A Bound for the Torsion in the
K-Theory
of Algebraic Integers
Christophe Soul´e
Received: November 6, 2002 Revised: May 30, 2003
A Kazuya Kato, qui marche sous la lune.
Abstract. Letmbe an integer bigger than one,Aa ring of algebraic integers, F its fraction field, and Km(A) the m-th Quillen K-group of A. We give a (huge) explicit bound for the order of the torsion subgroup ofKm(A) (up to small primes), in terms of m, the degree ofF overQ, and its absolute discriminant.
Let F be a number field, A its ring of integers and Km(A) the m-th Quillen
K-group ofA. It was shown by Quillen thatKm(A) is finitely generated. In this paper we shall give a (huge) explicit bound for the order of the torsion subgroup ofKm(A) (up to small primes), in terms ofm, the degree ofF over
Q, and its absolute discriminant.
Our method is similar to the one developed in [13] for F = Q. Namely, we reduce the problem to a bound on the torsion in the homology of the general linear group GLN(A). Thanks to a result of Gabber, such a bound can be obtained by estimating the number of cells of given dimension in any complex of free abelian groups computing the homology of GLN(A). Such a complex is derived from a contractibleCW-complex Wf on which GLN(A) with compact quotient. We shall use the construction of fW given by Ash in [1] . It consists of those hermitian metrics h on AN which have minimum equal to one and are such that their set M(h) of minimal vectors has rank equal to N in FN. To count cells in W /f GLN(A), one will exhibit an explicit compact subset of
AN⊗
The bound on the K-theory ofA implies a similar upper bound for the ´etale cohomology of Spec (A[1/p]) with coefficients in the positive Tate twists ofZp, for any (big enough) prime numberp.
However, this bound is quite large since it is doubly exponential both in m
and, in general, the discriminant of F. We expect the correct answer to be polynomial in the discriminant and exponential inm(see 5.1).
The paper is organized as follows. In Section 1 we prove a few facts on the geometry of numbers for A, including a result about the image of A∗ by the
regulator map (Lemma 3), which was shown to us by H. Lenstra. Using these, we study in Section 2 hermitian lattices overA, and we get a bound onM(h) whenhlies infW. The cell structure ofWfis described in Section 3. The main Theorems are proved in Section 4. Finally, we discuss these results in Section 5, where we notice that, because of the Lichtenbaum conjectures, a lower bound for higher regulators of number fields would probably provide much better upper bounds for the ´etale cohomology of Spec (A[1/p]). We conclude with the example ofK8(Z) and its relation to the Vandiver conjecture.
1 Geometry of algebraic numbers 1.1
LetF be a number field, andAits ring of integers. We denote byr= [F :Q] the degree of F over Q and by D = |disc (K/Q)| the absolute value of the discriminant of F over Q. Let r1 (resp. r2) be the number of real (resp.
complex) places of F. We haver=r1+ 2r2. We let Σ = Hom (F,C) be the
set of complex embeddings of F. These notations will be used throughout. Given a finite setX we let # (X) denote its cardinal.
1.2
We first need a few facts from the geometry of numbers applied to AandA∗.
The first one is the following classical result of Minkowski:
Lemma 1. LetLbe a rank one torsion-freeA-module. There exists a non zero element x∈L such that the submodule spanned byxinL has index
# (L/Ax)≤C1,
where
C1=
r!
rr·4
r2π−r2√D
in general, and C1= 1 whenAis principal.
Proof. TheA-moduleLis isomorphic to an idealIinA. According to [7], V
§4, p. 119, Minkowski’s first theorem implies that there existsx∈Ithe norm of which satisfies
Here |N(x)|= # (A/Ax) and N(I) = # (A/I), therefore # (I/Ax)≤C1. The
case whereAis principal is clear. q.e.d.
1.3
The family of complex embeddingsσ:F →C,σ∈Σ, gives rise to a canonical isomorphism of real vector spaces of dimensionr
F⊗QR= (CΣ)+,
where (·)+ denotes the subspace invariant under complex conjugation. Given
α∈F we shall write sometimes|α|σ instead of|σ(α)|.
Lemma 2. Given any element x= (xσ)∈ F⊗QR, there existsa ∈A such
that X
σ∈Σ
|xσ−σ(a)| ≤C2,
with
C2=
4r1πr2
rr−2r!
√ D
in general, and
C2= 1/2 if F =Q.
