Structure of the cuspidal rational torsion subgroup of J(1)(p(n))

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(1)J. London Math. Soc. (2) 82 (2010) 203–228. e 2010 London Mathematical Society. C. doi:10.1112/jlms/jdq013. Structure of the cuspidal rational torsion subgroup of J1 (pn ) Yifan Yang and Jeng-Daw Yu Abstract n. Let p be a prime and let J1 (p ) denote the Jacobian of the modular curve X1 (pn ). The Jacobian J1 (pn ) contains a Q-rational torsion subgroup generated by the cuspidal divisor classes [(a/pn ) − (∞)], where p a. In this paper, we determine the structure of the p-primary subgroup of this Q-rational torsion subgroup in the case where p is a regular prime.. 1. Introduction and statements of results Let Γ be a congruence subgroup of SL(2, Z). The modular curve X(Γ) and its Jacobian variety J(Γ) are very important objects in number theory. For instance, the problem of determining all possible structures of a (Q-)rational torsion subgroup of elliptic curves over Q is equivalent to that of determining whether the modular curves X1 (N ) have non-cuspidal rational points. Also, the celebrated theorem of Wiles and others shows that every elliptic curve over Q is a factor of the Jacobian J0 (N ). In the present paper, we are concerned with the arithmetic aspect of the Jacobian variety J1 (N ) of the modular curve X1 (N ). In particular, we will study the structure of the (Q-)rational torsion subgroup of J1 (N ). Recall that the modular curve X1 (N ) possesses a model over Q on which the cusp ∞ is a (Q-)rational point (see [11, Chapter 6] for details). Thus, if P is another rational cusp, then the image of P under the cuspidal embedding i∞ : X1 (N ) → J1 (N ) sending P to the divisor class [(P ) − (∞)] will be a rational point on J1 (N ). Moreover, according to a result of Manin [9], the point i∞ (P ) is of finite order. In other words, the rational torsion subgroup of J1 (N ) contains a subgroup generated by the image of rational cusps under i∞ . More generally, if D is a divisor of degree 0 defined over Q that is supported by cusps, then the divisor class of D gives us a rational torsion point on the Jacobian. We refer to the rational torsion subgroup arising in this way as the cuspidal rational torsion subgroup of J1 (N ). In general, it is believed that the cuspidal rational torsion subgroup should be the whole rational torsion subgroup. (For primes p, the conjecture was formally stated in [1, Conjecture 6.2.2]. The conjecture was verified for a few cases in the same paper.) Note that, for the case J0 (p), the Jacobian of X0 (p) of prime level p, Mazur [10, Theorem 1] has already shown that all rational torsion points are generated by the divisor class [(0) − (∞)]. On the aforementioned model of X1 (N ), all the cusps of type k/N with (k, N ) = 1 are rational over Q (see, for example, [12, Theorem 1.3.1]). Moreover, if the level N is relatively prime to 6, then these cusps are the only rational cusps. Since these cusps are precisely those lying over ∞ of X0 (N ), for convenience, we shall call them the ∞-cusps. Now suppose that we are given a divisor D of degree 0 supported by the ∞-cusps on X1 (N ). Then the order of the divisor class [D] in J1 (N ) is simply the smallest positive integer m such that mD is a principal divisor, that is, the divisor of a modular function on X1 (N ). Therefore, to determine the group Received 19 August 2008; revised 4 June 2009; published online 29 June 2010. 2000 Mathematics Subject Classification 11G18 (primary), 11F11, 14G35, 14H40 (secondary). The first author’s visit was supported by a fellowship of the Max-Planck-Institut. He was also partially supported by Grant 96-2628-M-009-014 of the National Science Council, Taiwan. The second author was supported in part by Professor N. Yui’s Discovery Grant from NSERC, Canada..

