A class number formula for higher derivatives of abelian L-functions

A. Hayward

Abstract

Gross and Rubin have made conjectures about special values of equivariant L-functions associated to abelian extensions of global fields. We describe a common refinement, due to Burns, and give evidence in favour of this conjecture for quadratic extensions and cyclotomic fields. We also note that the statement provides a new interpretation of further conjectures of Darmon and Gross.

1. Introduction

For K/k a finite abelian extension of global fields with Galois group G, and S and T finite disjoint sets of places of k such that S contains all infinite places, one defines an equivariant L-function Θ_{K/k,S,T} (s) for s ∈ ℂ, valued in ℂ[G]. When K = k, this is the (S-truncated, T -modified) zeta function of the field k. Dirichlet’s analytic class number formula tells about the properties of this zeta function at the point s = 0, specifically its order of vanishing and its leading term. This latter is the product of a transcendental “regulator” term, formed from the units of k, with h_{k,S,T} /w_{k,S,T}, a ratio of integer invariants related to the arithmetic in the field k.

Towards the end of the 1970s, Stark conjectured analogues of these properties for more general Lfunctions. In particular, for abelian extensions he proposed an integrality statement for Θ′_{K/k,S,T} (0). Of the work which followed this, we note the paper [Ru96] of Rubin, where he made a conjecture which extended Stark’s to higher orders of vanishing. Rubin’s conjecture has the property that it tends to strengthen as the order of vanishing increases; indeed for the zeroth derivative Θ_{K/k,S,T} (0), where it states Θ_{K/k,S,T} (0) ∈ ℤ[G], it follows easily from theorems of Deligne, Ribet and Weil. On the other hand, Gross in [Gr88] had made a conjecture of a different kind for this very element, in which he relates it to the class number h_{k,S,T} and a certain group-ring valued regulator. However, for higher orders of vanishing Gross’s conjecture becomes trivial.

In this paper we study a conjecture of Burns, 2.6 below, which unites these two approaches. It represents a strengthening of Rubin’s conjecture which is precisely in the spirit of Gross, and it specialises to Gross’s conjecture for the zeroth derivative Θ_{K/k,S,T} (0). The formulation was inspired by work in [Bu01], where it is shown that the Equivariant Tamagawa Number Conjecture, as formulated by Burns and Flach in [BF01], implies, for a certain class of extensions, a stronger variant of this Conjecture 2.6. The statement here proposes a generalisation of this to arbitrary abelian extensions.

In sections 2 to 4, we state the conjecture and give some elementary properties and special cases, including a proof for quadratic extensions. We then go on to use the theory of Dirichlet L-functions and cyclotomic fields to study the conjecture for certain extensions of number fields. In particular, we give evidence in the case of a real abelian extension of ℚ.

2000 Mathematics Subject Classification 11R42 (primary) 11R29 (secondary)

The author was funded during this work by an EPSRC doctoral training grant.

A. Hayward

We also discuss two conjectures bearing a family resemblance to Gross’s but concerning “minusunits” relative to a quadratic extension. These are due to Darmon ([Da95], dealing with an explicit “circular unit” related to first derivatives of L-functions) and Gross ([Gr88], Conjecture 8.8, which has more general hypotheses and concerns the values of the L-functions). In each context we interpret these conjectures as rather striking “base-change”-type statements for Burns’s conjecture, which transport it from an extension L̃/k to an extension L̃K/K, where K is a quadratic extension of k.

It is a pleasure to thank David Burns for his help and guidance while writing this paper, and Radan Kuc̆era for many useful comments.

2. Notation and Statement

2.1 Basic Setup Let F be a global field, S a nonempty finite set of places of F containing all archimedean places.

We define O_{F,S} := {α ∈ F : v(α) ⩾ 0 for all v ∉ S}, the ring of S-integers of F, and U_{F,S} = O_{F,S}^{×}, the S-units. The S-class group A_{F,S} is defined to be the Picard group of O_{F,S}, and fits into the exact sequence

0 ⟶ U_{F,S} ⟶ F^{×} ⟶ ⨁_{𝔭∉S}𝔭^{ℤ} ⟶ A_{F,S} ⟶ 0.  (1)

Now let T be a finite set of places of F, disjoint from S. The subgroup of U_{F,S} consisting of those S-units congruent to 1 modulo every prime in T is denoted U_{F,S,T}. The S ray-class group modulo T, denoted A_{F,S,T}, is the quotient of the group of fractional ideals of O_{F,S} prime to T by the subgroup of principal ideals with a generator congruent to 1 modulo each prime in T. The class groups fit into an exact sequence

0 ⟶ U_{F,S,T} ⟶ U_{F,S} ⟶ ∏_{𝔭∈T} 𝔽_{𝔭}^{×} ⟶ A_{F,S,T} ⟶ A_{F,S} ⟶ 0,  (2)

where 𝔽_{𝔭} denotes the residue field of F at 𝔭. For any finite place 𝔭 of F, we let N_{𝔭} be the size of 𝔽_{𝔭}. Define the S- and (S, T)- class numbers h_{F,S} = #A_{F,S}, h_{F,S,T} = #A_{F,S,T}. Then

h_{F,S,T} = h_{F,S} · \frac{∏_{𝔭∈T}(N𝔭 − 1)}{(U_{F,S} : U_{F,S,T})} = h_{F,S}(∏_{𝔭∈T} 𝔽_{𝔭}^{×} : Ũ_{F,S}),  (3)

where Ũ_{F,S} denotes the image of U_{F,S} in the residue fields.

For the rest of §2, we fix an abelian extension of global fields K/k with Galois group G, and a nonnegative integer r. Let S = S_{k} and T = T_{k} be finite sets of places of k, and define S_{K} and T_{K} to be the sets of places of K dividing places in S_{k} and T_{k} respectively. We will abbreviate U_{K,S_{K},T_{K}} to U_{K,S,T}, and do similarly for the class groups and class numbers. Let S_{1} be a subset of S. We assume S, S_{1} and T satisfy

Hypothesis 2.1. i) S contains all archimedean places of k;

ii) S contains the places which ramify in K/k;

iii) S_{1} consists of r places which split completely in K/k;

iv) #S ⩾ r + 1;

v) T ∩ S = ∅ and U_{K,S,T} is torsion-free.

We write #S = r + d + 1, so U_{k,S,T} is a free abelian group of rank r + d. Note that our setup closely follows [Ru96].

Abelian L-functions

2.2 The equivariant L-function For a finite unramified place v of k let Frob_{v} be the (arithmetic) Frobenius of the residue extension corresponding to w/v for a place w of K dividing v. Because K/k is abelian, this is a well-defined element of G.

For a character χ of the Galois group G of the extension K/k, write

e_{χ}(K/k) := \frac{1}{#G} ∑_{g∈G} χ(g)g^{−1}

for the corresponding idempotent in ℂ[G]. Define the S-truncated abelian (Artin) L-function of χ by

L_{K/k,S}(s, χ) = ∏_{v∉S_{k}}(1 − χ(Frob_{v})N_{v}^{−s})^{−1}.

The product converges for Re s > 1 and it is well-known that the function can be meromorphically extended to all of ℂ. The L-functions combine to give the S-truncated, T -modified equivariant L-function ℂ ⟶ ℂ[G], as defined, for example, in [Ta84], Chapter IV, §1:

Θ_{K/k,S,T} (s) := ( ∏_{t∈T_{k}}(1 − N_{t}^{1−s}Frob_{t}^{−1})) ∑_{χ∈Ĝ} L_{K/k,S_{k}}(s, χ^{−1})e_{χ}(K/k)
= ( ∏_{t∈T_{k}}(1 − N_{t}^{1−s}Frob_{t}^{−1}))( ∏_{v∉S_{k}}(1 − N_{v}^{−s}Frob_{v}^{−1}))^{−1}.  (4)

Because of the assumption that r places in S_{k} split completely in K/k, we see by Proposition I.3.4 of [Ta84] that each L_{K/k,S_{k}}(s, χ^{−1}) vanishes to order at least r at s = 0. Write e_{r} = ∑_{χ} e_{χ}, where the sum is over all those characters for which the order of vanishing is exactly r. The r-th term in the Taylor expansion is given by

Θ_{K/k,S,T}^{r} (0) := lim_{s→0} s^{−r}Θ_{K/k,S,T} (s).

It satisfies Θ_{K/k,S,T}^{r} (0) = Θ_{K/k,S,T}^{r} (0)e_{r}.

2.3 Special values and units We set Y_{S_{K}} := {∑_{v∈S_{K}} n_{v}v : n_{v} ∈ ℤ}, the free abelian group on S_{K}, and X_{S_{K}} := {∑_{v∈S_{K}} n_{v}v ∈ Y_{S_{K}} : ∑_{v∈S_{K}} n_{v} = 0} its augmentation subgroup.

Define absolute values at places v of K as follows:

|a|_{v} = \begin{cases}|a| & if K_{v} = ℝ, \\ |a|^{2} & if K_{v} = ℂ, \\ N_{v}^{−v(a)} & for v a finite place, \end{cases}

where the valuation v is normalised so that its image is ℤ.

For any ℤ[G]-module M and any ring R, RM := R ⊗_{ℤ}M will denote the R[G]-module obtained from M by extending scalars to R. The logarithmic regulator map is defined by:

\begin{aligned}λ_{S_{K}} : & U_{K,S} & ⟶ & ℝX_{S_{K}} \\ & u & ⟼ & − ∑_{v∈S_{K}} ln |u|_{v}v \end{aligned}

It is well-known that this induces an ℝ[G]-module isomorphism ℝU_{K,S} → ℝX_{S_{K}}. Its extension to a map ∧_{G}^{n}U_{K,S} ⟶ ℝ ∧_{G}^{n} X_{S_{K}} will be written λ_{S_{K}}^{(n)}.

Let us recall the analytic class number formula of Dirichlet. U_{k,S,T} has ℤ-rank r + d, the same as X_{S_{k}}. Choose a basis u_{1}, …, u_{r+d} for U_{k,S,T} modulo torsion. Order the elements of S_{k} as v_{1}, …, v_{r+d+1}; then v_{1} − v_{r+d+1}, …, v_{r+d} − v_{r+d+1} is a basis for X_{S_{k}}. The map λ_{S_{k}}^{(r+d)} gives us a real determinant with respect to these bases. The determinant may be calculated as

R_{k,S,T} = det ( − ln |u_{i}|_{v_{j}})_{1⩽i,j⩽r+d}.

The choice of the ordering of S_{k} only affects the determinant up to sign. In this paper we will choose to ignore systematically all questions of signs of regulators.

Dirichlet’s analytic class number formula (see [Gr88], (1.6)) says that the meromorphic function ζ_{k,S,T} has a zero of exact order #S_{k} − 1 at 0, and that the coefficient of the leading term in the Taylor expansion here is

−\frac{h_{k,S,T} |R_{k,S,T} |}{#(U_{k,S,T})_{tors}}.

We now relate Θ_{K/k,S,T}^{r} (0) to the S-units of K. Let W be an r-tuple (w_{1}, …, w_{r}) where w_{i} is a place of K chosen above v_{i} ∈ S_{1,k}. Define w_{i}^{∗} ∈ Hom_{G}(Y_{S_{K}}, ℤ[G]) on w′ ∈ S_{K} by w_{i}^{∗}(w′) = ∑_{γ} γ, summed over the elements γ of G with γw_{i} = w′. Set W^{∗} = w_{1}^{∗} ∧ ⋯ ∧ w_{r}^{∗} ∈ ∧_{G}^{r}Hom_{G}(Y_{S_{K}}, ℤ[G]). Then Remark 2 of [Po99], §1.6 shows that

W^{∗} ∘ λ_{S_{K}}^{(r)} : (ℂ ∧_{G}^{r} U_{K,S,T})e_{r} ⟶ ℂ[G]e_{r}

is a ℂ[G]-isomorphism. Hence there is a unique element^{1}

η = η_{K/k,S,T,r,W} ∈ ℂ ∧_{G}^{r} U_{K,S,T} e_{r}

such that W^{∗} ∘ λ_{S_{K}}^{(r)}(η) = Θ_{K/k,S,T}^{r} (0). If we choose another place w ∈ S_{K} − S_{1,K} and set b := (w_{1} − w) ∧ ⋯ ∧ (w_{r} − w), then we have

Θ_{K/k,S,T}^{r} (0) ∧_{G}^{r} X_{S_{K}} = Θ_{K/k,S,T}^{r} (0)ℤ[G]b, λ_{S_{K}}^{(r)}(η) = Θ_{K/k,S,T}^{r} (0)b.

We refer to [Ru96], Lemma 2.6(ii) and the proof of Proposition 2.4, for the proof of this.

We are interested in integrality properties of this η_{K/k,S,T,r,W}, which we will test using elements Φ ∈ ∧_{G}^{r}Hom_{ℤ[G]}(U_{K,S,T} , ℤ[G]). It will always suffice for our purposes to assume Φ is a primitive tensor ϕ_{1} ∧ ⋯ ∧ ϕ_{r} (or 1 ∈ ℤ[G] if r = 0), by linearity of our statements. Then Φ(u_{1} ∧ ⋯ ∧ u_{r}) means det(ϕ_{j}(u_{i}))_{i,j}. The element Φ induces a ℂ-linear map ℂ ∧_{G}^{r} U_{K,S,T} ⟶ ℂ[G], and we consider Φ(η) ∈ ℂ[G]. We propose to strengthen the following conjecture, which is Conjecture B′ of [Ru96]:

Conjecture 2.2 (Rubin). For every Φ ∈ ∧_{G}^{r}Hom_{ℤ[G]}(U_{K,S,T} , ℤ[G]), we have Φ(η) ∈ ℤ[G].

2.4 Formulation of the conjecture Let aug : ℤ[G] ⟶ ℤ be the augmentation homomorphism, and write I_{G} for its kernel, the augmentation ideal of ℤ[G]. Assume Rubin’s conjecture holds. Burns’s conjecture puts further conditions on the group ring element Φ(η), by proposing a congruence for Φ(η) modulo I_{G}^{d+1}.

For G any abelian group and M, N any ℤ[G]-modules, one may make the group Hom_{G}(M, N) := Hom_{ℤ[G]}(M, N) into a ℤ[G]-module with G-action given by (gα)(m) = gα(m) for g ∈ G, α ∈ Hom_{G}(M, N), m ∈ M. There is a canonical isomorphism of abelian groups

\begin{matrix}Hom_{ℤ}(M, ℤ) & ⟶ & Hom_{G}(M, ℤ[G]) \\ ϕ^{1} & ⟼ & (x ↦ ∑_{g∈G} ϕ^{1}(g^{−1}x)g). \end{matrix}  (5)

We write ϕ ↦ ϕ^{1} for the inverse of this isomorphism.

1Rubin in [Ru96] denotes this by ε instead of η. Throughout, we will omit subscripts from η which are clear from the context.

In §1.2 of [Ru96], Rubin observes that for any ℤ-module M and any n > 0, every h ∈ Hom_{ℤ}(M, ℤ) induces a homomorphism

\begin{aligned}h̃ : & ∧_{ℤ}^{n}M & ⟶ & ∧_{ℤ}^{n−1}M \\ & m_{1} ∧ ⋯ ∧ m_{n} & ⟼ & ∑_{i=1}^{n}(−1)^{i+1}h(m_{i})m_{1} ∧ ⋯ ∧ m̂_{i} ∧ ⋯ ∧ m_{r}, \end{aligned}

where ˆ means “omit”. This construction can be iterated to obtain

\begin{matrix}∧^{r}Hom_{ℤ}(M, ℤ) & ⟶ & Hom_{ℤ}(∧^{n}M, ∧^{n−r}M) \\ h_{1} ∧ ⋯ ∧ h_{n} & ⟼ & h̃_{1} ∘ ⋯ ∘ h̃_{n}. \end{matrix}

If Φ = ϕ_{1} ∧ ⋯ ∧ ϕ_{r} ∈ ∧_{ℤ[G]}^{r}Hom_{ℤ[G]}(U_{K,S,T} , ℤ[G]), we define Φ̃ to be the map from ∧_{ℤ}^{r+d}U_{k,S,T} to ∧_{ℤ}^{d}U_{k,S,T} thus obtained from ϕ_{1}^{1}, …, ϕ_{r}^{1}.

