Local Leopoldt’s Problem

for Rings of Integers in Abelian p-Extensions

of Complete Discrete Valuation Fields

M. V. Bondarko

Received: April 12, 2000 Revised: December 8, 2000

Communicated by A. S. Merkurev

Abstract. Using the standard duality we construct a linear embedding of an associated module for a pair of ideals in an extension of a Dedekind ring into a tensor square of its fraction field. Using this map we investigate properties of the coefficient-wise multiplication on associated orders and modules of ideals. This technique allows to study the question of determining when the ring of integers is free over its associated order. We answer this question for an Abelian totally wildly ramified p-extension of complete discrete valuation fields whose different is generated by an element of the base field. We also determine when the ring of integers is free over a Hopf order as a Galois module.

1991 Mathematics Subject Classification: 11S15, 11S20, 11S31 Keywords and Phrases: Complete discrete valuation (local) fields, additive Galois modules, formal groups, associated Galois modules

Introduction

0.1. Additive Galois modules and especially the ring of integers of local fields are considered from different viewpoints. Starting from H. Leopoldt [L] the ring of integers is studied as a module over its associated order. To be precise, if K is an extension of a local field k with Galois group being equal to G and 𝔒_{K} is the ring of integers of K, then 𝔒_{K} is considered as a module over 𝔄_{K/k}(𝔒_{K}) = {λ ∈ k[G], λ𝔒_{K} ⊂ 𝔒_{K}}. One of the main questions is to determine when 𝔒_{K} is free as an 𝔄_{K/k}(𝔒_{K})- module. Another related problem is to describe explicitly the ring 𝔄_{K/k}(𝔒_{K}) (cf. [Fr], [Chi], [CM]).

This question was actively studied by F. Bertrandias and M.-J. Ferton (cf. [Be], [B-F], [F1-2]) and more recently by M. J. Taylor, N. Byott and G. Lettl (cf. [T1], [By], [Le1]). In particular, G. Lettl proved that if K/ℚ_{p} is Abelian and k ⊂ K, then 𝔒_{K} ≈ 𝔄_{K/k}(𝔒_{K}) (cf. [Le1]). The proof was based on the fact that all Abelian extensions of ℚ_{p} are cyclotomic. So the methods of that paper and of most of preceding ones are scarcely applicable in more general situations. M. J. Taylor [T1] considers intermediate extensions in the tower of Lubin-Tate extensions. He proves that for some of these extensions 𝔒_{K} is a free 𝔄_{K/k}(𝔒_{K})- module. Taylor considers a formal Lubin-Tate group F (X, Y) over the ring of integers 𝔬 of a local field k. Let π be a prime element of the field k and T_{m} be equal to Ker[π^{m}] in the algebraic closure of the field k. For 1 ≤ r ≤ m, let L be equal to k(T_{m+r}), and let K be equal to k(T_{m}), . Lastly, let q be the cardinality of the residue field of k. Taylor proves that

(1) the ring 𝔒_{L} is a free 𝔄_{L/K}(𝔒_{L})-module and any element of L whose valuation is equal to q^{r} − 1 generates it, and

(2) 𝔄_{L/K}(𝔒_{L}) = 𝔒_{K} + ∑_{i=0}^{q^{r}−2} 𝔒_{K}σ_{i}, where σ_{i} ∈ K[G] and are described explicitly (cf. the details in [T1], subsection 1.4). This result was generalized to relative formal Lubin-Tate groups in the papers [Ch] and [Im]. Results of these papers were proved by direct computation. So these works do not show how one can obtain a converse result, i.e., how to find all extensions, that fulfill some conditions on the Galois structure of the ring of integers. To the best of author’s knowledge the only result obtained in this direction was proved in [By1] and refers only to cyclotomic Lubin-Tate extensions.

0.2. In the examples mentioned above the associated order is also a Hopf order in the group algebra (i.e., it is an order stable under comultiplication). Several authors are interested in this situation. The extra structure allows to deal with wild extensions as if they were tame (in some sense). This is why in this situation, following Childs, one speaks of “taming wild extensions by Hopf orders”. In the paper [By1], Byott proves that the associated order can be a Hopf order only in the case when the different of the extension is generated by an element of the smaller field. The present paper is also dedicated to extensions of this sort. More precisely, Theorem 4.4 of the paper [By1] implies that the order 𝔄_{K/k}(𝔒_{K}) can be a Hopf order (in the case when the ring 𝔒_{K} is 𝔬[G]- indecomposable) only if K/k fulfills the stated condition on the different and 𝔒_{K} is free over 𝔄_{K/k}(𝔒_{K}). Our main Theorem settles completely the question to determine when the order 𝔄_{K/k}(𝔒_{K}) is a Hopf order. It also describes completely Hopf orders that can be obtained as associated Galois orders. We shall also prove in subsection 3.4 that under the present assumptions if 𝔒_{K} is free over 𝔄_{K/k}(𝔒_{K}), then 𝔄_{K/k}(𝔒_{K}) is a Hopf order and determine when the inverse different of an Abelian totally ramified p-extension of a complete discrete valuation field is free over its associated order (cf. [By1] Theorem 3.10).

In the first section we study a more general situation. We consider a Galois extension K/k of fraction fields of Dedekind rings, with Galois group G. We prove a formula for the module

ℭ_{K/k}(I_{1}, I_{2}) = Hom_{𝔬}(I_{1}, I_{2}) ,

where I_{1}, I_{2} are fractional ideals of K (this set also can be defined for ideals that are not G-stable) in Theorem 1.3.1. We introduce two submodules of ℭ_{K/k}(I_{1}, I_{2}):

\begin{matrix}𝔄_{K/k}(I_{1}, I_{2}) = \{f ∈ k[G]|f(I_{1}) ⊂ I_{2}\} \\ 𝔅_{K/k}(I_{1}, I_{2}) = Hom_{𝔬[G]}(I_{1}, I_{2}), . \end{matrix}

These modules coincide in the case when K/k is Abelian (cf. Proposition 1.4.2). We call these modules the associated modules for the pair I_{1}, I_{2}. In case I_{1} = I_{2} we call the associated module associated order. In subsection 1.5 we define a multiplication on the modules of the type ℭ_{K/k}(I_{1}, I_{2}) and show the product of two such modules lies in some third one. We have also used this multiplication in the study of the decomposability of ideals in extensions of complete discrete valuation fields with inseparable residue field extension (cf. [BV]). Starting from the second section we consider totally wildly ramified extensions of complete discrete valuation fields with residue field of characteristic p with the restriction on the different:

𝔇_{K/k} = (δ), δ ∈ k  (∗).

This second section is dedicated to the study of conditions ensuring that the ring of integers 𝔒_{K} is free over its associated order 𝔄_{K/k}(𝔒_{K}) or 𝔅_{K/k}(𝔒_{K}). Let n = [K : k]. We prove the following statement.

Proposition. If in the associated order 𝔄_{K/k}(𝔒_{K}) (𝔅_{K/k}(𝔒_{K}) resp.) there is an element ξ which maps some (and so, any) element a ∈ 𝔒_{K} with valuation equal to n − 1 onto a prime element of the ring 𝔒_{K}, then 𝔒_{K} ≈ 𝔄_{K/k} (resp. 𝔅_{K/k}) and besides that 𝔄_{K/k} (resp. 𝔅_{K/k}) has a “power” base (in the sense of

multiplication ∗^{Δ}, cf. the second section), which is constructed explicitly using the element ξ.

The converse to this statement is also proved in the case when 𝔄_{K/k}(𝔒_{K}) (resp. 𝔅_{K/k}(𝔒_{K})) is indecomposable (cf. the Theorems 2.4.1, 2.4.2). An important part of our reasoning is due to Byott (cf. [By1]).

0.3. The third, fourth and fifth sections are dedicated to proving a more explicit form of the condition of the second section for the Abelian case. The main result of the paper is the following one. We will call an Abelian p-extension of complete discrete valuation fields of characteristic 0 almost maximally ramified if its degree divides the different.

Theorem A. Let K/k be an Abelian totally ramified p-extension of p-adic fields, which in case char k = 0 is not almost maximally ramified, and suppose that the different of the extension is generated by an element in the base field (see (*)). Then the following conditions are equivalent:

1. The extension K/k is Kummer for a formal group F, that there exists a formal group F over the ring of integers 𝔬 of the field k, a finite torsion subgroup T of the formal module F (𝔐_{𝔬}) and a prime element π_{0} of k such that K = k(x), where x is a root of the equation P (X) = π_{0}, where

P (X) = ∏_{t∈T}(X −_{F} t) .

2. The ring 𝔒_{K} is isomorphic to the associated order 𝔄_{K/k}(𝔒_{K}) as an 𝔬[G]- module.

Remark 0.3.1. Besides proving Theorem A we will also construct the element ξ explicitly and so describe 𝔄_{K/k}(𝔒_{K}) (cf. the theorems of §2).

Remark 0.3.2. IF k is of characteristic 0 the fact that 𝔒_{K} is indecomposable as an 𝔬[G]-module if and only if K/k is not almost maximally ramified, was proved in [BVZ]. The case of almost maximally ramified extensions is well understood (cf., for example, [Be]). It is obvious, that in the case char k = p the ring 𝔒_{K} is indecomposable as an 𝔬[G]-module, because in this case the algebra k[G] is indecomposable.

Remark 0.3.3. In the paper [CM] rings of integers in Kummer extensions for formal groups are also studied as modules over their associated orders. In that paper Kummer extensions are defined with the use of homomorphisms of formal groups. For extensions obtained in this way freeness of the ring of integers over its associated orders is proved. Childs and Moss also use some tensor product to prove their results. Yet their methods seem to be inapplicable for proving inverse results. Our notion of a Kummer extension for a formal group is essentially equivalent to the one in [CM]. We however only use one formal group and do not impose finiteness restriction on its height. Besides we consider also the equal characteristic case i.e., char k = char \overline{k} = p. Using the methods presented here one can prove that we can actually take the formal group F in Theorem A of finite height. Yet such a restriction does not seem to be natural.

Remark 0.3.4. Theorem A shows which Hopf orders can be associated to Galois orders for some Abelian extensions. The papers of mathematicians that “tame wild extensions by Hopf orders” do not show that their authors know or guess that such an assertion is valid. In the third section we study the fields that are Kummer in the sense of part 1 of Theorem A and deduce 2 from 1. In the fourth section we prove that we can suppose the coefficient b_{1} in ξ = δ^{−1} ∑ b_{σ}σ to be equal to 0. Further, if b_{1} is equal to 0, then we show that there exists a formal group F over 𝔬, such that for σ ∈ G the b_{σ}, form a torsion subgroup in the formal module F (𝔐_{𝔬}), i.e.,

b_{σ} +_{F} b_{τ} = b_{στ} .

In the fifth section we prove that if b_{σ} +_{F} b_{τ} is indeed equal to b_{στ}, then K/k is

Kummer for the group F.

0.4. This paper is the first in a series of papers devoted to associated orders and associated modules. The technique introduced in this paper (especially the map ϕ and the multiplication ∗ defined below) turns out to be very useful in studying The Galois structure of ideals. It allows the author to prove in another paper some results about freeness of ideals over their associated orders in extensions that do not fulfill the condition on the different (*). In particular, using Kummer extensions for formal groups we construct a wide variety of extensions in which some ideals are free over their associated orders. These examples are completely new. We also calculate explicitly the Galois structure of all ideals in such extensions. Such a result is very rare. In a large number of cases the necessary and sufficient condition for an ideal to be free over its associated order is found. The author is deeply grateful to professor S. V. Vostokov for his help and advice. The work paper is supported by the Russian Fundamental Research Foundation N 01-00-000140.

§1 General results

Let 𝔬 be a Dedekind ring, k be the fraction field of the ring 𝔬, K/k be a Galois extension with Galois group equal to G, 𝔇 = 𝔇_{K/k} be the different of the extension K/k, n = [K : k], tr be the trace operator in K/k. We also define some associated modules. For I_{1} and I_{2} G-stable ideals in K put 𝔅_{K/k}(I_{1}, I_{2}) = Hom_{𝔬[G]}(I_{1}, I_{2}). For I_{1}, I_{2} not being G-stable one can define

𝔅_{K/k}(I_{1}, I_{2}) = {f ∈ Hom_{𝔬[G]}(K, K), f(I_{1}) ⊂ I_{2}}.

For arbitrary I_{1}, I_{2} ⊂ K we define 𝔄_{K/k}(I_{1}, I_{2}) = {f ∈ k[G]|f(I_{1}) ⊂ I_{2}}, ℭ_{K/k}(I_{1}, I_{2}) = Hom_{𝔬}(I_{1}, I_{2}). Obviously, any 𝔬-linear map from I_{1} into I_{2} can be extended to a k-linear homomorphism from K into K. The dimension of Hom_{k}(K, K) over k is equal to n^{2}. Now we consider the group algebra K[G]. This algebra acts on K and the statement in (Bourbaki, algebra, §7, no. 5) implies that a non-zero element of K[G] corresponds to a non-zero map from K into K. Besides the dimension of K[G] over k is also equal to n^{2}. It follows that any element of Hom_{k}(K, K) can be expressed uniquely as an element of K[G]. So we reckon ℭ_{K/k}(I_{1}, I_{2}) being embedded in K[G].

