On the Milnor K-Groups of

Complete Discrete Valuation Fields

Jinya Nakamura

Received: April 5, 2000

Communicated by A. Merkurjev

Abstract. For a discrete valuation field K, the unit group K^{×} of K has a natural decreasing filtration with respect to the valuation, and the graded quotients of this filtration are given in terms of the residue field. The Milnor K-group Kq^{M}(K) is a generalization of the unit group, and it also has a natural decreasing filtration. However, if K is of mixed characteristics and has an absolute ramification index greater than one, the graded quotients of this filtration are not yet known except in some special cases.

The aim of this paper is to determine them when K is absolutely tamely ramified discrete valuation field of mixed characteristics (0, p >

2) with possibly imperfect residue field.

Furthermore, we determine the kernel of the Kurihara’s Kq^{M}- exponential homomorphism from the differential module to the Milnor K-group for such a field.

1991 Mathematics Subject Classification: 19D45, 11S70 Keywords and Phrases: The Milnor K-group, Complete Discrete Valuation Field, Higher Local Class Field Theory

1 Introduction

For a ring R, the Milnor K-group of R is defined as follows. We denote the unit group of R by R^{×}. Let J(R) be the subgroup of the q-fold tensor product of R^{×} overℤ generated by the elements a_{1} ⊗⋯⊗a_{q}, where a_{1}, …, a_{q} are elements of R^{×} such that a_{i} + a_{j} = 0 or 1 for some i ≠ j. Define

Kq^{M}(R) = (R^{×} ⊗_{ℤ} ⋯ ⊗_{ℤ} R^{×})/J(R).

We denotes the image of a_{1} ⊗ ⋯ ⊗ a_{q} by {a_{1}, …, a_{q}}.

Now we assume K is a discrete valuation field. Let v_{K} be the normalized valuation of K. Let O_{K}, F and 𝔪_{K} be the valuation ring, the residue field and the valuation ideal of K, respectively. There is a natural filtration on K^{×} defined by

U_{K}^{i} = \begin{cases}O_{K}^{×} & for i = 0 \\ 1 + 𝔪_{K}^{i} & for i ≥ 1. \end{cases}

We know that the graded quotients U_{K}^{i}/U_{K}^{i+1} are isomorphic to F^{×} if i = 0 and F if i ≥ 1. Similarly, there is a natural filtration on Kq^{M}(K) defined by

U^{i}Kq^{M}(K) = {{x_{1}, …, x_{q}} ∈ Kq^{M}(K) | x_{1} ∈ U_{K}^{i}, x_{2}, …, x_{q} ∈ K^{×}} .

Let gr^{i} Kq^{M}(K) = U^{i}Kq^{M}(K)/U^{i+1}Kq^{M}(K) for i ≥ 0. gr^{i} Kq^{M}(K) are determined in the case that the characteristics of K and F are both equal to 0 in [5], and in the case that they are both nonzero in [2] and [9]. If K is of mixed characteristics (0, p), gr^{i} Kq^{M}(K) is determined in [3] in the range 0 ≤ i ≤ e_{K}p/(p − 1), where e_{K} = v_{K}(p). However, gr^{i} Kq^{M}(K) still remains mysterious for i > ep/(p − 1). In [16], Kurihara determined gr^{i} Kq^{M}(K) for all i if K is absolutely unramified, i.e., v_{K}(p) = 1. In [13] and [19], gr^{i} Kq^{M}(K) is determined for some K with absolute ramification index greater than one.

The purpose of this paper is to determine gr^{i} Kq^{M}(K) for all i and a discrete valuation field K of mixed characteristics (0, p), where p is an odd prime and p ∤ e_{K}. We do not assume F to be perfect. Note that the graded quotient gr^{i} Kq^{M}(K) is equal to gr^{i} Kq^{M}(K̂), where K̂ is the completion of K with respect to the valuation, thus we may assume that K is complete under the valuation.

Let F be a field of positive characteristic. Let Ω_{F}^{1} = Ω_{F/ℤ}^{1} be the module of absolute differentials and Ω_{F}^{q} the q-th exterior power of Ω_{F}^{1} over F. As in [7], we define the following subgroups of Ω_{F}^{q}. Z_{1}^{q} = Z_{1}Ω_{F}^{q} denotes the kernel of d : Ω_{F}^{q} → Ω_{F}^{q+1} and B_{1}^{q} = B_{1}Ω_{F}^{q} denotes the image of d : Ω_{F}^{q−1} → Ω_{F}^{q}. Then there is an exact sequence

0 ⟶ B_{1}^{q} ⟶ Z_{1}^{q} ⟶^{C} Ω_{F}^{q} ⟶ 0,

where C is the Cartier operator defined by

\begin{matrix}x^{p} \frac{dy_{1}}{y_{1}} ∧ … \frac{dy_{q}}{y_{q}} ⟼ x\frac{dy_{1}}{y_{1}} ∧ … \frac{dy_{q}}{y_{q}} , \\ B_{1}^{q} → 0. \end{matrix}

The inverse of C induces the isomorphism

\begin{matrix}C^{−1} : Ω_{F}^{q} & ⟶^{≅} Z_{1}^{q}/B_{1}^{q} \\ x\frac{dy_{1}}{y_{1}} ∧ … ∧ \frac{dy_{q}}{y_{q}} & ⟼ x^{p} \frac{dy_{1}}{y_{1}} ∧ … ∧ \frac{dy_{q}}{y_{q}} \end{matrix}  (1)

The Milnor K-Groups 153

for x ∈ F and y_{1}, …, y_{q} ∈ F^{×}. For i ≥ 2, let B_{i}^{q} = B_{i}Ω_{F}^{q} (resp. Z_{i}^{q} = Z_{i}Ω_{F}^{q} ) be the subgroup of Ω_{F}^{q} defined inductively by

\begin{matrix}B_{i}^{q} ⊃ B_{i−1}^{q}, C^{−1} : B_{i−1}^{q} ⟶^{≅} B_{i}^{q}/B_{1}^{q} \\ (resp. Z_{i}^{q} ⊂ Z_{i−1}^{q}, C^{−1} : Z_{i−1}^{q} ⟶^{≅} Z_{i}^{q}/B_{1}^{q}). \end{matrix}

Let Z_{∞}^{q} be the intersection of all Z_{i}^{q} for i ≥ 1. We denote Z_{i}^{q} = Ω_{F}^{q} for i ≤ 0. The main result of this paper is the following

Theorem 1.1. Let K be a discrete valuation field of characteristic zero, and F the residue field of K. Assume that p = char(F) is an odd prime and e = e_{K} = v_{K}(p) is prime to p. For i > ep/(p − 1), let n be the maximal integer which satisfies i − ne ≥ e/(p − 1) and let s = v_{p}(i − ne), where v_{p} is the p-adic order. Then

gr^{i} Kq^{M}(K) ≅ Ω_{F}^{q−1}/B_{s+n}^{q−1}.

Corollary 1.2. Let U^{i}(Kq^{M}(K)/p^{m}) be the image of U^{i}Kq^{M}(K)

in Kq^{M}(K)/p^{m}Kq^{M}(K) for m ≥ 1 and gr^{i}(Kq^{M}(K)/p^{m}) =

U^{i}(Kq^{M}(K)/p^{m})/U^{i+1}(Kq^{M}(K)/p^{m}). Then

gr^{i}(Kq^{M}(K)/p^{m}) ≅ \begin{cases}Ω_{F}^{q−1}/B_{s+n}^{q−1} & ( if m > s + n) \\ Ω_{F}^{q−1}/Z_{m−n}^{q−1} & ( if m ≤ s + n, i − en ≠ \frac{e}{p−1}) \\ Ω_{F}^{q−1}/(1 + aC)Z_{m−n+1}^{q−1} & ( if m ≤ s + n, i − en = \frac{e}{p−1}) \end{cases}

where a is the residue class of p/π^{e} for a fixed prime element π of K.

Remark 1.3. If 0 ≤ i ≤ ep/(p − 1), gr^{i} Kq^{M}(K) is known by [3].

To show (1.1), we use the (truncated) syntomic complexes with respect to O_{K} and O_{K}/pO_{K}, which were introduced in [11]. In [12], it was proved that there exists an isomorphism between some subgroup of the q-th cohomology group of the syntomic complex with respect to O_{K} and some subgroup of Kq^{M}(K)ˆ which includes the image of U^{1}Kq^{M}(K) (cf. (2.1)). On the other hand, the cohomology groups of the syntomic complex with respect to O_{K}/pO_{K} can be calculated easily because O_{K}/pO_{K} depends only on F and e. Comparing these two complexes, we have the exact sequence (2.4)

H^{1}(𝕊_{q}) ⟶ Ω̂_{A/ℤ}^{q−1}/pdΩ̂_{A/ℤ}^{q−2} ⟶^{exp_{p}} Kq^{M}(K)ˆ

as an long exact sequence of syntomic complexes, where 𝕊_{q} is the truncated translated syntomic complex with respect to O_{K}/pO_{K}, hat means the p-adic completion, and exp_{p} is the Kurihara’s Kq^{M}-exponential homomorphism with respect to p. For more details, see Section 2. The left hand side of this exact sequence is determined in (2.6), and we have (1.1) by calculating these groups and the relations explicitly.

In Section 2, we see the relations between the syntomic complexes mentioned above, the Milnor K-groups, and the differential modules. The method of the proof of (1.1) is mentioned here. Note that we do not assume p ∤ e in this section and we get the explicit description of the cohomology group of the syntomic complex with respect to O_{K}/pO_{K} which was used in the proof of (1.1) without the assumption p ∤ e. In Section 3, we calculate differential module of O_{K}. We calculate the kernel of the Kq^{M}-exponential homomorphism (4) explicitly in Section 4, 5, 6 and 7. In Section 8, we show Theorem 1.1 and Corollary 1.2. In Section 9, we have an application related to higher local class field theory.

Notations and Definitions. All rings are commutative with 1. For an element x of a discrete valuation ring, x̄ means the residue class of x in the residue field. For an abelian group M and positive integer n, we denote M/p^{n} = M/p^{n}M and M̂ = \underleftarrow{lim}_{n}M/p^{n}. For a subset N of M, ⟨N⟩ means the subgroup of M generated by N. For a ring R, let Ω_{R}^{1} = Ω_{R/ℤ}^{1} be the absolute differentials of R and Ω_{R}^{q} the q-th exterior power of Ω_{R}^{1} over R for q ≥ 2. We denote Ω_{R}^{0} = R and Ω_{R}^{q} = 0 for negative q. If R is of characteristic zero, let

ℨ_{n}Ω̂_{R}^{q} = Ker (Ω̂_{R}^{q} ⟶^{d} Ω̂_{R}^{q+1}/p^{n})

for positive n. For an element ω ∈ Ω̂_{R}^{q}, let v_{p}(ω) be the maximal n which satisfies ω ∈ ℨ_{n}Ω̂_{R}^{q}. For n ≤ 0, let ℨ_{n}Ω̂_{R}^{q} = Ω̂_{R}^{q}. Let ℨ_{∞}Ω̂_{R}^{q} be the intersection of ℨ_{n}Ω̂_{R}^{q} of all n ≥ 0. All complexes are cochain complexes. For a morphism of non-negative complexes f: C^{·} → D^{·}, [f: C^{·} → D^{·}] and

[\begin{matrix}C^{0} & ⟶^{d} & C^{1} & ⟶^{d} & C^{2} & ⟶^{d} & … \\ ↓^{f} & & ↓^{f} & & ↓^{f} \\ D^{0} & ⟶^{d} & D^{1} & ⟶^{d} & D^{2} & ⟶^{d} & … \end{matrix} ]

both denote the mapping fiber complex with respect to the morphism f, namely, the complex

(C^{0} ⟶^{d} C^{1} ⊕ D^{0} ⟶^{d} C^{2} ⊕ D^{1} ⟶^{d} …),

where the leftmost term is the degree-0 part and where the differentials are defined by

\begin{matrix}C^{i} ⊕ D^{i−1} & ⟶ C^{i+1} ⊕ D^{i} \\ (a, b) & ⟼ (da, f(a) − db). \end{matrix}

Acknowledgements. I would like to express my gratitude to Professor Kazuya Kato, Professor Masato Kurihara and Professor Ivan Fesenko for their valuable advice. I also wish to thank Takao Yamazaki for many helpful comments.

In [20], I.Zhukov calculated the Milnor K-groups of multidimensional complete fields in a different way. He gives an explicit description by using topological generators. In [8], B.Kahn also calculated K_{2}(K) of local fields with perfect residue fields without an assumption p ∤ e_{K}.

2 Exponential homomorphism and syntomic cohomology

Let K be a complete discrete valuation field of mixed characteristics (0, p). Assume that p is an odd prime. Let A = O_{K} be the ring of integers of K and F the residue field of K. Let A_{0} be the Cohen subring of A with respect to F, namely, A_{0} is a complete discrete valuation ring under the restriction of the valuation of A with the residue field F and p is a prime element of A_{0} (cf. [4], IX, Section 2). Let K_{0} be the fraction field of A_{0}. Then K/K_{0} is finite and totally ramified extension of extension degree e = e_{K}. We denote e′ = ep/(p − 1). Let π be a prime element of K and fix it. We further assume that F has a finite p-base and fix their liftings 𝕋 ⊂ A_{0}. We can take the frobenius endomorphism f of A_{0} such that f(T) = T^{p} for T ∈ 𝕋 (cf. [12] or [17]). Let U^{i}Kq^{M}(A) be the subgroup defined by the same way of U^{i}Kq^{M}(K), namely,

U^{i}Kq^{M}(A) = ⟨{x_{1}, …, x_{q}} ∈ Kq^{M}(A) | x_{1} ∈ U_{K}^{i}, x_{2}, …, x_{q} ∈ A^{×}⟩ .

Let U^{i}Kq^{M}(K)ˆ (resp. U^{i}Kq^{M}(A)ˆ ) be the closure of the image of U^{i}Kq^{M}(K) (resp. U^{i}Kq^{M}(A)) in Kq^{M}(K)ˆ (resp. Kq^{M}(A)ˆ ). Note that gr^{i} Kq^{M}(K) ≅ gr^{i} Kq^{M}(K)ˆ for i > 0.

At first, we introduce an isomorphism between U^{1}Kq^{M}(K)ˆ and a subgroup of the cohomology group of the syntomic complex with respect to A. For further details, see [12]. Let B = A_{0}[[X]], where X is an indeterminate. We extend the operation of the frobenius f on B by f(X) = X^{p}. We define I and J as follows.

J = Ker (B ⟶^{X↦π} A)
I = Ker (B ⟶^{X↦π} A ⟶^{mod p} A/p) = J + pB.

Let D and J ⊂ D be the PD-envelope and the PD-ideal with respect to B → A, respectively ([1],Section 3). Let I ⊂ D be the PD-ideal with respect to B → A/p. D is also the PD-envelope with respect to B → A/p. Let J^{[q]} and I^{[q]} be their q-th divided powers. Notice that I^{[1]} = I, J^{[1]} = J and I^{[0]} = J^{[0]} = D. If q is an negative integer, we denote J^{[q]} = I^{[q]} = D. We define the complexes 𝕁^{[q]} and 𝕀^{[q]} as

\begin{aligned}𝕁^{[q]} & = (J^{[q]} →^{d} J^{[q−1]} ⊗_{B} Ω̂_{B}^{1} →^{d} J^{[q−2]} ⊗_{B} Ω̂_{B}^{2} ⟶ ⋯) \\ 𝕀^{[q]} & = (I^{[q]} →^{d} I^{[q−1]} ⊗_{B} Ω̂_{B}^{1} →^{d} I^{[q−2]} ⊗_{B} Ω̂_{B}^{2} ⟶ ⋯) , \end{aligned}

where Ω̂_{B}^{q} is the p-adic completion of Ω_{B}^{q}. The leftmost term of each complex is the degree 0 part. We define 𝔻 = 𝕀^{[0]} = 𝕁^{[0]}. For 1 ≤ q < p, let S(A, B)(q) and S′(A, B)(q) be the mapping fibers of

\begin{matrix}𝕁^{[q]} & ⟶^{1−f_{q}} 𝔻 \\ 𝕀^{[q]} & ⟶^{1−f_{q}} 𝔻, \end{matrix}

respectively, where f_{q} = f/p^{q}. S(A, B)(q) is called the syntomic complex of A with respect to B, and S′(A, B)(q) is also called the syntomic complex of A/p with respect to B (cf. [11]). We notice that

H^{q}(S(A, B)(q))

= Ker ((D⊗Ω̂_{B}^{q}) ⊕ (D⊗Ω̂_{B}^{q−1}) → (D⊗Ω̂_{B}^{q+1}) ⊕ (D⊗Ω̂_{B}^{q})) ; Im ((J⊗Ω̂_{B}^{q−1}) ⊕ (D⊗Ω̂_{B}^{q−2}) → (D⊗Ω̂_{B}^{q}) ⊕ (D⊗Ω̂_{B}^{q−1})), (2)

where the maps are the differentials of the mapping fiber. If q ≥ p, we cannot define the map 1 − f_{q} on 𝕁^{[q]} and 𝕀^{[q]}, but we define H^{q}(S(A, B)(q)) by using (2) in this case. This is equal to the cohomology of the mapping fiber of

σ_{>q−3}𝕁^{[q]} ⟶^{1−f_{q}} σ_{>q−3}𝔻,

where σ_{>n}C^{·} means the brutal truncation for a complex C^{·}, i.e., (σ_{>n}C^{·})^{i} is C^{i} if i > n and 0 if i ≤ n. Let U^{1}(D ⊗ Ω̂_{B}^{q−1}) be the subgroup of D ⊗ Ω̂_{B}^{q−1} generated by XD ⊗ Ω̂_{B}^{q−1}, (X^{e})^{[m]}D ⊗ Ω̂_{B}^{q−1} for all m ≥ 1 and D ⊗ Ω̂_{B}^{q−2} ∧ dX. Let U^{1}H^{q}(S(A, B)(q)) be the subgroup of H^{q}(S(A, B)(q)) generated by the image of (D ⊗ Ω̂_{B}^{q}) ⊕ U^{1}(D ⊗ Ω̂_{B}^{q−1}). Then there is a result of Kurihara:

Theorem 2.1 (Kurihara, [12]). A and B are as above. Then

U^{1}H^{q}(S(A, B)(q)) ≅ U^{1}Kq^{M}(A)ˆ .

Furthermore, we have the following

Lemma 2.2. A and K are as above. Assume that A has the primitive p-th roots of unity. Then

(i) The natural map Kq^{M}(A)ˆ → Kq^{M}(K)ˆ is an injection.

(ii) U^{1}H^{q}(S(A, B)(q)) ≅ U^{1}Kq^{M}(A)ˆ ≅ U^{1}Kq^{M}(K)ˆ .

Remark 2.3. When F is separably closed, this lemma is also the consequence of the result of Kurihara [14]. But even if F is not separably closed, calculation goes similarly to [14].

Proof of Lemma 2.2. The first isomorphism of (ii) is (2.1). The natural map

U^{1}Kq^{M}(A)ˆ → U^{1}Kq^{M}(K)ˆ

is a surjection by the definition of the filtrations and the fact that we can define an element {1+π^{i}a_{1}, a_{2}, …, a_{q−1}, π} as an element of Kq^{M}(A)ˆ by using Dennis-Stain Symbols, see [17]. Thus we only have to show (i). Let ζ_{p} be a primitive p-th root of unity and fix it. Let μ_{p} be the subgroup of A^{×} generated by ζ_{p}. For n ≥ 2, see the following commutative diagram.

\begin{matrix}Kq−1^{M}(K)/p & ⟶^{\{∗,ζ_{p}\}} & Kq^{M}(A)/p^{n−1} & ⟶^{p} & Kq^{M}(A)/p^{n} \\ ^{=}↓ & & ↓ & & ↓ \\ Kq−1^{M}(K)/p & ⟶^{\{∗,ζ_{p}\}} & Kq^{M}(K)/p^{n−1} & ⟶^{p} & Kq^{M}(K)/p^{n} \end{matrix} ; ⟶ Kq^{M}(A)/p ⟶ 0 ; ↓ ; ⟶ Kq^{M}(K)/p ⟶ 0.  (3)

The bottom row are exact by using Galois cohomology long exact sequence with respect to the Bockstein

⋯ → H^{q−1}(K, ℤ/p(q)) → H^{q}(K, ℤ/p^{n−1}(q)) → H^{q}(K, ℤ/p^{n}(q)) → …

and

Kq^{M}(K)/p^{n} ≅ H^{q}(K, ℤ/p^{n}(q))

by [3]. The map {∗, ζ_{p}} in the top row is well-defined if Kq^{M}(A)/p^{n−1} → Kq^{M}(K)/p^{n−1} is injective, and the top row are exact except at Kq^{M}(A)/p^{n−1}. Using the induction on n, we only have to show the injectivity of Kq^{M}(A)/p → Kq^{M}(K)/p. We know the subquotients of the filtration of Kq^{M}(K)/p by [3] and we also know the subquotients of the filtration of Kq^{M}(A)/p using the isomorphism U^{1}H^{q}(S(A, B)(q)) ≅ U^{1}Kq^{M}(A)ˆ in [12] and the explicit calculation of H^{q}(S(A, B)(q)) by [14] except gr^{0}(Kq^{M}(A)/p). Natural map preserves filtrations and induces isomorphisms of subquotients. Thus U^{1}(Kq^{M}(A)/p) → U^{1}(Kq^{M}(K)/p) is an injection. Lastly, the composite map of the natural maps

Kq^{M}(F)/p ↠ gr^{0}(Kq^{M}(A)/p) → gr^{0}(Kq^{M}(K)/p) →^{≅} Kq^{M}(F)/p ⊕ Kq−1^{M}(F)/p

is also an injection. Hence Kq^{M}(A)/p → Kq^{M}(K)/p is injective.

