The Hebrew University of Jerusalem

The Institute of Mathematics

On Nilpotency Problems in Metabelian Extensions of Local Fields

Sefi Ladkani December 1996

This work was carried out under the supervision of Prof. Ehud de-Shalit of the Hebrew University of Jerusalem.

I would like to express my deep gratitude to Prof. de-Shalit for his guidance.

Abstract

Let k be a local field. Abelian Local Class Field Theory (LCFT) makes a correspondence between abelian extensions L/k of k and closed subgroups of the completion k̂^{×} of the multiplicative group of the field. Lubin and Tate (see [L–T] and [Iw, ch. 4] for an exposition) have constructed a onedimensional commutative group law, enabling us to define an 𝔬_{k}-module structure on the maximal ideal of a completion of the algebraic closure of k. The fields generated by torsion points of this module play important role in the development of explicit LCFT.

Recently, De-Shalit and Koch, in [dS–K], developed an analogous theory for metabelian extensions. They introduced the groups 𝔊_{d}(k, ϕ) and 𝔊(k, ϕ), which in their theory are analogous to the group k̂^{×}, and defined a correspondence between metabelian extensions of k and closed subgroups of 𝔊(k, ϕ).

This work deals with nilpotency problems concerning metabelian extensions. A nilpotent (finite) group G is characterized by the fact that its lower and upper central series start with G (or 1) and reach 1 (or G).

We bound the center of the group 𝔊_{d}(k, ϕ) from above and from below. This enables us to show that the group 𝔊(k, ϕ), which correspond to the maximal metabelian of k, is centerless. We also show that the center of a dense subgroup of 𝔊_{d}(k, ϕ) is isomorphic to 𝔽_{q}^{×}. This is done by considering the action of k̂^{×} on power series 𝔽_{q^{d}}[[X]]^{×} and showing that the only power series which are fixed by every u ∈ U_{k} are the constant series. Another use of this fact is in our proof that the groups 𝔊_{d}(k, ϕ) cannot be represented as a semi-direct product of k̂^{×} and 𝔽_{q^{d}}[[X]]^{×}.

Using similar techniques we make some steps toward computing the second element of the upper central series of 𝔊_{d}(k, ϕ). It follows from our results that this series terminates at a rather small subgroup of 𝔊_{d}(k, ϕ).

Given a group G, one can search for maximal pro-nilpotent (inverse limit of nilpotent groups) subgroups of G, and also for the maximal pronilpotent quotient of G. Given a Galois extension L/K, this correspond to the question of finding maximal pro-nilpotent extensions L/M and the maximal pro-nilpotent extension N/K for K ⊆ M, N ⊆ L.

In analogy to Kummer extensions, we construct a family of metabelian extensions which are not nilpotent when the residue field of k is of order q > 2. The idea is to add to the fields k(ω_{n}), which contain the roots π^{n+1} · ω_{n} = 0, new roots π^{n+1} · α_{n} = a for a ∈ k, and to consider the resulting extensions k(ω_{n}, α_{n})/k, as it is done when adding n-th roots to a field containing the n-th roots of unity.

We deduce that for a non-ramified extension K/k, the metabelian extension (Kk_{π}^{0})^{ab}/K is not pro-nilpotent. In the case q = 2, the extensions k(ω_{n}, α_{n})/k are nilpotent, and their nilpotency class can be arbitrarily large.

Finally, we discuss some aspects of the lower central series. We define, for a closed subgroup H ≤ U_{k}, subgroups B_{d,α}(H) of 𝔽_{q^{d}}[[X]]^{×}, and relate them to nilpotent extensions of L in (k_{d}^{nr}k_{π})^{ab}, where L is a class field to P × H for P ≤ ⟨π^{d}⟩ℤ̂. This relation enables us to prove that for a nonramified extension K/k, the extension (Kk_{π})^{ab}/(Kk_{π}^{0}) is pro-nilpotent, and the maximal pro-nilpotent extension of K contained in (Kk_{π})^{ab} equals the maximal pro-nilpotent extension of k_{π,p}k^{nr} contained in (Kk_{π})^{ab}, where k_{π,p}k^{nr} is the class field of V, the cyclic subgroup of (q −1)-th roots of unity in U_{k}.

Contents

I Background 1

1 Local Fields 1 2 Abelian Local Class Field Theory 3 3 Coleman’s Norm Operator 8 4 Power Series over Fields 10 5 The Groups 𝔊_{d}(k, ϕ) 16 6 Aim of the Work 23

II Upper Central Series 27

7 Computation of Z(𝔊_{d}(k, ϕ)) 27 8 Computation of Z_{2}(𝔊_{d}(k, ϕ)) 33 9 Computation of Z(𝔊(k, ϕ)) 37

III Explicit Constructions 40

10 Definition of the Extensions k(α_{n}, ω_{n})/k 40 11 The Groups Gal(k(α_{n}, ω_{n})/k) 43 12 Central Series of Gal(k(α_{n}, ω_{n})/k) 49

IV Lower Central Series 54

13 The Groups B_{d,α}(H) 54 14 The Extensions N_{d,α}(H) 59 15 Applications to the Extensions (Kk_{π})^{ab}/K 62

References 69

Part I

Background

1 Local Fields

In this section, we briefly review the basic facts on local fields. For a detailed discussion, including proofs, we refer the reader to [Iw, chapters 1,2].

Basic Definitions

Let k be a field. A valuation on k is a function

ν : k^{×} ⟶ ℝ ∪ {∞}

satisfying

ν(0) = +∞ , ν(x) ∈ ℝ  for x ∈ k^{×}
ν(xy) = ν(x) + ν(y)  ∀x, y ∈ k
ν(x + y) ≥ min{ν(x), ν(y)}  ∀x, y ∈ k

Let

𝔬 = {x ∈ k : ν(x) ≥ 0}
𝔭 = {x ∈ k : ν(x) > 0}

𝔬 is a local ring whose maximal ideal is 𝔭. The field 𝔣 = 𝔬/𝔭 is called the residue field of k.

Throughout, we assume that the image ν(k^{×}) is a discrete subgroup of ℝ. In such case we can normalize ν such that ν(k^{×}) = ℤ.

Let (k, ν) be a field with a normalized valuation. An element π ∈ k for which ν(π) = 1 is called a prime of k. We have 𝔭 = π𝔬, and every ideal of 𝔬 is of the form 𝔭^{n} = π^{n}𝔬 for n ≥ 1. Thus 𝔬 is a principal domain.

Let

U = {x ∈ k : ν(x) = 0}
U_{n} = 1 + 𝔭^{n}

U = 𝔬 \ 𝔭 is the set of invertible elements of the ring 𝔬, and k^{×} = ⟨π⟩ × U. U_{n} are multiplicative subgroups of U, and they form a filtration of U. We have

U/U_{1} ≅ 𝔣^{×}

canonically, and

U_{n}/U_{n+1} ≅ 𝔣^{+}

non-canonically.

The metric defined by

d(x, y) = e^{−ν(x−y)}

equips the groups k^{+} and k^{×} with the topology it induces. In this topology, the family {𝔭^{n}} is a basis of neighborhoods of 0 (which are also subgroups of k^{+}), and the family {U_{n}} is a basis of neighborhoods of 1 in the group k^{×}.

A field (k, ν) is complete if the metric defined above is complete (i.e. every Cauchy sequence converges). In a complete field, a necessary and sufficient condition for the convergence of an infinite sum (product) is the convergence of the entries in the sum (product) to 0 (1).

Definition. A field (k, ν) is called local field if (k, ν) is complete and the residue field 𝔣 is finite.

Let (k, ν) be a local field with residue field 𝔽_{q}. Then U is a compact group. It contains the cyclic group V of the (q − 1)-th roots of unity. In fact, for every u ∈ U, the limit

lim_{n→∞} u^{q^{n}}

exists and equals the root whose image mod 𝔭 equals u mod 𝔭, and we have U = V × U_{1}.

Ramification

Let k′/k be a finite extension. There is a unique extension of ν to k′, denoted by ν′. Denote by 𝔬′, 𝔭′, 𝔣′ the local ring, etc. of k′. The ramification index e(k′/k) and residue degree f(k′/k) are defined by

e(k′/k) = [ν′(k′^{×}) : ν(k^{×})]
f(k′/k) = [𝔣′ : 𝔣]

If e(k′/k) = 1, we call the extension k′/k non-ramified. If f(k′/k) = 1, we call it totally ramified. We have the relation

[k′ : k] = e(k′/k)f(k′/k)

An infinite extension K/k is non-ramified if every finite extension k′/k contained in K/k is non-ramified. The concept of being totally ramified for such extensions is defined similarly.

For every d ≥ 1 we have a unique non-ramified extension of degree d over k, denoted by k_{d}^{nr}. It is Galois over k, and we have

Gal(k_{d}^{nr}/k) ≅ Gal(𝔽_{q^{d}}/𝔽_{q}) ≅ ℤ/dℤ

The maximal non-ramified extension over k, denoted by k^{nr}, is the union of the fields k_{d}^{nr}. Its residue field equals 𝔽̄_{q}, the algebraic closure of 𝔽_{q}, and

Gal(k^{nr}/k) ≅ Gal(𝔽̄_{q}/𝔽_{q}) ≅ ℤ̂

The Frobenius automorphism ϕ of k^{nr}/k is the one corresponding to the Frobenius automorphism of 𝔽̄_{q}/𝔽_{q}. It is characterized by

ϕ(x) ≡ x^{q} (mod 𝔭_{k^{nr}})

and generates (topologically) Gal(k^{nr}/k).

2 Abelian Local Class Field Theory

As in the previous section, we only define the basic concepts and give the main results, omitting their proofs. For details, we refer the reader to [Iw, ch. 4].

Formal Group Laws

Let R be a commutative ring with 1, and consider the ring of formal power series over R, denoted by R[[X]]. Addition and multiplication of power series are defined as usual. We can also define a composition of a power series ξ with a power series η ∈ XR[[X]]; if ξ = ∑_{i=0}^{∞} a_{i}X^{i}, define

ξ ∘ η = ∑_{i=0}^{∞} a_{i}η^{i}

and this is well-defined due to our assumption that η ∈ XR[[X]]. The composition of power series in more than one variable is defined similarly.

Definition. A one-dimensional commutative formal group law over R is a power series F (X, Y) ∈ R[[X, Y]] satisfying

F (X, Y) ≡ X + Y (mod deg 2)
F (F (X, Y), Z) = F (X, F (Y, Z))
F (X, Y) = F (Y, X)

where by (mod deg 2) we mean that the difference between the expressions in both sides of the equation does not have any terms of degree less than 2 (the degree of a monomial cX^{i}Y^{j} is i + j).

It can be shown that every such a formal group law has the form

F (X, Y) = X + Y + ∑_{i,j≥1} c_{ij}X^{i}Y^{j}

We think of F as a group law in the following sense; let f, g ∈ XR[[X]], then we can define “addition” ∔ by

f ∔ g = F (f(X), g(X))

and this definition makes XR[[X]] a commutative group.

A morphism f from one group law F to another group law G is a power series f ∈ XR[[X]] satisfying

f(F (X, Y)) = G(f(X), f(Y))

If f is invertible (with respect to the composition), it is called an isomorphism.

Lubin–Tate Power Series

In [L–T], Lubin and Tate constructed a family of formal group laws. They used such group laws to define an 𝔬-module structure on 𝔭_{Ω} (see below). In this approach to local class field theory, the study of the fields defined by considering the π-torsion points is a crucial step towards the (constructive) definition of the Artin map.

Let (k, ν) be a local field, and let K be the completion of the maximal non-ramified extension k^{nr} of k. Denote by 𝔬_{K}, 𝔭_{K} the ring of integers of K and its maximal ideal.

Let π ∈ K be a prime element. Denote by F_{π} the set of power series h ∈ 𝔬_{K}[[X]] for which

h(X) ≡ πX (mod deg 2)
h(X) ≡ X^{q} (mod 𝔭_{K})

Let ϕ be the extension of the Frobenius automorphism to K. For a power series F (X, Y) ∈ 𝔬_{K}[[X, Y]], denote by F^{ϕ}(X, Y) the power series obtained by applying ϕ on its coefficients.

Let h ∈ F_{π}. There is a unique formal group law F_{h}(X, Y) such that

h ∘ F_{h}(X, Y) = F_{h}^{ϕ}(h(X), h(Y))

Moreover, for every a ∈ 𝔬, there is a unique power series [a]_{h} ∈ 𝔬_{K}[[X]] satisfying

[a]_{h}(X) ≡ aX (mod deg 2)
h ∘ [a]_{h} = [a]_{h}^{ϕ} ∘ h

It follows that [a]_{h} is an endomorphism of the group law F_{h}, and the map a ↦ [a]_{h} is an injective ring homomorphism:

[ab]_{h} = [a]_{h} ∘ [b]_{h}
[a + b]_{h} = F_{h}([a]_{h}(X), [b]_{h}(X))

We note also that if k′/k is a finite non-ramified extension and π ∈ k′, h ∈ 𝔬_{k′}[[X]], then F_{h} and all [a]_{h} have their coefficients in 𝔬_{k′}. In particular, if π ∈ k, then F_{h} ∈ 𝔬[[X, Y]] and [a]_{h} ∈ 𝔬[[X]].

If π, π′ are two prime elements of K and h ∈ F_{h}, h′ ∈ F_{h′}, then there is an isomorphism θ ∈ 𝔬_{K}[[X]] between the formal groups F_{h} and F′_{h}. If, h, h′ ∈ F_{π} (same π) and π ∈ k′, then the isomorphism θ is actually in 𝔬_{k′}[[X]].

Torsion Points

Let π ∈ K, h ∈ F_{h}. Fix an algebraic closure Ω of k, and let μ be an extension of ν to Ω. Let Ω̄ be a completion of Ω, with valuation μ̄. As usual, let 𝔭Ω̄ denote the maximal ideal of 𝔬Ω̄.

We can define an 𝔬-module structure on 𝔭Ω̄ by

α ∔ β = F_{h}(α, β)  α, β ∈ 𝔭Ω̄
a · α = [a]_{h}(α)  a ∈ 𝔬, α ∈ 𝔭Ω̄

Let n ≥ −1. The π^{n+1}-torsion submodule W_{n} (we omit the reference to h, but these modules do depend on h) is defined by

W_{n} = {x ∈ 𝔭Ω̄ : π^{n+1} · x = 0}

(W_{−1} = 0). It can be shown that

W_{n} = {x ∈ 𝔭Ω̄ : (ϕ^{n}h ∘ ϕ^{n−1}h ∘ ⋯ ∘ ϕh ∘ h)(x) = 0}

and that these modules are finite, |W_{n}| = q^{n+1}.

Let W̃_{n} = W_{n} \ W_{n−1}. Any element ω_{n} ∈ W̃_{n} generates W_{n} as an 𝔬module, that is, W_{n} = 𝔬 · ω_{n}, and there is an isomorphism W_{n} ≅ 𝔬/𝔭^{n+1} defined by a ↦ a · ω_{n} (non-canonically).

The group of module automorphisms of W_{n} is denoted by Aut W_{n}. Every u ∈ U defines an automorphism ω ↦ u · ω. In fact, the automorphisms defined in this way are all possible automorphisms, and we have an isomorphism of groups

U/U_{n+1} ≅ Aut W_{n}  (2.1)

induced by mapping u into the automorphism it defines.

The fields k^{nr}(W_{n}) do not depend on the particular choice of π and h ∈ F_{π}.

Let k′/k be non-ramified. If π ∈ k′, h ∈ F_{π} ∩ 𝔬_{k′}[[X]], then the fields k′(W_{n}) do not depend on h. The extensions k′(W_{n})/k′ are Galois, totally ramified of degree (q−1)q^{n}. If ω_{n} is a generator of W_{n}, then k′(W_{n}) = k′(ω_{n}) and ω_{n} is a prime in that extension.

The homomorphism

Gal(k′(W_{n})/k′) ⟶ Aut W_{n}

defined by restricting σ ∈ Gal(k′(W_{n})/k′) to W_{n} is an isomorphism. Combining this and (2.1) we get that

Gal(k′(W_{n})/k′) ≅ U/U_{n+1}  (2.2)

Consider now the module W_{∞} = ⋃_{n=−1}^{∞} W_{n} and the field k′(W_{∞}) (which is the union of the fields k′(W_{n})). It is clear that

W_{∞} = {x ∈ 𝔭Ω̄ : there is an n ≥ −1 for which π^{n+1} · x = 0}

The following diagram, consisting of the isomorphisms in (2.2), is commutative, for m ≥ n;

Gal(k′(W_{m})/k′) ⟶^{≅} U/U_{m+1}
res↓  ↓
Gal(k′(W_{n})/k′) ⟶^{≅} U/U_{n+1}

where the left vertical arrow is the restriction map, and the right vertical arrow is the canonical map. Therefore, we get an isomorphism

Gal(k′(W_{∞})/k′) = \underleftarrow{lim}_{n} Gal(k′(W_{n})/k′) ≅ \underleftarrow{lim}_{n}U/U_{n+1} = U

One can verify that the explicit form of this homomorphism, denoted by δ : U → Gal(k′(W_{∞})/k′), is given by

δ(u)(ω) = u · ω

for every torsion point ω ∈ W_{∞}.

Main Results of Abelian Local Class Field Theory

Let π ∈ k, h ∈ F_{π} ∩ 𝔬[[X]]. Denote by k_{π} the field k(W_{∞}) (which does not depend on h), and let L_{k} = k^{nr}k_{π}.

The extension L_{k}/k is abelian, and the Frobenius automorphism ϕ of k^{nr}/k can be uniquely extended to an automorphism ψ_{π} of L_{k}, whose fixed field contains k_{π}.

Definition. The Artin map ρ_{k} : k̂^{×} → Gal(L_{k}/k) is defined by

ρ_{k}(a) = ψ_{π}^{ν}δ(u^{−1})

for a = π^{ν}u, ν ∈ ℤ̂, u ∈ U.

One can show that this definition of ρ_{k} does not depend on the particular choice of π ∈ k. Definition. Let F/k be an extension. The norm group N(F/k) is

N(F/k) = ⋂_{\begin{matrix}k⊆k′⊆F \\ [k′:k]<∞ \end{matrix}} N_{k′/k}(k̂′^{×})

where N_{k′/k} is the usual norm from k′^{×} to k^{×} (extended to the completions). Theorem. Let k be a local field. Denote by k^{ab} the maximal abelian extension over k. Then:

1. For every prime π ∈ k,

k^{ab} = k^{nr}k_{π}

2. The correspondence

F/k ⟼ N(F/k)

is one-to-one and onto from the set of abelian extensions F/k to the set of closed subgroups of k̂^{×}. If k′/k is abelian and finite, then

[k′ : k] = [k̂^{×} : N(k′/k)]

(the first equality of LCFT).

3. For every abelian extension F/k, we have

ρ_{k}(N(F/k)) = Gal(k^{ab}/F)

and this induces an isomorphism

k̂^{×}/N(F/k) ≅ Gal(F/k)

For a proof, we refer the reader to [Iw, ch. 6].

3 Coleman’s Norm Operator

Fix a non-ramified extension k′/k, with residue field 𝔣′. Let π ∈ k, h ∈ F_{π} ∩ 𝔬_{k}[[X]]. For α ∈ 𝔭Ω̄, and g ∈ 𝔬_{k′}((X)), define

g(X ∔ α) = g ∘ F_{h}(X, α)

Coleman defined in [Co] the norm operator

N_{h} : 𝔬_{k′}((X))^{×} ⟶ 𝔬_{k′}((X))^{×}

by

N_{h}(g) ∘ h = ∏_{ω∈W_{0}} g(X ∔ ω)

This operator is well defined, and satisfies the following properties:

N_{h}(g_{1}g_{2}) = N_{h}(g_{1})N_{h}(g_{2})
N_{h}(g) ≡ g^{ϕ} (mod 𝔭_{k′})
N_{h}(g) ≡ 1 (mod 𝔭_{k′}^{i+1})  if g ≡ 1 (mod 𝔭_{k′}^{i})

One can define the n-th iteration of N_{h}, by induction:

N_{h}^{n+1}(g) = N_{h}(N_{h}^{n}(g))

We have that

N_{h}^{n+1}(g) ≡ ϕN_{h}^{n}(g) (mod 𝔭_{k′}^{n+1})

thus the limit

N_{h}^{∞}(g) = lim_{n→∞} ϕ^{−n}N_{h}^{n}(g)

exists and satisfies

N_{h}(N_{h}^{∞}(g)) = ϕN_{h}^{∞}(g)

Definition. A power series g ∈ 𝔬_{k′}((X))^{×} which satisfies

N_{h}(g) = ϕg

is called a Coleman power series.

The set of all Coleman power series is a subgroup of 𝔬_{k′}((X))^{×} which will be denoted by C(k′, h).

