\documentclass{amsart} %\hfuzz=5pt %\parskip 6 pt %\addtolength{\textheight}{.3in} %\addtolength{\textwidth}{.2in} \usepackage{amssymb} \parindent 0cm \def\C{{ \mathbb C }} \def\R{{ \mathbb R }} \def\Z{{ \mathbb Z }} \def\Q{{\mathbb Q}} \def\N{{\mathbb N}} \def\Pr{{\mathbb P}} \def\D{{\mathfrak D}} \def\F{{\mathbb F}} \def\1{{\bf 1}} \def\G{{\mathcal G}} \def\a{{\mathcal Q}} \def\o{{\mathcal O}} \def\J{{\mathcal J}} \def\f{{\mathcal F}} \def\b{{\mathcal B}} \def\D{{\mathcal D}} \def\M{{\mathcal M}} \def\si{\Sigma} \def\Si{{\Sigma_I}} \def\k{\overline{k}} \def\U{{\mathcal U}} \def\Gam{G_{\si_I}} \def\m{{\mathfrak m}} \def\P{{\mathfrak P}} \def\p{{\mathfrak p}} \def\cha{{\boldsymbol \Delta}} \def\q{{\mathfrak q}} \def\l{{\ell}} \def\r{{\mathop{\rm r}}} \def\d{{\mathop{\rm d}}} \def\Iw#1#2{{{\text I }_{#1}({#2})}} \def\iw#1#2{{{\text i}_{#1}({#2})}} \def\hm{\hat{\m}} \def\w{{w_H}} \def\half{{\frac{1}{2}}} \def\Tr{{ \mathop{{\rm {Tr}}} }} \def\rd{{ \mathop{{\rm {rd}}} }} \def\Gal{{ \mathop{{\rm {Gal}}} }} \def\GG{G_{\overline{k}}} \def\Irr{{ \mathop{{\rm {Irr}}} }} \def\Cl{\text{Cl}} \def\Hom{{ \mathop{{\rm {Hom}}} }} \def\disc{{ \mathop{{\rm {disc}}} }} \def\rk{{ \mathop{{\rm {rk}}} }} \def\Ind{{ \mathop{{\rm {Ind}}} }} \def\Im{{ \mathop{{\rm {Im}}} }} \def\Reg{{\chi_{\mbox{reg}}}} \def\Res{\mbox{Res}} \def\mod{\text{mod}} \def\Gl{\text{Gl}} \def\cd{\text{cd}} \def\T{{ \mathop{{\rm {Tor}}} }} \def\L{\text{L}} \def\X{{ \mathop{{\rm {X}}} }} \def\d{{ \mathop{{\rm {d}}} }} \def\r{{ \mathop{{\rm {r}}} }} \def\SS{{ \mathop{{\rm {S}}} }} \def\AA{{ \mathop{{\rm {A}}} }} \setcounter{tocdepth}{2} \setcounter{part}{0} \setcounter{section}{0} \setcounter{subsection}{0} \newtheorem{Theorem}{Theorem} \newtheorem{Lemma}[Theorem]{Lemma} \newtheorem{Proposition}[Theorem]{Proposition} \newtheorem{Corollary}[Theorem]{Corollary} \newtheorem{Definition}[Theorem]{Definition} \theoremstyle{remark} \newtheorem{rem}{Remark}[section] \newtheorem{Example}{Example}[section] \newtheorem{Hypotheses}{Hypotheses} \newtheorem{Conjecture}{Conjecture} \newtheorem{Question}{Question} \newcommand{\thmref}[1]{Theorem~\ref{#1}} \newcommand{\secref}[1]{\S\ref{#1}} \newcommand{\lemref}[1]{Lemma~\ref{#1}} \newcommand{\propref}[1]{Proposition~\ref{#1}} \begin{document} \author{Christian Maire } \title{On the $\Z_l$-rank of abelian extensions with restricted ramification} \date{\today} \thanks{2000 {\em Mathematics Subject Classification.} Primary: 11R37, 11S37. {\em Key words and phrases.} $S$-ramification, $\Z_\l$-rank, Schanuel $\l$-adic Conjecture.} \begin{abstract} For a prime number $\l$, we consider $\l$-extensions $k_S$ of number fields $k$ unramified outside a finite set $S\subset S_\l$ of places of $k$ and, we study the $\Z_\l$-rank of the abelian part of $k_S/k$ and the number of relations of the Galois group $G_S:=\Gal(k_S/k)$.\end{abstract} \maketitle \section*{Introduction} Let $k$ be a number field. In class field theory, the study of $\l$-extension of $k$ with restricted ramification at a finite set $S$ is very classical. Many authors have contributed to unterstand the invariants of $k$ attached to such extensions (e.g Shafarevich \cite{sa}, Koch \cite{ko1}, \cite{ko2}, Haberland \cite{ha} ...). However in the wild ramification case, the standard results are essentially with the assumption that $S$ contains the set $S_\l$ of all places of $k$ above $\l$. This case is related to the famous Leopoldt's Conjecture. Let us explain now this conjecture more precisely: The Galois group over $k$ of the maximal abelian $\l$-extension of $k$ unramified outside $S_\l$ has a ${\Z_\l}$-rank equals to ${r_2+1+\delta}$ where $r_2$ is the number of complex places of $k$, and $\delta$ is an non-negative integer, which is zero under Leopoldt's Conjecture. In particular for all number fields $k$, there exists at least one $\Z_\l$-extension : namely the cyclotomic extension $k^\infty$. Here we ought to mention Brumer's Theorem \cite{br} (which is a $\l$-adic version of a Theorem of Baker) that proves Leopoldt's Conjecture for abelian fields and a result of Waldschmidt \cite{wal2} which proves ``one half" of Leopoldt's Conjecture. A nice description of all these facts can be found, for instance, in the book of Neukirch-Schmidt-Wingberg \cite{nsw} and the book of Gras \cite{gr1}. Let $G_S$ be the Galois group of $k_S/k$, and let $G_S^{ab}$ be its abelianization. Now concerning the $\Z_\l$-rank of $G_S^{ab}$ few results have been proved when $S$ is a proper, nonempty subset of $S_\l$. One of our main results (Theorem \ref{sup}) shows that for a number field $k$, the choice of $S$ is very important for the $\Z_\l$-rank. In particular, it is possible to find examples with $S$ small and $G_{S}^{ab}$ {\em infinite} and on the other hand for the same field $k$ and prime $\l$, one can find $S$ very large such that $G_{S}^{ab}$ is {\em finite}. To make this statement precise, we have to introduce some definitions. For a prime $\l$ and a number field $k$, let $\iw{\l}{k}$ be the smallest value of $|S|/|S_\l|$ such that $G_S^{ab}$ is infinite, and define $\Iw{\l}{k}$ to be the smallet rational number $c$, such that for all $|S|$ with $|S|/|S_\l|\geq c$, $G_S^{ab}$ is infinite. \begin{Theorem}\label{sup} For all prime $\l$, $$\sup_{ \mbox{{\small number fields }} k} \Iw{\l}{k}-\iw{\l}{k} =1.$$ \end{Theorem} An other quantity of interest is the cohomological dimension of the Galois groups of the extensions which appear. For example it is well-known that $\k(\l)/k^\infty$ is free (for $\l >2$), where $\k(\l)$ is the maximal $\l$-extension of $k$. Denote by $\widetilde{k_S}$ be the compositum of all $\Z_\l$-extensions of $k$ contained in $k_S$. Before giving some examples with $S\subsetneqq S_\l$ for which $\k(\l)/\widetilde{k_S}$ is free, we give a definition: \begin{Definition}\label{aaa} Let $k/N$ be a finite Galois extension of number fields with Galois group $G$. Let $\p$ be a prime of $N$. Define $S_{\bar{\p}}=\{\P \subset \o_k, \P\cap \o_N =\p\}=\{ \P^g, g\in G\}$. \end{Definition} First let us give a result for extensions of quadratic imaginary fields: \begin{Theorem} \label{imaginary} Let $N$ be an imaginary quadratic field, $\p$ be a prime of $N$ and $k/N$ be a finite Galois extension. Let $k_{S_{\bar{\p}}}$ be the maximal $\l$-extension of $k$ unramified outside $S_{\bar{\p}}$. Then $\Gal(\k(\l)/\widetilde{k_{S_{\bar{\p}}}})$ is free. \end{Theorem} For the real quadratic case, we prove: \begin{Theorem}[Under Schanuel $\l$-adic Conjecture (see \S 2.3.1)]\label{dimension3} Let $f(x) \in \Z[x]$ be an irreducible polynomial of degree 4, such that: \begin{enumerate} \item The discriminant $d$ of $f$ is a fundamental discriminant of a real quadratic field; $N=\Q(\sqrt{d})$; \item The roots of $f$ are not real. \end{enumerate} It is well-known that the Galois closure $k$ of $f$ over $\Q$ is of degree 24, and $\Gal(k/N)\simeq \AA_4$ (see \cite{kon}). Choose a prime number $\l$ such that $\left(\frac{d}{\l}\right)=1$, and fix a prime $\p$ of $N$ dividing $\l$. Let $k_{S_{\overline{\p}}}$ be the maximal $\l$-extension of $k$ unramified outside $S_{\bar{\p}}$. Then $\Gal(\k(\l)/\widetilde{k_{S_{\bar{\p}}}})$ is free. \end{Theorem} The organization of this paper is as follows. After a preliminary section, in the second section we study the $\Z_\l$-rank of $G_S^{ab}$ by using techniques based on linear representations inspired by \cite{ja}. In the next section, the nice situation where $\p$ is of degree 1 ($[N_\p:\Q_\l]=1$) is studied in detail, and many examples are given. The section 2 finishes with the proof of Theorems \ref{sup}, \ref{imaginary} and \ref{dimension3}. In the last section, we are interested in the group $G_S$, in particular we consider the situation in which it is possible to give the exact value of the minimal number of the relations of $G_S$. The case where $\cd(G_S)=1$ is studied. The results are in connection with the chapter 3 of the book of Gras \cite{gr1}. This work was supported by a Postdoctoral fellowship position in {\em Mathematical Sciences Research Institute}\footnote{Research at MSRI is supported in part by NSF grant DMS-9701755} at Berkeley and, by {\em CNRS} and {\em department A2X of Bordeaux I}. I thank G. Gras and J.