Next, we introduce Kq^{M}-exponential homomorphism and consider the kernel. By [17], there is the Kq^{M}-exponential homomorphism with respect to η for q ≥ 2 and η ∈ K such that v_{K}(η) ≥ 2e/(p − 1) defined by

\begin{matrix}exp_{η} : Ω̂_{A}^{q−1} & ⟶ Kq^{M}(K) \\ ˆ a\frac{db_{1}}{b_{1}} ∧ ⋯ ∧ \frac{db_{q−1}}{b_{q−1}} & ⟼ \{exp(ηa), b_{1}, …, b_{q−1}\} \end{matrix}  (4)

for a ∈ A, b_{1}, …, b_{q−1} ∈ A^{×}. Here exp is

exp(X) = ∑_{n=0}^{∞} \frac{X^{n}}{n!} .

We use this Kq^{M}-exponential homomorphism only in the case η = p in this paper. On the other hand, there exists an exact sequence of complexes

0 → [\begin{matrix}σ_{>q−3}𝕁^{[q]} \\ ↓ 1−f_{q} \\ σ_{>q−3}𝔻 \end{matrix}] → [\begin{matrix}σ_{>q−3}𝕀^{[q]} \\ ↓ 1−f_{q} \\ σ_{>q−3}𝔻 \end{matrix}] → [\begin{matrix}σ_{>q−3}𝕀^{[q]} & /σ_{>q−3}𝕁^{[q]} \\ & ↓ \\ & 0 \end{matrix}] → 0. (5)

[σ_{>q−3}𝕀^{[q]}/σ_{>q−3}𝕁^{[q]} → 0] is none other than the complex σ_{>q−3}𝕀^{[q]}/σ_{>q−3}𝕁^{[q]}.

We denote the complex [σ_{>q−3}𝕀^{[q]} →^{1−f_{q}} σ_{>q−3}𝔻][q − 2] by 𝕊_{q}. It is the mapping fiber complex

[\begin{matrix}I^{[2]} ⊗ Ω̂_{B}^{q−2} & ⟶^{d} & I ⊗ Ω̂_{B}^{q−1} & ⟶^{d} & D ⊗ Ω̂_{B}^{q} & ⟶^{d} & … \\ ↓^{1−f_{q}} & & ↓^{1−f_{q}} & & ↓^{1−f_{q}} \\ D ⊗ Ω̂_{B}^{q−2} & ⟶^{d} & D ⊗ Ω̂_{B}^{q−1} & ⟶^{d} & D ⊗ Ω̂_{B}^{q} & ⟶^{d} & … \end{matrix} ] . (6)

Taking cohomology, we have the following

Proposition 2.4. A, B and K are as above. Then Kq^{M}-exponential homomorphism with respect to p factors through Ω̂_{A}^{q−1}/pdΩ_{A}^{q−2} and there is an exact sequence

H^{1}(𝕊_{q}) ⟶^{ψ} Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2} ⟶^{exp_{p}} Kq^{M}(K)ˆ .

Proof. See the cohomological long exact sequence with respect to the exact sequence (5). The q-th cohomology group of the left complex of (5) is equal to H^{q}(S(A, B)(q)), thus the sequence

H^{1}(𝕊_{q}) →^{ψ} H^{1}((σ_{>q−3}𝕀^{[q]}/σ_{>q−3}𝕁^{[q]})[q − 2]) → H^{q}(S(A, B)(q))

is exact.  Here we denote the first map by ψ.  The complex

(σ_{>q−3}𝕀^{[2]}/σ_{>q−3}𝕁^{[2]})[q − 2] is

((I^{[2]} ⊗ Ω̂_{B}^{q−2})/(J^{[2]} ⊗ Ω̂_{B}^{q−2}) → (I ⊗ Ω̂_{B}^{q−1})/(J ⊗ Ω̂_{B}^{q−1}) → 0 → ⋯) .

(I ⊗ Ω̂_{B}^{q−1})/(J ⊗ Ω̂_{B}^{q−1}) is the subgroup of (D ⊗ Ω̂_{B}^{q−1})/(J ⊗ Ω̂_{B}^{q−1}) = A ⊗ Ω̂_{B}^{q−1}. The image of I ⊗Ω̂_{B}^{q−1} in A⊗Ω̂_{B}^{q−1} is equal to pA⊗Ω̂_{B}^{q−1}. Thus (I ⊗Ω̂_{B}^{q−1})/(J ⊗ Ω̂_{B}^{q−1}) = pA ⊗ Ω̂_{B}^{q−1}. The image of

(I^{[2]} ⊗ Ω̂_{B}^{q−2})/(J^{[2]} ⊗ Ω̂_{B}^{q−2}) ⟶^{d} pA ⊗ Ω̂_{B}^{q−1}

is equal to the image of I^{2} ⊗ Ω̂_{B}^{q−2}. By I = (p) + J, d(I^{2} ⊗ Ω̂_{B}^{q−2}) is equal to d(J^{2} ⊗ Ω̂_{B}^{q−2}) + pd(J ⊗ Ω̂_{B}^{q−2}) + p^{2}d(Ω̂_{B}^{q−2}). By the exact sequence

0 ⟶ J ⟶ B ⟶ A ⟶ 0,

we have an exact sequence

(J /J^{2}) ⊗ Ω̂_{B}^{q−2} ⟶^{d} A ⊗ Ω̂_{B}^{q−1} ⟶ Ω̂_{A}^{q−1} ⟶ 0.  (7)

Thus the image of d(J^{2} ⊗ Ω̂_{B}^{q−2}) in A ⊗ Ω̂_{B}^{q−1} is zero. A ⊗ Ω̂_{B}^{q−1} is torsion free, thus

\frac{pA ⊗ Ω̂_{B}^{q−1}}{pd(J ⊗ Ω̂_{B}^{q−2}) + p^{2}dΩ̂_{B}^{q−2}} ≅^{p^{−1}} \frac{A ⊗ Ω̂_{B}^{q−1}}{d(J ⊗ Ω̂_{B}^{q−2}) + pdΩ̂_{B}^{q−2}} ≅ Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}. (8)

Hence we have H^{1}((σ_{>q−3}𝕀^{[2]}/σ_{>q−3}𝕁^{[2]})[q − 2]) ≅ Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}. By chasing the connecting homomorphism Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2} → H^{q}(S(A, B)(q)), we can show that the image is contained by U^{1}H^{q}(S(A, B)(q)) and the composite map

Ω̂_{A}^{q−1} → Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2} → U^{1}H^{q}(S(A, B)(q)) →^{≅} U^{1}Kq^{M}(K)ˆ

is equal to exp_{p}. We got the desired exact sequence.

Remark 2.5. By [3], there exist surjections

\begin{matrix}Ω_{F}^{q−2} ⊕ Ω_{F}^{q−1} & ⟶ gr^{i} Kq^{M}(K) \\ (x\frac{dy_{1}}{y_{1}} ∧ ⋯ ∧ \frac{dy_{q−2}}{f_{q−2}} , 0) & ⟼ \{1 + π^{i}x̃, ỹ_{1}, …, ỹ_{q−2}, π\} \\ (0, x\frac{dy_{1}}{y_{1}} ∧ ⋯ ∧ \frac{dy_{q−1}}{f_{q−1}} ) & ⟼ \{1 + π^{i}x̃, ỹ_{1}, …, ỹ_{q−1}\} \end{matrix}  (9)

for i ≥ 1, where x ∈ F, y_{1}, …, y_{q−1} ∈ F^{×} and where x̃, ỹ_{1}, …, ỹ_{q−1} are their liftings to A. If i ≥ e + 1, then we can construct all elements of gr^{i} Kq^{M}(K) as the image of exp_{p}, namely,

{ω ∈ Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2} | exp_{p}(ω) ∈ U^{i}Kq^{M}(K)ˆ } ⟶^{exp_{p}} gr^{i} Kq^{M}(K)
\frac{π^{i−1}}{p} a\frac{db_{1}}{b_{1}} ∧ ⋯ ∧ \frac{db_{q−2}}{b_{q−2}} ∧ dπ ⟼ \{exp(π^{i}a), b_{1}, …, b_{q−2}, π\}
= {1 + π^{i}a, b_{1}, …, b_{q−2}, π}
\frac{π^{i}}{p} a\frac{db_{1}}{b_{1}} ∧ ⋯ ∧ \frac{db_{q−1}}{b_{q−1}} ⟼ \{exp(π^{i}a), b_{1}, …, b_{q−1}\}
= {1 + π^{i}a, b_{1}, …, b_{q−1}}.

Thus U^{e+1}Kq^{M}(K)ˆ is contained by the image of exp_{p}. On the other hand, (2.4) says the kernel of the Kq^{M}-exponential homomorphism is ψ(H^{1}(𝕊_{q})). Recall that the aim of this paper is to determine gr^{i} Kq^{M}(K) for all i, but we already know them in the range 0 ≤ i ≤ e′ in [3]. Thus if we want to know gr^{i} Kq^{M}(K) for all i, we only have to know ψ(H^{1}(𝕊_{q})). We determine H^{1}(𝕊_{q}) in the rest of this section, and ψ(H^{1}(𝕊_{q})) in Section 4, 5, 6 and 7.

To determine H^{1}(𝕊_{q}), we introduce a filtration into it. Let 0 ≤ r < p and s ≥ 0 be integers. Recall that B = A_{0}[[X]]. For i ≥ 0 and s ≥ 0, let fil^{i}(I^{[r]} ⊗ Ω̂_{B}^{s}) be the subgroup of I^{[r]} ⊗ Ω̂_{B}^{s} generated by the elements

\begin{matrix}\{X^{n}(X^{e})^{[j]}ω | n + ej ≥ i, n ≥ 0, j ≥ r, ω ∈ D ⊗ Ω̂_{B}^{s}\} \\ ∪ \{X^{n−1}(X^{e})^{[j]}ω ∧ dX | n + ej ≥ i, n ≥ 1, j ≥ r, ω ∈ D ⊗ Ω̂_{B}^{s−1}\} . \end{matrix}

The homomorphism 1 − f_{r+s} : I^{[r]} ⊗ Ω̂_{B}^{s} → D ⊗ Ω̂_{B}^{s} preserves filtrations. Thus we can define the following complexes

fil^{i}(σ_{>q−3}𝕀^{[q]})[q − 2]

= (fil^{i} (I^{[2]} ⊗ Ω̂_{B}^{q−2}) → fil^{i}(I ⊗ Ω̂_{B}^{q−1}) → fil^{i}(D ⊗ Ω̂_{B}^{q}) → …)

fil^{i}(σ_{>q−3}𝔻)[q − 2]

= (fil^{i} (D ⊗ Ω̂_{B}^{q−2}) → fil^{i}(D ⊗ Ω̂_{B}^{q−1}) → fil^{i}(D ⊗ Ω̂_{B}^{q}) → …)
fil^{i} 𝕊_{q} = [fil^{i}(σ_{>q−3}𝕀^{[q]})[q − 2] ⟶^{1−f_{q}} fil^{i}(σ_{>q−3}𝔻)[q − 2]]
gr^{i}(σ_{>q−3}𝕀^{[r]})[q − 2] = \frac{fil^{i}(σ_{>q−3}𝕀^{[r]})[q − 2]}{fil^{i+1}(σ_{>q−3}𝕀^{[r]})[q − 2]} for r = 0, q
gr^{i} 𝕊_{q} = [gr^{i}(σ_{>q−3}𝕀^{[q]})[q − 2] ⟶^{1−f_{q}} gr^{i}(σ_{>q−3}𝔻)[q − 2]] .

Note that if i ≥ 1, 1 − f_{q}: gr^{i}(σ_{>q−3}𝕀^{[q]})[q − 2] → gr^{i}(σ_{>q−3}𝔻)[q − 2] is none other than 1 because f_{q} takes the elements to the higher filters. fil^{i} 𝕊_{q} forms the filtration of 𝕊_{q} and we have the exact sequences

0 ⟶ fil^{i+1} 𝕊_{q} ⟶ fil^{i} 𝕊_{q} ⟶ gr^{i} 𝕊_{q} ⟶ 0

for i ≥ 0. This exact sequence of complexes give a long exact sequence

⋯ → H^{n}(fil^{i+1} 𝕊_{q}) → H^{n}(fil^{i} 𝕊_{q}) → H^{n}(gr^{i} 𝕊_{q}) → H^{n+1}(fil^{i+1} 𝕊_{q}) → … (10)

Furthermore, we have the following

Proposition 2.6. {H^{1}(fil^{i} 𝕊_{q})}_{i} forms the finite decreasing filtration of H^{1}(𝕊_{q}). Denote fil^{i} H^{1}(𝕊_{q}) = H^{1}(fil^{i} 𝕊_{q}) and gr^{i} H^{1}(𝕊_{q}) = fil^{i} H^{1}(𝕊_{q})/ fil^{i+1} H^{1}(𝕊_{q}). Then

gr^{i}H^{1}(𝕊_{q}) =
0  ( if i > 2e)
X^{2e−1}dX ∧ (Ω̂_{A_{0}}^{q−3}/p)  ( if i = 2e)
X^{i} (Ω̂_{A_{0}}^{q−2}/p) ⊕ X^{i−1}dX ∧ (Ω̂_{A_{0}}^{q−3}/p)  ( if e < i < 2e)
X^{e} (Ω̂_{A_{0}}^{q−2}/p) ⊕ X^{e−1}dX ∧ (ℨ_{1}Ω̂_{A_{0}}^{q−3}/p^{2}Ω̂_{A_{0}}^{q−3})  ( if i = e, p | e)
X^{e−1}dX ∧ (ℨ_{1}Ω̂_{A_{0}}^{q−3}/p^{2}Ω̂_{A_{0}}^{q−3})  ( if i = e, p ∤ e)

(X^{i} \frac{(p^{Max(η′_{i}−v_{p}(i),0)}Ω̂_{A_{0}}^{q−2} ∩ ℨ_{η_{i}} Ω̂_{A_{0}}^{q−2}) + p^{2}Ω̂_{A_{0}}^{q−2}}{p^{2}Ω̂_{A_{0}}^{q−2}} )

⊕ (X^{i−1}dX ∧ \frac{ℨ_{η_{i}} Ω̂_{A_{0}}^{q−3} + p^{2}Ω̂_{A_{0}}^{q−3}}{p^{2}Ω̂_{A_{0}}^{q−3}} )

( if 1 ≤ i < e)
0  ( if i = 0),

where η_{i} and η′_{i} be the integers which satisfy p^{η_{i}−1}i < e ≤ p^{η_{i}}i and p^{η′_{i}−1}i−1 < e ≤ p^{η′_{i}}i − 1 for each i.

To prove (2.6), we need the following lemmas. Lemma 2.7. For ω ∈ D ⊗ Ω̂_{B}^{q} and n ≥ 0,

v_{p}(f^{n}(ω)) ≥ v_{p}(ω) + nq.  (11)

In particular, if ω ∈ Ω̂_{A_{0}}^{q}, then

v_{p}(f^{n}(ω)) = v_{p}(ω) + nq.  (12)

Proof. ω ∈ D ⊗ Ω̂_{B}^{q} can be rewrite as ω = ∑_{i} a_{i}ω_{i}, where a_{i} ∈ D and ω_{i} are the canonical generators of Ω̂_{B}^{q}, which are

ω_{i} = \frac{dT_{1}}{T_{1}} ∧ ⋯ ∧ \frac{dT_{q}}{T_{q}}

for T_{1}, …, T_{q} ∈ 𝕋∪{X}. Canonical generators have the property f(ω_{i}) = p^{q}ω_{i}, thus we have (11). Furthermore, if ω ∈ Ω̂_{A_{0}}^{q}, then a_{i} ∈ A_{0} and we have v_{p}(f(a_{i})) = v_{p}(a_{i}). Thus (12) follows. Lemma 2.8. If 1 ≤ r < p, s ≥ 0 and i > er, then there exists a homomorphism

∑_{m=0}^{∞} f_{r+s}^{m} : fil^{i}(D ⊗ Ω̂_{B}^{s}) ⟶ fil^{i}(I^{[r]} ⊗_{B} Ω̂_{B}^{s})

This is the inverse map of 1−f_{r+s}, hence 1−f_{r+s} : fil^{i}(I^{[r]}⊗Ω̂_{B}^{s}) → fil^{i}(D⊗Ω̂_{B}^{s}) is an isomorphism.

Documenta Mathematica 5 (2000) 151–200

Proof. By i > er, fil^{i}(I^{[r]} ⊗ Ω̂_{B}^{s}) = fil^{i}(D ⊗ Ω̂_{B}^{s}) because X^{i} = r!X^{i−er}(X^{e})^{[r]}. All elements of fil^{i} D ⊗ Ω̂_{B}^{s} can be written as the sum of the elements of the form X^{n}(X^{e})^{[j]}ω, where ω ∈ D ⊗ Ω̂_{B}^{s} and n + ej ≥ i. Now r < p, thus (X^{e})^{[r]} = X^{er}/r! in D, hence we may assume j ≥ r. The image of X^{n}(X^{e})^{[j]}ω is

∑_{m=0}^{∞} f_{r+s}^{m}(X^{n}(X^{e})^{[j]}ω) = ∑_{m=0}^{∞} \frac{(p^{m}j)!}{p^{rm}(j!)}X^{np^{m}}(X^{e})^{[p^{m}j]} \frac{f^{m}(ω)}{p^{sm}}.

Here, f^{m}(ω) is divisible by p^{sm} by (11). The coefficients (p^{m}j)!/p^{rm}(j!) are p-integers for all m and if j ≥ 1 then the sum converges p-adically. If j = r = 0, n ≥ 1 says that the order of the power of X is increasing. This also means the sum converges p-adically in D ⊗_{B} Ω̂_{B}^{s}. The image is in fil^{i}(I^{[r]} ⊗_{B} Ω̂_{B}^{s}) because p^{m}j ≥ r for all m, thus the map is well-defined. Obviously, ∑_{m=0}^{∞} f_{r+s}^{m} is the inverse map of 1 − f_{r+s}.

Lemma 2.9. Let i ≥ 1 and e ≥ 1 be integers. For each n ≥ 0, let m_{n} (resp. m′_{n}) be the maximal integer which satisfies ip^{n} ≥ m_{n}e (resp. ip^{n} − 1 ≥ m′_{n}e). Then

Min {v_{p}(m_{n}!) + m_{n} − n}_{n}

= \begin{cases}1 − η_{i} ≤ 0 & ( when n = η_{i} − 1, if η_{i} ≥ 1) \\ v_{p}(m_{0}!) + m_{0} ≥ 1 & ( when n = 0, if η_{i} = 0) \end{cases}

Min {v_{p}(m′_{n}!) + m′_{n} − n}_{n}

= \begin{cases}1 − η′_{i} ≤ 0 & ( when n = η′_{i} − 1, if η′_{i} ≥ 1) \\ v_{p}(m′_{0}!) + m′_{0} ≥ 1 & ( when n = 0, if η′_{i} = 0), \end{cases}

where η_{i} and η′_{i} are as in (2.6).

Proof. By the definition of {m_{n}}_{n}, m_{n+1} is greater than or equal to pm_{n}. Thus v_{p}(m′_{n+1}!) ≥ v_{p}(pm′_{n}!) and

\begin{matrix}v_{p}(m_{n+1}!) + m_{n+1} − (n + 1) − (v_{p}(m_{n}!) + m_{n} − n) \\ = v_{p}(m_{n+1}!) − v_{p}(m_{n}!) + m_{n+1} − m_{n} − 1 \end{matrix}  (13)

is greater than zero if m_{n} > 0. On the other hand, η_{i} is the number which has the property that if n < η_{i}, then m_{n} = 0 and m_{η_{i}} ≥ 1. Thus the value of (13) is less than zero if and only if n < η_{i}. Hence the minimum of v_{p}(m_{n}!) + m_{n} − n is the value when n = η_{i} − 1 if η > 0 and n = 0 if η_{i} = 0. The rest of the desired equation comes from the same way.

Proof of Proposition 2.6. At first, we show that {H^{1}(fil^{i} 𝕊_{q})}_{i} forms the finite decreasing filtration of H^{1}(𝕊_{q}). See

gr^{i} 𝕊_{q} =

[\begin{matrix}gr^{i} I^{[2]} ⊗ Ω̂_{B}^{q−2} & ⟶^{d} & gr^{i} I ⊗ Ω̂_{B}^{q−1} & ⟶^{d} & gr^{i} D ⊗ Ω̂_{B}^{q} & ⟶^{d} & ⋯ \\ ↓^{1−f_{q}} & & ↓^{1−f_{q}} & & ↓^{1−f_{q}} \\ gr^{i} D ⊗ Ω̂_{B}^{q−2} & ⟶^{d} & gr^{i} D ⊗ Ω̂_{B}^{q−1} & ⟶^{d} & gr^{i} D ⊗ Ω̂_{B}^{q} & ⟶^{d} & ⋯ \end{matrix} ] .

