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-groupKM
q (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 KM
q -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 MilnorK-group, Complete Discrete Val-uation Field, Higher Local Class Field Theory
1 Introduction
For a ringR, the MilnorK-group ofRis defined as follows. We denote the unit group of R byR×. Let J(R) be the subgroup of theq-fold tensor product of R×overZgenerated by the elementsa
1⊗· · ·⊗aq, wherea1, . . . , aqare elements ofR× such thata
i+aj= 0 or 1 for some i6=j. Define
KM
We denotes the image ofa1⊗ · · · ⊗aq by{a1, . . . , aq}.
Now we assume K is a discrete valuation field. Let vK be the normalized valuation of K. Let OK, F and mK be the valuation ring, the residue field and the valuation ideal ofK, respectively. There is a natural filtration onK×
defined by
UKi = (
O×K fori= 0
1 +mi
K fori≥1. We know that the graded quotients Ui
K/U i+1
K are isomorphic to F× ifi = 0 andF ifi≥1. Similarly, there is a natural filtration onKM
q(K) defined by
UiKM
q(K) = n
{x1, . . . , xq} ∈KMq(K) x1∈U
i
K, x2, . . . , xq ∈K× o
.
Let griKM
q(K) = UiKMq(K)/Ui+1KMq(K) for i ≥ 0. griKMq(K) are deter-mined 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), griKM
q(K) is determined in [3] in the range 0 ≤i ≤ eKp/(p−1), where eK = vK(p). However, griKMq(K) still remains mysterious fori > ep/(p−1). In [16], Kurihara determined griKM
q(K) for all
i ifK is absolutely unramified, i.e.,vK(p) = 1. In [13] and [19], griKM
q(K) is determined for someK with absolute ramification index greater than one.
The purpose of this paper is to determine griKM
q(K) for alliand a discrete valuation fieldK of mixed characteristics (0, p), wherepis an odd prime and
p ∤ eK. We do not assume F to be perfect. Note that the graded quotient griKM
q(K) is equal to griKMq ( ˆK), where ˆKis the completion ofKwith respect to the valuation, thus we may assume thatK is complete under the valuation.
Let F be a field of positive characteristic. Let Ω1
F = Ω1F/Z be the module
of absolute differentials and ΩqF the q-th exterior power of Ω1
F over F. As in [7], we define the following subgroups of ΩqF. Z1q =Z1Ω
q
F denotes the kernel of
d: ΩqF →Ωq+1F and B1q =B1Ω q
F denotes the image ofd: Ω q−1 F →Ω
q
F. Then there is an exact sequence
0−→Bq1−→Z1q−→C ΩqF −→0,
where C is the Cartier operator defined by
xpdy1 y1 ∧
. . .dyq yq 7−→
xdy1 y1 ∧
. . .dyq yq
,
B1q→0.
The inverse of C induces the isomorphism C−1: ΩqF −→∼= Z1q/B
q 1
xdy1 y1 ∧
. . .∧dyq yq 7−→
xpdy1 y1 ∧
. . .∧dyq yq
forx∈F andy1, . . . , yq ∈F×. Fori≥2, letBqi =BiΩ q
F (resp. Z q i =ZiΩ
q F ) be the subgroup of ΩqF defined inductively by
Biq⊃B q i−1, C
−1 :Biq−1
∼
=
−→Biq/B q 1 (resp. Ziq ⊂Z
q
i−1, C−1:Z q i−1
∼
=
−→Ziq/B q 1). LetZq
∞ be the intersection of allZ
q
i fori≥1. We denoteZ q i = Ω
q
F fori≤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=eK =vK(p)is prime top. Fori > ep/(p−1), letnbe the maximal integer
which satisfies i−ne≥e/(p−1)and lets=vp(i−ne), wherevp is thep-adic
order. Then
griKM
q(K)∼= Ω q−1 F /B
q−1 s+n. Corollary 1.2. Let Ui(KM
q(K)/pm) be the image of UiKMq(K)
in KM
q(K)/pmKMq(K) for m ≥ 1 and gri(KMq(K)/pm) =
Ui(KM
q(K)/pm)/Ui+1(KMq(K)/pm). Then
gri(KM
q(K)/pm)∼=
ΩqF−1/B q−1
s+n (if m > s+n)
ΩqF−1/Zmq−−1n (if m≤s+n, i−en6= p−e1) ΩqF−1/(1 +aC)Zmq−−1n+1 (if m≤s+n, i−en= e
p−1)
wherea is the residue class ofp/πe for a fixed prime elementπ ofK.
Remark 1.3. If 0≤i≤ep/(p−1), griKM
q (K) is known by [3].
To show (1.1), we use the (truncated) syntomic complexes with respect to
OK and OK/pOK, 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 OK and some subgroup of
KM
q(K)ˆ which includes the image ofU1KMq(K) (cf. (2.1)). On the other hand, the cohomology groups of the syntomic complex with respect toOK/pOK can be calculated easily because OK/pOK depends only onF ande. Comparing these two complexes, we have the exact sequence (2.4)
H1(Sq)−→ΩˆA/q−1Z/pdΩˆqA/−2Z expp
−−−→KM
q (K)ˆ
In Section 2, we see the relations between the syntomic complexes mentioned above, the MilnorK-groups, and the differential modules. The method of the proof of (1.1) is mentioned here. Note that we do not assumep∤ein this section and we get the explicit description of the cohomology group of the syntomic complex with respect toOK/pOKwhich was used in the proof of (1.1) without the assumption p ∤ e. In Section 3, we calculate differential module of OK. We calculate the kernel of theKM
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
xof a discrete valuation ring, ¯xmeans the residue class ofxin the residue field. For an abelian groupM and positive integern, we denote M/pn =M/pnM and ˆM = lim←−nM/pn. For a subsetN of M, hNi means the subgroup of M generated by N. For a ringR, let Ω1
R= Ω1R/Z be the absolute differentials of
R and ΩqR theq-th exterior power of Ω1
R overR forq≥2. We denote Ω0R=R and ΩqR= 0 for negativeq. IfRis of characteristic zero, let
ZnΩˆqR= Ker
ˆ
ΩqR−→d Ωˆq+1R /pn
for positive n. For an element ω ∈ ΩˆqR, let vp(ω) be the maximal n which satisfiesω∈ZnΩˆqR. Forn≤0, letZnΩˆqR= ˆΩqR. LetZ∞ΩˆqR be the intersection of ZnΩˆqR of all n≥0. All complexes are cochain complexes. For a morphism of non-negative complexesf:C·
→D·, [f:C·
→D·] and
C0 d
−−−−→ C1 d
−−−−→ C2 d
−−−−→ . . .
yf
yf
yf
D0 d
−−−−→ D1 d
−−−−→ D2 d
−−−−→ . . .
both denote the mapping fiber complex with respect to the morphism f, namely, the complex
(C0−→d C1⊕D0−→d C2⊕D1−→d . . .),
where the leftmost term is the degree-0 part and where the differentials are defined by
Ci⊕Di−1−→Ci+1⊕Di
(a, b)7−→(da, f(a)−db).
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.
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 = OK be the ring of integers of K and F the residue field ofK. Let A0 be the Cohen subring ofA with respect to F, namely, A0 is a complete discrete valuation ring under the restriction of the valuation of A with the residue fieldF and pis a prime element ofA0 (cf. [4], IX, Section 2). Let K0 be the fraction field of A0. Then K/K0 is finite and totally ramified extension of extension degree e = eK. We denote
e′ =ep/(p−1). Letπbe a prime element ofK and fix it. We further assume
that F has a finite p-base and fix their liftings T ⊂ A0. We can take the frobenius endomorphism f ofA0 such thatf(T) =Tp forT ∈ T(cf. [12] or [17]). Let UiKM
q(A) be the subgroup defined by the same way of UiKMq(K), namely,
UiKM
q(A) = D
{x1, . . . , xq} ∈KMq(A) x1∈U
i
K, x2, . . . , xq ∈A× E
.
LetUiKM
q(K)ˆ (resp. UiKMq(A)ˆ) be the closure of the image ofUiKMq(K) (resp.
UiKM
q(A)) inKMq(K)ˆ (resp. KMq(A)ˆ). Note that griKMq (K)∼= griKMq(K)ˆ for
i >0.
