# Quadratic extension of quadratic extensions

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.

## 1 Answer

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?

• Thank you so much. Almost. I got $cos(\pi /6)+isin(\pi /6)=\frac{1}{2}*\sqrt{3} + \frac{1}{2}$. But here I'm a little confused because this would mean that $d'=\frac{1}{4}$.? – Matriz Dec 7 '14 at 20:23
• You forgot the $i$. – Arthur Dec 7 '14 at 20:40
• Thank you a lot for your help. – Matriz Dec 7 '14 at 20:55
• @Arthur Apologize but for the second part the field is $\mathbb Q(e^\frac{2\pi}{3})(\sqrt{d'})$, not $\mathbb Q(\sqrt{d'})$, So any element of the former looks like $a+b\sqrt{-3}+\left(a+b\sqrt{-3}\right)\sqrt{d'}$, am I wrong ? and an element of $\mathbb Q(e^\frac{\pi}{6})$ is of the form $a'+b'\sqrt{3}+ib'$, and if I plug in $-\frac14$ for $d'$, should I get equality for appropriate $a,b,a',b'$ ? – inequal Dec 7 '14 at 21:50
• @inequal No, the answer $d' = \frac{1}{4}$ is incorrect. A hint to find a correct $d'$: What is the square of $e^{\frac{\pi i}{6}}$? – Arthur Dec 7 '14 at 21:53