Different groups of circular units of a compositum of real quadratic fields

LXVII.2 (1994)

Different groups of circular units of a compositum of real quadratic fields

by

Radan Kuˇ cera (Brno)

1. Introduction. There are many different definitions of the group of circular units of a real abelian field. The aim of this paper is to study their relations in the special case of a compositum k of real quadratic fields such that −1 is not a square in the genus field K of k in the narrow sense.

The reason why fields of this type are considered is as follows. In such a field it is possible to define a group C of units (slightly bigger than Sinnott’s group of circular units) such that the Galois group acts on C/(±C ² ) trivially (see [K, Lemma 2]).

Due to this key property we can easily compare different groups of cir- cular units (see the conclusion of this paper).

2. The group C and the Sinnott group C ⁰ . Let k be a compositum of quadratic fields and suppose −1 is not a square in the genus field K of k in the narrow sense. This condition can be written equivalently as follows: either 2 does not ramify in k and k = Q( √

d ₁ , . . . , √

d _s ), where d 1 , . . . , d s with s ≥ 1 are square-free positive integers all congruent to 1 modulo 4, or 2 ramifies in k and there is a unique x ∈ {2, −2} such that k = Q( √

d ₁ , . . . , √

d _s ), where d ₁ , . . . , d _s with s ≥ 1 are square-free positive integers such that d i ≡ 1 (mod 4) or d i ≡ x (mod 8) for each i ∈ {1, . . . , s}.

In the former case, let

J = {p ∈ Z : p ≡ 1 (mod 4), |p| is a prime ramifying in k}, and, in the latter case, let

J = {x} ∪ {p ∈ Z : p ≡ 1 (mod 4), |p| is a prime ramifying in k}.

For any p ∈ J, let

n _{p} =

 |p| if p is odd, 8 if p is even.

[123]

For any S ⊆ J let (by convention, an empty product is 1) n _S = Y

p∈S

n _{p} , ζ _S = e ^2πi/n

, Q ^S = Q(ζ _S ), K _S = Q( √

p : p ∈ S).

It is easy to see that K J = K and that n J is the conductor of k. Let us define

ε S =

 



1 if S = ∅,

√ 1

p N _Q

/K

(1 − ζ S ) if S = {p}, N _Q

/K

(1 − ζ _S ) if #S > 1,

k _S = k ∩ K _S and η _S = N _K

_/k

(ε _S ) for any S ⊆ J. It is easy to see that ε _S and η S are units in K S and k S , respectively.

For any p ∈ J let σ _p be the non-trivial automorphism in Gal(K _J /K _J\{p} ).

Then G = Gal(K _J /Q) can be considered as a (multiplicative) vector space over F 2 with F 2 -basis {σ p : p ∈ J}. Let

X = {ξ ∈ b G : ξ(σ) = 1 for all σ ∈ Gal(K _J /k)},

where b G is the character group of G. Then X can be viewed also as the group of all Dirichlet characters corresponding to k. For any χ ∈ X let

S χ = {p ∈ J : χ(σ p ) = −1}.

Let C be the group generated by −1 and by {η _S ^σ : S ⊆ J, σ ∈ G}.

Let C ⁰ be the Sinnott group of circular units of k, i.e., the group of units in the group generated by −1 and

{N _Q

/Q

∩k (1 − ζ _S ) ^σ : σ ∈ G, S ⊆ J, S 6= ∅}

(see [L]). When we speak about a basis of a group of units we always have in mind a basis of the non-torsion part.

Proposition 1. The set {η _S

: χ ∈ X, χ 6= 1} is a basis of C and [E : C] =  Y

χ∈X χ6=1

(2 · [k : k S

])



· [k : Q] ^−[k:Q]/2 · h,

where h is the class number of k and E is the full group of units in k. The set

{η _S

: χ ∈ X, #S _χ > 1} ∪ {η _S ²

: χ ∈ X, #S _χ = 1}

is a basis of C ⁰ and [C : C ⁰ ] = 2 ^a , where a = #{p ∈ J : √

p ∈ k} = #{χ ∈ X : #S _χ = 1}.

P r o o f. The results concerning C were proved in [K, Theorem 1]. It was proved in [K, Section 4] that C ⁰ is generated by

{−1} ∪ {η S : S ⊆ J, #S > 1} ∪ {η ² _{p} : p ∈ J, p > 0, √ p ∈ k}.

It was shown in [K, proof of Lemma 5] that for any S ⊆ J such that S 6= S _χ for all χ ∈ X there are a T ∈ Z satisfying

η _S = ± Y

T (S

η ^a _T

.

But η _T is totally positive if #T > 1 (it is a norm from an imaginary abelian field to a real one) while η ^1+σ _{p}

= −1 for any p ∈ J such that √

p ∈ k due to [K, Lemma 1]. Thus a _{p} is even for all such p and the proposition follows.

3. The groups defined by Hasse, Leopoldt, Gras and Gillard.

To define all groups we are interested in we shall follow Gillard’s paper [G]. Let F be a real abelian field. Let ξ be a non-principal Q-irreducible Q-character on Gal(F/Q) with kernel denoted by ker ξ (i.e., ξ is the sum of all linear characters Gal(F/Q) → C ^× with kernel equal to ker ξ). Let F _ξ denote the subfield of F corresponding to ker ξ, f ξ the conductor of F ξ and G _ξ = Gal(F _ξ /Q). It is easy to see that G _ξ is a cyclic group. Let ζ _n = e ^2πi/n for any positive integer n. Then we define

θ ξ = Y

σ

(ζ 2f

− ζ _2f ⁻¹

) ^{¯ σ}

where the product is taken over all σ ∈ Gal(Q(ζ _f

) ⁺ /F _ξ ) and σ means an extension of σ to Q(ζ _2f

). Thus θ _ξ is well-defined up to sign and

(−1) ^s(ξ) θ _χ ² = N _Q(ζ

_)/F

(1 − ζ _f

) ∈ F _ξ , where s(ξ) = [Q(ζ _f

) ⁺ : F _ξ ]. For any α ∈ G _ξ fix some √

(θ ² _ε ) ^α and denote it by θ _ξ ^α . This definition can be extended to α ∈ Z[G ξ ] by linearity.