(14)

If i ≥ 1, all vertical arrows of (14) are equal to 1. Thus they are injections by the definition of the filtration. Especially, the injectivity of the first vertical arrow gives H^{0}(gr^{i} 𝕊_{q}) = 0, this means

0 ⟶ H^{1}(fil^{i+1} 𝕊_{q}) ⟶ H^{1}(fil^{i} 𝕊_{q}) ⟶ H^{1}(gr^{i} 𝕊_{q})  (15)

is exact. If i = 0, the first vertical arrow of (14) is 1−f_{q}: p^{2}Ω̂_{A_{0}}^{q−2} → Ω̂_{A_{0}}^{q−2}. This is also injective because of the invariance of the valuation of A_{0} by the action of f. Thus the exact sequence (15) also follows when i = 0. Hence {H^{1}(fil^{i} 𝕊_{q})}_{i} forms a decreasing filtration of H^{1}(𝕊_{q}).

Next we calculate H^{1}(gr^{i} 𝕊_{q}). If i > 2e, fil^{i} 𝕊_{q} is acyclic by (2.8). Thus we only consider the case i ≤ 2e. Furthermore, if i ≥ 1, we may consider that H^{1}(gr^{i} 𝕊_{q}) is the subgroup of gr^{i} D ⊗ Ω̂_{B}^{q−2} because of the injectivity of the vertical arrows of (14).

Let i = 2e. Then gr^{2e} 𝕊_{q} is

[\begin{matrix}X^{2e}Ω̂_{A_{0}}^{q−2} ⊕ pX^{2e−1}dX ∧ Ω̂_{A_{0}}^{q−3} & ⟶^{d} & X^{2e}Ω̂_{A_{0}}^{q−1} ⊕ X^{2e−1}dX ∧ Ω̂_{A_{0}}^{q−2} & ⟶^{d} & ⋯ \\ ↓^{1} & ↓^{1} \\ X^{2e}Ω̂_{A_{0}}^{q−2} ⊕ X^{2e−1}dX ∧ Ω̂_{A_{0}}^{q−3} & ⟶^{d} & X^{2e}Ω̂_{A_{0}}^{q−1} ⊕ X^{2e−1}dX ∧ Ω̂_{A_{0}}^{q−2} & ⟶^{d} & ⋯ \end{matrix}] .

The second vertical arrow is a surjection, thus

H^{1}(gr^{2e} 𝕊_{q}) ≅ X^{2e−1}dX ∧ (Ω̂_{A_{0}}^{q−3}/p).  (16)

Let e < i < 2e. Then gr^{2e} 𝕊_{q} is

[\begin{matrix}pX^{i}Ω̂_{A_{0}}^{q−2} ⊕ pX^{i−1}dX ∧ Ω̂_{A_{0}}^{q−3} & ⟶^{d} & X^{i}Ω̂_{A_{0}}^{q−1} ⊕ X^{i−1}dX ∧ Ω̂_{A_{0}}^{q−2} & ⟶^{d} & ⋯ \\ ↓^{1} & ↓^{1} \\ X^{i}Ω̂_{A_{0}}^{q−2} ⊕ X^{i−1}dX ∧ Ω̂_{A_{0}}^{q−3} & ⟶^{d} & X^{i}Ω̂_{A_{0}}^{q−1} ⊕ X^{i−1}dX ∧ Ω̂_{A_{0}}^{q−2} & ⟶^{d} & ⋯ \end{matrix} ] .

The second vertical arrow is also a surjection, thus

H^{1}(gr^{i} 𝕊_{q}) ≅ X^{i}(Ω̂_{A_{0}}^{q−2}/p) ⊕ X^{i−1}dX ∧ (Ω̂_{A_{0}}^{q−3}/p).  (17)

Let i = e. Then gr^{e} 𝕊_{q} is

[\begin{matrix}pX^{e}Ω̂_{A_{0}}^{q−2} ⊕ p^{2}X^{e−1}dX ∧ Ω̂_{A_{0}}^{q−3} & ⟶^{d} & X^{e}Ω̂_{A_{0}}^{q−1} ⊕ pX^{e−1}dX ∧ Ω̂_{A_{0}}^{q−2} & ⟶^{d} & ⋯ \\ ↓^{1} & ↓^{1} \\ X^{e}Ω̂_{A_{0}}^{q−2} ⊕ X^{e−1}dX ∧ Ω̂_{A_{0}}^{q−3} & ⟶^{d} & X^{e}Ω̂_{A_{0}}^{q−1} ⊕ X^{e−1}dX ∧ Ω̂_{A_{0}}^{q−2} & ⟶^{d} & ⋯ \end{matrix} ] .

The second vertical arrow is not a surjection. For an element X^{e}ω ∈ X^{e}Ω̂_{A_{0}}^{q−2}, d(X^{e}ω) is included in X^{e}Ω̂_{A_{0}}^{q−1} ⊕ pX^{e−1}dX ∧ Ω̂_{A_{0}}^{q−2} if and only if p | e or p | ω. For an element X^{e−1}ω ∧ dX ∈ X^{e−1}dX ∧ Ω̂_{A_{0}}^{q−3}, d(X^{e−1}ω ∧ dX) is included in X^{e}Ω̂_{A_{0}}^{q−1} ⊕ pX^{e−1}dX ∧ Ω̂_{A_{0}}^{q−2} if and only if p | dω. Thus we have

H^{1}(gr^{e} 𝕊_{q}) ≅ \begin{cases}X^{e}(Ω̂_{A_{0}}^{q−2}/p) ⊕ X^{e−1}dX ∧ (ℨ_{1}Ω̂_{A_{0}}^{q−3}/p^{2}Ω̂_{A_{0}}^{q−3}) & ( if p | e) \\ X^{e−1}dX ∧ (ℨ_{1}Ω̂_{A_{0}}^{q−3}/p^{2}Ω̂_{A_{0}}^{q−3}) & ( if p ∤ e). \end{cases}

(18)

Let 1 ≤ i < e. Then gr^{i} 𝕊_{q} is

[\begin{matrix}p^{2}X^{i}Ω̂_{A_{0}}^{q−2} ⊕ p^{2}X^{i−1}dX ∧ Ω̂_{A_{0}}^{q−3} & ⟶^{d} & pX^{i}Ω̂_{A_{0}}^{q−1} ⊕ pX^{i−1}dX ∧ Ω̂_{A_{0}}^{q−2} & ⟶^{d} & ⋯ \\ ↓^{1} & ↓^{1} \\ X^{i}Ω̂_{A_{0}}^{q−2} ⊕ X^{i−1}dX ∧ Ω̂_{A_{0}}^{q−3} & ⟶^{d} & X^{i}Ω̂_{A_{0}}^{q−1} ⊕ X^{i−1}dX ∧ Ω̂_{A_{0}}^{q−2} & ⟶^{d} & ⋯ \end{matrix}].

The image of X^{i}ω ∈ X^{i}Ω̂_{A_{0}}^{q−2} is included in pX^{i}Ω̂_{A_{0}}^{q−1} ⊕ pX^{i−1}dX ∧ Ω̂_{A_{0}}^{q−2} if and only if p | iω and p | dω, and the image of X^{i−1}dX ∧ ω ∈ X^{i−1}dX ∧ Ω̂_{A_{0}}^{q−3} is included in pX^{i}Ω̂_{A_{0}}^{q−1} ⊕ pX^{i−1}dX ∧ Ω̂_{A_{0}}^{q−2} if and only if p | dω. Thus

H^{1}(gr^{i} 𝕊_{q}) ≅

X^{i} (ℨ_{1}Ω̂_{A_{0}}^{q−2}/p^{2}Ω̂_{A_{0}}^{q−2}) ⊕ X^{i−1}dX ∧ (ℨ_{1}Ω̂_{A_{0}}^{q−3}/p^{2}Ω̂_{A_{0}}^{q−3})  ( if p | i)
X^{i}(pΩ̂_{A_{0}}^{q−2}/p^{2}Ω̂_{A_{0}}^{q−2}) ⊕ X^{i−1}dX ∧ (ℨ_{1}Ω̂_{A_{0}}^{q−3}/p^{2}Ω̂_{A_{0}}^{q−3})  ( if p ∤ i).

(19)

If i = 0, we need more calculation. The complex gr^{0} 𝕊_{q} is

[\begin{matrix}p^{2}Ω̂_{A_{0}}^{q−2} & ⟶^{d} & pΩ̂_{A_{0}}^{q−1} & ⟶^{d} & Ω̂_{A_{0}}^{q} & ⟶^{d} & ⋯ \\ ↓^{1−f_{q}} & ↓^{1−f_{q}} & ↓^{1−f_{q}} \\ Ω̂_{A_{0}}^{q−2} & ⟶^{d} & Ω̂_{A_{0}}^{q−1} & ⟶^{d} & Ω̂_{A_{0}}^{q} & ⟶^{d} & ⋯ \end{matrix} ] .

We introduce a p-adic filtration to gr^{0} 𝕊_{q} as follows.

fil_{p}^{m}(gr^{0} 𝕊_{q}) = [\begin{matrix}p^{2+m}Ω̂_{A_{0}}^{q−2} & ⟶^{d} & p^{1+m}Ω̂_{A_{0}}^{q−1} & ⟶^{d} & p^{m}Ω̂_{A_{0}}^{q} & ⟶^{d} & ⋯ \\ ↓^{1−f_{q}} & & ↓^{1−f_{q}} & & ↓^{1−f_{q}} \\ p^{m}Ω̂_{A_{0}}^{q−2} & ⟶^{d} & p^{m}Ω̂_{A_{0}}^{q−1} & ⟶^{d} & p^{m}Ω̂_{A_{0}}^{q} & ⟶^{d} & ⋯ \end{matrix} ] .

Then, for all m ≥ 0,

gr_{p}^{m}(gr^{0} 𝕊_{q}) = [\begin{matrix}Ω_{F}^{q−2} & ⟶^{0} & Ω_{F}^{q−1} & ⟶^{0} & Ω_{F}^{q} & ⟶ & ⋯ \\ ↓_{− C^{−1}} & ↓_{− C^{−1}} & ↓_{1−C^{−1}} \\ Ω_{F}^{q−2} & ⟶^{d} & Ω_{F}^{q−1} & ⟶^{d} & Ω_{F}^{q} & ⟶ & ⋯ \end{matrix} ] .  (20)

The injectivity of the leftmost vertical arrow of (20) says that

H^{0}(gr_{p}^{m}(gr^{0} 𝕊_{q})) = 0

for all m ≥ 0. Thus {H^{1}(fil_{p}^{m}(gr^{0} 𝕊_{q}))}_{m} is a decreasing filtration of H^{1}(gr^{0} 𝕊_{q}). On the other hand, the intersection of the image of − C^{−1} : Ω_{F}^{q−1} → Ω_{F}^{q−1} and the image of d: Ω_{F}^{q−2} → Ω_{F}^{q−1} = B_{1}^{q−1} is {0} by (1). Thus we also have H^{1}(gr_{p}^{m}(gr^{0} 𝕊_{q})) = 0 for all m ≥ 0. Hence we have H^{1}(gr^{0} 𝕊_{q}) = 0.

We already have known H^{1}(gr^{i} 𝕊_{q}) for all i ≥ 0, but the third arrow of (15) is not surjective in general. So we must know the image of H^{1}(fil^{i} 𝕊_{q}) → H^{1}(gr^{i} 𝕊_{q}). Let i ≥ 1 and let x be an element of fil^{i} D ⊗ Ω̂_{B}^{q−2} which represents an element of H^{1}(gr^{i} 𝕊_{q}). H^{1}(fil^{i} 𝕊_{q}) is

H^{1} [\begin{matrix}fil^{i} I^{[2]} ⊗ Ω̂_{B}^{q−2} & ⟶^{d} & fil^{i} I ⊗ Ω̂_{B}^{q−1} & ⟶^{d} & fil^{i} D ⊗ Ω̂_{B}^{q} & ⟶^{d} & ⋯ \\ ↓^{1−f_{q}} & & ↓^{1−f_{q}} & & ↓^{1−f_{q}} \\ fil^{i} D ⊗ Ω̂_{B}^{q−2} & ⟶^{d} & fil^{i} D ⊗ Ω̂_{B}^{q−1} & ⟶^{d} & fil^{i} D ⊗ Ω̂_{B}^{q} & ⟶^{d} & ⋯ \end{matrix} ] .

Now the second vertical arrow is an injection. Thus x also represents the element of H^{1}(fil^{i} 𝕊_{q}) if and only if

∑_{n=0}^{∞} f_{q}^{n}(dx) ∈ fil^{i} I ⊗ Ω̂_{B}^{q−1}.  (21)

The elements of H^{1}(gr^{i} 𝕊_{q}) are represented by two types of the elements of D ⊗ Ω̂_{B}^{q−2}, these are X^{i}ω for ω ∈ Ω̂_{A_{0}}^{q−2} and X^{i−1}dX ∧ ω for ω ∈ Ω̂_{A_{0}}^{q−3}. Thus we must know the condition when (21) follows for these elements.

At first, we calculate X^{i}ω for ω ∈ Ω̂_{A_{0}}^{q−2}.

∑_{n=0}^{∞} f_{q}^{n}(dX^{i}ω)

= ∑_{n=0}^{∞} f_{q}^{n}(X^{i}dω + iX^{i−1}dX ∧ ω)

= ∑_{n=0}^{∞} (\frac{1}{p^{nq}} X^{ip^{n}}f^{n}(dω) + \frac{ip^{n}}{p^{nq}} X^{ip^{n}−1}dX ∧ f^{n}(ω))

= ∑_{n=0}^{∞} (\frac{m_{n}!}{p^{n}}X^{ip^{n}−m_{n}e}(X^{e})^{[m_{n}]} \frac{f^{n}(dω)}{p^{n(q−1)}}

+\frac{i(m′_{n}!)}{p^{n}}X^{ip^{n}−1−m′_{n}e}(X^{e})^{[m′_{n}]}dX ∧ \frac{f^{n}(ω)}{p^{n(q−2)}} ) .

Here m_{n} and m′_{n} be the maximal integers which satisfy ip^{n} − m_{n}e ≥ 0 and ip^{n} − 1 − m′_{n}e ≥ 0. Note that f^{n}(dω) and f^{n}(ω) can be divided by p^{n(q−1)} and p^{n(q−2)}, respectively, by (11). Furthermore, v_{p}(f^{n}(dω)/p^{n(q−1)}) = v_{p}(dω) and v_{p}(f^{n}(ω)/p^{n(q−2)}) = v_{p}(ω) by (12). To be included in I ⊗ Ω̂_{B}^{q−1}, the sum of the p-adic order and divided power degree must be greater than or equal to 1, i.e., v_{p}(m_{n}!) − n + m_{n} + v_{p}(dω) ≥ 1 and v_{p}(i) + v_{p}(m′_{n}!) − n + m′_{n} + v_{p}(ω) ≥ 1 must be satisfied for all n. We already know the minimal of v_{p}(m_{n}!) − n + m_{n} and v_{p}(m′_{n}!) − n + m′_{n} by (2.9), thus ∑_{n=0}^{∞} f_{q}^{n}(dX^{i}ω) belongs to I⊗_{B}^{q−1} if and only if

\begin{cases}no condition & ( if e + 1 ≤ i) \\ v_{p}(i) + v_{p}(ω) ≥ 1 & ( if e = i) \\ v_{p}(dω) ≥ η_{i} and v_{p}(i) + v_{p}(ω) ≥ η′_{i} & ( if 1 ≤ i < e). \end{cases}  (22)

Next, we calculate X^{i−1}dX ∧ ω for ω ∈ Ω̂_{A_{0}}^{q−3}.

∑_{n=0}^{∞} f_{q}^{n}(dX^{i−1}dX ∧ ω)

= ∑_{n=0}^{∞} f_{q}^{n}(X^{i−1}dX ∧ dω)

= ∑_{n=0}^{∞} (\frac{p^{n}}{p^{nq}} X^{ip^{n}−1}dX ∧ f^{n}(dω))

= ∑_{n=0}^{∞} (\frac{(m′_{n}!)}{p^{n}}X^{ip^{n}−1−m′_{n}e}(X^{e})^{[m′_{n}]}dX ∧ \frac{f^{n}(dω)}{p^{n(q−2)}}) .

To be included in I ⊗ Ω̂_{B}^{q−1}, v_{p}(m′_{n}!) − n + m′_{n} + v_{p}(dω) ≥ 1 must be satisfied for all n. As the same way as above, ∑_{n=0}^{∞} f_{q}^{n}(X^{i−1}dX ∧ ω) belongs to I⊗_{B}^{q−1} if and only if

\begin{cases}no condition & ( if e + 1 ≤ i) \\ v_{p}(dω) ≥ η′_{i} & ( if 1 ≤ i ≤ e). \end{cases}  (23)

For ω ∈ Ω̂_{A_{0}}^{q−1}, the condition v_{p}(ω) ≥ n means ω ∈ p^{n}Ω̂_{A_{0}}^{q−1} and v_{p}(dω) ≥ n means ω ∈ ℨ_{n}Ω̂_{A_{0}}^{q−1}. Thus, by (16), (17), (18), (19), (22) and (23), we get (2.6).

3 Differential modules and filtrations

Let K, A, A_{0}, K_{0} and B are as in Section 2. We assume that p ∤ e = e_{K}, i.e., K/K_{0} is tamely totally ramified extension from here. Let k be the constant field of K (cf. [18]), i.e., k is the complete discrete valuation subfield of K with the restriction of the valuation of K, algebraically closed in K, and the residue field of k is the maximal perfect subfield of F. Then there exists a prime element of K such that π is the element of k. Let k_{0} = K_{0} ∩ k. Then π is algebraic over k_{0} and we get Ω̂_{O_{k_{0}}}^{1} = 0, where O_{k_{0}} is the ring of integers of

k_{0}. Thus π^{e−1}dπ = 0 in Ω̂_{A}^{1} by taking the differential of the minimal equation of π over k_{0}.

By the equation π^{e−1}dπ = 0 in Ω̂_{A}^{1}, we have

\begin{matrix}Ω̂_{A}^{q} & ≅ & ( ⨁_{i_{1}<i_{2}<···<i_{q}} A\frac{dT_{i_{1}}}{T_{i_{1}}} ∧ ⋯ ∧ \frac{dT_{i_{q}}}{T_{i_{q}}} ) \\ ⊕ ( ⨁_{i_{1}<i_{2}<···<i_{q−1}} A/(π^{e−1})\frac{dT_{i_{1}}}{T_{i_{1}}} ∧ ⋯ ∧ \frac{dT_{i_{q−1}}}{T_{i_{q−1}}} ∧ dπ) , \end{matrix}  (24)

where {T_{i}} = 𝕋. We introduce a filtration on Ω̂_{A}^{q} as follows. Let

fil^{i} Ω̂_{A}^{q} = \begin{cases}Ω̂_{A}^{q} & ( if i = 0) \\ π^{i}Ω̂_{A}^{q} + π^{i−1}dπ ∧ Ω̂_{A}^{q−1} & ( if i ≥ 1). \end{cases}

The subquotients are

gr^{i} Ω̂_{A}^{q} = fil^{i} Ω̂_{A}^{q}/ fil^{i+1} Ω̂_{A}^{q}

= \begin{cases}Ω_{F}^{q} & ( if i = 0 or i ≥ e) \\ Ω_{F}^{q} ⊕ Ω_{F}^{q−1} & ( if 1 ≤ i < e), \end{cases}

where the map is

\begin{matrix}π^{i}Ω̂_{A}^{q} ∋ π^{i}ω ⟼ ω̄ ∈ Ω_{F}^{q} \\ π^{i−1}dπ ∧ Ω̂_{A}^{q−1} ∋ π^{i−1}dπ ∧ ω ⟼ ω̄ ∈ Ω_{F}^{q−1}. \end{matrix}

Let fil^{i}(Ω̂_{A}^{q}/pdΩ̂_{A}^{q−1}) be the image of fil^{i} Ω̂_{A}^{q} in Ω̂_{A}^{q}/pdΩ̂_{A}^{q−1}. Then we have the following

Proposition 3.1. For j ≥ 0,

gr^{j} (Ω̂_{A}^{q}/pdΩ̂_{A}^{q−1}) = \begin{cases}Ω_{F}^{q} & (j = 0) \\ Ω_{F}^{q} ⊕ Ω_{F}^{q−1} & (1 ≤ j < e) \\ Ω_{F}^{q} /B_{l}^{q} & (e ≤ j), \end{cases}

where l be the maximal integer which satisfies j − le ≥ 0.

