Let $K/F$ be a normal algebraic extension and let $L = (K/F)^{\mathrm{sep}}$ be the subfield of elements of $K$ which are separable over $F$ (this is also called the separable closure of $F$ in $K$). Is $L/F$ necessarily normal?
I really do not have a clue about this question so any hints will be appreciated.
On
yes, consider any embedding f of L into algebraic closure of K.f can be extended to embedding of K into algebraic closure.as K is normal that embedding maps K into itself. let,x be any element of L then f(x) is in K.as x is separable implies f(x) is also separable so f(x) belongs to L so any embedding of L into algebraic closure is automorphism of L.So L is normal.
Take $a\in L$ and let $p$ be its minimal polynomial. Since $a\in K$ and $K$ is normal, $p$ is a product of linear factors : $$p=\prod_{j=1}^n(X-a_j),$$ (where $a=a_1$) and by definition of $L$ all the $a_j$'s are distinct. They have the same minimal polynomial $p$, so all of them are in $L$; hence $L$ is normal.