Local Leopoldt’s Problem for Rings of Integers in Abelian p-Extensions of Complete Discrete Valuation Fields

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 eld. Using this map we investi- gate properties of the coecien t-wise multiplication on associated orders and modules of ideals. This technique allows to study the question of de- termining when the ring of integers is free over its associated order. We answer this question for an Abelian totally wildly ramied p-extension of complete discrete valuation elds whose dieren t is generated by an element of the base eld. We also determine when the ring of integers is free over a Hopf order as a Galois module.

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 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) 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 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 ): 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: This second section is dedicated to the study of conditions ensuring that the ring of integers O K is free over its associated order We prove the following statement.
Proposition.If in the associated order ) 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 ∆ 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 2. The ring O K is isomorphic to the associated order Remark 0.3.1.Besides proving Theorem A we will also construct the element ξ explicitly and so describe -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 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 Documenta Mathematica 5 (2000) 657-693 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 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, tr be the trace operator in K/k.We also define some associated modules.
For I 1 and For arbitrary I 1 , I 2 ⊂ K we define 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 where ψ = σ ψ σ and ψ σ is the projection on the coordinate σ, defined by the equality 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) It is clear that φ(α) as a function acts on K as follows: (2) Besides that, the map φ may be expressed through ψ σ in the form 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: Documenta Mathematica 5 (2000) 657-693 We define a pairing , ⊗ on K ⊗ k K, that takes its values in k: 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: The pairing , K [G] is also non-degenerate because it is a direct sum of n = [K : k] non-degenerate pairings 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 and the equality (4) is proved.
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 (this is dual of I 1 for the bilinear trace form on K).For the associated modules the following equality holds: Proof.First we show that φ( , then for any z ∈ I 1 we have, according to the definition (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) This map is correctly defined since For a o-module M we denote by M the module of o-linear functions from M into o.It is clear that We identify the ideal I 2 with the o-module I * 2 via: Obviously f a (z) = tr(az) for any z ∈ I 2 .
In a completely analogous way we identify I * 1 with I * 1 Using these identifications and the fact that I 1 and I * 2 are projective o-modules we obtain an isomorphism (7) 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 (I * 2 ⊗ o I 1 ) (cf. ( 6)), and so, according to the isomorphism (7), it corresponds to an element Then ( 7) implies that the functional h α f , that corresponds to the element α f , is defined in the following way: On the other hand, from the definition of θ f (cf.( 5)) we obtain: and for any f in C K/k (I 1 , I 2 ) we have found its preimage in I 2 ⊗ I * 1 , i.e., 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: where X is the dual module to X in K with respect to the pairing K × K → k defined by the trace tr.
have unique extensions to k-linear maps from K to K. To be more precise, if f : 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.
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 For the reverse inclusion, let f belong to We take an element x that generates a normal base of the field K over k.Then there exists an element We obtain that k-homomorphisms f and g coincide on the basic elements and so 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 On the other hand, From the G-invariance of the element α it follows that 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 : It follows that Now using the fact that φ is a bijection we obtain We apply to both sides of the equality the map that obviously is an homomorphism.We have 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 Proof.Let f be equal to φ(α), g be equal to φ(β), where α, β x i u j σ(y i v j )σ.
On the other hand, 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 So from the Theorem 1. Proposition 1.7.2.Let f belong to B K/k (I 1 , I 2 ) and let g belong to From Proposition 1.5 we deduce that f and g belong to phi(K and this implies that Proof.From Proposition 1.7 it follows that 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 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 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 algebras with a unit with respect to the multiplication (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 ∆ * g maps D into D, and so it also maps o into itself since D = δO K .We obtain that ∆ * 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 ∆ * 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 So, according to proposition 1.6.1, and we obtain f with respect to the multiplication ∆ * .
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 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( Thus tr x = εδ.Further, if we multiply the element x by ε −1 ∈ o * , then we get the element a.

Documenta Mathematica 5 (2000) 657-693
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.
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).

