A note on Sinnott’s index formula

LXXXII.1 (1997)

A note on Sinnott’s index formula

by

Kazuhiro Dohmae (Tokyo)

0. Introduction. Let k be an imaginary abelian number field with exactly two ramified primes. The letters E and C denote the group of units and the group of circular units in k respectively. Sinnott’s index formula for this case is the following (see Proposition 4.1, Theorem 4.1 and Theorem 5.1 in [6]).

Theorem A (Sinnott). Let k be an imaginary abelian number field with conductor m = p

^e₁¹

p

^e₂²

, where p

1

and p

2

are distinct prime numbers and both e

1

and e

2

are positive integers. Denote by k

i

(i = 1, 2) the maximal subfield of k which is unramified outside p

i

∞. Let G be the Galois group of k over Q. Further , T

pi

and D

pi

denote the inertia group and the decompo- sition group of p

i

in G (i = 1, 2). Then the group C has finite index in E, and

(1) [E : C] = [k

1

: Q][k

2

: Q]

[k : Q] · 2

^−g0

· 2

^{ε1[G:Dp1]+ε2[G:Dp2]+δ1+δ2−1}

· Qh

⁺

, where Q is the unit index of k, h

⁺

the class number of the maximal real subfield of k and g

⁰

some rational integer. Moreover , ε

i

and δ

i

are defined by

ε

i

=  0 if k

3−i

is imaginary, 1 otherwise,

δ

i

=  0 if k

3−i

is real and [D

pi

: T

pi

] is odd , 1 otherwise,

for i = 1, 2. Finally, the rational integer g

⁰

satisfies µ ≤ g

⁰

≤ ν, where µ = ]{1 ≤ i ≤ 2 : k

i

is imaginary},

ν = ]{1 ≤ i ≤ 2 : [k

i

: Q] is even}.

In general, the formula (1) contains the unknown factor 2

^−g0

. But if both k

1

and k

2

are imaginary, then we have µ = ν = 2 and g

⁰

= 2. Hence,

1991 Mathematics Subject Classification: Primary 11R27.

[57]

in this case, (1) reads

(2) [E : C] = [k

1

: Q][k

²

: Q]

[k : Q] · 2

⁻¹

· Qh

⁺

.

In a previous paper [1], we gave another proof of (2) by constructing a system of fundamental circular units (i.e., a basis of the free part of C) of k.

It is the main purpose of this note to prove the following completely explicit version of Theorem A.

Theorem B. Let the notation be as in Theorem A. Then (3) [E : C] = [k

1

: Q][k

2

: Q]

[k : Q] · 2

^{ε1[G:Dp1]+ε2[G:Dp2]−1}

· Qh

⁺

.

The proof includes the explicit construction of a system of fundamental circular units of k. By the way, comparing (1) with (3), we obtain

(4) g

⁰

= δ

1

+ δ

2

.

Kuˇ cera kindly wrote me a direct proof of the inequality g

⁰

≥ δ

₁

+ δ

2

. But I have never found any direct proof of (4) .

In the last section, we also mention our result on a real abelian number field with exactly two ramified primes.

1. Notation. Let k be an imaginary abelian number field of conductor m = p

^e₁¹

p

^e₂²

. We note that k is a subfield of the mth cyclotomic field K = Q(ζ

m

), where ζ

m

= e

^2π

√−1/m

. Let N be the subgroup of T = (Z/mZ)

^×

which corresponds to Gal(K/k) under the natural isomorphism

(Z/mZ)

^×

3 t mod m 7→ (t, K) ∈ Gal(K/Q),

where the automorphism (t, K) maps ζ

m

to ζ

_m^t

. Throughout this paper, we use the following symbols:

• q

₁

= p

^e₁¹

, q

2

= p

^e₂²

;

• ζ = ζ

_m

;

• K

₁

= Q(ζ

q1

), K

2

= Q(ζ

q2

);

• k

₁

= k ∩ K

1

, k

2

= k ∩ K

2

;

• W is the group of roots of unity in k;

• D(q

₁

) = h−1, N

K1/k1

(1 − ζ

^aq2

) : 1 ≤ a < q

1

∧ (a, p

₁

) = 1i;