Suppose that for any such ξ 6= 1 we have an ideal I ξ ⊆ Z[G ξ ]. Then we can consider the group Q

ξ6=1 {±θ ^α _ξ : α ∈ I _ξ }. For some special choices of I _ξ we obtain the following interesting groups. The Leopoldt group of formal cyclotomic units C ⁽⁰⁾ is obtained if I ξ is the ideal generated by

γ _ξ = Y

p|n

(1 − σ ^n/p ),

where σ is a generator of the cyclic group G _ξ of order n, and p in the product runs through all primes dividing n. We obtain the Hasse group C ⁽¹⁾ if I _ξ is the augmentation ideal of Z[G ξ ] (i.e., I ξ is generated by σ − 1, where σ denotes a generator of G _ξ ). We get the Gillard group C ⁽²⁾ if

I _ξ = {α ∈ Z[G _ξ ] : θ ^α _ξ is a unit in F }

and the Gras group C ⁽³⁾ (for F not necessarily cyclic) if I _ξ = {α ∈ Z[G _ξ ] : θ _ξ ^α is a unit in F _ξ }.

Finally, the Leopoldt group of cyclotomic units H is the intersection E ∩C ⁽⁴⁾ , where C ⁽⁴⁾ is obtained if I _ξ = Z[G _ξ ].

Now, consider these groups for F being our field k. So we need not distinguish between linear characters and Q-irreducible Q-characters. For any χ ∈ X, χ 6= 1, the field F _χ is a quadratic subfield of k _S

. The conductor of F χ is f χ = n S

, so ζ f

= ζ S

. Moreover, s(χ) = ¹ ₄ ϕ(f χ ) is odd if and only if S _χ = {p} and p = 2 or |p| = p ≡ 5 (mod 8) or if S _χ = {p, q} and p 6= q are odd and negative. If S _χ = {p} then p > 0, k _S

= K _S

= F _χ = Q( √

p) and

(−1) ^s(χ) θ ² _χ = N _Q

/F

(1 − ζ S

) = √ p · ε S

. On the other hand, if #S _χ > 1 then

(1) (−1) ^s(χ) θ ² _χ = N _Q

/F

(1 − ζ _S

) = N _k

_/F

(η _S

).

Fix some σ _χ ∈ Gal(k _S

/Q) \ Gal(k _S

/F _χ ) for any χ ∈ X, χ 6= 1. Then Gal(F _χ /Q) = {1, σ _χ | _F

}. It is easy to see that C ⁽⁰⁾ = C ⁽¹⁾ is generated by

−1 and by

{θ _χ ^1−σ

: χ ∈ X, χ 6= 1}

and that this set is a basis because the number of elements involved is precisely the Z-rank. If S χ = {p} then

(θ _χ ² ) ^1−σ

= ( √

p ε _{p} ) ^1−σ

= pε ² _{p}

( √

p ε _{p} ) ^1+σ

= ε ² _{p} = η _{p} ²

by [K, Lemma 1] and because K _S

= k _S

. Let us concentrate on the case where #S χ > 1. Then

(θ _χ ² ) ^1−σ

= N _k

_/F

(η S

) ^1−σ

= N _k

_/F

(η _S

) ²

N _k

_/Q (η S

) = N _k

_/F

(η S

) ² , because N _k

_/Q (η _S

) = N _Q

/Q (1 − ζ _S

) = 1. Therefore (recall that θ _χ can be outside of k _S

and that θ χ ^1−σ

is determined only up to sign in this case) (2) θ _χ ^1−σ

= ±N _k

_/F

(η _S

).

Let σ ∈ Gal(k S

/F χ ), so χ(σ)=1. Choose T ⊆ S χ such that σ= Q

p∈T σ p | k

. Then

1 = χ(σ) = Y

p∈T

χ(σ p ) = (−1) ^#T ,

and

η _S ^1−σ

χ

= η _S ^1−Π

^σ

χ

= Y

p∈T

(η ^1+σ _S

χ

) ^Π

(−σ

) . Of course,

η _S ^1+σ

χ

= N _K

_/K

Sχ\{p}

(η _S

) = N _k

_K

Sχ\{p}

/K

(η _S

) ^[K

^:k

^K

^]

= (±η 1−Frob(|p|,k

)

S

\{p} ) ^[K

^:k

^K

^]

by [K, Lemma 4], because k S

∩ K _S

_\{p} = k _S

_\{p} . Therefore η _S ^1−σ

= ± Y

R(S

η ^2a _R

for suitable integers a _R due to Lemma 3 of [K]. So

θ ^1−σ _χ

= ±N _k

_/F

(η _S

) (3)

= ± Y

σ∈Gal(k

/F

)

η _S ^σ

= η _S ^[k

^:F

^] ·



± Y

R(S

η ^2b _R



for suitable integers b _R . But {η _S

: χ ∈ X, χ 6= 1} is a basis of C and if some η _R is not in this basis then it can be written as a combination of η _R

, where R ⁰ ( R (see [K, Theorem 1 and the proof of Lemma 5]). We have proved the following

Proposition 2. The set {θ χ ^1−σ

: χ ∈ X, χ 6= 1} is a basis of C ⁽⁰⁾ = C ⁽¹⁾ ⊆ C and

[C : C ⁽⁰⁾ ] = Y

χ∈X χ6=1

[k _S

: F _χ ] = Y

χ∈X χ6=1

1

2 [k _S

: Q]
.

