# An algebraic number is not a root of unity?

This problem is related to my study of the Burau representation of the braid group $B_3$: I was trying to show that certain "congruence subgroups" are of infinite index.

There is an approach that boils it down to the following question:

Let $\xi$ be a primitive root of unity of degree $2n>12$ (just in case). Then the roots $\lambda$ of $\chi(\lambda):=\lambda^2+(\xi^2-\xi+1)\lambda+\xi^2$ are not roots of unity.

I was particularly interested in the cases $n=7$ or $9$. In both cases, $\deg\Phi_{2n}=6$ (here $\Phi$ is the cyclotomic polynomial) and, hence, $\mathbb{Q}[\lambda]$ has degree at most $12$. There are finitely many (although quite a few) cyclotomic polynomials $\Phi_k$ with $\deg\Phi_k\le12$. For each such polynomial, one computes the $\lambda$-resultant $R(\xi)$ of $\Phi_k$ and $\chi$ and checks that $R(\xi)\ne0\bmod\Phi_{2n}(\xi)$.

Clearly, this approach should work for any given $n$, but I have no idea how this can be done "in general". So, here is the question:

Is there a smarter way to prove that an algebraic number is not a root of unity?

• This isn't a way of testing algebraic numbers for being roots of unity, but a little bit of algebra reveals that this problem is exactly equivalent to, writing $\xi = exp(2\pi i\ m/n),$ showing that there do not exist $p, q$ with $$\cos(\pi\ m/n) + \cos(\pi\ p/q) = 1/2.$$ Feb 28, 2014 at 10:47
• If $\lambda '$ is a root of $\chi$ then $\mathbb{Q}(\xi,\lambda ')$ has degree at most 2 over $\mathbb{Q}(\xi)$ so $\lambda '$ would have to be either a power of $\xi$ or a primitive $4n$-th root of unity. But the minimal polynomial of the latter is given by $\lambda^2-\xi$ hence this is not the case. So what remains to show is that no power of $\xi$ is a root of this polynomial. Feb 28, 2014 at 10:52