index of the group of cyclotomic units in the full unit group
Let ${K}_{n}=\mathbb{Q}({\zeta}_{{p}^{n}})$ where ${\zeta}_{{p}^{n}}$ is a primitive ${p}^{n}$th root of unity^{}, let ${h}_{n}$ be the class number^{} of ${K}_{n}$ and let ${\mathcal{O}}_{n}={\mathcal{O}}_{{K}_{n}}$ be the ring of integers^{} in ${K}_{n}$. Let ${E}_{n}={\mathcal{O}}_{n}^{\times}$ be the group of units in ${K}_{n}$. The cyclotomic units are a subgroup^{} ${C}_{n}$ of ${E}_{n}$ which satisfy:

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The elements of ${C}_{n}$ are defined analytically.

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The subgroup ${C}_{n}$ is of finite index in ${E}_{n}$. Furthermore, the index is ${h}_{n}^{+}$: Let ${E}_{n}^{+}$ be the group of units in ${K}_{n}^{+}$ and let ${C}_{n}^{+}={C}_{n}\cap {E}_{n}^{+}$. Then $[{E}_{n}^{+}:{C}_{n}^{+}]={h}_{n}^{+}$. Moreover, it can be shown that $[{E}_{n}:{C}_{n}]=[{E}_{n}^{+}:{C}_{n}^{+}]$ because ${E}_{n}={\mu}_{{p}^{n}}{E}_{n}^{+}$ (this is exercise 8.5 in [1]).

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The subgroups ${C}_{n}$ behave “well” in towers. More precisely, the norm of ${C}_{n+1}$ down to ${K}_{n}$ is ${C}_{n}$. This follows from the fact that the norm of ${\zeta}_{{p}^{n+1}}$ down to ${K}_{n}$ is ${\zeta}_{{p}^{n}}$.
Definition 1.
Let $p$ be prime and let $n\mathrm{\ge}\mathrm{1}$. Let ${\zeta}_{{p}^{n}}$ be a primitive ${p}^{n}$th root of unity.

1.
The cyclotomic unit group ${C}_{n}^{+}\subset {K}_{n}^{+}=\mathbb{Q}{({\zeta}_{{p}^{n}})}^{+}$ is the group of units generated by $1$ and the units
$${\xi}_{a}={\zeta}_{{p}^{n}}^{(1a)/2}\frac{1{\zeta}_{{p}^{n}}^{a}}{1{\zeta}_{{p}^{n}}}=\pm \frac{\mathrm{sin}(\pi a/{p}^{n})}{\mathrm{sin}(\pi /{p}^{n})}$$ with $$ and $\mathrm{gcd}(a,p)=1$.

