Prove any unit $u$ in $\mathbb{Z}[\zeta_{p}]$ is a product of a root of unity and real unit.

Question

Given $\zeta_{p}$ is a $p^{th}$ root of unity for some odd prime $p$ I was able to show for any unit $u$ that $\frac{u}{\overline{u}}=\pm \zeta_{p}^{k} $. Thus we have $u=\pm \zeta_{p}^{k} \overline{u}$. The problem is that I can't seem to prove $\overline{u}$ is a real unit (which I suppose would mean $u$ is a unit in $\mathbb{R} \cap \mathbb{Z}[\zeta_{p}]$). I tried by contradiction using the fact $\omega^{k}+\overline{\omega^{k}}$ is a real number and thought when substituted with $\frac{u}{\overline{u}}$ and $\frac{\overline{u}}{u}$ respectfully I'd see their sum equal a complex number. If anything it showed $u$ should be a complex number. Any insight or hints would be greatly appreciated!

Answer 1

In your formula for $u/\bar u$, you must show that only the sign + can occur. At least two kinds of proofs can be found in textbooks, but they both use a zest of arithmetic in a hidden way:

1) Directly for $\mathbf Q(\zeta_p)$, on using divisibility by $(1-\zeta_p)$ in the ring $\mathbf Z[\zeta_p]$. See Washington, prop. 1.5, or Marcus, ex. 12 of chapter 2.

2) Via the algebraic theory of CM-fields which provides, with obvious notations, the formula $Q_K:= [E:WE^+]= 1$ or $2$ for any CM-field $K$ (Wash., thm. 4.12). But for $K=\mathbf Q(\zeta_n)$, the choice between the values 1 and 2 needs arithmetic arguments: the same proof as in 1) gives $Q=1$when $n$ is a prime power; when $n$ is not a prime power, an additional argument modulo some power of $2$ gives $Q=2$.