Computing the Galois group of polynomials $x^n-a \in \mathbb{Q}[x]$

Question

I have some problems with this exercise. I don't know if it can be done.

Consider the polynomial $ x^n - a \in \mathbb{Q}[x]$. Can I compute the Galois group of this over $\mathbb{Q}$?

Maybe having a nice basis. The splitting field is given by $\mathbb{Q}(\zeta_n,\alpha)$, where $\zeta_n$ is a primitive root of unity, and $\alpha$ is some number such that $\alpha^n = a $. First of all, I want a good basis for the splitting field. In the sense that the minimal polynomials of the adjoined elements are different (in this case the computation of the Galois group is very simple).

For example, an easy case it's when $a>0$, then $a^{\frac{1}{n}} \in \mathbb{R}$, so clearly the minimal polynomials of $a^{\frac{1}{n}}$, respectively $\zeta_n$ are distinct, and I'm done. If $n$ is odd, then it's also easy, since one root it's also real, for example $x^3+3,a=-3$, the real root is $ \root 3 \of { - 3} = - \root 3 \of 3 $, so I can consider the splitting field as $\mathbb{Q}(-\root 3 \of 3 , \zeta_3 )=\mathbb{Q}(\root 3 \of 3 , \zeta_3 )$. The difficult case is when $n$ is even and $a<0$, for example $x^8+20$ or $x^4+20$. In some cases, as in the second, there are particular cases since there exist algorithms for the Galois group of quartics, but in general it can be done?

Remark. I'm searching a good basis for the splitting field. In the sense that the minimal polynomials of the adjoined elements, are different, since in this case the computation of the Galois group is very simple.

Answer 1

Let $K=\mathbb{Q}(\zeta_n)$ and $L=\mathbb{Q}(\sqrt[n]{a})$. Then the splitting field of $X^n-a$ is given by $F=KL$ and has degree $$ [F:\mathbb{Q}]=[KL:\mathbb{Q}]=\frac{[K:\mathbb{Q}][L:\mathbb{Q}]}{[K\cap L:\mathbb{Q}]}=\frac{\phi(n)n}{2^s}, $$ where $s\ge 0$ satisfies $2^s\mid \phi(n)$ and $K\cap L=\mathbb{Q}(\sqrt[2^s]{a})$. So the point is that we need to determine the intersection $\mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\sqrt[n]{a})$. This intersection is often not just $\mathbb{Q}$; note for example that $\sqrt{p}\in \mathbb{Q}(\zeta_p)$ for all primes $p\equiv 1 \bmod 4$. A complete answer for this (and the whole question) has been published by Jacobson and Velez in "The Galois group of a radical extension of the rationals", in 1990. Let $G$ be the Galois group of $X^n-a$. It is obvious that $G$ embeds naturally into the semidirect product of the Galois group $(\mathbb{Z}/n)^{\times}$ of $X^n-1$ and the automorphism group $\operatorname{Aut}_{\mathbb{Q}}(\mathbb{Q}(\sqrt[n]{a}))\cong \mathbb{Z}/n$.

Proposition [Jacobson, Velez]: One has $G\cong \mathbb{Z}/n \rtimes (\mathbb{Z}/n)^{\times}$ if and only if $n$ is odd, or $n$ is even and $\sqrt{a}\not\in \mathbb{Q}(\zeta_n)$.

Special attention is needed for the case $n=2^k$; see also this MO-question.

Answer 2

The minimal polynomial of $\zeta_n$ and $\sqrt[n]a$ are almost always different. In fact, every root in $\mathbb C$ of the $n$th cyclotomic polynomial has absolute value $1$, whereas $\sqrt[n]a$ usually has not. The only exceptions are $a=1$ and $a=-1$. For $a=1$, the splitting field is simply $\mathbb Q[\zeta_n]$. For $a=-1$, a solution to $x^n+1=0$ is also a solution to $x^{2n}-1=0$, hence the splitting field is simple $\mathbb Q[\zeta_{2n}]$.

Answer 3

A few remarks on this specific galois group: Consider the field extension $\mathbb{Q}(\zeta)|\mathbb{Q}$ and $\zeta$ a primitive root of $X^{n} - 1 \in \mathbb{Q}[X]$. A general fact of cyclotomic field extensions is that in this case: $\mathrm{Gal}(\mathbb{Q}(\zeta)|\mathbb{Q})\cong (\mathbb{Z}/n\mathbb{Z})^{*}$ whereas $|(\mathbb{Z}/n\mathbb{Z})^{*}|=\varphi(n)$. Let $a \in \mathbb{Q}$ such that $\sqrt[n]{a} \notin \mathbb{Q} $ then the splitting field of $X^n - a$ is as mentioned above given by $$ \mathrm{SF}(X^n - a)=\mathbb{Q}(\zeta, \sqrt[n]{a}).$$ Furthermore the automorphism group of $ \mathbb{Q}(\sqrt[n]{a})$ is cyclic of order $[\mathbb{Q}(\sqrt[n]{a}):\mathbb{Q}]=n$ hence $\mathrm{Aut}_{\mathbb{Q}}(\mathbb{Q}(\sqrt[n]{a})) \cong \mathbb{Z}/n\mathbb{Z}$. Note that both groups embed into the full galois group and moreover $\mathbb{Q}(\sqrt[n]{a}) \cap \mathbb{Q}(\zeta)= \mathbb{Q} $. One could then suggest (following theorems of galois theory) that the full galois groups is given by their product but this fails if we consider $\mathrm{Gal}(X^4 - 5) \cong \mathrm{D}_{4} $ which is not a direct product of cyclic groups. However if $ n=p $ is prime then $(\mathbb{Z}/p\mathbb{Z})^{*}=\mathbb{Z}/(p-1)\mathbb{Z}\triangleleft \mathrm{G}$ and since $|\mathrm{G}|=p(p-1)$ we can apply Schur-Zassenhaus because for all primes $p$ $$ \mathrm{gcd}(|(\mathbb{Z}/p\mathbb{Z})^{*}|, [G:(\mathbb{Z}/p\mathbb{Z})^{*}]) =\mathrm{gcd}(p-1,p)=1. $$ This yields a subgroup $\mathrm{U} \subseteq \mathrm{G}$ such that $\mathrm{G}\cong (\mathbb{Z}/p\mathbb{Z})^{*}\ltimes \mathrm{U} $ and $|\mathrm{U}|=p $ hence $\mathrm{U}\cong \mathbb{Z}/p\mathbb{Z}$ therefore $$\mathrm{Gal}(X^p-a) \cong \mathbb{Z}/(p-1)\mathbb{Z} \ltimes \mathbb{Z}/p\mathbb{Z}. $$