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.

  • $\begingroup$ Is your field $Q$ the field of rational numbers or any field? $\endgroup$ – Felipe Monteiro Dec 2 at 23:35
  • $\begingroup$ @FelipeMonteiro Rational numbers $\endgroup$ – Sorfosh Dec 2 at 23:36
  • $\begingroup$ I feel like this is written up in several algebra textbooks. Perhaps in Lang? Or Dummitt and Foote? $\endgroup$ – Matthew Leingang Dec 2 at 23:38
  • $\begingroup$ This is Kummer theory, see e.g. Lang's "Algebra", VIII, 8. $\endgroup$ – 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 $\text{Gal}(\mathbb{Q}(\alpha)/F ) \simeq \mathbb{Z}_n$.

  • $\begingroup$ Why would $Q(\alpha)$ be the splitting field of $x^n-\alpha$? $Q(\alpha)$ might not have all roots of unity. $\endgroup$ – Sorfosh Dec 7 at 23:52

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