Proof. Define a norm onF⊗QRby the formula
kxk=X σ∈Σ
|xσ|.
The additive group A is a lattice in F ⊗QR, and we let µ1, . . . , µr be its successive minima. In particular, there exista1, . . . , ar∈Asuch thatkaik=µi, 1≤i≤r, and{a1, . . . , ar} are linearly independent overZ. Anyx∈F⊗QR can be written
x= r X
i=1
λiai, λi∈R.
Letni∈Zbe such that|ni−λi| ≤1/2, for alli= 1, . . . , r, and
a= r X
i=1
niai.
Clearly
kx−ak ≤
r X
i=1
|λi−ni| kaik ≤ 1
2(µ1+· · ·+µr)≤
r
On the other hand, we know from the product formula that, given any a ∈ A− {0},
Y
σ∈Σ
|σ(a)| ≥1. (2)
By the inequality between arithmetic and geometric means this implies
kak=X σ∈Σ
|σ(a)| ≥r ,
hence
µi≥r for all i= 1, . . . , r . (3)
Minkowski’s second theorem tells us that
µ1. . . µr≤2rW2−r2
√
D (4)
([7], Lemma 2, p. 115), where W is the euclidean volume of the unit ball for
k · kin F⊗QR. (Note that the covolume ofAis
√
D.) The volumeW is the euclidean volume of those elements (xi, zj)∈Rr1×Cr2 such that
r1
X
i=1
|xi|+ 2 r2
X
j=1
|zj| ≤1.
One finds ([7], Lemma 3, p. 117)
W = 2r14−r2(2π)r2/r!. (5)
From (3) and (4) we get
µr≤2rW
√
D2−r2r−(r−1). (6)
The lemma follows from (1), (5) and (6). q.e.d.
1.4
We also need a multiplicative analog of Lemma 2. Let R(F) be the regulator ofF, as defined in [7] V,§1, p. 109. Lets=r1+r2−1.
Lemma 3. Let (λσ), σ ∈ Σ be a family of positive real numbers such that λσ = λσ when σ is the complex conjugate of σ. There exists a unit u∈ A∗
such that
Sup σ∈Σ
(λσ|u|σ)≤C3
à Y
σ∈Σ
λσ !1/r
with
C3= exp (s(4r(log 3r)3)s−12r2−1R(F)).
Proof. We follow an argument of H. Lenstra. Let H ⊂ Rr1+r2 be the s
-dimensional hyperplane consisting of vectors (x1, . . . , xr1+r2) such thatx1+x2+
· · ·+xr1+r2 = 0. Choose a subset{σ1, . . . , σr1+r2} ⊂Σ such thatσ1, . . . , σr1
are the real embeddings of Σ and σi 6=σj if i=6 j. Given λ= (λσ)σ∈Σ as in
the lemma, we let
ρ(λ) = (log(λσ1), . . . ,log(λσr1),2 log(λσr1 +1), . . . ,2 log(λσr1 +r2)).
Ifu∈A∗is a unit, andλ= (|u|σ), we haveρ(λ)∈H. We get this way a lattice
L={ρ(|u|σ), u∈A∗}
in H.
Define a normk · konH by the formula
k(xi)k= Sup ( Sup
1≤i≤r1
|xi|, Sup r1+1≤i≤r1+r2
|xi|/2).
It is enough to show that, for any vectorx∈H there exists a∈Lsuch that
kx−ak ≤log(C3).
According to [14], Cor. 2, p. 84, we have (when r≥2)
kak ≥ε
where
ε=r−1(log(3r))−3,
for any a∈ L− {0}. Therefore, using Minkowski’s second theorem as in the proof of Lemma 2 we get that, for any x∈H there exists a∈Lwith
kx−ak ≤s2s−1ε1−sWvol (H/L),
whereW is the euclidean volume of the unit ball fork · k, where we identifyH
with Rs by projecting on the first scoordinates. ClearlyW ≤2s+r2−1 when,
by definition (loc.cit.), vol (H/L) is equal toR(F). The lemma follows.
1.5
We now give an upper bound for the constantC3 of Lemma 3.
Lemma 4. The following inequality holds
Proof. Letκbe the residue ats= 1 of the zeta function of F. According to [11], Cor. 3, p. 333, we have
κ≤2r+1Daa1−r whenever 0< a≤1. Takinga= log(D)−1 we get
κ≤e2r+1log(D)r−1. (7)
On the other hand
κ= 2r1(2π)r2 h(F)R(F)
w(F)√D , (8)
whereh(F) is the class number ofF andw(F) the number of roots of unity in
F ([7], Prop. 13, p. 300). Sinceh(F)≥1 we get
R(F)≤w(F) 2−(r1+r2)π−r2e2r+1√Dlog(D)r−1.
