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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?

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  • $\begingroup$ I admit I am not sure what you're getting at with your use of the word "why." Does that mean that you won't accept a counterexample as an answer? Is it signaling that you're looking for some kind of intuition about such a counterexample? $\endgroup$ Commented May 2, 2011 at 16:44
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    $\begingroup$ @Qiaochu: Both. A counterexample would be great, but I'm also looking for intuition. I don't understand what is the impediment for the situation to be symmetrical. $\endgroup$ Commented May 2, 2011 at 23:48

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The intuition I can provide is that in an algebraic extension $F\subset K$ of characteristic $p$ , the field $K$ will be a separable extension of its purely inseparable extension $K_p$ over $F$ whenever $trdeg_{\mathbb F_p} F\leq 1$. This is why my counterexamples on MathOverflow here and here have $F=\mathbb F_p(u,v)$ , a purely transcendental extension of $\mathbb F_p$ of degree two.

I can't prove my intuition but it implies that Zev's example won't work (he didn't say it did, so he made no mistake), and here is a proof that indeed it doesn't.

The key point is that is that if $\mathbb F_q$ is a finite field of characteristic $p$, the Frobenius map $\mathbb F_q \to \mathbb F_q: x\mapsto x^p$ is surjective and this implies that for $f(T)\in \mathbb F_q [T]$ we have $ (f(T))^{1/p}\in \mathbb F_q [T^{1/p}]$, because of the freshman's dream $(a+bT+\ldots)^{1/p}=a^{1/p}+b^{1/p}T^{1/p}+\ldots $ !
This implies that

$$(\mathbb F_q(T))^{p^{-\infty}}=\mathbb F_q(T,T^{p^{-1}},T^{p^{-2}},\ldots)$$

And now the rest is easy. If $F= \mathbb F_p(T)$ and $K=(\mathbb F_{p^2}(T))^{p^{-\infty}}=\mathbb F_{p^2}(T,T^{p^{-1}},T^{p^{-2}},\ldots \quad )$ we have $K_p=\mathbb F_p(T,T^{p^{-1}},T^{p^{-2}},\ldots \quad )$ and $K$ is a separable extension of the perfect closure $K_p$ of $F$ in $K$ . Indeed $K$ is just the simple separable extension $K=K_p(g)$ of $K_p$, where $g \in \mathbb F_{p^2}$ is any element such that $\mathbb F_{p^2}=\mathbb F_{p}(g)$. [ The element $g$ is of course separable over the (perfect !) field $\mathbb F_p$, so it is a fortiori separable over $K_p$].

Edit What I called my intuition above has now been proved by ulrich on MathOverflow here. He proves that if $trdeg_{\mathbb F_p} F\leq 1$, any algebraic extension $F\subset K$ has the property that $K$ is separable over its purely inseparable extension $K_p$ over $F$.

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This is Prop 3.23 in Karpilovsky's Field Theory: $K/K_p$ is separable if and only if $K=K_pK_s$.

The proof requires these facts:

  1. Given fields $A\subseteq B\subseteq C$, then $A/C$ is separable (resp. purely inseparable) iff $A/B$ and $B/C$ separable (resp. purely inseparable).

  2. Given $A/B$, then $A/A_s$ is purely inseparable.

  3. If $A/C$ is separable and $B/C$ is any extension for which $A$ and $B$ are contained in some common field, then $AB/B$ is separable.

If $K/K_p$ is separable, then $K/K_pK_s$ is separable (clear because $K_pK_s\supseteq K_p$) and $K/K_pK_s$ is purely inseparable (combine facts 1 and 2). Conversely if $K=K_pK_s$, then by fact 3, because $K_s/K$ is separable, then $K=K_pK_s/K_p$ is separable.

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    $\begingroup$ Why doesn't this always happen? $\endgroup$ Commented May 1, 2011 at 23:46
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    $\begingroup$ @Bruno: That is an excellent question - I can't think of any examples off the top of my head. $\endgroup$ Commented May 1, 2011 at 23:54
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    $\begingroup$ All I can think of is just trying to pick the crappiest fields possible - maybe $F=\mathbb{F}_p(T)$ and $$K=\mathbb{F}_{p^2}(T)^{p^{-\infty}}=\{\alpha\in\overline{\mathbb{F}_{p^2}(T)}\mid \alpha^{p^n}\in \mathbb{F}_{p^2}(T)\text{ for some }n\}$$ but I'm not sure if this works. $\endgroup$ Commented May 2, 2011 at 0:04
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    $\begingroup$ Not crappy enough: see my answer. $\endgroup$ Commented Jun 26, 2011 at 17:54
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Counterexamples and intuition have been provided at MO.

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Assume $K=F(x)$ and let $f\in F[X]$ be the minimal polynomial of $x$ over $F$. One then knows that $f(X)=g(X^{p^m})$ for some $m\in\mathbb{N}$ and a separable polynomial $g\in F[X]$. Thus $K_s =F(x^{p^m})$.

If there exists an intermediate field $L$ such that $L/F$ is purely inseparable and $K/L$ is separable, then $[K:L]=[K_s:F]$ since $L$ and $K_s$ are linearly disjoint over $F$.

Consequently the minimal polynomial $h\in L[X]$ of $x$ over $L$ has the same degree as $g$.

Now $f=hh_1$ in $L[X]$. Since $x$ is a simple root of $h$ put a $p^m$-fold root of $f$ one concludes $f=h^{p^m}h_2$. Comparing degrees yields $h^{p^m}=g(X^{p^m})$. This shows that a necessary condition for $L$ to exist is that the coefficients of $g$ are $p^m$-th powers in $K$.

I guess this is sufficient too - have no time to think about it now. And I guess that what I just wrote down is what mephisto meant ...

By the way: if $K/F$ is normal, then $L$ exists: it is the fixed field of $\mathrm{Aut (K/F)}$.

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