Published in J. Math. Sci. Univ. Tokyo 10 (2003), no. 1, pp. 119–138.

GROSS’ CONJECTURE FOR EXTENSIONS

RAMIFIED OVER THREE POINTS OF ℙ^{1}

Michael Reid

June 6, 2002

Abstract. B. Gross has formulated a conjectural generalization of the class number formula. Suppose L/K is an abelian extension of global fields with Galois group G. A generalized Stickelberger element θ ∈ ℤ[G] is constructed from special values of L-functions at s = 0. Gross’ conjecture then predicts some I-adic information about θ, where I ⊆ ℤ[G] is the augmentation ideal. In this paper, we prove (under a mild hypothesis) the conjecture for the maximal abelian extension of the rational function field 𝔽_{q}(X) that is unramified outside a set of three degree 1 places.

1. Introduction

Let K be a global field, i.e. either a finite extension of ℚ, or a finite, separable extension of 𝔽_{q}(X). Let S be any finite, non-empty set of places of K. We also require, in the number field case, that S contains all the archimedean places. We consider the S-zeta function

ζ_{S}(s) = ∑_{𝔞⊆O_{S}} (N𝔞)^{−s} ,

where the summation ranges over all ideals 𝔞 in the ring O_{S} of S-integers. This series definition makes sense only for ℜ(s) > 1, where the series converges absolutely. However, it is well-known that this function has an analytic continuation to the whole complex plane, with only a simple pole at s = 1.

In the Taylor series expansion for ζ_{S}(s), at s = 0, the leading term has a very simple form. Specifically, we have

ζ_{S}(s) = −\frac{h_{S}R_{S}}{w_{S}} s^{n} + O(s^{n+1}) near s = 0,  (1.1)

where h_{S} is the class number of O_{S}, R_{S} is the S-regulator, w_{S} is the number of roots of unity in O_{S} (which does not actually depend upon S) and n = #S − 1, which is the rank of the unit group, O_{S}^{∗}, by Dirichlet’s unit theorem.

Typeset by AMS-TEX

2000 Mathematics Subject Classification. 11R58 (11G40).

In order to state Gross’ conjecture, we need a formula such as (1.1) above, but which has no denominator. Towards this end, we introduce an auxiliary set, T, of places of K. For now, we only insist that T is finite and disjoint from S, but we will make further restrictions below. We introduce the modified zeta function

ζ_{S,T} (s) = ζ_{S}(s) ∏_{𝔮∈T} (1 − N𝔮^{1−s}) .

The auxiliary factors are all regular and non-zero at s = 0. We will see that the leading term of the Taylor series of ζ_{S,T} (s) also has a simple form, namely

ζ_{S,T} (s) = (−1)^{#T −1} \frac{h_{S,T} R_{S,T}}{w_{S,T}} s^{n} + O(s^{n+1}) near s = 0.  (1.2)

Each term in this formula has an interpretation analogous to the corresponding term in formula (1.1). Here h_{S,T} is the order of the ray class group modulo T (i.e. modulo the conductor which is the product of primes in T), R_{S,T} is the regulator of U_{S,T}, the subgroup of units that are ≡ 1 mod T (i.e. are congruent to 1 modulo every place in T), and w_{S,T} is the number of roots of unity in U_{S,T}, or equivalently, the order of its torsion subgroup.

To prove formula (1.2) above, note that we have an exact sequence

1 ⟶ U_{S,T} ⟶ U_{S} ⟶ ∏_{𝔮∈T} κ(𝔮)^{∗} ⟶ Cl_{S,T} ⟶ Cl_{S} ⟶ 1

where κ(𝔮) is the residue field at 𝔮, Cl_{S} is the class group of O_{S}, and Cl_{S,T} is the ray class group modulo T. This shows that (U_{S} : U_{S,T})\frac{h_{S,T}}{h_{S}} = ∏_{𝔮∈T} (N𝔮 − 1). Moreover, we have (U_{S} : U_{S,T}) = \frac{R_{S,T}}{R_{S}} · \frac{w_{S}}{w_{S,T}} . These last two equations, along with (1.1) and the definition of ζ_{S,T} give (1.2).

Thus, to achieve a formula without a denominator, we need only insist that w_{S,T} = 1, which imposes only mild conditions on T. Henceforth we make this assumption on T.

Gross’ Conjecture

Let L be a finite abelian extension of K with Galois group G. Consider S as in the previous section, but now we will impose another condition, namely that S contains all places of K that ramify in L. For any character χ ∈ Ĝ, we define the L-function

L_{S}(χ, s) = ∑_{𝔞⊆O_{S}} χ̃(𝔞)(N𝔞)^{−s},

where χ̃ is defined by χ̃(𝔭) = χ(φ_{𝔭}) for prime ideals, and extended to integral ideals by multiplicativity. Here φ_{𝔭} denotes the Frobenius element at 𝔭. We define modified Lfunctions in a manner analogous to the modified zeta functions. Specifically, if T is finite and disjoint from S, we define

L_{S,T} (χ, s) = L_{S}(χ, s) ∏_{𝔮∈T} (1 − χ(φ_{𝔮})(N𝔮)^{1−s}) .

GROSS’ CONJECTURE FOR EXTENSIONS RAMIFIED OVER THREE POINTS OF ℙ^{1} 3

Next we introduce the opposite Stickelberger element, θ̄_{S,T}. This is the element of ℂ[G] characterized by the property that

χ(θ̄_{S,T}) = L_{S,T} (χ, 0) for all characters χ ∈ Ĝ.

We use this terminology because it seems traditional to call the Stickelberger element the element that, under χ, maps to the value of the L-function for the conjugate character, χ̄. There is also the opposite Stickelberger element θ̄_{S}, which is characterized by the property that

χ(θ̄_{S}) = L_{S}(χ, 0) for all characters χ ∈ Ĝ.

