Rational Torsion on the Generalized Jacobian

of a Modular Curve With Cuspidal Modulus

Takao Yamazaki^{1} and Yifan Yang^{2}

Received: June 22, 2016 Revised: August 16, 2016

Communicated by Don Blasius

Abstract. We consider the generalized Jacobian J̃_{0}(N) of a modular curve X_{0}(N) with respect to a reduced divisor given by the sum of all cusps on it. When N is a power of a prime ≥ 5, we exhibit that the group of rational torsion points J̃_{0}(N)(ℚ)_{Tor} tends to be much smaller than the classical Jacobian.

2010 Mathematics Subject Classification: Primary 14H40; Secondary 11G16, 11F03, 14G35. Keywords and Phrases: Generalized Jacobian, torsion points, modular units, cuspidal divisor class.

1 Introduction

1.1

Let N be a natural number and let X_{0}(N) be the modular curve with respect to Γ_{0}(N) = \{( \begin{matrix}a & b \\ c & d \end{matrix}) ∈ SL_{2}(ℤ) | c ≡ 0 mod N\}, which we regard as a smooth projective curve over ℚ. Its Jacobian variety J_{0}(N) is an important object in arithmetic geometry and is intensively studied by many authors. By the Mordell-Weil theorem, the group J_{0}(N)(ℚ) of ℚ-rational points on J_{0}(N) is finitely generated, and hence its torsion subgroup J_{0}(N)(ℚ)_{Tor} is finite. The torsion subgroup contains an important subgroup 𝒞 (N) generated by classes of ℚ-rational divisors of degree 0 with support on cusps of X_{0}(N), called the ℚ-rational cuspidal divisor class group. (By Manin’s theorem [4], divisors with support on cusps are of finite order in J_{0}(N).) Ogg [6] conjectured and later Mazur [5] proved that when N = p is a prime, the two groups 𝒞 (p) and J_{0}(p)(ℚ)_{Tor} coincide and are cyclic of order (p − 1)/(p − 1, 12). For general cases, it is still an open problem whether the two groups are equal, although the works of Lorenzini [3] and Ling [2] have given a partially affirmative answer to the problem. We summarize the results mentioned above in the theorem below. Here for m ∈ ℤ_{>0}, we say two abelian groups are isomorphic up to m-torsion if they become isomorphic after tensoring with ℤ[1/m].

1Supported by JSPS KAKENHI Grant (15K04773).

2Supported by Grant 102-2115-M-009-001-MY4 of the Ministry of Science and Technology, Taiwan (R.O.C.).

Documenta Mathematica 21 (2016) 1669–1690

Theorem 1.1.1 Let p be a prime number and set a := (p − 1)/(p − 1, 12), b := (p + 1)/(p + 1, 12). Let n be a positive integer.

1. If N = p, then J_{0}(p)(ℚ)_{Tor} = 𝒞 (p) and is a cyclic group of order a. (Mazur [5, Theorem 1].)

2. Suppose p ≢ 11 mod 12. If p ≥ 5 and N = p^{n}, then the three groups J_{0}(p^{n})(ℚ)_{Tor}, 𝒞 (p^{n}), and (ℤ/aℤ)^{n} × (ℤ/bℤ)^{n−1} are isomorphic up to 2ptorsion. (Lorenzini [3, Theorem 4.6].)

3. The previous statement (2) holds without the assumption p ≢ 11 mod 12 but up to 6p-torsion (Ling [2, Theorem 4]).

4. Assume that p ≥ 5. If n is even, then

𝒞 (p^{n}) ≃ (ℤ/aℤ)^{n} × (ℤ/bℤ)^{n−1} × ∏_{i=n/2}^{n−2} ℤ/p^{i}ℤ × ∏_{i=(n/2)+1}^{n−1} ℤ/p^{i}ℤ.

If n is odd, then

𝒞 (p^{n}) ≃ (ℤ/aℤ)^{n} × (ℤ/bℤ)^{n−1} × ∏_{i=(n+1)/2}^{n−2} ℤ/p^{i}ℤ × ∏_{i=(n+1)/2}^{n−1} ℤ/p^{i}ℤ.

In particular, the order of 𝒞 (p^{n}) is a^{n}b^{n−1}p^{k_{n}}, where

k_{n} = \begin{cases}(n − 2)(3n − 2)/4, & ifn is even, \\ (n − 1)(3n − 5)/4, & ifn is odd. \end{cases}

(Ling [2, Theorem 1].)

Remark 1.1.2 Recently, Ohta [7] proved that 𝒞 (N) and J_{0}(N)(ℚ)_{Tor} are isomorphic up to 2-torsion when N is the product of distinct primes different from 3.

Let C_{0}(N) be the closed subset of X_{0}(N) consisting of all cusps. We regard C_{0}(N) as an effective reduced divisor on X_{0}(N). In this paper, we consider the generalized Jacobian J̃_{0}(N) of X_{0}(N) with modulus C_{0}(N) in the sense of Rosenlicht-Serre [8]. It should be as important as J_{0}(N) in arithmetic geometry of modular curves, but somehow J̃_{0}(N) has not been studied much. We are interested in the group of ℚ-rational points J̃_{0}(N)(ℚ) on J̃_{0}(N). Although it is not finitely generated (unless N = 1), its torsion subgroup J̃_{0}(N)(ℚ)_{Tor} is finite. In this paper we observe that J̃_{0}(N)(ℚ)_{Tor} is unexpectedly smaller than J_{0}(N)(ℚ)_{Tor} by proving the following result, which shows a sharp contrast with Theorem 1.1.1. (For example, Mazur’s theorem shows that the cardinality of J_{0}(p)(ℚ)_{Tor} grows linearly as p increases, but on the contrary, our result shows that the cardinality of J̃_{0}(p)(ℚ)_{Tor} remains the same for all p.)

Theorem 1.1.3 Let p be a prime number and n be a positive integer.

1. If N = p, then J̃_{0}(p)(ℚ)_{Tor} is a cyclic group of order 2.

2. Suppose p ≢ 11 mod 12. If p ≥ 5 and N = p^{n}, then J̃_{0}(p^{n})(ℚ)_{Tor} is isomorphic to the trivial group up to 2p-torsion.

3. The previous statement (2) holds without the assumption p ≢ 11 mod 12 but up to 6p-torsion.

4. Assume that p ≥ 5. Suppose that the conjecture J_{0}(p^{n})(ℚ)_{Tor} = 𝒞 (p^{n}) is true. If n is even, then

J̃_{0}(p^{n})(ℚ)_{Tor} ≃ ∏_{i=0}^{(n/2)−1} ℤ/(2p^{i}ℤ) × ∏_{i=1}^{n/2} ℤ/(2p^{i}ℤ).

If n is odd, then

J̃_{0}(p^{n})(ℚ)_{Tor} ≃ ∏_{i=0}^{(n−1)/2} ℤ/(2p^{i}ℤ) × ∏_{i=1}^{(n−1)/2} ℤ/(2p^{i}ℤ).

This result is actually a consequence of our main theorem (Theorem 1.3.1) below. However, before we state our main result, let us pause here to recall some basic facts about generalized Jacobian (cf. [8]).

