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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

r

i=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

)

c

mod (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−3

such that B

ipj

≡ 0 mod p

2j+1

for 0 ≤ j < u

i

and B

ipui

6≡ 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

)

ci

mod (1 − ζ

p

)

ci+1

with c

i

= i + (p − 1)u

0i

for 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

p

the p-adic valuation normalized by v

p

(p) = 1.

1991 Mathematics Subject Classification: 11R21, 11R29.

[249]

(2)

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+1

of 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

i

π

ci

mod π

ci+1

, i = 1, . . . , r, with c

i

= i +

2d1

ϕ(p

n+1

)u

i

for some rational integer u

i

≥ 0.

Here d = [Q(ζ

pn+1

)

+

: K

n+1

], r = rank E

Kn+1

=

2d1

ϕ(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

i

and hence u

i

depends 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+1

and E

0

a 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+1

and E

0

is 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

T

be a finite Galois extension over Q which is unramified at p and K

V

a finite Galois extension which is totally ramified at p. Let K = K

T

K

V

, [K

T

: Q] = d

K

and [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

, . . . , ℘

g

be the primes of K

T

over p, ℘

0i

(i = 1, . . . , g) the primes of K over ℘

i

, and ℘ the unique prime of K

V

over p. Then ℘

i

= ℘

0eiK

and (p) = ℘

1

. . . ℘

g

= (℘

01

. . . ℘

0g

)

eK

= ℘

eK

in the ring O

K

of 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.

(3)

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+ T

and [E

K+

: E] < ∞. Then there exists a basis {η

1

, . . . , η

r

} of E

mpδeK

E

K+

T

mod- ulo E

K+

T

such that

η

i

≡ a

i

+ b

i

π

Kci

mod π

Kci+1

, i = 1, . . . , r = rank E

K+

/E

K+ T

,

where a

i

and b

i

belong to the ring O

KT

of integers of K

T

, a

i

6≡ 0 modulo any prime over p, b

i

6≡ 0 mod p and c

i

is a natural number such that c

i

6≡

0 mod e

K

.

Further , let u

i

(E) = 

c

eKi

 (Gauss symbol) and E

0

be 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

T

such that η

a

∈ E

K+

for some natural number a. Then, for any σ ∈ Gal(K/K

T

), (η

σ

/η)

a

=

T

a

)

σ

a

= η

a

a

= 1. So, η

σ−1

is an ath root of unity. Moreover, when K is a real or CM-field, η

σ−1

is 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

K

is 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−1

X

eK−1

+ . . . + c

01

X + c

00

∈ Z[X] be the minimal polynomial of π

K

which is of Eisenstein type. That is, c

0eK−1

≡ . . . ≡ c

00

≡ 0 mod p and c

00

6≡ 0 mod p

2

. We write c

0i

= −pc

i

(i = 0, 1, . . . , e

K

− 1). Then

(1) π

KeK

≡ pc

0

mod π

KeK+1

and p ≡ b

0

π

eKK

mod π

eKK+1

,

where b

0

is the natural number such that c

0

b

0

≡ 1 mod p and 1 ≤ b

0

≤ p−1.

(4)

For any z ∈ O

K

\ K

T

which is prime to p, we define

c(z) = max{0 ≤ c ∈ Z : z ≡ x + yπ

cK

mod π

Kc+1

with 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

0

and y

0

give c(z) (z ∈ O

K

\ K

T

), it is clear that x

0

6≡ 0 mod ℘

i

(i = 1, . . . , g) and y

0

6≡ 0 mod p.

If c(z) is a multiple of e

K

, then writing c(z) = ce

K

with a natural number c, we have

z ≡ x

0

+ y

0

π

KceK

≡ x

0

+ y

0

p

c

c

c0

mod 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

T

which is prime to p,

c(z) = max{0 ≤ c ∈ Z : z ≡ x + yπ

Kc

mod π

c+1K

for 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

0

and y

0

be elements in O

KT

giving c(z). Then x

0

6≡ 0 mod ℘

i

(i = 1, . . . , g) and y

0

6≡ 0 mod p, and further , x

0

and y

0

are uniquely determined by z modulo π

c(z)+1K

and p, respectively.

