= Solution
First suppose $\operatorname{char}k\ne2$. 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. Thus
$$
A=k[t_1,\ldots,t_n]/(t_1^2+\cdots+t_n^2)
$$
is a domain of dimension $n-1$. Its gradient is $(2t_1,\ldots,2t_n)$, so the <Jacobian criterion> shows that the only singular point is the origin. Its codimension is $n-1\ge2$, 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 \b[$\boxed{X\text{ is normal}}$], 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.
Back to article page
