IndexOfTheGroupOfCyclotomicUnitsInTheFullUnitGroup 2006-02-28 13:31:16 2006-02-28 13:31:16 Theorem cyclotomic units % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newtheorem{thm1}{Theorem} \newtheorem{defn1}{Definition} \newtheorem{prop1}{Proposition} \newtheorem{lemma1}{Lemma} \newtheorem{cor1}{Corollary} \theoremstyle{definition} \newtheorem{exa}{Example} \newtheorem{remark}{Remark} % Some sets \newcommand{\Nats}{\mathbb{N}} \newcommand{\Ints}{\mathbb{Z}} \newcommand{\Reals}{\mathbb{R}} \newcommand{\Complex}{\mathbb{C}} \newcommand{\Rats}{\mathbb{Q}} \newcommand{\Gal}{\operatorname{Gal}} \newcommand{\Cl}{\operatorname{Cl}} Let $K_n=\Rats(\zeta_{p^n})$ 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: \begin{itemize} \item The elements of $C_n$ are defined analytically. \item 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 \cite{wash}). \item 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 $\zeta_{p^n}$. \end{itemize} \begin{defn1} Let $p$ be prime and let $n\geq 1$. Let $\zeta_{p^n}$ be a primitive $p^n$th root of unity. \begin{enumerate} \item The cyclotomic unit group $C_n^+\subset K_n^+=\Rats(\zeta_{p^n})^+$ is the group of units generated by $-1$ and the units $$\xi_a=\zeta_{p^n}^{(1-a)/2}\frac{1-\zeta_{p^n}^a}{1-\zeta_{p^n}}=\pm \frac{\sin(\pi a/p^n)}{\sin(\pi/p^n)}$$ with $1<a<\frac{p^n}{2}$ and $\gcd(a,p)=1$. \item The cyclotomic unit group $C_n\subset K_n=\Rats(\zeta_{p^n})$ is the group generated by $\zeta_{p^n}$ and the cyclotomic units $C_n^+$ of $K_n^+$. \end{enumerate} \end{defn1} \begin{remark} Let $\sigma_a:\zeta_{p^n}\to \zeta_{p^n}^a$ be an element of $\Gal(K_n/\Rats)$. Then: $$\xi_a=\zeta_{p^n}^{(1-a)/2}\frac{1-\zeta_{p^n}^a}{1-\zeta_{p^n}}= \frac{(\zeta_{p^n}^{-1/2}(1-\zeta_{p^n}))^{\sigma_a}}{\zeta_{p^n}^{-1/2}(1-\zeta_{p^n})}.$$ \end{remark} \begin{remark} \label{rem1} Let $g$ be a primitive root modulo $p^n$. Let $a\equiv g^r \mod p^n$. Then one can rewrite $\xi_a$ as: $$\xi_a= \prod_{i=0}^{r-1}\xi_g^{\sigma_g^i}.$$ In particular $\xi_g$ generates $C_n^+/\{\pm 1\}$ as a module over $\Ints[\Gal(\Rats(\zeta_{p^n})^+/\Rats)]$. \end{remark} 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 $[K_n:\Rats]=2d$. Since $K_n$ is a totally imaginary field and by Dirichlet's unit theorem the free rank of $E_n$ is $r_1+r_2-1=d-1$. On the other hand, $[K_n^+:\Rats]=d$ and $K_n^+$ is totally real, thus the free rank of $E_n^+$ is also $d-1$. 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. \begin{thm1}[\cite{wash},Thm. 8.2] \label{cycloindex} Let $p$ be a prime and $n\geq 1$. Let $h^+_{n}$ be the class number of $\Rats(\zeta_{p^n})^+$. The cyclotomic units $C_{n}^+$ of $\Rats(\zeta_{p^n})^+$ are a subgroup of finite index in the full unit group $E_{n}^+$. Furthermore: $$h^+_{n}=[E_{n}^+:C^+_{n}]=[E_n:C_n].$$ \end{thm1} In the proof of the previous theorem one calculates the regulator of the units $\xi_a$ in terms of values of Dirichlet L-functions with even characters. In particular, one calculates: $$R(\{\xi_a\})=\pm\prod_{\chi\neq \chi_0}\frac{1}{2}\tau(\chi)L(1,\overline{\chi})=h^+_{n}\cdot R^+$$ where in the last equality one uses the properties of Gauss sums and the class number formula in terms of Dirichlet L-functions evaluated at $s=1$. This yields that $R(\{\xi_a\})$ in non-zero, 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 $C_n^+$. \begin{thebibliography}{00} \bibitem{wash} L. C. Washington, {\em Introduction to Cyclotomic Fields}, Second Edition, Springer-Verlag, New York. \end{thebibliography}