# How to show that a subfield of a Galois extension with Galois group $S_n$ has only trivial automorphisms.

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 $n\ge 4$.

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Your last sentence does nor make sense: the Galois group cannot be generated by elements of the field! – Mariano Suárez-Alvarez Mar 9 '12 at 15:32

Hint for part a): Show first that $\alpha$ is the only zero of $f(x)$ in $\mathbb{Q}(\alpha)$.

Hint for part b): You may have the right idea, but you phrased it a bit funnily. If the minimal polynomial were of the form $x^n-q$, then that would also have to be $f(x)$ up to a scalar multiple. Hence you get the splitting field of $f(x)$ by adjoining to $\mathbb{Q}$ an $n$th root of $q$ and a primitive $n$th root of unity. This does, indeed, lead to a splitting field of a wrong degree (or to a Galois group of wrong order).

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Well any automorphism has to act transitively on the roots, so we cannot have $\sigma(\alpha)=\alpha$. If $\alpha'$ were another zero of $f(x)$ in $\mathbb{Q}(\alpha)$, then since the automorphisms act transitively, $\mathbb{Q}(\alpha)$ would have to contain all of the roots and thus $\alpha$ would have to be a primitive root of the extension $K$. Is this argument correct so far? I am not entirely sure why this would be a contradiction, but it seems close. – Lepidopterist Mar 9 '12 at 15:45
@jaywendt: It is false that "any automorphism has to act transitively on the roots". (This is the same error in thinking you committed in your previous question). If the roots are $\alpha_1,\ldots,\alpha_n$, then the action of an element $\sigma\in S_n$ is given naturally by $\alpha_i\mapsto \alpha_{\sigma(i)}$. There are plenty of elements of $S_n$ that don't act transitively on $\{1,2,\ldots,n\}$. The entire group acts transitively, but its elements don't have to. – Arturo Magidin Mar 9 '12 at 16:57

$a)$ Let $\sigma \neq Id \in Aut(\mathbb{Q}(\alpha))$.So $\sigma(\alpha) \neq \alpha$ is a root of $f$ and $\sigma(\alpha) \in \operatorname{Aut}(\mathbb{Q}(\alpha))$. But since $G=Gal(K/\mathbb{Q})=S_n$ so exist $\tau \in G$ such that $\tau(\alpha)=\alpha$ but $\tau(\sigma(\alpha)) \neq \sigma(\alpha)$. Which is a contradiction since $\tau(\alpha)=\alpha \implies \tau(\sigma(\alpha))=\sigma(\alpha)$ as $\sigma(\alpha)\in \mathbb{Q}(\alpha)$

$b)$ Since of $G=S_n$ so, $f$ is irreducible over $\mathbb{Q}[x]$ of degree $n$. But if $\alpha^n=a \in \mathbb{Q}$ then clearly $f(x)=c(x^n-a) \implies |G| \leq n^2 \implies n!\leq n^2 \implies n<4$ a contradiction!

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