I'm trying to proof this proposition (below), I already find a proof in a pdf file but it was difficult to understand so I though would be better to do it myself but I got stuck trying to do it. Please help me, please!
Proposition: If $[\mathbb K:\mathbb F]<\infty$ and it is Galois, then it is normal and separable.
Proof: We need to show that every irreducible $p(x)\in\mathbb F$ with one root $\alpha\in\mathbb K,$ splits over $\mathbb K$. (This is for normality) and that for every $\alpha\in\mathbb K $ is separable.
So let $p(x)\in\mathbb F$ be irreducible with one root $\alpha\in\mathbb K$ and let $\alpha\in\mathbb K. $
By hypothesis, $G(\mathbb K:\mathbb F)^+=\sigma(\mathbb F)≈\mathbb F$ and also we have that $\mathbb K$ is a splitting field of a separable $p(x)\in\mathbb F.$
Now let $\sigma\in G(\mathbb K:\mathbb F)^+$ and consider the polynomial $x-\alpha\in\mathbb K.$ Thus $\sigma(x-\alpha)=x-\sigma(\alpha)$.
So we can get something like this $f(x)=\prod_{\sigma\in G(\mathbb K:\mathbb F)^+}(x-\alpha)=\prod_{\sigma\in G(\mathbb K:\mathbb F)^+}(x-\sigma(\alpha))$.
Now $f(x)\in\mathbb K$ and splits on it too, can I considerer $f$ to be the same irreducible polynomial as $p(x)?.$ If I can consider $f(x)=p(x),$ then $\mathbb K:\mathbb F$ would be normal.
The only thing to show now would be to proof that each $\sigma_i(\alpha)$ is different to then be able to conclude that each root of $f(x)$ it's different on $\mathbb F.$ How can I do this?
Any kind of help would be very appreciated :)
This is a not so uncommon mistake. If $G = \{\sigma_1, \dots, \sigma_n\}$ then $\sigma_1(\alpha), \dots, \sigma_n(\alpha)$ are not all distinct. If they were then the minimal polynomial of $\alpha$ would have $n$ roots: $\sigma_1(\alpha), \dots, \sigma_n(\alpha)$. Consequentially, $[\mathbf{F}(\alpha) : \mathbf{F}]$ would be $n$, which is true for most $\alpha$ but not all $\alpha$.
Define the subgroup $G_\alpha = \{\sigma \in G : \sigma(\alpha) = \alpha\}$, called the stabilizer of $\alpha$. Also define $G\cdot\alpha = \{\sigma(\alpha) : \sigma \in G\}$, called the orbit of $\alpha$. The set $G\cdot\alpha$ consists of the distinct values of $\sigma(\alpha)$ as $\sigma$ runs over all the elements of $G$.
The minimal polynomial of $\alpha$ is
$$ f_\alpha(x) := \prod_{\beta \in G \cdot \alpha} (x - \beta). $$
The polynomial given by
$$ \prod_{\sigma \in G} (x - \sigma(\alpha)) $$
is called the characteristic polynomial of $\alpha$ and is equal to $f_\alpha(x)^{[\mathbf{K} : \mathbf{F}(\alpha)]}$.
Why is $f_\alpha$ the minimal polynomial of $\alpha$?
First, note that $f_\alpha(\alpha) = 0$, which follows since $\alpha = \operatorname{id}(\alpha) \in G \cdot \alpha$.
Second, note that $f_\alpha \in \mathbf{F}[x]$, which follows since for all $\tau \in G$ we have $$\tau \cdot (G \cdot \alpha) = \{\tau(\beta) : \beta \in G \cdot \alpha\} = \{\tau\sigma(\alpha) : \sigma \in G\} = \{\sigma(\alpha) : \sigma \in G\} = G\cdot \alpha.$$ Hence, $$\tau \cdot f_\alpha(x) = \tau \cdot \prod_{\beta \in G \cdot \alpha} (x - \beta) = \prod_{\beta \in G \cdot \alpha} (x - \tau(\beta)) = \prod_{\gamma \in \tau\cdot(G \cdot \alpha)} (x - \gamma) = \prod_{\gamma \in G \cdot \alpha} (x - \gamma) = f_\alpha(x). $$ Therefore $f_\alpha(x) \in \mathbf{K}^G[x] = \mathbf{F}[x]$. Here $\mathbf{K}^G$ is the fixed field of $G$ and equals $\mathbf{F}$ since $\mathbf{K}/\mathbf{F}$ is Galois.