Definition 2.3. Let r ⩽ n be nonnegative integers. Define a set of permutations

\left[{n \atop r}\right] := \{σ ∈ S_{n} : \begin{matrix}σ(1) < σ(2) < … < σ(r) and \\ σ(r + 1) < σ(r + 2) < … < σ(n) \end{matrix} \} .

Note that the cardinality of this set is the binomial coefficient \binom{n}{r}. Each element corresponds to choosing a subset of r elements from {1, …, n}, and associates to it a sign, sign(σ).

Lemma 2.4. We have the formula

Φ̃(u_{1} ∧ ⋯ ∧ u_{r+d}) = ∑_{σ∈\left[{r+d \atop r}\right]} sign(σ) det (ϕ_{j}^{1}(u_{σ(i)}))_{1⩽i,j⩽r}u_{σ(r+1)} ∧ ⋯ ∧ u_{σ(r+d)}.

The proof is routine.

Now, following Gross, we define a group ring-valued regulator. Let the places in S_{k} − S_{1,k} be denoted v′_{1}, …, v′_{d+1}. For each v′_{i}, local class field theory gives us a local reciprocity map

f_{v′_{i}} : k^{×} ⟶ G,

coming from the reciprocity map in the local extension K_{w}/k_{v′_{i}} for a place w of K above v′_{i}. We compose this with the isomorphism

\begin{matrix}G & ⟶ & I_{G}/I_{G}^{2} \\ g & ⟼ & g − 1 \end{matrix}

to get a homomorphism to the additive group I_{G}/I_{G}^{2}. We now define the Gross-style regulator homomorphism (cf. [Gr88], (2.2)):

Reg_{G} = Reg_{K/k,S,r}^{(v′_{1},...,v′_{d})} : ∧_{ℤ}^{d}U_{k,S,T} ⟶  ℤ[G]/I_{G}^{d+1}
u_{1} ∧ ⋯ ∧ u_{d} ⟼ det (f_{v′_{j}}(u_{i}) − 1)_{1⩽i,j⩽d}.

We will vary the subscripts of Reg according to any clarification needed in context. Note that here we have chosen to exclude v′_{d+1}. So we need:

Proposition 2.5. The homomorphism above does not depend on the choice of which of the v′_{i} to exclude, or the ordering of the v′_{i}, up to sign.

Proof. By the product formula of global class field theory, we have ∏_{v∈S} f_{v}(x) = 1 for all x ∈ O_{k,S}^{×}. Because v ∈ S_{1} split completely in K/k, f_{v}(x) = 1 for these v. Hence

∏_{v′∈S−S_{1}} f_{v′}(x) = 1.

Now choose j ∈ {1, …, d}. In the determinant Reg_{G}(u_{1} ∧⋯∧u_{d}), adding every other column to the column corresponding to v′_{j} and using the product formula shows that the i-th entry in column j is congruent to −(f_{v′_{d+1}}(u_{i})−1) (mod I_{G}^{2}). So the determinant becomes −Reg_{G}(u_{1}∧⋯∧u_{d}) calculated with respect to the places v′_{1}, …, v′_{j−1}, v′_{d+1}, v′_{j+1}, …, v′_{d}. Reordering these can only change the sign again.

Let u_{1}, …, u_{r+d} be a ℤ-basis for U_{k,S,T}. We set Reg_{G}^{Φ} = Reg_{G}(Φ̃(u_{1} ∧ ⋯ ∧ u_{r+d})), and note that this is independent of the choice of basis up to sign. We have by Lemma 2.4 that

Reg_{G}^{Φ} = ∑_{σ∈\left[{r+d \atop r}\right]} sign(σ) det (ϕ_{j}^{1}(u_{σ(i)}))_{1⩽i,j⩽r} det (f_{v′_{j}}(u_{σ(r+i)}) − 1)_{1⩽i,j⩽d},

where \left[{r+d \atop r}\right] was defined in Definition 2.3. The conjecture we will discuss is as follows.

Conjecture 2.6 (Burns). Let K/k, S ⊇ S_{1}, T, r satisfy Hypothesis 2.1. Assume that Rubin’s conjecture holds for this data, so that for every Φ ∈ ∧_{ℤ[G]}^{r}Hom_{ℤ[G]}(U_{K,S,T} , ℤ[G]), we have Φ(η) ∈ ℤ[G]. Then this element satisfies

Φ(η) ≡ ±h_{k,S,T} Reg_{G}^{Φ} (mod I_{G}^{d+1}).

Note that this conjecture implies Φ(η) ∈ I_{G}^{d}, an “order of vanishing” statement which generalises [Gr88] (4.2) (via Proposition 3.9 below). For more on the formulation of Conjecture 2.6, including a method for specifying the sign in the congruence, see [Bu03].

3. Basic properties of the conjecture

3.1 Varying the data Firstly, we wish to check that Conjecture 2.6 will remain true if we lower the top field K. We note a useful result about the unit groups.

Lemma 3.1. For any K/k, S, T such that K/k is Galois and U_{K,S,T} is torsion-free, the quotient U_{K,S,T} /U_{k,S,T} is also torsion-free.

Proof. Suppose u ∈ U_{K,S,T} is such that u^{n} ∈ U_{k,S,T} for some n > 0. This means that for all σ ∈ Gal(K/k), we have (u^{n})^{σ−1} = 1. But this is (u^{σ−1})^{n}. Hence for all σ ∈ Gal(K/k), u^{σ−1} is a torsion element of U_{K,S,T} and so is 1. Hence u ∈ k as required.

Proposition 3.2. Let L/K/k be a tower of finite extensions, with L/k and K/k abelian with groups Γ and G = Γ/H respectively. If Conjecture 2.6 holds for L/k, S ⊇ S_{1}, T then it holds for K/k, S ⊇ S_{1}, T.

Proof. It is clear, using Proposition IV.1.8 of [Ta84], that η_{K/k} = (∧^{r}N_{L/K})η_{L/k}.

The inclusion U_{K,S,T} ↪ U_{L,S,T} and the ℤ[G]-module isomorphism

\begin{matrix}ℤ[Γ]^{H} & ⟶^{∼} & ℤ[G] \\ N_{L/K} & ⟼ & 1 \end{matrix}

induce a surjective map

Hom_{Γ}(U_{L,S,T} , ℤ[Γ]) ⟶ Hom_{G}(U_{K,S,T} , ℤ[G])

whereby each ϕ in the second group can be lifted to a ϕ̂ in the first in such a way that the projection of ϕ̂(u) to ℤ[G] is ϕ(N_{L/K}u) for all u ∈ U_{L,S,T}. This follows by applying [Ru96], Proposition 1.1, to the exact sequence of ℤ-torsion free Γ-modules given by Lemma 3.1, and using [Ru96], diagram (16).

Take Φ ∈ ∧_{G}^{r}Hom_{G}(U_{K,S,T} , ℤ[G]) and lift it to Φ̂ ∈ ∧_{Γ}^{r}Hom_{Γ}(U_{L,S,T} , ℤ[Γ]) componentwise. Now

Φ̂(η_{L/k}) ≡ ±h_{k,S,T} Reg_{Γ}(˜Φ̂(u)) (mod I_{Γ}^{d+1}),

where u = u_{1} ∧ ⋯ ∧ u_{r+d}, u_{i} a ℤ-basis of U_{k,S,T}. Passing to the quotient in this congruence, and noting that ˜ Φ̂ = Φ̃, we get

Φ(η_{K/k}) ≡ ±h_{k,S,T} Reg_{G}(Φ̃(u)) (mod I_{G}^{d+1}),

as required.

We now look at enlarging T.

Proposition 3.3. Suppose Conjecture 2.6 holds for K/k, S ⊇ S_{1}, T, and v is a place of k not in S or T. Set T′ = T ∪ {v}. Then Conjecture 2.6 holds for K/k, S ⊇ S_{1}, T′.

Proof. The definition of Θ_{K/k,S,T} shows that η_{T′} = η_{T}^{1−N_{v}Frob_{v}^{−1}}.

We will adapt [Po02], proof of Proposition 5.3.1. Let ϕ′_{1}, …, ϕ′_{r} be in Hom_{G}(U_{K,S,T′}, ℤ[G]) and set Φ′ = ϕ′_{1} ∧ ⋯ ∧ ϕ′_{r}. Popescu proves that there exist ϕ_{i} ∈ Hom_{G}(U_{K,S,T} , ℤ[G]), α_{i} ∈ ℤ[G] and ϕ_{0} ∈ Hom_{G}(U_{K,S,T′}, ℤ[G]) such that ϕ′_{i} = ϕ_{i} + α_{i}ϕ_{0} for all i = 1, …, r. Let δ_{v} = 1 − N_{v}Frob_{v}^{−1}, then it is clear that the map δ_{v}ϕ_{0} : x ↦ ϕ_{0}(δ_{v}x) is in Hom_{G}(U_{K,S,T} , ℤ[G]). Popescu shows that

Φ′(η_{T′}) = Ψ(η_{T}),  (6)

where Ψ ∈ ∧_{G}^{r}Hom_{G}(U_{K,S,T} , ℤ[G]) is given by

δ_{v}(ϕ_{1} ∧ ⋯ ∧ ϕ_{r}) + ∑_{i=1}^{r} α_{i}ϕ_{1} ∧ ⋯ ∧ ϕ_{i−1} ∧ δ_{v}ϕ_{0} ∧ ϕ_{i+1} ∧ ⋯ ∧ ϕ_{r}.  (7)

Now let u = u_{1} ∧ ⋯ ∧ u_{r+d} be the wedge of a basis of U_{k,S,T} and u′ = u′_{1} ∧ ⋯ ∧ u′_{r+d} similarly for U_{k,S,T′}. We have (U_{k,S,T} : U_{k,S,T′})u = u′ in ∧_{ℤ}^{r+d}U_{k,S,T}. Apply Ψ̃ to both sides of this equality. We note that (δ_{v}ϕ_{0})^{1}(u′_{i}) = aug(δ_{v})ϕ_{0}^{1}(u′_{i}) since u′_{i} ∈ U_{k,S,T′}. Note also that aug(δ_{v}) = −(N_{v} − 1). By the form of (7), this shows that

(U_{k,S,T} : U_{k,S,T′})Ψ̃(u) = −(N_{v} − 1)Φ̃′(u′) in ∧_{ℤ}^{d}U_{k,S,T}.

But (U_{k,S,T} : U_{k,S,T′}) divides (N_{v} − 1). Since the group in which the equality holds is torsion-free, we may cancel (U_{k,S,T} : U_{k,S,T′}) from both sides. Furthermore, by (3),

h_{k,S,T′} = \frac{(N_{v} − 1)}{(U_{k,S,T} : U_{k,S,T′})}h_{k,S,T} .

So we have h_{k,S,T} Ψ̃(u) = −h_{k,S,T′} Φ̃′(u′) in ∧_{ℤ}^{d}U_{k,S,T}. Applying Reg_{G} to this and using (6) gives the result.

We next look at changing S. We will use:

Lemma 3.4. Let k, S, T be such that U_{k,S,T} is torsion-free. Suppose u_{1}, …, u_{n} is a ℤ-basis for U_{k,S,T}. Let v be a place of k not in S or T. Take u′ ∈ U_{k,S∪{v},T} such that v(u′) is minimal positive. Then u_{1}, …, u_{n}, u′ is a basis for U_{k,S∪{v},T}.

Proof. Let u ∈ U_{k,S∪{v},T}. Then there exists a ∈ ℤ such that v(u) = av(u′). Then u/u′^{a} ∈ U_{k,S,T}, so we see u_{1}, …, u_{n}, u′ generates U_{k,S∪{v},T}. Linear independence follows from considering the valuations at v.

Lemma 3.5. Using the notation of the previous Lemma, we have

h_{k,S∪{v},T} · v(u′) = h_{k,S,T} .

Proof. Write S′ = S ∪ {v} for short. The result follows from the analytic class number formula as follows. If n = #S − 1, we have

ζ_{k,S′,T} (s) ≡ h_{k,S′,T} R_{k,S′,T} s^{n+1} (mod s^{n+2}),
ζ_{k,S,T} (s) ≡ h_{k,S,T} R_{k,S,T} s^{n} (mod s^{n+1}),

and the leading terms are related by

h_{k,S′,T} R_{k,S′,T} = lim_{s→0} \frac{1 − N_{v}^{−s}}{s} h_{k,S,T} R_{k,S,T} .

Hence h_{k,S′,T} R_{k,S′,T} = (ln N_{v})h_{k,S,T} R_{k,S,T}. On the other hand, the definition of the regulator, and the fact that v(u_{i}) = 0 for i = 1, …, n shows that R_{k,S′,T} = ln |u′|_{v}R_{k,S,T} = v(u′)(ln N_{v})R_{k,S,T}. This gives the result.

Proposition 3.6. Let K/k, S, T, r be data satisfying Hypothesis 2.1, and let v be a place of k not in S or T and which splits completely in K/k. Set S′ = S ∪ {v} and S′_{1} = S_{1} ∪ {v}. Suppose Conjecture 2.6 holds for K/k, S′ ⊇ S′_{1}, T, r + 1. Then Conjecture 2.6 holds for K/k, S ⊇ S_{1}, T, r.

Proof. Choose bases as in Lemma 3.4. Note that “n”= r + d in the notation of that lemma, and define u_{r+d+1} = u′. We choose w above v to go into W. By [Ru96], Proposition 5.2, we have η_{S} = w̌(η_{S′}), where w̌ ∈ Hom_{G}(U_{K,S′,T} , ℤ[G]) is defined by

w̌(u) = ∑_{g∈G} w(g^{−1}u)g.

Take Φ ∈ ∧_{G}^{r}Hom_{G}(U_{K,S,T} , ℤ[G]). The hypothesis that U_{K,S,T} and U_{K,S′,T} are torsion-free implies that the map Hom_{G}(U_{K,S′,T} , ℤ[G]) → Hom_{G}(U_{K,S,T} , ℤ[G]) (restriction) is surjective (cf. [Ru96], Proposition 1.1 (ii)). So we may lift Φ componentwise to Φ′ ∈ ∧_{G}^{r}Hom_{G}(U_{K,S′,T} , ℤ[G]). Then

Φ(η_{S}) = (Φ′ ∘ w̌)(η_{S′}) ≡ ±h_{k,S′,T} Reg_{G}(Φ̃( ˜w̌(u_{1} ∧ ⋯ ∧ u′))) (mod I_{G}^{d+1}),

and ˜ w̌(u_{1} ∧ ⋯ ∧ u′) = ∑_{i=1}^{r+d+1}(−1)^{i+1}w(u_{i})u_{1} ∧ ⋯ ∧ u_{i−1} ∧ u_{i+1} ∧ ⋯ ∧ u_{r+d+1}. But u_{1}, …, u_{r+d} are S-units so this collapses to ±w(u′)u_{1} ∧ ⋯ ∧ u_{r+d}. Now because v splits completely in K/k, we have w(u′) = v(u′). Hence

Φ(η_{S}) ≡ ±v(u′)h_{k,S′,T} Reg_{G}(Φ̃(u_{1} ∧ ⋯ ∧ u_{r+d})) (mod I_{G}^{d+1}),

which by Lemma 3.5 is what we want.