1.2

Let C be a smooth projective geometrically connected curve over a field k, and J the Jacobian variety of C. We give ourselves distinct closed points P_{0}, …, P_{n} ∈ C. We assume that P_{n} is a k-rational point. We consider the generalized Jacobian J̃ of C with modulus D = P_{0} + ⋯ + P_{n}. There is an exact sequence

0 → 𝔾_{m} → ⨁_{i=0}^{n} Res_{k(P_{i})/k}𝔾_{m} → J̃ → J → 0

of commutative algebraic groups over k. Here Res_{k(P_{i})/k} denotes the Weil restriction. We have Res_{k(P_{n})/k}𝔾_{m} = 𝔾_{m} since P_{n} is a k-rational point. As we have H^{1}(k, Res_{k(P_{i})/k}𝔾_{m}) = 0 (by Hilbert 90 and Shapiro’s lemma), it induces exact sequences of abelian groups

0 → ⨁_{i=0}^{n−1} k(P_{i})^{×} →^{ι} J̃(k) → J(k) → 0  (1)

and

0 → ⨁_{i=0}^{n−1} μ(k(P_{i})) → J̃(k)_{Tor} →^{ρ} J(k)_{Tor} →^{δ} ⨁_{i=0}^{n−1} k(P_{i})^{×} ⊗ ℚ/ℤ,  (2)

where we denote by μ(F) the group of all roots of unity in F for a field F.

Remark 1.2.1 Consider an effective divisor D′ which has the same support as D (that is, D′ = ∑_{i=0}^{n} a_{i}P_{i} with a_{i} ∈ ℤ_{>0}). One can consider the generalized Jacobian J′ of C with modulus D′. Then there is a canonical surjection J′ → J̃ whose kernel is unipotent. In particular, when k is of characteristic zero, we have an isomorphism J′(k)_{Tor} → J̃(k)_{Tor} and hence there is nothing new in our problem.

1.3

We return to the setting in §1.1. Let p be a prime number and let n be a positive integer. Then C_{0}(p^{n}) consists of n+1 points P_{0}, …, P_{n}, which we will arrange in such a way that the residue field ℚ(P_{i}) of P_{i} is the cyclotomic field ℚ(μ_{p^{d(i)}}) of degree p^{d(i)} with d(i) := min(i, n − i) for each i = 0, …, n. (See §3.1 for more details.) In particular, P_{0} and P_{n} are ℚ-rational. Then the map δ in (2) for (k, C, D) = (ℚ, X_{0}(p^{n}), C_{0}(p^{n})) reads

δ : J_{0}(p^{n})(ℚ)_{Tor} → ⨁_{i=0}^{n−1} ℚ(μ_{p^{d(i)}})^{×} ⊗ ℚ/ℤ, d(i) = min(i, n − i). (3)

Thus, Theorem 1.1.3 follows from Theorem 1.1.1 and the following theorem, which is the main result of this article.

Theorem 1.3.1 Let p ≥ 5 be a prime number and let n be a positive integer. Then the restriction of the map (3) to 𝒞 (p^{n}) is injective.

The proof of Theorem 1.3.1 will occupy almost all of the rest of this article and will be completed in §6. On the other hand, Theorem 1.3.1 does not admit a naive generalization to other values of level N. Indeed, in the last section §7 we shall observe the following result:

Proposition 1.3.2 Let p, q be two distinct prime numbers. (Then C_{0}(pq) consists of four ℚ-rational points.) If p ≡ q ≡ 1 mod 12, then the kernel of the restriction to 𝒞 (pq) of

δ : J_{0}(pq)(ℚ)_{Tor} → (ℚ^{×})^{3} ⊗ ℚ/ℤ

is a cyclic group of order (p − 1)(q − 1)/24.

In view of Ohta’s result (see Remark 1.1.2), we find that J̃_{0}(pq)(ℚ)_{Tor} is isomorphic to a cyclic group of order (p − 1)(q − 1)/3 up to 2-torsion. This shows another sharp contrast with Theorem 1.1.3. We are not able (but hoping) to find more conceptual reason for such difference.

Notation

Let \overline{ℚ} be an algebraic closure of ℚ and fix an embedding \overline{ℚ} ↪ ℂ. For m ∈ ℤ_{>0}, we set ζ_{m} := e^{2πi/m} ∈ \overline{ℚ}^{×} and μ_{m} := \{ζ_{m}^{k} | k ∈ ℤ\} ⊂ \overline{ℚ}^{×}. For an abelian group A, we write A_{Tor} for the subgroup of torsion elements of A. For a field F, we write μ(F) = (F^{×})_{Tor}.

2 Torsion rational points on generalized Jacobian

2.1

In this section, we use the notations introduced in §1.2. We always assume P_{n} is k-rational. We will give an explicit description of the map δ in (2) in Lemma 2.3.1 below. Take x ∈ ⨁_{i=0}^{n−1} k(P_{i})^{×}, m ∈ ℤ_{>0} and a ∈ J(k) such that ma = 0. Then by definition we have δ(a) = x ⊗ \frac{1}{m} if there is a lift ã ∈ J̃(k) of a such that ι(x) = mã, where ι is the map appearing in (1).

2.2

We recall some basic facts about the relative Picard group and generalized Jacobian (cf. [8, Chapter V]). Denote by K the function field of C. For a closed point P on C, we write K_{P} for the completion of K at P, O_{P} for the ring of integers in K_{P}, t_{P} ∈ O_{P} for a (fixed) uniformizer, U_{P} := (1 + t_{P} O_{P})^{×} for the group of principal units in O_{P}, and k(P) := O_{P} /t_{P} O_{P} for the residue field at P. Let U := C \ |D| be the open complement of the divisor D = P_{0} + ⋯ + P_{n}. Let us consider the abelian group

Div(C, D) := Div(U) ⊕ ⨁_{i=0}^{n}(K_{P_{i}}^{×}/U_{P_{i}}).

We have a canonical map

K^{×} → Div(C, D), f ↦ (div_{U}(f); (f mod U_{P_{i}})_{i=0}^{n})),

whose cokernel is by definition the relative Picard group Pic(C, D) of C relative to D. We also have a commutative diagram with exact rows

[figure: 0; K^{×}; Div(C, D); α; Pic(C, D); \overline{α}; 0; k^{×}; K^{×}; Div(C); Pic(C); 0,]

where α is a canonical surjection given by

(E; (f_{i} mod U_{P_{i}})_{i=0}^{n}) ↦ E + ∑_{i=0}^{n} ord_{P_{i}}(f_{i})P_{i},

and \overline{α} is induced by α. Combined with isomorphisms

ker(α) ≅ ⨁_{i=0}^{n}(O_{P_{i}}^{×}/U_{P_{i}}) ≅ ⨁_{i=0}^{n} k(P_{i})^{×},

⨁_{i=0}^{n} k(P_{i})^{×}/(k^{×}) ≅ ⨁_{i=0}^{n−1} k(P_{i})^{×}, (c_{i})_{i=0}^{n} mod (k^{×}) ↦ (c_{i}/c_{n})_{i=0}^{n−1},

where (k^{×}) is the image of the diagonal map k^{×} → ⊕_{i} k(P_{i})^{×}, we obtain an exact sequence

0 → ⨁_{i=0}^{n−1} k(P_{i})^{×} → Pic(C, D) → Pic(C) → 0.  (4)

On the other hand, there are canonical isomorphisms

J(k) ≅ ker[Pic(C) →^{deg} ℤ],
J̃(k) ≅ ker[Pic(C, D) →^{\overline{α}} Pic(C) →^{deg} ℤ].

