# How do I show that the Galois group of an algebraic number over the field containing roots of unity is cyclic?

Suppose that I have a complex number $\alpha\not\in\mathbb Q$ such that $\alpha^n\in\mathbb Q$. I take the extension $\mathbb Q(\alpha)$. Suppose that $\mathbb Q(\alpha)$ is Galois. Clearly, this extension need not be cyclic. However, I believe that if I consider the extension $F$ of $\mathbb Q$ where $F$ contains all the roots of unity in $\mathbb Q(\alpha)$, then the extension $[\mathbb Q(\alpha):F]$ becomes cyclic. I've seen this used in a few places, but can't seem to rigorously find why this is the case. I can see that $\alpha$ satisfies $x^n-\alpha^n$, but how do I know the minimal polynomial? I guessed that the roots of the minimal polynomial might be $\alpha\zeta_n$, where $\zeta_n$ is a primitive root of unity, but even then I'm not sure. I'd appreciate some help. Thanks.

• You probably want to say that $F(\alpha)/F$ is cyclic, not $\mathbb Q(\alpha)/F$ because in the latter, $\mathbb Q(\alpha)$ does not necessarily contain $F$. – Ravi Nov 23 '16 at 6:10
• Yes, but I think I may as well suppose $\mathbb Q(\alpha)$ is Galois over $\mathbb Q$. I've edited it. – adrija Nov 23 '16 at 6:13
• I don't think you should assume that because that will lead to you quickly coming to the conclusion that $n=2$ from your opening sentence. – Ravi Nov 23 '16 at 6:16
• How so? I didn't say $\alpha$ is real, it may as well be a primitive $n^{th}$ root. – adrija Nov 23 '16 at 6:20

First some definitions and notations: $a\in \mathbf Q^*, K=\mathbf Q(\alpha)$, where $\alpha$ is a root of $g_n (X)=X^n - a$. You suppose that $K/ \mathbf Q$ is normal, and you ask whether a certain subfield $F$ of $K$, containing enough roots of unity, is such that $K/F$ is a cyclic extension. Let us consider two cases according to the (ir)reducibility of $g_n (X)=X^n - a$. In the following, just to avoid petty trouble with $2$ , all rational primes which enter the game will be odd (but this not a mathematical restriction).
1) If $g_n (X)$ is irreducible, the normality hypothesis implies that $K$ contains the group $\mu_n$ of $n$-th roots of unity. Take $F=\mathbf Q (\mu_n)$ ; then $K=F(\alpha)$ is a simple Kummer extension, hence cyclic (see e.g. S. Lang's "Algebra", chap.8, §8), of degree dividing $n$
2) If $g_n (X)$ is reducible, a general criterion in op. cit., chap.8, §9, asserts that there exists a prime divisor $p$ of $n$ such that $a\in \mathbf Q ^{*p}$. Let $p^r$ the maximal power such that $a\in \mathbf Q^{*p^r}$. Then $\beta :=\alpha^{p^r}$ is a root of $g_m(X)=X^m - a$, where $n=mp^r$. If $\beta$ were a $p$-th power in $\mathbf Q(\beta)$, say $\beta=\gamma^{p^r}$, taking norms in $\mathbf Q(\beta) / \mathbf Q$ would give that $a= N(\pm\gamma) ^{p^{r+1}}$ (here we use that $p$ is odd), which contradicts the maximality of $r$. Hence, according to the same criterion op. cit., $X^{p^r} – a$ is irreducible over $\mathbf Q(\beta)$. The same argument as in 1) then shows that $K$ is cyclic over $F_p := \mathbf Q(\beta, \mu_{p^r})$, of degree dividing $p^r$. Repeat this process for all prime divisors $q$ of $n$ wich share the same propery as $p$ above, and take $F$ to be the compositum of all the $F_q$ ‘s. Then $Gal(K/F)$, being the intersection of all the $Gal(K/F_q)$'s, is cyclic