Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 134 / 2 / iii

Because $C\cong {\mathbb{P}}^1$ and $\mathrm{deg}{\mathcal{O}}_C(C)=C^2=0$, its normal line bundle is ${\mathcal{O}}_C(C)\cong {\mathcal{O}}_{{\mathbb{P}}^1}$. Thus every ${\mathcal{O}}_C(mC)$ is trivial and has $H^1=0$. The divisor restriction exact sequence gives, for every $m\ge 1$,The maps $H^1(X,(m-1)C)\to H^1(X,mC)$ are surjective. Their nonnegative finite dimensions eventually stabilize, so for every sufficiently large $m$ restriction $H^0(X,mC)\to H^0(C,{\mathcal{O}}_C)=k$ is surjective. A lift of $1$ is nowhere zero along $C$, and the canonical section of $mC$ is nowhere zero outside $C$. Together they generate ${\mathcal{O}}_X(mC)$. Therefore $C$ is semiample.

For those same large $m$, the exact sequence of global sections givesThus $h^0(X,mC)=m+O(1)$, and the Iitaka dimension is exactly one:Equivalently, its basepoint-free multiple defines a morphism to a curve: it is nonconstant because its sections grow, and cannot have two-dimensional image because $(mC{)}^2=0$. This is the semiampleness of a square-zero rational curve; it uses no characteristic-zero vanishing theorem.