Blow-up of a three-dimensional variety at a node

Question

Let $X$ be a three-dimensional variety over $\mathbb{C}$ with a nodal singularity at a point, say $P$. Is the exceptional divisor of the blow-up of $X$ at $P$ isomorphic to a smooth quadric in $\mathbb{P}^3$? I read this statement in an article, but am not able to find a proof.

Answer 1

There is no need to take analytic coordinates.

There is a general principle for blowing up the solution set of a hypersurface $f(x_1,\dots,x_n)=0$ at a point $x_1,....,x_n=0$.

Recall that blowing up involves introducing projective coordinates $(y_1: \dots :y_n)$ in addition to our original ones, satisfying the relations $x_i y_j = y_i x_j$, and taking the closure of our original variety, minus the origin.

This closure will satisfy the equation $f(x_1,\dots,x_n)=0$. However, this equation is not very useful, as it vanishes when the $x_i$ vanish. To imrprove this equation, we can multiply it by $\frac{y_1}{x_1}$, and then clear the denominator using $y_1 x_j = x_1 y_j$. In other words, we take each monomial in the $x_i$s and replace one of the $x$s by a $y$. This produces an honest polynomial as long as the constant term vanishes. In fact, we can repeat it $k$ times, replacing $k$ $x_i$s in each monomial with a $y_i$, as long as the monomials in $f$ have degree at least $k$.

When we do this, we obtain an equation satisfied by the blow-up. As soon as we do this the maximal number of times, we obtain an equation with an $n$-dimensional solution set with an $n-1$-dimensional space of solutions over $0$, hence one in which the nonzero solutions are dense, which therefore must be the blowup.

The exceptional divisor is then calculated by setting the $x_i$s to $0$. In other words, we take the lowest-degree terms in the $x_i$ and then swap the $x_i$s for $y_i$s.

So if the lowest degree terms form a nondegenerate quadratic form, the exceptional divisor is the solution set of a nondegenerate quadratic form - a smooth hypersurface.