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I've been self-studying inseparable extensions and there's something that seems obvious to everybody but not to me. Let's clear out some definitions that are not so universal:

Let $K$ be a field and $f\in K[X]$ a polynomial. We say it is separable if all its roots are distinct over its splitting field. We say it is purely inseparable if it has only one root over its splitting field, and this root is multiple.

A field extension $F\subset K$ is purely inseparable (resp. separable) if every element of $K$ is the root of a purely inseparable (resp. *separable) polynomial in $F[X]$.

A field K is perfect if every irreducible polynomial over K is separable.

Let $F\subset K$ be a field extension of characteristic $p$. The perfect closure, or purely inseparable closure of $F$ in $K$ is the greatest intermediate field that is purely inseparable over $F$. We will denote it by $K^p_F$

We can prove that $K^p_F$ consists of all the elements $\alpha\in K$ such that there is $n\in \mathbb{N}$ such that $\alpha^{p^n}\in F$.

Another property that may be useful is that an irreducible polynomial over a field of characteristic $p$ splits on its splitting field like this: $$f(X)= a_0(X-a_1)^{p^n} \dots (X-a_r)^{p^n}$$ where the $a_1,\dots,a_r$ are pairwise distinct.

Now, why is it true that $K^p_F$ is a perfect field? Please try to prove it only using these definitions and properties.

EDIT: I have also proved that

A field is perfect iff every finite extension is separable, iff every algebraic extension is separable.

which may come in handy.

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I sense this is a silly question that arises from my lack of intuition on these matters of inseparability. – Bruno Stonek May 1 '11 at 19:16
up vote 4 down vote accepted

The answer to your question is, that it isn't necessarily true that $K_F^p$ is a perfect field. For example, let $F$ be any non-perfect field (e.g. $\mathbb{F}_p(T)$), and let $K=F$, so that we must have $K_F^p=F$.

When $K=\bar{F}$, an algebraic closure of $F$, then $K_F^p$ is the smallest perfect subfield of $K$ containing $F$, hence the name "perfect closure". This is Lemma 3.16 in Karpilovsky's Field Theory (for some reason I couldn't find this in a more standard reference like Lang). As you mention, $K_F^p$ consists of those $\alpha\in K$ such that $\alpha^{p^n}\in F$ for some $n$, and note that the field that Karpilovsky refers to as the perfect closure is $$F^{p^{-\infty}}=\{a\in\bar{F}\mid a^{p^n}\in F \text{ for some }n\geq0\}.$$

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Why is it true that "all subextensions $L/F_p$ of $\bar{F}/F_p$ are separable"? – Bruno Stonek May 1 '11 at 19:45
@Bruno: I have edited. – Zev Chonoles May 1 '11 at 20:17
@Zev: Sadly, I can't find the book you mention... – Bruno Stonek May 1 '11 at 20:58
@Bruno: I have also found a reference for the above proposition in Bourbaki (Prop 3 in the link). It is stated in more generality (it works for all rings of characteristic $p$). – Zev Chonoles May 1 '11 at 21:07
Oh, but I found it as lemma 2.24 in Karpilovsky's "Topics in Field Theory" which is, ahem, findable on the web. It is quite clear there. Thank you. – Bruno Stonek May 1 '11 at 21:07

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