ON THE ℤ_{l}-RANK OF ABELIAN EXTENSIONS WITH RESTRICTED RAMIFICATION

CHRISTIAN MAIRE

Abstract. For a prime number ℓ, we consider ℓ-extensions k_{S} of number fields k unramified outside a finite set S ⊂ 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).

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 [18], Koch [7], [8], Haberland [5] ...). 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 + δ where r_{2} is the number of complex places of k, and δ 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^{∞}. Here we ought to mention Brumer’s Theorem [1] (which is a ℓ-adic version of a Theorem of Baker) that proves Leopoldt’s Conjecture for abelian fields and a result of Waldschmidt [19] 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 [12] and the book of Gras [3]. 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 1) 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} 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 finite. To make this statement precise, we have to introduce some definitions. For a prime ℓ and a number field k, let i_{ℓ}(k) be the smallest value of |S|/|S_{ℓ}| such that G_{S}^{ab} is infinite, and define I_{ℓ}(k) to be the smallet rational number c, such that for all |S| with |S|/|S_{ℓ}| ≥ c, G_{S}^{ab} is infinite.

Theorem 1. For all prime ℓ,

sup_{number fields k} I_{ℓ}(k) − i_{ℓ}(k) = 1.

Date: December 12, 2000.

2000 Mathematics Subject Classification. Primary: 11R37, 11S37. Key words and phrases.

S-ramification, ℤ_{ℓ}-rank, Schanuel ℓ-adic Conjecture.

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 \overline{k}(ℓ)/k^{∞} is free (for ℓ > 2), where \overline{k}(ℓ) is the maximal ℓ-extension of k. Denote by k̃_{S} be the compositum of all ℤ_{ℓ}-extensions of k contained in k_{S}. Before giving some examples with S ⫋ S_{ℓ} for which \overline{k}(ℓ)/k̃_{S} is free, we give a definition:

Definition 2. Let k/N be a finite Galois extension of number fields with Galois group G. Let 𝔭 be a prime of N. Define S𝔭̄ = {𝔓 ⊂ O_{k}, 𝔓 ∩ O_{N} = 𝔭} = {𝔓^{g}, g ∈ G}.

First let us give a result for extensions of quadratic imaginary fields:

Theorem 3. Let N be an imaginary quadratic field, 𝔭 be a prime of N and k/N be a finite Galois extension. Let k_{S𝔭̄} be the maximal ℓ-extension of k unramified outside S𝔭̄. Then Gal(\overline{k}(ℓ)/k̃_{S𝔭̄}) is free.

For the real quadratic case, we prove:

Theorem 4 (Under Schanuel ℓ-adic Conjecture (see §2.3.1)). Let f(x) ∈ ℤ[x] be an irreducible polynomial of degree 4, such that:

1. The discriminant d of f is a fundamental discriminant of a real quadratic field; N = ℚ(\sqrt{d});

2. The roots of f are not real. It is well-known that the Galois closure k of f over ℚ is of degree 24, and Gal(k/N) ≃ A_{4} (see [9]). Choose a prime number ℓ such that (\frac{d}{ℓ}) = 1, and fix a prime 𝔭 of N dividing ℓ. Let k_{S\overline{𝔭}} be the maximal ℓ-extension of k unramified outside S𝔭̄. Then Gal(\overline{k}(ℓ)/k̃_{S𝔭̄}) is free.

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 [6]. 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 1, 3 and 4. 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 [3]. This work was supported by a Postdoctoral fellowship position in Mathematical Sciences Research Institute^{1} at Berkeley and, by CNRS and 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.

1. Preliminaries

1.1. Notations. We fix a prime number ℓ and an algebraic closure ℚ̄ 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:

1Research at MSRI is supported in part by NSF grant DMS-9701755

- k_{𝔭} the completion of k in 𝔭; N𝔭 the absolute norm, e_{𝔭} and f_{𝔭} the ramification index and the residue degree of 𝔭 in k/ℚ respectively;

- [k_{𝔭} : ℚ_{ℓ}] = e_{𝔭} · f_{𝔭} be the degree of 𝔭;

- O_{𝔭} the completion at 𝔭 of the ring of the integers O_{k} of k;

- U_{𝔭} the units group of O_{𝔭};

- U_{𝔭}^{i} the units of U_{𝔭} congruent to 1 modulo 𝔭^{i};

- G_{𝔭} the pro-ℓ-absolute Galois group of k_{𝔭}.

- E_{k} the units group of k;

- E_{k}^{T} the T -units group of k (in the classical sense), where T is a finite set of places of k;

- k_{S}^{1} = {x ∈ k^{×}, ∀𝔭 ∈ S, v_{𝔭}(x − 1) ≥ 1};

- E_{S}^{T} = k_{S}^{1} ∩ E_{k}^{T} and E_{S} = k_{S}^{1} ∩ E_{k};

- Δ_{S} = {x ∈ k^{×}, (x) = Q^{ℓ}, x ∈ k_{𝔭}^{ℓ}, ∀𝔭 ∈ S}/k^{×}^{ℓ}

- k^{∞} the ℤ_{ℓ} cyclotomic extension of k;

- k_{S} the maximal ℓ-extension of k unramified outside S (archimedean places are decomposed);

- k_{S}^{T} the maximal subextension of k_{S} in which all places of T split totally;

- G_{S} = Gal(k_{S}/k) and G_{S}^{ab} = G_{S}/[G_{S}, G_{S}];

- \overline{k} = ℚ̄, and \overline{k}(ℓ) the maximal ℓ-extension of k contained in \overline{k};

- G\overline{k} = Gal(\overline{k}/k) the absolute Galois group of k. If A is a ℤ_{ℓ} (resp. ℤ) module of finite type:

- Tor(A) the ℤ_{ℓ}-torsion part of A (resp. ℤ);

- d_{ℓ}(A) = d(A) the 𝔽_{ℓ}-rank of A;

- rk_{ℤ_{ℓ}} (resp. rk_{ℤ}), the ℤ_{ℓ}-rank of A (resp. ℤ-rank).

1.2. Leopoldt’s Conjecture. Fix now an algebraic closure \overline{ℚ_{ℓ}} of ℚ_{ℓ}, and for each place 𝔭_{i} (i = 1, ⋯, s) of S, we fix σ_{i} an imbedding of ℚ in k_{𝔭_{i}} ⊂ \overline{ℚ_{ℓ}}. Now consider the following homomorphism Θ_{S} of groups:

\begin{matrix}Θ_{S} : k_{S}^{1} & → & ∏_{𝔭∈S} \overline{ℚ_{ℓ}} \\ x & ↦ & (σ_{1}(x), ⋯, σ_{s}(x)) \end{matrix}

and let L_{S} be the Logarithmic map:

\begin{matrix}L_{S} : ∏_{𝔭∈S} U_{𝔭}^{1} & → & ∏_{𝔭∈S} k_{𝔭} ≃ ℚ_{ℓ}^{∑_{𝔭} e_{𝔭}f_{𝔭}} \\ (x_{1}, ⋯, x_{s}) & ↦ & (log_{ℓ}(x_{1}), ⋯, log_{ℓ}(x_{s})) \end{matrix}

Note that the kernel of the logarithmic map is finite. We can now present Leopoldt’s Conjecture:

Conjecture 1. If S = S_{ℓ}, then

rk_{ℤ}E_{k} = rk_{ℤ_{ℓ}} \overline{Θ_{S}(E_{S})}.

Or equivalently (L_{S}(\overline{Θ_{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}(Θ_{S}(E_{S}))) is in the hyperplane of ℚ_{ℓ}^{n} defined by the equation: x_{1} + ⋯ + x_{n} = 0.

