# Self intersection of blown up points and the lines which they lie on

I'm currently trying to understand the process of blowing-up, and a few things strike me as a little difficult to get an intuitive understanding of what's happening.

The current problem is on self intersection.

Why does the exceptional divisor I get by blowing up at a point $\mathbb{P}^1(\mathbb{C})\times\mathbb{P}^1(\mathbb{C})$ have self intersection -1? A mathematically abstract answer would be nice, but an included intuitive explanation of negative self-intersection would really take the cake.

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See the answers to this question: mathoverflow.net/questions/16788/… for an explanation of this. –  Emerton Sep 19 '11 at 2:51
Ah cheers Emerton, I did a search but obviously my keywords weren't exhaustive enough. –  philiph Sep 19 '11 at 6:25

If $\pi : \tilde{S} \to S$ is the blowup of $S$ (a smooth projective surface over $\mathbb{C}$) at $p$ and $E$ is the exceptional divisor then $E$ has self intersection $-1$; i.e. $E^2 = -1$. The way to see this is as follows:
1) If $C$ is a smooth curve passing through $p$ with multiplicity $m$, and $\tilde{C}$ is the strict transform of $C$, then $\pi^*C = \tilde{C} + mE$. (working this out in the example of blowing up the node of $y^2 = x^3 + x^2$ is helpful and the general argument is more or less the same).
2)If $D$ is any divisor on $S$ then $E.\pi^*(D) = 0$. This is because we can compute intersection numbers using any divisor in the linear equivalence class of $D$.
3) Now suppose that $C$ is a curve passing through $p$ with multiplicity one so that $\tilde{C} = \pi^*C - E$. Now intersect with $E$ to get $E^2 = -1$.