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Suppose $f\in\mathbb{Q}[x]$ has degree $n$, and let $K$ be the splitting field of $f$ over $\mathbb{Q}$. Suppose the Galois group of $K/\mathbb{Q}$ is $S_n$ with $n\geq 3$. It is not hard to show that $f$ is irreducible and that for any root $\alpha$ of $f(x)$ the only automorphism of $\mathbb{Q}(\alpha)$ is the identity. Now, I am trying to understand why if $n\geq 4$ then $\alpha^n\not\in\mathbb{Q}$.

If $\alpha\in\mathbb{Q}$ then $\alpha$ satisfies a polynomial $x^n-a/b$, where $a/b\in\mathbb{Q}$ and we have that $f(x)$ divides this polynomial as it is the minimal polynomial for $\alpha$. However, how should I proceed to arrive at a contradiction?

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One way to proceed is the following: Not only is the only automorphism of $\mathbb Q(\alpha)$ the identity, the only automorphism of $\mathbb Q(\alpha,\beta)$ for $\alpha,\beta$ roots of $f(x)$ are the identity and the one swapping $\alpha$ and $\beta$.

On the other hand, if the roots were of the form $\alpha\zeta_n^i$ for $\zeta_n$ a $n$-th root of unity and $1\leq i \leq n$ (as they would have to be if $\alpha^n \in \mathbb Q$), then $\mathbb Q(\alpha,\alpha\zeta_n)$ is Galois and contains all the roots and so has many automorphisms (for $n\geq 3)$.

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