(2) 204. YIFAN YANG AND JENG-DAW YU. structure of the cuspidal rational torsion subgroup of J1 (N ) generated by the ∞-cusps, it is vital to study the group of modular units on X1 (N ) having divisors supported on the ∞-cusps. (In the literature, if a modular function f on a congruence subgroup Γ has a divisor supported on cusps, then f is called a modular unit.) In a series of papers [3–7], Kubert and Lang studied the group of modular units on X(N ) and X1 (N ). For the curves X1 (N ), in [8, Chapter 3] they showed that all modular units on X1 (N ) with divisors supported on the ∞-cusps are products of a certain class of Siegel functions (see Subsection 3.1 for details). Furthermore, in [7] they also determined the order of the torsion subgroup of J1 (pn ) generated by the ∞-cusps for the case p is a prime greater than 3. (The case N = p was first obtained in [2]). Then Yu [18] gave a formula for all positive integers N . (Note that all the results mentioned above dealt with modular units with divisors supported on the cusps lying over 0 of X0 (N ), instead of the ∞-cusps, but it is easy to translate the results using the Atkin–Lehner involution N0 −1 0 .) In a very recent paper [17], we applied Yu’s divisor class number formula to determine an explicit basis for the group of modular units on X1 (N ) with divisors supported on the ∞-cusps for any positive integer N . As applications, we used the basis to compute the group structure of the cuspidal rational torsion subgroup of J1 (N ) with divisors supported on the ∞-cusps. A remarkable discovery is that, when p is a regular prime, the structure of the p-primary subgroup of the cuspidal rational torsion subgroup of J1 (pn ) seems to follow a simple pattern. (Recall that an odd prime p is said to be regular if p does not divide the numerators of the Bernoulli numbers B2 , B4 , . . . , Bp−3 .) More precisely, let p be a prime, let n be a positive integer and let C1∞ (pn ) be the subgroup of J1 (pn ) generated by the ∞-cusps. Consider the endomorphism [p] : C1∞ (pn ) → C1∞ (pn ) defined by multiplication by p. Define the p-rank of C1∞ (pn ) to be the integer k such that the kernel of [p] has pk elements. Assume that p is a regular prime. Then the p-rank of C1∞ (pn ) is 1 (p − 1)pn−2 − 1 2 for prime power pn 8 with n 2. More precisely, the number of copies of Z/p2k Z in the primary decomposition of C1∞ (pn ) is given by 1 (p − 1)2 pn−k−2 − 1 if p = 2 and k n − 3, 2 1 (p − 1)2 pn−k−2 − 1 if p 3 and k n − 2, 2 1 (p − 5) if p 5 and k = n − 1, 2 0 else. Conjecture (Yang [17]).. and the number of copies of Z/p2k−1 Z is given by 1 if p = 2 and k n − 3, 1 if p = 3 and k n − 2, 1 if p 5 and k n − 1, 0 otherwise. Example. For the primes p = 2, 3, 5, the above conjecture asserts that the p-parts of C1∞ (pn ) follow the pattern depicted in Table 1. Here the notation (pe1 )n1 . . . (pek )nk means that the primary decomposition of C1∞ (pn ) contains ni copies of Z/pei Z..

(3) CUSPIDAL RATIONAL TORSION SUBGROUP OF J1 (pn ). 205. Table 1. The p-primary part of C1∞ (pn ). pn. p-primary subgroups. 24 25 26 27. (2) (2)(22 )(23 ) (2)(22 )3 (23 )(24 )(25 ) (2)(22 )7 (23 )(24 )3 (25 )(26 )(27 ). 33 34 35 36. (3)(32 ) (3)(32 )5 (33 )(34 ) (3)(32 )17 (33 )(34 )5 (35 )(36 ) (3)(32 )53 (33 )(34 )17 (35 )(36 )5 (37 )(38 ). 52 53 54 55. (5) (5)(52 )7 (53 ) (5)(52 )39 (53 )(54 )7 (55 ) (5)(52 )199 (53 )(54 )39 (55 )(56 )7 (57 ). The main purpose of the present paper is to prove this conjecture. Theorem A.. The conjecture is true.. We note that the assumption that p is a regular prime is crucial in the proof of Theorem A. This assumption is used to establish an exact formula for the p-rank of C1∞ (pn ) and to determine the kernel of the homomorphism C1∞ (pn ) → C1∞ (pn−1 ) induced from the covering X1 (pn ) → X1 (pn−1 ). At present, we do not know how to extend our method to the case of irregular primes. On the other hand, it is possible to obtain a similar result for modular curves X1 (pn q m ), where q is another prime, under the assumption that the product 1 B2,χ p 4 χ of generalized Bernoulli numbers B2,χ associated with even Dirichlet characters χ mod pq m is a p-unit. For example, following the argument in the present paper, we can show that the 2primary subgroup of the torsion subgroup of J1 (3 · 2n ) generated by the ∞-cusps is isomorphic to n−2 n−k−2 (Z/22k Z)2 . k=1. However, we will not pursue this direction here because it does not constitute a significant extension of Theorem A and the proof of some key lemmas in these cases is much more complicated than the prime power cases. (For instance, it takes more than one page just to describe the basis for the group of modular units on X1 (pn q m ).) Remark. Note that in a recent work [13], Sun considered the -adic evaluation v (|C1∞ (N pn )|) for a prime = p. His result showed that there exists an integer ν such that v (|C1∞ (N pn )|) = τ pn−1 + ν for all sufficiently large n, where (p − 1)φ(N/v (N ) )(v (N )−1 − 1) τ= 0. if | N, if N..

