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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?

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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 $$\cos(\pi\ m/n) + \cos(\pi\ p/q) = 1/2.$$ –  drvitek Feb 28 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. –  Sebastian Schoennenbeck Feb 28 at 10:52
In your examples, you take $n=7$ or $n=3^2$: since it makes things easier, I am wondering if it is enough for you that $n$ be a prime-power or whether you might want to take things like $n=60. –  Filippo Alberto Edoardo Feb 28 at 10:52
@drvitek Of course, but this looks even worse (= transcendental) –  Alex Degtyarev Feb 28 at 10:53
@@GerryMyerson The paper that we're looking for is Newman's "Some Results on Roots of Unity, with an Application to a Diophantine Problem" in which he characterizes all rational $x, y$ with $\sin(\pi x)\sin(\pi y) = 1/4$, which is equivalent via prosthaphaeresis (cosine-addition, but I just wanted an excuse to use that word) to the equation I gave (which is in fact missing some 2's). In particular, none of the solutions have denominators greater than 12, so it looks like you're good. I have access to the PDF; please let me know if you can't get a hold of it. –  drvitek Feb 28 at 11:54
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3 Answers

up vote 7 down vote accepted

Assume that the roots $\zeta$ and $\xi^2/\zeta$ of your equation are roots of unity. Then by Vieta Theorem $\zeta+\xi^2/\zeta=-(\xi^2-\xi+1)$; equivalently sum of 5 roots of unity $\zeta+\xi^2/\zeta+\xi^2-\xi+1$ vanishes. All such relations are classified in Theorem 6 of paper by Conway and Jones. This implies the result.

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This is probably the simplest answer. Thanks! –  Alex Degtyarev Mar 2 at 9:49
You are very welcome! –  Victor Ostrik Mar 2 at 16:07
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You are interested in the solutions of the bivariate equation $V\colon f(X,Y)=X^2+X(Y^2-X+1)+Y^2=0$ in roots of unity, a sort of question known as "Manin-Mumford problem". The answer is known in the greater generality of subvarieties of algebraic groups — except for finitely many points, they are explained by the existence of "torsion subvarieties".

In the case you're interested in, the answer has been proved by Ihara, Serre, Tate, see Lang, Division points on curves. The proof is quite elementary. Let $z=(\xi,\eta)$ be a point of $V$ where $\xi,\eta$ are roots of unity; let $n$ be the lcm of their orders. Then, for every integer $d\geq2$ which is prime to $n$, there exists $\sigma\in\mathop{\rm Gal}(\overline{\mathbf Q}/\mathbf Q)$ such that $\sigma(z)=z^d$. Then, $z$ is a common root of $f(X,Y)$ and of $f(X^d,Y^d)$. By Bézout's theorem, the locus defined by these two equations is either equal to $V$ (if $f(X,Y)$ divides $f(X^d,Y^d)$), or is finite, of cardinality $\leq \deg(f)^2 d=9d$. In the first case, $f$ (which is irreducible) would be the difference of two monomials, which it isn't. So we are in the latter case. Since all conjugates of $z$ belong to this finite set, one gets the inequality $\phi(n)\leq 9d$.

By the prime number theorem, it is possible to choose $d$ quite small with respect to $n$, say $d\ll \log(n)$. (The worst case is when $n=p_1\dots p_r$, the product of the $r$ first primes—then, $d=p_{r+1}$ is possible, and (PNT), one has $p_{r+1}\sim r\log(r)$ while $\log(n)=\sum_{i\leq r} \log(p_i)\sim r\log(r)$ as well, so that $p_{r+1} \approx \log(n)$.) Similarly, $\phi(n)\gg n^{1-\epsilon}$, hence $n^{1-\epsilon}\ll 9 \log(n)$. This bounds $n$ from above, and it remains to check every possible $z$ in a finite set.

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I assume that $n=p$ be a prime and $p>3$: indeed, if $p=3$ then your statement is false because $\xi^2-\xi+1=0$. Your polynomial is $$ \chi(\lambda)=\lambda^2+(\xi-1)^2\lambda+\lambda\xi+\xi^2\in\mathbb{Z}[\xi] $$ where $\xi$ is a primitive $2p$-th root of unity. Let $\zeta$ be a root of $\chi$ and assume it is a root of unity: hence, as Sebastian Schoennenbeck observed, either $\zeta\in\mathbb{Q}(\xi)=\mathbb{Q}(\mu_{2p})$ or it lies in the quadratic extension $\mathbb{Q}(\mu_{4p})$. The latter is not the case since the minimal polynomial of $\zeta$ over $\mathbb{Q}(\mu_{2p})$ would be $\lambda^2-\xi$, therefore $\zeta\in \langle\xi\rangle$, the cyclic group of order $2p$ generated by $\xi$ inside $\mathbb{Z}[\xi]^\times$. To show that this is not the case, observe that all roots of unity in $\mathbb{Z}[\xi]$ reduce to $\pm 1$ modulo the maximal ideal $\mathfrak{p}\subseteq \mathbb{Z}[\xi]$ which lies above $p$ and contains (is in fact generated by) $\xi-1$ (see Proposition 2.8 of Washingtons's Introduction to Cyclotomic Fields). If we reduce the polynomial$\mod\mathfrak{p}$ we thus find $$ \bar{\chi}(\lambda)=\lambda^2\pm \lambda+1 $$ and neither $1$ nor $-1$ are roots of either of these polynomials unless $p=3$ which we excluded. Hence, no roots of unity can be a root of $\chi$.

The argument above extends immediately to the case when $n=p^e$ is a prime power but does not extend to the case where $n=p^aq^b$ is composite because $\xi-1$ becomes a unit.

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