1.1. We consider the G-Galois algebra K ⊗_{k} K. It is easily seen that the tensor product K ⊗_{k} K is isomorphic to a direct sum of n copies of K as a k-algebra. Being more precise, let K_{σ}, σ ∈ G denote a field, isomorphic to K.

Lemma 1.1.1. There is an isomorphism of K-algebras

ψ : K ⊗_{k} K → ∑_{σ∈G} K_{σ},

where ψ = ∑_{σ} ψ_{σ} and ψ_{σ} is the projection on the coordinate σ, defined by the equality

ψ_{σ}(x ⊗ y) = xσ(y) ∈ K_{σ}.

The proof is quite easy, you can find it, for example, in ”Algebra” of Bourbaki. Also see 1.2 below. Now we construct a map ϕ from the G-Galois algebra K ⊗_{k} K into the group algebra K[G]:

(1)  \begin{matrix}ϕ : K ⊗_{k} K → K[G] \\ α = ∑_{i} x_{i} ⊗ y_{i} → ϕ(α) = ∑_{i} x_{i} (∑_{σ∈G} σ(y_{i})σ) . \end{matrix}

It is clear that ϕ(α) as a function acts on K as follows:

(2)  ϕ(α)(z) = ∑_{i} x_{i}tr(y_{i}z), z ∈ K.

Besides that, the map ϕ may be expressed through ψ_{σ} in the form

(3)  ϕ(α) = ∑_{σ} ψ_{σ}(α)σ, α ∈ K ⊗_{k} K.

Proposition 1.1.2. The map ϕ is an isomorphism of k-vector spaces between K ⊗_{k} K and K[G].

Cf. the sketch of the proof in Remark 1.2 below.

1.2 Pairings on K ⊗_{k} K and K[G]. There is a natural isomorphism of the k-space K and its dual space of kfunctionals:

\begin{matrix}K → K̂ \\ a → f_{a}(b) = tr(ab). \end{matrix}

We define a pairing ⟨, ⟩_{⊗} on K ⊗_{k} K, that takes its values in k:

\begin{matrix}(K ⊗_{k} K) × (K ⊗_{k} K) → k \\ ⟨a ⊗ b, c ⊗ d⟩_{⊗} → (f_{a} ⊗ f_{c})(b ⊗ d) = (tr ac)(tr bd). \end{matrix}

This pairing is correctly defined and is non-degenerate. The second fact is easily proved with the use of dual bases. We also define a pairing on K[G] that takes its values in k:

\begin{matrix}K[G] × K[G] → k \\ α = ∑_{σ} a_{σ}σ, β = ∑_{σ} b_{σ}σ → ∑_{σ} tr a_{σ}b_{σ} = ⟨α, β⟩_{K[G]}. \end{matrix}

The pairing ⟨, ⟩_{K[G]} is also non-degenerate because it is a direct sum of n = [K : k] non-degenerate pairings

\begin{matrix}K × K → k \\ (a, b) → tr ab. \end{matrix}

We check that for any two x, y the following equality is fulfilled:

(4)  ⟨x, y⟩_{⊗} = ⟨ϕ(x), ϕ(y)⟩_{K[G]}.

Indeed, it follows from linearity that it is sufficient to prove that for x = a ⊗ b, y = c ⊗ d. In that case we have

⟨a ⊗ b, c ⊗ d⟩_{⊗} = tr ac tr bd,

and besides that

⟨ϕ(a ⊗ b), ϕ(c ⊗ d)⟩_{K[G]}

= ⟨a ∑_{σ} σ(b)σ, c ∑_{σ} σ(d)σ⟩_{K[G]}

= ∑_{σ} tr(acσ(bd)) = ∑_{σ} ∑_{τ} τ(acσ(bd))

= ∑_{τ} ∑_{σ} τ(ac)τσ(bd) = tr ac tr bd.

and the equality (4) is proved.

Remark 1.2. It follows from (4) that the map ϕ from (1) is an injection. Indeed, let ϕ(α) be equal to 0 for α ∈ K ⊗_{k} K, then ⟨ϕ(α), ϕ(β)⟩_{K[G]} = 0 for any b ∈ K ⊗_{k} K. So, according to (4), ⟨α, β⟩_{⊗} = 0 for any β ∈ K ⊗_{k} K. From the non-degeneracy of the pairing ⟨, ⟩_{⊗} it follows that α = 0. That implies Ker(ϕ) = 0. Using the equality of dimensions we can deduce that ϕ is an isomorphism.

1.3 Modules of homomorphisms for a pair of ideals. Let I_{1}, I_{2} be fractional ideals of the field K.

Theorem 1.3.1. Let ϕ be the bijection from (1) and let I_{1}^{∗} = 𝔇^{−1}I_{1}^{−1} (this is dual of I_{1} for the bilinear trace form on K). For the associated modules the following equality holds:

ℭ_{K/k}(I_{1}, I_{2}) = ϕ(I_{2} ⊗_{𝔬} I_{1}^{∗}) .

Proof. First we show that ϕ(I_{2} ⊗_{𝔬} I_{1}^{∗}) ⊂ ℭ_{K/k}(I_{1}, I_{2}). If x ∈ I_{2}, y ∈ I_{1}^{∗}, then for any z ∈ I_{1} we have, according to the definition (2):

ϕ(x ⊗ y)(z) = x tr(yz) ∈ I_{2},

since x ∈ I_{2} and tr(yz) ∈ 𝔬, which follows from the definition of I_{1}^{∗} and of the different 𝔇. Conversely, let f ∈ ℭ_{K/k}(I_{1}, I_{2}). We define a map θ_{f} and show that it is sent onto f by ϕ. Let:

(5)  \begin{matrix}θ_{f} : I_{2}^{∗} ⊗_{𝔬} I_{1} → 𝔬 \\ x ⊗ y → tr(xf(y)). \end{matrix}

This map is correctly defined since x ∈ I_{2}^{∗} = 𝔇^{−1}I_{2}^{−1} and f(y) ∈ I_{2}, thus

tr(xf(y)) ∈ 𝔬.

For a 𝔬-module M we denote by M̂ the module of 𝔬-linear functions from M into 𝔬. It is clear that

(6)  θ_{f} ∈ { \atop (I_{2}^{∗} ⊗_{𝔬} I_{1})}.

We identify the ideal I_{2} with the 𝔬-module Î_{2}^{∗} via:

\begin{matrix}I_{2} → Î_{2}^{∗} \\ a → f_{a} = ∑_{σ∈G} σ(a)σ. \end{matrix}

Obviously f_{a}(z) = tr(az) for any z ∈ Î_{2}. In a completely analogous way we identify I_{1}^{∗} with Î_{1}^{∗} Using these identifications and the fact that I_{1} and I_{2}^{∗} are projective 𝔬-modules we obtain an isomorphism

(7)  \begin{matrix}I_{2} ⊗_{𝔬} I_{1}^{∗} → { \atop (I_{2}^{∗} ⊗_{𝔬} I_{1})} \\ a ⊗ b → h_{a,b}, \end{matrix}

where h_{a,b}(x ⊗ y) = ⟨a ⊗ b, x ⊗ y⟩_{⊗} = (tr ax)(tr by) for all x in I_{2}^{∗} and y in I_{1}.

The map θ_{f} in (5) lies in { \atop (I_{2}^{∗} ⊗_{𝔬} I_{1})} (cf. (6)), and so, according to the isomorphism (7), it corresponds to an element α_{f} in I_{2} ⊗_{𝔬} I_{1}^{∗}, i.e.,

α_{f} = ∑_{i} a_{i} ⊗ b_{i}, a_{i} ∈ I_{2}, b_{i} ∈ I_{1}^{∗}.

Then (7) implies that the functional h_{α_{f}}, that corresponds to the element α_{f}, is defined in the following way:

h_{α_{f}} (x ⊗ y) = ⟨α_{f}, x ⊗ y⟩_{⊗} = ∑_{i} tr a_{i}x tr b_{i}y.

On the other hand, from the definition of θ_{f} (cf. (5)) we obtain:

h_{α_{f}} (x ⊗ y) = θ_{f}(x ⊗ y) = tr(xf(y)).

It follows that f(y) = ∑_{i} a_{i} tr(b_{i}y). So we have

f = ϕ(∑ a_{i} ⊗ b_{i}),

and for any f in ℭ_{K/k}(I_{1}, I_{2}) we have found its preimage in I_{2} ⊗ I_{1}^{∗}, i.e.,

ℭ_{K/k}(I_{1}, I_{2}) ⊂ ϕ(I_{2} ⊗_{𝔬} I_{1}^{∗}).

The theorem is proved. □

Remark 1.3.2. In the same way as above we can prove, that if we replace the ideals I_{1}, I_{2} by two arbitrary free 𝔬-submodules X and Y of K of dimension n, then we will obtain the following formula:

ℭ_{K/k}(X, Y) = ϕ(Y ⊗ X̂),

where X̂ is the dual module to X in K with respect to the pairing K × K → k defined by the trace tr.

Remark 1.3.3. All elements in ℭ_{K/k}(I_{1}, I_{2}), 𝔄_{K/k}(I_{1}, I_{2}), 𝔅_{K/k}(I_{1}, I_{2}) have unique extensions to k-linear maps from K to K. To be more precise, if f : I_{1} → I_{2} is an 𝔬-homomorphism, then for all x ∈ K we can assume f(x) = αf(a) if x = αa, α ∈ k, a ∈ I_{1}. It is easily seen that the map we obtained in this way is a correctly defined k-linear homomorphism from K into K.

1.4. Now we compare the modules 𝔄 and 𝔅.

Proposition 1.4.

(8)  𝔄_{K/k}(I_{1}, I_{2}) = 𝔅_{K/k}(I_{1}, I_{2})

if and only if K/k is an Abelian extension.

Proof. 1. Let K/k be an Abelian extension. To verify the equality (8) in this case, let first f belong to 𝔄_{K/k}(I_{1}, I_{2}), then f is an 𝔬-homomorphism from I_{1} into I_{2}. Besides that, f commutes with all elements of G since G is an Abelian group, i.e., σf(a) = f(σ(a)) for all σ ∈ G and a ∈ I_{1}. So we obtain that f is an 𝔬[G]-homomorphism from I_{1} into I_{2}, thus f ∈ 𝔅_{K/k}(I_{1}, I_{2}). For the reverse inclusion, let f belong to 𝔅_{K/k}(I_{1}, I_{2}). Then f induces an 𝔬[G]-homomorphism from K into K. We take an element x that generates a normal base of the field K over k. Then there exists an element g ∈ k[G] such that f(x) = g(x), to be more precise, if f(x) = ∑_{σ} a_{σ}σ(x), a_{σ} ∈ k then we take g = ∑ a_{σ}σ. We consider the 𝔬[G]-homomorphism g from K into K. Since G is an Abelian group, f(σ(x)) = σ(f(x)) = σ(g(x)) = g(σ(x)) for any σ ∈ G. We obtain that k-homomorphisms f and g coincide on the basic elements and so f = g ∈ k[G] and f(I_{1}) ⊂ I_{2}, i.e., f ∈ 𝔄_{K/k}(I_{1}, I_{2}).

2. Now we suppose that 𝔄_{K/k}(I_{1}, I_{2}) = 𝔅_{K/k}(I_{1}, I_{2}) and check that K/k is an Abelian extension. Indeed, since k𝔄_{K/k}(I_{1}, I_{2}) = k[G], 𝔄_{K/k}(I_{1}, I_{2}) contains elements of the form aσ, where a ∈ k^{∗}, for any σ ∈ G. It follows from our assumption that aσ ∈ 𝔅_{K/k}(I_{1}, I_{2}), and so aσ is an 𝔬[G]-homomorphism. We obtain that G commutes with all elements σ ∈ G. Proposition is proved. □

Proposition 1.5. If we assume the action of G on K ⊗_{k} K to be diagonal, then

𝔅_{K/k}(I_{1}, I_{2}) = ϕ((I_{2} ⊗_{o} I_{1}^{∗})^{G}) .

Proof. Let α belong to (I_{2} ⊗_{𝔬} I_{1}^{∗})^{G}. We have to show that ϕ(a) is an 𝔬[G]- homomorphism from I_{1} into I_{2}. This means that

σϕ(a)(z) = ϕ(a)(σ(z))

for all z ∈ I_{1}. Let α be equal to ∑ a_{i} ⊗ b_{i}, a_{i} ∈ I_{2}, b_{i} ∈ I_{1}^{∗}. Then from the definition of ϕ (cf. (2)) it follows that

\begin{aligned}ϕ(a)(σ(z)) & = ∑_{i} a_{i} tr(b_{i}σ(z)) \\ & = ∑_{i} a_{i} tr(σ^{−1}(b_{i})σ(z)) = ϕ (∑_{i} a_{i} ⊗ σ^{−1}(b_{i})) (z). \end{aligned}

On the other hand,

\begin{matrix}σϕ(a)(z) = σ (∑ a_{i} tr(b_{i}z)) \\ = ∑(σ(a_{i}) tr(b_{i}z)) = ϕ (∑ σ(a_{i}) ⊗ b_{i}) (z). \end{matrix}

From the G-invariance of the element α it follows that

ϕ (∑ a_{i} ⊗ σ^{−1}(b_{i})) = ϕ (∑ σ(a_{i}) ⊗ σ(σ^{−1}(b_{i}))) = ϕ (∑ σ(a_{i}) ⊗ b_{i}) .

