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 $n\ge 4$.
 A: 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).
A: $a)$ 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!