Let η ∈ E

K

\ K

T

. Let η ≡ x + yπ

c(η)K

mod π

Kc(η)+1

be 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

T

and 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

1

and η

2

be elements in E

K

\ K

T

such that c(η

1

) < c(η

2

). Then c(η

1

η

2

) = c(η

1

) and f (η

1

η

2

) = f (η

1

).

(3) Let η

1

, . . . , η

s

be elements in E

K

\ K

T

such 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

V

is imaginary, then π

KJ

≡ −π

K

mod π

K2

.

(5)

P r o o f. By the definition of π

K

, we have (π

KJ

) = (π

K

). So, there exists u ∈ E

KV

such that π

KJ

= π

K

u. We have π

K

= (π

KJ

)

J

= (π

K

u)

J

= π

K

uu

J

. Thus, uu

J

= 1.

First, we assume u 6∈ K

T

. Let u ≡ a + bπ

c(u)K

mod π

c(u)+1K

according to Lemma 1.1. For any σ ∈ Gal(K/K

V

), u

σ

= u, π

Kσ

= π

K

and u ≡ a

σ

+ b

σ

π

Kc(u)

mod π

Kc(u)+1

. Because O

KV

/(π

K

) = Z/(p), we can always write u ≡ a + bπ

cK

mod π

c+1K

where a, b ∈ Z are prime to p. Then 1 = uu

J

a

2

mod π

K

, so a ≡ ±1 mod p. Since π

KJ

≡ aπ

K

mod π

2K

, we have u

J

≡ a + ba

c(u)

π

c(u)K

mod π

c(u)+1K

. By Lemma 1.2, f (u

−1

) = f (u

J

) ≡ ba

c(u)−1

mod p and f (u

−1

) = −f (u) ≡ −ba

−1

mod 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

KJ

6= π

K

by our assumption, so that u = −1. The proof is complete.

Lemma 1.4. For any η ∈ E

K

\ K

T

we have c(η

pδ

) > e

K

/(p − 1). If c(η) > e

K

/(p − 1), then c(η

pa

) = c(η) + ae

K

and f (η

pa

) = b

a0

f (η) for all natural numbers a. Here δ and b

0

are as in the introduction.

P r o o f. Let η ≡ x + yπ

Kc(η)

mod π

c(η)+1K

according to Lemma 1.1. Then there exists y

1

∈ O

K

such that η = x + y

1

π

c(η)K

and y

1

≡ y mod π

K

. So,

η

p

= x

p

+ px

p−1

y

1

π

c(η)K

(2)

+

 p 2



x

p−2

y

12

π

2c(η)K

+ . . . +

 p p − 1



xy

p−11

π

K(p−1)c(η)

+ y

1p

π

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

K

p − 1 , pc(η)

 . Further,

c(η

p2

) ≥ min

 e

K

p − 1 , pc(η

p

)



≥ min

 e

K

p − 1 , p min

 e

K

p − 1 , pc(η)



= min

 e

K

p − 1 , p

2

c(η)

 . For all natural numbers a, we get by induction

c(η

pa

) ≥ min

 e

K

p − 1 , p

a

c(η)



.

(6)

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

K

p − 1 . When e

K

/(p − 1) < c(η), we have

e

K

+ c(η) < pc(η) and η

p

≡ x

p

+ px

p−1

y

1

π

Kc(η)

mod π

eKK+c(η)+1

. Now from (1), we have

(3) η

p

≡ x

p

+ b

0

x

p−1

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

2

π

c(ηK p)

mod π

Kc(ηp)+1

according to Lemma 1.1. Then

x

2

≡ η

p

≡ x

p

+ b

0

x

p−1

eKK+c(η)

mod π

KeK+c(η)+1

and

x

2

− x

p

≡ b

0

x

p−1

KeK+c(η)

mod π

eKK+c(η)+1

.

Take a prime ℘

0i

dividing π

K

of K such that y 6≡ 0 mod ℘

0i

. Then, from the above, e

K

+c(η) = v

0

i

(b

0

x

p−1

KeK+c(η)

) = v

0

i

(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

K

for all natural numbers a.

Furthermore, from (3),

f (η

p

) ≡ b

0

x

p−1

y x

p

≡ b

0

y

x ≡ b

0

f (η) mod p.