-F. Jaulent for discussions in this topic, F. Hajir and N. Yui for their comments and remarks. \section{Preliminaries} \subsection{Notations} We fix a prime number $\l$ and an algebraic closure $\bar{\Q}$ of $\Q$. For a number field $k$, $S_\l$ denotes the set of places of $k$ above $\l$, $S$ is a non-empty subset of $S_\l$, and $\p$ is a prime ideal of the ring of integers $\o_k$ of $k$. Some standard notations are as follows: \begin{enumerate} \item[-] $k_\p$ the completion of $k$ in $\p$; $N\p$ the absolute norm, $e_\p$ and $f_\p$ the ramification index and the residue degree of $\p$ in $k/\Q$ respectively; \item[-] $[k_\p:\Q_\l]=e_\p\cdot f_\p$ be the degree of $\p$; \item[-] $\o_\p$ the completion at $\p$ of the ring of the integers $\o_k$ of $k$; \item[-] $U_\p$ the units group of $\o_\p$; \item[-] $U_\p^i$ the units of $U_\p$ congruent to 1 modulo $\p^i$; \item[-] $\G_\p$ the pro-$\l$-absolute Galois group of $k_\p$. \item[-] $E_k$ the units group of $k$; \item[-] $E_k^T$ the $T$-units group of $k$ (in the classical sense), where $T$ is a finite set of places of $k$; \item[-] $k_S^1=\{ x \in k^\times,\ \forall \p \in S, v_\p(x-1)\geq 1\}$; \item[-] $E_S^T=k_S^1\cap E_k^T$ and $E_S=k_S^1\cap E_k$; \item[-] $\cha_S=\{x \in k^\times, (x)=\a^\l, x\in k_\p^\l, \forall \p \in S\}/{k^\times}^\l$ \item[-] $k^\infty$ the $\Z_\l$ cyclotomic extension of $k$; \item[-]$k_S$ the maximal $\l$-extension of $k$ unramified outside $S$ (archimedean places are decomposed); \item[-] $k_S^T$ the maximal subextension of $k_S$ in which all places of $T$ split totally; \item[-] $G_S=\Gal(k_S/k)$ and ${G_S}^{ab}=G_S/[G_S,G_S]$; \item[-] $\k=\bar{\Q}$, and $\k(\l)$ the maximal $\l$-extension of $k$ contained in $\k$; \item[-] $\GG=\Gal(\k/k)$ the absolute Galois group of $k$. \end{enumerate} If $A$ is a $\Z_\l$ (resp. $\Z$) module of finite type: \begin{enumerate} \item[-] $\T(A)$ the $\Z_\l$-torsion part of $A$ (resp. $\Z$); \item[-] $d_\l(A)=d(A)$ the $\F_\l$-rank of $A$; \item[-] $\rk_{\Z_\l}$ (resp. $\rk_\Z$), the $\Z_\l$-rank of $A$ (resp. $\Z$-rank). \end{enumerate} \subsection{Leopoldt's Conjecture} Fix now an algebraic closure $\overline{\Q_\l}$ of $\Q_\l$, and for each place $\p_i$ ($i=1,\cdots, s$) of $S$, we fix $\sigma_i$ an imbedding of $\Q$ in $k_{\p_i}\subset \overline{\Q_\l}$. Now consider the following homomorphism $\Theta_S$ of groups: \begin{center} \begin{math} \begin{array}{rcl} \Theta_S : k_S^1 & \rightarrow & \prod_{\p \in S} \overline{\Q_\l}\\ x & \mapsto & \left(\sigma_1(x), \cdots, \sigma_s(x)\right) \end{array} \end{math} \end{center} and let $\L_S$ be the Logarithmic map: \begin{center} \begin{math} \begin{array}{rcl} \L_S : \prod_{\p \in S} U_\p^1 & \rightarrow & \prod_{\p \in S} k_\p\simeq {\Q_\l}^{\sum_\p e_\p f_\p}\\ \left(x_1, \cdots, x_s\right)& \mapsto & \left(\log_\l(x_1), \cdots, \log_\l(x_s)\right) \end{array} \end{math} \end{center} Note that the kernel of the logarithmic map is finite. We can now present Leopoldt's Conjecture: \begin{Conjecture} If $S=S_\l$, then $$\rk_{\Z}E_k=\rk_{\Z_\l}\overline{\Theta_S(E_S)}.$$ Or equivalently $(\L_S(\overline{\Theta_S(E_S)}))$ is a $\Z_\l$-lattice of $\Q_\l^n$ with rank $r_1+r_2-1$. If $\l$ splits totally in $k/\Q$, then $\L_S(\Theta_S(E_S)))$ is in the hyperplane of $\Q_\l^n$ defined by the equation: $x_1+\cdots+x_n=0$. \end{Conjecture} \subsection{On the $\Z_\l$-rank of $G_S^{ab}$} In order to prove Theorem \ref{sup}, we need to consider subextensions where places split completely. \begin{Theorem}\label{rang} Let $T$ and $S$ be two finite sets of finite places of $k$ ($S\subset S_\l$). Let $k_S^T$ be the maximal $\l$-extension of $k$ unramified outside $S$ in which all places of $T$ split. Put $G_S^T=\Gal(k_S^T/k)$. Then we have the following isomorphism of $\Z_\l$-modules: $${G_S^T}^{ab}\simeq \Z_\l^t\times \T ({G_S^T}^{ab}),$$ where $t=\sum_{\p \in S}[k_\p:\Q_\l]-\rk_{\Z_\l}\overline{\Theta_S(E_S^T)}=\sum_{\p \in S}[k_\p:\Q_\l]-\rk_{\Z_\l}\L_S(\overline{\Theta_S(E_S^T)})$. \end{Theorem} \begin{proof} We want to determine $t$. By Class field theory ${k_S^T}^{ab}$ is associated to the following sub-group of ideles: $$k^\times\prod_{\p \notin S\cup T}U_{\p}\prod_{\p\in T} k_\p^\times.$$ Remark that archimedean places play no role for the determination of $t$. Let $n\geq 1$ be an integer and consider the following exact sequence: \begin{math} \begin{array}{rl} \displaystyle{1\longrightarrow \frac{k^\times \prod_{\p\notin S\cup T}U_p\prod_{\p\in S}U_\p^1\prod_{\p\in T}k_\p^\times}{k^\times \prod_{\p \notin S\cup T}U_\p \prod_{\p \in S}U_\p^n \prod_{\p\in T}k_\p^\times}}&\displaystyle{\longrightarrow \frac{J_k}{k^\times \prod_{\p \notin S\cup T}U_\p \prod_{\p \in S}U_\p^n \prod_{\p\in T}k_\p^\times}}\\ &\displaystyle{\longrightarrow \frac{J_k}{k^\times \prod_{\p\notin S\cup T}U_p\prod_{\p\in S}U_\p^1\prod_{\p\in T}k_\p^\times}\longrightarrow 1} \end{array} \end{math} Now $\displaystyle{\frac{J_k}{k^\times \prod_{\p\notin S\cup T}U_p\prod_{\p\in S}U_\p^1\prod_{\p\in T}k_\p^\times}}$ corresponds to the maximal abelian extension ${k_S^T}^{tame}$ of $k$, unramified outside $S$, tamely ramified for all places of $S$ and totally decomposed for places of $T$; it is finite and does not depend on $n$. Now ${k_S^T}^{ab}$ is the inductive limit of extensions $k_n$ of $k$ associated to : $k^\times \prod_{\p \notin S\cup T}U_\p \prod_{\p \in S}U_\p^n \prod_{\p\in T}k_\p^\times$. So we have: $$1\longrightarrow \lim_{\stackrel{\leftarrow}{n}}A_n\longrightarrow \Gal({k_S^T}^{ab}/k) \longrightarrow \Gal({k_S^T}^{tame}/k)\longrightarrow 1,$$ where $A_n=\displaystyle{\frac{k^\times \prod_{\p\notin S\cup T}U_p\prod_{\p\in S}U_\p^1\prod_{\p\in T}k_\p^\times}{k^\times \prod_{\p \notin S\cup T}U_\p \prod_{\p \in S}U_\p^n \prod_{\p\in T}k_\p^\times}}$, and so $\displaystyle{t=\rk_{\Z_\l} \lim_{\stackrel{\leftarrow}{n}}A_n}$. It is easy to see that we have the following continuous isomorphism of groups: \begin{eqnarray} A_n\simeq \frac{\prod_{\p \in S}U_\p^1}{\Theta_S(E_S^T)\prod_{\p \in S}U_p^n}, \end{eqnarray} where restrictions are compatible. So $$\lim_{\stackrel{\leftarrow}{n}}A_n \simeq \lim_{\stackrel{\leftarrow}{n}}\frac{\prod_{\p \in S}U_\p^1}{\Theta_S(E_S^T)\prod_{\p \in S}U_p^n}.$$ Now for each integer $n$, $\Theta_S(E_S^T)\prod_{\p \in S}U_\p^n$ is of finite index in $\prod_{\p \in s}U_\p^1$, which is a pro-$\l$-group, and so these sub-groups are closed and we obtain (see for example \cite{se2} lemma 1 page 3) $$\lim_{\stackrel{\leftarrow}{n}}\frac{\prod_{\p \in S}U_\p^1}{\Theta_S(E_S^T)\prod_{\p \in S}U_p^n}\simeq \frac{\prod_{\p\in S}U_\p^1}{\bigcap_n (\Theta_S(E_S^T)\prod_{\p \in S}U_\p^n)}.$$ The conclusion comes from the observation that $$\bigcap_n \Theta_S(E_S^T)\prod_{\p \in S}U_\p^n=\overline{\Theta_S(E_S)}.