Proposition 3.7. Suppose Conjecture 2.6 holds for K/k, S ⊇ S_{1}, T, r. Let v be a place of k not in S or T, and set S′ = S ∪ {v}. Then Conjecture 2.6 holds for K/k, S′ ⊇ S_{1}, T, r.

Proof. Again we have n = r+d and define u_{r+d+1} = u′. We note that, because S satisfies Hypothesis 2.1, v is unramified in K/k. Therefore the Artin symbol at v can be calculated by the formula

f_{v}(x) = (Frob_{v})^{v(x)},

for all x in k^{×}. The definition of Θ_{K/k,S,T} shows that η_{S′} = η_{S}^{1−Frob_{v}}. Take Φ = ϕ_{1} ∧ ⋯ ∧ ϕ_{r} ∈ ∧_{G}^{r}Hom_{ℤ[G]}(U_{K,S′,T} , ℤ[G]), then Conjecture 2.6 for S′ asks for:

Φ(η_{S′}) ≡ ±h_{k,S′,T} Reg_{G}(Φ̃(u_{1} ∧ ⋯ ∧ u_{r+d+1})) (mod I_{G}^{d+2}).

We choose the places v_{1}, …, v_{d+1} appearing in Reg_{G} by taking the set S−S_{1} of places not designated as splitting, excluding one place, then adding v = v_{d+1}. We have^{2}

Reg_{G}(Φ̃(u_{1} ∧ ⋯ ∧ u_{r+d+1})) = ∑_{σ∈\left[{r+d+1 \atop r}\right]} sign(σ) det (ϕ_{j}^{1}(u_{σ(i)})) det (f_{v_{j}}(u_{σ(r+i)}) − 1).

2for the definition of \left[{r+d+1 \atop r}\right] see Definition 2.3.

Let us consider two cases of σ. If r + d + 1 ∈ {σ(1), …, σ(r)} then in the corresponding term, the column in the second determinant corresponding to v is all zeros, since u_{1}, …, u_{r+d} are S-units. The other possibility is that r + d + 1 = σ(r + d + 1). Then this same column is all zeros apart from the bottom-right entry, which is

f_{v}(u_{r+d+1}) − 1 = (Frob_{v})^{v(u′)} − 1 ≡ v(u′)(Frob_{v} − 1) (mod I_{G}^{2}).

Hence Reg_{K/k,S′}(Φ̃(u_{1} ∧ ⋯ ∧ u_{r+d+1})) = ±v(u′)(Frob_{v} − 1)Reg_{K/k,S}(Φ̃(u_{1} ∧ ⋯ ∧ u_{r+d})), where in the second expression we may consider Φ as being restricted to U_{K,S,T}. So Conjecture 2.6 for S′ now reads, using Lemma 3.5,

(1 − Frob_{v})Φ(η_{S}) ≡ ±(Frob_{v} − 1)h_{k,S,T} Reg_{G}(Φ̃(u_{1} ∧ ⋯ ∧ u_{r+d})) (mod I_{G}^{d+2}).

Therefore if Conjecture 2.6 holds for S, it holds for S′.

On the other hand, the proof shows that we have the following, possibly weaker, implication in the other direction.

Proposition 3.8. Suppose Conjecture 2.6 holds for K/k, S ∪ {v} ⊇ S_{1}, T, r. Then we have, for the data K/k, S ⊇ S_{1}, T, r,

(1 − Frob_{v})Φ(η) ≡ (1 − Frob_{v})(±h_{k,S,T} Reg_{G}^{Φ}) (mod I_{G}^{d+1}),

in the notation of Conjecture 2.6. That is, we obtain the image of the congruence in the next level of the augmentation filtration under multiplication by (1 − Frob_{v}).

3.2 Special cases We study the behaviour of the conjecture for some interesting special cases of the data.

Proposition 3.9. Suppose r = 0 and K/k, S ⊇ ∅, T, 0 satisfies Hypothesis 2.1, ie. we designate no places as splitting in K/k. Then Conjecture 2.6 is equivalent to Conjecture 4.1 of [Gr88], up to sign.

Proof. The element η ∈ ℂ ∧_{G}^{0} U_{K,S,T} e_{r} = ℂ[G]e_{r} is characterised by η = Θ_{K/k,S,T} (0). Taking Φ = 1 ∈ ℤ[G], Conjecture 2.6 now reads:

Θ_{K/k,S,T} (0) ≡ ±h_{k,S,T} Reg_{G}(u_{1} ∧ ⋯ ∧ u_{d}) (mod I_{G}^{d+1}).

This is a sign-indifferent version of Gross’s conjecture for the extension K/k and sets S and T.

Proposition 3.10. Conjecture 2.6 holds when more than r places in S split completely.

Proof. We adapt the method of [Ru96], Proposition 3.1.

Note that all the S-truncated L-functions corresponding to nontrivial characters vanish to order greater than r at s = 0 (see [Ta84], Proposition I.3.4). If #S > r + 1 then this is true for the trivial character too, and so η is the identity. On the other hand, if #S > r + 1 then our Gross-style regulators in Conjecture 2.6 can be calculated with respect to a totally split place and so are all zero. Hence Conjecture 2.6 says 0 ≡ 0.

Now assume #S = r+1. Let u_{1}, …, u_{r} be a ℤ-basis for U_{k,S,T}. In his proof of [Ru96], Proposition 3.1, Rubin shows that, by the analytic class number formula,

η = \frac{h_{k,S,T}}{#G^{r}} u_{1} ∧ ⋯ ∧ u_{r},

(for which we might have to invert a unit u_{1} to get the sign right). We apply Φ = ϕ_{1} ∧ ⋯ ∧ ϕ_{r} ∈ ∧_{G}^{r}Hom_{G}(U_{K,S,T} , ℤ[G]) to η. Note that, because u_{i} ∈ k, ϕ_{j}(u_{i}) = ϕ_{j}^{1}(u_{i})N_{G}, where N_{G} = ∑_{g∈G} g. We obtain

Φ(η) = ±\frac{h_{k,S,T}}{(#G)^{r}} N_{G}^{r} det(ϕ_{j}^{1}(u_{i})) = ±\frac{h_{k,S,T}}{#G} N_{G} det(ϕ_{j}^{1}(u_{i})).

Rubin (loc. cit.) argues by class field theory that #G | h_{k,S,T}, so this is an element of ℤ[G]. Since N_{G} has augmentation #G, reducing this equation mod I_{G} gives us Conjecture 2.6.

Corollary 3.11. If k/k, S, T, r satisfy Hypothesis 2.1, then Conjecture 2.6 is true for this data.

Proof. This is because all places in S split completely.

Proposition 3.12. Assume K/k, S ⊇ S_{1}, T, r satisfy Hypothesis 2.1 and furthermore that #S ⩾ r + 2. Assume Conjecture 2.2 holds for this data. Then we have Φ(η) ∈ I_{G}.

Proof. This is [Bu01], Theorem 4.4 (iii). We reproduce the proof. Since ζ_{k,S} vanishes to order r + 1 at s = 0, Θ_{K/k,S,T}^{r} (0) lies in ℂI_{G}. Hence N_{K/k}η = 1, and so Φ(η) ∈ ℂI_{G}. Now if Conjecture 2.2 holds, we have Φ(η) ∈ ℂI_{G} ∩ ℤ[G] = I_{G}, as required.

4. Quadratic Extensions

In this section we take K/k, S, T, r with K/k quadratic with group G generated by τ. We will assume, using Proposition 3.10, that exactly r places S_{1} split in K/k.

Perhaps the most involved arguments of Rubin in [Ru96] and Gross in [Gr88] are to verify their respective conjectures in this situation. We adapt their methods to prove:

Theorem 4.1. Let K/k, S ⊇ S_{1}, T, r be data satisfying Hypothesis 2.1, with [K : k] = 2. Then Conjecture 2.6 holds.

Remark 4.2. This result provides a new proof of the validity of Gross’s conjecture ([Gr88], Conjecture 4.1) for quadratic extensions. Its proof avoids the technicalities and special cases considered by Gross in [Gr88], §6, by using the extra functorial properties of Conjecture 2.6 with respect to increase of S. For comparison, note that §4.2 below corresponds to the known case “n = 0” of Gross’s conjecture ([Gr88] (4.3)), and that in §4.3 below the sign of the regulator is irrelevant.

4.1 Cohomology of U_{K,S,T} Let u_{1}, …, u_{r+d+r} be a basis of U_{K,S,T} such that u_{1}, …, u_{r+d} is a basis of U_{k,S,T}, which is possible by Lemma 3.1. The relevant structure of this basis is closely related to the Galois cohomology of the G-module U_{K,S,T}. Our first result in this direction is the following (cf. [Ru96], Theorem 3.5, proof).

Lemma 4.3. If H^{1}(G, U_{K,S,T}) ≠ 0 then we can assume N_{K/k}u_{r+d+1} = 1.

Proof. Take u ∈ U_{K,S,T} representing a nontrivial element of H^{1}(G, U_{K,S,T}) = U_{K,S,T}^{−} /U_{K,S,T}^{1−τ}, where U_{K,S,T}^{−} is the set of (S,T)-units of norm 1. Write u = ϵ ∏_{i} u_{r+d+i}^{α_{i}}, where ϵ ∈ U_{k,S,T}, and write ϵ_{i} for the norm u_{r+d+i}^{1+τ} ∈ U_{k,S,T}. Then u_{r+d+i}^{1−τ} = u_{r+d+i}^{2}ϵ_{i}^{−1}. Therefore we can assume each α_{i} is 0 or 1. But they cannot all be 0. Hence u can go into a basis of U_{K,S,T}.

Lemma 4.4. We have

\frac{#Ĥ^{0}(G, U_{K,S,T})}{#H^{1}(G, U_{K,S,T})} = 2^{d}.

Proof. Note the left-hand side is the Herbrand quotient h(U_{K,S,T}) of the ℤ[G]-module U_{K,S,T} in the sense of [Se79], Ch. VIII, §4. The composite isomorphism of ℚ[G]-modules

ℚU_{K,S,T} ≅ ℚX_{S_{K}} ≅ ℚ[G]^{r} ⊕ ℚ^{d}

implies that there is an injection of U_{K,S,T} into ℤ[G]^{r} ⊕ ℤ^{d} with finite cokernel. Then h(U_{K,S,T}) = h(ℤ[G])^{r}h(ℤ)^{d} = 2^{d} as required.

Lemma 4.5. If H^{1}(G, U_{K,S,T}) = 0, we can assume that u_{i} = N_{K/k}u_{r+d+i} for i = 1, …, r.

Proof. Write Nu_{r+d+j} = ∏_{i=1}^{r+d} u_{i}^{α_{ji}} for j = 1, …, r. We may perform the following operations on the r × (r + d) matrix (α_{ji}): elementary column operations, which correspond to swapping and multiplying the units in the basis of U_{k,S,T}, and elementary row operations, which correspond to swapping and multiplying the units u_{r+d+1}, …, u_{r+d+r}.

Thus we can put (α_{ji}) into diagonal form with integers a_{1}, …, a_{r} on the diagonal. Now suppose some a_{i} is even, so Nu_{r+d+i} = ϵ^{2} for some ϵ in U_{k,S,T}. Then u_{r+d+i}ϵ^{−1} is a unit in K − k with norm 1. But the group U_{K,S,T}^{1−τ} ⊆ U_{k,S,T} U_{K,S,T}^{2} . So u_{r+d+i}ϵ^{−1} represents a nontrivial element of

U_{K,S,T}^{−} /U_{K,S,T}^{1−τ} = H^{1}(G, U_{K,S,T}).

This is a contradiction to our assumption. We conclude that each a_{i} is odd. Now replacing u_{r+d+i} by u_{r+d+i}u_{i}^{−[a_{i}/2]} gives us the result.

In our situation, noting that S contains a place not splitting in K, Lemma 3.4 of [Ru96] states:

Lemma 4.6 (Rubin). i) h_{k,S,T} | h_{K,S,T} ii) #H^{1}(G, U_{K,S,T}) | h_{k,S,T}

iii) If Ĥ^{0}(G, U_{K,S,T}) and H^{1}(G, U_{K,S,T}) are trivial then h_{k,S,T} ≡ h_{K,S,T} /h_{k,S,T} (mod 2).

We write ϵ_{−} = u_{r+d+1} ∧ ⋯ ∧ u_{r+d+r}. In proving his conjecture for quadratic extensions ([Ru96], proof of Proposition 3.5), Rubin uses the analytic class number formula to express η in terms of this element; we make extensive use of his formulae, which are quoted below.

4.2 The case d = 0 We assume d = 0, that is, #S = r + 1, and show that the congruence in Conjecture 2.6 holds. We require:

Φ(η) ≡ ±h_{k,S,T} | \begin{matrix}ϕ_{1}^{1}(u_{1}) & ⋯ & ϕ_{r}^{1}(u_{1}) \\ ⋮ & ... & ⋮ \\ ϕ_{1}^{1}(u_{r}) & ⋯ & ϕ_{r}^{1}(u_{r}) \end{matrix} | (mod I_{G}).  (8)

Note that this is an equality in ℤ[G]/I_{G} ≅ ℤ.

Assume first that H^{1}(G, U_{K,S,T}) = 0. Then Rubin shows in [Ru96], proof of Proposition 3.5, that

η = ± (\frac{h_{K,S,T}}{h_{k,S,T}} \frac{(1 − τ)}{2} ± h_{k,S,T} \frac{(1 + τ)}{2}) ϵ_{−}.

Lemma 4.6 shows the ℚ[G]-factor here lies in ℤ[G], and we note its augmentation is ±h_{k,S,T}. Now we can assume u_{r+d+i}^{1+τ} = u_{i} for i = 1, …, r by Lemma 4.5, so

ϕ(u_{r+d+i}) = ϕ^{1}(u_{r+d+i}) − τϕ^{1}(u_{r+d+i}) + τϕ^{1}(u_{i}) ≡ ϕ^{1}(u_{i}) (mod I_{G}).

Hence (8) is satisfied in this case.

Now assume H^{1}(G, U_{K,S,T}) ≠ 0. Let ū_{1}, …, ū_{r} ∈ U_{K,S,T} such that N_{K/k}ū_{i} is a basis for NU_{K,S,T}. Set ϵ_{+} = ū_{1} ∧ ⋯ ∧ ū_{r}. Rubin shows (loc. cit.) that

η = ±\frac{h_{k,S,T}}{#Ĥ^{0}(G, U_{K,S,T})} \frac{(1 + τ)}{2}ϵ_{+} ± \frac{h_{K,S,T}}{h_{k,S,T}} \frac{(1 − τ)}{2}ϵ_{−},

where again the ℚ[G]-factors are in ℤ[G]. By Lemma 4.3 we can assume N_{K/k}u_{r+d+1} = 1, and then

ϕ_{j}(u_{r+d+1}) = ϕ_{j}^{1}(u_{r+d+1}) + τϕ_{j}^{1}(u_{r+d+1}^{τ}) ≡ ϕ_{j}^{1}(N_{K/k}u_{r+d+1}) = 0 (mod I_{G}),

so Φ(ϵ_{−}) ≡ 0 (mod I_{G}). On the other hand, ϕ_{j}(ū_{i}) ≡ ϕ_{j}^{1}(N_{K/k}ū_{i}) (mod I_{G}) and the index of the group generated by the N_{K/k}ū_{i} in U_{k,S,T} is #Ĥ^{0}(G, U_{K,S,T}). This shows that (8) holds in this case too.

This verifies Theorem 4.1 in the case d = 0.

