If we want to apply our equivalent formulation of the Gross-Kuz’min conjecture to construct fields in which the Gross-Kuz’min conjecture holds, we have two major obstacles:
We do not have a canonical extension Ω+E0/K∞ such that ΩE0 = ΩEΩ+E0. We can still find aZrp extension with this property but it depends on the choice of thep-units we take as generators for the radical.
If we want to verify that (U∞/E∞W)(T∗) is finite it is no longer sufficient to consider ΩE0/ΩE. In fact we have seen above that, if the Gross conjecture is true, there is a non-trivial action of φ(W) on ΩE and it is not clear how one can describe the radical of ΩE/ΩZE and how W acts on it.
Chapter 6
A conditional proof of Leopoldt’s conjecture
The results in this chapter stem from joint work with Preda Mih˘ailescu torwards a proof of Leopoldt’s conjecture. Preda Mih˘ailescu recently gave a much simpler proof of Theorem 6.2.1 not using group cohomology but a simple counting argument instead. We still provide a full proof here as the author is convinced that the analysis of the cohomology groups gives additional insight.
Throughout this section we will assume thatp >2 andKis aCM number field.
LetK∞/Kbe a CM Zp-extension. Let An be the p-class group of the field Kn and defineA∞ = lim∞←nAn. It is well known that A−∞ is a finitely generated Λ-torsion module and pseudo isomorphic to a module of the form
k
M
i=1
Λ/peiM
s
M
j=1
Λ/fj(T)dj
for irreducible distinguished polynomialsfj(T). As in Chapter 1 we define µ(A−∞) = Pk
i=1ei and λ(A−∞) =Ps
j=1deg(fj(T))dj. We will refer to the following conjecture asµ= 0 conjecture:
Conjecture 6.0.1. For any CM Zp-extension of a number field K we have µ(A−∞) = 0.
If K∞ is the cyclotomic Zp-extension and if ζp ∈ K it is well known [Wash, Proposition 13.24] thatµ(A∞) = 0 if and only ifµ(A−∞) = 0. Note that the cyclotomic Zp-extension is the onlyCM Zp-extension of K, if K satisfies Leopoldt’s conjecture (we give the precise statement below). To abbreviate notation we will also write µ andµ− forµ(A∞) and µ(A−∞), respectively.
The second conjecture we want to consider in this chapter is the Leopoldt conjec-ture:
Conjecture 6.0.2 (Leopoldt’s conjecture). Let E be the units of K and E their p-adic closure in U. Then Zp-rank(E) =Z-rank(E).
115
This conjecture can be reformulated as follows.
Conjecture 6.0.3 (Leopoldt’s conjecture – second statement). K admits exactly r2+ 1 independent Zp-extensions, where r2 denotes the number of pairs of complex conjugate embeddings of K.
It is easy to show that each number fields has at least r2+ 1 independent Zp -extensions. Thus, to prove Leopoldt’s conjecture it suffices to prove thatr2+ 1 is an upper bound.
Both conjectures, the Leopoldt conjecture and the µ = 0 conjecture have the property that they remain false under finite extensions of the base field. These results are folklore and were already known by Iwasawa [Iw 2].
Lemma 6.0.4. Let L/K be a finite Galois extension of CM number fields. Assume that the Leopoldt conjecture or the µ= 0 conjecture does not hold for K. Then they do not hold forL.
To analyze the Leopoldt conjecture more carefully we need the following reformu-lation of the Leopoldt conjecture.
Lemma 6.0.5. Let K be aCM number field. Then the Leopoldt conjecture fails for Kif and only if K admits at least twoCM Zp-extensions.
Proof. LetU be the local units ofKas defined in the introduction andE the closure of the image of the group of units of K in U. LetM be the maximal p-ramified, p-abelian extension ofKandHthep-Hilbert class field ofK. Then by [Wash, Corollary 13.6] and the definition ofU
Gal(M/H)∼=U/E.