This means that f (η

pa

) = b

a0

f (η) for all natural numbers a. The proof is complete.

Lemma 1.5. Let η

1

, . . . , η

s

be elements in E

K

\ K

T

such 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

s

are Z-independent.

P r o o f. Suppose that η

1

, . . . , η

s

are Z-dependent, that is, η

e11

. . . η

ses

= 1 with some e

1

, . . . , e

s

∈ Z. We may assume e

i

6= 0 for all i.

Let e

i

= a

i

p

bi

(Z 3 a

i

6≡ 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

i

e

K

and f (η

iei

) = a

i

f (η

ipbi

) = b

b0i

a

i

f (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

s

with some t (1 ≤ t ≤ s).

From our assumption, X

t i=1

f (η

iei

) = X

t i=1

b

b0i

a

i

f (η

i

) 6≡ 0 mod p.

(7)

Now, c(η

eii

) = c(η

i

) + b

i

e

K

= c(η

i

) + βe

K

for all i = 1, . . . , t. So, from Lemma 1.2, c( Q

t

i=1

η

iei

) = c(η

1

) + βe

K

< c(η

ejj

) for all t + 1 ≤ j ≤ s. There- fore, c( Q

s

i=1

η

iei

) = c( Q

t

i=1

η

eii

) = c(η

1

)+βe

K

. This contradicts Q

s

i=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

V

is imaginary, and f (η)

J

= f (η) if K

V

is real.

P r o o f. First, we assume that K

V

is imaginary. Let η ≡ x + yπ

c(η)K

modπ

c(η)+1K

according 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)

J

mod p.

When K

V

is real, η = η

J

≡ x

J

+y

J

π

Kc(η)

mod π

c(η)+1K

. Thus, f (η) = f (η

J

) ≡ (y/x)

J

≡ f (η)

J

as 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

mpδeK

E

K+

T

/E

K+

T

which satisfies some conditions on the Fermat quotient and the level.

When K

T

is 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

KT

and E

K[K++:KT+]

⊂ Ker(N

+

) · E

K+

T

, where N and N

+

is the norm map from K to K

T

and from K

+

to K

T+

, respectively.

The next lemma is due to Washington [6].

Lemma 2.1. Let E be a subgroup of E

K

of finite index and let η be a non-torsion element of E. If v

0i

(log

p

η) is sufficiently large for all primes

0i

(i = 1, . . . , g) then η is a pth power in E. Here, we consider v

0i

(log

p

η) and log

p

η 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

Z

E

K

) such that {η, u

2

, . . . , u

r

} generates a subgroup E

0

of E of

(8)

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 ℘

0ci

for 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

p

η) and the level of η.

Lemma 2.2. Let η be any element of E

K

\ K

T

and ℘

0i

(i = 1, . . . , g) the prime of K over p. Suppose N (η) = 1 and c(η) > e

K

/(p − 1). Then

v

0

i

(log

p

η) ≥ 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)K

mod π

Kc(η)+1

according to Lemma 1.1. Let c = c(η). Fix any prime ℘

0i

| p. From the assumption, 1 = N (η) ≡ x

eK

mod π

Kc

. Observe that the π

K

-order of x

eK

− 1 is a multiple of e

K

and c is not a multiple of e

K

by Lemma 1.1. Thus x

eK

≡ 1 mod π

Kc+1

. As c > e

K

/(p − 1), we have v

0i

(log

p

x

eK

) ≥ c + 1 (see Lemma 5.5 of Washington [7]).

Thus, v

0i

(e

K

) + v

0i

(log

p

x) ≥ c + 1. From v

0i

(e

K

) = v

p

(e

K

)e

K

, we obtain v

0i

(log

p

x) ≥ c+1−v

p

(e

K

)e

K

. There exists y

1

∈ O

K

such that η = x+y

1

π

cK

and y

1

≡ y mod π

K

. Then, since log

p

η = log

p

x + log

p

(1 + y

1

π

cK

/x), we have

v

0i

(log

p

η) ≥ min



v

0i

(log

p

x), v

0i

 log

p

 1 + y

1

x 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

T

such that c(η) = c} ⊂ O

KT

/(p).