Then (1) is deduced from (4) by restriction. We also obtain

J(k)_{Tor} ≅ Pic(C)_{Tor}, J̃(k)_{Tor} ≅ Pic(C, D)_{Tor}.

2.3

We are now ready to describe explicitly the map δ from (2).

Lemma 2.3.1 Let E = ∑_{i=0}^{n} a_{i}P_{i} ∈ Div^{0}(C) be a degree zero divisor supported on D. Suppose that its class [E] in J(k) is killed by m ∈ ℤ_{>0} so that there is f ∈ K^{×} such that div_{C}(f) = mE. Define

E := ((\frac{f}{t_{P_{n}}^{ma_{n}}})(P_{n})(\frac{t_{P_{i}}^{ma_{i}}}{f})(P_{i}))_{i=0}^{n−1} ∈ ⨁_{i=0}^{n−1} k(P_{i})^{×}.

Then we have

δ([E]) = E ⊗ \frac{1}{m} in ⨁_{i=0}^{n−1} k(P_{i})^{×} ⊗ ℚ/ℤ.

(Note that E ⊗ \frac{1}{m} does not depend on the choices of t_{P_{i}} and f.)

Proof. We use the fact recalled in §2.1. Put Ẽ := (0; (t_{P_{i}}^{a_{i}})_{i=0}^{n}) ∈ Div(C, D) so that α(Ẽ) = E. It suffices to prove that the class of mẼ in Pic(C, D) is the same as ι(E) ∈ J̃(k) ⊂ Pic(C, D) (see (1) for the map ι). By definition, ι(E) is given by the class of (0; (f^{−1}t_{P_{i}}^{ma_{i}})_{i=0}^{n}). Since div_{C}(f) = mE, we have div_{U}(f) = 0, and hence the class of (0; (f^{−1}t_{P_{i}}^{ma_{i}})_{i=0}^{n}) agrees with that of (0; (t_{P_{i}}^{ma_{i}})_{i=0}^{n}) = mẼ in Pic(C, D). We are done. □

3 Preliminaries on modular curves

3.1

We return to the setting in §1.1. We take an integer N > 1 and consider the modular curve X_{0}(N). Recall that C_{0}(N) denotes the set of cusps on X_{0}(N) so that we have a canonical bijection C_{0}(N)(ℂ) ≅ Γ_{0}(N)\ℙ^{1}(ℚ). For each divisor d > 0 of N, there is a unique Q_{d} ∈ C_{0}(N) such that the set of ℂ-rational points lying over Q_{d} is given by Γ_{0}(N)-orbits of a/d ∈ ℙ^{1}(ℚ) with a ∈ ℤ, (a, d) = 1. We call Q_{d} ∈ C_{0}(N) the cusp of level d. The residue field of Q_{d} is ℚ(ζ_{m}) with m = (d, N/d), hence the degree of Q_{d} is ϕ(m), where ϕ denotes the Euler function. The classes of 0 and ∞ ∈ ℙ^{1}(ℚ) are ℚ-rational and are of level 1 and N respectively. We define

\begin{aligned}𝒟(N) & := \{E ∈ Div^{0}(X_{0}(N)) | |E| ⊂ C_{0}(N)\} \\ & = ⟨Q_{d} − ϕ((d, N/d))Q_{N} | d|N⟩, \end{aligned}

𝒫(N) := 𝒟(N) ∩ div(ℚ(X_{0}(N))^{×}),  (5)

𝒞 (N) := 𝒟(N)/𝒫(N).

As we recalled in the introduction, 𝒞 (N) is a subgroup of J_{0}(N)(ℚ)_{Tor}, hence finite.

3.2

We will use the Dedekind eta function to construct modular functions needed for our purpose. Here let us recall some well-known properties of the Dedekind eta function η(τ), where as usual τ is a variable on the upper half plane ℍ. We shall make use of the standard identification X_{0}(N)(ℂ) ≅ Γ_{0}(N)\(ℍ ∪ ℙ^{1}(ℚ)).

Proposition 3.2.1 ([1, Proposition 3.2.1]) Let N be a positive integer. The product h = ∏_{δ|N} η(δτ)^{r_{δ}}, r_{δ} ∈ ℤ, is a modular function on X_{0}(N) if and only if the following conditions are satisfied:

1. ∑_{δ|N} r_{δ} = 0,

2. ∏_{δ|N} δ^{r_{δ}} is the square of a rational number,

3. ∑_{δ|N} r_{δ}δ ≡ 0 mod 24, and

4. ∑_{δ|N} r_{δ}(N/δ) ≡ 0 mod 24.

We also remark that, if these conditions are satisfied, then h is defined over ℚ (see [1, p. 32, Remarque]).

Lemma 3.2.2 ([1, Proposition 3.2.8]) Let N be a positive integer. Let d and δ be positive divisors of N. Then the order of η(δτ) at a cusp of level d is a_{N}(d, δ)/24, where

a_{N}(d, δ) := \frac{N}{(d, N/d)} \frac{(d, δ)^{2}}{dδ} .

In particular, if g(τ) = ∏_{δ|N} η(δτ)^{r_{δ}} is an eta-product satisfying the conditions in Proposition 3.2.1, then

divg = \frac{1}{24} ∑_{d,δ|N} r_{δ}a_{N}(d, δ)Q_{d}.

When N = p^{n} is a prime power, the orders of η(p^{k}τ) at cusps can be summarized as follows.

Corollary 3.2.3 Let p^{n} be a prime power.

1. If m ≥ n/2, then the order of η(p^{k}τ) at a cusp of level p^{m} is

\begin{cases}p^{k}/24, & ifk ≤ m, \\ p^{2m−k}/24, & ifk > m. \end{cases}

2. If m < n/2, then the order of η(p^{k}τ) at a cusp of level p^{m} is

\begin{cases}p^{n−k}/24, & ifm ≤ k, \\ p^{n+k−2m}/24, & ifm > k. \end{cases}

In order to obtain the Fourier expansion of an eta-product at a cusp, we should need the following transformation formula for the Dedekind eta function.

Lemma 3.2.4 ([10, pp. 125–127]) For

γ = (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ SL_{2}(ℤ),

the transformation formula for η(τ) is given by, for c = 0,

η(τ + b) = e^{πib/12}η(τ),

and, for c > 0,

η(γτ) = ε(a, b, c, d)\sqrt{\frac{cτ + d}{i}}η(τ)

with

ε(a, b, c, d) = \begin{cases}(\frac{d}{c}) i^{(1−c)/2}e^{πi(bd(1−c^{2})+c(a+d))/12}, & ifc is odd, \\ (\frac{c}{d}) e^{πi(ac(1−d^{2})+d(b−c+3))/12}, & ifd is odd, \end{cases}

where (\frac{d}{c}) is the Jacobi symbol.

4 Cuspidal divisor class group

4.1

In §§4–6, we consider the case N = p^{n}, where p is a prime greater than or equal to 5 and n is a positive integer. We describe the group of modular units on X_{0}(p^{n}) that gives us 𝒫(p^{n}) (see (5) for its definition).

Proposition 4.1.1 Let p^{n} be a prime power with p ≥ 5 prime and n ∈ ℤ_{>0}. Then the group 𝒫(p^{n}) is generated by the divisors of

f(τ) = (\frac{η(pτ)}{η(τ)})^{24/(p−1,12)} , g_{k}(τ) = \frac{η(p^{k+2}τ)}{η(p^{k}τ)}, k = 0, …, n − 2.

