$f$ is irreducible $\iff$ $G$ act transitively on the roots

Let $F$ be a field and $f\in F[X]$ a separable polynomial. Let $K_f$ be the splitting field of $f$ and $G_f=\text{Gal}(K_f/F)$ its Galois group. Show that $f$ is irreducible $\iff$ $G_f$ acts transitively on the roots on $f$.

My work

$\implies :$ Let $f\in F[x]$ irreducible, $\alpha \in K_f$ a root and $\alpha _1,...,\alpha _n$ the other roots of $f$ (so $K_f=F(\alpha ,\alpha _1,...,\alpha _n)$). What I have to show (I think) is that for all $i$ there is a $\sigma \in G_f$ s.t. $\sigma \alpha =\alpha _i.$

Q1) Is it right ?

Q2) Is $f$ the minimal polynomial of $\alpha $ ? (To me it is, but my teacher said that it can be not the case, but I don't understand why).

Since $\alpha $ and $\alpha _1$ have the same minimal polynomial, there is an isomorphism $\sigma :\alpha \longmapsto \alpha _1$. We can observe that $\sigma : F(\alpha )\longrightarrow K_f$. Now $F(\alpha ,\alpha _1)/F(\alpha )$ is algebraic, then we can extend $\sigma $ to a morphism $F(\alpha ,\alpha _1)\longrightarrow K_f$. Now, we can continue like that and extend $\sigma $ as $\psi: K_f\longrightarrow K_f$ s.t. $\psi(\alpha )=\alpha _1$.

Q3) The fact that $\psi(\alpha )=\alpha _1$ comes from the fact that $\psi|_{F(\alpha )}=\sigma $, right ?

The bijectivity is obvious since it's a field homomorphism and that $\sigma (K_f)=K_f$.

Q4) How to conclude that $G_f$ act transitively on the roots ?

I don't have particular problem for the converse.

Solution 1:

Proposition. Let $P\in K[X]\setminus K$ be separable over $K$ and $L$ be a splitting field of $P$ over $K$. Let $\mathcal{R}_L(P)$ be the roots of $P$ in $F$.

(i) $\textrm{Gal}(L/K)$ acts on $\mathcal{R}_L(P)$.

(i) $P$ is irreducible over $K$ if and only if $\textrm{Gal}(L/K)$ acts transitively on $\mathcal{R}_L(P)$.

Proof. $L/K$ is a galois extension, since $L$ is a splitting field of a separable polynomial over $K$. Let: $$G_P:=\textrm{Gal}(L/K).$$ (i) There exists $\lambda\in K^\times$ such that: $$P=\lambda\prod_{\eta\in\mathcal{R}_L(P)}(X-\eta).$$ Let $g\in G_p$ and $\eta\in\mathcal{R}_L(P)$, since $P\in K[X]$ and $L^{G_P}=K$, one has: $$P(g(\eta))=g(P(\eta))=g(0)=0.$$ Therefore, $g(\eta)\in\mathcal{R}_L(P)$ and the following map is well-defined: $$\left\{\begin{array}{ccc}G_P & \rightarrow & \mathfrak{S}(\mathcal{R}_L(P))\\g & \mapsto & g_{\vert\mathcal{R}_L(P)}\end{array}\right..$$ $G_P$ acts on $\mathcal{R}_L(P)$.

(ii) Let $\{\mathcal{O}_i\}_{i\in\{1,\ldots,k\}}$ be the pairwise distinct orbits of the action of $G_P$ on $\mathcal{R}_L(P)$. One has: $$\mathcal{R}_L(P)=\coprod_{i=1}^k\mathcal{O}_i.$$ Therefore, one derives: $$P=\lambda\prod_{\eta\in\mathcal{R}_L(P)}(X-\eta)=\lambda\prod_{i=1}^k\prod_{\eta\in\mathcal{O}_i}(X-\eta).$$ Notice that for all $i\in\{1,\ldots,k\}$, $\displaystyle\prod_{\eta\in\mathcal{O}_i}(X-\eta)$ is irreducible over $K$, since it is the minimal polynomial of an element of $\mathcal{O}_i$. Hence, $P$ is irreducible if and only if $k=1$ if and only if there is only one orbit of the action of $G_P$ on $\mathcal{R}_L(P)$ if and only if $G_P$ acts transitively on $\mathcal{R}_L(P)$. $\Box$

N.B. If you need it, I can explain why $\displaystyle\prod_{\eta\in\mathcal{O}_i}(X-\eta)$ is the minimal polynomial of an element of $\mathcal{O}_i$.

To answer precisely to your question :

Q1) You're right, this means that there is only one orbit of the action of $G_P$ on $\mathcal{R}_L(P)$.

Q2) You're teacher seems to be wrong, if $P$ is irreducible and $P(\alpha)=0$, $P$ is the minimial polynomial of $\alpha$. However, if $P$ isn't irreducible this isn't true! Indeed, let $K:=\mathbb{Q}$ and $P:=X^4-1$, then $L:=\mathbb{Q}(i)$ is a splitting field of $X^4-1$ over $\mathbb{Q}$. Howeover, $X^4-1$ is not the minimal polynomial of $i$ over $\mathbb{Q}$.