Let $L:=\mathbb{Q}(e^{\frac{2 \pi i}{3}},\sqrt2)$.
First I want to determine $[\mathbb{Q}(e^{\frac{2 \pi i}{3}},\sqrt2) : \mathbb{Q}]$.
For this I am trying to find a $\mathbb{Q}$-basis of $\mathbb{Q}(e^{\frac{2 \pi i}{3}})$.
Elements in $\mathbb{Q}(e^{\frac{2 \pi i}{3}})$ have the form $a+be^{\frac{2 \pi i}{3}}$, thus the wanted basis is $\{1, e^{\frac{2 \pi i}{3}} \}$.
Next, I want to find a $\mathbb{Q}(e^{\frac{2 \pi i}{3}})$-basis of $\mathbb{Q}(e^{\frac{2 \pi i}{3}},\sqrt2)$.
For that I write $\mathbb{Q}(e^{\frac{2 \pi i}{3}},\sqrt2)$ as $\mathbb{Q}(e^{\frac{2 \pi i}{3}})(\sqrt2)$. Elements in it have the form $x+y\sqrt2$ with $c,d \in \mathbb{Q}(e^{\frac{2 \pi i}{3}})$ This means they have the form $(a+be^{\frac{2 \pi i}{3}})+(c+de^{\frac{2 \pi i}{3}})\sqrt2$. And so the $\mathbb{Q}$-basis of $\mathbb{Q}(e^{\frac{2 \pi i}{3}},\sqrt2)$ is $\{ 1,\sqrt2,e^{2 \pi i/3},\sqrt2 e^{2 \pi i/3} \}$.
Question: Is my calculation right?
The extension $\mathbb{Q}(e^{\frac{2\pi i}{3}}) / \mathbb{Q}$ has degree $3$, so the basis needs $3$ elements.
The field $\mathbb{Q}(e^{\frac{2 \pi i}{3}})$ also contains $e^{\frac{2\pi i}{3}} \cdot e^{\frac{2 \pi i}{3}} = e^{\frac{4 \pi i}{3}}$ and this is not of the form $a+b e^{\frac{2\pi i}{3}}$ with $a,b \in \mathbb{Q}$