# The sum of unity roots

Let $p\not=2$ be a prime. Suppose that $\zeta_1,\cdots,\zeta_n$ are $p$-th unity roots. If their sum $S$ is an integer, show that $S$ is congruent to $n$ modulo $p$.

I don't know how to deal with such problems. Generally, is it necessary that $\zeta_i-1\in pO$, where $O$ is the algebraic integer ring of $\mathbb{Q}(\xi)$ and $\xi$ is a primitive $p$-th root?

$\newcommand{\ze}{\zeta}$ $\newcommand{\Z}{\Bbb Z}$ So, each $\ze_i=\eta^{a_i}$ where $\eta=\exp(2\pi i/p)$ and $a_i$ is an integer. We may assume that $0\le a_i<p$. Our task is to prove that if $S=\sum_i (\ze_i-1)\in\Z$ then $S$ is a multiple of $p$. But $S$ is a multiple of $\eta-1$ in the cyclotomic ring $\Z[\eta]$. So all one has to prove is that $(\eta-1)\Z[\eta]\cap\Z\subseteq p\Z$. Indeed here, both sides are equal, and this is a standard result in cyclotomic fields.
• The standard result is that $\eta-1$ generates the unique prime ideal above $p$ in $\Bbb Z[\eta]$. Jul 15, 2017 at 5:21
• And a quick way to see that is to see look at the norm of $\eta - 1$ down to $\mathbb{Q}$. The norm is $\prod_{i=1}^{p-1} (\eta^i - 1)$, and which is $\prod_{i=1}^{p-1} (\eta^i x^i - 1) = x^{p-1} + x^{p-2} + ... + x + 1$ evaluated at $1$, so it equals $p$. Since it has prime norm, the absolute norm of the ideal $(\eta - 1)$ is prime, so quotienting the ring of integers by this ideal gives a field, so $(\eta - 1)$ is a prime ideal (and we see it divides $p$). Jul 15, 2017 at 11:16