Quadratic extension of quadratic extensions

Question

I need help for the following exercise:

The field $\mathbb{Q}(e^{\frac{2 \pi i}{3}})$ is a quadratic extension of $\mathbb{Q}$ and $\mathbb{Q}(e^{\frac{\pi i}{6}})$ is a quadratic extension of $\mathbb{Q}(e^{\frac{2 \pi i}{3}})$. Exhibit elements $d \in \mathbb{Q}$ and $d' \in \mathbb{Q}(e^{\frac{2 \pi i}{3}})$, such that $\mathbb{Q}(e^{\frac{2 \pi i}{3}})=\mathbb{Q}(\sqrt{d})$ and $\mathbb{Q}(e^{\frac{\pi i}{6}})$=$\mathbb{Q}(e^{\frac{2 \pi i}{3}})(\sqrt{d'})$.

I have absolutely no clue how to solve this. I'm very grateful for any tips.

Answer 1

For the first one: $e^{\frac{2 \pi i}{3}} = \cos(\frac{2 \pi}{3}) + i \sin(\frac{2 \pi}{3}) = - \frac{1}{2} + \frac{1}{2} \sqrt{-3}$, so $\mathbb{Q}(e^{2 \pi i/3}) = \mathbb{Q}(\sqrt{-3})$. Can you do the second one now?