The proof of this proposition will be given in §4.2. We first deduce a corollary that will be used later. For i = 0, …, n, we write P_{i} := Q_{p^{i}} for the cusp of level p^{i} (see §3.1).

Corollary 4.1.2 Let p, n be as in Proposition 4.1.1.

1. The divisors div(f), div(g_{0}), …, div(g_{n−2}) form a free ℤ-basis of 𝒫(p^{n}).

2. Let c_{0}, …, c_{n−2} ∈ ℤ and write

div(fg_{0}^{c_{0}}g_{1}^{c_{1}} … g_{n−2}^{c_{n−2}}) = ∑_{i=0}^{n} s_{i}P_{i}, s_{i} ∈ ℤ

in 𝒟(p^{n}). Then we have (s_{0}, s_{1}, …, s_{n}) = (p − 1)/(p − 1, 12).

Proof. (1) This is an immediate consequence of Proposition 4.1.1, since 𝒫(p^{n}) is a free ℤ-module of rank n (as it is a finite index subgroup of 𝒟(p^{n}).)

(2) Put a := (p − 1)/(p − 1, 12). By using Corollary 3.2.3 we first see that the order of g_{k} at any cusp is divisible by a for k = 0, …, n − 2. Hence, by (1), it suffices to show the statement for c_{0} = ⋯ = c_{n−2} = 0, which again follows from Corollary 3.2.3. □

In the proof of Proposition 4.1.1, we use the following elementary lemma. We omit its proof.

Lemma 4.1.3 Let L_{0} ⊂ ℝ^{n+1} be the lattice of rank n generated by the vectors of the form (0, …, 1, −1, 0, …, 0). Let L_{1} be a sublattice of L_{0} of the same rank generated by v_{1}, …, v_{n} ∈ L_{1}. Let v_{n+1} = (c_{1}, …, c_{n+1}) be any vector such that ∑_{i} c_{i} ≠ 0, and M be the (n + 1) × (n + 1) matrix whose ith row is v_{i}. Then we have

(L_{0} : L_{1}) = |(∑_{i=1}^{n+1} c_{i})^{−1} det M| .

4.2 Proof of Proposition 4.1.1

Let L_{0} be the lattice of rank n in ℤ^{n+1} = ⨁_{i=0}^{n} ℤe_{i} generated by vectors of the form (0, …, 1, −1, 0, …, 0). Recall that D_{i} := P_{i} − ϕ((p^{i}, p^{n−i}))P_{n} (i = 0, …, n − 1) form a ℤ-basis of 𝒟(p^{n}). Consider the natural group homomorphism λ : 𝒟(p^{n}) → ℤ^{n+1} defined by

λ(D_{i}) = ϕ((p^{i}, p^{n−i}))e_{i} − ϕ((p^{i}, p^{n−i}))e_{n} (i = 0, …, n − 1).

Let L_{1} = λ(𝒟(p^{n})) be the image of 𝒟(p^{n}) under λ. It is a sublattice of L_{0}. Let 𝒟′ be the group generated by the divisors of f and g_{k} and L_{2} be the image of 𝒟′ under λ. Then to show that the divisors of f and g_{k} generate 𝒫(p^{n}), it suffices to show that the index of L_{2} in L_{1} is equal to the divisor class number given in Part (4) of Theorem 1.1.1. To show that this indeed holds, we form 3 square matrices M, U, and V of dimension n + 1. The first matrix M = (M_{ij})_{i,j=0}^{n} is defined by

M_{ij} = the order of η(p^{i}τ) at cusps of level p^{j}

= \begin{cases}p^{i}/24, & if j ≥ n/2 and i ≤ j, \\ p^{2j−i}/24, & if j ≥ n/2 and i > j, \\ p^{n−i}/24, & if j < n/2 and i ≥ j, \\ p^{n+i−2j}/24, & if j < n/2 and i < j. \end{cases}

The second matrix U is a diagonal matrix with the diagonal entries being ϕ((p^{i}, p^{n−i})), i = 0, …, n. The third matrix V is

V = (\begin{matrix}−c & c & 0 & 0 & ⋯ & ⋯ & 0 \\ −1 & 0 & 1 & 0 & ⋯ & ⋯ & 0 \\ 0 & −1 & 0 & 1 & ⋯ & ⋯ & 0 \\ ⋮ & ⋮ & & & & ⋮ & ⋮ \\ 0 & ⋯ & ⋯ & 0 & −1 & 0 & 1 \\ 1 & ⋯ & ⋯ & ⋯ & 1 & 1 & 1 \end{matrix}) , c = \frac{24}{(p − 1, 12)}.

That is, if we associate to an eta-product ∏_{i=0}^{n} η(p^{i}τ)^{r_{i}} a vector (r_{0}, …, r_{n}), then the first n rows of V are the vectors corresponding to the functions f and g_{k}, while the last row of V consists of 1’s. Then the first n rows of the matrix V MU are precisely λ(divf) and λ(divg_{k}), k = 0, …, n − 2. Now we claim that (see Theorem 1.1.1 for the definition of a and b)

1. det V = 24(−1)^{n}(n + 1)/(p − 1, 12),

2. det M = \begin{cases}(ab)^{n}p^{(n−1)(3n−1)/4}/24, & if n is odd, \\ (ab)^{n}p^{n(3n−4)/4}/24, & if n is even. \end{cases}

3. det U = ∏_{i=0}^{n} ϕ((p^{i}, p^{n−i})), and

4. the sum of the entries in the last row of V MU is (n + 1)p^{n−1}(p + 1)/24.

Assuming that the claims are true for the moment, let us complete the proof of the proposition. It is clear that

(L_{0} : L_{1}) = ∏_{i=0}^{n} ϕ((p^{i}, p^{n−i})) = det U.  (6)

By Lemma 4.1.3, we have

(L_{0} : L_{2}) = C^{−1}| det(V MU)|,

where C is the sum of the entries in the last row of V MU. By the four claims above,

det(V MU) = (−1)^{n} \frac{(n + 1)}{(p − 1, 12)}(ab)^{n}(det U) × \begin{cases}p^{(n−1)(3n−1)/4}, & if n is odd, \\ p^{n(3n−4)/4}, & if n is even, \end{cases}

and C = (n + 1)p^{n−1}(p + 1)/24. It follows that

(L_{0} : L_{2}) = \frac{24(ab)^{n} det U}{(p + 1)(p − 1, 12)} × \begin{cases}p^{(n−1)(3n−5)/4}, & if n is odd, \\ p^{(n−2)(3n−2)/4}, & if n is even. \end{cases}

Recall that the number b is defined to be (p + 1)/(p + 1, 12). Also we may check case by case that (p − 1, 12)(p + 1, 12) = 24. Therefore, the expression above can also be written as

(L_{0} : L_{2}) = a^{n}b^{n−1}(det U) × \begin{cases}p^{(n−1)(3n−5)/4}, & if n is odd, \\ p^{(n−2)(3n−2)/4}, & if n is even. \end{cases}

