TITLE: Splitting field of a separable polynomial is separable
QUESTION [20 upvotes]: Probably a stupid question, but..
Why is the splitting field of a separable polynomial necessarily separable?
Thanks.
Follow up question
Show that if $F$ is a splitting field over $K$ for $P \in K[X]$, then $[F:K] \leq n!$
I've proven this by induction on $n$, but I'm convinced there's a more algebraic approach ($n!$ screams $S_n$). 
If I knew $P$ were separable, then I'd know $F$ was Galois, so $ |\mbox{Gal}(F/K)| = [F:K] $. Considering the action of $ \mbox{Gal}(F/K) $ on the roots of $P$, we'd get an injective homomorphism into $ S_n $, giving the result. 
But $P$ is not given to be separable, so $P$ could have repeated roots when factorised in $F$. If $G$ is any finite group of $K$-automorphisms of $L$, then I know that $[F:K] \leq |G| $. Considering the action of $G$ on the set of roots of $P$, say $\Omega$, we get an injective homomorphism of $G$ into $S_{|\Omega|}$, where $|\Omega| \leq n$ (I think). I have a feeling I'm wrong here, since I think $G$ is forced to be $\mbox{Aut}(F/K)$. Any advice would be appreciated.

REPLY [7 votes]: $\DeclareMathOperator{\aut}{Aut}$ $\newcommand{\vp}{\varphi}$Here is a short proof.
Let $E$ be the splitting field of a separable polynomial $p(x)$ over a field $F$. We need to show that $E:F$ is a separable extension.
Claim 1. $[E:F]=|\aut(E:F)|$.
Proof. Say $\deg p=n$ and let $a_1, \ldots, a_n$ be all the roots of $p$ in $E$. Write $F_i$ to denote $F(a_1, \ldots, a_i)$. We will show that each $F$-embedding of $F_i$ in $E$ admits exactly $[F_{i+1}:F_i]$ extensions to an embedding of $F_{i+1}$ into $E$.
Suppose $\vp:F_i\to E$ be an $F$-embedding of $F_i$ in $E$ and let $d_i=[F_{i+1}:F_i]$. We need to show that there are exactly $d_i$ different extensions of $\vp$ to an embedding of $F_{i+1}$ into $E$. Let $p_i(x)\in F_i[x]$ be the minimal polynomial of $a_i$ over $F_i$. Since $p_i$ and $p$ both have $a_{i+1}$ has a root, we have $p_i|p$ and therefore $p_i$ has $d_i$ distinct roots in $E$. Now we may extend $\vp$ to $F_{i+1}$ by sending $a_{i+1}$ to any root of $p_i$ we wish. Further, all extensions of $\vp$ send $a_{i+1}$ to some root of $p_i$. Lastly, any extension of $\vp$ is determined completely by its action on $a_{i+1}$. Therefore there are precisely $d_i=[F_{i+1}:F_i]$ different extensions of $\vp$ to $F_{i+1}$.
Thus there are at least $[F_n:F_{n-1}]\cdots [F_1:F]=[E:F]$ distinct elements in $\aut(E:F)$. There cannot be any more and thus we are done. $\blacksquare$
Claim 2. The fixed field in $E$ of $\aut(E:F)$ is $F$.
Proof. Let $M$ be the fixed field of $\aut(E:F)$. Thus there is a natural bijection between $\aut(E:M)$ and $\aut(E:F)$. Note that $p(x)$ is a separable polynomial over $M$ whose splitting field is $E$. Therefore, by Claim 1, we have $|\aut(E:M)|=[E:M]$. This shows that $|\aut(E:F)|=[E:M]$. In the light of Claim 1, this is only possible if $M=F$. $\blacksquare$
Claim 3. $E$ is separable over $F$.
Proof. Let $\alpha\in E$ be chosen arbitrarily and $f(x)$ be the minimal polynomial of $\alpha$ over $F$. We need to show that $f(x)$ is separable over $F$. Note that $E$ is normal over $F$ and therefore completely splits $f$. Write $f=(x-\beta_1)^{r_1}\cdots (x-\beta_k)^{r_k}$ in $E[x]$, where $\beta_i$'s are pairwise distinct. If $\vp\in \aut(E:F)$, then we see that
$$f=(x-\beta_1)^{r_1}\cdots (x-\beta_k)^{r_k}= (x-\vp(\beta_1))^{r_1}\cdots (x-\vp(\beta_k))^{r_k}$$
which lets us conclude that $\vp$ permutes $\{\beta_1, \ldots, \beta _k\}$. Since $\vp$ was arbitrary in $\aut(E:F)$, we conclude that $\aut(E:F)$ acts on $\{\beta_1, \ldots, \beta_k\}$. Therefore the symmetric functions on $\beta_i$'s remain fixed by $\aut(E:F)$. Applying Claim 2, we see that each symmetric function in $\beta_i$'s is a member of $F$. Therefore the polynomial $(x-\beta_1)\cdots (x-\beta_k)$ is in $F[x]$. Since $f(x)$ was irreducible over $F$, we are forced to have $f(x)=(x-\beta_1)\cdots (x-\beta_k)$ and we have shown that $\alpha$ is separable over $F$. $\blacksquare$