Since the degree overQ ofQ(√n
1) isϕ(n), whereϕis the Euler function, we must have
ϕ(w(F))≤r .
Whenn=ptis an odd prime power we have
ϕ(pt) = (p−1)pt−1≥pt/2.
Therefore
w(F)≤2r2.
Since
2−(r1+r2)π−r22r=
µ 2
π
¶r2
≤1
and 4e≤11, the lemma follows.
2 Hermitian lattices 2.1
Anhermitian latticeM = (M, h) is a torsion freeA-moduleM of finite rank, equipped with an hermitian scalar product hon M ⊗ZC which is invariant under complex conjugation. In other words, if we let Mσ =M ⊗AC be the complex vector space obtained from M by extension of scalars viaσ∈Σ,his given by a collection of hermitian scalar productshσ onMσ,σ∈Σ, such that
hσ(x, y) =hσ(x, y) wheneverxandy are inM. We shall also write
and
kxkσ= p
hσ(x).
Lemma 5. Let M be an hermitian lattice of rankN. Assume thatM contains N vectors e1, . . . , eN which are F-linearly independent in M ⊗AF and such
that
keik ≤1
for alli= 1, . . . , N. Then there exist a direct sum decomposition M =L1⊕ · · · ⊕LN
where each Li has rank one and contains a vectorfi such that
# (Li/A fi)≤C1
and
kfik ≤(i−1)C1C2+C11/rC3.
Here C1, C2, C3 are the constants defined in Lemmas 1, 2, 3 respectively.
Proof. We proceed by induction onN. WhenN = 1, Lemma 1 tells us that
L1=M containsx1 such that
# (L1/A x1)≤C1.
Let us write
x1=α e1
withα∈F∗. Using Lemma 3, we can chooseu∈A∗ such that
Sup σ∈Σ|
u α|σ ≤C3
à Y
σ
|α|σ !1/r
=C3N(α)1/r≤C3C11/r.
The lemma follows withf1=u x1.
Assume now that N ≥2, and let L1 =M ∩F e1 in M ⊗AF. As above, we choose f1=a11e1in L1 with [L1:A f1]≤C1and
Sup σ∈Σ|
a11|σ≤C3C11/r.
The quotient M′ = L/L
1 is torsion free of rank N −1. We equip M′ with
the quotient metric induced by h, we let p:M →M′ be the projection, and
e′
i=p(ei),i= 2, . . . , N. Clearly
We assume by induction thatM′ can be written
M′ =L′2⊕ · · · ⊕L′N and that L′
i contains a vectorfi′ such that
ni= # (L′i/A fi′)≤C1
with
f′
i = X
2≤j≤i
aije′j, (9)
aij ∈F, and, for allσ∈Σ,
|aij|σ ≤C1C2 if 2≤j < i≤N ,
and
|aii|σ ≤C11/rC3, 2≤i≤N .
Let s:M′ →M be any section of the projectionp. From (9) it follows that
there exists µi∈F such that
s(f′
i)− X
2≤j<i
aijej =µie1.
Applying Lemma 2, we can choosebi∈A such that X
σ∈Σ
¯ ¯ ¯ ¯µnii −bi
¯ ¯ ¯ ¯ σ
≤C2.
Definet:M′ →M by the formulae
t(x) =s(x)−a(x)bie1
wheneverx∈L′
i, hence
nix=a(x)fi′,
for some a(x)∈A, 2≤i≤N. If we takefi=t(fi′), we get
fi=s(fi′)−nibie1=
X
2≤j<i
aijej+ (µi−nibi)e1
and, for allσ∈Σ,
|µi−nibi|σ≤niC2≤C1C2.
We defineai1=µi−nibi andLi=t(L′i). Then
satisfies our induction hypothesis:
Lemma 6. LetI⊂Abe a nontrivial ideal. There exists a set of representatives R ⊂Aof Amodulo I such that, for any xinR,
Proof. According to the proof of Lemma 2, theZ-moduleAcontains a basis ofrelementse1, . . . , er such that and we can chooseRamong those
2.3
such that Lcontains a vector e2 with
# (L/A e2)≤n1n2,
Proof. The algebraic content of this lemma is [9], Lemma 1.7, p. 12. To control the norms in this proof we first define an isomorphism
ui:Li→Ii
whereIi is an ideal ofA. Ifx∈Li,ui(x)∈A is the unique element such that
nix=ui(x)fi, i= 1,2. In particularni=ui(fi).