(4) 206. YIFAN YANG AND JENG-DAW YU. This formula in particular implies that if 2 N , then the -adic valuation of |C1∞ (N pn )| eventually becomes a constant for all sufficiently large n. This result is comparable to Washington’s theorem [14] on the boundedness of the -part of ideal class groups in a Zp extension of an abelian number field. The rest of the article is organized as follows. In Section 2, we describe our strategy in proving Theorem A. We will show that Theorem A will follow immediately from five properties of the divisor groups, namely, Propositions 1–5. In Section 3, we review our basis for the group of modular units on X1 (N ), which constitutes the cornerstone of our argument. In Section 4, we study the natural maps between the cuspidal groups in different levels. We then give the proof of the five propositions in Section 5.. 2. Outline of proof of Theorem A In this section, we first collect all the notation and conventions used throughout the paper. We then describe our strategy in proving Theorem A. Our arguments depend crucially on our explicit knowledge on the basis for the group of modular units on X1 (N ), which will be reviewed in Subsection 3.2. 2.1. Notation and conventions Let p be a prime. We fix an integer α that generates (Z/pn+1 Z)× / ± 1 for all integers n 0. Explicitly, for p = 2, we choose α = 3, and for an odd prime p, we let α be an integer that generates (Z/pZ)× but satisfies αp−1 ≡ 1 mod p2 . For n 0, we define Xn = the modular curve X1 (pn+1 ), Cn = the set of cusps of Xn lying over ∞ of X0 (pn+1 ), that is, the set of ∞-cusps, φn = |Cn | = φ(pn+1 )/2 = pn (p − 1)/2, Pn,k = the cusp αk /pn+1 in Cn , Dn = the group of divisors of degree 0 on Xn having support on Cn , Fn = the group of modular units on Xn having divisors supported on Cn , Pn = div Fn , the subgroup of principal divisors on Xn having support on Cn , Cn = Dn /Pn , the rational torsion subgroup of J1 (pn+1 ) generated by Cn , πn = the canonical homomorphism from Dn to Dn−1 induced from the covering Xn → Xn−1 , ιn = the embedding Dn−1 → Dn defined by ιn (P ) = p. . Q.. Q: πn (Q)=P. Note that Pn,k and Pn,m represent the same cusp on Xn if and only if k ≡ m mod φn . Then we have Cn = {Pn,k : k = 0, . . . , φn − 1}, and πn (Pn,k ) = Pn−1,k ,. ιn (Pn−1,k ) = p. p−1 . Pn,k+hφn−1 .. h=0. Since we are mainly interested in the orders of a function at the ∞-cusps, for a modular function f on Xn , we introduce the notation div∞ denoting the Cn -part div∞ f = ordf (P )P P ∈Cn. of the divisor of f ..

