# $|\mathbb{Q}(e^{2\pi i/n}) : \mathbb{Q}(\cos(2\pi/n)| = 1 \text{ or } 2$

I was trying to follow a sentence in a paper, which says that:

$|\mathbb{Q}(e^{2\pi i/n}) : \mathbb{Q}(\cos(2\pi/n)| = 1 \text{ or } 2$

because $e^{2\pi i/n}$ is a root of $x^2 - 2\cos(2\pi/n) + 1$.

I know that, for a simple extension $F \subseteq F(a)$, $|F(a) : F| =$ degree of minimal polynomial of a over F.

I also know that, because $|\mathbb{Q}(e^{2\pi i/p}) : \mathbb{Q}(\cos(2\pi/p)|$ is finite, $\mathbb{Q}(e^{i2\pi/n})$ is a simple extension of $\mathbb{Q}(\cos(2\pi i/n))$.

i.e. $\mathbb{Q}(e^{i2\pi/n}) = \mathbb{Q}(a,\;\cos(2\pi i/n))$ for some $a \in \mathbb{Q}(e^{i2\pi/n})$.

However, to find the possible degrees of the extension, you'd have to find a polynomial satisfied by $a$, rather than by $e^{2i\pi/n}$, so I don't think the statement follows immediately?

Any help on understanding this is appreciated.

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Simply let $a=e^{i2\pi/n}$. Then $\frac{a+a^{-1}}2=\cos(2\pi/n)$ implies $\mathbb Q(a, \cos(2\pi/n))=\mathbb Q(e^{i2\pi/n})$.
In general if $L,M$ are extensions of $K$ and $M\subseteq L$ and $L=K(\alpha)$, then also $L=M(\alpha)$.
Thanks. Gosh, don't know how I missed that...I already had that $cos(2i\pi/n) = (e^{2i\pi/n} + e^{2i\pi/n})/2$ but somehow didn't piece it together. I was anticipating that it'd be hard to find such an $a$ to make it a simple extension. Thanks! –  maliky0_o Nov 17 '12 at 16:28