**Proof.**
Assume $X$ is weakly normal. Since a weakly normal scheme is seminormal, we see that (1) holds (by our definition of weakly normal schemes). In particular $A$ is reduced. Let $p, z, w$ be as in (2). Choose $x, y \in A$ such that $z^ p x = w^ p$ and $zy = pw$. Then $p^ p x = y^ p$. The ring map $A \to C = A[t]/(t^ p - x, pt - y)$ induces a universal homeomorphism on spectra. The normalization $X^\nu $ of $X$ is the spectrum of the integral closure $A'$ of $A$ in the total ring of fractions of $A$, see Lemma 29.54.3. Note that $a = w/z \in A'$ because $a^ p = x$. Hence we have an $A$-algebra homomorphism $A \to C \to A'$ sending $t$ to $a$. At this point the defining property $X = X^{wn} = X^{X^\nu /wn}$ of being weakly normal tells us that $C \to A'$ maps into $A$. Thus we find $a \in A$ as desired.

Conversely, assume (1) and (2). Let $A'$ be as in the previous paragraph. We have to show that $X^{X^\nu /wn} = X$. By construction in the proof of Lemma 29.55.1, the scheme $X^{X^\nu /wn}$ is the spectrum of the subring of $A'$ which is the union of the subrings $A[a_1, \ldots , a_ n] \subset A'$ such that for $i = 1, \ldots , n$ we have

$a_ i^2, a_ i^3 \in A[a_1, \ldots , a_{i - 1}]$, or

there exists a prime number $p$ with $pa_ i, a_ i^ p \in A[a_1, \ldots , a_{i - 1}]$.

Then we can use (1) and (2) to inductively see that $a_1, \ldots , a_ n \in A$; we omit the details. Consequently, we have $X = X^{X^\nu /wn}$ and hence $X$ is weakly normal.
$\square$

## Comments (0)