I came across an old exam problem and I wonder if my solution is correct.
Let $L=\mathbb{Q}(\omega)$, where $\omega=e^{\frac{2\pi i}{6}}$ is a primitive sixth root of unity:
a) determine $[L:\mathbb{Q}]$
b) prove that $t^{3}-2$ is irreducible over $L$
c) decide if the field $\mathbb{Q}(\omega, \sqrt[3]{2})$ is normal over $\mathbb{Q}$
Here is what I think
a) consider the polynomial $x^{6}-1\in\mathbb{Q}[x]$ then $\frac{x^{6}-1}{(x-1)(x+1)}=x^4+x^2+1$ which is irreducible over $\mathbb{Q}$ by the rational root theorem, hence $[L:\mathbb{Q}]=4$
b) $t^{3}-2=(x-\sqrt[3]{2})(x-\varepsilon\sqrt[3]{2})(x-\varepsilon^2\sqrt[3]{2})$ where $\varepsilon=e^{\frac{2\pi i}{3}}=-\frac{1}{2}+\frac{\sqrt{3}i}{2}$
I guess I need to show that neither of the above roots is contained in $\mathbb{Q}(\omega)$ but I don't see any good and fast method for proving it.
We have
$$\omega^2=e^{\frac{2\pi i}{3}}$$
$$\omega^3=e^{\frac{4\pi i}{3}}$$
Hence $t^{3}-2$ will be irreducible if $\sqrt[3]{2}\not\in \mathbb{Q}(\omega)$
If $\sqrt[3]{2}\in \mathbb{Q}(\omega)$ then $\sqrt[3]{2}=a+b\omega+c\omega^2+c\omega^3$, but then $a=0$ as $a\in\mathbb{Q}$ but how to check the rest?
c) if I can prove b) then I think that $\mathbb{Q}(\omega, \sqrt[3]{2})\supseteq \mathbb{Q}$ is normal as $\mathbb{Q}(\omega, \sqrt[3]{2})\supseteq \mathbb{Q}$ will be the splitting field for $t^{3}-2$.
Any help appreciated! Thank you.