LXXXV.3 (1998)
Index for subgroups of the group of units in number fields
by
Tsutomu Shimada (Yamato)
We define a sequence of rational integers u
i(E) for each finite index subgroup E of the group of units in some finite Galois number fields K in which prime p ramifies. For two subgroups E
0⊂E of finite index in the group of units of K we prove the formula v
p([E : E
0]) = P
ri=1
{u
i(E
0) − u
i(E)}.
This is a generalization of results of P. D´enes [3], [4] and F. Kurihara [5].
Introduction. Let p be an odd prime number, Q and Z the field of rational numbers and the ring of rational integers, respectively. For each unit ε of Q(ζ
p) which is not in Z, there exist rational integers a, b and c satisfying ε ≡ a + b(1 − ζ
p)
cmod (1 − ζ
p)
c+1, ab 6≡ 0 mod p and c 6≡ 0 mod (p − 1), where c is uniquely determined by ε. P. D´enes [2] de- fined the p-character of the Bernoulli numbers to be the rational inte- gers u
2, u
4, . . . , u
p−3such that B
ipj≡ 0 mod p
2j+1for 0 ≤ j < u
iand B
ipui6≡ 0 mod p
2ui+1, where i = 2, 4, . . . , p − 3, and proved the following results under the assumption that the p-character of the Bernoulli numbers exists:
Theorem A. There exists a basis {θ
2, θ
4, . . . , θ
p−3} for the group of units of Q(ζ
p)
+modulo {±1} such that
θ
i≡ a
i+ b
i(1 − ζ
p)
cimod (1 − ζ
p)
ci+1with c
i= i + (p − 1)u
0ifor some integer 0 ≤ u
0i≤ u
i.
Theorem B. We have
v
p(h(Q(ζ
p)
+)) =
p−3
X
i=2, even
(u
i− u
0i).
Here h(∗) denotes the class number of a field ∗ and v
pthe p-adic valuation normalized by v
p(p) = 1.
1991 Mathematics Subject Classification: 11R21, 11R29.
[249]
L. C. Washington [6] showed that u
i= v
p(L
p(1, ω
i)), i = 2, 4, . . . , p − 3, and then proved D´enes’ assumption stated above, where ω is the Teichm¨ uller character. Furthermore, Washington gave simple proofs of the theorems above. In [5], F. Kurihara generalized the results above to a subfield K
n+1of Q(ζ
pn+1)
+, 0 ≤ n ∈ Z, and showed the following two theorems.
Theorem C. Let E be a subgroup of E
Kn+1, the group of units of K
n+1, such that [E
Kn+1: E] is finite. Then there exists a basis {θ
1, . . . , θ
r} for E modulo {±1} such that
θ
ipn≡ a
i+ b
iπ
cimod π
ci+1, i = 1, . . . , r, with c
i= i +
2d1ϕ(p
n+1)u
ifor some rational integer u
i≥ 0.
Here d = [Q(ζ
pn+1)
+: K
n+1], r = rank E
Kn+1=
2d1ϕ(p
n+1) − 1 and π is the image of (1 − ζ
pn+1)(1 − ζ
p−1n+1) by the norm from Q(ζ
pn+1)
+to K
n+1. Now since c
iand hence u
idepends only on E, it is denoted by c
i(E) and u
i(E), respectively.
Theorem D. Let E be a subgroup of E
Kn+1and E
0a subgroup of E.
Suppose that the index [E
Kn+1: E
0] is finite. Then v
p([E : E
0]) =
X
r i=1{u
i(E
0) − u
i(E)}.
Considering the case where K
n+1= Q(ζ
pn+1)
+, E = E
Kn+1and E
0is the group of cyclotomic units in the sense of Sinnott, we see that Theorem D is a generalization of Theorem B.
Our aim is to prove similar results in some other number fields: the composite of two Galois extensions of finite degree over Q, one unramified at p and the other totally ramified.
