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Let $X: xz=y^2 \subset \mathbb{A}^3$ be a surface with ordinary double point. It is claimed that there exists a resolution $f:Y \to X$ for which the exceptional divisor is a curve $E \cong \mathbb{P}^1, E^2=-2$.

I don't know how to show the above claim. Is this resolution the blowup at the origin? Then how to show $E \cong \mathbb{P}^1, E^2= -2$?

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Yes you can resolve the singularity by a single blowing-up. That the exceptional divisor is $\mathbb P^1$ can be seen directly from the computation. The $2$ in the RHS of $E^2=-2$ is the dimension of the tangent space of $X$ at the singular point minus $1$. –  Cantlog Sep 25 '13 at 20:52
    
I don’t understand what you said about self intersection of exceptional divisor. Could you explain more about that? –  Li Yutong Sep 26 '13 at 1:21
    
You have $E^2=-(d-1)$, where $d$ is the dimension of the tangent space of the singular point. Unfortunately I don't know a simple proof. –  Cantlog Sep 26 '13 at 5:55
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One place to look for guidance is Miles Reid, "Chapters on algebraic surfaces", Chapter 4, Exercises 1--2 (in the version arxiv.org/pdf/alg-geom/9602006). –  Asal Beag Dubh Sep 26 '13 at 8:07

1 Answer 1

The resolution is indeed obtained by blowing up the origin. By blowing up, one replaces the origin by a copy of $\mathbb{P}^2$, called an exceptional surface. You will see that the intersection of your surface with this exceptional one will be a smooth plane quadric curve (where the role of the plane is taken by the exceptional surface. Does this make sense to you?). It is well known and also not hard to show that such a curve is a $\mathbb{P}^1$.

About the self intersection, the following way is an alternative. Let us complete our surfaces to projective ones and call your nodal surface $C$. It is in this case the same equation, but now in $\mathbb{P}^3$.

After blowing up $\mathbb{P}^3$ we obtain a strict transform of your surface, lets call it $\widetilde C$.

Are you familiar with a canonical divisor? One shows through a local calculation in $\mathbb{P}^3$ followed by adjunction that the canonical divisor on $\widetilde C$ is simply the pullback of the canonical divisor on $C$. In particular we will find $K_{\widetilde C}.E = 0$ (here $K_{\widetilde C}$ is the canonical divisor).

Now we use the genus formula (equivalently adjunction): $$ g(E) = 1 + (K_{\widetilde C}.E + E^2)/2 $$ and $E^2 = -2$ follows.

I hope im not using terms that are unfamiliar to you.

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