TITLE: Separable closure is a field
QUESTION [8 upvotes]: Let $K\subset L$ a field extension. Define
$K_s=\{x\in L : x \text{ is algebraic and separable over } K\}$
Show that $K_s$ is a subfield of $F$.
Attempt:
I tried to prove that $K(\alpha,\beta)/K$ is a separable extension when $\alpha,\beta$ are separable and algebraic (I know that $K(\alpha,\beta)/K$ is algebraic when $\alpha,\beta$ are algebraic). Then $\alpha\pm\beta, \alpha^{-1},\alpha\beta, \alpha/\beta$ are separable then $K_s$ is a field.
I have no idea how to prove that $K(\alpha)/K$ is separable if $\alpha$ is separable. Also, if I prove that, how to conclude? I tried to prove the following "If $K/F$  is separable and $L/K$  is separable then $L/F$  is separable"

REPLY [2 votes]: As Ewan Delanoy, first note the following result

Lemma Let F be a field and $f(x)\in F[x]$ be an irreducible polynomial with $deg f(x)\geqslant 1$.Then (1) if $char(F)=0$, then $f(x)$ has no multiple roots; (2) if $char(F)=p>0$, then $f(x)$ has multiple roots iff $f(x)=g(x^p)$ for some $g(x)\in F[x]$.

Then the follwing steps will answer your question:

Step 1 Any algebraic extension of an characteristic zero field is separable.

By above lemma.

Step 2 Let $E/F$ be a field extension with $char(F)=p>0$, then an element $\alpha\in E$ is separable over $F$ iff $F(\alpha)=F(\alpha^p)$.

By disproof and above lemma

Step 3 If $\alpha$ is separable over $F$, $\beta$ is separable over $F(\alpha)$, then $\beta$ is separable over $F$.

Assume that $char(F)=p>0$, then show that $F(\beta)=F(\beta^p)$.

Step 4 Let $E/F$ be an algebraic extension and $S\subseteq E$ be a subset, then $F(S)/F$ is separable iff every element of $S$ is separable.

By inductive step to the case of Step 3.