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\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  $<I,I^\prime>$ 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}
<I,I^\prime>&=&<\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: $<I,I>=|H/G\backslash H|$.

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:
$$|H/G\backslash H|=<I,I>=\sum_\psi a_\psi^2.$$
The first result follows.

2) By looking at the scalar products $<I,I>$, $<I,I^\prime>$ and $<I^\prime,I^\prime>$, 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}


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\end{thebibliography}

\


\begin{sc}
\noindent Christian Maire\\
Laboratoire A2X\\
Universit\'e Bordeaux I\\
Cours de la Lib\'eration\\
33405 Talence Cedex France \\
\end{sc}
\verb?maire@math.u-bordeaux.fr?

\

Current address:

 
\begin{sc}
\noindent  Mathematical Sciences Research Institute \\
1000 Centennial Drive\\
Berkeley, Califoria 94720 \\
USA\\
\end{sc}
\verb?maire@msri.org?
\end{document}
