# Curve $C$ on a rational surface $S$, $h^0(S,\mathcal{O}_S(C))=1$ iff $C$ is an exceptional divisor

Let $$S$$ be a smooth rational surface over $$\Bbb{C}$$ with a birational morphism $$f:S\to\Bbb{P}^2$$.

It is well know that that $$f$$ is a composition of isomorphisms and blowups, so essentially $$S$$ is a sequence of blowups of $$\Bbb{P}^2$$.

I'm trying to prove the following claim (I think it's true, but I'm not sure):

A curve $$C\subset S$$ is such that $$h^0(S,\mathcal{O}_S(C))=1\Leftrightarrow$$ $$C$$ is an exceptional divisor from one of the blowups.

Here's my attempt:

First, if $$E$$ is an exceptional divisor, assume by contradiction that there is a nonconstant $$g\in k(S)$$ such that $$\text{div}(g)+E\geq 0$$. Then $$E$$ is the only pole of $$g$$. Since $$f^*:k(\Bbb{P}^2)\to k(S)$$ is an isomorphism, there is some function $$h\in k(\Bbb{P}^2)$$ with $$f^*(h)=g$$ and whose only pole is $$f(E)$$, which is a point in $$\Bbb{P}^2$$ (absurd).

Conversely let $$C$$ such that $$h^0(S,\mathcal{O}_S(C))=1$$. If $$C$$ is not an exceptional divisor, then there is a curve $$C'\subset\Bbb{P}^2$$ such that $$C$$ is the strict transform of $$C'$$. Taking $$g\in k(\Bbb{P}^2)$$ with $$C'$$ as a pole, then $$f^*(g)=g\circ f\in k(S)$$ has $$C$$ as a pole, which contradicts $$h^0(S,\mathcal{O}_S(C))=1$$.

Does this make any sense?

Certainly it is the case that if $$C$$ is the exceptional curve of a blowup, then $$h^0(S,O_S(C))=1$$.
But the converse is false. The simplest counterexample is to let $$f \colon S \rightarrow \mathbf P^2$$ be the blowup of 2 points, and let $$C$$ be the strict transform of the line joining the points. Then $$C$$ is a $$(-1)$$-curve, but it is not an exceptional curve of $$f$$.
In this example, $$C$$ is not an exceptional curve of the given morphism, but there is another morphism from $$S$$ for which $$C$$ is exceptional. However, it is also possible to find a curve $$C$$ on a surface $$S$$ as in your question for which $$h^0(S,O_S(C))=1$$ but $$C$$ cannot be contracted by any morphism to a smooth surface. For this, 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.
• Just one question about your last sentence. Let $F\in \Bbb{C}[x,y,z]$ be the polynomial defining the cubic through the 9 points and let $g:=\frac{x^3}{F}\in k(\Bbb{P}^2)$. Then $f^*(g)\in k(S)$ would be a nonconstant element in $H^0(S,\mathcal{O}_S(C))$, so that $h^0(S,\mathcal{O}_S(C))>1$. What am I getting wrong? May 11, 2021 at 0:26
• The rational function $f^*(g)$ is not an element of $H^0(S,O_S(C))$, because it also has poles along the exceptional curves. May 11, 2021 at 8:12