Separable closure

Remarks

The set $L_s$ is a subfield of $L$ containing $K$, the extension $L_s/K$ is separable, and every separable subextension of $L/K$ lies in $L_s$. The extension $L/L_s$ is purely inseparable. The field $K^{\mathrm{sep}}$ is separably closed, and it is unique up to an isomorphism over $K$.

Example: let $K = \Q$ and $L = \overline{\Q}$. Check: in characteristic $0$ every irreducible polynomial has distinct roots, so every element of $\overline{\Q}$ is separable and $\Q^{\mathrm{sep}} = \overline{\Q}$.

Non-example: let $K = \F_p(t)$ and $L = K(s)$ with $s^p = t$. The element $s$ is not in the separable closure of $K$ in $L$, since its minimal polynomial $x^p - t = (x - s)^p$ has a repeated root; in fact $L_s = K$.