Lemma 3.1. Let g ∈ C(k′, h), u ∈ U. Then

g ∘ [u]_{h} ∈ C(k′, h)

Proof. Since u acts as a permutation on W_{0}, and [u] commutes with F_{h}, we have

N_{h}(g ∘ [u]) ∘ h = ∏_{ω∈W_{0}} g ∘ [u] ∘ F_{h}(X, ω) = ∏_{ω∈W_{0}} g ∘ F_{h}([u](X), u · ω) =

∏_{ω∈W_{0}} g ∘ F_{h}(X, ω) ∘ [u] = N_{h}(g) ∘ h ∘ [u] = N_{h}(g) ∘ [u]^{ϕ} ∘ h

hence

N_{h}(g ∘ [u]) = N_{h}(g) ∘ [u]^{ϕ}

and the lemma follows.

Theorem. There is a split short exact sequence

1 ⟶ 1 + π𝔬_{k′}[[X]] ⟶ 𝔬_{k′}((X))^{×} ⟶^{N_{h}^{∞}} C(k′, h) ⟶ 1

The splitting i : C(k′, h) ⟶ 𝔬_{k′}((X))^{×} is the identity homomorphism.

For the proof, see [Co].

Corollary. Reduction modulo 𝔭_{k′} is an isomorphism

C(k′, h) ≅ 𝔣′((X))^{×}

Thus for any ξ ∈ 𝔣′((X))^{×} there is a unique lifting to Coleman power series g_{ξ} ∈ C(k′, h) such that

g_{ξ} mod 𝔭_{k′} = ξ

Normed sequences

For n ≥ 0, denote by k′_{n} the field k′(W_{n}).

Definition. A sequence (β_{n})_{n≥0}, β_{n} ∈ k′_{n} is called normic if, for every m ≥ n,

N_{k′_{m}/k′_{n}}(β_{m}) = β_{n}

Due to the transitivity of the norm, it is sufficient to take m = n + 1 in the above definition.

Let (ω_{n})_{n≥0} be a sequence such that

π · ω_{0} = 0
π · ω_{n+1} = ω_{n}  for every n ≥ 0

Then for every n ≥ 0, ω_{n} is a generator of the module W_{n}.

Lemma 3.2. Let g ∈ C(k′, h), and let, for every n ≥ 0,

β_{n} = g^{ϕ^{−n}}(ω_{n})

Then (β_{n})_{n≥0} is normic.

Proof. This is a straightforward computation:

N_{k′_{n+1}/k′_{n}}(β_{n+1}) = ∏_{σ∈Gal(k′_{n+1}/k′_{n})} σ(g^{ϕ^{−(n+1)}}(ω_{n+1}))

= ∏_{u∈U_{n}/U_{n+1}} g^{ϕ^{−(n+1)}}(u · ω_{n+1})

= ∏_{ω∈W_{0}} g^{ϕ^{−(n+1)}}(ω_{n+1} ∔ ω)

= N_{h}(g^{ϕ^{−(n+1)}}) ∘ h(ω_{n+1}) = g^{ϕ^{−n}} ∘ ω_{n} = β_{n}

This converse of this lemma is also true (although it is much harder to prove), see [Co]. We thus have

Theorem. There is a one-to-one and onto correspondence between Coleman power series and normic sequences, given by

g ⟼ (g^{ϕ^{−n}}(ω_{n}))_{n≥0}

Definition. A power series h ∈ F_{π} is called normic if

X ∈ C(k′, h)

It follows that for a normic power series, the sequence (ω_{n})_{n≥0} is normic (since it corresponds to X).

4 Power Series over Fields

The rings of power series 𝔽_{q^{d}}[[X]] and 𝔽̄_{q}[[X]] play a major role in the sequel. To provide a better understanding of these rings, the basic facts concerning power series over fields are presented in this section.

Let F be a field, and F ((X)) be the field of formal Laurent series over F, that is, the set

{ξ = ∑_{i=−n}^{∞} a_{i}X^{i} : n ∈ ℤ, a_{i} ∈ F }

where addition and multiplication are defined in the ordinary way.

Define a valuation v on F ((X)) by

v( ∑_{i=−n}^{∞} a_{i}X^{i}) = min {i ∈ ℤ : a_{i} ≠ 0} .

This valuation induces a complete metric on F ((X)). The ring F [[X]] of formal power series over F is exactly the ring of Laurent series with nonnegative valuation, and the group of units equals the multiplicative group F [[X]]^{×} of invertible power series.

It follows that a basis of open neighborhoods of 1 in F [[X]]^{×} is given by the family of subgroups {1 + X^{n}F [[X]]}.

Composition

The composition of power series over F is defined exactly as in the case of a general commutative ring R. We have the associative law

ξ ∘ (η ∘ ζ) = (ξ ∘ η) ∘ ζ

for any ξ ∈ F [[X]] and η, ζ ∈ XF [[X]], and also the distributive laws

(ξ_{1} + ξ_{2}) ∘ η = ξ_{1} ∘ η + ξ_{2} ∘ η
(ξ_{1}ξ_{2}) ∘ η = (ξ_{1} ∘ η)(ξ_{2} ∘ η)

Composition is a continuous operation, as the following lemma indicates.

Lemma 4.1. Let ξ_{n} → ξ be a convergent sequence of power series and η_{n} → η be a convergent sequence of power series having no constant term. Then

ξ_{n} ∘ η_{n} ⟶ ξ ∘ η

Proof. Write

\begin{aligned}ξ ∘ η − ξ_{n} ∘ η_{n} & = ξ ∘ η − ξ ∘ η_{n} + ξ ∘ η_{n} − ξ_{n} ∘ η_{n} \\ & = ξ ∘ η − ξ ∘ η_{n} + (ξ − ξ_{n}) ∘ η_{n} \end{aligned}  (4.1)

Let m ≥ 0. Since ξ_{n} → ξ, we have N_{1} such that v(ξ − ξ_{n}) ≥ m for every n > N_{1}. The power series η_{n} have no constant terms, hence

v((ξ − ξ_{n}) ∘ η_{n}) ≥ m  (4.2)

for every n > N_{1}.

Now, write ξ = ∑_{i=0}^{∞} a_{i}X^{i}. It follows that

ξ ∘ η − ξ ∘ η_{n} = ∑_{i=1}^{∞} a_{i}(η^{i} − η_{n}^{i})  (4.3)

(the constant terms a_{0} cancel each other). Since η_{n} → η, we have N_{2} such that v(η − η_{n}) ≥ m for every n > N_{2}. From

η^{i} − η_{n}^{i} = (η − η_{n}) ∑_{j=0}^{i−1} η^{i−1−j}η_{n}^{j}

and (4.3), we get that for every n > N_{2},

v(ξ ∘ η − ξ ∘ η_{n}) ≥ m

Combining this with (4.1) and (4.2), we conclude that for n > max{N_{1}, N_{2}},

v(ξ ∘ η − ξ_{n} ∘ η_{n}) ≥ m

Raising to p^{n}-th powers

Assume now that char F = p.

Lemma 4.2. The mapping ξ ↦ ξ^{p} is a continuous ring homomorphism of F [[X]].

Proof. We start by proving that this mapping is a ring homomorphism. It is clear that (ξη)^{p} = ξ^{p}η^{p}. Moreover, we have

(ξ + η)^{p} = ξ^{p} + η^{p} + ∑_{i=1}^{p−1} \binom{p}{i}ξ^{i}η^{p−i}

Since \binom{p}{i} are divisible by p for 1 ≤ i ≤ p − 1 and char F = p, the sum in the RHS of the equation is 0, hence (ξ + η)^{p} = ξ^{p} + η^{p} and ξ ↦ ξ^{p} is indeed a ring homomorphism.

Regarding continuity, it suffices to show that if ξ_{n} → 0, then ξ_{n}^{p} → 0. But this is clear, as v(ξ_{n}) ≥ m yields v(ξ_{n}^{p}) ≥ pm.

Lemma 4.3. let ξ ∈ 1 + XF [[X]]. Then lim_{n→∞} ξ^{p^{n}} = 1.

Proof. We prove by induction on n ≥ 0, that

ξ^{p^{n}} ∈ 1 + X^{p^{n}}F [[X]] .

This is clear for n = 0. Write ξ^{p^{n}} = 1 + X^{p^{n}}ξ′. Then

ξ^{p^{n+1}} = (ξ^{p^{n}})^{p} = (1 + X^{p^{n}}ξ′)^{p} = 1 + X^{p^{n+1}}ξ′^{p}

and the assertion follows.

Since {1 + X^{m}F [[X]]} is a basis of neighborhoods of 1, it follows that lim_{n→∞} ξ^{p^{n}} exists and equal to 1.

If F is also a finite field, we can say even more.

Lemma 4.4. Assume also that F is finite, and let ξ ∈ F [[X]]^{×}. Then

lim_{n→∞} ξ^{(|F |−1)p^{n}} = 1

Proof. Since F is finite, the multiplicative group F^{×} is cyclic of order |F |−1. Thus a^{|F |−1} = 1 for every a ∈ F^{×}.

Write ξ as a_{0} + Xξ_{1}. The constant term of ξ^{|F |−1} is a_{0}^{|F |−1} = 1, so that ξ^{|F |−1} ∈ 1 + XF [[X]]. Now the result follows from the previous lemma.

Endomorphisms of F [[X]]^{×}

We state and prove a criterion for the surjectivity of certain endomorphisms of F [[X]]^{×}. To make the exposition as general as possible, we have chosen to use the language of group rings.

Let G = Aut F, and let ℤ[G] be its group ring over ℤ. Let χ = ∑_{ψ∈G} c_{ψ}ψ be an element of ℤ[G], such that c_{ψ} = 0 for every ψ ∈ G except for a finite number of ψ-s.

χ gives rise to the endomorphisms of the multiplicative and the additive groups of F, respectively:

χ^{×} : F^{×} → F^{×}
χ^{+} : F^{+} → F^{+}

defined by

χ^{×}(x) = ∏_{ψ∈G} ψ(x)^{c_{ψ}}
χ^{+}(x) = ∑_{ψ∈G} c_{ψ}ψ(x)

where multiplication of y ∈ F by c ∈ ℤ is understood as a shorthand for writing y + y + ⋯ + y where the addition is done c times.

Since there is only finite number of c_{ψ} ≠ 0, the above endomorphisms are indeed well-defined.

In addition to these endomorphisms, χ gives rise to an endomorphism of the multiplicative group of F [[X]] as described below.

Let ξ ∈ F^{×} be an invertible power series. Write

ξ = ∑_{i=0}^{∞} a_{i}X^{i}

For ψ ∈ G, define ψ(ξ) as the power series obtained from ξ by applying ψ on the coefficients:

ψ(ξ) = ∑_{i=0}^{∞} ψ(a_{i})X^{i}  (4.4)

For general χ = ∑_{ψ∈G} c_{ψ}ψ, define

χ^{[]} : F^{×} ⟶ F^{×}

by

χ^{[]}(ξ) = ∏_{ψ∈G} ψ(ξ)^{c_{ψ}}

where ψ(ξ) is defined in (4.4). Again, χ^{[]} is well-defined as all but a finite number of the elements in the product are equal to 1.

Proposition 4.5. Let χ ∈ ℤ[G] be as above, and suppose that both χ^{×} and χ^{+} are surjective endomorphisms.

Then the endomorphism χ^{[]} is also surjective.

Proof. Write U = F [[X]]^{×}, U_{n} = 1 + X^{n}F [[X]]. Since U_{n} are subgroups which are invariant under coefficient-wise endomorphisms, we have χ^{[]}(U_{n}) ⊆ U_{n} (recall that χ^{[]} is a product of the coefficient-wise automorphisms ψ).

From the formulae

∏_{ψ} ψ(a_{0} + …)^{c_{ψ}} = (∏_{ψ} ψ(a_{0})^{c_{ψ}}) + …

∏_{ψ} ψ(1 + a_{n}X^{n} + …)^{c_{ψ}} = 1 + (∑_{ψ} c_{ψ}ψ(a_{n}))X^{n} + …

we get that the following diagrams commute:

\begin{matrix}U & ⟶^{χ^{[]}} & U \\ ^{p_{0}}↓ & & ↓^{p_{0}} \\ U/U_{1} & ⟶^{χ^{×}} & U/U_{1} \end{matrix}  (4.5)

\begin{matrix}U_{n} & ⟶^{χ^{[]}} & U_{n} \\ ^{p_{n}}↓ & & ↓^{p_{n}} \\ U_{n}/U_{n+1} & ⟶^{χ^{+}} & U_{n}/U_{n+1} \end{matrix}  (4.6)

where the vertical mappings p_{n} are defined by

p_{0}(a_{0} + a_{1}X + …) = a_{0}
p_{n}(1 + a_{n}X^{n} + …) = a_{n}  ,  n ≥ 0

and the quotients U/U_{1}, U_{n}/U_{n+1} are identified with F^{×}, F^{+}.

Let η ∈ U. By (4.5) and the assumption that χ^{×} is surjective, we can find ξ_{1} ∈ U such that

η_{1} = \frac{η}{χ^{[]}(ξ_{1})} ∈ U_{1}

Assume that we have been able to find ξ_{1}, …, ξ_{n} such that

η_{n} = \frac{η}{χ^{[]}(ξ_{1} · · · · · ξ_{n})} ∈ U_{n}  (4.7)

Then, by (4.6) and the assumption that χ^{+} is surjective, we can find ξ_{n+1} ∈ U_{n} such that

η_{n+1} = \frac{η_{n}}{χ^{[]}(ξ_{n+1})} ∈ U_{n+1}

Consider now the product ∏_{n=1}^{∞} ξ_{n}. As ξ_{n+1} ∈ U_{n}, this product converges to ξ ∈ U. Moreover, by (4.7) it follows that η = χ^{[]}(ξ). As η was arbitrary, we deduce that χ^{[]} is surjective.

In the following proposition, we list two corollaries of proposition 4.5 which will be important in the sequel.

Proposition 4.6. Let ϕ be the Frobenius automorphism of 𝔽̄_{q}, and let d, e ≥ 1. Then the following endomorphisms of 𝔽̄_{q}[[X]]^{×} are surjective:

ξ ⟼ \frac{ϕ^{d}ξ}{ξ}  (4.8)
ξ ⟼ ∏_{i=0}^{e−1} ϕ^{i·d}ξ  (4.9)

Proof. These mappings correspond to the elements A = ϕ^{d} − 1 and B = 1+ϕ^{d}+⋯+ϕ^{(e−1)d} in the group ring ℤ[Gal(𝔽̄_{q}/𝔽_{q})]. By the last proposition, it is sufficient to check the surjectivity of the endomorphisms defined by these elements on 𝔽̄_{q}^{×} and 𝔽̄_{q}^{+}.

For every x ∈ 𝔽̄_{q}, ϕ(x) = x^{q}. Therefore, A and B induce on 𝔽̄_{q}^{×} the mappings

x ↦ x^{q^{d}−1}
x ↦ x^{1+q^{d}+···+q^{(e−1)d}}

whose surjectivity follows from the fact that 𝔽̄_{q} is algebraically closed. Similarly, the mappings on 𝔽̄_{q}^{+} induced by A and B are

x ↦ x^{q^{d}} − x
x ↦ x + x^{q^{d}} + ⋯ + x^{q^{(e−1)d}}

Again, these mappings are surjective since 𝔽̄_{q} is algebraically closed.

5 The Groups 𝔊_{d}(k, ϕ)

In this section we define, for every d ≥ 1, the groups 𝔊_{d}(k, ϕ), and prove some of their basic properties. These groups were introduced in [dS–K] to describe the Galois groups Gal((k_{d}^{nr}k_{π})^{ab}/k). Their role in the theory of metabelian extensions is analogous to the role of the group k̂^{×} in the abelian theory.

We prove here new result about the action of U on power series in 𝔽_{q^{d}}((X)) through composition with {u}. This result will be used in the computation of the upper central series of the groups 𝔊_{d}(k, ϕ). A nice application of this result, on the H^{2} class corresponding to the extension 𝔊_{d}(k, ϕ) of 𝔽_{q^{d}}[[X]]^{×} by k̂^{×}, is given here. Namely, we show that this class is not trivial.

Setup

Let (k, ν) be a local field, with 𝔬 its ring of integers, 𝔭 the maximal ideal in 𝔬, and 𝔽_{q} the residue field. Let U denote the group of units in 𝔬.

Let ϕ be the Frobenius automorphism on k^{nr}/k, and fix once and for all an extension of ϕ to the separable closure of k, k^{sep}/k. Call this extension also ϕ. The fixed field of ϕ in k^{ab} is a maximal totally ramified abelian extension, hence it equals k_{π} for some prime π ∈ k.

Let h(X) ∈ F_{π} be a normic Lubin-Tate power series. As before, we have the formal group law F_{h}(X, Y) ∈ 𝔬[[X, Y]] which satisfies

h ∘ F_{h}(X, Y) = F_{h}^{ϕ}(h(X), h(Y)

and have the automorphisms [a]_{h} ∈ 𝔬[[X]] for which

[a]_{h}(X) ≡ aX (mod deg 2)
h ∘ [a]_{h} = [a]_{h} ∘ h

In the sequel, we omit the subscript h, meaning that h has been fixed once and for all.

For each a ∈ 𝔬, we define {a} to be the reduction of [a] modulo 𝔭. Since 𝔬/𝔭 ≅ 𝔽_{q}, we certainly have {a} ∈ 𝔽_{q}[[X]]. By [π] = h ≡ X^{q} (mod 𝔭), we get that {π} = X^{q}.

Lemma 5.1. If u ∈ U_{n}, then {u} ≡ X (mod deg q^{n}).

Proof. Write u = 1 + aπ^{n} for a ∈ 𝔬. Then

[u] = F_{h}(X, [aπ^{n}]) = X + [aπ^{n}] + ∑_{i,j≥1} c_{ij}X^{i}[aπ^{n}]^{j}

Reducing modulo 𝔭, and using {π^{n}} = X^{q^{n}}, we get that

{u} = X + {a}(X^{q^{n}}) + ∑_{i,j≥1} c_{ij}X^{i}({a}(X^{q^{n}}))^{j}

As {a} is divisible by X, {a}(X^{q^{n}}) is divisible by X^{q^{n}}. We therefore see that the right sum is divisible by X^{q^{n}+1}, hence

{u} ≡ X + āX^{q^{n}} (mod deg q^{n} + 1)

(ā = a mod 𝔭) and the lemma follows.

Remark. The proof of the lemma shows also that if u ∉ U_{n+1}, then

{u} ≢ X (mod deg q^{n} + 1) .

Lemma 5.2. The mapping U ⟶ (𝔽_{q}[[X]], ∘) defined by

u ⟼ {u}

is a continuous injection.

Proof. We already know that {uv} = {u} ∘ {v} for u, v ∈ U. To prove that the mapping is an injection, take u ≠ 1. Then u ∉ U_{n+1} for some n ≥ 0, and by the above remark,

{u} ≢ X (mod deg q^{n} + 1)

therefore {u} ≠ X.

The continuity of this map follows directly from the last lemma, which ensures that elements in smaller neighborhoods of 1 are mapped to power series in smaller neighborhoods of 1.

Definition of 𝔊_{d}(k, ϕ)

Let k̂^{×} be the completion of k^{×} defined by

k̂^{×} = ⟨π⟩ℤ̂ × U

(π is a fixed prime element of k). The group k̂^{×} acts on the group of power series 𝔽̄_{q}[[X]]^{×} by

ξ^{a} = ϕ^{−ν}ξ ∘ {u}  (5.1)

where a = π^{ν}u, ν ∈ ℤ̂, u ∈ U.

Let b = π^{μ}v. Then

\begin{aligned}(ξ^{a})^{b} & = ϕ^{−μ}(ϕ^{−ν}ξ ∘ \{u\}) ∘ \{v\} = ϕ^{−μ−ν}ξ ∘ \{u\} ∘ \{v\} = \\ & = ϕ^{−(μ+ν)}ξ ∘ \{u · v\} = ξ^{ab} \end{aligned}

so that (5.1) is indeed a group action.

The following proposition shows that there are not too many power series which are fixed by U.

Proposition 5.3. Let ξ ∈ 𝔽_{q^{d}}((X)), and let H ≤ U be a subgroup of finite index. Assume that for every u ∈ H,

ξ ∘ {u} = ξ

Then ξ is a constant series.

Proof. Denote by k′ the field k_{d}^{nr}. Let g_{ξ} ∈ C(k′, h) be the lifting of g to a Coleman power series. Then by lemma 3.1, g_{ξ} ∘ [u] is also a Coleman power series, and from

g_{ξ} ∘ [u] ≡ ξ ∘ {u} (mod 𝔭_{k′})

we deduce that it is the lifting of ξ ∘ {u} to C(k′, h). As ξ = ξ ∘ {u} for every u ∈ H, and the lifting is unique, we get that

g_{ξ} ∘ [u] = g_{ξ}

for every u ∈ H.

