# Galois group acts transitively over irreducible factors

My question arises from this.

I'm wondering that if $K$ is a field, $f \in K[X]$ is a separable and not necessarily irreducible polynomial and $E$ is the splitting of $f$ over $K$, then $Gal(E/K)$ acts transitively on the roots of any irreducible factor of $f$.

I think it is possible to make a similar argument to the given in the other question. If $g$ is an irreducible factor, and $\alpha,\beta$ are roots of $g$, we can still make the isomorphism $\sigma:K(\alpha)/K \to K(\beta)/K$ with $\sigma(\alpha)=\beta$, as $g = Irr(\alpha,K) = Irr(\beta,K)$. Then, as $E/K$ is still a normal extension containing $K(\alpha)$ and $K(\beta)$, $\sigma$ can be extended to an automorphism $\overline{\sigma}:E/K \to E/K$, so that $\overline{\sigma} \in Gal(E/K)$ and $\overline{\sigma}(\alpha)=\beta$. Is this right?

• Nice LaTeX use, but FYI it is generally frowned upon to use external links, in case they go bad. Please put all such information in the question itself! – The Count Dec 1 '16 at 21:06
• Yes, it's right. Do you even need $f$ to be separable? – Prahlad Vaidyanathan Dec 5 '16 at 5:07
• I think so, just to be able to talk about "Galois extension", @Prahlad. – Maximator Dec 5 '16 at 18:03
• @Maximator: You don't need the extension to be Galois for this result. – Prahlad Vaidyanathan Dec 11 '16 at 19:11