If $f(x) = x^3 - 2 $, why is its splitting field $\mathbb{Q}(\sqrt{2},\omega)$? I don't know how to calculate the splitting field of polynomials that involves unity roots. What is the procedure?
I did read in some books that its splitting field is $\mathbb{Q}(\sqrt{2},\omega)$ where $\omega$ is one root of $x^2 + x + 1 = 0$. 
But I can't understand what is $\omega$ and why it is this the splitting field of this polynomial.
Thanks a lot.
 A: Suppose $\omega$ is a root of $x^2 + x + 1$. This means:
$\omega^2 + \omega + 1 = 0$, or put another way:
$\omega^2 = -\omega - 1$.
If we multiply both sides by $\omega$, we get:
$\omega^3 = -\omega^2 - \omega$, and using $\omega^2 = -\omega - 1$, we have:
$\omega^3 = -(-\omega - 1) - \omega = \omega - \omega + 1 = 1$.
So $\omega$ is a cube root of $1$. Specifically, the complex numbers:
$\omega = -\dfrac{1}{2} \pm i\dfrac{\sqrt{3}}{2}$ are our roots, as you can verify using the quadratic formula.
It follows that $(\sqrt[3]{2}\omega^k)^3 = (\sqrt[3]{2})^3(\omega^k)^3 = 2(\omega^3)^k = 2$ for $k = 0,1,2$, so $\sqrt[3]{2}\omega^k$ for $k = 0,1,2$ must give our $3$ roots of $x^3 - 2$.
So $x^3 - 2$ definitely splits in $\Bbb Q(\sqrt[3]{2},\omega)$, and it doesn't split in $\Bbb Q(\sqrt[3]{2})$ (this is a subfield of the reals, and $\omega \not\in \Bbb R$).
Since $x^2 + x + 1$ doesn't have any roots in $\Bbb Q(\sqrt[3]{2})$, it follows it is irreducible over that field (because it is only a quadratic), and thus $[\Bbb Q(\sqrt[3]{2},\omega):\Bbb Q(\sqrt[3]{2})] = 2$.
This doesn't leave any room for intermediate fields between them.
A: Let $\zeta$ be a primitive third root of unity.
Given any choice of $\sqrt[3]{2}$, it's easy to see that the three roots of $x^3 - 2$ will all be $\zeta^n \sqrt[3]{2}$, for $n=0,1,2$.
Consequently, the splitting field contains $\frac{\zeta \sqrt[3]{2}}{\sqrt[3]{2}} = \zeta$. Since adjoining $\zeta$ gives a field with all three roots, the splitting field must be $\mathbb{Q}(\sqrt[3]{2}, \zeta)$.
The rest of the problem is just recognizing that $\zeta^2 + \zeta + 1 = 0$.
