Showing $\operatorname{Spec} k[x,y,z]/(x^2 + y^2+z^2)$ is normal 
Let $k$ be a field of characteristic not $2$ and algebraically closed. I would like to show that $\operatorname{Spec} k[x,y,z]/(x^2 + y^2+z^2)$ is normal. 

I would appreciate any help/hint. Thank you!
 A: Normal is equivalent to showing that the ring $k[x,y,z]/(x^2+y^2+z^2)$ is integrally closed. Normal is also equivalent to the Serre conditions $R_1$ and $S_2$, which are that the singular locus is at least codimension 2 (this is $R_1$) and that a rational function on an open subset has no poles in codimension one, then it is in fact regular (this is $S_2$). Proving integrally closed is probably not too hard, but it will likely be somewhat long/tedious, although I'm sure someone on this site has a wonderful trick for it. I'll instead explain why this variety satisfies $R_1$ and $S_2$ and is therefore normal.
Since our variety $X\subset \mathbb{A}^3$ is a hypersurface ($x^2+y^2+z^2$ is irreducible), condition 2 is automatic (this works for any local complete intersection inside a smooth variety). So normal is equivalent to the singular locus being codimension 2 (in this case, zero-dimensional). We can compute the singular locus from the Jacobian. This matrix is $$\begin{pmatrix} 2x & 2y & 2z \end{pmatrix}$$
and since $\operatorname{char} k\neq 2$, this matrix is rank zero exactly when $x=y=z=0$, which is a single point. So we've shown that the singular locus is codimension two, and therefore this variety is normal.
A: I think this is a standard HW problem in Hartshorne (somewhere in Chapter 2). A step by step guided proof based on other problems can be found in a series of problems in Ravi Vakil's book (page 162). 
If you still cannot do it, there is in fact an identical post addressing your question posted a while ago. Hope this 25 cents helps. 
