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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.

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    What is $Q(\cos(2\pi/n))$? – 2017-02-13
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    @R_D, $Q$ is the rationals, $Q(\alpha)$ is the smallest field containing $Q$ and $\alpha$. – 2017-02-13
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    Automorphisms of your field are restrictions of automorphisms of the cyclotomic field. – 2017-02-13
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    I think Kaj Hansen's way (implicit in Gerry's comment) is the way to do this. It might be possible to go via the route of 1) $\cos(2m\pi/n)\in\Bbb{Q}(\cos(2\pi/n))$: by trig identities there exists a polynomial $T_m(x)$ with integer coefficients such that $\cos mx= T_m(\cos x)$. 2) If $mm'\equiv1\pmod n$ then using $T_{m'}(\cos(2\pi m/n))=\cos(2\pi /n)$ gives the reverse inclusion. 3) showing that $\cos(2\pi/n)$ and $\cos(2\pi m/n)$ are zeros of the same polynomial (modify the polynomial $T_n(x)$ to fit this). Proving that this polynomial is irreducible does not seem trivial though :-( – 2017-02-13

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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 $K$.

Notice that 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's the splitting field for the associated cyclotomic polynomial), it suffices to find its automorphisms and restrict down to the field in question. 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 determined by $\cos(2i \pi/n) \mapsto \cos(2i \pi k/n)$.

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    That makes sense. I was misunderstanding what ζ^k_n meant. – 2017-02-13
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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.

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    I've actually been trying to use that, especially since I already know the automorphisms for ω. I just haven't figured out how to use that to find the automorphism for cos. Everything I do ends up with cos^m(2π/n). – 2017-02-13
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    @Will, I suspect Kummer knew that $2\cos(2\pi/n)$ is an algebraic integer. If not Kummer, then Dirichlet, or Dedekind, or some other 19th century figure. – 2017-02-13
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    @Gerry understood. It was late, and this 1933 reference appeared in a new article in the M.A.A. Monthly, Chebyshev Polynomials and the Minimal Polynomial of $\cos ( 2 \pi / n) .$ The other ingredient for me is finding Gauss method for cyclotomy in Cox, Galois Theory, then tables in an 1875 book by Reuschle. https://books.google.com/books?id=wt7lgfeYqMQC&pg=PR1&lpg=PR1&dq=reuschle++tafeln+complexer+primzahlen&source=bl&ots=VGZFPrfUBn&sig=MlQ667PqXaQ9rAvLWkG3_F1rwsk&hl=en&sa=X&ved=0ahUKEwiIwtSvm9TQAhUJ-2MKHXJIA_kQ6AEIODAE#v=onepage&q=reuschle%20%20tafeln%20complexer%20primzahlen&f=false – 2017-02-13