As an exercise I am trying to do the following:
Let $L$ be the splitting field of $x^3-3\in \mathbb{Q}[X]$ over $\mathbb{Q}$.
I want to determine $Gal(L/\mathbb{Q})$, all subgroups of $Gal(L/\mathbb{Q})$ and all intermediate fields of $L/\mathbb{Q}$.
My attempt: Let $\xi:= e^{\frac{2 \pi i}{3}}$, solving $x^3-3=0$ yields $3^{1/3}$,$3^{1/3} \xi$,$3^{1/3} \xi^2$ . Thus $L=\mathbb{Q}(3^{1/3}, \xi)$
I did calculate $[\mathbb{Q}(3^{1/3},\xi):\mathbb{Q}]=6$
Since $Gal(L/\mathbb{Q})=6$, we have either $Gal(L/\mathbb{Q}) \cong \mathbb{Z}/6$ or $Gal(L/\mathbb{Q}) \cong S_3$
How do I find out which of these two it is?
My idea would be to consider that for $\tau \in\Gamma$, we have that $\tau(3^{1/3})\in \{3^{1/3},3^{1/3} \xi,3^{1/3} \xi^2\} $ and $\tau(\xi) \in \{\xi,\xi^2\}$. So I guess $Gal(L/\mathbb{Q}) \cong S_3$ (is there a better argumentation for that?) Well, by this I get that
$\tau_1=Id$
$\tau_2: a \mapsto a, \xi \mapsto \xi^2$
$\tau_3: a \mapsto a\xi, \xi \mapsto \xi$
$\tau_4: a \mapsto a \xi, \xi \mapsto \xi^2$
$\tau_5: a \mapsto a\xi^2, \xi \mapsto \xi$
$\tau_6: a \mapsto a\xi^2, \xi \mapsto \xi^2$
Now I want to find all subgroups of $Gal(L/\mathbb{Q})$:
$S_3,\{Id\}, \{Id,\tau_2\},\{Id,\tau_3\},\{Id,\tau_4\}, \{Id,\tau_5, \tau_6\}$
Well thats how far I got. I know that I should somehow use the subgroups of $Gal(L\mathbb{Q})$ to determine the intermediate fields, but can't figure out how.