U has Zp-rank 2r2 = [K : Q] and E has the rank r2 −1−δ, where δ denotes the Leopoldt defect. δ is a non-negative integer and vanishes if and only if the Leopoldt conjecture is true forK. As usual, let jdenote the complex conjugation then E1−j is a finite group and we see thatZp-rank(U1−j/(U1−j∩E)) =Zp-rank(U1−j) =r2. As p6= 2 there is a decomposition U =U1−j ⊕U1+j and we see that there are exactly 1 +δ independent Zp-extensions that are fixed by U1−j. Let M be the compositum of theseZp-extensions then Gal(M/H) is annihilated by (1−j). LetM+⊂M be the maximal subextension of Mthat is abelian over K+, the maximal real subfield ofK.
Then Gal(M+/K+) hasZp-rank 1 +δ and all Zp-extensions inM+Kare indeedCM extensions.
Now we have all ingredients to prove Lemma 6.0.4.
Proof of Lemma 6.0.4. Assume first that the Leopoldt conjecture is false forK. Then K admits at least two CM Zp-extension (one of these extensions is the cyclotomic Zp-extension). AsL/K is finite the same holds forL.
Assume now that the µ= 0 conjecture is false for Kand recall thatH−∞=HA
+∞
∞ . We have a short exact sequence
0→Gal(H−∞(K)L∞/L∞)→Gal(H−∞(K)L∞/K∞)→Gal(L∞/K∞)→0
6.1. RADICALS AND THEIR COHOMOLOGIES 117 As Gal(H−∞(K)/K∞) is a quotient of Gal(H−∞(K)L∞/K∞) and Gal(H−∞(K)/K∞) has infinite p-rank, the same holds for Gal(H−∞(K)L∞/K∞). The term Gal(L∞/K∞) is finite by assumption. Therefore, Gal(H−∞(K)L∞/L∞) has a positive µ-invariant. As Gal(H−∞(K)L∞/L∞) is a quotient of Gal(H−∞(L)/L∞) the same follows for the group Gal(H−∞(L)/L∞).
In view of Lemma 6.0.4 we impose the following conditions on our base field K:
Kis aCM field and Galois overQ.
ζp∈K.
The cyclotomicZp-extension ofKis totally ramified at all ideals abovep.
6.1 Radicals and their cohomologies
Vlad Cri¸san developed in his thesis the theory of projective radicals forZp-free Galois extensions F/K∞ of finite rank [Cr]. In the following we will show that it is also possible to construct projective radicals for extensions of µ-type, i.e. for extensions of finite exponent but infinite rank.
Assume that ζpk ∈ K. Let L/K be a finite Kummer extension of exponent pk. Let K0 ⊂ K be a subfield such that L/K0 and K/K0 are Galois. There is a natural action of Γ = Gal(K/K0) on X = Gal(L/K). Let B be the Kummer-radical of L/K. Then Γ acts naturally on B as well. We write X ↔ B to indicate that B is the Kummer-radical for an extension with Galois group X and vice versa. There is a canonical non-degenerate Kummer pairing
h·,·i:B×X →µpk
such that hρ, xi = x(ρ1/pk)
ρ1/pk . Consider the canonical restriction of the cyclotomic character
χ: Γ→(Z/pkZ)×, γ(ζpk) =ζpχ(γ)k .
Note thathγρ, xi=hρ, χ(γ)γ−1xi (this is a simple reformulation of the equivariance of the Kummer pairing [Gu, Theorem 1.26]). To simplify notation we write γ∗ = χ(γ)γ−1. Thus, ∗ induces an involution on the group ring (Z/pk)[Γ]. With these definitions, we have the following relations:
Lemma 6.1.1. Let f ∈(Z/pk)[Γ]. Thenf:X→X induces a group homomorphism and we obtain
|X|=|f X| · |X[f]|=|B|=|B[f∗]| · |f∗B|, (6.1)
B[f∗]↔X/f X (6.2)
B/f∗B ↔X[f]. (6.3)
Proof. Since X ∼= B as Zp-modules, we obviously have |X| = |B|. The other two equalities in (6.1) follow directly from the isomorphism theorem for finite modules.