At first, we introduce an isomorphism betweenU1KM
q(K)ˆ and a subgroup of the cohomology group of the syntomic complex with respect toA. For further details, see [12]. Let B =A0[[X]], whereX is an indeterminate. We extend the operation of the frobenius f onB by f(X) =Xp. We define I and J as follows.
J = KerB−−−→X7→π A
I = KerB−−−→X7→π A−−−−−→modp A/p=J +pB.
LetDandJ ⊂Dbe the PD-envelope and the PD-ideal with respect toB →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 toB→A/p. LetJ[q]andI[q] be theirq-th divided powers. Notice thatI[1]=I, J[1] =J and I[0] =J[0]=D. Ifqis an negative integer, we denoteJ[q]=I[q]=D. We define the complexes
J[q] andI[q] as
J[q]=
J[q]→d J[q−1]⊗
B ˆ Ω1B
d
→J[q−2]⊗
B ˆ
Ω2B−→ · · ·
I[q]=
I[q]→d I[q−1]⊗
B ˆ Ω1B
d
→I[q−2]⊗
B ˆ
Ω2B−→ · · ·
,
where ˆΩqB is thep-adic completion of ΩqB. 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) andS′(A, B)(q) be the mapping fibers of
respectively, wherefq =f /pq. S(A, B)(q) is called the syntomic complex ofA with respect toB, andS′(A, B)(q) is also called the syntomic complex ofA/p
with respect to B (cf. [11]). We notice that
Hq(S(A, B)(q))
= Ker
(D⊗ΩˆqB)⊕(D⊗Ωˆ q−1
B )→(D⊗Ωˆ q+1
B )⊕(D⊗Ωˆ q B)
Im(J⊗ΩˆqB−1)⊕(D⊗ΩˆqB−2)→(D⊗ΩˆqB)⊕(D⊗ΩˆqB−1)
, (2)
where the maps are the differentials of the mapping fiber. Ifq≥p, we cannot define the map 1−fq onJ[q] andI[q], but we define Hq(S(A, B)(q)) by using (2) in this case. This is equal to the cohomology of the mapping fiber of
σ>q−3J[q] 1
−fq
−→ σ>q−3D,
where σ>nC· means the brutal truncation for a complexC·, i.e., (σ>nC·)i is
Ci ifi > nand 0 if i≤n. Let U1(D⊗Ωˆq−1
B ) be the subgroup of D⊗Ωˆ q−1 B generated byXD⊗ΩˆqB−1, (Xe)[m]D⊗Ωˆ
q−1
B for allm≥1 andD⊗Ωˆ q−2 B ∧dX. Let U1Hq(S(A, B)(q)) be the subgroup of Hq(S(A, B)(q)) generated by the image of (D⊗ΩˆqB)⊕U1(D⊗Ωˆ
q−1
B ). Then there is a result of Kurihara: Theorem 2.1 (Kurihara, [12]). AandB are as above. Then
U1Hq(S(A, B)(q))∼=U1KM
q(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 mapKM
q(A)ˆ→KMq (K)ˆis an injection. (ii) U1Hq(S(A, B)(q))∼=U1KM
q(A)ˆ∼=U1KMq(K)ˆ.
Remark 2.3. WhenF is separably closed, this lemma is also the consequence of the result of Kurihara [14]. But even ifF 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
U1KM
q(A)ˆ→U1K
M
q(K)ˆ
p-th root of unity and fix it. Let µp be the subgroup of A× generated byζp. Forn≥2, see the following commutative diagram.
KM
q−1(K)/p
{∗,ζp}
−−−−→ KM
q(A)/pn−1 p
−−−−→ KM
q(A)/pn =
y
y
y
KM
q−1(K)/p
{∗,ζp}
−−−−→ KM
q(K)/pn−1 p
−−−−→ KM
q(K)/pn
−−−−→ KM
q(A)/p −−−−→ 0
y
−−−−→ KM
q(K)/p −−−−→ 0.
(3)
The bottom row are exact by using Galois cohomology long exact sequence with respect to the Bockstein
· · · →Hq−1(K,Z/p(q))→Hq(K,Z/pn−1(q))→Hq(K,Z/pn(q))→. . .
and
KM
q(K)/pn∼=Hq(K,Z/pn(q))
by [3]. The map {∗, ζp} in the top row is well-defined if KMq(A)/pn−1 →
KM
q(K)/pn−1 is injective, and the top row are exact except at KMq(A)/pn−1. Using the induction onn, we only have to show the injectivity ofKM
q(A)/p→
KM
q(K)/p. We know the subquotients of the filtration of KMq(K)/p by [3] and we also know the subquotients of the filtration ofKM
q(A)/pusing the iso-morphismU1Hq(
S(A, B)(q))∼=U1KM
q(A)ˆ in [12] and the explicit calculation of Hq(
S(A, B)(q)) by [14] except gr0(KM
q(A)/p). Natural map preserves fil-trations and induces isomorphisms of subquotients. Thus U1(KM
q (A)/p) →
U1(KM
q (K)/p) is an injection. Lastly, the composite map of the natural maps
KM
q(F)/p։gr0(KMq(A)/p)→gr0(KMq(K)/p)
∼
=
→KM
q(F)/p⊕KMq−1(F)/p is also an injection. HenceKM
q(A)/p→KMq(K)/pis injective. Next, we introduceKM
q-exponential homomorphism and consider the kernel. By [17], there is theKM
q-exponential homomorphism with respect toηforq≥2 andη∈K such thatvK(η)≥2e/(p−1) defined by
expη : ˆΩqA−1−→K
M
q(K)ˆ
adb1 b1 ∧ · · · ∧
dbq−1
bq−1 7−→ {exp(ηa), b1, . . . , bq−1}
(4)
fora∈A, b1, . . . , bq−1∈A×. Here exp is
exp(X) =
∞
X
n=0
Xn
We use this KM
q-exponential homomorphism only in the case η = p in this paper. On the other hand, there exists an exact sequence of complexes
0→
σ>q−3J[q]
↓1−fq σ>q−3D
→
σ>q−3I[q]
↓1−fq σ>q−3D
→
σ>q−3I[q]/σ>q−3J[q]
↓
0
→0. (5)
[σ>q−3I[q]/σ>q−3J[q] →0] is none other than the complexσ>q−3I[q]/σ>q−3J[q]. We denote the complex [σ>q−3I[q] 1
−fq
→ σ>q−3D][q−2] bySq. It is the map-ping fiber complex
I[2]⊗Ωˆq−2 B
d
−−−−→ I⊗ΩˆqB−1 −−−−→d D⊗ΩˆqB −−−−→d . . .
y1−fq
y1−fq
y1−fq
D⊗ΩˆqB−2 −−−−→d D⊗ΩˆBq−1 −−−−→d D⊗ΩˆqB −−−−→d . . .
. (6)
Taking cohomology, we have the following
Proposition 2.4. A, B and K are as above. Then KM
q-exponential
homo-morphism with respect topfactors throughΩˆqA−1/pdΩqA−2and there is an exact sequence
H1(Sq)−→ψ ΩˆAq−1/pdΩˆqA−2−−−→expp KM
q (K)ˆ.
Proof. See the cohomological long exact sequence with respect to the exact sequence (5). Theq-th cohomology group of the left complex of (5) is equal to
Hq(S(A, B)(q)), thus the sequence
H1(Sq) ψ
→H1((σ>q−3I[q]/σ>q−3J[q])[q−2])→Hq(S(A, B)(q))
is exact. Here we denote the first map by ψ. The complex (σ>q−3I[2]/σ>q−3J[2])[q−2] is
(I[2]
⊗ΩˆqB−2)/(J[2]
⊗ΩˆqB−2)→(I⊗ΩˆBq−1)/(J⊗ΩˆBq−1)→0→ · · ·.