Now we introduce some notations. Let K
Tbe a finite Galois extension over Q which is unramified at p and K
Va finite Galois extension which is totally ramified at p. Let K = K
TK
V, [K
T: Q] = d
Kand [K
V: Q] = e
K. We fix an embedding of K into C, the field of complex numbers. Let J denote the restriction of the complex conjugation to K. Let K
+and K
T+be the fixed field of hJi in K and K
T, respectively. Let ℘
1, . . . , ℘
gbe the primes of K
Tover p, ℘
0i(i = 1, . . . , g) the primes of K over ℘
i, and ℘ the unique prime of K
Vover p. Then ℘
i= ℘
0eiKand (p) = ℘
1. . . ℘
g= (℘
01. . . ℘
0g)
eK= ℘
eKin the ring O
Kof integers of K. Let E
∗be the group of units of a field ∗. Let δ be the least natural number a that satisfies e
K/(p − 1) < p
a−1.
Throughout this paper, we assume the following three conditions:
(A
1) ℘ is a principal ideal.
(A
2) The exponent of the torsion part of E
K+/E
K+T
is prime to p.
(A
3) The Leopoldt conjecture is valid for K and p.
Then we may write ℘ = (π
K) with some π
K∈ O
KV. Let m be the exponent in (A
2).
Our main result is the following:
Theorem E. Let E be a subgroup of E
K+such that E ⊃ E
K+ Tand [E
K+: E] < ∞. Then there exists a basis {η
1, . . . , η
r} of E
mpδeKE
K+T
mod- ulo E
K+T
such that
η
i≡ a
i+ b
iπ
Kcimod π
Kci+1, i = 1, . . . , r = rank E
K+/E
K+ T,
where a
iand b
ibelong to the ring O
KTof integers of K
T, a
i6≡ 0 modulo any prime over p, b
i6≡ 0 mod p and c
iis a natural number such that c
i6≡
0 mod e
K.
Further , let u
i(E) =
ceKi
(Gauss symbol) and E
0be a subgroup of E such that E
0⊃ E
K+T
and [E
K+: E
0] < ∞. Then v
p([E : E
0]) =
X
r i=1{u
i(E
0) − u
i(E)}.
Remark 1. Let η be any element of E
K+\ K
Tsuch that η
a∈ E
K+for some natural number a. Then, for any σ ∈ Gal(K/K
T), (η
σ/η)
a=
T(η
a)
σ/η
a= η
a/η
a= 1. So, η
σ−1is an ath root of unity. Moreover, when K is a real or CM-field, η
σ−1is real, hence ±1 and (η
2)
σ−1= 1. Thus, η
2∈ E
K+T
and m equals 1 or 2, hence in this case it is prime to p, i.e. (A
2) is valid.
Remark 2. When K = Q(ζ
apn) with a prime to p, then K
T= Q(ζ
a).
Note that the condition (A
3) is valid by the theorem of A. Brumer [1]. Let E = E
K+and E
0= C
K+E
K+T
, where C
Kis the group of cyclotomic units and C
K+= C
K∩ K
+. Then, since v
p([E : E
0]) = v
p(h(K
+)/h(K
T+)), we get a generalization of Theorem B:
v
p(h(Q(ζ
apn)
+)/h(Q(ζ
a)
+)) = X
r i=1{u
i(E
0) − u
i(E)}.
1. The Fermat quotient and the level of unit. Let the notations be as in the introduction. Note that K
V= Q(π
K). Let f (X) = X
eK+ c
0eK−1X
eK−1+ . . . + c
01X + c
00∈ Z[X] be the minimal polynomial of π
Kwhich is of Eisenstein type. That is, c
0eK−1≡ . . . ≡ c
00≡ 0 mod p and c
006≡ 0 mod p
2. We write c
0i= −pc
i(i = 0, 1, . . . , e
K− 1). Then
(1) π
KeK≡ pc
0mod π
KeK+1and p ≡ b
0π
eKKmod π
eKK+1,
where b
0is the natural number such that c
0b
0≡ 1 mod p and 1 ≤ b
0≤ p−1.
For any z ∈ O
K\ K
Twhich is prime to p, we define
c(z) = max{0 ≤ c ∈ Z : z ≡ x + yπ
cKmod π
Kc+1with some x, y ∈ O
KT}.
It can be easily seen that 1 ≤ c(z) < ∞. Furthermore, we define c(z) = ∞ for z ∈ O
KT.
When x
0and y
0give c(z) (z ∈ O
K\ K
T), it is clear that x
06≡ 0 mod ℘
i(i = 1, . . . , g) and y
06≡ 0 mod p.
