Why are K3 surfaces minimal?

Question

I need to prove that all K3 surfaces are minimal surfaces, so that every birational map between K3 surfaces is an isomorphism. I've started to read Beauville's book on complex algebraic surfaces: there it says that the fact that K3 surfaces are minimal comes from the definition ($K=0$ and $H^{1,0}(X)=0$), but I can't see why. Do you know the proof or where I can find it?

Answer 1

What Beauville says is that "[c]learly, $K\equiv 0$ implies that they are minimal."

A surface $S$ is minimal if any birational morphism $S\to S'$ to any other surface $S'$ is an isomorphism. Thus, suppose that some K3, say $S,$ is not minimal. This means, by definition, that there exists a birational morphism $S\to S'$ which is not an isomorphism. By Theorem II.11 of Beauville, such a morphism can be factored as a (finite, nonzero) sequence of blowups at a point, and by Proposition II.3(iv), $K_S$ is a nonzero effective linear combination of pullbacks of the exceptional divisors of the successive blowups plus the strict transform of $K_{S'}$. Since $\operatorname{Pic}(S)=\operatorname{Pic}(S')\oplus \mathbb Z^m,$ where $m$ is the number of blowups, this gives $K_S\neq 0$ (i.e., we cannot cancel by linear equivalence) thus the contradiction, hence the K3 must be minimal.