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.
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3$\begingroup$ For me this follows quickly from the definitions of "nodal singularity" and "blow-up". Are you using different definitions? I would say a scheme has a nodal singularity if it is locally isomorphic to the vanishing locus of a single equation whose lowest-degree term is a nondegenerate quadratic form. $\endgroup$– Will SawinFeb 11, 2021 at 0:14
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$\begingroup$ @WillSawin It is the same definition. Can you suggest a reference or write down the idea in the answer. $\endgroup$– JanaFeb 11, 2021 at 0:25
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$\begingroup$ In local analytic coordinates, the variaty has the form $x^2+y^2+z^2+w^4=0$ in a 4-dimensional polydisc. Blowup the origin and take the strict transform, the exceptional divisor is given by the same equation in $\mathbb P^3$, which is a smooth quadric. $\endgroup$– AG learnerFeb 11, 2021 at 3:16
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$\begingroup$ @AGlearner: $w^4$ should be $w^2$. $\endgroup$– SashaFeb 11, 2021 at 4:09
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$\begingroup$ Oops! Sorry for the typo. $\endgroup$– AG learnerFeb 11, 2021 at 4:13
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1 Answer
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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.
answered Feb 11, 2021 at 19:14