Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 152 / 3 / c / Solution

The divisor $D_1$ is a fiber of the ruling and has $D_1^2=0$. Hence $(k{D}_1{)}^2=0$ for every $k$. An ample divisor on a complete surface has positive self-intersection, so no $\mathcal{O}(k{D}_1)$ with $k\ge 0$ is ample. These results are summarized by multiples of a fiber on the first Hirzebruch surface.