Thus ϕ(α)(σ(z)) = σϕ(α)(z) for any σ ∈ G, i.e., ϕ(α) ∈ 𝔅_{K/k}(I_{1}, I_{2}). Conversely, let f belong to 𝔅_{K/k}(I_{1}, I_{2}), then f ∈ ℭ_{K/k}(I_{1}, I_{2}) and so, according to Theorem 1.3.1, there is an α ∈ I_{2} ⊗ I_{1}^{∗}, such that f = ϕ(α). It remains to check that α is G-invariant. We use the fact that f is an Ghomomorphism, i.e., σf(z) = f(σz) for all σ ∈ G and z ∈ I_{1}. We obtain an equality σϕ(α)(z) = ϕ(α)(σz). By writing the left and the right side of the equality as above we obtain for α = ∑ a_{i} ⊗ b_{i}:

\begin{matrix}σϕ(α)(z) = ϕ (∑ σa_{i} ⊗ σ(σ^{−1}b_{i})) (z) \\ ϕ(α)(σz) = ϕ (∑ a_{i} ⊗ σ^{−1}b_{i}) (z). \end{matrix}

It follows that

ϕ (∑ σa_{i} ⊗ σ(σ^{−1}b_{i})) = ϕ (∑ a_{i} ⊗ σ^{−1}b_{i}) .

Now using the fact that ϕ is a bijection we obtain

∑ σa_{i} ⊗ σ(σ^{−1}b_{i}) = ∑ a_{i} ⊗ σ^{−1}b_{i}.

We apply to both sides of the equality the map

\begin{matrix}1 ⊗ σ : K ⊗_{k} K → K ⊗_{k} K \\ a ⊗ b → a ⊗ σ(b) \end{matrix},

that obviously is an homomorphism. We have

∑ σa_{i} ⊗ σb_{i} = ∑ a_{i} ⊗ b_{i},

i.e., σ(α) = α. □

1.6 The multiplication ∗ on K[G]. On the algebra K ⊗_{k} K there is a natural multiplication: (a⊗b)·(c⊗d) = ac⊗bd. Using it and the bijection ϕ we define a multiplication on K[G]. To be more precise, if f, g ∈ K[G], then we define

f ∗ g = ϕ(ϕ^{−1}(f) · ϕ^{−1}(g)) ∈ K[G].  (9)

Proposition 1.6.1. If f = ∑_{σ} a_{σ}σ and g = ∑_{σ} b_{σ}σ, then

f ∗ g = ∑_{σ} a_{σ}b_{σ}σ.

Proof. Let f be equal to ϕ(α), g be equal to ϕ(β), where α, β ∈ K ⊗_{k} K. If α = ∑ x_{i} ⊗ y_{i}, β = ∑ u_{j} ⊗ v_{j}, then from the definition of ϕ (cf. (1)) we obtain

ϕ(α) = ∑ x_{i} ∑_{σ} σy_{i}σ, ϕ(β) = ∑ u_{j} ∑_{σ} σv_{j}σ.

It follows that

∑_{σ} a_{σ}b_{σ}σ = ∑_{i,j} x_{i}u_{j}σ(y_{i}v_{j})σ.

On the other hand,

f ∗ g = ϕ(αβ) = ϕ (∑_{i,j}(x_{i}u_{i} ⊗ y_{i}v_{j})) = ∑_{i,j} x_{i}u_{j}σ(y_{i}v_{j})σ

and we obtain the proof of Proposition. □

Remark 1.6.2. The formula from Proposition 1.6.1 will be used further as an another definition of the multiplication ∗.

1.7 Multiplication on associated modules. Now we consider the multiplication (9) on the different associated modules. Here we will see appearing the different which we will suppose to be induced from the base field in the following sections.

Proposition 1.7.1. Let f belong to ℭ_{K/k}(I_{1}, I_{2}), and let g belong to ℭ_{K/k}(I_{3}, I_{4}). Then

f ∗ g ∈ ℭ_{K/k}(I_{1}I_{3}𝔇, I_{2}I_{4}).

Proof. It is clear that ϕ^{−1}(f) and ϕ^{−1}(g) belong to I_{2} ⊗_{o} I_{1}^{∗} and I_{4} ⊗ I_{3}^{∗} respectively. So we obtain that ϕ^{−1}(f)ϕ^{−1}(g) lies in the product

\begin{matrix}(I_{2} ⊗_{o} I_{1}^{∗})(I_{4} ⊗_{o} I_{3}∗) = I_{2}I_{4} ⊗_{o} (I_{1}^{∗}I_{3}^{∗}) \\ = I_{2}I_{4} ⊗_{o} 𝔇^{−2}I_{1}^{−1}I_{3}^{−1} = I_{2}I_{4} ⊗_{o} (𝔇I_{1}I_{3})^{∗}. \end{matrix}

So from the Theorem 1.3.1 it follows that f ∗g belongs to ℭ_{K/k}(I_{1}I_{3}𝔇, I_{2}I_{4}). □

Now we study the multiplication ∗ on the modules 𝔅_{K/k}.

Proposition 1.7.2. Let f belong to 𝔅_{K/k}(I_{1}, I_{2}) and let g belong to 𝔅_{K/k}(I_{3}, I_{4}), then

f ∗ g ∈ 𝔅_{K/k}(I_{1}I_{3}𝔇, I_{2}I_{4}).

Proof. Since 𝔅_{K/k}(I, J) is a submodule in ℭ_{K/k}(I, J), from Proposition 1.7 it follows that f ∗ g ∈ ℭ_{K/k}(I_{1}I_{3}𝔇, I_{2}I_{4}). From Proposition 1.5 we deduce that f and g belong to phi(K ⊗_{k} K^{G}), g ∈ ϕ(K ⊗_{k} K^{G}). So f ∗ g ∈ ϕ(K ⊗_{k} K^{G}), and this implies that

f ∗ g ∈ ℭ_{K/k}(I_{1}I_{3}𝔇, I_{2}I_{4}) ∩ ϕ(K ⊗_{k} K^{G}) = 𝔅_{K/k}(I_{1}I_{3}𝔇, I_{2}I_{4}).

□

Proposition 1.7.3. Let f belong to 𝔄_{K/k}(I_{1}, I_{2}) and LET g belong to 𝔄_{K/k}(I_{3}, I_{4}), then

f ∗ g ∈ 𝔄_{K/k}(I_{1}I_{3}𝔇, I_{2}I_{4}).

Proof. From Proposition 1.7 it follows that

f ∗ g ∈ ℭ_{K/k}(I_{1}I_{3}𝔇, I_{2}I_{4})  (10)

since 𝔄_{K/k}(I_{1}, I_{2}) and 𝔄_{K/k}(I_{3}, I_{4}) are submodules OF ℭ_{K/k}(I_{1}, I_{2}) and ℭ_{K/k}(I_{3}, I_{4}) respectively. From the definition of 𝔄_{K/k} it follows that f and g belong to k[G]. So the coefficients of f and g lie in k, and from Proposition 1.6.1 it follows that f ∗ g also belongs to k[G]. Then (10) implies that

f ∗ g ∈ ℭ_{K/k}(I_{1}I_{3}𝔇, I_{2}I_{4}) ∩ k[G] = 𝔄_{K/k}(I_{1}I_{3}𝔇, I_{2}I_{4}),

and thus the proposition is proved. □

§2 Isomorphism of rings of integers of totally wildly ramified extensions of complete discrete valuation fields with their associated orders.

2.1. Let K/k be a totally wildly ramified Galois extension of a complete discrete valuation field with residue field of characteristic p. Let 𝔇 be the different of the extension and let 𝔒_{K} be the ring of integers of the field K. From this moment and up to the end of the paper we will suppose the condition (*) of the introduction to be fulfilled, i.e., that 𝔇 = (δ), with δ ∈ k. We will write 𝔄_{K/k}(𝔒_{K}), 𝔅_{K/k}(𝔒_{K}) instead of 𝔄_{K/k}(𝔒_{K}, 𝔒_{K}), 𝔅_{K/k}(𝔒_{K}, 𝔒_{K}). We denote prime elements of the fields k and K by π_{0} and π respectively, and their maximal ideals by 𝔐_{𝔬} and 𝔐.

Proposition 2.1. The modules ℭ_{K/k}(𝔒_{K}), 𝔄_{K/k}(𝔒_{K}), 𝔅_{K/k}(𝔒_{K}) are 𝔬algebras with a unit with respect to the multiplication

f ∗^{Δ}g = δf ∗ g = δ^{−1}ϕ(ϕ^{−1}(δf)ϕ^{−1}(δg))

(cf. (9)). The unit is given by δ^{−1}tr.

The motivation for the above definition is given by Theorem 2.4.1 below.

Proof. Let f and g belong to ℭ_{K/k}(𝔒_{K}), then according to Proposition 1.7,

the product f ∗ g maps the different 𝔇 into the ring 𝔒_{K}. It follows that f ∗^{Δ}g maps 𝔇 into 𝔇, and so it also maps 𝔬 into itself since 𝔇 = δ𝔒_{K}. We obtain that ∗^{Δ} defines a multiplication on the each of the modules associated to 𝔒_{K}. Now we consider the element δ^{−1} tr and prove that it is the unit for the multiplication ∗^{Δ} in each of these modules. It is clear that δ^{−1} tr maps 𝔒_{K} into itself and that δ^{−1} tr belongs to k[G], so δ^{−1} tr belongs to 𝔄_{K/k}(𝔒_{K}). Besides that, δ^{−1} tr commutes with all elements of G and so δ^{−1} tr lies in 𝔅_{K/k}(𝔒_{K}). Let now f belong to K[G], then

f = ∑_{σ} a_{σ}σ, a_{σ} ∈ K, δ^{−1} tr = ∑_{σ} δ^{−1}σ.

So, according to proposition 1.6.1,

f ∗ (δ^{−1} tr) = ∑_{σ} δ^{−1}a_{σ}σ

and we obtain

f ∗^{Δ}(δ^{−1} tr) = ∑ a_{σ} = f.

Since 𝔄_{K/k}(𝔒_{K}), 𝔅_{K/k}(𝔒_{K}) are 𝔬-submodules in ℭ_{K/k}(𝔒_{K}), δ^{−1} tr ∈ 𝔄_{K/k}(𝔒_{K}), 𝔅_{K/k}(𝔒_{K}), δ^{−1} tr is also an identity in 𝔄_{K/k}(𝔒_{K}), 𝔅_{K/k}(𝔒_{K})

with respect to the multiplication ∗^{Δ}.

2.2. Let as before n be equal to [K : k].

Lemma 2.2.1. Let x be an element of the ring 𝔒_{K} whose valuation equals n − 1, i.e., v_{K}(x) = n − 1. Then

v_{k}(tr x) = v_{k}(δ) .

In particular, there is an element a in 𝔒_{K} with v_{K}(a) = n − 1 and such that tr a = δ.

Proof. Let 𝔐 and 𝔐_{𝔬} be the maximal ideals of K and k respectively. From the definition of the different and surjectivity of the trace operator it follows that tr(𝔐^{−1}𝔇^{−1}) = 𝔐_{𝔬}^{−1}. Moreover any element of 𝔐^{−1}𝔇^{−1}, that does not belong to 𝔇^{−1}, has a non-integral trace. So the trace of the element z = π_{0}^{−1}δ^{−1}x is equal to

tr z = π_{0}^{−1}δ^{−1} tr x = π_{0}^{−1}ε, ε ∈ 𝔬^{∗}.

Thus tr x = εδ. Further, if we multiply the element x by ε^{−1} ∈ 𝔬^{∗}, then we get the element a. □

Lemma 2.2.2. In the ring 𝔒_{K} we can choose a basis a_{0}, a_{1}, …, a_{n−2}, a, where a is as in Lemma 2.2.1, and the a_{i} for 0 ≤ i ≤ n − 2 are such that v_{K}(a_{i}) = i, and satisfy tr a_{i} = 0.

Proof. The kernel Ker tr(𝔒_{K}) has 𝔬-rank equal to n−1. Let x_{0}, …, x_{n−2} be an 𝔬-basis of Ker tr(𝔒_{K}). Along with the element a they form a 𝔬-base of the ring 𝔒_{K}. By elementary operations in Ker tr(𝔒_{K}) we can get from x_{0}, …, x_{n−2} a set of elements with pairwise different valuations. Their valuations have to be less than n − 1. Indeed, otherwise by subtracting from the element x_{0} of valuation n − 1 an element a of the same valuation multiplied by a coefficient in 𝔬^{∗} we can obtain an element of 𝔐^{n}, which is impossible. □

2.3. Let a be an element of 𝔒_{K} with valuation equal to n−1, where n = [K : k], and let

tr a = δ,  (12)

where δ is a generator of the different 𝔇_{K/k} (cf. Lemma 2.2.1).

Proposition 2.3.1. 1. The module 𝔄_{K/k}(𝔒_{K})(a) mod 𝔐^{n} is a subring with an identity in 𝔒_{K} mod 𝔐^{n} (with standard multiplication).

2. The multiplication ∗^{Δ} in 𝔄_{K/k}(𝔒_{K}) (cf. (11)) induces the standard multiplication in the ring 𝔄_{K/k}(𝔒_{K})(a) mod 𝔐^{n}, i.e.,

f ∗^{Δ}g(a) ≡ f(a)g(a) mod 𝔐^{n}.