1.3. On the ℤ_{ℓ}-rank of G_{S}^{ab}. In order to prove Theorem 1, we need to consider subextensions where places split completely.

Theorem 5. Let T and S be two finite sets of finite places of k (S ⊂ 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} ≃ ℤ_{ℓ}^{t} × Tor(G_{S}^{T}^{ab}),

where t = ∑_{𝔭∈S}[k_{𝔭} : ℚ_{ℓ}] − rk_{ℤ_{ℓ}} \overline{Θ_{S}(E_{S}^{T})} = ∑_{𝔭∈S}[k_{𝔭} : ℚ_{ℓ}] − rk_{ℤ_{ℓ}}L_{S}(\overline{Θ_{S}(E_{S}^{T})}).

Proof. We want to determine t. By Class field theory k_{S}^{T}^{ab} is associated to the following sub-group of ideles:

k^{×} ∏_{𝔭∉S∪T} U_{𝔭} ∏_{𝔭∈T} k_{𝔭}^{×}.

Remark that archimedean places play no role for the determination of t. Let n ≥ 1 be an integer and consider the following exact sequence:

\begin{aligned}1 ⟶ \frac{k^{×} ∏_{𝔭∉S∪T} U_{p} ∏_{𝔭∈S} U_{𝔭}^{1} ∏_{𝔭∈T} k_{𝔭}^{×}}{k^{×} ∏_{𝔭∉S∪T} U_{𝔭} ∏_{𝔭∈S} U_{𝔭}^{n} ∏_{𝔭∈T} k_{𝔭}^{×}} & ⟶ \frac{J_{k}}{k^{×} ∏_{𝔭∉S∪T} U_{𝔭} ∏_{𝔭∈S} U_{𝔭}^{n} ∏_{𝔭∈T} k_{𝔭}^{×}} \\ & ⟶ \frac{J_{k}}{k^{×} ∏_{𝔭∉S∪T} U_{p} ∏_{𝔭∈S} U_{𝔭}^{1} ∏_{𝔭∈T} k_{𝔭}^{×}} ⟶ 1 \end{aligned}

Now \frac{J_{k}}{k^{×} ∏_{𝔭∉S∪T} U_{p} ∏_{𝔭∈S} U_{𝔭}^{1} ∏_{𝔭∈T} k_{𝔭}^{×}} corresponds to the maximal abelian exten-

sion 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^{×} ∏_{𝔭∉S∪T} U_{𝔭} ∏_{𝔭∈S} U_{𝔭}^{n} ∏_{𝔭∈T} k_{𝔭}^{×}. So we have:

1 ⟶ lim_{n^{←}} A_{n} ⟶ Gal(k_{S}^{T}^{ab}/k) ⟶ Gal(k_{S}^{T}^{tame}/k) ⟶ 1,

where A_{n} = \frac{k^{×} ∏_{𝔭∉S∪T} U_{p} ∏_{𝔭∈S} U_{𝔭}^{1} ∏_{𝔭∈T} k_{𝔭}^{×}}{k^{×} ∏_{𝔭∉S∪T} U_{𝔭} ∏_{𝔭∈S} U_{𝔭}^{n} ∏_{𝔭∈T} k_{𝔭}^{×}}, and so t = rk_{ℤ_{ℓ}} lim_{n^{←}} A_{n}.

It is easy to see that we have the following continuous isomorphism of groups:

A_{n} ≃ \frac{∏_{𝔭∈S} U_{𝔭}^{1}}{Θ_{S}(E_{S}^{T}) ∏_{𝔭∈S} U_{p}^{n}} , (1)

where restrictions are compatible. So

lim_{n^{←}} A_{n} ≃ lim_{n^{←}} \frac{∏_{𝔭∈S} U_{𝔭}^{1}}{Θ_{S}(E_{S}^{T}) ∏_{𝔭∈S} U_{p}^{n}} .

Now for each integer n, Θ_{S}(E_{S}^{T}) ∏_{𝔭∈S} U_{𝔭}^{n} is of finite index in ∏_{𝔭∈s} U_{𝔭}^{1}, which is a pro-ℓ-group, and so these sub-groups are closed and we obtain (see for example [16] lemma 1 page 3)

lim_{n^{←}} \frac{∏_{𝔭∈S} U_{𝔭}^{1}}{Θ_{S}(E_{S}^{T}) ∏_{𝔭∈S} U_{p}^{n}} ≃ \frac{∏_{𝔭∈S} U_{𝔭}^{1}}{⋂_{n}(Θ_{S}(E_{S}^{T}) ∏_{𝔭∈S} U_{𝔭}^{n})}.

The conclusion comes from the observation that

⋂_{n} Θ_{S}(E_{S}^{T}) ∏_{𝔭∈S} U_{𝔭}^{n} = \overline{Θ_{S}(E_{S})}.

1.4. On the rank and relations of G_{S}. We recall the following result:

Proposition 6. Let k be a number field, ℓ be a prime number and S ⊂ 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

1. d(G_{S}) = d_{ℓ}Δ_{S} + ∑_{𝔭∈S} U_{𝔭}^{1} − d_{ℓ}E_{k}.

2.

r(G_{S}) − d(G_{S}) ≤ d_{ℓ}E_{k} − δ_{k} + δ_{S} − ∑_{𝔭∈S}[k_{𝔭} : ℚ_{ℓ}],

where δ_{S} = 1 if k contains μ_{ℓ} and S is empty, 0 otherwise.

3. There exists the following surjective homomorphism:

Δ_{S} ↠ (ker(H^{2}(G_{S}, 𝔽_{ℓ}) → ⊕_{𝔭∈S^{∗}}H^{2}(G_{𝔭}, 𝔽_{ℓ})))^{∗} ,

where S^{∗} is S minus one place if k contains μ_{ℓ}.

4. r(G_{S_{ℓ}}) − d(G_{S_{ℓ}}) = −(r_{2} + 1).

Proof. cf. [5], [7], [8]. For ℓ = 2 in the point 4), see [14].

1.5. First examples.

Theorem 7. 1) The group G_{S}^{ab} is finite in the following cases:

(i) k is totally real, S does not contain all places above ℓ and Leopoldt’s Conjecture is true for k and ℓ.

(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.

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 : ℓO_{k} = 𝔭_{1} ⋯ 𝔭_{n}. Put S = {𝔭_{1}, ⋯, 𝔭_{i}}. If i ≥ 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.

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^{∞} of k. In this extension, all places above ℓ are infinitely ramified. So k_{S} ∩ k^{∞} is finite over k, and the same is for G_{S}^{ab}. 1 ii) We use Theorem 5 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 ≤ r_{2} − 1. So here i − r ≥ 1.

3) By computation, we find d(G_{S}) = 0.

2. On the group G_{S}^{ab}

Before to precise the situation, we recall some facts about linear representations of finite groups.

2.1. Linear representations. Let G be a finite group and let M be a ℚ[G]- module. We will identify the following modules:

1. the ℚ[G]-module M

2. the ℚ_{ℓ}[G] module ℚ_{ℓ} ⊗_{ℚ} M

3. the ℂ_{ℓ}[G]-module ℂ_{ℓ} ⊗_{ℚ} M.