If c(z) is a multiple of e
K, then writing c(z) = ce
Kwith a natural number c, we have
z ≡ x
0+ y
0π
KceK≡ x
0+ y
0p
cc
c0mod π
KceK+1,
which contradicts the maximality of c(z). Thus, we get c(z) 6≡ 0 mod e
K. To sum up, we have the following:
Lemma 1.1. For any z ∈ O
K\ K
Twhich is prime to p,
c(z) = max{0 ≤ c ∈ Z : z ≡ x + yπ
Kcmod π
c+1Kfor some x, y ∈ O
KT} is a natural number which depends only on z (it does not depend on the choice of π
K) and c(z) 6≡ 0 mod e
K. Let x
0and y
0be elements in O
KTgiving c(z). Then x
06≡ 0 mod ℘
i(i = 1, . . . , g) and y
06≡ 0 mod p, and further , x
0and y
0are uniquely determined by z modulo π
c(z)+1Kand p, respectively.
Let η ∈ E
K\ K
T. Let η ≡ x + yπ
c(η)Kmod π
Kc(η)+1be a congruence giving c(η) according to Lemma 1.1. Then, in the following, we call c(η) the level of η and (y/x) mod p ∈ O
KT/(p) the (generalized) Fermat quotient of η and we denote the latter by f (η). Of course they are uniquely determined by η.
In the rest of this section, we present several elementary properties of the level and the Fermat quotient.
The next lemma immediately follows from the definitions of the Fermat quotient and the level.
Lemma 1.2. (1) For any η ∈ E
K\ K
Tand any natural number a prime to p, we have c(η
J) = c(η), c(η
a) = c(η), f (η
a) = af (η) and c(η
−1) = c(η), f (η
−1) = −f (η).
(2) Let η
1and η
2be elements in E
K\ K
Tsuch that c(η
1) < c(η
2). Then c(η
1η
2) = c(η
1) and f (η
1η
2) = f (η
1).
(3) Let η
1, . . . , η
sbe elements in E
K\ K
Tsuch that c(η
1) = . . . = c(η
s) and f (η
1) + . . . + f (η
s) 6≡ 0 mod p. Then
c(η
1. . . η
s) = c(η
1) and f (η
1. . . η
s) = f (η
1) + . . . + f (η
s).
Lemma 1.3. If K
Vis imaginary, then π
KJ≡ −π
Kmod π
K2.
P r o o f. By the definition of π
K, we have (π
KJ) = (π
K). So, there exists u ∈ E
KVsuch that π
KJ= π
Ku. We have π
K= (π
KJ)
J= (π
Ku)
J= π
Kuu
J. Thus, uu
J= 1.
First, we assume u 6∈ K
T. Let u ≡ a + bπ
c(u)Kmod π
c(u)+1Kaccording to Lemma 1.1. For any σ ∈ Gal(K/K
V), u
σ= u, π
Kσ= π
Kand u ≡ a
σ+ b
σπ
Kc(u)mod π
Kc(u)+1. Because O
KV/(π
K) = Z/(p), we can always write u ≡ a + bπ
cKmod π
c+1Kwhere a, b ∈ Z are prime to p. Then 1 = uu
J≡ a
2mod π
K, so a ≡ ±1 mod p. Since π
KJ≡ aπ
Kmod π
2K, we have u
J≡ a + ba
c(u)π
c(u)Kmod π
c(u)+1K. By Lemma 1.2, f (u
−1) = f (u
J) ≡ ba
c(u)−1mod p and f (u
−1) = −f (u) ≡ −ba
−1mod p. This means a
c(u)≡ −1 mod p, so that a ≡ −1 mod p and the lemma is proved in this case.
Secondly, we assume u ∈ K
T. Then u ∈ K
T∩ K
V= Q and u = ±1.
Now, π
KJ6= π
Kby our assumption, so that u = −1. The proof is complete.
Lemma 1.4. For any η ∈ E
K\ K
Twe have c(η
pδ) > e
K/(p − 1). If c(η) > e
K/(p − 1), then c(η
pa) = c(η) + ae
Kand f (η
pa) = b
a0f (η) for all natural numbers a. Here δ and b
0are as in the introduction.
