Math. Ann. (2006) 336: 639–657 DOI 10.1007/s00208-006-0015-9

Mathematische Annalen

Elliptic units and sign functions Hassan Oukhaba

Received: 22 February 2005 / Revised: 23 August 2005 / Published online: 28 July 2006 © Springer-Verlag 2006

Abstract In the first part of this paper we give a new definition of the elliptic analogue of Sinnott’s group of circular units. In this we essentially use the ideas discussed in Oukhaba (in Ann Inst Fourier, 55(33):753–772, 2005). In the second part of the paper we are interested in computing the index of this group of elliptic units. This question is closely related to the behaviour of the universal signed ordinary distributions introduced in loc. cit. Such distributions have a natural resolution discovered by Anderson. Consequently, we can apply Ouyang’s general index formula and the powerful Anderson’s theory of double complex to make the computations Mathematics Subject Classification (2000)

11G16

1 Introduction Let k ⊂ C be a imaginary quadratic field of discriminant Dk < 0. Let m be an integral ideal of k prime to 6. Let km ⊂ C be the ray class field of k modulo m. In [4] we defined a certain group of elliptic units Em ⊂ km and computed its index in Ok×m , the group of units of km. We obtained an index-formula similar to that obtained by Sinnott in [9] for his group of circular units of a cyclotomic field. It is also similar to the formulas proved by Yin in positive characteristic, cf. [12, 13]. To define his units, Yin used the torsion points of normalized Drinfel’d modules of rank one.This gave us precious informations for the construction of

H. Oukhaba (B) Laboratoire de mathématique, 16 Route de Gray, 25030 Besançon Cedex, France e-mail: [email protected]

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Em. Moreover, many decisive steps in our index computations are borrowed from the method of Sinnott–Yin. Let us denote by deg(m) the number of prime ideals p of k such that p divides m. Unfortunately, when deg(m) ≥ 3, we succeeded in computing [Ok×m : Em] only if the ideal class number of k is odd. Indeed, in Sect. 7 of [4] we had to compute some determinant which we interpreted in cohomological terms. To compute such cohomology we employed an inductive method due to Sinnott and Yin, which works only if k has an odd ideal class number. In this paper we propose a new approach of the construction of Em based on the material introduced in [5]. The fundamental idea is the concept of sign function. Let us recall what this means. Let H = k(1) ⊂ C be the Hilbert class field of k. Let μH be the group of roots of unity of H. Let 6 be the multiplicative group of elements of k that are prime to 6. Then a sign function of k is a surjective homomorphism s : 6 −→ μH satisfying s(ζ ) = ζ for all ζ ∈ μH ∩ k and such that s(u) = s(v) for all prime to 6 algebraic integers u, v ∈ k with u ≡ v modulo 12. Let Ok be the ring of integers of k. Then, an ideal a of Ok is said to be primitive if it is not of the form ta , with 2 ≤ t ∈ N and a ⊂ Ok . In the sequel we will exclusively use the following sign function sgn defined for all λ ∈ Ok ∩ 6 , such that the ideal λOk is primitive, by the formula sgn(λ) = λ

η2 (Ok ) , η2 (λOk )

(1)

where a −→ η(a) is the eta function on primitive ideals of Ok prime to 6, cf. [2] definition 8. Let us recall the relation (2π )12 η(a)24 = (−1)d−1 N(a)12 (¯a), where a¯ is the image of a by the complex conjugation, N(a) is the cardinal of the ring √ Ok /a and d is the unique square-free positive integer satisfying k = Q( −d). If L is a lattice of C then (L) is the discriminant of the elliptic curve determined by L. Lemma 14 of [2] implies in particular that sgn is indeed a sign function in the sens given above. We use sgn to define the signed universal ordinary distributions. These objects are what we need to adapt the theory of double complex, introduced by Anderson in [1] and improved by Ouyang in [6] and [7], in order to compute the determinant (R− : U − ) (see formula (24) below) and consequently to handle the index [Ok×m : Em] in full generality. The function sgn is also used to define the algebraic integers m(c), associated to any integral ideal c of k prime to 6N(m). These numbers are roots of the well known Ramachandra invariants. The norm relations which we recall below in Theorem 2.1 give a concrete example of a signed distribution. They may be considered as a refinement of the norm formulas proved by Robert in [8] Théorème 2.

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2 Elliptic units In this section we introduce the elliptic units which will serve to define the group Em. To manipulate these units we will use a family of finite abelian extensions of k called the narrow ray class fields with respect to sgn, defined as follows. Let f be an integral ideal of k prime to 6 and let k12f ⊂ C be the ray class field of k modulo 12f, then consider the subgroup Rf = {(xOk , k12f/k) | x ∈ 6 , x ≡ 1 mod× f and sgn(x) = 1} of Gal(k12f/k), where (xOk , k12f/k) is the k-automorphism of k12f associated to the fractional ideal xOk by the Artin map. Then we denote by Kf the subfield of k12f fixed by Rf. We call Kf the narrow ray class field of k modulo f, with respect to sgn. It is easy to check that kf ⊂ Kf and Gal(Kf/kf) is isomorphic to μH (resp. μH /μk ) if f = (1) (resp. f = (1)). Moreover, Gal(Kf/K(1) ) (Ok /f)× , the multiplicative group of Ok /f. To define Em we shall use the eta-quotients introduced by Hajir and Villegas in [2]. Let us recall some properties of these eta-quotients. Let a be a primitive ideal of Ok prime to 6. Let λ ∈ 6 ∩ Ok and denote by σλ the automorphism of kab /k associated to λOk by the Artin map, where kab is the maximal abelian extension of k in C. Then by Proposition 10 (i) of [2] we have η2 (Ok ) ∈ kab η2 (a)

and

 η2 (O ) σλ −1 k = sgn(λ)N(a)−1 . 2 η (a)

