I was thinking about the following:
Let $L/K$ be a finite extension and $K_s=\{a \in L : a \text{ is separable over } K\}$ Is $L/K_s$ purely inseparable?
I will start with the definiton: The algebraic extension $L/K$ is called purely inseparable if every element $a \in L\setminus K$ is inseparable.
Now if $a \in L\setminus K_s$ would be separable, then $a \in (K_s)_s:=\{a \in L : a \text{ is separable over } K_s \}$. This would mean $(K_s)_s(a^p)=(K_s)_s(a)$ but since $K \subset K_s \subset (K_s)_s$ this would mean that $K(a^p)=K(a)$, this is a contradiction to $a \in L \setminus K_s$.
I am not really sure if, the argumentation is correct. I think I am confusing something with something.
I don't understand how you go from $(K_s)_s(a^p) = (K_s)_s(a)$ to $K(a^p) = K(a)$.
If you want to prove that $L/K_s$ is purely insparable, consider $a \in L\backslash K_s$, then its minimal polynomial over $K$ can be written as $f(X) = g(X^p)$ with $g$ irreducible. Repeating the operation, we deduce that $f(X) = h(X^q)$ where $q$ is a power of $p$ and $h$ is irreducible and separable.
You deduce that $a^q \in K_s$ since its minimal polynomial is the separable polynomial $h$. Therefore, the polynomial $p(X) = X^q - a^q$ has coefficients in $K_s$ and $p(a) = 0$. Notice that $p(X) = (X - a)^q$ because $q$ is a power of $p$, thus the minimal polynomial of $a$ in $K_s$ is of the form $(X - a)^{q'}$ with $q' \leqslant q$.
$a \notin K_s$ by hypothesis so $q' > 1$. The minimal polynomial of $a$ has one unique multiple root, which is $a$ itself. $L/K_s$ is purely inseparable.