Let χ be a ℚ_{ℓ}-irreducible character of G. One knows that χ is equal to a sum of ℂ_{ℓ}-irreducible characters ψ. More precisely recall the definition of the Schur index m_{ψ} of a ℂ_{ℓ}-irreducible character ψ : m_{ψ} = min_{k}[k : ℚ_{ℓ}(ψ)], the minimum being over all fields k realizing ψ and where ℚ_{ℓ}(ψ) is the extension of ℚ_{ℓ} generated by the values of ψ. We know some properties of m_{ψ}, e.g. m_{ψ} divides the degree n_{ψ} of ψ. For more details see [2] or [17]. So if χ is a ℚ_{ℓ}-irreducible character, one has

χ = m_{χ} (ψ_{1} + ⋯ + ψ_{n})

where the characters ψ_{i} are ℂ_{ℓ}-irreducible conjugates (with the same degree n_{χ}), and moreover m_{χ} = m_{ψ_{1}} = ⋯ = m_{ψ_{n}} and where finally n = [ℚ_{ℓ}(χ) : ℚ_{ℓ}]. The character χ is of degree m_{χ}n_{χ}[ℚ_{ℓ}(χ) : ℚ_{ℓ}]. If G is abelian then m_{ψ}n_{ψ} = 1, for all ℂ_{ℓ}-irreducible characters ψ.

Definition 8. For a finite group G, Irr(ℂ_{ℓ}) denotes the set of ℂ_{ℓ}-irreducible characters of G.

2.1.1. Regular representation. The regular representation of G is the representation of G acting on itself by translation. The character χreg of this representation is known:

χreg = ∑_{ψ∈Irr(ℂ_{ℓ})} n_{ψ}ψ = 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 χreg: this can be seen by transitivity formula of induction:

Ind_{{1}}^{G}1 = Ind_{H}^{G}(Ind_{{1}}^{H}1)
= Ind_{H}^{G}(χreg(H))
= Ind_{H}^{G}(1 + ⋯)
= Ind_{H}^{G}(1) + ⋯

To conclude this part, let us give :

Lemma 9. Let G be a finite group. Let H and H′ be two subgroups of G with the same order. Let I = Ind_{H}^{G}1 and I′ = Ind_{H′}^{G}1. Consider now the decomposition of I and I′ as a sum of ℂ_{ℓ}-irreducible characters ψ: I = ∑_{ψ} a_{ψ}ψ, and I′ = ∑_{ψ} a′_{ψ}ψ. As I and I′ appear in χreg, one has: a_{ψ}, a′_{ψ} ≤ n_{ψ}. Then

1. H ◁ G if and only if a_{ψ} = n_{ψ}, ∀ψ ∈ Irr(ℂ_{ℓ});

2. If H ◁ G, then I = I′ if and only if H = H′.

Proof. Let us calculate the scalar product < I, I′ > in G: To do this we use [17]. Let S be a representing system on the double coset H/G\H′:

G = ⋃_{s∈S}^{·} HsH′.

For s ∈ S, denote H_{s}′ = sH′s^{−1} ∩ H, and consider the following representation 1_{s} of H_{s}′:

∀x ∈ H_{s}′, 1_{s}(x) = 1_{H′}(s^{−1}xs),

where 1_{H′} is the unit representation of H′. One has in fact 1_{s} = 1_{H_{s}}′. We get:

(2) Res_{H}(Ind_{H′}^{G}1) = ∑_{s∈S} Ind_{H_{s}′}^{H}(1_{s})

where Res_{H} is the restriction to H. By using the Frobenius formula, one has:

< I, I′ > = < 1_{H}, Res_{H}(Ind_{H′}^{G}1) >_{H}
= < 1_{H}, ∑_{s∈S} Ind_{H_{s}′}^{H}(1_{s}) >_{H}
= ∑_{s∈S} < 1_{H}, Ind_{H_{s}′}^{H}(1_{s}) >_{H}
= ∑_{s∈S} < Res_{H_{s}}′(1_{H}), 1_{s} >_{H_{s}}′
= |S|

By a similary computation: < I, I >= |H/G\H|. We can now prove the first point:

1) One knows that H ◁ G if and only if |G/H| = |H/G\H|. But |G/H| is of order deg(I), i.e

|G/H| = ∑_{ψ} a_{ψ}n_{ψ}.

But we have also the identity:

|H/G\H| =< I, I >= ∑_{ψ} a_{ψ}^{2}.

The first result follows.

2) By looking at the scalar products < I, I >, < I, I′ > and < I′, I′ >, one obtains I = I′ if and only if

∑_{ψ} a_{ψ}^{2} = ∑_{ψ} a_{ψ}a′_{ψ} = ∑_{ψ} (a′_{ψ})^{2} = |S|.

It suffices then to show that for H ≠ H′, |S| ≠ ∑_{ψ} a_{ψ}^{2}. Suppose H ≠ H′. There exists h′ ∈ H and ∉ H; a representing system of G/H could be written {1, h′, ⋯ }. Now two elements of G which are in the same class in G/H are in the same class in H/G\H′. Moreover 1 and h′ are also in the same class in H/G\H′. So |S| ≤ |G|/|H| − 1. To conclude our proof it suffices to note that (by using the fact that H ◁ G)

∑_{ψ} a_{ψ}^{2} = ∑_{ψ} a_{ψ}n_{ψ} = deg(I) = |G|/|H|.

2.2. 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 ⊂ O_{k}, 𝔓 ∩ O_{N} ∈ S_{N}}. Then G acts on U_{S}^{1} := ∏_{𝔭∈S} U_{𝔭}^{1} and, Θ_{S} and L_{S} ∘ Θ_{S} are homomorphisms of ℚ_{ℓ}[G]-modules. With these facts, we can give some informations on the character of Θ_{S}(E_{S}) and so on the ℤ_{ℓ}-rank of G_{S}^{ab}.

Definition 10. Denote by χ_{S} the character of Θ_{S}(E_{S}).

2.2.1. Character of units. Let 𝔭_{∞_{1}}, ⋯, 𝔭_{∞_{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 [4]):

Proposition 11. 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:

χ(E_{k}^{T}) = χ(E_{S}^{T}) = ∑_{i=1,··· ,n} Ind_{D_{𝔭∞_{i}}}^{G} (1) + ∑_{𝔭∈T} Ind_{D_{𝔭}}^{G}1 − 1,

where D_{𝔭} is the decomposition group of 𝔭 in k/N.

2.2.2. Character of U_{S}^{1}.

Lemma 12.

χ(U_{S}^{1}) = ∑_{𝔭∈S_{N}}[N_{𝔭} : ℚ_{ℓ}]χreg.

Proof. Fix 𝔭 ∈ S_{N}. Let 𝔓|𝔭. Note first that for i ≥ 1 the two modules ℚ_{ℓ} ⊗_{ℤ_{ℓ}} U_{k_{𝔓}}^{1} and ℚ_{ℓ} ⊗_{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} ≃ O_{k_{𝔓}}. Thus for large i, we get:

∏_{𝔓∈S} U_{𝔓}^{i} ≃ ∏_{𝔓∈S} O_{k_{𝔓}}.

One then has the following isomorphisms of G-modules:

ℚ_{ℓ} ⊗_{ℤ_{ℓ}} ∏_{𝔓|𝔭} U_{k_{𝔓}}^{1} ≃ ℚ_{ℓ} ⊗ ∏_{𝔓|𝔭} U_{k_{𝔓}}^{i}
≃ ℚ_{ℓ} ⊗_{ℤ_{ℓ}} ∏_{𝔓|𝔭} O_{k_{𝔓}}
≃ ∏_{𝔓|𝔭} ℚ_{ℓ} ⊗_{ℤ_{ℓ}} k_{𝔓}
≃ ∏_{𝔓|𝔭} k_{𝔓}
= N_{𝔭} ⊗_{N} k
≃ N_{𝔭} ⊗_{N} N[G]
≃ N_{𝔭}[G]
≃ ℚ_{ℓ}[G] ⊕ ⋯ ⊗ ℚ_{ℓ}[G] = [N_{𝔭} : ℚ_{ℓ}]ℚ_{ℓ}[G].

