# Why isn't the perfect closure separable?

This question has escalated from math.stackexchange. I'm doing this because it has been a while since the question has been open, receiving no satisfactory answers, even when subject to a bounty. I hope it is adequate for MO, I apologize if it is not.

Let $F\subset K$ be an algebraic extension of fields. By taking the separable closure $K_s$, we obtain a tower $F\subset K_s \subset K$ such that $F\subset K_s$ is separable and $K_s\subset K$ is purely inseparable.

Wikipedia, following Isaacs, Algebra, A Graduate Course p.301, says:

On the other hand, an arbitrary algebraic extension $F\subset K$ may not possess an intermediate extension $E$ that is purely inseparable over $F$ and over which $K$ is separable.

The question is: why? And more explicitly, had I not seen this this soon, I would surely have conjectured that the perfect closure, $K_p$, which satisfies $F\subset K_p$ purely inseparable, also satisfied $K_p\subset K$ separable... But why doesn't it?

To clarify, I'm looking both for a counterexample and for intuition regarding the impediment of the situation to be symmetrical.

ADDED: After posting the link to this answer on Math.SE, Georges Elencwajg kindly answered there also, providing further intuition on this subject.

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– Kevin Buzzard May 27 '11 at 13:01
There is a paper "Balanced Field Extensions" by Joe Lipman in Amer. Math. Monthly 73 (1966), pp. 373-374 which discusses the possibility of such a reversed intermediate tower of field extensions, if I recall this correctly. – KConrad May 27 '11 at 23:52

A counterexample
Take as ground field $F=\mathbb F_2(u,v)$ and consider the polynomial $f(X)=X^6+uvX^2+u\in F[X]$. This polynomial is irreducible by Eisenstein. Let $F\subset K$ be the extension obtained by adjoining a root $a$ of $f(X)$ to F, so that $K=F[a]$, $[K:F]=6$ and $f(a)=0$ .The element $a^2\in K$ has minimal polynomial over $F$ the separable (because of degree $3$) polynomial $g(X)=X^3+uvX+u$ .
From this follows that the separable closure of $F$ inside $K$ is $F^{sep}=F[a^2]$ and that the extension $F^{sep}=F[a^2] \subset K=F[a]$ is purely inseparable, as expected.

And now comes the surprise: there are no elements in $K$ purely inseparable over $F$ !(except the elements of $F$, of course) and so $K$ is not separable over the purely inseparable closure $F^{perf}=F$ of $F$. The proof that an element of $r\in K$ purely inseparable over $F$ belongs to $F$ is easy, once you realize that such an element must satisfy $r^2\in F$ . [Expand $r$ in the $F$-basis $a^i, 0\leq i\leq5$ and calculate $r^2$ (easy in characteristic $2$ !), remembering that $f(a)=0$.You will see that $r$ was already in $F$.]
(Edit: Kevin's comment and link have reminded that I had already given analogous examples in characteristic $p>2$, and completely forgotten about them! The two answers now cover all positive characteristics.)

A positive result
If $F\subset K$ is normal , then indeed $K$ is separable over $F^{perf}$, the extensions $F^{sep}$ and $F^{perf}$ are linearly disjoint over $F$ and the canonical map $F^{sep} \otimes_F F^{perf} \to K$ is an isomorphism. (As an aside observe that this is one of the rare cases where the tensor product of two fields is a field)

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In the special case that your extension $K/F$ is of the form $K=F(\text{root of }f)$ for some irreducible polynomial $f(x)$, then there is some nice intuition as follows. If $f$ is separable, then $K/F$ is separable, and nothing interesting is going on. If $f$ is inseparable, then that means $\gcd(f(x),f'(x))$ is not $1$. But $f(x)$ is irreducible, and the degree of $f'(x)$ is smaller than that of $f(x)$, so we must have $f'(x)\equiv 0$. This is only possible if $\operatorname{char}K=p$ and $f(x)=g(x^p)$ for some prime $p$ (just look termwise at the derivative and this should be obvious). Since $f$ is irreducible, so is $g$. Now, we continue this process to the polynomial $g$, until we have written $f(x)=h(x^{p^n})$ for some $n\geq 0$ and some separable polynomial $h$. Then the field extension is $F\subseteq F(\text{root of }h)\subseteq F((\text{root of }h)^{1/p^n})=K$, where the first part is separable, and the second is purely inseparable.

With the above proof in mind, the intuition you are looking for might be the following. We know purely inseparable extensions come from taking $p$th roots in characteristic $p$. In the whole extension, you may have taken the $p$th root of something not in the ground field, and so you'd better take the separable closure before looking for the purely inseparable part of your field extension.

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Could you please explain why the perfect closure of $F$ in $K$ is $F$? I certainly believe you, but I can't find an argument. – Georges Elencwajg May 27 '11 at 23:44
Ah, I see you have edited your answer by adding an indeterminate to your example . Do you still believe that your preceding version was correct? – Georges Elencwajg May 28 '11 at 16:13
Thanks! I don't believe either example anymore. – John Pardon May 28 '11 at 17:45
Dear unknown, congratulations on your sense of fair-play. And welcome to MathOverflow! – Georges Elencwajg May 28 '11 at 18:21