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

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} $ 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 $. Well first of all, I want a $\underline{"good basis"}$ 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 one easy case, it's when $a>0$, then $(a)^{\frac{1}{n}} \in \mathbb{R}$ , so clearly the minimal polynomial of $(a)^{\frac{1}{n}} , \zeta_n$ are distincs, 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 $, 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?

$\underline{Remark}$ I'm searching a $\underline{"good basis"}$ 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.

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