Degree of extension - Is this $\phi(n)$ or $\phi(n)/2$?

Question

Let $a=\cos(2\pi/n) $.

I have shown that $Q(a)/Q$ is a Galois extension and now I want to show that $[Q(a):Q]=\phi (n)/2$.

I have done the following:

It holds that $|Gal(\mathbb{Q}(a)/\mathbb{Q})|=|(\mathbb{Z}/n\mathbb{Z})^{\times}|$.

We also have that $\mathbb{Q}(a)$ is the splitting field of the cyclotomic polynomial is $\Phi_n$.

Does it follow from that that $[\mathbb{Q}(a):\mathbb{Q}]=\deg \Phi_n=\phi (n)$ ?

But I have found $\phi(n)$ and at the exercise statement it is divided by $2$.

Have I done something wrong?

Answer 1

$\mathbb{Q}(a)$ is not the splitting field of the cyclotomic polynomial $\Phi_n$ because $\mathbb{Q}(a) \subseteq \mathbb{R}$.

$2a = \omega + \bar \omega$, where $\omega = \exp(2\pi i/n)$. Therefore, $\mathbb{Q}(a)$ is in the fixed field induced by conjugation, which defines a subgroup of order $2$ of the Galois group and so $[\mathbb{Q}(a):\mathbb{Q}] \le \phi (n)/2$.

Answer 2

Let $ ζ= e^{2\pi i\over n}$. It follows that Q(ζ):Q has degree φ(n) as the minimum polynomial of ζ is $Φ_n(x)$, which is irreducible.

Now consider $2α=ζ+ζ^{-1}= 2cos({2\pi \over n})$ and since $Q(2\alpha)=Q(\alpha)$, consider the tower of extensions: $ Q \subset Q(a) \subset Q(\zeta).$ The degree of $[Q(\alpha):Q]$ can't be φ(n) as that'd imply that $Q(\alpha)=Q(\zeta)$, but $Q(\alpha)$ is a real extension, while $Q(\zeta)$ is not.

What you can do instead is use the tower law to conclude that $[Q(\alpha):Q] = \frac{[Q(\zeta):Q]}{[Q(\zeta):Q(\alpha)]}.$ The numerator is $\phi(n)$, and then show the denominator is 2. The denominator can't be 1. Look at $x^2 - (\zeta+\zeta^{-1})x + 1$, which is a polynomial over $Q(\alpha)$ that sends $\zeta$ to 0.