Let $F$ be a field of characteristic $0$ and let $f(x)\in F[x]$, and let $G$ be the Galois group of $f(x)$ over $F$.
Now, if $f(x)$ is irreducible, I know that $G$ acts transitively on the roots of $f(x)$. $(*)$
Suppose that $f(x)$ is not irreducible, and let $g(x)$ be an irreducible factor of $f(x)$. Does $G$ act transitively on the roots of $g(x)$?
I think we can show this by a restriction homomorphism but I'm not sure. My idea is this:
If $K$ is the splitting field of $f(x)$ and $L$ is the splitting field of $g(x)$ then $G = \text{Gal}(K/F)$ and $L$ is an intermediate field of $K/F$. Let $H = \text{Gal}(L/F)$. Define a map $$\phi : G\rightarrow H$$ $$\sigma \mapsto \sigma|_L$$
Now this map is surjective and so the image of $\phi$ is $H$ and so $\text{im}\phi$ acts transitively on the roots of $g(x)$ by $(*)$. But since $\phi$ is just a restriction, $G$ must also act transitively on the roots of $g(x)$.
It's quite a wordy proof but is the idea correct?
Best Answer
That looks good. You can potentially cut down on the wordiness of the argument by "going in reverse" logically, by extending maps instead of restricting maps:
We know that the Galois group of $g$, which you call $\operatorname{Gal}(L/F)$, acts transitively on the roots of $g$. Since $L$ is a subfield of $K$ (which is an algebraic extension of $F$), every $F$-automorphism of $L$ can be extended to an $F$-automorphism of $K$ per the isomorphism extension theorem. Clearly, these extensions are elements of $\operatorname{Gal}(K/F)$ as this is the set of all $F$-automorphisms of $K$. Thus, we can conclude that $\operatorname{Gal}(K/F)$ acts transitively on the roots of $g$.