(2)

Thus η2 (Ok )/η2 (a) ∈ K(1) and K(1) is the extension of H generated by all these quotients. Let a, b be primitive ideals of Ok , prime to 6Dk and such that N(a) and N(b) are coprime. Then ab is primitive. Moreover, the quotient η(a, b) =

η(a)η(b) η(Ok )η(ab)

(3)

satisfies η(a, b) ∈ K(1)

and

η(a, b)σλ −1 = sgn(λ) 2 (N(a)−1)(N(b)−1) , 1

(4)

thanks to Theorem 19(i) of [2]. In fact η(a, b) is a unit in K(1) . To go further we need some more notation. If a is an ideal of Ok then a is uniquely written in the form a = ta , with 1 ≤ t ∈ N and a is a primitive ideal of Ok . We will denote a by pr(a). The order of Z/a ∩ Z will be denoted by ea. If L is any finite abelian extension of k and b is a fractional ideal of k prime to the conductor of L/k then we denote by (b, L/k) the k-automorphism of L associated to b by the Artin map. In [5] we introduced the following new invariants.

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Definition 2.1 For all ideals m = (1) and c of Ok prime to 6, such that c is prime to 6N(m) we define     ¯ −2π f N(c), m¯c η2 pr(mc) m(c) = sgn(c1 ) emc

  em and  m(c) = m(c)

where c1 = N(c)/ec and z −→ f(z, m¯c) is the Klein form associated to m¯c considered as a lattice of C, cf. [3] formula (2.8). Let Cl(m) be the ideal class group of k moduo m. Let C be the class of c. Then we have m(c)12em = (−1)d−1 ϕ(C)m, where ϕ(C)m is the Ramachandra–Robert invariant associated to C. In [5] §3 m(c) ∈ Km. Moreover we we proved that m(c) ∈ K(m) , where m = N(m), and  have (5) m(1)(c, K(m) /k) = m(c). Let us consider a principal ideal c = λOk , where λ ∈ Ok is prime  to 6N(m).   Then pr(c) is generated by μ = λ/c1 . In addition, since f N(c), m¯c = λ¯ f λ, m , formula (5) implies m(1)

(c, K(m) /k)−1

  f λ, m 1 η2 (μpr(m)) ¯  sgn(λ/μ) =  2 ¯ f 1, m μ η (pr(m))   −m f λ, m = sgn(λ) . f 1, m

Let us fix α ∈ Ok such that Im(α) > 0, Ok = Zα⊕Z and m = Zm1 α⊕Zem, where m1 = N(m)/em. Suppose we have λ ≡ 1 modulo m, that is λ = 1 + xm1 α + yem with x, y ∈ Z. Then we have m(1)(c, K(m) /k)−1 = sgn(λ)−m (−1)xy+x+y e−iπ x/em ,

(6)

thanks to the transformation law of the Klein form, cf. formula (K3) in [3]. Theorem 2.1 Suppose we have m = nq, where q is a prime ideal dividing m. Then we have ⎧   em ⎪ sgn(eq)  n(1) en ⎪ ⎨    (1−(q,Kn /k)−1 ) em en  m(1) = NKm /Kn  n(1) ⎪   2 ⎪ ⎩sgn(e ) e η (Ok ) eq q q η2 (pr(q¯ ))

if q|n, if q  n and n = (1)

(7)

if n = (1)

Proof we refer the reader to the proof of Theorem 3.2 of [5].

 

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3 The group Em We are ready now to define the group Em. We will denote the ring of integers of a number field by OF , its group of units by OF× and the torsion subgroup of OF× by μF . Also we let wF be the order of μF . If X, Y are subgroups of the multiplicative group F × then by XY we shall always mean the subgroup of F × generated by X and Y. Thus XY = {xy, with x ∈ X and y ∈ Y}. Moreover, if n is a positive integer then we put X n = {xn , with x ∈ X}. E ∩ H, where  E is the group of units of K(1) If m = (1) then we let Em =  generated by μH = μK(1) and by all η(a, b), a and b being primitive ideals of Ok , prime to 6Dk and such that N(a) and N(b) are coprime. Let us remark that  E is a galois module because we have η(a, b)(c, K(1) /k) =

η(ac, b) , η(b, c)

for any primitive ideal c of Ok prime to 6Dk N(ab), thanks to Proposition 10(i) of [2]. If m = (1) then we define P ⊂ K(m) , where m = N(m), to be the galois module generated by μK(m) = μk(m) and by n(1), for all the integral ideals n = (1) dividing m (it is also useful to put P = 1 in case m = (1)). Then we let × C = P ∩ OK (m)

and  C= EC.

The group  C may be considered as the analogue of the extended group of cyclotomic units constructed by Yin in [13] definition 2.5. The group Em is the intersection C ∩ km. (8) Em =  Lemma 3.1 Let m be the group of elements x of Km such that xn ∈ Ok×m for some positive integer n. Then we have [ m : Ok×m ] = wH /wk if deg(m) ≤ 1, and C em [ m : Ok×m ] = wH otherwise. Moreover we have m = Ok×m  Proof Since Km/km is cyclic we see that for x ∈ Km, wH

× if m = (1)) x ∈ m ⇐⇒ xwH ∈ Ok×m (resp. x wk ∈ OH

Let j be a generator of J = Gal(Km/km) and consider the group homomorphism ϕ : m −→ μH defined by ϕ(x) = xj /x. We have ker(ϕ) = Ok×m . Let us choose λ ∈ Ok prime to N(m) and such that (λOk , Km/k) = j. If deg(m) ≥ 2 then  m(1) ∈ m. Moreover, we have ϕ( m(1)) = sgn(λ)−m1 ,

