prove $2^{1/4}$ not in $Q(\xi)$

Question

$\xi$ is a root of unity and $f(x)\in Q[x]$ ,prove that $f(\xi) \ne 2^{1/4}$ for all $f(x)$ and $\xi$

I know it means $2^{1/4} \notin Q(\xi)$.

If $i \in Q(\xi)$ I can prove that by considering the Galois group of $X^4 -2=0$ as it is not cyclic but generally it will not happen. Can anyone give some help ?

Answer 1

The Galois group of $\Bbb Q(\xi)/\Bbb Q$ is Abelian. If $\alpha\in \Bbb Q(\xi)$ then $\Bbb Q(\alpha)$ is a Galois extension, with Abelian Galois group over $\Bbb Q$. But $\Bbb Q(\sqrt[4]2)$ is not Galois over $\Bbb Q$.