The element θ̄_{S} nominally lives in ℂ[G]. However, it actually lives in ℚ[G]; in the number field case, this is a consequence of a famous theorem of Seigel, in the function field case, it follows from Weil’s work. Moreover, with our assumptions on T, θ̄_{S,T} lies in ℤ[G]. In the function field case, this again follows from Weil’s work. In the number field case, it requires the p-adic congruences first proved by Barsky and Cassou-Noguès, and later by Deligne and Ribet.

Before introducing the Gross regulator, it is worthwhile to revisit the usual regulator. We order the places in S as v_{0}, v_{1}, …, v_{n}. Let u_{1}, u_{2}, …, u_{n} be a ℤ-basis of the torsionfree group U_{S,T}, and consider the (n + 1) × n matrix (log |u_{j}|_{v_{i}}). We obtain the usual regulator by deleting an arbitrary row of this matrix, and taking the absolute value of the determinant. The Gross regulator is intended as an algebraic analogue, however, there is no algebraic analogue of the absolute value. For definiteness, we delete the 0-th row (corresponding to the place v_{0}). Then we can say that

det (log |u_{j}|_{v_{i}})_{1≤i,j,≤n} = ±R_{S,T}

with the realization that the choice of sign depends upon the ordering of places in S, as well as the orientation of ℤ-basis of U_{S,T}. We denote this determinant above as det_{ℝ}(λ_{S,T}); then we have

ζ_{S,T} (s) = ±h_{S,T} det_{ℝ}(λ_{S,T})s^{n} + O(s^{n+1}) near s = 0.

Let L/K be an abelian extension, unramified outside of S, with Galois group G. Consider the (n + 1) × n matrix (r_{v_{i}}(u_{j})), where r_{v_{i}} is the local reciprocity map at the place v_{i} for the extension L/K. Now r_{v}(u_{j}) = 1 for v ∉ S, since L/K is unramified outside of S, so the product formula tells us that the product of the elements in any given column is the identity element of G. Using the isomorphism G → I/I^{2} as an analogue of the logarithm, we define the Gross regulator to be

det_{G}(λ_{S,T}) = det (r_{v_{i}}(u_{j}) − 1)_{1≤i,j,≤n} mod I^{n+1}.

This definition also depends upon the ordering of places in S, as well as the orientation of ℤ-basis of U_{S,T}, which may effect a change in sign. Otherwise, it is well-defined, modulo I^{n+1}.

Now we may state Gross’ conjecture. Conjecture 1.3. (Gross’ Conjecture) With the notation above, we have

θ̄_{S,T} ≡ ±h_{S,T} det_{G}(λ_{S,T}) mod I^{n+1},

where the sign is chosen to agree with the sign in

ζ_{S,T} (s) = ±h_{S,T} det_{ℝ}(λ_{S,T})s^{n} + O(s^{n+1}) near s = 0.

The conjecture behaves well with respect to various functorialities, which we note here.

Proposition 1.4. Suppose that Gross’ conjecture holds for the extension L/K with respect to S and T. Let L′/K be a subextension of L/K, S′ a superset of S and T′ a superset of T.

(a) Gross’ conjecture holds for L′/K with respect to S and T.

(b) Gross’ conjecture holds for L/K with respect to S′ and T.

(c) Gross’ conjecture holds for L/K with respect to S and T′.

Part (a) is more or less obvious from the functoriality of the Stickelberger element and the local reciprocity maps that comprise the Gross regulator. Parts (b) and (c) are also straightforward. We refer the reader to Aoki’s paper [1] where the proofs are given in some detail.

As Gross notes, the functoriality of part (a) allows a formulation for infinite extensions by passing to the limit. The conjecture for an infinite extension is equivalent to it holding for every finite subextension.

Proposition 1.5. Gross’ conjecture holds for an extension L/K if and only if it holds for each subextension of prime power degree.

We will prove this proposition in the next section.

We enumerate here some cases of Gross’ conjecture that have been proven. Theorem 1.6. Gross’ conjecture holds in the following cases.

(a) S is a singleton.

(b) L/K is a constant field extension of function fields.

(c) L/K is a p-power extension, or pro-p, and K is a function field of characteristic p.

(d) L/K is an elementary abelian p-extension, and the class number of K is not divisible by p.

(e) K = ℚ.

(f) S contains a place that splits completely in L.

Proof. Part (a) is immediate from the formulation of the conjecture. In this case, the conjecture becomes the usual class number formula.

For (b), the extension is unramified, so we may reduce S to a singleton. Then it holds from part (a) and also 1.4(b).

Part (c) was proved by Tan [11].

Part (d) was proved by Lee [7].

Part (e) was proved by Aoki [1].

For part (f), the Stickelberger element is 0, because all the L-functions for L/K vanish at s = 0. On the other hand, the Gross regulator vanishes modulo I^{n+1}, because the row of the matrix corresponding to the split place also vanishes. □

Statement of result

Let K = 𝔽_{q}(X) be the rational function field over the finite field with q elements. We will consider abelian extensions of K that are unramified outside a set S of three degree 1 places of K. Let K_{S} be the maximal abelian extension of K that is unramified outside of S.

Theorem 1.7. If T is any set of places, whose greatest common divisor of their degrees is relatively prime to q − 1, then Gross’ conjecture holds for K_{S}/K.

Remark. The set T should be considered secondary in the conjecture, so the restriction placed on it is only a minor issue. Nonetheless, it would be of some interest to eliminate this hypothesis.

2. Group rings

In this section we develop some general results about group rings and their augmentation ideals. Let G be a group, ℤ[G] the group ring, and I_{G} ⊆ ℤ[G] its augmentation ideal. Our goal is a moderate understanding of congruences modulo I_{G}^{3}, so that we may verify the congruence of the conjecture. Of course, congruences modulo I_{G}^{2} are well-understood, so we use that as our starting point.

In the following, when the group G is understood, we will write simply I rather than the more cumbersome I_{G}. The following lemma, whose proof we omit, is entirely elementary.

Lemma 2.1. (a) If a, b ∈ I and a ≡ a′ mod I^{2} and b ≡ b′ mod I^{2}, then ab ≡ a′b′ mod I^{3}.

