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I am trying to learn about Kummer Rings, and in particular what makes $n=3,4,6$ so special. (That is the Gaussian and Eisenstein integers)

The only $\theta\in [0,\frac{\pi}{2}]$ which are rational multiples of $\pi$ for which $\cos(\theta)\in \mathbb{Q}$ are $\theta=\frac{\pi}{2},\frac{\pi}{3}$ which corresponds exactly to $n=4,6$ in $\frac{2\pi}{n}$.

Can someone give me an explanation for why $\cos(\theta)$ is rational only in these cases? Also, can we go the other way, and use some nice property of the Kummer Rings to show that $\cos(2\pi/n)$ is rational if and only if $n=1,2,3,4,6$?

Thanks,

Edit: As pointed out by Qiaochu, what I previously wrote above was certainly not the norm.

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    I wish the cited wikipedia article gave a source for calling these rings "Kummer rings". I acknowledge that Kummer worked on them (as did others before and since), but every number theorist I know calls them **cyclotomic integer rings** or something close to that.2011-07-03

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In a number field $K$, the norm of an element $N_{K/\mathbb{Q}}(a) = N(a)$ can be given various equivalent definitions, one of which is that it is the determinant of the linear map $x \mapsto ax$ acting on $K$ regarded as a vector space over $\mathbb{Q}$. If $\sigma_i : K \to \mathbb{C}$ denote the complex embeddings of $K$, then we also have

$N(a) = \prod_i \sigma_i(a).$

The norm is always rational.

If $K$ has degree $n$, then it has $n$ complex embeddings (for example by the primitive element theorem); in particular, fixing a basis of $K$ and expressing $a$ in it, the norm is a homogeneous polynomial of degree $n$. It is a quadratic form if and only if $n = 2$.

Now the degree of $\mathbb{Q}(\zeta_m)$ is equal to $\varphi(m)$, which is equal to $2$ if and only if $m = 3, 4, 6$. In other words, these are the only cyclotomic fields which give quadratic extensions. This is related to the crystallographic restriction theorem.

Yes, you can use cyclotomic fields to prove that $\cos \frac{2\pi}{m}$ is rational only when $m = 1, 2, 3, 4, 6$. Once you know that $\mathbb{Q}(\zeta_m)$ has degree $\varphi(m)$ (but this is not trivial), you can show that that $\mathbb{Q}(\zeta_m + \zeta_m^{-1})$ is a subfield of index $2$, hence is $\mathbb{Q}$ if and only if $\varphi(m) \le 2$.

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    @Eric: if $K$ is a Galois extension with Galois group $G$, the norm can be written $N(a) = \prod_{g \in G} ga$. Here the Galois group is cyclic of order $4$ generated by $\zeta_5 \mapsto \zeta_5^2$ (a primitive root modulo $5$). This statement about the Galois group turns out to be equivalent to the statement that the corresponding cyclotomic polynomial is irreducible (together with the existence of primitive roots). By the way, you only need to go up to $\zeta_5^3$. This kind of material should be covered in any good book on algebraic number theory / Galois theory.2011-07-04