5
$\begingroup$

Let $K$ be the splitting field of $f(x)\in\mathbb{Q}[x]$ over $\mathbb{Q}$, of degree $n$, and suppose that $\operatorname{Gal}(K/\mathbb{Q})=S_n$.

It is easy to show that his implies that $f$ is irreducible, since the Galois group acts transitively on the roots.

I would like to show now that

a) if $f(\alpha)=0,$ then $\operatorname{Aut}(\mathbb{Q}(\alpha))$ is trivial,

and

b) if $n\ge 4$, then $\alpha^n$ cannot be rational.

I think I see b), since if this were so $x^n-q$ would be a minimal polynomial for $\alpha$ over $\mathbb{Q}$ for some rational $q$, and it seems rather clear that there are only two possibilities for the Galois group in this case, a cyclic group or a group generated by an $n$-th root of $q$ and an $n$-th root of unity, neither of which can be isomorphic to $S_n$ when $n\ge 4$.

  • 3
    Your last sentence does nor make sense: the Galois group cannot be generated by elements of the field!2012-03-09

2 Answers 2