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

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 $G$ embedds naturally into the semidirect product of the Galois group $(\mathbb{Z}/n)^{\times}$ of $X^n-1$ and the automorphism group $Aut_{\mathbb{Q}}(\mathbb{Q}(\sqrt[n]{a})\cong \mathbb{Z}/n$.

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 $\sqrt{a}\not\in \mathbb{Q}(\zeta_n)$.

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

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 $\mathbb Q[\zeta_{2n}]$.

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 $$\mathrm{Gal}(X^p-a) \cong \mathbb{Z}/(p-1)\mathbb{Z} \ltimes \mathbb{Z}/p\mathbb{Z}. $$