• D(q

₂

) = h−1, N

_K2/k2

(1 − ζ

^q1b

) : 1 ≤ b < q

2

∧ (b, p

₂

) = 1i;

• D(m) = h−1, N

_K/k

(1 − ζ

^x

) : 1 ≤ x < m ∧ (x, p

1

) = (x, p

2

) = 1i;

• D = D(q

1

)D(q

2

)D(m);

• C = D ∩ E is the group of circular units in k;

• T

₁

= {a (= a mod m) ∈ T : a ≡ 1 mod q

2

};

• T

2

= {b (= b mod m) ∈ T : b ≡ 1 mod q

1

};

• T

₁⁰

= {a ∈ T

1

: there exists b ∈ T

2

such that ab ∈ N };

• T

₂⁰

= {b ∈ T

2

: there exists a ∈ T

1

such that ab ∈ N };

• J = −1 mod m ∈ T ;

• J

₁

≡ −1 mod q

₁

∧ J

₁

≡ 1 mod q

₂

;

• J

₂

≡ 1 mod q

₁

∧ J

₂

≡ −1 mod q

₂

.

Lemma 1. With the above notation we have:

(1) Gal(K

1

/k

1

) = T

₁⁰

, (2) Gal(K

2

/k

2

) = T

₂⁰

, (3) Gal(k/k

1

) = T

₁⁰

T

2

/N , (4) Gal(k/k

2

) = T

1

T

₂⁰

/N . P r o o f. Easy.

R e m a r k. By the statements (1) and (2) of Lemma 1, we can see that k

i

is imaginary if and only if J

i

6∈ T

_i⁰

(i = 1, 2).

In order to prove the theorem, we have to consider the following four cases separately:

I. Both k

1

and k

2

are imaginary (⇔ J

1

6∈ T

₁⁰

∧ J

₂

6∈ T

₂⁰

);

II. k

1

is imaginary and k

2

is real (⇔ J

1

6∈ T

₁⁰

∧ J

₂

∈ T

₂⁰

);

II

⁰

. k

1

is real and k

2

is imaginary (⇔ J

1

∈ T

₁⁰

∧ J

₂

6∈ T

₂⁰

);

III. Both k

1

and k

2

are real (⇔ J

1

∈ T

₁⁰

∧ J

₂

∈ T

₂⁰

).

But case I was treated in [1], and cases II and II

⁰

are similar. So it is sufficient to consider cases II and III. In case II, we use the following symbols:

• a

₁

(= 1), a

2

, . . . , a

r1

a system of representatives for T

1

/hJ

1

iT

₁⁰

;

• b

₁

(= 1), b

2

, . . . , b

r2

a system of representatives for T

2

/T

₂⁰

;

• d

₁

(= 1), d

2

, . . . , d

s

a system of representatives for T

₁⁰

T

₂⁰

/N ;

• Y = {a

_i1

b

i2

d

j

: 1 ≤ i

1

≤ r

₁

∧ 1 ≤ i

₂

≤ r

₂

∧ 1 ≤ j ≤ s};

• Y

⁰

= Y − {1, a

2

, . . . , a

r1

, b

2

, . . . , b

r2

};

• M = {a

i1

q

2

: 2 ≤ i

1

≤ r

1

} ∪ {q

1

b

i2

: 2 ≤ i

2

≤ r

2

} ∪ Y

⁰

;

• M

⁰

= M ∪ {0}.

And, in case III, we use the following ones:

• a

₁

(= 1), a

2

, . . . , a

r1

a system of representatives for T

1

/T

₁⁰

;

• b

₁

(= 1), b

2

, . . . , b

r2

a system of representatives for T

2

/T

₂⁰

;

• d

₁

(= 1), d

2

, . . . , d

s

a system of representatives for T

₁⁰

T

₂⁰

/hJ iN ;

• Y = {a

_i1

b

i2

d

j

: 1 ≤ i

1

≤ r

₁

∧ 1 ≤ i

₂

≤ r

₂

∧ 1 ≤ j ≤ s};

• Y

⁰

= Y − {1, a

2

, . . . , a

r1

, b

2

, . . . , b

r2

};

• M = {a

_i1

q

2

: 2 ≤ i

1

≤ r

₁

} ∪ {q

₁

b

i2

: 2 ≤ i

2

≤ r

₂

} ∪ Y

⁰

;

• M

⁰

= M ∪ {0}.