Proof. Let f and g belong to 𝔄_{K/k}(𝔒_{K}). Then the preimages ϕ^{−1}(δf), ϕ^{−1}(δg) with respect to the bijection ϕ belong to 𝔒_{K} ⊗_{o} 𝔒_{K}, since

𝔇 ⊗_{o} 𝔇^{−1} = δ𝔒_{K} ⊗_{o} δ^{−1}𝔒_{K} = 𝔒_{K} ⊗_{o} 𝔒_{K}.

We prove that

ϕ^{−1}(δa) = x ⊗ 1 + y,  (13)

where x ∈ 𝔒_{K} and y ∈ 𝔒_{K} ⊗ 𝔐. Indeed, ϕ^{−1}(δa) = ∑ a_{i} ⊗ b_{i}. If b_{i} ∈ 𝔐, then a_{i} ⊗ b_{i} ∈ 𝔒_{K} ⊗ 𝔐, otherwise b_{i} = c_{i} +d_{i}, where c_{i} ∈ 𝔬, d_{i} ∈ 𝔐 and so a_{i} ⊗b_{i} = a_{i} ⊗c_{i} +a_{i} ⊗d_{i} = c_{i}x_{i} ⊗1+y_{i}, where c_{i}x_{i} ∈ 𝔒_{K}, since c_{i} ∈ 𝔬 and y_{i} ∈ 𝔒_{K} ⊗ 𝔐. So (13) follows. Similarly

ϕ^{−1}(δg) = x′ ⊗ 1 + y′,  (14)

where x′ ∈ 𝔒_{K}, y′ ∈ 𝔒_{K} ⊗ 𝔐. Thus

\begin{matrix}ϕ^{−1}(δf)ϕ^{−1}(δg) = (x ⊗ 1 + y)(x′ ⊗ 1 + y′) \\ = xx′ ⊗ 1 + y((x′ ⊗ 1) + y′) + (x ⊗ 1)y′ \\ = xx′ ⊗ 1 + z, where z ∈ 𝔒_{K} ⊗ 𝔐. \end{matrix}  (15)

We consider the action of the element f ∈ 𝔄_{K/k}(𝔒_{K}) on the element a with valuation equal to n − 1. From (13) we obtain f = δ^{−1}ϕ(x ⊗ 1 + y), where x ∈ 𝔒_{K}, y ∈ 𝔒_{K} ⊗ 𝔐. Then from the definition of the map ϕ we have:

f(a) = δ^{−1}ϕ(x ⊗ 1 + y)(a)
= δ^{−1}(ϕ(x ⊗ 1)(a) + ϕ(y)(a))
= δ^{−1}x tr a + δ^{−1}ϕ(y)(a).

We show that δ^{−1}ϕ(y)(a) ∈ 𝔐^{n}, i.e.,

f(a) = δ^{−1}x tr a + z, where z ∈ 𝔐^{n}.  (16)

Indeed, let y be equal to ∑ a_{i} ⊗ b_{i}, then from the definition of ϕ we deduce that

ϕ(y)(a) = ∑ a_{i} tr(b_{i}a).

Moreover tr b_{i}a ∈ 𝔇𝔐_{𝔬}, thus δ^{−1}a_{i} tr(b_{i}a) ∈ 𝔐^{n}, i.e., z = δ^{−1}ϕ(y)(a) ∈ 𝔐^{n} and we obtain (16). Our assumptions imply that tr a = δ, so

f(a) ≡ x mod 𝔐^{n}

and similarly

g(a) ≡ x′ mod 𝔐^{n},

where x′ is the element from (14). Then

f(a)g(a) ≡ xx′ mod 𝔐^{n}.

On the other hand from the definition of the multiplication f ∗^{Δ}g (cf. (11)) it follows that

f ∗^{Δ}g(a) = δ^{−1}ϕ(ϕ^{−1}(δf)ϕ^{−1}(δg))(a).

Using this and keeping in mind (15) and (16) we obtain

f ∗^{Δ}g(a) = δ^{−1}ϕ(xx′ ⊗ 1 + z)(a) ≡ xx′ mod 𝔐^{n}.

So we have the congruence

f ∗^{Δ}g(a) ≡ f(a)g(a) mod 𝔐^{n}.

Also, the element δ^{−1} tr in 𝔄_{K/k}(𝔒_{K}) gives us an identity element in the ring 𝔄_{K/k}(𝔒_{K})(a) mod 𝔐^{n}, since δ^{−1} tr a = 1 (cf. (12)). □

Remark 2.3.2. For any other element a′ in the ring 𝔒_{K} with valuation equal to n − 1 we have 𝔄_{K/k}(𝔒_{K})(a′) ≡ ε𝔄_{K/k}(𝔒_{K})(a) mod 𝔐^{n}, ε ∈ 𝔬^{∗}.

Remark 2.3.3. A similar statement also holds for the module 𝔅_{K/k}(𝔒_{K})(a) mod 𝔐^{n}.

2.4. Now we formulate the statements which we will begin to prove in the next subsection. We will investigate the following condition: in the order 𝔄_{K/k}(𝔒_{K}) (resp. 𝔅_{K/k}(𝔒_{K}) ) there exists an element ξ such that

ξ(a) = π,  (17)

where π is a prime element of the field K and a is some element with valuation equal to n − 1.

Theorem 2.4.1. If in the ring 𝔄_{K/k}(𝔒_{K}) (resp. 𝔅_{K/k}(𝔒_{K})) the condition (17) is fulfilled, then the element ξ generates a “power” basis of 𝔄_{K/k}(𝔒_{K})

(𝔅_{K/k}(𝔒_{K}) resp.) over 𝔬 with respect to the multiplication ∗^{Δ} (cf. 11), i.e.,

𝔄_{K/k}(𝔒_{K}) = ⟨ξ^{0}, ξ^{1}, …, ξ^{n−1}⟩,

where ξ^{0} = δ^{−1} tr is the unit and ξ^{i} = ξ∗^{Δ}ξ^{i−1}.

Theorem 2.4.2. 1. If for the ring 𝔄_{K/k}(𝔒_{K}) (resp. 𝔅_{K/k}(𝔒_{K})) the condition (17) is fulfilled, then the ring 𝔒_{K} is a free 𝔄_{K/k}(𝔒_{K})-module (resp. free 𝔅_{K/k}(𝔒_{K})-module).

2. If the ring 𝔒_{K} is a free module over the ring 𝔄_{K/k}(𝔒_{K}) (resp. 𝔅_{K/k}(𝔒_{K})) and if moreover the order 𝔄_{K/k}(𝔒_{K}) (resp. 𝔅_{K/k}(𝔒_{K})) is indecomposable (i.e does not contain non-trivial idempotents), then for the ring 𝔄_{K/k}(𝔒_{K}) (resp. 𝔅_{K/k}(𝔒_{K})) the condition (17) and so also the assertions of the theorem 2.4.1 are fulfilled.

Remark 2.4.3. If ξ maps some element with valuation equal to n − 1 onto an element with valuation equal to 1, then ξ also maps any other element with valuation equal to n − 1 onto an element with valuation equal to 1. Indeed, if a ∈ K, v_{k}(a) = n − 1 and v_{K}(ξ(a)) = 1, then any other element a′ ∈ 𝔒_{K}, for which v_{K}(a′) = n − 1, is equal to εa + b, where ε ∈ 𝔬^{∗}, b ∈ 𝔐^{n}, so b = π_{0}b′, where π_{0} is a prime element in k. Besides that b′ ∈ 𝔒_{K} and we obtain ξ(b) = π_{0}ξ(b′) and this implies v_{K}(ξ(b)) ≥ n. We also have v_{K}(ξ(εa)) = v_{K}(εξ(a)) = v_{K}(ε) + 1 = 1 i.e., ε ∈ 𝔬^{∗}, and so v_{K}(ξ(a′)) = 1.

Remark 2.4.4. Any element ξ in 𝔄_{K/k}(𝔒_{K}) (resp. in 𝔅_{K/k}(𝔒_{K})) that fulfills the condition (17) generates a power base of the 𝔬-module 𝔄_{K/k}(𝔒_{K}) (resp.

𝔅_{K/k}(𝔒_{K})) with respect to the multiplication ∗^{Δ}.

2.5 Proof of Theorem 2.4.1 and of the first part of Theorem 2.4.2.. We take the element a in the ring 𝔒_{K} such that v_{K}(a) = n − 1 and tr a = δ (cf. (12)). By assumption we have ξ(a) = π, where π is a prime element of the field K. We check that

ξ^{i}(a) ≡ π^{i} mod 𝔐_{K}^{n}.  (18)

Indeed, if i = 0, then for the usual product ξ^{0}(a)δ^{−1} tr(a) = 1. Let further ξ^{i−1}(a) be equal to π^{i−1} mod 𝔐_{K}^{n}, then according to Proposition 2.3.1 we have

π^{i} ≡ ξ^{i−1}(a)ξ(a) ≡ (ξ^{i−1}∗^{Δ}ξ)(a) ≡ ξ^{i}(a) mod 𝔐_{K}^{n}

and the congruence (18) is proved. This equality implies that ξ^{i}(a), 0 ≤ i ≤ n − 1, generate 𝔒_{K}, i.e., 𝔒_{K} = 𝔬ξ^{0}(a) ⊕ ⋯ ⊕ 𝔬ξ^{n−1}(a). Now we show that

𝔄_{K/k}(𝔒_{K}) = ⟨ξ^{0}, …, ξ^{n−1}⟩.

If there exists an element η ∈ 𝔄_{K/k}(𝔒_{K}) that does not belong to a 𝔬-module ⟨ξ^{0}, …, ξ^{n−1}⟩, then η(a) = b ∈ 𝔒_{K}. So we obtain

η(a) = ∑_{i=0}^{n−1} α_{i}ξ^{i}(a), α_{i} ∈ 𝔬,

i.e.,

(η − ∑ α_{i}ξ^{i})(a) = 0  (19)

Now we show that

η = ∑ α_{i}ξ^{i}.  (20)

Indeed the spaces k𝔄_{K/k}(𝔒_{K}) and k⟨ξ^{0}, …, ξ^{n−1}⟩ have equal dimensions and so coincide. It follows from (19) that

η − ∑ α_{i}ξ^{i} ∈ 𝔄_{K/k}(𝔒_{K}) ⊂ k𝔄_{K/k}(𝔒_{K}) = k⟨ξ^{0}, …, ξ^{n−1}⟩,

and so

η − ∑ α_{i}ξ^{i} = ∑_{i=0}^{n−1} α′_{i}ξ^{i},

where α′_{i} ∈ k. Since α′_{i} ∈ k and the valuations of the elements ξ^{i}(a), 0 ≤ i ≤ n− 1 are pairwise non-congruent mod n (cf. (16)), the valuations v_{K}((α′_{i}ξ^{i})(a)) are also pairwise non-congruent mod n, and so ∑ α′_{i}ξ^{i}(a) ≠ 0 if not all the α′_{i} are equal to 0. This reasoning proves (20). So we have obtained that the ring 𝔒_{K} is a free 𝔄_{K/k}(𝔒_{K})-module:

𝔒_{K} = 𝔄_{K/k}(𝔒_{K})(a)

and we have proved Theorem 2.4.1 and the first part of Theorem 2.4.2.

Lemma 2.6. Let x be an element of the ring 𝔒_{K} such that tr x = 0, then for any f ∈ 𝔄_{K/k}(𝔒_{K}) and g ∈ 𝔅_{K/k}(𝔒_{K}) the following equalities hold:

tr f(x) = 0 = tr g(x) = 0.

Proof. If f ∈ 𝔄_{K/k}(𝔒_{K}), then f = ∑_{σ∈G} a_{σ}σ, a_{σ} ∈ k, and so

tr f(x) = tr(∑ a_{σ}σ(x)) = ∑ a_{σ} tr σ(x) = 0.

If g ∈ 𝔅_{K/k}(𝔒_{K}), then tr g(x) = g(tr(x)) = g(0) = 0, since g is an 𝔬[G]- homomorphism. □

2.7. Proof of necessity in Theorem 2.4.2. Let 𝔒_{K} be a free 𝔄_{K/k}(𝔒_{K})-module and assume the order 𝔄_{K/k}(𝔒_{K}) IS indecomposable. We prove that in the order 𝔄_{K/k}(𝔒_{K}) there exists an element ξ that fulfills the condition (17) of 2.4. We take elements a_{0}, …, a_{n−2} such that v_{K}(a_{i}) = i and such that tr a_{i} = 0 (cf. Lemma 2.2.2 and also [By1]). We take further an element a with valuation equal to n − 1. Let χ : 𝔒_{K} → 𝔄_{K/k}(𝔒_{K}) be an isomorphism of 𝔬[G]-modules, then

𝔄_{K/k}(𝔒_{K}) = ⟨χ(a_{0}), …, χ(a)⟩_{𝔬}.

Thus, in particular, there exist α_{i}, α ∈ 𝔬 such that

1 = α_{0}χ(a_{0}) + ⋯ + α_{n−2} + αχ(a).