For studying C ⁽²⁾ and C ⁽³⁾ we need to know when θ χ ∈ k and θ χ ∈ F χ , respectively. We shall suppose that #S _χ > 1, because θ _χ is not a unit if

#S _χ = 1. If s(χ) is odd then −θ ² _χ = N _Q

/F

(1 − ζ _S

) > 0, so θ _χ is pure imaginary and θ _χ 6∈ k. Suppose now that s(χ) is even. Recall that χ can be considered as an even Dirichlet character modulo f _χ = n _S

. We need to distinguish two cases.

First, suppose that n _S

is odd. Let q = min S _χ and write |q| − 1 = 2 ^b · c with c odd. Let ψ be a Dirichlet character modulo |q| of order 2 ^b , so ψ(−1) =

−1, and let

A = {a ∈ Z : 1 ≤ a ≤ f _χ , χ(a) = 1, (ψ(a) = 1 or Im ψ(a) > 0)}.

It is easy to see that for any σ ∈ Gal((Q ^S

) ⁺ /F χ ) there is precisely one

a ∈ A such that σ is the restriction to (Q ^S

) ⁺ of the automorphism of Q ^S

sending ζ S

to ζ _S ^a

. Therefore θ _χ = Y

a∈A

(ξ ^a − ξ ^−a ),

where ξ = ζ _S ^(1+f

^)/2 . We want to prove that θ χ ∈ F χ . Choose any σ ∈ Gal(Q ^S

/F _χ ). If y is determined by σ(ζ _S

) = ζ _S ^y

χ

then we define A ₁ = {a ∈ A : ψ(ay) = 1 or Im ψ(ay) > 0}, A ₂ = {a ∈ A : ψ(ay) = −1 or Im ψ(ay) < 0}.

Because χ(y) = 1 and χ(−1) = 1, the mapping f : A → A defined by f (a) ≡ ay (mod f _χ ) if a ∈ A ₁ , and f (a) ≡ −ay (mod f _χ ) if a ∈ A ₂ , is a permutation. Therefore

σ(θ _χ ) = Y

a∈A

(ξ ^ay − ξ ^−ay )

=  Y

a∈A

(ξ ^{f (a)} − ξ ^{−f (a)} )  Y

a∈A

(−1)(ξ ^{f (a)} − ξ ^{−f (a)} )



= (−1) ^#A

· θ χ .

It is easy to see that #A = ¹ ₄ ϕ(f χ ) = s(χ) and that #{a ∈ A : ψ(a) = ψ(a ₀ )} = 2 ^1−b s(χ) for any fixed a ₀ ∈ A. But A ₂ is a disjoint union of such sets involving some a ₀ , so 2 ^1−b s(χ) | (#A ₂ ). If q < 0 then |q| ≡ 3 (mod 4), so b = 1, s(χ) | (#A 2 ) and #A 2 is even. If q > 0 then also q ⁰ = min(S χ \ {q}) >

q > 0 (recall #S _χ > 1) and q ⁰ ≡ 1 (mod 4). Thus 2 ^1−b s(χ) = 2 ^−1−b ϕ(f χ ) = c q ⁰ − 1

2

Y

p∈S

\{q,q

}

ϕ(p) ≡ 0 (mod 2) and #A ₂ is again even. We have proved that θ _χ ∈ F _χ if n _S

is odd and either #S _χ > 2 or S _χ = {p, q} with p > 0 and q > 0.

Now, suppose that n _S

is even. Then n _S

= 8n for some odd n > 1 and s(χ) = ϕ(n). Directly from the definition we have

(4) θ _χ = Y

a

(ζ _16n ^a − ζ _16n ^−a ),

where the product is taken over all integers a satisfying 0 < a < 16n and χ(a) = 1 which are congruent to 1 or to 5 modulo 16. It is easy to see that there is y ≡ 5 (mod 16) such that χ(y) = 1. Let σ ∈ Gal(Q(ζ 16n )/Q) be determined by ζ _16n ^σ = ζ _16n ^y . Then σ ∈ Gal(Q(ζ _16n )/F _χ ) and

θ _χ ^σ−1 =

 Y

0<a<16n χ(a)=1 a≡5,9 (mod 16)

(ζ _16n ^a − ζ _16n ^−a )

 Y

0<a<16n χ(a)=1 a≡1,5 (mod 16)

(ζ _16n ^a − ζ _16n ^−a )

 ₋₁

.

Of course, a ≡ 9 (mod 16) if and only if a ± 8n ≡ 1 (mod 16), so θ _χ ^σ−1 = (−1) #{a∈Z:0<a<16n, χ(a)=1, a≡9 (mod 16)} = (−1) ^ϕ(n)/2 .

Consider any automorphism τ ∈ Gal(Q(ζ 16n )/F χ ) and let x ∈ Z be such that ζ _16n ^τ = ζ _16n ^x , so χ(x) = 1. If x ≡ 1 (mod 4) then there is j ∈ {1, . . . , 4}

satisfying 5 ^j x ≡ 1 (mod 16), so y ^j x ≡ 1 (mod 16) and σ ^j τ acts on θ χ

identically, because it only permutes the terms in the product (4). Thus in this case

θ ^{τ −1} _χ = θ ^σ _χ

^{τ −1} θ _χ ^σ−1
_−(σ

_+...+1)τ

= (−1) ^jϕ(n)/2 .

