Let $E/\ F$ be a fine extension of fields that $E = F[\alpha]$, show that if the minimal polynomial $f (x)$ of $\alpha$...

Question

Let $E/\ F$ be a fine extension of fields that $E = F[\alpha]$, show that if the minimal polynomial $f (x)$ of $\alpha$ has $deg (f)$ roots in $E $ then $E$ is of Galois

$E=F[\alpha]$

Let $deg(f)=n$ and $A = \{a_1,a_2...a_n\}$ be the set of all roots other than $f$ it is obvious that

$F[a_1] =F[a_2]= ... F[a_n]$

Let's define

$\sigma : \sigma(\alpha) = a_i$ for some $i \in \{1...n\} $

How can I show that this actually defines an F-automorphism?

The greatest doubt is how to show that if $ e \in E $ is the root of a polynomial $g(x) \in F[x]$ then $\sigma(e)$ is also root of $g$, how to show this?

Answer 1

A field is Galois if it is separable and normal. If the degree of the minimal polynomial $f$ of $\alpha$ has $def( f) $ distinct roots $f$ is separable by definition. $E$ us normal since it is the splitting field of $f$.