If $E$ is a curve on a surface $X$ with $E^2=0$, then $h^0(X,\mathcal{O}_X(E))\geq 2$?

Question

Let $X$ be a smooth rational algebraic surface over $\Bbb{C}$ such that $-K_X$ is effective.

Let $E\subset X$ be a curve such that $E^2=0$.

Is it necessarily true that $h^0(X,\mathcal{O}_X(E))\geq 2$?

Here's where I'm at:

By Serre duality $h^2(X,\mathcal{O}_X(E))=h^0(X,K_X-E)=0$, since $E-K_F$ is effective. Since $X$ is rational, $\chi(X)=1$ and by Riemann-Roch, $h^0(X,\mathcal{O}_X(E))-h^1(X,\mathcal{O}_X(E))=1-\frac{1}{2}E\cdot K_X$.

Since $-K_X$ is effective and $E^2=0$, then $-E\cdot K_X\geq 0$. If $-E\cdot K_X>0$, then necessarily $h^0(X,\mathcal{O}_X(E))\geq 2$.

Now I don't know what to do when $E\cdot K_X=0$.

Answer 1

This is not true, and the counterexample is in fact the same one as for your previous question: let $S$ be the blowup of $\mathbf P^2$ in 9 general points, and let $C$ be the proper transform of the unique cubic through those 9 points.

Then $C^2=0$ and $h^0(S,C)=1$.