Define a sequence of generators (ω_{n}) by

π · ω_{0} = 0
π · ω_{n+1} = ω_{n}

By the correspondence between Coleman power series and norm-compatible sequences, we get that the sequence (β_{n}) defined by

β_{n} = g_{ξ}^{ϕ^{−n}}(ω_{n})

is normic. From now on, we consider the subsequence (γ_{n}) obtained from (β_{n}) by taking the values appearing at places which are integral multiples of d. We have

γ_{n} = g_{ξ}^{ϕ^{−dn}}(ω_{dn}) = g_{ξ}(ω_{dn})

(recall that g_{ξ} ∈ 𝔬_{k′}[[X]] on which ϕ^{d} acts trivially), and

(g_{ξ} ∘ [u])(ω_{dn}) = g_{ξ}([u]ω_{dn}) = g_{ξ}(δ(u)(ω_{dn})) = δ(u)(g_{ξ}(ω_{dn})) = δ(u)(γ_{n}) (5.2)

where δ is the isomorphism between U and Gal(k′(W_{∞})/k′).

Since H as a finite index in U, there is N ≥ 1 such that U_{d(N+1)} ≤ H. Let L = k′(W_{dN}). By (5.2) we get that

σ(γ_{n}) = γ_{n}

for every automorphism σ ∈ Gal(k′(W_{∞})/L), thus γ_{n} ∈ L for every n ≥ N.

Let n ≥ N. Since γ_{n+1} ∈ L and [k′(W_{d(n+1)}) : k′(W_{dn})] = q^{d}, we get that

γ_{n} = N_{k′(W_{d(n+1)})/k′(W_{dn})}(γ_{n+1}) = γ_{n+1}^{q^{d}}

thus

γ_{n} = γ_{n+m}^{q^{d}^{m}}  (5.3)

for every m ≥ 0.

The residue field of L is of order q^{d} (L is a totally ramified extension of k_{d}^{nr}). Therefore, for every x ∈ 𝔬_{L}, the sequence (x^{q^{dm}})_{m≥0} converges to a (q^{d} − 1)-th root of unity (or to zero). By (5.3) we deduce that γ_{n}, for every n ≥ N, is a root of unity (or 0). Therefore γ_{n} ∈ k′ and

γ_{n} = v ∈ V_{q^{d}−1} ∪ {0}

for every n ≥ 0. It follows that β_{n} ∈ k′ and β_{n} = v^{q^{nmodd}}

The power series corresponding to such normic sequences are exactly the constant series. Hence g_{ξ} is a constant series, thus ξ ∈ 𝔽_{q^{d}}^{×}.

Definition. Let A, B be commutative groups, and assume that there is an action of A on B, that is, there is a homomorphism ψ : A → Aut B. Denote by b^{a} the action of a ∈ A on b ∈ B.

Define the semi-direct product A ⋉ B by

A ⋉ B = {(a, b) : a ∈ A, b ∈ B}  (5.4)

where multiplication of (a, b) and (a′, b′) is defined as

(a, b)(a′, b′) = (aa′, b + b′^{a})  (5.5)

(we use multiplicative notation for A, and additive notation for B).

It is easy to verify that this definition satisfies the associative law, the identity is (1, 0), and the inverse of (a, b) is (a^{−1}, −b^{a^{−1}}).

By taking in the above definition A = k̂^{×}, B = 𝔽̄_{q}[[X]]^{×}, using the action defined in (5.1), we get a semi-direct product

k̂^{×} ⋉ 𝔽̄_{q}[[X]]^{×}

Definition. Let d ≥ 1. The group 𝔊_{d}(k, ϕ) is the subgroup of k̂^{×}⋉𝔽̄_{q}[[X]]^{×} defined by

𝔊_{d}(k, ϕ) = \{(a, ξ) ∈ k̂^{×} ⋉ 𝔽̄_{q}[[X]]^{×} : a = π^{ν}u, \frac{ϕ^{d}ξ}{ξ} = \frac{\{u\}(X)}{X}\}

To prove that these are subgroups, let (a, ξ), (b, η) ∈ 𝔊_{d}(k, ϕ), a = π^{ν}u, b = π^{μ}v. Then

(a, ξ)(b, η) = (ab, ξ · ϕ^{−ν}η ∘ {u})

and

\frac{ϕ^{d}(ξ · ϕ^{−ν}η ∘ \{u\})}{ξ · ϕ^{−ν}η ∘ \{u\}} = \frac{ϕ^{d}ξ}{ξ} · \frac{ϕ^{d}ϕ^{−ν}η ∘ \{u\}}{ϕ^{−ν}η ∘ \{u\}} = \frac{ϕ^{d}ξ}{ξ} · (ϕ^{−ν} \frac{ϕ^{d}η}{η} ∘ \{u\}) =

= \frac{\{u\}(X)}{X} · (\frac{\{v\}(X)}{X} ∘ \{u\}) = \frac{\{u\}(X)}{X} · \frac{\{vu\}(X)}{\{u\}(X)} = \frac{\{uv\}(X)}{X}

(note the use of the fact that ϕ acts trivially on {u} ∈ 𝔽_{q}[[X]]^{×}).

Proposition 5.4. There is a short exact sequence

1 → 𝔽_{q^{d}}[[X]]^{×} → 𝔊_{d}(k, ϕ) → k̂^{×} → 1  (5.6)

Proof. Define the mappings i : 𝔽_{q^{d}}[[X]]^{×} → 𝔊_{d}(k, ϕ), p : 𝔊_{d}(k, ϕ) → k̂^{×} by

i(ξ) = (1, ξ)
p((a, ξ)) = a

Since ξ ∈ 𝔽_{q^{d}}[[X]]^{×}, ϕ^{d}ξ/ξ = 1 = {1}(X)/X, hence (1, ξ) ∈ 𝔊_{d}(k, ϕ). Therefore i is well defined, and is clearly an injection.

Take a ∈ k̂^{×} and write it as a = π^{ν}u for ν ∈ ℤ̂, u ∈ U. By proposition 4.6, the mapping ξ ↦ ϕ^{d}ξ/ξ is surjective in 𝔽̄_{q}[[X]]^{×}, hence there is ξ ∈ 𝔽̄_{q}[[X]]^{×} for which

\frac{ϕ^{d}ξ}{ξ} = \frac{\{u\}(X)}{X}

It follows that for such ξ, (a, ξ) ∈ 𝔊_{d}(k, ϕ), thus p is surjective.

Now, Im i = Ker p follows from the fact that (1, ξ) ∈ 𝔊_{d}(k, ϕ) for ξ ∈ 𝔽̄_{q}[[X]]^{×} if and only if ξ ∈ 𝔽_{q^{d}}[[X]]^{×}.

The last proposition states that 𝔊_{d}(k, ϕ) is an extension of k̂^{×} by 𝔽_{q^{d}}[[X]]^{×}. By definition, the group 𝔊_{d}(k, ϕ) is embedded in the semi-direct product k̂^{×} ⋉ 𝔽̄_{q}[[X]]^{×}. It seems natural to ask whether there is an isomorphism

𝔊_{d}(k, ϕ) ≅ k̂^{×} ⋉ 𝔽_{q^{d}}[[X]]^{×}

preserving the first coordinate.

The answer to this question is negative, and the precise statement is given in the proposition below , which is an application of proposition 5.3.

Proposition 5.5. The sequence in (5.6) is not split.

Proof. Assume the contrary, i.e. that there is an injection j : k̂^{×} → 𝔊_{d}(k, ϕ) such that j ∘ p is the identity map on k̂^{×}.

Let

j(π^{d}) = (π^{d}, ξ)
j(u) = (u, η_{u})  for every u ∈ U

Since j is an injection and k̂^{×} is abelian, the images j(π^{d}) and j(u) commute for every u ∈ U. Carrying out the multiplications, we get

(π^{d}u, ξ · ϕ^{−d}η_{u}) = (uπ^{d}, η_{u} · (ξ ∘ {u}))

or equivalently,

\frac{ϕ^{−d}η_{u}}{η_{u}} = \frac{ξ ∘ \{u\}}{ξ}  (5.7)

for every u ∈ U.

Since (u, η_{u}) ∈ 𝔊_{d}(k, ϕ), we have

\frac{ϕ^{d}η_{u}}{η_{u}} = \frac{\{u\}(X)}{X} ∈ 𝔽_{q}[[X]]^{×}

and therefore

\frac{ϕ^{−d}η_{u}}{η_{u}} = ϕ^{−d} \frac{η_{u}}{ϕ^{d}η_{u}} = \frac{η_{u}}{ϕ^{d}η_{u}} = \frac{X}{\{u\}(X)}

Substituting this in (5.7), we get that

\frac{X}{\{u\}(X)} = \frac{ξ ∘ \{u\}}{ξ}

for every u ∈ U, or equivalently,

(ξ · X) ∘ {u} = ξ · X  (5.8)

for every u ∈ U.

Observe that ξ ∈ 𝔽_{q^{d}}[[X]]^{×}. By proposition 5.3, the only power series ζ ∈ 𝔽_{q^{d}}((X))^{×} for which ζ ∘ {u} = ζ for every u ∈ U are the constant series. But ξ · X cannot be constant, a contradiction.

This contradiction implies that our initial assumption of existence of such a j was false, hence the sequence (5.6) is not split.

Definition of 𝔊(k, ϕ)

Let d, e ≥ 1, and consider the mappings

ψ_{de,d} : 𝔊_{de}(k, ϕ) ⟶ 𝔊_{d}(k, ϕ)

defined by

ψ_{de,d} : (a, ξ) ⟼ (a, ∏_{i=0}^{e−1} ϕ^{i·d}ξ)

Since

\frac{ϕ^{de}(ξ)}{ξ} = ∏_{i=0}^{e−1} \frac{ϕ^{d(i+1)}ξ}{ϕ^{di}ξ} = \frac{ϕ^{d}(∏_{i=0}^{e−1} ϕ^{di}ξ)}{∏_{i=0}^{e−1} ϕ^{di}ξ} ,  (5.9)

we get that

\frac{ϕ^{de}ξ}{ξ} = \frac{\{u\}(X)}{X}

if and only if

\frac{ϕ^{d}(∏_{i=0}^{e−1} ϕ^{di}ξ)}{∏_{i=0}^{e−1} ϕ^{di}ξ} = \frac{\{u\}(X)}{X}

hence the mappings ψ_{de,d} are well-defined.

Using proposition 4.6, we prove Proposition 5.6. The mappings ψ_{de,d} are surjective. Proof. Take (a, η) ∈ 𝔊_{d}(k, ϕ). By proposition 4.6, since η ∈ 𝔽̄_{q}[[X]]^{×}, there is ξ ∈ 𝔽̄_{q}[[X]]^{×} such that

η = ∏_{i=0}^{e−1} ϕ^{i·d}ξ .

By (5.9) we get that

\frac{ϕ^{de}ξ}{ξ} = \frac{ϕ^{d}η}{η}

hence (a, ξ) ∈ 𝔊_{de}(k, ϕ) and ψ_{de,d}(a, ξ) = (a, η). Definition.

𝔊(k, ϕ) = \underleftarrow{lim} 𝔊_{d}(k, ϕ)

where the inverse limit is taken with respect to the homomorphisms ψ_{de,d} for every d, e ≥ 1.

For the proof of the following theorem, see [dS–K]. Theorem. Let k be a local field.

1. A correspondence

F/k ⟼ M(F/k)

between metabelian extensions (soluble extensions of degree 2) and closed subgroups of 𝔊(k, ϕ) can be defined. This correspondence is one-to-one and onto, and

[F : k] = [𝔊(k, ϕ) : M(F/k)]

when [F : k] is finite.

2. There is a mapping

ı : 𝔊(k, ϕ) ⟼ Gal((k^{ab})^{ab}/k)

which induces an isomorphism

𝔊(k, ϕ)/M(F/k) ≅ Gal(F/k)

for every metabelian extension F/k.

3. Let d ≥ 1. The following diagram is commutative:

𝔽_{q^{d}}[[X]]^{×}  ⟶  𝔊_{d}(k, ϕ)  ⟶  k̂^{×}
↓  ↓^{ı}  ↓^{ρ_{k}}
Gal((k_{π}k_{d}^{nr})^{ab}/k^{ab}) ⟶ Gal((k_{π}k_{d}^{nr})^{ab}/k) ⟶ Gal(k^{ab}/k)

where the top horizontal arrows are those of (5.6).

6 Aim of the Work

This work deals with nilpotency problems concerning metabelian extensions of a local field k. Before stating the objectives of this work, we briefly review some of the basic concepts of nilpotency and pro-nilpotency. Definition. Let G be a (topological) group, and let H, K ≤ G be subgroups. The commutator group [H, K] of H,K is

[H, K] = ⟨xyx^{−1}y^{−1} : x ∈ H, y ∈ K⟩

where by ⟨A⟩ we mean the closed subgroup of G generated by A.

Definition. Let G be a group. The lower central series of G is defined as

G^{0} = G
G^{n+1} = [G^{n}, G] for n ≥ 0

This is a decreasing sequence of closed subgroups of G.

Definition. A group G is nilpotent if there is an n for which G^{n} = 1. The least such n is called the nilpotency class of G.

In analogy to the lower central series, one can also define the upper central series. Let Z(G) denote the center of a group G. Then the groups Z_{n}(G) are defined inductively as

Z_{0}(G) = 1
Z_{n+1}(G)/ Z_{n}(G) = Z(G/ Z_{n}(G))

If G is finite, then a well-known result states that a necessary and sufficient condition for G to be nilpotent is that there exists an n such that Z_{n}(G) = G.

The Galois groups of (possibly infinite) field extensions are pro-finite groups, that is, inverse limits of finite groups. The notion of pro-nilpotent groups is defined similarly.

Definition. Let G be a group. G is pro-nilpotent if there exist nilpotent groups {H_{i}} (and mappings f_{ij} : H_{j} → H_{i}) such that

G = \underleftarrow{lim} H_{i}

Here is a characterization of pro-nilpotent groups by their lower central series.

Lemma 6.1. Let G be a group. Then G is pro-nilpotent if and only if

⋂_{n≥1} G^{n} = 1

Proof. The groups G/G^{n} are nilpotent (since (G/G^{n})^{n} = 1). It follows that if the intersection of the groups G^{n} is 1, then

G ≅ \underleftarrow{lim} G/G^{n}

is pro-nilpotent.

Conversely, assume that G = \underleftarrow{lim} H_{i}. For every n ≥ 1, the inverse limits \underleftarrow{lim} H_{i}^{n} are well-defined, since the homomorphisms

f_{ij} : H_{j} ⟶ H_{i}

map commutators to commutators, thus f_{ij}(H_{j}^{n}) ⊆ H_{i}^{n}. This argument shows also that these limits can be embedded canonically in G.

Since multiplication on G is defined by coordinates, we get that

G^{n} ⊆ \underleftarrow{lim} H_{i}^{n}

Now, take g in the intersection of all G^{n}. Its i-th coordinate lies in the intersection of all H_{i}^{n}, which is 1, as H_{i} is nilpotent. Therefore g = 1 and the lemma follows.

Lemma 6.2. Let G be a compact, pro-nilpotent group. Let N ◁ G be a normal open subgroup. Then G/N is nilpotent.

Proof. Since the groups G/G^{n} are nilpotent, it in enough to show that G^{n} ≤ N for some n.

Assume the contrary. Then for every n ≥ 1, we can find x_{n} ∈ G^{n} such that

x_{n} ∈ G^{n} , x_{n} ∉ N

Since G is compact, there is a subsequence (x_{n_{k}}) of (x_{n}) which converges to x ∈ G. As G^{n} form a decreasing sequence of closed subgroups, we must have x = 1, thus x ∈ N. But this is a contradiction, since x is the limit point of x_{n_{k}} ∉ N, and N is open.

Corollary. Let G be compact, pro-nilpotent, and let N ◁ G be a normal closed subgroup of finite index. Then G/N is nilpotent

Proof. The conditions on N imply that N is open in G.

Definition. Let E/F be Galois extension of fields. E/F is called pronilpotent if Gal(E/F) is pro-nilpotent.

Since Gal(E/F) is the inverse limit of Gal(E′/F) where [E′ : F] < ∞ contained in E, we get that E/F is pro-nilpotent if and only if every E′/F, for [E′ : F] < ∞ contained in E, is nilpotent.

We ask the following basic questions:

• Let k be a local field. Is (k^{ab})^{ab}/k pro-nilpotent ?

More generally, we seek for a description of the maximal pro-nilpotent extension N of k contained in (k^{ab})^{ab}/k.

• We can also consider the maximal nilpotent extension of class n over k. A description of the maximal 2-step nilpotent over k can be found in [dS–K].

• A similar question is to find large pro-nilpotent extensions (k^{ab})^{ab}/M.

In part II we analyze the upper central series of the groups 𝔊_{d}(k, ϕ), which correspond to the extensions (k_{d}^{nr}k_{π})^{ab}/k.

In section 7, we bound Z_{1}(𝔊_{d}(k, ϕ)) from above and from below, and then we show that the center of a dense subgroup of 𝔊_{d}(k, ϕ) is isomorphic to 𝔽_{q}^{×}. This is done by using proposition 5.3.

In section 8, we use this approach to compute Z_{2}(𝔊_{d}(k, ϕ)). Again, we show that for the dense subgroup of 𝔊_{d}(k, ϕ) mentioned above, the next group in the upper central series is rather small.

To conclude this part, we prove in section 9 that the group 𝔊(k, ϕ) is centerless.

In part III we construct a family of metabelian extensions, which are not nilpotent if the order of the residue field of k is greater than 2. This enables us to show that for every non-ramified extension K/k, the extension (Kk_{π}^{0})^{ab}/K is not pro-nilpotent, where k_{π}^{0}/k is an extension of order q − 1, generated by the root π · ω_{0} = 0. This is an example for a relatively small extension which is not pro-nilpotent.

In section 10 we define these extensions, while in section 11 we compute their Galois groups over k and show that they can be described as extensions of Aut W_{n} by W_{m}, for 0 ≤ m ≤ n. In section 12 we show that such groups are not nilpotent when q > 2, and their nilpotency class is m + 2 (in particular, it can be arbitrarily large), when q = 2.

In part IV we analyze the lower central series. In section 13 we define, for d ≥ 1, and a closed subgroup H ≤ U, a sequence of subgroups B_{d,α}(H) of 𝔽_{q^{d}}[[X]]^{×}, and prove their important properties.

In section 14, we relate these subgroups to nilpotent extensions over a field L for which

Gal(k^{ab}/L) = P × H

where P ≤ ⟨π^{d}⟩ℤ̂. The main result of this section is that

Gal((k_{d}^{nr}k_{π})^{ab}/N_{L}) ≅ B_{d,ω}(H)

for the maximal pro-nilpotent extension N_{L} over L contained in (k_{d}^{nr}k_{π})^{ab}.

In section 15 we use the results of the previous sections, for the specific cases H = U_{1} and H = U. We prove that B_{d,ω}(U_{1}) = 1 for every d ≥ 1, and deduce that for every non-ramified extension K/k, the extension (Kk_{π})^{ab}/(Kk_{π}^{0}) is pro-nilpotent (note that the construction in part III implies that (Kk_{π})^{ab}/K is not pro-nilpotent).

Later, we show that for such K, the maximal pro-nilpotent extension over K contained in (Kk_{π})^{ab} equals the maximal pro-nilpotent extension over k_{π,p}k^{nr} in (Kk_{π})^{ab}, where k_{π,p}k^{nr} is the class field of V, the group of (q − 1)-th roots of unity in U. We also give a description of the groups B_{d,ω}(V), which correspond to such maximal pro-nilpotent extensions.

Part II

Upper Central Series

7 Computation of Z(𝔊_{d}(k, ϕ))

In this section we present some results concerning Z(𝔊_{d}(k, ϕ)). We do not have yet an exact description of the center. However, we do have results that enable us to “bound” it from above and from below. In particular, as we see later, we can compute Z(𝔊(k, ϕ)) using these results.

Throughout this section, we fix d ≥ 1 and consider the group 𝔊_{d}(k, ϕ).

Upper Bounds

We start by defining a group which will serve as an upper bound for the center.

Definition.

ℌ_{d}(k, ϕ) = {(a, ξ) ∈ 𝔊_{d}(k, ϕ) : a ∈ ⟨π^{d}⟩ℤ̂, ξ ∈ 𝔽_{q}[[X]]^{×}}

It is straightforward to verify that ℌ_{d}(k, ϕ) is actually a subgroup of 𝔊_{d}(k, ϕ). The following propositions shed some light on the structure of ℌ_{d}(k, ϕ).

Proposition 7.1. ℌ_{d}(k, ϕ) ≅ ℤ̂ × 𝔽_{q}[[X]]^{×}.

Proof. Take two elements (π^{dν}, ξ), (π^{dμ}, η) in ℌ_{d}(k, ϕ). Then, since ϕ acts trivially on 𝔽_{q}[[X]]^{×}, we have

(π^{dν}, ξ)(π^{dμ}, η) = (π^{d(ν+μ)}, ξη)  (7.1)

Thus the mapping (π^{dν}, ξ) ↦ (ν, ξ) is a homomorphism from ℌ_{d}(k, ϕ) to ℤ̂ × 𝔽_{q}[[X]]^{×}. It is clear that this homomorphism is injective. It is also surjective since every ν ∈ ℤ̂ and ξ ∈ 𝔽_{q}[[X]]^{×} define an element (π^{dν}, ξ) ∈ ℌ_{d}(k, ϕ).