R e m a r k. We note that Y is a system of representatives for T /hJ iN .

Further, the cardinality of M is equal to

¹₂

[k : Q]−1, which is the unit rank

of k.

Now let l : k

^×

→ R[G] be the G-module homomorphism defined by l : k

^×

3 α 7→ X

σ∈G

log |α

^σ

| · σ

⁻¹

∈ R[G].

It is easy to see that Ker l ∩ E = Ker l ∩ C = W . Hence [E : C] = [l(E) : l(C)].

Furthermore, any finite subset {v

1

, . . . , v

r

} of C is a system of fundamental circular units of k if and only if {l(v

1

), . . . , l(v

r

)} is a Z-basis of l(C).

For any a, b, y ∈ T , we define circular units v

aq2

, v

q1b

, v

y

by v

aq2

= N

K1/k1

 1 − ζ

^aq2

1 − ζ

^q2



, v

q1b

= N

K2/k2

 1 − ζ

^q1b

1 − ζ

^q1



, v

y

= N

K/k

(1 − ζ

^y

).

Then we notice the following facts:

(1) If a ≡ a

⁰

mod T

2

N , then v

aq2

= v

a⁰q2

. (2) If b ≡ b

⁰

mod T

1

N , then v

q1b

= v

q1b⁰

. (3) If y ≡ y

⁰

mod N , then v

y

= v

y⁰

.

Let C

⁰

be the subgroup of C generated by {v

x

: x ∈ M }. Later, we shall see that l(C

⁰

) has finite index in l(C) and l(E). Hence we have

[E : C] = [l(E) : l(C)] = [l(E) : l(C

⁰

)]

[l(C) : l(C

⁰

)] .

2. Computation of [l(C) : l(C

⁰

)]. First, from the definition of C

⁰

, we can easily see the following lemma.

Lemma 2. l(C

⁰

) is generated by {l(v

x

) : x ∈ M }.

We choose two integers l

1

and l

2

such that

l

1

≡ 1 mod q

₁

, p

1

l

1

≡ 1 mod q

₂

, p

2

l

2

≡ 1 mod q

1

, l

2

≡ 1 mod q

2

. Then l

1

∈ T

₂

and l

2

∈ T

₁

. We define f

1

, g

1

, f

2

and g

2

by

f

1

= [hl

1

iT

₂⁰

: T

₂⁰

], g

1

= [T

2

: hl

1

iT

₂⁰

], f

2

= [hl

2

iT

₁⁰

: T

₁⁰

], g

2

= [T

1

: hl

2

iT

₁⁰

].

Let {1, s

2

, . . . , s

g1

} and {1, t

₂

, . . . , t

g2

} be systems of representatives for T

2

/hl

1

iT

₂⁰

and T

1

/hl

2

iT

₁⁰

respectively.

Proposition 3. In case II, l(C) is generated by l(v

a_i1q2

) (2 ≤ i

1

≤ r

₁

), l(v

y

) (y ∈ Y

⁰

) and

1

2

l(v

q1l1

), . . . ,

¹₂

l(v

_q

1l^f1−1₁

),

l(v

q1s

),

¹₂

(l(v

q1s

) − l(v

q1sl1

)), . . . ,

¹₂

(l(v

_q

1sl₁^f1−2

) − l(v

_q

1sl^f1−1₁

))

(s = s

2

, . . . , s

g1

).

Moreover , [l(C) : l(C

⁰

)] = 2

^r2−g1

.

Proposition 4. In case III, l(C) is generated by l(v

y

) (y ∈ Y

⁰

)

and

1

2

l(v

q1l1

), . . . ,

¹₂

l(v

_q

1l^f1−1₁

), l(v

q1s

),

¹₂

(l(v

q1s

) − l(v

q1sl1

)), . . . ,

¹₂

(l(v

_q

1sl₁^f1−2

) − l(v

_q

1sl^f1−1₁

))

(s = s

2

, . . . , s

g1

) and

1

2

l(v

l2q2

), . . . ,

¹₂

l(v

_lf2−1 2 q2

), l(v

tq2

),

¹₂

(l(v

tq2

) − l(v

tl2q2

)), . . . ,

¹₂

(l(v

_tlf2−2

2 q2

) − l(v

_tlf2−1 2 q2

))

(t = t

2

, . . . , t

g2

).

