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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.

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Do you mean the polynmial $x^n-a \in \mathbb{Q}[x]$? If not then your "polynomial" is a fixed number. –  Fly by Night Sep 29 '12 at 20:38
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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 $\mathbb Q[\zeta_{2n}]$.

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