The minimal polynomial can't have multiple roots (in my proof) I have a question about the roots of the minimal polynomial $f_\alpha$, I can't see why $f_\alpha$ is separable, i.e. all it's roots must be different on a splitting field $\mathbf K$. I know it's roots are $\sigma_i(\alpha),$ but What is the proof to know that all the $\sigma_i's$ are different?
Proposition: If $\mathbf K:\mathbf F<\infty,$ and is Galois, then it is normal and separable.
Proof:
(summary of t.gunn's proof  :)
Let $\alpha\in\mathbf K.$
The minimal polynomial of $\alpha$ is
$$ f_\alpha(x) := \prod_{\beta \in G \cdot \alpha} (x - \beta). $$
Indeed:
First, note that $f_\alpha(\alpha) = 0$, which follows since $\alpha = \operatorname{id}(\alpha) \in G \cdot \alpha$.


*

*Second, note that $f_\alpha \in \mathbf{F}[x]$, 

*Third note that $f_\alpha$ is minimal. Indeed if $f(\alpha) = 0$ then $f(\sigma(\alpha)) = \sigma(f(\alpha)) = \sigma(0) = 0$ for all $\sigma \in G$. Thus $\sigma(\alpha)$ is a root for all $\sigma \in G$. Thus $f_\alpha \mid f$.
Finally, we note that $f_\alpha$ splits over $\mathbf{K}$ and is separable, by construction.
 A: Did you listen when we said in your previous questions that $K/F$ is Galois iff $F = K^G$ where $G= Gal(K/F)$ is a finite group of automorphisms of $K$ ? 
For $\alpha \in K$, it means that the polynomial with distinct roots $f(x) = \prod_{\beta \in G (\alpha)} (x-\beta) \in K[x]$ has coefficients in the fixed field, ie. $f \in F[x]$, therefore it is the minimal polynomial of $\alpha$. 
$G( \alpha)= \{ \beta \in K, \exists \sigma \in G, \sigma(\alpha) = \beta\}$.
$K^G = \{ \alpha \in K, \forall \sigma \in G, \sigma(\alpha) = \alpha\}$.
$f(x) =\prod_{\beta \in G (\alpha)} (x-\beta)= \sum_{n=0}^d c_n x^n, \quad \sum_{n=0}^d \sigma(c_n) x^n = \prod_{\beta \in G (\alpha)} (x-\sigma(\beta))= f(x)$.
$f \in F[x] \land f(\alpha) = 0 \land \sigma \in Gal(K/F) \implies f(\sigma(\alpha)) =\sigma(f(\alpha)) = 0$.



*

*Show the main statement when $K = F(\alpha)$ and use induction. It is equivalent to $|Gal(K/F)| = [K:F]$.

*Because $x^p-1 \equiv (x-1)^p \bmod p$ it means $\zeta_p$ doesn't exist in $\mathbb{F}_p$ and $x^p-t^p$ is the non-separable minimal polynomial of $t$ over $\mathbb{F}_p(t^p)$ so that $\mathbb{F}_p(t)/\mathbb{F}_p(t^p)$ is a non-separable finite extension. 

*If $E=F(\alpha)$ where the minimal polynomial $f$ of $\alpha$ is separable then its normal closure (the splitting field of $f$) is Galois. 