(b) Suppose G has exponent n. Then n annihilates I^{r}/I^{r+1} for every positive r. □

We will use these two throughout the computation without mention.

Lemma 2.2. Let g ∈ G have order dividing n. If n is odd, then n(g − 1) ≡ 0 mod I^{3}, and if n = 2m is even, then n(g − 1) ≡ m(g − 1)^{2} mod I^{3}.

Proof. We have n(g − 1) = [n − (1 + g + g^{2} + ⋯ + g^{n−1})](g − 1) ≡ \frac{n(1−n)}{2}(g − 1)^{2} mod I^{3}. If n is odd, this is 0 mod I^{3}, because n(g − 1) ∈ I^{2}. If n = 2m is even, then this is m(g − 1)^{2} − mn(g − 1)^{2} ≡ m(g − 1)^{2} mod I^{3}. □

Lemma 2.3. Let G be the direct product of two cyclic groups of order n. If n is odd, then ∑_{γ∈G}(γ − 1) ≡ 0 mod I^{3}. If n is even, then ∑_{γ∈G}(γ − 1) ≡ (σ − 1)^{2} + (σ − 1)(τ − 1) + (τ − 1)^{2} mod I^{3}, where σ and τ generate the 2-torsion subgroup of G. Proof. Let g, h be generators of the two cyclic components. Then

∑_{γ∈G}(γ − 1) = (1 + g + g^{2} + ⋯ + g^{n−1} − n)(1 + h + h^{2} + ⋯ + h^{n−1} − n)

+ n(1 + g + g^{2} + ⋯ + g^{n−1} − n) + n(1 + h + h^{2} + ⋯ + h^{n−1} − n),

and note that

(1 + g + g^{2} + ⋯ + g^{n−1} − n)(1 + h + h^{2} + ⋯ + h^{n−1} − n)

≡ (\frac{n(n−1)}{2}(g − 1)) (\frac{n(n−1)}{2}(h − 1)) mod I^{3}.

Also, n kills I^{2}/I^{3}, so that n(1 + g + g^{2} + ⋯ + g^{n−1} − n) ≡ n(\frac{n(n−1)}{2}(g − 1)) mod I^{3}. Moreover, n(g − 1) ≡ −\frac{n(n−1)}{2}(g − 1)^{2} mod I^{3}, as in the proof of Lemma 2.1, and since n(n − 1) kills I^{2}/I^{3}, this is also ≡ \frac{n(n−1)}{2}(g − 1)^{2} mod I^{3}. Thus n(1 + g + g^{2} + ⋯ + g^{n−1} − n) ≡ (\frac{n(n−1)}{2})^{2}(g − 1)^{2} mod I^{3}, and similarly, n(1 + h + h^{2} + ⋯ + h^{n−1} − n) ≡ (\frac{n(n−1)}{2})^{2}(h − 1)^{2} mod I^{3}, so that

∑_{γ∈G}(γ − 1) ≡ (\frac{n(n−1)}{2})^{2} ((g − 1)^{2} + (g − 1)(h − 1) + (h − 1)^{2}) mod I^{3}.

If n is odd, then (\frac{n(n−1)}{2})^{2} is divisible by n, and therefore kills I^{2}/I^{3}, so the sum in question vanishes modulo I^{3}.

On the other hand, if n = 2m is even, then (\frac{n(n−1)}{2})^{2} ≡ m^{2} mod n, so that

\begin{aligned}∑_{γ∈G}(γ − 1) & ≡ m^{2}((g − 1)^{2} + (g − 1)(h − 1) + (h − 1)^{2}) \\ & ≡ (g^{m} − 1)^{2} + (g^{m} − 1)(h^{m} − 1) + (h^{m} − 1)^{2} mod I^{3}. \end{aligned}

The proof concludes by noting that g^{m} and h^{m} generate the 2-torsion in G. □

Remark. In the lemma, if n is divisible by 4, then the sum ∑_{G}(g − 1) ≡ 0 mod I^{3}, since (\frac{n(n−1)}{2})^{2} is divisible by n. However, we will not need this below.

Let G_{1} [respectively G_{2}] be a finite group of exponent n_{1} [respectively n_{2}], where (n_{1}, n_{2}) = 1, and let G = G_{1} × G_{2}. Let π_{i} : G → G_{i} be the natural projection and ι_{i} : G_{i} → G the inclusions defined by ι_{1}(g) = (g, 1) and ι_{2}(g) = (1, g). Let I_{G} ⊆ ℤ[G] denote the augmentation ideal, and similarly for I_{G_{i}} ⊆ ℤ[G_{i}].

Lemma 2.4. If η ∈ I_{G}, then η ≡ ι_{1}(π_{1}(η)) + ι_{2}(π_{2}(η)) mod I_{G}^{r} for any r. Proof. First consider an element of the form η = g − 1, for some g ∈ G, and let g_{i} = ι_{i}(π_{i}(g)) for i = 1, 2, so that g = g_{1}g_{2}. Then we have η − ι_{1}(π_{1}(η)) − ι_{2}(π_{2}(η)) = (g_{1} − 1)(g_{2} − 1), so we must show that this element is in I_{G}^{r}. Since n_{1} and n_{2} are relatively prime, we may write a_{1}n_{1} + a_{2}n_{2} = 1 for some integers a_{1} and a_{2}. Let

θ = a_{1}(n_{1} − (1 + g_{1} + g_{1}^{2} + ⋯ + g_{1}^{n_{1}−1})) + a_{2}(n_{2} − (1 + g_{2} + g_{2}^{2} + ⋯ + g_{2}^{n_{2}−1})),

which is visibly an element of I_{G}. Since g_{i} − 1 annihilates 1 + g_{i} + g_{i}^{2} + ⋯ + g_{i}^{n_{i}−1}, we have θ(g_{1} − 1)(g_{2} − 1) = (a_{1}n_{1} + a_{2}n_{2})(g_{1} − 1)(g_{2} − 1) = (g_{1} − 1)(g_{2} − 1). Thus, for any r, we have (g_{1} − 1)(g_{2} − 1) = θ^{r−2}(g_{1} − 1)(g_{2} − 1) ∈ I_{G}^{r}, as desired. This proves the result for elements of the form η = g − 1. The result now holds for arbitrary η ∈ I_{G} by ℤ-linearity. □

