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?


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    $\begingroup$ 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. $\endgroup$ Commented Mar 10, 2013 at 11:45

1 Answer 1


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 $5$.


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