On the Milnor K-Groups of Complete Discrete Valuation Fields

For a discrete valuation eld K, the unit group K of K has a natural decreasing ltration with respect to the valuation, and the graded quotients of this ltration are given in terms of the residue eld. The Milnor K-group K M q (K) is a generalization of the unit group, and it also has a natural decreasing ltration. However, if K is of mixed characteristics and has an absolute ramication index greater than one, the graded quotients of this ltration are not yet known except in some special cases. The aim of this paper is to determine them when K is absolutely tamely ramied discrete valuation eld of mixed characteristics (0;p > 2) with possibly imperfect residue eld. Furthermore, we determine the kernel of the Kurihara's K M q - exponential homomorphism from the dieren tial module to the Milnor K-group for such a eld.


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 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 We know that the graded quotients U i K /U i+1 K are isomorphic to F × if i = 0 and F if i ≥ 1.Similarly, there is a natural filtration on K M q (K) defined by Let gr i K M q (K) = U i K M q (K)/U i+1 K M q (K) for i ≥ 0. gr i K M q (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 K M q (K) is determined in [3] in the range 0 ≤ i ≤ e K p/(p − 1), where e K = v K (p).However, gr i K M q (K) still remains mysterious for i > ep/(p − 1).In [16], Kurihara determined gr i K M q (K) for all i if K is absolutely unramified, i.e., v K (p) = 1.In [13] and [19], gr i K M q (K) is determined for some K with absolute ramification index greater than one.
The purpose of this paper is to determine gr i K M q (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 K M q (K) is equal to gr i K M q ( K), where K 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 1 F = Ω 1 F/Z be the module of absolute differentials and Ω q F the q-th exterior power of Ω 1 F over F .As in [7], we define the following subgroups of Ω q F .Z q 1 = Z 1 Ω q F denotes the kernel of d : Ω q F → Ω q+1 F and B q 1 = B 1 Ω q F denotes the image of d : Ω q−1 F → Ω q F .Then there is an exact sequence 0 −→ B q 1 −→ Z q 1 C −→ Ω q F −→ 0, where C is the Cartier operator defined by x p dy 1 y 1 ∧ . . .dy q y q −→ x dy 1 y 1 ∧ . . .dy q y q , B q 1 → 0. The inverse of C induces the isomorphism x dy 1 y 1 ∧ . . .∧ dy q y q −→ x p dy 1 y 1 ∧ . . .∧ dy q y q (1) Documenta Mathematica 5 (2000) 151-200 for x ∈ F and y 1 , . . ., y q ∈ F × .For i ≥ 2, let B q i = B i Ω q F (resp.Z q i = Z i Ω q F ) be the subgroup of Ω q F defined inductively by Let Z q ∞ be the intersection of all Z q i for i ≥ 1.We denote Z q i = Ω q F 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 Corollary 1.2.Let U i (K M q (K)/p m ) be the image of U i K M q (K) in K M q (K)/p m K M q (K) for m ≥ 1 and gr i (K M q (K)/p m ) = U i (K M q (K)/p m )/U i+1 (K M q (K)/p m ).Then 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 K M q (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 K M q (K)ˆwhich includes the image of U 1 K M q (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) s 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 K M q -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 K M q -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, x 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 M = 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 1 R = Ω 1 R/Z be the absolute differentials of R and Ω q R the q-th exterior power of Ω 1 R over R for q ≥ 2. We denote Ω 0 R = R and Ω q R = 0 for negative q.If R is of characteristic zero, let Ωq R of all n ≥ 0. All complexes are cochain complexes.For a morphism of non-negative complexes f : both denote the mapping fiber complex with respect to the morphism f , namely, the complex where the leftmost term is the degree-0 part and where the differentials are defined by 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 .

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 K M q (A) be the subgroup defined by the same way of U i K M q (K), namely, At first, we introduce an isomorphism between U 1 K M q (K)ˆand a subgroup of the cohomology group of the syntomic complex with respect to A. For further details, see [12].
, 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.