Corollary 2.5. Let η ∈ ℤ[G]. Then η ∈ I_{G}^{r} if and only if π_{1}(η) ∈ I_{G_{1}}^{r} and π_{2}(η) ∈ I_{G_{2}}^{r}. Proof. The “only if” part is trivial, and the “if” part is immediate for r = 1. Now suppose that r > 1, and that η ∈ ℤ[G] has the property that π_{i}(η) ∈ I_{G_{i}}^{r} for i = 1, 2. Then η ∈ I_{G}, so from Lemma 1, η ≡ ι_{1}(π_{1}(η)) + ι_{2}(π_{2}(η)) mod I_{G}^{r}. Finally, ι_{i}(π_{i}(η)) ∈ ι_{i}(I_{G_{i}}^{r}) ⊆ I_{G}^{r}, for i = 1, 2, so that η ∈ I_{G}^{r}, as desired. □

Proof of Proposition 1.5. This follows easily from Corollary 2.5 and the functoriality of both sides of the conjecture, as in 1.4(a). □

3. Extensions ramified over three points of ℙ^{1}

Let K = 𝔽_{q}(X) be the rational function field over a finite field with q elements, and S = {0, 1, ∞} a set of three degree 1 places of K. Also let p be the characteristic of K. Let K_{S} be the maximal abelian extension of K that is unramified outside of S. The Galois group Gal(K_{S}/K) factors canonically as the product of its prime-to-p part and its p-power part. These correspond to linearly disjoint subextensions, K_{S}^{(p)}/K and K_{S}^{(not p)}/K. From Corollary 2.5, to verify Gross’ conjecture for K_{S}/K, it is sufficient to verify it for K_{S}^{(p)}/K and K_{S}^{(not p)}/K. Tan’s result (1.6(c)) handles the extension K_{S}^{(p)}/K. Thus our task is to verify Gross’ conjecture for K_{S}^{(not p)}/K; in fact, we will do slightly more. Let K_{S}^{tame} be the maximal abelian extension of K that is unramified outside of S and is only tamely ramified. Then clearly K_{S}^{(not p)} ⊆ K_{S}^{tame}. We will verify Gross’ conjecture for K_{S}^{tame}/K.

Proposition 3.1. K_{S}^{tame} = \overline{𝔽}_{q}( \sqrt[q−1]{X}, \sqrt[q−1]{X − 1}).

Proof. Let J_{K} denote the idele group of K. We have a short exact sequence

1 ⟶ ∏_{v∈S} κ(v)^{∗}/image of(𝔽_{q}^{∗}) ⟶ J_{K}/(K^{∗} · ∏_{v∉S} U_{v} × ∏_{v∈S} U_{v}^{1}) ⟶ J_{K}/(K^{∗} · ∏_{v} U_{v}) ⟶ 1.

Moreover, J_{K}/(K^{∗} · ∏_{v} U_{v}) = Pic(K) ≅ ℤ, since Pic^{0}(K) is trivial. The middle term of the exact sequence corresponds to the extension K_{S}^{tame}/K, under classfield theory, so we get another short exact sequence

1 ⟶ ∏_{v∈S} κ(v)^{∗}/image of(𝔽_{q}^{∗}) ⟶ Gal(K_{S}^{tame}/K) ⟶ ℤ̂ ⟶ 0,

where the quotient ℤ̂ is the Galois group of the constant field extension \overline{𝔽}_{q}(X)/K. This sequence splits (non-canonically). It is easy to see that ∏_{v∈S} κ(v)^{∗}/image of(𝔽_{q}^{∗}) ≅ (𝔽_{q}^{∗})^{2}, and then it is straightforward to identify K( \sqrt[q−1]{X}, \sqrt[q−1]{X − 1}) as a complement. □

In order to prove Theorem 1.7, we will use the following calculation.

Theorem 3.2. Let L = 𝔽_{q^{m}}( \sqrt[q−1]{X}, \sqrt[q−1]{X − 1}), S = \{0, 1, ∞\} as above, and let T contain a single place of degree d. Then Gross’ conjecture holds for L/K when both sides are multiplied by (q^{d} − 1)/(q − 1), in other words, we have the congruence

(1 + q + q^{2} + ⋯ + q^{d−1})θ̄_{S,T} ≡ (1 + q + q^{2} + ⋯ + q^{d−1})h_{S,T} det_{G}(λ_{S,T}) mod I^{n+1}.

Before proving Theorem 3.2, we will show how to prove Theorem 1.7 from it.

Proof of Theorem 1.7. As remarked above, we need only prove Gross’ conjecture for K_{S}^{tame}/K. Moreover, from 2.5, we may work one prime at a time.

Let ℓ be a prime. If ℓ does not divide q − 1, then the ℓ-power part of K_{S}^{tame}/K is the constant field extension 𝔽_{q^{ℓ^{∞}}} (X)/K, and Gross’ conjecture is known for constant field extensions.

Now suppose that ℓ divides q − 1. By hypothesis, T contains a place whose degree is not divisible by ℓ. Let 𝔞 ∈ T be such a place, of degree d, and let T_{0} = {𝔞}. Also let L_{m} = 𝔽_{q^{m}}( \sqrt[q−1]{X}, \sqrt[q−1]{X − 1}). Theorem 3.2 shows that Gross’ conjecture for L_{m}/K holds when multiplied by (1 + q + q^{2} + ⋯ + q^{d−1}). The functoriality of the conjecture then implies that the same is true for any subextension of L_{m}/K, in particular, the ℓ-power part of it, L_{m}^{(ℓ)}/K. But then Gal(L_{m}^{(ℓ)}/K) is an ℓ-group, so I^{n}/I^{n+1} is ℓ^{∞}-torsion. However, 1 + q + q^{2} + ⋯ + q^{d−1} ≡ d ≢ 0 mod ℓ, because ℓ divides q − 1, but not d. In particular, multiplication by (1 + q + q^{2} + ⋯ + q^{d−1}) is an automorphism of I^{n}/I^{n+1}. Therefore, the conjecture itself holds for L_{m}^{(ℓ)}/K. Moreover, the compositum of L_{m}^{(ℓ)} over all m is the ℓ-power part of K_{S}^{tame}, since the compositum of all such L_{m} is K_{S}^{tame}. Thus the conjecture holds for the ℓ-power part of K_{S}^{tame}/K, with respect to T_{0}, and therefore also with respect to T. This proves the theorem. □

