Finding the $6$ $\mathbb{Q}$ Automorphisms of $\mathbb{Q}[\sqrt[3]{2}, \omega]$ Let $F = \mathbb{Q}$ and consider $\mathbb{Q}[\sqrt[3]{2}, \omega] = E$, a $6$ degree extension of $F$ that splits $p(x) = x^3 - 2$.  I am trying to establish that there are $6$ $F$ automorphisms of $E$.  That is, I'm trying to show that $|G[E:F]| = 6$ where if $g \in G[E:F]$, then $g$ is an automorphism of $E$ s.t. $f(q) = q$ for all $q \in F$.
Attempt:


*

*I know that if $g \in G[E:F]$, then $g$ must permute the roots of any member of $F[x]$.

*Then since $p(x) = x^3 - 2 \in F[x]$ has the roots $(\sqrt[3]{2})$,  $(\sqrt[3]{2}) (e^{2 \pi i/3})$, and $(\sqrt[3]{2}) (e^{4 \pi i/3})$, it must be that any $g \in G[E:F]$ permutes the elements of $R_p = \{\sqrt[3]{2}, (\sqrt[3]{2}) (e^{2 \pi i/3}), (\sqrt[3]{2}) (e^{4 \pi i/3})\}$.

*For ease of notation let $r = \sqrt[3]{2}$ and $\omega = (e^{2 \pi i/3})$.  Then $R_p = \{r, r\omega, r\omega^2\}$ and we aim to discover the $6$ members of $G[E:F]$ that must permute the members of $R_p$.

*Now my first candidate function was $g_1: E \rightarrow E$ s.t. $g_1(x) = x$ if $x \in F$ else $g_1(x) = \omega x$.  This would at least permute the elements of $R_p$ since $g(r) = \omega r$, $g(r \omega) = r \omega^2$ and $g_1(r \omega^2) = r \omega^3 = r$.

*The problem with this is that $g_1$ would fail to be an automorphism, for multiplication is not preserved.  That is, if $e_1, e_2 \in E - F$ then $g_1(e_1 e_2) = \omega e_1 e_2 \ne \omega e_1 \omega e_2 = g_1(e_1) g_1 (e_2)$.
So how can I find the $6$ members of $G[E:F]$?

There is an old trick for this.
Let $K=\mathbb{Q}(\sqrt[3]{2},\omega)$. We see that $K$ is the splitting field of $x^3-2$, and thus $K/\mathbb{Q}$ is Galois. Thus, $|\text{Gal}(K/\mathbb{Q})|=6$. Now, since $K$ is generated by $\sqrt[3]{2}$ and $\omega$ any $\sigma\in\text{Gal}(K/\mathbb{Q})$ is determined by its values on the pair. But, since $\sigma$ permutes the roots of the minimal polynomial of an algebraic number you have that 
$$\sigma:\qquad\sqrt[3]{2}\mapsto \omega^k\sqrt[3]{2},\text{ }k=0,1,2\qquad \omega\mapsto \omega^j,\text{ }j=1,2$$
Thus, there are six possible $\mathbb{Q}$-algebra maps $K\to K$. Namely, 
$$\sigma_{i,j}:(\sqrt[3]{2},\omega)\mapsto (\omega^k\sqrt[3]{2},\omega^j)$$
for $i\in \{0,1,2\}$ and $j\in\{1,2\}$. Since there MUST be six automorphisms of $K/\mathbb{Q}$, all six of these candidate morphisms must actually be automorphisms. Thus, $\text{Gal}(K\mathbb{Q})=\{\sigma_{i,j}\}$.