Combining this with (6), we find that

(L_{1} : L_{2}) = a^{n}b^{n−1} × \begin{cases}p^{(n−1)(3n−5)/4}, & if n is odd, \\ p^{(n−2)(3n−2)/4}, & if n is even, \end{cases}

which agrees with the class number given in Part (4) of Theorem 1.1.1. Therefore, we conclude that the divisors of f and g_{k}, k = 0, …, n − 2 generate 𝒫(p^{n}). It remains to prove that the four claims are true. Claims (1) and (3) are obvious. To prove Claim (2), we start by giving examples. Consider the case n = 5. The matrix M in this case is

\frac{1}{24} (\begin{matrix}p^{5} & p^{3} & p & 1 & 1 & 1 \\ p^{4} & p^{4} & p^{2} & p & p & p \\ p^{3} & p^{3} & p^{3} & p^{2} & p^{2} & p^{2} \\ p^{2} & p^{2} & p^{2} & p^{3} & p^{3} & p^{3} \\ p & p & p & p^{2} & p^{4} & p^{4} \\ 1 & 1 & 1 & p & p^{3} & p^{5} \end{matrix})

We subtract the second column from the first column, the third column from the second column, the fourth column from the fifth column, and then the fifth column from the last column. The matrix becomes

\frac{1}{24} (\begin{matrix}p^{3}(p^{2} − 1) & p(p^{2} − 1) & p & 1 & 0 & 0 \\ 0 & p^{2}(p^{2} − 1) & p^{2} & p & 0 & 0 \\ 0 & 0 & p^{3} & p^{2} & 0 & 0 \\ 0 & 0 & p^{2} & p^{3} & 0 & 0 \\ 0 & 0 & p & p^{2} & p^{2}(p^{2} − 1) & 0 \\ 0 & 0 & 1 & p & p(p^{2} − 1) & p^{3}(p^{2} − 1) \end{matrix}) ,

with the determinant unchanged. Thus,

det M = \frac{1}{24^{6}} p^{14}(p^{2} − 1)^{5} = \frac{1}{24}(ab)^{5}p^{14}.

In general, if n is an odd integer greater than 3, then a similar matrix manipulation (subtracting the second column from the first column, the third column from the second column, etc.) will produce a matrix of the form

\frac{1}{24} (\begin{matrix}A_{1} & B_{1} & 0 \\ 0 & A_{2} & 0 \\ 0 & B_{2} & A_{3} \end{matrix}) ,

where A_{1} is an upper-triangular matrix of dimension (n−1)/2 whose diagonal entries are p^{n−2}(p^{2} − 1), …, p^{(n−1)/2}(p^{2} − 1), A_{3} is a lower-triangular matrix of the same dimension whose diagonal entries are p^{(n−1)/2}(p^{2} − 1), …, p^{n−2}(p^{2} − 1),

A_{2} = (\begin{matrix}p^{(n+1)/2} & p^{(n−1)/2} \\ p^{(n−1)/2} & p^{(n+1)/2} \end{matrix}) ,

and B_{i} are some immaterial (n − 1)/2-by-2 matrices. It follows that

\begin{aligned}det M & = \frac{1}{24^{n+1}} p^{2((n−1)/2+(n+1)/2+···+(n−2))}(p^{2} − 1)^{n−1}(p^{n+1} − p^{n−1}) \\ & = \frac{1}{24}(ab)^{n}p^{(n−1)(3n−1)/4}. \end{aligned}

This proves Claim (2) for the case of odd n. The proof of the case of even n is similar. For the case n = 4, we have

M = \frac{1}{24} (\begin{matrix}p^{4} & p^{2} & 1 & 1 & 1 \\ p^{3} & p^{3} & p & p & p \\ p^{2} & p^{2} & p^{2} & p^{2} & p^{2} \\ p & p & p & p^{3} & p^{3} \\ 1 & 1 & 1 & p^{2} & p^{4} \end{matrix}) .

Subtracting the second column from the first column, the third column from the second column, the fourth column from the last column, and then the third column from the fourth column, we obtain the matrix

\frac{1}{24} (\begin{matrix}p^{2}(p^{2} − 1) & p^{2} − 1 & 1 & 0 & 0 \\ 0 & p(p^{2} − 1) & p & 0 & 0 \\ 0 & 0 & p^{2} & 0 & 0 \\ 0 & 0 & p & p(p^{2} − 1) & 0 \\ 0 & 0 & 1 & p^{2} − 1 & p^{2}(p − 1) \end{matrix}) ,

whose determinant is

\frac{1}{24^{5}} p^{8}(p^{2} − 1)^{4} = \frac{1}{24}(ab)^{4}p^{8}.

In general, a similar matrix manipulation yields a matrix of the form

\frac{1}{24} (\begin{matrix}A_{1} & B_{1} & 0 \\ 0 & p^{n/2} & 0 \\ 0 & B_{2} & A_{2} \end{matrix}) ,

where A_{1} is an upper-triangular matrix of dimension n/2 with the diagonal entries being p^{n−2}(p^{2} − 1), …, p^{n/2−1}(p^{2} − 1) and A_{2} is a lower-triangular matrix of dimension n/2 with the diagonals being p^{n/2−1}(p^{2} − 1), …, p^{n−2}(p^{2} − 1). Therefore,

\begin{aligned}det M & = \frac{1}{24^{n+1}} p^{2((n/2−1)+n/2+···+(n−2))+n/2}(p^{2} − 1)^{n} \\ & = \frac{1}{24}(ab)^{n}p^{n(3n−4)/4}. \end{aligned}

This completes the proof of Claim (2). To prove Claim (4), we first observe that since the last row of V consists of 1’s, the sum of the entries in the last row of V MU is simply the sum of all entries in MU. Now the (i, j)-entry of MU is the order of η(p^{i−1}τ) at a cusp of level p^{j−1} times the number of such cusps. Therefore, the sum of the entries in the ith row of MU is the degree of divη(p^{i−1}τ), which is equal to

\frac{1}{24}(SL(2, ℤ) : Γ_{0}(p^{n})) = \frac{1}{24}p^{n−1}(p + 1).

(In general, the degree of the divisor of a modular form of weight k on Γ_{0}(N) is k(SL(2, ℤ) : Γ_{0}(N))/12. Here the weight of the Dedekind eta function is 1/2.) Hence the sum of the entries in the last row of V MU is (n + 1)p^{n−1}(p + 1)/24. This completes the proof of the proposition. □

5 Leading Fourier coefficients of modular units at cusps

5.1

In this section, we work out the leading Fourier coefficients of the modular functions f(τ) and g_{k} defined in Proposition 4.1.1 at cusps. To speak of such coefficients, we first need to choose uniformizers at cusps. Then the coefficients are canonically defined as an element of the residue fields of cusps, but we can calculate it after base change to ℂ. As cusps of the same level are Galois conjugates, for our purpose, we only need to calculate the leading coefficient at one of the ℂ-valued points of cusps of each level. In general, to define a local uniformizer at a cusp α ∈ ℙ^{1}(ℚ) of X_{0}(N), we choose an element σ in GL^{+}(2, ℚ) such that σ∞ = α. Let h be the smallest positive number such that

σ (\begin{matrix}1 & h \\ 0 & 1 \end{matrix}) σ^{−1} ∈ Γ_{0}(N).

Then a local uniformizer at α is

q_{α} = e^{2πiσ^{−1}τ/h}.

