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

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 $a$. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)