Proposition 7.2. ℌ_{1}(k, ϕ) is a maximal abelian subgroup of 𝔊_{1}(k, ϕ).

Proof. Eq. (7.1) shows that ℌ_{1}(k, ϕ) is abelian. To prove maximality, take an arbitrary element (a, ξ) ∈ 𝔊_{1}(k, ϕ) not belonging to ℌ_{1}(k, ϕ). Write a as π^{ν}u, where u ≠ 1. Then

(1, 1 + X)(a, ξ) = (a, (1 + X) · ξ)
(a, ξ)(1, 1 + X) = (a, ξ · (1 + {u}))

As {u} ≠ X, the above equations show that (a, ξ) does not commute with (1, 1 + X) ∈ ℌ_{1}(k, ϕ).

Remark. By metabelian local class field theory, one has

ℌ_{1}(k, ϕ) ≅ Gal(k_{π}^{ab}/k_{π}) .

We now show that ℌ_{d}(k, ϕ) serves as an upper bound for the center of 𝔊_{d}(k, ϕ).

Proposition 7.3. Z(𝔊_{d}(k, ϕ)) ⊆ ℌ_{d}(k, ϕ)

Proof. The idea of the proof is rather simple; we take an element in the center, and see what conditions it satisfies. We prove that these conditions imply that the element is contained in ℌ_{d}(k, ϕ).

Let (a_{0}, ξ_{0}) ∈ Z(𝔊_{d}(k, ϕ)) be an element in the center. Write a_{0} = π^{ν_{0}}u_{0} where ν_{0} ∈ ℤ̂ and u_{0} ∈ U. We have, for every (a, ξ) ∈ 𝔊_{d}(k, ϕ),

(a_{0}, ξ_{0})(a, ξ) = (a, ξ)(a_{0}, ξ_{0})

Rewriting this equation after carrying out the multiplications for a = π^{ν}u (ν ∈ ℤ̂, u ∈ U), we get

(a_{0}a, ξ_{0} · (ϕ^{−ν_{0}}ξ ∘ {u_{0}})) = (aa_{0}, ξ · (ϕ^{−ν}ξ_{0} ∘ {u}))

or equivalently,

\frac{ϕ^{−ν_{0}}ξ ∘ \{u_{0}\}}{ξ} = \frac{ϕ^{−ν}ξ_{0} ∘ \{u\}}{ξ_{0}}  (7.2)

Equation (7.2) holds for every element (a, ξ). We take a = 1 and a = π and examine the constraints imposed on (a_{0}, ξ_{0}).

Take a = 1, so that ν = 0 and u = 1. From the definition of 𝔊_{d}(k, ϕ), we see that in order to have (1, ξ) ∈ 𝔊_{d}(k, ϕ), ξ ∈ 𝔽̄_{q}[[X]]^{×} must satisfy

\frac{ϕ^{d}ξ}{ξ} = 1

or equivalently, ξ ∈ 𝔽_{q^{d}}[[X]]^{×}. Substituting in (7.2), we get

\frac{ϕ^{−ν_{0}}ξ ∘ \{u_{0}\}}{ξ} = 1  (7.3)

Substitute ξ = 1 + X in (7.3). We get {u_{0}} = X, thus u_{0} = 1 (recall that the map u ↦ {u} is an injection).

Rewriting (7.3), we obtain

\frac{ϕ^{−ν_{0}}ξ}{ξ} = 1 .

As this is true for every ξ ∈ 𝔽_{q^{d}}[[X]]^{×}, we must have ν_{0} ∈ dℤ̂.

Now write ν_{0} = dμ_{0} for μ_{0} ∈ ℤ̂. Eq. (7.2) gets the form

\frac{ϕ^{−dμ_{0}}ξ}{ξ} = \frac{ϕ^{−ν}ξ_{0} ∘ \{u\}}{ξ_{0}}  (7.4)

for every (a, ξ) ∈ 𝔊_{d}(k, ϕ).

Take a = π, so that u = 1 and ν = 1. Similar reasoning as in the case of a = 1 shows that (π, ξ) ∈ 𝔊_{d}(k, ϕ) for every ξ ∈ 𝔽_{q^{d}}[[X]]^{×}. Substituting in (7.4), we see that

1 = \frac{ϕ^{−1}ξ_{0}}{ξ_{0}}

hence ξ_{0} ∈ 𝔽_{q}[[X]]^{×}.

We have shown that for every element (a_{0}, ξ_{0}) in the center, a_{0} ∈ ⟨π^{d}⟩ℤ̂ and ξ_{0} ∈ 𝔽_{q}[[X]]^{×}, or in other words,

Z(𝔊_{d}(k, ϕ)) ⊆ ℌ_{d}(k, ϕ) .

Remark. In the proof we used only the commutativity of (a_{0}, ξ_{0}) and (a, ξ) where a ∈ ⟨π⟩. This corresponds to commutativity with all automorphisms σ ∈ Gal((k_{π}k_{d}^{nr})^{ab}/k_{π}).

We see from the proof of the preceding proposition, that a necessary and sufficient condition for an element (a_{0}, ξ_{0}) ∈ ℌ_{d}(k, ϕ), a_{0} = π^{dμ_{0}}, to be in Z(𝔊_{d}(k, ϕ)) is

\frac{ϕ^{−dμ_{0}}ξ}{ξ} = \frac{ξ_{0} ∘ \{u\}}{ξ_{0}}  (7.5)

for every (u, ξ) ∈ 𝔊_{d}(k, ϕ).

We now prove that the center of a dense subgroup of 𝔊_{d}(k, ϕ) is rather small.

Definition.

𝔉_{d}(k, ϕ) = {(a, ξ) ∈ 𝔊_{d}(k, ϕ) : a = π^{ν}u where ν ∈ ℤ, u ∈ U}

𝔉_{d}(k, ϕ) is a subgroup of 𝔊_{d}(k, ϕ). It is also dense since ⟨π⟩^{ℤ} is dense in ⟨π⟩ℤ̂.

Proposition 7.4. Z(𝔊_{d}(k, ϕ)) ∩ 𝔉_{d}(k, ϕ) = {(1, c) : c ∈ 𝔽_{q}^{×}} Proof. Let (a_{0}, ξ_{0}) be in the subgroup defined by the LHS. We already know that a_{0} = π^{dμ_{0}} where μ_{0} ∈ ℤ̂ and ξ_{0} ∈ 𝔽_{q}[[X]]^{×}.

Our assumption is that dμ_{0} ∈ ℤ is an integer. Since both (a_{0}, ξ_{0}) and (a_{0}, ξ_{0})^{−1} lie in the center, we may assume without loss of generality that μ_{0} ≥ 0. Let (u, ξ) ∈ 𝔊_{d}(k, ϕ) be arbitrary. Since

\frac{ϕ^{d}ξ}{ξ} = \frac{\{u\}(X)}{X} ∈ 𝔽_{q}[[X]] ,

we have

\frac{ϕ^{dn}ξ}{ξ} = \frac{\{u\}(X)^{n}}{X^{n}} , \frac{ϕ^{−dn}ξ}{ξ} = \frac{X^{n}}{\{u\}(X)^{n}}

for every n ≥ 0. Rewriting (7.5), we obtain

\frac{X^{μ_{0}}}{\{u\}(X)^{μ_{0}}} = \frac{ξ_{0} ∘ \{u\}}{ξ_{0}}

and this is equivalent to

(X^{μ_{0}}ξ_{0}) ∘ {u} = X^{μ_{0}}ξ_{0} ,  (7.6)

this equation being true for every u ∈ U.

Proposition 5.3 states that the only power series λ ∈ 𝔽_{q}[[X]] that satisfy

λ ∘ {u} = λ

for every u ∈ U, are the constant series. Applying this to our case, we get that μ_{0} must be 0 and ξ_{0} ∈ 𝔽_{q}^{×}.

On the other hand, it is easy to see that elements of the form (1, c), c ∈ 𝔽_{q}^{×}, commute with every (a, ξ) ∈ 𝔊_{d}(k, ϕ).

Corollary. Z(𝔉_{d}(k, ϕ)) = {(1, c) : c ∈ 𝔽_{q}^{×}}

Proof. It suffices to show that

Z(𝔉_{d}(k, ϕ)) = Z(𝔊_{d}(k, ϕ)) ∩ 𝔉_{d}(k, ϕ)

The inclusion ⊇ follows from the fact that 𝔉_{d}(k, ϕ) is a subgroup 𝔊_{d}(k, ϕ). The reverse inclusion follows from the fact that an element which commutes with every element in a dense subgroup of a given group, lies in the center of that group.

Corollary. 𝔊_{d}(k, ϕ)′ ∩ Z(𝔊_{d}(k, ϕ)) = {(1, c) : c ∈ 𝔽_{q}^{×}}

Proof. Let (a, ξ) ∈ 𝔊_{d}(k, ϕ)′. Then a = 1 and the result follows from the previous proposition.

Lower Bound

We already know that the center of the dense subgroup 𝔉_{d}(k, ϕ) is very small. We now show that this is not true for 𝔊_{d}(k, ϕ). The reason for that is the fact that power series ξ ∈ 𝔽̄_{q}[[X]]^{×} for which there is a ∈ k̂^{×} with (a, ξ) ∈ 𝔊_{d}(k, ϕ) have their coefficients in a field much smaller than 𝔽̄_{q}. This enables us to find a large family of homomorphisms whose action on such power series is trivial. These homomorphisms correspond to elements in the center, as described below.

We begin by examining such power series ξ more carefully.

Notation.

𝔽_{q^{d(q−1)p^{∞}}} = ⋃_{n=0}^{∞} 𝔽_{q^{d(q−1)p^{n}}}

Lemma 7.5. If ξ ∈ 𝔽̄_{q}[[X]]^{×} satisfies

\frac{ϕ^{d}ξ}{ξ} = \frac{\{u\}(X)}{X}

for a unit u ∈ U, then

ξ ∈ 𝔽_{q^{d(q−1)p^{∞}}} [[X]]^{×}

More precisely, if

ξ = ∑_{i=0}^{∞} a_{i}X^{i} ,

then for every n ≥ 0 and i < p^{n}, we have a_{i} ∈ 𝔽_{q^{d(q−1)p^{n}}}.

Proof. {u}(X)/X has a non-zero constant term, namely u mod 𝔭, thus it belongs to 𝔽_{q}[[X]]^{×}. By lemma 4.4, we have

lim_{n→∞} (\frac{\{u\}(X)}{X})^{(q−1)p^{n}} = 1

From the proof of that lemma, we deduce also that

(\frac{\{u\}(X)}{X})^{(q−1)p^{n}} ≡ 1 (mod X^{p^{n}})

for every n ≥ 0.

Since ϕ^{d}ξ/ξ ∈ 𝔽_{q}[[X]], we have

\frac{ϕ^{d(q−1)p^{n}}ξ}{ξ} = (\frac{ϕ^{d}ξ}{ξ} )^{(q−1)p^{n}} = (\frac{\{u\}(X)}{X})^{(q−1)p^{n}} ≡ 1 (mod X^{p^{n}})

Multiplying by ξ, we get

ϕ^{d(q−1)p^{n}}ξ ≡ ξ (mod X^{p^{n}})

As ϕ^{d(q−1)p^{n}} operates on ξ coefficient-wise, by examining the coefficients we conclude that for every i < p^{n},

ϕ^{d(q−1)p^{n}}(a_{i}) = a_{i}

or in other words, a_{i} lies in the fixed field of ϕ^{d(q−1)p^{n}}.

ϕ(x) = x^{q} is the Frobenius automorphism of 𝔽_{q}, therefore ϕ^{d(q−1)p^{n}}(x) = x^{q^{d(q−1)p^{n}}} is the Frobenius automorphism of 𝔽_{q^{d(q−1)p^{n}}}, and 𝔽_{q^{d(q−1)p^{n}}} is its fixed field. Hence, for every i < p^{n}, a_{i} ∈ 𝔽_{q^{d(q−1)p^{n}}} and

ξ ∈ 𝔽_{q^{d(q−1)p^{∞}}} [[X]]^{×}

Corollary. If (a, ξ) ∈ 𝔊_{d}(k, ϕ), then ξ ∈ 𝔽_{q^{d(q−1)p^{∞}}} [[X]]^{×}.

Proof. Write a = π^{ν}u. Then ϕ^{d}ξ/ξ = {u}(X)/X and the result follows from the proposition.

Lemma 7.6. Let ϕ be the Frobenius automorphism on 𝔽̄_{q}/𝔽_{q}. Then a necessary and sufficient condition for ν ∈ ℤ̂ to satisfy ϕ^{dν} ∈ Gal(𝔽̄_{q}/𝔽_{q^{d(q−1)p^{∞}}}) is

ν ∈ (q − 1)ℤ̂ ∩ ∏_{l≠p} ℤ_{l}  (7.7)

where ℤ̂ ≅ ∏_{l} ℤ_{l} canonically (l runs over all primes).

Proof. The statement ϕ^{dν} ∈ Gal(𝔽̄_{q}/𝔽_{q^{d(q−1)p^{∞}}}) is equivalent to the statement that ϕ^{dν} is the identity on every subfield 𝔽_{q^{d(q−1)p^{n}}} (n ≥ 0).

Now, ϕ^{dν} is the identity on 𝔽_{q^{d(q−1)p^{n}}} if and only if (q − 1)p^{n} | ν. Thus an equivalent condition on ν is to be divisible by every (q − 1)p^{n}.

Assume first that ν is divisible by every (q − 1)p^{n}. Since ν is divisible by q − 1, we surely have ν ∈ (q − 1)ℤ̂. Now, view ν as a vector (ν_{l})_{l} where l runs over all rational primes. As p^{n} | ν for every n ≥ 0, we must have ν_{p} = 0. Combining these conditions, we finally get

ν ∈ (q − 1)ℤ̂ ∩ ∏_{l≠p} ℤ_{l}

The above arguments can be reversed, hence the condition in (7.7) is also sufficient.

We can now prove

Proposition 7.7.

Z(𝔊_{d}(k, ϕ)) ∩ {(a, ξ) ∈ 𝔊_{d}(k, ϕ) : a = π^{dν} for ν ∈ (q − 1)ℤ̂ ∩ ∏_{l≠p} ℤ_{l}} =

{(a, c) : a = π^{dν} for ν ∈ (q − 1)ℤ̂ ∩ ∏_{l≠p} ℤ_{l} , c ∈ 𝔽_{q}^{×}} (7.8)

Proof. Let (a_{0}, ξ_{0}) be an element in the LHS of (7.8). By proposition 7.3, a_{0} = π^{dν_{0}} for ν_{0} ∈ ℤ̂, and ξ_{0} ∈ 𝔽_{q}[[X]]^{×}.

(a_{0}, ξ_{0}) satisfies (7.5), that is, for each (u, ξ) ∈ 𝔊_{d}(k, ϕ) where u ∈ U, the following equality holds:

\frac{ϕ^{−dν_{0}}ξ}{ξ} = \frac{ξ_{0} ∘ \{u\}}{ξ_{0}}  (7.9)

Take (u, ξ) ∈ 𝔊_{d}(k, ϕ). By lemma 7.5, ξ ∈ 𝔽_{q^{d(q−1)p^{∞}}} [[X]]^{×}. Now, lemma 7.6 states that ϕ^{dν_{0}} is the identity of 𝔽_{q^{d(q−1)p^{∞}}}, so that the LHS of (7.9) equals 1. Thus

1 = \frac{ξ_{0} ∘ \{u\}}{ξ_{0}}

for every u ∈ U. Proposition 5.3 now yields that ξ_{0} ∈ 𝔽_{q}^{×} is a constant series.

As (7.9) is also a sufficient condition for being in the center, we see that elements belonging to the RHS of (7.8) lie in the center.

Corollary. Z(𝔊_{d}(k, ϕ)) contains the subgroup

{(a, c) ∈ 𝔊_{d}(k, ϕ) : a = π^{dν} for ν ∈ (q − 1)ℤ̂ ∩ ∏_{l≠p} ℤ_{l} , c ∈ 𝔽_{q}^{×}}

Proof. Clear.

8 Computation of Z_{2}(𝔊_{d}(k, ϕ))

In this section, we present some results regarding the computation of the second element in the upper central series of 𝔊_{d}(k, ϕ). Throughout this section we keep d ≥ 1 fixed.

Proposition 8.1. Let G be a group and let z ∈ Z_{2}(G). Then the mapping

ρ_{z}(g) = [z, g]

is a homomorphism ρ_{z} : G ⟶ G′ ∩ Z(G).

Proof. Let ḡ be the image of g under the canonical homomorphism from G to G/ Z(G). Since

Z_{2}(G)/ Z(G) = Z(G/ Z(G))

and z ∈ Z_{2}(G), we have z̄ḡ = ḡz̄ for every g ∈ G, or equivalently, \overline{[z, g]} = 1, thus ρ_{z}(g) ∈ Z(G). Since ρ_{z}(g) is a commutator, it is surely in G′.

Now,

\begin{aligned}[z, gh] & = zghz^{−1}h^{−1}g^{−1} = \\ & = zgz^{−1}g^{−1}gzhz^{−1}h^{−1}g^{−1} = [z, g][z, h]^{g} = [z, g][z, h] \end{aligned}

because [z, h] ∈ Z(G) is invariant under conjugation. Similarly,

[z, g^{−1}] = zg^{−1}z^{−1}g = (g^{−1}zgz^{−1})^{−1} = [z, g]^{−g^{−1}} = [z, g]^{−1}

so that ρ_{z} : G ⟶ G′ ∩ Z(G) is a homomorphism.

Fix (a_{0}, ξ_{0}) ∈ Z_{2}(𝔊_{d}(k, ϕ)). The previous proposition states that there is a homomorphism

ρ_{(a_{0},ξ_{0})} : 𝔊_{d}(k, ϕ) ⟶ 𝔊_{d}(k, ϕ)′ ∩ Z(𝔊_{d}(k, ϕ))

From the corollary after proposition 7.4, the image of this homomorphism is contained in the subgroup {(1, c) : c ∈ 𝔽_{q}^{×}}. We can therefore write

ρ_{(a_{0},ξ_{0})}(a, ξ) = (1, c_{(a,ξ)})

for every (a, ξ) ∈ 𝔊_{d}(k, ϕ).

Let us now exploit some of the properties of the homomorphism c : (a, ξ) ↦ c_{(a,ξ)}. Recall that by definition of ρ,

[(a_{0}, ξ_{0}), (a, ξ)] = (1, c_{(a,ξ)})

Carrying out the multiplications, substituting a_{0} = π^{ν_{0}}u_{0} and a = π^{ν}u where u_{0}, u ∈ U, we get

\frac{ξ_{0}}{ϕ^{−ν}ξ_{0} ∘ \{u\}} · \frac{ϕ^{−ν_{0}}ξ ∘ \{u_{0}\}}{ξ} = c_{(a,ξ)}  (8.1)

Lemma 8.2. Let (1, ξ) ∈ 𝔊_{d}(k, ϕ). Then c_{(1,ξ)} = 1.

Proof. As a = 1, equation (8.1) has the form

\frac{ϕ^{−ν_{0}}ξ ∘ \{u_{0}\}}{ξ} = c_{(1,ξ)}  (8.2)

Substitute ξ = X + 1 in (8.2). We get

\frac{1 + \{u_{0}\}(X)}{1 + X} = c_{(1,1+X)}  (8.3)

From (8.3) we deduce that c_{(1,1+X)} = 1 and therefore {u_{0}} = X. Rewriting (8.2), we see that

\frac{ϕ^{−ν_{0}}ξ}{ξ} = c_{(1,ξ)}

Now let ξ be arbitrary and let ξ_{1} = 1 + Xξ. Substituting, we get

c_{(1,1+Xξ)} = \frac{ϕ^{−ν_{0}}ξ_{1}}{ξ_{1}} = \frac{1 + Xϕ^{−ν_{0}}ξ}{1 + Xξ} = \frac{1 + c_{(1,ξ)}Xξ}{1 + Xξ}

hence 1 = c_{(1,1+Xξ)} = c_{(1,ξ)}.

Corollary. c_{(a,ξ)} does not depend on ξ.

Proof. Let (a, ξ), (a, ξ′) be two elements of 𝔊_{d}(k, ϕ). Then

(1, \frac{ξ′}{ξ})(a, ξ) = (a, ξ′)

and therefore

c_{(1, \frac{ξ′}{ξ})}c_{(a,ξ)} = c_{(a,ξ′)}

and the result follows from c_{(1, \frac{ξ′}{ξ})} = 1.

Corollary. Let (a_{0}, ξ_{0}) ∈ Z_{2}(𝔊_{d}(k, ϕ)). Then a_{0} ∈ ⟨π^{d}⟩ℤ̂.

