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.
[Math] Does an inseparable extension have a purely inseparable element
abstract-algebragalois-theory
Related Solutions
It really does seem to be wrong. There are examples in an earlier MSE posting. Your question boils down to this: if, in an extension $L\supset K$, the field $L$ has no elements purely inseparable over $K$, is $L$ separable over $K$? And the answer, grâce à Joe Lipman, is no.
Let $K$ be a field. Then for each element $\alpha\in K$ we have the evaluation map $K[x]\to K$ which is $K$-linear and sends $x\mapsto\alpha$. We write $f(\alpha)$ for the image of a polynomial $f\in K[x]$.
We say that $\alpha$ is a root of $f$ provided $f(\alpha)=0$. By the division algorithm, this is equivalent to saying that $x-\alpha$ divides $f(x)$ in $K[x]$.
If now $F\subset K$ is a subfield, then we can regard $F[x]$ as a subring of $K[x]$, and the evaluation map restricts to a map $F[x]\to K$, which is $F$-linear and sends $x\mapsto\alpha$. Thus for $f\in F[x]$ we have $\alpha\in K$ is a root of $f$ provided $f(\alpha)=0$, or equivalently $x-\alpha$ divides $f(x)$ in $K[x]$.
That answers (1) and (2).
(3). You are correct that the definition from the book is a bit ambiguous. An algebraic element $\alpha\in K$ is purely inseparable over the subfield $F$ provided the minimal polynomial $m:=\mathrm{min}(F,\alpha)$ has only one root in every field extension of $K$. In this case it follows that $m=(x-\alpha)^d\in K[x]$ for some $d$, so that $m$ actually splits over $K$, and conversely if $m=(x-\alpha)^d\in K[x]$, then $\alpha$ is purely inseparable over $F$.
In summary, $\alpha\in K$ is purely inseparable over $F$ provided the minimal polynomial $\mathrm{min}(F,\alpha)$ factors as $(x-\alpha)^d$ in $K[x]$.
This is different to saying that $m$ has only the one root $\alpha$ in $K$, since it may not split over $K$. For example, let $\alpha=\sqrt[3]2\in\mathbb R$, and take $F=\mathbb Q$ and $K=\mathbb Q(\alpha)$. Then the minimal polynomial $m=x^3-2$ has only one root in $K$ but $\alpha$ is separable over $\mathbb Q$.
Best Answer
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 fi eld $E'/E$ 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 $K/F$.