$$ \end{proof} \subsection{On the rank and relations of $G_S$} We recall the following result: \begin{Proposition}\label{gen_rel} Let $k$ be a number field, $\l$ be a prime number and $S\subset S_\l$. Let $k_S$ be the maximal $\l$-extension of $k$ unramified outside $S$, $G_S=\Gal(k_S/k)$, and let $d(G_S)$ be the $\l$-rank of $G_S$ and $r(G_S)$ be the minimal number of relations of $G_S$. Then \begin{enumerate} \item $\displaystyle{d(G_S)=d_\l \cha_S + \sum_{\p \in S} U_\p^1 - d_\l E_k}.$ \item $$r(G_S)-d(G_S) \leq d_\l E_k -\delta_k+\delta_S -\sum_{\p \in S} [k_\p:\Q_\l],$$ where $\delta_S=1$ if $k$ contains $\mu_\l$ and $S$ is empty, $0$ otherwise. \item There exists the following surjective homomorphism: $$\cha_S \twoheadrightarrow \left(\ker (H^2(G_S,\F_\l)\rightarrow \oplus_{\p\in S^*} H^2(\G_\p,\F_\l))\right)^*,$$ where $S^*$ is $S$ minus one place if $k$ contains $\mu_\l$. \item $r(G_{S_\l})-d(G_{S_\l})=-(r_2+1)$. \end{enumerate} \end{Proposition} \begin{proof} cf. \cite{ha}, \cite{ko1}, \cite{ko2}. For $\l=2$ in the point 4), see \cite{sch}. \end{proof} \subsection{First examples} \begin{Theorem} 1) The group $G_S^{ab}$ is finite in the following cases: \begin{enumerate} \item[(i)] $k$ is totally real, $S$ does not contain all places above $\l$ and Leopoldt's Conjecture is true for $k$ and $\l$. \item[(ii)] $k$ is not an imaginary quadratic field, $\l$ splits in $k/\Q$ and $S=\{\p\}$, where $\p$ is a prime of $k$ above $\l$. \end{enumerate} 2) Let $k$ be a totally imaginary field of degree $n$ over $\Q$; $n=2r_2$. Let $\l$ be a prime which splits completely in $k/\Q$ : $\l \o_k =\p_1\cdots \p_n$. Put $S=\{\p_1,\cdots, \p_i\}$. If $i\geq r_2$, $G_S^{ab}$ is infinite. 3) Consider $k=\Q(\sqrt{17})$, $S=\{\p\}$, where $\p$ is a prime above $2$. Then $G_S$ is trivial. \end{Theorem} \begin{proof} 1 i) As $k$ is totally real, $k$ has exactly one $\Z_\l$-extension (by Leopoldt's Conjecture): it is the cyclotomic extension $k^\infty$ of $k$. In this extension, all places above $\l$ are infinitely ramified. So $k_S\cap k^\infty$ is finite over $k$, and the same is for $G_S^{ab}$. 1 ii) We use Theorem \ref{rang} to note that the abelian part is finite. 2) The number of $\Z_\l$-extensions contained in $G_S^{ab}$ is $i-r$, where $r\leq r_2-1$. So here $i-r\geq 1$. 3) By computation, we find $d(G_S)=0$. \end{proof} \section{On the group $G_S^{ab}$} Before to precise the situation, we recall some facts about linear representations of finite groups. \subsection{Linear representations} Let $G$ be a finite group and let $M$ be a $\Q[G]$-module. We will identify the following modules: \begin{enumerate} \item the $\Q[G]$-module $M$ \item the $\Q_\l[G]$ module $\Q_\l\otimes_{\Q}M$ \item the $\C_\l[G]$-module $\C_\l\otimes_{\Q}M$. \end{enumerate} Let $\chi$ be a $\Q_\l$-irreducible character of $G$. One knows that $\chi$ is equal to a sum of $\C_\l$-irreducible characters $\psi$. More precisely recall the definition of the Schur index $m_\psi$ of a $\C_\l$-irreducible character $\psi$ : $m_\psi=\min_k[k:\Q_\l(\psi)]$, the minimum being over all fields $k$ realizing $\psi$ and where $\Q_\l(\psi)$ is the extension of $\Q_\l$ generated by the values of $\psi$. We know some properties of $m_\psi$, e.g. $m_\psi$ divides the degree $n_\psi$ of $\psi$. For more details see \cite{cr} or \cite{se3}. So if $\chi$ is a $\Q_\l$-irreducible character, one has $$\chi=m_\chi\left(\psi_1 +\cdots +\psi_n \right)$$ where the characters $\psi_i$ are $\C_\l$-irreducible conjugates (with the same degree $n_\chi$), and moreover $m_\chi =m_{\psi_1}=\cdots=m_{\psi_n}$ and where finally $n=[\Q_\l(\chi):\Q_\l]$. The character $\chi$ is of degree $m_\chi n_\chi[\Q_\l(\chi):\Q_\l]$. If $G$ is abelian then $m_\psi n_\psi=1$, for all $\C_\l$-irreducible characters $\psi$. \begin{Definition} For a finite group $G$, $\Irr(\C_\l)$ denotes the set of $\C_\l$-irreducible characters of $G$. \end{Definition} \subsubsection{Regular representation} The regular representation of $G$ is the representation of $G$ acting on itself by translation. The character $\Reg$ of this representation is known: $$\Reg=\sum_{\psi \in \Irr(\C_\l)}n_\psi \psi=\Ind_{\{1\}}^G\1,$$ where $\1$ is the character of the unit (or trivial) representation of $G$ (Here $\Ind_{\{1\}}^G\1$ means the character of the induced representation from $\{1\}$ to $G$ of the trivial representation). Now if $H$ is a subgroup of $G$, then $\Ind_{H}^G\1$ is contained in $\Reg$: this can be seen by transitivity formula of induction: \begin{center} \begin{math} \begin{array}{rcl} \Ind_{\{1\}}^G\1 &=&\Ind_H^G(\Ind_{\{1\}}^H\1)\\ &=&\Ind_H^G(\Reg(H))\\ &=&\Ind_H^G(\1+\cdots)\\ &=& \Ind_H^G(\1)+ \cdots \end{array} \end{math} \end{center} To conclude this part, let us give : \begin{Lemma}\label{egalite} Let $G$ be a finite group. Let $H$ and $H^\prime$ be two subgroups of $G$ with the same order. Let $I=\Ind_H^G\1$ and $I^\prime=\Ind_{H^\prime}^G\1$. Consider now the decomposition of $I$ and $I^\prime$ as a sum of $\C_\l$-irreducible characters $\psi$: $I=\sum_\psi a_\psi \psi,$ and $I^\prime=\sum_\psi a_\psi^\prime \psi.$ As $I$ and $I^\prime$ appear in $\Reg$, one has: $a_\psi, a_\psi^\prime \leq n_\psi.$ Then \begin{enumerate} \item $H \lhd G$ if and only if $a_\psi=n_\psi$, $\forall \psi \in \Irr(\C_\l)$; \item If $H \lhd G$, then $I=I^\prime$ if and only if $H=H^\prime$. \end{enumerate} \end{Lemma} \begin{proof} Let us calculate the scalar product $$ in $G$: To do this we use \cite{se3}. Let $S$ be a representing system on the double coset $H/G\backslash H^\prime$: $$G=\bigcup_{s\in S}^\cdot H s H^\prime.$$ For $s\in S$, denote ${H_s}^\prime=s{H^\prime}s^{-1} \cap H$, and consider the following representation $\1_s$ of ${H_s}^\prime$: $$\forall x \in {H_s}^\prime, \ \1_s(x)=\1_{H^\prime}(s^{-1}xs),$$ where $\1_{H^\prime}$ is the unit representation of $H^\prime$. One has in fact $\1_s=\1_{{H_s}^\prime}$. We get: \begin{eqnarray} \Res_H(\Ind_{H^\prime}^G\1)=\sum_{s\in S}\Ind_{{H_s}^\prime}^H(\1_s) \end{eqnarray} where $\Res_H$ is the restriction to $H$. By using the Frobenius formula, one has: \begin{center} \begin{math} \begin{array}{rcl} &=&<\1_H,\Res_H(\Ind_{H^\prime}^G\1)>_H\\ &=& <\1_H, \sum_{s\in S}\Ind_{{H_s}^\prime}^H(\1_s)>_H\\ &=& \sum_{s\in S} <\1_H, \Ind_{{H_s}^\prime}^H(\1_s)>_H\\ &=& \sum_{s\in S}<\Res_{{H_s}^\prime}(\1_H),\1_s>_{{H_s}^\prime}\\ &=& |S| \end{array} \end{math} \end{center} By a similary computation: $=|H/G\backslash H|$. We can now prove the first point: 1) One knows that $H \lhd G$ if and only if $|G/H|=|H/G\backslash H|$. But $|G/H|$ is of order $\deg(I)$, i.e $$|G/H|=\sum_\psi a_\psi n_\psi.$$ But we have also the identity: $$|H/G\backslash H|==\sum_\psi a_\psi^2.$$ The first result follows. 2) By looking at the scalar products $$, $$ and $$, one obtains $I=I^\prime$ if and only if $$\sum_\psi a_\psi^2=\sum_\psi a_\psi a_\psi^\prime=\sum_\psi {(a_\psi^\prime)}^2=|S|.$$ It suffices then to show that for $H\not= H^\prime$, $|S|\not=\sum_\psi a_\psi^2$. Suppose $H\not= H^\prime$. There exists $h^\prime \in H$ and $\notin H$; a representing system of $G/H$ could be written $\{1, h^\prime, \cdots \}$. Now two elements of $G$ which are in the same class in $G/H$ are in the same class in $H/G\backslash H^\prime$. Moreover $1$ and $h^\prime$ are also in the same class in $H/G\backslash H^\prime$. So $|S|\leq |G|/|H|-1$. To conclude our proof it suffices to note that (by using the fact that $H \lhd G$) $$\sum_\psi a_\psi^2 =\sum_\psi a_\psi n_\psi=\deg(I)=|G|/|H|.