Let $S$ be a *uniruled* surface, ie admits a dominant map $ f:X \times \mathbb{P}^1$. Why then it's canonical divisor $\omega\_X$ cannot be trivial? Motivation: I want to understand why $K3$ surfaces cannot be uniruled?



$K3$ surfaces can't be uniruled
The comment by user @naf is completely correct: there is a simpler argument than the argument I sketched. However, the argument I sketched gives a stronger result, which Mumford conjectured is sharp.


Because of the semipositivity, for every smooth, projective variety $S$ over a characteristic zero field $k$, for a geometric generic point $x$ of $S$, the tangent direction at $x$ of any rational curve contained in $X$ and containing $x$ is contained in the stalk of the following coherent sheaf, $$
\mathcal{F}:=\bigcap \text{Ker}\left( T\_{S/k} \xrightarrow{\phi} \bigotimes^r \Omega\_{S/k}\right),
$$ where the intersection is over all integers $r\geq 0$ and all $\mathcal{O}\_S$-module homomorphisms $\phi:T\_{S/k} = \Omega^\vee\_{S/k} \to \bigotimes^r \Omega\_{S/k}$. In particular, for a nonzero, skew-symmetric section $\phi:T\_{S/k} \to \Omega\_{S/k}$ in dimension $2$, the kernel is the zero sheaf. Thus, there is no rational curve containing a geometric generic point on a surface with positive $h^0(S,\omega\_{S/k})$. The same argument proves that $S$ is non-uniruled (for $S$ of arbitrary dimension) if there exists an integer $m>1$ such that $h^0(S,\omega^{\otimes m}\_{S/k})$ is nonzero, i.e., if the $m^{\text{th}}$ plurigenus is nonzero


Mumford conjectured that the sheaf above, $\mathcal{F}$, is (generically) the kernel of the derivative map for the "rational quotient", i.e., the quotient of $S$ by all free rational curves. This quotient is only regular on a dense Zariski open subset of $S$, which we may as well take to be an open on which the quotient morphism is smooth to its image and on which $\mathcal{F}$ is locally free. Mumford conjectures that the restriction of $\mathcal{F}$ to this open equals the kernel of the derivative map of this smooth morphism.


Mumford's Conjecture is wide open; it is not even clear that the sheaf $\mathcal{F}$ is involutive. By the Rationally Connected Fibration Theorem, we do know that Mumford's conjecture is equivalent to the "Uniruledness Conjecture" (sometimes also attributed to Mumford): $S$ is uniruled if and only if the $m^{\text{th}}$ plurigenus equals zero for every $m>0$.


