Rational first chern class of algebraic variety with zero Kodaira dimension.

Question

Let $X$ be a compact Kahler algebraic variety which has zero Kodaira dimension. Then the integral first chern class vanishes? What about rational first chern class?

Answer 1

No, the first Chern class need not vanish rationally (in particular, not integrally either).

The basic point is that Kodaira dimension is a birational invariant, whereas the first Chern class (or its negative, the canonical divisor class) is a biregular invariant. So you can change one without changing the other.

Here's an example. Let $X$ be a $K3$ surface. Then $\kappa(X)=0$ and $c_1(X)=0$. Now blow up a point on $X$, to give a smooth surface $Y$. The blowup map $\pi:Y \rightarrow X$ is birational, so $\kappa(Y)=\kappa(X)=0$. But the formula for the canonical class of a blowup says

$$ K_Y = \pi^*K_X +E = E$$

where $E$ is the exceptional curve of the blowup. So $c_1(Y)=-K_Y=-E$, which is a non-torsion class in $H^2(Y)$.

However, if $X$ is minimal, meaning that $K_X$ is a nef line bundle, and $\kappa(X)=0$, then it is a theorem of Kawamata that $mK_X$ is trivial for some natural number $m$. Hence $c_1(X)$ must be torsion, that is, it must vanish rationally. To see that this is not true integrally, take $X$ to be an Enriques surface: this is a minimal surface on which $K_X$ is nontrivial, but $2K_X$ is trivial.