$$ \end{proof} \subsection{Presentation of the situation} Let $N$ be a number field and let $k/N$ be a finite Galois extension with group $G$. Let $S_N$ be a set of places of $N$ above $\l$ and let $S$ be the set of places of $k$ above $S_N$: $S=\{\P \subset \o_k, \P\cap \o_N \in S_N\}$. Then $G$ acts on $U_S^1:=\prod_{\p\in S} U_\p^1$ and, $\Theta_S$ and $\L_S\circ\Theta_S$ are homomorphisms of $\Q_\l[G]$-modules. With these facts, we can give some informations on the character of $\Theta_S(E_S)$ and so on the $\Z_\l$-rank of $G_S^{ab}$. \begin{Definition} Denote by $\chi_{S}$ the character of ${\Theta_S(E_S)}$. \end{Definition} \subsubsection{Character of units} Let $\p_{\infty_1}, \cdots,\p_{\infty_n}$ be the archimedean places of $N$; $n={r_1(N)+r_2(N)}$. We recall Dirichlet's Theorem for representations (whose proof can be found in \cite{gr2}): \begin{Proposition}\label{unites} Let $k/N$ be a finite Galois extension of number fields with group $G$. Let $T$ and $S$ be two finite sets of primes of $k$ on which $G$ acts respectively. The group of $T$-units of $k$ tensored by $\Q$ is a $\Q[G]$-module with character: $$\chi( E_k^T )= \chi(E_S^T)=\sum_{i=1,\cdots, n} \Ind_{D_{\p_{\infty_i}}}^G(\1) +\sum_{\p\in T}\Ind_{D_\p}^G\1 -\1 ,$$ where $D_\p$ is the decomposition group of $\p$ in $k/N$. \end{Proposition} \subsubsection{Character of $U_S^1$} \begin{Lemma} $$\chi( U_S^1 )= \sum_{\p \in S_N}[N_\p:\Q_\l]\Reg.$$ \end{Lemma} \begin{proof} Fix $\p \in S_N$. Let $\P|\p$. Note first that for $i\geq 1$ the two modules $\Q_\l\otimes_{\Z_\l} U_{k_\P}^1$ and $\Q_\l\otimes_{\Z_\l} U_{k_\P}^i$ are isomorphic because $U_{k_\P}^1/U_{k_\P}^i$ is finite. Moreover for large $i$, one has the isomorphsim of $\Z_\l[G]$-modules via the logarithm : $U_{k_\P}^i \simeq \o_{k_\P}$. Thus for large $i$, we get: $$\prod_{\P \in S} U_\P^i\simeq\prod_{\P \in S} \o_{k_\P}.$$ One then has the following isomorphisms of $G$-modules: \begin{math} \begin{array}{rcl} \Q_\l\otimes_{\Z_\l}\prod_{\P|\p}U_{k_\P}^1&\simeq & \Q_\l \otimes\prod_{\P|\p} U_{k_\P}^i \\ &\simeq &\Q_\l\otimes_{\Z_\l}\prod_{\P|\p}\o_{k_\P}\\ &\simeq&\prod_{\P|\p}\Q_\l\otimes_{\Z_\l}k_\P\\ &\simeq &\prod_{\P|\p}k_\P\\ &= &N_\p\otimes_{N} k\\ &\simeq & N_\p\otimes_N N[G]\\ &\simeq & N_\p[G]\\ &\simeq & \Q_\l[G]\oplus \cdots \otimes \Q_\l[G]=[N_\p:\Q_\l]\Q_\l[G]. \end{array} \end{math} \end{proof} \subsubsection{Upper bound for $\chi_S$} It comes immediately the following: \begin{Proposition}\label{majoration} One has for the character $\chi_S$ of the $\Q_\l[G]$-module $\Q_\l\otimes_{\Z_\l} \Theta_S(E_S)$: $$\chi_S \leq \sum_{\p \in S_N}[N_\p:\Q_\l]\Reg\wedge \left(\sum_{i=1,\cdots, r(N)} \Ind_{D_{\infty_i}}^G(\1) -\1\right).$$ \end{Proposition} \subsection{Lower bound for $\chi_S$ and Schanuel $\l$-adic Conjecture} \subsubsection{Conjectural results} The $\l$-adic independance conjecture of Schanuel is at the center of the problem: {\em Schanuel $\l$-adic Conjecture.} {\it Let $x_1,\cdots,x_r \in \overline{\Q_\l}$ be a family of algebraic numbers, such that the family ${\log(x_i)}_{i=1,\cdots,r}$ is independent over $\Q$. Then the transcendance degree over $\Q$ of $\Q(\log_\l(x_1),\cdots,\log_\l(x_r))$ is $r$.} Under this Conjecture we can prove the following result which can be found, for example, in \cite{ja}: \begin{Proposition}\label{scha}Let $M_0$ be a $\Q[G]$-module of finite type included in $k_S^1$. Let $m=\dim_{\Q_\l}(\Reg\wedge \chi(M_0))$. With Schanuel's Conjecture one has: $$\rk_{\Z_\l}(\L_S(\Theta_S(M_0)))\geq m.$$ \end{Proposition} \begin{rem} Proposition \ref{scha} shows that Schanuel's Conjecture implies Leopoldt's Conjecture. \end{rem} To prove the Proposition \ref{scha} we need a Lemma: \begin{Lemma}[\cite{ja}]\label{restriction}Let $\varepsilon \in M_0$. There exists a $\Q[G]$-submodule $M$ of $k_S^1$ with $\chi(M)=\Reg$ and such that the principal $\Q[G]$-module $\Q[G]\cdot \varepsilon$ is a submodule of $M$. \end{Lemma} \begin{proof} Let $\q$ be a prime of $N$ which splits completely in $k/N$. Take $T$ to be the set of primes of $k$ above $\q$ and consider $E_S^T$. The character of $\Q\otimes E_S^T$ contains $\Reg$. Let $M$ be the maximal principal submodule of $\Q\otimes (E_S^T +M_0)$ containing $\varepsilon$. Then the character of $M$ is $\chi(E_S^T +M_0)\wedge \Reg=\Reg$. \end{proof} \begin{proof}(Proposition \ref{scha}) It suffices to prove it for $S_N=\{\p\}$. Let $S=\{ \P_1,\cdots, \P_s\}$ be the primes of $k$ above $\p$. Recall that a $\Q[G]$-module is principal if and only if its character appears in $\Reg$. Let $\chi=\Reg\wedge \chi(M_0)$. It is the character of a $\Q[G]$-principal module $M_1$, thus there exists $\varepsilon \in M_0$ such that $M_1=\Q[G]\cdot \varepsilon$. By Lemma \ref{restriction}, there exists a $\Q[G]$-module $M$ containing $M_1$ and with representation $\Reg$. We will show that $\overline{\Theta_S(M)}$ has maximal $\Q_\l$-rank $|G|$, and this would imply that $\overline{\Theta_S(M_1)}$ has $\Q_\l$-rank $m$. Let $(x_1,\cdots,x_n)$ be a $\Q$-basis of $M$ such that $x_i=x^{g_i}$, with $g_i\in G$. Suppose the existence of a linear relation on $\Z_\l$ between the images of $x_i$ by $\L_S\circ \Theta_S$. We may write such a linear relation by: $$\sum_{i=1,\cdots,m}a_i\L_S(\Theta_S(x_i))=0,$$ with $a_i\in \Z_\l$. Then one has: $$\forall \sigma_j, \sum_{i=1,\cdots,m}a_i\log_\l(\sigma_j(x_i))=0.$$ Therefore we obtain: \begin{Lemma}\label{lemme1} $$\forall g\in G, \sum_{i=1,\cdots,m}a_i\log_\l(\sigma_1({x_i}^g))=0.$$ \end{Lemma} \begin{proof} Let $g \in G$. If $g$ is in the decomposition group of $\P_1$ in $N/k$, then there exists $g^\prime \in \Gal(k_{\P_1}/N_\p)$ such that $\sigma_1({y}^g)={\sigma_1(y)}^{g^\prime}, \forall y \in k$. Then \begin{center} \begin{math} \begin{array}{rcl} \sum_i a_i\log_\l(\sigma_1({x_i}^g))&=&\sum_i a_i\log_\l(\sigma_1({x_i}))^{g^\prime}\\ &=& \sum_i a_i (\log_\l(\sigma_1({x_i})))^{g^\prime}\\ &=& (\sum_i a_i (\log_\l(\sigma_1({x_i})))^{g^\prime}\\ &=& 0 \end{array} \end{math} \end{center} Now as $|S_N|=1$, $G$ acts transitively on $S$. Consequently $\sigma_1(y^g)=g^\prime\cdot \sigma_{i(g)}(y)$ where $g^\prime$ is in $\Gal(k_{\P_{i(g)}}/N_\p)$, and we conclude our proof as before. \end{proof} Now consider the square matrix $\M$ of size $|G|\times |G|$: ${(\log(\sigma_1({x_i}^g)))}_{g\in G,i=1,\cdots,|G|}$. The determinant of this matrix is zero (by Lemma \ref{lemme1}). It is a polynomial on $\l$-adic logarithms of independent algebraic numbers. If we show that this polynomial is not the zero polynomial then we obtain a contradiction to Schanuel's Conjecture. \begin{Lemma} Put $X_g=x^g$, for $g\in G$. Let $\M^\prime$ be the matrix ${(X_{g\cdot g^\prime})}_{g,g^\prime \in G}$ and let $P({(X_g)}_{g\in G})$ be the determinant of this matrix. One has: $\det(\M)=P({(x^g)}_g)$. Then $P$ is not the zero polynomial. \end{Lemma} We can assume that $\M^\prime$ can be written: \begin{center} \begin{math} \left( \begin{array}{cccc} X_1 & X_\sigma &\cdots &X_\tau\\ X_{\sigma^{-1}}&X_1 &\cdots&X_{\sigma^{-1}\tau}\\ \cdots &\cdots&\cdots&\cdots\\ X_{\tau^{-1}}&\cdots&\cdots&X_1 \end{array} \right) \end{math} \end{center} Then the determinant of $\M$ is a monic polynomial in $\X_1$ of degree $|G|$. \end{proof} \subsubsection{Unconditionnal results} First recall the $\l$-adic version of Baker's Theorem proved by Brumer \cite{br} on the independance of logarithms: \begin{Proposition} Let ${(\alpha_i)}_{i=1,\cdots,r}$ be a family of non zero algebraic numbers such that there are linearely multiplicatively independent. Look at these elements in $\overline{\Q_\l}$. Then the family ${(\log_\l \alpha_i)}_{i=1,\cdots,r}$ is $\overline{\Q}$-independent. \end{Proposition} As noted in \cite{ja}, we have: \begin{Proposition}\label{jaulent} For all noetherian $\Z[G]$-module $M$ of $k_S^1$, the character of the $\Q_\l[G]$-module $\Q_\l\otimes_{\Z} \Theta_S(M)$ contains all $\Q_\l$-irreducible character $\chi$ represented in $\chi(M)$. More precisely: \begin{center} \begin{math} \begin{array}{rcl} \chi(\Theta_S (M))&\geq & \displaystyle{\sum_{\chi \leq \chi(M)}} \chi\\ &\geq &\displaystyle{ \sum_{\stackrel{\psi \leq \chi(M)}{\psi \in \Irr (\C_\l)}} m_\psi \psi} \end{array} \end{math} \end{center} \end{Proposition} \begin{proof} Let $M_0$ be a principal $\Q[G]$-sub module of $M$ with character $\chi(M_0)$: $M_0=\Q_\l[G]\cdot x$, with $x\in M_0$. Let $\chi$ be a $\Q_\l$-irreducible character dividing $\chi(M_0)$. The values of $\chi$ are in $\bar{\Q}$. One has \begin{eqnarray}\label{nonnul}\sum_{g\in G} \chi(g^{-1})g\cdot x\not=1.