4.3 The case d > 0 Now we assume d > 0, ie. #S > r + 1. For d > 0, I_{G}^{d}/I_{G}^{d+1} is a group of order 2, so the congruence statement in Conjecture 2.6 only concerns in which power of the augmentation ideal the terms lie. We have (1 − τ)^{n} = 2^{n−1}(1 − τ) for n > 0. Note that the map

\begin{matrix}I_{G}^{d}/I_{G}^{d+1} & ⟶ & I_{G}^{d+1}/I_{G}^{d+2} \\ x & ⟼ & (1 − τ)x \end{matrix}

is a bijection. We have the following freedom to increase S:

Lemma 4.7. Let K/k, S, T, r satisfy Hypothesis 2.1 with K/k a quadratic extension. Assume d = #S − r − 1 > 0. Let v be a place of k not in S or T, and set S′ = S ∪ {v}, S′_{1} = S_{1} ∪ {v}. Then either of the following conditions implies Conjecture 2.6 for K/k, S, T, r.

i) v splits in K/k, and Conjecture 2.6 holds for K/k, S′ ⊇ S′_{1}, T, r + 1; ii) v is inert in K/k, and Conjecture 2.6 holds for K/k, S′ ⊇ S_{1}, T, r.

Proof. This follows from Propositions 3.6 and 3.8, given the structure of the augmentation filtration.

Rubin shows in [Ru96], proof of Proposition 3.5, that for the case K/k quadratic and d > 0 we have

η = ±2^{d} \frac{h_{K,S,T}}{h_{k,S,T}} \frac{(1 − τ)}{2}ϵ_{−}.

We note that 2^{d−1}(1 − τ) ∈ I_{G}^{d}.

It turns out that we only need to consider the congruence statement under the following cohomological assumption^{3}.

Lemma 4.8. Suppose d > 0 and H^{1}(G, U_{K,S,T}) = 0. Then the congruence of Conjecture 2.6 is implied by the following statement:

2^{d−1}(1 − τ)\frac{h_{K,S,T}}{h_{k,S,T}} ≡ h_{k,S,T} Reg_{G}(u_{r+1} ∧ ⋯ ∧ u_{r+d}) (mod I_{G}^{d+1}).  (9)

Proof. We apply Lemma 4.5 to assume that u_{i} = N_{K/k}u_{r+d+i} for i = 1, … r. Then ϕ_{j}(u_{r+d+i}) ≡ ϕ_{j}^{1}(u_{i}) (mod I_{G}). Thus

Φ(η) ≡ 2^{d−1}(1 − τ)\frac{h_{K,S,T}}{h_{k,S,T}} | \begin{matrix}ϕ_{1}^{1}(u_{1}) & ⋯ & ϕ_{r}^{1}(u_{1}) \\ ⋮ & ... & ⋮ \\ ϕ_{1}^{1}(u_{r}) & ⋯ & ϕ_{r}^{1}(u_{r}) \end{matrix} | (mod I_{G}^{d+1})

For the right-hand side of the congruence in Conjecture 2.6, we note that if any u_{i}, 1 ⩽ i ⩽ r, appears in the argument of Reg_{G}, then the corresponding term is 0 (because u_{i} is a norm from K, and therefore in the kernel of all the local reciprocity maps). So the right-hand side collapses to a

3In fact it is easy to show that both sides of the congruence vanish if this assumption is not satisfied.

single term:

h_{k,S,T} | \begin{matrix}ϕ_{1}^{1}(u_{1}) & ⋯ & ϕ_{r}^{1}(u_{1}) \\ ⋮ & ... & ⋮ \\ ϕ_{1}^{1}(u_{r}) & ⋯ & ϕ_{r}^{1}(u_{r}) \end{matrix} | Reg_{G}(u_{r+1} ∧ ⋯ ∧ u_{r+d}).

This gives the result.

Next we identify a condition for the non-vanishing of the regulator. We need an auxiliary lemma. For Tate’s theory of the group cohomology of finite cyclic groups, we refer to [Se79], Ch. VIII, §4.

Lemma 4.9. Suppose h_{k,S} = 1. Then there is an isomorphism

\frac{U_{k,S} ∩ NK^{×}}{NU_{K,S}} ≅ A_{K,S}^{G}.

Proof. (cf. [Gr88] p191) We have the exact sequence (from (1)):

0 ⟶ K^{×}/U_{K,S} ⟶ ⨁_{𝔓∉S_{K}} 𝔓^{ℤ} ⟶ A_{K,S} ⟶ 0.

Considering the decompositions of primes shows that H^{1}(G, ⨁_{𝔓∉S_{K}} 𝔓^{ℤ}) = 0. Then taking cohomology gives an exact sequence

0 ⟶ H^{0}(G, K^{×}/U_{K,S}) ⟶ ⨁_{𝔭∉S_{k}} 𝔭^{ℤ} ⟶^{0} A_{K,S}^{G} ⟶ H^{1}(G, K^{×}/U_{K,S}) ⟶ 0.

Observe that k^{×}/U_{k,S} injects into H^{0}(G, K^{×}/U_{K,S}), and (1) for k shows that it surjects onto ⨁_{𝔭∉S_{k}} 𝔭^{ℤ}, because the S-class group of k is trivial. This is why the map marked 0 is zero. Therefore we have A_{K,S}^{G} ≅ H^{1}(G, K^{×}/U_{K,S}).

On the other hand, applying Hilbert’s Theorem 90 ([Se79], Ch. X, Proposition 2) and Tate cohomology to the short exact sequence

0 ⟶ U_{K,S} ⟶ K^{×} ⟶ K^{×}/U_{K,S} ⟶ 0

gives an exact sequence

0 ⟶ H^{1}(G, K^{×}/U_{K,S}) ⟶ Ĥ^{0}(G, U_{K,S}) ⟶ Ĥ^{0}(G, K^{×}).

Therefore

H^{1}(G, K^{×}/U_{K,S}) ≅ ker(U_{k,S}/NU_{K,S} ⟶ k^{×}/NK^{×}) = U_{k,S} ∩ NK^{×}/NU_{K,S}.

This completes the proof.

Lemma 4.10. Suppose h_{K,S,T} = 1. Then Reg_{G}(u_{r+1} ∧ ⋯ ∧ u_{r+d}) ≡ / 0 (mod I_{G}^{d+1}).

Proof. By (3) and Lemma 4.6 (part (i)), we have that h_{k,S} = 1, and then Corollary 2 of [Ri65] shows that H^{1}(G, U_{K,S}) = 0. Therefore Ĥ^{0}(G, U_{K,S}) is a 2-torsion group with 2^{d} elements by Lemma 4.4, and hence is isomorphic to (ℤ/2ℤ)^{d}.

Similarly to [Gr88] p191, we define a homomorphism

f : U_{k,S} ⟶ G^{S−S_{1}} ≅ (ℤ/2ℤ)^{d+1}

by the local reciprocity maps f_{v}. Then, by the product formula,

im f ⊆ V := {(g_{v})_{v∈S−S_{1}} : ∏ g_{v} = 1} ≅ (ℤ/2ℤ)^{d}.

Now u ∈ U_{k,S} is in ker f if and only if u is a local norm at all the places in S − S_{1}, and we note it is automatically a norm at all other places. Since K/k is cyclic, u is a local norm everywhere if and only if it is a global norm. So ker f = U_{k,S} ∩ NK^{×}. Since h_{K,S,T} = 1, A_{K,S}^{G} = 0 and so Lemma 4.9 shows that ker f = NU_{K,S}. On the other hand, Ĥ^{0}(G, U_{K,S}) = U_{k,S}/NU_{K,S} ≅ (ℤ/2ℤ)^{d}, so we have f(U_{k,S}) = V.

We note that the form of our regulator and the choice of our unit basis show that the nonvanishing of our regulator

Reg_{G}(u_{r+1} ∧ ⋯ ∧ u_{r+d}) ≡ / 0 (mod I_{G}^{d+1})

is equivalent to saying f(U_{k,S,T}) = V.

The reduction map U_{K,S} ⟶ ∏_{𝔓∈T_{K}} 𝔽_{𝔓}^{×} is surjective, by (2) for K and the assumption that h_{K,S,T} = 1. Also, the norm in an extension of finite fields is surjective. Hence ker f = NU_{K,S} surjects onto ∏_{𝔭∈T} 𝔽_{𝔭}^{×}. The sequence (2) for k shows that this latter is isomorphic by the reduction map to U_{k,S}/U_{k,S,T}. So every element of U_{k,S} can be written as the product of something in ker f by something in U_{k,S,T}, which gives our result.

Now consider A_{K,S,T}, the S_{K} ray class group modulo T_{K}. Let S′_{K} = {w_{1}, …, w_{n}} be a set of primes of O_{K,S} coprime to T_{K} whose classes generate this group. Set S′ to be the set of places of k lying below these. If S contains S′, then h_{K,S,T} = 1. On the other hand:

Lemma 4.11. If h_{K,S,T} = 1, then the congruence of Conjecture 2.6 holds.

Proof. By Lemma 4.6 (part (i)) we have that h_{k,S,T} = 1, which also shows by part (ii) of the same lemma that H^{1}(G, U_{K,S,T}) = 0. So by Lemma 4.8 it is sufficient to show (9) holds. Lemma 4.10 shows that the right-hand side of (9) is not zero. On the other hand h_{K,S,T} /h_{k,S,T} = 1 so the left-hand side is not zero either.

Now by Lemma 4.7, we may assume S′ ⊆ S, increasing r by the number of split primes in S′ −S. Then h_{K,S,T} = 1, so Lemma 4.11 implies that the congruence of Conjecture 2.6 holds.

We have verified Theorem 4.1 in all cases.

5. Real abelian extensions of ℚ

In real abelian extensions of ℚ, the infinite place splits and the Stark unit is known to be essentially a cyclotomic element. In this section we show that Conjecture 2.6 can be verified (up to a factor of 2 on each side) using the theory of cyclotomic elements.

5.1 Determination of the special unit Suppose F is a totally real, nontrivial, abelian extension of ℚ with group G and conductor m. We consider Conjecture 2.6 for the extension F/ℚ. By the Kronecker-Weber theorem, F is contained in ℚ(ζ_{m}).

We set S_{ℚ} = {p | m}∪{∞}, S_{1,ℚ} = {∞} and r = 1 in the notation of §2, noting that the infinite place does indeed split completely in the extension F/ℚ because F is real. Let ∞_{F} be the infinite place of F induced by the embedding

\begin{matrix}ℚ(ζ_{m}) & ⟶ & ℂ \\ ζ_{m} & ⟼ & e^{2πi/m}. \end{matrix}

Set β = 1 − ζ_{m} and, in the notation of §2, W = (∞_{F}).

Lemma 5.1. W^{∗}(λ_{S_{F}} (N_{ℚ(ζ_{m})/F}β)) = 2Θ_{F/ℚ,S_{ℚ},∅}^{1}(0).

Proof. We have

W^{∗}(λ_{S_{F}} (N_{ℚ(ζ_{m})/F}β)) = − ∑_{σ∈G} ln |σN_{ℚ(ζ_{m})/F}β|σ^{−1}.

On the other hand, the value at s = 0 of the L-function of an even Dirichlet character χ defined modulo m is given by:

L(0, χ) = 0,  L′(0, χ) = −\frac{1}{2} ∑_{i=1}^{m−1} χ(i) ln |1 − ζ_{m}^{i}|,  (10)

which holds whether or not m is the conductor of χ. The result follows easily by combining these formulae.

Let T = T_{ℚ} be as required by Hypothesis 2.1, ie. T contains a prime of odd residue characteristic. Then the T -correction factor is

δ_{T} = ∏_{v∈T}(1 − N_{v}Frob_{v}^{−1}), ie. Θ_{F/ℚ,S,T}^{1}(0) = δ_{T} Θ_{F/ℚ,S,∅}^{1}(0).

We have W^{∗}(λ_{S_{F}} (δ_{T} N_{ℚ(ζ_{m})/F}β)) = 2Θ_{F/ℚ,S,T}^{1}(0), so (in the notation of §2) δ_{T} N_{ℚ(ζ_{m})/F}β = 2η_{F/ℚ} in ℂU_{F,S,T}.

Because of this, we wish to study the properties of the cyclotomic elements (1 − ζ_{m}). The next lemma summarises their well-known distribution properties.

Lemma 5.2. For each positive integer n, set ζ_{n} = e^{2πi/n}, and define the norm element of the integral group ring of Γ_{n} := Gal(ℚ(ζ_{n})/ℚ) by N_{n} := ∑_{g∈Γ_{n}} g. Take positive integers p, f, r such that p is prime, f > 1 and p ∤ f. By linear disjointness of ℚ(ζ_{p^{r}}) and ℚ(ζ_{f}) as extensions of ℚ, there is a natural inclusion Γ_{p^{r}} ↪ Γ_{p^{r}f}. Let σ_{a,b} denote the automorphism of ℚ(ζ_{b}) sending ζ_{b} to ζ_{b}^{a} for a coprime to b. Then we have:

i) (1 − ζ_{p^{r}f})^{N_{p^{r}}} = (1 − ζ_{f})^{(1−σ_{p,f}^{−1} )}; ii) (1 − ζ_{p^{r}})^{N_{p^{r}}} = p; iii) If two distinct primes divide n, then (1 − ζ_{n})^{N_{n}} = 1

These well-known facts follow from the factorisation of X^{p^{r}} − 1 ∈ ℂ[X].

5.2 Relations between determinants of certain matrices The following linear algebra result will be useful in the following section. Fix a commutative ring with 1, and call it R. Let B be a finite set of positive integers, and for each i, j ∈ B with i ≠ j fix a_{ij} ∈ R. For each I ⊆ B, let A^{I} be the square matrix indexed by I with (i, j)-th entry a_{ij} for i ≠ j and − ∑_{k∈I-{j}} a_{ik} for i = j, so that A^{I} is a matrix with row-sum zero. Let A_{i}^{I} be the (i, i)-th minor determinant of A^{I} for i ∈ I.

Proposition 5.3. For each i ∈ B,

∑_{{i}⊆I⊆B} A_{i}^{I} ∏_{j∈B−I} ∑_{k∈I} a_{jk} = 0.

Proof. The proof uses trees in an analogous way to [GK03], proof of Theorem 8. If J is a finite set, then a tree T on J consists of the set of vertices J and edges between them which form a connected graph with no loops. A choice of a vertex r ∈ J to be the “root” \sqrt{T} of T induces a direction on each edge such that the out-degree of r is 0 and the out-degree of all other vertices is 1. For a directed tree T on a subset of B, define A(T) := ∏_{(i→j)∈T} a_{ij}.

Since the row-sums of A^{I} are zero, the Kirchhoff-Tutte theorem (see [Tu48], or [GK02], Theorem 4) states that:

A_{i}^{I} = (−1)^{#I−1} ∑_{\begin{matrix}T tree on I: \\ \sqrt{T} =i \end{matrix}} A(T).

We also note that

∏_{j∈B−I} ∑_{k∈I} a_{jk} = ∑_{f:(B−I)→I} ∏_{j∈B−I} a_{j,f(j)}.

Hence the left-hand side in the desired equality is

∑_{\{i\}⊆I⊆B} ∑_{\begin{matrix}T_{I} tree on I: \\ \sqrt{T} =i \end{matrix}} ∑_{f_{I}:(B−I)→I}(−1)^{#I−1}A(T_{I}) ∏_{j∈B−I} a_{j,f_{I}(j)}.

For each tree T on B with root i ∈ B, we calculate the coefficient of A(T) in the above sum. If V is a subset of the set of vertices of in-degree 0 in T, removing the vertices V and the edges attached to them gives a tree T_{I} on B − V =: I. Defining f_{I} : (B − I) → I by the relation (j → f_{I}(j)) ∈ T, the pair (T, V) corresponds bijectively to the index (I, T_{I}, f_{I}) from the sum, whose term is (−1)^{#I}A(T). But given T, there are as many V with #I even as #I odd. Hence the term for T is 0.