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thanks to (6). But sgn(λ)−m1 is a generator of μH . Thus Im(ϕ) = μH if deg(m) ≥ 2. If deg(m) = 1 then for all σ ∈ Gal(Km/k) we have  m(1)σ −1 ∈ m and

  ϕ  m(1)σ −1 = (sgn(λ)−m1 )σ −1 .

w

This shows that μHk ⊂ Im(ϕ). To prove the equality let us write m = pe , where p is a prime ideal and e is a positive integer. Then the ramification index of p in Km/k is N(pe ) − N(pe−1 ). This forces the order of ϕ(x), for x ∈ m, to be at most equal to wH /wk . Indeed, the order of ϕ(x) is equal to the degree of the extension km(x)/km in which the ideal p does not ramify. Thus in the w case deg(m) = 1 we have Im(ϕ) = μHk . Now suppose m = (1). Let p1 , . . . , pt be prime ideals of Ok all prime to 6, and n1 , . . . , nt , be rational integers such that wk = ni (N(pi ) − 1). Let us assume that p1 , . . . , pu are primitive for some positive integer u ≤ t and (if necessary) pu+1 , . . . , pt have residual degree 2. The formula (4) then implies ⎛ ϕ⎝

u 

⎞   η pi , pj ⎠ = sgn(λ)wk .

i,j=1



This proves the lemma in all cases. Proposition 3.1 We have em

[μkm C

 wkm /wk : μkm Em ] = wkm em

if deg(m) ≤ 1 otherwise.

em × C em =  E and μkm Em = E ∩ H. Since (1) =  EOH Proof If m = (1) then μkm  × we have [ E :  E ∩ H] = [ (1) : OH ] = wH /wk , thanks to lemma 3.1. Let us suppose m = (1) and consider the two groups em Cem /μkm Em A = μkm 

and

B = μkm P Eem /μkm [P E ∩ km]em

The inclusion  C ⊂ P E induces an injective homomorphism A → B. The group B is cyclic generated by the class of  m(1). Indeed, formulas (4) and (6) may be used to prove that the following three elements of P E N(m)

(m(1)) N(n) , n(1)

γm(c) = m(1)N(c)−(c, K(m) /k)

and

η(a, b)m(1)e(a,b) ,

where e(a, b) = 12 (N(a) − 1)(N(b) − 1)N(m), are in fact in km. Now, let us compute the order d of the class of  m(1) in B. The previous remark clearly shows that d divides wkm . Indeed, recall that wkm = wKm may be written as a finite

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sum na(N(a) − 1), na ∈ Z, for some ideals a of Ok prime to 6N(m) and such that (a, Km/k) = 1, cf. [11] Chapter IV-1. Therefore we have  m(1)wkm =



γm(a)na

em

.

a

By its very definition d is the least positive integer such that, (m(1))d = ζ x, where x ∈ km and ζ is a root of unity in K(m) . This is equivalent to saying that ¯

(m(1))d((λOk , K(m) /k)−1) = ζ λλ−1 , for all λ ∈ Ok prime to 6N(m) and λ ≡ 1 modulo m. Let us take λ = 1+12em and apply formula (6). We obtain 1 = ζ 24em . Furthermore, if we take λ = 1+12m1 α, where α is chosen as above, that is, α ∈ Ok with Im(α) > 0, Ok = Zα ⊕ Z and m = Zm1 α ⊕ Zem, then we find ¯ e−12π id/em = ζ 12m1 (α+α) .

Since α ∈ pr(m) its trace α + α¯ is in m∗2 Z, where m∗2 is the product of the ramified prime numbers p in k/Q and such that p|pr(m). Thus m1 m∗2 divides d. Let us apply the same technique to a λ ∈ Ok prime to 6 and λ ≡ 1 modulo mOk then we find sgn(λ)−md = 1. This forces d to be a multiple of wH since sgn(λ)−m takes all the roots of unity in H. Thus wH m1 m∗2 divides d. But wkm = wH m1 m∗2 , as one may prove using the famous lemma 5 of [8]. So we have proved that d = wkm . If deg(m) ≥ 2 then m(1) is a unit and then we have A B. If deg(m) = 1 then m(1) is not a unit but the elements (m(1))(c, K(m) /k) m(1) are units. In particular the cokernel of the injective map A → B has order dividing wk . Moreover the group C is generated by these quotients and by μK(m) . This implies that A is annihilated by wkm /wk . The proof of the proposition is now complete.  Recall that we want the index [Ok×m : Em]. But the identity [Ok×m : Em] =

em Cem ][μkm  Cem : μkm Em ] [ m : μkm  , em [ m : Ok×m ][Em : μkm Em ]

(9)

together with Lemma 3.1 and Proposition 3.1 clearly show that this is equivalent to compute the index [ m : μkm  Cem ].

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 em ] 4 The index [m : μkm C Cem ] we will follow the method of Sinnott– To compute the index [ m : μkm  Yin developed in [9, 10, 12, 13]. Let us put G = Gal(Km/k), R = Z[G] and In = Gal(Km/Kn) for n|m. If X is a subset of G then we write s(X) for the sum σ in R of all σ ∈ X. Also we will denote by |X| = #X the cardinal of X. If p is a prime ideal dividing m we let Tp ⊂ G (resp. Fp ∈ G) be the inertia group (resp. any Frobenius automorphism) for p. We set p∗ = Fp−1 s(Tp)/|Tp|. Let V be the R-submodule of Q[G] generated by αn = s(In)



(1 − p∗ )

p|n

for n|m and n = (1) (resp. V = 0 if m = (1)). Also we need the following element of R[G] ω = wH em