(I⊗ΩˆqB−1)/(J⊗ΩˆqB−1) is the subgroup of (D⊗ΩˆBq−1)/(J⊗ΩˆBq−1) =A⊗ΩˆqB−1. The image ofI⊗ΩˆqB−1inA⊗ΩˆqB−1is equal topA⊗ΩˆqB−1. Thus (I⊗ΩˆqB−1)/(J⊗
ˆ
ΩqB−1) =pA⊗ΩˆqB−1. The image of
(I[2]⊗ΩˆqB−2)/(J[2]⊗ΩˆqB−2)−→d pA⊗ΩˆqB−1 is equal to the image of I2
⊗ΩˆqB−2. ByI= (p) +J,d(I2
⊗ΩˆqB−2) is equal to
d(J2
⊗ΩˆqB−2) +pd(J ⊗ΩˆBq−2) +p2d( ˆΩq−2
we have an exact sequence
(J/J2)⊗ΩˆBq−2−→d A⊗ΩˆqB−1−→ΩˆqA−1−→0. (7) Thus the image ofd(J2
⊗ΩˆqB−2) inA⊗ΩˆBq−1is zero. A⊗ΩˆqB−1is torsion free, thus
pA⊗ΩˆqB−1
pd(J ⊗ΩˆqB−2) +p2dΩˆq−2 B
p−1
∼
= A⊗Ωˆ q−1 B
d(J ⊗ΩˆqB−2) +pdΩˆqB−2 ∼= ˆΩ q−1 A /pdΩˆ
q−2 A . (8)
Hence we have H1((σ
>q−3I[2]/σ>q−3J[2])[q−2])∼= ˆΩqA−1/pdΩˆ q−2
A . By chasing the connecting homomorphism ˆΩqA−1/pdΩˆ
q−2
A →Hq(S(A, B)(q)), we can show that the image is contained byU1Hq(
S(A, B)(q)) and the composite map ˆ
ΩAq−1→ΩˆAq−1/pdΩˆqA−2→U1Hq(S(A, B)(q))→∼= U1KM
q(K)ˆ is equal to expp. We got the desired exact sequence.
Remark 2.5. By [3], there exist surjections ΩqF−2⊕ΩqF−1−→griKM
q(K)
xdy1 y1 ∧ · · · ∧
dyq−2
fq−2
,0
7−→ {1 +πix,˜ y˜1, . . . ,y˜q−2, π}
0, xdy1 y1 ∧ · · · ∧
dyq−1
fq−1
7−→ {1 +πix,˜ y˜1, . . . ,y˜q−1}
(9)
for i≥1, wherex∈F,y1, . . . , yq−1 ∈F× and where ˜x,y˜1, . . . ,y˜q−1 are their liftings toA. If i≥e+ 1, then we can construct all elements of griKM
q (K) as the image of expp, namely,
n
ω∈ΩˆqA−1/pdΩˆqA−2
expp(ω)∈UiKMq(K)ˆ oexpp
−→griKM
q(K)
πi−1
p a db1
b1 ∧ · · · ∧
dbq−2
bq−2 ∧
dπ7−→ {exp(πia), b1, . . . , bq−2, π} ={1 +πia, b1, . . . , bq−2, π}
πi
pa db1
b1 ∧ · · · ∧
dbq−1
bq−1 7−→ {
exp(πia), b
1, . . . , bq−1} ={1 +πia, b
1, . . . , bq−1}. ThusUe+1KM
q(K)ˆ is contained by the image of expp. On the other hand, (2.4) says the kernel of the KM
q-exponential homomorphism is ψ(H1(Sq)). Recall that the aim of this paper is to determine griKM
q(K) for alli, but we already know them in the range 0≤i≤e′ in [3]. Thus if we want to know griKM
q(K) for alli, we only have to knowψ(H1(Sq)). We determineH1(Sq) in the rest of this section, andψ(H1(S
To determine H1(S
q), we introduce a filtration into it. Let 0 ≤ r < p and s ≥ 0 be integers. Recall that B = A0[[X]]. For i ≥ 0 and s ≥ 0, let fili(I[r]
⊗Ωˆs
B) be the subgroup ofI[r]⊗ΩˆsB generated by the elements n
Xn(Xe)[j]ω
n+ej≥i, n≥0, j≥r, ω∈D⊗ΩˆsB o
∪nXn−1(Xe)[j]ω∧dX
n+ej≥i, n≥1, j≥r, ω∈D⊗ΩˆsB−1 o
.
The homomorphism 1−fr+s:I[r]⊗ΩˆsB→D⊗ΩˆsB preserves filtrations. Thus we can define the following complexes
fili(σ>q−3I[q])[q−2]
= fili (I[2]⊗ΩˆqB−2)→fil i
(I⊗ΩˆqB−1)→fil i
(D⊗ΩˆqB)→. . .
fili(σ>q−3D)[q−2]
= fili (D⊗ΩˆBq−2)→fili(D⊗ΩˆqB−1)→fili(D⊗ΩˆqB)→. . .
filiSq=
fili(σ>q−3I[q])[q−2] 1−fq
−→ fili(σ>q−3D)[q−2]
gri(σ>q−3I[r])[q−2] = fil i
(σ>q−3I[r])[q−2] fili+1(σ>q−3I[r])[q−2]
forr= 0, q
griSq=
gri(σ>q−3I[q])[q−2] 1−fq
−→ gri(σ>q−3D)[q−2]
.
Note that if i≥1, 1−fq: gri(σ>q−3I[q])[q−2]→gri(σ>q−3D)[q−2] is none other than 1 because fq takes the elements to the higher filters. filiSq forms the filtration ofSq and we have the exact sequences
0−→fili+1Sq −→filiSq −→griSq −→0
fori≥0. This exact sequence of complexes give a long exact sequence
· · · →Hn(fili+1Sq)→Hn(filiSq)→Hn(griSq)→Hn+1(fili+1Sq)→. . . (10)
Furthermore, we have the following
Proposition 2.6. {H1(filiSq)}i forms the finite decreasing
filtra-tion of H1(Sq). Denote filiH1(Sq) = H1(filiSq) and griH1(Sq) = filiH1(Sq)/fili+1H1(S
griH1(Sq) =
To prove (2.6), we need the following lemmas. Lemma 2.7. For ω∈D⊗ΩˆqB andn≥0, the canonical generators of ˆΩqB, which are
Proof. Byi > er, fili(I[r]
⊗Ωˆs B) = fil
i(D
⊗Ωˆs
B) because Xi =r!Xi−er(Xe)[r]. All elements of filiD⊗Ωˆs
B can be written as the sum of the elements of the form Xn(Xe)[j]ω, where ω
∈ D ⊗Ωˆs
B and n+ej ≥ i. Now r < p, thus (Xe)[r]=Xer/r! in D, hence we may assumej
≥r. The image ofXn(Xe)[j]ω is
∞
X
m=0
fr+sm (Xn(Xe)[j]ω) =
∞
X
m=0 (pmj)!
prm(j!)X npm
(Xe)[pmj]f m(ω)
psm .
Here, fm(ω) is divisible by psm by (11). The coefficients (pmj)!/prm(j!) are
p-integers for allmand ifj≥1 then the sum convergesp-adically. Ifj =r= 0,
n≥1 says that the order of the power ofX is increasing. This also means the sum convergesp-adically inD⊗BΩˆsB. The image is in fil
i (I[r]⊗
BΩˆsB) because
pmj ≥rfor allm, thus the map is well-defined. Obviously, P∞
m=0fr+sm is the inverse map of 1−fr+s.
Lemma 2.9. Let i≥1 and e≥1 be integers. For each n≥0, let mn (resp.
m′
n) be the maximal integer which satisfies ipn≥mne(resp. ipn−1≥m′ne).
Then
Min{vp(mn!) +mn−n}n =
(
1−ηi≤0 (whenn=ηi−1, if ηi≥1)
vp(m0!) +m0≥1 (whenn= 0, if ηi= 0) Min{vp(m′n!) +m′n−n}n
= (
1−η′
i≤0 (whenn=ηi′−1, if ηi′≥1)
vp(m′
0!) +m′0≥1 (whenn= 0, if η′i= 0),
whereηi andηi′ are as in (2.6).
Proof. By the definition of{mn}n,mn+1is greater than or equal topmn. Thus
vp(m′
n+1!)≥vp(pm′n!) and
vp(mn+1!) +mn+1−(n+ 1)−(vp(mn!) +mn−n) =vp(mn+1!)−vp(mn!) +mn+1−mn−1
(13)
is greater than zero ifmn>0. On the other hand,ηi is the number which has the property that ifn < ηi, thenmn = 0 andmηi ≥1. Thus the value of (13)
Proof of Proposition 2.6. At first, we show that{H1(filiS
q)}i forms the finite decreasing filtration ofH1(S
q). See Ifi≥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 givesH0(griS
is also injective because of the invariance of the valuation ofA0by the action of
f. Thus the exact sequence (15) also follows wheni= 0. Hence{H1(filiSq)}i forms a decreasing filtration ofH1(S
q). Next we calculate H1(griS
q). If i >2e, filiSq is acyclic by (2.8). Thus we only consider the case i ≤ 2e. Furthermore, if i ≥ 1, we may consider that
H1(griS
q) is the subgroup of griD⊗ΩˆqB−2 because of the injectivity of the vertical arrows of (14).
