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If we blow up the projective plane in $3$ points in general position, then we get a del Pezzo surface of degree $3$. What happens if the points are all on one line (i.e. collinear)?

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  • $\begingroup$ @K02 that looks like an answer to me - please consider recording it as such below. $\endgroup$
    – KReiser
    Nov 27, 2023 at 22:03

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@KReiser suggested I record my comments as an actual answer. I have deleted my comments and put them here.

Antiampleness of the canonical divisor fails.

First off, let's recll some facts. For a blow-up, $Bl_p: S \rightarrow \mathbb{P}^2$, at one point $p\in \mathbb{P}^2$, the pullback of a curve $\pi^* C$ is linearly equivalent to $\widehat{C}+m E$ where $\widehat{C}$ is the strict transform and $m$ is the order of vanishing/multiplciity of $C$ along the point $p$ that you blew-up. Now you are doing this at three points so let $\pi:=Bl_{p_1,p_2,p_3}:S\to \mathbb{P}^2$ be the blow-up at three colinear points.

Let $\ell$ be the line through the three points. If you blow-up the three collinear points, then the strict transform $\widehat{\ell}$ is $\widehat{\ell}\sim\pi^* \ell-\pi^* E_1-\pi^* E_2-E_3$ for $E_i$ the exceptional divisors and $\pi: S \rightarrow \mathbb{P}^2$ the composition of the blow-ups. The canonical divisor is $\pi^* K_{\mathbb{P}^2}+\pi^* E_1+\pi^* E_2+E_3$. It is not anti-ample because when restricted to $\widehat{\ell}$, you get $\widehat{\ell} \cap-K_S=3-1-1-1=0$. A degree 0 line bundle on a rational curve $\widehat{\ell}$ has no chance of being ample.

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