TITLE: First examples in Galois theory
QUESTION [8 upvotes]: I'm studying Field Theory and after studying theorems and problems about extensions, splitting fields, etc... I'm starting with the first theorems of the Galois Theory itself. In order to see if I understand such theorems, I'm trying to prove the first examples in Galois Theory such that the 

Galois group of $x^3-2\in \mathbb Q[x]$ is the group of symmetries of the triangle.

I know that the roots of the equation of $x^3-2=0$ are $2^{1/3},2^{1/3}w,2^{1/3}w^{2}$, where $w$ is a root of the irreducible polynomial $x^2+x+1$ over $\mathbb Q(2^{1/3})$. Thus we write $x^3-2=(x-2^{1/3})(x-2^{1/3}w)(x-2^{1/3}w^{2})$, whence $E=\mathbb Q(2^{1/3},w)$. It follows that $[E:\mathbb Q]=6$. Since E is a splitting field and therefore normal we have also $G(E/\mathbb Q)=6$, then we have six automorphisms of $E$. 
I'm stuck here, I can't go further, I need help please.
Thanks a lot.

REPLY [7 votes]: The easiest way to solve this particular problem is to note that $G(E/\mathbb{Q})$ is nonabelian and has order $6$, from which the answer follows immediately.  Below is a constructive method to determine this, which should hopefully also give you insight into how to approach future problems of this nature.
Let $\sigma\in G(E/\mathbb{Q})$.  Since $E=\mathbb{Q}(2^{1/3},\omega)$, it suffices to define $\sigma$'s action on $E$ by defining its action on $\omega$ and $2^{1/3}$. 
First, we consider $\sigma(2^{1/3})$. $\sigma$ must send conjugates to conjugates - that is, if $x\in E$, the minimal polynomial of $x$ must be the same as the minimal polynomial of $\sigma(x)$. Therfore, we must have $\sigma(2^{1/3})=2^{1/3}$, $\omega 2^{1/3}$, or $\omega^2 2^{1/3}$.  Similarly, $\sigma(\omega)$ must go to $\omega$ or $\omega^2$.  (Note that $\sigma(\omega)$ can't go to $\omega^3=1$, since this would not fix $\mathbb{Q}$.)
Intuitively we see it's best to define $$\rho:\left\{\begin{array}{l}2^{1/3}\mapsto \omega 2^{1/3}\\\omega\rightarrow\omega\end{array}\right. \text{  and  }\tau:\left\{\begin{array}{l}2^{1/3}\mapsto 2^{1/3}\\\omega\rightarrow\omega^2\end{array}\right.$$  We can easily verify that $\rho$ and $\tau$ are indeed automorphisms of $E$ fixing $\mathbb{Q}$, and therefore that $\langle \sigma,\tau \rangle\leqslant G(E/\mathbb{Q})$.  Indeed, by design we can write any admissible $\sigma\in G(E/\mathbb{Q})$ in terms of $\sigma$ and $\tau$, so we expect $\langle \sigma,\tau \rangle = G(E/\mathbb{Q})$.  We confirm by matching the presentation of $\langle \sigma,\tau \rangle$ to $$D_3=\langle r,s|r^3,s^2,r^s=r^{-1}\rangle$$ that $\langle \sigma,\tau \rangle = G(E/\mathbb{Q})=D_3$.