Proof. If 1 ≤ j < e, gr^{j} Ω̂_{A}^{q} = gr^{j}(Ω̂_{A}^{q}/pdΩ̂_{A}^{q−1}) because pdΩ̂_{A}^{q−1} ⊂ fil^{e} Ω̂_{A}^{q}. Assume that j ≥ e and let l as above. By (24) and pi^{e}dπ = 0, Ω̂_{A}^{q−1} is generated by the elements pπ^{i}dω for 0 ≤ i < e and ω ∈ Ω̂_{A_{0}}^{q−1}. By [7] (Cor. 2.3.14), pπ^{i}dω ∈ fil^{e(1+n)+i} Ω̂_{A}^{q} if and only if the residue class of p^{−n}dω belongs to B_{n+1}. Thus gr^{j}(Ω̂_{A}^{q}/pdΩ̂_{A}^{q−1}) ≅ Ω_{F}^{q} /B_{l}^{q}.

168 Jinya Nakamura

We need the lemma in the following sections. Lemma 3.2. (i) For n ≥ 0, there exist maps

f_{q}^{n} = \frac{f^{n}}{p^{nq}} : Ω̂_{A_{0}}^{q} ⟶ ℨ_{n}Ω̂_{A_{0}}^{q}.

(ii) For n ≥ 1,

ℨ_{n}Ω̂_{A_{0}}^{q} = (∑_{l=0}^{n−1} p^{l}f_{q}^{n−l}Ω̂_{A_{0}}^{q}) + p^{n}Ω̂_{A_{0}}^{q} + ℨ_{∞}Ω̂_{A_{0}}^{q}.

(iii) For n ≥ 1,

⨁_{i=0}^{n−1} \frac{d}{p^{i}} ∘ f_{q−1}^{i}: (Ω̂_{A_{0}}^{q−1}/ℨ_{1}Ω̂_{A_{0}}^{q−1})^{⊕n} ⟶ Ω̂_{A_{0}}^{q}/p ≅ Ω_{F}^{q}

is injective and the image is B_{n}Ω_{F}^{q}. (iv) For any n ≥ 0,

Ω̂_{A_{0}}^{q}/ℨ_{1}Ω̂_{A_{0}}^{q} ⟶^{f_{q}^{n}} ℨ_{n}Ω̂_{A_{0}}^{q}/(ℨ_{n+1}Ω̂_{A_{0}}^{q} + ℨ_{n}Ω̂_{A_{0}}^{q} ∩ pΩ̂_{A_{0}}^{q})

is an isomorphism.

(v) For any n ≥ 0,

(Ω̂_{A_{0}}^{q}/p) ⊕ (Ω̂_{A_{0}}^{q−1}/ℨ_{1}Ω̂_{A_{0}}^{q−1})^{⊕n} \underrightarrow{f_{q}^{n}⊕⨁_{i=0}^{n−1} \frac{d}{p^{i}} ∘f_{q−1}^{i}} \frac{ℨ_{n}Ω̂_{A_{0}}^{q}}{ℨ_{n}Ω̂_{A_{0}}^{q} ∩ pΩ̂_{A_{0}}^{q}}

is an isomorphism. Proof. (i) By (12), f^{n}(ω) belongs to p^{nq}Ω̂_{A_{0}}^{q}. Ω̂_{A_{0}}^{q} is p-torsion free, thus f_{q}^{n} is well-defined as the map to Ω̂_{A_{0}}^{q}. Furthermore,

d(f_{q}^{n}(ω)) = \frac{1}{p^{nq}} f^{n}(dω) = p^{n}f_{q+1}^{n}(dω),

thus f_{q}^{n}(ω) ∈ ℨ_{n}Ω̂_{A_{0}}^{q}.

(ii) For 0 ≤ l ≤ n − 1, the image of the natural injection

\frac{ℨ_{n}Ω̂_{A_{0}}^{q} ∩ p^{l}Ω̂_{A_{0}}^{q}}{(ℨ_{n}Ω̂_{A_{0}}^{q} ∩ p^{l+1}Ω̂_{A_{0}}^{q}) + (ℨ_{∞}Ω̂_{A_{0}}^{q} ∩ p^{l}Ω̂_{A_{0}}^{q})}

⟶ \frac{p^{l}Ω̂_{A_{0}}^{q}}{p^{l+1}Ω̂_{A_{0}}^{q} + (ℨ_{∞}Ω̂_{A_{0}}^{q} ∩ p^{l}Ω̂_{A_{0}}^{q})} ≅ Ω_{F}^{q} /Z_{∞}Ω_{F}^{q}

Documenta Mathematica 5 (2000) 151–200

is coincide with Z_{n−l}Ω_{F}^{q} /Z_{∞}Ω_{F}^{q} by [7] (Cor. 3.2.14). The image of p^{l}f_{q}^{n−l}Ω̂_{A_{0}}^{q} is also Z_{n−l}Ω_{F}^{q} /Z_{∞}Ω_{F}^{q} for all l, thus the natural projection

(∑_{l=0}^{n−1} p^{l}f_{q}^{n−l}Ω̂_{A_{0}}^{q}) ⟶ \frac{ℨ_{n}Ω̂_{A_{0}}^{q}}{p^{n}Ω̂_{A_{0}}^{q} + ℨ_{∞}Ω̂_{A_{0}}^{q}}

is surjective. Hence we have (ii).

(iii) The following diagram commute

(Ω̂_{A_{0}}^{q−1}/ℨ_{1}Ω̂_{A_{0}}^{q−1})^{⊕n} \underrightarrow{⨁_{i=0}^{n−1} \frac{d}{p^{i}} ∘f_{q−1}^{i}} Ω̂_{A_{0}}^{q}/p
≅↓  ≅↓
(Ω_{F}^{q−1}/Z_{1}^{q−1})^{⊕n}  \underrightarrow{⨁_{i=0}^{n−1} C^{−i} d} Ω_{F}^{q} .

The image of the bottom arrow is B_{n}^{q}.

(iv) The image of ℨ_{n}Ω̂_{A_{0}}^{q}/(ℨ_{n}Ω̂_{A_{0}}^{q} ∩ pΩ̂_{A_{0}}^{q}) under the isomorphism Ω̂_{A_{0}}^{q}/p → Ω_{F}^{q} is Z_{n}^{q} by [7] (Cor. 3.2.14). (iv) follows from the diagram

Ω̂_{A_{0}}^{q}/ℨ_{1}Ω̂_{A_{0}}^{q} ⟶^{f_{q}^{n}} ℨ_{l}Ω̂_{A_{0}}^{q}/(ℨ_{n+1}Ω̂_{A_{0}}^{q} + ℨ_{n}Ω̂_{A_{0}}^{q} ∩ pΩ̂_{A_{0}}^{q})
≅↓  ≅↓
Ω_{F}^{q} /Z_{1}^{q}  ⟶^{C^{−n}}  Z_{n}^{q}/Z_{n+1}^{q}.

(v) The image of

(Ω̂_{A_{0}}^{q−1}/ℨ_{1}Ω̂_{A_{0}}^{q−1})^{⊕n} \underrightarrow{⨁_{i=0}^{n−1} \frac{d}{p^{i}} ∘f_{q−1}^{i}} Ω̂_{A_{0}}^{q}/p ≅ Ω_{F}^{q}

is B_{n}^{q} by (iii), and the image of the composite

Ω̂_{A_{0}}^{q}/p ⟶^{f_{q}^{n}} Ω̂_{A_{0}}^{q}/p ≅ Ω_{F}^{q} → Ω_{F}^{q} /B_{n}^{q}

is Z_{n}^{q}/B_{n}^{q}. Hence we get (v).

4 The image of H^{1}(𝕊_{q})

We assume p ∤ e. We further assume that there exists the prime element π of K such that π^{e} = p. If there does not exist such π, we replace K by K(p^{\frac{1}{e}}). Note that the extension K(p^{\frac{1}{e}})/K is unramified of degree prime to p. In this section, we calculate ψ(H^{1}(𝕊_{q})) explicitly. We need some preparations.

Let N_{0}^{q} be the subset of Ω̂_{A_{0}}^{q} such that the canonical map N_{0}^{q} → Ω_{F}^{q} \ Z_{1}^{q} is an injection, the image generates Ω_{F}^{q} /Z_{1}^{q} and have the property

If ω̄ + C^{−1}ω̄ = 0, then dω = 0.  (25)

We can take such N_{0}^{q} because of the following Lemma 4.1. Take x ∈ Ω̂_{F}^{q}. If x + C^{−1} x = 0 then there exists ω ∈ Ω̂_{A_{0}}^{q} such that ω̄ = x and dω = 0.

Proof. x can be written as

x = ∑_{τ} x_{τ} τ,

where τ runs through the canonical generators (cf. in the proof of (2.7)) and x_{τ} ∈ F. The assumption x + C^{−1} x = 0 means that x_{τ} + x_{τ}^{p} = 0 for all τ, thus x_{τ} ∈ E for all τ, where E is the maximal perfect subfield of F. The canonical generators have the fixed lifts denoted by τ̃, and we can take lifts of x_{τ}, denoted by x̃_{τ}, in the ring of Witt vectors with coefficients in E, denotes W (E). Fix an inclusion W (E) → A_{0}. Let

ω = ∑_{τ} x̃_{τ} τ̃.

Then dω = 0 in Ω̂_{A_{0}}^{q} because dx̃_{τ} = 0 in Ω̂_{A_{0}}^{q} and ω̄ = x. This ω is the desired one.

For any q, l ≥ 0, let N_{l}^{q} = f_{q}^{l}(N_{0}^{q}) as a subset of Ω̂_{A_{0}}^{q} and let

\begin{matrix}N_{∞}^{q} = ℨ_{∞}Ω̂_{A_{0}}^{q} \ (ℨ_{∞}Ω̂_{A_{0}}^{q} ∩ pΩ̂_{A_{0}}^{q}), \\ N_{f}^{q} = ⋃_{l≥0} N_{l}^{q}, N^{q} = N_{f}^{q} ∪ N_{∞}^{q}. \end{matrix}

Then, by (3.2,iv), N^{q} generates Ω̂_{A_{0}}^{q}/p and ω ≠ 0 in Ω̂_{A_{0}}^{q}/p for all ω ∈ N^{q}. Furthermore, by using (3.2,v) and the isomorphism

\frac{ℨ_{n−1}Ω̂_{A_{0}}^{q}}{ℨ_{n−1}Ω̂_{A_{0}}^{q} ∩ pΩ̂_{A_{0}}^{q}} ⟶^{p} \frac{ℨ_{n}Ω̂_{A_{0}}^{q} ∩ pΩ̂_{A_{0}}^{q}}{ℨ_{n}Ω̂_{A_{0}}^{q} ∩ p^{2}Ω̂_{A_{0}}^{q}} ,

we have

\begin{matrix}⟨f_{q}^{n}N^{q} ∪ ⋃_{m=0}^{n−1} \frac{d}{p^{m}} f_{q−1}^{m}N_{0}^{q−1}⟩ = \frac{ℨ_{n}Ω̂_{A_{0}}^{q}}{ℨ_{n}Ω̂_{A_{0}}^{q} ∩ pΩ̂_{A_{0}}^{q}} , \\ ⟨pf_{q}^{n−1}N^{q} ∪ ⋃_{m=0}^{n−2} p\frac{d}{p^{m}} f_{q−1}^{m}N_{0}^{q−1}⟩ = \frac{ℨ_{n}Ω̂_{A_{0}}^{q} ∩ pΩ̂_{A_{0}}^{q}}{ℨ_{n}Ω̂_{A_{0}}^{q} ∩ p^{2}Ω̂_{A_{0}}^{q}} . \end{matrix}  (26)

Thus the union of the sets of the left hand side of (26) generates ℨ_{n}Ω̂_{A_{0}}^{q}/(ℨ_{n}Ω̂_{A_{0}}^{q} ∩ p^{2}Ω̂_{A_{0}}^{q}). If q < 0 then let N_{l}^{q} = ∅.

Let S_{i,1}^{0}, S_{i,1}^{1}, S_{i,2}^{0} and S_{i,2}^{1} be the subsets of D ⊗ Ω̂_{B}^{q−2} defined as follows.

S_{i,1}^{0} = \begin{cases}∅ & (i = 0, e or i ≥ 2e) \\ X^{i}N^{q−2} & (if e < i < 2e) \\ ∅ & (if 1 ≤ i < e, η_{i} − v_{p}(i) ≥ 1) \\ X^{i} (f_{q−2}^{η_{i}}N^{q−2} ∪ ⋃_{m=0}^{η_{i}−1} \frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3}) & (if 1 ≤ i < e, η_{i} − v_{p}(i) ≤ 0), \end{cases}

S_{i,1}^{1} = \begin{cases}∅ & (i = 0 or i ≥ e ) \\ ∅ & (1 ≤ i < e, η_{i} − v_{p}(i) ≥ 2) \\ X^{i} (pf_{q−2}^{η_{i}−1}N^{q−2} ∪ ⋃_{m=0}^{η_{i}−2} p\frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3}) & (1 ≤ i < e, η_{i} − v_{p}(i) ≤ 1), \end{cases}

S_{i,2}^{0} = \begin{cases}∅ & (i = 0 or i > 2e) \\ X^{i−1}dX ∧ N^{q−3} & (e < i ≤ 2e) \\ X^{e−1}dX ∧ (f_{q−3}^{1}N^{q−3} ∪ dN_{0}^{q−4}) & (i = e) \\ X^{i−1}dX ∧ (f_{q−3}^{η_{i}}N^{q−3} ∪ ⋃_{m=0}^{η_{i}−1} \frac{d}{p^{m}} f_{q−4}^{m}N_{0}^{q−4}) & (1 ≤ i < e), \end{cases}

S_{i,2}^{1} = \begin{cases}∅ & (i = 0 or i > e) \\ X^{e−1}dX ∧ pN^{q} & (if i = e) \\ X^{i−1}dX ∧ (pf_{q−3}^{η_{i}−1}N^{q−3} ∪ ⋃_{m=0}^{η_{i}−2} p\frac{d}{p^{m}} f_{q−4}^{m}N_{0}^{q−4}) & (if 1 ≤ i < e). \end{cases}

Let S_{i,1} = S_{i,1}^{0} ∪S_{i,1}^{1}, S_{i,2} = S_{i,2}^{0} ∪S_{i,2}^{1}, S_{i} = S_{i,1} ∪S_{i,2} and S the union of all S_{i}. By the above definitions, S_{i} generates gr^{i} H^{1}(𝕊_{q}), hence S generates H^{1}(𝕊_{q}).

The following lemma is useful to calculate ψ.

Lemma 4.2. If 1 ≤ i < e then the minimal value of v_{K}(π^{ip^{n}}/p^{n+1}) = ip^{n} − e(n + 1) is

ip^{η_{i}−1} − eη_{i}  ( when n = η_{i} − 1; if e′ < ip^{η_{i}} < ep)
ip^{η_{i}} − e(η_{i} + 1)  ( when n = η_{i}; if e < ip^{η_{i}} < e′)
ip^{η_{i}−1} − eη_{i}  ( when n = η_{i} − 1, η_{i}; if < ip^{η_{i}} = e′)

and if e < i then the minimal value of v_{K}(π^{ip^{n}}/p^{n+1}) is i − e.

Proof. Lemma follows from the definition of η_{i}.

Remark 4.3. Method of calculation of ψ. In (2.6) and in the definition of S, we use elements of D ⊗ Ω̂_{B}^{q−2}, which is the degree zero part of the complex σ_{>q−3}𝔻[q − 2], to represent elements of H^{1}(𝕊_{q}). Chasing the complex (6) and the map (8), ψ is the composite of

\begin{matrix}D ⊗ Ω̂_{B}^{q−2} ⟶^{d} D ⊗ Ω̂_{B}^{q−1} \underrightarrow{∑_{n≥0} f_{q}^{n}} I ⊗ Ω̂_{B}^{q−1} ⟶^{I→pA} pA ⊗ Ω̂_{B}^{q−1} \\ ⟶^{p^{−1}} A ⊗ Ω̂_{B}^{q−1} ⟶^{dX=dπ} Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}. \end{matrix}

Thus, for ω ∈ Ω̂_{A_{0}}^{q−2} (resp. ω ∈ Ω̂_{A_{0}}^{q−3}) and i ≥ 1,

\begin{matrix}ψ(X^{i}ω) = ∑_{n≥0} (\frac{i}{p^{n+1}} π^{ip^{n}} \frac{dπ}{π} ∧ f_{q−2}^{n}(ω) + \frac{1}{p^{n+1}} π^{ip^{n}}f_{q−1}^{n}(dω)) \\ (resp. ψ (X^{i} \frac{dX}{X} ∧ ω) = ∑_{n≥0} \frac{1}{p^{n+1}} π^{ip^{n}} \frac{dπ}{π} ∧ f_{q−2}^{n}(dω)) . \end{matrix}  (27)

Here, to avoid the complication of notations, we use

X^{i} \frac{dX}{X} ( resp. π^{i} \frac{dπ}{π} )

which only denotes the meaning of X^{i−1}dX (resp. π^{i−1}dπ) when i ≥ 1. By using (4.2), n = η_{i} − 1 or n = η_{i} is the number at which the value v_{K} of the coefficients of dπ in (27) is the minimal. If X^{i}ω ∈ S (resp. X^{i−1}dX ∧ω ∈ S) for ω ∈ Ω̂_{A_{0}}^{q−2} (resp. ω ∈ Ω̂_{A_{0}}^{q−3}), then ω has the property (22) (resp. (23)). Under this condition, the right hand side of (27) belongs to Ω̂_{A}^{q−1}. Furthermore, by π^{e−1}π = 0, if η_{i} ≥ 1 then

ψ(X^{i}ω) = \frac{i}{p^{η_{i}}} π^{ip^{η_{i}−1}} \frac{dπ}{π} ∧ f_{q−2}^{η_{i}−1}(ω) + \frac{i}{p^{η_{i}+1}} π^{ip^{η_{i}}} \frac{dπ}{π} ∧ f_{q−2}^{η_{i}}(ω)

+ ∑_{n≥0} (\frac{1}{p^{n+1}} π^{ip^{n}}f_{q−1}^{n}(dω)) ,

ψ(X^{i} \frac{dX}{X} ∧ ω) = \frac{1}{p^{η_{i}}} π^{ip^{η_{i}−1}} \frac{dπ}{π} ∧ f_{q−2}^{η_{i}−1}(dω)\frac{1}{p^{η_{i}+1}} π^{ip^{η_{i}}} \frac{dπ}{π} ∧ f_{q−2}^{η_{i}}(dω),

and if η_{i} = 0 then

\begin{aligned}ψ(X^{i}ω) & = \frac{i}{p}π^{i} \frac{dπ}{π} ∧ ω + ∑_{n≥0} (\frac{1}{p^{n+1}} π^{ip^{n}}f_{q−1}^{n}(dω)) , \\ ψ(X^{i} \frac{dX}{X} ∧ ω) & = \frac{1}{p}π^{i} \frac{dπ}{π} ∧ dω. \end{aligned}

Note that if η_{i} ≥ 1,

v_{K} (\frac{1}{p^{η_{i}}} π^{ip^{η_{i}−1}}) − v_{K} (\frac{1}{p^{η_{i}+1}} π^{ip^{η_{i}}} ) \begin{cases}< 0 & ( if e′ < ip^{η_{i}} < ep) \\ > 0 & ( if e < ip^{η_{i}} < e′) \\ = 0 & ( if ip^{η_{i}} = e′). \end{cases} (28)

By the definition, S generates H^{1}(𝕊_{q}). But ψ: H^{1}(𝕊_{q}) → Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2} has the kernel in general. The following lemma compute some subset of this kernel.

Lemma 4.4. (i) S_{2e}, S_{e} ⊂ Ker ψ.

(ii) If e < i < 2e then ψ(S_{i,2}^{0} \ (X^{i−1}dX ∧ N_{0}^{q−3})) = 0. If 1 ≤ i < e then ψ(S_{i,2}^{0} \ (X^{i−1}dX ∧ f_{q−3}^{η_{i}}N_{0}^{q−3})) = 0 and ψ(S_{i,2}^{1} \ (X^{i−1}dX ∧ pf_{q−3}^{η_{i}−1}N_{0}^{q−3})) = 0.

(iii) If e < i < 2e and p ∤ i, then ψ(S_{i,2}) ⊂ ⟨ψ(S_{i,1})⟩.

(iv) If e′ < i < 2e and p | i, then

ψ(S_{i,2}) ⊂ ⟨ψ ( ⋃_{1≤j<e} S_{j,1}^{1})⟩ .

(v) Let 1 ≤ i < e, s = η_{i} + v_{p}(i) and i_{0} = i/p^{v_{p}(i)}. If e′ < ip^{η_{i}} < ep and s ≥ 2, then

ψ(S_{i,2}) ⊂ ⟨( ⋃_{1≤j<e} S_{j,1}) ∪ S_{ip^{η_{i}} −e,1}⟩ .