Next, we choose an idealJ1in the class ofI1 which is prime toI2. According
to [9], proof of Lemma 1.8, we can choose
J1=
The direct sum of the inverses ofv1 andu2is an isomorphism
ϕ:J1⊕I2−→∼ L1⊕L2=M .
Since J1 and I2 are prime to each other we have an exact sequence (as in [9]
loc.cit.)
and mapL toM by the composite map
L−→i J1⊕I2
ϕ
−→M
wherei(x) = (x,−x). We choose
so thatϕ◦i(e2) = (n2v1(f1), x0u2(f2)) has norm
kϕ◦i(e2)k ≤n2kf1k+C2
µ
r+ 3 4
¶
nr1kf2k.
Furthermore we have isomorphisms
L⊕A(−→i,s)J1⊕I2
ϕ
−→M
and
# (L/A e2) = # (J1I2/A e2)≤# (J1/A x0)×# (I2/A n2) =n1n2.
q.e.d.
2.4
Proposition 1. Let M be a rank N hermitian free A-module such that its unit ball contains a basis of M ⊗AF. Then M has a basis (e1, . . . , eN) such
that
keik ≤Bi
with Bi= (i−1)C2+C3,i= 1, . . . , N, whenA is principal and
Bi= (1 +C1C2)(N C2+C3)
µ 1 +C2
r+ 3 4
¶log2(N)+2
C12(r+1)N/i
in general. Here log2(N)is the logarithm of N in base 2.
Proof. WhenA is principal,C1 = 1 and Proposition 1 follows from Lemma
5.
In general Lemma 5 tells us that
M =L1⊕ · · · ⊕LN andLi contains a vectorfi with # (Li/A fi)≤C1and
kfik ≤C1((i−1)C2+C3)≤C1(N C2+C3).
Let k > 1 be an integer and λ > 0 be a real number. We shall prove by induction N that, ifM has a decomposition as above with
# (Li/A fi)≤k and
kfik ≤k λ , thenM has a basis (e1, . . . , eN) such that
keik ≤λ µ
1
k+C2
¶ µ
1 +C2r+ 3
4 ¶t
for alli= 1, . . . , N, where t≥1 is such that
ThereforeM satisfies the induction hypothesis (10). Since
1 + 2 +· · ·+ 2t= 2t+1−1≤2Ni
andt≤log2(N) + 1, Proposition 1 follows by takingk=C1 and
λ=C1(N C2+C3)
2.5
LetM be a rankN hermitian freeA-module. We let
m(h) = Inf{h(x), x∈M − {0}}
be the minimum value ofhonM − {0} and
M(h) ={x∈M/h(x) =m(h)}
be the (finite) set of minimal vectors ofM. LetωN be the standard volume of the unit ball inRN.
Proposition 2. LetM = (M, h)be as above. Assume thatm(h) = 1and that M(h)spans theF-vector spaceM ⊗AF. Then M has a basis f1, . . . , fN such
that anyx∈M(h)is of the form
x= N X
i=1
yifi
with X
σ∈Σ
|yi|2σ≤Ti,
Ti=rrNC32rN+2γN
Y
j6=i
Bj2,
and
γ= 4r1+r2ω−2r1/N
N ω
−2r2/N
2N D .
Proof. From Proposition 1 we know that M has a basis (e1, . . . , eN) with
keik ≤Bi. Letx∈M(h) be a minimal vector and (xi) its coordinates in the basis (ei).
Fixi∈ {1, . . . , N} andσ∈Σ. Consider the square matrix
Hi= (hσ(vk, vℓ)), wherevk=ek ifk6=iandvi=x. Furthermore, let
Hσ = (hσ(ek, eℓ)). Since
|xi|2σ = det(Hi) det(Hσ)−1 the Hadamard inequality implies
|xi|2σ≤hσ(x) Y
j6=i
For any unitu∈A∗ we can replacee
According to Icaza [4], Theorem 1, there existsz∈Lsuch that Y
From this it follows that Y
σ∈Σ
det(Hσ)−1≤(r C32)rNγN (14)
and Proposition 2 follows from (13) and (14).
