Suppose that $\alpha \in C$ with $\alpha^n \in Q$ such that $Q[\alpha]:Q$ is Galois. Suppose that $\alpha \in C$ with $\alpha^n \in Q$ such that $Q[\alpha]:Q$ is Galois. Now Let F be the field containing $Q$ generated by all the roots of unity in $Q[\alpha]$ prove $Gal(Q[\alpha]:F)$ is cyclic. I have no clue how to start this problem. All i know is that $f(x)=x^n-a$ has $\alpha$ as a root for some $a \in Q$. We don't know if it is irreducible, if it was then $Q(\alpha)$ would have the nth root of unity. I am not sure how to proceed.
 A: Suppose that $\alpha^n \in \mathbb{Q}$ and that this $n$ is the smallest possible with this property. First note that, as you take the extension $\mathbb{Q}(\alpha) / F$, you are getting the splitting field of the polynomial $f(x) = x^n - \alpha$ over $F$, since the roots of this polynomial over $\mathbb{C}$ are given by the set:
$$
S = \{ \alpha \cdot \xi^i : \xi \text{ is a primitive n-th root of unity and } 0 \leq i \leq n-1\}
$$
Then you need to define this morphism of groups:
$$
\psi : \text{Gal}(\mathbb{Q}(\alpha)/F ) \rightarrow \{ \xi \in \mathbb{C} : \xi^n = 1 \} ( \simeq \mathbb{Z}_n )
$$
defined in each element $\sigma \in \text{Gal}(\mathbb{Q}(\alpha)/F )$ by:
$$
\psi(\sigma) \doteq \frac{\sigma(\alpha)}{\alpha}.
$$
You can prove that $\psi$ is well defined, and that $\psi$ is an injection. Then, in particular, you get that
$$
\text{Gal}(\mathbb{Q}(\alpha)/F ) \simeq \mathbb{Z}_d
$$
with $d \mid n$. Indeed, the polynomial $f(x)$ do not need to be irreducible. That's the case if and only if you get $\text{Gal}(\mathbb{Q}(\alpha)/F ) \simeq \mathbb{Z}_n$.