\end{eqnarray} Let $(x_1,\cdots,x_r)$ be a $\Q$-basis of $M_0$. Look at the projection on $\chi$ of the image of $x$ by $\L_S(\Theta_S)$: If it is trivial, it implies that $$\ \L_S(\Theta_S(\sum_{g\in G}\chi(g^{-1})g\cdot x))=0.$$ Look at this vector equality on the first component to obtain: $$ \log_\l(\sigma_1(\sum_{g\in G}\chi(g^{-1})g\cdot x))=0.$$ Now we write $g\cdot x$ in the $\Q$-basis $(x_1,\cdots,x_r)$ to obtain $$\sum_{i=1,\cdots,r} a_i \log_\l(\sigma_1(x_i))=0,$$ where $a_i$ are algebraic over $\Q$. Moreover this relation is not trivial by (\ref{nonnul}). Then Theorem of Baker-Brumer yields a contradiction. So the character $\chi$ appears in $\chi_S$.\end{proof} To finish, in our case we can apply a $\l$-adic version of Theorem 1.16 of Chapter 1 of \cite{wal} (I thank M. Waldschmidt for telling me the existence of this version): \begin{Proposition}\label{wal} Let ${(x_{i,j})}_{\stackrel{i=1,\cdots, d}{j=1,\cdots, l}}$ be a family of algebraic numbers. Let $l(\cdot)=\L_S (\sigma_1(\cdot))$ be the logarithmic on $\overline{\Q_\l}$. Assume that for all ${(t_i)}_{i=1,\cdots,d} \in \Z^d\backslash \{0\}$ and ${(s_i)}_{i=1,\cdots,l} \in \Z^l\backslash \{0\}$ we have: $$\sum_{i,j} t_i s_j x_{i,j} \not= 0.$$ Then the rank of the matrix ${(l(x_{i,j})))}_{\stackrel{i=1,\cdots, d}{j=1,\cdots, l}}$ is at least $\displaystyle{\frac{l d}{l+d}}$. \end{Proposition} As consequence of Proposition \ref{wal}, we have: \begin{Theorem} \label{wal1} Let $M$ be a $\Q[G]$-module of finite type included in $k_S^1$. Let $\chi$ be a $\Q$-character of $\Q[G]$-module such that $\chi \leq \Reg$. If $m \cdot \chi \leq \chi(M)$, then $$\rk_{\Z_\l} \L_S(\Theta_S(M))\geq \frac{m \cdot n_\chi}{m + n_\chi}.$$ \end{Theorem} \begin{proof} As $m\cdot \chi \leq Reg$, there exists $x_1,\cdots, x_m$ in $M$ such that: \begin{enumerate} \item $\chi(\Q[G]\cdot x_i)=\chi$; \item $\Q[G]\cdot x_i=\Q< x_i^{(1)},\cdots, x_i^{(d)}>$, where $d=n_\chi$. \item The elements $x_i^{(j)}$ can be chosen as to be the Galois conjugates of $x_i$: There exists $s_1,\cdots,s_d \in G$ such that $$\forall i,j \ ,\ x_i^{(j)}=s_j\cdot x_i={x_i}^{s_j}.$$ \item The family ${(x_i^{(j)})}_{\stackrel{i=1,\cdots,m}{j=1,\cdots,d}}$ is linearly $\Q$-independants. \end{enumerate} Put $X_i=\L_S(\Theta_S(x_i))$. We will prove that the $\Q_\l$-rank of $\{ X_1,\cdots, X_m\}$ is at least $r=\displaystyle{\frac{m \cdot n_\chi}{m + n_\chi}}$. Suppose that the rank is less than $r-1$. That means that for all sub-families $\{X_{i_1},\cdots, X_{i_s}\}$ with $s\leq r-1$, there exists ${(a_j)}_j \in \Q_\l^{r-1}$ such that $$\sum_{j=1}^{r-1} a_j X_{i_j}=0.$$ Using the same method as for the proof of Lemma \ref{lemme1}, one obtains: \begin{eqnarray}\label{rrr}\forall g\in G,\ \sum_{j=1}^{r-1} a_j\log_\l (\sigma_1(g\cdot x_{i_j}))=0.\end{eqnarray} Now look at the matrix $\widetilde{\M}$ of size $d\times m$ defined by: $\widetilde{\M}_{j,i}={(\log_\l(\sigma_1(x_i^{(j)})))}_{\stackrel{j=1,\cdots,d}{i=1,\cdots,m}}$: its rank is at least $r$ by Proposition \ref{wal}. Moreover $\widetilde{\M}$ is a sub-matrix of ${\M}= {(\log_\l(\sigma_1(g\cdot x_i)))}_{\stackrel{g\in G}{i=1,\cdots,d}}$ (of size $|G|\times m$) with rank at most $r-1$ by (\ref{rrr}). And so we obtain a contradiction. \end{proof} \section{The degree one case} In all this section, $k/N$ is a finite Galois extension and $\p$ a prime of $N$ of degree 1. We recall: \begin{Definition} Let $k/N$ be a Galois extension of number fields with group $G$ and $\p \subset \o_k$. Define $S_{\bar{\p}}=\{ \p^g, g\in G\}$. \end{Definition} We first will show what Schanuel Conjecture, Baker-Brumer Theorem and Theorem \ref{wal1} imply for the ${\Z_\l}$-rank of $ G_{S_{\bar{\p}}}^{ab}$. After this, we will give some examples (\S 3.3 and 3.4) and conclude by the proof of Theorem \ref{sup} (\S 3.5). \subsection{With Schanuel's Conjecture} \begin{Proposition}\label{schanuel} Let $k/N$ be a Galois extension of number fields with group $G$, $\p$ be a prime of $N$ of degree $1$. Let $M_0$ be a $\Q[G]$-module of finite type included in $k_{S_{\bar{\p}}}^1$. Then Schanuel's conjecture implies: $\chi(\Theta_{S_{\bar{\p}}}(M_0))=\Reg\wedge \chi(M_0)$. \end{Proposition} \begin{proof} Consequence of Proposition \ref{scha}. \end{proof} \begin{Theorem} Let $k/N$ be a Galois extension of number fields, $\p$ be a prime of $N$ of degree $1$. Assume $[N:\Q]>2$ and $r_2(N)>0$. Then Schanuel's Conjecture implies that $G_{S_{\bar{\p}}}^{ab}$ is finite. \end{Theorem} \begin{proof} Application of Proposition \ref{schanuel} and Theorem \ref{rang}. \end{proof} \subsection{Unconditional results} \begin{Theorem}\label{corollaire1} Let $k/N$ be an {\bf abelian} extension of number fields with group $G$. Let $\l$ be a prime such that there exists $\p\in N$ above $\l$ of degree $1$. Then $$G_{S_{\bar{\p}}}^{ab}\simeq \Z_\l^t\times \T(G_{S_{\bar{\p}}}^{ab}),$$ where $t=|G|-|\psi \in \Irr(\C_\l), \psi \leq \chi(E_k)|$. In particular if $N$ is totally real, $k$ is totally complex and if all groups of decomposition of archimedean places are equals, then $t=|H|/2$. But if $[N:\Q]>2$ and if $r_2(N)>0$, then $t=0$.\end{Theorem} \begin{proof} This is a direct application of Propositions \ref{majoration} and \ref{jaulent}. \end{proof} To finish, we give some applications of Theorem \ref{wal1}: \begin{Theorem} Let $N$ be a number field $k/N$ be a finite Galois extension with group $G$. Put $n=r_1(N)+r_2(N)$. Assume that $n\geq 2$ and that all decomposition groups for archimedean places are trivial in $k/N$. Let $\p$ be a prime of $\o_N$ with degree $1$. Then $|G| < \displaystyle{\frac{1+\sqrt{3+ 4 n}}{2}}$ implies $G_{S_{\bar{\p}}}^{ab}$ finite. \end{Theorem} \begin{proof} By hypothesis, in the extension $k/N$ the character of the units contains $(n_1-1)\Reg$. By Theorem \ref{wal1}, one has: $$\rk_{\Z_\l} \L_S(\Theta_S(E_S))\geq \frac{|G| (n-1)}{|G|+n-1}.$$ Now by Theorem \ref{rang}, the $\Z_\l$-rank of $G_{S_{\bar{\p}}}^{ab}$ is trivial if and only if $$ |G|-\rk_{\Z_\l} \L_S(\Theta_S(E_S))<1.$$ And in particular this is the case, if $|G| < \displaystyle{\frac{1+\sqrt{3+ 4 n}}{2}}$. \end{proof} \subsection{Compositum of $\Z_\l$-extensions} \begin{Definition} Let $K/k$ be a pro-$\l$-extension of a number field $k$. We denote by $\widetilde{K}$ the compositum of $\Z_\l$-extensions of $k$ contained in $K$ and by $\widetilde{\Gal(K/k)}$ the Galois group $\Gal(\widetilde{K}/k)$. In particular $\widetilde{\Gal(K/k)}\simeq \Z_\l^t$. \end{Definition} We have immediately these properties: \begin{Proposition}\label{xxx} 1) If $S \cap S^\prime =\emptyset$, then $\rk_{\Z_\l}\widetilde{\Gal(k_{S} k_{S^\prime}/k)}=\rk_{\Z_\l}\widetilde{\Gal(k_S/k)}+\rk_{\Z_\l}\widetilde{\Gal(k_{S^\prime}/k)}$. 