Average of sum of unit roots is an algebraic integer Let $\alpha_1,\ldots, \alpha_n$ be roots of unity, and let $a=\frac{1}{n}\sum\alpha_i$. Then if $a$ is an algebraic integer, we have either $a=0$ or $a=\alpha_1=\dots=\alpha_n$. Why?
 A: 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.
A: I don't think we need Kronecker? First notice $|a|\le 1$, with equality iff all the $\alpha_i$s coincide, so WLOG assume $|a|<1$. Let $a'$ be any conjugate of $a$. WLOG $a$ is $1/n$ times the sum of $n$ $k$-th roots of unity; and so $a'$ must also be $1/n$ times the sum of $n$ $k$-th roots of unity, since the conjugates of $k$-th roots of unity are still $k$-th roots of unity. Immediately we get $|a'|\le 1$. Now let 
$a^m + r_{m-1}a^{m-1}+\cdots + r_1a^1+r_0=0$
be a minimal polynomial for $a$. WLOG (if $a\neq 0$) we may assume $r_0\neq 0$. By Vieta, we have $\pm r_0$ equals the product of $a$ with all its conjugates, counting multiplicities. Thus $|r_0|<1$ and so $r_0=0$ (because it's an integer), our desired contradiction.