Leti= 2e. Then gr2eS
The second vertical arrow is a surjection, thus
H1(gr2eSq)∼=X2e−1dX∧( ˆΩqA−03/p). (16)
The second vertical arrow is also a surjection, thus
H1(griSq)∼=Xi( ˆΩqA−02/p)⊕X
i−1dX
Leti=e. Then greS
The second vertical arrow is not a surjection. For an elementXeω
∈XeΩˆq−2
Ifi= 0, we need more calculation. The complex gr0S q is
We introduce ap-adic filtration to gr0S
Then, for allm≥0,
The injectivity of the leftmost vertical arrow of (20) says that
H0(grm (15) is not surjective in general. So we must know the image of H1(filiS
q)→
Now the second vertical arrow is an injection. Thus x also represents the element ofH1(filiSq) if and only if
q) are represented by two types of the elements of
D⊗ΩˆqB−2, these are Xiω forω ∈Ωˆ q−2 A0 andX
i−1dX∧ω forω ∈Ωˆq−3 A0 . Thus
Here mn and m′n be the maximal integers which satisfy ipn−mne ≥ 0 and
ipn
−1−m′
ne≥0. Note thatfn(dω) andfn(ω) can be divided bypn(q−1)and
pn(q−2), respectively, by (11). Furthermore, vp(fn(dω)/pn(q−1)) =vp(dω) and
vp(fn(ω)/pn(q−2)) = v
p(ω) by (12). To be included in I⊗ΩˆqB−1, the sum of thep-adic order and divided power degree must be greater than or equal to 1, i.e.,vp(mn!)−n+mn+vp(dω)≥1 andvp(i) +vp(m′n!)−n+m′n+vp(ω)≥1 must be satisfied for alln. We already know the minimal ofvp(mn!)−n+mn andvp(m′
n!)−n+m′n by (2.9), thus P∞
n=0fqn(dXiω) belongs toI⊗ q−1 B if and only if
no condition ( ife+ 1≤i)
vp(i) +vp(ω)≥1 ( ife=i)
vp(dω)≥ηi andvp(i) +vp(ω)≥ηi′ ( if 1≤i < e).
(22)
Next, we calculateXi−1dX
∧ω forω∈ΩˆqA−03.
∞
X
n=0
fn
q(dXi−1dX∧ω) =
∞
X
n=0
fn
q(Xi−1dX∧dω)
=
∞
X
n=0 pn
pnqX
ipn−1dX
∧fn(dω)
=
∞
X
n=0
(m′n!)
pn X
ipn−1−m′
ne(Xe)[m′n]dX∧f
n(dω)
pn(q−2)
.
To be included inI⊗ΩˆqB−1,vp(m′
n!)−n+m′n+vp(dω)≥1 must be satisfied for alln. As the same way as above,P∞
n=0fqn(Xi−1dX∧ω) belongs toI⊗ q−1 B if and only if
(
no condition ( ife+ 1≤i)
vp(dω)≥η′
i ( if 1≤i≤e).
(23)
Forω ∈ΩˆqA−01, the conditionvp(ω)≥n meansω ∈pnΩˆq−1
A0 and vp(dω)≥n
means ω ∈ ZnΩˆqA−01. Thus, by (16), (17), (18), (19), (22) and (23), we get
(2.6).
3 Differential modules and filtrations
LetK,A, A0,K0 andB are as in Section 2. We assume thatp∤e=eK, i.e.,
K/K0 is tamely totally ramified extension from here. Letk be the constant field of K (cf. [18]), i.e., k is the complete discrete valuation subfield of K
residue field of k is the maximal perfect subfield of F. Then there exists a prime element ofKsuch that πis the element ofk. Letk0=K0∩k. Thenπ is algebraic overk0 and we get ˆΩ1Ok0 = 0, whereOk0 is the ring of integers of k0. Thusπe−1dπ= 0 in ˆΩ1
Aby taking the differential of the minimal equation ofπoverk0.
By the equationπe−1dπ= 0 in ˆΩ1
A, we have ˆ
ΩqA∼=
M
i1<i2<···<iq AdTi1
Ti1
∧ · · · ∧dTiq Tiq
⊕
M
i1<i2<···<iq−1
A/(πe−1)dTi1 Ti1
∧ · · · ∧dTiq−1 Tiq−1
∧dπ
,
(24)
where{Ti}=T. We introduce a filtration on ˆΩqA as follows. Let
filiΩˆqA= (
ˆ
ΩqA ( ifi= 0)
πiΩˆq
A+πi−1dπ∧Ωˆ q−1
A ( ifi≥1). The subquotients are
griΩˆqA= fil iˆ
ΩqA/fil i+1ˆ
ΩqA =
(
ΩqF ( ifi= 0 ori≥e) ΩqF ⊕Ω
q−1
F ( if 1≤i < e), where the map is
πiΩˆqA∋πiω7−→ω¯∈ΩqF
πi−1dπ∧ΩˆqA−1∋πi−1dπ∧ω7−→ω¯ ∈ΩqF−1.
Let fili( ˆΩqA/pdΩˆqA−1) be the image of filiΩˆqA in ˆΩqA/pdΩˆqA−1. Then we have the following
Proposition 3.1. For j≥0,
grjΩˆqA/pdΩˆqA−1=
ΩqF (j = 0) ΩqF⊕ΩqF−1 (1≤j < e) ΩqF/Blq (e≤j),
wherel be the maximal integer which satisfiesj−le≥0. Proof. If 1 ≤ j < e, grjΩˆq
A = grj( ˆΩ q A/pdΩˆ
q−1
A ) because pdΩˆ q−1 A ⊂ fil
eΩˆq A. Assume that j ≥ e and let l as above. By (24) and piedπ = 0, ˆΩq−1
A is generated by the elements pπidω for 0
≤i < e and ω ∈ ΩˆqA−01. By [7] (Cor. 2.3.14),pπidω
∈file(1+n)+iΩˆqAif and only if the residue class ofp−ndωbelongs to Bn+1. Thus grj( ˆΩq
A/pdΩˆ q−1 A )∼= Ω
q F/B
We need the lemma in the following sections. Lemma 3.2. (i) For n≥0, there exist maps
fqn =
fn
pnq : ˆΩ q
A0 −→ZnΩˆ
q A0.
(ii) Forn≥1,
ZnΩˆqA0 =
n−1 X
l=0
plfn−l q Ωˆ
q A0
! +pnΩˆq
A0+Z∞Ωˆ
q A0.
(iii) Forn≥1,
n−1 M
i=0
d pi ◦f
i q−1:
ˆ
ΩqA−01/Z1ΩˆqA−01
⊕n
−→ΩˆqA0/p∼= ΩqF
is injective and the image isBnΩqF.
(iv) For anyn≥0,
ˆ
ΩqA0/Z1ΩˆqA0
fn q
−→ZnΩˆqA0/(Zn+1
ˆ
ΩqA0+ZnΩˆqA0∩p
ˆ ΩqA0)
is an isomorphism.
(v) For anyn≥0,
ˆ
ΩAq0/p⊕ΩˆqA−01/Z1ΩˆqA−01
⊕n fqn⊕Ln−1i=0 d pi◦f
i q−1
−−−−−−−−−−−−→ ZnΩˆ
q A0
ZnΩˆqA0∩pΩˆ
q A0
is an isomorphism.
Proof. (i) By (12),fn(ω) belongs topnqΩˆq A0. ˆΩ
q
A0 isp-torsion free, thusf
n q is well-defined as the map to ˆΩqA0. Furthermore,
d(fqn(ω)) = 1
pnqf
n(dω) =pnfn q+1(dω), thusfn
q(ω)∈ZnΩˆqA0.