(5) CUSPIDAL RATIONAL TORSION SUBGROUP OF J1 (pn ). 207. Finally, the generalized Bernoulli numbers Bk,χ associated with a Dirichlet character χ mod N , not necessarily primitive, are defined by the series N χ(r)tert r=1. eN t. −1. =. ∞ k=0. Bk,χ. tk . k!. In particular, we have B2,χ = N. N . χ(r)B2. r=1. 2. N r r 1 r =N + χ(r) − . N N2 N 6 r=1. Here B2 (x) = {x}2 − {x} + 1/6 and {x} denotes the fractional part of a real number x. The readers should be mindful that our definition differs from some other authors’ definition. See the remark following Theorem E for details. 2.2. Outline of proof of Theorem A In this section, we will describe our strategy in proving Theorem A. Intuitively, just by looking at Table 1, one immediately realizes that if the conjecture is to hold, then the p-primary subgroup of Cn / ker[p2 ] must have the same structure as that of Cn−1 , where [p2 ] denotes the multiplication-by-p2 homomorphism for an additive group, and one expects that there should be a canonical isomorphism between the p-primary subgroups of the two groups. The only sensible candidate for such an isomorphism is the one induced by the covering Xn → Xn−1 . To establish this isomorphism, we first show that πn induces an isomorphism between the p-part of Cn = Dn /Pn and that of πn (Dn )/πn (Pn ) = Dn−1 /πn (Pn ). We then show that the kernel of [p2 ] of the latter group is Pn−1 /πn (Pn ), and thereby establish the isomorphism. The following diagram illustrate the relations between various groups: Cn = Dn /Pn / ker[p2 ]. πn. - Dn−1 /πn (Pn ). p-part . / ker[p2 ]. ? ? Cn / ker[p2 ] p-part- Cn−1 = Dn−1 /Pn−1 Now assume that the isomorphism between the p-parts of Cn / ker[p2 ] and Cn−1 is established. This would show if the p-part of Cn−1 is (Z/pei Z)ri , then the p-part of Cn is (Z/pZ)s1 × that 2 s2 ei +2 Z)ri for some non-negative integers s1 and s2 . If we can determine the (Z/p Z) × (Z/p p-ranks of Cn−1 and Cn and the index of πn (Pn ) in Pn−1 , this will yield information about s1 + s2 and s1 + 2s2 , respectively, which in turn will give us the exact values of s1 and s2 . Finally, if we know the structure of C0 (C1 for p = 3 and C2 for p = 2), then the structure of the p-primary subgroup of Cn is determined for all n. In summary, to establish Theorem A, it suffices to prove the following propositions. Proposition 1. If p is a regular prime, then p does not divide |C0 |. Also, if p = 2 and p = 3, then p |C1 |, and if p = 2, then p |C2 |. Proposition 2. Let p be a regular prime. If pn+1 5, then the p-rank of Cn is pn−1 (p − 1)/2 − 1..

(6) 208. YIFAN YANG AND JENG-DAW YU. Proposition 3. For all primes p, we have πn (Pn ) ⊂ Pn−1 , and the index of πn (Pn ) in n−1 Pn−1 is pp (p−1)−3 if pn+1 5. Moreover, the structure of the factor group Pn−1 /πn (Pn ) is given by n−1. (Z/p2 Z)p. (p−1)/2−2. × (Z/pZ).. Proposition 4. Assume that p is a regular prime. Then the p-part of Cn = Dn /Pn is isomorphic to the p-part of Dn−1 /πn (Pn ). Proposition 5. Let p be a prime. Then the kernel of the multiplication-by-p2 endomorphism [p2 ] of Dn−1 /πn (Pn ) is Pn−1 /πn (Pn ). Remark. We remark that the assumption that p is a regular prime is crucial in the proof of Propositions 1, 2 and 4. In fact, the assumption is a necessary and sufficient condition for the three propositions. For example, by carefully examining the proof of Proposition 2, one sees that if p is an irregular prime, then the p-rank of Cn is strictly greater than pn−1 (p − 1)/2 − 1. The proof of these propositions will be postponed until Section 5. Here let us formally complete the proof of Theorem A, assuming the