Now we return to prove Theorem 3.2. Let L = 𝔽_{q^{m}}( \sqrt[q−1]{X}, \sqrt[q−1]{X − 1}), and G = Gal(L/K). Let E_{0} = 𝔽_{q^{m}}(X) and E_{1} = 𝔽_{q}( \sqrt[q−1]{X}, \sqrt[q−1]{X − 1}) be the given intermediate fields, and G_{i} = Gal(L/E_{i}), so that G ≅ G_{0} × G_{1}.

The group G_{1} is cyclic of order m, with a canonical generator, the “arithmetic” Frobenius, F, which acts on constants via the q-th power map.

Standard Kummer theory shows that G_{0} ≅ (𝔽_{q}^{∗})^{2}. For α, β ∈ 𝔽_{q}^{∗}, let τ(α, β) ∈ G_{0} be the element that acts via

τ(α, β)( \sqrt[q−1]{X}) = α \sqrt[q−1]{X} and τ(α, β)( \sqrt[q−1]{X − 1}) = β \sqrt[q−1]{X − 1}

and acts trivially on constants. For notational convenience, we will let δ(α, β) denote the group ring element τ(α, β) − 1.

For a place v ∉ S of K, let φ_{v} denote its Frobenius element for the extension L/K. This is characterized by the condition that φ_{v}(x) ≡ x^{Nv} mod v for all x that are integral at v. We will only need to determine these Frobenius elements for degree 1 places.

Lemma 3.3. Let f(X) ∈ 𝔽_{q}[X] be a monic irreducible polynomial of degree d. Then f(X) divides X^{1+q+q^{2}+···+q^{d−1}} − (−1)^{d}f(0) and also (X − 1)^{1+q+q^{2}+···+q^{d−1}} − (−1)^{d}f(1).

Proof. Let α be a root of f(X) in 𝔽_{q^{d}}; then the other roots are α^{q}, α^{q^{2}}, α^{q^{3}}, …, α^{q^{d−1}}. The product of these roots is (−1)^{d}f(0). Therefore, over 𝔽_{q^{d}}, f(X) divides

X^{1+q+q^{2}+···+q^{d−1}} − α^{1+q+q^{2}+···+q^{d−1}} = X^{1+q+q^{2}+···+q^{d−1}} − (−1)^{d}f(0).

The same divisibility then holds in 𝔽_{q}[X], and the second divisibility follows in a similar way. □

Lemma 3.4. If 𝔞 ∉ S is a degree d place, which corresponds to a monic irreducible polynomial f(X), its Frobenius element is given by F^{d}τ((−1)^{d}f(0), (−1)^{d}f(1)). In particular, the Frobenius element of the degree 1 place at b ≠ 0, 1 is given by F τ(b, b − 1).

Proof. For a constant α ∈ \overline{𝔽}_{q}, we have φ_{𝔞}(α) ≡ α^{N𝔞} = α^{q^{d}} mod 𝔞, and since φ_{𝔞}(α) is also a constant, we have φ_{𝔞}(α) = α^{q^{d}}. Furthermore,

φ_{𝔞}( \sqrt[q−1]{X}) ≡ (X^{1+q+q^{2}+···+q^{d−1}})( \sqrt[q−1]{X}) ≡ (−1)^{d}f(0)( \sqrt[q−1]{X}) mod 𝔞.

Since φ_{𝔞}( \sqrt[q−1]{X}) is a constant times \sqrt[q−1]{X}, this congruence is in fact an equality. Similarly, φ_{𝔞}( \sqrt[q−1]{X − 1}) = (−1)^{d}f(1)( \sqrt[q−1]{X − 1}). This proves the first statement, and the second is a special case. □

Next we determine the Stickelberger element. Let T be a singleton {𝔞} where 𝔞 ∉ S is a place of degree d which corresponds to the monic irreducible polynomial f(X).

Proposition 3.5. θ̄_{S,T} = (1−q^{d}φ_{𝔞})(1+∑_{b≠0,1}φ_{b})+F^{2}(1+qF +q^{2}F^{2}+⋯+q^{d−1}F^{d−1}) ∑_{g∈G_{0}}g.

Proof. Let η = (1−qφ_{𝔞})(1+∑_{b≠0,1} φ_{b})+F^{2}(1+qF +q^{2}F^{2}+⋯+q^{d−1}F^{d−1}) ∑_{g∈G_{0}} g, and let χ be a character of G. We aim to show that χ(η) = L_{S,T} (χ, 0). Note that L_{S}(χ, s) = ∑_{n=0}^{∞} q^{−ns} ∑_{deg 𝔟=n} χ̃(𝔟), where the inner sum is over effective divisors of degree n with support outside of S. In general, if χ is non-trivial on the subgroup corresponding to the maximal constant field extension, then this expression is a polynomial in q^{−s}, of degree 2g − 2 + deg 𝔣, where g is the genus of the base field and 𝔣 is the conductor of χ (see [12, Chapter VII, Theorem 6]). In our situation, we have g = 0 and deg 𝔣 = 3, so the inner sum vanishes for n ≥ 2. Therefore, if χ is non-trivial on G_{0}, then L_{S}(χ, s) = 1 + q^{−s} ∑_{b≠0,1} χ(φ_{b}). Now, still assuming that χ is non-trivial on G_{0}, we have

L_{S,T} (χ, 0) = (1 − qχ(φ_{𝔞}))(1 + ∑_{b≠0,1} χ(φ_{b})) = χ(η),

since χ(∑_{G_{0}} g) = 0.