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 If q is an negative integer, we denote J [q] = I [q] = D. We define the complexes J [q] and I [q] as where Ωq B is the p-adic completion of Ω q B .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 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 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 where σ >n C • means the brutal truncation for a complex C • , i.e., ( Then there is a result of Kurihara: Theorem 2.1 (Kurihara, [12]).A and B are as above.Then 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 K M q (A)ˆ→ K M q (K)ˆis an injection.
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 s 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 K M q (A)ˆby using Dennis-Stain Symbols, see [17].Thus we only have to show (i).Let ζ p be a primitive Documenta Mathematica 5 (2000) 151-200 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.
The bottom row are exact by using Galois cohomology long exact sequence with respect to the Bockstein q (K)/p n−1 is injective, and the top row are exact except at K M q (A)/p n−1 .Using the induction on n, we only have to show the injectivity of K M q (A)/p → K M q (K)/p.We know the subquotients of the filtration of K M q (K)/p by [3] and we also know the subquotients of the filtration of K M q (A)/p using the isomorphism U 1 H q (S(A, B)(q)) ∼ = U 1 K M q (A)ˆin [12] and the explicit calculation of H q (S(A, B)(q)) by [14] except gr 0 (K M q (A)/p).Natural map preserves filtrations and induces isomorphisms of subquotients.Thus U 1 (K M q (A)/p) → U 1 (K M q (K)/p) is an injection.Lastly, the composite map of the natural maps q−1 (F )/p is also an injection.Hence K M q (A)/p → K M q (K)/p is injective.Next, we introduce K M q -exponential homomorphism and consider the kernel.By [17], there is the K M q -exponential homomorphism with respect to η for q ≥ 2 and η ∈ K such that v K (η) ≥ 2e/(p − 1) defined by We use this K M q -exponential homomorphism only in the case η = p in this paper.On the other hand, there exists an exact sequence of complexes [σ >q−3 I [q] /σ >q−3 J [q] → 0] is none other than the complex σ >q−3 We denote the complex [σ >q−3 Taking cohomology, we have the following Proposition 2.4.A, B and K are as above.Then K M q -exponential homomorphism with respect to p factors through Ωq−1 A /pdΩ q−2 A and there is an exact sequence 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 is exact.
Here we denote the first map by ψ.
Remark 2.5.By [3], there exist surjections for i ≥ 1, where x ∈ F , y 1 , . . ., y q−1 ∈ F × and where x, ỹ1 , . . ., ỹq−1 are their liftings to A. If i ≥ e + 1, then we can construct all elements of gr i K M q (K) as the image of exp p , namely, Thus U e+1 K M q (K)ˆis contained by the image of exp p .On the other hand, (2.4) says the kernel of the K M q -exponential homomorphism is ψ(H 1 (S q )).Recall that the aim of this paper is to determine gr i K M q (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 K M q (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 The homomorphism 1 − f r+s : for r = 0, q 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 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 where η i and η i be the integers which satisfy p ηi−1 i < e ≤ p ηi i and To prove (2.6), we need the following lemmas.
Lemma 2.7.For ω ∈ D ⊗ Ωq B and n ≥ 0, In particular, if ω ∈ Ωq A0 , then Proof.ω ∈ D ⊗ Ωq B can be rewrite as ω = i a i ω i , where a i ∈ D and ω i are the canonical generators of Ωq B , which are 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 ∈ Ωq A0 , then a i ∈ A 0 and we have B can be written as the sum of the elements of the form X n (X e ) [j] ω, where ω ∈ D ⊗ Ωs B 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 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 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 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 ηi ≥ 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 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 is exact.If i = 0, the first vertical arrow of ( 14) . 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 ⊗ Ωq−2 B because of the injectivity of the vertical arrows of (14).
Let i = 2e.Then gr The second vertical arrow is a surjection, thus Let e < i < 2e.Then gr 2e S q is The second vertical arrow is also a surjection, thus Let i = e.Then gr e S q is The second vertical arrow is not a surjection.For an element Thus we have if and only if p | iω and p | dω, and the image of If i = 0, we need more calculation.The complex gr 0 S q is We introduce a p-adic filtration to gr 0 S q as follows.
