Statement) Let $K$ be a Galois extension over a field $F \ \text{s.t.} \ \vert G(K/F) \vert =[K;F] =n $
Say $ G(K/F) = \{ \sigma_1(=id) , \sigma_2,…, \sigma_n \}$
Take a element of $\alpha \in K \setminus F$, whose $deg(irr(\alpha, F))= n $
Then $\text{irr}(\alpha,F) = \prod_{i=1}^n (x-\sigma_i(\alpha))$
In my guess, the above statement is true. Considering the properties of the Galois group, Because the $\forall \sigma_i$ which are the elements of the $G(K/F)$, $\alpha$ and $\sigma_i(\alpha)$ are conjugates sharing the same irreducible polynomials.
Is my thought right?
Any help would be appreciated. Thanks.
Best Answer
$\newcommand{\mc}{\mathcal}$ You are right.
Let $G$ be the Galois group of $K:F$. For any $\alpha\in K$,define
$$ p_\alpha(x)=\prod_{\beta\in O_\alpha}(x-\beta) $$ where $O_\alpha$ is the $G$-orbit of $\alpha$.
Claim. Then for all $\alpha\in K$, $p_\alpha(x)$ lies in $F[x]$ and is irreducible over $F$.
Proof. Let $\alpha\in K$. The coefficients of $p_\alpha(x)$ are fixed by each member of $G$ and are thus in the fixed field of $G$, which is $F$. To prove irreducibility of $p_\alpha(x)$ over $F$, assume on the contrary that $p_\alpha(x)$ is reducible over $F$. Thus there exists a proper subset $O$ of $O_\alpha$ such that $p(x):=\prod_{\beta\in O}(x-\beta)$ lies in $F[x]$. But then $p(x)$ is fixed by each member of $G$, and consequently we have $\sigma(\beta)\in O$ for all $\beta\in O$. But this would mean that the $G$-orbit of an element of $O$ is not $O_\alpha$, which is a contradiction.
So we see that the degree of the minimal polynomial of any $\alpha\in K$ is same as the the size of the $G$-orbit of $\alpha$. So if $\deg(\text{irr}(\alpha, F)) = n= |G|$, we must have that the orbit of $\alpha$ has size $n$ and consequently the minimal polynomial of $\alpha$ is $$ \prod_{\sigma\in G} (x-\sigma(\alpha)) $$