Lemma 9.14.6. Let $E/F$ be an algebraic field extension. There exists a unique subextension $F \subset E_{sep} \subset E$ such that $E_{sep}/F$ is separable and $E/E_{sep}$ is purely inseparable.

** Any algebraic field extension is uniquely a separable field extension followed by a purely inseparable one. **

**Proof.**
If the characteristic is zero we set $E_{sep} = E$. Assume the characteristic if $p > 0$. Let $E_{sep}$ be the set of elements of $E$ which are separable over $F$. This is a subextension by Lemma 9.12.13 and of course $E_{sep}$ is separable over $F$. Given an $\alpha $ in $E$ there exists a $p$-power $q$ such that $\alpha ^ q$ is separable over $F$. Namely, $q$ is that power of $p$ such that the minimal polynomial of $\alpha $ is of the form $P(x^ q)$ with $P$ separable algebraic, see Lemma 9.12.1. Hence $E/E_{sep}$ is purely inseparable. Uniqueness is clear.
$\square$

## 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Your email address will not be published.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 (2)

Comment #826 by Johan Commelin on

Comment #5488 by Théo de Oliveira Santos on

There are also: