# Blow-up of projective plane in three collinear points is not a del Pezzo surface

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)?

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

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.