On the other hand, if x ≡ −1 (mod 4) then we can consider τ ⁰ ∈ Gal(Q(ζ _16n )/F _χ ) satisfying ζ _16n ^τ

= ζ _16n ^−x (recall that χ(−1) = 1 because F χ is real). Because

(ζ _16n ^a − ζ _16n ^−a ) ^τ = −(ζ _16n ^a − ζ _16n ^−a ) ^τ

and there is an even number of terms in the product (4), we have θ _χ ^τ = θ ^τ _χ

. We have proved that θ χ ∈ F χ if and only if ϕ(n) is divisible by 4. If

#S _χ > 2 then there are at least two different primes dividing n and 4 | ϕ(n).

If S _χ = {2, p} then √

2p ∈ k, so p > 0 and n = p ≡ 1 (mod 4), hence again 4 | ϕ(n). Finally, if S χ = {−2, p} then √

−2p ∈ k, so p < 0 and n = −p ≡ 3 (mod 4), in which case 4 does not divide ϕ(n). We shall prove that in the last case even θ _χ 6∈ k. Indeed, if τ ∈ Gal(Q(ζ _16n )/F _χ ( √

2)) and if x ∈ Z satisfies ζ _16n ^τ = ζ _16n ^x then x ≡ ±1 (mod 8) and θ ^{τ −1} _χ = 1 by the previous computation. But this means that θ _χ ∈ F _χ ( √

2). So √

2 ∈ F _χ (θ _χ ) ⊆ K _J (θ _χ ) but √

2 6∈ K _J , because √

−1 6∈ K _J and √

−2 ∈ K _J in this case. Thus K J 6= K J (θ χ ), which implies θ χ 6∈ k ⊆ K J .

Proposition 3. Let J ⁺ = {p ∈ J : p > 0} and J ⁻ = {p ∈ J : p < 0}.

Then the set

{θ _χ : χ ∈ X, #S _χ ≥ 2, S _χ ⊆ J ⁺ if #S _χ = 2}

∪ {θ ^1−σ _χ

: χ ∈ X, #S χ = 1 or 2, S χ ⊆ J ⁻ if #S χ = 2}

is a basis of C ⁽²⁾ = C ⁽³⁾ . The set

{θ χ : χ ∈ X, [k S

: Q] > 2} ∪ {θ _χ ^1−σ

: χ ∈ X, [k S

: Q] = 2}

is a basis of C ⁽²⁾ ∩C. Moreover , [C ⁽²⁾ : C ⁽⁰⁾ ] = 2 ^b and [C ⁽²⁾ : C ⁽²⁾ ∩C] = 2 ^c , where

b = #{χ ∈ X : #S χ ≥ 2, S χ ⊆ J ⁺ if #S χ = 2},

c = #{χ ∈ X : #S _χ ≥ 2, [k _S

: Q] = 2, S _χ ⊆ J ⁺ if #S _χ = 2}.

P r o o f. Let χ ∈ X, χ 6= 1. We have shown in the previous computation that θ _χ ∈ k if and only if θ _χ ∈ F _χ , and that this is the case if and only if

#S _χ > 1 and S _χ ⊆ J ⁺ if #S _χ = 2. Thus C ⁽²⁾ = C ⁽³⁾ . If #S _χ > 1 then θ ² _χ = ±θ ^1−σ χ

by (1) and (2), so a basis of C ⁽²⁾ can have the above described form.

If k _S

= F _χ , then θ χ ^1−σ

= ±η _S

by (2). If k _S

6= F _χ then #S _χ > 1 and (5) θ _χ = ±η _S ^[k

^:F

^]/2

χ

· Y

R(S

η ^b _R

for suitable b _R ∈ Z by (3). But {η _S

: χ ∈ X, χ 6= 1} is a basis of C by Proposition 1, hence

{θ _χ : χ ∈ X, #S _χ ≥ 2, k _S

6= F _χ , S _χ ⊆ J ⁺ if #S _χ = 2}

∪ {θ ^1−σ _χ

: χ ∈ X, χ 6= 1, (k _S

= F _χ or #S _χ = 1

or (#S _χ = 2 and S _χ ⊆ J ⁻ ))}

is a basis of C ⁽²⁾ ∩ C, because if χ ∈ X satisfies k _S

= F _χ , then S _χ

6⊆ S _χ for any χ ⁰ ∈ X such that 1 6= χ ⁰ 6= χ. Of course, if #S _χ = 1 then k _S

= F _χ . If #S χ = 2 and S χ ⊆ J ⁻ , then again k S

= F χ . It is clear that k S

= F χ

if and only if [k _S

: Q] = 2. Hence this basis is of the stated form and the proposition follows.

Let us study Leopoldt’s group H now. We have seen that θ χ ∈ E for any χ ∈ X such that #S _χ > 2 or such that #S _χ = 2 and S _χ ⊆ J ⁺ . Moreover, if S χ = {p} then θ χ has non-zero |p|-adic valuation. Therefore H is generated by −1 and

{θ χ : χ ∈ X, #S χ ≥ 2, S χ ⊆ J ⁺ if #S χ = 2}

∪ {θ _χ ^1−σ

: χ ∈ X, #S χ = 1} ∪ n Y

χ∈X

θ ^a _χ

∈ k : a χ ∈ Z o

,

where X ₁ = {χ ∈ X : #S _χ = 2, S _χ ⊆ J ⁻ }, because θ ^1+σ _χ

is a root of unity for χ ∈ X ₁ . Thus we need to find when Q

χ∈X

θ χ ^a

∈ k for a _χ ∈ Z.