Furthermore, let

j = \begin{cases}i_{0}p^{\frac{s}{2}} & ( if s is even ), \\ i_{0}p^{\frac{s−1}{2}} & ( if s is odd ). \end{cases}

Then

\begin{matrix}ψ(S_{i,2}^{0}) ⊂ ⟨ψ(S_{j,1})⟩ & if 3η_{i} ≥ v_{p}(i), \\ ψ(S_{i,2}^{1}) ⊂ ⟨ψ(S_{j,1})⟩ & if 3η_{i} ≥ v_{p}(i) + 2. \end{matrix}

Proof. (i) Take X^{2e−1}dX ∧ ω ∈ S_{2e,2}. Then

ψ (X^{2e} \frac{dX}{X} ∧ ω) = \frac{1}{p}π^{2e} \frac{dπ}{π} ∧ dω = 0

by (4.3). Next, take X^{e−1}dX ∧ ω ∈ S_{e,2}. By the definition of S_{e,2}, such an ω can be divided by p. Thus, by using (4.3), we get

ψ (X^{e} \frac{dX}{X} ∧ ω) = \frac{1}{p}π^{e} \frac{dπ}{π} ∧ p\frac{dω}{p} = 0.

(ii) At first, let e < i < 2e. When we take X^{i−1}dX ∧ω from S_{i,2}^{0} \(X^{i−1}dX ∧ N_{0}^{q−3}), then ω has the property v_{p}(dω) ≥ 1. Thus by using (4.3),

ψ (X^{i} \frac{dX}{X} ∧ ω) = \frac{1}{p}π^{i} \frac{dπ}{π} ∧ p\frac{dω}{p} = 0.

Next let 1 ≤ i < e. When we take X^{i−1}dX ∧ ω from

(S_{i,2}^{0} \ (X^{i} \frac{dX}{X} ∧ f_{q−3}^{η_{i}}N_{0}^{q−3})) ∪ (S_{i,2}^{1} \ (X^{i} \frac{dX}{X} ∧ pf_{q−3}^{η_{i}−1}N_{0}^{q−3})) ,

ω has the property v_{p}(dω) ≥ η_{i} + 1. Thus ψ(X^{i−1}dX ∧ ω) = 0 by using (4.3).

(iii) In this case, S_{i,2} = X^{i−1}dX ∧N^{q−3} and S_{i,1} = X^{i}N^{q−2}. For an element X^{i−1}dX ∧ ω ∈ S_{i,2} with ω ∈ N^{q−3}, there exists X^{i}dω ∈ S_{i,1} because dω ∈ N_{∞}^{q−2}, and

d (X^{i} \frac{dX}{X} ∧ ω) = d (\frac{X^{i}dω}{i} ) .

This means ψ(X^{i−1}dX ∧ ω) = ψ(X^{i}dω)/i. Thus ψ(S_{i,2}) ⊂ ⟨ψ(S_{i,1})⟩.

(iv) Take an element X^{i−1}dX ∧ ω ∈ S_{i,2} with ω ∈ N^{q−3}. Let j = j_{0} = i − e and j_{l} = j_{l−1}p − e for j ≥ 1. Then, {j_{l}}_{l} have the property

\begin{matrix}p ∤ j_{l}, \\ \frac{e}{p − 1} < j_{0} < j_{1} < j_{2} < … \end{matrix}

by p | i and i > e′. Let L be the minimal integer such that j_{L} ≥ 2e/p. Then η_{j_{l}} = 1 for all 0 ≤ l ≤ L. There exist the elements X^{j_{l}}pf_{q−2}^{l}(dω) ∈ S_{j_{l},1}^{1} because S_{j_{l},1}^{1} = X^{j_{l}}N^{q−2} and pf_{q−2}^{l}(dω) ∈ N_{∞}^{q−2}. Thus the element, denoted by Y,

Y = ∑_{l=0}^{L} \frac{(−1)^{l}}{j_{l}} X^{j_{l}}pf_{q−2}^{l}(dω)

exists in ⟨⋃_{k=1}^{e−1} S_{k,1}^{1}⟩. By (4.3), ψ(X^{i−1}dX ∧ω) = π^{i−e−1}dπ ∧dω. On the other hand,

ψ(Y) = ∑_{l=0}^{L} ((−1)^{l}π^{j_{l}} \frac{dπ}{π} ∧ f_{q−2}^{l}(dω) + (−1)^{l} \frac{1}{p}π^{j_{l}p} \frac{dπ}{π} ∧ f_{q−2}^{l+1}(dω))

= ∑_{l=0}^{L} ((−1)^{l}π^{j_{l}} \frac{dπ}{π} ∧ f_{q−2}^{l}(dω) + (−1)^{l}π^{j_{l+1}} \frac{dπ}{π} ∧ f_{q−2}^{l+1}(dω))

= π^{j_{0}} \frac{dπ}{π} ∧ dω.

The third equation follows from j_{L+1}−1 ≥ e−1. Hence ψ(X^{i−1}dX∧ω) = ψ(Y) because j_{0} = i − e, and we get (iv).

(v) Now S_{i,2}^{0} and S_{i,2}^{1} are

\begin{aligned}S_{i,2}^{0} & = X^{i} \frac{dX}{X} ∧ (f_{q−3}^{η_{i}}N^{q−3} ∪ ⋃_{m=0}^{η_{i}−1} \frac{d}{p^{m}} f_{q−4}^{m}N_{0}^{q−4}) , \\ S_{i,2}^{1} & = X^{i} \frac{dX}{X} ∧ (pf_{q−3}^{η_{i}−1}N^{q−3} ∪ ⋃_{m=0}^{η_{i}−2} p\frac{d}{p^{m}} f_{q−4}^{m}N_{0}^{q−4}) . \end{aligned}

By (ii), we only have to calculate the element of

X^{i} \frac{dX}{X} ∧ f_{q−3}^{η_{i}}N_{0}^{q−3}, X^{i} \frac{dX}{X} ∧ pf_{q−3}^{η_{i}−1}N_{0}^{q−3}

to show (v). If e′ < ip^{η_{i}} < ep and s ≥ 2, then

X^{i} \frac{dX}{X} ∧ f_{q−3}^{η_{i}}ω = π^{ip^{η_{i}−1}} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−1}(dω)

+ π^{ip^{η_{i}} −e} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}}(dω),

X^{i} \frac{dX}{X} ∧ pf_{q−3}^{η_{i}−1}ω = π^{ip^{η_{i}−1}} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−2}(dω)

+ π^{ip^{η_{i}} −e} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−1}(dω).

The first terms of the right hand side come from

S_{ip^{η_{i}−1}+e,1} ⊃X^{ip^{η_{i}−1}+e}N_{∞}^{q−2}

∋ X^{ip^{η_{i}−1}+e}f_{q−2}^{2η_{i}−1}(dω) ⟶^{ψ} eπ^{ip^{η_{i}−1}} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−1}(dω),

S_{ip^{η_{i}−1}+e,1} ⊃X^{ip^{η_{i}−1}+e}N_{∞}^{q−2}

∋ X^{ip^{η_{i}−1}+e}f_{q−2}^{2η_{i}−2}(dω) ⟶^{ψ} eπ^{ip^{η_{i}−1}} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−2}(dω).

On the other hand, the second terms of the right hand side are, if ip^{η_{i}} ≥ 2e then vanished. If ip^{η_{i}} < 2e, then by using the same argument of (iv) with j_{0} = ip^{η_{i}} − e and

Y = \begin{cases}∑_{l=0}^{L} \frac{(−1)^{l}}{j_{l}} X^{j_{l}}pf_{q−2}^{l}(f_{q−2}^{2η_{i}}(dω)) & ( the first case ) \\ ∑_{l=0}^{L} \frac{(−1)^{l}}{j_{l}} X^{j_{l}}pf_{q−2}^{l}(f_{q−2}^{2η_{i}−1}(dω)) & ( the second case ), \end{cases}

we get

ψ(S_{i,2}) ⊂ ⟨( ⋃_{1≤j<e} S_{j,1}) ∪ S_{ip^{η_{i}} −e,1}⟩ .

Next, we do not assume e < ip^{η_{i}} < e′ and s ≥ 2. In this case, we have to show

\begin{matrix}ψ (X^{i} \frac{dX}{X} ∧ f_{q−3}^{η_{i}}N_{0}^{q−3}) ⊂ ψ(S_{j,1}) & if 3η_{i} ≥ v_{p}(i), \\ ψ (X^{i} \frac{dX}{X} ∧ pf_{q−3}^{η_{i}−1}N_{0}^{q−3}) ⊂ ψ(S_{j,1}) & if 3η_{i} ≥ v_{p}(i) + 2. \end{matrix}

Take ω ∈ N_{0}^{q−3}. Then, by (4.3),

ψ (X^{i} \frac{dX}{X} ∧ f_{q−3}^{η_{i}}(ω))

= π^{ip^{η_{i}−1}} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−1} (dω) + π^{ip^{η_{i}} −e} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}} (dω) ,

ψ (X^{i} \frac{dX}{X} ∧ pf_{q−3}^{η_{i}−1}(ω))

= π^{ip^{η_{i}−1}} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−2} (dω) + π^{ip^{η_{i}} −e} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−1} (dω) .

On the other hand, there exist elements

ω′_{1} = f_{q−2}^{2η_{i}− \frac{s}{2}} (dω)  ( if s is even and 3η_{i} ≥ v_{p}(i)),
ω′_{2} = pf_{q−2}^{2η_{i}− \frac{s+1}{2}} (dω) ( if s is odd and 3η_{i} ≥ v_{p}(i)),
ω′_{3} = f_{q−2}^{2η_{i}− \frac{s}{2} −1}(dω) ( if s is even and 3η_{i} ≥ v_{p}(i) + 2),
ω′_{4} = pf_{q−2}^{2η_{i}− \frac{s+1}{2} −1}(dω) ( if s is odd and 3η_{i} ≥ v_{p}(i) + 2)

in Ω̂_{A_{0}}^{q−2} because the conditions are, 2η_{i} ≥ s/2 if and only if 3η_{i} ≥ v_{p}(i) when s is even, 2η_{i} ≥ (s+1)/2 if and only if 3η_{i} ≥ v_{p}(i) when s is odd, 2η_{i} ≥ (s/2)+1 if and only if 3η_{i} ≥ v_{p}(i) + 2 when s is even, and 2η_{i} ≥ ((s + 1)/2) + 1 if and only if 3η_{i} ≥ v_{p}(i) + 2 when s is odd. The image of ψ of an each element is

ψ(X^{j}ω′_{1}) = i_{0}π^{i_{0}p^{s−1}} \frac{dπ}{π} ∧ f_{q−2}^{\frac{s}{2} −1}(ω′_{1}) + \frac{i_{0}}{p} π^{i_{0}p^{s}} \frac{dπ}{π} ∧ fq−2^{\frac{s}{2}}(ω′_{1})

= i_{0}π^{ip^{η_{i}−1}} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−1}(dω) + i_{0}π^{ip^{η_{i}} −e} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}}(dω)

= i_{0}ψ (X^{i} \frac{dX}{X} ∧ f_{q−3}^{η_{i}}(ω)) ,

ψ(X^{j}ω′_{2}) = \frac{i_{0}}{p} π^{i_{0}p^{s−1}} \frac{dπ}{π} ∧ fq−2^{\frac{s−1}{2}} (ω′_{2}) + \frac{i_{0}}{p^{2}} π^{i_{0}p^{s}} \frac{dπ}{π} ∧ fq−2^{\frac{s+1}{2}} (ω′_{2})

= i_{0}π^{ip^{η_{i}−1}} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−1}(dω) + i_{0}π^{ip^{η_{i}} −e} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}}(dω)

= i_{0}ψ (X^{i} \frac{dX}{X} ∧ f_{q−3}^{η_{i}}(ω)) ,

ψ(X^{j}ω′_{3}) = i_{0}π^{i_{0}p^{s−1}} \frac{dπ}{π} ∧ f_{q−2}^{\frac{s}{2} −1}(ω′_{3}) + \frac{i_{0}}{p} π^{i_{0}p^{s}} \frac{dπ}{π} ∧ fq−2^{\frac{s}{2}}(ω′_{3})

= i_{0}π^{ip^{η_{i}−1}} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−2}(dω) + i_{0}π^{ip^{η_{i}} −e} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−1}(dω)

= i_{0}ψ (X^{i} \frac{dX}{X} ∧ pf_{q−3}^{η_{i}−1}(ω)) ,

ψ(X^{j}ω′_{4}) = \frac{i_{0}}{p} π^{i_{0}p^{s−1}} \frac{dπ}{π} ∧ fq−2^{\frac{s−1}{2}} (ω′_{4}) + \frac{i_{0}}{p^{2}} π^{i_{0}p^{s}} \frac{dπ}{π} ∧ fq−2^{\frac{s+1}{2}} (ω′_{4})

= i_{0}π^{ip^{η_{i}−1}} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−2}(dω) + i_{0}π^{ip^{η_{i}} −e} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−1}(dω)

= i_{0}ψ (X^{i} \frac{dX}{X} ∧ pf_{q−3}^{η_{i}−1}(ω)) .

Compare the definition of S_{j,1} with the condition of ω′_{1}, …, ω′_{4}. If s is even and 3η_{i} ≥ v_{p}(i) then

X^{j}ω′_{1} = \begin{cases}X^{j} \frac{d}{p^{2η_{i}− \frac{s}{2}}} f_{q−2}^{2η_{i}− \frac{s}{2}} (dω) ∈ X^{j} \frac{d}{p^{2η_{i}− \frac{s}{2}}} f_{q−2}^{2η_{i}− \frac{s}{2}} N_{0}^{q−3} & (if η_{i} − \frac{s}{2} ≤ η_{j} − 1), \\ X^{j}f_{q−2}^{η_{j}}f_{q−2}^{2η_{i}− \frac{s}{2} −η_{j}}(dω) ∈ X^{j}f_{q−2}^{η_{j}}N_{∞}^{q−2} & (if η_{i} − \frac{s}{2} ≥ η_{j}). \end{cases}

Thus X^{j}ω′_{1} ∈ S_{i,1}^{0}. By the similar way, we have

X^{j}ω′_{2} ∈ S_{j,1}^{1} ( if s is odd and 3η_{i} ≥ v_{p}(i)),
X^{j}ω′_{3} ∈ S_{j,1}^{0} ( if s is even and 3η_{i} ≥ v_{p}(i) + 2),
X^{j}ω′_{4} ∈ S_{j,1}^{1} ( if s is odd and 3η_{i} ≥ v_{p}(i) + 2).

The claim (v) was proved.

Remark 4.5. Let S_{i,2}^{′0} (resp. S_{i,2}^{′1}) be the subset of S_{i,2}^{0} (resp. S_{i,2}^{1} ) defined as follows.

S_{i,2}^{′0} = \begin{cases}X^{i} \frac{dX}{X} ∧ N_{0}^{q−3} & (e < i ≤ e′, p | i) \\ X^{i} \frac{dX}{X} ∧ f_{q−3}^{η_{i}}N_{0}^{q−3} & 1 ≤ i < e, e < ip^{η_{i}} ≤ e′, 3η_{i} < v_{p}(i)) \\ ∅ & ( otherwise ), \end{cases}

S_{i,2}^{′1} = \begin{cases}X^{i} \frac{dX}{X} ∧ pf_{q−3}^{η_{i}−1}N_{0}^{q−3} & (1 ≤ i < e, e < ip^{η_{i}} ≤ e′, 3η_{i} < v_{p}(i) + 2) \\ ∅ & ( otherwise ). \end{cases}

Remark that if 1 ≤ i < e satisfies v_{p}(i) + η_{i} = 1 and e′ < ip^{η_{i}} < ep, then v_{p}(i) = 0, e/(p − 1) < i < e and η_{i} = 1. Thus this i satisfies neither 3η_{i} < v_{p}(i) + 2 nor 3η_{i} < v_{p}(i). Let S′_{i,2} = S_{i,2}^{′0} ∪ S_{i,2}^{′1}. Then by (4.4), ψ(H^{1}(𝕊_{q})) is generated by

( ⋃_{1≤i<2e} S_{i,1}) ∪ ( ⋃_{1≤i<2e} S′_{i,2}) .

We need some modification of generators of ψ(H^{1}(𝕊_{q})) as follows.

Let the index sets Λ_{0} and Λ_{1} be

Λ_{0} = {i ; 1 ≤ i < e, e′ < ip^{η_{i}} < 2e, η_{i} = v_{p}(i)} ,  (29)

Λ_{1} = {i ; 1 ≤ i < e, e′ < ip^{η_{i}} < 2e, η_{i} = v_{p}(i) + 1, p ∤ (i + e)} .  (30)

Let Λ = Λ_{0} ∪ Λ_{1}. For i ∈ Λ, let

\begin{matrix}s = η_{i} + v_{p}(i), \\ i′ = i/p^{v_{p}(i)}, \\ i_{0} = i′p^{s−1}, \\ i_{l} = i_{l−1}p − e for l ≥ 1, \\ L = Min\{l ; i_{l} ≥ 2e/p\}. \end{matrix}  (31)

{i_{l}}_{l} are monotonely increasing, thus we can take such L. Note that p ∤ i′, η_{i_{l}} = 1 for 0 ≤ l ≤ L and p ∤ i_{l} for l ≥ 1. If i ∈ Λ_{0} then let g_{i,0} be

g_{i,0}(X^{i}ω) = \frac{1}{i′} X^{i}ω − \frac{1}{i_{0} + e}X^{i_{0}+e}f_{q−2}^{η_{i}−1}(ω) + ∑_{l=1}^{L−1} \frac{(−1)^{l}}{i_{l}} pX^{i_{l}}f_{q−2}^{η_{i}+l−1}(ω)

for ω ∈ ℨ_{η_{i}} Ω̂_{A_{0}}^{q−2}. This function satisfies g_{i,0}(ω) ≡ (1/i′)X^{i}ω modulo fil^{i+1} H^{1}(𝕊_{q}), thus we can replace S_{i,1}^{0} by g_{i,0}(S_{i,1}^{0}) to generate ψ(H^{1}(𝕊_{q})). When i ∈ Λ_{1}, then let g_{i,1} be

g_{i,1}(X^{i}pω) = \frac{1}{i′} X^{i}pω − \frac{1}{i_{0} + e}X^{i_{0}+e}f_{q−2}^{η_{i}−1}(ω) + ∑_{l=1}^{L−1} \frac{(−1)^{l}}{i_{l}} pX^{i_{l}}f_{q−2}^{η_{i}+l−1}(ω)

for pω ∈ pΩ̂_{A_{0}}^{q−2}∩ℨ_{η_{i}} Ω̂_{A_{0}}^{q−2}. This function satisfies g_{i,1}(pω) ≡ (1/i′)X^{i}pω modulo fil^{i+1} H^{1}(𝕊_{q}), thus we can replace S_{i,1}^{1} by g_{i,1}(S_{i,1}^{1}) to generate ψ(H^{1}(𝕊_{q})).

5 Explicit Calculation, Case (a)

We compute ψ(S_{i,1}), ψ(S′_{i,2}), ψ(g_{i,0}S_{i,1}^{0}) and ψ(g_{i,1}S_{i,1}^{1}) explicitly in Section 5, 6 and 7.

Define the index sets as

Γ_{a} = \{i | 1 ≤ i < e, \frac{e}{p − 1} < ip^{η_{i}−1} < e\} ∪ \{i | e′ < i < 2e\} ,

Γ_{b} = {i | 1 ≤ i < e, e < ip^{η_{i}} < e′} ∪ {i | e < i < e′} ,

Γ_{c} = \{\frac{e}{p − 1}, e′\} .

These sets are disjoint to each other, and Γ_{a} ∪ Γ_{b} ∪ Γ_{c} is coincide with {i; 1 ≤ i < 2e, i ≠ e}. Λ is the subset of Γ_{a}. In this section, we compute ψ(S_{i,1} ∪ S′_{i,2}) for i ∈ Γ_{a} \ Λ, ψ(g_{i,0}(S_{i,1}^{0}) ∪ S_{i,1}^{1} ∪ S′_{i,2}) for i ∈ Λ_{0} and ψ(g_{i,1}(S_{i,1}^{1}) ∪ S′_{i,2}) for i ∈ Λ_{1}. We compute ψ(S_{i,1} ∪ S′_{i,2}) when i ∈ Γ_{b} in Section 6 and when i ∈ Γ_{c} in Section 7.

At first, we compute ψ when i ∈ Γ_{a} and 1 ≤ i < e. Let e/(p − 1) < j < e, s = v_{p}(j) + 1 and j_{0} = j/p^{s−1}. Then the integers i which satisfy ip^{η_{i}−1} = j are

(i, η_{i}) = (j_{0}, s), (j_{0}p, s − 1), …, (j_{0}p^{s−1}, 1).

Let i = j_{0}p^{t}. Then i ∈ Γ_{a} for all t. Notice that if i ∈ Γ_{a} and i < e then there exists e/(p − 1) < j < e such that ip^{η_{i}−1} = j.

If t < \frac{s−1}{2} then S_{i,1} = ∅.

If t = (s − 1)/2 and p | (j + e), then i ∈ Λ_{1}, S_{i,1}^{0} = ∅ and

S_{i,1}^{1} = X^{i} (pf_{q−2}^{η_{i}−1}N^{q−2} ∪ ⋃_{m=0}^{η_{i}−2} p\frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3}) .