The ring 𝔄_{K/k}(𝔒_{K}) is indecomposable by assumption, and so, according to the Krull-Schmidt Theorem, it is a local ring. We obtain that one of the χ(a_{i}) OR χ(a) has to be invertible in the ring 𝔄_{K/k}(𝔒_{K}). The elements χ(a_{i}) cannot be invertible since, according to Lemma 2.6, 𝔄_{K/k}(𝔒_{K})(a_{i}) ∈ Ker tr 𝔒_{K}. Thus χ(a) is invertible and it follows that 𝔄_{K/k}(𝔒_{K})(a) = 𝔒_{K}. We obtain that there exists a ξ in 𝔄_{K/k}(𝔒_{K}) such that ξ(a) = π. Theorem 2.4.2 is proved.

Remark 2.7. The corresponding reasoning for the ring 𝔅_{K/k}(𝔒_{K}) almost literally repeats the one we used above.

§3 Kummer extensions for formal groups. The proof of sufficiency in Theorem A.

Starting from this section we assume that the extension K/k is Abelian.

3.1. We denote the valuation on k by v_{0}. We also denote by v_{0} the valuation on K that coincides with v_{0} on k. We suppose that the field k fulfills the conditions of §2 and that F is some formal group over the ring 𝔬 (the coefficients of the series F (X, Y) may, generally speaking, lie, for example, in the ring of integers of some smaller field). On the maximal ideal 𝔐_{𝔬} of the ring 𝔬 we introduce a structure of a formal ℤ_{p}-module using the formal group F by letting for x, y ∈ 𝔐_{𝔬} and α ∈ ℤ_{p}

x +_{F} y = F (x, y),

αx = [α](x).

We denote the ℤ_{p}-module obtained in this way by F (𝔐_{𝔬}). Let T be a finite torsion subgroup in F (𝔐_{𝔬}) and let n = card T be the cardinality of the group T. Obviously, n is a power of p. We construct the following series:

P (X) = ∏_{t∈T}(X −_{F} t).  (21)

Remark 3.1.1. The constant term of the series P (X) is equal to zero, the coefficient at X^{n} is invertible in 𝔬, and the coefficients at the powers, not equal to n, belong to the ideal 𝔐_{𝔬}.

Lemma 3.1.2. Let a be a prime element of the field k and K = k(x) be the extension obtained from k by adjoining the roots of the equation P (X) = a, where the series P is as in (21). Then the extension K/k is a totally ramified Abelian extension of degree n and the different 𝔇 of the extension K/k is generated by an element of the base field, i.e., 𝔇 = (δ), δ ∈ k. The ramification jumps of the extension K/k are equal to h_{l} = nv_{0}(t_{l}) − 1, t_{l} ∈ T.

Proof. Using the Weierstrass Preparation Lemma we decompose the series P (X) − a into a product

P (X) − a = cf(X)ε(X),

where ε(X) ∈ 𝔬[[X]]^{∗} is an invertible series (with respect to multiplication), f(X) is A unitary polynomial, c ∈ 𝔬. Then, according to Remark 3.1.1, f(X) is an Eisenstein polynomial of degree n. The series P (X) − a has the same roots (we consider only the roots of P (X) − a with positive valuation) as the polynomial f(X), and so there are exactly n roots. It is obvious that if P (x) = a, then

P (x +_{F} τ) = ∏_{t∈T}(x −_{F} t +_{F} τ) = ∏_{t∈T}(x −_{F} t) = P (x) = a, τ ∈ T

and so the roots of P (X) − a and f(X) are exactly the elements x +_{F} τ, τ ∈ T.

Thus we proved that all roots of f(X) lie in K and are all distinct. It follows that K/k is a Galois extension. We denote the Galois group of the extension

K/k by G. Obviously if σ_{1}, σ_{2} ∈ G, σ_{1}(x) = x +_{F} t_{1}, σ_{2}(x) = x +_{F} t_{2}, then

σ_{2}(σ_{1}(x)) = σ_{2}(x +_{F} t_{1}) = σ_{2}(x) +_{F} t_{1} = x +_{F} t_{2} +_{F} t_{1} (as F (X, Y) is defined

over 𝔬, t_{1} ∈ T). Since the addition +_{F} is commutative, the extension K/k is

Abelian. We also have that ∏ σ(x) = a, v_{k}(a) = 1, and so v_{0}(x) = \frac{e(K/k)}{n}. This implies that the extension K/k is totally ramified and that x is a prime element in K. Now we compute the ramification jumps of the extension K/k. Let F (X, Y) = X +Y +∑_{i,j>0} a_{ij}X^{i}Y^{j} be the formal group law, then we have

x − σ_{l}(x) = x − (x +_{F} t_{l}) = t_{l} + ∑_{i,j>0} a_{ij}x^{i}t^{j} = t_{l}ε_{t},

where ε_{t} is a unit of the ring 𝔒_{K}. It follows that the ramification jumps of the extension K/k are equal to nv_{k}(t_{l}) − 1. Hence the exponent of the different is equal to ∑_{t_{l}∈T} (h_{l} + 1) = n ∑_{t_{l}∈T} v_{k}(t_{l}) (cf., for example, [Se], Ch. 4, Proposition 4) and so v_{k}(𝔇) ≡ 0 mod n. □

3.2. Before beginning the proof of Theorem A we prove that the first condition of Theorem A is equivalent to a weaker one.

Proposition 3.2. Let a belong to 𝔬, v_{k}(a) = ns + 1, where 0 ≤ s < min_{t∈T} v_{k}(t). Then the extension k(x)/k, where x is A root of the equation P (X) = a, has the same properties as the extensions from the first condition of Theorem A.

Proof. We consider the series

F_{s}(X, Y) = π_{0}^{−s}F (π_{0}^{s}X, π_{0}^{s}Y).

It is easily seen that F_{s} also defines a formal group law and the elements of T_{s} = {π_{0}^{−s}t, t ∈ T } form some torsion subgroup in the formal module F_{s}(𝔐_{𝔬}). Indeed, if F (X, Y) = ∑ a_{ij}X^{i}Y^{j}, then F_{s}(X, Y) = ∑ π_{0}^{s(i+j−1)}a_{ij}X^{i}Y^{j}, and so the coefficients of F_{s}(X, Y) are integral. Since π_{0}^{−s}X +_{F_{s}} π_{0}^{−s}Y = π_{0}^{−s}(X +_{F}Y),

F_{s} indeed defines an associative and commutative addition. Besides that if

u_{1}, u_{2} ∈ T_{s}, then u_{1} +_{F_{s}} u_{2} = π_{0}^{−s}(π_{0}^{s}u_{1} +_{F} π_{0}^{s}u_{2}) ∈ T_{s} and so T_{s} is indeed a

subgroup in F (𝔐_{𝔬}). Now we compute the series P_{F_{s}}(X) for the formal group F_{s}. We obtain:

P_{F_{s}}(X) = ∏_{t∈T}(X −_{F_{s}} π_{0}^{−s}t) = ∏_{t}(π_{0}^{−s}(π_{0}^{s}X) −_{F_{s}} π_{0}^{−s}t)

= ∏_{t} π_{0}^{−s}(π_{0}^{s}X −_{F} t) = π_{0}^{−sn}P_{F} (π_{0}^{s}X).

Thus the equation P_{f}(X) = a is equivalent to P_{F_{s}}(π_{0}^{−s}X) = aπ_{0}^{−sn}. Besides that v(π_{0}^{−sn}a) = v(a) − sn, i.e., π_{0}^{−sn}a is a prime element in k. Now it remains to note that a root of the equation P_{f}(X) = a can be obtained by multiplication of a root of the equation P_{F_{s}}(Y) = aπ_{0}^{−sn} by π_{0}^{s}, and so the extensions obtained by adjoining the roots of these equations coincide. □

3.3 The proof of Theorem A: 1 ⟹ 2. Let K/k be an extension obtained by adjoining the roots of the equation P (X) = π_{0}. So K = k(x) for some root x. We consider the maximal ideal 𝔐_{⊗} of the tensor product 𝔒_{K} ⊗ 𝔒_{K}. Obviously 𝔐_{⊗} = 𝔒_{K} ⊗ 𝔐 + 𝔐 ⊗ 𝔒_{K}. We also have 𝔐_{⊗}^{i} = ∑_{j=0}^{i} 𝔐^{j} ⊗ 𝔐^{i−j}, i > 0, and so it is easily seen that ∩_{i>0}𝔐_{⊗}^{i} = 0. It follows that we can introduce ℤ_{p}-module structure F (𝔐_{⊗}) on 𝔐_{⊗}. We consider the element

α = x ⊗ 1 −_{F} 1 ⊗ x ∈ 𝔐_{⊗} .

We can define ξ in the following way:

ξ = δ^{−1}ϕ(α).  (22)

We check the following properties of the element ξ:

1. ξ ∈ 𝔄_{K/k}(𝔒_{K})

2. If y belongs to 𝔒_{K} and v(y) = n − 1, then ξ(y) is a prime element in K. For (1): Let X −_{F} Y be equal to ∑ b_{ij}X^{i}Y^{j}, then the equalities

ϕ(α) = ∑ b_{ij}ϕ(x^{i} ⊗ 1^{i}) ∗ ϕ(1^{j} ⊗ x^{j})

= ∑_{σ∈G, i,j} b_{ij}x^{i}σ(x^{j})σ = ∑(x −_{F} σx)σ = ∑ t_{σ}σ ∈ k[G]

follow from the definitions of ϕ and ∗. It also follows from Theorem 1.3.1 that ξ belongs to ℭ_{K/k}(𝔒_{K}). Thus ξ ∈ k[G] ∩ ℭ_{K/k}(𝔒_{K}) = 𝔄_{K/k}(𝔒_{K}). For (2): It easily seen that

(x ⊗ 1) −_{F} (1 ⊗ x) = x ⊗ 1 + y,  (23)

where y ∈ 𝔐 ⊗ 𝔐. Indeed,

x ⊗ 1 −_{F} 1 ⊗ x = x ⊗ 1 + ∑_{i≥1,j≥0} b_{ij}x^{i} ⊗ x^{j}.

Since b_{ij} ∈ 𝔬, b_{ij}x^{i} ⊗ x^{j} ∈ 𝔒_{K} ⊗ 𝔐 and we obtain (23). We can assume that tr(xa) = δ (cf. Remark 2.4.3). Then, as in Proposition 2.3.1, we have

ξ(a) = δ^{−1}ϕ(x ⊗ 1 + y)(a) ≡ x mod 𝔐^{n}.

Since v(x) = 1, we have proved the property 2.

3.4. Now we construct explicitly a basis of an associated order for extensions that fulfill the condition 1 of Theorem A. We have proved above that we can take the element ξ to be equal to δ^{−1} ∑ t_{σ}σ. Then it follows from Theorem 2.4.1 that

𝔄_{K/k}(𝔒_{K}) = ⟨δ^{−1} ∑_{σ} t_{σ}^{i}σ, i = 0, …, n − 1⟩.

Now suppose that K is generated by a root of the equation P (X) = b, where v_{k}(b) = sn + 1, s < min v_{0}(t_{l}). In 3.2 we proved that K may be generated by a root of the equation P_{s}(X) = bπ_{0}^{−sn}, and so it follows in this case that

𝔄_{K/k}(𝔒_{K}) = ⟨δ^{−1}π_{0}^{−si}t_{σ}^{i}σ, i = 0, …, n − 1⟩.

Proposition 3.4.1. Suppose that the extension K/k fulfills the condition 1 of Theorem A. Then 𝔄_{K/k}(𝔒_{K}) is a Hopf order in the group ring k[G] with respect to the standard Hopf structure.

Proof. 𝔄_{K/k}(𝔒_{K}) is an order in the group ring. It is easily seen, that if f = ∑_{σ∈G} c_{σ}σ ∈ 𝔄_{K/k}(𝔒_{K}), then ∑ c_{σ}σ^{−1} ∈ 𝔄_{K/k}(𝔒_{K}). Indeed, we have ϕ^{−1}(f) ∈ δ^{−1}𝔒_{K} ⊗_{o} 𝔒_{K}. Now consider the 𝔬-linear map i : K ⊗_{k} K → K ⊗_{k} K, that maps x ⊗ y into y ⊗ x. Obviously, i(ϕ^{−1}(f)) ∈ δ^{−1}𝔒_{K} ⊗_{o} 𝔒_{K}. Besides, we have ϕ(x ⊗ y) = ∑_{σ∈G} xσ(y), ϕ(y ⊗ x) = ∑_{σ∈G} yσ(x) = ∑_{σ∈G} σ(xσ^{−1}(y). Thus we have

∑ c_{σ}σ^{−1} = ϕ(i(ϕ^{−1}(f))) ∈ 𝔄_{K/k}(𝔒_{K}).

Thus it is sufficient to prove that for any f ∈ 𝔄_{K/k}(𝔒_{K}) we have

Δ(f) ∈ 𝔄_{K/k}(𝔒_{K}) ⊗ 𝔄_{K/k}(𝔒_{K}) ,

, where Δ(∑ c_{σ}σ) = ∑ c_{σ}σ ⊗ σ. Theorem 2.4.1 implies immediately that it is sufficient to check this assertion for f = ξ^{l}, l ≥ 0, the power is taken with respect to multiplication ∗^{Δ}. We consider the polynomial

J(X) = ∏_{t∈T \\{0\}} \frac{t − X}{t} .