5.2

We return to the case N = p^{n}, where p^{n} is a prime power with p ≥ 5. For convenience, our choice of a cusp α_{m} ∈ ℙ(ℚ) of level p^{m} is

α_{m} = \begin{cases}1/p^{m}, & if m ≥ n/2, \\ −1/p^{m}, & if m < n/2. \end{cases}  (7)

Then we can choose σ_{m} to be

σ_{m} = \begin{cases}( \begin{matrix}1 & 0 \\ p^{m} & 1 \end{matrix} ) , & if m ≥ n/2, \\ ( \begin{matrix}0 & −1 \\ p^{n} & 0 \end{matrix} ) ( \begin{matrix}1 & 0 \\ p^{n−m} & 1 \end{matrix} ) = ( \begin{matrix}−p^{n−m} & −1 \\ p^{n} & 0 \end{matrix} ) , & if m < n/2. \end{cases}  (8)

We summarize our discussion as a lemma:

Lemma 5.2.1 With the choice of σ_{m} given in (8), the local uniformizer at the cusps α_{m} in (7) is

e^{2πiσ_{m}^{−1}τ}

for each m.

5.3

In the following lemma, we adopt the following notation

f(τ)| (\begin{matrix}a & b \\ c & d \end{matrix}) := f (\frac{aτ + b}{cτ + d}) ,

which is slightly different from the usual meaning of the slash operator.

Lemma 5.3.1 Let p be an odd prime.

1. If k ≤ m, then

η(p^{k}τ)| (\begin{matrix}1 & 0 \\ p^{m} & 1 \end{matrix}) = e^{2πi/8}e^{−2πip^{m−k}/24}\sqrt{\frac{p^{m}τ + 1}{i}}η(p^{k}τ).

2. If k ≥ m, then

η(p^{k}τ)| (\begin{matrix}1 & 0 \\ p^{m} & 1 \end{matrix}) = e^{2πip^{k−m}/24}\sqrt{\frac{p^{m}τ + 1}{p^{k−m}i}}η (\frac{p^{m}τ + 1}{p^{k−m}}) .

Proof. If k ≤ m, we have

p^{k} \frac{τ}{p^{m}τ + 1} = (\begin{matrix}1 & 0 \\ p^{m−k} & 1 \end{matrix}) p^{k}τ.

Hence, by Lemma 3.2.4, we find that

\begin{aligned}η(p^{k}τ)| (\begin{matrix}1 & 0 \\ p^{m} & 1 \end{matrix}) & = ε(1, 0, p^{m−k}, 1)\sqrt{\frac{p^{m}τ + 1}{i}}η(p^{k}τ) \\ & = i^{(1−p^{m−k})/2}e^{2πip^{m−k}/12}\sqrt{\frac{p^{m}τ + 1}{i}}η(p^{k}τ), \end{aligned}

which yields the first statement of the lemma. If k ≥ m, we have

p^{k} \frac{τ}{p^{m}τ + 1} = (\begin{matrix}p^{k−m} & −1 \\ 1 & 0 \end{matrix}) \frac{p^{m}τ + 1}{p^{k−m}}.

By Lemma 3.2.4 again, we have

\begin{aligned}η(p^{k}τ)| (\begin{matrix}1 & 0 \\ p^{m} & 1 \end{matrix}) & = ε(p^{k−m}, −1, 1, 0)\sqrt{\frac{p^{m}τ + 1}{p^{k−m}i}}η (\frac{p^{m}τ + 1}{p^{k−m}}) \\ & = e^{2πip^{k−m}/24}\sqrt{\frac{p^{m}τ + 1}{p^{k−m}i}}η (\frac{p^{m}τ + 1}{p^{k−m}}) . \end{aligned}

This proves the lemma. □

Remark 5.3.2 From Lemmas 5.2.1 and 5.3.1, we can easily deduce the orders of η(p^{k}τ) at each cusp, recovering the results in Corollary 3.2.3.

5.4

We now use Lemma 5.3.1 to obtain the leading coefficients of modular functions at cusps. Here for an odd prime p, we let

p^{∗} = e^{2πi(p−1)/4}p,  \sqrt{p^{∗}} = e^{2πi(p−1)/8}\sqrt{p}.  (9)

Proposition 5.4.1 Assume that p^{n} is a prime power with p ≥ 5 and n ≥ 1. Let

f(τ) = (\frac{η(pτ)}{η(τ)})^{24/(p−1,12)} , g_{k}(τ) = \frac{η(p^{k+2}τ)}{η(p^{k}τ)}, k = 0, …, n − 2,

be the modular functions defined in Proposition 4.1.1.

1. If m ≥ n/2, then the leading Fourier coefficients of f(τ) and g_{k}(τ) with respect to the local uniformizer at 1/p^{m} chosen using (8) are

1  for f(τ),
1  for g_{k}(τ) withk ≤ m − 2,
(−1)^{(p−1)/2}e^{−2πiab/p}/\sqrt{p^{∗}} for g_{m−1}(τ),
e^{−2πiab/p^{k+2−m}}/p  for g_{k}(τ) withk ≥ m.

2. If m < n/2, then the leading Fourier coefficients of f(τ) and g_{k}(τ) with respect to the local uniformizer at −1/p^{m} chosen using (8) are

p^{−12/(p−1,12)} for f(τ) whenm = 0,
e^{2πia/p^{m}}  for f(τ) whenm ≥ 1,
1/p  for g_{k}(τ) withk ≥ m,
e^{2πiab/p}/\sqrt{p^{∗}} for g_{m−1}(τ),
e^{2πiab/p^{m−k}} for g_{k}(τ) withk ≤ m − 2.

Proof. Consider the function f(τ) first. Since n is assumed to be at least 1, when m ≥ n/2, we have m ≥ 1 and the first part of Lemma 5.3.1 applies. We find that

\frac{η(pτ)}{η(τ)}| (\begin{matrix}1 & 0 \\ p^{m} & 1 \end{matrix}) = e^{2πip^{m−1}(p−1)/24} \frac{η(pτ)}{η(τ)}.

It follows that

f(τ)| (\begin{matrix}1 & 0 \\ p^{m} & 1 \end{matrix}) = f(τ)

and the leading Fourier coefficient is 1. Similarly, if k is less than or equal to m − 2, then the leading Fourier coefficient of g_{k}(τ) at the cusp 1/p^{m} is 1. If k = m − 1, then k < m and k + 2 = m + 1 > m. By Lemma 5.3.1, we have

η(p^{m+1}τ)| (\begin{matrix}1 & 0 \\ p^{m} & 1 \end{matrix}) = e^{2πip/24}\sqrt{\frac{p^{m}τ + 1}{pi}}η (\frac{p^{m}τ + 1}{p}) ,

η(p^{m−1}τ)| (\begin{matrix}1 & 0 \\ p^{m} & 1 \end{matrix}) = e^{2πi/8}e^{−2πip/24}\sqrt{\frac{p^{m}τ + 1}{i}}η(p^{m−1}τ),

and the leading coefficient of g_{m−1}(τ) at 1/p^{m} is

\begin{aligned}\frac{1}{\sqrt{p}}e^{2πi(p/12−1/8+1/24p)} & = \frac{1}{\sqrt{p}}e^{2πi(p−1)/8}e^{−2πi(p^{2}−1)/24p} \\ & = \frac{(−1)^{(p−1)/2}}{\sqrt{p^{∗}}} e^{−2πiab/p}. \end{aligned}

(Here we remind the reader that the leading coefficient of η((cτ + d)/e) is e^{2πid/24e}.) When k ≥ m, by Part (2) of Lemma 5.3.1,

\frac{η(p^{k+2}τ)}{η(p^{k}τ)}| (\begin{matrix}1 & 0 \\ p^{m} & 1 \end{matrix}) = \frac{1}{p}e^{2πip^{k−m}(p^{2}−1)/24} \frac{η((p^{m}τ + 1)/p^{k+2−m})}{η((p^{m}τ + 1)/p^{k−m})},

whose leading Fourier coefficient is

\frac{1}{p}e^{2πi(1−p^{2})/24p^{k+2−m}} = \frac{1}{p}e^{−2πiab/p^{k+2−m}}.