Proof. Write a_{0} = π^{ν_{0}}u_{0} for ν_{0} ∈ ℤ̂, u_{0} ∈ U. In the proof of lemma 8.2 we saw that {u_{0}} = X, thus u_{0} = 1. Moreover, by that lemma we also get that c_{(1,ξ)} = 1 for every (1, ξ) ∈ 𝔊_{d}(k, ϕ). Therefore

\frac{ϕ^{−ν_{0}}ξ}{ξ} = 1

for every ξ ∈ 𝔽_{q^{d}}[[X]]^{×}, hence ν_{0} ∈ dℤ̂.

Remark. One can also prove that c_{(1,ξ)} = 1 by metabelian class field theory, as follows. We have a commutative diagram

𝔽_{q^{d}}[[X]]^{×}  ⟶  𝔊_{d}(k, ϕ)  ⟶  k̂^{×}
↓  ↓  ↓
Gal((k_{π}k_{d}^{nr})^{ab})/k^{ab}) ⟶ Gal((k_{π}k_{d}^{nr})^{ab}/k) ⟶ Gal(k^{ab}/k)

Since k^{ab}/k is the maximal abelian extension over k, we must have that 𝔊_{d}(k, ϕ)′ = {(1, ξ) : ξ ∈ 𝔽_{q^{d}}[[X]]^{×}}. Now, as c : 𝔊_{d}(k, ϕ) → 𝔽_{q}^{×} is an homomorphism into an abelian group, its kernel contains the commutator subgroup.

Since c_{(a,ξ)} does not depend on ξ, we can regard c as an homomorphism c : k̂^{×} → 𝔽_{q}^{×}, and rewrite (8.1) as

\frac{ϕ^{−dμ_{0}}ξ}{ξ} · \frac{ξ_{0}}{ϕ^{−ν}ξ_{0} ∘ \{u\}} = c_{a}  (8.4)

for (π^{dμ_{0}}, ξ_{0}) ∈ Z_{2}(𝔊_{d}(k, ϕ)) and every (a, ξ) ∈ 𝔊_{d}(k, ϕ), where a = π^{ν}u.

Lemma 8.3. c : U → 𝔽_{q}^{×} is continuous.

Proof. Let u_{n} → u. For each n, choose ξ_{n} such that (u_{n}, ξ_{n}) ∈ 𝔊_{d}(k, ϕ). Since 𝔽̄_{q}[[X]]^{×} is compact, we can find a convergent subsequence of {ξ_{n}}. To simplify notation, we call it again {ξ_{n}}.

Since ξ_{n} → ξ, and since u ↦ {u} is continuous, we get that

\frac{\{u\}(X)}{X} = lim_{n→∞} \frac{\{u_{n}\}(X)}{X} = lim_{n→∞} \frac{ϕ^{d}ξ_{n}}{ξ_{n}} = \frac{ϕ^{d}ξ}{ξ}

therefore (u, ξ) ∈ 𝔊_{d}(k, ϕ).

Now, by (8.4),

c_{u} = \frac{ϕ^{−dμ_{0}}ξ}{ξ} · \frac{ξ_{0}}{ξ_{0} ∘ \{u\}} = lim_{n→∞} \frac{ϕ^{−dμ_{0}}ξ_{n}}{ξ_{n}} · lim_{n→∞} \frac{ξ_{0}}{ξ_{0} ∘ \{u_{n}\}} = lim_{n→∞} c_{u_{n}}

Lemma 8.4. U_{1} ≤ Ker c.

Proof. Let u ∈ U_{1}. Then u^{q^{n}} → 1. But since c(u) ∈ 𝔽_{q}^{×},

c(u^{q^{n}}) = c(u)^{q^{n}} = c(u)

and by continuity of c, c(u) = c(1) = 1.

As in the case of the center, we can now prove that the intersection of Z_{2}(𝔊_{d}(k, ϕ)) and the dense subgroup 𝔉_{d}(k, ϕ) is quite small.

Proposition 8.5. Z_{2}(𝔊_{d}(k, ϕ)) ∩ 𝔉_{d}(k, ϕ) ⊆ {(1, x) : x ∈ 𝔽_{q^{d}}^{×}}

Proof. Rewriting (8.4) for (u, ξ) when u ∈ U_{1}, and using lemma 8.4, we get

\frac{ϕ^{−dμ_{0}}ξ}{ξ} · \frac{ξ_{0}}{ξ_{0} ∘ \{u\}} = 1

But this is precisely the condition in (7.5).

Now the proposition follows by repeating the argument in the proof of proposition 7.4, replacing U with the group U_{1} when using proposition 5.3.

Corollary. Z_{2}(𝔊_{1}(k, ϕ)) ∩ 𝔉_{1}(k, ϕ) = Z(𝔉_{1}(k, ϕ)).

Proof. Apply the proposition with d = 1, and use the corollary after proposition 7.4.

9 Computation of Z(𝔊(k, ϕ))

In this section, we compute Z(𝔊(k, ϕ)) using the results we have obtained on Z(𝔊_{d}(k, ϕ)). As 𝔊(k, ϕ) is defined as an inverse limit of 𝔊_{d}(k, ϕ), considerations involving such limits come into play.

We start by computing the inverse limit of the groups ℌ_{d}(k, ϕ).

Proposition 9.1. \underleftarrow{lim}_{d}ℌ_{d}(k, ϕ) = 1.

Proof. First we show that the inverse limit is well defined. To do this, we prove that for every d, e ≥ 1, we have

ψ_{de,d}(ℌ_{de}(k, ϕ)) ⊆ ℌ_{d}(k, ϕ)

Take an element (a, ξ) ∈ ℌ_{de}(k, ϕ). We have to verify that

(a, ∏_{i=0}^{e−1} ϕ^{di}ξ) ∈ ℌ_{d}(k, ϕ) ,

but this is clear from the definition, since ⟨π^{de}⟩ℤ̂ ⊆ ⟨π^{d}⟩ℤ̂ and ξ ∈ 𝔽_{q}[[X]]^{×}.

Now, take an element (a_{d}, ξ_{d})_{d} ∈ \underleftarrow{lim}_{d}ℌ_{d}(k, ϕ), and fix d ≥ 1. By definition of the maps ψ_{de,d}, we have (for the first coordinate)

a_{de} = a_{d}

for every e ≥ 1, where a_{de} ∈ ⟨π^{de}⟩ℤ̂. Thus

a_{d} ∈ ⋂_{e≥1}⟨π^{de}⟩ℤ̂ = {1}

Considering the second coordinate, we have, for every e ≥ 1,

ξ_{d} = ∏_{i=0}^{e−1} ϕ^{di}ξ_{de} = ξ_{de}^{e}  (9.1)

where the latter equality follows from the fact that ξ_{de} ∈ 𝔽_{q}[[X]]^{×}. Let n ≥ 0, and choose e = (q − 1)p^{n}. Substituting for e in (9.1), we get that

ξ_{d} = ξ_{d(q−1)p^{n}}^{(q−1)p^{n}}

Since η^{(q−1)p^{n}} ∈ 1 + X^{p^{n}}𝔽_{q}[[X]] for every η ∈ 𝔽_{q}[[X]]^{×}, we see that ξ_{d} ∈ 1 + X^{p^{n}}𝔽_{q}[[X]]. As this is true for every n ≥ 0, we finally get ξ_{d} = 1.

Now, since a_{d} = 1 and ξ_{d} = 1 for every d ≥ 1, the proposition follows.

We now prove a simple proposition which connects the center of an inverse limit of groups with the inverse limit of the centers.

Proposition 9.2. Let I be a partially ordered set, and let {G_{i}} be a family of groups indexed by I. Suppose that we have homomorphisms

f_{ij} : G_{i} ⟶ G_{j}

for every i > j, and define \underleftarrow{lim} G_{i} in the usual manner. Set also Z_{i} = Z(G_{i}). If the above homomorphisms are surjective then we have the following:

(i) f_{ij}(Z_{i}) ⊆ Z_{j} for every i > j. Hence \underleftarrow{lim} Z_{i} is defined and can be canonically embedded in \underleftarrow{lim} G_{i}.

(ii) Z(\underleftarrow{lim} G_{i}) = \underleftarrow{lim} Z_{i}.

Proof. Fix i > j, and let z_{i} ∈ Z_{i}. To show that f_{ij}(z_{i}) ∈ Z_{j}, take g_{j} ∈ G_{j}. Since f_{ij} is surjective, we can find g_{i} ∈ G_{i} for which g_{j} = f_{ij}(g_{i}). Now,

f_{ij}(z_{i})g_{j} = f_{ij}(z_{i})f_{ij}(g_{i}) = f_{ij}(z_{i}g_{i}) = f_{ij}(g_{i}z_{i}) = g_{j}f_{ij}(z_{i}) (9.2)

where we used the fact that z_{i} commutes with g_{i}. Since (9.2) is true for every g_{j} ∈ G_{j}, we have f_{ij}(z_{i}) ∈ Z_{j}.

We may now define \underleftarrow{lim} Z_{i} using the maps f_{ij}. In particular, (z_{i})_{i} ∈ \underleftarrow{lim} Z_{i} can be viewed as an element of \underleftarrow{lim} G_{i} (since z_{i} ∈ G_{i} and f_{ij}(z_{i}) = z_{j} for every i > j).

To prove the second assertion of the proposition, take z ∈ Z(\underleftarrow{lim} G_{i}), and write it as z = (z_{i})_{i}, where z_{i} ∈ G_{i}. Since the maps f_{ij} are surjective, the projections from \underleftarrow{lim} G_{i} to each G_{i} are surjective. Take i ∈ I and g_{i} ∈ G_{i}. We can find g ∈ \underleftarrow{lim} G_{i} whose i-th coordinate is g_{i}. Since zg = gz, we surely have z_{i}g_{i} = g_{i}z_{i}. As this is true for every g_{i} ∈ G_{i} and every i ∈ I, we get z_{i} ∈ Z_{i} for every i, hence

Z(\underleftarrow{lim} G_{i}) ⊆ \underleftarrow{lim} Z_{i}

To prove the opposite inclusion, take z = (z_{i})_{i} ∈ \underleftarrow{lim} Z_{i} and let g = (g_{i})_{i} ∈ \underleftarrow{lim} G_{i} be arbitrary. Since z_{i}g_{i} = g_{i}z_{i} for every i, we get zg = gz, hence z ∈ Z(\underleftarrow{lim} G_{i}) and the proposition is proven.

With proposition 9.2 at hand, we can now state and prove the main theorem of this section.

Theorem 9.3. Z(𝔊(k, ϕ)) = 1.

Proof. 𝔊(k, ϕ) = \underleftarrow{lim}_{d}𝔊_{d}(k, ϕ) is defined using the homomorphisms

ψ_{de,d} : 𝔊_{de}(k, ϕ) → 𝔊_{d}(k, ϕ)

which from proposition 5.6 are known to be surjective.

Applying proposition 9.2, we get that

Z(𝔊(k, ϕ)) = \underleftarrow{lim}_{d} Z(𝔊_{d}(k, ϕ))

But from proposition 7.3 we have Z(𝔊_{d}(k, ϕ)) ⊆ ℌ_{d}(k, ϕ), and proposition 9.1 states that \underleftarrow{lim}_{d}ℌ_{d}(k, ϕ) = 1.

Thus \underleftarrow{lim}_{d} Z(𝔊_{d}(k, ϕ)) = 1, and

Z(𝔊(k, ϕ)) = 1 .

Corollary. The center of Gal((k^{ab})^{ab}/k) is trivial.

Proof. Gal((k^{ab})^{ab}/k) = 𝔊(k, ϕ).

Part III

Explicit Constructions

In this part, we will construct a family of extensions of a local field k which are metabelian and non-nilpotent (in the case where the order of the residue field of k is greater than 2). We later use these extensions to show that some metabelian extensions are not pro-nilpotent over k.

The construction is analogous to the construction of Kummer extensions over a field F. When constructing Kummer extensions, we first attach an n-th root of unity, ζ_{n}, to F, and then look at the extension F (α_{n}, ζ_{n})/F where α_{n}^{n} = a for a ∈ F. Here we replace the operation of raising to the n-th power by applying an endomorphism [π^{n+1}] derived from a Lubin-Tate power series, in a way described below.

This analogy suggests the name “Kummer-Lubin-Tate” for the extensions that we construct.

10 Definition of the Extensions k(α_{n}, ω_{n})/k

As before, (k, ν) is a local field with a ring of integers 𝔬, a maximal ideal 𝔭 and residue field 𝔽_{q}. Let π ∈ k be a prime element, and h(X) be a Lubin-Tate power series with respect to π (see section 2), that is,

h(X) ≡ πX  (mod deg 2)  (10.1)
h(X) ≡ X^{q}  (mod 𝔭)  (10.2)

Assume also that h(X) is a polynomial of degree q in 𝔬[X].

Let F_{h}(X, Y) ∈ 𝔬[[X, Y]] be the group law that corresponds to h. Recall that for every a ∈ 𝔬 we have a power series [a] which is an endomorphism of F_{h}. By h ∈ 𝔬[[X]] and (10.1) we deduce that [π] = h, and therefore [π^{n}] = h ∘ ⋯ ∘ h (composition n times).

Let Ω be an algebraic closure of k, and let μ be an extension of ν to Ω. Let Ω̄ be a completion of Ω with valuation μ̄, and 𝔭Ω̄ be the maximal ideal in Ω̄ (i.e. the set of elements with positive valuation). Recall that we can define an 𝔬-module structure on 𝔭Ω̄. Let W_{n}, for n ≥ 1, be the π^{n+1}-torsion submodules.

From now on we assume that ω_{n}, for n ≥ 0, have been fixed once and for all as generators of the corresponding modules W_{n}.

Throughout this part, we fix a prime element a ∈ k. For every n ≥ 0, we choose α_{n} ∈ Ω such that

[π^{n+1}](α_{n}) = a  (10.3)

Such a choice can be made, since Ω is algebraically closed, and [π^{n}] are polynomials.

First we show that α_{n} ∈ 𝔭_{Ω}. In order to do that, we need a general lemma on the roots of Eisenstein polynomials.

Lemma 10.1. Let f ∈ 𝔬[X] be an Eisenstein polynomial of degree m, f = X^{m} + c_{m−1}X^{m−1} + ⋯ + c_{1}X + c_{0}, and let α be a root of f in Ω.

Then the extension k(α)/k is totally ramified,

[k(α) : k] = m

and α is a prime in k(α).

Proof. It is known that an Eisenstein polynomial is irreducible, therefore [k(α) : k] = m. Since N_{k(α)/k}(α) = ±c_{0}, we see that

mμ(α) = μ(c_{0})  (10.4)

(because conjugate elements over k have the same valuation).

Now, the fact that f is Eisenstein implies that ν(c_{0}) = 1, hence μ(k^{×}) = μ(c_{0})ℤ. It is also evident that μ(k(α)^{×}) ⊆ μ(α)ℤ. Combining these and (10.4), we get for the ramification index e(k(α)/k),

[k(α) : k] ≥ e(k(α)/k) = [μ(k(α)^{×}) : μ(k^{×})] ≥

≥ [μ(α)ℤ : μ(c_{0})ℤ] = \frac{μ(c_{0})}{μ(α)} = m = [k(α) : k]

so that k(α)/k is totally ramified of degree m, and all inequalities are in fact equalities. In particular, as μ(k(α)^{×}) = μ(α)ℤ, we also get that α is a prime element of k(α).

Lemma 10.2. Let n ≥ 0. Then the extension k(α_{n})/k is totally ramified of degree q^{n+1}, and α_{n} is a prime in that extension.

Proof. It suffices to check that the polynomial

[π^{n+1}](X) − a

is Eisenstein of degree q^{n+1}.

Recall that [π^{n+1}] = h ∘ h ∘ ⋯ ∘ h. The degree of h is q, thus the degree of the n + 1-th composition is q^{n+1}. In addition, as h is divisible by X, the same is true for the composition. By (10.2) we also get that

[π^{n+1}](X) ≡ X^{q^{n+1}} (mod 𝔭)

We still have to show that the constant term of [π^{n+1}] − a is a prime of k. But this is clear, as a is a prime and [π^{n+1}] is divisible by X.

Lemma 10.3. Let α_{n} be defined as above and let σ ∈ Gal(Ω/k). Then σ(α_{n}) is another solution of (10.3) and α_{n} . − σ(α_{n}) ∈ W_{n}.

Proof. Since [π^{n+1}] − a has its coefficients in k, it is clear that the image under σ of a solution to (10.3) is also a solution of that equation. Now,

π^{n+1} · (α_{n} . − σ(α_{n})) = π^{n+1} · α_{n} . − π^{n+1} · σ(α_{n}) = a . − a = 0

hence α_{n} . − σ(α_{n}) ∈ W_{n}.

Proposition 10.4. Let m, n ≥ 0 and assume that m ≥ n. Then the extension k(α_{n}, ω_{m})/k is Galois.

Proof. It suffices to show that k(α_{n}, ω_{m}) is left invariant under arbitrary σ ∈ Gal(Ω/k). For such σ we have σ(ω_{m}) ∈ W_{m}.

By the previous lemma, if we let ω = σ(α_{n}) . − α_{n}, then ω ∈ W_{n}. Since m ≥ n, it follows that W_{n} ⊆ W_{m} ⊆ k(ω_{m}). In particular, ω ∈ k(ω_{m}). As k(α_{n}, ω_{m})/k is finite, it is also closed in Ω̄, hence

σ(α_{n}) = α_{n} ∔ ω ∈ k(α_{n}, ω_{m})

and the proposition follows.

We shall examine later in detail the Galois groups of the extensions k(α_{n}, ω_{n})/k for every n ≥ 0, and show that these extensions are examples of a metabelian non-nilpotent extensions. At the moment, we present a simple result for the case n = 0, whose proof does not involve any group considerations.

Proposition 10.5. The extension k(α_{0}, ω_{0})/k is totally ramified of degree q(q − 1).

Proof. We already know that the extension k(ω_{0})/k is totally ramified of degree q − 1 and that ω_{0} is prime. Thus

μ(ω_{0}) = \frac{1}{q − 1}  (10.5)

By lemma 10.2, k(α_{0})/k is totally ramified of degree q and α_{0} is prime element. Hence

μ(α_{0}) = \frac{1}{q}  (10.6)

Now take ω_{0}/α_{0} ∈ k(α_{0}, ω_{0}). By (10.5) and (10.6),

μ(\frac{ω_{0}}{α_{0}} ) = (\frac{1}{q − 1} − \frac{1}{q}) = \frac{1}{q(q − 1)}

thus e(k(α_{0}, ω_{0})/k) ≥ q(q − 1).

Since [k(ω_{0}) : k] = q − 1 and [k(α_{0}) : k] = q, we must have [k(α_{0}, ω_{0}) : k] ≤ q(q − 1). Combining these inequalities, we get

q(q − 1) ≥ [k(α_{0}, ω_{0}) : k] ≥ e(k(α_{0}, ω_{0})/k) ≥ q(q − 1)

and all inequalities are in fact equalities. We deduce that the extension k(α_{0}, ω_{0})/k is totally ramified of degree q(q − 1).

11 The Groups Gal(k(α_{n}, ω_{n})/k)

In this section we describe the groups Gal(k(α_{n}, ω_{n})/k) for n ≥ 0.

We use the construction defined in section 5, (5.4) and (5.5) for the groups A = Aut W_{n} and B = W_{m} where m ≤ n (note that by Aut W_{n} we mean automorphisms that preserve not only the additive structure of W_{n}, but also its 𝔬-module structure.)

Definition. Let m, n ≥ 0, m ≤ n. The group Aut W_{n} ⋉ W_{m} is defined as

Aut W_{n} ⋉ W_{m} = {(ε, ω) : ε ∈ Aut W_{n}, ω ∈ W_{m}}

where multiplication of two elements (ε, ω) and (ε′, ω′) is defined as

(ε, ω)(ε′, ω′) = (εε′, ω ∔ ε(ω′))

Note that since W_{m} is invariant under Aut W_{n}, the semi-direct product is well-defined.

Proposition 11.1. Let n ≥ 0. There is an injective group homomorphism

ρ_{n} : Gal(k(α_{n}, ω_{n})/k) ⟶ Aut W_{n} ⋉ W_{n}

Proof. Let σ ∈ Gal(k(α_{n}, ω_{n})/k). Let ε_{σ} be the restriction of σ to W_{n}, and let ω_{σ} = σ(α_{n}) . − α_{n}. Define ρ_{n}(σ) = (ε_{σ}, ω_{σ}).

It is well known that ε_{σ} ∈ Aut W_{n}. Moreover, by lemma 10.3, ω_{σ} ∈ W_{n}, therefore ρ_{n}(σ) = (ε_{σ}, ω_{σ}) ∈ Aut W_{n} ⋉ W_{n}.

