f(x) is irreducible over K if and only if Gal(F/K) acts transitively on the roots of f(x).

Question

Consider the following theorem from book John A.Beachy Abstract Algebra. Proposition 8.6.2.

I don't understand why it can be extended to an automorphism of the splitting field $F$. Can someone explain that part?

Answer 1

This follows from the following more general theorem:

Theorem. Let $\sigma : K \to K^\prime$ be an automorphism of fields, and let $f \in K[X]$. Let $f^\sigma \in K^\prime[X]$ be the polynomial induced by $\sigma$. Let $F$ be a splitting field of $f$ over $K$, and let $F^\prime$ be a splitting field of $f^\sigma$ over $K^\prime$. Then there exists a field automorphism $\tau : F \to F^\prime$ such that $\tau\big|_K = \sigma$.

Proof. We induct on $[F : K]$. If $[F : K] = 1$, then let $\tau = \sigma$. If $[F : K] > 1$, then $f$ does not split over $K$, so $f$ has an irreducible factor $g \in K[X]$ with $\deg(g) \geq 2$. Let $g^\sigma \in K^\prime[X]$ be the polynomial induced by $\sigma$. If $\alpha \in F$ is a root of $g$ and $\alpha^\prime \in F^\prime$ is a root of $g^\sigma$, then there is an automorphism $\rho : K(\alpha) \to K^\prime(\alpha)$ given by composing the automorphisms $$ K(\alpha) \cong K[X]/(g) \cong K^\prime[X]/(g^\sigma) \cong K^\prime(\alpha^\prime). $$ Thus, $\rho(\alpha) = \alpha^\prime$, and $\rho\big|_F = \sigma$. Since $$ [F : K] = [F : K(\alpha)][K(\alpha) : K] > [F : K(\alpha)], $$ we may, by induction, extend $\rho$ to an automorphism $\tau : F \to F^\prime$. But then $\tau\big|_K = \sigma$, so we are done.