2.2.3. Upper bound for χ_{S}. It comes immediately the following:

Proposition 13. One has for the character χ_{S} of the ℚ_{ℓ}[G]-module ℚ_{ℓ}⊗_{ℤ_{ℓ}}Θ_{S}(E_{S}):

χ_{S} ≤ ∑_{𝔭∈S_{N}}[N_{𝔭} : ℚ_{ℓ}]χreg ∧ ( ∑_{i=1,··· ,r(N)} Ind_{D_{∞_{i}}}^{G} (1) − 1) .

2.3. Lower bound for χ_{S} and Schanuel ℓ-adic Conjecture.

2.3.1. Conjectural results. The ℓ-adic independance conjecture of Schanuel is at the center of the problem: Schanuel ℓ-adic Conjecture. Let x_{1}, ⋯, x_{r} ∈ \overline{ℚ_{ℓ}} be a family of algebraic numbers, such that the family log(x_{i})_{i=1,··· ,r} is independent over ℚ. Then the transcendance degree over ℚ of ℚ(log_{ℓ}(x_{1}), ⋯, log_{ℓ}(x_{r})) is r. Under this Conjecture we can prove the following result which can be found, for example, in [6]:

Proposition 14. Let M_{0} be a ℚ[G]-module of finite type included in k_{S}^{1}. Let m = dim_{ℚ_{ℓ}}(χreg ∧ χ(M_{0})). With Schanuel’s Conjecture one has:

rk_{ℤ_{ℓ}}(L_{S}(Θ_{S}(M_{0}))) ≥ m.

Remark 2.1. Proposition 14 shows that Schanuel’s Conjecture implies Leopoldt’s Conjecture.

To prove the Proposition 14 we need a Lemma:

Lemma 15 ([6]). Let ε ∈ M_{0}. There exists a ℚ[G]-submodule M of k_{S}^{1} with χ(M) = χreg and such that the principal ℚ[G]-module ℚ[G] · ε is a submodule of M.

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 ⊗ E_{S}^{T} contains χreg. Let M be the maximal principal submodule of ℚ ⊗ (E_{S}^{T} + M_{0}) containing ε. Then the character of M is χ(E_{S}^{T} + M_{0}) ∧ χreg = χreg.

Proof. (Proposition 14) It suffices to prove it for S_{N} = {𝔭}. Let S = {𝔓_{1}, ⋯, 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 χreg. Let χ = χreg∧χ(M_{0}). It is the character of a ℚ[G]-principal module M_{1}, thus there exists ε ∈ M_{0} such that M_{1} = ℚ[G] · ε. By Lemma 15, there exists a ℚ[G]-module M containing M_{1} and with representation χreg. We will show that \overline{Θ_{S}(M)} has maximal ℚ_{ℓ}-rank |G|, and this would imply that \overline{Θ_{S}(M_{1})} has ℚ_{ℓ}-rank m. Let (x_{1}, ⋯, x_{n}) be a ℚ-basis of M such that x_{i} = x^{g_{i}}, with g_{i} ∈ G. Suppose the existence of a linear relation on ℤ_{ℓ} between the images of x_{i} by L_{S} ∘Θ_{S}. We may write such a linear relation by:

∑_{i=1,··· ,m} a_{i}L_{S}(Θ_{S}(x_{i})) = 0,

with a_{i} ∈ ℤ_{ℓ}. Then one has:

∀σ_{j}, ∑_{i=1,··· ,m} a_{i} log_{ℓ}(σ_{j}(x_{i})) = 0.

Therefore we obtain:

Lemma 16.

∀g ∈ G, ∑_{i=1,··· ,m} a_{i} log_{ℓ}(σ_{1}(x_{i}^{g})) = 0.

Proof. Let g ∈ G. If g is in the decomposition group of 𝔓_{1} in N/k, then there exists g′ ∈ Gal(k_{𝔓_{1}}/N_{𝔭}) such that σ_{1}(y^{g}) = σ_{1}(y)^{g′}, ∀y ∈ k. Then

∑_{i} a_{i} log_{ℓ}(σ_{1}(x_{i}^{g})) = ∑_{i} a_{i} log_{ℓ}(σ_{1}(x_{i}))^{g′}
= ∑_{i} a_{i}(log_{ℓ}(σ_{1}(x_{i})))^{g′}
= (∑_{i} a_{i}(log_{ℓ}(σ_{1}(x_{i})))^{g′}
= 0

Now as |S_{N}| = 1, G acts transitively on S. Consequently σ_{1}(y^{g}) = g′ · σ_{i(g)}(y) where g′ is in Gal(k_{𝔓_{i(g)}}/N_{𝔭}), and we conclude our proof as before.

Now consider the square matrix M of size |G| × |G|: (log(σ_{1}(x_{i}^{g})))_{g∈G,i=1,··· ,|G|}. The determinant of this matrix is zero (by Lemma 16). 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.

Lemma 17. Put X_{g} = x^{g}, for g ∈ G. Let M′ be the matrix (X_{g·g′})_{g,g′∈G} and let P ((X_{g})_{g∈G}) be the determinant of this matrix. One has: det(M) = P ((x^{g})_{g}). Then P is not the zero polynomial.

We can assume that M′ can be written:

( \begin{matrix}X_{1} & X_{σ} & ⋯ & X_{τ} \\ X_{σ^{−1}} & X_{1} & ⋯ & X_{σ^{−1}τ} \\ ⋯ & ⋯ & ⋯ & ⋯ \\ X_{τ^{−1}} & ⋯ & ⋯ & X_{1} \end{matrix} )

Then the determinant of M is a monic polynomial in X_{1} of degree |G|.

2.3.2. Unconditionnal results. First recall the ℓ-adic version of Baker’s Theorem proved by Brumer [1] on the independance of logarithms:

Proposition 18. Let (α_{i})_{i=1,··· ,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} α_{i})_{i=1,··· ,r} is \overline{ℚ}-independent.

As noted in [6], we have:

Proposition 19. For all noetherian ℤ[G]-module M of k_{S}^{1}, the character of the ℚ_{ℓ}[G]-module ℚ_{ℓ} ⊗_{ℤ} Θ_{S}(M) contains all ℚ_{ℓ}-irreducible character χ represented in χ(M). More precisely:

\begin{aligned}χ(Θ_{S}(M)) & ≥ & ∑_{χ≤χ(M)}χ \\ & ≥ & ∑_{ψ∈Irr(ℂ_{ℓ})^{ψ≤χ(M)}} m_{ψ}ψ \end{aligned}

Proof. Let M_{0} be a principal ℚ[G]-sub module of M with character χ(M_{0}): M_{0} = ℚ_{ℓ}[G] · x, with x ∈ M_{0}. Let χ be a ℚ_{ℓ}-irreducible character dividing χ(M_{0}). The values of χ are in ℚ̄. One has

∑_{g∈G} χ(g^{−1})g · x ≠ 1. (3)

Let (x_{1}, ⋯, x_{r}) be a ℚ-basis of M_{0}. Look at the projection on χ of the image of x by L_{S}(Θ_{S}): If it is trivial, it implies that

L_{S}(Θ_{S}(∑_{g∈G} χ(g^{−1})g · x)) = 0.

Look at this vector equality on the first component to obtain:

log_{ℓ}(σ_{1}(∑_{g∈G} χ(g^{−1})g · x)) = 0.

Now we write g · x in the ℚ-basis (x_{1}, ⋯, x_{r}) to obtain

∑_{i=1,··· ,r} a_{i} log_{ℓ}(σ_{1}(x_{i})) = 0,

where a_{i} are algebraic over ℚ. Moreover this relation is not trivial by (3). Then Theorem of Baker-Brumer yields a contradiction. So the character χ appears in χ_{S}.

