Average of sum of unit roots is an algebraic integer

Solution 1:

Suppose that $\alpha_1, \dots, \alpha_n$ are $k$th roots of unity. Since all conjugates of the $\alpha_i$ are again $k$th roots of unity, for any conjugate $a'$ of $a$, we have the (complex) norm of $a'$ is bounded: $$ |a'| = \frac{1}{n} \sum \left |\zeta_k^{t_i} \right| \leq \frac{1}{n} \sum 1 = 1. $$ But any nonzero algebraic integer whose conjugates are all bounded by 1 must be a root of unity. So $a = 0$ or $a$ is also a ($k$th) root of unity.

If $a \neq 0$, then the triangle inequality $$ |a| = \frac{1}{n} | \alpha_i | \leq \frac{1}{n} \sum |\alpha_i| = 1 $$ is an equality. By a simple induction argument, this implies $\alpha_i = \pm \alpha_j$ for $1 \leq i,j \leq n$.

Write $\alpha_1 = \zeta_k^\ell$, so each $\alpha_i = \pm \zeta_k^\ell$. Reordering (and potentially swapping signs) if necessary, we may assume $\alpha_1 = \dots = \alpha_m = \zeta_k^\ell$ and $\alpha_{m+1} = \dots = \alpha_n = - \zeta_k^\ell$ with $\frac{n}{2} \leq m \leq n$.

Then $$\begin{align*} a &= \frac{1}{n} \left(\sum_{i=1}^m \zeta_k^\ell - \sum_{i = m+1}^n \zeta_k^\ell \right)\\ &= \frac{2m - n}{n} \zeta_k^\ell \end{align*}$$

But $a$ is an algebraic integer, hence $\frac{2m-n}{n} \in \mathbf{Z}$. Thus $m = \frac{n}{2}$ or $m = n$, giving $a = 0$ or $a = \alpha_1 = \dots = \alpha_n$, respectively.