If is $L/K$ a normal extension, then it follows that $K_s/K$ is a normal extension.
Definition of normal extension:
Let $L/K$ be an algebraic extension and $\overline{L}$ be a algebraic closure of $L$, the following are equivalent:
- L is a splitting field of $p \in K[x]$ over $K$
- For every injective $K-$Homomorphism $\phi: L \rightarrow \overline {L} $ is $\phi(L)=L$
- Every irreducible polynomial from $K[x]$ with a root in $L$, splits over $L$
If an algebraic field extension $L/K$ satisfies one of the above condition (and thus all of the above conditions), then it is called a normal extension.
Let $L/K$ be an algebraic extension. Then $K_s:=\{a \in L : a \text{ is separable over } K \}$
An Element $a \in L$ is called separable if its minimal polynomial $m(a,K)$ is separable over $K$ and $a$ is algebraic over $K$. A polynomial $P \in K[x]$ is called separable, if every irreducible factor of $P$ in a splitting field of $P$ over $K$ only has simple roots.
My attempt: I think the easiest approach would be to show that every irreducible polynomial from $K[x]$ with a root in $K_s$ splits over $K_s$. So let $p \in K[x]$ be irreducible and assume that $p$ has a root in $K_s$. Thus, we can write $p(x)=q(x)(x-r)$ for some polynomial $q$
Now there are two options: $q$ has root in $K_s$
$q$ has no root in $K_s$
Assume $q$ has no root in $K_s$ Now I don't know how to continue.