Finally, we note that $f_\alpha$ splits over $\mathbf{K}$ and is separable, by construction.
The proof that $$\prod_{\sigma \in G} (x - \sigma(\alpha)) = f_\alpha(x)^{[\mathbf{K} : \mathbf{F}(\alpha)]}.$$ Only read if interested.
The Galois Correspondence Theorem says:
The correspondence theorem tells us that $|G_\alpha| = [\mathbf{K} : \mathbf{K}^{G_\alpha}]$. But we know that $G_\alpha$ is exactly the group that fixed $\alpha$. Thus $G_\alpha$ fixes $\mathbf{F}(\alpha)$, i.e. $$|G_\alpha| = [\mathbf{K} : \mathbf{F}(\alpha)]. \tag{1}$$
Decompose $G$ as a disjoint union of cosets of $G_\alpha$: $$G = \sigma_1G_\alpha \cup \cdots \cup \sigma_rG_\alpha \tag{2}$$ Then $\sigma_1(\alpha), \dots, \sigma_r(\alpha)$ are all distinct because otherwise $\sigma_i(\alpha) = \sigma_j(\alpha)$ implies $\sigma_j^{-1}\sigma_i(\alpha) = \alpha$ implies $\sigma_j^{-1}\sigma_i \in G_\alpha$ implies $\sigma_iG_\alpha = \sigma_jG_\alpha$. Since the cosets are disjoint, this forces $i = j$.
We also know that $\{\sigma_1(\alpha), \dots, \sigma_r(\alpha)\} = G \cdot \alpha$. Certainly $$\{\sigma_1(\alpha), \dots, \sigma_r(\alpha)\} \subseteq \{\sigma(\alpha) : \sigma \in G\} = G \cdot \alpha \tag{3}$$ and if $\sigma \in G$ then we can write $\sigma = \sigma_i\tau$ for some $i \in \{1,\dots,r\}$ and $\tau \in G_\alpha$ by $(2)$ and then $\sigma(\alpha) = \sigma_i(\tau(\alpha)) = \sigma_i(\alpha)$ (note: $\tau(\alpha) = \alpha$ by definition of $G_\alpha$). Thus $$G \cdot \alpha \subseteq \{\sigma_1(\alpha), \dots, \sigma_r(\alpha)\}. \tag{4}$$ Combining $(3)$ and $(4)$ gives us $$ \{\sigma_1(\alpha), \dots, \sigma_r(\alpha)\} = G \cdot \alpha \tag{5}. $$
Therefore \begin{align} \prod_{\sigma \in G} (x - \sigma(\alpha)) &= \prod_{i = 1}^r \prod_{\tau \in G_\alpha} (x - \sigma_i\tau(\alpha)) \tag{by (2)} \\ &= \prod_{i = 1}^r \prod_{\tau \in G_\alpha} (x - \sigma_i(\alpha)) \tag{def. of $G_\alpha$} \\ &= \left( \prod_{i = 1}^r (x - \sigma_i(\alpha)) \right)^{|G_\alpha|} \\ &= \left( \prod_{\beta \in G \cdot \alpha} (x - \beta) \right)^{|G_\alpha|} \tag{by (5)} \\ &= f_\alpha(x)^{|G_\alpha|} \tag{def. of $f_\alpha(x)$} \\ &= f_\alpha(x)^{[\mathbf{K} : \mathbf{F}(\alpha)]}. \tag{by (1)} \end{align}