P r o o f. Let η ≡ x + yπ
Kc(η)mod π
c(η)+1Kaccording to Lemma 1.1. Then there exists y
1∈ O
Ksuch that η = x + y
1π
c(η)Kand y
1≡ y mod π
K. So,
η
p= x
p+ px
p−1y
1π
c(η)K(2)
+
p 2
x
p−2y
12π
2c(η)K+ . . . +
p p − 1
xy
p−11π
K(p−1)c(η)+ y
1pπ
pc(η)K. Since the π
K-orders of terms on the right hand side are
0, e
K+ c(η), e
K+ 2c(η), . . . , e
K+ (p − 1)c(η) and pc(η), it follows that
c(η
p) ≥ min{e
K+ c(η), pc(η)} ≥ min
e
Kp − 1 , pc(η)
. Further,
c(η
p2) ≥ min
e
Kp − 1 , pc(η
p)
≥ min
e
Kp − 1 , p min
e
Kp − 1 , pc(η)
= min
e
Kp − 1 , p
2c(η)
. For all natural numbers a, we get by induction
c(η
pa) ≥ min
e
Kp − 1 , p
ac(η)
.
Since c(η
pδ−1) ≥ e
K/(p − 1), (2) means that
c(η
pδ) ≥ min{e
K+ c(η
pδ−1), pc(η
pδ−1)} > e
Kp − 1 . When e
K/(p − 1) < c(η), we have
e
K+ c(η) < pc(η) and η
p≡ x
p+ px
p−1y
1π
Kc(η)mod π
eKK+c(η)+1. Now from (1), we have
(3) η
p≡ x
p+ b
0x
p−1yπ
KeK+c(η)mod π
KeK+c(η)+1. So, we conclude that c(η
p) ≥ c(η) + e
K.
Suppose c(η
p) > c(η)+e
K. Let η
p≡ x
2+y
2π
c(ηK p)mod π
Kc(ηp)+1according to Lemma 1.1. Then
x
2≡ η
p≡ x
p+ b
0x
p−1yπ
eKK+c(η)mod π
KeK+c(η)+1and
x
2− x
p≡ b
0x
p−1yπ
KeK+c(η)mod π
eKK+c(η)+1.
Take a prime ℘
0idividing π
Kof K such that y 6≡ 0 mod ℘
0i. Then, from the above, e
K+c(η) = v
℘0i
(b
0x
p−1yπ
KeK+c(η)) = v
℘0i
(x
2−x
p). This is a multiple of e
K, so that c(η) is also a multiple of e
K. That is a contradiction. Therefore, c(η
p) = c(η) + e
K. Inductively, we obtain c(η
pa) = c(η) + ae
Kfor all natural numbers a.
Furthermore, from (3),
f (η
p) ≡ b
0x
p−1y x
p≡ b
0y
x ≡ b
0f (η) mod p.
This means that f (η
pa) = b
a0f (η) for all natural numbers a. The proof is complete.
Lemma 1.5. Let η
1, . . . , η
sbe elements in E
K\ K
Tsuch that c(η
1) = . . . = c(η
s) > e
K/(p − 1) and {f (η
1), . . . , f (η
s)} is an F
p-independent sys- tem. Then η
1, . . . , η
sare Z-independent.
P r o o f. Suppose that η
1, . . . , η
sare Z-dependent, that is, η
e11. . . η
ses= 1 with some e
1, . . . , e
s∈ Z. We may assume e
i6= 0 for all i.
Let e
i= a
ip
bi(Z 3 a
i6≡ 0 mod p, 0 ≤ b
i∈ Z, i = 1, . . . , s). Then from Lemmas 1.2 and 1.4 we have
c(η
iei) = c(η
ipbi) = c(η
i) + b
ie
Kand f (η
iei) = a
if (η
ipbi) = b
b0ia
if (η
i).
We denote by β the minimum of {b
1, . . . , b
s} and assume, without loss of generality, β = b
1= . . . = b
t< b
t+1, . . . , b
swith some t (1 ≤ t ≤ s).
From our assumption, X
t i=1f (η
iei) = X
t i=1b
b0ia
if (η
i) 6≡ 0 mod p.
Now, c(η
eii) = c(η
i) + b
ie
K= c(η
i) + βe
Kfor all i = 1, . . . , t. So, from Lemma 1.2, c( Q
ti=1
η
iei) = c(η
1) + βe
K< c(η
ejj) for all t + 1 ≤ j ≤ s. There- fore, c( Q
si=1
η
iei) = c( Q
ti=1
η
eii) = c(η
1)+βe
K. This contradicts Q
si=1
η
iei= 1 (whose level is ∞) and the lemma is proved.