To finish, in our case we can apply a ℓ-adic version of Theorem 1.16 of Chapter 1 of [20] (I thank M. Waldschmidt for telling me the existence of this version):

Proposition 20. Let (x_{i,j})_{j=1,··· ,l^{i=1,··· ,d}} be a family of algebraic numbers. Let l(·) =

L_{S}(σ_{1}(·)) be the logarithmic on \overline{ℚ_{ℓ}}. Assume that for all (t_{i})_{i=1,··· ,d} ∈ ℤ^{d}\\{0\} and (s_{i})_{i=1,··· ,l} ∈ ℤ^{l}\{0} we have:

∑_{i,j} t_{i}s_{j}x_{i,j} ≠ 0.

Then the rank of the matrix (l(x_{i,j})))_{j=1,··· ,l^{i=1,··· ,d}} is at least \frac{ld}{l + d}.

As consequence of Proposition 20, we have:

Theorem 21. Let M be a ℚ[G]-module of finite type included in k_{S}^{1}. Let χ be a ℚ-character of ℚ[G]-module such that χ ≤ χreg. If m · χ ≤ χ(M), then

rk_{ℤ_{ℓ}}L_{S}(Θ_{S}(M)) ≥ \frac{m · n_{χ}}{m + n_{χ}} .

Proof. As m · χ ≤ Reg, there exists x_{1}, ⋯, x_{m} in M such that:

1. χ(ℚ[G] · x_{i}) = χ;

2. ℚ[G] · x_{i} = ℚ < x_{i}^{(1)}, ⋯, x_{i}^{(d)} >, where d = n_{χ}.

3. The elements x_{i}^{(j)} can be chosen as to be the Galois conjugates of x_{i}: There exists s_{1}, ⋯, s_{d} ∈ G such that

∀i, j , x_{i}^{(j)} = s_{j} · x_{i} = x_{i}^{s_{j}}.

4. The family (x_{i}^{(j)})_{j=1,··· ,d^{i=1,··· ,m}} is linearly ℚ-independants.

Put X_{i} = L_{S}(Θ_{S}(x_{i})). We will prove that the ℚ_{ℓ}-rank of {X_{1}, ⋯, X_{m}} is at least

r = \frac{m · n_{χ}}{m + n_{χ}}.

Suppose that the rank is less than r − 1. That means that for all sub-families {X_{i_{1}}, ⋯, X_{i_{s}}} with s ≤ r − 1, there exists (a_{j})_{j} ∈ ℚ_{ℓ}^{r−1} such that

∑_{j=1}^{r−1} a_{j}X_{i_{j}} = 0.

Using the same method as for the proof of Lemma 16, one obtains:

∀g ∈ G, ∑_{j=1}^{r−1} a_{j} log_{ℓ}(σ_{1}(g · x_{i_{j}})) = 0. (4)

Now look at the matrix M̃ of size d×m defined by: M̃_{j,i} = (log_{ℓ}(σ_{1}(x_{i}^{(j)})))_{i=1,··· ,m^{j=1,··· ,d}}:

its rank is at least r by Proposition 20. Moreover M̃ is a sub-matrix of M = (log_{ℓ}(σ_{1}(g · x_{i})))_{i=1,··· ,d^{g∈G}} (of size |G| × m) with rank at most r − 1 by (4). And so

we obtain a contradiction.

3. 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:

Definition 22. Let k/N be a Galois extension of number fields with group G and 𝔭 ⊂ O_{k}. Define S𝔭̄ = {𝔭^{g}, g ∈ G}.

We first will show what Schanuel Conjecture, Baker-Brumer Theorem and Theorem 21 imply for the ℤ_{ℓ}-rank of G_{S𝔭̄}^{ab}. After this, we will give some examples (§3.3 and 3.4) and conclude by the proof of Theorem 1 (§3.5).

3.1. With Schanuel’s Conjecture.

Proposition 23. 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_{S𝔭̄}^{1}. Then Schanuel’s conjecture implies: χ(Θ_{S𝔭̄}(M_{0})) = χreg ∧ χ(M_{0}).

Proof. Consequence of Proposition 14.

Theorem 24. 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_{S𝔭̄}^{ab} is finite.

Proof. Application of Proposition 23 and Theorem 5.

3.2. Unconditional results.

Theorem 25. Let k/N be an abelian extension of number fields with group G. Let ℓ be a prime such that there exists 𝔭 ∈ N above ℓ of degree 1. Then

G_{S𝔭̄}^{ab} ≃ ℤ_{ℓ}^{t} × Tor(G_{S𝔭̄}^{ab}),

where t = |G| − |ψ ∈ Irr(ℂ_{ℓ}), ψ ≤ χ(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.

Proof. This is a direct application of Propositions 13 and 19.

To finish, we give some applications of Theorem 21:

Theorem 26. 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 ≥ 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| < \frac{1 + \sqrt{3 + 4n}}{2} implies G_{S𝔭̄}^{ab} finite.

Proof. By hypothesis, in the extension k/N the character of the units contains (n_{1} − 1)χreg. By Theorem 21, one has:

rk_{ℤ_{ℓ}}L_{S}(Θ_{S}(E_{S})) ≥ \frac{|G|(n − 1)}{|G| + n − 1}.

Now by Theorem 5, the ℤ_{ℓ}-rank of G_{S𝔭̄}^{ab} is trivial if and only if

|G| − rk_{ℤ_{ℓ}}L_{S}(Θ_{S}(E_{S})) < 1.

And in particular this is the case, if |G| < \frac{1 + \sqrt{3 + 4n}}{2}.

3.3. Compositum of ℤ_{ℓ}-extensions.

Definition 27. Let K/k be a pro-ℓ-extension of a number field k. We denote by K̃ the compositum of ℤ_{ℓ}-extensions of k contained in K and by { \atop Gal(K/k)} the Galois group Gal(K̃/k). In particular { \atop Gal(K/k)} ≃ ℤ_{ℓ}^{t}.

We have immediately these properties:

Proposition 28. 1) If S ∩ S′ = ∅, then rk_{ℤ_{ℓ}} { \atop Gal(k_{S}k_{S′}/k)} = rk_{ℤ_{ℓ}} { \atop Gal(k_{S}/k)} +

rk_{ℤ_{ℓ}} { \atop Gal(k_{S′}/k)}.

2) If S ≠ S_{ℓ}, then rk_{ℤ_{ℓ}} { \atop Gal(k^{∞}k_{S}/k)} = 1 + rk_{ℤ_{ℓ}} { \atop Gal(k_{S}/k)}.

We can present:

Theorem 29. 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 k̃_{S_{ℓ}} = k^{∞} · ∏_{𝔭∈S_{ℓ}} k̃_{S𝔭̄} 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∞_{i}}}^{G} (1) = Ind_{D_{𝔭∞_{j}}}^{G} (1).

Moreover if for every ℂ_{ℓ}-irreducible characters ψ dividing Ind_{D_{𝔭∞}}^{G} (1), ψ divises it deg(ψ) times (which is equivalent to D_{𝔭_{∞}} is distinguished in G for an archimedean place 𝔭_{∞} cf. lemma 9) then k is a CM-field. Or, (iii) N is a imaginary quadratic field and [k : N] = 2.

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 23), Proposition 28 and Theorem 5, the number of ℤ_{ℓ}-extensions coming from the compositum of the fields k_{S𝔭̄} with the cyclotomic extension k^{∞}, is:

r_{1} + 2r_{2} − [N : ℚ] · dim_{ℚ_{ℓ}} ((−1 + ∑_{i} Ind_{D_{𝔭∞_{i}}}^{G} (1)) ∧ χreg) + 1.

So we want:

(5) r_{1} + r_{2} = [N : ℚ] · dim_{ℚ_{ℓ}} ((−1 + ∑_{i} Ind_{D_{𝔭∞_{i}}}^{G} (1)) ∧ χreg) .