L (0, χ¯ )eχ ,

χ =1 χ (J)=1

where χ runs over the nontrivial characters of G which are trivial on the group J = Gal(Km/km), s −→ L(s, χ¯ ) is the primitive L function associated to the conjugate character χ¯ of χ and eχ ∈ C[G] is the idempotent associated to χ . × −→ The relation between P, V and ω is given map l : Km by theσ logarithm × 2 −1 R[G], defined for x ∈ Km by l(x) = − σ ∈G ln(|x | )σ . Recall that l is a R× = μKm = μkm . Let l∗ = (1 − e1 )l, homomorphism with the property ker l ∩ OK m where e1 is the idempotent associated to the trivial character of G. Proposition 4.1 We have Pem ⊂ Km and l∗ (Pem ) = ωV. Proof The proposition is trivial if m = (1). In the  caseem = (1) the order of μK(m) divides emwKm . Also we know from §2 that n(1) m ∈ Km. On the other  e  hand we have the equality l∗ n(1) m = ωαn, for all (1) = n|m. To prove this claim we have to check that   e  ρχ l∗ n(1) m = ρχ (ωαn) (10) for all the characters χ of G. Here ρχ is the ring homomorphism C[G] −→ C defined for σ ∈ G by ρχ (σ ) = χ (σ ). Let us recall that zeχ = ρχ (z)eχ for all z ∈ C[G]. Now the definition of ωαn, the relation    w NKm /km  n(1) = −  n(1) H and the fact that  n(1) ∈ Kn imply that the two sides of (10) vanish if χ is non trivial on J = Gal(Km/km) or on Gal(Km/Kn). The same conclusion is true if χ

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is the trivial character of G. Thus we may suppose that χ is a non trivial character of Gal(kn/k). But in this case (10) is easily derived from the Kronecker limit formula as stated in [4] formula (5).   Proposition 4.2 Let a be a primitive ideal of Ok prime to 6. Let τ be any extension of (a, H/k) to Km. Then we have eml∗



η2 (Ok ) η2 (a)



  = s I(1) ω (1 − τ ) .

Proof We have to show that   2      η (Ok ) ρχ eml∗ = ρχ s I(1) ω (1 − τ ) 2 η (a)

(11)

for all nontrivial character χ of G. This equality is already true if χ is not trivial on Gal(Km/H). On the other hand (11), for χ trivial on Gal(Km/H), is equivalent to the Kronecker limit formula, cf. [4] formula (5).   For an ideal f of Ok prime to 6 we define Qf to be the galois submodule of × K(1) generated by μH = μK(1) and by λq := eq

η2 (Ok ) η2 (pr(¯q))

for the prime ideals q dividing f, if any. Let us recall that λq generates in OK(1) × the ideal qOK(1) , cf. [2] Proposition 10(iv). By definition Ef = Qf ∩ OK . Let us (1)  choose an ideal m of Ok prime to 6Dk N(m) and such that Gal(H/k) is generated by the Frobenius automorphisms (p, H/k), p|m (we may take m = (1) if Gal(H/k) = 1). Then we have Em = μH  Ewk . × . Then we have Let us define P = μkm Pwk em Qemm and C = P ∩ OK (m)

C = μkm  Cwk em because any x ∈ P is coprime to any y ∈ Qm . Let us put r = [km : k] − 1 and e+ = s(J)/|J|. Then we have 

    −r    Cem = w−r m : μkm  k m : C = wk l( m) : l(C ) .

(12)

Now consider the R-vector space Y = (1 − e1 )e+ R[G]. Since Y is a G-module its dimension as a R-vector space is equal to the number of the characters χ of G such that ρχ (Y) = 0, that is the non trivial characters χ which are trivial on J.

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Thus we have dimR Y = r. Moreover, the R-modules l( m), l(C ) and l∗ (P ) are lattices of Y. Recall that by definition, a lattice L in a finitely generated F-vector space E (with F = Q or R), is a free Z-module of rank dimF E, such that the F-vector space generated by L is equal to E. If M and N are two lattices in E we define their Sinnott’s index (M : N) to be | det γ | of any R-linear endomorphism of E satisfying γ (M) = N. The properties of (M : N) are summarized in [10] Sect. 1 pp 187–188. If A is any R-submodule of R[G] then we put A0 = (1 − e1 )A. It is the kernel in A of the multiplication by s(G). Let U  = wk V + s(I(1) )R.

U = V + s(I(1) )R and

(13)

Since e+ R0 , e+ U0 and e+ U0 are also lattices of Y the index [l( m) : l(C )] is equal to the product [l( m) : l(C )] = (l( m) : e+ R0 )(e+ R0 : e+ U0 ) (e+ U0 : e+ U0 )(e+ U0 : l(C )).

(14)

Lemma 4.1 We have +

+

(l( m) : e R0 )(e

U0

  w r wk h(km)  H × m : Okm : l (P )) = em wkm h |J| ∗



r

where h(km) (resp. h) is the ideal class number of km (resp. k). Proof We have l∗ (P ) = ωe+ U0 , thanks to Proposition 4.1 and Proposition 4.2. Multiplication by ω is a R-linear map of Y sending e+ U0 onto l∗ (P ). Thus (e+ U0 : l∗ (P )) equals 

ρχ (ω) = (wH em)r

χ =1 χ (J)=1



L (0, χ¯ ) = (wH em)r

χ =1 χ (J)=1

wk h(km)Reg(km) . wkm h

The last equality being the analytic class number formula, cf. [4] formula (4). Here Reg(km) is the regulator of km. We have Reg(km)|J|r = (e+ R0 : l(Ok×m )). This completes the proof of the lemma.   − deg(m)

Lemma 4.2 We have [l∗ (P ) : l(C )] = h[Km : K(1) ]wk

.

mm

and w = wkm . Then by rewriting the proof of Proposition Proof Let n = 8 in [4] one obtains the formula − deg(m)

[(P )12wh ∩ k : (Qn)12whem ∩ k][l∗ (P ) : l(C )] = h[Km : K(1) ]wk

.