Moreover , [l(C) : l(C

⁰

)] = 2

^{r1+r2−g1−g2}

.

R e m a r k. In the next section, we shall see that l(C

⁰

) has finite index in l(E). Hence

rank

_Z

l(C

⁰

) = rank

_Z

l(C) = rank

_Z

l(E).

On the other hand, we notice that the cardinality of every system of gener- ators for l(C

⁰

) or l(C) stated in the above propositions is equal to the unit rank of k. This implies that these systems of generators are bases.

We can prove Propositions 3 and 4 in a similar fashion. So we only prove Proposition 3. Let L be the subgroup of R[G] generated by the elements stated in the proposition.

First we prove L ⊂ l(C). Fix a b

i2

(1 ≤ i

2

≤ r

₂

). Then X

1≤i1≤r1

1≤j≤s

l(v

a_i1b_i2dj

)

= X

1≤i1≤r1

1≤j≤s

l(N

_K/k

(1 − ζ

^ai1bi2dj

))

= 1 2

X

1≤i1≤r1

1≤j≤s

l(N

K/k

(1 − ζ

^ai1bi2dj

)) + 1 2

X

1≤i1≤r1

1≤j≤s

l(N

K/k

(1 − ζ

^J1ai1bi2dj

))

= 1 2

X

0≤h1≤1 1≤i1≤r1

1≤j≤s

l(N

_K/k

(1 − ζ

^J1h1ai1bi2dj

))

= 1 2 l

 Y

0≤h1≤1 1≤i1≤r1

1≤j≤s

N

K/k

(1 − ζ

^J1h1ai1bi2dj

)

 .

Now, by Lemma 1(4), we have Y

0≤h1≤1 1≤i1≤r1

1≤j≤s

N

K/k

(1 − ζ

^J1h1ai1bi2dj

) = N

k/k2

(N

K/k

(1 − ζ

^bi2

)).

So we get X

1≤i1≤r1

1≤j≤s

l(v

a_i1b_i2dj

)

= 1

2 l(N

k/k2

(N

K/k

(1 − ζ

^bi2

)))

= 1

2 l(N

_K2/k2

(N

_K/K2

(1 − ζ

^bi2

)))

= 1 2 l

 N

K2/k2

 1 − ζ

^q1bi2

1 − ζ

^q1l1bi2



= 1 2 l

 N

K2/k2

 1 − ζ

^q1bi2

1 − ζ

^q1



− 1 2 l

 N

K2/k2

 1 − ζ

^q1l1bi2

1 − ζ

^q1



= 1

2 (l(v

q1b_i2

) − l(v

q1l1b_i2

)).

Hence

1

2 (l(v

q1b_i2

) − l(v

q1l1b_i2

)) ∈ l(C).

From this, we can easily deduce that L ⊂ l(C).

Next we prove l(C) ⊂ L. For this purpose, it is sufficient to prove that l(v

y

) ∈ L for y ∈ Y − Y

⁰

, because it is obvious that l(v

q1b_i2

) ∈ L for 2 ≤ i

2

≤ r

₂

. Fix an a

i1

(2 ≤ i

1

≤ r

₁

). Then

X

1≤i2≤r2

1≤j≤s

l(v

a_i1b_i2dj

) = l(N

k/k1

(N

K/k

(1 − ζ

^ai1

)))

= l(N

K1/k1

(N

K/K1

(1 − ζ

^ai1

)))

= l

 N

K1/k1

 1 − ζ

^ai1q2

1 − ζ

^ai1q2l2



= l

 N

K1/k1

 1 − ζ

^ai1q2

1 − ζ

^q2



− l

 N

K1/k1

 1 − ζ

^ai1q2l2

1 − ζ

^q2



= l(v

a_i1q2

) − l(v

a_i1q2l2

) ∈ L.