On the other hand, if χ is trivial on G_{0}, it is easy to calculate

L_{S,T} (χ, s) = (1 − χ(F)q^{−s})^{2} (1 + χ(F)q^{1−s} + (χ(F)q^{1−s})^{2} + ⋯ + (χ(F)q^{1−s})^{d−1}) ,

so that

L_{S,T} (χ, 0) = (1 − χ(F))^{2}(1 + qχ(F) + q^{2}χ(F^{2}) + ⋯ + q^{d−1}χ(F^{d−1})).

Moreover, since χ is trivial on G_{0}, we have χ(φ_{𝔞}) = χ(F^{d}) and χ(φ_{b}) = χ(F), so that

χ(η) = (1 − q^{d}χ(F^{d}))(1 + (q − 2)χ(F))

+ (q − 1)^{2}χ(F^{2})((1 + qχ(F) + q^{2}χ(F^{2}) + ⋯ + q^{d−1}χ(F^{d−1})),

which equals the value of L_{S,T} (χ, 0) above. This shows that χ(η) = L_{S,T} (χ, 0) for all χ ∈ Ĝ, so η = θ̄_{S,T}, as claimed. □

Now we attempt to compute the Gross regulator. In general, we do not know enough units to compute it exactly, but we can compute some multiple of it. The unit group U_{S} is generated by 𝔽_{q}^{∗}, X and X − 1. We easily identify two independent units in U_{S,T}, namely, u_{0} = X^{1+q+q^{2}+···+q^{d−1}}/((−1)^{d}f(0)) and u_{1} = (X −1)^{1+q+q^{2}+···+q^{d−1}}/((−1)^{d}f(1)). For notational convenience, we let α denote (−1)^{d}f(0) and β = (−1)^{d}f(1). Let V be the group generated by u_{0} and u_{1}. The index (U_{S} : V) is (1 + q + q^{2} + ⋯ + q^{d−1})(q^{d} − 1), whereas (U_{S} : U_{S,T}) = (q^{d} − 1)/h_{S,T}. Therefore (U_{S,T} : V) = (1 + q + q^{2} + ⋯ + q^{d−1})h_{S,T}, whence

(1 + q + q^{2} + ⋯ + q^{d−1})h_{S,T} det_{G}(λ_{S,T}) ≡ det ( \begin{matrix}r_{0}(u_{0}) − 1 & r_{0}(u_{1}) − 1 \\ r_{1}(u_{0}) − 1 & r_{1}(u_{1}) − 1 \end{matrix} ) mod I^{3}, (3.6)

where r_{0} and r_{1} are the local reciprocity maps at 0 and 1 respectively.

Proposition 3.7. We have the following values of the local reciprocity maps:

\begin{aligned}r_{0}(u_{0}) & = F^{−(1+q+q^{2}+···+q^{d−1})}τ((−1)^{d}α^{−1}, (−1)^{d}) \\ r_{0}(u_{1}) & = τ((−1)^{d}β^{−1}, 1) \\ r_{1}(u_{0}) & = τ(1, α^{−1}) \\ r_{1}(u_{1}) & = F^{−(1+q+q^{2}+···+q^{d−1})}τ(1, (−1)^{d}β^{−1}) \end{aligned}

Proof. On constants, r_{0}(u_{0}) acts as F^{− ord_{0}(u_{0})} = F^{−(1+q+q^{2}+···+q^{d−1})}. For its action on \sqrt[q−1]{X} and \sqrt[q−1]{X − 1}, we use the “tame symbol” (see Serre [9, Chapter XIV, Proposition 8]). We calculate

(u_{0}, X)_{0} = (−1)^{1+q+q^{2}+···+q^{d−1}}u_{0}/X^{1+q+q^{2}+···+q^{d−1}} ≡ (−1)^{d}α^{−1} mod 𝔭_{0}

and

(u_{0}, X − 1)_{0} = (−1)^{0}u_{0}^{0}/(X − 1)^{1+q+q^{2}+···+q^{d−1}} ≡ (−1)^{d} mod 𝔭_{0}.

Thus r_{0}(u_{0}) = F^{−(1+q+q^{2}+···+q^{d−1})}τ((−1)^{d}α^{−1}, (−1)^{d}), as claimed. The other values are similar computations. □

In order to simplify (3.6) above, we need a lemma.

Lemma 3.8. For x, y ∈ 𝔽_{q}^{∗}, we have δ(x, 1)δ(1, y) ≡ δ(y, 1)δ(1, x) mod I^{3}.

Proof. We use the fact that 𝔽_{q}^{∗} is cyclic; let t be a generator. Then t^{m} = x and t^{n} = y for some integers m, n. Then we have δ(x, 1)δ(1, y) ≡ mδ(t, 1)nδ(1, t) mod I^{3}, and similarly, δ(y, 1)δ(1, x) ≡ nδ(t, 1)mδ(1, t) mod I^{3}. □

Proposition 3.9. We have

(1 + q + q^{2} + ⋯ + q^{d−1})h_{S,T} det_{G}(λ_{S,T}) ≡ (1 + q + q^{2} + ⋯ + q^{d−1})^{2}(F − 1)^{2}
+ (1 + q + q^{2} + ⋯ + q^{d−1})(F − 1)δ((−1)^{d}α, β)
+ dδ(−1, −1)δ(1, (−1)^{d}β) mod I^{3}.

Proof. From 3.7, we have

r_{0}(u_{0}) − 1 ≡ −(1 + q + q^{2} + ⋯ + q^{d−1})(F − 1) − δ((−1)^{d}α, (−1)^{d}) mod I^{2},

as well as r_{0}(u_{1}) − 1 ≡ −δ((−1)^{d}β, 1) mod I^{2}, r_{1}(u_{0}) − 1 ≡ −δ(1, α) mod I^{2}, and

r_{1}(u_{1}) − 1 ≡ −(1 + q + q^{2} + ⋯ + q^{d−1})(F − 1) − δ(1, (−1)^{d}β) mod I^{2}.

