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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.

Bruno Joyal
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Ben Perez
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  • Numbers of the form $u_k=(\zeta^k-\zeta^{-k})/(\zeta-\zeta^{-1})$ with $k=2,3,(p-1)/2$ are easily seen to be units in the ring $\Bbb{Z}[\zeta+\zeta^{-1}]$. I may be wrong, but I think they form a basis for the free part of the unit group. I don't remember the argument, and may be it only showed that the units $u_k$ are independent? – Jyrki Lahtonen Oct 04 '14 at 19:04
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    Dear @Jyrki: I don't believe these units (called "cyclotomic units") form a basis, but they are indeed independent (this follows from the nonvanishing of Dirichlet $L$-functions at $s=1$, and from the formula for these $L$-values in terms of logs of these units). In fact, I think that the index of the subgroup they generate is a subtle invariant, closely related to the class number of the cyclotomic field. – Bruno Joyal Oct 26 '14 at 22:27
  • Thanks for the extras @Bruno. Chance for me to learn something! – Jyrki Lahtonen Oct 27 '14 at 05:26

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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!

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Bruno Joyal
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  • But, by Washington's "Introduction to Cyclotomic Fields", Corollary 4.13. Let $E,E^+$ be the group of units in $\mathbb{Q}(\zeta)$ and its maximal real subfield, then $[E:\left<\zeta\right>E^+]$ should be $2$ when $n$ is not a prime power. What's wrong? – S.Gau at Math Feb 11 '15 at 08:11
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    I guess the $s=\zeta^a$ should be $s=\pm\zeta^a$. – S.Gau at Math Feb 11 '15 at 08:33
  • ($\zeta$ is $p$-th root of unity) $u^p$ is a unit. Suppose $s=-\zeta^a$. Since $u^p=t$ for some $t\in\mathbb{Z}$, $p$ divides $u^p$, a contradiction. – defacto Jan 16 '22 at 22:24