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) ≥ 2. That means in particular that N is not an imaginary quadratic field. Suppose moreover that at least one archimedean place 𝔭_{∞_{i}} of N splits in k/N. As Ind_{D_{𝔭∞}}^{G} (1) contains the unit representation 1 and that r_{1}(N) + r_{2}(N) ≥ 2, we have

dim_{ℚ_{ℓ}} ((−1 + ∑ Ind_{D_{𝔭∞_{i}}}^{G} (1)) ∧ χreg) = |G|.

So (5) becomes:

\begin{aligned}r_{1} + r_{2} & = & [N : ℚ] · |G| \\ & = & [k : ℚ] = r_{1} + 2r_{2} \end{aligned}

and this is a contradiction. So N is totally real and k totally complex. The equality (5) now becomes:

\begin{aligned}\frac{|G| · [N : ℚ]}{2} & = & r_{2} \\ & = & [N : ℚ] · dim_{ℚ_{ℓ}} (χ(E_{S}) ∧ χreg) . \end{aligned}

Now, as N has at least two archimedean places,

(−1 + ∑ Ind_{D_{𝔭∞_{i}}}^{G} (1)) ∧ χreg = (∑ Ind_{D_{𝔭∞_{i}}}^{G} (1)) ∧ χreg.

Recall that Ind_{D_{∞_{i}}}^{G} (1)) is in χreg and is of dimension |G|/2. Suppose that there exist i and j, such that

Ind_{D_{𝔭∞_{i}}}^{G} (1)) ≠ Ind_{D_{𝔭∞_{j}}}^{G} (1);

One would have

dim_{ℚ_{ℓ}} ((∑ Ind_{D_{𝔭∞_{i}}}^{G} (1)) ∧ χreg) > |G|/2,

and so a contradiction. If all irreducible characters ψ dividing Ind_{D_{𝔭∞}}^{G} (1), occur with multiplicity deg(ψ), by Lemma 9, one has D_{𝔭_{∞_{i}}} = D_{𝔭_{∞_{j}}} ∀i, j. And so k is a CM field with k^{+} = k^{D_{𝔭∞_{i}}}, for one i.

3) Now if N an imaginary quadratic field, the equality (5) becomes

r_{2} = [N : ℚ](|G| − 1),

and the last case occurs.

3.4. 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} ≥ 2, and let K be a ℤ_{ℓ}-extension of k; Γ = Gal(K/k). Let S ⊂ 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 ∪_{n}k_{n} = K with [k_{n} : k] = ℓ^{n}, and S_{n} = {𝔓 ⊂ O_{k_{n}}, 𝔓 ∩ O_{k} ∈ S}. Also put X_{∞,S} = lim_{n^{←}} X_{n,S} (the restriction map is

the norm). Let Λ = ℤ_{ℓ}[[T]]. We know that X_{∞,S} is a finitely generated Λ-module pseudoisomorphic to

Λ^{r} ⊕ (⊕_{i}Λ/ℓ^{n_{i}}) ⊕ (⊕_{i}Λ/f_{i}) ,

f_{i} being polynomials of ℤ_{ℓ}[T]. For more details see, for example, [21]. We will determine r in three cases:

3.4.1. 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 χ(E_{k_{n}}) = (r_{1} + r_{2})χreg(G_{n}) − 1 and so by Theorem 25, 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.

3.4.2. 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^{∞} be the cyclotomic ℤ_{ℓ}-extension of k. Then for all n, Gal(k_{n}/k^{+}) ≃ ℤ/ℓ^{n}ℤ × ℤ/2ℤ. By using Theorem 25, one obtains

X_{n,S} ≃ ℤ_{ℓ}^{ℓ^{n}} × (finite).

So in this case r = 1.

3.4.3. 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} ≃ ℤ_{ℓ} × (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_{2ℓ^{n}} = ⟨x, y, x^{l^{n}} = 1, y^{2} = 1, yxy = x^{−1}⟩. This group has (ℓ^{n} − 1)/2 + 2 ℂ_{ℓ}- irreducible representations. Two of them are of degree 1: 1 and ψ_{0}(x) = 1 ψ_{0}(y) = −1, and (ℓ^{n} − 1)/2 representations ρ_{h} are of degree 2 (for h = 1, ⋯, (ℓ^{n} − 1)/2) :

ρ_{h}(x) = ( \begin{matrix}ζ^{−h} & 0 \\ 0 & ζ^{h} \end{matrix} ) , ρ_{h}(y) = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} )

where ζ is a ℓ^{n}-primitive root of 1. Now we can note that ∑_{h} ρ_{h} is a ℚ_{ℓ}-irreducible character of degree ℓ^{n} − 1 (The Schur index of ρ_{h} is 1). For an archimedean place 𝔭_{∞} of k^{+} the decomposition group D_{𝔭_{∞}} = ⟨x^{a}y⟩ in Gal(k_{n}/k^{+}) is of order 2. As ℓ is odd, the subgroups of order 2 of D_{2ℓ^{n}} are conjugated and so:

χ(E_{k_{n}}) = [k^{+} : ℚ] (∑_{h} ρ_{h} + 1) − 1.

One then has χ(Θ_{S_{n}}(E_{k_{n}})) = 1 + ∑_{h} ρ_{h} or χ(Θ_{S_{n}}(E_{k_{n}})) = 1 + 2 ∑_{h} ρ_{h}. In the first case X_{n,S} = ℤ_{ℓ}^{ℓ^{n}} × (finite) and in the second X_{n,S} = ℤ_{ℓ} × (finite). Thus Gal(k_{n,S_{n}}^{ab}/k_{∞}) ≃ ℤ_{ℓ}^{ℓ^{n}−1} ×(finite) in the first case, and Gal(k_{n,S_{n}}^{ab}/k_{∞}) is finite in the second case. Now we have to remark that Gal(k_{n,S_{n}}^{ab}/k_{∞}) ≃ X_{∞,S}/ω_{n} where ω_{n} = (1 + T)^{ℓ^{n}} − 1 (cf. [21] for example) and this implies:

rk_{ℤ_{ℓ}}Gal(k_{n,S_{n}}^{ab}/k_{∞}) = r · ℓ^{n} + a,

with a ∈ ℕ. Only the second case is possible. In conclusion we have for all n, χ(Θ_{S_{n}}(E_{k_{n}})) = 1 + 2 ∑_{h} ρ_{h} and r = 0.

Remark 3.1. Schanuel’s Conjecture implies in the last example that χ(Θ_{S_{n}}(E_{k_{n}})) = 1 + 2 ∑_{h} ρ_{h}.

3.5. Proof of Theorem 1. We recall the definition of i_{ℓ}(k) and I_{ℓ}(k):

Definition 30. (i) i_{ℓ}(k) = min_{S⊂S_{ℓ}}\{\frac{|S|}{|S_{ℓ}|}, G_{S}^{ab} is infinite\};

(ii) I_{ℓ}(k) = min c, such that ∀S ⊂ S_{ℓ}, \frac{|S|}{|S_{ℓ}|} ≥ c ⟹ G_{S}^{ab} is infinite.

