How to show the degree $[\Bbb Q(\alpha _n+\frac{1}{\alpha_n}) : \Bbb Q]$ is $\phi(n)/2$, where $\alpha_n$ is a primitive $n$-th root of unity?

Question

How to show the degree $[\Bbb Q(\alpha _n+\frac{1}{\alpha_n}) : \Bbb Q]$ is $\phi(n)/2$, where $\alpha_n$ is a primitive $n$-th root of unity?

I already know that $[\Bbb Q(\alpha _n ):\Bbb Q]$ is Euler-phi function $\phi(n)$. But, I have no clue what to do next? Is it possible to find the irreducible polynomial for $\alpha _n +\frac{1}{\alpha_n}$ over $\Bbb Q$?

Answer 1

For any $k,\gcd(n,k)=1$ : $$ \mathbb{Q}(\zeta_n)= \mathbb{Q}(\cos(2\pi k/n))(i\sin(2\pi k/n))$$ $i \sin(2\pi k/n)$ is a root of $x^2+1-\cos^2(2\pi k/n)$ a polynomial of degree $2$ over $\mathbb{Q}(\cos(2\pi k/n))$ and so $$[ \mathbb{Q}(\zeta_n): \mathbb{Q}(\cos(2\pi k/n))] \le 2$$ Since it is real $\mathbb{Q}(\cos(2\pi k/n)) \ne \mathbb{Q}(\zeta_n)$ and : $$[ \mathbb{Q}(\zeta_n): \mathbb{Q}(\cos(2\pi k/n))] = 2, \qquad \phi(n) = [ \mathbb{Q}(\zeta_n): \mathbb{Q}] = [ \mathbb{Q}(\zeta_n): \mathbb{Q}(\cos(2 \pi k/n))][ \mathbb{Q}(\cos(2\pi k/n)): \mathbb{Q}]$$ $$[ \mathbb{Q}(\cos(2\pi k/n)): \mathbb{Q}] = \phi(n)/2$$

Answer 2

If you let $\beta_n=\alpha_n+\frac{1}{\alpha_n}$. Then, clearly $\mathbb Q(\beta_n)\subseteq \mathbb Q(\alpha_n)$, since $\beta_n=\mathbb Q(\alpha_n)$.

You first have to show it is a strict subfield - that $\mathbb Q(\alpha_n)\neq \mathbb Q(\beta_n)$. That can be seen by noting that $\beta_n$ is real, and $\alpha_n$ is not, for $n>2$.

An alternative approach is to show that the automorphism of $\mathbb Q(\alpha_n)$ sending $\alpha_n$ to $\alpha_n^{-1}$ fixes $\beta_n$ and thus all of $\mathbb Q(\beta_n)$, but does not fix $\alpha_n$ unless $n=2$.

But if $\alpha_n+\frac{1}{\alpha_n}=\beta_n$ then $x^2-\beta_nx +1$ has $\alpha_n$ as a root, so $\alpha_n$ is at most degree $2$ over $\mathbb Q(\beta_n)$.

So we get $$\phi(n)=[\mathbb Q(\alpha_n):\mathbb Q]=[\mathbb Q(\alpha_n):\mathbb Q(\beta_n)][\mathbb Q(\beta_n):\mathbb Q]=2[\mathbb Q(\beta_n):\mathbb Q]$$

The minimal polynomial for $\beta_n$ can be computed in terms of Chebyshev polynomials, but you have to deal with repeated roots and the like.