5.3 A congruence statement for cyclotomic elements

Let m > 1 and write m = p_{1}^{a_{1}}p_{2}^{a_{2}} … p_{d+1}^{a_{d+1}}. Write β_{m} = 1−ζ_{m} for the associated cyclotomic element.

If p | m, and 𝔭 is a place of ℚ(ζ_{m}) above p, we let f_{p}(x) denote the Artin symbol (x, ℚ(ζ_{m})_{𝔭}/ℚ_{p}) for all nonzero x ∈ ℤ. It is a simple exercise in the global class field theory of cyclotomic fields to show the following

Lemma 5.4. If j ≠ i, f_{p_{j}}(p_{i})^{−1} is given by the automorphism σ_{p_{i},p_{j}^{a_{j}}} of ℚ(ζ_{m}) defined by:

ζ_{p_{j}^{a_{j}}} ↦ ζ_{p_{j}^{a_{j}}}^{p_{i}}
ζ_{p_{k}^{a_{k}}} ↦ ζ_{p_{k}^{a_{k}}}  for k ≠ j.

In the notation of the previous section, we set B = {1, …, d + 1} and a_{ij} = f_{p_{j}}(p_{i}) − 1 ∈ ℤ[Γ], where Γ := Gal(ℚ(ζ_{m})/ℚ). Then there are defined certain elements A_{i}^{I} ∈ ℤ[Γ] for i ∈ I ⊆ B. Set S = {p | m} ∪ {∞}, a finite set of places of ℚ. We prove the following congruence statement for β_{m}:

Proposition 5.5. For all ϕ : U_{ℚ(ζ_{m}),S} ⟶ ℤ[Γ], we have

ϕ(β_{m}) ≡ ∑_{i=1}^{d+1} ϕ^{1}(p_{i})A_{i}^{B} (mod I_{Γ}^{d+1}).

Proof. (cf. [Da95], Theorem 4.2) By induction on d + 1. If d + 1 = 1 we have m = p_{1}^{a_{1}}. Then

ϕ(β_{m}) = ∑_{g∈Γ_{m}} ϕ^{1}(g^{−1}β_{m})g ≡ ∑_{g∈Γ_{m}} ϕ^{1}(g^{−1}β_{m}) (mod I_{Γ}),

and this is ϕ^{1}(N_{p_{1}^{a_{1}}} β_{m}) = ϕ^{1}(p_{1}) by Lemma 5.2. Hence the claim is true for d + 1 = 1.

Now assume it is true for d + 1 = 1, 2, …, n. Set d + 1 = n + 1 > 1. For I ⊆ {1, …, d + 1}, write Γ_{I} = ∏_{i∈I}(ℤ/p_{i}^{a_{i}}ℤ)^{∗} ↪ Γ_{m}, Γ_{i} = Γ_{{i}} and m(I) = ∏_{i∈I} p_{i}^{a_{i}}. We have the following equality in ℤ[Γ]:

∑_{g_{1}∈Γ_{1}} … ∑_{g_{d+1}∈Γ_{d+1}} ϕ^{1}((g_{1}^{−1} … g_{d+1}^{−1})β_{m})(g_{1} − 1) … (g_{d+1} − 1)

= ∑_{g_{1}∈Γ_{1}} … ∑_{g_{d+1}∈Γ_{d+1}} ϕ^{1}((g_{1}^{−1} … g_{d+1}^{−1})β_{m}) ∑_{I⊆B}(−1)^{d+1−#I} ∏_{i∈I} g_{i}
= ∑_{I⊆B}(−1)^{d+1−#I} ∑_{g∈Γ_{I}} ϕ^{1}(g^{−1} ∏_{j∉I} N_{p_{j}^{a_{j}}} β_{m})g  (11)

We recall (Lemma 5.2) that, if I ≠ ∅, we have ∏_{j∉I} N_{p_{j}^{a_{j}}} β_{m} = ∏_{j∉I}(1 − σ_{p_{j},m(I)}^{−1})β_{m(I)}, and we note

that σ_{p_{j},m(I)} ∈ Γ_{I}. For I = ∅, ∏_{j} N_{p_{j}^{a_{j}}} β_{m} = 1 by Lemma 5.2. So (11) is equal to

∑_{∅≠I⊆B}(−1)^{d+1−#I}ϕ_{(I)}(β_{m(I)}) ∏_{j∉I}(1 − σ_{p_{j},m(I)}^{−1}),

where ϕ_{(I)} means the ℤ[Γ_{I}]-homomorphism U_{ℚ(ζ_{m(I)}),S} ⟶ ℤ[Γ_{I}] associated to the restriction of ϕ^{1} to U_{ℚ(ζ_{m(I)}),S}.

Lemma 5.4 shows that σ_{p,m(I)} = ∏_{i∈I} f_{p_{i}}(p)^{−1} for p ∤ m(I), and our induction hypothesis gives ϕ_{(I)}(β_{m(I)}) ∈ I_{Γ}^{#I−1} for I ≠ B. So if we reduce our equality modulo I_{Γ}^{d+1}, we obtain

0 ≡ ∑_{I⊆B}(−1)^{d+1−#I}ϕ_{(I)}(β_{m(I)}) ∏_{j∉I} ( − ∑_{k∈I}(f_{p_{k}}(p_{j}) − 1))
≡ ∑_{I⊆B} ϕ_{(I)}(β_{m(I)}) ∏_{j∉I} ( ∑_{k∈I} a_{jk}) (mod I_{Γ}^{d+1}).

By the induction hypothesis, ϕ_{(I)}(β_{m(I)}) ≡ ∑_{i∈I} ϕ^{1}(p_{i})A_{i}^{I} (mod I_{Γ}^{#I}) if I ≠ B. Therefore

0 ≡ ϕ(β_{m}) + ∑_{i=1}^{d+1} ϕ^{1}(p_{i}) ∑_{{i}⊆I⊊B} A_{i}^{I} ∏_{j∈B−I} ( ∑_{k∈I} a_{jk}).

Now Proposition 5.3 shows that the i-th term of the sum is −ϕ^{1}(p_{i})A_{i}^{B}, as required.

5.4 The congruence statement for a real abelian extension of ℚ Let F/ℚ be a finite, real, abelian extension. Let G be the Galois group and m = p_{1}^{a_{1}} … p_{d+1}^{a_{d+1}} the conductor of this extension. Recall from §5.1 that

2η_{F/ℚ} = δ_{T_{ℚ}}N_{ℚ(ζ_{m})/F}β_{m},

where β_{m} = (1 − ζ_{m}). We set S_{ℚ} = S = {∞} ∪ {p | m}, S_{1} = {∞}, r = 1, and T to be as required by Hypothesis 2.1.

Set B = {1, …, d + 1}, and define a_{ij} ∈ ℤ[Gal(ℚ(ζ_{m})/ℚ)] as in §5.3. Then a_{ij} ↦ f_{p_{j}}(p_{i}) − 1 ∈ ℤ[G] under the natural projection to ℤ[G]. We relate the projections Ā_{i}^{I} of the A_{i}^{I} determinants from Proposition 5.5 to our group ring-valued regulators. We must choose an ordered set of d places with respect to which to calculate, and we choose p_{1}, …, p_{d}.

Proposition 5.6. Let σ ∈ S_{d+1} (the symmetric group on B) such that σ(1) < … < σ(d). Then

Reg_{G}(p_{σ(1)} ∧ ⋯ ∧ p_{σ(d)}) = sign(σ)Ā_{σ(d+1)}^{B}.

Proof. If d + 1 = 1 then both sides are the determinants of 0 × 0 matrices and so are 1. So we suppose d + 1 > 1. First assume σ = id. The regulator is, by the product rule, the determinant of (the projection of)

( \begin{matrix}c_{1} & a_{12} & ⋯ & a_{1d} \\ a_{21} & c_{2} & ⋯ & a_{2d} \\ ⋮ & ⋮ & ... & ⋮ \\ a_{d1} & a_{d2} & ⋯ & c_{d} \end{matrix} ) ,

with c_{j} = − ∑_{k} a_{jk}, where k runs over {1, …, d + 1}-{j}. This is what we need.

Now assume σ ≠ id. Let b = σ(d + 1). In the matrix defining the regulator, add the other columns to the column for the place p_{b}. This gives, by the product rule,

−Reg_{G}^{(1,2,...,b−1,n+1,b+1,...,d)}(p_{σ(1)} ∧ ⋯ ∧ p_{σ(d)}),

where the upper d-tuple is the ordered set of places with respect to which the regulator is calculated. This ordered set differs from the ordered set (σ(1), …, σ(d)) by the permutation σ ∘ (n + 1 b). So

Reg_{G}^{(1,...,d)}(p_{σ(1)} ∧ ⋯ ∧ p_{σ(d)}) = sign(σ)Reg_{G}^{(σ(1),...,σ(d))}(p_{σ(1)} ∧ ⋯ ∧ p_{σ(d)}).

Now the result follows from the case σ = id by renaming the primes p_{σ(i)} to p_{i}.

Now Proposition 5.5 and a “lowering the top field” argument entirely analogous to the proof of Proposition 3.2 together prove the following:

Proposition 5.7. For all ϕ : U_{F,S} ⟶ ℤ[G], we have

ϕ(N_{ℚ(ζ_{m})/F}β_{m}) ≡ (−1)^{d} ∑_{i=1}^{d+1}(−1)^{i+1}ϕ^{1}(p_{i})Reg_{G}(p_{1} ∧ ⋯ ∧ p̂_{i} ∧ ⋯ ∧ p_{d+1}) (mod I_{G}^{d+1}),

where “ˆ ” means “omit”.

Remark 5.8. If F/ℚ is cyclic of prime-power degree, and all the p_{i} are totally tamely ramified in F, then this result can be deduced from Theorem 1 of [GK03].

Finally, we deduce a T -modified version. Let δ_{T} = ∏_{v∈T} (1 − N_{v}Frob_{v}^{−1}).

Lemma 5.9. Let u_{1}, …, u_{d+1} be a ℤ-basis for U_{ℚ,S,T}. Then we have the following equality in ∧_{ℤ}^{d+1}U_{ℚ,S}:

2u_{1} ∧ ⋯ ∧ u_{d+1} = (U_{ℚ,S} : U_{ℚ,S,T})p_{1} ∧ ⋯ ∧ p_{d+1}.

Proof. Write u_{j} = ± ∏_{i=1}^{d+1} p_{i}^{c_{ji}} for 1 ⩽ j ⩽ d + 1 and a d + 1 square matrix C = (c_{ji}) over ℤ. The d + 2 square matrix of relations between the d + 2 generators −1, p_{1}, …, p_{d+1} of U_{ℚ,S}/U_{ℚ,S,T} is of the form

( \begin{matrix}2 & 0 \\ ? & C \end{matrix} )

where 2 is the top-left entry. Hence the index (U_{ℚ,S} : U_{ℚ,S,T}) = 2 det C. On the other hand we have

u_{1} ∧ ⋯ ∧ u_{d+1} = (det C)p_{1} ∧ ⋯ ∧ p_{d+1} + X

where X is a sum of terms of the form (−1) ∧ x, so that 2X = 0. Multiplying this by 2 gives the stated result.

Now we can show:

Theorem 5.10. Let F/ℚ be a real abelian extension with Galois group G and conductor m, r = 1, S_{1} = {∞}, S = {∞} ∪ {p | m}, T ⊈ {2} a finite nonempty set of primes of ℚ disjoint from S. Then F/ℚ, S ⊇ S_{1}, T, r satisfies Hypothesis 2.1. In this case the congruence of Conjecture 2.6 is satisfied up to a factor of 2. That is, for all ϕ ∈ Hom_{ℤ[G]}(U_{F,S,T} , ℤ[G]) we have

2ϕ(η) ≡ 2(±h_{ℚ,S,T}Reg_{G}^{ϕ}) (mod I_{G}^{d+1}).

Proof. Let ϕ ∈ Hom_{ℤ[G]}(U_{F,S,T} , ℤ[G]). Applying Proposition 5.7 to the map x ↦ ϕ(x^{δ_{T}}) and using Lemma 5.9 shows

2ϕ(δ_{T} N_{ℚ(ζ_{m})/F}β_{m}) ≡ 2\frac{aug(δ_{T})}{(U_{ℚ,S} : U_{ℚ,S,T})}Reg_{G}(ϕ̃(u_{1} ∧ ⋯ ∧ u_{d+1})) (mod I_{G}^{d+1}).

Finally, equation (3) for ℚ shows that h_{ℚ,S,T} = (−1)^{#T} aug(δ_{T})/(U_{ℚ,S} : U_{ℚ,S,T}).

Remark 5.11. In particular, if G is of odd order then Conjecture 2.6 holds.

6. Base change via a conjecture of Darmon

We now move on to studying what happens when we make a quadratic extension of the base field k in Conjecture 2.6.

6.1 Quadratic extension of the base field for L-functions Let K/k and L̃/k be linearly disjoint finite abelian extensions of global fields. Assume [K : k] = 2. Write L = L̃K. Hence we have the following diagram of fields:

[figure: K; L; k; L̃]

L/k is Galois with group Gal(L/k) = Gal(L̃/k) × Gal(K/k). Let ω be the nontrivial character of Gal(K/k), and let G := Gal(L/K) = Gal(L̃/k). Let S = S_{k} and T = T_{k} be disjoint finite sets of places of k with S nonempty and containing all infinite places. The Euler factors defining L functions in the extensions L̃/k and L/K are related as follows. Let v be a finite prime of k, then for each place w of K lying over v we have a Frobenius element Frob_{w} ∈ Gal(L/K) ↪ Gal(L/k). We compare these with Frob_{v} ∈ Gal(L/k) to obtain:

∏_{\begin{matrix}w place of K \\ w|v \end{matrix}}(1 − N_{w}^{s}Frob_{w}^{−1}) = (1 − N_{v}^{s}Frob_{v}^{−1})(1 − ω(v)N_{v}^{s}Frob_{v}^{−1}).  (12)

This follows by considering each Euler factor in the three cases ω(v) = 1 (v splits in K/k), ω(v) = −1 (v is inert) and ω(v) = 0 (v ramifies). We see that the L-functions satisfy the following base-change factorisation when passing from L/K to L/k:

Θ_{L/K,S_{K},T_{K}}(s) = Θ_{L/K/k,S_{k},T_{k}}(s, ω)Θ_{L̃/k,S_{k},T_{k}}(s),  (13)

where Θ_{L/K/k,S,T} (s, ω) is the twisted Stickelberger function defined as

( ∏_{t∈T_{k}}(1 − ω(t)N_{t}^{1−s}Frob_{t}^{−1})) ∑_{χ∈Ĝ} L_{L/k,S_{k}}(s, ωχ^{−1})e_{ωχ}(L/k).

= ( ∏_{t∈T_{k}}(1 − ω(t)N_{t}^{1−s}Frob_{t}^{−1}))( ∏_{v∉S_{k}}(1 − ω(v)N_{v}^{−s}Frob_{v}^{−1}))^{−1}.

The validity of equation (13) follows from the lemma and the definition (4) in the region of convergence Re s > 1 and then everywhere by meromorphic continuation.

We will also use the notation

δ_{T}^{ω} := ∏_{v∈T_{k}}(1 − ω(v)N_{v}Frob_{v}^{−1})

for the relative T -modification factor at s = 0.

6.2 The circular unit Here we show how the “circular unit” defined in [Da95] corresponds to the change in L-functions which results from raising the base field from ℚ to a linearly disjoint real quadratic field.