If a

i1

b

i2

d

j

6= a

_i1

, then a

i1

b

i2

d

j

∈ Y

⁰

by the definition of Y

⁰

. Hence l(v

a_i1

) ∈ L. Now we fix a b

i2

(1 ≤ i

2

≤ r

₂

). Then, as we have seen above,

X

1≤i1≤r1

1≤j≤s

l(v

a_i1b_i2dj

) = 1

2 (l(v

q1b_i2

) − l(v

q1l1b_i2

)).

If a

i1

b

i2

d

j

6= b

_i2

, then l(v

a_i1b_i2dj

) ∈ L. Hence l(v

b_i2

) ∈ L. We have thus proved that l(C) ⊂ L.

Finally, by computing the determinant of the transition matrix, we can see that

[l(C) : l(C

⁰

)] = 1

1 2

f1−1

·

¹₂

f1−1

g1−1

= 2

^f1g1−g1

= 2

^r2−g1

. This completes the proof of Proposition 3.

3. Computation of [l(E) : l(C

⁰

)]. For each t ∈ T , we let c

0

= 1,

c

tq2

=  1 if t ∈ T

₂

N, 0 otherwise, c

tq1

=  1 if t ∈ T

1

N,

0 otherwise, c

t

= n 1 if t ∈ N ,

0 otherwise.

Further, we define b

0,t

= 2,

b

aq2,t

= (c

(aq2)t

+ c

J (aq2)t

) − (c

tq2

+ c

J tq2

) (a ∈ T ), b

q1b,t

= (c

_(q1b)t

+ c

_{J (q1b)t}

) − (c

tq1

+ c

J tq1

) (b ∈ T ),

b

y,t

= (c

yt

+ c

J yt

) − 1

[k : k

2

] (c

ytl1q1

+ c

J ytl1q1

)

− 1

[k : k

1

] (c

ytl2q2

+ c

J ytl2q2

) + 2

[k : Q] (y ∈ T ).

We denote by σ

t

(t ∈ T ) the automorphism of k over Q which is the image of (t, K) under the canonical surjection

Gal(K/Q) → G = Gal(k/Q).

We omit the proof of the next lemma because it is just the same as those of Lemmas 3.5 and 3.6 in [1].

Lemma 5. Let Q be the unit index of k and h

⁺

the class number of k

⁺

. Then

[l(E) : l(C

⁰

)] = 1

[k : Q] · |det(b

_x,t

)

x∈M⁰,t∈Y

| · Qh

⁺

.

In what follows, we compute the determinant of the matrix (b

x,t

)

x∈M⁰,t∈Y

. We define

A = {J

₁^h1

a

i1

q

2

: 0 ≤ h

1

≤ 1 ∧ 1 ≤ i

₁

≤ r

₁

}, B = {q

1

b

i2

: 1 ≤ i

2

≤ r

₂

} in case II, and

A = {a

i1

q

2

: 1 ≤ i

1

≤ r

₁

}, B = {q

1

b

i2

: 1 ≤ i

2

≤ r

₂

} in case III. Then it is easy to see that

X

y∈A

c

yt

= X

y∈B

c

yt

= 1 for any t ∈ T .

Lemma 6. (1) For any x ∈ A, there exist rational numbers α

i1

(1 ≤ i

1

≤ r

1

) such that

c

xt

+ c

J xt

= α

1

(c

0

+ c

0

) +

r1

X

i1=2

α

i1

(c

(a_i1q2)t

+ c

J (a_i1q2)t

) for any t ∈ T .

(2) For any x ∈ B, there exist rational numbers β

i2

(1 ≤ i

2

≤ r

₂

) such that

c

xt

+ c

J xt

= β

1

(c

0

+ c

0

) +

r2

X

i2=2

β

i2

(c

(q1b_i2)t

+ c

J (q1b_i2)t

) for any t ∈ T .

P r o o f. As cases II and III are similar, we prove the lemma for case II.

If x ∈ M , there is nothing to prove. So we suppose that x 6∈ M . (1) Since

c

_(J1a

i1q2)t

+ c

_{J (J1a}

i1q2)t

= c

_(J1a

i1q2)t

+ c

_(a

i1q2)t

= c

_(a

i1q2)t

+ c

_{J (a}

i1q2)t

, it suffices to consider the case x = q

2

. Let us define

A

0

= {a

i1

q

2

: 1 ≤ i

1

≤ r

₁

}, A

1

= {J

1

a

i1

q

2

: 1 ≤ i

1

≤ r

₁

}.

