Prove there is no such nth root of unity $\zeta$ such that $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{Q}(\zeta) \quad$

Question

I'm trying to do the above problem. My approach is to use the fact that $\mathbb{Q(\zeta)}$ is the fixed subfield of the normal subgroup $A_3$ of $S_3$ and then since $A_3$ has no subgroup of the form $\{1, \tau\}$ for some transposition $\tau$, then $\mathbb{Q}(\sqrt[3]{2}) \subseteq \mathbb{Q}(\zeta)$ cannot happen since $\mathbb{Q}(\sqrt[3]{2})$ is the fixed subfield for the subgroup $\{1, \tau\}$. I'm not sure if this correct but I think the proof is supposed to be along these lines. Thanks.

Answer 1

I don't see why $\mathbb{Q}(\zeta)$ should be the fixed field of $A_3$. The degree of $\mathbb{Q}(\zeta)$ over $\mathbb{Q}$ can be much bigger than three.

Here's a hint: If $\zeta$ is a root of unity, then $\mathbb{Q}(\zeta)$ is an abelian extension of $\mathbb{Q}$. That is, it is Galois with an abelian Galois group.

Now recall that every subgroup of an abelian group is a normal subgroup. What does this say about the subfields of $\mathbb{Q}(\zeta)$?