Solution
= Solution
The divisor $D_1$ is a fiber of the ruling and has $D_1^2=0$. Hence $(kD_1)^2=0$ for every $k$. An ample divisor on a complete surface has positive self-intersection, so no $\mathcal O(kD_1)$ with $k\geq0$ is ample. These results are summarized by <multiples of a fiber on the first Hirzebruch surface>.