Galois group of $K/\mathbb{Q}$ where $K $ is splitting field of $x^{12} -3 $ I am trying to find the Galois group of $K/\mathbb{Q}$ where $K$ is the splitting field of $x^{12}-3$ and how it acts on the roots of $x^{12} - 3$.
I was able to show that the degree of the Galois extension $K/\mathbb{Q}$ is 24 and I know that the roots of the given polynomial are $3^{1/12}, w_{12}3^{1/12}, w^2_{12}3^{1/12}, ... ,w^{11}_{12}3^{1/12}$ where $w_{12}^{12} = 1$. However, from here I am lost, I don't know how deduce the structure of the Galois group and how it acts on the roots. Thanks for any hint or help on this problem!
 A: First of all let $\zeta_{12}$ be a 12-th root of unity. Then it's not hard to prove that $\mathbb{Q}(\zeta_{12}) = \mathbb{Q}(\sqrt{3},i)$. Thus we have that:
$$\mathbb{Q}\left(\zeta_{12}, \sqrt[12]{3}\right) = \mathbb{Q}\left(\sqrt{3},i,\sqrt[12]{3}\right) = \mathbb{Q}\left(\sqrt[12]{3},i\right)$$
Now as $\mathbb{Q}\left(\sqrt[12]{3}\right)$ is a real subfield we have that $i \not \in \mathbb{Q}\left(\sqrt[12]{3}\right)$ and thus $\left[\mathbb{Q}\left(\sqrt[12]{3},i\right):\mathbb{Q}\left(\sqrt[12]{3}\right)\right]=2$. From here we get the Galois group is of order $24$. This seems to be the easiest way, as proving that $x^6 - \sqrt{3}$ is irreducible in $\mathbb{Q}\left(\zeta_{12}\right)$ isn't that easy (at least for me). Moreover writing the splitting field as $\mathbb{Q}\left(\sqrt[12]{3},i\right)$ gives us two linearly independent subfields $\mathbb{Q}\left(\sqrt[12]{3}\right)$ and $\mathbb{Q}\left(i\right)$ and this can help us with determining the automorphisms.
Now because of the linear independence we can easily write all $24$ automorphisms. They are all of the form:
$$\sigma_{a,b}: \begin{array}{lr}
       \sqrt[12]{3} \to \zeta_{12}^a\sqrt[12]{3}\\
       i \to (-1)^bi
     \end{array}$$
Now note that $\sigma_{1,0}$ has order $12$ and $\sigma_{0,1}$ has order $2$. Also we have:
$$ \sigma_{0,1}\circ \sigma_{1,0} \circ \sigma_{0,1}(i) = \sigma_{1,0}^{-1}(i)$$
$$ \sigma_{0,1}\circ \sigma_{1,0} \circ \sigma_{0,1}\left(\sqrt[12]{3}\right) = \sigma_{0,1}\left(\zeta_{12}\sqrt[12]{3}\right) = \sigma_{0,1}\left(\left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right)\sqrt[12]{3}\right) $$
$$= \left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right)\sqrt[12]{3} = \zeta_{12}^{11}\sqrt[12]{3} = \sigma_{11,0}\left(\sqrt[12]{3}\right) = \sigma_{1,0}^{-1}\left(\sqrt[12]{3}\right)$$
Thus we get the relation $\sigma_{0,1}\circ \sigma_{1,0} \circ \sigma_{0,1}= \sigma_{1,0}^{-1}$ and this is exactly the relation that characterizes the dihedral group of order $24$. Thus we have:
$$\operatorname{Gal}(K/\mathbb{Q}) \cong D_{12}$$