For comparison with [Da95], we assume the following hypothesis for the rest of §6.

Hypothesis 6.1 (Darmon’s setup). Let N and S be coprime integers with N > 1 and S > 1. Let ω be a primitive, quadratic, even Dirichlet character defined modulo N. Set K = ℚ(ζ_{N})^{ker ω}, a real quadratic field, and call its nontrivial automorphism τ. Let L̃ be a real subfield of ℚ(ζ_{S}), normal over ℚ. Write L = L̃K.

Hence we are in the situation of §6.1 with the further assumptions that k = ℚ and that L̃ and K are totally real and have coprime conductors. We define the set S_{ℚ} from the integer S in the obvious way: S_{ℚ} = {p | S, ∞}.

All the characters of these extensions come from even Dirichlet characters because the fields are totally real. Since, by (10), L-functions of even characters vanish at s = 0, differentiating (13) twice shows that we have the following equality in ℂ[G]

Θ_{L/K,S_{K},T_{K}}^{2}(0) = \frac{1}{2}Θ′′_{L/K,S_{K},T_{K}}(0) = Θ′_{L/K/ℚ,S_{ℚ},T_{ℚ}}(0, ω)Θ′_{L̃/ℚ,S_{ℚ},T_{ℚ}}(0).  (14)

We now relate the base-change factor Θ′_{L/K/ℚ,S_{ℚ},T_{ℚ}}(0, ω) to the circular unit defined in [Da95], §4. This is the following element of K_{S} := K(ζ_{S}).

α_{S} := ∏_{σ∈Gal(ℚ(ζ_{NS})/ℚ(ζ_{S}))}σ(ζ_{NS} − 1)^{ω(σ)} ∈ U_{K_{S}}.

Write ∞_{L} for the place of L corresponding to the embedding of L into ℝ given by ζ_{NS} ↦ e^{2πi/NS}, and \overline{∞_{L}} its conjugate by τ.

Lemma 6.2. Θ′_{L/K/ℚ,S_{ℚ},∅}(0, ω) · (∞_{L} − \overline{∞_{L}}) = \frac{1}{2}λ_{S_{L}}(N_{K_{S}/L}α_{S}).

Proof. Because NS is not a prime power, (ζ_{NS} −1) is a global unit in ℚ(ζ_{NS}). Hence λ_{S_{L}}(N_{K_{S}/L}α_{S}) is zero outside the archimedean places. Now

λ_{S_{L}}(N_{K_{S}/L}α_{S}) = − ∑_{γ∈Gal(L/ℚ)}ln |γ^{−1}N_{K_{S}/L}α_{S}|γ∞_{L} = − ∑_{γ∈G} ln |γ^{−1}N_{K_{S}/L}α_{S}|γ(∞_{L} − \overline{∞_{L}}),

since the generator of Gal(K/ℚ) inverts α_{S}. For χ a character of L/K, it suffices to prove

2Θ′_{L/K/ℚ,S_{ℚ},∅}(0, ω)e_{ωχ}(L/ℚ) = − ∑_{γ∈G} ln |γ^{−1}N_{K_{S}/L}α_{S}|ωχ(γ)e_{ωχ}(L/ℚ).

This is easy to show using the formulas (10) for values of Dirichlet L-series at s = 0, as in Lemma 5.1.

Definition 6.3. We set η_{ω} := δ_{T}^{ω}N_{K_{S}/L}α_{S}.

Then Θ′_{L/K/ℚ,S_{ℚ},T_{ℚ}}(0, ω) · (∞_{L} − \overline{∞_{L}}) = \frac{1}{2}λ_{S_{L}}(η_{ω}).

6.3 Calculation of η By Hypothesis 6.1, Gal(L/ℚ) = Gal(L̃/ℚ)×Gal(K/ℚ) and the restriction map Gal(L/K) = G ⟶ Gal(L̃/ℚ) is an isomorphism. Let #S_{s} = #S_{split} be the number of primes p dividing S with ω(p) = 1, and #S_{i} = #S_{inert} the number of p with ω(p) = −1.

We consider Conjecture 2.6 for the extensions L/K and L̃/ℚ in turn. We then show that the results and conjecture of [Da95] relate them.

– L/K: Since K is a real quadratic field, there are two infinite places of K, which we call ∞_{K} and \overline{∞_{K}}. Because L is also real, these split completely in L/K. To avoid confusion, we write r′ and d′ for the “r” and “d” of §2 for the extension L/K. We take r′ = 2 and S_{1,K} = \{∞_{K}, \overline{∞_{K}}\}. The element η_{L/K} ∈ ℂ ∧_{G}^{2} U_{K,S_{K},T_{K}}e_{2} is defined by:

Θ_{L/K,S_{K},T_{K}}^{2}(0)(∞_{L} − w_{0}) ∧ (\overline{∞_{L}} − w_{0}) = λ_{S_{L}}^{(2)}(η_{L/K})

for some finite place w_{0} of S_{L}.

– L̃/ℚ: In the notation of §2 we take r = 1 and S_{1,ℚ} = {∞}. Then d = #S_{s} + #S_{i}. The element η_{L̃/ℚ} ∈ ℂU_{ℚ,S_{ℚ},T_{ℚ}}e_{1}(L̃/ℚ) is defined by

Θ_{L̃/ℚ,S_{ℚ},T_{ℚ}}^{1}(0)(∞L̃ − v_{0}) = λ_{SL̃}(η_{L̃/ℚ}),

for some finite place v_{0} ∈ SL̃.

The results of §6.2 allow us to express η_{L/K} in terms of η_{ω} and η_{L̃/ℚ}. We have:

2Θ_{L/K,S_{K},T_{K}}^{2}(0)(∞_{L} − w_{0}) ∧ (\overline{∞_{L}} − w_{0}) =

Θ_{L/K/ℚ,S_{ℚ},T_{ℚ}}^{1}(0, ω)(∞_{L} − \overline{∞_{L}}) ∧ Θ′_{L̃/ℚ,S_{ℚ},T_{ℚ}}(0)(∞_{L} + \overline{∞_{L}} − 2w_{0}). (15)

We now calculate λ_{S_{L}}(η_{L̃/ℚ}). Let us review the various identifications and inclusions. Along with the canonical identification of Galois groups G = Gal(L̃/ℚ), we define homomorphisms UL̃ ↪ U_{L} (inclusion) and

\begin{aligned}i_{L/L̃} : & Y_{SL̃} & ⟶ & Y_{S_{L}} \\ & v & ⟼ & w + w̄, \end{aligned}

where w is a place of L chosen arbitrarily above the place v of L̃, and w̄ = w^{τ}. With these maps, the following diagrams commute:

\begin{matrix}Gal(L̃/ℚ) × Y_{SL̃} & ⟶ & Y_{SL̃} & UL̃ & ⟶^{λ_{SL̃}} & ℝX_{SL̃} \\ ↓ & □ & ↓i_{L/L̃} & ↓ & □ & ↓i_{L/L̃} \\ G × Y_{S_{L}} & ⟶ & Y_{S_{L}}, & U_{L} & ⟶^{λ_{S_{L}}} & ℝX_{S_{L}}. \end{matrix}  (16)

Hence λ_{S_{L}}(η_{L̃/ℚ}) = Θ′_{L̃/ℚ,S_{ℚ},T_{ℚ}}(0)(∞_{L} + \overline{∞_{L}} − w_{1} − \overline{w_{1}}), where w_{1} is a place of L chosen above v_{0}. If we assume #S_{i} ≠ 0, we can choose v_{0} such that w_{1} = \overline{w_{1}}. Then setting w_{0} = w_{1} in (15) shows that

λ_{S_{L}}^{(2)}(η_{ω} ∧ η_{L̃/ℚ}) = 4Θ_{L/K,S_{K},T_{K}}^{2}(0)(∞_{L} − w_{0}) ∧ (\overline{∞_{L}} − w_{0}),

and hence 4η_{L/K} = η_{ω} ∧ η_{L̃/ℚ}.

In [Da95], Darmon makes a congruence conjecture for his circular unit. We propose to interpret this as a base-change statement for Conjecture 2.6, under the following assumptions:

Hypothesis 6.4. i) #S_{i} ≠ 0 ii) For every place p in T_{ℚ}, we have ω(p) = 1.

For a group U on which τ acts, we define U^{−} = {u ∈ U : u^{τ} = u^{−1}}.

Proposition 6.5. Assuming Hypothesis 6.4, δ_{T}^{ω}(U_{L,S}^{−}) ⊆ U_{L,S,T}^{−}. Therefore η_{ω} ∈ U_{L,S,T}^{−}.

Proof. Let x ∈ U_{L,S}^{−} and y = x^{δ_{T}^{ω}}. Let v ∈ T split into w, w̄ in K. Then δ_{T}^{ω} contains a factor (1−N_{w}Frob_{w}^{−1}) by (12). Hence w(y−1) > 0. But we also have w̄(y−1) = w(y^{−1}−1) = w((1−y)/y) = w(1 − y) > 0. Therefore y ≡ 1 (mod t) for all t ∈ T_{K} as required. Setting x = N_{K_{S}/L}α_{S} proves the second assertion.

6.4 Indices of minus units Let K/k be a quadratic Galois extension of global fields with Galois group generated by τ. For this section we only need to assume that S is a finite, nonempty set of places of k containing all infinite places, and that T is any finite disjoint set of places of k.

Lemma 6.6. (U_{K,S} : U_{K,S,T}) = (U_{k,S} : U_{k,S,T})(U_{K,S}^{1−τ} : U_{K,S,T}^{1−τ})

Proof. Consider the following commutative diagram, in which the rows are exact.

0 ⟶ U_{k,S,T} ⟶ U_{K,S,T} ⟶^{1−τ} U_{K,S,T}^{1−τ} ⟶ 0
↓ ↓ ↓
0 ⟶ U_{k,S} ⟶ U_{K,S} ⟶^{1−τ} U_{K,S}^{1−τ} ⟶ 0,

The vertical arrows are inclusions. Applying the snake lemma, we obtain the exact sequence

0 ⟶ U_{k,S}/U_{k,S,T} ⟶ U_{K,S}/U_{K,S,T} ⟶^{1−τ} U_{K,S}^{1−τ} /U_{K,S,T}^{1−τ} ⟶ 0.

This shows the result.

We consider the subgroup U_{K,S,T}^{−} := {u ∈ U_{K,S,T} : u^{τ} = u^{−1}} of “minus S-units” in U_{K,S,T}. This contains U_{K,S,T}^{1−τ}. The quotient is, by Tate’s finite group cohomology ([Se79], Chapter VIII),

U_{K,S,T}^{−} /U_{K,S,T}^{1−τ} = H^{1}(⟨τ⟩, U_{K,S,T}).

Lemma 6.7. Suppose h_{k,S} = 1. Then:

i) H^{1}(⟨τ⟩, U_{K,S,∅}) = 0. ii) (U_{K,S}^{1−τ} : U_{K,S,T}^{1−τ}) = (U_{K,S}^{−} : U_{K,S,T}^{−})#H^{1}(⟨τ⟩, U_{K,S,T}).

Proof. Corollary 2 of [Ri65] shows that H^{1}(⟨τ⟩, U_{K,S,∅}) embeds into the S-class group of k, which is trivial. This shows (i), and (ii) follows immediately.

Finally we adapt the method of [Ta84] (§II.2 and Theorem IV.5.4) to show the following. Let n denote the number of places of S which split in K/k.

Lemma 6.8.

(U_{K,S,T} : U_{k,S,T} U_{K,S,T}^{−}) = \frac{2^{n}#(U_{K,S,T}^{−} ∩ \{±1\})}{#H^{1}(⟨τ⟩, U_{K,S,T})}

Proof. The sequence

0 ⟶ U_{k,S,T} U_{K,S,T}^{−} ⟶ U_{K,S,T} ⟶^{1−τ} \frac{U_{K,S,T}^{1−τ}}{(U_{K,S,T}^{−})^{2}} ⟶ 0

is exact. This shows that (U_{K,S,T} : U_{k,S,T} U_{K,S,T}^{−}) = (U_{K,S,T}^{1−τ} : (U_{K,S,T}^{−})^{2}). We also have:

(U_{K,S,T}^{−} : (U_{K,S,T}^{−})^{2}) = (U_{K,S,T}^{−} : U_{K,S,T}^{1−τ})(U_{K,S,T}^{1−τ} : (U_{K,S,T}^{−})^{2}).

The first factor on the right is #H^{1}(⟨τ⟩, U_{K,S,T}). The factor on the left can be calculated from the standard decomposition of the finitely generated abelian group U_{K,S,T}^{−}. The free rank of this group is n, and the cokernel of squaring on the torsion part has the same size as the kernel, which is U_{K,S,T}^{−} ∩ {±1}. The second factor on the right is what we want to calculate. Hence

2^{n}#(U_{K,S,T}^{−} ∩ {±1}) = #H^{1}(⟨τ⟩, U_{K,S,T})(U_{K,S,T} : U_{k,S,T} U_{K,S,T}^{−}),

as required.

6.5 Darmon’s conjecture We return to the situation of Hypothesis 6.1. We first state Darmon’s conjecture in our notation.

Write Γ := Gal(K(ζ_{S})/K). For each prime l_{i}|S such that ω(l_{i}) = 1, l_{i} splits into two distinct places λ_{i} and λ̄_{i} in K. Darmon claims that U_{K,S,∅}^{−} is a free ℤ-module ([Da95], §4), but this is not in fact the case since it contains −1, so actually U_{K,S,∅}^{−} ≅ ℤ/2ℤ × ℤ^{#S_{s}+1}. Taking either T = ∅ or T such that U_{K,S,T} is torsion-free, choose a basis u_{1}, …, u_{#S_{s}+1} for a maximal free subgroup of U_{K,S,T}^{−}, which will have index 2 if T = ∅. Following Darmon, we define a regulator

R_{S,T} := ∑_{i=1}^{#S_{s}+1}(−1)^{i+1}u_{i} ⊗ det(f_{λ_{k}}(u_{j}) − 1)_{k,j} ∈ U_{K,S,T} ⊗ \frac{I_{Γ}^{#S_{s}}}{I_{Γ}^{#S_{s}+1}} ,  (17)

where, in the matrix, k runs from 1 to #S_{s} and j runs from 1 to #S_{s} + 1, omitting i. Note that R_{S,∅} might depend upon the choice of maximal free subgroup if the torsion element −1 is not in the kernel of the local Artin maps.

We state Darmon’s conjecture [Da95], Conjecture 4.3, under the ring automorphism involution of ℤ[Γ] given by g ↦ g^{−1}, which amounts to a sign change in the statement, and then ignoring all issues of sign.

Conjecture 6.9 (Darmon). We have the following equality in U_{K(ζ_{S}),S} ⊗ I_{Γ}^{#S_{s}}/I_{Γ}^{#S_{s}+1}:

∑_{σ∈Γ} σ^{−1}α_{S} ⊗ σ = ±2^{#S_{i}+1}h_{K,S}R_{S,∅}.

We consider a T -modified version. This will fit with our general framework, and avoids the problem of torsion in the unit group. We assume Hypothesis 6.4 (ii) which implies that each v in T splits into w and w̄ in K, with N_{w} = N_{v}. Then by (3), we have the following:

h_{ℚ,S,T} = h_{ℚ,S} · \frac{∏_{v∈T} (N_{v} − 1)}{(U_{ℚ,S} : U_{ℚ,S,T})}, h_{K,S,T} = h_{K,S} · \frac{∏_{v∈T} (N_{v} − 1)^{2}}{(U_{K,S} : U_{K,S,T})} ,

where we note h_{ℚ,S} = 1. The quotient is

\frac{h_{K,S,T}}{h_{ℚ,S,T}} = h_{K,S} · \frac{∏_{v∈T} (N_{v} − 1)}{(U_{K,S}^{1−τ} : U_{K,S,T}^{1−τ})} = h_{K,S} · \frac{∏_{v∈T} (N_{v} − 1)}{(U_{K,S}^{−} : U_{K,S,T}^{−})#H^{1}(⟨τ⟩, U_{K,S,T})},  (18)

using Lemmas 6.6 and 6.7 (ii).