2.6
To count the number of vectors inM(h) using Proposition 2 we shall apply the following lemma :
Lemma 8. The number of elementsa inA such that
X
σ∈Σ
is at most
B(T) = Sup (Tr/22r+3,1).
Proof. When r2 > 0, this follows from [7], V § 1, Theorem 0, p. 102, by
noticing that one can takeC3= 2r+3in loc.cit. Whenr2= 0, the argument is
similar.
3 Reduction theory 3.1
Fix an integerN ≥2. Let
Γ = GLN(A) and
G= GLN(F⊗QR).
On the standard latticeL0=AN consider the hermitian metrich0 defined by
h0(x, y) =
X
σ∈Σ
N X
i=1
xiσyiσ
for all vectorsx= (xiσ) andy= (yiσ) inL0⊗ZC= (CN)Σ. Anyg∈Gdefines an hermitian metric h=g(h0) onL0by the formula
g(h0) (x, y) =h0(g(x), g(y)).
LetK be the stabilizer ofh0 andGandX =K\G . We can view eachh∈X
as a metric onL0.
Following Ash [1], we say that a finite subset M ⊂ L0 is well-roundedwhen
it spans the F-vector spaceL0⊗AF. We letWf⊂X be the space of metrics
h such that m(h) = 1 and M(h) is well-rounded. Given a well-rounded set
M ⊂L0 we letC(M)⊂Wfbe the set of metricshsuch that
• h(x) = 1 for allx∈M
• h(x)>1 for allx∈L0−(M ∪ {0}).
As explained in [1], proof of (iv), pp. 466-467,C(M) is either empty or topo-logically a cell, and the family of closed cellsC(M) gives a Γ-invariant cellular decomposition of fW, such that
C(M) = a M′⊃M
C(M′).
3.2
Proposition 3. i) For any integerk≥0, the number of cells of codimension k inWf is at most
c(k, N) = µ
a(N)
N+k
¶
where
a(N) = 2N(r+3)
ÃN Y
i=1
Ti !r/2
,
andTi is as in Proposition 2.
ii)Given a cell inWf, its number of codimension one faces is at mosta(N)N+1.
Proof. Let Φ be the set of vectors x = (xi) in AN such that, for all i =
1, . . . , N, X
σ∈Σ
|xi|2σ≤Ti.
Given h∈Wf, Proposition 2 says that we can find a basis (fi) ofL0such that
any x in M(h) has its coordinates (xi) bounded as above. If γ ∈ Γ is the matrix mapping the standard basis ofAN to (fi), this means thatM(γ(h)) =
γ−1(M(h)) is contained in Φ.
LetC(M) be a nonempty closed cell of codimensionk inWf. For anyx∈L0,
the equation h(x) = 1 defines a real affine hyperplane in the set of N ×N
hermitian matrices with coefficients in (F ⊗QC)+. The equationsh(x) = 1,
x∈M, may not be linearly independent, but, sinceC(M) has codimensionk,
M has at leastN+k elements. And sinceM ⊂M(h) for someh∈fW, there existsγ∈Γ such thatγ−1(M) is contained in Φ. Therefore, modulo the action
of Γ, there are at most³card(Φ)N+k ´ cellsC(M) of codimensionk. From Lemma 7 we know that
card(Φ)≤a(N),
therefore i) follows.
To prove ii), consider a cellC(M) and a codimension one faceC(M′) ofC(M).
We can writeM′=M∪ {x}for some vectorxand there existsγ∈Γ such that
γ(M′)⊂Φ. Since M is well-rounded, the matrix γ is entirely determined by
the set of vectors γ(M), i.e. there are at most card(Φ)N matricesγ such that
γ(M)⊂Φ. Sinceγ(x)∈Φ, there are at most card(Φ)N+1 vectorsxas above.
q.e.d.
3.3
Lemma 9. Letγ∈Γ− {1} andpbe a prime number such that γp= 1. Then
Proof. Since γ is non trivial we have P(γ) = 0 where P is the cyclotomic polynomial
P(x) =Xp−1+Xp−2+
· · ·+ 1.
If F does not contain the p-th roots of one,P is irreducible, and therefore it divides the characteristic polynomial of the matrixγ overF, hencep−1≤N. Otherwise,F containsQ(µp), which is of degreep−1, thereforep−1≤r.