Now it follows that

(1 + q + q^{2} + ⋯ + q^{d−1})h_{S,T} det_{G}(λ_{S,T})

≡ (1 + q + q^{2} + ⋯ + q^{d−1})^{2}(F − 1)^{2}
+ (1 + q + q^{2} + ⋯ + q^{d−1})(F − 1)[δ((−1)^{d}α, (−1)^{d}) + δ(1, (−1)^{d}β)]

+ δ((−1)^{d}α, (−1)^{d})δ(1, (−1)^{d}β) − δ((−1)^{d}β, 1)δ(1, α) mod I^{3},

from equation (3.6) above. We also have

(F − 1)[δ((−1)^{d}α, (−1)^{d}) + δ(1, (−1)^{d}β)] ≡ (F − 1)δ((−1)^{d}α, β) mod I^{3},

and, using the lemma,

\begin{aligned}δ((−1)^{d}α, (−1)^{d})δ(1, (−1)^{d}β) & ≡ [δ((−1)^{d}, (−1)^{d}) + δ(α, 1)]δ(1, (−1)^{d}β) \\ & ≡ dδ(−1, −1)δ(1, (−1)^{d}β) + δ((−1)^{d}β, 1)δ(1, α) mod I^{3}, \end{aligned}

from which the result follows. □

We now return to consider the Stickelberger element. To simplify the expression for this, we need a few easy lemmas.

Lemma 3.10. We have ∑_{g∈G_{0}}(g − 1) ≡ δ(−1, 1)^{2} + δ(1, −1)δ(−1, −1) mod I^{3}.

Proof. If q is even, then the left hand side vanishes modulo I^{3}, from Lemma 2.3, while the right hand side vanishes because −1 = 1. If q is odd, then it also follows from Lemma 2.3, by noting that τ(−1, 1) and τ(1, −1) generate the 2-torsion in G_{0}, and also that δ(−1, 1)δ(1, −1) + δ(1, −1)^{2} ≡ δ(1, −1)δ(−1, −1) mod I^{3}. □

Lemma 3.11. For any x ∈ 𝔽_{q}^{∗}, we have (q − 1)δ(x, 1) ≡ δ(x, 1)δ(−1, 1) mod I^{3}, and (q − 1)δ(1, x) ≡ δ(1, x)δ(1, −1) mod I^{3}.

Proof. We prove the first statement; the second is similar. If q is even, then the left hand side vanishes modulo I^{3} by Lemma 2.2, while the right hand side vanishes because −1 = 1. If q is odd, then Lemma 2.2 shows that (q − 1)δ(x, 1) ≡ \frac{q−1}{2}δ(x, 1)^{2} ≡ δ(x, 1)δ(x^{(q−1)/2}, 1) mod I^{3}. Thus we must show that δ(x, 1)δ(−x^{(q−1)/2}, 1) ≡ 0 mod I^{3}. If x is a square in 𝔽_{q}, say x = y^{2}, then δ(x, 1)δ(−x^{(q−1)/2}, 1) = δ(y^{2}, 1)δ(−1, 1) ≡ 2δ(y, 1)δ(−1, 1) ≡ δ(y, 1)δ(1, 1) = 0 mod I^{3}. On the other hand, if x is not a square, then δ(−x^{(q−1)/2}, 1) = δ(1, 1) = 0. □

Now we return our attention to the Stickelberger element.

Proposition 3.12. We have the congruence

θ̄_{S,T} ≡ (1 + q + q^{2} + ⋯ + q^{d−1})(F − 1)^{2} + (F − 1)δ((−1)^{d}α, β) + δ(α, β)δ(−1, −1) mod I^{3}.

Proof. We begin with the expression in Proposition 3.5, and write

(1 − q^{d}φ_{𝔞})(1 + ∑_{b≠0,1} φ_{b})

= (q − 1)(1 − q^{d}) − q^{d}(q − 1)(φ_{𝔞} − 1) + (1 − q^{d}) ∑_{b≠0,1}(φ_{b} − 1)

− q^{d}(φ_{𝔞} − 1) ∑_{b≠0,1}(φ_{b} − 1)

and now consider each term individually. The term (q − 1)(1 − q^{d}) does not require any simplification.

Firstly, we have

−q^{d}(q − 1)(φ_{𝔞} − 1) = −q^{d}(q − 1)(F^{d}τ(α, β) − 1)
= −q^{d}(q − 1)((F^{d} − 1)δ(α, β) + (F^{d} − 1)
+ δ(α, 1)δ(1, β) + δ(α, 1) + δ(1, β))
≡ −q^{d}(q − 1)((F^{d} − 1) + δ(α, 1) + δ(1, β)) mod I^{3},

because (q − 1) kills δ(α, β), δ(α, 1) and δ(1, β), modulo I^{2}. Also, from 3.11, we have

−q^{d}(q − 1)δ(α, 1) ≡ −q^{d}δ(α, 1)δ(−1, 1) ≡ −δ(α, 1)δ(−1, 1) ≡ δ(α, 1)δ(−1, 1) mod I^{3}

and similarly, −q^{d}(q − 1)δ(1, β) ≡ δ(1, β)δ(1, −1) mod I^{3}. Expand (F^{d} − 1) in powers of (F − 1) to see that (F^{d} − 1) ≡ d(F − 1) + \frac{d(d−1)}{2}(F − 1)^{2} mod I^{3}. Therefore we get

− q^{d}(q − 1)(φ_{𝔞} − 1) ≡ −dq^{d}(q − 1)(F − 1) − \frac{d(d−1)}{2}q^{d}(q − 1)(F − 1)^{2}

+ δ(α, 1)δ(−1, 1) + δ(1, β)δ(1, −1) mod I^{3} (3.13)

