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

For $D=k{D}_1$, the lattice polytope of a toric divisor consists of $(m_1,m_2)$ satisfyingThus $m_2=0$ and $-k\le m_1\le 0$. Its lattice points index a basis of global sections, so

For every maximal cone, one endpoint of the interval of lattice points from part a realizes the corresponding Cartier datum. The toric basepoint-free criterion therefore makes $\mathcal{O}(k{D}_1)$ basepoint-free for $k\ge 0$. Equivalently, part d identifies it with the pullback of the globally generated bundle ${\mathcal{O}}_{{\mathbb{P}}^1}(k)$.

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.

Let $E_1$ be the invariant point divisor of ${\mathbb{P}}^1$ corresponding to the ray ${\mathbb{R}}_{\ge 0}$. Under $(u_1,u_2)\mapsto u_1$, the ray $(1,0)$ maps primitively to that ray, while no other ray maps into its interior. The toric pullback formula givesequivalently $D_1={\pi }^{-1}(E_1)$ as divisors.

The assertion is false. The Picard group of the Hirzebruch surface is freely generated by the negative section $S$ and a fiber $F=D_1$. Pullbacks from ${\mathbb{P}}^1$ form only the subgroup $\mathbb{Z}F$. For example, $\mathcal{O}(S)$ cannot be a pullback: its restriction to a fiber has degree $S\cdot F=1$, whereas every pullback from the base restricts trivially to every fiber.