If p is a prime ideal dividing n then we denote by xp a generator of the ideal ph . By Lemma 2 of [4] we know that (P )12wh ∩ k and (Qn)12whem ∩ k are equal to m the subgroup of k× generated by x12we , p|n. This proves the lemma.  p

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Lemma 4.3 Let D be the subgroup of Gal(H/k) generated by the Frobenius automorphisms at the prime ideals dividing m and put e = [Gal(H/k) : D]. Then we have +

+

[e U0 : e

U0 ]

 1 = (wk )[km :k]−e

if m = (1) if m = (1)

Proof The case m = (1) is obvious. So, let us suppose m = (1). We have the isomorphism e+ U0 /e+ U0 e+ V/M, with M = e+ V ∩ e+ U0 . We claim that M = wk e+ V. To see this it is enough to show that e+ s(I(1) )R0 ∩ e+ V ⊂ wk e+ V. Let Jm be the ideal of R generated by (1 − σ ) for the automorphisms σ ∈ Gal(Km/H D ). If x ∈ V then there exists a positive integer n such that nx ∈ Jm. Thus e+ x ∈ e+ s(I(1) )R0 ∩ e+ V imply that the class of e+ x is torsion in e+ s(I(1) )R0 /e+ s(I(1) )Jm because n|I(1) |e+ x ∈ e+ s(I(1) )Jm. But this factor group is torsion free. Thus e+ x ∈ e+ s(I(1) )Jm ⊂ wk e+ V. To conclude the proof of the lemma we have just to check that e+ V is a free abelian group of rank [km : k]−e. This may be done by the use of character theory, cf. [12] Lemma 3.1.   Lemma 4.4 If A is a R-module then we denote by A− the kernel in A of the multiplication by (#J)e+ = s(J). Then we have (e+ R0 : e+ U0 ) =

⎧ ⎨1

if m = (1)

1 ⎩ [Km : K(1) ](R− : U − )

if m = (1).

Proof Let us suppose m = (1). The properties of Sinnott’s index, cf. [10] Sect. 1 pp 187–188, give (R0 : U0 ) = (e+ R0 : e+ U0 )((R0 )− : (U0 )− ) (R : U) = (R0 : U0 )(s(G)R : s(G)U). On the other hand we have (s(G)R : s(G)U) = [s(G)Z : [Km : K(1) ]s(G)Z] = [Km : K(1) ]. Moreover, (R0 )− = R− and (U0 )− = U − . Thus we have (e+ R0 : e+ U0 ) =

(R : U) [Km : K(1) ](R− : U − )

Since the Galois group of Km/K(1) is the direct product of the inertia groups Tp, for the prime ideals p|m, we have (R : U) = 1 exactly as in [9] Sect. 6. Let us give an indication of the proof here. If p is a prime ideal of Ok such that p|m then we define Up to be the R-submodule of Q[G] generated by s(Tp) and by (1 − p∗ ). Let p1 , . . . , pt , t = deg(m), be the prime divisors of m. Let U0 = R and if i ∈ {1, . . . , t} let Ui be the R-module j≤i Upj . Since Ui is a lattice of Q[G] for all i the index (Ui : Ui+1 ) is well defined. Let p = pi+1 and ep = s(Tp)/|Tp|.

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Then we have            Ui : Ui+1 = 1 − ep Ui : 1 − ep Ui+1 ker 1 − ep |U : ker 1 − ep |U i

 i+1

The relation Ui+1 = Ui Up implies in particular that (1 − ep)Ui = (1 − ep)Ui+1 . Moreover, if A is any R-submodule of Q[G] we have     ker 1 − ep |A = ATp = a ∈ A| σ a = a, for all σ ∈ Tp . T

T

p Now the inclusion Ui+1 ⊂ Ui p is easily seen to be a consequence of Ui+1 = Ui Up. As we have already pointed out we have

Tp ∩



Tpj = {1}

j≤i

This is what we need to prove by induction on j ∈ {0, . . . , i} that Uj is a free T

Tp-module, see [9] Proposition 5.2, and then Uip = s(Tp)Ui ⊂ Ui+1 . Thus we   have (Ui : Ui+1 ) = 1 and (R : U) = (U0 : Ut ) = i (Ui : Ui+1 ) = 1 too. Corollary 4.1 We have ⎧ h(H) ⎪ ⎪ ⎪ ⎪ ⎨ −e wH erm wk h(km) C em ] = (R− : U − )[ m : μkm  wkm ⎪ ⎪ wH erm ⎪ ⎪ ⎩wk2−e−deg(m) h(km) wkm

if m = (1) if deg(m) ≤ 2 if deg(m) ≥ 3

Proof One has just to use the formulas (12) and (14), and the lemmas 4.1, 4.2, 4.3 and 4.4.   5 Calculation of (R− : U − ) by Ouyang’s method In this last section we compute the index (R− : U − ) in the case m = (1). We recall that Sinnott and Yin had to compute similar indices respectively in [9] §5 and [12] Sect. 6. Their method may be successfully applied to our case, but only if the ideal class number h of k is odd. Let us briefly explain this method. Since R is a free Z[G]-module we have R− = (1 − j)R and (R− : U − ) = ((1 − j)R : (1 − j)U)((1 − j)U : U − ). But (1 − j)U ⊂ U − and U − /(1 − j)U = H 1 (J, U). If h is odd we are able to determine the structure of this cohomology group and also its order. Indeed, let us set, for n ≥ 1 and i ∈ {0, . . . , deg(m)},   T Ani = H n J, Ui i ,

