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.

• What is the definition of Galois extension you are using? – Ennar Jul 30 '17 at 0:17
• @Ennar $\mathbf K:\mathbf F<\infty$ is Galois if $G(\mathbf K:\mathbf F)^+=\sigma(\mathbf F)$, where $\sigma$ is the monomorphism between the fields – user441848 Jul 30 '17 at 0:20
• I'm not familiar with the notation, do you mean that the fixed field of automorphism group of $K$ over $F$ is $\sigma(F)$? – Ennar Jul 30 '17 at 0:24
• $$\prod_{\beta\in G\cdot\alpha}(x-\beta)\neq \prod_{\sigma\in G}(x-\sigma(\alpha))$$ – Ennar Jul 30 '17 at 1:02
• You know that $I = \{1\} = \{1,1,1\}$, right? If I wrote $\prod_{i\in I}(x-i)$ it would mean $x-1$, not $(x-1)(x-1)(x-1)$. I will once again say, if $\alpha\in F$, then $\sigma_1(\alpha)=\sigma_2(\alpha)=\ldots=\sigma_n(\alpha) = \alpha$, $G\cdot\alpha = \{\sigma_1(\alpha),\sigma_2(\alpha),\ldots,\sigma_n(\alpha)\} = \{\alpha\}$ and finally, $$\prod_{\beta\in G\cdot\alpha}(x-\beta) = x - \alpha \neq (x-\alpha)^n = \prod_{\sigma\in G} (x-\sigma(\alpha)).$$ – Ennar Jul 30 '17 at 1:34

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.

• why did you mention the bullets? – user441848 Jul 30 '17 at 3:19
• This polynomial $f(x) = \prod_{\beta \in G (\alpha)} (x-\beta) \in K[x]$ has different roots, why?. I already know that $f\in F[x],$ and that $f$ is the minimal polynomial of $\alpha$. Why did you write that as 'conclusion'? – user441848 Jul 30 '17 at 3:28
• @Annet. Are you serious ? $G(\alpha)$ is a finite set, we pick $\beta$ only once. – reuns Jul 30 '17 at 3:32
• then why the product $\prod$ notation? – user441848 Jul 30 '17 at 3:50
• Of course I'm serious 😳 – user441848 Jul 30 '17 at 3:51