Group of Units in Cyclotomic Integers

Question

I'm trying to show that for any $p$-th root of unity $\zeta$, where $p$ is an odd prime, we have $\mathbb{Z}[\zeta]^{\times} = \left<\zeta\right>\mathbb{Z}[\zeta + \zeta^{-1}]^{\times}$. Obviously the $\left<\zeta\right>$ factor comes as a result of Dirichlet's unit theorem. However, I'm struggling to show that $\mathbb{Z}[\zeta + \zeta^{-1}]^\times$ is a free abelian group of rank $r + s - 1$, where $r$ is the number of real embeddings of $\mathbb{Z}[\zeta]$ and $s$ is the number of complex conjugate pairs of embeddings.

Answer 1

This is true more generally if $\zeta$ is a primitive $n$-th root of unity, where $n$ is odd (not necessarily prime).

Let $u \in \mathbb Z[\zeta]$. Let $\overline{u}$ denote the complex conjugate of $u$ (remark that this is independent of any choice of embedding of $\mathbb Q(\zeta)$ in $\mathbb C$, because $\mathbb Q(\zeta)$ is a CM field - complex conjugation always acts via $\zeta \mapsto \zeta^{-1}$).

Let $s= u/\overline{u}$.

Remark that under any embedding of $\mathbb Q(\zeta)$ in $\mathbb C$, $s$ is mapped to a complex number of absolute value $1$.

It follows that $s$ is a root of unity contained in $\mathbb Q(\zeta)$, hence $s=\zeta^a$ for some $a \in \mathbb Z/n\mathbb Z$.

Let $b \in \mathbb Z/n\mathbb Z$ be such that $2b=-a$, which exists because $n$ is odd. Let $v = \zeta^b u$. Then

$$\overline{v} = \zeta^{-b} \overline{u} = \zeta^{-b} u/s = \zeta^{-b - a} u = \zeta^b u = v.$$

Hence $v$ is totally real, hence $v \in \mathbb Z[\zeta + \zeta^{-1}]^\times$. Since $u = \zeta^{-b}v$, it follows that $\mathbb Z[\zeta]^\times = \left<\zeta\right> \mathbb Z[\zeta + \zeta^{-1}]^\times$.

Remark: This calculation is very Galois-cohomological in flavor. I'd be curious to see a proof of this result using only Galois cohomology!