Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 139 / 2 / vii / Solution

The intersection calculation in part (vi)(a) gives $F^2=0$. Part (vi)(b), with its necessary $F\ne 0$ qualification, excludes every nonzero fixed part. Thus $F=0$ and $|D|$ is movable. Part (v) gives $\dim |D|\ge 1$, and $D^2=0$ by hypothesis. Applying part (ii) provesThus a nontrivial nef divisor of square zero on a K3 surface defines a fibration over a curve; the key step is fixed-part elimination on a K3 surface.