This completes the proof of the case of m ≥ n/2. We next consider the case α_{m} = −1/p^{m} with m < n/2. Recall that the choice of σ_{m} is given in (8). Noticing that

η(p^{k}τ)| (\begin{matrix}0 & −1 \\ p^{n} & 0 \end{matrix}) = η(−1/p^{n−k}τ) = \sqrt{\frac{p^{n−k}τ}{i}} η(p^{n−k}τ),

we have

g_{k}(τ)|σ_{m} = \frac{η(p^{k+2}τ)}{η(p^{k}τ)}| (\begin{matrix}0 & −1 \\ p^{n} & 0 \end{matrix}) |σ_{n−m} = \frac{1}{pg_{n−k−2}(τ)}|σ_{n−m}.

Thus, using the results in Part (1), we find that the leading coefficients of g_{k}(τ) at the cusp α_{m} = −1/p^{m} are

\begin{cases}1/p, & if k ≥ m, \\ e^{2πiab/p}/\sqrt{p^{∗}}, & if k = m − 1, \\ e^{2πiab/p^{m−k}} & if k ≤ m − 2. \end{cases}

Finally, for the function f(τ), we have

f(τ)|σ_{m} = \frac{1}{p^{12/(p−1,12)}} (\frac{η(p^{n−1}τ)}{η(p^{n}τ)})^{24/(p−1,12)} |σ_{n−m}.

When m ≥ 1, by Part (2) of Lemma 5.3.1,

\begin{matrix}(\frac{η(p^{n−1}τ)}{η(p^{n}τ)})^{24/(p−1,12)} |σ_{n−m} \\ = p^{12/(p−1,12)} (\frac{η((p^{n−m}τ + 1)/p^{m−1})}{η((p^{n−m}τ + 1)/p^{m})})^{24/(p−1,12)} . \end{matrix}

Documenta Mathematica 21 (2016) 1669–1690

Thus, the leading coefficient of f(τ) at α_{m} is

(e^{2πip^{−m}(p−1)/24})^{24/(p−1,12)} = e^{2πia/p^{m}}.

When m = 0, by Part (1) of Lemma 5.3.1,

(\frac{η(p^{n−1}τ)}{η(p^{n}τ)})^{24/(p−1,12)} |σ_{n} = (e^{2πi(1−p)/24} \frac{η(p^{n−1}τ)}{η(p^{n}τ)})^{24/(p−1,12)} .

Thus, the leading coefficient of f(τ) at α_{0} is p^{−12/(p−1,12)}. This completes the proof of the proposition. □

6 Proof of Theorem 1.3.1

6.1

We keep to assume N = p^{n}, where p ≥ 5 is a prime and n is a positive integer. As in §4, we write P_{i} := Q_{p^{i}} for the cusp of level p^{i} for i = 0, …, n (see §3.1). The residue field of P_{i} is given by

ℚ(P_{i}) = ℚ(ζ_{p^{d(i)}}), d(i) = min(i, n − i).

Our task is to show the injectivity of the composition map

𝒞 (p^{n}) ↪ J_{0}(p^{n})(ℚ)_{Tor} →^{δ} ⨁_{i=0}^{n−1} ℚ(P_{i})^{×} ⊗ ℚ/ℤ,  (10)

where δ is the map from (2).

Lemma 6.1.1 Let p ≥ 3 be a prime and m ∈ ℤ_{>0}. Then the maps

ℚ/ℤ → ℚ^{×} ⊗ ℚ/ℤ, x ↦ p ⊗ x,
ℚ/ℤ → ℚ(ζ_{p^{m}})^{×} ⊗ ℚ/ℤ, x ↦ \sqrt{p^{∗}} ⊗ x,

are (split) injections. (See (9) for the definition of \sqrt{p^{∗}}.)

Proof. Splitting is automatic because ℚ/ℤ is injective. The first statement follows from the elementary fact that ℚ^{×} is the direct sum of {±1} and the free abelian group on the set of all prime numbers. From this, the second statement is reduced to showing

ker[ℚ^{×} ⊗ (ℚ/ℤ) → ℚ(ζ_{p^{m}})^{×} ⊗ (ℚ/ℤ)] = \{0, p^{∗} ⊗ \frac{1}{2}\}.  (11)

The right hand side is contained in the left, because \sqrt{p^{∗}} ∈ ℚ(ζ_{p^{m}}). Thus it suffices to show the left hand side of (11) is of order 2. Put ℚ/ℤ(1) := μ(\overline{ℚ}). In terms of Galois cohomology, this group can be rewritten as

ker[H^{1}(ℚ, ℚ/ℤ(1)) → H^{1}(ℚ(ζ_{p^{m}}), ℚ/ℤ(1))], which is then identified with

H^{1}(G, H^{0}(ℚ(ζ_{p^{m}}), ℚ/ℤ(1))), G = Gal(ℚ(ζ_{p^{m}})/ℚ)

by the inflation-restriction sequence. Note that H^{0}(ℚ(ζ_{p^{m}}), ℚ/ℤ(1)) = μ_{2p^{m}}. Since G ≅ (ℤ/p^{m}ℤ)^{×} is a cyclic group, the order of H^{1}(G, μ_{2p^{m}}) agrees with that of the Tate cohomology

Ĥ^{0}(G, μ_{2p^{m}}) := \frac{\{x ∈ μ_{2p^{m}} | σ(x) = x for all σ ∈ G\}}{\{∏_{σ∈G} σ(x) | x ∈ μ_{2p^{m}}\}}.

Direct computation shows that Ĥ^{0}(G, μ_{2p^{m}}) is of order two. This completes the proof of the lemma. □

6.2

Let Λ_{i} be the subgroup of ℚ(P_{i})^{×}/μ(ℚ(P_{i})) generated by p (resp. \sqrt{p^{∗}}) for i = 0 (resp. for i = 1, …, n − 1). We also let

Λ := ⨁_{i=0}^{n−1} Λ_{i} ⊂ ⨁_{i=0}^{n−1} ℚ(P_{i})^{×}/μ(ℚ(P_{i})).

Proposition 5.4.1 and Lemma 2.3.1 show that the map δ in (10) factors as

\xymatrix{𝒞 (p^{n}) \ar[r]^{δ̃} \ar[dr]_{(10)} & Λ ⊗ ℚ/ℤ \ar@{^{(}->}[d] \\  & ⨁_{i=0}^{n−1} ℚ(P_{i})^{×} ⊗ ℚ/ℤ,}  (12)

where the right vertical injection is provided by Lemma 6.1.1. We are reduced to showing the injectivity of δ̃. We choose a uniformizer t_{P_{i}} described in Lemma 5.2.1 at each cusp P_{i}. Using them, we define a homomorphism

Δ : 𝒫(p^{n}) → Λ  (13)

by, for any h ∈ K^{×} such that div(h) is supported on C_{0}(p^{n}),

Δ(div(h)) = ((\frac{h}{t_{P_{n}}^{ord_{Pn} (h)}} )(P_{n})(\frac{t_{P_{i}}^{ord_{P_{i}} (h)}}{h})(P_{i}))_{i=0}^{n−1} ∈ Λ ⊂ ⨁_{i=0}^{n−1} ℚ(P_{i})^{×}/μ(ℚ(P_{i})).