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

f(x) = x(x^{ℓ} − x + 1) ∈ 𝔽_{ℓ}[x] (6)

which is the decomposition into irreducible polynomials in 𝔽_{ℓ}[x]. And this guarantees the irreducibility of f in ℚ(x). Let N = ℚ(δ), where theta is a root of f. Then N is of degree ℓ + 1 on ℚ and totally imaginary. The factorization (6) also gives us the factorization of ℓO_{N}:

ℓO_{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 5)

[N_{𝔭_{2}}/ℚ_{ℓ}] − (r_{1}(k) + r_{2}(k) − 1 + |T |) = ℓ − (ℓ + 1)/2 ≥ 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 ≥ n_{0}. For a field k_{n}, we have: |T (k_{n})| = ℓ^{n}, |S(k_{n})| ≤ ℓ^{n_{0}}, and

ℓ^{n} + 1 ≤ |S_{ℓ}(k_{n})| = |T (k_{n})| + |S(k_{n})| ≤ 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

i_{ℓ}(k_{n}) ≤ \frac{ℓ^{n_{0}}}{ℓ^{n} + 1}.

Now we apply Proposition 25 in the extension k_{n}/N with S_{N} = T = {𝔭_{1}}. The ℤ_{ℓ}-rank of k_{n}^{ab}_{S𝔭̄_{1}} /k_{n} is trivial, and so

I_{ℓ}(k_{n}) > \frac{ℓ^{n}}{|S_{ℓ}(k_{n})|} ≥ \frac{ℓ^{n}}{ℓ^{n} + ℓ^{n_{0}}} .

Letting n → ∞, 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.

3.6. Proof of Theorems 3 and 4. We first prove:

Theorem 31. Let k be a number field (for ℓ > 2, assume that k is totally complex) and S ⊂ S_{ℓ} such that:

1. For all 𝔭 ∉ S, rk_{ℤ_{ℓ}}G_{S}^{{𝔭}}^{ab} < rk_{ℤ_{ℓ}}G_{S}^{ab} (⇔ rk_{ℤ_{ℓ}}Θ_{S}(E_{S}^{{𝔭}}) > rk_{ℤ_{ℓ}}Θ_{S}(E_{S}));

2. For all S′ ⊊ S, rk_{ℤ_{ℓ}}G_{S}^{ab} > rk_{ℤ_{ℓ}}G_{S′}^{ab}.

Then Gal(\overline{k}(ℓ)/k̃_{S}) is free.

Proof. Let k̃_{S} (resp. k̃_{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}).

Lemma 32. With the conditions 1 and 2 of Theorem 31, one has for all finite places 𝔭 of k, [k̃_{S}_{𝔭} : k_{𝔭}] = ℓ^{∞}.

Proof. For a finite place 𝔭 of k, denote by D_{𝔭} the decomposition group of 𝔭 in Gal(k̃_{S}/k) ≃ 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 k̃_{S}^{D_{𝔭}} the subfield of k̃_{S} fixed by D_{𝔭}. Note that k̃_{S}^{D_{𝔭}} is included in k_{S}^{{𝔭}} and

k̃_{S}^{{𝔭}} ⊂ k̃_{S}.

Now assume that for 𝔭 ∉ S, D_{𝔭} is trivial. Then k̃_{S} = k̃_{S}^{D_{𝔭}} ⊂ k_{S}^{{𝔭}}. Also k̃_{S} ⊂ k̃_{S}^{{𝔭}}

and the equality k̃_{S} = k̃_{S}^{{𝔭}} yealds. This implies that for 𝔭 ∉ S, rk_{ℤ_{ℓ}}G_{S}^{{𝔭}}^{ab} = rk_{ℤ_{ℓ}}G_{S}^{ab} which contradicts the condition 1. Now for 𝔭 ∈ S, as we have:

k ⊂ k̃_{S}^{\{𝔭\}} ⊂ { \atop k_{S\\{𝔭\}}} ⊂ k̃_{S},

the conclusion is the same as for 𝔭 ∉ S.

Now we will apply a result which can be found in [16]: the ℓ-cohomological dimension Gal(\overline{k}/k̃_{S}) is less than 1. In consequence, we can now apply Proposition 2 page 85 of [16] and so cd_{ℓ}(Gal(\overline{k}(ℓ)/k̃_{S})) = 1.

3.6.1. Proof of Theorem 3. If S𝔭̄ = 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 5, we have rk_{ℤ_{ℓ}}Gal(K/N) = 1, and for 𝔭′ ≠ 𝔭, rk_{ℤ_{ℓ}}Gal(K^{{𝔭′}}/N) = 0. Then for all 𝔭 ∈ O_{N}, [K_{𝔭} : N_{𝔭}] = ∞. By base change, the same is true for k_{S𝔭̄}/k. Then the end of the proof of Theorem 31 can be applied.

3.6.2. Proof of Theorem 4. Let f(x) ∈ 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 S_{4}, and moreover, k/ℚ(\sqrt{d}) is unramified everywhere, with Galois group G = A_{4}. Let 𝔭|ℓ be a prime of N := ℚ(\sqrt{d}) which is of degree one. Put S_{N} = \{𝔭\} and S = S𝔭̄ = {𝔓 ∈ O_{k}, 𝔓 ∩ O_{N} = 𝔭}. We will study the action of A_{4}. The group A_{4} has three irreducible characters of degree one (1, ψ_{1}, ψ_{2}) plus one irreducible character ψ of degree 3. As the archimedean places are unramified in k/N, we have:

χ(E_{k}) = 2(1 + ψ_{1} + ψ_{2} + ψ) − 1 = 1 + 2ψ_{1} + 2ψ_{2} + 2ψ.

Now by Proposition 14, one has:

χ_{S} = 1 + ψ_{1} + ψ_{2} + 2ψ,

and so

rk_{ℤ_{ℓ}}G_{S}^{ab} = 3.

Now we prove that conditions 1 and 2 of Theorem 31 are satisfied.

1) First a lemma:

Lemma 33. Let 𝔮 be a prime of k prime to S. Let T = \{𝔮\} and \overline{T} = \{𝔮^{s}, s ∈ G\}.

Then rk_{ℤ_{ℓ}}G\overline{T} ; S^{ab} < rk_{ℤ_{ℓ}}G_{S}^{ab} if and only if rk_{ℤ_{ℓ}}G_{S}^{T}^{ab} < rk_{ℤ_{ℓ}}G_{S}^{ab}.

Proof. For a finite place 𝔮 of k, denote by D_{𝔮} the decomposition group of 𝔮 in k̃_{S}/k. First we have :

ℚ ⊂ N ⊂ k ⊂ k̃\overline{T} ; S ⊂ k̃_{S}^{T} ⊂ k̃_{S}^{D_{𝔮}} ⊂ k̃_{S}.

Note that k̃_{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\overline{T} ; S^{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 k̃_{S} = k̃_{S}^{T}. The decomposition group of 𝔮 in k̃_{S}/k is then trivial and so the decomposition group of

all conjugate of 𝔭 by G are trivial. Consequently, k̃_{S} ⊂ k\overline{T} ; S and so k̃_{S} ⊂ k̃\overline{T} ; S, giving

k̃_{S} = k̃\overline{T} ; S .

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 ≤ χreg, implying ψ ≤ Ind_{D_{𝔭}}^{G}. And so for all prime 𝔭 of N, we have (by using Proposition 11):

χ(ET̄ ; S) ≥ χreg.

By Proposition 14,

Θ_{S}(E\overline{T} ; S) = χreg,

and so by Theorem 5

rk_{ℤ_{ℓ}}G\overline{T} ; S^{ab} = 0.

By Lemma 33, one has the first condition of Theorem 31.

2) To prove the second condition, we use a method similar to that of Lemma 33. For a prime 𝔮 in S, denote by k_{S\{𝔮}} the maximal ℓ-extension of k, unramified outside

S\\{𝔮\} and let { \atop k_{S\\{𝔮\}}} be the compositum of ℤ_{ℓ}-extension included in k_{S\\{𝔮\}}. We have

ℚ ⊂ N ⊂ k ⊂ { \atop k_{S\\{𝔮\}}} ⊂ k̃_{S}^{I_{𝔮}} ⊂ k̃_{S},

where I_{𝔮} is the ramification group of 𝔮 in k̃_{S}/k. Now I_{𝔮} is a closed subgroup of Gal(K_{S}/k) ≃ 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 ∈ G, I_{𝔮^{s}} is also trivial, and so k̃_{S}/k is unramified everywhere which is impossible. The second condition of Theorem 31 is then verified.

