Computing $h^0(\mathbb{F}_n, \mathcal{O}_{\mathbb{F}_n}(aC_0+bF))$ in a Hirzebruch surface Let $\mathbb{F}_n=\mathbb{P}(\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(-n))$, $n\neq 1$ be a Hirzebruch surface. There exists a section $C_0$ of the natural projection to $\mathbb{P}^1$ such that $C_0^2=-n$. On the other hand, we call $F$ to a fiber of this projection. We have that:
$$
\mathrm{Pic}(\mathbb{F}_n)=\mathbb{Z}\langle C_0\rangle\oplus\mathbb{Z}\langle F\rangle.
$$
In other words, any divisor is linearly equivalent to $aC_0+bF$ for some integers $a,b$.
My question is, can we compute $h^0(\mathbb{F}_n, \mathcal{O}_{\mathbb{F}_n}(aC_0+bF))$?
 A: Firs, notice that if $a<0$, then $(aC_0+bF)\cdot F=a<0$ and since $F$ is nef, the $h^0=0$. So, we may assume $a\geq 0$. Next intersecting with $C_0$, we see that if $b<an$, then $h^0(aC_0+bF)=h^0((a-1)C_0+bF)$. So, we only need to understand the $h^0$ when $a\geq 0, b\geq an$. We use induction on $a$, if $a=0$, we see that $h^0(bF)=b+1$. Next use the exact sequence $0\to (a-1)C_0+bF\to aC_0+bF\to \mathcal{O}_{C_0}(b-an)\to 0$. An easy induction will give the $h^1$ to be zero and thus we get $h^0(aC_0+bF)=(a+1)(b+1)-n\frac{a(a+1)}{2}$ when $a\geq0, b\geq an$.
A: A more direct way is to compute first the pushforward of the line bundle with respect to the projection $p \colon \mathbb{F}_n \to \mathbb{P}^1$. Assume $a \ge 0$. One has
$$
p_*\mathcal{O}(aC_0 + bF) \cong
\mathcal{O}(b) \otimes Sym^a(\mathcal{O} \oplus \mathcal{O}(-n)) \cong
\mathcal{O}(b) \oplus \mathcal{O}(b-n) \oplus \dots \oplus \mathcal{O}(b-an)
$$
and since
$$
H^0(\mathbb{F}_n, \mathcal{O}(aC_0 + bF)) \cong 
H^0(\mathbb{P}^1, p_*\mathcal{O}(aC_0 + bF)),
$$
it remains to apply the standard formula for the global sections of line bundles on $\mathbb{P}^1$.
