundle?

I am trying to the affirmative answer using Nakai's criterion, which says that a line bundle $L$ on a compact surface is ample iff $(L,L)>0$ and $(L,D) > 0$ for every effective divisor $D$ on $X$.

By Riemann-Roch we have that $$ \dim H^0(X, O_X(D)) + \dim H^0(X, O_X(-D) \otimes K_X) - \dim H^1 (X, O_X(D) ) = \chi(X) + 1/2( (D, D) - (D,K_X) )$$

I'm not sure where to go from here, or if RR helps at all.

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  • $\begingroup$ As in your last question, you need more assumptions. Try the case of $\Bbb P^2$ first and see what happens. (This is the same example as last time.) $\endgroup$
    – KReiser
    Commented Jan 9, 2023 at 16:20
  • $\begingroup$ @KReiser the self-intersection is negative then though, right? $\endgroup$
    – hunter
    Commented Jan 9, 2023 at 16:34
  • $\begingroup$ The intersection product on $\Bbb P^2$ is $\mathcal{O}(a).\mathcal{O}(b)=ab$, right? Or am I misremembering? $\endgroup$
    – KReiser
    Commented Jan 9, 2023 at 16:40

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The answer is no. For example, the canonical divisor of $\mathbb{P}^2$ is $K=\mathcal{O}(-3)$, and so $K^2=9$. However, Nakai-Moishezon's criterion says that if $D$ is a divisor on a smooth surface such that $D^2>0$ and for any curve $C$, the intersection $DC$ is also positive, then $D$ is ample. See, for instance, Hartshorne's Book.

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