4 The main results 4.1
For any integern >0 and any finite abelian group Awe let cardn(A) be the largest divisor of the integer #(A) such that no primep≤ndivides cardn(A). LetN ≥2 be an integer. We keep the notation of§3 and we let
e
w= dim(X)−N =r1
N(N+ 1) 2 +r2N
2
−N
be the dimension ofWf. For anyk≤wewe define
h(k, N) =a(N)(N+1)c(we−k−1,N),
wherec(·, N) anda(N) are defined in Proposition 3.
Theorem 1. The torsion subgroup of the homology of GLN(A)is bounded as follows
card1+sup(r,N)Hk(GLN(A),Z)tors≤h(k, N).
Proof. We know from [1] thatWfis contractible and the stabilizer of anyh∈
f
W is finite. From Lemma 9 it follows that, moduloS1+sup(r,N), the homology
of Γ = GLN(A) is the homology of a complex (C·, ∂), where Ck is the free abelian group generated by a set of Γ-representatives of those k-dimensional cells c in fW such that the stabilizer ofc does not change its orientation ([2], VII). According to Proposition 3, the rank ofCk is at mostc(we−k, N) and any cell ofWf has at mosta(N)N+1 faces. Theorem 1 then follows from a general
result of Gabber ([13], Proposition 3 and equation (18)).
4.2
For any integerm≥1 let
k(m) =h(m,2m+ 1).
Theorem 2. The following inequality holds
cardsup(r+1,2m+2)Km(A)tors≤k(m).
Proof. As in [13], Theorem 2, we consider the Hurewicz map
H:Km(A)→Hm(GL(A),Z),
the kernel of which lies inSn,n≤(m+ 1)/2. Since, according to Maazen and Van der Kallen,
Hm(GL(A),Z) =Hm(GLN(A),Z) whenN ≥2m+ 1, Theorem 2 is a consequence of Theorem 1.
4.3
Letpbe an odd prime andn≥2 an integer. For anyν≥1 denote byZ/pν(n) the ´etale sheaf µ⊗pνn on Spec(A[1/p]), and let
H2(Spec(A[1/p]),Zp(n)) = lim←− ν
H2(Spec(A[1/p]),Zpν(n)).
From [12], we know that this group is finite and zero for almost all p.
Theorem 3. The following inequality holds
Y
p≥4n−1
p≥r+2
cardH2(Spec(A[1/p]),Zp(n))≤k(2n−2).
Proof. According to [12], the cokernel of the Chern class
cn,2:K2n−2(A)→H2(Spec(A[1/p]),Zp(n))
lies in Sn+1 for all p. Furthermore, Borel proved that K2m−2(A) is finite.
Therefore Theorem 3 follows from Theorem 2.
4.4
By Lemmas 1 to 7 and Propositions 1 to 3, the constant k(m) is explicitly bounded in terms ofm, r andD. We shall now simplify this upper bound.
Proposition 4. i) log logk(m)≤220m4log(m)r4r√D log(D)r−1
ii) IfF has class number one,
iii) If F=Q(√−D)is imaginary quadratic
log logk(m)≤1120m4log(m) log(D) ;
if furthermoreF has class number one
log logk(m)≤510m4log(m) log(D).
iv)When F=Qandm≥9
log logk(m)≤8m4log(m) ;
furthermore
log logk(7)≤40 545
and
log logk(8)≤70 130.
Proof. By definition
k(m) =h(m,2m+ 1) =a(N)(N+1)c(we−m−1,N) withN = 2m+ 1 and
c(we−m−1, N) = µ
a(N)
N+we−m−1 ¶
.
Since
N+we−m−1 = r1
N(N+ 1) 2 +r2N
2−m−1
≤ 2r m2+ 3r m+r−2m−1 and sincea(2m+ 1) is very big, we get
log logk(m) ≤ (2r m2+ 3r m+r−2m−1) loga(2m+ 1) + log(2m+ 2) + log loga(2m+ 1)
≤ r(2m2+ 3m+ 1) loga(2m+ 1). (15) From Proposition 3 and Proposition 2 we get
a(N) = 2N(r+3) ÃN
Y
i=1
Ti !r/2
, (16)
and
N Y
i=1
Ti= (rrNγNC32rN+2)N
N Y
i=1
According to Proposition 1
is much bigger thanC2. Therefore
log(N C2+C3)≤log(N) + log(C3∗). (19)
from which it follows that
X1 ≤ 208 log(m)m4r4r
√
D log(D)r−1
whenm≥2 andr≥2.
