Why isn't the perfect closure separable? 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?
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 $F$.
Hint: let $g(X)$ be irreducible and separable; let $f(X)=g(X^p)$; adjoin one root of f to your field; study the extension.