In the end we investigate the action of J on the Fermat quotient of a real unit.
Lemma 1.6. For any η ∈ E
K+\ K
T, f (η)
J= (−1)
c(η)f (η) if K
Vis imaginary, and f (η)
J= f (η) if K
Vis real.
P r o o f. First, we assume that K
Vis imaginary. Let η ≡ x + yπ
c(η)Kmodπ
c(η)+1Kaccording to Lemma 1.1. From Lemmas 1.2 and 1.3,
η = η
J≡ x
J+ y
J(−1)
c(η)π
Kc(η)mod π
c(η)+1K. Therefore,
f (η) = f (η
J) ≡ y
J(−1)
c(η)x
J≡ (−1)
c(η)y x
J≡ (−1)
c(η)f (η)
Jmod p.
When K
Vis real, η = η
J≡ x
J+y
Jπ
Kc(η)mod π
c(η)+1K. Thus, f (η) = f (η
J) ≡ (y/x)
J≡ f (η)
Jas desired.
2. A basis of units modulo units of K
T+. Let the notation be as before. In this section, we shall prove the existence of a set of representatives of a basis of E
mpδeKE
K+T
/E
K+T
which satisfies some conditions on the Fermat quotient and the level.
When K
Tis imaginary, let
O
KT/(p) = (O
KT/(p))
+⊕ (O
KT/(p))
−be the decomposition associated with (1 + J)/2 and (1 − J)/2. Then it is easy to see that
(i) dim
Fp(O
KT/(p))
+= dim
Fp(O
KT/(p))
−= d
K/2.
(ii) E
KeK⊂ Ker(N ) · E
KTand E
K[K++:KT+]⊂ Ker(N
+) · E
K+T
, where N and N
+is the norm map from K to K
Tand from K
+to K
T+, respectively.
The next lemma is due to Washington [6].
Lemma 2.1. Let E be a subgroup of E
Kof finite index and let η be a non-torsion element of E. If v
℘0i(log
pη) is sufficiently large for all primes
℘
0i(i = 1, . . . , g) then η is a pth power in E. Here, we consider v
℘0i(log
pη) and log
pη in the localization of K with respect to ℘
0i.
P r o o f. If η is not a pth power in E, then we can take u
2, . . . , u
r∈ E
(r = rank
ZE
K) such that {η, u
2, . . . , u
r} generates a subgroup E
0of E of
finite index prime to p. Let R
p(∗) be the p-adic regulator of ∗ (see Washing- ton [7]). From our assumption, R
p(E
0) ≡ 0 mod ℘
0cifor all ℘
0i| p, where c is sufficiently large. Now,
R
p(E
0) = [E
K: E][E : E
0]R
p(E
K) 6= 0
by our assumption (A
3). So, v
p(R
p(E
0)) = v
p([E
K: E]) + v
p(R
p(E
K)).
The right hand side depends only on K and E. But the left hand side is sufficiently large. That is a contradiction and the proof is complete.
Next we prove a relation between v
℘0i(log
pη) and the level of η.
Lemma 2.2. Let η be any element of E
K\ K
Tand ℘
0i(i = 1, . . . , g) the prime of K over p. Suppose N (η) = 1 and c(η) > e
K/(p − 1). Then
v
℘0i
(log
pη) ≥ min{c(η) + 1 − v
p(e
K)e
K, c(η)} for all ℘
0i.
P r o o f. Let η ≡ x+yπ
c(η)Kmod π
Kc(η)+1according to Lemma 1.1. Let c = c(η). Fix any prime ℘
0i| p. From the assumption, 1 = N (η) ≡ x
eKmod π
Kc. Observe that the π
K-order of x
eK− 1 is a multiple of e
Kand c is not a multiple of e
Kby Lemma 1.1. Thus x
eK≡ 1 mod π
Kc+1. As c > e
K/(p − 1), we have v
℘0i(log
px
eK) ≥ c + 1 (see Lemma 5.5 of Washington [7]).
