Show that the surface $xy = z^2$ in $\mathbb A^3$ is normal, working in characteristic $2$. I want to prove that the surface $S : xy = z^2$ in $\mathbb A^3$ is normal. In characteristic $\ne 2$, this is easy:

*

*Notice that $S \cong \mathbb A^2 / G$, where $G = \{ 1, -1 \}$ acts on $\mathbb A^2$ by scalar multiplication.


*Since $\mathbb A^2$ is a normal surface, then so is the orbit space $\mathbb A^2 / G$.
But this does not work in characteristic $2$. How could I solve this problem?
 A: $\def\Frac{\operatorname{Frac}}\def\fieldchar{\operatorname{char}}$
We'll show that $k[x,y,z]/(xy-z^2)$ is integrally closed.
We see that $K=\Frac (k[x,y,z]/(xy-z^2))$ is a degree-two extension of $k(x,y)$, so we can write any element of $K$ as $u+vz$ for $u,v\in k(x,y)$.
Assume $u+vz$ is integral over $k[x,y,z]/(xy-z^2)$, and let $Q(T)$ be a monic polynomial with coefficients from $k[x,y,z]/(xy-z^2)$ which is satisfied by $u+vz$.
We may assume $Q(T)$ has coefficients from $k[x,y]$: define an automorphism $\overline{-}$ of $k[x,y,z]/(xy-z^2)$ by $z\mapsto -z$ and replace $Q(T)$ by $Q(T)\overline{Q(T)}$ to get a monic polynomial with coefficients from $k[x,y]$ satisfied by $u+vz$.
If $v=0$, then $Q(T)$ shows that $u\in k(x,y)$ is integral over $k[x,y]$, so $u\in k[x,y]$ since $k[x,y]$ is integrally closed.
If $v\neq 0$, then $P(T)=T^2-2uT-(v^2xy-u^2)$ is the minimal polynomial of $u+vz$ as an element of $\Frac(k[x,y,z]/(xy-z^2))$ over $k(x,y)$, and therefore it must divide $Q(T)$ in $k(x,y)[T]$.
By Gauss' lemma, this means that $P(T)$ actually has coefficients in $k[x,y]$.
Now if $\fieldchar k\neq 2$, we are quickly done: $2u\in k[x,y]$ implies $u\in k[x,y]$, and thus $v^2xy$ (hence $v^2$ hence $v$) must also be in $k[x,y]$, so $k[x,y,z]/(xy-z^2)$ is integrally closed.
(This generalizes to $k[x_i,z]/(z^2-f)$ being normal for $k$ not of characteristic $2$ and $f\in k[x_i]$ squarefree.)
If the characteristic of $k$ is $2$, things are a little more involved. We may start by assuming $v\neq 0$ in our element $u+vz\in K$ integral over $k[x,y]$.
Our goal is to show that if $v^2xy+u^2=r \in k[x,y]$, then $u,v\in k[x,y]$.
Write $u=f/g$ and $v=p/q$ in lowest terms, and after expanding the equality, we get $$g^2p^2xy=q^2(rg^2-f^2).$$
Now I claim $g=q$ up to a multiplicative constant. Letting $\alpha$ be an irreducible factor of $g$ appearing with exponent $d$ in the factorization of $g$, $\alpha^{2d}$ must divide $q^2$ because $g$ and $rg^2-f^2$ are relatively prime. Therefore $\alpha^d$ must divide $q$. Conversely, if $\beta$ is an irreducible factor of $q$ appearing with exponent $e$ in the factorization of $q$, $\beta^{2e}$  must divide $g^2xy$ because $p$ and $q$ are relatively prime. Since $x$ and $y$ are coprime irreducibles, $\beta$ can only divide at most one of them to order at most one, so $\beta^{2d-1}$ must divide $g^2$, and therefore $\beta^d$ must divide $g$.
Since $k[x,y]$ is a domain, our equation reduces to $$cp^2xy=rg^2-f^2$$ for $c\in k^\times$.
Now we need to do a little manipulation with $r$.
Let $r=r_{00}+r_{10}+r_{01}+r_{11}$ denote the decomposition in to polynomials where all entries in $r_{ij}$ have all exponents of $x$ equal to $i$ mod $2$ and exponents of $y$ equal to $j$ mod $2$.
Since the LHS of our equation consists only of terms of type $11$ and $f^2$ as well as $g^2$ consists only of terms of type $00$, we see that $r_{01}=r_{10}=0$, $r_{11}g^2=cp^2xy$, and $r_{00}g^2=f^2$.
By coprimality of $f$ and $g$, we have either $g=1$ or $r_{00}=0$.
By coprimailty of $g$ and $p$, we have either $g=1$ or $r_{11}=0$ (here's where we use that $xy$ is squarefree - any factor of $g$ divides the RHS to order $2$, and thus divide $p^2$ to order $1$, which means that it must divide $p$, contradicting coprimality).
So either $g=1$ or $r=0$, and in both cases we get $u,v\in k[x,y]$.
A: I will rewrite KReiser's answer for the benefit of other people who, like me, might have trouble reading long paragraphs of mathematical prose. No new mathematical content is here. All the ideas are due to KReiser.
Let $A = \Bbbk[x,y]$ and $K = \Bbbk(x,y)$. Denote by $A', K'$ the ring extensions formed by adjoining a square root of $xy$ called $z$. Then $K'$ is the field of fractions of $A'$, and we have to show that $A'$ is integrally closed in $K'$.
Let $w \in K'$ be integral over $A'$. Then,

*

*$w = u + vz$, for some $u, v \in K$, because $K'/K$ is a quadratic extension.

*$w^2 = u^2 + v^2 xy \in K$, because we are working in characteristic $2$.

*$w^2 \in A$, because $A$ is a UFD, hence integrally closed in $K$.

Write $u, v$ as irreducible fractions: $u = p/q$ and $v = r/s$. Then,

*

*$w^2 q^2 s^2 = p^2 s^2 + q^2 r^2 xy \in Aq^2 s^2$

*$\gcd(p,q) = 1 \implies q^2 \mid s^2 \implies q \mid s$

*$\gcd(r,s) = 1 \implies s^2 \mid q^2 xy \implies s \mid q$. (We use here the fact that $xy$ is square-free.)

*Since $q, s$ are associates, we may assume that $q = s$. Therefore, $p^2 + r^2 xy \in Aq^2$. Write
$$
p^2 + r^2xy = aq^2.
$$
Note $p^2$, $q^2$, and $r^2$ are each in the subring $B = \Bbbk[x^2, y^2]$. From the way elements of $\Bbbk[x,y]$ look, we have a direct sum decomposition
$$
A = B \oplus Bx \oplus By \oplus Bxy.
$$
Write $a = b_0 + b_1x + b_2y + b_3xy$, where each $b_i$ is in $B$, so
$$
p^2 + r^2xy = (b_0 + b_1x + b_2y + b_3xy)q^2 = b_0q^2 + b_1q^2x + b_2q^2y + b_3q^2xy.
$$
Since $q^2 \in B$ and $B$ is a ring,
comparing the left and right sides above with the direct sum decomposition of $A$, we get
$$
p^2 = b_0q^2, \ \ b_1 = 0, \ \ b_2 = 0, \ \ r^2 = b_3q^2.
$$
Since $q$ is relatively prime to $p$ in the UFD $A$,
from the relation $p^2 = b_0q^2$ we see that $q$ is a unit, so
$u = p/q$ and $v = r/q$ are each in $A$. (We don't need the other three relations above.) Thus $w' = u + vz \in A[z]$, so $A' \subset A[z] \subset A'$, which implies $A' = A[z]$.