First, suppose that χ ∈ X ₁ and that S _χ = {p, q} with p and q odd. Then

θ _χ = Y

1≤a≤pq (

)=(

)=1

(ξ ^a − ξ ^−a ) ∈ K _{p,q} ,

where ξ = ζ _{p,q} ^(1+pq)/2 . The complex conjugation on K _{p,q} is σ _p σ _q , so θ ^σ _χ

^σ

= Y

1≤a≤pq (

)=(

)=1

(ξ ^−a − ξ ^a ) = −θ χ ,

because |p| ≡ |q| ≡ 3 (mod 4). Hence if σ = Q

p∈S σ p ∈ Gal(K J /k) for some S ⊆ J, then

θ ^σ _χ =

 θ _χ if S _χ ∩ S = ∅,

−θ χ if S χ ⊆ S

(it is clear that #(S _χ ∩ S) = 1 is not possible because Q( √

pq) = F _χ ⊆ k).

Now, suppose that χ ∈ X ₁ and that S _χ = {−2, q}. Then θ _χ 6∈ K _J but θ χ ∈ K J ( √

2). It is clear that K J ( √

2) = K J ( √

−1) in this case. So we need to extend our automorphisms σ _p to K _J ( √

−1): for any p ∈ J let σ _p ⁰ be the non-trivial automorphism in Gal(K _J ( √

−1)/K _J\{p} ( √

−1)), and let σ ₋₁ be the non-trivial automorphism in Gal(K _J ( √

−1)/K _J ). Then ζ _{−2} ^σ

= ζ _{−2} , ζ _{−2} ^σ

= ζ _{−2} ⁵ and ζ _{−2} ^σ

= ζ _{−2} ³ ,

so θ χ ^σ

= θ χ ^σ

= −θ χ , while θ ^σ χ

= θ χ , due to the computations preceding Proposition 3.

Suppose that σ = Q

p∈S σ p ∈ Gal(K J /k) for some S ⊆ J. Then we have two extensions of σ to K J ( √

−1), namely σ ⁰ and σ ⁰ σ −1 , where σ ⁰ = Q

p∈S σ _p ⁰ , and

(6)

 Y

χ∈X

θ ^a _χ

 _σ

= (−1) ^Σ

^a

Y

χ∈X

θ _χ ^a

,

 Y

χ∈X

θ ^a _χ

 _σ

= (−1) ^Σ

^a

Y

χ∈X

θ _χ ^a

, where X ₂ = {χ ∈ X ₁ : −2 ∈ S _χ }.

Consider the equivalence relation on J ⁻ defined by p ∼ q if and only if √

pq ∈ k.

Let us show that if p 6= q then p ∼ q if and only if there is χ ∈ X 1 such that S _χ = {p, q}. Indeed, if χ ∈ X ₁ and S _χ = {p, q}, then Q( √

pq) = F _χ ⊆ k, so p ∼ q. On the other hand, if √

pq ∈ k for p, q ∈ J ⁻ , p 6= q, then χ ∈ b G defined by

χ(σ t ) =

 −1 if t ∈ {p, q}, 1 if t ∈ J \ {p, q}, satisfies χ(σ) = 1 for any σ ∈ Gal(K _J /Q( √

pq)), hence χ ∈ X and S _χ = {p, q}. It is easy to see that if

σ = Y

p∈S

σ p ∈ Gal(K J /k),

then for any class T ∈ J ⁻ /∼ either T ⊆ S or T ∩ S = ∅. If X ₂ = {χ ∈ X ₁ :

−2 ∈ S χ } is not empty, fix χ 0 ∈ X 2 . Then (6) implies that θ χ θ χ

∈ k for any

χ ∈ X ₂ . For any class T ∈ (J ⁻ \ {−2})/∼ satisfying #T > 1, fix χ _T ∈ X ₁

such that S χ

⊆ T . Then (6) implies that θ χ θ χ

∈ k for any χ ∈ X 1 , where T ∈ (J ⁻ \ {−2})/∼ satisfies S _χ ⊆ T . Hence we need only find when

Y

T ∈(J

\{−2})/∼

#T >1

θ _χ ^a

∈ k,

where a T ∈ Z.

Let J 0 be the union of all T ∈ (J ⁻ \ {−2})/∼ such that #T > 1. If J ₀ = ∅ then X ₁ = X ₂ , #X ₂ ≤ 1 and H = C ⁽²⁾ . Suppose that J ₀ 6= ∅. Then

∼ can be considered as an equivalence relation on J 0 and θ χ ∈ K J

for any χ ∈ X ₁ \ X ₂ . So (6) implies that

Y

T ∈J

/∼

θ ^a _χ

∈ k if and only if X

T ∈J

/∼

T ⊆S

a _T ≡ 0 (mod 2)

for all S ⊆ J ₀ such that Y

p∈S

σ _p ∈ Gal(K _J

/k _J

).

Choose S ₁ , . . . , S _l ⊆ J ₀ such that the restrictions of τ ₁ = Y

p∈S

σ _p , . . . , τ _l = Y

p∈S

σ _p

form a basis of the (multiplicative) vector space Gal(K _J

/k _J

) over F ₂ . We shall prove that the equations

(7) X

T ∈J

/∼

T ⊆S

x _T = 0, i = 1, . . . , l,

over F 2 are linearly independent. Indeed, suppose that there is L ⊆ {1, . . . , l}

such that

#{i ∈ L : T ⊆ S _i } ≡ 0 (mod 2)

for all T ∈ J ₀ /∼. Now, for any p ∈ J ₀ there is T ∈ J ₀ /∼ such that p ∈ T . But for any i ∈ {1, . . . , l}, we have p ∈ S _i if and only if T ⊆ S _i . Therefore

#{i ∈ L : p ∈ S i } is even for all p ∈ J 0 . Thus Y

i∈L

τ _i = Y

p∈J

σ ^{#{i∈L:p∈S} _p

^} = 1.