2) If $S\not= S_\l$, then $\rk_{\Z_\l}\widetilde{\Gal(k^\infty k_S/k)}=1+\rk_{\Z_\l} \widetilde{\Gal(k_S/k)}$. \end{Proposition} We can present: \begin{Theorem} \label{main theorem}Let $k/N$ be a Galois extension of number fields with group $G$. Let $\l$ be a prime number totally decomposed in $N/\Q$. Assume that Schanuel Conjecture is true. Then $\widetilde{k_{S_\l}}=k^\infty \cdot \prod_{\p \in S_\l}\widetilde{k_{S_{\bar{\p}}}}$ if and only if: (i) $k$ is totally real. Or, (ii) $N$ is totally real, $k$ is totally imaginary and for all $i,j$, $\Ind_{D_{\p_{\infty_i}}}^G(\1)=\Ind_{D_{\p_{\infty_j}}}^G(\1)$. Moreover if for every $\C_\l$-irreducible characters $\psi$ dividing $\Ind_{D_{\p_{\infty}}}^G(\1)$, $\psi$ divises it $\deg(\psi)$ times (which is equivalent to $D_{\p_\infty}$ is distinguished in $G$ for an archimedean place $\p_\infty$ cf. lemma \ref{egalite}) then $k$ is a CM-field. Or, (iii) $N$ is a imaginary quadratic field and $[k:N]=2$. \end{Theorem} \begin{proof} By Leopoldt's Conjecture (which is a consequence of Schanuel's Conjecture) the $\Z_\l$-rank of $G_{S_\l}^{ab}$ is $r_2+1$. By Schanuel's Conjecture (Proposition \ref{schanuel}), Proposition \ref{xxx} and Theorem \ref{rang}, the number of $\Z_\l$-extensions coming from the compositum of the fields $k_{S_{\bar{\p}}}$ with the cyclotomic extension $k^\infty$, is: $$r_1+2r_2-[N:\Q]\cdot\dim_{\Q_\l}\left((-\1+\sum_{i}\Ind_{D_{\p_{\infty_i}}}^G(\1))\wedge\Reg\right)+1.$$ So we want: \begin{eqnarray}\label{11}r_1+ r_2=[N:\Q]\cdot\dim_{\Q_\l}\left((-\1+\sum_{i}\Ind_{D_{\p_{\infty_i}}}^G(\1))\wedge\Reg\right).\end{eqnarray} 1) The first case is trivial. 2) Let $(r_1,r_2)$ be the signature of $k$. Assume that $r_2>0$ and that $r_1(N)+r_2(N)\geq 2$. That means in particular that $N$ is not an imaginary quadratic field. Suppose moreover that at least one archimedean place $\p_{\infty_i}$ of $N$ splits in $k/N$. As $\Ind_{D_{\p_\infty}}^G(\1)$ contains the unit representation $\1$ and that $r_1(N)+r_2(N)\geq 2$, we have $$\dim_{\Q_\l}\left((-\1+\sum\Ind_{D_{\p_{\infty_i}}}^G(\1))\wedge\Reg\right)=|G|.$$ So (\ref{11}) becomes: \begin{center} \begin{math} \begin{array}{rcl} r_1+r_2&=&[N:\Q]\cdot |G|\\ &=&[k:\Q]=r_1+2r_2 \end{array} \end{math} \end{center} and this is a contradiction. So $N$ is totally real and $k$ totally complex. The equality (\ref{11}) now becomes: \begin{center} \begin{math} \begin{array}{rcl} \displaystyle{\frac{|G|\cdot [N:\Q]}{2}}&=&r_2\\ &=&[N:\Q]\cdot\dim_{\Q_\l}\left(\chi(E_S)\wedge\Reg\right). \end{array} \end{math} \end{center} Now, as $N$ has at least two archimedean places, $$(-\1+\sum\Ind_{D_{\p_{\infty_i}}}^G(\1))\wedge\Reg=(\sum\Ind_{D_{\p_{\infty_i}}}^G(\1))\wedge\Reg.$$ Recall that $\Ind_{D_{\infty_i}}^G(\1))$ is in $ \Reg$ and is of dimension $|G|/2$. Suppose that there exist $i$ and $j$, such that $$\Ind_{D_{\p_{\infty_i}}}^G(\1))\not=\Ind_{D_{\p_{\infty_j}}}^G(\1);$$ One would have $$\dim_{\Q_\l}\left((\sum\Ind_{D_{\p_{\infty_i}}}^G(\1))\wedge\Reg\right) >|G|/2,$$ and so a contradiction. If all irreducible characters $\psi$ dividing $\Ind_{D_{\p_\infty}}^G (\1)$, occur with multiplicity $\deg(\psi)$, by Lemma \ref{egalite}, one has $D_{\p_{\infty_i}}=D_{\p_{\infty_j}}$ $\forall i,j$. And so $k$ is a CM field with $k^+=k^{D_{\p_{\infty_i}}}$, for one $i$. 3) Now if $N$ an imaginary quadratic field, the equality (\ref{11}) becomes $$r_2=[N:\Q](|G|-1),$$ and the last case occurs. \end{proof} \subsection{Behavior of the $\Z_\l$-rank of $G_S^{ab}$ in a $\Z_\l$-extension} We will look at the following situation. Let us fixe $k$ be a number field such that $r_1+r_2 \geq 2$, and let $K$ be a $\Z_\l$-extension of $k$; $\Gamma=\Gal(K/k)$. Let $S\subset S_\l$ be a set of places of $k$. Denote by $k_S^{ab}$ the maximal abelian $\l$-extension of $k$ unramified outside $S$ ; $\X_S=G_S^{ab}=\Gal(k_S^{ab}/k)$. More generally put $\X_{n,S}=\Gal(k_{n, S_n}^{ab}/k_n)$ where $\cup_n k_n=K$ with $[k_n:k]=\l^n$, and $S_n=\{ \P \subset \o_{k_n}, \P\cap \o_k \in S\}$. Also put $\displaystyle{\X_{\infty,S}=\lim_{\stackrel{\leftarrow}{n}} \X_{n,S}}$ (the restriction map is the norm). Let $\Lambda=\Z_\l[[T]]$. We know that $\X_{\infty,S}$ is a finitely generated $\Lambda$-module pseudo-isomorphic to $$\Lambda^r \oplus \left(\oplus_i \Lambda/ \l^{n_i} \right) \oplus \left(\oplus_i \Lambda / f_i\right),$$ $f_i$ being polynomials of $\Z_\l[T]$. For more details see, for example, \cite{wa}. We will determine $r$ in three cases: \subsubsection{ }Consider the following case: $S=\{\p\}$ with $\p$ of degree 1. We look at the action of $G_n=\Gal(k_n/k)$. As in a $\Z_\l$-extension, archimedean places are decomposed, one has $\chi(E_{k_n})=(r_1+r_2)\Reg(G_n) -\1$ and so by Theorem \ref{corollaire1}, for all $n$ the $\Z_\l$-rank of $\X_{n,S}$ is trivial. The group $\X_{n,S}$ is then finite and so $r=0$. \subsubsection{ } Assume that $k$ is a CM field. Denote by $k^+$ the maximal real subfield of $k$. Take $\p$ in $k^+$ of degree 1. Take for $S$ all the places of $k$ above $\p$, and $K=k^\infty $ be the cyclotomic $\Z_\l$-extension of $k$. Then for all $n$, $\Gal(k_n/k^+)\simeq \Z/\l^n\Z \times \Z/2\Z$. By using Theorem \ref{corollaire1}, one obtains $$\X_{n,S}\simeq {\Z_\l}^{\l^n}\times \mbox{(finite)}.$$ So in this case $r=1$. \subsubsection{ } Take $\l>2$. Assume that $k$ is a CM field (non quadratic). Denote by $k^+$ the maximal real subfield of $k$. Let $\p$ in $k^+$ of degree 1, and for $S$ all the places of $k$ above $\p$. Here we have $\X_{S}\simeq \Z_\l \times \mbox{(finite)}$. Take then $K$ as being the $\Z_\l$-extension unramified outside $S$. The extensions $k_n/k^+$ are Galois (because $\rk_{\Z_\l}\X_{S}=1$) with a non abelian group (because $k^+$ is totally real and so $G_{\{\p\}}^{ab}(k^+)$ is finite) isomorphic to the diedral group $D_{2\l^n}=\langle x,y, x^{l^n}=1, y^2=1, yxy=x^{-1}\rangle$. This group has $(\l^n-1)/2+2$ $\C_\l$-irreducible representations. Two of them are of degree 1: $\1$ and $\psi_0(x)=1$ $\psi_0(y)=-1$, and $(\l^n-1)/2$ representations $\rho_h$ are of degree $2$ (for $h=1,\cdots, (\l^n-1)/2$) : \begin{center} \begin{math} \rho_h(x)=\left(\begin{array}{cc} \zeta^{-h} & 0\\ 0 &\zeta^h \end{array} \right) , \ \rho_h(y)=\left(\begin{array}{cc} 0 & 1\\ 1 &0\end{array} \right) \end{math}\end{center} where $\zeta$ is a $\l^n$-primitive root of $1$. Now we can note that $\sum_h \rho_h$ is a $\Q_\l$-irreducible character of degree $\l^n-1$ (The Schur index of $\rho_h$ is 1). For an archimedean place $\p_\infty$ of $k^+$ the decomposition group $D_{\p_\infty}=\langle x^ay\rangle$ in $\Gal(k_n/k^+)$ is of order $2$. As $\l$ is odd, the subgroups of order $2$ of $D_{2\l^n}$ are conjugated and so: $$\chi(E_{k_n})=[k^+:\Q]\left(\sum_{h} \rho_h +\1\right) -\1.