(ii) For 0≤l≤n−1, the image of the natural injection ZnΩˆqA0∩plΩˆqA0
(ZnΩˆqA0∩p
l+1Ωˆq
A0) + (Z∞
ˆ
ΩqA0∩plΩˆq A0)
−→ p
lΩˆq A0 pl+1Ωˆq
A0+ (Z∞Ωˆ
q A0∩p
lΩˆq A0)
∼
is coincide withZn−lΩqF/Z∞ΩFq by [7] (Cor. 3.2.14). The image ofplfqn−lΩˆ q A0
is alsoZn−lΩqF/Z∞ΩqF for alll, thus the natural projection n−1
X
l=0
plfqn−lΩˆ q A0
!
−→ ZnΩˆ
q A0 pnΩˆq
A0+Z∞Ωˆ
q A0
is surjective. Hence we have (ii). (iii) The following diagram commute
ˆ
ΩqA−01/Z1ΩˆqA−01
⊕n Ln−1i=0 pid◦f i q−1
−−−−−−−−−→ ΩˆqA0/p ∼=
y
∼
= y
ΩqF−1/Z1q−1⊕n
Ln−1 i=0 C
−id
−−−−−−−−→ ΩqF.
The image of the bottom arrow isBq n. (iv) The image ofZnΩˆqA0/(ZnΩˆ
q A0∩pΩˆ
q
A0) under the isomorphism ˆΩ
q A0/p→
ΩqF isZnq by [7] (Cor. 3.2.14). (iv) follows from the diagram ˆ
ΩqA0/Z1ΩˆqA0 fqn
−−−−→ ZlΩˆqA0/(Zn+1ΩˆqA0+ZnΩˆqA0∩pΩˆqA0)
∼
= y
∼
= y
ΩqF/Z1q −−−−→C−n Zq
n/Z q n+1. (v) The image of
ˆ
ΩqA−01/Z1ΩˆqA−01
⊕n Ln−1i=0 pid◦f i q−1
−−−−−−−−−→ΩˆqA0/p∼= ΩqF isBqn by (iii), and the image of the composite
ˆ ΩqA0/p f
n q
−→ΩˆqA0/p∼= ΩqF →Ω q F/B
q n isZq
n/Bqn. Hence we get (v). 4 The image of H1(Sq)
We assumep∤e. We further assume that there exists the prime element πof
K such thatπe=p. If there does not exist suchπ, we replace K by K(p1 e).
Note that the extensionK(p1e)/K is unramified of degree prime top. In this
section, we calculateψ(H1(S
q)) explicitly. We need some preparations. LetN0q be the subset of ˆΩ
q
A0 such that the canonical mapN
q 0 →Ω
q F \Z
q 1 is an injection, the image generates ΩqF/Z1q and have the property
Lemma 4.1. Take x∈ΩˆqF. If x+ C−1x= 0 then there exists ω ∈ΩˆqA0 such that ω¯ =xanddω= 0.
Proof. xcan be written as
x=X
τ
xττ,
where τ runs through the canonical generators (cf. in the proof of (2.7)) and
xτ∈F. The assumptionx+ C−1x= 0 means thatxτ+xpτ = 0 for allτ, thus
xτ∈Efor allτ, whereE is the maximal perfect subfield ofF. The canonical generators have the fixed lifts denoted by ˜τ, and we can take lifts ofxτ, denoted by ˜xτ, in the ring of Witt vectors with coefficients inE, denotes W(E). Fix an inclusionW(E)→A0. Let
ω=X
τ ˜
xττ .˜
Thendω= 0 in ˆΩqA0 becausedx˜τ = 0 in ˆΩqA0 and ¯ω=x. Thisω is the desired one.
For anyq, l≥0, letNlq=fl q(N
q
0) as a subset of ˆΩ q
A0 and let N∞q =Z∞ΩˆqA0\(Z∞ΩˆqA0∩pΩˆqA0), Nfq =[
l≥0
Nlq, Nq =Nfq∪N∞q .
Then, by (3.2,iv), Nq generates ˆΩq
A0/p and ω 6= 0 in ˆΩ
q
A0/p for all ω ∈ N
q. Furthermore, by using (3.2,v) and the isomorphism
Zn−1ΩˆqA0 Zn−1ΩˆqA0∩p
ˆ ΩqA0
p
−→ ZnΩˆ
q A0∩pΩˆ
q A0
ZnΩˆqA0∩p
2Ωˆq A0
,
we have *
fqnNq∪ n−1
[
m=0
d pmf
m q−1N
q−1 0
+
= ZnΩˆ q A0
ZnΩˆqA0∩pΩˆ
q A0
,
*
pfqn−1Nq∪ n−2
[
m=0
p d pmf
m q−1N
q−1 0
+ = Zn
ˆ
ΩqA0∩pΩˆqA0 ZnΩˆqA0∩p
2Ωˆq A0
.
(26)
Thus the union of the sets of the left hand side of (26) generates ZnΩˆqA0/(ZnΩˆ
q A0∩p
2Ωˆq
A0). Ifq <0 then letN
LetS0
i,1,Si,11 ,Si,20 andSi,21 be the subsets ofD⊗Ωˆ q−2
B defined as follows.
Si,10 =
The following lemma is useful to calculateψ.
Lemma 4.2. If 1 ≤i < e then the minimal value of vK(πip
Proof. Lemma follows from the definition ofηi.
Thus, forω∈ΩˆqA−02 (resp. ω∈ΩˆqA−03) andi≥1,
Here, to avoid the complication of notations, we use
XidX this condition, the right hand side of (27) belongs to ˆΩqA−1. Furthermore, by
Lemma 4.4. (i) S2e, Se⊂Kerψ. (ii) If e < i < 2e then ψ(S0
i,2 \(Xi−1dX ∧N q−3
0 )) = 0. If 1 ≤ i < e
then ψ(S0
i,2 \ (Xi−1dX ∧ f ηi
q−3N q−3
0 )) = 0 and ψ(Si,21 \ (Xi−1dX ∧
pfηi−1
q−3 N q−3 0 )) = 0.
(iii) Ife < i <2eandp∤i, thenψ(Si,2)⊂ hψ(Si,1)i.
(iv) Ife′ < i <2eandp
|i, then
ψ(Si,2)⊂
*
ψ
[
1≤j<e
Sj,11
+
.
(v) Let 1 ≤i < e, s= ηi+vp(i) and i0 =i/pvp(i). If e′ < ipηi < ep and
s≥2, then
ψ(Si,2)⊂ *
[
1≤j<e
Sj,1
∪Sipηi−e,1 +
.
Furthermore, let
j= (
i0p
s
2 (ifs is even), i0p
s−1
2 (ifs is odd).
Then
ψ(Si,20 )⊂ hψ(Sj,1)i if 3ηi≥vp(i),
ψ(Si,21 )⊂ hψ(Sj,1)i if 3ηi≥vp(i) + 2.
Proof. (i) TakeX2e−1dX
∧ω∈S2e,2. Then
ψ
X2edX X ∧ω
= 1
pπ
2edπ
π ∧dω= 0
by (4.3). Next, takeXe−1dX
∧ω ∈Se,2. By the definition of Se,2, such an ω
can be divided byp. Thus, by using (4.3), we get
ψ
XedX X ∧ω
=1
pπ
edπ
π ∧p dω
p = 0.
(ii) At first, lete < i <2e. When we takeXi−1dX
∧ωfromS0
i,2\(Xi−1dX∧
N0q−3), thenω has the propertyvp(dω)≥1. Thus by using (4.3),
ψ
XidX X ∧ω
=1
pπ
idπ
π ∧p dω
By (ii), we only have to calculate the element of
XidX X ∧f
ηi
q−3N q−3 0 , Xi
dX X ∧pf
ηi−1
q−3 N q−3 0
to show (v). Ife′< ipηi< epand s≥2, then
XidX X ∧f
ηi
q−3ω=π
ipηi−1dπ π ∧f
2ηi−1
q−2 (dω) +πipηi−edπ
π ∧f
2ηi
q−2(dω),
XidX X ∧pf
ηi−1
q−3 ω=πip
ηi−1dπ π ∧f
2ηi−2
q−2 (dω) +πipηi−edπ
π ∧f
2ηi−1
q−2 (dω). The first terms of the right hand side come from
Sipηi−1+e,1⊃Xipηi −1+e
N∞q−2
∋Xipηi−1+ef2ηi−1
q−2 (dω) ψ
−→eπipηi−1dπ π ∧f
2ηi−1
q−2 (dω),
Sipηi−1+e,1⊃Xip ηi−1
+eNq−2
∞
∋Xipηi−1+ef2ηi−2
q−2 (dω) ψ
−→eπipηi−1dπ π ∧f
2ηi−2
q−2 (dω). On the other hand, the second terms of the right hand side are, if ipηi ≥2e
then vanished. If ipηi < 2e, then by using the same argument of (iv) with j0=ipηi−eand
Y = (PL
l=0 (−1)l
jl X
jlpfl
q−2(f 2ηi
q−2(dω)) ( the first case ) PL
l=0 (−1)l
jl X
jlpfl
q−2(f 2ηi−1
q−2 (dω)) ( the second case ), we get
ψ(Si,2)⊂ *
[
1≤j<e
Sj,1
∪Sipηi−e,1 +
.
