# 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.

• Is your field $Q$ the field of rational numbers or any field? – Felipe Monteiro Dec 2 at 23:35
• @FelipeMonteiro Rational numbers – Sorfosh Dec 2 at 23:36
• I feel like this is written up in several algebra textbooks. Perhaps in Lang? Or Dummitt and Foote? – Matthew Leingang Dec 2 at 23:38
• This is Kummer theory, see e.g. Lang's "Algebra", VIII, 8. – nguyen quang do Dec 4 at 9:25

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 $$\text{Gal}(\mathbb{Q}(\alpha)/F ) \simeq \mathbb{Z}_n$$.
• Why would $Q(\alpha)$ be the splitting field of $x^n-\alpha$? $Q(\alpha)$ might not have all roots of unity. – Sorfosh Dec 7 at 23:52