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Comrades,

Let $S$ be a non-ruled, minimal (smooth projective complex algebraic) surface. Let $K$ be a canonical divisor of $S$ and $H$ a hyperplane section of $S$ (for your favorite embedding).

Suppose I know that $K^2 > 0$. Then supposedly $S$ being non-ruled forces $H.K > 0$. Is this correct or should it only say $H.K \geq 0$? Why is it correct?

This is at the beginning of chapter 9 in beauville's surfaces book. I can say for sure that $H.K \geq 0$ since a surface is ruled iff there exists a non-exceptional curve $C \subseteq S$ satisfying $C.K < 0$.

Thanks in advance.

edited, i said something stupid in original post.

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2 Answers

up vote 5 down vote accepted

$K^2 >0$ and $S$ non ruled implies that $S$ is of general type. In particular $mK$ is effective for some $m \geq 1$.

Since $H$ is very ample it follows $mKH >0$, that is $KH >0$.

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This actually holds in general, that is, if $d=\dim X$ is arbitrary. At least in characteristic $0$. $X$ being minimal is defined by $K_X$ being nef, for the classical notion of minimal surfaces this is equivalent to being a non-birationally-ruled surface (strictly speaking $\mathbb P^2$ is not ruled, but you would surely want to avoid it, one could also say non-ruled, non-rational).

If $X$ is minimal (i.e., $K_X$ is nef), then $K_X^d>0$ implies that $K_X$ is big, i.e., $X$ is of general type. Then by the Basepoint-free Theorem (see Theorem 3.3 in Kollár-Mori: Birational Geometry of Algebraic Varieties) $|mK_X|$ is basepoint-free, so for any $0<j<d$, $K^j_X$ maybe represented by an effective cycle and hence $H^{d-j}\cdot K_X^j>0$.

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