Galois group of $f := X^6 - 6 ∈ \Bbb Q[X]$

Question

A quick sanity check:

A splitting field for $f$ over $\Bbb Q$ is $L := \Bbb Q(\zeta_3, \sqrt[6]{6})$. It is of degree 12 over $\Bbb Q$, so Gal$(f)$ will be a group of order 12. An automorphism of $L$ must send $\sqrt[6]{6}$ to $± \zeta_3^k \sqrt[6]{6}$ for some $k ∈ \{0,1,2\}$, but it must also send $\zeta_3 $ to $\zeta_3^k $. Does this provide enough information to determine the Galois group?

Answer 1

I think rather than $\zeta_3$ a root of $X^2 + X + 1$ you need $\zeta_6$ a root of $X^2 - X + 1$.

You will have the complex conjugation automorphism $\tau$.

And 6 different maps $\sigma \sqrt[6]{6} = \zeta_6^r \sqrt[6]{6}$ for $r$ being $1$ up to $6$.

The complex conjugation $\tau$ map interacts with the $\sigma$ maps by reversing the cycle, similar to the dihedral group.