$$ One then has $\chi(\Theta_{S_n}(E_{k_n}))=\1 +\sum_{h} \rho_h$ or $\chi (\Theta_{S_n}(E_{k_n}))=\1 +2\sum_{h} \rho_h$. In the first case $\X_{n,S}=\Z_\l^{\l^n}\times \mbox{(finite)}$ and in the second $\X_{n,S}=\Z_\l\times \mbox{(finite)}$. Thus $\Gal(k_{n,S_n}^{ab}/k_{\infty})\simeq \Z_\l^{\l^n-1}\times \mbox{(finite)}$ in the first case, and $ \Gal(k_{n,S_n}^{ab}/k_{\infty})$ is finite in the second case. Now we have to remark that $\Gal(k_{n,S_n}^{ab}/k_{\infty})\simeq \X_{\infty,S}/\omega_n$ where $\omega_n ={(1+T)}^{\l^n}-1$ (cf. \cite{wa} for example) and this implies: $$\rk_{\Z_\l} \Gal(k_{n,S_n}^{ab}/k_\infty)=r \cdot \l^n +a,$$ with $a\in\N$. Only the second case is possible. In conclusion we have for all $n$, $\chi (\Theta_{S_n}(E_{k_n}))=\1 +2\sum_{h} \rho_h$ and $r=0$. \begin{rem} Schanuel's Conjecture implies in the last example that $\chi (\Theta_{S_n}(E_{k_n})) = \1 +2\sum_{h} \rho_h$. \end{rem} \subsection{Proof of Theorem \ref{sup}} We recall the definition of $\iw{\l}{k}$ and $\Iw{\l}{k}$: \begin{Definition} \begin{enumerate} \item[(i)] $\iw{\l}{k}=\displaystyle{\min_{S\subset S_\l}}\{\displaystyle{\frac{|S|}{|S_\l|}}, G_S^{ab} \ \text{ is infinite}\}$; \item[(ii)] $\Iw{\l}{k}=\min c$, such that $\forall S \subset S_\l$, $\displaystyle{\frac{|S|}{|S_\l|}}\geq c \Longrightarrow $ $G_S^{ab}$ is infinite. \end{enumerate} \end{Definition} In the previous sections, we have gathered much data concerning these quantities. a) First assume $\l>2$. Consider the polynomial $f(x)= x^{\l+1}-x^2+x+\l$. We can note that all the roots of $f$ are complex. Moreover, if we look at $f$ modulo $\l$, we find \begin{eqnarray}\label{reduction}f(x)=x{(x^\l-x+1)} \ \in \F_\l[x] \end{eqnarray} which is the decomposition into irreducible polynomials in $\F_\l[x]$. And this guarantees the irreducibility of $f$ in $\Q(x)$. Let $N=\Q(\delta)$, where $\\theta$ is a root of $f$. Then $N$ is of degree $\l+1$ on $\Q$ and totally imaginary. The factorization (\ref{reduction}) also gives us the factorization of $\l\o_N$: $$\l\o_N=\p_1 \p_2,$$ where $\p_1$ is of degree 1, and $\p_2$ of degree $\l$. Put $T=\{\p_1\}$ and $S=\{\p_2\}$. Consider now, the maximal $\l$-extension $k_S^T$, $S$-ramified and $T$-decomposed of $k$; $G_S^T=\Gal(k_S^T/k)$. Then the $\Z_\l$-rank of ${G_S^T}^{ab}$ is at least (by Theorem \ref{rang}) $$[N_{\p_2}/\Q_\l]-(r_1(k)+r_2(k)-1+|T|)=\l-(\l+1)/2\geq 1.$$ Then let $K$ be a $\Z_\l$-extension of $N$ included in $k_S^T$. Denote now by $k_n$ the subfield of $K$ defined by $[k_n:N]=\l^n$. As the class group of $N$ is finite, there exists $n_0$, such that $\p_2$ is totally ramified in $K/k_{n_0}$. Now we consider $n\geq n_0$. For a field $k_n$, we have: $|T(k_n)|=\l^n$, $|S(k_n)|\leq \l^{n_0}$, and $$\l^n +1 \leq |S_\l(k_n)|=|T(k_n)|+|S(k_n)|\leq \l^n+\l^{n_0}.$$ The field $k_n$ admits $K/k_n$ as a $\Z_\l$-extension which is ramified only at $S(k_n)$, and so $$\iw{\l}{k_n}\leq \frac{\l^{n_0}}{\l^n+1}.$$ Now we apply Proposition \ref{corollaire1} in the extension $k_n/N$ with $S_N=T=\{\p_1\}$. The $\Z_\l$-rank of ${k_n^{ab}}_{S_{\bar{\p_1}}}/k_{n}$ is trivial, and so $$\Iw{\l}{k_n}> \frac{\l^n}{|S_\l(k_n)|} \geq \frac{\l^n}{\l^n+\l^{n_0}}.$$ Letting $n \rightarrow \infty$, we have the result. b) For $\l=2$ we can apply the same argument as above by taking for example $f(x)=x^4+x^2+x+2$. \subsection{Proof of Theorems \ref{imaginary} and \ref{dimension3}} We first prove: \begin{Theorem}\label{dimension2} Let $k$ be a number field (for $\l>2$, assume that $k$ is totally complex) and $S\subset S_\l$ such that: \begin{enumerate} \item For all $\p \notin S$, $\rk_{\Z_\l} {G_S^{\{\p\}}}^{ab} < \rk_{\Z_\l} G_S^{ab}$ ($\Leftrightarrow \rk_{\Z_\l} \Theta_S(E_S^{\{\p\}})>\rk_{\Z_\l} \Theta_S(E_S)$); \item For all $S^\prime \varsubsetneq S$, $\rk_{\Z_\l} G_{S}^{ab} > \rk_{\Z_\l}{G_{S^\prime}}^{ab}$. \end{enumerate} Then $\Gal(\k(\l)/\widetilde{k_S})$ is free. \end{Theorem} \begin{proof} Let $\widetilde{k_S}$ (resp. $\widetilde{k_S^T}$, for a finite set $T$ of places of $k$) be the compositum of all $\Z_\l$-extensions of $k$ contained in $k_S$ (resp. in $k_S^T$). \begin{Lemma}\label{infini2} With the conditions 1 and 2 of Theorem \ref{dimension2}, one has for all finite places $\p$ of $k$, $[\widetilde{k_S}_\p:k_\p]=\l^\infty$. \end{Lemma} \begin{proof} For a finite place $\p$ of $k$, denote by $D_\p$ the decomposition group of $\p$ in $\Gal(\widetilde{k_S}/k)\simeq Z_\l^t$. As $D_\p$ is a closed subgroup of $\Z_\l^t$, it is trivial, or contains a subgroup isomorphic to $\Z_\l$. We will show that $D_\p$ can not be trivial. Denote by $\widetilde{k_S}^{D_\p}$ the subfield of $\widetilde{k_S}$ fixed by $D_\p$. Note that $\widetilde{k_S}^{D_\p}$ is included in $k_S^{\{\p\}}$ and $\widetilde{k_S^{\{\p\}}}\subset \widetilde{k_S}$. Now assume that for $\p \notin S$, $D_\p$ is trivial. Then $\widetilde{k_S}=\widetilde{k_S}^{D_\p}\subset k_S^{\{\p\}}$. Also $\widetilde{k_S} \subset \widetilde{k_S^{\{\p\}}}$ and the equality $\widetilde{k_S} =\widetilde{k_S^{\{\p\}}}$ yealds. This implies that for $\p \notin S$, $\rk_{\Z_\l} {G_S^{\{\p\}}}^{ab} = \rk_{\Z_\l} G_S^{ab}$ which contradicts the condition 1. Now for $\p \in S$, as we have: $$k\subset \widetilde{k_S^{\{\p\}}}\subset \widetilde{k_{S\backslash \{\p\}}} \subset \widetilde{k_S},$$ the conclusion is the same as for $\p \notin S$. \end{proof} Now we will apply a result which can be found in \cite{se2}: the $\l$-cohomological dimension $\Gal(\k/\widetilde{k_S})$ is less than 1. In consequence, we can now apply Proposition 2 page 85 of \cite{se2} and so $\cd_\l(\Gal(\k(\l)/\widetilde{k_S}))=1$. \end{proof} \subsubsection{Proof of Theorem \ref{imaginary}} If $S_{\bar{\p}}=S_\l$ this is well-known. Let $K$ be the maximal abelian $\l$-extension of $N$ unramified outside $\p$, and for a prime $\p'$ different from $\p$, let $K^{\{\p'\}}$ be the maximal abelian $\l$-extension of $N$ contained in $K$ and in which $\p'$ splits completely. By Theorem \ref{rang}, we have $\rk_{\Z_\l} {\Gal(K/N)}=1$, and for $\p' \not= \p$, $\rk_{\Z_\l} \Gal(K^{\{\p'\}}/N) =0$. Then for all $\p \in \o_N$, $[{K}_\p:N_\p]=\infty$. By base change, the same is true for $k_{S_{\bar{\p}}}/k$. Then the end of the proof of Theorem \ref{dimension2} can be applied. \subsubsection{Proof of Theorem \ref{dimension3}} Let $f(x)\in \Z[x]$ be an irreducible polynomial of degree $4$, such that the discriminant $d$ of $f$ is a fundamental discriminant of a real quadratic field: for example $f(x)=x^4-x+1$ with $d=229$. It is well know that the Galois closure $k$ of $f$ is of degree 24 over $\Q$ with Galois group $\SS_4$, and moreover, $k/\Q(\sqrt{d})$ is unramified everywhere, with Galois group $G=\AA_4$. Let $\p |\l$ be a prime of $N:=\Q(\sqrt{d})$ which is of degree one. Put $S_N=\{\p\}$ and $S=S_{\bar{\p}}=\{\P\in \o_k, \P\cap \o_N =\p\}$. We will study the action of $\AA_4$. The group $\AA_4$ has three irreducible characters of degree one ($\1, \psi_1,\psi_2$) plus one irreducible character $\psi$ of degree $3$. As the archimedean places are unramified in $k/N$, we have: $$\chi(E_k)= 2 (\1+\psi_1+\psi_2+\psi)-\1 = \1 +2\psi_1+2\psi_2 +2\psi.$$ Now by Proposition \ref{scha}, one has: $$\chi_S=\1+\psi_1+\psi_2+2\psi,$$ and so $$\rk_{\Z_\l} G_S^{ab}=3.$$ Now we prove that conditions 1 and 2 of Theorem \ref{dimension2} are satisfied. 1) First a lemma: \begin{Lemma} \label{lemme3} Let $\q$ be a prime of $k$ prime to $S$. Let $T=\{\q\}$ and $\overline{T}=\{\q^s, s\in G\}$. Then $\rk_{\Z_\l} {G_S^{\overline{T}}}^{ab} < \rk_{\Z_\l} G_S^{ab}$ if and only if $\rk_{\Z_\l} {G_S^T}^{ab} < \rk_{\Z_\l} G_S^{ab}$. \end{Lemma} \begin{proof} For a finite place $\q$ of $k$, denote by $D_\q$ the decomposition group of $\q$ in $\widetilde{k_S}/k$. First we have : $$\Q \subset N\subset k \subset \widetilde{k_S^{\overline{T}}} \subset \widetilde{k_S^{T}}\subset \widetilde{k_S}^{D_\q}\subset \widetilde{k_S}.