Lemma 6.10. Under hypothesis 6.4,

\frac{h_{K,S,T}}{h_{ℚ,S,T}} #H^{1}(⟨τ⟩, U_{K,S,T})

is an integer.

Proof. For T empty, this is clear. Now let T = {v_{1}, …, v_{n}} and choose a place w_{i} of K above each v_{i}. Let K(w_{i}) be the residue field of K at w_{i}. Then the natural sequence

0 ⟶ U_{K,S,T}^{−} ⟶ U_{K,S}^{−} ⟶ ⨁_{i=1}^{n} K(w_{i})^{×}

is exact. For if u ∈ U_{K,S}^{−} reduces to 1 modulo each w_{i}, then \overline{w_{i}}(u − 1) = w_{i}(ū − 1) = w_{i}((1 − u)/u) = w_{i}(1 − u) > 0, as in Proposition 6.5. Hence u ∈ U_{K,S,T} ∩ U_{K,S}^{−} = U_{K,S,T}^{−}.

This shows that (U_{K,S}^{−} : U_{K,S,T}^{−}) | ∏_{v∈T} (N_{v} − 1), and by (18) this gives the result.

We propose the following slight modification of Darmon’s conjecture:

Conjecture 6.11. Assume T satisfies Hypothesis 6.4 (ii). Then we have the following equality in U_{K(ζ_{S}),S,T} ⊗ I_{Γ}^{#S_{s}}/I_{Γ}^{#S_{s}+1}:

∑_{σ∈Γ} σ^{−1}α_{S}^{δ_{T}^{ω}} ⊗ σ = ±2^{#S_{i}} \frac{h_{K,S,T}}{h_{ℚ,S,T}} #H^{1}(⟨τ⟩, U_{K,S,T})#(U_{K,S,T})_{tors}R_{S,T} .  (19)

If we put T = ∅ in this statement, then h_{ℚ,S,T} = 1, #H^{1}(⟨τ⟩, U_{K,S,T}) = 1 (by Lemma 6.7 (i)), and #(U_{K,S,T})_{tors} = 2. Hence we recover Conjecture 6.9. On the other hand, let us look at how Conjecture 6.11 varies when we replace T by T ∪ {v}. If T is empty, then for the comparison statement we will have to assume that the regulator in Conjecture 6.9 is calculated with respect to a maximal free subgroup of U_{K,S} which contains U_{K,S,{v}}. Examining how the various factors change on increase of T shows that Conjecture 6.11 behaves well, and that it follows from Conjecture 6.9 when U_{K,S,T}^{−} can be embedded in a maximal free submodule of U_{K,S}^{−}.

The consequence of Darmon’s conjecture that we wish to use is the following.

Proposition 6.12. Assume the setup of Hypothesis 6.1 and let T satisfy Hypothesis 6.4 (ii) with U_{L,S,T} torsion-free. Then Conjecture 6.11 implies that for each ϕ ∈ Hom_{G}(U_{L,S,T} , ℤ[G]), we have

ϕ(η_{ω}) ≡ ±2^{#S_{i}} \frac{h_{K,S,T}}{h_{ℚ,S,T}} #H^{1}(⟨τ⟩, U_{K,S,T})ϕ^{1}(R_{S,T}) (mod I_{G}^{#S_{s}+1}).

Proof. Denote as usual G = Gal(L/K). We apply the natural projection U_{K(ζ_{S}),S} ⊗ ℤ[Γ] ⟶ U_{K(ζ_{S}),S}⊗ℤ[G], which maps the left-hand side of (19) to ∑_{σ∈G} σ^{−1}(δ_{T}^{ω}N_{K(ζ_{S})/L}α_{S})⊗σ in U_{L,S}⊗ℤ[G]. Then Conjecture 6.11 implies the following equality in U_{L,S} ⊗ I_{G}^{#S_{s}}/I_{G}^{#S_{s}+1}:

∑_{σ∈G} σ^{−1}η_{ω} ⊗ σ = ±2^{#S_{i}} \frac{h_{K,S,T}}{h_{ℚ,S,T}} #H^{1}(⟨τ⟩, U_{K,S,T})R_{S,T} ,

with η_{ω} from Definition 6.3.

Recall the isomorphism (5). Applying the homomorphism

ϕ^{1} ⊗ id : U_{L,S,T} ⊗ I_{G}^{#S_{s}}/I_{G}^{#S_{s}+1} ⟶ I_{G}^{#S_{s}}/I_{G}^{#S_{s}+1}

gives the stated result.

6.6 Factorisation of the regulator We assume T is such that U_{L,S,T} is torsion-free, and that Hypothesis 6.4 is satisfied. We let u_{1}, …, u_{#S_{s}+1} be a basis for U_{K,S,T}^{−} and u_{#S_{s}+2}, …, u_{2+d′} be a basis for U_{ℚ,S,T}. Then these u_{i} form a basis for U_{ℚ,S,T}U_{K,S,T}^{−}. The index of this group in U_{K,S,T} was calculated in Lemma 6.8. We calculate the regulator from Conjecture 2.6 for these u_{i}. Let Φ = ϕ_{1}∧ϕ_{2} ∈ ∧_{G}^{2}Hom_{ℤ[G]}(U_{L,S,T} , ℤ[G]). Let R_{S,T} be the regulator defined in §6.5 in terms of the u_{i}. Write u_{ℚ} = u_{#S_{s}+2} ∧ ⋯ ∧ u_{2+d′}.

Recall that for each prime l_{i}|S such that ω(l_{i}) = 1, l_{i} splits into distinct places λ_{i}, λ̄_{i} in K. The other #S_{i} primes dividing S are inert in K/ℚ, and will be denoted q_{1}, …, q_{#S_{i}}. For reference, we summarise S_{ℚ} and S_{K}:

S_{ℚ} = {∞, l_{1}, …, l_{#S_{s}}, q_{1}, …, q_{#S_{i}}},
S_{1,ℚ} = {∞}, r = 1, #S_{ℚ} = r + d + 1, so d = #S_{s} + #S_{i};
S_{K} = \{∞_{L}, \overline{∞_{L}}, λ_{1}, …, λ_{#S_{s}}, λ̄_{1}, …, λ̄_{#S_{s}}, q_{1}, …, q_{#S_{i}}\},
S_{1,K} = \{∞_{L}, \overline{∞_{L}}\}, r′ = 2, #S_{K} = r′ + d′ + 1, so d′ = 2#S_{s} + #S_{i}.

Proposition 6.13. We have the following equality in ℤ[G]/I_{G}^{d′+1}:

Reg_{L/K}(Φ̃(u_{1} ∧ ⋯ ∧ u_{2+d′})) = ±2^{#S_{s}} | \begin{matrix}2^{#S_{i}−1}ϕ_{1}^{1}(R_{S,T}) & 2^{#S_{i}−1}ϕ_{2}^{1}(R_{S,T}) \\ Reg_{L̃/ℚ}(ϕ̃_{1}(u_{ℚ})) & Reg_{L̃/ℚ}(ϕ̃_{2}(u_{ℚ})) \end{matrix} | .

Proof. We saw in Lemma 2.4 that

Φ̃(u_{1} ∧ ⋯ ∧ u_{2+d′}) = ∑_{σ∈\left[{2+d′ \atop 2}\right]} sign(σ) | \begin{matrix}ϕ_{1}^{1}(u_{σ(1)}) & ϕ_{2}^{1}(u_{σ(1)}) \\ ϕ_{1}^{1}(u_{σ(2)}) & ϕ_{2}^{1}(u_{σ(2)}) \end{matrix} | u_{σ(3)} ∧ ⋯ ∧ u_{σ(2+d′)}.  (20)

The terms u_{σ(3)}∧⋯∧u_{σ(2+d′)} are made by choosing two of the u_{i} for the integer determinant. So each σ excludes 0, 1 or 2 units of the U_{K,S,T}^{−} basis from the wedge of units. Let m_{σ} = #(σ({3, 4, …, 2 + d′}) ∩ {1, …, #S_{s} + 1}) be the number of minus-units included in u_{σ(3)} ∧ ⋯ ∧ u_{σ(2+d′)} in the term corresponding to σ, so m_{σ} = #S_{s} − 1, #S_{s} or #S_{s} + 1.

We calculate our matrix with respect to the following places of K, using Hypothesis 6.4 to exclude q_{#S_{i}}:

λ_{1}, …, λ_{#S_{s}}, λ̄_{1}, …, λ̄_{#S_{s}}, q_{1}, …, q_{#S_{i}−1}.

This means we have the determinant of the following d′×d′ matrix to calculate for Reg_{L/K}(u_{σ(3)}∧ ⋯ ∧ u_{σ(2+d′)}):

\begin{aligned}& λ_{1}, …, λ_{#S_{s}} & λ̄_{1}, …, λ̄_{#S_{s}} & q_{1}, …, q_{#S_{i}−1} \\ u_{σ(3)} ; ⋮ ; u_{σ(m_{σ}+2)} & f_{λ_{j}}(u_{i}) − 1 & fλ̄_{j}(u_{i}) − 1 & f_{q_{j}}(u_{i}) − 1 \\ u_{σ(m_{σ}+3)} ; ⋮ ; u_{σ(2+d′)} & f_{λ_{j}}(u_{i}) − 1 & fλ̄_{j}(u_{i}) − 1 & f_{q_{j}}(u_{i}) − 1 \end{aligned}  (21)

where the units at the top are in U_{K,S,T}^{−} and the units at the bottom are in U_{ℚ,S,T}. We will distinguish between the cases where m_{σ} takes the different values.

First consider the case m_{σ} = #S_{s} + 1. We may add the column for λ_{j} to the column for \overline{λ_{j}} for j = 1, …, #S_{s} without altering the value of the determinant. The (i, j)-th entry in the top-centre (#S_{s} + 1) × (#S_{s}) block is then congruent mod I_{G}^{2} to f_{λ_{j}}(ū_{i}u_{i}) − 1 = 0. On the other hand, for each q_{j}, the local extension K_{q_{j}}/ℚ_{q_{j}} has degree 2, and f_{q_{j}}(u_{i}) only depends on the norm of u_{i} in this local extension. If u_{i} ∈ U_{K,S,T}^{−} then u_{i}^{1+τ} = 1. Hence f_{q_{j}}(u_{i}) − 1 = 0 for these u_{i}. Therefore the entire top-right (#S_{s} + 1) × (#S_{s} + #S_{i} − 1) block is zero. Hence there are at most #S columns which are nonzero in their first #S + 1 rows. Therefore the determinant is zero.

Now in (21) we subtract the column for λ̄_{j} from the column for λ_{j} for j = 1, …, #S_{s} to show the determinant is the same as the determinant of the following matrix.

\begin{aligned}& λ_{1}, …, λ_{#S_{s}} & λ̄_{1}, …, λ̄_{#S_{s}} & q_{1}, …, q_{#S_{i}−1} \\ u_{σ(3)} ; ⋮ ; u_{σ(m_{σ}+2)} & f_{λ_{j}}(u_{i}) − fλ̄_{j}(u_{i}) & f_{λ_{j}}(u_{i}) − 1 & f_{q_{j}}(u_{i}) − 1 \\ u_{σ(m_{σ}+3)} ; ⋮ ; u_{σ(2+d′)} & 0 & fλ̄_{j}(u_{i}) − 1 & f_{q_{j}}(u_{i}) − 1 \end{aligned}  (22)

If m_{σ} = #S_{s} − 1, the first #S_{s} columns have all zeros except perhaps in the first #S_{s} − 1 rows. Therefore the determinant is again zero.

We are left with the case m_{σ} = #S_{s}. In this case the matrix (22) is block-upper-triangular. Let us consider the top-left (#S_{s})×(#S_{s}) block first. We note that f_{λ_{j}}(u_{σ(i)})−fλ̄_{j}(u_{σ(i)}) ≡ 2(f_{λ_{j}}(u_{σ(i)})−1) (mod I_{G}^{2}). So the top-left block has determinant 2^{#S_{s}} det(f_{λ_{j}}(u_{σ(i)}) − 1)_{i,j}. Note the relation to the regulator R_{S,T} of (17).

Now we calculate the determinant of the bottom-right block. We have fλ̄_{j}(u) = f′_{l_{j}}(u), f_{q_{j}}(u) = f′_{q_{j}}(u)^{2} for each j and each u appearing, where the f′ denote the local symbols coming from the extension L̃/ℚ. So the bottom-right block is 2^{#S_{i}−1}Reg_{L̃/ℚ}(u_{σ(m_{σ}+3)} ∧ ⋯ ∧ u_{σ(2+d′)}).

Referring back to (20), the only terms which appear in the sum after applying Reg_{L/K} are those for σ such that σ(1) ⩽ #S_{s} + 1 and σ(2) > #S_{s} + 1. We put this in correspondence with a pair (i, j), 1 ⩽ i ⩽ #S_{s} + 1, 1 ⩽ j ⩽ d + 1 such that σ(1) = i, σ(2) = #S_{s} + 1 + j. Then one may check that sign(σ) = (−1)^{#S_{s}}(−1)^{i+1}(−1)^{j+1}.

Putting all this together with (20) gives the stated result, with sign (−1)^{#S_{s}} on the right.

6.7 Base change for the congruence We are now ready to show the base change statement for Conjecture 2.6.

Theorem 6.14. We use the setup of Hypothesis 6.1, assume Hypothesis 6.4 and use the definition of u_{i} from §6.6.

Assume Conjecture 2.6 holds for the extension L̃/ℚ, ie. that

ϕ(η_{L̃/ℚ}) ≡ ±h_{ℚ,S,T}Reg_{L̃/K}(ϕ̃(u_{#S_{s}+2} ∧ ⋯ ∧ u_{2+d′})) (mod I_{G}^{d+1}),

for all ϕ ∈ Hom_{G}(U_{L̃,S,T} , ℤ[G]). Assume also that the modified Darmon conjecture 6.11 holds. Then Conjecture 2.6 holds for the extension L/K up to a power of 2. Explicitly, for all Φ ∈ ∧_{G}^{2}Hom_{G}(U_{L,S,T} , ℤ[G]), we have

4 · 2^{#S_{s}}Φ(η_{L/K}) ≡ ±4 · 2^{#S_{s}}h_{K,S,T} Reg_{L/K}(Φ̃(ϵ_{1} ∧ ⋯ ∧ ϵ_{2+d′})) (mod I_{G}^{d′+1}),

where the ϵ_{i} form a ℤ-basis for U_{K,S,T}.

