3
$\begingroup$

First of all, I would like to apologize if my question is stupid or a well known fact.

Let $F$ be a rational surface with $K_F^2=5$ and $f: F\rightarrow \mathbb{P_k^2}$ be a birational morphism contracting four exceptional curves. If $\mathbb{k}$ is not necessarily algebraically closed, is it true that $F$ is a del Pezzo surface of degree 5?

Thanks.

$\endgroup$
1
  • 2
    $\begingroup$ Even if $k$ is algebraically closed, the surface $F$ does not need to be del Pezzo: it is not dP if three of the blown-up points are collinear. $\endgroup$ Mar 10, 2013 at 11:45

1 Answer 1

5
$\begingroup$

The surface, as a surface defined over $\mathbb{k}$, is a del Pezzo surface if and only if it is a del Pezzo surface, viewed as a surface defined over $\overline{\mathbb{k}}$. So your surface is a del Pezzo surface of degree $5$ if and only if your birational morphism contracts exactly four $(-1)$-curves, and if the image of the four points are such that no $3$ are collinear.

Note that if the four curves are all defined over $\mathbb{k}$, then the surface is unique, up to isomorphism, since the points can be chosen to be $[1:0:0]$, $[0:1:0]$, $[0:0:1]$, $[1:1:1]$. However, there are different del Pezzo surfaces of degree $5$, not isomorphic over $\mathbb{k}$ (but only one isomorphism class over $\overline{\mathbb{k}}$), since you can blow-up points not defined over $\overline{\mathbb{k}}$, and the Picard group over $\mathbb{k}$ can be smaller than $5$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.