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

 A: 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.