Proof. Write u = u_{1} ∧ ⋯ ∧ u_{2+d′}, ϵ = ϵ_{1} ∧ ⋯ ∧ ϵ_{2+d′}. Set Φ = ϕ_{1} ∧ ϕ_{2}, and recall from §6.3 that 4η_{L/K} = η_{ω} ∧ η_{L̃/ℚ}. The conjectures tell us, using Proposition 6.12,

4Φ(η_{L/K}) = | \begin{matrix}ϕ_{1}(η_{ω}) & ϕ_{2}(η_{ω}) \\ ϕ_{1}(η_{L̃/ℚ}) & ϕ_{2}(η_{L̃/ℚ}) \end{matrix} | ≡ ± | \begin{matrix}2^{#S_{i}} \frac{h_{K,S,T}}{h_{ℚ,S,T}} #H^{1}ϕ_{1}^{1}(R_{S,T}) & 2^{#S_{i}} \frac{h_{K,S,T}}{h_{ℚ,S,T}} #H^{1}ϕ_{2}^{1}(R_{S,T}) \\ h_{ℚ,S,T}Reg_{L̃/ℚ}(ϕ̃_{1}(u_{ℚ})) & h_{ℚ,S,T}Reg_{L̃/ℚ}(ϕ̃_{2}(u_{ℚ})) \end{matrix} |

modulo I_{G}^{d+1+#S_{s}+1}, where #H^{1} = #H^{1}(⟨τ⟩, U_{K,S,T}) and R_{S,T} is calculated with respect to the ℤ-basis u_{1}, …, u_{#S_{s}+1} of U_{K,S,T}^{−}. Noting that d′ = d + #S_{s}, this is

≡ ±2^{#S_{i}}h_{K,S,T} #H^{1} | \begin{matrix}ϕ_{1}^{1}(R_{S,T}) & ϕ_{2}^{1}(R_{S,T}) \\ Reg_{L̃/ℚ}(ϕ̃_{1}(u_{ℚ})) & Reg_{L̃/ℚ}(ϕ̃_{2}(u_{ℚ})) \end{matrix} | (mod I_{G}^{d′+2}).

Hence by Proposition 6.13, we have 4·2^{#S_{s}}Φ(η_{L/K}) ≡ ±h_{K,S,T} ·2#H^{1}Reg(Φ̃(u)) (mod I_{G}^{d′+1}). Now we know from Lemma 6.8 that (U_{K,S,T} : U_{ℚ,S,T}U_{K,S,T}^{−}) = 2^{#S_{s}+1}/#H^{1}, so u = (2^{#S_{s}+1}/#H^{1})ϵ. This gives the result.

Note that if #G is odd, this last congruence is the full statement of Conjecture 2.6.

Remark 6.15. Using the method of [Da95], Lemma 8.1, it is possible to prove that ϕ(N_{K(ζ_{S})/L}α_{S}) ∈ I_{G}^{#S_{s}} for all ϕ ∈ Hom_{G}(U_{L,S}^{−}, ℤ[G]), without assuming the validity of Darmon’s conjecture. It then follows that ϕ′(η_{ω}) ∈ I_{G}^{#S_{s}} for all ϕ′ ∈ Hom_{G}(U_{L,S,T} , ℤ[G]). Thus if Conjecture 2.6 holds for L̃/ℚ, then for L/K we have 4Φ(η_{L/K}) ∈ I_{G}^{d′}, for all Φ ∈ ∧_{G}^{2}Hom_{G}(U_{L,S,T} , ℤ[G]), consistent with the “order of vanishing” implied by Conjecture 2.6.

7. Base change via Gross’s conjecture on the L-functions of tori

In §8 of [Gr88], Gross makes a conjecture motivated by considering algebraic tori. Similarly to Darmon’s later conjecture, this involves a quadratic extension of the base field and consideration of the “minus-units” in this extension. It also involves a “Θ” element which is twisted by the nontrivial character of the extension. In the previous section we saw that Darmon’s conjecture, which was related to the first derivative of the relative factor in (14), gave us a base change property for Conjecture 2.6 where the order of vanishing, r, increased by 1. Gross’s conjecture, by contrast, concerns the value (zero-th derivative) of the relative factor and correspondingly it gives us a base change property where r doesn’t change.

7.1 Setup and calculation of η Our setup is as follows. Let k be a global field and L̃/k and K/k linearly disjoint abelian extensions, with [K : k] = 2. Let ω : Gal(K/k) ⟶ {±1} be the nontrivial character of K/k. Setting L = L̃K, we are in the situation of §6.1. Write G = Gal(L/K) = Gal(L̃/k). Let S = S_{k} be a set of places of k containing all infinite places and all places ramifying in L/k. That is, both L̃/k and K/k are unramified outside S. Take T = T_{k} such that U_{L,S,T} is torsion-free. We define n to be the number of places in S splitting in the quadratic extension K/k, and refer to the other #S − n inert or ramified places as nonsplit. We write τ for the nontrivial automorphism of this extension. This is the situation of §8 of [Gr88] except we have an unfortunate clash of notation, summarised in the following table.

Gross’s notation L K χ σ ; Our notation K L̃ ω τ

Assume there are r places S_{1,k} in S_{k} splitting completely in L̃/k. Then all the places above these in K split completely in L/K, and there are at least r of them, so the two sets of data L/K, S_{K} ⊇ S_{1,K}, T_{K}, r and L̃/k, S_{k} ⊇ S_{1,k}, T_{k}, r both satisfy Hypothesis 2.1. Differentiating the base-change factorisation of the L-functions (13) r times and evaluating at s = 0 gives

Θ_{L/K,S_{K},T_{K}}^{r}(0) = Θ_{L/K/k,S_{k},T_{k}}(0, ω)Θ_{L̃/k,S_{k},T_{k}}^{r}(0).  (23)

The base-change factor Θ_{L/K/k,S_{k},T_{k}}(0, ω) lies in ℤ[G] by the argument following [Gr88] (8.7), where the corresponding element is denoted θ_{G}(χ). Gross’s tori conjecture concerns this element, and we will show that its validity would imply that Conjecture 2.6 for L/K, S_{K} ⊇ S_{1,K}, T_{K}, r (weakened by powers of 2, similarly to the case of Darmon’s conjecture) follows from the conjecture for L̃/k, S_{k} ⊇ S_{1,k}, T_{k}, r.

First note that we may assume that the r places in S_{1,k} are nonsplit in K/k, since otherwise more than r places in S_{K} split in L/K and Conjecture 2.6 already holds for L/K, S_{K}, T_{K}, r by Proposition 3.10. We also impose the following assumption, which is the same as Hypothesis 6.4 (i).

Hypothesis 7.1. There is a place in S_{k} − S_{1,k} which is nonsplit in K/k. That is, d ⩾ n.

Let v_{0} be such a place. Write S_{1,k} = {v_{1}, …, v_{r}}. Choose w_{i} a place of L above v_{i} for i = 0, 1, …, r. Set b_{L} = (w_{1} − w_{0}) ∧ ⋯ ∧ (w_{r} − w_{0}). Write w̃_{i} for the place of L̃ induced by w_{i} for i = 0, …, r, and set bL̃ = (w̃_{1} − w̃_{0}) ∧ ⋯ ∧ (w̃_{r} − w̃_{0}). Then with these choices of the W in the definition of η, we have that η_{L̃/k} ∈ ℂ ∧_{ℤ[G]}^{r} U_{L̃,S,T} is defined by the equation

λL̃(η_{L̃/k}) = Θ_{L̃/k,S_{k},T_{k}}^{r}(0)bL̃.

The commutative diagrams (16) hold here (with k instead of ℚ). Since v_{0}, …, v_{r} are all nonsplit in K/k, there is a unique w_{i} over w̃_{i} for i = 0, …, r. Hence

λ_{L}(η_{L̃/k}) = Θ_{L̃/k,S_{k},T_{k}}^{r}(0)((2w_{1} − 2w_{0}) ∧ ⋯ ∧ (2w_{r} − 2w_{0})) = 2^{r}Θ_{L̃/k,S_{k},T_{k}}^{r}(0)b_{L}.

Therefore we have, using (23), λ_{L}(η_{L̃/k}^{Θ_{L/K/k,S,T} (0,ω)}) = 2^{r}Θ_{L/K,S_{K},T_{K}}^{r}(0)b_{L}. It follows that

η_{L/K} = \frac{1}{2^{r}} η_{L̃/k}^{Θ_{L/K/k,S,T} (0,ω)}.  (24)

7.2 Regulator calculations

Keeping the assumptions of the previous section, we now go on to study the regulators involved in the various conjectures. Break up S_{k} as follows:

S_{k} = \{ \overbrace{v_{1}, …, v_{r},}^{S_{1,k}, nonsplit in K} \overbrace{v_{r+1}, …, v_{r+n}}^{n split in K}, \overbrace{v_{r+n+1}, …, v_{r+d+1}}^{d−n+1 nonsplit in K} \}.

Note that d−n+1 > 0 by Hypothesis 7.1. Choose v′_{i} to be place of K above v_{i} for i = 1, …, r+d+1. Choose a ℤ-basis μ_{1}, …, μ_{n} of U_{K,S,T}^{−}. Then we can define a minus-unit regulator R_{G}^{−} ∈ ℤ[G]/I_{G}^{n+1} by the determinant of the n × n matrix with (i, j)-th entry f_{v′_{r+j}}(μ_{i}) − 1 for 1 ⩽ i, j ⩽ n. This is denoted det_{G}(λ_{τ}) in [Gr88].

We choose a ℤ-basis u_{1}, …, u_{r+d+n} for U_{K,S,T} such that u_{1+n}, …, u_{r+d+n} is a basis for U_{k,S,T}, which is possible by Lemma 3.1.

The analogue of Proposition 6.13 in this situation is the following.

Proposition 7.2. We have the following in ℤ[G]/I_{G}^{d+n+1}:

Reg_{L/K}(Φ̃(u_{1} ∧ ⋯ ∧ u_{r+d+n})) = ±2^{d−n}R_{G}^{−}Reg_{L̃/k}(Φ̃(u_{1+n} ∧ ⋯ ∧ u_{r+d+n}))(U_{K,S,T}^{−} : U_{K,S,T}^{1−τ}).

Proof. The regulator on the left is

∑_{σ∈\left[{r+d+n \atop r}\right]} sign(σ) det (ϕ_{j}^{1}(u_{σ(i)}))_{1⩽i,j⩽r}Reg_{L/K}(u_{σ(r+1)} ∧ ⋯ ∧ u_{σ(r+d+n)}),  (25)

where, after manipulations as in the proof of Proposition 6.13, Reg_{L/K}(u_{σ(r+1)} ∧ ⋯ ∧ u_{σ(r+d+n)}) is seen to be the determinant in ℤ[G]/I_{G}^{d+n+1} of the matrix

v′_{r+1}, …, v′_{r+n} v̄′_{r+1}, …, v̄′_{r+n} v′_{r+n+1}, …, v′_{r+n+d}
u_{σ(r+1)} ; ⋮ ; u_{σ(r+m_{σ})}  f_{v′_{j}}(u_{i}^{1−τ}) − 1 f_{v′_{j}}(u_{i}^{τ}) − 1  f_{v′_{j}}(u_{i}) − 1

u_{σ(r+m_{σ}+1)} ; ⋮ ; u_{σ(r+d+n)}  0  f_{v_{j}}(u_{i}) − 1  2(f_{v_{j}}(u_{i}) − 1)

in which m_{σ} = #σ({r+1, …, r+d+n})∩{1, …, n}. Now if m_{σ} < n then this determinant is clearly 0. So for nonzero terms in the sum (25) we must have m_{σ} = n, ie. σ(r + 1) = 1, …, σ(r + n) = n. Then u_{σ(r+1)}^{1−τ}, …, u_{σ(r+n)}^{1−τ} is a ℤ-basis for U_{K,S,T}^{1−τ} and so the determinant of the top left n × n block is (U_{K,S,T}^{−} : U_{K,S,T}^{1−τ})R_{G}^{−}. The determinant of the bottom-right d × d block is 2^{d−n}Reg_{L̃/k}(u_{σ(r+n+1)} ∧ ⋯ ∧ u_{σ(r+d+n)}).

Note that for such σ, the map σ ∘ (1 2 … r + n)^{r} is a permutation of {n + 1, …, r + d + n} of the form n + k ↦ n + σ′(k) for σ′ ∈ \left[{r+d \atop r}\right]. We have

Reg_{L/K}(Φ̃(u_{1} ∧ ⋯ ∧ u_{r+d+n})) = (U_{K,S,T}^{−} : U_{K,S,T}^{1−τ})R_{G}^{−} ∑_{σ′}((−1)^{r(r+n+1)}sign(σ′)

det (ϕ_{j}^{1}(u_{n+σ′(i)}))_{1⩽i,j⩽r}Reg_{L̃/k}(u_{n+σ′(r+1)} ∧ ⋯ ∧ u_{n+σ′(r+d)})),

which gives the result.

7.3 Gross’s conjecture on the L-functions of tori We will now state Conjecture 8.8 of [Gr88]. The analytic class number formula makes it possible to calculate the coefficient of the leading term of the Taylor expansion of L_{K/k,S,T} (s, ω) at s = 0, as in [Ta84], Chapter II, §2. It is m_{ω}R^{−}, where R^{−} is a logarithmic regulator calculated with respect to bases of the minus-parts of U_{K,S,T} and X_{S_{K}}, and

m_{ω} = ±\frac{h_{K,S,T}}{h_{k,S,T}} 2^{#S−n−1}(U_{K,S,T}^{−} : U_{K,S,T}^{1−τ}).

The reader is warned that the factor 2^{#S−n−1} is missing in equation (8.5) of [Gr88].

Lemma 7.3. Assuming Hypothesis 7.1, m_{ω} is an integer.

Proof. The hypothesis shows that #S − n − 1 ⩾ 0. Also K/k is a quadratic extension unramified outside S such that at least one place in S is inert. Therefore Lemma 4.6 (i) shows that h_{k,S,T} divides h_{K,S,T}. This gives the result.

We can now state Gross’s tori conjecture, which in our setup, assuming Hypothesis 7.1 in order to have the conclusion of Lemma 7.3, states the following

Conjecture 7.4 (Gross). We have

Θ_{L/K/k,S,T} (0, ω) ≡ m_{ω}R_{G}^{−} (mod I_{G}^{n+1}).

7.4 Base change Theorem 7.5. Let L̃/k, K/k be finite linearly disjoint abelian extensions of a global field k, with [K : k] = 2. Set L = L̃K. Assume S = S_{k}, T = T_{k} are such that L/k is unramified outside S_{k} and U_{L,S,T} is torsion-free. Let S_{1} ⊆ S be a set of r places which split in L̃/k but not in K/k. Assume Hypothesis 7.1 for these data.

Assume that Conjecture 2.6 holds for L̃/k, S, T, r and that Conjecture 7.4 holds. Then for all Φ ∈ ∧_{ℤ[G]}^{r}Hom_{ℤ[G]}(U_{L,S,T} , ℤ[G]) we have

2^{r}Φ(η_{L/K}) ≡ ±2^{r}h_{K,S,T} Reg_{G}^{Φ} (mod I_{G}^{d+n+1}).

That is, the conclusion of Conjecture 2.6 for L/K, S_{K}, T_{K}, r holds with a factor of 2^{r} on each side. Proof. By (24) we have 2^{r}Φ(η_{L/K}) = Θ_{L/K/k,S,T} (0, ω)Φ(η_{L̃/k}). Multiplying the congruences of Conjecture 7.4 and Conjecture 2.6 gives

2^{r}Φ(η_{L/K}) ≡ ±\frac{h_{K,S,T}}{h_{k,S,T}} 2^{r+d−n}(U_{K,S,T}^{−} : U_{K,S,T}^{1−τ})R_{G}^{−} ·
h_{k,S,T} Reg_{L̃/k}(Φ̃(u_{1+n} ∧ ⋯ ∧ u_{r+d+n})) (mod I_{G}^{d+n+2})
≡ ±2^{r}h_{K,S,T} Reg_{L/K}(Φ̃(u_{1} ∧ ⋯ ∧ u_{r+d+n})) (mod I_{G}^{d+n+1}),

by the regulator calculation in Proposition 7.2.

Note that if r = 0 then this shows that, under Hypothesis 7.1, Gross’s conjecture on tori actually gives a base-change property with no weakening factor for Conjecture 4.1 in [Gr88].

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A. Hayward hayward@mth.kcl.ac.uk Department of Mathematics, King’s College London, Strand, London, WC2R 2LS