Secondly, we have

\begin{aligned}(1 − q^{d}) ∑_{b≠0,1}(φ_{b} − 1) & = (1 − q^{d}) ∑_{b≠0,1}((F − 1)δ(b, b − 1) + (F − 1) + δ(b, b − 1)) \\ & ≡ (q − 2)(1 − q^{d})(F − 1) + (1 − q^{d}) ∑_{b≠0,1} δ(b, b − 1), \end{aligned}

because (q − 1)δ(b, b − 1) ∈ I^{2}. Moreover, we have ∑_{b≠0,1} δ(b, b − 1) ≡ δ(−1, 1) mod I^{2}, from Wilson’s Theorem, and because (q − 1) kills I_{G_{0}}^{2}/I_{G_{0}}^{3}, we have

(1 − q^{d}) ∑_{b≠0,1} δ(b, b − 1) ≡ (1 − q^{d})δ(−1, 1)
≡ −(1 + q + q^{2} + ⋯ + q^{d−1})δ(−1, 1)^{2}
≡ dδ(−1, 1)^{2} mod I^{3},

14 MICHAEL REID

also using 3.11. Therefore we have

(1 − q^{d}) ∑_{b≠0,1}(φ_{b} − 1) ≡ (q − 2)(1 − q^{d})(F − 1) + dδ(−1, 1)^{2} mod I^{3}.  (3.14)

Thirdly, we have

(φ_{𝔞} − 1) ∑_{b≠0,1}(φ_{b} − 1) ≡ [d(F − 1) + δ(α, β)] ∑_{b≠0,1}[(F − 1) + δ(b, b − 1)]
≡ [d(F − 1) + δ(α, β)][(q − 2)(F − 1) + δ(−1, 1)]
≡ d(q − 2)(F − 1)^{2} − (F − 1)δ((−1)^{d}α, β)
+ δ(α, β)δ(−1, 1) mod I^{3}.

Multiply by −q^{d} to get

− q^{d}(φ_{𝔞} − 1) ∑_{b≠0,1}(φ_{b} − 1) ≡ −dq^{d}(q − 2)(F − 1)^{2} + (F − 1)δ((−1)^{d}α, β)

+ δ(α, β)δ(−1, 1) mod I^{3}. (3.15)

Lastly, since ∑_{g∈G_{0}} g ≡ (q − 1)^{2} + δ(−1, 1)^{2} + δ(1, −1)δ(−1, −1) mod I^{3}, we get

F^{2}(1 + qF + q^{2}F^{2} + ⋯ + q^{d−1}F^{d−1}) ∑_{g∈G_{0}} g

≡ (q − 1)^{2}F^{2}(1 + qF + q^{2}F^{2} + ⋯ + q^{d−1}F^{d−1})

+ (1 + q + q^{2} + ⋯ + q^{d−1})(δ(−1, 1)^{2} + δ(1, −1)δ(−1, −1)) mod I^{3}.

We also have

(1 + q + q^{2} + ⋯ + q^{d−1})(δ(−1, 1)^{2} + δ(1, −1)δ(−1, −1))

≡ d(δ(−1, 1)^{2} + δ(1, −1)δ(−1, −1)) mod I^{3}.

Expand (q − 1)^{2}F^{2}(1 + qF + q^{2}F^{2} + ⋯ + q^{d−1}F^{d−1}) in ascending powers of (F − 1) to see that it is congruent to c_{0} + c_{1}(F − 1) + c_{2}(F − 1)^{2} modulo I^{3}, where the coefficients are c_{0} = (q − 1)(q^{d} − 1), c_{1} = (d + 1)q^{d+1} − (d + 2)q^{d} − q + 2, and

c_{2} = 1 + q + q^{2} + ⋯ + q^{d−1} − \frac{d(d+3)}{2}q^{d} + \frac{d(d+1)}{2}q^{d+1}.

Putting these together, we obtain

F^{2}(1 + qF + q^{2}F^{2} + ⋯ + q^{d−1}F^{d−1}) ∑_{g∈G_{0}} g

≡ c_{0} + c_{1}(F − 1) + c_{2}(F − 1)^{2} + d(δ(−1, 1)^{2} + δ(1, −1)δ(−1, −1)) mod I^{3}. (3.16)

Now add together equations (3.13) through (3.16) to obtain

θ̄_{S,T} ≡ (1 + q + q^{2} + ⋯ + q^{d−1})(F − 1)^{2} + (F − 1)δ((−1)^{d}α, β) + δ(α, β)δ(−1, 1) + δ(α, 1)δ(−1, 1) + δ(1, β)δ(1, −1) + dδ(1, −1)δ(−1, −1).

Finally, we have

δ(α, β)δ(−1, 1)+δ(α, 1)δ(−1, 1) + δ(1, β)δ(1, −1) + dδ(1, −1)δ(−1, −1)
≡ δ(1, β)δ(−1, 1) + δ(1, β)δ(1, −1) + dδ(1, −1)δ(−1, −1)
≡ δ(1, β)δ(−1, −1) + dδ(1, −1)δ(−1, −1)
≡ δ(1, (−1)^{d}β)δ(−1, −1) mod I^{3},

which proves the required congruence. □

Finally, we complete the proof of Theorem 3.2.

Proof of Theorem 3.2. Start with the congruence of Proposition 3.12, and multiply by (1 + q + q^{2} + ⋯ + q^{d−1}). Note also that

(1 + q + q^{2} + ⋯ + q^{d−1})δ(1, (−1)^{d}β)δ(−1, −1) ≡ dδ(1, (−1)^{d}β)δ(−1, −1) mod I^{3},

because (q − 1) kills I_{G_{0}}^{2}/I_{G_{0}}^{3}. Now the theorem follows by comparing with the congruence in Proposition 3.9. □

Acknowledgments

This paper grew out of, and considerably extends, results obtained in my Ph.D. thesis [8]. I would like to give my heartiest thanks to my advisor, Michael Rosen, for all his guidance and encouragement. I am also grateful to David Hayes for many helpful conversations and insights.

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Old address:  Current address:
Michael Reid  Michael Reid
Department of Mathematics  Department of Mathematics
University of Arizona  University of Central Florida
Tucson, AZ 85721  Orlando, FL 32816
U.S.A.  U.S.A.

reid@math.arizona.edu                          reid@cflmath.com