Thus, v
℘0i(e
K) + v
℘0i(log
px) ≥ c + 1. From v
℘0i(e
K) = v
p(e
K)e
K, we obtain v
℘0i(log
px) ≥ c+1−v
p(e
K)e
K. There exists y
1∈ O
Ksuch that η = x+y
1π
cKand y
1≡ y mod π
K. Then, since log
pη = log
px + log
p(1 + y
1π
cK/x), we have
v
℘0i(log
pη) ≥ min
v
℘0i(log
px), v
℘0ilog
p1 + y
1x π
cK≥ min{c + 1 − v
p(e
K)e
K, c}.
The lemma is proved.
For any natural number c, we define
F
Kc= {f (η) : η ∈ E
K+\ K
Tsuch that c(η) = c} ⊂ O
KT/(p).
Lemma 2.3. (I) If K
Tand K
Vare imaginary, then F
Kc⊂ (O
KT/(p))
+if c is even, and F
Kc⊂ (O
KT/(p))
−if c is odd. Moreover , dim
FpF
Kc≤
12d
K. (II) If K
Tis imaginary and K
Vis real, then F
Kc⊂ (O
KT/(p))
+for all c, and dim
FpF
Kc≤
12d
K.
(III) If K
Tis real and K
Vis imaginary, then e
Kis even and c(η) is even for all η ∈ E
K+\ K
T. Obviously, F
Kc⊂ (O
KT/(p)) = (O
KT/(p))
+and dim
FpF
Kc≤ d
K.
(IV) If K
Tand K
Vare real, then F
Kc⊂ (O
KT/(p)) = (O
KT/(p))
+and dim
FpF
Kc≤ d
K.
P r o o f. (I) Clearly, K is imaginary. Let η ∈ E
K+\ K
Tand η ≡ x +
yπ
Kc(η)mod π
Kc(η)+1according to Lemma 1.1. Let c = c(η). Since K
Vis imag-
inary, the statement follows from Lemma 1.6 and (i).
(II) In this case, our statement follows easily from Lemma 1.6 and (i).
(III) Let η ∈ E
K+\ K
Tand η ≡ x + yπ
Kc(η)mod π
c(η)+1Kaccording to Lemma 1.1. Let c = c(η). Because the order of J is 2, e
Kis clearly even.
From Lemma 1.3,
η = η
J≡ x
J+ y
J(−1)
cπ
Kcmod π
c+1K.
Here, x
J= x and y
J= y because K
Tis real. So, x + yπ
Kc≡ x + y(−1)
cπ
cKmod π
Kc+1. This means that c is even.
(IV) It is clear.
Remark 3. We have r = rank
Z(E
K+/E
K+T
) = rank
ZE
K+− rank
ZE
K+T
. Hence we easily observe that:
r =
12d
K(e
K− 1) and dim
FpF
Kc≤
12d
Kin the case (I) or (II).
r = d
K(
12e
K− 1) and dim
FpF
Kc≤ d
Kin the case (III).
r = d
K(e
K− 1) and dim
FpF
Kc≤ d
Kin the case (IV).
Theorem 2.4. Let E ⊃ E
K+T
be a subgroup of E
K+of finite index. Let r = rank
Z(E
K+/E
K+T
). Then there exists a set of representatives {η
1, . . . , η
r} of a basis of E
mpδeKE
K+T
/E
K+T
such that (1) c(η
i) > e
K/(p − 1) (i = 1, . . . , r).
(2) N
+(η
i) = 1 (i = 1, . . . , r).
(3) c
1≤ c
2≤ . . . ≤ c
rwhere c
i= c(η
i).
(4) Let S
j= {η
i: c(η
i) ≡ j mod e
K} (1 ≤ j < e
Kand j is even only if K
Tis real and K
Vis imaginary). Then ]S
j=
12d
K(d
Kresp.) when K
Tis imaginary (resp. real) and {f (η
i) : η
i∈ S
j} is an F
p-independent system for each j which defines S
j.
P r o o f. Let {ξ
1, . . . , ξ
r}, ξ
i∈ E, be a set of representatives of a basis of E
mE
K+T
/E
K+T
. Observe that E
mE
K+ T/E
K+T
is torsion-free. From Lemma 1.4, c(ξ
ipδ) > e
K/(p − 1). From (ii), as [K
+: K
T+] = e
Kor e
K/2, we have ξ
ipδeK∈ Ker(N
+) · E
K+T
. Therefore,
ξ
piδeK= η
iu
iwith some η
i∈ Ker(N
+) ∩ E and u
i∈ E
K+T
, 1 ≤ i ≤ r.