The multiplication
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 where where Documenta Mathematica 5 (2000) 657-693 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 Then from the definition of the map φ we have: We show that δ On the other hand from the definition of the multiplication f Using this and keeping in mind ( 15) and ( 16) we obtain So we have the congruence Also, the element δ −1 tr in A K/k (O K ) gives us an identity element in the ring Remark 2.3.2.For any other element a in the ring O K with valuation equal to n − 1 we have Remark 2.3.3.A similar statement also holds for the module 2.4.Now we formulate the statements which we will begin to prove in the next subsection.
We will investigate the following condition: 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 17) is fulfilled, then the element ξ generates a "power" basis of ) over o with respect to the multiplication ∆ * (cf.11), i.e., where ξ 0 = δ −1 tr is the unit and and if moreover the order does not contain non-trivial idempotents), then for the ring the condition (17) and so also the assertions of the theorem 2.4.1 are fulfilled.We also have v with respect to the multiplication ∆ * .
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 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 n K , then according to Proposition 2.3.1 we have and the congruence ( 18) is proved.This equality implies that ξ i (a), 0 . Now we show that If there exists an element 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 and so 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: 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: 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 : Thus, in particular, there exist α i , α ∈ o such that 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, 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 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: 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 ] * 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 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 Abelian.We also have that σ(x) = a, v k (a) = 1, and so v 0 (x) = 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 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 It is easily seen that F s also defines a formal group law and the elements of F s indeed defines an associative and commutative addition.Besides that if ).Now we compute the series P Fs (X) for the formal group F s .We obtain: Thus the equation P f (X) = a is equivalent to P Fs (π −s 0 X) = aπ −sn 0 .Besides that v(π −sn 0 a) = v(a) − sn, i.e., π −sn 0 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 Fs (Y ) = aπ −sn 0 by π s 0 , and so the extensions obtained by adjoining the roots of these equations coincide.

The proof of Theorem
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 We can define ξ in the following way: We check the following properties of the element ξ: follow from the definitions of φ and * .It also follows from Theorem 1.3.1 that where y ∈ M ⊗ M. Indeed, and we obtain (23).
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 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 −sn 0 , and so it follows in this case that 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 Thus it is sufficient to prove that for any , where ∆( 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 ∆ * .We consider the polynomial 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 3.1.2imply that J(X) The fact that A K/k (O K ) is an o-algebra with a unit with respect to the multiplication where the powers are taken with respect to ∆ * 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 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 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.
From the definition of φ it follows that it is equal to ψ 1 (δ −1 α).We have We decompose the element y in the base of O K ⊗ M: 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 (a σ − a 1 )σ.

A preliminary group law.
We consider the expansion 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 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): 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 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
. We introduce the following notation: We consider the series f S (X), whose coefficients are found from the equation 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 Documenta Mathematica 5 (2000) 657-693 4.5.In the Abelian group G we choose a family of subgroups 1 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 We introduce the following notation: 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 is fulfilled for all i = p rs , j = s − p rs , 2 ≤ s ≤ m, then it is also fulfilled for 0 ≤ i ≤ p rm − 1, 0 ≤ j ≤ m − 1 and for i = p rm , 0 ≤ j ≤ m − p rm − 1.
2. If (41) is fulfilled for i = p rs , j = s − p rs , 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 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 < 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 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 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 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 + 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: (we assume here that the group G acts only on the first component of the tensor product).
The remaining terms indeed have greater ni + j since (i 0 , p) = 1 and so the valuation of τ s+1 π i0 s+1 − π i0 s+1 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 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 .Documenta Mathematica 5 (2000) 657-693

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, 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 .
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.

) 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.
) is fulfilled.Thus we obtainf bp w = f b p w , b < p, f (p−1)p w f p w = f p w = f (σ p p w )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 p w = f (σ (pc−1)p w σ p w ), §5The proof of necessity in Theorem A

F 1 ⊗
z. Using the formula for ψ 1 and the fact that b 1 = 0 we obtain that0 = 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:

a
i0j0 (τ s+1 π i0 s+1 − π i0 s+1 ) ⊗ π j0 + 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.Documenta Mathematica 5 (2000) 657-693