Let $k$ be a field of characteristic $\neq 2$, and $n\geq m\geq 3$. Then $k[x_1,\ldots,x_n]/\langle x_1^2+\ldots+x_m^2\rangle$ is integrally closed.

Question

I wish to show that $B:=k[x_1,\ldots,x_n]/\langle x_1^2+\ldots+x_m^2\rangle$ is a normal domain -- that is to say, it is integrally closed in its ring of fractions. Unfortunately, I don't know of any theorems about quotients of normal domains being normal. The best I can say so is that $B$ is an integral domain, that is to say, $x_1^2+\ldots+x_m^2$ is irreducible.

Answer 1

$B$ integrally closed in it's ring of fractions is equivalent to $\operatorname{Spec} B$ being normal. This can be checked by the two Serre conditions, $R_1$ and $S_2$ (see for example http://stacks.math.columbia.edu/tag/031O).

$S_2$ is easy in this case, as it is true for any locally complete intersection inside a smooth variety. $\operatorname{Spec} B$ is a hypersurface in affine space, so we are done with this.

$R_1$ is really the meat of this question. $R_1$ means that we want $\operatorname{Spec} B$ to be regular in codimension 1- ie, the set of all singular points must be contained in some closed set of codimension 2. Where are the singular points? They're exactly the places where the Jacobian has non-maximal rank. Let's compute:

$$J=\begin{pmatrix} 2x_1 & 2x_2 & \cdots & 2x_m & 0 & \cdots & 0 \end{pmatrix}$$

Because this matrix is $1\times n$ and 2 is a unit, the set of singular points is all the points in $\operatorname{Spec} B$ such that $x_1=x_2=\cdots=x_m=0$. This has codimension $m-1$, which is at least 2 as $m\geq 3$. So $R_1$ and $S_2$ are satisfied.