We have J(0) = 1, J(t) = 0 for t ∈ T \ {0}. The standard formula for the valuation of the different (cf. [Se]) and the last assertion of Lemma 3.1.2 imply that J(X) ∈ δ^{−1}𝔬[X]. The fact that 𝔄_{K/k}(𝔒_{K}) is an 𝔬-algebra with a unit with respect to the multi-

plication ∗^{Δ} implies that for f_{1}, f_{2} ∈ 𝔄_{K/k}(𝔒_{K})⊗𝔄_{K/k}(𝔒_{K}) their product f_{1}∗f_{2} which is defined coefficient-wise lies in δ^{−2}𝔄_{K/k}(𝔒_{K}) ⊗ 𝔄_{K/k}(𝔒_{K}). Thus for the element I = ∑_{τ,σ∈G} t_{στ^{−1}}στ belongs to δ^{2}𝔄_{K/k}(𝔒_{K}) ⊗ 𝔄_{K/k}(𝔒_{K}) since

I = ∑_{τ,σ∈G}(t_{σ} −_{F} t_{τ}) = δ^{2} ∑_{i,j≥0} b_{ij}ξ^{i} ⊗ ξ^{j},

where the powers are taken with respect to ∗^{Δ} and b_{ij} are the coefficients in the expansion of the formal difference X −_{F} Y into the powers of X and Y.

We also obtain that the element

L = Δ(tr) = ∑_{σ∈G} σ ⊗ σ = J(I)

belongs to δ𝔄_{K/k}(𝔒_{K}) ⊗ 𝔄_{K/k}(𝔒_{K}). Then we have the equality

Δ(ξ^{l}) = (ξ^{l} ⊗ tr) ∗ L ∈ 𝔄_{K/k}(𝔒_{K}) ⊗ 𝔄_{K/k}(𝔒_{K})

since tr ∈ δ𝔄_{K/k}(𝔒_{K}). □

In a future paper we will prove a similar statement for an extension that fulfills the second condition of Theorem A not supposing K/k to be Abelian.

§4 Construction of a formal group

4.1. We suppose that in the associated order 𝔄_{K/k}(𝔒_{K}) there exists an element ξ such that it maps an element a ∈ 𝔒_{K}, v_{K}(a) = n − 1 into a prime element π of the field K, i.e.,

ξ(a) = π  (24)

(cf. Lemma 2.2.1 and Theorem 2.4.2). We choose an element a such that tr a = δ (cf. Lemma 2.2.1). Let

\begin{matrix}ψ_{1} : K ⊗_{k} K → K \\ x ⊗ y → xy \end{matrix}  (25)

be the map from 1.1, and let ϕ be the bijection between K ⊗_{k} K and K[G], that was defined in subsection 1.1.

Lemma 4.1.1. We can choose an element ξ in 𝔄_{K/k}(𝔒_{K}) fulfilling (4) so that 1. ϕ^{−1}(δξ) = π ⊗ 1 − 1 ⊗ π + z, where z belongs to the maximal ideal 𝔐 of the ring 𝔒_{K}.

2. In the expansion ξ = ∑_{σ∈G} a_{σ}σ, we have a_{1} = 0 and δa_{σ} ∈ 𝔐_{𝔬} for σ ≠ 1, where 𝔐_{𝔬} is the maximal ideal in 𝔬.

Proof. Theorem 1.3.1 implies that the preimage ϕ^{−1}(ξ) belongs to 𝔒_{K} ⊗ 𝔇^{−1}. From our assumptions we also have 𝔇 = (δ), δ ∈ k (cf. (*)), so ϕ^{−1}(δξ) ∈ 𝔒_{K} ⊗ 𝔒_{K}. As in 2.3, we can prove that

ϕ^{−1}(δξ) = x ⊗ 1 + y,

where x = ξ(a)δ tr(a)^{−1} = π, y ∈ 𝔒_{K} ⊗ 𝔐. It follows that

ϕ^{−1}(δψ) = π ⊗ 1 + y, y ∈ 𝔒_{K} ⊗ 𝔐.

We compute the coefficient at 1 = id_{K} of the element ξ = ϕ(δ^{−1}(π ⊗ 1 + y)). From the definition of ϕ it follows that it is equal to ψ_{1}(δ^{−1}α). We have

ψ_{1}(δ^{−1}(π ⊗ 1 + y)) = δ^{−1}(π + ψ_{1}(y)).  (26)

We decompose the element y in the base of 𝔒_{K} ⊗ 𝔐:

y = ∑_{i≥0, j≥1} a_{ij}(π^{i} ⊗ π^{j}), a_{ij} ∈ 𝔬.

Then

ψ_{1}(∑ a_{ij}π^{i} ⊗ π^{j}) = ∑_{ij} a_{ij}π^{i+j}  (27)

If ξ = ∑ a_{σ}σ, a_{σ} ∈ k, then ψ_{1}(δ^{−1}α) = a_{1} ∈ k.

This implies that in order for the element (26) to lie in 𝔄_{K/k}(𝔒_{K}) it is necessary that the coefficient a_{01} in (27) is congruent with −1 mod π_{0}, where π_{0} is the prime element of k. It follows that y = −1 ⊗ π + z where z ∈ 𝔐 ⊗ 𝔐. The first claim of the lemma is proved. Now we consider

ξ′ = ξ − ψ_{1}(ξ) tr = ∑_{σ∈G}(a_{σ} − a_{1})σ = ∑_{σ∈G, σ≠1}(a_{σ} − a_{1})σ.

We have ψ_{1}(ξ′) = 0. Besides that, δa_{1} = π + ∑ a_{ij}π^{i+j}, a_{1} ∈ k, and it follows that δa_{1} ∈ 𝔐_{𝔬}. Thus a_{1} tr ∈ 𝔐_{0}𝔄_{K/k}(𝔒_{K}), and it follows that ξ′ ∈ 𝔄_{K/k}(𝔒_{K}) and ξ′(a) ≡ ξ(a) mod 𝔐^{n}. So ξ may be replaced by ξ′. It remains to prove that in the expansion ξ = ∑ a_{σ}σ all δa_{σ} lie in 𝔐. From the definition of ψ_{σ} we have

δa_{σ} = ψ_{σ}(δξ) = ψ_{σ}(π ⊗ 1) = ψ_{σ}(1 ⊗ π) + ψ_{σ}(z) = π − σ(π) + ψ_{σ}(z) ∈ 𝔐 (28)

since z ∈ 𝔐 ⊗ 𝔐. Yet a_{σ} ∈ k and it follows that a_{σ} ∈ 𝔐 ∩ k = 𝔐_{𝔬}. □

Corollary 4.1.2. We introduce the following notation: δξ = ∑_{σ} b_{σ}σ. If ϕ^{−1}(δξ) = π ⊗ 1 − 1 ⊗ π + ∑_{1≤i,j≤n} a_{ij}π^{i} ⊗ π^{j}, a_{ij} ∈ 𝔬, then we have

b_{σ} = π − σ(π) + ∑ a_{ij}π^{i}σπ^{j}  (29)

and

b_{σ^{−1}τ} = σπ − τπ + ∑ a_{ij}σπ^{i}τπ^{j}.  (30)

(the last statement follows from (29) and an obvious equality σb_{σ^{−1}τ} = b_{σ^{−1}τ}).

4.2 A preliminary group law. We consider the expansion

ϕ^{−1}(δξ) = π ⊗ 1 − 1 ⊗ π + ∑_{1≤i,j<n} a_{ij}π^{i} ⊗ π^{j}, a_{ij} ∈ 𝔬.  (31)

We replace π ⊗ 1 in this decomposition by X, and 1 ⊗ π by Y and decompose π^{i} ⊗ π^{j} into the product (π^{i} ⊗ 1)(1 ⊗ π^{j}). Then from the expansion (31) we obtain a polynomial in two variables of degree not greater than n; we denote it by

R(X, Y) = X − Y + ∑_{1≤i,j≤n} a_{ij}X^{i}Y^{j}, a_{ij} ∈ 𝔬.  (32)

Now we identify π ⊗ 1 with π, and 1 ⊗ π with Z and obtain from (31) a polynomial in Z of degree not greater than n. We denote it by s(Z):

s(Z) = π − Z + ∑_{1≤j≤n}( ∑_{1≤i≤n} a_{ij}π^{i})Z^{j}.  (33)

All the coefficients of this polynomial except the coefficient at Z belong to 𝔐, and the coefficient at Z is invertible, so the polynomial s(Z) is invertible in 𝔒_{K}[[Z]] with respect to composition. We denote the inverse to s by s^{−1}(Z). It follows from (28) and (31) that

b_{σ} = δa_{σ} = (π − σ(π)) + ∑_{1≤i,j≤n} a_{ij}π^{i}σπ^{j}.

So, keeping in mind (33), we obtain

s(σπ) = b_{σ}, s^{−1}(b_{σ}) = σπ.

Besides that, (32) and (30) imply:

G(b_{σ}, b_{τ}) = R(s^{−1}(b_{σ}), s^{−1}(b_{τ})) = b_{σ^{−1}τ} ,

we also know that the series G(X, Y) = R(s^{−1}(X), s^{−1}(Y)) ∈ 𝔒_{K}[[X, Y]] and

G(b_{σ}, b_{τ}) ≡ (b_{σ} − b_{τ}) mod 𝔐𝔒_{K}[[X, Y]].

So the series G(b_{σ}, b_{τ}) is invertible with respect to composition as a series in b_{τ}. We denote this inverse series by H(b_{σ}, b_{τ}) ∈ 𝔒_{K}[[b_{σ}, b_{τ}]]. We also note that H(b_{σ}, b_{τ}) = b_{στ}. We introduce the following notation:

M(X) = ∏_{σ∈G}(X − b_{σ}).  (34)

Consider the reduction of the series H(X, Y) modulo the ideal (M(X), M(Y)). We obtain a polynomial J(X, Y), whose degree in each variable is less than n. Since M(b_{σ}) = 0,

J(b_{σ}, b_{τ}) = b_{στ} .  (35)

Proposition 4.2. J(X, Y) ∈ 𝔬[[X, Y]].

Proof. We denote by f(X, Y) the interpolation polynomial whose degree in each variable is less than n and that fulfills the system of relations

f(b_{σ}, b_{τ}) = b_{στ} , σ, τ ∈ G.  (36)

The polynomial J also fulfills the system of equalities (36) and so

f(X, Y) = J(X, Y).

Since f ∈ k[X, Y], J ∈ 𝔒_{K}[X, Y], the coefficients of J belong to k ∩ 𝔒_{K} = 𝔬. □

4.3 The main statement concerning finite torsion submodules of formal modules over formal groups. Let G be an Abelian group and suppose that for any σ ∈ G there is an element b_{σ} ∈ 𝔐_{𝔬} chosen, and suppose that b_{1} = 0.

Theorem 4.3.1. The following conditions are equivalent 1 There exists a formal group F (X, Y) over the ring of integers 𝔬_{E} of some extension E of the field k such that b_{σ} +_{F} b_{τ} = b_{στ} for all σ, τ ∈ G.

2. There exists a formal group F (X, Y) over the ring 𝔬, fulfilling the same conditions: b_{σ} +_{F} b_{τ} = b_{στ}.

3. The coefficients of the interpolation polynomial f(X, Y), of degree less than n in X and Y and such that for σ, τ ∈ G

f(b_{σ}, b_{τ}) = b_{στ}

belong to 𝔬.

Remark 4.3.2. This Theorem gives us a schematic description of finite subsets of 𝔐_{𝔬} that are finite groups with respect to an addition defined by some formal group with integral coefficients. Additive Galois modules are not mentioned in the stating of this theorem and so it can be used without them.

Proof of Theorem 4.3.1: 2 ⟹ 1 ⟹ 3. The condition 2 obviously implies 1. Now we prove that 1 implies 3. We consider the reduction of F (X, Y) modulo the ideal (M(X), M(Y)) (cf. (34)) and denote it by F_{red}(X, Y). It follows from the condition 1 that F_{red}(b_{σ}, b_{τ}) = b_{στ}, since M(b_{σ}) = 0. So F_{red}(X, Y) coincides with the interpolation polynomial f(X, Y), whose coefficients belong to k. Yet F_{red}[X, Y] ∈ 𝔬_{E}[X, Y], so the coefficients of f(X, Y) lie in 𝔬 = k ∩𝔬_{E}. It remains to prove that 3 implies 2.

4.4 Some universal formal group laws. We construct a formal group law in the same way as Hazewinkel (cf. [Ha], Ch. I, §3, subsection 3.1). Consider the ring of polynomials ℤ_{p}[S_{2}, S_{3}, …, S_{n}] = ℤ_{p}[S] ⊂ ℚ_{p}[S]. We introduce the following notation:

σ : ℚ_{p}[S] → ℚ_{p}[S], S_{i} → S_{i}^{p}.