But this means that L = ∅ because τ ₁ , . . . , τ _l is a basis. The equations in (7) are then linearly independent. So there are l classes C 1 , . . . , C l ∈ J 0 /∼

such that (7) is equivalent to

(8) x _C

= X

T ∈R

b _T,i x _T , i = 1, . . . , l,

for suitable elements b T,i ∈ F 2 , where R = (J 0 /∼) \ {C 1 , . . . , C l }. Thus Q

T ∈J

/∼ θ ^a _χ

with a _T ∈ Z is in k if and only if x _T = a _T + 2Z is a solution

of (8), where we have identified F 2 = Z/2Z. Therefore {θ _χ : χ ∈ X, #S _χ ≥ 2, S _χ ⊆ J ⁺ if #S _χ = 2}

∪ {θ _χ ^1−σ

: χ ∈ X, #S χ = 1} ∪ {θ χ θ χ

: χ ∈ X 2 }

∪ {θ _χ θ _χ

: χ ∈ X ₁ \ X ₂ , T ∈ (J ⁻ \ {−2})/∼ with S _χ ⊆ T, χ 6= χ _T }

∪ n

θ χ

Y l i=1

θ ^b _χ

: T ∈ R o

∪ {θ _χ ²

: i = 1, . . . , l}

is a basis of H, where each element b _T,i ∈ F ₂ , used in (8), is understood as the integer 0 or 1.

Proposition 4. Let J ⁻ = {p ∈ J : p < 0}, J ₀ = {p ∈ J ⁻ \ {−2} :

√ pq ∈ k for some q ∈ J ⁻ \ {−2} with q 6= p} and d = #{χ ∈ X : #S _χ = 2, S χ ⊆ J ⁻ }. Let d 0 = 1 if there is an odd p ∈ J ⁻ such that √

−2p ∈ k, and d ₀ = 0 otherwise. Then

[H : C ⁽²⁾ ] = 2 ^d−d

[K _J

: k _J

] . Moreover , H ∩ C = C ⁽²⁾ ∩ C.

P r o o f. The former equality can be obtained directly by comparing the basis of C ⁽²⁾ (see Proposition 3) with the basis of H described above. To prove the latter equality, let us compare the basis of H with the basis of C (see Proposition 1). If #S _χ = 1 then θ χ ^1−σ

= ±η _S

, which is an element (up to sign) of both bases. So we need to find when

ε = Y

χ∈X

#S

>1

θ _χ ^c

∈ k

with c χ ∈ Z is an element of C. We shall prove that ε ∈ C if and only if c _χ [k _S

: F _χ ] is even for all χ ∈ X with #S _χ > 1.

Fix some linear ordering ≺ on X such that S _χ ⊆ S _ψ ⇒ χ ≺ ψ

for any χ, ψ ∈ X. As we mentioned in the proof of Proposition 1, for any S ⊆ J such that S 6= S _χ for all χ ∈ X, there are a _T ∈ Z satisfying

η S = ± Y

T (S

η ^a _T

.

Therefore (3) implies that for any χ ∈ X such that #S χ > 1, θ ² _χ = ±η ^[k _S

^:F

^] · Y

ψ∈X\{1}

ψ≺χ

η ^2b _S

ψ

for suitable integers b χ,ψ . Thus, with respect to the basis of C, ε ² has the following form:

ε ² = Y

χ∈X

#S

>1

θ _χ ^2c

= ± Y

χ∈X

#S

>1



η _S ^c

^[k

^:F

^]

χ

· Y

ψ∈X\{1}

ψ≺χ

η _S ^2c

^b

ψ

 .

It is easy to see that ε ∈ C if and only if the exponent of η S

in this expression is even for each ψ ∈ X \ {1}. This exponent is

X

χ∈X ψ≺χ

2c _χ b _χ,ψ or c _ψ [k _S

: F _ψ ] + X

χ∈X ψ≺χ

2c _χ b _χ,ψ

depending on whether #S _ψ = 1 or #S _ψ > 1. Hence ε ∈ C if and only if c χ [k S

: F χ ] is even for all χ ∈ X with #S χ > 1.

Now we can use the basis of H described before the proposition to obtain the following basis of H ∩ C:

{θ χ : χ ∈ X, #S χ ≥ 2, k S

6= F χ } ∪ {θ _χ ^1−σ

: χ ∈ X, #S χ = 1}

∪ {θ _χ ² : χ ∈ X, #S χ ≥ 2, k S

= F χ }, because k _S

= F _χ for any χ ∈ X ₁ . But that is (maybe, up to some signs) the basis of C ⁽²⁾ ∩ C given in Proposition 3.

4. Sinnott’s group of square roots and Washington’s group. Let C ₁ ⁰ be the group defined in [S, p. 209], namely

C ₁ ⁰ = {ε ∈ E : ε ² ∈ C ⁰ }.

Similarly, define

C 1 = {ε ∈ E : ε ² ∈ C}.

Finally, let C ⁰⁰ be the group of cyclotomic units defined in [W, p. 143], namely the intersection of E and the group of cyclotomic units in the small- est cyclotomic field containing k.

Proposition 5. C ₁ = C ₁ ⁰ .

P r o o f. Because C ⁰ ⊆ C, we have C ₁ ⁰ ⊆ C ₁ directly from the definitions.

