8

How to show that blow-up of $\mathbb P^2$ in 2 points is isomorphic to blowup of quadric in 1 point? I think it is a standard fact, but Google can not help me with it.

Update: I found an answer here: J. Harris "Algebraic Geometry: A First Course", example 7.22.

evgeny
  • 3,971
  • 2
    You can look at http://math.stanford.edu/~vakil/02-245/sclass13A.pdf Also the book Algebraic Surfaces and Holomorphic Vector Bundles by Robert Friedman Chapter 5. – Krish Mar 25 '15 at 16:08
  • 6
    As projective toric varieties, the polytope for ${\Bbb P}^1$ is a line, the polytope for ${\Bbb P}^1\times {\Bbb P}^1$ is a square. The polytope for ${\Bbb P}^2$ is a triangle. Cutting a corner is blowing up a point. Cutting two corners from the triangle gives you the same polytope as cutting one corner off the square. – Jan-Magnus Økland Mar 25 '15 at 16:19
  • @Jan-MagnusØkland: although I don't know toric geometry, it sounds very nice, thank you! – evgeny Apr 24 '15 at 08:03

1 Answers1

5

The following is presumably the argument in Harris:

It's a classical fact that a smooth quadric surface $Q$ in $\mathbf{P}^{3}$ is doubly-ruled. Let $p$ be an arbitrary point on $Q$, and $H$ a hyperplane not containing $p$. Every line through $p$ intersects $H$ exactly once. Moreover, every line through $p$ either is one of the two rulings of $Q$ through $p$, or else (because $Q$ is a quadric) intersects $Q$ exactly twice, at $p$, and at one other point.

Let $\overline{Q}$ be the blow-up of $Q$ at $p$. Each ruling of $Q$ has self-intersection $0$, so the proper transforms of the rulings through $p$ have self-intersection $-1$ in $\overline{Q}$.

Projection away from $p$ induces a map $\overline{Q} \to H$, bijective away from the proper transforms of the rulings through $p$, and collapsing (i.e., blowing down) the two rulings.

That is, blowing up $Q$ at $p$ is the same as blowing up $H$ at the two points corresponding to the rulings through $p$.

Andrew D. Hwang
  • 82,708