Lemma 2.3. (I) If K

T

and K

V

are imaginary, then F

Kc

⊂ (O

KT

/(p))

+

if c is even, and F

Kc

⊂ (O

KT

/(p))

if c is odd. Moreover , dim

Fp

F

Kc

12

d

K

. (II) If K

T

is imaginary and K

V

is real, then F

Kc

⊂ (O

KT

/(p))

+

for all c, and dim

Fp

F

Kc

12

d

K

.

(III) If K

T

is real and K

V

is imaginary, then e

K

is even and c(η) is even for all η ∈ E

K+

\ K

T

. Obviously, F

Kc

⊂ (O

KT

/(p)) = (O

KT

/(p))

+

and dim

Fp

F

Kc

≤ d

K

.

(IV) If K

T

and K

V

are real, then F

Kc

⊂ (O

KT

/(p)) = (O

KT

/(p))

+

and dim

Fp

F

Kc

≤ d

K

.

P r o o f. (I) Clearly, K is imaginary. Let η ∈ E

K+

\ K

T

and η ≡ x +

Kc(η)

mod π

Kc(η)+1

according to Lemma 1.1. Let c = c(η). Since K

V

is imag-

inary, the statement follows from Lemma 1.6 and (i).

(9)

(II) In this case, our statement follows easily from Lemma 1.6 and (i).

(III) Let η ∈ E

K+

\ K

T

and η ≡ x + yπ

Kc(η)

mod π

c(η)+1K

according to Lemma 1.1. Let c = c(η). Because the order of J is 2, e

K

is clearly even.

From Lemma 1.3,

η = η

J

≡ x

J

+ y

J

(−1)

c

π

Kc

mod π

c+1K

.

Here, x

J

= x and y

J

= y because K

T

is real. So, x + yπ

Kc

≡ x + y(−1)

c

π

cK

mod 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

Z

E

K+

− rank

Z

E

K+

T

. Hence we easily observe that:

r =

12

d

K

(e

K

− 1) and dim

Fp

F

Kc

12

d

K

in the case (I) or (II).

r = d

K

(

12

e

K

− 1) and dim

Fp

F

Kc

≤ d

K

in the case (III).

r = d

K

(e

K

− 1) and dim

Fp

F

Kc

≤ d

K

in 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

mpδeK

E

K+

T

/E

K+

T

such that (1) c(η

i

) > e

K

/(p − 1) (i = 1, . . . , r).

(2) N

+

i

) = 1 (i = 1, . . . , r).

(3) c

1

≤ c

2

≤ . . . ≤ c

r

where c

i

= c(η

i

).

(4) Let S

j

= {η

i

: c(η

i

) ≡ j mod e

K

} (1 ≤ j < e

K

and j is even only if K

T

is real and K

V

is imaginary). Then ]S

j

=

12

d

K

(d

K

resp.) when K

T

is 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

m

E

K+

T

/E

K+

T

. Observe that E

m

E

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

K

or e

K

/2, we have ξ

ipδeK

∈ Ker(N

+

) · E

K+

T

. Therefore,

ξ

piδeK

= η

i

u

i

with 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

mpδeK

E

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

T

is real and K

V

is 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}

(10)

∪ {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

i

b

α0i

f (η

i

) with some a

i

∈ Z,

where α

i

=

e1

K

(c(η

s+1

)−c(η

i

)). We have c(η

aiipαi

) = c(η

s+1

) and f (η

aiipαi

) = a

i

b

α0i

f (η

i

). Then

f

 Y

1≤i≤s, ci≡l mod eK

η

iaipαi



= X

1≤i≤s, ci≡l mod eK

a

i

b

α0i

f (η

i

) = f (η

s+1

).

So, letting

η

0s+1

= η

s+1

 Y

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

1

, . . . , η

r

} as desired.

Note that, in this theorem, the sum of ]S

j

for 1 ≤ j < e

K

(j is even when K

T

is real and K

V

is imaginary) is equal to r by Remark 3.

3. A formula for index of subgroups. Let E and E

0

be 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

0

to be

12

d

K

if K

T

is imaginary (in the case (I) or (II) in Lemma 2.3) and d

K

if K

T

is 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

l

in the case (III) as follows:

B

l

= 

l

2

− 1 

d

0

+ 1, . . . ,

2l

d

0

for 1 < l < e

K

and l even.

