Calculating a minimal polynomial over $\mathbb{Q}(\sqrt[3]{2})$

Question

Let $L_1=\mathbb{Q}(\omega\sqrt[3]{2})$ where $\omega=e^\frac{2\pi i}{3}$ and $L_2=\mathbb{Q}(\sqrt[3]{2})$.

I want to calculate $[L_1L_2:L_2]$, that it is the degree of the minimal polynomial over $L_2$ with root $\omega\sqrt[3]{2}$.

I think it must be 2, but all my efforts to prove this has been failed!

Answer 1

Calculate the minimal polynomial of $\omega$ over the rationals, and see how this is related to the polynomial you seek.

Answer 2

Show that $p_{\omega,\mathbb{Q}} = X^2+X+1$ is irreducible over $\mathbb{Q}$ and that $\omega$ is a root of $p_{\omega,\mathbb{Q}}$. Let $\alpha = \sqrt[3]{2}$.

Also keep in mind that if $[\mathbb{Q}[\omega]:\mathbb{Q}] = 2$, $[\mathbb{Q}[\alpha]:\mathbb{Q}] = 3$ where $GDC(2,3) = 1$ and $[\mathbb{Q}[\omega,\alpha]:\mathbb{Q} [\alpha]] = r$ , $[\mathbb{Q}[\omega,\alpha]:\mathbb{Q} [\omega]] = s$. What can you tell about $r$ and $s$? And $p_{\omega,\mathbb{Q}}$? Draw a diagram.