Then

X

y∈A0

(c

yt

+ c

J yt

) = X

y∈A

(c

yt

+ c

J yt

) − X

y∈A1

(c

yt

+ c

J yt

).

On the other hand, we can see that X

y∈A1

(c

yt

+ c

J yt

) = X

y∈A1

(c

_{(J y)t}

+ c

_{J (J y)t}

) = X

y∈A0

(c

yt

+ c

J yt

).

Hence X

y∈A0

(c

yt

+ c

J yt

) = 1 2

X

y∈A

(c

yt

+ c

J yt

) = 1

2 (1 + 1) = 1

2 (c

0

+ c

0

).

Consequently, we obtain c

q2t

+ c

J q2t

= 1

2 (c

0

+ c

0

) −

r1

X

i1=2

(c

(a_i1q2)t

+ c

J (a_i1q2)t

).

(2) From x 6∈ M , we have x = q

1

. Since

r2

X

i2=1

(c

(q1b_i2)t

+ c

J (q1b_i2)t

) = 1 + 1 = c

0

+ c

0

, we get

c

q1t

+ c

J q1t

= (c

0

+ c

0

) −

r2

X

i2=2

(c

(q1b_i2)t

+ c

J (q1b_i2)t

).

Using the above lemma, we can deduce that

|det(b

x,t

)

x∈M⁰,t∈Y

| = r

1

r

2

· |det(c

xt

+ c

J xt

)

x∈M⁰,t∈Y

|

by the same argument as in the proof of Lemma 3.8 in [1]. So, in order to accomplish our purpose, we have to compute the determinant of the matrix (c

xt

+ c

J xt

)

x∈M⁰,t∈Y

.

For each t ∈ Y , there exists exactly one u ∈ Y such that tu ∈ hJ iN , because Y is a system of representatives for T /hJ iN . We denote this u by t

⁰

. Then the map Y 3 t 7→ t

⁰

∈ Y is bijective.

Lemma 7. Suppose x ∈ M , t ∈ Y and c

xt⁰

+ c

J xt⁰

6= 0. Then:

(1) if x ∈ Y

⁰

, then t = x;

(2) if x = a

i1

q

2

(2 ≤ i

1

≤ r

₁

), then t ∈ Y

⁰

or t = a

i1

; (3) if x = q

1

b

i2

(2 ≤ i

2

≤ r

₂

), then t ∈ Y

⁰

or t = b

i2

. P r o o f. (1) Straightforward.

(2) If c

_(a

i1q2)t⁰

+ c

_{J (a}

i1q2)t⁰

6= 0, then a

i1

t

⁰

∈ hJ iT

2

N . Hence there is a y = b

i2

J

^h

d

j

such that a

i1

yt

⁰

= a

i1

b

i2

J

^h

d

j

t

⁰

∈ hJ iN , and so we have a

i1

b

i2

d

j

t

⁰

∈ hJ iN . Therefore t = a

i1

b

i2

d

j

, and we obtain t = a

i1

or t ∈ Y

⁰

.

(3) In case III, the proof is similar to (2). So we consider case II. If c

(q1b_i2)t⁰

+c

J (q1b_i2)t⁰

6= 0, then b

_i2

t

⁰

∈ hJ iT

₁

N . Hence there is a w = J

₁^h1

a

i1

d

j

such that wb

i2

t

⁰

= J

₁^h1

a

i1

b

i2

d

j

t

⁰

∈ hJ iN , and so a

_i1

b

i2

J

₂^h1

d

j

t

⁰

∈ hJ iN . As

J

2

∈ T

₂⁰

, there exists a d

j⁰

such that J

₂^h1

d

j

≡ d

_j⁰

mod hJ iN , and then a

i1

b

i2

d

j⁰

t

⁰

∈ hJ iN . Therefore t = b

_i2

or t ∈ Y

⁰

.

From this, we can obtain the following conclusion.

Proposition 8. Let Q be the unit index of k and h

⁺

the class number of k

⁺

.

