$f$ irreducible iff Galois group acts transitively on roots

Question

Let $f\in{Q[t]}$ be of degree $>0$, with no repeated roots, and let L be the splitting field for F. Show that $f\in{Q}$ (rationals) is irreducible iff for any $a,b\in{Root_f(L)}$ there is $G\in{Gal(L,Q)}$ such that $G(a)=b$.

A proof that piqued my interest was the simple proof in Theorem 24.3 here. However, I do not understand when they write:

Let $g(r)=h(s)=0$, where $r,s\in{Q^a}$ (algebraic closure of rationals). Let $G\in{Gal_f}$ be a permutation which maps $r$ to $s$. We may assume that $g(x)$ is irreducible. But then $s$ has to be a root of $g(x)$.

Specifically, what I do not understand is how one can be certain such a mapping in $Gal_f$ exists/why this then implies $s$ must be a root of $g(x)$?

Answer 1

1. Why the algebraic closure: if we fix an algebraic closure, then things become more canonical, e.g. there is a unique splitting field (inside the algebraic closure) for any polynomial, unique normal closure, etc. Most importantly, a polynomial $f$ has exactly $\deg f$ roots (up to multiplicity).

2. Why such a mapping exists: by assumption, since $r$ and $s$ are roots of $f$.

3. Why must $s$ be a root of $g$: Let $g = \sum a_i t^i$. We know that $g(r) = 0$, so $\sum a_i r^i = 0$. Now, $G$ is a field homomorphism, so $G(\sum a_i r^i) = \sum a_i G(r)^i$, but the latter is just $g(G(r))$, i.e. $g(s)$, so $g(s) = 0$.

Answer 2

Let $\alpha,\beta$ be roots of $f$. Since $f$ is irreducible, $[\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\beta):\mathbb{Q}]=\deg f$, so there exist an isomorphism $\varphi:\mathbb{Q}(\alpha)\to\mathbb{Q}(\beta)$, sending $\alpha$ to $\beta$ and fixing $\mathbb{Q}$. This isomorphism can be uniquely extended to an automorphism $\sigma$ on $\Omega_{\mathbb{Q}}^f$, and so we have $\sigma\in\text{Gal}(f)$ with $\sigma(\alpha)=\sigma(\beta)$.