Simple question about surfaces of general type.

Question

I have found that if $S$ is a minimal not ruled surface $\chi_{top}(S)\ge0$. On some book i saw that in the case of surfaces of general type the inequality is strictly. Why?

Answer 1

This is true for any smooth surface of general type. Here's one argument.

First, if $S$ is not minimal, then its minimal model $S_m$ is smooth, minimal, of general type, and $\chi(S_m)<\chi(S)$. So it is enough to prove the statement for minimal surfaces of general type.

If $S$ is minimal, then $K_S$ is nef. But since $S$ is general type too, $K_S$ is also big. A nef divisor on a surface is big if and only if its selfintersection is $>0$, so we get $c_1(S)^2=K_S^2>0$.

Now the Bogomolov–Miyaoka–Yau inequality says that $\chi(S) \geq \frac13 K_S^2>0$.