Let $E$ be the splitting field of $x^3 - 2$ over $\mathbb{Q}$.
I proved that $E = \mathbb{Q}(\sqrt{2},\omega)$. Where $\omega$ is a primitive root of unity.
And I also know that $G := \operatorname{Gal}(E/ \mathbb{Q}) =S_3 $.
I find that $\mathbb{Q}(\sqrt{2},\omega)=\{a+b\sqrt{2}+c\omega+d\omega^2+e\sqrt{2}\omega+f\sqrt{2}\omega^2 : a,b,c,d,e,f \in \mathbb{Q}\}.$
A little intermediate question I have is if does that mean that the dimension of $\mathbb{Q}(\sqrt{2},\omega)$ over $\mathbb{Q}$ is $6$ right? (I also know this because [$E:\mathbb{Q}]=|\operatorname{Gal}(E/\mathbb{Q})|=|S_3|=6$).
So finding the dimension of $\mathbb{Q}(\alpha_1,\alpha_2,...,\alpha_n)$ over $\mathbb{Q}$ where $\alpha_i$ are specific and known, is as easy as count how many of the products of the $\alpha_i$ are in a different fields and express $\mathbb{Q}(\alpha_1,\alpha_2,...,\alpha_n)$ as a combination linear of these elements over $\mathbb{Q}$. Just as done with $\mathbb{Q}(\sqrt{2},\omega)$. Am I right?...
So let $H \subset G$ be, for example, the 3-cycle $\{\operatorname{id},(123),(132)\}$. In order to find $E^H$ we see that if $x \in \mathbb{Q}(\sqrt{2},\omega)=\{a+b\sqrt{2}+c\omega+d\omega^2+e\sqrt{2}\omega+f\sqrt{2}\omega^2 : a,b,c,d,e,f \in \mathbb{Q}\}$ say
$x=a+b\sqrt{2}+c\omega+d\omega^2+e\sqrt{2}\omega+f\sqrt{2}\omega^2$ for specific $a,b,c,d,e,f \in \mathbb{Q}$
we must have $b=c=d$, $e=f$
since $(123)$ permutes every element of $\sqrt{2},\omega,\omega^2.$
So $E^H = \{a+b(\sqrt{2}+\omega+\omega^2)+e(\sqrt{2}\omega+\sqrt{2}\omega^2) : a,b,e \in \mathbb{Q}\} =\mathbb{Q}(\sqrt{2}+\omega+\omega^2,\sqrt{2}\omega+\sqrt{2}\omega^2)$
Is my reasoning correct?
Thank you very much in advance.
For a start, a $\Bbb Q$-basis of $E$ is $1$, $\sqrt[3]2$, $\sqrt[3]4$, $\omega$, $\omega\sqrt[3]2$, $\omega\sqrt[3]4$.
Let $H=\{\text{id},(123),(132)$. Then $E^H$ is the subset of $E$ fixed by $(123)$.
If we label the roots of $x^3-2=0$ as $\alpha_1=\sqrt[3]2$, $\alpha_2=\omega\sqrt[3]2$, $\alpha_3=\omega^2\sqrt[3]2$, then $\sigma=(123)$ has $\sigma(\alpha_1)=\alpha_2$, $\sigma(\alpha_2)=\alpha_3$ and $\sigma(\alpha_3)=\alpha_1$. This means that $$\sigma(\sqrt[3]2)=\omega\sqrt[3]2$$ and $$\sigma(\omega)=\sigma(\alpha_2/\alpha_1)=\alpha_3/\alpha_2=\omega.$$ Therefore \begin{align} &\sigma(a+b\sqrt[3]{2}+c\sqrt[3]{4}+d\omega+e\omega\sqrt[3]{2}+f\omega\sqrt[2]{4})\\ &=a+b\omega\sqrt[3]{2}+c\omega^2\sqrt[3]{4}+d\omega+e\omega^2\sqrt[3]{2}+f\sqrt[2]{4}\\ &=a+b\omega\sqrt[3]{2}+c(-1-\omega)\sqrt[3]{4}+d\omega+e(-1-\omega)\sqrt[3]{2}+f\sqrt[2]{4}\\ \end{align} Therefore $\sigma$ fixes this element iff $b=-e$, $c=-c+f$, $e=b-e$ and $f=-c$. These imply $b=c=e=f=0$ and so the element is $a+b\omega$. Thus $E^H=\Bbb Q(\omega)$.
There are some obvious subfields of $E$, namely $E$, $\Bbb Q$ $\Bbb Q(\omega)$, $\Bbb Q(\sqrt[3]2)$, $\Bbb Q(\omega\sqrt[3]2)$, $\Bbb Q(\omega^2\sqrt[3]2)$. From Galois theory, do we expect any more?