Let $X$ be a relatively minimal rational elliptic surface over a number field $k$, with elliptic fibration $\pi:X\to\Bbb{P}^1$.
It is well known that the anticanonical divisor $-K_X$ is linearly equivalent to a general fiber $F$ and $\chi(0)=1$, where $\chi(D):=h^0(D)-h^1(D)+h^2(D)$ and $h^i(D):=\dim_kH^0(X,\mathcal{O}_X(D))$ for any divisor $D$ in $X$.
Since the linear system $|F|$ defines the fibration $\pi:X\to\Bbb{P}^1$, I should expect $h^0(F)=2$, but I still don't now how to prove it.
Since $F^2=0$ and $F\cdot K_X=-F^2=0$, by Riemann-Roch for surfaces $\chi(F)=1$.
By Serre duality, $h^2(F)=h^0(K_X-F)=h^0(-2F)=0$, since $F$ is effective. Consequently $h^0(F)-h^1(F)=1$, but I don't know how to treat $h^1(F)$.
Again by duality, $h^1(F)=h^1(-2F)$, by I don't think this helps.
Any suggestions?
You can use the exact sequence $$ 0 \to \mathcal{O}_X \to \mathcal{O}_X(F) \to \mathcal{O}_F(F) \to 0. $$ The cohomology of the first term have dimensions $(1,0,0)$, because these are birational invariants and $X$ is rational. Furthermore, $$ \mathcal{O}_F(F) \cong \mathcal{N}_{F/X} \cong \mathcal{O}_F $$ because $F$ is the fiber of a morphism. Since this is an elliptic curve, its cohomology has dimensions $(1,1,0)$. Now from the exact sequence it follows that the cohomology of $\mathcal{O}_X(F)$ have dimensions $(2,1,0)$.