Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 21 / 3

We first prove the local algebra fact behind regularity in codimension one. Let $(A,\mathfrak{m})$ be a one-dimensional normal Noetherian local domain. Choose $0\ne a\in \mathfrak{m}$. Its radical is $\mathfrak{m}$, so some ${\mathfrak{m}}^n$ lies in $(a)$; take minimal $n$. If $n=1$, then $\mathfrak{m}=(a)$. Otherwise choose $b\in {\mathfrak{m}}^{n-1}\setminus (a)$ and put $z=b/a$ in the fraction field. We have $z\notin A$ but $z\mathfrak{m}\subseteq A$.

If $z\mathfrak{m}\subseteq \mathfrak{m}$, express multiplication by $z$ on a finite generating set of $\mathfrak{m}$. The determinant trick gives a monic polynomial in $z$ annihilating $\mathfrak{m}$; as this ideal is nonzero in a domain, the polynomial itself vanishes. Normality would then give $z\in A$, a contradiction. Hence there is $u\in \mathfrak{m}$ with $zu$ a unit of $A$. Every $v\in \mathfrak{m}$ satisfies $v=u(zv)/(zu)$, proving $\mathfrak{m}=(u)$. The maximal ideal therefore has one generator in dimension one, so $A$ is a regular local ring, indeed a discrete valuation ring. This proves one-dimensional normal local rings are discrete valuation rings directly.

Now a codimension-one point of the normal variety has such a local ring, and a codimension-zero local ring is a field. Both are regular. Since the algebraically closed ground field is perfect, regularity and smoothness agree for this finite-type variety, and the Jacobian criterion makes the singular locus closed. It contains no codimension-zero or codimension-one point. Therefore every irreducible component of the singular locus has codimension at least two, givingFor $d\le 1$ this means the singular locus is empty. In particular the singular set itself is the required closed subset, not merely a set contained in some larger one.

First suppose $\mathrm{char}k\ne 2$. The quadratic form has rank $n>2$, so it is irreducible: a reducible homogeneous quadratic would be a product of two linear forms, whose associated symmetric matrix has rank at most two. Thusis a domain of dimension $n-1$. Its gradient is $(2t_1,\dots ,2t_n)$, so the Jacobian criterion shows that the only singular point is the origin. Its codimension is $n-1\ge 2$, giving regularity at all codimension-one points, the $(R_1)$ condition.

At every localization the polynomial ambient ring is regular and hence a Cohen-Macaulay ring. The nonzero quadratic equation is a nonzero divisor; quotienting by it decreases both depth and dimension by one. Hence the hypersurface local rings remain Cohen-Macaulay, so their depth is their dimension and in particular at least $\min (2,\dim )$. This verifies $(S_2)$. The Serre criterion for normality now yields $\boxed{X\text{ is normal}}$, establishing normality of a quadratic cone with all its hypotheses checked.

In characteristic two, $t_1^2+\cdots +t_n^2=(t_1+\cdots +t_n{)}^2$. The reduced zero variety is therefore the hyperplane $t_1+\cdots +t_n=0$, isomorphic to affine $(n-1)$-space, and its polynomial coordinate ring is integrally closed. Thus the reduced variety is normal in this case too. If one instead keeps the nonreduced hypersurface scheme defined by the squared equation, its coordinate ring has a nonzero square-zero class and is not normal. The variety interpretation in the question requires the reduced zero set.