Then, for all m ≥ 0, The injectivity of the leftmost vertical arrow of (20) says that H 0 (gr m p (gr 0 S q )) = 0 for all m ≥ 0. Thus {H 1 (fil m p (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 (1).Thus we also have H 1 (gr m p (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 . 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 The elements of H 1 (gr i S q ) are represented by two types of the elements of D ⊗ Ωq−2 B , these are X i ω for ω ∈ Ωq−2 A0 and X i−1 dX ∧ ω for ω ∈ Ωq−3 A0 .Thus we must know the condition when (21) follows for these elements.

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 Documenta Mathematica 5 (2000) 151-200 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 Ω1 O k 0 = 0, where O k0 is the ring of integers of k 0 .Thus π e−1 dπ = 0 in Ω1 A by taking the differential of the minimal equation of π over k 0 .
By the equation π e−1 dπ = 0 in Ω1 A , we have where {T i } = T.We introduce a filtration on Ωq A as follows.Let The subquotients are where the map is Let fil i ( Ωq A /pd Ωq−1 A ) be the image of fil i Ωq A in Ωq A /pd Ωq−1 A .Then we have the following Proposition 3.1.For j ≥ 0, where l be the maximal integer which satisfies j − le ≥ 0.
(v) For any n ≥ 0, Proof.(i) By (12), f n (ω) belongs to p nq Ωq A0 .Ωq A0 is p-torsion free, thus f n q is well-defined as the map to Ωq A0 .Furthermore, F for all l, thus the natural projection The image of the bottom arrow is

14). (iv) follows from the diagram
, and the image of the composite 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 1 e ).Note that the extension K(p 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 q 0 be the subset of Ωq A0 such that the canonical map N q 0 → Ω q F \ Z q 1 is an injection, the image generates Ω q F /Z q 1 and have the property If ω + C −1 ω = 0, then dω = 0. (25) We can take such N q 0 because of the following Documenta Mathematica 5 (2000) 151-200 Lemma 4.1.Take x ∈ Ωq F .If x + C −1 x = 0 then there exists ω ∈ Ωq A0 such that ω = x and dω = 0.
Proof.x can be written as 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 p τ = 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 τ , and we can take lifts of x τ , denoted by xτ , in the ring of Witt vectors with coefficients in E, denotes W (E). Fix an inclusion W (E) → A 0 .Let Then dω = 0 in Ωq A0 because dx τ = 0 in Ωq A0 and ω = x.This ω is the desired one.
For any q, l ≥ 0, let N q l = f l q (N q 0 ) as a subset of Ωq A0 and let Then, by (3.2,iv), N q generates Ωq A0 /p and ω = 0 in Ωq A0 /p for all ω ∈ N q .Furthermore, by using (3.2,v) and the isomorphism we have Thus the union of the sets of the left hand side of (26) generates Let S 0 i,1 , S 1 i,1 , S 0 i,2 and S 1 i,2 be the subsets of D ⊗ Ωq−2 B defined as follows.
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 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 ⊗ Ωq−2 B , 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 Thus, for ω ∈ Ωq−2 A0 (resp.ω ∈ Ωq−3 A0 ) and i ≥ 1, Here, to avoid the complication of notations, we use 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 ∈ Ωq−2 A0 (resp.ω ∈ Ωq−3 A0 ), then ω has the property (22) (resp.( 23)).Under this condition, the right hand side of (27) belongs to Ωq−1 A .Furthermore, by and if By the definition, S generates H 1 (S q ).But ψ : H 1 (S q ) → Ωq−1 A /pd Ωq−2 A has the kernel in general.The following lemma compute some subset of this kernel.
(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 ) .
(v) Now S 0 i,2 and S 1 i,2 are By (ii), we only have to calculate the element of . If e < ip ηi < ep and s ≥ 2, then The first terms of the right hand side come from On the other hand, the second terms of the right hand side are, if ip ηi ≥ 2e then vanished.If ip ηi < 2e, then by using the same argument of (iv) with j 0 = ip ηi − e and Y = L l=0 Next, we do not assume e < ip ηi < e and s ≥ 2. In this case, we have to show

Documenta Mathematica 5 (2000) 151-200
Take ω ∈ N q−3 0 .Then, by (4.3), On the other hand, there exist elements A0 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 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 Thus X j ω 1 ∈ S 0 i,1 .By the similar way, we have The claim (v) was proved.