For X^{i}pω ∈ S_{i,1}^{1}, ψ(g_{i,1}(X^{i}pω)) is

\begin{aligned}ψ(g_{i,1}(X^{i}pω)) & = ∑_{n≥0} \frac{pπ^{ip^{n}}}{i′p^{n+1}} f_{q−1}^{n}(dω) \\ & − ∑_{n≥0} \frac{p^{η_{i}−1}}{(i_{0} + e)p^{n+1}} π^{(i_{0}+e)p^{n}}f_{q−1}^{η_{i}+n−1}(dω) \\ & + ∑_{l=1}^{L−1} ∑_{n≥0} \frac{(−1)^{l}p^{η_{i}+l−1}}{i_{l}p^{n}} π^{i_{l}p^{n}}f_{q−1}^{η_{i}+l+n−1}(dω) \end{aligned}  (32)

by using (4.3) and the same kind of calculation in (4.4,iv) with the notation (31). If pω ∈ pf_{q−2}^{η_{i}−1}N_{∞}^{q−2} or pω ∈ p\frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3} for some m, then ψ(g′_{i}(X^{i}pω)) = 0 by (32). If pω ∈ pf_{q−2}^{η_{i}−1}N_{f}^{q−2}, then take l ≥ 0 and f_{q−2}^{l}ω′ ∈ N_{l}^{q−2} for ω′ ∈ N_{0}^{q−2} such that ω = f_{q−2}^{η_{i}+l−1}(ω′). For this ω′, we have

ψ(g_{i,1}(X^{i}pf_{q−2}^{η_{i}+l−1}(ω′))) ≡ \begin{cases}\frac{p^{l}}{i′} π^{ip^{η_{i}−1}}f_{q−1}^{2η_{i}+l−2}(dω′) & ( if η_{i} ≠ 1) \\ \frac{ep^{l}}{i(i+e)}π^{i}f_{q−1}^{l}(dω′) & ( if η_{i} = 1) \end{cases}
mod fil^{ip^{η_{i}−1}+el+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}).

Now ip^{η_{i}−1} = j and η_{i} = s − t = (s + 1)/2,

ψ(g_{i,1}(X^{i}pf_{q−2}^{η_{i}+l−1}(ω′))) ≡ \begin{cases}\frac{p^{l}}{i′} π^{j}f_{q−1}^{s+l−1}(dω′) & ( if s − t > 1) \\ \frac{ep^{l}}{i(i+e)}π^{j}f_{q−1}^{l}(dω′) & ( if t = 0, s = 1) \end{cases}
mod fil^{j+el+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}).

(33)

If s = 1 and p | (j + e), then t can be taken only 0. S_{i,1}^{0} = ∅. This i is not in Λ_{1}, hence we compute S_{i,1}^{1} without g_{i,1}. Now S_{i,1}^{1} = X^{i}pN^{q−2}. For X^{i}pω ∈ X^{i}pN^{q−2},

\begin{aligned}ψ(X^{i}pω) & = iπ^{i−1}dπ ∧ ω + iπ^{ip−e−1}dπ ∧ f_{q−2}(ω) \\ & + ∑_{n≥0} \frac{1}{p^{n}} π^{ip^{n}} ∧ f_{q−1}^{n}(dω) \\ & ≡ iπ^{i−1}dπ ∧ ω + π^{i} ∧ dω \\ & mod fil^{i+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). \end{aligned}  (34)

If t = s/2, then i ∈ Λ_{0} and

S_{i,1}^{0} = X^{i} (f_{q−2}^{η_{i}}N^{q−2} ∪ ⋃_{m=0}^{η_{i}−1} p\frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3}) .

For X^{i}ω ∈ S_{i,1}^{0}, ψ(g_{i,0}(X^{i}ω)) is

\begin{aligned}ψ(g_{i,0}(X^{i}ω)) & = ∑_{n≥0} \frac{π^{ip^{n}}}{i′p^{n+1}} f_{q−1}^{n}(dω) \\ & − ∑_{n≥0} \frac{1}{(i_{0} + e)p^{n+1}} π^{(i_{0}+e)p^{n}}f_{q−1}^{n}(df_{q−2}^{η_{i}−1}(ω)) \\ & + ∑_{l=1}^{L−1} ∑_{n≥0} \frac{(−1)^{l}p^{η_{i}+l−1}}{i_{l}p^{n}} π^{i_{l}p^{n}}f_{q−1}^{η_{i}+l+n−1}(dω). \end{aligned}  (35)

If ω ∈ f_{q−2}^{η_{i}}N_{∞}^{q−2} or ω ∈ \frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3} for some m, then ψ(g_{i,0}(X^{i}ω)) = 0 by (35). If ω ∈ f_{q−2}^{η_{i}}N_{f}^{q−2}, then take l ≥ 0 and f_{q−2}^{l}ω′ ∈ N_{l}^{q−2} for ω′ ∈ N_{0}^{q−2} such that ω = f_{q−2}^{η_{i}+l}(ω′). For this ω′, we have

\begin{aligned}ψ(g_{i,0}(X^{i}f_{q−2}^{η_{i}+l}(ω′))) & ≡ \frac{p^{l}}{i′} π^{ip^{η_{i}−1}}f_{q−1}^{2η_{i}+l−1}(dω′) \\ & ≡ \frac{p^{l}}{i′} π^{j}f_{q−1}^{s+l−1}(dω′) \\ & mod fil^{j+el+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). \end{aligned}  (36)

S_{i,1}^{1} is

S_{i,1}^{1} = X^{i} (pf_{q−2}^{η_{i}−1}N^{q−2} ∪ ⋃_{m=0}^{η_{i}−2} p\frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3}) .

For X^{i}pω ∈ S_{i,1}^{1}, ψ(X^{i}pω) is

ψ(X^{i}pω) = ∑_{n≥0} \frac{p^{η_{i}−1}}{p^{n}} π^{ip^{n}}f_{q−1}^{n} (\frac{dω}{p^{η_{i}−1}} ) .  (37)

Documenta Mathematica 5 (2000) 151–200

Thus if ω ∈ f_{q−2}^{η_{i}−1}N_{∞}^{q−2} or ω ∈ \frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3} for some m, then ψ(X^{i}pω) = 0. If ω ∈ f_{q−2}^{η_{i}−1}N_{f}^{q−2}, let l ≥ 0 and f_{q−2}^{l}ω′ ∈ N_{l}^{q−2} for ω′ ∈ N_{0}^{q−2} such that ω = f_{q−2}^{η_{i}+l}(ω′). For this ω′, we have

\begin{aligned}ψ(X^{i}pf_{q−2}^{η_{i}+l−1}(ω′)) & ≡ p^{l}π^{ip^{η_{i}−1}}f_{q−1}^{2η_{i}+l−2}(dω′) \\ & ≡ p^{l}π^{j}f_{q−1}^{s+l−2}(dω′) \\ & mod fil^{j+el+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). \end{aligned}  (38)

If t > s/2, then i ∉ Λ and

S_{i,1}^{0} = X^{i} (f_{q−2}^{η_{i}}N^{q−2} ∪ ⋃_{m=0}^{η_{i}−1} p\frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3}) .

For X^{i}ω ∈ S_{i,1}^{0}, ψ(X^{i}ω) is

ψ(X^{i}ω) = ∑_{n≥0} \frac{p^{η_{i}}}{p^{n+1}} π^{ip^{n}}f_{q−1}^{n} (\frac{dω}{p^{η_{i}}} ) .  (39)

If ω ∈ f_{q−2}^{η_{i}}N_{∞}^{q−2} or ω ∈ \frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3} for some m, then ψ(X^{i}ω) = 0. If ω ∈ f_{q−2}^{η_{i}}N_{f}^{q−2}, then take l ≥ 0 and f_{q−2}^{l}ω′ ∈ N_{l}^{q−2} for ω′ ∈ N_{0}^{q−2} such that ω = f_{q−2}^{η_{i}+l}(ω′). For this ω′, we have

\begin{aligned}ψ(X^{i}f_{q−2}^{η_{i}+l}(ω′)) & ≡ p^{l}π^{ip^{η_{i}−1}}f_{q−1}^{2η_{i}+l−1}(dω′) \\ & ≡ p^{l}π^{j}f_{q−1}^{2s−2t+l−1}(dω′) \\ & mod fil^{j+el+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). \end{aligned}  (40)

When we take X^{i}pω ∈ S_{i,1}^{1}, by the same calculation of the case t = s/2, we have

\begin{aligned}ψ(X^{i}pf_{q−2}^{η_{i}+l−1}(ω′)) & ≡ p^{l}π^{ip^{η_{i}−1}}f_{q−1}^{2η_{i}+l−2}(dω′) \\ & ≡ p^{l}π^{j}f_{q−1}^{2s−2t+l−2}(dω′) \\ & mod fil^{j+el+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). \end{aligned}  (41)

When i ∈ Γ_{a} and i < e then S′_{i,2} = ∅.

Next, we compute i ∈ Γ_{a} and e < i. In this case S′_{i,2} = ∅, thus we only have to compute S_{i,1}. Let e/(p − 1) < j < e and i = j + e. Then S_{i,1}^{1} = ∅ and S_{i,1}^{0} = X^{i}N^{q−2}. For an element X^{i}ω ∈ S_{i,1}^{0},

ψ(X^{i}ω) = iπ^{i−e} \frac{dπ}{π} ∧ ω + ∑_{n≥0} \frac{1}{p^{n+1}} π^{ip^{n}}f_{q−1}^{n}(dω).

If p | i, then the first term of the right hand side is zero. Hence if ω = f_{q−2}^{l}(ω′) then

\begin{aligned}ψ(X^{i}ω) & ≡ p^{l}π^{i−e}f_{q−1}^{l}(dω′) \\ & mod fil^{j+el+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}) \end{aligned}  (42)

and if ω ∈ N_{∞}^{q−2} then ψ(X^{i}ω) = 0. If p ∤ i, then

\begin{aligned}ψ(X^{i}ω) & ≡ iπ^{i−e} \frac{dπ}{π} ∧ ω + π^{i−e}dω \\ & mod fil^{j+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). \end{aligned}  (43)

We have computed all S_{i} or substitutes of S_{i} for i ∈ Γ_{a} as above. Next, we construct the sets {M^{j}}_{j≥0} which are rearrangements of the generators of ψ(H^{1}(𝕊_{q})). The law of rearrangement is, for example, as follows. See (33). For an element g_{i,1}(X^{i}pf_{q−2}^{η_{i}+l−1}(ω′)), the image of ψ is

\begin{aligned}ψ(g_{i,1}(X^{i}pf_{q−2}^{η_{i}+l−1}(ω′))) & ≡ \frac{p^{l}}{i′} π^{j}f_{q−1}^{s+l−1}(dω′) \\ & mod fil^{j+el+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}) \end{aligned}

when s − t > 1. Thus this element goes to gr^{j+el}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}) and it seems non-zero. So we put g_{i,1}(X^{i}pf_{q−2}^{η_{i}+l−1}(ω′)) into M^{j+el}. We will know its image is really non-zero in Section 8 but now we do not know it is true or not. We construct the set M^{j+el} by, roughly speaking, the set of the elements which come to gr^{j+el}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}) and seem non-zero. The real definition of M^{∗} is as follows.

Use (33), (34), (36), (38), (40), (41), (42) and (43) to define M^{j+el} for e/(p − 1) < j < e and l ≥ 0. Let e/(p − 1) < j < e, s = v_{p}(j) + 1 and l ≥ 0. If s = 1 then let

M^{j+el} =

g_{j,1}(X^{j}pN_{0}^{q−2}) ∪ X^{j+e}N^{q−2}  (p ∤ (j + e) and l = 0) … (33),(43)
g_{j,1}(X^{j}pf_{q−2}^{l}N_{0}^{q−2})  (p ∤ (j + e) and l ≥ 1) … (33)
X^{j}pN^{q−2} ∪ X^{j+e}N_{0}^{q−2}  (p | (j + e) and l = 0) … (34),(42)
X^{j+e}f_{q−2}^{l}N_{0}^{q−2}  (p | (j + e) and l ≥ 1) … (42).

(44)

By (3.1), gr^{j}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}) ≅ Ω_{F}^{q−1} ⊕ Ω_{F}^{q−2} and gr^{j+el}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}) ≅ Ω_{F}^{q−1}/B_{l}^{q−1} for l ≥ 1. The image of M^{j} is, if p ∤ (j + e),

\begin{matrix}Ω_{F}^{q−2}/Z_{1}^{q−2} ⟶^{ψ∘g_{j,1}X^{j}p} gr^{j}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}) ⟶^{≅} Ω_{F}^{q−1} ⊕ Ω_{F}^{q−2} \\ ω ⟼ ψ ∘ g_{j,1}X^{j}pω = \frac{e}{j + e}π^{j}dω ⟼ (\frac{e}{j + e}dω̄, 0) \end{matrix}

The Milnor K-Groups 183

and

\begin{matrix}Ω_{F}^{q−2} ⟶^{ψX^{j+e}} gr^{j}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}) ⟶^{≅} Ω_{F}^{q−1} ⊕ Ω_{F}^{q−2} \\ ω ⟼ ψ ∘ X^{j+e}ω = π^{j}dω + (j + e)π^{j} \frac{dπ}{π} ∧ ω ⟼ (dω̄, iω̄). \end{matrix}

Thus we get

ψ(x) ≠ 0 for x ∈ M^{j} in gr^{j}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}),
gr^{j}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2})/⟨ψ(M^{j})⟩ ≅ Ω_{F}^{q−1}/B_{1}^{q−1}.

(45)

The case s = 1 and p | (j + e) goes similarly to the case above. If l ≥ 1, the image of M^{l+el} in gr^{j+el}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}) ≅ Ω_{F}^{q−1}/B_{l}^{q−1} is

Ω_{F}^{q−2}/Z_{1}^{q−2} ∋ x ⟼ C^{−l} dx ∈ Ω_{F}^{q−1}/B_{l}^{q−1}

and hence non-zero. Thus ψ(x) ≠ 0 for x ∈ M^{j+el} in gr^{j+el}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}),

gr^{j+el}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2})/⟨ψ(M^{j+el})⟩ ≅ Ω_{F}^{q−1}/B_{l+1}^{q−1}.

(46)

If s is even and s ≥ 2, let

\begin{matrix}M^{j} = & X^{j+e}N^{q−2} … (43) \\ ∪ g_{j_{0}p^{\frac{s}{2}} ,0} (X^{j_{0}p^{\frac{s}{2}}} fq−2^{\frac{s}{2}}N_{0}^{q−2}) … (36) \\ ∪ X^{j_{0}p^{\frac{s}{2}}} pf_{q−2}^{\frac{s}{2} −1}N_{0}^{q−2} … (38) \\ ∪ ( ⋃_{s/2<t≤s−1} X^{j_{0}p^{t}}f_{q−2}^{s−t}N_{0}^{q−2} ∪ X^{j_{0}p^{t}}pf_{q−2}^{s−t−1}N_{0}^{q−2}) … (40),(41), \end{matrix} (47)

and

\begin{aligned}M^{j+el} & = \\ & g_{j_{0}p^{\frac{s}{2}} ,0} (X^{j_{0}p^{\frac{s}{2}}} f_{q−2}^{\frac{s}{2} +l}N_{0}^{q−2}) … (36) \\ & ∪ X^{j_{0}p^{\frac{s}{2}}} pf_{q−2}^{\frac{s}{2} +l−1}N_{0}^{q−2} … (38) \\ & ∪ ( ⋃_{s/2<t≤s−1} X^{j_{0}p^{t}}f_{q−2}^{s−t+l}N_{0}^{q−2} ∪ X^{j_{0}p^{t}}pf_{q−2}^{s−t+l−1}N_{0}^{q−2}) … (40),(41) \end{aligned} (48)

for l ≥ 1. The image of (43) is the image of

Ω_{F}^{q−1} ∋ x ⟼^{(d,i)} (dx, ix) ∈ Ω_{F}^{q−1} ⊕ Ω_{F}^{q−2}

Documenta Mathematica 5 (2000) 151–200

184 Jinya Nakamura and the image of (36), (38) and (40) is

Ω_{F}^{q−1}/Z_{1}^{q−1} ⊕ Ω_{F}^{q−1}/Z_{1}^{q−1} ⊕ ⨁_{s/2<t≤s−1}(Ω_{F}^{q−1}/Z_{1}^{q−1} ⊕ Ω_{F}^{q−1}/Z_{1}^{q−1})

\underrightarrow{C^{−(s−1)d} ⊕ C^{−(s−2)d} ⊕ ⨁_{s/2<t≤s−1}(C^{−(2s−2t−1)} d⊕C^{−(2s−2t−2)})} B_{s}^{q−1} ⊕ 0 ⊂ Ω_{F}^{q−1} ⊕ Ω_{F}^{q−2}.

Thus we get

ψ(x) ≠ 0 for x ∈ M^{j} in gr^{j}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}),
gr^{j}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2})/⟨ψ(M^{j})⟩ ≅ Ω_{F}^{q−1}/B_{s}^{q−1}.

(49)

If l ≥ 1, the image of M^{j+el} is

Ω_{F}^{q−1}/Z_{1}^{q−1} ⊕ Ω_{F}^{q−1}/Z_{1}^{q−1} ⊕ ⨁_{s/2<t≤s−1}(Ω_{F}^{q−1}/Z_{1}^{q−1} ⊕ Ω_{F}^{q−1}/Z_{1}^{q−1})

\underrightarrow{C^{−(s+l−1)} d⊕C^{−(s+l−2)} ⊕ ⨁_{s/2<t≤s−1}(C^{−(2s−2t+l−1)} ⊕ C^{−(2s−2t+l−2)})} B_{s+l}^{q−1}/B_{l}^{q−1} ⊂ Ω_{F}^{q−1}/B_{l}^{q−1}.

Thus we get

ψ(x) ≠ 0 for x ∈ M^{j+el} in gr^{j+el}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}),
gr^{j+el}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2})/⟨ψ(M^{j+el})⟩ ≅ Ω_{F}^{q−1}/B_{s+l}^{q−1}.

(50)

If s is odd and s ≥ 3, let

\begin{matrix}M^{j} = & X^{j+e}N^{q−2} … (43) \\ ∪ g_{j_{0}p^{\frac{s−1}{2}} ,1} (X^{j_{0}p^{\frac{s−1}{2}}} pf_{q−2}^{\frac{s−1}{2} −1}N_{0}^{q−2}) … (33) \\ ∪ ( ⋃_{s/2<t≤s−1} X^{j_{0}p^{t}}f_{q−2}^{s−t}N_{0}^{q−2} ∪ X^{j_{0}p^{t}}pf_{q−2}^{s−t−1}N_{0}^{q−2}) … (40),(41), \end{matrix} (51)

and

\begin{aligned}M^{j+el} & = \\ & g_{j_{0}p^{\frac{s−1}{2}} ,1} (X^{j_{0}p^{\frac{s−1}{2}}} pf_{q−2}^{\frac{s−1}{2} +l−1}N_{0}^{q−2}) … (33) \\ & ∪ ( ⋃_{s/2<t≤s−1} X^{j_{0}p^{t}}f_{q−2}^{s−t+l}N_{0}^{q−2} ∪ X^{j_{0}p^{t}}pf_{q−2}^{s−t+l−1}N_{0}^{q−2}) … (40),(41) \end{aligned} (52)

for l ≥ 1. By the similar calculation as the case s is even, we get the same results (49) and (50).

By the definition of M^{j+el},

( ⋃_{i∈Γ_{a}\Λ} S_{i,1}^{0} ∪ S_{i,1}^{1}) ∪ ( ⋃_{i∈Λ_{0}} g_{i,0}S_{i,1}^{0} ∪ S_{i,1}^{1}) ∪ ( ⋃_{i∈Λ_{1}} g_{i,1}S_{i,1}^{1}) ∪ ( ⋃_{i∈Γ_{a}} S′_{i,2})

is equal to the union of M^{j+el} for all e/(p − 1) < j < e and all l ≥ 0.

6 Explicit Calculation, Case (b)

In this section, we compute ψ(S_{i,1}) and ψ(S′_{i,2}) for i ∈ Γ_{b}.

At first, we compute ψ when i ∈ Γ_{b} and 1 ≤ i < e. Let e < j < e′, s = v_{p}(j) and j_{0} = j/p^{s}. Then the integers i which satisfy ip^{η_{i}} = j are

(i, η_{i}) = (j_{0}, s), (j_{0}p, s − 1), …, (j_{0}p^{s−1}, 1).

Let i = j_{0}p^{t}. Then i ∈ Γ_{b} for all t. Notice that if i ∈ Γ_{b} and i < e then there exists e < j < e′ such that ip^{η_{i}} = j. But if s = 0 then there is no i ∈ Γ_{b} such that i < e and ip^{η_{i}} = j. Thus we assume s ≥ 1 to calculate when i < e.

If t < \frac{s−1}{2} then S_{i,1} = ∅.

If t = (s − 1)/2, then S_{i,1}^{0} = ∅ and

S_{i,1}^{1} = X^{i} (pf_{q−2}^{η_{i}−1}N^{q−2} ∪ ⋃_{m=0}^{η_{i}−2} p\frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3}) .