To evaluateX2 first notice that
by [10], II, (1.5), Remark, hence
By the Stirling formula and Lemma 1, ifr≥2,
Furthermore (18) becomes
5 Discussion 5.1
The upper bound in Theorem 2 and Propostition 4 seems much too large. When m = 0, cardK0(A)tors is the class number h(F), which is bounded as
follows:
h(F)≤α√Dlog(D)r−1, (25)
for some constant α(r) [11], Theorem 4.4, p. 153. Furthermore, when F =
Q, m = 2n−2 and n is even, the Lichtenbaum conjecture predicts that cardK2n−2(Z) is the order of the numerator of Bn/n, where Bn is the n-th Bernoulli number. The upper bound
Bn≤n!≈nn
suggests, since the denominator ofBn/n is not very big, that cardKm(Z)tors
should be exponential in m. We are thus led to the following:
Conjecture. Fix r ≥ 1. There exists positive constants α, β, γ such that, for any number fieldF of degreer onQ,
cardKm(A)tors≤αexp(βmγlogD).
Furthermore, we expect thatγdoes not depend onr.
5.2
As suggested by A. Chambert-Loir, it is interesting to consider the analog in positive characteristic of the conjecture above. LetX be a smooth connected projective curve of genusg over the finite field withqelements,ζX(s) its zeta function and
P(t) =
2g Y
i=1
(1−αit),
whereαiare the roots of Frobenius acting on the fistℓ-adic cohomology group of X. Whenn > 1, it is expected that the finite group K2n−2(X) has order
the numerator ofζX(1−n), i.e. P(qn−1). Since|αi|=q1/2 for alli= 1· · ·2g, we get
P(qn−1)
≤(1 +qn−1/2)2g
≤q2ng.
5.3
The upper bound fork(m) in Proposition 4 i) is twice exponential inD. One exponential is due to our use of Lemma 3, whereC3is exponential inD. Maybe
this can be improved in general, and not only whens= 0.
The exponential in D occuring in Proposition 4 ii) might be due to our use of the geometry of numbers. Indeed, if one evaluates the class number h(F) by applying naively Minkowski’s theorem (Lemma 1), the bound one gets is exponential inD; see however [8], Theorem 6.5., for a better proof.
5.4
One method to prove (25) consists in combining the class number formula (see (7) and (8)) with a lower bound for the regulator R(F). This suggests replacing the arguments of this paper by analytic number theory, to get good upper bounds for ´etale cohomology.
More precisely, let n ≥ 2 be an integer, and let ζF(1−n)∗ be the leading coefficient of the Taylor series ofζF(s) ats= 1−n. Lichtenbaum conjectured that
ζF(1−n)∗=±2r1R2n−1(F)
Y
p
cardH2(Spec(A[1/p]),Zp(n))
Y
p
cardH1(Spec(A[1/p]),Zp(n))tors
, (26)
whereR2n−1(F) is the higher regulator for the groupK2n−1(F). The equality
(26) is known up a power of 2 whenF is abelian overQ[5], [6], [3].
The order of the denominator on the right-hand side of (26) is easy to evaluate, as well asζF(1−n)∗ (since it is related by the functional equation toζF(n)).
Problem. Can one find a lower bound forR2n−1(F)?
If such a problem could be solved, the equality (26) is likely to produce a much better upper bound for ´etale cohomology than Theorem 3. Zagier’s conjecture suggests that this problem could be solved if one knew that the values of the
n-logarithm onF areQ-linearly independent.
5.5
To illustrate our discussion, letF =Qandn= 5. Then we have
H2(Spec(Z[1/p]),Zp(5))/p=C(p−5),
whereCis the class group ofQ(√p1) modulop, andC(i)is the eigenspace ofC
that C(p−5) = 0 when pis odd. It is true when p≤ 4.106. Theorem 3 and
Proposition 4 tell us that Y
p
H2(Spec(Z[1/p],Zp(5))≤k(8)≤exp exp (70130).
If one could find either a better upper bound for the order ofK8(Z) or a good
lower bound for R9(Q), this would get us closer to the expected vanishing of
C(p−5).
Notice that, using knowledge onK4(Z), Kurihara has proved thatC(p−3)= 0.
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Christophe Soul´e CNRS and IHES 35 route de Chartres 91440 Bures-sur-Yvette France
