Find the Galois group of $\Bbb Q(\zeta_n,\sqrt[n]{2})/\Bbb Q(\zeta_n)$ Question 
Find the Galois group of $\Bbb Q(\zeta_n,\sqrt[n]{2})/\Bbb Q(\zeta_n)$, where $\zeta_n$ is a primitive $n$-th root of $1$. 
Attempt
We observe that $\Bbb Q(\zeta_n,\sqrt[n]{2})$ is the splitting field of $x^n-2\in \Bbb Q(\zeta_n)$ (which is separable) and that $x^n-2$ is irreducible over $\Bbb Q(\zeta_n)$ so $[\Bbb Q(\zeta_n,\sqrt[n]{2}):\Bbb Q(\zeta_n)]=n$. Every $ \sigma\in \operatorname{Gal}(\Bbb Q(\zeta_n,\sqrt[n]{2})/\Bbb Q(\zeta_n))$ is determined by its image on $\sqrt[n]{2}$. In particular $\sigma(\sqrt[n]{2})\in\{\sqrt[n]{2} \zeta_n^{i}\},0\leq i\leq n.$ Since $ \tau:\sqrt[n]{2}\to \sqrt[n]{2}\zeta_n \in \operatorname{Gal}(\Bbb Q(\zeta_n,\sqrt[n]{2})/\Bbb Q(\zeta_n))$ we have that $\operatorname{Gal}(\Bbb Q(\zeta_n,\sqrt[n]{2})/\Bbb Q(\zeta_n))=c_n$
Is this correct? How can I justify that $x^n-2$ is irreducible in $\Bbb Q(\zeta_n)$?
 A: It's not necessarily the case that $x^n-2$ is irreducible over $\Bbb Q(\zeta_n)$. Take $n=8$. Then $\sqrt2=e^{\pi i/4}+e^{-\pi i/4}
=\zeta_8+\zeta_8^7$. Therefore
$$x^8-2=(x^4-\zeta_8-\zeta_8^7)(x^4+\zeta_8+\zeta_8^7).$$
A: See Lang's "Algebra", chap. VIII, §9, thm.16 for the following irreducibility criterion over an arbitrary field $K$ of the polynomial  $X^n -a, n\ge 2, a\in K^*$: a sufficient condition is that for all primes $p \mid n$ one has $a\notin K^p$, and if $4\mid n$ then $a\notin -4K^4$. In your case here, the criterion shows that $X^n - 2$ is irreducible over $\mathbf Q$ for all $n\ge2$; over $\mathbf Q(\zeta_n)$ if $i$ (sin$\frac {\pi}{4})^{1/2} \notin \mathbf Q(\zeta_n)$ - a not very tractable condition, but notice that $i\in \mathbf Q(\zeta_n)$ if $4\mid n$.
Actually, by Kummer theory, a cyclic extension of degree $n$ over $K=\mathbf Q(\zeta_n)$ is of the form $K(a^{1/n})$, where $\bar a \in K^*/{K^*}^n$  has order $n$ exactly. As I understand it, you want to take $a\in {\mathbf Q}^*$, so a necessary preliminary is to determine the kernel $\Delta$ of the natural map $ {\mathbf Q}^*/{{\mathbf Q}^*}^n \to K^*/{K^*}^n$, which is obviously equal to  $({\mathbf Q}^*\cap {K^*}^n)/{K^*}^n$. This $\Delta$ is a finite group (why ?), and it remains only to pick up a class $\bar a$ of order $n$ in ${\mathbf Q}^*/{{\mathbf Q}^*}^n$ minus $\Delta$. 
NB. It would be more convincing to give an explicit description of $\Delta$. This could actually be done, but at the price of heavy cohomological calculations (as always with the prime $2$).