Note that the image of Δ is contained in Λ by Propositions 4.1.1 and 5.4.1. Note also that, unlike δ, this map depends on our choice of uniformizers t_{P_{i}}. Recall that a = (p − 1)/(p − 1, 12). Put a′ = 12/(p − 1, 12).

Lemma 6.2.1 1. The map Δ is injective.

2. The cokernel of Δ is a cyclic group of order a′ generated by the class of λ := p ∈ Λ_{0} ⊂ Λ.

3. There exist c_{0}, …, c_{n−2} ∈ ℤ such that a′λ = Δ(div(fg_{0}^{c_{0}} … g_{n−2}^{c_{n−2}})). Here f, g_{0}, …, g_{n−2} are functions introduced in Proposition 4.1.1.

Proof. Recall from Corollary 4.1.2 (1) that div(f), div(g_{0}), …, div(g_{n−2}) form a ℤ-basis of 𝒫(p^{n}). Let us take a ℤ-basis of Λ = ⊕_{i=0}^{n−1}Λ_{i} given by the generators λ = p ∈ Λ_{0} and \sqrt{p^{∗}} ∈ Λ_{i} for i = 1, …, n − 1. Then Proposition 5.4.1 shows that the map Δ is represented by

(\begin{matrix}a′ & 0 & 0 & 0 & … & 0 \\ 1 & 1 & 0 & 0 & … & 0 \\ 1 & 2 & 1 & 0 & … & 0 \\ 1 & 2 & 2 & 1 & … & 0 \\ ⋮ & & & & & ⋮ \\ 1 & 2 & … & 2 & 2 & 1 \end{matrix})

from which the lemma follows. □

6.3 Proof of Theorem 1.3.1

Recall from Lemma 6.2.1 (1) that the map Δ defined in (13) is injective. Since 𝒫(p^{n}) is of finite index in 𝒟(p^{n}) (see (5)), Δ has a unique extension

Δ̃ : 𝒟(p^{n}) → Λ ⊗ ℚ.

which is also injective. By Lemma 2.3.1, we have a commutative diagram with exact rows

\xymatrix{0 \ar[r] & 𝒫(p^{n}) \ar[d]^{Δ} \ar[r] & 𝒟(p^{n}) \ar[d]^{Δ̃} \ar[r] & 𝒞 (p^{n}) \ar[d]^{δ̃} \ar[r] & 0 \\ 0 \ar[r] & Λ \ar[r] & Λ ⊗ ℚ \ar[r] & Λ ⊗ ℚ/ℤ \ar[r] & 0.}

where δ̃ is from (12). We get an exact sequence

0 → ker(δ̃) → Coker(Δ) →^{ψ} Coker(Δ̃).  (14)

It remains to show ψ is injective. In view of Lemma 6.2.1 (2), this amounts to showing that the image of λ = p ∈ Λ_{0} ⊂ Λ in Coker(Δ̃) has order a′. Let b > 0 be a divisor of a′ such that the image of bλ vanishes in Coker(Δ̃). This means that div(fg_{0}^{c_{0}} … g_{n−2}^{c_{n−2}}) = (a′/b)E for some E ∈ 𝒟(p^{n}), where c_{0}, …, c_{n−2} ∈ ℤ are taken from Lemma 6.2.1 (3). Since a and a′ are relatively prime to each other, Corollary 4.1.2 (2) shows that b must be a′. This completes the proof. □

7 The case of N = pq

7.1

In this section, we present an outline of the proof of Proposition 1.3.2. Since the proof goes similarly with Theorem 1.3.1, we will be brief and omit details. Let N = pq where p, q are two distinct prime numbers such that p ≡ q ≡ 1 mod 12. We use Takagi’s result [9, Theorem 5.1] that determines the order of 𝒞 (N) for any square free N. Here we state it in the special case N = pq, p, q ≡ 1 mod 12:

Proposition 7.1.1 (Takagi) The order of 𝒞 (pq) is given by 4abc, where

a = \frac{(p − 1)(q + 1)}{24}, b = \frac{(p + 1)(q − 1)}{24}, c = \frac{(p − 1)(q − 1)}{24}.

7.2 Proof of Proposition 1.3.2

There are exactly four cusps on X_{0}(pq), all of which are ℚ-rational. Their levels are 1, p, q, pq (see §3.1). We give their names as follows:

C_{0}(pq) = {P_{0} = Q_{1}, P_{1} = Q_{p}, P_{2} = Q_{q}, P_{3} = Q_{pq}}

so that we have a ℤ-basis of 𝒟(pq) given by

D_{1} = P_{0} − P_{3}, D_{2} = P_{1} − P_{3}, D_{3} = P_{2} − P_{3}.

Using Proposition 7.1.1, one sees that the group 𝒫(pq) is generated by the divisors of

f_{1} = \frac{η(τ)η(qτ)}{η(pτ)η(pqτ)}, f_{2} = \frac{η(τ)η(pτ)}{η(qτ)η(pqτ)}, f_{3} = \frac{η(τ)η(pqτ)}{η(pτ)η(qτ)}

by an argument similar to Proposition 4.1.1. Lemma 3.2.2 shows that the divisors of the functions f_{1}, f_{2}, f_{3} are respectively given by

a(D_{1} − D_{2} + D_{3}), b(D_{1} + D_{2} − D_{3}), c(D_{1} − D_{2} − D_{3}).

Let m ∈ {1, p, q, pq} and put m′ = pq/m. Choose integers b and d such that dm′ − bm = 1. Put

σ_{m} := (\begin{matrix}m′ & b \\ pq & dm′ \end{matrix}) .

Then σ_{m} normalizes Γ_{0}(pq) and satisfies σ_{m}∞ = 1/m. We may take e^{2πiσ_{m}^{−1}τ} as a local uniformizer at the cusp of level m (cf. Lemma 5.2.1). With this choice, the leading coefficients of f_{1}, f_{2}, f_{3}, up to ±1 signs, are given by the following table:

P_{0} P_{1} P_{2} P_{3}
f_{1} p 1 p 1
f_{2} q q 1 1
f_{3} 1 1 1 1

Finally, by using Lemma 2.3.1, we find the kernel of 𝒞 (pq) ↪ J_{0}(pq)(ℚ)_{Tor} →^{δ} (ℚ^{×})^{3} ⊗ ℚ/ℤ is a cyclic group of order c generated by the class of D_{1} − D_{2} − D_{3}. □

Acknowledgement. The authors would like to thank the anonymous referee for the detailed comments.

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Takao Yamazaki Mathematical Institute Tohoku University Aoba, Sendai 980-8578 Japan ytakao@math.tohoku.ac.jp Yifan Yang Department of Applied Mathematics National Chiao Tung University Hsinchu 300 Taiwan yfyang@math.nctu.edu.tw