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 ηi−1 = j.
by using (4.3) and the same kind of calculation in (4.4,iv) with the notation (31 , then take l ≥ 0 and (ω ).For this ω , we have If s = 1 and p | (j + e), then t can be taken only 0. ( for some m, then ψ(g i,0 (X i ω)) = 0 by , then take l ≥ 0 and f l q−2 ω ∈ N q−2 l for ω ∈ N q−2 0 such that ω = f ηi+l q−2 (ω ).For this ω , we have (37) Documenta Mathematica 5 (2000) 151-200 , let l ≥ 0 and f l q−2 ω ∈ N q−2 l for ω ∈ N q−2 0 such that ω = f ηi+l q−2 (ω ).For this ω , we have If t > s/2, then i ∈ Λ and , then take l ≥ 0 and f l q−2 ω ∈ N q−2 l for ω ∈ N q−2 0 such that ω = f ηi+l q−2 (ω ).For this ω , we have When we take X i pω ∈ S 1 i,1 , by the same calculation of the case t = s/2, we have 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 1 i,1 = ∅ and S 0 i,1 = X i N q−2 .For an element If p | i, then the first term of the right hand side is zero.Hence if ω = f l q−2 (ω ) then ψ(X i ω) ≡ p l π i−e f l q−1 (dω ) mod fil j+el+1 ( Ωq−1 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 ηi+l−1 q−2 (ω )), the image of ψ is when s − t > 1.Thus this element goes to gr j+el ( Ωq−1 A /pd Ωq−2 A ) and it seems non-zero.So we put g i,1 (X i pf ηi+l−1 q−2 (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 ( Ωq−1 A /pd Ωq−2 A ) and seem non-zero.The real definition of M * is as follows.
(44) and Thus we get 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 ( Ωq−1 and hence non-zero.Thus If s is even and s ≥ 2, let and for l ≥ 1.The image of (43) is the image of and the image of (36), ( 38) and ( 40) is Thus we get Thus we get If s is odd and s ≥ 3, let and 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 , is equal to the union of M j+el for all e/(p − 1) < j < e and all l ≥ 0.
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 ηi = j.But if s = 0 then there is no i ∈ Γ b such that i < e and ip ηi = j.Thus we assume s ≥ 1 to calculate when i < e.
x → (C −s dx, j 0 C −s x) ( if e < j < e , l = 0) Ω q−1 F /B q−1 s+l ( if e < j < e , l ≥ 1), where s = v p (j) and j 0 = j/p s .gr e/(p−1)+el ( Ωq−1 A pd Ωq−2 A )/ ψ(M e/(p−1)+el ) gr el ( Ωq−1 A pd Ωq−2 A )/ ψ(M el ) ∼ = gr el ( Ωq−1 A pd Ωq−2 A ) ∼ = Ω q−1 F /B q−1 l ( 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 K M q (K) ∼ = gr n ( Ωq−1 A /pd Ωq−2 A )/ ψ(M n ) .Let n > e/(p − 1).Then (78), (79), ( 80) and (81) say gr n ( Ωq−1 A pd Ωq−2 A )/ ψ(M n ) ∼ = Ω q−1 F /B q−1 s +1+k , where s = v p (n − ke).Hence we have gr n+e K M q (K) ∼ = Ω q−1 F /B q−1 s +1+k 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 , 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 K M q (K) −→ gr im K M q (L) 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 q−1 FL /B l Ω q−1 FL ∼ = Ω q−1 FK /B l Ω q−1 FK ⊗ F p l K 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 q F /B l Ω q F −→ Ω q F /B l Ω q F are injective for all q and l because, if F /F is separable extension, then Ω q F = F ⊗ F Ω q F and if F /F is purely transcendental extension F = F (T ) then Ω q F = (F ⊗ F Ω q F ) ⊕ (F ⊗ F Ω q−1 F ∧ dT ).Hence we get Theorem (1.1) by taking inductive limit.