Elliptic units and sign functions

651

where Ui is defined in the proof of Lemma 4.4 and Ti is the subgroup of G  generated by the inertia groups Tpr , for r > i. By definition Tdeg( m) = 1. Recall that p1 , . . . , pdeg(m) are the prime ideals of Ok that divide m. We have U0 = R and Udeg(m) = U. The group Gal(Km/K(1) ) acts trivially on Ani , see [12] Lemma 6.2. Therefore, since Gal(Km/H) is generated by J and by Gal(Km/K(1) ), we deduce that Ani is a Z[Gal(H/k)]-module. Exactly as in [12] Proposition 6.3 one may prove the existence of an exact sequence of R-modules Fp i    0 −→ Ani−1 / 1 − Fpi −→ Ani −→ An+1 −→ 0, i−1

(15)

where Fpi is the Frobenius automorphism associated to pi in Km/k. Now an easy computation gives An0 = 0 if n is odd and An0 (Z/wk Z)[Gal(H/k)] if n is even. From which we deduce that Ani is annihilated by some power wdki of wk , thanks to (15). Let us suppose that h is odd. Then the gcd(wH , wdki ) = wk ( if √ wk = 2 then i = −1 ∈ H, otherwise 2 = [k(i) : k] divides h). Since wH Ani = 0 we see that Ani is a (Z/wk Z)[Gal(H/k)]-module. In fact we have, for all i, n ≥ 1, 2i−1   Ani (Z/wk Z) G(i) ,

(16)

where G(i) = Gal(H/k)/Li and Li =< τp1 , . . . , τpi > is the subgroup of Gal(H/k) generated by the automorphisms τpr = (pr , H/k), for r ≤ i. Indeed, on one hand the group An1 satisfies the property (16), thanks to the exact sequence (15). On the other hand, if Ani−1 satisfies (16) then Fp i  2i−2     (i)

Z) G . Ani−1 / 1 − Fpi An+1 (Z/w k i−1

(17)

If, in addition, h is odd then the group Li acts trivially on Ani and (15) splits, see Fp i , which clearly [12] Lemma 6.5. In other words Ani Ani−1 /(1 − Fpi ) ⊕ (An+1 i−1 ) n shows that Ai satisfies (16). In particular we get (U − : (1 − j)U) = #H 1 (J, U) = #A1t = we2 k

t−1

,

(18)

where t = deg(m) and e = [Gal(H/k) :< τp1 , . . . , τpt >]. We still have to compute the index ((1 − j)R : (1 − j)U), which we decompose as a product ((1 − j)R : (1 − j)U) =

t−1 

((1 − j)Ui : (1 − j)Ui+1 ).

i=0

Following Sinnott, cf. [9] formulas (6.8) and (6.10), one may prove the following:  T     (Ui : Ui+1 ) = (Ui )Tp : Ui+1 p = #Bi / 1 − Fp Bi ,

652

H. Oukhaba

where  = (1 − j), p = pi+1 and Bi is the group (Ui )Tp /s(Tp)Ui . We have B0 = 0 and then ((1 − j)R : (1 − j)U) = 1 in case t = 1. If t ≥ 2 and i > 0 then we have T

Bi H 1 (Tp, UiJ ) H 1 (J, Ui p ). See [9] formula (6.12) for a proof of these two isomorphisms. Let us remark that Ui is a free JTp-module provided that 0 ≤ i < t − 1, cf. loc. cit. Proposition 5.2. In particular Bi = 0 and ((1 − j)Ui : (1 − j)Ui+1 ) = 1 when 0 ≤ i < t − 1. Let us suppose 0 < i = t − 1. Since Bi A1t−1 the isomorphism (16) (proved under the hypothesis 2  h) gives  2t−2 Bt−1 /(1 − Fpt )Bt−1 (Z/wk Z)[G(t) ] .

(19)

Consequently we obtain the formula  −

−

(R : U ) =

w−e k

t−2 w−e2 k

if t = 1 if t ≥ 2 and 2  h.

(20)

Quite recently Ouyang developed in [7] a new approach of this subject. In what follows we propose a method of computing (R− : U − ) largely inspired from Ouyang’s approach. This will allow us to obtain (R− : U − ) in full generality. First we prove that the R-module U is isomorphic to the universal sgn0 (m) of level m introduced in [5] §4. Let us recall that U 0 (m) distribution Usgn sgn is defined as the quotient 0 (m) = Usgn

0sgn (m) D0 (m)

,

where 0sgn (m) is the free abelian group on ϒm = f∈m Gal(Kf/k) and m is the set of the integral ideals n|m such that the gcd(n, m/n) = (1). Moreover D0 (m) is the R-submodule of 0sgn (m) generated by the sums S0 (f, pu , σ ) = σ − σ (p, Kf/k)−1 − s(Kfpu , σ ), where f, pu ∈ m are such that p is a prime ideal not dividing f, σ ∈ Gal(Kf/k) and s(Kfpu , σ ) is the sum of all the automorphisms of Kfpu /k whose restriction to Kf is equal to σ . Let us consider the surjective R-homomorphism F : 0sgn (m) −→ U which σ αn,  σ being associates to a σ ∈ Gal(Kn/k), with n ∈ m, its image F(σ ) =  any extension of σ to Km. One may easily check that D0 (m) ⊂ ker(F). But U 0 (m) are Z-free and have the same rank, cf. [5] Theorem 4.1. Thus the and Usgn quotient ker(F)/D0 (m) is trivial and then F induces the desired isomorphism 0 (m) U. Usgn

Elliptic units and sign functions

653

0 (m), analogue of In [5] we introduced a certain resolution (Lm, d1 ) of Usgn Anderson’s resolution of the universal ordinary distribution Ur of level r ∈ N∗ , discovered by Anderson in [6], Appendix. In [7], Ouyang uses Anderˆ i (I, Ur ), where son’s resolution to produce spectral sequences convergent to H −1 I = Gal(Q(ζr )/Q(ζr + ζr )) and ζr is a primitive r-th root of 1. This allows him to compute these cohomology groups. In [5] we applied the method of 0 (m)). We point out that all the specˆ i (J, Usgn Ouyang to compute the groups H tral sequences used in this context are obtained from some double complexes. In fact, it is Anderson who invented in [1] the theory of double complex for computing the sign-cohomology groups of the universal ordinary distribution. Ouyang, by using Anderson’s resolution, greatly improved and simplified the method of Anderson. By definition Lm is the free abelian group on the symbols

