Question

Let $p$ be a rational prime and $\zeta_p$ a primitive $p$th root of unity.

What do we know about the set ${z\in\mathbb{Q}(\zeta_p):|z|=1}$?

Answer 1

For any $u \in L^\times = \mathbf{Q}(\zeta_p)^\times$, letting $z=u/\bar{u}$, we have $|z|=1$. The converse is true : let $G=\mathbf{Z}/2\mathbf{Z}$ act on $L^\times$ by complex conjugation, then $G$ is the Galois group of $L$ over $K = L \cap \mathbf{R}$. By Hilbert 90, $H^1(G,L^\times) = {1}$, which says precisely that any $z \in L^\times$ such that $|z|^2 = z \bar{z}=1$ is of the form $z=u/\bar{u}$ for some $u \in L^\times$.

Let $U \subset L^\times$ be the subgroup of elements $z$ satisfying $|z|=1$. Then the map $u \mapsto u/\bar{u}$ induces an isomorphism $U \cong L^\times / K^\times$. In particular $U$ is not of finite type.