Let σ, τ be two automorphisms of k(α_{n}, ω_{n})/k. Then

ε_{στ} = (στ) ⇂_{W_{n}}= ε_{σ}ε_{τ}

and

στ(α_{n}) . − α_{n} = στ(α_{n}) . − σ(α_{n}) ∔ σ(α_{n}) . − α_{n} =

σ(τ(α_{n}) . − α_{n}) ∔ (σ(α_{n}) . − α_{n}) = ε_{σ}(ω_{τ}) ∔ ω_{σ} (11.1)

hence ρ_{n}(στ) = ρ_{n}(σ)ρ_{n}(τ).

Assume that ε_{σ} = 1 and ω_{σ} = 0. Then σ(ω_{n}) = ω_{n} and σ(α_{n}) = α_{n}, thus σ is the identity on k(α_{n}, ω_{n}). Hence ρ_{n} is injective.

We will now determine the image of ρ_{n}. In order to do that, we need a general lemma on Galois groups.

Let L/F be a finite Galois extension of fields. Let F ⊆ K ⊆ L be intermediate field and assume that K/F is also Galois. Write G = Gal(L/F), H = Gal(L/K), Δ = Gal(K/F), and assume that H is abelian. We will use additive notation when referring to the group H.

Let σ ∈ H, and δ ∈ Δ. Let δ̃ be an extension of δ to L/F. Define

⟨δ, σ⟩ = δ̃σδ̃^{−1}  (11.2)

Lemma 11.2. The pairing defined in (11.2) satisfies the following properties:

1. ⟨δ, σ⟩ ∈ H

2. The definition in (11.2) does not depend on the particular choice of an extension δ̃ of δ.

3. ⟨δ, σ_{1} + σ_{2}⟩ = ⟨δ, σ_{1}⟩ + ⟨δ, σ_{2}⟩

4. ⟨δ_{1}δ_{2}, σ⟩ = ⟨δ_{1}, ⟨δ_{2}, σ⟩⟩

Proof. Let σ ∈ H, δ ∈ Δ, and choose an extension δ̃ of δ to L/F.

1. Let x ∈ K. Then, since σ fixes K pointwise,

δ̃σδ̃^{−1}(x) = δ(δ^{−1}(x)) = x

thus ⟨δ, σ⟩ ∈ H.

2. Let δ_{1}, δ_{2} be two extensions of δ. As δ_{1}(x) = δ_{2}(x) = δ(x) for every x ∈ K, we have δ_{2}δ_{1}^{−1} ∈ H.

We already know that δ_{1}σδ_{1}^{−1} ∈ H. Since H is abelian, the elements δ_{2}δ_{1}^{−1} and δ_{1}σδ_{1}^{−1} commute. Therefore,

δ_{2}σδ_{2}^{−1} = (δ_{2}δ_{1}^{−1})(δ_{1}σδ_{1}^{−1})(δ_{2}δ_{1}^{−1})^{−1} = δ_{1}σδ_{1}^{−1}

and ⟨δ, σ⟩ does not depend on the particular choice of an extension of δ.

3. This is clear, as

δ̃σ_{1}σ_{2}δ̃^{−1} = (δ̃σ_{1}δ̃^{−1})(δ̃σ_{2}δ̃^{−1})

4. If δ̃_{1} and δ̃_{2} are extensions of δ_{1},δ_{2} resp., then δ̃_{1}δ̃_{2} is an extension of δ_{1}δ_{2}, and the result is clear from the definition (and the independence on the extension).

In our setting, G = Gal(k(α_{n}, ω_{n})/k), H = Gal(k(α_{n}, ω_{n})/k(ω_{n})) and Δ = Gal(k(ω_{n})/k).

Lemma 11.3. The extension k(α_{n}, ω_{n})/k(ω_{n}) is abelian.

Proof. Let σ ∈ Gal(k(α_{n}, ω_{n})/k(ω_{n})) and consider ω_{σ} defined in the proof of proposition 11.1.

Let σ, τ ∈ Gal(k(α_{n}, ω_{n})/k(ω_{n})). By (11.1),

ω_{στ} = σ(ω_{τ}) ∔ ω_{σ}

but since σ acts trivially on W_{n}, we actually have

ω_{στ} = ω_{τ} ∔ ω_{σ}

hence the mapping σ ↦ ω_{σ} is a homomorphism. It is injective, since if ω_{σ} = 0, then σ(α_{n}) = α_{n} and σ is the identity on k(α_{n}, ω_{n}).

Now the lemma follows, as Gal(k(α_{n}, ω_{n})/k(ω_{n})) is embedded in the abelian group W_{n}.

We will now use the pairing defined in (11.2). First, we connect this pairing to ρ_{n}.

Lemma 11.4. Let σ ∈ Gal(k(α_{n}, ω_{n})/k(ω_{n})), δ ∈ Gal(k(ω_{n})/k). Then

ω_{⟨δ,σ⟩} = δ(ω_{σ})

Proof. Let δ̃ be an extension of δ to k(α_{n}, ω_{n}). By (11.1)

ω_{δ̃σδ̃^{−1}} = δ̃(ω_{σδ̃^{−1}}) ∔ ωδ̃ = δ̃(σ(ωδ̃^{−1}) ∔ ω_{σ}) ∔ ωδ̃  (11.3)

Since σ acts trivially on W_{n}, we have σ(ωδ̃^{−1}) = ωδ̃^{−1}. We also have

δ̃(ωδ̃^{−1}) = δ̃(δ̃^{−1}(α_{n}) . − α_{n}) = . − ωδ̃

Substituting this in (11.3), we get

ω_{δ̃σδ̃^{−1}} = δ̃(ω_{σ}) = δ(ω_{σ})

Lemma 11.5. Let W be a subgroup of the additive group of W_{n}, and assume also that W is invariant under Gal(k(ω_{n})/k).

Then there exists −1 ≤ m ≤ n such that W = W_{m}.

Proof. By local class field theory, there is a one-to-one correspondence between Gal(k(ω_{n})/k) and U/U_{n+1}. Thus if ω ∈ W, then u · ω ∈ W for every u ∈ U.

We will show that W is an 𝔬-submodule of W_{n}. Let u_{1} = 1 + π^{2} and u_{2} = 1 + π + π^{2} be two units, and let ω ∈ W. Since W is invariant under the Galois group and since it is an additive subgroup, we deduce that

π · ω = u_{2} · ω . − u_{1} · ω ∈ W

and therefore ⟨π⟩ · ω ⊆ W (apply the same argument inductively). As U · ω ⊆ W, we conclude that 𝔬 · ω ∈ W.

Now the result follows from the fact that the only submodules of W_{n} are W_{m} for m ≤ n.

Lemma 11.6. There exists −1 ≤ m ≤ n for which

Gal(k(α_{n}, ω_{n})/k(ω_{n})) ≅ W_{m}

Proof. In the proof of lemma 11.3 we saw that there is an injection

Gal(k(α_{n}, ω_{n})/k(ω_{n})) ⟶ W_{n}

defined by σ ↦ ω_{σ}.

The image of that injection is a subgroup of the additive group of W_{n}. By lemma 11.4, it is also invariant under every δ ∈ Gal(k(ω_{n})/k). Finally, by lemma 11.5, there exists −1 ≤ m ≤ n such that this image equals W_{m}.

Definition. A group G is of type Γ_{n,m} if there exists a commutative diagram

0 ⟶ W_{m} ⟶  G  ⟶ Aut W_{n} ⟶ 1
↓  ↓  ↓
0 ⟶ W_{n} ⟶^{i} Aut W_{n} ⋉ W_{n} ⟶^{p} Aut W_{n} ⟶ 1

(11.4)

where the middle vertical arrow is an injection, the other vertical arrows are the canonical embeddings, and the horizontal maps i and p are defined by

i : ω ⟼ (1, ω)
p : (ε, ω) ⟼ ε

Groups of type Γ_{n,m} can be considered as subgroups of Aut W_{n} ⋉ W_{n}. A typical element in such group can thus be viewed as a pair (ε, ω) for ε ∈ Aut W_{n} and ω ∈ W_{n}. The commutativity of the above diagram implies that elements of the form (1, ω) belong to a group of type Γ_{n,m} if and only if ω ∈ W_{m}. It also implies that for every ε ∈ Aut W_{n}, there is an ω ∈ W_{n} such that (ε, ω) lies in the group.

Proposition 11.7. The extension k(α_{n}, ω_{n})/k is Galois. Its Galois group is of type Γ_{n,m}, where −1 ≤ m ≤ n.

Proof. By proposition 10.4, k(α_{n}, ω_{n})/k is Galois. Let us compute Im ρ_{n} for

ρ_{n} : Gal(k(α_{n}, ω_{n})/k) ⟶ Aut W_{n} ⋉ W_{n}

defined in proposition 11.1. Since ρ_{n} is injective, we will get an isomorphism between the Galois group and Im ρ_{n}.

By lemma 11.6, there is an −1 ≤ m ≤ n such that the elements of the form (1, ω) which lie in Im ρ_{n} are exactly the ones for which ω ∈ W_{m}.

The mapping δ ↦ ε_{δ} is an isomorphism between Gal(k(ω_{n})/k) and Aut W_{n}. Since any such δ can be extended to k(α_{n}, ω_{n}), we get that for every ε ∈ Aut W_{n}, there is an ω(ε) such that (ε, ω(ε)) ∈ Im ρ_{n}.

Combining these facts, we get a short exact sequence

0 ⟶ W_{m} ⟶^{i′} Im ρ_{n} ⟶^{p′} Aut W_{n} ⟶ 1

with respect to the maps

i′ : ω ⟼ (1, ω)
p′ : (ε, ω) ⟼ ε

As Im ρ_{n} is a subgroup of Aut W_{n}⋉W_{n}, it is easy to see that the diagram in (11.4) commutes.

The value of m depends on the choice of a in the construction of the fields k(α_{n}, ω_{n}). The remainder of this section is devoted to the understanding of this dependency.

Let 𝔭_{n} be the maximal ideal of k(ω_{n}). Denote by π^{i} · 𝔭_{n} the set

π^{i} · 𝔭_{n} = {π^{i} · x : x ∈ 𝔭_{n}}

for i ≥ 0. Since 𝔭_{n} is topologically closed, the sets π^{i} · 𝔭_{n} are also closed and contained in 𝔭_{n}.

Lemma 11.8. k ∩ π^{n+1} · 𝔭_{n} ⊆ 𝔭^{2}.

Proof. Let β ∈ 𝔭_{n}. Recall that [π^{n+1}] ∈ X𝔬[[X]], and by the proof of lemma 10.2,

[π^{n+1}](X) ≡ X^{q^{n+1}} (mod 𝔭)

Therefore, π^{n+1} · β = πγ + β^{q^{n+1}} where γ ∈ 𝔭_{n}.

Since k(ω_{n})/k is totally ramified of degree (q − 1)q^{n}, we have μ(β) ≥ 1/(q − 1)q^{n} (recall that μ is an extension of the valuation ν to the algebraic closure Ω).

Now, as μ(γ) > 0, we get μ(πγ) > 1. We also have

μ(β^{q^{n+1}}) ≥ \frac{q^{n+1}}{(q − 1)q^{n}} > 1 .

It follows that μ(π^{n+1} · β) > 1. In particular, if π^{n+1} · β ∈ k, then its valuation is at least 2, and it is contained in 𝔭^{2}.

Proposition 11.9. Let m be the integer for which

a ∈ π^{n−m} · 𝔭_{n} \ π^{n−m+1} · 𝔭_{n} .

Then 0 ≤ m ≤ n, and Gal(k(α_{n}, ω_{n})/k) is of type Γ_{n,m}.

Proof. Since π^{0} · 𝔭_{n} = 𝔭_{n}, we have a disjoint union decomposition

𝔭 = ⋃_{i=0}^{∞} (k ∩ (π^{i} · 𝔭_{n} \ π^{i+1} · 𝔭_{n}))

But by the lemma, π^{n+1} · 𝔭_{n} ⊆ 𝔭^{2}, hence

𝔭 \ 𝔭^{2} ⊆ ⋃_{i=0}^{n} (k ∩ (π^{i} · 𝔭_{n} \ π^{i+1} · 𝔭_{n}))

It follows that if a is a prime of k, then m defined as above satisfies 0 ≤ m ≤ n.

First, assume that Gal(k(α_{n}, ω_{n})/k) is of type Γ_{n,m} (proposition 11.7 ensures that this is the form of the Galois group). Then ω_{σ} ∈ W_{m} for every automorphism σ ∈ Gal(k(α_{n}, ω_{n})/k(ω_{n})). Thus

π^{m+1} · (σ(α_{n}) . − α_{n}) = 0

for every such σ, or equivalently, since σ commutes with [π^{m+1}] ∈ 𝔬[[X]],

σ(π^{m+1} · α_{n}) = π^{m+1} · α_{n}

thus π^{m+1} · α_{n} ∈ k(ω_{n}).

By lemma 10.2, k(α_{n})/k is a totally ramified extension of degree q^{n+1} and α_{n} is a prime in that extension. In particular, α_{n} ∈ 𝔭Ω̄, and π^{m+1} · α_{n} ∈ 𝔭Ω̄. Therefore π^{m+1} · α_{n} ∈ 𝔭_{n}, and

a = π^{n+1} · α_{n} = π^{n−m} · (π^{m+1} · α_{n}) ∈ π^{n−m} · 𝔭_{n}

For the opposite direction, assume that a = π^{n−m} · β for β ∈ 𝔭_{n}. Then

π^{n+1} · α_{n} = π^{n−m} · (π^{m+1} · α_{n}) = a = π^{n−m} · β

hence β . − π^{m+1}α_{n} ∈ W_{n−m−1}. Since β ∈ 𝔭_{n} and W_{n−m−1} ⊆ W_{n} ⊆ 𝔭_{n}, we have

π^{m+1}α_{n} ∈ 𝔭_{n}

Thus for every σ ∈ Gal(k(α_{n}, ω_{n})/k(ω_{n})),

σ(π^{m+1} · α_{n}) = π^{m+1} · α_{n}

Again, since σ commutes with [π^{m+1}] ∈ 𝔬[[X]], this is equivalent to

π^{m+1} · ω_{σ} = π^{m+1}(σ(α_{n}) . − α_{n}) = 0

Therefore ω_{σ} ∈ W_{m} for every such σ.

The proposition now follows from the combination of the above arguments and the special form of the Galois group; this group is of type Γ_{n,m′} where m′ is characterized by the fact that

Gal(k(α_{n}, ω_{n})/k(ω_{n})) ≅ W_{m′}

Assume that a ∈ π^{n−m} · 𝔭_{n} \ π^{n−m+1} · 𝔭_{n}. The second argument above implies that the image of the subgroup Gal(k(α_{n}, ω_{n})/k(ω_{n})) is contained in W_{m}, hence m′ ≤ m. But the first argument implies that if the Galois group is of type Γ_{n,m′}, then a ∈ π^{n−m′} · 𝔭_{n}, hence we must have m = m′.

12 Central Series of Gal(k(α_{n}, ω_{n})/k)

The groups Gal(k(α_{n}, ω_{n})/k) are of type Γ_{n,m} for 0 ≤ m ≤ n. We will now compute the upper and lower central series of these groups. The nature of the results we shall obtain depends on the value of q, the order of the residue field of k. It seems appropriate to distinguish between two cases: the case where q > 2 and the case where q = 2.

The case q > 2

Let G be of type Γ_{n,m} where n ≥ 0 and 0 ≤ m ≤ n (in the case where m = −1 we have merely G = Aut W_{n} ≅ U/U_{n+1} which is an abelian group and the upper (lower) series are trivial).

Lemma 12.1. Let ω ∈ W_{m}, and assume that ε(ω) = ω for every ε ∈ Aut W_{m}. Then ω = 0.

Proof. There is an isomorphism Aut W_{m} ≅ Gal(k(ω_{m})/k) defined by σ ↦ ε_{σ} (the restriction of σ to W_{m}). Thus the condition in the lemma can be reformulated as σ(ω) = ω for every σ ∈ Gal(k(ω_{m})/k). Therefore, ω ∈ k ∩ W_{m}.

If ω generates the module W_{i} (i ≥ 0), then it is well known that k(ω) = k(W_{i}), an extension of degree (q−1)q^{i} > 1 over k. It follows that ω generates W_{−1} = 0, i.e. ω = 0. (Note that the assumption q > 2 is necessary for this implication to hold).

Lemma 12.2. Let G be of type Γ_{n,m}. Then

(i) Z(G) ⊆ {(ε, ω) : ε acts trivially on W_{m}}

(ii) If (1, ω) ∈ Z(G), then ω = 0.

Proof. Let (ε, ω) ∈ Z(G). For every (ε′, ω′) ∈ G, we have

(ε, ω)(ε′, ω′) = (ε′, ω′)(ε, ω)

Carrying out the multiplication, we get the constraints:

εε′ = ε′ε
ω ∔ ε(ω′) = ω′ ∔ ε′(ω)  (12.1)

where the first constraint is always satisfied, as Aut W_{n} is an abelian group. Take ε′ = 1 and ω′ ∈ W_{m} arbitrary. Substituting in (12.1), we see that

ε(ω′) = ω′

hence ε acts trivially on W_{m}.

For the second claim, assume that (1, ω) ∈ Z(G). Then by (12.1),

ω = ε′(ω)  (12.2)

for every (ε′, ω′) ∈ G. Since the projection from G to Aut W_{n} is surjective, we can take in (12.2) any ε′ ∈ Aut W_{n}, and deduce, by lemma 12.1, that ω = 0.

Lemma 12.3. Let G be of type Γ_{n,m} as before. Then

G′ = {(1, ω) : ω ∈ W_{m}} ≅ W_{m}

Proof. Consider a commutator [(ε, ω), (ε′, ω′)]. Its first coordinate equals [ε, ε′] = 1, since Aut W_{n} is abelian. Therefore,

G′ ⊆ {(1, ω) : ω ∈ W_{m}}

To prove the opposite inclusion, note that the group {(1, ω) ∈ G} is normal in G. This follows from the fact that this group is the kernel of the homomorphism G → Aut W_{n} defined by (ε, ω) ↦ ε. The image of this homomorphism is the abelian group Aut W_{n}, hence we get that

{(1, ω) ∈ G} ⊆ G′

Remark. In the proof of this lemma we did not use the assumption q > 2. Hence it is also true for q = 2.

Lemma 12.4. Z_{2}(G) = Z(G)

Proof. Let z ∈ Z_{2}(G). By proposition 8.1, [z, g] ∈ G′ ∩ Z(G) for every g ∈ G. Applying this to our case, using the previous lemmas, we see that if [z, g] = (ε, ω), then since [z, g] ∈ G′ we must have ε = 1, and as [z, g] ∈ Z(G), we must have ω = 0. Hence [z, g] = 1 for every g ∈ G, and therefore z ∈ Z(G).

Corollary. The upper central series of G gets fixed after the second stage.

Proof. This is clear from Z_{2}(G) = Z(G).

We prove symmetrical result for the lower central series.

Lemma 12.5. G^{2} = G^{1}

Proof. Let (1, ω) ∈ G^{1}, (ε′, ω′) ∈ G. Then

[(ε′, ω′), (1, ω)] = (1, ε′(ω) . − ω)

Denote by W the subgroup

W = ⟨ε(ω) . − ω : ε ∈ Aut W_{n}, ω ∈ W_{m}⟩

Let ε ∈ Aut W_{n}. There is u_{ε} ∈ U such that ε(ω) = u_{ε} · ω for every ω ∈ W_{n}. Thus

W = ⟨(u − 1) · ω : u ∈ U, ω ∈ W_{m}⟩

Since q > 2, we can find a unit u ∈ U \U_{1} for which u−1 ∈ U. It follows that [u − 1] acts on W_{m} as a permutation, thus (u − 1) · W_{m} = W_{m} and the lemma follows.

Proposition 12.6. Let G be of type Γ_{n,m}, n ≥ m ≥ 0. Then G is metabelian and non-nilpotent.

Proof. Consider the normal subgroup G′. Since G′ ≅ W_{m}, we get that both G′ and G/G′ are abelian, hence G is metabelian.

If G were nilpotent, then we would have Z_{n}(G) = G for some n ≥ 1. However, we already know that Z_{n}(G) = Z(G) for every n ≥ 1. As G is non-abelian (G′ ≠ 1), we have Z_{n}(G) = Z(G) ≠ G , a contradiction.

Corollary. Let n ≥ 0. The extensions k(α_{n}, ω_{n})/k are metabelian and non-nilpotent. The maximal abelian extension over k contained in k(α_{n}, ω_{n}) equals k(ω_{n}).

Proof. Since G = Gal(k(α_{n}, ω_{n})/k) is of type Γ_{n,m} for 0 ≤ m ≤ n, the first claim follows immediately from the previous proposition. Moreover, since

Gal(k(α_{n}, ω_{n})/k(ω_{n})) ≅ {(1, ω) : ω ∈ W_{m}} = G′

we get that Gal(k(ω_{n})/k) is the maximal abelian quotient of G and the second claim follows.

We now state some consequences of the proposition related to the maximal abelian extensions of some fields.

Notation. Denote by k_{π}^{0} the extension k(ω_{0}), where ω_{0} is a generator of the module W_{0}.

