TITLE: Does an inseparable extension have a purely inseparable element?
QUESTION [6 upvotes]: Assume $K/F$ is an inseparable extension. Is it necessary that K contains an element $u \notin F$ that is purely inseparable over $F$?
I also posted in MO.

REPLY [12 votes]: The standard counterexample is to consider $F = \mathbf{F}_p(x,y)$ and $K = F(\alpha)$ where $\alpha$ is a root of $f = T^{p^2} + x T^p + y \in F[T]$
(where $f$ is irreducible by Gauss' Lemma).  Clearly $K/F$ is non-separable of degree $p^2$ and $F' := F(\alpha^p)$ is a degree-$p$ subextension that is separable over $F$, so this is the maximal separable subextension.  Hence, $[K:F]_s = p$ and so $[K:F]_i = p$.  To rule out the existence of an element of $K-F$ that is purely inseparable over $F$ it suffices to show that $F'$ is the unique degree-$p$ subextension of $K/F$.  
Suppose $E$ is another such subextension, so it must be purely inseparable over $F$ (as otherwise it would be separable and hence the composite $F'E$ would be separable yet has to exhaust $K$ for $F$-degree reasons).  By multiplicativity of separable degree in towers, it follows that $[K:E]_s=p$, so the degree-$p$ extension $K/E$ is separable.  But $K=E(\alpha)$, so the degree-$p$ minimal polynomial $g \in E[T]$ of $\alpha$ is separable. In a splitting field $E'/E$ of $f$ over $E$, 
$f$ is a product of $p$th powers of $p$ monic linear factors and $g$ is the product of just these monic linear factors (since $g|f$ in $E[T]$ and $g$ is separable over $E$ of degree $p$). 
Hence, in $E'[T]$ we must have $g^p=f$. But this latter equality then must hold in $E[T]$ as well. It then follows that $g = T^p + uT + v$ for some $u, v \in E$ satisfying $u^p = x, v^p = y$. But the extension $\mathbf{F}_p(x^{1/p}, y^{1/p}) \supset F$ has degree $p^2$ and so cannot lie inside the degree-$p$ extension $E$ of $F$. This is a contradiction, so no such $E \ne F'$ exists inside $K/F$.