On a Proof that the Splitting Field of a Separable Polynomial is Galois Prop.: If $f \in F[x]$ is separable, then the splitting field of $f$ over $F$ is a Galois extension of $F$.
Proof: By induction over $[E:F]$, where $E$ is the splitting field.
By previous results concerning finite extensions, we know it suffices to show that $[E:F] = |\text{Aut}(E/F)|$.
If $[E:F]=1$ there is nothing to do.
If $[E:F]>1$, then we can write $f=pq$ where $p,q \in F[x]$, $p$ irreducible and $\deg p>1$. Since $f$ is separable, $p$ is separable. Write
$$
p(x) := \prod_{i=1}^n (X-\alpha_i),
$$
where $\alpha_i \in E$ are different. Let $E_i := F(\alpha_i)$. Then $E$ is the splitting field of $f/(x-\alpha_1) \in E_1[x]$. Since $m := [E:E_1]<[E:F]$, the induction hypothesis tells us there are $m$ elements in $\text{Aut}(E/E_1)$, let's say $\text{Aut}(E/E_1) = \{\tau_1, \ldots, \tau_m\}$. There are also $n$ isomorphisms
\begin{align}
\sigma_i : E_1 &\to E_i \\
\alpha_1 &\mapsto \alpha_i.
\end{align}
Each combination $(\tau_j, \sigma_i)$ gives an element in $\text{Aut}(E/F)$. Hence there are $mn$ elements in $\text{Aut}(E/F)$. But $mn=[E:E_1][E_1:F]=[E:F]$. $\blacksquare$

Questions:
  
  
*
  
*Does $\alpha_i \in E\backslash F$ for all $i$? In fact, this is used to deduce $[E:E_1]<[E:F]$? I think the answer to both is yes and that $\alpha_i \not\in F$ follows from the irreducibility of $p$ over $F$ and from the fact that $e \in E\backslash F$ & $f\in F$ $\implies$ $ef \in E\backslash F$.
  
*To apply the induction hypothesis, we note that $E$ is the splitting field of $f/(x-\alpha_1) \in E_1[x]$. But is it not true that we have more simply that $E$ is the splitting field of $f \in E_1[x]$?
  
*When we say that each combination $(\tau_j, \sigma_i)$ gives an element in $\text{Aut}(E/F)$, what are those elements? I think it must be some compositions, but the domains and codomains of $\tau_j$ and $\sigma_i$ don't quite match...? Also why are those $mn$ elements different?

 A: *

*Any root of $p$ cannot be contained in $F$, since otherwise you would have that $(x-\alpha) \mid p$ for some $\alpha \in F$, $p(\alpha) = 0$.

*Think about what it means to solve this problem inductively. What we are doing is assuming that $[E:E_1] = |\operatorname{Aut}(E/E_1)|$. Why can we do that? 

*$\tau_j$ describes automorphisms of $E$ which leave $E_1$ fixed, and therefore fixes all conjugates of $a_1$ which is the set of all roots of $p$. $\sigma_i$ describes isomorphisms of $E_1$ onto $E_i$. We can extend $\sigma_i$ naturally onto an action of $E$, which does not affect elements orthogonal to $\operatorname{span}\{\alpha_1,\dots,\alpha_n\}$ (viewing $E$ as a vector space over $F$). Then $\tau_j$ and $\sigma_i$ commute as automorphisms of $E$. Furthermore, any automorphism of $E$ which leaves $F$ fixed can be described by its action on basis elements of $E$. I don't know of a simple way to justify this, but we can have a basis $B$ of $E$ which contains powers of $\alpha_1$ and elements orthogonal to $\alpha_1$. Then $\{\tau_j\}$ describes all possible actions of $B \setminus \{\alpha_1^k\}$ and $\{\sigma_i\}$ describes all possible actions on $\alpha_1$, so the set of compositions $\{\tau_j \sigma_i\}$ describes all automorphisms of $E$ fixing $F$, so we have that the number of automorphisms is equal to $[E:F]$.