2.
The cyclotomic unit group ${C}_{n}\subset {K}_{n}=\mathbb{Q}({\zeta}_{{p}^{n}})$ is the group generated by ${\zeta}_{{p}^{n}}$ and the cyclotomic units ${C}_{n}^{+}$ of ${K}_{n}^{+}$.
Remark 1.
Let ${\sigma}_{a}:{\zeta}_{{p}^{n}}\to {\zeta}_{{p}^{n}}^{a}$ be an element of $\mathrm{Gal}({K}_{n}/\mathbb{Q})$. Then:
$${\xi}_{a}={\zeta}_{{p}^{n}}^{(1a)/2}\frac{1{\zeta}_{{p}^{n}}^{a}}{1{\zeta}_{{p}^{n}}}=\frac{{({\zeta}_{{p}^{n}}^{1/2}(1{\zeta}_{{p}^{n}}))}^{{\sigma}_{a}}}{{\zeta}_{{p}^{n}}^{1/2}(1{\zeta}_{{p}^{n}})}.$$ 
Remark 2.
Let $g$ be a primitive root^{} modulo ${p}^{n}$. Let $a\equiv {g}^{r}mod{p}^{n}$. Then one can rewrite ${\xi}_{a}$ as:
$${\xi}_{a}=\prod _{i=0}^{r1}{\xi}_{g}^{{\sigma}_{g}^{i}}.$$ 
In particular ${\xi}_{g}$ generates ${C}_{n}^{+}/\{\pm 1\}$ as a module over $\mathbb{Z}[\mathrm{Gal}(\mathbb{Q}{({\zeta}_{{p}^{n}})}^{+}/\mathbb{Q})]$.
Notice that in order to show that the index of ${C}_{n}$ in ${K}_{n}$ is finite it suffices to show that the index of ${C}_{n}^{+}$ in ${K}_{n}^{+}$ is finite. Indeed, let $[{K}_{n}:\mathbb{Q}]=2d$. Since ${K}_{n}$ is a totally imaginary field and by Dirichlet’s unit theorem the free rank of ${E}_{n}$ is ${r}_{1}+{r}_{2}1=d1$. On the other hand, $[{K}_{n}^{+}:\mathbb{Q}]=d$ and ${K}_{n}^{+}$ is totally real, thus the free rank of ${E}_{n}^{+}$ is also $d1$. Therefore the free rank of ${E}_{n}^{+}$ and ${E}_{n}$ are equal. As we claimed before, the index $[{E}_{n}^{+}:{C}_{n}^{+}]$ is rather interesting to us.
Theorem 1 ([1],Thm. 8.2).
Let $p$ be a prime and $n\mathrm{\ge}\mathrm{1}$. Let ${h}_{n}^{\mathrm{+}}$ be the class number of $\mathrm{Q}\mathit{}{\mathrm{(}{\zeta}_{{p}^{n}}\mathrm{)}}^{\mathrm{+}}$. The cyclotomic units ${C}_{n}^{\mathrm{+}}$ of $\mathrm{Q}\mathit{}{\mathrm{(}{\zeta}_{{p}^{n}}\mathrm{)}}^{\mathrm{+}}$ are a subgroup of finite index in the full unit group ${E}_{n}^{\mathrm{+}}$. Furthermore:
$${h}_{n}^{+}=[{E}_{n}^{+}:{C}_{n}^{+}]=[{E}_{n}:{C}_{n}].$$ 
In the proof of the previous theorem one calculates the regulator^{} of the units ${\xi}_{a}$ in terms of values of Dirichlet Lfunctions with even characters^{}. In particular, one calculates:
$$R(\{{\xi}_{a}\})=\pm \prod _{\chi \ne {\chi}_{0}}\frac{1}{2}\tau (\chi )L(1,\overline{\chi})={h}_{n}^{+}\cdot {R}^{+}$$ 
where in the last equality one uses the properties of Gauss sums and the class number formula^{} in terms of Dirichlet Lfunctions evaluated at $s=1$. This yields that $R(\{{\xi}_{a}\})$ in nonzero, therefore the index in ${E}_{n}^{+}$ is finite and moreover
$${h}_{n}^{+}=\frac{R(\{{\xi}_{a}\})}{{R}^{+}}=[{E}_{n}^{+}:{C}_{n}^{+}]=[{E}_{n}:{C}_{n}].$$ 
An immediate consequence of this is that if $p$ divides ${h}_{n}^{+}$ then there exists a cyclotomic unit $\gamma \in {C}_{n}^{+}$ such that $\gamma $ is a $p$th power in ${E}_{n}^{+}$ but not in ${C}_{n}^{+}$.
References
 1 L. C. Washington, Introduction to Cyclotomic Fields^{}, Second Edition, SpringerVerlag, New York.
Title  index of the group of cyclotomic units in the full unit group 

Canonical name  IndexOfTheGroupOfCyclotomicUnitsInTheFullUnitGroup 
Date of creation  20130322 15:42:49 
Last modified on  20130322 15:42:49 
Owner  alozano (2414) 
Last modified by  alozano (2414) 
Numerical id  4 
Author  alozano (2414) 
Entry type  Theorem 
Classification  msc 11R18 