We consider the series f_{S}(X), whose coefficients are found from the equation

f_{S}(X) = g(X) + ∑_{i≥1} \frac{S_{p^{i}}}{p} σ_{∗}^{i}f_{S}(X^{p^{i}}), g(X) = X + ∑_{i≥2} S_{i}X^{i} − ∑_{i≥1} S_{p^{i}}X^{p^{i}},

where σ_{∗}^{i}f_{S} is the series obtained from f_{S} by applying the homomorphism σ^{i} to the coefficients, for i > n we reckon S_{i} being equal to 0. It is easily seen that f_{S} = X + ∑_{i>1} a_{i}X^{i}, a_{i} ∈ Q[S]. Then the Hazewinkel’s Functional Equation Lemma implies that

F_{S}(X, Y) = f_{S}^{−1}(f_{S}(X) + f_{S}(Y))

is a formal group law over ℤ_{p}[S]. Besides that

\begin{matrix}F_{S}[X, Y] ≡ & X + Y + S_{m}v_{p}(m)^{−1}B_{m}(X, Y) \\ mod (S_{2}, …, S_{m−1}, deg(m + 1)), 2 ≤ m ≤ n, \end{matrix}  (37)

where v_{p}(m) = p if m = p^{r}, r ∈ ℤ, r > 0, else v_{p}(m) = 1, and

B_{m}(X, Y) = X^{m} + Y^{m} − (X + Y)^{m}.

We note that the series F_{S} is a formal group law also in the case char k = p (in that case we should compute the coefficients of v_{p}(m)^{−1}B_{m}(X, Y) ’formally’ in ℤ). This formal group differs from Hazewinkel’s only because S_{i} = 0 for i > n. Now we modify the formal group law we have obtained. To be more precise, we make the following change of variables. Now we define some values r_{m} in the following way. Let r^{m} be equal to s if m = p^{s}m_{0} and (m_{0}, p) = 1, m_{0} > 1 or m_{0} = p. Then

F_{S}(X, Y) = X + Y + ∑_{2≤m≤n} d_{m}X^{p^{rm}} Y^{m−p^{rm}} + summands of other degrees.

We consider the ring

ℤ_{p}[V] = ℤ_{p}[V_{2}, …, V_{n}], V_{i} = d_{i}.  (38)

Lemma 4.4. We can express the variables S_{i} as polynomials in V_{i} with integral coefficients and so obtain a new formal group law over ℤ_{p}[V].

Proof. We have the equality V_{2} = S_{2} Besides that, for i > 2 the equality V_{i} = S_{i} + f_{i}(S_{2}, …, S_{i−1}), f_{i} ∈ ℤ_{p}[S_{2}, …, S_{i−1}] is fulfilled (cf. (37)). It follows that we can express S_{i} as a polynomial in d_{i}, S_{2}, …, S_{i−1} with integral coefficients. Making all such changes we obtain an expression of S_{i} as a polynomial in V_{2}, …, V_{i−1}. Lemma is proved. □

We note that the formal group law we constructed has the form

F_{V} (X, Y) = X + Y + ∑_{2≤m≤n} V_{m}X^{p^{rm}} Y^{m−p^{rm}}
+ summands of other degrees.

(39)

4.5. In the Abelian group G we choose a family of subgroups 1 = G_{0} ⊂ G_{1} ⊂ G_{2} ⊂ ⋯ ⊂ G_{l} = G, where n = p^{l} and the cardinality of G_{i} is equal to p^{i}. In each subgroup G_{i} we choose an element σ_{p^{i−1}} such that the coset σ_{p^{i−1}} mod G_{i−1} generates the cyclic group G_{i}/G_{i−1} of cardinality p. We obtain a set of generators σ_{1}, σ_{p}, …, σ_{p^{l−1}} for the group G. By induction on the cardinality of G it can be easily proved that any element σ of G can be expressed uniquely in the form

σ = ∏_{0≤i≤l−1} σ_{p^{i}}^{c_{i}}, where 0 ≤ c_{i} ≤ p − 1.  (40)

We introduce the following notation:

c_{σ} = c_{0} + pc_{1} + ⋯ + p^{l−1}c_{l−1} ∈ ℤ, 0 ≤ c_{σ} ≤ n − 1.

We obtain a one-to-one correspondence σ → c_{σ}. We also use the inverse notation: σ_{c} = σ ⟺ c = c_{σ}. We will construct the desired formal group by induction On the m-th step we ’get rid’ of the variable V_{m} and adjoin the m-th relation. First we prove a simple lemma about relations in an arbitrary Abelian group.

Lemma 4.5. Let H be an Abelian group, f be a map from G into H, f(1) = 1_{H}. Then the following statements are fulfilled

1. If the relation

f(σ_{i})f(σ_{j}) = f(σ_{i}σ_{j})  (41)

is fulfilled for all i = p^{r_{s}}, j = s − p^{r_{s}}, 2 ≤ s ≤ m, then it is also fulfilled for 0 ≤ i ≤ p^{r_{m}} − 1, 0 ≤ j ≤ m − 1 and for i = p^{r_{m}}, 0 ≤ j ≤ m − p^{r_{m}} − 1.

2. If (41) is fulfilled for i = p^{r_{s}}, j = s − p^{r_{s}}, 2 ≤ s ≤ n, then it is also fulfilled for all 0 ≤ i < n, 0 ≤ j < n.

Proof. 1. First we prove by induction that (41) is fulfilled for 0 ≤ i ≤ p^{t}, 0 ≤ j ≤ m − 1, for all 0 ≤ t ≤ r_{m}, i.e the restriction f_{t} of f onto G_{t} is a group homomorphism and F_{tx} = \{f_{i} = f(σ_{i}), i = p^{t}x+u, 0 ≤ u ≤ p^{t}−1\}, 0 < x < \frac{m}{p^{t}} are cosets modulo the subgroup F_{t} = f(G_{t}). This is fulfilled for t = 0 since f(1_{G}) = 1_{H}. We suppose that such statement is fulfilled for t = w. Now we prove it for t = w + 1. Since F_{wx} are cosets modulo the subgroup F_{w}, it is sufficient to check that (41) is fulfilled for i = p^{w}a, j = p^{w}b, 0 < a ≤ p − 1, 0 < b < \frac{m}{p^{w}}. If 0 < b ≤ p, then r_{bp^{w}} = w, and so for i = (b − 1)p^{w}, j = p^{w} the relation (41) is fulfilled. Thus we obtain

f_{bp^{w}} = f_{p^{w}}^{b}, b < p, f_{(p−1)p^{w}}f_{p^{w}} = f_{p^{w}} = f(σ_{p^{w}}^{p})

and f_{w+1} is a group homomorphism. Besides that, for b > p, (b, p) = 1 we also have r_{bp^{w}} = w, and so f_{(b−1)p^{w}}f_{p^{w}} = f_{bp^{w}}. For b = pc we have:

f_{(pc−1)p^{w}}f_{p^{w}} = f_{(pc−p)p^{w}}f_{p^{w}}^{p} = f(σ_{(pc−1)p^{w}}σ_{p^{w}}), since σ_{p^{w}}^{p} ∈ G_{w}. It follows that (41) is fulfilled indeed for all i = p^{w}a, j = p^{w}b, 0 < a ≤ p − 1, 0 < b < \frac{m}{p^{w}}, and so F_{w+1x}, 0 < x < \frac{m}{p^{w+1}} indeed form cosets modulo F_{w+1}. The relation (41) for i = p^{w}, w = r_{m} 0 ≤ j ≤ m − p^{w} − 1 is also sufficient to prove for j = p^{w}b, since F_{wx} are cosets modulo F_{w}. We argue in the same way as in the previous reasoning. We have again the equality r_{bp^{w}} = w for 0 < b ≤ p. It follows that if \frac{m}{p^{w}} ≤ p, then we obtain the desired assumption. If \frac{m}{p^{w}} > p, then for \frac{m}{p^{w}} ≥ b > p, (b, p) = 1 also r_{bp^{w}} = w, and so f_{(b−1)p^{w}}f_{p^{w}} = f(σ_{(b_{1})p^{w}}σ_{p^{w}}). For p | b this is also fulfilled as f_{(p−1)p^{w}} = f_{p^{w}}^{p−1}. It follows that for pc + p − 1 < \frac{m}{p^{w}} we have

f_{(pc+p−1)p^{w}}f_{p^{w}} = f_{pc}f_{p^{w}}^{p} = f(σ_{(pc+p−1)p^{w}}σ_{p^{w}}),

since σ_{p^{w}}^{p} ∈ G_{w}.

2. In the proof of the first part from the relation (41) for i = p^{r_{s}}, j = s−p^{r_{s}}, 2 ≤ s ≤ m we deduced that F_{t}, p^{t} | m is a group homomorphism, the same proof for m = n gives us the desired statement. □

4.6 The proof of 3 ⟹ 2 in Theorem 4.3.1. In the beginning we are in the following situation. We have a formal group F_{V} (X, Y), for which the following relations are fulfilled: F (b_{1}, b_{σ}) = b_{σ} for all σ ∈ G, since b_{1} = 0 from our assumptions. We describe the second step. We take the generator σ_{1} of the group G_{1} and try to fit the relation

F_{V} (b_{σ_{1}}, b_{σ_{1}}) = b_{σ_{1}^{2}}.  (42)

The interpolation polynomial f(X, Y) from our condition 3 can be written in the form f(X, Y) = X +Y +XY ψ(X, Y), where ψ(X, Y) ∈ 𝔬[X, Y]. We denote the ring of all series in 𝔬[[V_{i}]], that converge at all (integral) values of V_{i}, by 𝔬̃[[V_{i}]], and its ideal, consisting of series with coefficients in 𝔐_{𝔬}, by 𝔐̃_{𝔬}[[V_{i}]]. We have b_{σ_{1}^{2}} = b_{σ_{1}} + b_{σ_{1}} + b_{σ_{1}}^{2}c, c ∈ 𝔬. On the other hand, if the relation (42) is fulfilled, then

b_{σ_{1}^{2}} = b_{σ_{1}} +_{F} b_{σ_{1}} = b_{σ_{1}} + b_{σ_{1}} + b_{σ_{1}}^{2}V_{2} + b_{σ_{1}}^{3}(…), (…) ∈ 𝔬̃[[V_{2}, V_{3}, …, V_{n}]].

We obtain

V_{2} + b_{σ_{1}}(…) = const ∈ 𝔬

where (…) ∈ 𝔬̃[[V_{2}, V_{3}, …, V_{n}]]. Using the last relation we express V_{2} in V_{3}, …, V_{n}, i.e., V_{2} = c + g(V_{3}, V_{4}, …, V_{n}), g ∈ 𝔐̃_{𝔬}[[V_{3}, ˙,V_{n}]]. This is possible, since in this relation the coefficient at the first power of V_{2} is invertible, and coefficients at all other powers contain b_{σ_{1}}. We denote the formal group we obtained in this way by F_{V}^{2} (X, Y). It depends on variables V_{3}, …, V_{n} and fits the relations:

\begin{matrix}F_{V}^{2} (b_{1}, b_{σ}) = b_{σ}, σ ∈ G \\ F_{V}^{2} (b_{σ_{1}}, b_{σ_{1}}) = b_{σ_{1}^{2}}. \end{matrix}

The second step is ended. Now we describe the m-th step. We suppose that at the m−1-step we obtained a formal group F_{V}^{m−1}(X, Y) that depends on the variables V_{m}, V_{m+1}, …, V_{n} and fits the relations:

b_{σ_{i}} +_{F} b_{σ_{j}} = b_{σ_{i}σ_{j}}  (43)

for 0 ≤ i ≤ p^{r_{m}} − 1, 0 ≤ j ≤ m − 1 and i = p^{r_{m}}, 0 ≤ j ≤ m − p^{r_{m}} − 1. Thus we have (m − 1)(p^{r_{m}} + 1) + 1 relations. We denote by A_{m−1} the set of points (b_{σ_{i}}, b_{σ_{j}}) in (43). We need some lemma about interpolation. We introduce the following notation:

χ_{m−1} = F_{V}^{m−1}(X, Y) − f(X, Y),  (44)

where f(X, Y) is an interpolation polynomial from the formula (36). The relations for f(X, Y) imply that χ_{m−1}(b_{σ_{i}}, b_{σ_{j}}) = 0 for the indices (i, j) mentioned in (43).

Lemma 4.6.1. Let R be an integral domain, b_{σ_{i}} be a set of elements in R, I be an ideal IN R[X, Y], consisting of all polynomials that take the value zero at all pairs (b_{σ_{i}}, b_{σ_{j}}) in A_{m−1}. Then I = (M, N, L), where M = ∏_{0≤i≤p^{rm}} (X − b_{i}), N = ∏_{0≤j≤m−1}(Y − b_{σ_{j}}), L = ∏_{0≤i≤p^{rm} −1}(X − b_{i}) ∏_{0≤j≤m−p^{rm} −1}(Y − b_{σ_{j}}).

