# The only two rational values for cosine and their connection to the Kummer Rings

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.

• That is not the norm in $\mathbb{Z}[\zeta_n]$. – Qiaochu Yuan Jul 3 '11 at 22:32
• @Qiaochu: What would the norm be in the general Kummer Rings? – Eric Naslund Jul 3 '11 at 22:39
• 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. – Pete L. Clark Jul 3 '11 at 23:31

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 $\varphi(m) \le 2$.

• Does this "crystallographic Restriction" have to do with the fact that $\mathbb{Z}[\zeta_n]$ has finitely many units if and only if $n=1,2,3,4,6$ by Dirichlet's unit theorem? – Eric Naslund Jul 3 '11 at 22:43
• Well, yes, but that's not how I would think about it. The only number fields $K$ such that $\mathcal{O}_K$ has finitely many units satisfy $r + s = 1$ where $r$ is the number of real embeddings and $s$ the number of pairs of complex embeddings, hence either $r = 1, s = 0$ and $K = \mathbb{Q}$ or $r = 0, s = 1$ and $K$ is an imaginary quadratic number field. This is the important property, as it implies that $\mathcal{O}_K$ naturally embeds into $\mathbb{C}$ as a discrete subgroup, so in particular the units of $\mathcal{O}_K$ act on a lattice. – Qiaochu Yuan Jul 3 '11 at 22:50
• Crystallographic restriction is, to my mind, really a theorem about elements of finite order in $\text{GL}_n(\mathbb{Z})$. The point is that an element of order exactly $m$ has the property that its characteristic polynomial is a product of cyclotomic polynomials, at least one of which must be $\Phi_m$, which has degree $\varphi(m)$, hence $\varphi(m) \le n$. Thus in $3$ dimensions we can still only have $m = 1, 2, 3, 4, 6$ but in $4$ dimensions we can have $m = 1, 2, 3, 4, 5, 6, 8$, and so forth. – Qiaochu Yuan Jul 3 '11 at 22:53
• One more question: I just want to understand this example: What is the norm in $\mathbb{Z}[\zeta_5]$? So we can write each element as $a+b\zeta_5+c\zeta_5^2+d\zeta_5^3+e\zeta_5^4$. What will the polynomial for the norm be? How does it relate to the cyclotomic polynomial $\Phi_5(X)=X^4+X^3+X^2+X^1+1$? Thanks again for your help! This answer has been very useful! – Eric Naslund Jul 4 '11 at 16:26
• @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. – Qiaochu Yuan Jul 4 '11 at 16:44