Note that the extension k_{π}^{0} depends only on π, and not on the particular choice of the Lubin-Tate series h(X).

Theorem 12.7. Let K/k be a non-ramified extension. Then the extension (Kk_{π}^{0})^{ab}/K is not pro-nilpotent.

Proof. Let h(X) ∈ 𝔬[[X]] a Lubin-Tate power series over k with respect to π, which is also a polynomial of degree q.

Consider the extension K(α_{0}, ω_{0})/K. By proposition 10.5, the extension k(α_{0}, ω_{0})/k is totally ramified, therefore

Gal(K(α_{0}, ω_{0})/K) ≅ Gal(k(α_{0}, ω_{0})/k)

is of type Γ_{0,0}.

We deduce that K(α_{0}, ω_{0})/K is a finite extension, which is abelian over K(ω_{0}) = Kk_{π}^{0}, by lemma 11.3, and not nilpotent, by proposition 12.6.

Since this extension is finite and is contained in (Kk_{π}^{0})^{ab}, we get that (Kk_{π}^{0})^{ab}/K is not pro-nilpotent.

Corollary. Let K/k be a non-ramified extension.

Then the extensions (Kk_{π})^{ab}/K and (k^{ab})^{ab}/K are not pro-nilpotent.

Proof. These extensions contain the extension (Kk_{π}^{0})^{ab}/K.

The case q = 2

Again, let G be of type Γ_{n,m} for n ≥ m ≥ 0. The order of G equals (q − 1)q^{n} · q^{m+1} which is a power of 2 (since q = 2). In particular, G is a finite p-group (p = 2), and therefore nilpotent. The following proposition is an immediate consequence of the nilpotency of G.

Proposition 12.8. Assume that q = 2, and let n ≥ 0. Then the extensions k(α_{n}, ω_{n})/k are metabelian and nilpotent.

Proof. For the proof that these extensions are metabelian, cf. the proof of proposition 12.6. The nilpotency follows from the nilpotency of the groups of type Γ_{n,m} in this case.

We will now compute the lower central series. As a consequence, we will determine the nilpotency class of the group G.

Proposition 12.9. Let G be as before, and let i ≥ 1. Then the i-th group of the lower central series of G equals

G^{i} = {(1, ω) : ω ∈ W_{m+1−i}}

In particular, G^{m+2} = 1 and G is of nilpotency class m + 2.

Proof. By lemma 12.3, the proposition is true for i = 1. We continue by induction on i; we assume that the proposition is true for i and prove it for i + 1.

Let (1, ω) ∈ G^{i}, (ε′, ω′) ∈ G. Then

[(ε′, ω′), (1, ω)] = (1, ε′(ω) . − ω)

and by the induction hypothesis, ω ∈ W_{m+1−i}.

To determine the image of ε(ω) . − ω for ω ∈ W_{m+1−i}, we use an argument similar to the one in the proof of lemma 12.5. Write ε(ω) as u · ω for u ∈ U. Since q = 2, we have U = U_{1} and therefore u − 1 ∈ 𝔭 for every u ∈ U. It follows that

u · ω . − ω = (u − 1) · ω ∈ 𝔭 · ω

and we deduce that this subgroup is contained in 𝔭 · W_{m+1−i} = W_{m−i}.

Take now u = 1 + π. As ω runs over W_{m+1−i}, (u − 1) · ω runs over all elements of W_{m−i}, and the induction step is complete.

Since W_{0} ≠ W_{−1} = 0, we get that G^{m+1} ≠ 1, G^{m+2} = 1. Hence G’s nilpotency class equals m + 2.

Part IV

Lower Central Series

13 The Groups B_{d,α}(H)

As before, let (k, ν) be a local field, with residue field 𝔽_{q} = 𝔬/𝔭 and prime element π, and let h(X) ∈ 𝔬[[X]] be a Lubin-Tate power series with respect to π. Throughout this section, we fix d ≥ 1.

Let H ≤ U be a closed subgroup of U. We define, by transfinite induction, a sequence of subgroups of 𝔽_{q^{d}}[[X]]^{×}.

Definition. The groups B_{d,α}(H) are defined as follows:

1. For α = 1, define

B_{d,1}(H) = 𝔽_{q^{d}}[[X]]^{×}

2. For every ordinal α, define

B_{d,α+1}(H) = ⟨\frac{ξ ∘ \{u\}}{ξ} : ξ ∈ B_{d,α}, u ∈ H⟩

where by ⟨A⟩ we mean the closure of the subgroup generated by the set A.

3. If α is a limit ordinal (not a successor), define

B_{d,α}(H) = ⋂_{β<α} B_{d,β}(H)

Lemma 13.1. The groups B_{d,α}(H) satisfy the following properties:

(i) B_{d,α}(H_{1}) ⊆ B_{d,α}(H_{2}) for H_{1} ≤ H_{2}.

(ii) If u ∈ U and ξ ∈ B_{d,α}(H), then ξ ∘ {u} ∈ B_{d,α}(H).

(iii) B_{d,β}(H) ⊆ B_{d,α}(H) for α < β.

Proof. (i) This is clear for α = 1. We proceed by transfinite induction. For successors α + 1, since B_{d,α}(H_{1}) ⊆ B_{d,α}(H_{2}) by the induction hypothesis and H_{1} ≤ H_{2}, we get that the set of topological generators of B_{d,α+1}(H_{1}) is contained in the set of topological generators of B_{α+1}(H_{2}), thus B_{d,α+1}(H_{1}) ⊆ B_{d,α+1}(H_{2}).

For limit ordinals α the induction step is clear, as B_{d,α}(H) is defined as the intersection of all previous B_{d,β}(H).

(ii) Again, this is proved by transfinite induction. The statement is clear for α = 1, since B_{d,1}(H) = 𝔽_{q^{d}}[[X]]^{×}.

Assume the statement is true for an ordinal α. To prove it for α + 1, we first note that since the mapping η ↦ η ∘ {u} is a continuous multiplicative endomorphism of 𝔽_{q^{d}}[[X]]^{×}, it suffices to show that η ∘ {u} ∈ B_{d,α+1}(H) for a generator η of B_{d,α+1}(H) and u ∈ U.

Take such a generator η, and write it as η = (ξ ∘ {v})/ξ for v ∈ H, ξ ∈ B_{d,α}(H). Then

η ∘ \{u\} = \frac{ξ ∘ \{v\} ∘ \{u\}}{ξ ∘ \{u\}} = \frac{(ξ ∘ \{u\}) ∘ \{v\}}{ξ ∘ \{u\}}

which is another generator of B_{d,α+1}(H) since ξ ∘ {u} ∈ B_{d,α}(H) by the induction hypothesis.

The induction step for limit ordinals α is obvious; if ξ ∈ B_{d,α}(H), then for every β < α we have ξ ∈ B_{d,β}(H), thus by our hypothesis, ξ ∘ {u} ∈ B_{d,β}(H) for every u ∈ U, hence ξ ∘ {u} ∈ B_{d,α}(H) for every u ∈ U.

(iii) It is enough to show B_{d,α+1}(H) ⊆ B_{d,α}(H). Again, we may limit our attention to generators of B_{d,α+1}. Such a generator has the form

\frac{ξ ∘ \{u\}}{ξ}

for ξ ∈ B_{d,α}(H), u ∈ H. This quotient lies in B_{d,α}(H) since ξ ∘ {u} ∈ B_{d,α}(H) by the previous statement.

We shall now show that we may consider only ordinals α ≤ ω. First, a lemma on compact groups.

Lemma 13.2. Let G be a commutative compact group, and A be a closed subgroup. Let {B_{n}}_{n≥1} a decreasing sequence of closed subgroups (that is, B_{n} ⊆ B_{m} if n > m).

Then

A ( ⋂_{n≥1} B_{n}) = ⋂_{n≥1} AB_{n}

Proof. Denote by B the intersection of all B_{n}-s. The inclusion ⊆ is trivial, as B ⊆ B_{n} for every n ≥ 1.

To prove the reverse inclusion, let x ∈ ⋂_{n≥1} AB_{n}. For every n ≥ 1, write x as x = a_{n}b_{n} for a_{n} ∈ A, b_{n} ∈ B_{n}. Since G is compact, we can find a subsequence {n_{l}}_{l≥1} such that the two sequences a_{n_{l}} and b_{n_{l}} converge.

Write a = lim_{l→∞} a_{n_{l}}, b = lim_{l→∞} b_{n_{l}}. As A is closed, we get that a ∈ A, as a limit of points in A. Since B_{n_{l}} contains B_{n} for n ≥ n_{l}, we see that all but finite number of elements of the sequence {b_{n_{l}}} belong to B_{n_{l}}, hence b ∈ B_{n_{l}}. As this is true for every l ≥ 1, we obtain b ∈ B.

Now, ab = lim_{l→∞} a_{n_{l}}b_{n_{l}} = x, hence x ∈ AB, as required.

Proposition 13.3. Let H ≤ U. Then B_{d,ω+1}(H) = B_{d,ω}(H).

Proof. For r ≥ 1, denote by I_{r} the subgroup 1 + X^{r}𝔽_{q^{d}}[[X]] of 𝔽_{q^{d}}[[X]]^{×}. The sequence {I_{r}}_{r≥1} forms a filtration of 𝔽_{q^{d}}[[X]]^{×}.

The quotient 𝔽_{q^{d}}[[X]]^{×}/I_{r} is nothing but the multiplicative group of the ring 𝔽_{q^{d}}[X]/(X^{r}). In particular, as 𝔽_{q^{d}} is a finite field, this group is finite.

Fix r ≥ 1 and consider the subgroups B_{d,α}(H)I_{r}/I_{r}. They form a decreasing sequence of subgroups of the finite group 𝔽_{q^{d}}[[X]]^{×}/I_{r}. Therefore there is an integer l_{r} ≥ 1 for which B_{d,l_{r}}(H)I_{r}/I_{r} = B_{d,l_{r}+1}(H)I_{r}/I_{r}. It follows that for every ordinal α ≥ l_{r},

B_{d,α}(H)I_{r} = B_{d,l_{r}}(H)I_{r}

and in particular, B_{d,ω+1}(H)I_{r} = B_{d,ω}(H)I_{r} (since both subgroups equal B_{d,l_{r}}(H)).

We deduce that B_{d,ω}(H) ⊆ B_{d,ω+1}(H)I_{r} for every r ≥ 1. Therefore

B_{d,ω}(H) ⊆ ⋂_{r≥1} B_{d,ω+1}(H)I_{r} = B_{d,ω+1}(H)

where the last equality follows from lemma 13.2 and ∩_{r≥1}I_{r} = 1. As the reverse inclusion is obvious, the proposition follows.

The next result connects the group B_{d,α}(H_{1}H_{2}), for two subgroups H_{1}, H_{2} of U, with the groups associated with H_{1} and H_{2}.

Lemma 13.4. Let H_{1}, H_{2} be subgroups of U. Then for every α ≤ ω,

B_{d,α}(H_{1}H_{2})B_{d,2}(H_{2}) = B_{d,α}(H_{1})B_{d,2}(H_{2})

Proof. As H_{1}H_{2} ≥ H_{2}, the inclusion ⊇ is clear. To prove the reverse inclusion, it is enough to show that B_{d,α}(H_{1}H_{2}) ⊆ B_{d,α}(H_{1})B_{d,2}(H_{2}).

Again, we use (transfinite) induction. The statement is clear for α = 1, as both sides equal 𝔽_{q^{d}}[[X]]^{×}. Assume now that it is true for an ordinal α, and take a generator η of B_{d,α+1}(H_{1}H_{2}). Write η as

η = \frac{ξ ∘ \{u\}}{ξ}  (13.1)

for ξ ∈ B_{d,α}(H_{1}H_{2}) and u ∈ H_{1}H_{2}.

Write u = u_{1}u_{2} with u_{1} ∈ H_{1}, u_{2} ∈ H_{2}. By the induction hypothesis, we can write ξ = ξ_{1}ξ_{2} for ξ_{1} ∈ B_{d,α}(H_{1}), ξ_{2} ∈ B_{d,2}(H_{2}). We can therefore rewrite (13.1) as

\begin{aligned}η & = \frac{(ξ_{1}ξ_{2}) ∘ \{u\}}{ξ_{1}ξ_{2}} = \frac{ξ_{1} ∘ \{u_{1}u_{2}\}}{ξ_{1}} · \frac{ξ_{2} ∘ \{u\}}{ξ_{2}} = \\ & = \frac{(ξ_{1} ∘ \{u_{2}\}) ∘ \{u_{1}\}}{ξ_{1} ∘ \{u_{2}\}} · \frac{ξ_{1} ∘ \{u_{2}\}}{ξ_{1}} · \frac{ξ_{2} ∘ \{u\}}{ξ_{2}} \end{aligned}  (13.2)

Now, as ξ_{1} ∈ B_{d,α}(H_{1}), we get that ξ_{1} ∘ {u_{2}} ∈ B_{d,α}(H_{1}) and therefore the leftmost element in the product belongs to B_{d,α+1}(H_{1}). Since ξ_{1} ∈ 𝔽_{q^{d}}[[X]]^{×} and u_{2} ∈ H_{2}, it follows that the middle element belongs to B_{d,2}(H_{2}). Finally, since ξ_{2} ∈ B_{d,2}(H_{2}) and u ∈ U, we get by lemma 13.1 that the rightmost element also belongs to B_{d,2}(H_{2}), and η lies in B_{d,α+1}(H_{1})B_{d,2}(H_{2})

If α = ω is a limit ordinal, then by the induction hypothesis and lemma 13.2 for the compact group 𝔽_{q^{d}}[[X]]^{×},

B_{d,α}(H_{1}H_{2}) = ⋂_{β<α} B_{d,β}(H_{1}H_{2}) ⊆ ⋂_{β<α} B_{d,β}(H_{1})B_{d,2}(H_{2}) =

( ⋂_{β<α} B_{d,β}(H_{1})) B_{d,2}(H_{2}) = B_{d,α}(H_{1})B_{d,2}(H_{2})

Proposition 13.5. Let H_{1},H_{2} be subgroups of U. Then

B_{d,ω}(H_{1}H_{2}) = B_{d,ω}(H_{1})B_{d,ω}(H_{2})

Proof. First, note that the inclusion ⊇ follows from the fact that H_{1}H_{2} ≥ H_{1}, H_{1}H_{2} ≥ H_{2}; the group on the LHS contains both groups in the RHS, thus it contains their product. It remains to show the opposite inclusion.

We start by proving that for every n, m ≥ 2,

\begin{aligned}B_{d,ω}(H_{1}H_{2}) & ⊆ B_{d,ω}(H_{1}) · B_{d,ω}(H_{2}) · (B_{d,m}(H_{1}) ∩ B_{d,n}(H_{2})) \\ & · (B_{d,2}(H_{1}) ∩ B_{d,n+1}(H_{2})) \end{aligned}  (13.3)

We prove this by double induction; we assume (13.3) for a pair (n, m) and prove it for (n, m + 1). Then we prove it for (n + 1, 2) assuming it for (n, m) for every m < ω. This double induction is actually ordinary transfinite induction for ordinals smaller than ω · ω.

First, we prove (13.3) for n = 2, m = 2. This is merely a double application of lemma 13.4. By this lemma, we get

B_{d,ω}(H_{1}H_{2}) ⊆ B_{d,ω}(H_{1})B_{d,2}(H_{2})

and by exchanging the roles of H_{1}, H_{2}, we also get

B_{d,ω}(H_{1}H_{2}) ⊆ B_{d,ω}(H_{2})B_{d,2}(H_{1})

to that

B_{d,ω}(H_{1}H_{2}) ⊆ B_{d,ω}(H_{1})B_{d,2}(H_{2}) ∩ B_{d,ω}(H_{2})B_{d,2}(H_{1})  (13.4)

Assume now that x ∈ B_{d,ω}(H_{1})B_{d,2}(H_{2}) ∩ B_{d,ω}(H_{2})B_{d,2}(H_{1}). Then we can write x as

x = y_{1}z_{2} = y_{2}z_{1}

for y_{1} ∈ B_{d,ω}(H_{1}), y_{2} ∈ B_{d,ω}(H_{2}), z_{1} ∈ B_{d,2}(H_{1}), z_{2} ∈ B_{d,2}(H_{2}). Hence

B_{d,2}(H_{2}) ∋ \frac{z_{2}}{y_{2}} = \frac{z_{1}}{y_{1}} ∈ B_{d,2}(H_{1})

so that these quotients lie in the intersection of the two groups. Therefore,

x = y_{1} · y_{2} · \frac{z_{2}}{y_{2}} ∈ B_{d,ω}(H_{1})B_{d,ω}(H_{2}) (B_{d,2}(H_{1}) ∩ B_{d,2}(H_{2}))

Combining this and (13.4), we finally get

B_{d,ω}(H_{1}H_{2}) ⊆ B_{d,ω}(H_{1})B_{d,ω}(H_{2}) (B_{d,2}(H_{1}) ∩ B_{d,2}(H_{2}))

and (13.3) follows for n = 2, m = 2.

Assume now that (13.3) is true for (n, m). Since B_{d,ω+1}(H) = B_{d,ω}(H) for every H, it is enough to show that elements of the form ξ ∘ {u}/ξ for ξ ∈ B_{d,ω}(H), u ∈ H_{1}H_{2} lie in the RHS of (13.3).

Since B_{d,ω}(H_{i}) (i = 1, 2) are closed under composition with {u}, it is sufficient to show that

\frac{ξ ∘ \{u\}}{ξ} ∈ (B_{d,m+1}(H_{1}) ∩ B_{d,n}(H_{2})) (B_{d,2}(H_{1}) ∩ B_{d,n+1}(H_{2}))

for ξ ∈ (B_{d,m}(H_{1}) ∩ B_{d,n}(H_{2})) (B_{d,2}(H_{1}) ∩ B_{d,n+1}(H_{2})).

Write such ξ as ξ = ξ_{1}ξ_{2} for ξ_{1} ∈ B_{d,m}(H_{1}) ∩ B_{d,n}(H_{2}), ξ_{2} ∈ B_{d,2}(H_{1}) ∩ B_{d,n+1}(H_{2}). Write u = u_{1}u_{2} for u_{1} ∈ H_{1}, u_{2} ∈ H_{2}. Then by (13.2),

\frac{ξ ∘ \{u\}}{ξ} = \frac{(ξ_{1} ∘ \{u_{2}\}) ∘ \{u_{1}\}}{ξ_{1} ∘ \{u_{2}\}} · \frac{ξ_{1} ∘ \{u_{2}\}}{ξ_{1}} · \frac{ξ_{2} ∘ \{u\}}{ξ_{2}}

Since ξ_{1} ∈ B_{d,m}(H_{1}) ∩ B_{d,n}(H_{2}), we get that the leftmost element in the product lies in B_{d,m+1}(H_{1}) ∩ B_{d,n}(H_{2}) (because u_{1} ∈ H_{1}), the middle element lies in B_{d,m}(H_{1}) ∩ B_{d,n+1}(H_{2}) (because u_{2} ∈ H_{2}). Similarly, we get that the rightmost element lies in B_{d,2}(H_{1}) ∩ B_{d,n+1}(H_{2}). Therefore the product lies in

(B_{d,m+1}(H_{1}) ∩ B_{d,n}(H_{2})) (B_{d,2}(H_{1}) ∩ B_{d,n+1}(H_{2}))

and (13.3) is also true for (n, m + 1).

Assume now that (13.3) is true for all pairs (n, m) where n is fixed and m < ω. Then

B_{d,ω}(H_{1}H_{2}) ⊆
⋂_{m≥2}B_{d,ω}(H_{1})B_{d,ω}(H_{2}) (B_{d,m}(H_{1}) ∩ B_{d,n}(H_{2})) (B_{d,2}(H_{1}) ∩ B_{d,n+1}(H_{2}))
=B_{d,ω}(H_{1})B_{d,ω}(H_{2}) (B_{d,ω}(H_{1}) ∩ B_{d,n}(H_{2})) (B_{d,2}(H_{1}) ∩ B_{d,n+1}(H_{2}))
=B_{d,ω}(H_{1})B_{d,ω}(H_{2}) (B_{d,2}(H_{1}) ∩ B_{d,n+1}(H_{2}))

by lemma 13.2, and therefore (13.3) is true for (n + 1, 2).

Consider (13.3) for (n, m). Write it as

\begin{aligned}B_{d,ω}(H_{1}H_{2}) & ⊆ B_{d,ω}(H_{1})B_{d,ω}(H_{2}) (B_{d,m}(H_{1}) ∩ B_{d,n}(H_{2})) · \\ & · (B_{d,2}(H_{1}) ∩ B_{d,n+1}(H_{2})) ⊆ B_{d,m}(H_{1})B_{d,n+1}(H_{2}) \end{aligned}

By taking the intersection over all m we get

B_{d,ω}(H_{1}H_{2}) ⊆ B_{d,ω}(H_{1})B_{d,n+1}(H_{2})

for every n ≥ 2. By taking intersection over all n, we finally get

B_{d,ω}(H_{1}H_{2}) ⊆ B_{d,ω}(H_{1})B_{d,ω}(H_{2})

as required.