Next, we do not assume e < ipηi < e′ ands ≥2. In this case, we have to
show
ψ
XidX X ∧f
ηi
q−3N q−3 0
⊂ψ(Sj,1) if 3ηi≥vp(i),
ψ
XidX X ∧pf
ηi−1
q−3 N q−3 0
Takeω∈N0q−3. Then, by (4.3), On the other hand, there exist elements
ψ(Xjω4′) =
i0
pπ
i0ps−1dπ π ∧f
s−1 2
q−2(ω′4) +
i0
p2π i0psdπ
π ∧f s+1
2
q−2(ω4′) =i0πip
ηi−1dπ π ∧f
2ηi−2
q−2 (dω) +i0πip
ηi−edπ
π ∧f
2ηi−1
q−2 (dω) =i0ψ
XidX X ∧pf
ηi−1
q−3 (ω)
.
Compare the definition ofSj,1with the condition ofω1′, . . . , ω4′. Ifsis even and 3ηi≥vp(i) then
Xjω′1= (
Xj d p2ηi−s2f
2ηi−s2
q−2 (dω)∈Xjp2ηid−s2f
2ηi−s2
q−2 N q−3
0 ifηi−s2 ≤ηj−1,
Xjfηj
q−2f
2ηi−s2−ηj
q−2 (dω)∈Xjf ηj
q−2N∞q−2 ifηi−s2 ≥ηj. ThusXjω′
1∈Si,10 . By the similar way, we have
Xjω2′ ∈Sj,11 ( ifsis odd and 3ηi≥vp(i)),
Xjω3′ ∈Sj,10 ( ifsis even and 3ηi≥vp(i) + 2),
Xjω′
4∈Sj,11 ( ifsis odd and 3ηi≥vp(i) + 2). The claim (v) was proved.
Remark 4.5. LetS′0
i,2 (resp. S
′1
i,2) be the subset ofSi,20 (resp. S1i,2 ) defined as follows.
Si,2′0 =
Xi dX X ∧N
q−3
0 (e < i≤e′, p|i)
Xi dX X ∧f
ηi
q−3N q−3
0 1≤i < e, e < ipηi ≤e′, 3ηi< vp(i))
∅ ( otherwise ),
Si,2′1 = (
Xi dX X ∧pf
ηi−1
q−3 N q−3
0 (1≤i < e, e < ipηi≤e′, 3ηi< vp(i) + 2)
∅ ( otherwise ).
Remark that if 1 ≤ i < e satisfies vp(i) +ηi = 1 and e′ < ipηi < ep, then
vp(i) = 0, e/(p−1) < i < e and ηi = 1. Thus this i satisfies neither 3ηi <
vp(i) + 2 nor 3ηi < vp(i). LetSi,2′ =S
′0
i,2∪S
′1
i,2. Then by (4.4), ψ(H1(Sq)) is generated by
[
1≤i<2e
Si,1
∪
[
1≤i<2e
Si,2′
.
We need some modification of generators ofψ(H1(S
q)) as follows. Let the index sets Λ0and Λ1be
Λ0={i; 1≤i < e, e′ < ipηi <2e, η
Λ1={i; 1≤i < e, e′< ipηi<2e, η
i =vp(i) + 1, p∤(i+e)}. (30) Let Λ = Λ0∪Λ1. Fori∈Λ, let
s=ηi+vp(i),
i′=i/pvp(i), i0=i′ps−1,
il=il−1p−eforl≥1,
L= Min{l ; il≥2e/p}.
(31)
{il}l are monotonely increasing, thus we can take such L. Note that p ∤ i′,
ηil= 1 for 0≤l≤Landp∤ilforl≥1. Ifi∈Λ0then letgi,0 be
gi,0(Xiω) = 1
i′X
iω
−i 1
0+e
Xi0+efηi−1
q−2 (ω) + L−1 X
l=1 (−1)l
il
pXilfηi+l−1
q−2 (ω)
for ω ∈ ZηiΩˆ
q−2
A0 . This function satisfies gi,0(ω) ≡ (1/i
′)Xiω modulo fili+1H1(Sq), thus we can replace S0
i,1 by gi,0(Si,10 ) to generate ψ(H1(Sq)). Wheni∈Λ1, then letgi,1 be
gi,1(Xipω) = 1
i′X
ipω
−i 1
0+eX
i0+efηi−1
q−2 (ω) + L−1 X
l=1 (−1)l
il pX
ilfηi+l−1
q−2 (ω)
forpω∈pΩˆqA−02∩ZηiΩˆ
q−2
A0 . This function satisfiesgi,1(pω)≡(1/i
′)Xipωmodulo fili+1H1(Sq), thus we can replaceS1
i,1 bygi,1(Si,11 ) to generateψ(H1(Sq)). 5 Explicit Calculation, Case (a)
We computeψ(Si,1),ψ(S′
i,2),ψ(gi,0Si,10 ) andψ(gi,1Si,11 ) explicitly in Section 5, 6 and 7.
Define the index sets as Γa =
i
1≤i < e,
e p−1 < ip
ηi−1< e
∪ni e
′< i <2eo,
Γb =ni
1≤i < e, e < ip
ηi < e′o
∪ni
e < i < e
′o,
Γc = e
p−1, e
′
.
These sets are disjoint to each other, and Γa∪Γb∪Γcis coincide with{i; 1≤
i <2e, i6=e}. Λ is the subset of Γa. In this section, we computeψ(Si,1∪Si,2′ ) for i∈Γa\Λ,ψ(gi,0(Si,10 )∪Si,11 ∪Si,2′ ) for i∈Λ0 andψ(gi,1(Si,11 )∪Si,2′ ) for
At first, we computeψ wheni∈Γa and 1≤i < e. Lete/(p−1)< j < e,
s=vp(j) + 1 andj0=j/ps−1. Then the integersiwhich satisfyipηi−1=jare (i, ηi) = (j0, s),(j0p, s−1), . . . ,(j0ps−1,1).
Leti=j0pt. Theni∈Γa for allt. Notice that ifi∈Γa andi < ethen there existse/(p−1)< j < esuch that ipηi−1=j.
Ift < s−21 thenSi,1=∅.
Ift= (s−1)/2 and p|(j+e), theni∈Λ1,S0
i,1=∅and
Si,11 =Xi pf ηi−1
q−2 Nq−2∪ ηi−2
[
m=0
p d pmf
m q−3N
q−3 0
!
.
ForXipω
∈S1
i,1,ψ(gi,1(Xipω)) is
ψ(gi,1(Xipω)) =X n≥0
pπipn
i′pn+1f n q−1(dω)
−X
n≥0
pηi−1
(i0+e)pn+1
π(i0+e)pnfηi+n−1
q−1 (dω)
+ L−1 X
l=1 X
n≥0
(−1)lpηi+l−1 ilpn
πilpnfηi+l+n−1
q−1 (dω)
(32)
by using (4.3) and the same kind of calculation in (4.4,iv) with the nota-tion (31). If pω ∈ pfηi−1
q−2 N∞q−2 or pω ∈ ppdmfqm−3N q−3
0 for some m, then
ψ(g′
i(Xipω)) = 0 by (32). If pω ∈ pf ηi−1
q−2 N q−2
f , then take l ≥ 0 and
fl
q−2ω′ ∈ N q−2
l for ω′ ∈ N q−2
0 such that ω = f ηi+l−1
q−2 (ω′). For this ω′, we have
ψ(gi,1(Xipfηi+l−1
q−2 (ω′)))≡ (pl
i′πip ηi−1
f2ηi+l−2
q−1 (dω′) ( ifηi6= 1) epl
i(i+e)π ifl
q−1(dω′) ( ifηi= 1) mod filipηi−1+el+1( ˆΩqA−1/pdΩˆ
q−2 A ). Nowipηi−1=j andη
i=s−t= (s+ 1)/2,
ψ(gi,1(Xipfηi+l−1
q−2 (ω′)))≡ (pl
i′πjf
s+l−1
q−1 (dω′) ( ifs−t >1) epl
i(i+e)πjfql−1(dω′) ( ift= 0, s= 1) mod filj+el+1( ˆΩqA−1/pdΩˆqA−2).