4. On the group G_{S}

4.1. The exact value of the number of relations of G_{S}.

4.1.1. Relations and torsion. Leopoldt’s conjecture says that for S = S_{ℓ}, the ℤ_{ℓ}- rank of \overline{Θ_{S}(E_{S})} is exactly the ℤ-rank of the group of units of k. As noted in [11], this result has a link with the Schur multiplicator of G_{S}:

Proposition 34. Leopoldt’s Conjecture is true for (ℓ, k) if and only if H_{2}(G_{S_{ℓ}}, ℤ_{ℓ}) is trivial.

Proof. Proposition 34 is a direct consequence of the following Lemma:

Lemma 35 (see [13] for example). Let G be a pro-ℓ-group. Then

d(H_{2}(G, ℤ_{ℓ})) = r(G) − d(Tor(G)) = r(G) − d(G) + rk_{ℤ_{ℓ}}G^{ab}.

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.

The triviality of this cohomology group, allows us to see d(Tor(G_{S_{ℓ}})) as the number of relations of G_{S_{ℓ}}:

Theorem 36. Leopoldt’s Conjecture is equivalent to

r(G_{S_{ℓ}}) = d(Tor(G_{S_{ℓ}})).

4.1.2. General case.

Theorem 37. Assume S ⊂ S_{ℓ} (not empty). If rk_{ℤ_{ℓ}} \overline{Θ_{S}(E_{S})} = rk_{ℤ}E_{k}, one has: 1. r(G_{S}) = d(G_{S}) − ∑_{𝔭∈S}[k_{𝔭} : ℚ_{ℓ}] + r_{1} + r_{2} − 1 = d(Tor(G_{S}));

2. H_{2}(G_{S}, ℚ_{ℓ}/ℤ_{ℓ}) ≃ Tor(G_{S}).

Proof. By Propositions 35 and 6, one has:

0 ≤ d(H_{2}(G_{S}, ℤ_{ℓ})) ≤ r(G_{S}) − d(G_{S}) + rk_{ℤ_{ℓ}}G_{S}^{ab} ≤ rk_{ℤ}E_{k} − rk_{ℤ_{ℓ}} \overline{Θ_{S}(E_{S})}. (7)

The first assertion is then proved. For the second assertion, we use a result of Nguyen [11]. First by the inequality (7), H^{2}(G_{S}, ℚ_{ℓ}/ℤ_{ℓ}) = H_{2}(G_{S}, ℤ_{ℓ})^{∗} = 0. So the exact sequence

1 ⟶ ℤ_{ℓ} ⟶^{ℓ^{m}} ℤ_{ℓ} ⟶ ℤ/ℓ^{m}ℤ ⟶ 1,

gives for large m

H_{2}(G_{S}, ℤ/ℓ^{m}ℤ) ≃ Tor(G_{S}).

By passing to inductive limit, we have the result.

We can now give some examples.

4.1.3. Examples. 1) Let k be a quadratic field and S ⊂ S_{ℓ} (S not empty). Then r(G_{S}) = d(Tor(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 σ_{1} and σ_{2} be the two archimedean places of N. Suppose α ∈ N\N^{2} and β ∈ N\N^{2} have negative absolute norm satisfying σ_{1}(α) > 0, σ_{2}(β) > 0. Put k = N(\sqrt{α}, \sqrt{β}) ; k is a totally complex biquadratic extension over N. Put for the two cases S = {𝔓 ∈ k, 𝔓|𝔭}. Then one has:

r(G_{S}) = d(Tor(G_{S})) = \{ \begin{matrix}d(G_{S}) − (r_{2} + 1), if S = S_{ℓ} \\ d(G_{S}) − 1, else \end{matrix}

In the two cases, G_{S}^{ab} is infinite.

Proof. 1) Direct application of Theorem 37.

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 23. So we have:

χ_{S} = χreg ∧ χ(E_{S}) = χ(E_{S}) = χreg − 1.

ii) Denote N_{0} = N, N_{1} = N(\sqrt{α}), N_{2} = N(\sqrt{β}) and N_{3} = N(\sqrt{α · β}). Let φ_{i} be the ℂ_{ℓ}-irreducible character of G = Gal(k/N) such that the kernel of φ_{i} defined N_{i}. Then

χ(E_{S}) = φ_{0} + φ_{1} + φ_{2}.

And so χ_{S} = χ(E_{S}), by Proposition 19. In the two cases, Theorem 37 can be applied.

4.2. Free pro-ℓ-groups.

4.2.1. Classical results.

Proposition 38. The pro-ℓ-group G_{S_{ℓ}} is free if and only if one of the following is satisfied:

1. There exists a place 𝔭 of S_{ℓ} such that μ_{ℓ} ⊂ k_{𝔭}, then μ_{ℓ} ⊂ k, Δ_{S_{ℓ}} = 0 and |S_{ℓ}| = 1.

2. For all places 𝔭 of S_{ℓ}, k_{𝔭} does not contain μ_{ℓ}, and Δ_{S_{ℓ}} = 0. In this case, G_{S_{ℓ}} is the free pro-ℓ-group with r_{2} + 1 generators.

Proof. One has r(G_{S_{ℓ}})−d(G_{S_{ℓ}}) = −(r_{2}+1). Moreover d(G_{S_{ℓ}}) = r_{2}+1+∑_{𝔭∈S_{ℓ}} δ_{𝔭}+ d(Δ_{S_{ℓ}}) − δ_{k}. So r(G_{S_{ℓ}}) = 0 if and only if δ_{k} = ∑_{𝔭∈S_{ℓ}} δ_{𝔭} + d(Δ_{S_{ℓ}}).

4.2.2. General case. We have:

Theorem 39. Let S ⊂ S_{ℓ}. Then G_{S} is pro-ℓ-free in the following situations: 1. μ_{ℓ} ⊂ k, S = {𝔭} and Δ_{S} = 0.

2. For all places 𝔭 of S, k_{𝔭} does not contain μ_{ℓ} and Δ_{S} = 0. In the two cases G_{S} is the free pro-ℓ-group with ∑_{𝔭∈S}[k_{𝔭} : ℚ_{ℓ}] + 1 − (r_{2} + r_{1}) generators.

Proof. Recall the surjective map of Proposition 6 :

Δ_{S} ↠ ker (H^{2}(G_{S}, 𝔽_{ℓ}) → ⨁_{𝔭∈S^{∗}} H^{2}(G_{𝔭}, 𝔽_{ℓ}))^{∗}.

Now ker (H^{2}(G_{S}, 𝔽_{ℓ}) → ⨁_{𝔭} H^{2}(G_{𝔭}, 𝔽_{ℓ})) = H^{2}(G_{S}, 𝔽_{ℓ}) in one of the following:

1. |S| = 1, if μ_{ℓ} ∈ k;

2. k_{𝔭} does not contain μ_{ℓ} for all places 𝔭 of S (in this case, H^{2}(G_{𝔭}, 𝔽_{ℓ}) = 0). These conditions plus triviality of Δ_{S} imply H^{2}(G_{S}, 𝔽_{ℓ}) = 1.

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Christian Maire Laboratoire A2X Université Bordeaux I Cours de la Libération 33405 Talence Cedex France

maire@math.u-bordeaux.fr

Current address: Mathematical Sciences Research Institute 1000 Centennial Drive Berkeley, Califoria 94720 USA

maire@msri.org