For X^{i}pω ∈ S_{i,1}^{1}, ψ(X^{i}pω) is

\begin{aligned}ψ(X^{i}pω) & = j_{0}π^{j_{0}p^{s−1}} \frac{dπ}{π} ∧ fq−2^{\frac{s−1}{2}} (ω) + j_{0}π^{j−e} \frac{dπ}{π} ∧ fq−2^{\frac{s+1}{2}} (ω) \\ & + ∑_{n≥0} \frac{1}{p^{n}} π^{ip^{n}}f_{q−1}^{n}(dω) \\ & ≡ j_{0}π^{j−e} \frac{dπ}{π} ∧ fq−2^{\frac{s+1}{2}} (ω) + π^{j−e}fq−2^{\frac{s+1}{2}} (\frac{dω}{p^{\frac{s−1}{2}}} ) \\ & mod fil^{j−e+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). \end{aligned}  (53)

Note that if X^{i}pω ∈ S_{i,1}^{1} then ω ∈ ℨ_{η_{i}−1}Ω̂_{A_{0}}^{q−2}.

If t = s/2, then

S_{i,1}^{0} = X^{i} (f_{q−2}^{η_{i}}N^{q−2} ∪ ⋃_{m=0}^{η_{i}−1} \frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3})

and

S_{i,1}^{1} = X^{i} (pf_{q−2}^{η_{i}−1}N^{q−2} ∪ ⋃_{m=0}^{η_{i}−2} p\frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3}) .

For X^{i}ω ∈ S_{i,1}^{0}, ψ(X^{i}ω) is

\begin{aligned}ψ(X^{i}ω) & ≡ j_{0}π^{j−e} \frac{dπ}{π} ∧ fq−2^{\frac{s}{2}}(ω) + π^{j−e}fq−1^{\frac{s}{2}} (\frac{dω}{p^{\frac{s}{2}}} ) \\ & mod fil^{j−e+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). \end{aligned}  (54)

For X^{i}pω ∈ S_{i,1}^{1}, ψ(X^{i}pω) is

ψ(X^{i}pω) = ∑_{n≥0} \frac{1}{p^{n}} π^{ip^{n}}f_{q−2}^{n}(dω).

Thus if X^{i}pω ∈ S_{i,1}^{1} \ X^{i}pf_{q−2}^{η_{i}−1}N_{f}^{q−2} then ψ(X^{i}pω) = 0. Take ω′ ∈ N_{0}^{q−2} such that ω = f_{q−2}^{η_{i}−1}f_{q−2}^{l}ω′. Then

\begin{aligned}ψ(X^{i}pω) & ≡ p^{l}π^{j−e}f_{q−1}^{s−1+l}dω′ \\ & mod fil^{j−e+el+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). \end{aligned}  (55)

If t > s/2, then

S_{i,1}^{0} = X^{i} (f_{q−2}^{η_{i}}N^{q−2} ∪ ⋃_{m=0}^{η_{i}−1} p\frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3})

and

S_{i,1}^{1} = X^{i} (pf_{q−2}^{η_{i}−1}N^{q−2} ∪ ⋃_{m=0}^{η_{i}−2} p\frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3}) .

For X^{i}ω ∈ S_{i,1}^{0}, ψ(X^{i}ω) is

ψ(X^{i}ω) = ∑_{n≥0} \frac{1}{p^{n+1}} π^{ip^{n}}f_{q−2}^{n}(dω).

Thus if X^{i}ω ∈ S_{i,1}^{0} \ X^{i}f_{q−2}^{η_{i}}N_{f}^{q−2} then ψ(X^{i}ω) = 0. Take ω′ ∈ N_{0}^{q−2} such that ω = f_{q−2}^{η_{i}}f_{q−2}^{l}ω′. Then

\begin{aligned}ψ(X^{i}ω) & ≡ p^{l}π^{j−e}f_{q−1}^{2s−2t+l}dω′ \\ & mod fil^{j−e+el+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). \end{aligned}  (56)

For X^{i}pω ∈ S_{i,1}^{1}, by the same calculation as in the case t = s/2,

\begin{aligned}ψ(X^{i}pω) & ≡ p^{l}π^{j−e}f_{q−1}^{2s−2t+l−1}dω′ \\ & mod fil^{j−e+el+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). \end{aligned}  (57)

The Milnor K-Groups 187

Next, we compute S′_{i,2} in the case i ∈ Γ_{b} and i < e. By (4.5), S_{i,2}^{′0} (resp. S_{i,2}^{′1}) exists when 3s/4 < t (resp. (3s − 2)/4 < t). If 3s/4 < t,

\begin{matrix}S_{i,2}^{′0} = X^{i} \frac{dX}{X} ∧ f_{q−3}^{η_{i}}N_{0}^{q−3} ∋ X^{i} \frac{dX}{X} ∧ f_{q−3}^{η_{i}}(ω) \\ ⟼ ψ(X^{i} \frac{dX}{X} ∧ f_{q−3}^{η_{i}}(ω)) \\ = π^{ip^{η_{i}−1}} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−1}(dω) + \frac{1}{p}π^{ip^{η_{i}}} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}}(dω) \\ ≡ π^{j−e} \frac{dπ}{π} ∧ f_{q−2}^{2s−2t}(dω) \\ mod fil^{j−e+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). \end{matrix}  (58)

If (3s − 2)/4 < t,

\begin{matrix}S_{i,2}^{′1} = X^{i} \frac{dX}{X} ∧ pf_{q−3}^{η_{i}−1}N_{0}^{q−3} ∋ X^{i} \frac{dX}{X} ∧ pf_{q−3}^{η_{i}−1}(ω) \\ ⟼ ψ(X^{i} \frac{dX}{X} ∧ f_{q−3}^{η_{i}−1}(ω)) \\ = π^{ip^{η_{i}−1}} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−2}(dω) + \frac{1}{p}π^{ip^{η_{i}}} \frac{dπ}{π} ∧ f_{q−2}^{2η_{i}−1}(dω) \\ ≡ π^{j−e} \frac{dπ}{π} ∧ f_{q−2}^{2s−2t−1}(dω) \\ mod fil^{j−e+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). \end{matrix}  (59)

Next, we compute S_{i,1} for i ∈ Γ_{b} and i > e. Let j = i. Now S_{i,1}^{1} = ∅ and S_{i,1}^{0} = X^{i}N^{q−2}. For an element X^{i}ω ∈ S_{i,1}^{0},

ψ(X^{i}ω) = iπ^{i−e−1}dπ ∧ ω + ∑_{n≥0} \frac{1}{p^{n+1}} π^{ip^{n}}f_{q−1}^{n}(dω).

If p | i, then the first term of the right hand side is zero. Hence if ω = f_{q−2}^{l}(ω′) then

\begin{aligned}ψ(X^{i}ω) & ≡ p^{l}π^{i−e}f_{q−1}^{l}(dω′) \\ & mod fil^{j−e+el+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}) \end{aligned}  (60)

and if ω ∈ N_{∞}^{q−2} then ψ(X^{i}ω) = 0. If p ∤ i, then

\begin{aligned}ψ(X^{i}ω) & ≡ iπ^{i−e} \frac{dπ}{π} ∧ ω + π^{i−e}dω \\ & mod fil^{j−e+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). \end{aligned}  (61)

For i ∈ Γ_{b} and i > e, S′_{i,2} is empty if p ∤ i. So assume p | i. Then, for

Documenta Mathematica 5 (2000) 151–200

X^{i−1}dX ∧ ω ∈ X^{i−1}dX ∧ N_{0}^{q−3} = S′_{i,2} = S_{i,2}^{′0},

\begin{aligned}X^{i} \frac{dX}{X} ∧ ω & = π^{i−e} \frac{dπ}{π} ∧ dω \\ & ≡ π^{j−e} \frac{dπ}{π} ∧ dω mod fil^{j−e+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). \end{aligned}  (62)

Use (53), (54), (55), (56), (57), (58), (59), (60), (61) and (62) to define M^{j+el} for e < j < e′ and l ≥ 0. Let e < j < e′, s = v_{p}(j). By (3.1),

gr^{j−e+el}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}) ≅ \begin{cases}Ω̂_{F}^{q−1} ⊕ Ω_{F}^{q−2} & (l = 0) \\ Ω̂_{F}^{q−1}/B_{l}^{q−1} & (l ≥ 1). \end{cases}

If s = 0 then let

\begin{matrix}M^{j−e} = X^{j}N^{q−2}, … (61) \\ M^{j−e+el} = ∅ \end{matrix}  (63)

for all l ≥ 0. The image of M^{j−e} is the image of

Ω_{F}^{q−2} ∋ x ⟼ (dx, jx) ∈ Ω_{F}^{q−1} ⊕ Ω_{F}^{q−2},

hence we get

ψ(x) ≠ 0 for x ∈ M^{j−e} in gr^{j−e}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}),
gr^{j−e}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2})/⟨ψ(M^{j−e})⟩
≅ Coker (Ω_{F}^{q−2} ∋ x ⟼ (dx, jx) ∈ Ω_{F}^{q−1} ⊕ Ω_{F}^{q−2})

(64)

and

ψ(x) ≠ 0 for x ∈ M^{j−e+el} in gr^{j−e+el}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}),
gr^{j−e+el}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2})/⟨ψ(M^{j−e+el})⟩ ≅ Ω_{F}^{q−1}/B_{l}^{q−1}

(65)

for l ≥ 1 because M^{j−e+el} = ∅.

Documenta Mathematica 5 (2000) 151–200

The Milnor K-Groups 189

If s is even and s ≥ 2, let

\begin{matrix}M^{j−e} = \\ X^{j_{0}p^{\frac{s}{2}}} (fq−2^{\frac{s}{2}}N^{q−2} ∪ ⋃_{m=0}^{(s/2)−1} \frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3}) … (54) \\ ∪ X^{j_{0}p^{\frac{s}{2}}} pf_{q−2}^{(s/2)−1}N_{0}^{q−2} … (55) \\ ∪ ⋃_{\frac{s}{2} <t≤s−1} (X^{j_{0}p^{t}}f_{q−2}^{s−t}N_{0}^{q−2} ∪ X^{j_{0}p^{t}}f_{q−2}^{s−t−1}N_{0}^{q−2}) … (56),(57) \\ ∪ ⋃_{\frac{3s}{4} <t≤s−1} (X^{j_{0}p^{t}} \frac{dX}{X} f_{q−3}^{s−t}N_{0}^{q−3}) … (58) \\ ∪ ⋃_{\frac{3s−2}{4} <t≤s−1} (X^{j_{0}p^{t}} \frac{dX}{X} pf_{q−3}^{s−t−1}N_{0}^{q−3}) … (59) \\ ∪ X^{j}N_{0}^{q−2} … (60) \\ ∪ X^{j} \frac{dX}{X} N_{0}^{q−3} … (62) \end{matrix}  (66)

and

\begin{matrix}M^{j−e+el} = \\ X^{j_{0}p^{\frac{s}{2}}} pf_{q−2}^{(s/2)−1}N_{l}^{q−2} … (55) \\ ∪ ⋃_{\frac{s}{2} <t≤s−1} (X^{j_{0}p^{t}}f_{q−2}^{s−t}N_{l}^{q−2} ∪ X^{j_{0}p^{t}}f_{q−2}^{s−t−1}N_{l}^{q−2}) … (56),(57) \\ ∪ X^{j}N_{l}^{q−2} … (60) \end{matrix}  (67)

for l ≥ 1. When l = 0, the image of (58), (59) and (62) is

( ⨁_{3s/4<t≤s−1} Ω_{F}^{q−3}/Z_{1}^{q−3}) ⊕ ( ⨁_{3s/4<t≤s−1} Ω_{F}^{q−3}/Z_{1}^{q−3}) ⊕ Ω_{F}^{q−3}/Z_{1}^{q−3}

⟶^{(⨁_{3s/4<t≤s−1} C^{−(2s−2t)} d)⊕(⨁_{(3s−2)/4<t≤s−1} C^{−(2s−2t−1)} d)⊕d}

0 ⊕ Bq−2_{\frac{s}{2}} ⊂ Ω_{F}^{q−1} ⊕ Ω_{F}^{q−2},

the image of (55), (56), (57) and (60) is

Ω_{F}^{q−2}/Z_{1}^{q−2} ⊕ ( ⨁_{\frac{s}{2} <t≤s−1} Ω_{F}^{q−2}/Z_{1}^{q−2} ⊕ Ω_{F}^{q−2}/Z_{1}^{q−2}) ⊕ Ω_{F}^{q−2}/Z_{1}^{q−2}

⟶^{C^{−(s−1)} d⊕(⨁_{\frac{s}{2} <t≤s−1} C^{−(2s−2t)} d⊕C^{−(2s−2t−1)})⊕d}

B_{s}^{q−1} ⊕ 0 ⊂ Ω_{F}^{q−1} ⊕ Ω_{F}^{q−2}.

Documenta Mathematica 5 (2000) 151–200

190 Jinya Nakamura Furthermore, the image of (54) modulo the image of the group generated by the other generators of M^{j} is

Ω_{F}^{q−2} ⊕ ( ⨁_{0≤m< \frac{s}{2}} Ω_{F}^{q−3}/Z_{1}^{q−3}) ⟶^{(C^{−s} d,j_{0} C^{−s})⊕(⨁_{0≤m< \frac{s}{2}} C^{−( \frac{s}{2} +m)} d)}

(C^{−s} d, j_{0} C^{−s})Ω_{F}^{q−2} + B_{s}^{q−2}/Bq−2_{\frac{s}{2}} ⊂ Ω_{F}^{q−1}/B_{s}^{q−1} ⊕ Ω_{F}^{q−2}/Bq−2_{\frac{s}{2}}.

Hence we get ψ(x) ≠ 0 for x ∈ M^{j−e} in gr^{j−e}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}), gr^{j−e}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2})/⟨ψ(M^{j−e})⟩ ≅ Coker (Ω_{F}^{q−2} ∋ x ⟼ (C^{−s} dx, j_{0} C^{−s} x) ∈ Ω_{F}^{q−1}/B_{s}^{q−1} ⊕ Ω_{F}^{q−2}/B_{s}^{q−2}) .

(68)

When l ≥ 1, the image of M^{j−e+el} is

Ω_{F}^{q−2}/Z_{1}^{q−2} ⊕ ( ⨁_{\frac{s}{2} <t≤s−1} Ω_{F}^{q−2}/Z_{1}^{q−2} ⊕ Ω_{F}^{q−2}/Z_{1}^{q−2}) ⊕ Ω_{F}^{q−2}/Z_{1}^{q−2}

⟶^{C^{−(s+l−1)} d⊕(⨁_{\frac{s}{2} <t≤s−1} C^{−(2s−2t+l)} d⊕C^{−(2s−2t+l−1)})⊕C^{−l} d}

B_{s+l}^{q−1}/B_{l}^{q−1} ⊂ Ω_{F}^{q−1}/B_{l}^{q−1}. Hence we get

ψ(x) ≠ 0 for x ∈ M^{j−e+el} in gr^{j−e+el}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}),
gr^{j−e+el}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2})/⟨ψ(M^{j−e+el})⟩ ≅ Ω_{F}^{q−1}/B_{s+l}^{q−1}.

(69)

If s is odd and s ≥ 1, let

\begin{matrix}M^{j−e} = X^{j_{0}p^{\frac{s−1}{2}}} (pfq−2^{\frac{s−1}{2}} N^{q−2} ∪ ⋃_{m=0}^{((s−1)/2)−1} p\frac{d}{p^{m}} f_{q−3}^{m}N_{0}^{q−3}) … (53) \\ ∪ ⋃_{\frac{s}{2} <t≤s−1} (X^{j_{0}p^{t}}f_{q−2}^{s−t}N_{0}^{q−2} ∪ X^{j_{0}p^{t}}f_{q−2}^{s−t−1}N_{0}^{q−2}) … (56), (57) \\ ∪ ⋃_{\frac{3s}{4} <t≤s−1} (X^{j_{0}p^{t}} \frac{dX}{X} f_{q−3}^{s−t}N_{0}^{q−3}) … (58) \\ ∪ ⋃_{\frac{3s−2}{4} <t≤s−1} (X^{j_{0}p^{t}} \frac{dX}{X} pf_{q−3}^{s−t−1}N_{0}^{q−3}) … (59) \\ ∪ X^{j}N_{0}^{q−2} … (60) \\ ∪ X^{j} \frac{dX}{X} N_{0}^{q−3} … (62) \end{matrix} (70)

and

\begin{aligned}M^{j−e+el} = & ⋃_{\frac{s}{2} <t≤s−1} (X^{j_{0}p^{t}}f_{q−2}^{s−t}N_{l}^{q−2} ∪ X^{j_{0}p^{t}}f_{q−2}^{s−t−1}N_{l}^{q−2}) … (56), (57) \\ & ∪ X^{j}N_{l}^{q−2} … (60) \end{aligned} (71)

for l ≥ 1. By the similar calculation to the case s is even, we get the same result (68) and (69).

By the definition of M^{j+el},

( ⋃_{i∈Γ_{b}} S_{i,1}^{0} ∪ S_{i,1}^{1}) ∪ ( ⋃_{i∈Γ_{b}} S′_{i,2})

is equal to the union of M^{j−e+el} for all e < j < e′ and all l ≥ 0.

7 Explicit Calculation, Case (c)

In this section, we compute ψ(S_{i,1}) and ψ(S′_{i,2}) for i ∈ Γ_{c}.

Γ_{c} has only two elements, e/(p − 1) and e′. At first let i = e/(p − 1). Then S_{i,1}^{0} = ∅, S′_{i,2} = ∅ and

S_{i,1}^{1} = X^{i}pN^{q−2}.

Note that this i has the property i = ip − e. Take X^{i}pω ∈ X^{i}pN^{q−2}, then

\begin{aligned}ψ(X^{i}pω) & = iπ^{i} \frac{dπ}{π} ∧ (ω + f_{q−2}(ω)) + π^{i}(dω + f_{q−1}(dω)) \\ & + ∑_{n≥2} \frac{1}{p^{n}} π^{ip^{n}}f_{q−1}^{n}(dω). \end{aligned}

If ω+f_{q−2}(ω) ≡ 0 mod p then the leftmost term of the right hand side vanishes. But ω + f_{q−2}(ω) ≡ 0 means ω̄ + C^{−1}ω̄ = 0 in Ω̂_{F}^{q−2}, thus dω = 0 hence ψ(X^{i}pω) = 0 by the property of N_{0}^{q−2}, see (25). So we get

ψ(X^{i}pω) \begin{cases}≡ iπ^{i} \frac{dπ}{π} ∧ (ω + f_{q−2}(ω)) + π^{i}(dω + f_{q−1}(dω)) \\ mod fil^{i+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}) \\ ( if ω + f_{q−2}(ω) ≢ 0 mod p) \\ = 0 ( if ω + f_{q−2}(ω) ≡ 0 mod p) \end{cases}  (72)

Next, let i = e′. Then S_{i,1}^{0} = X^{i}N_{q−2}, S_{i,2}^{′0} = X^{i−1}dX ∧ N_{0}^{q−3} and S_{i,1}^{1} = S_{i,2}^{′1} = ∅. For X^{i}ω ∈ X^{i}N_{q−2},

ψ(X^{i}ω) = ∑_{n≥0} \frac{1}{p^{n+1}} π^{ip^{n}}f_{q−1}^{n}(dω).

192 Jinya Nakamura Thus if ω ∈ N_{∞}^{q−2} then ψ(X^{i}ω) = 0 and if ω = f_{q−2}^{l}ω′ for ω′ ∈ N_{0}^{q−2} then

ψ(X^{i}f_{q−2}^{l}ω′) ≡ p^{l}π^{i−e}f_{q−1}^{l}(dω′) mod fil^{i−e+el+1}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}). (73)

For X^{i−1}dX ∧ ω ∈ X^{i−1}dX ∧ N_{0}^{q−3},

ψ(X^{i−1}dX ∧ ω) = π^{i−e} \frac{dπ}{π} ∧ dω.  (74)

Use (72), (73) and (74) to define M^{j+el} for j = e/(p − 1) and l ≥ 0.

M^{\frac{e}{p−1}} =X^{\frac{e}{p−1}} pN^{q−2} \ \{ω|ω + f_{q−2}ω ≡ 0 mod p\} … (72)

∪ X^{e′}N_{0}^{q−2} … (73)

∪ X^{e′} \frac{dX}{X} ∧ N_{0}^{q−3} … (74)

and let

M^{\frac{e}{p−1} +el} = X^{e′}N_{l}^{q−2} … (73).