Proof. Let ψ(X, Y) be an arbitrary polynomial in the ideal I. It is easily seen that it can be reduced uniquely modulo the ideal (M, N, L) to a polynomial ψ_{red}, since the higher coefficients of L, M, N are equal to 1. By its reduction we mean a polynomial that is congruent to it modulo (M, N, L) and has a non-zero coefficient at X^{i}Y^{j} only if (b_{σ_{i+1}}, b_{σ_{j+1}}) ∈ A_{m−1}. Now we have to prove that a polynomial ψ_{red} that belongs to I and contains non-zero coefficients only at powers mentioned above has to be equal to 0. It is sufficient to prove this statement for polynomials over the field R_{0} that is the fraction field of R since it does not depend on the ring. We note that if we take for all g(X, Y) ∈ R_{0}(X, Y) their values G(b_{σ_{i}}, b_{σ_{j}}), (b_{σ_{i}}, b_{σ_{j}}) in A_{m−1}, then we obtain a vector subspace of values in R_{0}^{(p^{rm} +1)(m−1)+1}. It is easily seen that any set of values corresponds to some polynomial in R_{0}(X, Y). A similar statement is obvious for polynomials in one variable and is easily carried on by induction to the case of any number of variables. It follows that the dimension of the space of values is equal to (p^{r_{m}} +1)(m−1)+1. The map R_{0}[X, Y] → C, where C is the (p^{r_{m}} + 1)(m − 1) + 1-dimensional space of values of polynomials in R_{0}[X, Y] in points of A_{m−1}, factorizes through the space I′ of reduction polynomials in R_{0}[X, Y]. So it follows from the equality of the dimensions that Ker(I′ → C) = {0} and thus ψ_{red} = 0. Lemma is proved. □

It is easily seen that we may also use this lemma for the ring of power series. Indeed, we can reduce series modulo ∏_{i<n}(X − b_{i}) ∏_{i<n}(Y − b_{i}) and obtain polynomials.

Now we make the m-th step of formal group construction. According to Lemma, the series from (44) can be represented in a form

χ_{m−1}(X, Y) = fM + gN + hL,  (45)

where f, g, h ∈ R[[X, Y]], and R = 𝔬̃[[V_{m}, …, V_{n}]]. We consider the following relation:

χ_{m−1}(b_{p^{rm}} , b_{m−p^{rm}}) = 0.

It follows from the definition of M and N that fM(b_{p^{rm}} , b_{m−p^{rm}}) = gN(b_{p^{rm}} , b_{m−p^{rm}}) = 0. We need h to be equal to 0. We consider h(X, Y) in the point (b_{p^{rm}} , b_{m−p^{rm}}). It is clear that h(b_{p^{rm}} , b_{m−p^{rm}}) ∈ R[b_{p^{rm}} , b_{m−p^{rm}}] ⊂ R. Besides that

h(b_{p^{rm}} , b_{m−p^{rm}}) ≡ absolute term of (X, Y) mod 𝔐̃_{𝔬}[[V_{m}, …, V_{n}]].

Since in M, N, L only higher coefficients are invertible, in the relation (45) the coefficient at X^{p^{rm}} Y^{m−p^{rm}} of the series χ_{m−1}(X, Y) is equal to

the absolute term of h(X, Y) + an element of 𝔐̃_{𝔬}[[V_{m}, …, V_{n}]].

On the other hand it is equal to V_{m} + const, const ∈ 𝔬, as χ_{m−1}(X, Y) = F_{V} (X, Y) − f(X, Y). Thus we obtain

V_{m} + c = h(b_{p^{rm}} , b_{m−p^{rm}}) + d,

where c ∈ 𝔬, d ∈ 𝔐̃_{𝔬}[[V_{m}, …, V_{n}]]. We chose V_{m} so that h(b_{p^{rm}} , b_{m−p^{rm}}) = 0 (in the formal group we replace V_{m} by a series in V_{m+1}, …, V_{n}). So we made the m-th step. Since all the formal group laws F_{V}^{m} are commutative and associative, we can apply to the group H = F_{V}^{m}(𝔐̃_{𝔬})[[V_{m+1}, …, V_{n}]], f : σ → b_{σ} the first part of Lemma 4.5 and so indeed after the m − 1-th step the group F_{V}^{m−1} fits the relations (43) that are necessary for the m-th step. From the second part of Lemma 4.5 it follows that after the last (n-th) step all the relations for b_{σ_{i}}, b_{σ_{j}} are fulfilled and all the variables V_{i} are got rid of and thus we obtained the desired formal group law F = F_{V}^{n} in 𝔬[[X, Y]].

Remark 4.6.2. The proof of Theorem 4.3.1 also implies that the formal group law F that fits the relations b_{σ} +_{F} b_{τ} = b_{στ} , σ, τ ∈ G and for which all the

S_{i}, i > n in expression F (X, Y) through F_{S}(X, Y), is unique. Besides that when we started instead of fixing S_{i} = 0, i > n we could demand S_{i}, i > n to be equal to arbitrary convergent series in S_{2}, …, S_{n} with integral coefficients and obtain a similar statement.

§5 The proof of necessity in Theorem A

Now we show that our extension is indeed Kummer for the formal group we constructed. We have a formal group F (X, Y) such that

F (b_{σ}, b_{τ}) = b_{στ} .  (46)

We introduce again the addition with the means of the formal group F on the maximal ideal 𝔐_{⊗} = 𝔒_{K} ⊗ 𝔐 + 𝔐 ⊗ 𝔒_{K} of the tensor product 𝔒_{K} ⊗ 𝔒_{K}. We denote the formal module obtained by F (𝔐_{⊗}). Now let ξ = ∑ a_{σ}σ act as in (4). Let α be equal to δϕ^{−1}(ξ). Our aim in the remaining subsections is the proof of the following statement.

Proposition 5.1. There are elements x and y in 𝔐 such that

α = x ⊗ 1 +_{F} 1 ⊗ y.

Here we show that the statement of Proposition implies theorem A. Suppose that 5.1 is fulfilled. We can express y as 1 ⊗ z, z ∈ 𝔐 and so α = x ⊗ 1 +_{F} 1 ⊗ z. Using the formula for ψ_{1} and the fact that b_{1} = 0 we obtain that

0 = b_{1} = ψ_{1}(α) = x +_{F} z. Thus z is equal to [−1]_{F} (x) ([−1]_{F} (x) is the inverse

to x in F (𝔐)) and it follows that α = x ⊗ 1 −_{F} 1 ⊗ x. Similarly

b_{σ} = ψ_{σ}(α) = ψ_{σ}(x ⊗ 1) −_{F} ψ_{σ}(1 ⊗ x) = x −_{F} σ(x).

We obtain that the conjugates of x in the extension K/k are exactly the elements of the form x +_{F} b_{σ}. Then ∏(x +_{F} b_{σ}) ∈ 𝔬 and v_{k}(∏(x +_{F} b_{σ})) = 1, thus the

extension K/k is Kummer for the formal group F and is generated by a root of the equation P (X) = w while v_{k}(w) = 1. Theorem A of the introduction is proved.

5.2. Now we start proving 5.1. Let n be equal to p^{l} = [K : k]. Consider a tower of intermediate extensions:

k = k_{0} ⊂ k_{1} ⊂ k_{2} ⊂ ⋯ ⊂ k_{l} = K, [k_{i} : k_{i−1}] = p.

We take representatives τ_{i} for the generators of the Galois groups G_{i} = Gal(k_{i}/k_{i−1}). For all τ ∈ G we prove the following equality:

τ(α) = α −_{F} b_{τ}  (47)

(we assume here that the group G acts only on the first component of the tensor product).

Indeed, for any z ∈ K we have

ϕ(τ(x) ⊗ y)(z) = τ(x) tr yz = τ(x tr yz) = τ(ϕ(x ⊗ y)(z)),

and so for each β ∈ K ⊗_{k} K the equality

ϕ((τ ⊗ 1)β(z)) = τ(ϕ(β)(z))  (48)

is also fulfilled. Thus we obtain

τ(δξ) = ∑_{σ} b_{σ}τσ = ∑_{σ}(b_{τσ} −_{F} b_{τ})σ_{τ} = σξ −_{F} (b_{τ} tr).

The formula (48) now implies (47).

5.3. Let 𝔬_{i} be the ring of integers of the field k_{i}. For each i : 0 ≤ i ≤ l we prove by induction the following lemma.

Lemma 5.3. There is an equality

α = x ⊗ 1 +_{F} y, y ∈ 𝔬_{i} ⊗ 𝔐  (49)

Proof. For i = l the claim is obvious. Let the claim be fulfilled for i = s + 1. We can expand the element y in the base of the formal module F (𝔬_{s+1} ⊗ 𝔐):

y = ∑_{0≤i 0≤j<n}_{(F)}a_{ij}π_{s+1}^{i} ⊗ π^{j},  (50)

where π_{s+1} is a prime element of the field k_{s+1}, the coefficients a_{ij} are either equal to 0, or a_{ij} ∈ 𝔬^{∗} (for j ≥ n we extract the prime element π_{0} of the field k from π^{n} and convert it into the first component). For the automorphism τ_{s+1} we have the equality (47):

τ_{s+1}(α) = α −_{F} b_{τ_{s+1}}.  (51)

Now we express α in the form

x ⊗ 1 +_{F} ∑_{0≤i 0≤j<n}_{(F)}b_{ij}π_{s}^{i} ⊗ π^{j} +_{F} ∑_{0≤i 0≤j<n}_{(F)}a_{ij}π_{s+1}^{i} ⊗ π^{j}.  (52)

We consequently convert the third summand in (52) into the first and the second, increasing the minimum of ni + j for i, j such that a_{ij} ≠ 0. Out of all pairs (i, j) for which a_{ij} ≠ 0 we chose a pair with the least i, which we denote by i_{0}, further out of all pairs (i_{0}, j) such that a_{i_{0}j} ≠ 0 we chose the pair with the least j and denote it by j_{0}. We call the corresponding ordering of (i, j) by order.

We have three cases. Case I. If j_{0} = 0, then

a_{i_{0}j_{0}}(π_{s+1}^{i_{0}} ⊗ π^{j_{0}}) = a_{i_{0}0}(π_{s+1}^{i_{0}} ⊗ 1)  (53)

and we import this term into the first summand in (52). It is clear that subtracting with respect to F the term a_{i_{0}0}(π_{s+1}^{i_{0}} ⊗ 1) gives only the terms of greater order (since X −_{F}Y = (X −Y)(1+r(X, Y)), r(X, Y) ∈ (X, Y)𝔬[[X, Y]])

and so we increase the value of ni_{0} + j_{0}.

Case II. Now let j_{0} be not equal to 0. If p | i_{0} then π_{s+1}^{i_{0}} = πs^{\frac{i_{0}}{p}} + r, where v_{s+1}(r) > i_{0} and \frac{i_{0}}{p} ∈ ℤ and thus

a_{i_{0}j_{0}}(π_{s+1}^{i_{0}} ⊗ π^{j_{0}}) = a_{i_{0}j_{0}}πs^{\frac{i_{0}}{p}} ⊗ π^{j_{0}} +_{F} terms of greater order.  (54)

It follows that we can import the term a_{i_{0}j_{0}}πs^{\frac{i_{0}}{p}} ⊗π^{j_{0}} into the second summand of the formula (52) and increase the minimum of ni + j again. Case III. It remains to consider the case (i_{0}, p) = 1, j_{0} ≠ 0, We consider

τ_{s+1}(α) −_{F} α = τ_{s+1}(x ⊗ 1) −_{F} (x ⊗ 1) +_{F} (τ_{s+1} −_{F} 1)a_{i_{0}j_{0}}(π_{s+1}^{i_{0}} ⊗ π^{j_{0}})

+_{F}(τ_{s+1} −_{F} 1)(terms of greater order).

We have:

\begin{matrix}τ_{s+1}(x ⊗ 1) −_{F} (x ⊗ 1) = (τ_{s+1}(x) −_{F} x) ⊗ 1 ∈ 𝔐 ⊗ 𝔬 \\ a_{τ_{s+1}} = a_{τ_{s+1}} ⊗ 1 ∈ 𝔐 ⊗ 𝔬. \end{matrix}  (55)

So the element (τ_{s+1} −_{F} 1)a_{i_{0}j_{0}}(π_{s+1}^{i_{0}} ⊗ π^{j_{0}}) +_{F} (τ_{s+1} −_{F} 1) (terms of greater order)

also belongs to 𝔐 ⊗ 𝔬. Yet that is impossible since

(τ_{s+1} −_{F} 1)a_{i_{0}j_{0}}(π_{s+1}^{i_{0}} ⊗ π^{j_{0}}) ≡ a_{i_{0}j_{0}}(τ_{s+1}π_{s+1}^{i_{0}} − π_{s+1}^{i_{0}}) ⊗ π^{j_{0}}

mod (terms of greater order).

The remaining terms indeed have greater ni + j since (i_{0}, p) = 1 and so the valuation of τ_{s+1}π_{s+1}^{i_{0}} − π_{s+1}^{i_{0}} in k_{s+1} is equal to i_{0} + h_{s+1,s+1}, where h_{s+1,s+1} is the ramification jump of τ_{s+1} in the field K_{s+1}. Thus we obtain

a_{i_{0}j_{0}}(τ_{s+1}π_{s+1}^{i_{0}} − π_{s+1}^{i_{0}}) ⊗ π^{j_{0}} +_{F} (terms of greater order),

and this sum cannot belong to 𝔐 ⊗ 𝔬 since j_{0} does not contain n from our assumptions (it can be easily proved by considering the expansion of an element of 𝔐 ⊗ 𝔬 in the base 𝔒_{K} ⊗ 𝔒_{K} over 𝔬.) So this case is impossible. Thus our assertion is valid for i = s. Lemma 5.3 is proved. □

Now by applying the lemma 5.3 for i = 0 we obtain α = x ⊗ 1 +_{F} y, y ∈ 𝔬 ⊗ 𝔐.

Theorem A is proved completely.

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Prof. M. Bondarko Department of Mathematics and Mechanics St. Petersburg State University Bibliotechnaya pl. 2, St. Petersburg, 198904, Russia m@vbond.usr.pu.ru 694

Documenta Mathematica 5 (2000)