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?


1 Answer 1


What you are trying to prove is not true.

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.

  • $\begingroup$ 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? $\endgroup$
    – rmdmc89
    May 11, 2021 at 0:26
  • $\begingroup$ 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. $\endgroup$ May 11, 2021 at 8:12

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