The question is in the headline above. I need to know this, because I want to show that $\mathbb{Q}(\sqrt[9]{2}, e^{\frac{2\pi i}{3}})/\mathbb{Q}$ is not a normal extension and it's quite obvious I think that $e^{\frac{2\pi i}{9}}$ couldn't be expressed by $\sqrt[9]{2}$, $e^{\frac{2\pi i}{3}}$ and the rational numbers using multiplication and addition.
If there won't be $\sqrt[9]{2}$ a generator, I would argue that if $e^{\frac{2\pi i}{9}}$ is in $\mathbb{Q}(e^{\frac{2\pi i}{3}})$ we have $\mathbb{Q}(e^{\frac{2\pi i}{9}}) \subseteq \mathbb{Q}(e^{\frac{2\pi i}{3}})$ and thereby $[\mathbb{Q}(e^{\frac{2\pi i}{9}}):\mathbb{Q}] \leq [\mathbb{Q}(e^{\frac{2\pi i}{3}}):\mathbb{Q}]$.
But the third and the ninth cyclotomic polynomials are of degree 2 and 6 and thus we have a contradiction.
Unfortunately I don't know if the cyclotomic polynomials are irreducible over $\mathbb{Q}(\sqrt[9]{2})$ too, so I couldn't transfer the above argument.
What can I do instead?
Thank you.
One way to do this is to use the Fundamental Theorem of Galois Theory and a bit of Group Theory.
Suppose that $K := {\mathbb Q}(\sqrt[9]{2},e^{2\pi i/3})$ is the splitting field of $x^9-2$ over ${\mathbb Q}$. So $|K:{\mathbb Q}|=18$ and the Galois group $G$ of the extension has order $18$.
Now ${\mathbb Q}(\sqrt[9]{2})$ contains only the one (real) $9$th root of $2$, because none of the other roots are real. So $K$ has nine distinct subfields of degree $9$ over ${\mathbb Q}$, one containing each $9$th root of $2$. So $G$ has $9$ subgroups of order $2$.
But $K$ also contains the $9$the cyclotomic field, which is a normal extension of ${\mathbb Q}$ of degree $6$ with Galois group cyclic of order $6$. So $G$ contains the cyclic group of order $6$ as a quotient group, which is not possible if it has $9$ Sylow $2$-subgroups.