Then R is the union of all B

l

.

We permute η

1

, . . . , η

r

and θ

1

, . . . , θ

r

such that c(η

i

) ≡ l mod e

K

for all

i ∈ B

l

and c(θ

j

) ≡ l mod e

K

for all j ∈ B

l

.

(11)

We let, for all j = 1, . . . , r, (5) θ

j

=

Y

r i=1

η

iajipeji

where a

ji

∈ Z is 0 or prime to p, and 0 ≤ e

ji

∈ Z.

Then

det(a

ji

p

eji

) = [E

mpδeK

E

K+ T

/E

K+

T

: (E

0

)

mpδeK

E

K+ T

/E

K+

T

]

= [E

m

E

K+ T

/E

K+

T

: (E

0

)

m

E

K+ T

/E

K+

T

]

= [E : E

0

] × (a natural number prime to p) since E

m

E

K+

T

/E

K+

T

and (E

0

)

m

E

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

0

be subgroups of E

K+

such that E ⊃ E

0

E

K+

T

and [E

K+

: E

0

] < ∞. Then v

p

(det(a

ji

p

eji

)) = v

p

([E : E

0

]).

Next we prove a formula for index [E : E

0

].

Theorem 3.2. Let E and E

0

be 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

K

n X

r

j=1

c(θ

j

) − X

r j=1

c(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

j

is non-empty and A

j

⊂ B

l

if j ∈ B

l

. Further,

f (θ

j

) = X

i∈Aj

f (η

iajipeji

) = X

i∈Aj

a

ji

b

e0ji

f (η

i

) for all j ∈ R.

Since {f (η

i

)}

i∈Bl

and {f (θ

j

)}

j∈Bl

are F

p

-independent systems, it follovs that

(6) det(a

0ji

b

e0ji

)

j,i∈Bl

6≡ 0 mod p for all l, where a

0ji

= a

ji

if i ∈ A

j

and 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

l

and P

l0

= {τ

l

∈ P

l

: τ

l

(j) ∈ A

j

for all j ∈ B

l

}.

Then, for each l,

(7) det(a

0ji

b

e0ji

)

j,i∈Bl

= X

τl∈Pl0



sgn(τ

l

) · Y

j∈Bl

a

l(j)

b

e0jτl(j)



.

(12)

From (6) and (7), we see that P

l0

is non-empty for every l. Let P be the set of all permutations on R and P

0

= {% ∈ P : %(j) ∈ A

j

for all j ∈ R}.

It is clear that any element of P

0

is 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

0

to B

l

is an element of P

l0

. So, we see that P

0

is 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=1

c(θ

j

) <

X

r j=1

c(η

a%(j)j%(j)pej%(j)

).

For % ∈ P

0

, by the definition, X

r j=1

c(θ

j

) = X

r j=1

c(η

a%(j)j%(j)pej%(j)

).

Since X

r j=1

c(η

a%(j)j%(j)pej%(j)

) = X

r j=1

{c(η

%(j)

) + e

K

e

j%(j)

} = X

r j=1

c(η

%(j)

) + e

K

X

r j=1

e

j%(j)

,

we have (9)

X

r j=1

e

j%(j)

= 1 e

K

n X

r

j=1

c(θ

j

) − X

r j=1

c(η

j

) o

for all % ∈ P

0

.

Similarly, from (8), (10)

X

r j=1

e

j%(j)

> 1 e

K

n X

r

j=1

c(θ

j

) − X

r j=1

c(η

j

) o

for all % ∈ P \ P

0

.

From (7) and (9), Y

l

det(a

0ji

b

e0ji

)

j,i∈Bl

= X

%∈P0



sgn(%) · Y

r j=1

a

j%(j)

b

e0j%(j)



= X

%∈P0



sgn(%) · b

Pr

j=1ej%(j)

0

·

Y

r j=1

a

j%(j)



= b

(1/eK)(

Pr

j=1c(¯θj)−Pr

j=1c(¯ηj))

0

· X

%∈P0



sgn(%) · Y

r j=1

a

j%(j)



.

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