Automorphism of $\mathbb Q(\cos(2\pi/n))$

Question

How can you show that any automorphism $\xi$ of $\mathbb Q(\cos(2\pi/n))$ is defined by $\xi(\cos(2\pi/n)) = \cos(2m\pi/n)$ for $\gcd (m,n)=1$?

It's simple to see that multiplication and addition are preserved, so it must be a homomorphism. I just can't figure out how to show that it is bijective as well, which must be linked to the fact that $m$ and $n$ are prime to each other.

Answer 1

A bit cleaner if you write $2 \cos \frac{2 \pi}{n}$ since $$ 2 \cos \frac{2 \pi}{n} = \omega + \frac{1}{\omega}, $$ where $\omega = e^{2 \pi i / n}$ is a root of unity. Lehmer pointed out, in 1933, that $2 \cos \frac{2 \pi}{n}$ is an algebraic integer.

Answer 2

Here's an indirect way of tackling the problem:

Let $F \subset L \subset K$ be a tower of algebraic extensions. The set of $F$-isomorphisms of $L$ will simply be the set of $F$-isomorphisms of $K$ restricted to $L$. This is a consequence of the isomorphism extension theorem: each isomorphism of $L$ extends to an isomorphism of $K$. Note further that any isomorphism for $K$ is determined by its action on a set of basis elements, and if $K$ is Galois, then $K$ contains all of the Galois conjugates of all of the basis elements. Thus, when $K$ is Galois, any automorphism of $L$ extends to an automorphism of $K$.

Here, we have a tower of fields $\mathbb{Q} \subset \mathbb{Q}( \cos(2 \pi/n)) \subset \mathbb{Q}(\zeta_n)$, where $\zeta_n$ is a primitive $n^\text{th}$ root of unity. Since $\mathbb{Q}(\zeta_n)$ is a Galois extension of the rationals (it is the splitting field for the associated cyclotomic polynomial), it suffices to find its automorphisms and restrict them to the intermediate field. The automorphisms of $\mathbb{Q}(\zeta_n)$ are determined by their action on $\zeta_n$ itself. Where can $\zeta_n$ be sent? The minimal polynomial for $\zeta_n$ is $\displaystyle \Phi_n(x) = \prod_{\gcd(k, n) = 1 \\ \ \ \ 1 \leq k \leq n} \left( x - e^{2i\pi k/n} \right)$, and since the Galois group of an irreducible polynomial acts transitively on its roots, it follows that we can send $\zeta_n \mapsto \zeta_n^k$ for any $1 \leq k \leq n$ such that $\gcd(k,n) = 1$.

Applying Euler's formula gives $\zeta_n = \cos(2\pi/n) + i \sin(2\pi/n)$ and $\zeta_n^k = \cos(2k \pi/n) + i \sin(2k \pi/n)$. Restricting the automorphism determined by $\zeta_n \mapsto \zeta_n^k$ to the field $\mathbb{Q}(\cos(2i\pi/n))$ gives the automorphism of that field determined by $\cos(2i \pi/n) \mapsto \cos(2i \pi k/n)$.