Let $E$ be the splitting field of $x^6-2$ over $\mathbb{Q}$. Show that $Gal(E/\mathbb{Q})\cong D_6$, the dihedral group of the regular hexagon.
I've shown that $E=\mathbb{Q}(\zeta_6, \sqrt[6]{2})$, where $\zeta_6$ is a (fixed) primitive sixth root of unity, and thus that $[E:\mathbb{Q}]=12$.
I'm getting a little mixed up working out the automorphisms, though. I know the Galois group is determined by the action on the generators $\zeta_6$ and $\sqrt[n]{2}$. So then the possibilities appear to be: \begin{align*}\sqrt[6]{2}&\mapsto\zeta_6^n\sqrt[6]{2}\;\;\;\mbox{ for } n=0,1,\ldots ,5 \\ \zeta_6&\mapsto \zeta_6^j\;\;\;\;\;\;\;\;\mbox{ for } j=1,5\,.\end{align*}
Does this make sense? Something doesn't quite feel right, but I'm not sure where the issue might be. I know that in some sense the generators are "independent", because I definitely can't get one generator from the other. (For example, it'd be different if we had fourth roots of unity because we could get $\sqrt{2}$ from both generators.)
Any help is appreciated