Here, {η
1, . . . , η
r} is also a set of representatives of a basis of the quotient E
mpδeKE
K+T
/E
K+T
that satisfies (1) and (2). And (3) is satisfied by an ap- propriate change of indices.
Now we define the condition (C
s) for 1 ≤ s < r : {f (η
i) : c(η
i) ≡
j mod e
K, 1 ≤ i ≤ s} is an F
p-independent system for all j (1 ≤ j < e
K,
j is even if K
Tis real and K
Vis imaginary). Clearly, (C
1) is true. Suppose
that (C
s) is valid. Let c(η
s+1) = l. If {f (η
i) : c(η
i) ≡ l mod e
K, 1 ≤ i ≤ s}
∪ {f (η
s+1)} is an F
p-independent system, then (C
s+1) is valid. If it is not F
p-independent, then
f (η
s+1) = X
1≤i≤s, ci≡l mod eK
a
ib
α0if (η
i) with some a
i∈ Z,
where α
i=
e1K
(c(η
s+1)−c(η
i)). We have c(η
aiipαi) = c(η
s+1) and f (η
aiipαi) = a
ib
α0if (η
i). Then
f
Y
1≤i≤s, ci≡l mod eK
η
iaipαi= X
1≤i≤s, ci≡l mod eK
a
ib
α0if (η
i) = f (η
s+1).
So, letting
η
0s+1= η
s+1Y
1≤i≤s, ci≡l mod eK
η
aiipαi −1,
we get c(η
s+10) > c(η
s+1). Now {η
1, . . . , η
s, η
s+10, η
s+2, . . . , η
r} is also a set of representatives that satisfies (1) and (2). By means of some permuta- tion of {η
s+10, η
s+2, . . . , η
r}, we may write it {η
s+1, η
s+2, . . . , η
r} again with c(η
s+1) ≤ . . . ≤ c(η
r). Then, further, we repeat the above procedure for η
s+1. Lemmas 2.1 and 2.2 imply that the procedure must stop after a fi- nite number of steps. Hence (C
s+1) becomes true. So, inductively, we get {η
1, . . . , η
r} as desired.
Note that, in this theorem, the sum of ]S
jfor 1 ≤ j < e
K(j is even when K
Tis real and K
Vis imaginary) is equal to r by Remark 3.
3. A formula for index of subgroups. Let E and E
0be subgroups of E
K+such that E ⊃ E
0⊃ E
K+T
and [E
K+: E
0] < ∞. Let {η
i} and {θ
i} be as in Theorem 2.4 for E and E
0, respectively.
For η ∈ E
K, let η denote η mod E
K+T
, and c(η) = c(η) and f (η) = f (η).
They are well defined because c(ηu) = c(η) and f (ηu) = f (η) for any u ∈ E
K+T
.
We define d
0to be
12d
Kif K
Tis imaginary (in the case (I) or (II) in Lemma 2.3) and d
Kif K
Tis real (in the case (III) or (IV) in Lemma 2.3).
We let R = {1, . . . , r} and B
l= {(l − 1)d
0+ 1, . . . , ld
0} (1 ≤ l < e
K), in the case (I), (II) or (IV) in Lemma 2.3. Moreover we define B
lin the case (III) as follows:
B
l=
l2
− 1
d
0+ 1, . . . ,
2ld
0for 1 < l < e
Kand l even.
Then R is the union of all B
l.
We permute η
1, . . . , η
rand θ
1, . . . , θ
rsuch that c(η
i) ≡ l mod e
Kfor all
i ∈ B
land c(θ
j) ≡ l mod e
Kfor all j ∈ B
l.
We let, for all j = 1, . . . , r, (5) θ
j=
Y
r i=1η
iajipejiwhere a
ji∈ Z is 0 or prime to p, and 0 ≤ e
ji∈ Z.