By (3.1),

gr^{e/(p−1)+el} Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2} ≅ \begin{cases}Ω_{F}^{q−1} ⊕ Ω_{F}^{q−2} & ( if l = 0) \\ Ω_{F}^{q−1}/B_{l}^{q−1} & ( if l ≥ 1). \end{cases}

When l = 0, the image of (73) and (74) is

Ω_{F}^{q−2}/Z_{1}^{q−2} ⊕ Ω_{F}^{q−3}/Z_{1}^{q−3} ⟶^{d⊕d} Ω_{F}^{q−1} ⊕ Ω_{F}^{q−2}

and the image of (72) modulo the subgroup generated by (73) and (74) is

Ω_{F}^{q−2}/Z_{1}^{q−2} ⟶^{((1+C^{−1})d, \frac{e}{p−1} (1+C^{−1}))}

Ω_{F}^{q−1}/B_{1}^{q−1} ⊕ Ω_{F}^{q−2}/B_{1}^{q−2} ≅ Ω_{F}^{q−1}/(1 + C)B_{1}^{q−1} ⊕ Ω_{F}^{q−2}/(1 + C)B_{1}^{q−2}. Here Ω_{F}^{q−1}/B_{1}^{q−1} ≅ Ω_{F}^{q−1}/(1 + C)B_{1}^{q−1} follows from C(B_{1}^{q−1}) = 0. Hence we get

ψ(x) ≠ 0 for x ∈ M^{e/(p−1)} in gr^{e/(p−1)}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}), gr^{e/(p−1)}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2})/⟨ψ(M^{e/(p−1)})⟩

≅ Coker (\begin{matrix}Ω_{F}^{q−2} & ⟶ Ω_{F}^{q−1}/(1 + C)B_{1}^{q−1} ⊕ Ω_{F}^{q−2}/(a + C)B_{1}^{q−2} \\ x & ⟼ ((1 + C)C^{−1}dx, \frac{e}{p − 1}(1 + C) C^{−1} x) \end{matrix}) .

(75)

When l ≥ 1, the image of (73) is

Ω_{F}^{q−2}/Z_{1}^{q−2} ⟶^{C^{−l} d} B_{l+1}^{q−1}/B_{l}^{q−1} ⊂ Ω_{F}^{q−1}/B_{l}^{q−1}.

Documenta Mathematica 5 (2000) 151–200

Hence we get

ψ(x) ≠ 0 for x ∈ M^{e/(p−1)+el} in gr^{e/(p−1)+el}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}),
gr^{e/(p−1)+el}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2})/⟨ψ(M^{e/(p−1)+el})⟩ ≅ Ω_{F}^{q−1}/B_{l+1}^{q−1}.

(76)

By the definition of M^{e/(p−1)+el},

( ⋃_{i∈Γ_{c}} S_{i,1}^{0} ∪ S_{i,1}^{1}) ∪ ( ⋃_{i∈Γ_{b}} S′_{i,2})

is equal to the union of M^{e/(p−1)+el} for all l ≥ 0.

8 The structure of the Milnor K-group

Proof of Theorem 1.1. At first, assume ζ_{p} ∈ K, there exists a prime element π of K such that π^{e} = p and the residue field F has a finite p-base. By the definition of M^{n}, the union of all M^{n} for n ≥ 1 and n/e ∉ ℤ generates ψ(H^{1}(𝕊_{q})). M^{el} for l ≥ 0 is not defined yet, so let M^{el} = ∅. Then M^{n} is defined for all n ≥ 1. There is map

⟨ ⋃_{n≥i} ψ(M^{n})⟩ / ⟨ ⋃_{n≥i+1} ψ(M^{n})⟩ ⟶ gr^{i}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2})  (77)

for each i ≥ 0. By the exact sequence of (2.4), if (77) are injective for all i ≥ 0 then

⟨ψ(M^{i})⟩ ⟶ gr^{i}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}) ⟶^{exp_{p}} gr^{i+e} Kq^{M}(K)

are also exact for all i ≥ 0. We already know ψ(x) ≠ 0 for x ∈ M^{i} in gr^{i}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}) and what is the group ⟨ψ(M^{i})⟩ in gr^{i}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2}) for all i ≥ 0 by (45), (46), (49), (50), (64), (65), (68), (69), (75) and (76). The results are as follows:

gr^{j+el}(Ω̂_{A}^{q−1}pdΩ̂_{A}^{q−2})/⟨ψ(M^{j+el})⟩ ≅ Ω_{F}^{q−1}/B_{s+l}^{q−1} ( if \frac{e}{p − 1} < j < e, l ≥ 0), (78)

where s = v_{p}(j) + 1.

gr^{j−e+el}(Ω̂_{A}^{q−1}pdΩ̂_{A}^{q−2})/⟨ψ(M^{j+el})⟩

≅ \begin{cases}Coker (\begin{matrix}Ω_{F}^{q−2} & → Ω_{F}^{q−1}/B_{s}^{q−1} ⊕ Ω_{F}^{q−2}/B_{s}^{q−2} \\ x & ↦ (C^{−s} dx, j_{0} C^{−s} x) \end{matrix}) & ( if e < j < e′, l = 0) \\ Ω_{F}^{q−1}/B_{s+l}^{q−1} & ( if e < j < e′, l ≥ 1), \end{cases}

(79) where s = v_{p}(j) and j_{0} = j/p^{s}.

gr^{e/(p−1)+el}(Ω̂_{A}^{q−1}pdΩ̂_{A}^{q−2})/⟨ψ(M^{e/(p−1)+el})⟩

≅ \begin{cases}Coker (\begin{matrix}Ω_{F}^{q−2} → Ω_{F}^{q−1}/(1 + C)B_{1}^{q−1} ⊕ Ω_{F}^{q−2}/(1 + C)B_{1}^{q−2} \\ x ↦ ((1 + C) C^{−1} dx, \frac{e}{p − 1}(1 + C) C^{−1} x) \end{matrix}) ( if l = 0) \\ Ω_{F}^{q−1}/B_{1+l}^{q−1} ( if l ≥ 1). \end{cases}

(80)

gr^{el}(Ω̂_{A}^{q−1}pdΩ̂_{A}^{q−2})/⟨ψ(M^{el})⟩ ≅ gr^{el}(Ω̂_{A}^{q−1}pdΩ̂_{A}^{q−2}) ≅ Ω_{F}^{q−1}/B_{l}^{q−1}( for l ≥ 0). (81)

Let n ≥ 1 and k be the integer which satisfies e/(p − 1) ≤ n − ke < e′. If 1 ≤ n ≤ e/(p − 1), then the results of (79) with l = 0 and (80) with l = 0 is coincide with the result of [3] by gr^{n+e} Kq^{M}(K) ≅ gr^{n}(Ω̂_{A}^{q−1}/pdΩ̂_{A}^{q−2})/⟨ψ(M^{n})⟩. Let n > e/(p − 1). Then (78), (79), (80) and (81) say

gr^{n}(Ω̂_{A}^{q−1}pdΩ̂_{A}^{q−2})/⟨ψ(M^{n})⟩ ≅ Ω_{F}^{q−1}/B_{s′+1+k}^{q−1},

where s′ = v_{p}(n − ke). Hence we have

gr^{n+e} Kq^{M}(K) ≅ Ω_{F}^{q−1}/B_{s′+1+k}^{q−1}

and we get Theorem (1.1) by shifting degrees.

We prove Theorem (1.1) in the case K does not contain primitive p-th roots of unity ζ_{p} or K does not contain a prime element π such that π^{e} = p as follows. Let L = K(ζ_{p}, \sqrt[e]{p}) and let m = [L : K]. Then p ∤ m and the extention L/K is unramified. By using standard norm argument, the composite map

gr^{i} Kq^{M}(K) ⟶ gr^{im} Kq^{M}(L) ⟶^{Norm} gr^{i} Kq^{M}(K)

is the multiplication by m, hence injective. Furthermore, F_{L}/F_{K} is a finite separable extension, where F_{L} (resp. F_{K}) is the residue field of L (resp. K), we get Ω_{F_{L}}^{q−1}/B_{l}Ω_{F_{L}}^{q−1} ≅ Ω_{F_{K}}^{q−1}/B_{l}Ω_{F_{K}}^{q−1} ⊗_{F_{K}^{p^{l}}} F_{L}. Thus Theorem (1.1) follows even if ζ_{p} ∉ K.

Lastly, do not assume that the residue field of K has a finite p-base. Then an inductive system of complete discrete valuation fields whose residue fields has a finite p-base and its limit is isomorphic to K exists by [9] Section 1.5. On the other hand, for a purely transcendental extension or a separable extension F′/F,

Ω_{F}^{q} /B_{l}Ω_{F}^{q} ⟶ Ω_{F′}^{q}/B_{l}Ω_{F′}^{q}

are injective for all q and l because, if F′/F is separable extension, then Ω_{F′}^{q} = F′ ⊗_{F} Ω_{F}^{q} and if F′/F is purely transcendental extension F′ = F (T) then Ω_{F′}^{q} = (F′ ⊗_{F} Ω_{F}^{q})⊕(F′ ⊗_{F} Ω_{F}^{q−1} ∧dT). Hence we get Theorem (1.1) by taking inductive limit.

To prove Corollary (1.2), we need the following

Lemma 8.1. Assume ζ_{p} ∈ K. Let V = Im({ζ_{p}, ∗}: Kq−1^{M}(K)/p → Kq^{M}(K)ˆ ). Then the sequence

\begin{matrix}0 ⟶ V ∩ U^{i}Kq^{M}(K)ˆ ∩ p^{n}Kq^{M}(K)ˆ ⟶ U^{i}Kq^{M}(K)ˆ ∩ p^{n}Kq^{M}(K) \\ ˆ ⟶ U^{i+e}Kq^{M}(K)ˆ ∩ p^{n+1}Kq^{M}(K)ˆ ⟶ 0 \end{matrix}

is exact for i > e/(p − 1).

Proof. Restricting the bottom row of (3) to the filtration of Kq^{M}(K), we have the exact sequence

0 ⟶ V ∩ U^{i}Kq^{M}(K)ˆ ⟶ U^{i}Kq^{M}(K)ˆ ⟶^{p} U^{i+e}Kq^{M}(K)ˆ ⟶ 0

and hence we get the exact sequence

\begin{matrix}0 ⟶ V ∩ U^{i}Kq^{M}(K)ˆ ∩ p^{n}Kq^{M}(K)ˆ ⟶ U^{i}Kq^{M}(K)ˆ ∩ p^{n}Kq^{M}(K) \\ ˆ ⟶ U^{i+e}Kq^{M}(K)ˆ ∩ p^{n+1}Kq^{M}(K)ˆ . \end{matrix}  (82)

We only have to show the surjectivity of the last arrow of (82). Take p^{n+1}x ∈ U^{i+e}Kq^{M}(K)ˆ ∩ p^{n+1}Kq^{M}(K)ˆ . By the surjectivity of the multiplication by p map U^{i}Kq^{M}(K)ˆ → U^{i+e}Kq^{M}(K), there exists y ∈ U^{i}Kq^{M}(K)ˆ such that p(y − p^{n}x) = 0. This y − p^{n}x is a p-torsion element of Kq^{M}(K)ˆ , thus y − p^{n}x ∈ V ⊂

U^{e/(p−1)}Kq^{M}(K)ˆ . Hence p^{n}x ∈ U^{e/(p−1)}Kq^{M}(K)ˆ because y ∈ U^{i}Kq^{M}(K)ˆ . Now

e/(p − 1) is prime to p, thus gr^{e/(p−1)} Kq^{M}(K)ˆ ≅ gr^{e/(p−1)}(Kq^{M}(K)/p^{n}) by [3], and p^{n}x goes to zero on this map. Hence we get p^{n}x ∈ U^{e/(p−1)+1}Kq^{M}(K)ˆ . Let j = (e/(p − 1)) + 1. By the definition, all rows and columns in the following commutative diagram are exact:

0  0
↓  ↓
0 ⟶ \frac{V ∩U_{∞}^{j}∩(p^{n})}{V ∩U_{∞}^{i}∩(p^{n})} ⟶ \frac{U_{∞}^{j}∩(p^{n})}{U_{∞}^{i}∩(p^{n})} ⟶^{p} \frac{U_{∞}^{j+e}∩(p^{n+1})}{U_{∞}^{i+e}∩(p^{n+1})}
↓  ↓  ↓
0 ⟶ \frac{V ∩U_{∞}^{j}}{V ∩U_{∞}^{i}} ⟶ \frac{U_{∞}^{j}}{U_{∞}^{i}}  ⟶^{p}  \frac{U_{∞}^{j+e}}{U_{∞}^{i+e}}  ⟶ 0
↓  ↓  ↓
0 ⟶ \frac{V_{p^{n}} ∩U_{p^{n}}^{j}}{V_{p^{n}} ∩U_{p^{n}}^{i}} ⟶ \frac{U_{p^{n}}^{j}}{U_{p^{n}}^{i}}  ⟶^{p}  \frac{U_{p^{n+1}}^{j+e}}{U_{p^{n+1}}^{i+e}} ⟶ 0
↓  ↓
0  0

where we denote V_{p^{n}} = Im(V → Kq^{M}(K)/p^{n}), U_{∞}^{m} = U^{m}Kq^{M}(K)ˆ , U_{p^{n}}^{n} = U^{n}(Kq^{M}(K)/p^{n}) and (p^{n}) = p^{n}Kq^{M}(K)ˆ only in this diagram. p^{n}x is in the middle group of the top row and goes to zero by multiplication by p. Thus there exists z ∈ V ∩U^{j}Kq^{M}(K)ˆ ∩p^{n}Kq^{M}(K)ˆ such that p^{n}x−z ≡ 0 modulo U^{i}Kq^{M}(K)ˆ . Furthermore, z ∈ p^{n}Kq^{M}(K)ˆ implies p^{n}x − z ∈ U^{i}Kq^{M}(K)ˆ ∩ p^{n}Kq^{M}(K)ˆ , thus

\begin{matrix}U^{i}Kq^{M}(K)ˆ ∩ p^{n}Kq^{M}(K) & ˆ⟶^{p} U^{i+e}Kq^{M}(K)ˆ ∩ p^{n+1}Kq^{M}(K) \\ ˆ p^{n}x − z & ⟼ p^{n+1}x − pz = p^{n+1}x. \end{matrix}

Hence surjectivity of the last arrow of (82) follows.

Corollary 8.2. All rows and columns are exact in the following commutative diagram :

0  0
↓  ↓
\frac{U^{i}K_{q}^{M} (K)ˆ ∩p^{n}K_{q}^{M} (K)}{ˆ U^{i+1}K_{q}^{M} (K)ˆ ∩p^{n}K_{q}^{M} (K)}ˆ ⟶ \frac{U^{i+e}K_{q}^{M} (K)ˆ ∩p^{n+1}K_{q}^{M} (K)}{ˆ U^{i+e+1}K_{q}^{M} (K)ˆ ∩p^{n+1}K_{q}^{M} (K)}ˆ ⟶ 0
↓  ↓
gr^{i} Kq^{M}(K)ˆ  ⟶^{p}  gr^{i+e} Kq^{M}(K)ˆ  ⟶ 0
↓  ↓
gr^{i}(Kq^{M}(K)/p^{n}) ⟶^{p} gr^{i+e}(Kq^{M}(K)/p^{n+1}) ⟶ 0
↓  ↓
0  0.

(83)

Proof. Exactness of the top row comes from (8.1).

Proof of Corollary 1.2. Denote Ker(gr^{i} Kq^{M}(K)ˆ → gr^{i} Kq^{M}(K)/p^{n+1}) by G_{i,n+1}. At first, we prove Corollary (1.2) for e′ < i ≤ e′ + e. Let s = v_{p}(i − e) and i_{0} = (i − e)/p^{s}. Then we know all gr^{i−e} Kq^{M}(K)ˆ and gr^{i−e}(Kq^{M}(K)/p^{n}) by [3], thus (83) is, if n ≤ s and i ≠ e′ + e then

0

↓

Z_{n}^{q−1} ⊕ Z_{n}^{q−2}  ⟶  G_{i,n+1}  ⟶ 0
↓  ↓
\frac{Ω_{F}^{q−1}/B_{s}^{q−1}⊕Ω_{F}^{q−2}/B_{s}^{q−2}}{(C^{−s} d,i_{0} C^{−s})Ω_{F}^{q−2}} ⟶ Ω_{F}^{q−1}/B_{s+1}^{q−1} ⟶ 0
↓  ↓
\frac{Ω_{F}^{q−1}}{Z_{n}^{q−1}} ⊕ \frac{Ω_{F}^{q−2}}{Z_{n}^{q−2}}  ⟶ gr^{i}(Kq^{M}(K)/p^{n+1}) ⟶ 0
↓  ↓
0  0

here all maps are natural maps, and if n ≤ s and i = e′ + e then

0

↓

(1 + aC)Z_{n}^{q−1} ⊕ (1 + aC)Z_{n}^{q−2} ⟶  G_{i,n+1}  ⟶ 0
↓  ↓
\frac{Ω_{F}^{q−1}/(1+aC)B_{s}^{q−1}⊕Ω_{F}^{q−2}/(1+aC)B_{s}^{q−2}}{((1+aC) C^{−s} d,i_{0}(1+aC) C^{−s})Ω_{F}^{q−2}} ⟶ Ω_{F}^{q−1}/B_{s+1}^{q−1} ⟶ 0
↓  ↓
\frac{Ω_{F}^{q−1}}{(1+aC)Z_{n}^{q−1}} ⊕ \frac{Ω_{F}^{q−2}}{(1+aC)Z_{n}^{q−2}}  ⟶ gr^{i}(Kq^{M}(K)/p^{n+1}) ⟶ 0
↓  ↓
0  0

where a is the residue class of p/π^{e}. We get (1.2) in this case by these diagrams. If n > s then gr^{i} Kq^{M}(K)ˆ → gr^{i}(Kq^{M}(K)/p^{n}) is an isomorphism, thus gr^{i+e} Kq^{M}(K)ˆ → gr^{i+e}(Kq^{M}(K)/p^{n+1}) is also an isomorphism.

By induction on i and calculating the diagram (83) for each case, we get (1.2).

9 An application

Theorem 9.1. Let K be a Henselian discrete valuation field of mixed characteristics (0, p > 2) with the residue field F. Assume p ∤ e and [F : F^{p}] = p^{q−1}, where e = v_{K}(p). Let L/K be a ferociously ramified cyclic extention of order p^{n} (i.e., the extention of the residue fields is inseparable of order p^{n}). Then p^{n} ≤ e′, where e′ = ep/(p − 1).

Remark 9.2. In [15] and [6], they give the upper bounds of such extensions. If K has the property p ∤ e, our bound is stricter than them (or equal to [6] if e is small).

Proof. We use the notation U_{p^{n}}^{i} = U^{i}(Kq^{M}(K)/p^{n}) for simplicity. The proof goes similarly to the argument of [15] Section 3. By the limit argument, we may assume F is a field of transcendental degree q − 1 over 𝔽_{p}. Then H^{q+1}(K, ℤ/p(q)) is non zero by [10] and furthermore we know that H^{q+1}(K, ℤ/p^{n}(q)) has an elements of order p^{n} by using Bockstein.

Let L/K be a cyclic extension of order p^{n} and let χ ∈ H^{1}(K, ℤ/p^{n}) be the character which coincide with L/K. Let ϕ_{χ} be the homomorphism

ϕ_{χ}: Kq^{M}(K)/p^{n} ⟶ H^{q+1}(K, ℤ/p^{n}(q))

which is induced by the pairing

H^{1}(K, ℤ/p^{n}) × Kq^{M}(K)/p^{n} ⟶ H^{q+1}(K, ℤ/p^{n}(q))

by using Kq^{M}(K)/p^{n} ≅ H^{q}(K, ℤ/p^{n}(q)). If L/K is ferociously ramified, by [15] Section 3, we know

ϕ_{χ} : U_{p^{n}}^{1} ⟶ H^{q+1}(K, ℤ/p^{n}(q))

is surjective and

ϕ_{χ}({1 + π^{i}x, y_{1}, …, y_{q−1}}) ∈ ϕ_{χ}(U_{p^{n}}^{i+1})  (84)

for any x, y_{1}, …, y_{q−1} ∈ O_{K}^{×} and i ≥ 1. Theorem (1.1) says that U_{p^{n}}^{e′+1} is generated by the elements of the form of the left hand side of (84), thus we get

ϕ_{χ}: U_{p^{n}}^{p}/U_{p^{n}}^{e′+1} ⟶ H^{q+1}(K, ℤ/p^{n}(q))

is defined and surjective. Furthermore, for any element {1 + π^{i}x, y_{1}, …, y_{q−2}, π} ∈ U_{p^{n}}^{1} for x, y_{1}, …, y_{q−2} ∈ O_{K}^{×} and i ≥ p, its order modulo U_{p^{n}}^{e′+1} is less than or equal to p^{l} by [3] Theorem 1.4, where l be the maximal integer which satisfies p^{l} ≤ e′. Thus he maximal order of the elements of U_{p^{n}}^{p} modulo U_{p^{n}}^{e′+1} is less than or equal to p^{l}. On the other hand, H^{q+1}(K, ℤ/p^{n}(q)) has a element of order p^{n}, thus n ≤ l. This is the inequality which we desired.

Note that there exists elements of U_{p^{n}}^{p}/U_{p^{n}}^{e′+1} of order p^{n}, for example, {1 + π^{p}T_{1}, T_{2}, …, T_{q−1}, π}, where {T_{1}, …, T_{q−1}} are the liftings of a p-base of F. Thus the maximal order of the elements of U_{p^{n}}^{p}/U_{p^{n}}^{e′+1} is p^{l}.

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Jinya Nakamura Graduate School of Mathematical Sciences University of Tokyo Komaba, Meguro-ku Tokyo, Japan jinya@ms357.ms.u-tokyo.ac.jp