(1) In case II, we have

[l(E) : l(C

⁰

)] = [k

1

: Q][k

2

: Q]

[k : Q] · 2

^r2−1

· Qh

⁺

. (2) In case III, we have

[l(E) : l(C

⁰

)] = [k

1

: Q][k

²

: Q]

[k : Q] · 2

^r1+r2−1

· Qh

⁺

. In particular , l(C

⁰

) and l(C) have finite indices in l(E).

4. Proof of Theorem B. First we consider case II. By Propositions 3 and 8(1), we have

[E : C] = [l(E) : l(C

⁰

)]

[l(C) : l(C

⁰

)] = [k

1

: Q][k

2

: Q]

[k : Q] · 2

^r2−1

2

^r2−g1

· Qh

⁺

= [k

1

: Q][k

2

: Q]

[k : Q] · 2

^g1−1

· Qh

⁺

.

Now, since σ

l1

is the inverse of the Frobenius automorphism for p

1

in k, we have

g

1

= [T : hl

1

iT

₁

N ] = [G : D

p1

].

Therefore

[E : C] = [k

1

: Q][k

²

: Q]

[k : Q] · 2

^[G:Dp1]−1

· Qh

⁺

.

Next we consider case III. By Propositions 4 and 8(2), we have [E : C] = [k

1

: Q][k

2

: Q]

[k : Q] · 2

^g1+g2−1

· Qh

⁺

. As also g

1

= [G : D

p1

] and g

2

= [G : D

p2

], we obtain

[E : C] = [k

1

: Q][k

²

: Q]

[k : Q] · 2

^{[G:Dp1]+[G:Dp2]−1}

· Qh

⁺

. This completes the proof of the theorem.

5. The real case. Let k be a real abelian number field with con-

ductor p

^e₁¹

p

^e₂²

. Let T

1

, T

2

, T

₁⁰

, T

₂⁰

, N be the same as in Section 1, and

{a

_i1

}, {b

_i2

}, {d

_j

} be systems of representatives for T

₁

/T

₁⁰

, T

2

/T

₂⁰

, T

₁⁰

T

₂⁰

/N re- spectively. Then

{v

_a

i1q2

: 2 ≤ i

1

≤ r

₁

} ∪ {v

_q1b

i2

: 2 ≤ i

2

≤ r

₂

} ∪ {v

_y

: y ∈ Y

⁰

} is a system of fundamental circular units of k, where

Y

⁰

= {a

i1

b

i2

d

j

: 1 ≤ i

1

≤ r

₁

∧ 1 ≤ i

₂

≤ r

₂

∧ 1 ≤ j ≤ s}

− {1, a

₂

, . . . , a

r1

, b

2

, . . . , b

r2

}.

We can deduce Sinnott’s index formula [E : C] = [k

1

: Q][k

²

: Q]

[k : Q] · 2

^[k:Q]−1

· h

by computing the regulator of the system of fundamental units. We omit the proof because it is very similar to that of the imaginary case.

Acknowledgments. I am grateful to Radan Kuˇ cera for many helpful comments and suggestions for improving an earlier version of this note. I would like to thank the referee for his careful reading of the manuscript and valuable suggestions.

References

[1] K. D o h m a e, On bases of groups of circular units of some imaginary abelian number fields, J. Number Theory 61 (1996), 343–364.

[2] R. G o l d and J. K i m, Bases for cyclotomic units, Compositio Math. 71 (1989), 13–28.

[3] R. K u ˇ c e r a, On bases of odd and even universal ordinary distributions, J. Number Theory 40 (1992), 264–283.

[4] —, On bases of the Stickelberger ideal and of the group of circular units of a cyclo- tomic field , ibid., 284–316.

[5] W. S i n n o t t, On the Stickelberger ideal and the circular units of a cyclotomic field , Ann. of Math. 108 (1978), 107–134.

[6] —, On the Stickelberger ideal and the circular units of an abelian field , Invent. Math.

62 (1980), 181–234.

[7] L. W a s h i n g t o n, Introduction to Cyclotomic Fields, Grad. Texts in Math. 83, Springer, New York, 1980.

Department of Mathematics Tokyo Metropolitan University Minami-Ohsawa 1-1, Hachioji-shi Tokyo 192-03, Japan

E-mail: dohmae@math.metro-u.ac.jp

Received on 26.9.1996

and in revised form on 7.4.1997 (3050)