Then
det(a
jip
eji) = [E
mpδeKE
K+ T/E
K+T
: (E
0)
mpδeKE
K+ T/E
K+T
]
= [E
mE
K+ T/E
K+T
: (E
0)
mE
K+ T/E
K+T
]
= [E : E
0] × (a natural number prime to p) since E
mE
K+T
/E
K+T
and (E
0)
mE
K+ T/E
K+T
are torsion-free and m is prime to p by our assumption (A
2). Consequently, we have:
Lemma 3.1. Let E and E
0be subgroups of E
K+such that E ⊃ E
0⊃ E
K+T
and [E
K+: E
0] < ∞. Then v
p(det(a
jip
eji)) = v
p([E : E
0]).
Next we prove a formula for index [E : E
0].
Theorem 3.2. Let E and E
0be subgroups of E
K+such that E ⊃ E
0⊃ E
K+T
and [E
K+: E
0] < ∞. Let {η
j} and {θ
j} be as in Theorem 2.4 for E and E
0, respectively. Then
v
p([E : E
0]) = 1 e
Kn X
rj=1
c(θ
j) − X
r j=1c(η
j) o
. P r o o f. By the properties of level given in Section 1,
c(θ
j) = min{c(η
iajipeji) : 1 ≤ i ≤ r}.
Define A
j= {i ∈ R : c(θ
j) = c(η
iajipeji)}, 1 ≤ j ≤ r. Clearly, A
jis non-empty and A
j⊂ B
lif j ∈ B
l. Further,
f (θ
j) = X
i∈Aj
f (η
iajipeji) = X
i∈Aj
a
jib
e0jif (η
i) for all j ∈ R.
Since {f (η
i)}
i∈Bland {f (θ
j)}
j∈Blare F
p-independent systems, it follovs that
(6) det(a
0jib
e0ji)
j,i∈Bl6≡ 0 mod p for all l, where a
0ji= a
jiif i ∈ A
jand a
0ji= 0 if i 6∈ A
j.
Now we define P
l(1 ≤ l < e
K, l is even in the case (III) in Lemma 2.3) to be the set of all permutations on B
land P
l0= {τ
l∈ P
l: τ
l(j) ∈ A
jfor all j ∈ B
l}.
Then, for each l,
(7) det(a
0jib
e0ji)
j,i∈Bl= X
τl∈Pl0
sgn(τ
l) · Y
j∈Bl
a
jτl(j)b
e0jτl(j).
From (6) and (7), we see that P
l0is non-empty for every l. Let P be the set of all permutations on R and P
0= {% ∈ P : %(j) ∈ A
jfor all j ∈ R}.
It is clear that any element of P
0is a product of τ
l∈ P
l0(1 ≤ l < e
K, l is even in the case (III) of Lemma 2.3) and the restriction of each % ∈ P
0to B
lis an element of P
l0. So, we see that P
0is non-empty. In general, for any % ∈ P , c(θ
j) ≤ c(η
a%(j)j%(j)pej%(j)) for all j ∈ R, while for each % ∈ P \ P
0, c(θ
j) < c(η
a%(j)j%(j)pej%(j)) with some j ∈ R. Therefore, for each % ∈ P \ P
0,
(8)
X
r j=1c(θ
j) <
X
r j=1c(η
a%(j)j%(j)pej%(j)).
For % ∈ P
0, by the definition, X
r j=1c(θ
j) = X
r j=1c(η
a%(j)j%(j)pej%(j)).
Since X
r j=1c(η
a%(j)j%(j)pej%(j)) = X
r j=1{c(η
%(j)) + e
Ke
j%(j)} = X
r j=1c(η
%(j)) + e
KX
r j=1e
j%(j),
we have (9)
X
r j=1e
j%(j)= 1 e
Kn X
rj=1
c(θ
j) − X
r j=1c(η
j) o
for all % ∈ P
0.
Similarly, from (8), (10)
X
r j=1e
j%(j)> 1 e
Kn X
rj=1
c(θ
j) − X
r j=1c(η
j) o
for all % ∈ P \ P
0.
From (7) and (9), Y
l
det(a
0jib
e0ji)
j,i∈Bl= X
%∈P0
sgn(%) · Y
r j=1a
j%(j)b
e0j%(j)= X
%∈P0
sgn(%) · b
Pr
j=1ej%(j)
0
·
Y
r j=1a
j%(j)= b
(1/eK)(Pr
j=1c(¯θj)−Pr
j=1c(¯ηj))
0
· X
%∈P0