Next we prove (6.2). Let L0 = Lf X be the field fixed by f X. Let Bf ⊂ B be the Kummer radical ofL0. For every f x∈f X and everyρ∈Bf we obtain
1 =hρ, f xi=hf∗ρ, xi
As this holds for allx∈X we see thatXacts trivially onK((f∗Bf)1/pk). The pairing is non-degenerate and we obtain that f∗Bf = {0} as well as Bf ⊂ B[f∗]. Let now ρ∈B[f∗] then
hρ, f xi=hf∗ρ, xi= 1.
So f X acts trivially on K(B[f∗]1/pk) which implies B[f∗]⊂Bf. Hence, Bf =B[f∗] and (6.2) follows.
For (6.3) define L0 = LX[f] and let Bf0 ⊂ B be the radical of L0/K. There is a natural homomorphism
B→L0/L0pk,
whose kernel is Bf0. Therefore, X[f]↔ B/Bf0. It remains to show that Bf0 = f∗B.
Letf∗ρ∈f∗B and x∈X[f] then
hf∗ρ, xi=hρ, f xi= 1.
As this holds for allρ ∈ B we see that X[f] acts trivially on K((f∗B)1/pk). Hence, f∗B ⊂Bf0. Let ˜L=K((f∗B)1/pk)⊂L0 and let Y ⊂X be the fixing group of ˜L. Let f∗ρ∈f∗B and x∈Y. Then
1 =hf∗ρ, xi=hρ, f xi
As this holds for allρ, the non-degeneracy of the pairing implies thatf x= 0. Hence, Y ⊂ X[f] and L0 = LX[f] ⊂ L˜ = LY. We showed already that ˜L ⊂ L0. Thus, we obtain equality and indeed (6.3).
Assume for the remainder of this section thatK/K0 is cyclic. Letγ be a generator.
Then we can write the algebraic norm as N =P[K:K0]
i=1 γi and ∆ =γ −1. The Tate cohomologies ofHbi(Γ, X), i= 0,1 are defined as usual:
Hb0(Γ, X) = X[∆]
N X , Hb1(Γ, X) = X[N]
∆X .
To use the correspondence we proved in Lemma 6.1.1 we define the cohomologies for B with respect to the twisted action:
Hb0(Γ, B) = B[∆∗]
N∗B , Hb1(Γ, B) =B[N∗]
∆∗B . With these definition, we deduce from Lemma 6.1.1
6.1. RADICALS AND THEIR COHOMOLOGIES 119 Corollary 6.1.2. We have the following relations:
Hb0(Γ, B)↔Hb1(Γ, X) and Hb1(Γ, B)↔Hb0(Γ, X). (6.4) Proof. Consider the fieldsL0 =LX[∆]⊆L00=LN X. We obtain
Gal(L00/L0)∼=X[∆]/N X ∼=Hb0(Γ, X). (6.5) In view of (6.2) we have,
Gal(L00/K)↔B[N∗].
There is a natural homomorphism
B→L0/L0p
k
. (6.6)
By (6.3) we know thatX[∆]↔B/∆∗B. Therefore, the kernel of the natural projec-tion (6.6) is ∆∗B and we obtain
Gal(L0/K)↔∆∗B.
Hence,
Gal(L00/L0)↔B[N∗]/∆∗B ∼=Hb1(Γ, B).
Together with (6.5) we obtainHb1(Γ, B)↔Hb0(Γ, X).
For the second implication consider L0=LX[N]⊆L00=L∆X. Then Gal(L00/L0)∼=X[N]/∆X=Hb1(Γ, X)
We know from (6.2) that Gal(L00/K)∼=X/∆X↔B[∆∗]. From (6.3) we have X[N]↔B/N∗B.