Suppose that ε ∈ C 1 . Then ε ∈ E and ε ² ∈ C. By comparing the bases of C ⁰ and C in Proposition 1, we see that there are ε ⁰ ∈ C ⁰ and S ⊆ {p ∈ J :

√ p ∈ k} such that

ε ² = ε ⁰ Y

p∈S

η _{p} .

But C ⁰ is generated by −1 and norms from imaginary abelian fields to real

ones, so ε ⁰ is totally positive or totally negative. If q ∈ S then

Y

p∈S

η _{p}

 _1−σ

= η _{q} ^1−σ

= −η _{q} ² < 0

by Lemma 1 of [K]. Of course, ε ² is totally positive. Therefore S = ∅ and ε ² = ε ⁰ ∈ C ⁰ . So ε ∈ C ₁ ⁰ and the proposition follows.

Lemma. Let S ⊆ J. If #S = 1 then η S is a cyclotomic unit in the n S -th cyclotomic field. If #S > 1 then η _S or −η _S is the square of a cyclotomic unit in the n _S -th cyclotomic field and √

η _S is in the maximal real subfield of K _S ( √

−1).

P r o o f. We shall distinguish two cases depending on the parity of n S . First, suppose that n S is odd. Let ξ = ζ _S ⁽¹⁺ⁿ

^)/2 ; then

α = N _Q

/K

(1 − ζ S ) = N _Q

/K

(−ξ)N _Q

/K

(ξ − ξ ⁻¹ ) = N _Q

/K

(ξ − ξ ⁻¹ ), where we have used the fact that N _Q

/K

(−ξ) is a totally positive root of unity. First, let S = {p}. Then

η S =

 ±1 if √ p 6∈ k,

√ 1

p α if √ p ∈ k.

Of course, p > 0 in the latter case, so p ≡ 1 (mod 4), α ^1+σ

= p (by Lemma 1 of [K]) and

η ² _S = α ^1−σ

=

p−1 Y

a=1

(ξ ^a − ξ ^−a ) ⁽

⁾ =

 ^(p−1)/2 Y

a=1

(ξ ^a − ξ ^−a ) ⁽

⁾

 ₂ , which is the square of a cyclotomic unit in the pth cyclotomic field.

Now, suppose that #S > 1 and that K _S is imaginary. Then α = N _K

S

/K

(N _Q

/K

(ξ − ξ ⁻¹ ))

= (−1) ^[Q

^{: K}

^] N _Q

/K

(ξ − ξ ⁻¹ ) ² .

Let τ ₀ , . . . , τ _l be a basis of the (multiplicative) vector space Gal(K _S /k _S ) over F 2 , where τ 0 is the complex conjugation. Let L be the subfield of K S

whose Galois group is generated by τ ₁ , . . . , τ _l . Then η S = N _K

S

/k

(α) = N _K

_/L (α) = (−1) ^[Q

^{: L]} N _Q

/L (ξ − ξ ⁻¹ ) ² . Therefore √

η _S ∈ L( √

−1) ⊆ K _S ( √

−1). Moreover, η _S is totally positive, so

√ η S is real. The lemma follows in this case because N _Q

/L (ξ − ξ ⁻¹ ) is a

cyclotomic unit in the n _S th cyclotomic field.

Now, suppose that #S > 1 and that K S is real. Then all p ∈ S are positive and

α = Y

a∈A

(ξ ^a − ξ ^−a ), where

A = 

a ∈ Z : 1 ≤ a ≤ n S , ^a _p

= 1 for all p ∈ S .

Choose q ∈ S and write q − 1 = 2 ^b · c with c odd. Let ψ be a Dirichlet character modulo q of order 2 ^b , so ψ(−1) = −1, and let

B = {a ∈ A : ψ(a) = 1 or Im ψ(a) > 0}.

Then A = B ∪ {n _S − a : a ∈ B} is a disjoint union, so α = (−1) ^#B Y

a∈B

(ξ ^a − ξ ^−a ) ² . Of course,

#B = 1

2 (#A) = 1 2

Y

p∈S

p − 1 2 is even. Let

β = Y

a∈B

(ξ ^a − ξ ^−a ).

We shall show that β ∈ K S , which means β = Y

a∈B

(ξ ^ay − ξ ^−ay ) for any y ∈ A.

Fix y ∈ A and define the mapping g : B → B by the following congruence modulo n _S : for any a ∈ B,

g(a) ≡

 ay if ψ(ay) = 1 or Im ψ(ay) > 0,

−ay if ψ(ay) = −1 or Im ψ(ay) < 0.

It is easy to see that g is a permutation and that Y

a∈B

(ξ ^ay − ξ ^−ay ) = (−1) ^#B

Y

a∈B

(ξ ^g(a) − ξ ^−g(a) ) = (−1) ^#B

β, where B ⁰ = {a ∈ B : g(a) ≡ −ay (mod n _S )}. We have

#{a ∈ A : ψ(a) = ψ(a 0 )} = 2 ^1−b (#A)

for any fixed a ₀ ∈ A. But B ⁰ is a disjoint union of such sets involving some a ₀ , so #B ⁰ is divisible by

2 ^1−b (#A) = c Y

p∈S\{q}

p − 1

2 ,

which is even. Thus β ∈ K S and η S = N _K

_/k

(α) = N _K

_/k

(β) ² . The lemma is proved if n _S is odd because β is a cyclotomic unit in the n _S th cyclotomic field.

Now, let us deal with the case of n S being even. If S = {−2} then η S = 1.

If S = {2} then η _S = −1 + √

2 or η _S = −1 depending on whether √ 2 ∈ k or not. It is easy to check that

−1 + √

2 = −ζ _{2} (1 − ζ _{2} )(1 − ζ _{2} ³ ) ⁻¹ is a cyclotomic unit in the eighth cyclotomic field.

