# question on Galois group

Let $f(x)\in F[x]$, let $E/F$ be a splitting field, and let $G=\text{Gal}(E/F)$ be the Galois group.

1) If $f(x)$ is irreducible, then $G$ acts transitively on the set of all roots of $f(x)$ (if $\alpha$ and $\beta$ are any two roots of $f(x)$ in $E$, there exists $\sigma \in G$ with $\sigma (\alpha)=\beta$)

I consider $\sigma : F\to F$ an isomorphism of field and \begin{align*} \sigma ^*:F[x]&\to F[x]\\ \sum_{i}r_ix^i&\mapsto\sum_{i}\sigma (r_i)x^i \end{align*} the corresponding isomorphism of ring. If $p(x)\in F[x]$, I denote $p^*(x)=\sigma ^*(p(x))$. Let $\alpha$ be a root of $f(x)$ and $\beta$ a root of $f^*(x)$. By a lemma of my course, there exists a unique automorphism $\hat\sigma :F(\alpha)\to F(\beta)$ such that $\hat \sigma(\alpha)=\beta$ and by definition of $\hat\sigma$, we have that $\hat\sigma \in G$.

Question: Is my prove correct ?

2) If $f(x)$ has no repeated roots and $G$ acts transively on the roots, then $f(x)$ is irreducible.

Hint: If $f(x)=g(x)h(x)$ then $\gcd(g(x),h(x))=1$; if $\alpha$ is a root of $g(x)$ such that $\sigma (\alpha)$ is a root of $h(x)$, then $\sigma (\alpha)$ is a common root of $g(x)$ and $h(x)$.

I don't know how to prove it.

• Take a look at my answer here, see if it helps. – Aaron Maroja Nov 23 '14 at 13:12
• Your first proof relies completely on "that lemma in your course" . Most probably most people can't know what lemma is that (though one can guess). For (2): you're only assuming we have a Galois Extension $\;E/F\;$ without this necessarily being the splitting field of $\;f\;$ over $\;F\;$ as in the first part, right? Otherwise the question is trivial. – Timbuc Nov 23 '14 at 13:13