14 The Extensions N_{d,α}(H)

We are now going to make a connection between the subgroups B_{d,α}(H) defined in the last section and maximal (pro-)nilpotent extensions of local fields.

We begin by establishing the necessary background on (pro-)nilpotent extensions, in the following few lemmas.

Lemma 14.1. Let L/K be an extension with Galois group G. For every n ≥ 1, let L_{n} be the extension defined by Gal(L/L_{n}) = G^{n} (recall that G^{n} is the closed n-th commutator subgroup of G).

Then the extension L_{n}/K is the maximal extension over K with nilpotency class n contained in L.

Proof. Gal(L_{n}/K) = G/G^{n}. This group is nilpotent of class n, as every n + 1-th commutator

[… [[x_{1}, x_{2}], x_{3}] …, x_{n+1}]  (14.1)

is the image of such a commutator in G. Since all such commutators in G lie in G^{n}, we must have that the commutator in (14.1) equals 1.

Assume that Gal(M/K) is nilpotent of class n. Then the image of every n + 1-th commutator in G as in (14.1) modulo Gal(L/M) is 1, meaning that all such commutators lie in Gal(L/M). Thus G^{n} ⊆ Gal(L/M), and M ⊆ L_{n}.

Lemma 14.2. Let L/K be as in the previous lemma. Denote by G^{ω} the intersection of the groups G^{n}, and by L_{ω} the union of all fields L_{n}.

Then Gal(L/L_{ω}) = G^{ω} and L_{ω}/K is the maximal pro-nilpotent extension over K contained in L.

Proof. The groups Gal(L_{n}/K) is nilpotent. Since

Gal(L_{ω}/K) = \underleftarrow{lim} Gal(L_{n}/K)

we get that Gal(L_{ω}/K) is pro-nilpotent.

Let M/K be pro-nilpotent in L. Let x ∈ M, and consider the normal closure of K(x)/K. As this is a finite extension in M and M/K is pronilpotent, it is nilpotent of some class n, hence x ∈ L_{n}. It follows that M ⊆ L_{ω}.

Notation. Let L/k be Galois, k ⊆ L ⊆ (k_{π}k_{d}^{nr})^{ab}, and let n ≥ 1. Denote by M_{d,n}(L) the maximal extension over L with nilpotency class n contained in (k_{π}k_{d}^{nr})^{ab}.

Denote by M_{d,ω}(L) the maximal pro-nilpotent extension over L contained in (k_{π}k_{d}^{nr})^{ab}

Remark. By the last lemma, M_{d,ω}(L) is the union of all M_{d,n}(L) for n ≥ 1.

As in the last section, we still keep d ≥ 1 fixed. By metabelian local class field theory, there is an isomorphism ı : 𝔊_{d}(k, ϕ) ⟶^{≅} Gal((k_{π}k_{d}^{nr})^{ab}/k). There is also an injection 𝔽_{q^{d}}[[X]]^{×} ⟶ 𝔊_{d}(k, ϕ), ξ ↦ (1, ξ). We denote, for B ⊆ 𝔽_{q^{d}}[[X]]^{×}, the image of this injection by 1 × B.

Definition. Let α be an ordinal. Define N_{d,α}(H) as the field contained in (k_{π}k_{d}^{nr})^{ab} for which

Gal((k_{π}k_{d}^{nr})^{ab}/N_{d,α}(H)) = ı(1 × B_{d,α}(H))

Remark. Since B_{d,ω}(H) is the intersection of all B_{d,n}(H), the field N_{d,ω}(H) is the union of the fields N_{d,n}(H).

The following lemma connects the extensions M_{d,n}(L) and N_{d,n}(H).

Lemma 14.3. Let H ≤ U, P ≤ ⟨π^{d}⟩ℤ̂ be closed subgroups. Let L be the field corresponding to the group P × H by local class field theory (that is, Gal(k^{ab}/L) = P × H).

Then

N_{d,n}(L) ⊆ M_{d,n}(H) ⊆ N_{d,n+1}(L)

Proof. The subgroup (k_{π}k_{d}^{nr})^{ab}/L of (k_{π}k_{d}^{nr})^{ab}/k corresponds, under the isomorphism

Gal((k_{π}k_{d}^{nr})^{ab}/k) → 𝔊_{d}(k, ϕ)

to the subgroup

G = {(a, ξ) ∈ 𝔊_{d}(k, ϕ) : a ∈ P × H}

of 𝔊_{d}(k, ϕ). By lemma 14.1, the group Gal((k_{π}k_{d}^{nr})^{ab}/M_{d,n}(L)), for n ≥ 1, correspond, under the same isomorphism, to the group G^{n}.

We prove by induction on n that

1 × B_{d,n+1}(H) ⊆ G^{n} ⊆ 1 × B_{d,n}(H)  (14.2)

It is evident that these inclusions imply the statement of the lemma.

Let (a, ξ), (b, η) ∈ G. Write a = π^{ν}u, b = π^{μ}v for ν, μ ∈ ℤ̂, u, v ∈ H. Then

[(a, ξ), (b, η)] = (1, \frac{ξ}{ϕ^{−μ}ξ ∘ \{v\}} · \frac{ϕ^{−ν}η ∘ \{u\}}{η})

This expression belongs to 𝔊_{d}(k, ϕ). Therefore the power series in the second coordinate belongs to 𝔽_{q^{d}}[[X]]^{×} = B_{d,1}(H). It follows that

G′ ⊆ 1 × B_{d,1}(H)  (14.3)

Now, take (1, η) ∈ G. Then η ∈ 𝔽_{q^{d}}[[X]]^{×}. As μ = 0 and v = 1, we get

[(a, ξ), (1, η)] = (1, \frac{ϕ^{−ν}η ∘ \{u\}}{η})  (14.4)

Since π^{ν} ∈ P and P ≤ ⟨π^{d}⟩, ϕ^{−ν} acts trivially on 𝔽_{q^{d}}[[X]]^{×}, thus the equation (14.4) gets the form

[(a, ξ), (1, η)] = (1, \frac{η ∘ \{u\}}{η})  (14.5)

The group generated by these elements is contained in G′. But this group is exactly 1 × B_{d,2}(H), and this and (14.3) imply (14.2) for n = 1.

Now assume (14.2) for n ≥ 1. G^{n+1} is generated by all the commutators [(a, ξ), (1, η)] for (1, η) ∈ G^{n}. By the induction hypothesis, η ∈ B_{d,n}(H), and by (14.5), such a commutator has its second coordinate in B_{d,n+1}(H).

If η ∈ B_{d,n+1}(H), then by the induction hypothesis (1, η) ∈ G^{n}. Let (η ∘ {u})/η be a generator of B_{d,n+2}(H). Choose (a, ξ) ∈ G where a = π^{ν}u for π^{ν} ∈ P (it is already known that such a choice is always possible). Then by (14.5), the commutator [(a, ξ), (1, η)] has its second coordinate equal to the given generator, and belongs, of course, to G^{n+1}. Hence the assertion follows.

Proposition 14.4. Let H ≤ U, P ≤ ⟨π^{d}⟩ℤ̂ be closed subgroups, and let L be the field corresponding to P × H by (abelian) local class field theory, as in the previous lemma.

Then N_{d,ω}(H) is the maximal pro-nilpotent extension over L contained in (k_{π}k_{d}^{nr})^{ab}.

Proof. By lemma 14.3, N_{d,n}(H) ⊆ M_{d,n}(L) for every n, thus N_{d,ω}(H) ⊆ M_{d,ω}(L). By the same lemma, N_{d,n+1}(H) ⊇ M_{d,n}(L) for every n, hence N_{d,ω}(H) ⊇ M_{d,ω}(L).

By lemma 14.2, M_{d,ω}(L) is the maximal pro-nilpotent extension over L contained in (k_{π}k_{d}^{nr})^{ab}, and the proposition follows.

Remark. By the last proposition, we see that the maximal pro-nilpotent extension over L in (k_{π}k_{d}^{nr})^{ab} does not depend on the group P.

15 Applications to the Extensions (Kk_{π})^{ab}/K

We will now use the results of the previous sections to describe the maximal pro-nilpotent extension over K contained in (Kk_{π})^{ab} for a non-ramified extension K/k.

We start by computing the group B_{d,ω}(U_{1}) for the subgroup U_{1} of the principal units in U.

Lemma 15.1. Let n ≥ 2. Then B_{d,n}(U_{1}) ⊆ 1 + X^{n}𝔽_{q^{d}}[[X]].

Proof. We prove this by induction on n. Let ξ ∈ 𝔽_{q^{d}}[[X]]^{×}, u ∈ U_{1}. Then {u} ≡ X (mod X^{2}). It follows that

\frac{ξ ∘ \{u\}}{ξ} ≡ \frac{ξ ∘ X}{ξ} = 1 (mod X^{2})

and the proof for n = 2 is complete.

Assume the statement of the lemma for n. Let ξ ∈ B_{d,n}(U_{1}), u ∈ U_{1}. By the induction hypothesis, ξ = 1 + X^{n}ξ_{1} for ξ_{1} ∈ 𝔽_{q^{d}}[[X]]^{×}. Hence

ξ ∘ {u} = 1 + {u}^{n} · (ξ_{1} ∘ {u})

Since u ∈ U_{1}, we have u ≡ X (mod X^{2}), hence {u}^{n} ≡ X^{n} (mod X^{n+1}). We get that the coefficient of X^{n} in ξ ∘ {u} equals the constant term of ξ_{1}.

As ξ = 1 + X^{n}ξ_{1}, the coefficient of X^{n} in ξ also equals the constant term of ξ_{1}. It follows that both ξ and ξ ∘ {u} have their X^{n} coefficient equal. Since they are both congruent to 1 modulo X^{n}, we conclude that

ξ ∘ {u} ≡ ξ (mod X^{n+1})

and therefore

\frac{ξ ∘ \{u\}}{ξ} ≡ 1 (mod X^{n+1})

Proposition 15.2. B_{d,ω}(U_{1}) = 1.

Proof. By the previous lemma,

B_{d,ω}(U_{1}) = ⋂_{n≥2} B_{d,n}(U_{1}) ⊆ ⋂_{n≥2}(1 + X^{n}𝔽_{q^{d}}[[X]]^{×}) = 1

Proposition 15.3. Let K/k be a finite non-ramified extension. Then the extension (Kk_{π})^{ab}/(Kk_{π}^{0}) is pro-nilpotent.

Proof. Write K = k_{d}^{nr}. As

Gal(k^{ab}/k_{d}^{nr}k_{π}^{0}) = ⟨π^{d}⟩ℤ̂ × U_{1}

we can apply proposition 14.4 and get that the maximal pro-nilpotent extension over k_{d}^{nr}k_{π}^{0} in (k_{d}^{nr}k_{π})^{ab} equals N_{d,ω}(U_{1}). But by the definition of the groups N_{d,ω}(H) and proposition 15.2,

Gal((k_{d}^{nr}k_{π})^{ab}/N_{d,ω}(U_{1})) = B_{d,ω}(U_{1}) = 1

hence N_{d,ω}(U_{1}) = (k_{d}^{nr}k_{π})^{ab}.

We can generalize proposition 15.3 and omit the assumption that K/k is finite. In order to do that, we need some lemmas.

Lemma 15.4. Let d_{1}, d_{2} ≥ 1. Then k_{d_{1}}^{nr}k_{d_{2}}^{nr} = k_{lcm(d_{1},d_{2})}^{nr}.

Proof. Write g = gcd(d_{1}, d_{2}), l = lcm(d_{1}, d_{2}). Since d_{1} | l, d_{2} | l, the field k_{l}^{nr} contains k_{d_{1}}^{nr} and k_{d_{2}}^{nr} and therefore contains their compositum.

By a similar argument, k_{g}^{nr} is contained in k_{d_{1}}^{nr} and k_{d_{2}}^{nr}. Now, [k_{d_{1}}^{nr} : k_{g}^{nr}] = d_{1}/g, [k_{d_{2}}^{nr} : k_{g}^{nr}] = d_{2}/g. Since d_{1}/g and d_{2}/g are relatively prime, the compositum satisfies

[k_{d_{1}}^{nr}k_{d_{2}}^{nr} : k_{g}^{nr}] = d_{1}d_{2}/g^{2} = l/g

thus [k_{d_{1}}^{nr}k_{d_{2}}^{nr} : k] = l and we get the required equality.

Proposition 15.5. Let K/k be non-ramified. Then K is a union of an increasing sequence of finite non-ramified extensions over k.

Proof. K is a compositum of finite non-ramified extensions k_{d}^{nr}. Order these extensions in a sequence {k_{d_{n}}^{nr}}, and define,

e_{1} = d_{1}
e_{2} = lcm(d_{1}, d_{2})
⋮
e_{n} = lcm(d_{1}, d_{2}, …, d_{n})

By the last lemma, K equals the compositum of {k_{e_{n}}^{nr}}, which is an increasing sequence of fields.

Proposition 15.6. Let {L_{n}} be an increasing sequence of fields. Let L = ∪_{n=1}^{∞}L_{n} and assume that the extensions L/L_{n} are abelian.

Then

L^{ab} = ⋃_{n=1}^{∞} L_{n}^{ab}

Proof. We prove that for every finite abelian extension M/L, there is an n for which M ⊆ L_{n}^{ab}.

Let M/L be such an extension. Since M/L is Galois and finite, there is a polynomial f(X) ∈ L[X] such that M is the splitting field of f(X) over L. As L is the union of all L_{n} and the number of coefficients in f(X) is finite, there is an n for which f(X) ∈ L_{n}[X].

Let M′ be the splitting field of f(X) over L_{n}. Then M = LM′ and

Gal(M/L ∩ M′) = Gal(M/L) × Gal(M/M′)
≅ Gal(M/L) × Gal(L/L ∩ M′)

Since M/L and L/L∩M′ are abelian (the Galois group of the latter extension is a subgroup of Gal(L/L_{n})), we get that Gal(M/L ∩ M′) is abelian. But L ∩ M′/L_{n} is finite, hence there is n′ for which L ∩ M′ ⊆ L_{n′}. We conclude that

Gal(M/L_{n′}) ⊆ Gal(M/L ∩ M′)

is abelian.

Theorem 15.7. Let K/k be non-ramified.

Then the extension (Kk_{π})^{ab}/(Kk_{π}^{0}) is pro-nilpotent.

Proof. Proposition 15.3 states that the theorem is true in the case that K/k is finite. In the general case, write K as the union of an increasing sequence of finite non-ramified extensions K_{n}/k.

By proposition 15.3, for every n ≥ 1,

Gal((K_{n}k_{π})^{ab}/(Kk_{π}^{0})) ⊆ Gal((K_{n}k_{π})^{ab}/(K_{n}k_{π}^{0}))

is pro-nilpotent.

The sequence L_{n} = K_{n}k_{π} satisfies the conditions of proposition 15.6 (take L = Kk_{π}), therefore

(Kk_{π})^{ab} = ⋃_{n=1}^{∞}(K_{n}k_{π})^{ab}

It follows that Gal((Kk_{π})^{ab}/(Kk_{π}^{0})) is the inverse limit of the pro-nilpotent groups Gal((K_{n}k_{π})^{ab}/(Kk_{π}^{0})), hence it is also pro-nilpotent.

Proposition 15.8. Let V be the group of the (q − 1)-th roots of unity in U. Then

B_{d,ω}(U) = B_{d,ω}(V)

Proof. By U = V × U_{1} and propositions 13.5 and 15.2,

B_{d,ω}(U) = B_{d,ω}(V)B_{d,ω}(U_{1}) = B_{d,ω}(V)

By local class field theory, k_{π}/k = U. Let L be the field for which k_{π}/L = V. From U/V ≅ U_{1} and the fact that U_{1} is a pro-p group, we get that L/k is a pro-p extension. Moreover, since V has an order relatively prime to p, it follows that L/k is the maximal pro-p extension contained in k_{π}. We denote L by k_{π,p}.

Theorem 15.9. Let K/k be non-ramified. Then the maximal pro-nilpotent extension over K contained in (Kk_{π})^{ab} equals the maximal pro-nilpotent extension over k_{π,p}k^{nr} contained in (Kk_{π})^{ab}.

Proof. Let k_{d}^{nr} ⊆ K, and apply proposition 14.4 for the fields K and k_{π,p}k^{nr}. We get that

M_{d,ω}(K) = N_{d,ω}(U)

and

M_{d,ω}(k_{π,p}k^{nr}) = N_{d,ω}(V)

hence, by proposition 15.8, the maximal pro-nilpotent extension over K contained in (k_{d}^{nr}k_{π})^{ab} equals the maximal pro-nilpotent extension over k_{π,p}k^{nr} contained in the same field.

Since (Kk_{π})^{ab} can be written as a union of fields (k_{d}^{nr}k_{π})^{ab} (cf. the proof of theorem 15.7), our theorem follows.

Corollary. The extensions k_{π,p}^{ab}/k and (k_{π,p}k^{nr})^{ab} ∩k_{π}^{ab}/k are pro-nilpotent.

Proof. Since

k_{π,p}^{ab} ⊆ (k_{π,p}k^{nr})^{ab} ∩ k_{π}^{ab}

it is enough to prove it for the latter extension. Now, (k_{π,p}k^{nr})^{ab} ∩ k_{π}^{ab} is abelian over k_{π,p}k^{nr} (and in particular, pro-nilpotent), contained in k_{π}^{ab}, and the corollary follows by taking K = k in theorem 15.9.

It may be interesting to describe explicitly the subgroups B_{d,ω}(V). Here is a result in this direction.

Proposition 15.10. Let v ∈ V be a generator. Let λ be the homomorphism

λ : 𝔽_{q^{d}}[[X]]^{×} ⟶ 𝔽_{q^{d}}[[X]]^{×}
λ : ξ ⟼ \frac{ξ ∘ \{v\}}{ξ}

Then for every n ≥ 0,

B_{d,n+1}(V) = Im λ^{n}

In particular, B_{d,ω}(V) = ⋂_{n≥1} Im λ^{n}.

Proof. We prove this by induction on n. The proposition is trivial for n = 0. For the induction step, we prove that

B_{d,n+1}(V) = λ(B_{d,n}(V))  (15.1)

Since (ξ∘{v})/ξ ∈ B_{d,n+1}(V) (this is one of the generators of B_{d,n+1}(V)), the inclusion ⊆ follows. Now, let

\frac{ξ ∘ \{v^{i}\}}{ξ}

be a generator of B_{d,n+1}(V), and write it as

\frac{ξ ∘ \{v^{i}\}}{ξ} = ∏_{j=0}^{i−1} \frac{(ξ ∘ \{v^{j}\}) ∘ \{v\}}{ξ ∘ \{v^{j}\}}

It follows that this generator belongs to λ(B_{d,n}(V)), and the opposite inclusion is proved.

Finally, by B_{d,n}(V) = Im λ^{n} and (15.1) we get B_{d,n+1}(V) = Im λ^{n+1}.

We can give also a topological description of B_{d,ω}(V).

Lemma 15.11. The set of functions {λ^{n}} is equicontinuous.

Proof. For every r ≥ 1, denote by I_{r} the group 1+X^{r}𝔽_{q^{d}}[[X]]. It is sufficient to prove that for every r ≥ 1 and n ≥ 1,

λ^{n}(I_{r}) ⊆ I_{r}

If ξ ∈ I_{r}, then since {v} is divisible by X, we have that ξ ∘ {v} ∈ I_{r}, hence λ(ξ) ∈ I_{r}. Therefore λ^{n}(ξ) ∈ I_{r} for every n ≥ 1.

Proposition 15.12. The group B_{d,ω}(V) equals the set of all limit points of the sequences {λ^{n}(ξ)}, ξ ∈ 𝔽_{q^{d}}[[X]]^{×}.

Proof. Let η be a limit point of a sequence {λ^{n}ξ}. There is a subsequence λ^{n_{k}}ξ → η, and since Im λ^{n} is a decreasing sequence of closed subgroups, η lies in their intersection, which equals B_{d,ω}(V) by proposition 15.10.

Conversely, let η ∈ B_{d,ω}(V). By proposition 15.10, for every n ≥ 1, there is ξ_{n} for which η = λ^{n}ξ_{n}. The sequence {ξ_{n}} has a convergent subsequence ξ_{n_{k}} → ξ, and by the equicontinuity of the functions λ^{n}, we have λ^{n_{k}}ξ → η. Therefore η is a limit point of a sequence {λ^{n}ξ}.

References

[Co] Coleman R., Division values in local fields, Inv. Math., 53 (1979), 91–116.

[dS–K] de-Shalit E., Koch H., Metabelian local class field theory, to appear in J. Crelle.

[Iw] Iwasawa K., Local Class Field Theory, Oxford University Press, New York, 1986.

[L–T] Lubin J., Tate J., Formal complex multiplication in local fields, Ann. Math., 81 (1965), 380–387.