## If the roots of a monic irreducible polynomial over ZZ all have modulus at most 1, then the roots are roots of unity

Let $p(x) \in \mathbb{Z}[x]$ be a monic irreducible polynomial and suppose that for each root $\zeta$ of $p(x)$, we have $||\zeta || \leq 1$, where $|| \cdot ||$ denotes complex modulus. Show that the roots of $p(x)$ are roots of unity.

Note that $p(x)$ factors as $p(x) = \prod (x - \zeta_i)$, where $\zeta_i$ ranges over the roots of $p$. Now the constant coefficient of $p(x)$ is $\prod \zeta_i$, and thus $||\prod \zeta_i|| = \prod ||\zeta_i|| \leq 1$. On the other hand, the constant coefficient is an integer, and so is either 0 or 1. If the constant coefficient is 0, then $x$ divides $p(x)$. So the constant coefficient of $p$ is 1, and we have $||\zeta_i|| = 1$ for all $i$.

By Lemma 11.8, the $\zeta_i$ are all roots of unity.