Lemma 38.38.6. Let $p$ be a prime number. An $\mathbf{F}_ p$-algebra $A$ is absolutely weakly normal if and only if it is perfect.

Proof. It is immediate from condition (2)(b) in Morphisms, Definition 29.47.1 that if $A$ is absolutely weakly normal, then it is perfect.

Assume $A$ is perfect. Suppose $x, y \in A$ with $x^3 = y^2$. If $p > 3$ then we can write $p = 2n + 3m$ for some $n, m > 0$. Choose $a, b \in A$ with $a^ p = x$ and $b^ p = y$. Setting $c = a^ n b^ m$ we have

$c^{2p} = x^{2n} y^{2m} = x^{2n + 3m} = x^ p$

and hence $c^2 = x$. Similarly $c^3 = y$. If $p = 2$, then write $x = a^2$ to get $a^6 = y^2$ which implies $a^3 = y$. If $p = 3$, then write $y = a^3$ to get $x^3 = a^6$ which implies $x = a^2$.

Suppose $x, y \in A$ with $\ell ^\ell x = y^\ell$ for some prime number $\ell$. If $\ell \not= p$, then $a = y/\ell$ satisfies $a^\ell = x$ and $\ell a = y$. If $\ell = p$, then $y = 0$ and $x = a^ p$ for some $a$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).