Degree $[\mathbb{Q}(α):\mathbb{Q}(α+α^{-1})]$

Question

$$\mathbb{Q}(\alpha) , \quad \alpha=e^{2\pi i\over n}$$

$$\mathbb{Q}\left(\cos\left({2\pi \over n}\right)\right)=\mathbb{Q}(\alpha+\alpha^{-1}),\qquad \cos\left({2\pi \over n}\right)= {1 \over 2}(\alpha+\alpha^{-1})$$

My question:

I am experiencing some problems and I have questions that

$$\quad[\mathbb{Q}(\alpha) : \mathbb{Q}(\alpha+\alpha^{-1})]=2 $$

and

$$[\mathbb{Q}(\alpha ):\mathbb{Q}] = \left[\mathbb{Q}\left(\sin\left({2\pi \over n}\right)\right):\mathbb{Q}\right].$$

It is true? I can't prove that.

Answer 1

Since $\mathbb{Q}(\alpha+\alpha^{-1})$ is a real field, clearly $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha+\alpha^{-1})]>1$. On the other hand, $\alpha$ is a root of $$x^2 - (\alpha+\alpha^{-1})x + 1.$$

The second statement is false: take $n=8$; then $[\mathbb{Q}(\alpha):\mathbb{Q}] = 4$ ($x^4+1$ is a minimal polynomial for $\alpha$), while $\left[\mathbb{Q}\left(\sin\left(\frac{2\pi}{8}\right)\right):\mathbb{Q}\right] = 2$ ($2x^2-1$ is a minimal polynomial for $\sin\left(\frac{2\pi}{8}\right) = \frac{\sqrt{2}}{2}$)