Blowup Computations

The doubled cone

The blowup of the affine plane at the origin, and self-intersection of the exceptional divisor.

(Taken from my notes on intersection theory)

Let $E$ be the exceptional divisor of $\mb P^2$ blown up at the origin $(0,0) =V(x, y)$. (Here we choose, arbitrarily, that the origin is the origin of the $z\neq 0$ affine patch.) Let $C=V(x)\subset \mathbb P^2$ and $D=V(y)\subset\mb P^2$. Then $C\cdot D = 1$ as two lines meet in $\mb P^2$ at exactly one point. Further consider $\tilde C$ and $\tilde D$, given by the strict transform of $C$ and $D$. The idea of this example is to argue that since $\pi$ is an isomorphism away from $E$, the intersection pairing of two divisors linearly equivalent $C$ and $D$ which do not meet $P$ is the same as the intersection pairing of the pre-image of those divisors under $\pi$, and then to use \cref{lem:linequivpullback} to compare those to the pre-images of $C$ and $D$ under $\pi$. Since the pre-image of $C$ and $D$ include $E$, we’ll then compute $\pi^{-1}(C)\cdot \pi^{-1}(D) = (\tilde C +E)\cdot (\tilde D+E)$ to deduce the intersection pairing of $E$ with itself.

Let $A$ denote the affine patch of $\mb P^2$ containing the origin (clearly the divisors cannot intersect away from the origin because the projection $\pi$ is an isomorphism away from the origin). Then the strict transform $\tilde C$ of $C$ can be described as a subset of $\mb A^2\times \mb P^1$ (with affine coordinates $(x, y)$ and projective coordinates $[z:w]$) as $\tilde C = V(x, xz-yw)$. Similarly, $\tilde D$ can be described as $\tilde D = V(y, xz-yw)$. In the affine patch $z\neq 0$ we may dehomogenize to obtain $\tilde C = V(x, x-yw) = V(yw, x-yw)$. This decomposes into two irreducible components; $V(w, x-yw)$ is the image of the strict transform $\tilde C$ under this dehomogenization and $V(y, x-yw)$ is the exceptional divisor. They intersect (transversely) at $V(x, y, w)$. When $w\neq 0$ there is no intersection; we dehomogenize to obtain $V(x, xz-y)$, which yields the single irreducible component $V(x,y)$ the exceptional divisor. A symmetric argument shows that $\tilde D$ intersects the exceptional divisor (transversely) at the point $V(x, y, z)$, and so the curves do not intersect. (They cannot intersect away from the exceptional divisor, and they do not intersect on the exceptional divisor). They are also effective divisors, because they are given by curves. Thus we obtain that $\tilde C\cdot \tilde D = 0$ by \cref{mult}.

Consider $\tilde C+E$ and $F = \pi^{-1}(V(x-1))$. By \cref{lem:linequivpullback}, $\pi^{-1}(C) = E + \tilde C$ is linearly equivalent to the line $\pi^{-1}(V(x-1))$, and $\pi^{-1}(D) = E+ \tilde D$ is linearly equivalent to $\pi^{-1}(V(y-1))$. Since $\pi$ is an isomorphism away from $E$ and neither $\pi^{-1}(V(x-1))$ nor $\pi^{-1}(V(y-1))$ meet $E$, we have that their intersection pairing is just the intersection pairing $V(x-1)\cdot V(y-1)$ in $\mb P^2$, which is one. This means that

But we already have that $\tilde C\cdot E = \tilde D \cdot E = 1$, and that $\tilde C\cdot \tilde D = 0$, so we obtain

or $0+1+1+E\cdot E =1$. Thus $E\cdot E=-1$.