Hence,N∗B is the kernel of the natural mapB →L0/L0p
k. Therefore, Gal(L0/K)↔N∗B.
We obtain
Gal(L00/L0)↔B[∆∗]/N∗B ∼=Hb0(Γ, B).
We want to apply these general results to indicate that we can define norm co-herent radicals forµ-type subfields ofH∞/K∞. LetHµbe the maximal subextension of H−∞ such that X = Gal(Hµ/K∞) is of finite exponent and does not contain a finite Λ-submodule. Let Mk be the maximal p-abelian p-ramified extension of Kk
and M−k the fixed field under the pluspart. We define Hk =Hµ∩M−k for all k and Xk= Gal(Hk/Kk). Note thatXk=X/ωkX. The multiplication with νk,k+v induces a homomorphism
ψk,k+v:Xk→Xk+v.
AsX does not contain νk,k+v-torsion this homomorphism is injective. Let φk,k+v:Xk+v →Xk
be the natural restriction homomorphism. Let
·νk,k+v:Xk+v →Xk+v
be the homomorphism onXk+vsendingxtoνk,k+vx. Clearly·νk,k+v =ψk,k+v◦φk,k+v. Asψk,k+v is injective and ker(φk,k+v) =ωkXk+v we see that ker(·νk,k+v) =ωkXk+v.
The exponent ofXkis uniformly bounded bypl. Assume that there is an indexn0
such thatKn contains µpl for alln≥n0. Note that ωn≡ −τpnω∗n mod pn, whereτ is a topological generator of Gal(K∞/K). So we can choosen0 large enough such that ωn ≡ −τpnω∗n mod pl for all n≥n0. Let Bn be the radical of Hn/Kn. As K∞/Kn
is disjoint toHn/Kn we see that the natural homomorhismBn→Bn0 is injective for alln0 ≥n≥n0.
Lemma 6.1.3. Forn0 ≥n > n0 we haveBn0[ωn] =Bn andBn=Nn0,n(Bn0).
Proof. Let ρ ∈ Kn0 be a representative of a class z in Bn0. As by Kummer duality Bn1−j0 = 1 and p 6= 2 we can assume that ρ1−j = 1. Then Nn0,n(ρ) ∈ Kn and Wn = Kn(Nn0,n(ρ)1/pl)/Kn is abelian over Kn. As Hn0/Kn is Galois we see that Wn ⊂ Hn0. Hence, Wn/Kn is unramified outside p and WnK∞ ⊂ Hµ. Therefore, Wn⊂Hn. This shows that Nn0,n(Bn0)⊂Bn for all possible choices ofnand n0.
We have already seen that ker(νn,n0 |Xn0) =ωnXn0. As νn,n0 =P
g∈Gal(Kn0/Kn)g it follows that Hb1(Gal(Kn0/Kn), Xn0) = 0. By Lemma 6.1.2 this implies that the groupHb0(Gal(Kn0/Kn), Bn0) is trivial. Thus,Bn0[ω∗n] =Nn∗0,n(Bn0). Asωn≡ −τpnωn∗ modpl and νn,n0 =P
g∈Gal(Kn0/Kn)g≡P
g∈Gal(Kn0/Kn)χ(g−1)g=νn,n∗ 0 we see that Bn0[ωn] =Bn0[ω∗n] =Nn0,n(Bn0).
Therefore,
Bn⊂Bn0[ωn] =Nn0,n(Bn0)⊂Bn
from which the claim is immediate.
Remark 6.1.4. The condition that Kn contains µpl for n large enough is trivially satisfied for the cyclotomic Zp-extension of a number field containing the p-th roots of unity. If K∞ is not the cyclotomic Zp-extension, i.e. a Leopoldt defect extension then the extensionK∞(ζpl)/K∞ is finite. So we can just replaceKby K(ζpl) andK∞
by K∞(ζpl).