[σ , g],

g ∈ m and

σ ∈ ϒm/g

If nσ (σ ) ∈ 0s (m/g) then some times we will write [x, g] instead of x = nσ [σ , g]. We equip Lm with the graduation deg([x, g]) = − deg(g). Let us recall the definition of d1 . If g ∈ m, f ∈ m/g and σ ∈ Gal(Kf/k) are given then we let 

0 d1 ([σ , g]) =

(−1)ω(p,g) [S0 (f, pˆ , σ ), g/pˆ ]

p|g

if g = (1) if g = (1).

where pˆ is the greatest power of p dividing m and ω(p, g) is defined as follows. If p1 , . . . , pt , t = deg(m), are the prime divisors of m and p = pi then ω(p, g) = #{j such that pj |g and j < i}. Then (Lm, d1 ) is a graded differential R-module. We refer the reader to the appendix of [6] for the proof that (Lm, d1 ) is acyclic in non-zero degree. The map [σ , (1)] −→ σ induces an isomorphism 0 (m). As Ouyang did in [7] we define a differential d on L H 0 (Lm, d1 ) Usgn 2 m to obtain a resolution of R. Indeed, if g, f and σ are as above then we put 

0 d2 ([σ , g]) =

(−1)ω(p,g)

p|g

  [σ , g/pˆ ] − [s(Kfpu , σ ), g/pˆ ]

if g = (1) if g = (1).

To check that (Lm, d2 ) is acyclic in non-zero degree one has just to repeat Anderson’s proof in [6]. Moreover the isomorphism H 0 (Lm, d2 ) R is straightforward. Now we prove a formula for the index (R− : U − ) analogue of the formula (4.1) of [7]. To this end we let Vm be the Q[G]-module freely generated as a Q-vector space by all [σ , g], g ∈ m and σ ∈ ϒm/g. Also we have to consider Ouyang’s connecting Q[G]-isomorphism m : (Vm, d1 ) −→ (Vm, d2 ) defined for σ ∈ Gal(Kn/k) by the formula

654

H. Oukhaba



m([σ , g]) =

(−1)

h∈n

deg(h)

   Fp−1 p|h

#Tp

[σ (n/h), g],

where σ (n/h) is the restriction of σ to Kn/h. Proposition 2.1 of [7] applied to m then gives I2 (R− : U − ) = , (21) I1 where Ij =

0 − #coker(H 0 (L− m, dj ) −→ H (Lm, dj ) )  − i (−1)i #torH 0 (L− m, dj ) i =0 #H (Lm, dj )

0 − Here the map H 0 (L− m, dj ) −→ H (Lm, dj ) is the one induced by the inclusion − Lm −→ Lm. The proof of (21) is exactly the same as the proof of Theorem 4.1 in [7]. Now we use spectral sequences to compute I1 and I2 . Let M = Z[J]/s(J)Z and let

(P, ∂) :

· · · −→ Pi −→ · · · −→ P1 −→ P0 −→ 0

be a projective resolution of the J-module M. For j ∈ {1, 2} fixed we use  (P,p∂)  p,q and (Lm, dj ) to define the double complex Cj , where Cj = HomJ Pq , Lm , equipped with the differentials ∂ and dj . Let Tot Cj be the total complex of Cj and let us consider the spectral sequence 

 p,q p  q E2 = Hdj ExtJ M, L•m ⇒ H p+q (Tot Cj )

 induced by the first filtration  Filp (Tot Cj ) = u≥p Cju,v of Tot Cj . Let Z0 ((Tot Cj )) be the set of cocycles in (Tot Cj )0 . Then we have a surjective map 1 : Z0 ((Tot Cj )0 ) −→ H 0 (Lm, dj )− defined by the formula ⎛ 1 ⎝



⎞

  φq ⎠ = φ0 (α) modulo dj L−1 m

q −q,q

where φq ∈ Cj and α is any element in the inverse image of 1 via the surjective map P0 −→ M. It is easy to see that 1 induces an isomorphism H 0 (Tot Cj ) −→ H 0 (Lm, dj )− , which we still denote 1 . We also have an isomorphism 2 : H 0 ( Fil0 (Tot Cj )) = HomJ (P0 , L0m) −→ (L0m)−1 defined by 2 (φ) = φ(α). But the following diagram is commutative

Elliptic units and sign functions

655

H 0 ( Fil0 (Tot Cj )) −−−−→ H 0 (Tot Cj ) ⏐ ⏐ ⏐ ⏐ ! 1 ! 2 (L0m)−1

−−−−→ H 0 (Lm, dj )−

  0,0 Therefore, since  E∞ = Im H 0 ( Fil0 (Tot Cj )) −→ H 0 (Tot Cj ) , we have the identity 1





     −  0,0 0 = Im H 0 L− , E∞ m, dj −→ H Lm, dj

from which we deduce     −    p,−p  0 #coker H 0 L− = . # E∞ m, dj −→ H Lm, dj

(22)

p =0

To determine H n (Tot Cj ) for all n we need consider  the spectral sequence associated to the second filtration  Filq (Tot Cj ) = v≥q Cju,v . We have 

p,q

q

E2 = ExtJ (M, H p (Lm, dj )) ⇒ H p+q (Tot Cj ),

Since H p (Lm, dj ) = 0 if p = 0 the spectrale sequence  E degenerates at  E2 . In particular we have H n (Tot Cj ) = ExtnJ (M, H 0 (Lm, dj )) and 

    p,q p  q p+q M, H 0 Lm, dj . E2 = Hdj ExtJ M, L•m ⇒ ExtJ

q ˆ q+1 (J, A) for all q > 0. We have If A is any J-module then ExtJ (M, A) = H proved in [5] Lemma 4.1 that the spectral sequence p