$$ Note that $\widetilde{k_S}/N$ is a Galois extension. It is clear that $\rk_{\Z_\l} {G_S^T}^{ab} < \rk_{\Z_\l} G_S^{ab}$ implies $\rk_{\Z_\l} {G_S^{\overline{T}}}^{ab} < \rk_{\Z_\l} G_S^{ab}$. We have only to prove the reciprocity: Assume that $\rk_{\Z_\l} {G_S^T}^{ab} = \rk_{\Z_\l} G_S^{ab}$. Then $\widetilde{k_S}=\widetilde{k_S^T}$. The decomposition group of $\q$ in $\widetilde{k_S}/ k$ is then trivial and so the decomposition group of all conjugate of $\p$ by $G$ are trivial. Consequently, $\widetilde{k_S} \subset k_S^{\overline{T}}$ and so $\widetilde{k_S}\subset \widetilde{k_S^{\overline{T}}}$, giving $\widetilde{k_S}= \widetilde{k_S^{\overline{T}}}$ . \end{proof} For a finite place $\p$ of $N$, the decomposition group $D_\p$ in $k/N$ is cyclic, because $\p$ is unramified. So the order of $D_\p$ is $1$, $2$ or $3$, which implies that the degree of $\Ind_{D_\p}^G \1$ is $12$, $6$ or $4$. Moreover $\Ind_{D_\p}^G \1 \leq \Reg$, implying $\psi \leq \Ind_{D_\p}^G $. And so for all prime $\p$ of $N$, we have (by using Proposition \ref{unites}): $$\chi(E_S^{\bar{T}}) \geq \Reg.$$ By Proposition \ref{scha}, $$\Theta_S(E_S^{\overline{T}})=\Reg,$$ and so by Theorem \ref{rang}$$\rk_{\Z_\l} {G_S^{\overline{T}}}^{ab}=0.$$ By Lemma \ref{lemme3}, one has the first condition of Theorem \ref{dimension2}. 2) To prove the second condition, we use a method similar to that of Lemma \ref{lemme3}. For a prime $\q$ in $S$, denote by $k_{S\backslash\{\q\}}$ the maximal $\l$-extension of $k$, unramified outside $S\backslash\{\q\}$ and let $\widetilde{k_{S\backslash\{\q\}}}$ be the compositum of $\Z_\l$-extension included in $k_{S\backslash\{\q\}}$. We have $$\Q \subset N \subset k \subset \widetilde{k_{S\backslash \{\q\}}}\subset \widetilde{k_S}^{I_\q} \subset \widetilde{k_S},$$ where $I_\q$ is the ramification group of $\q$ in $\widetilde{k_S}/k$. Now $I_\q$ is a closed subgroup of $\Gal(K_S/k)\simeq \Z_\l ^3$: it is trivial, or it contains a subgroup isomorphic to $\Z_\l$. As before, if $I_\q$ is trivial then for all conjugates $\q^s$, $s \in G$, $I_{\q^s}$ is also trivial, and so $\widetilde{k_S}/k$ is unramified everywhere which is impossible. The second condition of Theorem \ref{dimension2} is then verified. \section{ On the group $G_S$} \subsection{The exact value of the number of relations of $G_S$} \subsubsection{Relations and torsion} Leopoldt's conjecture says that for $S=S_\l$, the $\Z_\l$-rank of $\overline{\Theta_S(E_S)}$ is exactly the $\Z$-rank of the group of units of $k$. As noted in \cite{ng}, this result has a link with the Schur multiplicator of $G_S$: \begin{Proposition}\label{proposition6} Leopoldt's Conjecture is true for $(\l,k)$ if and only if $H_2(G_{S_\l},\Z_\l)$ is trivial. \end{Proposition} \begin{proof} Proposition \ref{proposition6} is a direct consequence of the following Lemma: \begin{Lemma}[see \cite{ro} for example] \label{relgen} Let $G$ be a pro-$\l$-group. Then $$d(H_2(G,\Z_\l))=r(G)-d(\T(G))=r(G)-d(G)+\rk_{\Z_\l} G^{ab}.$$ \end{Lemma} The difference $r(G_{S_\l})-d(G_{S_\l})$ is well-known and is equal to $-(r_2+1)$. But Leopoldt's Conjecture is true if and only if $\rk_{\Z_\l} G^{ab}=r_2+1$, i.e. if and only if $H_2(G,\Z_\l)$ is trivial. \end{proof} The triviality of this cohomology group, allows us to see $d(\T(G_{S_\l}))$ as the number of relations of $G_{S_\l}$: \begin{Theorem} Leopoldt's Conjecture is equivalent to $$r(G_{S_\l})= d(\T(G_{S_\l})).$$ \end{Theorem} \subsubsection{General case} \begin{Theorem}\label{rela2} Assume $S\subset S_\l$ (not empty). If $\rk_{\Z_\l} \overline{\Theta_S(E_S)} =\rk_{\Z}E_k$, one has: \begin{enumerate} \item $r(G_S)=d(G_S)-\sum_{\p \in S}[k_\p:\Q_\l] +r_1+r_2-1=d(\T(G_S))$; \item $H_2(G_S,\Q_\l/\Z_\l)\simeq \T(G_S)$. \end{enumerate} \end{Theorem} \begin{proof} By Propositions \ref{relgen} and \ref{gen_rel}, one has: \begin{eqnarray}\label{ine}0\leq d(H_2(G_S,\Z_\l))\leq r(G_S)-d(G_S)+\rk_{\Z_\l} G_S^{ab}\leq \rk_{\Z}E_k-\rk_{\Z_\l} \overline{\Theta_S(E_S)} .\end{eqnarray} The first assertion is then proved. For the second assertion, we use a result of Nguyen \cite{ng}. First by the inequality (\ref{ine}), $H^2(G_S,\Q_\l/\Z_\l)=H_2(G_S,\Z_\l)^*=0$. So the exact sequence $$1\longrightarrow \Z_\l \stackrel{\l^m}{\longrightarrow} \Z_\l \longrightarrow \Z/\l^m\Z \longrightarrow 1,$$ gives for large $m$ $$H_2(G_S,\Z/\l^m\Z)\simeq \T(G_S).$$ By passing to inductive limit, we have the result. \end{proof} We can now give some examples. \subsubsection{Examples} 1) Let $k$ be a quadratic field and $S\subset S_\l$ ($S$ not empty). Then $r(G_S)=d(\T(G_S))$. 2) Consider the two following situations: i) Let $N$ be an imaginary quadratic field, let $\p$ be a prime of $N$ above $\l$ and let $k/N$ be a finite Galois extension. Assume Schanuel's Conjecture. ii) Let $N$ be a real quadratic field and $\p$ be a prime of $N$. Let $\sigma_1$ and $\sigma_2$ be the two archimedean places of $N$. Suppose $\alpha \in N\backslash N^2$ and $\beta\in N\backslash N^2$ have negative absolute norm satisfying $\sigma_1(\alpha)>0$, $\sigma_2(\beta)>0$. Put $k=N(\sqrt{\alpha},\sqrt{\beta})$ ; $k$ is a totally complex biquadratic extension over $N$. Put for the two cases $S=\{\P \in k, \P|\p\}$. Then one has: \begin{center} \begin{math} r(G_S)=d(\T(G_S))=\left\{\begin{array}{l}d(G_S)-(r_2+1), \ \mbox{if } S=S_\l\\ d(G_S)-1, \ \mbox{else}\end{array}\right. \end{math} \end{center} In the two cases, $G_S^{ab}$ is infinite. \begin{proof} 1) Direct application of Theorem \ref{rela2}. 2) If $S=S_\l$ this is well-known. Let us show the second equality. To do this we use the action of $G=\Gal(k/N)$. i) We are exactly in the same context as in Proposition \ref{schanuel}. So we have: $$\chi_S=\Reg\wedge \chi(E_S)=\chi(E_S)=\Reg-\1.$$ ii) Denote $N_0=N$, $N_1=N(\sqrt{\alpha})$, $N_2=N(\sqrt{\beta})$ and $N_3=N(\sqrt{\alpha \cdot \beta})$. Let $\varphi_i$ be the $\C_\l$-irreducible character of $G=\Gal(k/N)$ such that the kernel of $\varphi_i$ defined $N_i$. Then $$\chi(E_S)=\varphi_0+\varphi_1+\varphi_2.$$ And so $\chi_S=\chi(E_S),$ by Proposition \ref{jaulent}. In the two cases, Theorem \ref{rela2} can be applied. \end{proof} \subsection{Free pro-$\l$-groups} \subsubsection{Classical results} \begin{Proposition} The pro-$\l$-group $G_{S_\l}$ is free if and only if one of the following is satisfied: \begin{enumerate} \item There exists a place $\p$ of $S_\l$ such that $\mu_\l\subset k_\p$, then $\mu_\l\subset k$, $\cha_{S_\l}=0$ and $|S_\l|=1$. \item For all places $\p$ of $S_\l$, $k_\p$ does not contain $\mu_\l$, and $\cha_{S_\l}=0$. \end{enumerate} In this case, $G_{S_\l}$ is the free pro-$\l$-group with $r_2+1$ generators. \end{Proposition} \begin{proof} One has $r(G_{S_\l})-d(G_{S_\l})=-(r_2+1)$. Moreover $d(G_{S_\l})=r_2+1+\sum_{\p \in S_\l}\delta_\p +d(\cha_{S_\l})-\delta_k$. So $r(G_{S_\l})=0$ if and only if $\delta_k=\sum_{\p \in S_\l}\delta_\p +d(\cha_{S_\l})$. \end{proof} \subsubsection{General case} We have: \begin{Theorem}\label{libre} Let $S\subset S_\l$. Then $G_S$ is pro-$\l$-free in the following situations: \begin{enumerate} \item $\mu_\l\subset k$, $S=\{\p\}$ and $\cha_S=0$. \item For all places $\p$ of $S$, $k_\p$ does not contain $\mu_\l$ and $\cha_S=0$. \end{enumerate} In the two cases $G_S$ is the free pro-$\l$-group with $\sum_{\p \in S} [k_\p:\Q_\l]+1-(r_2+r_1)$ generators. \end{Theorem} \begin{proof} Recall the surjective map of Proposition \ref{gen_rel} : $$\cha_S \twoheadrightarrow \ker {\left(H^2(G_S,\F_\l)\rightarrow \bigoplus_{\p \in S^*} H^2(\G_\p,\F_\l)\right)}^*.$$ Now $ \ker {\left(H^2(G_S,\F_\l)\rightarrow \bigoplus_\p H^2(\G_\p,\F_\l)\right)}=H^2(G_S,\F_\l)$ in one of the following: \begin{enumerate} \item $|S|=1$, if $\mu_\l \in k$; \item $k_\p$ does not contain $\mu_\l$ for all places $\p$ of $S$ (in this case, $H^2(\G_\p,\F_\l)=0$). \end{enumerate} These conditions plus triviality of $\cha_S$ imply $H^2(G_S,\F_\l)=1$. \end{proof} \begin{thebibliography}{99} \bibitem{br}A. 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