(33)
If s = 1 and p | (j+e), then t can be taken only 0. S0
i,1 = ∅. This i is not in Λ1, hence we compute S1
Xipω
∈XipNq−2,
ψ(Xipω) =iπi−1dπ∧ω+iπip−e−1dπ∧fq−2(ω)
+X n≥0
1
pnπ ipn
∧fqn−1(dω)
≡iπi−1dπ∧ω+πi∧dω
mod fili+1( ˆΩqA−1/pdΩˆqA−2).
(34)
Ift=s/2, theni∈Λ0 and
Si,10 =Xi f ηi
q−2Nq−2∪ ηi−1
[
m=0
p d pmf
m q−3N
q−3 0
!
.
ForXiω∈S0
i,1,ψ(gi,0(Xiω)) is
ψ(gi,0(Xiω)) =X n≥0
πipn i′pn+1f
n q−1(dω)
−X
n≥0 1 (i0+e)pn+1
π(i0+e)pnfn
q−1(df ηi−1
q−2 (ω))
+ L−1 X
l=1 X
n≥0
(−1)lpηi+l−1 ilpn
πilpnfηi+l+n−1
q−1 (dω).
(35)
If ω ∈fηi
q−2N∞q−2 or ω ∈ pdmfqm−3N q−3
0 for some m, then ψ(gi,0(Xiω)) = 0 by (35). Ifω∈fηi
q−2N q−2
f , then takel≥0 andfql−2ω′∈N q−2
l forω′∈N q−2 0 such that ω=fηi+l
q−2(ω′). For thisω′, we have
ψ(gi,0(Xifηi+l
q−2(ω′)))≡
pl
i′π
ipηi−1
f2ηi+l−1
q−1 (dω′)
≡ p
l
i′π
jfs+l−1 q−1 (dω
′)
mod filj+el+1( ˆΩqA−1/pdΩˆqA−2).
(36)
S1 i,1 is
Si,11 =Xi pf ηi−1
q−2 Nq−2∪ ηi−2
[
m=0
p d pmf
m q−3N
q−3 0
!
.
ForXipω
∈S1
i,1,ψ(Xipω) is
ψ(Xipω) =X n≥0
pηi−1 pn π
ipn fqn−1
dω
pηi−1
Thus if ω ∈fηi−1
q−2 N∞q−2 or ω ∈ pdmfqm−3N
q−3
0 for somem, then ψ(Xipω) = 0. If ω ∈ fηi−1
q−2 N q−2
f , let l ≥ 0 and fql−2ω′ ∈ N q−2
l for ω′ ∈ N q−2
0 such that
ω=fηi+l
q−2(ω′). For thisω′, we have
ψ(Xipfηi+l−1
q−2 (ω′))≡plπip
ηi−1
f2ηi+l−2
q−1 (dω′)
≡plπjfqs+l−1−2(dω′)
mod filj+el+1( ˆΩqA−1/pdΩˆ q−2 A ).
(38)
Ift > s/2, theni6∈Λ and
Si,10 =Xi f ηi
q−2Nq−2∪ ηi−1
[
m=0
p d pmf
m q−3N
q−3 0
!
.
ForXiω∈S0
i,1,ψ(Xiω) is
ψ(Xiω) =X n≥0
pηi pn+1π
ipn fqn−1
dω pηi
. (39)
If ω ∈ fηi
q−2N∞q−2 or ω ∈ pdmfqm−3N q−3
0 for some m, then ψ(Xiω) = 0. If
ω ∈ fηi
q−2N q−2
f , then takel ≥0 and fql−2ω′ ∈ Nlq−2 for ω′ ∈N q−2
0 such that
ω=fηi+l
q−2(ω′). For thisω′, we have
ψ(Xifηi+l
q−2(ω
′))
≡plπipηi−1f2ηi+l−1
q−1 (dω
′)
≡plπjfq2s−−12t+l−1(dω′)
mod filj+el+1( ˆΩqA−1/pdΩˆqA−2).
(40)
When we take Xipω ∈S1
i,1, by the same calculation of the case t =s/2, we have
ψ(Xipfηi+l−1
q−2 (ω′))≡plπip
ηi−1
f2ηi+l−2
q−1 (dω′)
≡plπjf2s−2t+l−2 q−1 (dω′) mod filj+el+1( ˆΩqA−1/pdΩˆ
q−2 A ).
(41)
Wheni∈Γa andi < ethenS′
i,2=∅.
Next, we computei∈Γa ande < i. In this caseS′
i,2=∅, thus we only have to compute Si,1. Let e/(p−1) < j < e and i = j+e. Then S1
i,1 = ∅ and
S0
i,1=XiNq−2. For an elementXiω∈Si,10 ,
ψ(Xiω) =iπi−edπ
π ∧ω+
X
n≥0 1
pn+1π ipnfn
Ifp|i, then the first term of the right hand side is zero. Hence ifω=fl q−2(ω′) then
ψ(Xiω)≡plπi−efql−1(dω′)
mod filj+el+1( ˆΩqA−1/pdΩˆqA−2) (42) and ifω∈Nq−2
∞ thenψ(Xiω) = 0. Ifp∤i, then ψ(Xiω)≡iπi−edπ
π ∧ω+π
i−edω mod filj+1( ˆΩqA−1/pdΩˆ
q−2 A ).
(43)
We have computed all Si or substitutes of Si for i ∈ Γa as above. Next, we construct the sets{Mj}j≥0 which are rearrangements of the generators of
ψ(H1(Sq)). The law of rearrangement is, for example, as follows. See (33). For an elementgi,1(Xipfηi+l−1
q−2 (ω′)), the image ofψis
ψ(gi,1(Xipfηi+l−1
q−2 (ω′)))≡
pl
i′π
jfs+l−1 q−1 (dω′) mod filj+el+1( ˆΩqA−1/pdΩˆ
q−2 A ) whens−t >1. Thus this element goes to grj+el( ˆΩq−1
A /pdΩˆ q−2
A ) and it seems non-zero. So we putgi,1(Xipfηi+l−1
q−2 (ω′)) intoMj+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 Mj+el by, roughly speaking, the set of the elements which come to grj+el( ˆΩqA−1/pdΩˆ
q−2
A ) and seem non-zero. The real definition ofM∗is as follows.
Use (33), (34), (36), (38), (40), (41), (42) and (43) to define Mj+el for
e/(p−1)< j < eandl≥0. Lete/(p−1)< j < e,s=vp(j) + 1 andl≥0. If
s= 1 then let
Mj+el =
gj,1(XjpNq−2
0 )∪Xj+eNq−2 (p∤(j+e) andl= 0) . . . (33),(43)
gj,1(Xjpfl q−2N
q−2
0 ) (p∤(j+e) andl≥1) . . . (33)
XjpNq−2
∪Xj+eNq−2
0 (p|(j+e) andl= 0) . . . (34),(42)
Xj+efl q−2N
q−2
0 (p|(j+e) andl≥1) . . . (42).
(44)
By (3.1), grj( ˆΩqA−1/pdΩˆ q−2 A ) ∼= Ω
q−1 F ⊕Ω
q−2
F and grj+el( ˆΩ q−1 A /pdΩˆ
q−2 A ) ∼= ΩqF−1/Blq−1forl≥1. The image ofMj is, ifp∤(j+e),
ΩqF−2/Z1q−2
ψ◦gj,1Xjp
−−−−−−→grj( ˆΩqA−1/pdΩˆAq−2)−→∼= ΩFq−1⊕ΩqF−2
ω7−→ψ◦gj,1Xjpω= e
j+eπ
jdω
7−→
e j+edω,¯ 0
and and hence non-zero. Thus
for l ≥ 1. By the similar calculation as the cases is even, we get the same results (49) and (50).
