Desingularization of ordinary double point of a surface 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 $E \cong \mathbb{P}^1, E^2= -2$?
 A: 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.