Hd

1

     ˆ p+q J, H 0 (Lm, d1 ) ˆ q J, L•m ⇒ H H

ˆ q (J, L•m)) = 0. If degenerates at the second term. Let us prove that Hd2 (H f ∈ m is such that f = (1) then Z[Gal(Kf/k)] is Z[J]-free. Moreover if q is ˆ q (J, Z[Gal(K(1) /k)]) = 0. We deduce from these two remarks that odd then H p ˆ q (J, Lm) = 0 if q is odd. If q is even then H p

ˆ q (J, Lpm) = H

"



g∈m deg(g)=−p

 A, g .

656

H. Oukhaba

  where A = Z/wk Z[Gal(H/k)] and A, g = {[x, g], x ∈ A}. Let us denote by d¯ 2 ˆ q (J, Lpm) induced by d2 . We have the differential on H 

0 d¯ 2 ([x, g]) =

if g = (1) if g = (1).

ω(p,g) [x, g/p ˆ] p|g(−1)

ˆ q (J, Lpm) be such Now let us fix a prime divisor q of m and let X = [Xg, g] ∈ H ˆ q (J, Lp−1 that d¯ 2 (X) = 0. Consider the element Y = [Yf, f] ∈ H m ) such that  Yf =

0 (−1)ω(q,f) Xf/pˆ

if q  f if q|f.

p ˆq (J, L•m)) = 0. Then we obviously have d¯ 2 (Y) = X. This proves that Hd2 (H p,q p,q   From the above we deduce that Er = Er+1 for all r ≥ 2 and all p, q satisfying p + q ≥ 1. This implies that for a given r ≥ 2



$  #  p,q Er , p + q ≤ 0 and (p, q) = (0, 0) ∪ tor  Er0,0

is a bounded complex of finite groups. In particular     p,q (−1)p+q #tor  Er0,0 # Er



=

p+q≤0 (p,q) =(0,0)

is independant of r ≥ 2. Now the formula (22) and the definition of Ij give Ij =



#tor  E20,0



  p,q (−1)p+q # E2  (−1)p =  p,0 p+q≤0 p =0 # E2 

(23)

q>0

p,q p ˆq If j = 2 and q > 0 then  E2 = Hd2 (H (J, L•m)) = 0. Thus we have I2 = 1. If j = 1 and q > 0 we have proved in [5] Theorem 4.2 the following result

 

#(

p,q E2 )

=

1 (wk )eC(p)

if q is even if q is odd,

where e is defined in Lemma 4.3 and C(p) is the binomial coefficient giving the number of subsets of cardinal −p in a set of cardinal deg(m). Therefore we have  w−e if deg(m) = 1 1 − − k (R : U ) = = (24) deg(m)−2 I1 (wk )−e2 if deg(m) ≥ 2, thanks to (21) and (23).

Elliptic units and sign functions

657

Theorem 5.1 We have  [Ok×m : Em] =

h(km) h(km)wak

if deg(m) ≤ 2 if deg(m) ≥ 3

where a = e(2deg(m)−2 − 1) + 2 − deg(m) if deg(m) ≥ 2 and a = 0 if deg(m) ≤ 1. Proof The theorem is a direct consequence of formula (9), Lemma 3.1, Proposition 3.1, Corollary 4.1 and formula (24)   Acknowledgements The author would like to express his sincere thanks to the referee for his valuable comments and suggestions to improve the manuscript.

References 1. Anderson, G.W.: A double complex for computing the sign-cohomology of the universal ordinary distribution. In Recent progress in algebra (Taejon/Seoul, 1997), pp. 1–27. Amer. Math. Soc. Providence, RI (1999) 2. Hajir, F., Villegas, F.R.: Explicit elliptic units. I. Duke Math. J. 90(3), 495–521 (1997) 3. Kubert, D.S.: Product formulae on elliptic curves. Invent. Math. 117(2), 227–273 (1994) 4. Oukhaba, H.: Index formulas for ramified elliptic units. Composit. Math. 137(1), 1–22 (2003) 5. Oukhaba, H.: Sign functions of imaginary quadratic fields and applications. Ann. Inst. Fourier 55(3), 753–772 (2005) 6. Ouyang, Y.: Group cohomology of the universal ordinary distribution. J. Reine Angew. Math. 537, 1–32, 2001. With an appendix by Greg W. Anderson 7. Ouyang, Y.: The universal norm distribution and Sinnott’s index formula. Proc. Amer. Math. Soc. 130(8), 2203–2213 (electronic), (2002) 8. Robert, G.: Unités elliptiques. Société Mathématique de France, Paris, 1973. Bull. Soc. Math. France, Mém. No. 36, Tome 101 9. Sinnott, W.: On the Stickelberger ideal and the circular units of a cyclotomic field. Ann. Math. (2) 108(1), 107–134, (1978) 10. Sinnott, W.: On the Stickelberger ideal and the circular units of an abelian field. Invent. Math. 62(2), 181–234, 1980/1981 11. Tate, J.: Les conjectures de Stark sur les fonctions L d’Artin en s = 0. Birkh¨auser Boston Inc, 1984. Lecture notes edited by Dominique Bernardi and Norbert Schappacher 12. Yin, L.: Index-class number formulas over global function fields. Composit. Math. 109(1), 49–66 (1997) 13. Yin, L.: On the index of cyclotomic units in characteristic p and its applications. J. Num. Theory 63(2), 302–324 (1997)