By the definition ofMj+el,
ForXiω
∈S0
i,1,ψ(Xiω) is
ψ(Xiω)≡j0πj−edπ
π ∧f s 2
q−2(ω) +πj−ef
s 2
q−1
dω ps2
mod filj−e+1( ˆΩAq−1/pdΩˆqA−2).
(54)
ForXipω∈S1
i,1,ψ(Xipω) is
ψ(Xipω) =X n≥0
1
pnπ ipn
fqn−2(dω).
Thus ifXipω
∈S1
i,1\Xipf ηi−1
q−2 N q−2
f thenψ(Xipω) = 0. Takeω′ ∈N q−2 0 such that ω=fηi−1
q−2 fql−2ω′. Then
ψ(Xipω)≡plπj−efqs−−11+ldω′
mod filj−e+el+1( ˆΩqA−1/pdΩˆ q−2 A ).
(55)
Ift > s/2, then
S0
i,1=Xi f ηi
q−2Nq−2∪ ηi−1
[
m=0
p d pmf
m q−3N
q−3 0
!
and
Si,11 =Xi pf ηi−1
q−2 Nq−2∪ ηi−2
[
m=0
p d pmf
m q−3N
q−3 0
!
.
ForXiω
∈S0
i,1,ψ(Xiω) is
ψ(Xiω) =X n≥0
1
pn+1π ipn
fqn−2(dω).
Thus ifXiω∈S0 i,1\Xif
ηi
q−2N q−2
f thenψ(Xiω) = 0. Takeω′∈N q−2
0 such that
ω=fηi
q−2fql−2ω′. Then
ψ(Xiω)≡plπj−efq2s−−12t+ldω′
mod filj−e+el+1( ˆΩqA−1/pdΩˆqA−2). (56)
ForXipω
∈S1
i,1, by the same calculation as in the caset=s/2,
ψ(Xipω)≡plπj−efq2s−−12t+l−1dω′
Next, we compute S′
i,2 in the case i ∈ Γb and i < e. By (4.5), S
′0
i,2 (resp.
S′1
i,2) exists when 3s/4< t(resp. (3s−2)/4< t). If 3s/4< t,
Si,2′0 =Xi
dX X ∧f
ηi
q−3N q−3 0 ∋Xi
dX X ∧f
ηi
q−3(ω)
7−→ψ(XidX
X ∧f
ηi
q−3(ω)) =πipηi−1dπ
π ∧f
2ηi−1
q−2 (dω) + 1
pπ
ipηidπ π ∧f
2ηi
q−2(dω)
≡πj−edπ π ∧f
2s−2t q−2 (dω)
mod filj−e+1( ˆΩAq−1/pdΩˆqA−2).
(58)
If (3s−2)/4< t,
Si,2′1 =Xi
dX X ∧pf
ηi−1
q−3 N q−3 0 ∋Xi
dX X ∧pf
ηi−1
q−3 (ω)
7−→ψ(XidX X ∧f
ηi−1
q−3 (ω)) =πipηi−1dπ
π ∧f
2ηi−2
q−2 (dω) + 1
pπ
ipηidπ π ∧f
2ηi−1
q−2 (dω)
≡πj−edπ π ∧f
2s−2t−1 q−2 (dω) mod filj−e+1( ˆΩAq−1/pdΩˆqA−2).
(59)
Next, we compute Si,1 for i∈Γb and i > e. Letj =i. NowS1i,1 =∅ and
S0
i,1=XiNq−2. For an elementXiω∈Si,10 ,
ψ(Xiω) =iπi−e−1dπ∧ω+X n≥0
1
pn+1π ipn
fqn−1(dω).
Ifp|i, then the first term of the right hand side is zero. Hence ifω=fql−2(ω′)
then
ψ(Xiω)≡plπi−efql−1(dω′)
mod filj−e+el+1( ˆΩAq−1/pdΩˆqA−2) (60) and ifω∈Nq−2
∞ thenψ(Xiω) = 0. Ifp∤i, then ψ(Xiω)≡iπi−edπ
π ∧ω+π
i−edω
mod filj−e+1( ˆΩAq−1/pdΩˆqA−2).
(61)
For i ∈ Γb and i > e, S′
Xi−1dX
∧ω∈Xi−1dX
∧N0q−3=S′
i,2=S
′0
i,2,
XidX
X ∧ω =π
i−edπ
π ∧dω
≡πj−edπ
π ∧dω mod fil
j−e+1( ˆΩq−1 A /pdΩˆ
q−2 A ).
(62)
Use (53), (54), (55), (56), (57), (58), (59), (60), (61) and (62) to defineMj+el fore < j < e′ andl
≥0. Lete < j < e′,s=vp(j). By (3.1),
grj−e+el( ˆΩqA−1/pdΩˆ q−2 A )∼=
( ˆ
ΩqF−1⊕Ω q−2
F (l= 0) ˆ
ΩqF−1/Blq−1 (l≥1).
Ifs= 0 then let
Mj−e=XjNq−2, . . . (61)
Mj−e+el=∅ (63)
for alll≥0. The image ofMj−eis the image of
ΩqF−2∋x7−→(dx, jx)∈ΩqF−1⊕ΩqF−2,
hence we get
ψ(x)6= 0 forx∈Mj−e in grj−e( ˆΩq−1 A /pdΩˆ
q−2 A ), grj−e( ˆΩqA−1/pdΩˆAq−2)/hψ(Mj−e)i
∼
= CokerΩqF−2∋x7−→(dx, jx)∈ΩqF−1⊕ΩqF−2
(64)
and
ψ(x)6= 0 forx∈Mj−e+el in grj−e+el( ˆΩq−1 A /pdΩˆ
q−2 A ), grj−e+el( ˆΩAq−1/pdΩˆqA−2)/hψ(Mj−e+el)i ∼= ΩqF−1/B
q−1 l
(65)
forl≥1 becauseMj−e+el =
Furthermore, the image of (54) modulo the image of the group generated by the other generators ofMj is
and
Mj−e+el= [
s 2<t≤s−1
Xj0ptfs−t
q−2N q−2 l ∪Xj0p
t
fqs−−2t−1Nlq−2 . . . (56), (57)
∪XjNlq−2 . . . (60)
(71) for l ≥ 1. By the similar calculation to the cases is even, we get the same result (68) and (69).
By the definition ofMj+el,
[
i∈Γb
S0i,1∪Si,11 !
∪ [
i∈Γb Si,2′
!
is equal to the union ofMj−e+el for alle < j < e′ and alll≥0.
7 Explicit Calculation, Case (c) In this section, we compute ψ(Si,1) andψ(S′
i,2) fori∈Γc.
Γc has only two elements,e/(p−1) and e′. At first leti=e/(p−1). Then Si,10 =∅,Si,2′ =∅and
Si,11 =XipNq−2.
Note that thisihas the propertyi=ip−e. TakeXipω
∈XipNq−2, then
ψ(Xipω) =iπidπ
π ∧(ω+fq−2(ω)) +π
i(dω+f
q−1(dω))
+X n≥2
1
pnπ ipnfn
q−1(dω).
Ifω+fq−2(ω)≡0 modpthen the leftmost term of the right hand side vanishes.
But ω +fq−2(ω) ≡ 0 means ¯ω +C−1ω¯ = 0 in ˆΩqF−2, thus dω = 0 hence
ψ(Xipω) = 0 by the property ofNq−2
0 , see (25). So we get
ψ(Xipω)
≡iπi dπ
π ∧(ω+fq−2(ω)) +π
i(dω+f
q−1(dω))
mod fili+1( ˆΩqA−1/pdΩˆqA−2) ( ifω+fq−2(ω)6≡0 modp)
= 0 ( if ω+fq−2(ω)≡0 modp)
(72)
Next, let i =e′. Then S0
i,1 =XiNq−2, S ′0
i,2 =Xi−1dX∧N q−3
0 and Si,11 =
S′1
i,2=∅. ForXiω∈XiNq−2, ψ(Xiω) =X
n≥0 1
pn+1π ipnfn