Now, suppose that #S > 1. Then ε _S = Y

a

(1 − ζ _S ^a ),

where the product is taken over all positive integers a < n S satisfying _|p| ^a

= 1 for all odd p ∈ S such that a ≡ ±1 (mod 8) if 2 ∈ S or a ≡ 1, 3 (mod 8) if −2 ∈ S. Let ξ = e ^πi/n

, so ξ ² = ζ S .

First, suppose that K _S is imaginary. Let τ be the complex conjugation on Q(ξ). Because n _S ≡ 8 (mod 16), we have

ε ^1+τ _S = Y

a

(1 − ζ _S ^a )(1 − ζ _S ^−a ) = Y

a

(−(ξ ^a − ξ ^−a ) ² ),

where the products are taken over all positive integers a < 2n S satisfying

a

|p|

= 1 for all odd p ∈ S such that a ≡ ±1 (mod 16) if 2 ∈ S or a ≡ 1, 3 (mod 16) if −2 ∈ S. The number of terms in these products is even, so ε ^1+τ _S = β ² , where

(9) β = Y

a

(ξ ^a − ξ ^−a ),

with a running through the same set as above. We need to prove that β ∈ (K _S ( √

−1)) ⁺ . For any σ ∈ Gal(Q(ξ)/K _S ( √

−1)) there is an integer y satisfying y ≡ 1 (mod 8) and ^y _p

= 1 for all odd p ∈ S such that ξ ^σ = ξ ^y . It is clear that if y ≡ 1 (mod 16) then σ only permutes the terms in the product (9), so β ^σ = β in this case. If y ≡ 9 (mod 16) then y ⁰ = y + n _S ≡ 1 (mod 16) and ξ ^y

= −ξ ^y . Moreover, ^y _p

= 1 for all odd p ∈ S, so β ^σ = Y

a

(ξ ^ay − ξ ^−ay ) = Y

a

(−(ξ ^ay

− ξ ^−ay

)) = β in this case, too. It is easy to see that β is real, so β ∈ (K _S ( √

−1)) ⁺ . It is clear that β is a cyclotomic unit in the n S th cyclotomic field and the lemma is proved in this case, since η _S = N _K

S

/k

(β ² ).

Finally, suppose that K S is real. Then all p ∈ S are positive and ε S = Y

a

(1 − ζ _S ^a )(1 − ζ _S ^−a ) = Y

a

(−(ξ ^a − ξ ^−a ) ² )

with the products taken over all positive integers a < n S such that a ≡ 1 (mod 8) and ^a _p

= 1 for all odd p ∈ S. The number of terms in these products is even, so ε _S = β ² , where

(10) β = Y

a

(ξ ^a − ξ ^−a ),

where a in the product runs through all integers satisfying 0 < a < 2n S

and a ≡ 1 (mod 16) such that ^a _p

= 1 for all odd p ∈ S. Let us show that β ∈ K _S . For any σ ∈ Gal(Q(ξ)/K _S ) there is an integer y satisfying ξ ^σ = ξ ^y such that y ≡ ±1 (mod 8) and ^y _p

= 1 for all odd p ∈ S. It is clear that if y ≡ 1 (mod 16) then σ only permutes the terms in the product (10). If y ≡ 9 (mod 16) then σ also changes the sign of each term in (10).

But the number of terms in the product (10) is even, so β ^σ = β in both previous cases. If y ≡ −1 (mod 8) then we have proved that β ^{τ σ} = β, where τ is the complex conjugation on Q(ξ). But β is real, so β ∈ K _S . It is clear that β is a cyclotomic unit in the n S th cyclotomic field and the lemma is proved.

Proposition 6. Let 2 ^g = [C ⁰ : {1, −1} × (C ₁ ⁰ ) ² ]. Then H ⊆ C ₁ ⁰ ⊆ C ⁰⁰ and

[C ₁ ⁰ : C ⁰ ] = 2 ^{[k:Q]−1−g} . Moreover , 2 ^g is a divisor of [K _J : k].

P r o o f. The fact that [C ⁰ : {1, −1} × (C ₁ ⁰ ) ² ] is a power of 2 follows from the inclusion C ⁰ ⊆ C ₁ ⁰ . By (1) and (2) we have θ ² _χ = ±θ χ ^1−σ

for any χ ∈ X such that #S _χ > 1. The form of the basis of H before Proposition 4 gives H ⊆ C 1 , because θ χ ^1−σ

∈ C by Proposition 2. But C 1 = C ₁ ⁰ by Proposition 5.

Let ε ∈ C ₁ ⁰ . Then ε ∈ E and ε ² ∈ C ⁰ . The Lemma gives that any element of the basis of C ⁰ given in Proposition 1 is (up to sign) the square of a cyclotomic unit in the n _J th cyclotomic field. Thus ε is a cyclotomic unit in this field and ε ∈ C ⁰⁰ .

The formula follows from

2 ^g · [C ₁ ⁰ : C ⁰ ] = [C ₁ ⁰ : {1, −1} × (C ₁ ⁰ ) ² ] = 2 ^[k:Q]−1 .

It remains to show that 2 ^g is a divisor of [K _J : k]. Let C ₀ ⁰ be the group of totally positive elements of C ⁰ . Proposition 1 gives that

{η _S

: χ ∈ X, #S _χ > 1} ∪ {η _S ²

: χ ∈ X, #S _χ = 1}

generates C ₀ ⁰ . The Lemma implies that √

ε ∈ (K J ( √

−1)) ⁺ for any ε ∈ C ₀ ⁰ . Of course,

(C ₁ ⁰ ) ² = {ε ∈ C ₀ ⁰ : √

ε ∈ k}.