truth of the propositions. Proof of Theorem A. By Propositions 4 and 5, when p is a regular prime, we have p-part of Cn / ker[p2 ] p-part of (Dn−1 /πn (Pn ))/ ker[p2 ] = p-part of (Dn−1 /πn (Pn ))/(Pn−1 /πn (Pn )) p-part of Dn−1 /Pn−1 = Cn−1 , Thus, if the structure of the p-part of Cn−1 is given by k . (Z/pei Z)ri ,. i=1. then, according to the structure theorem for finite abelian groups, the structure of the p-part of Cn is as follows: k (Z/pei +2 Z)ri (Z/pZ)s1 × (Z/p2 Z)s2 × i=1. for some non-negative integers s1 and s2 . Here the sum of ri is what we call the p-rank of Cn−1 , and the sum of s1 , s2 and ri is the p-rank of Cn . Using Proposition 2, we find that the integers s1 and s2 satisfy 1 s1 + s2 = pn−2 (p − 1)2 . (1) 2 On the other hand, by Propositions 3 and 4, we know that n−1. p-part of |Cn |/|Cn−1 | = |Pn−1 /πn (Pn )| = pp. (p−1)−3. ,. which, together with Proposition 2, implies that s1 + 2s2 = (pn−1 (p − 1) − 3) − 2(pn−2 (p − 1)/2 − 1) = pn−2 (p − 1)2 − 1. Combining this with (1), we get s1 = 1 and s2 = pn−2 (p − 1)2 /2 − 1. Finally, Proposition 1 shows that the p-part of C0 (C1 for p = 3 and C2 for p = 2) is trivial. Then an induction argument gives the claimed result..

(7) CUSPIDAL RATIONAL TORSION SUBGROUP OF J1 (pn ). 209. 3. Group of modular units on X1 (N ) In this section, we will introduce our basis for the group Fn , which is essential in our proof of Theorem A. The construction of our basis utilizes the Siegel functions. 3.1. Siegel functions The Siegel functions are usually defined as products of the Klein forms and the Dedekind eta function. For our purpose, we only need to know that they have the following infinite product representation. For a pair of rational numbers (a1 , a2 ) ∈ Q2 \Z2 and τ ∈ H, set z = a1 τ + a2 , qτ = e2πiτ and qz = e2πiz . Then the Siegel function G(a1 ,a2 ) (τ ) satisfies G(a1 ,a2 ) (τ ) = −e2πia2 (a1 −1)/2 qτB(a1 )/2 (1 − qz ). ∞ . (1 − qτn qz )(1 − qτn /qz ),. n=1. where B(x) = x − x + 1/6 is the second Bernoulli polynomial. To construct modular units on X1 (N ) with divisors supported on the ∞-cusps, we consider a special class of Siegel functions. Given a positive integer N and an integer a not congruent to 0 mod N , we set 2. Ea(N ) (τ ) = −G(a/N,0) (N τ ) = q N B(a/N )/2. ∞ . (1 − q (n−1)N +a )(1 − q nN −a ),. n=1 (N ). where q = e2πiτ . If the integer N is clear from the context, then we write Ea in place of Ea . We now review the properties of Ea . The material is mainly taken from [16]. For more details see [16]. In the first lemma, we describe two simple, but yet very important, relations between Siegel functions of two different levels. Lemma 6. Let M and N be two positive integers. Assume that N = nM for some integer n. Let a be an integer not congruent to 0 mod N . Then (N ) (τ ) = Ea(M ) (nτ ). Ena. (2). Moreover, for all integers a with 0 < a < M , we have. n−1 a kM + a , N B2 = M B2 N M. (3). k=0. and consequently n−1 . (N ). EkM +a (τ ) = Ea(M ) (τ ).. (4). k=0. (N ). Proof. Relation (2) follows trivially from the definition of Eg . Property (3) can be verified by a direct computation. Relation (4) is an immediate consequence of (3) and the definition (N ) of Ea . The next lemma gives the transformation law for Ea under the action of matrices in Γ0 (N ). Lemma 7 [16, Corollary 2]. satisfy. For integers g not congruent to 0 mod N , the functions Eg Eg+N = E−g = −Eg .. (5).

(8) 210. Moreover, let γ =. YIFAN YANG AND JENG-DAW YU. . a b cN d. . ∈ Γ0 (N ). For c = 0, we have Eg (τ + b) = eπibN B(g/N ) Eg (τ ),. and, for c > 0, we have Eg (γτ ) = (a, bN, c, d)eπi(g where (a, b, c, d) =. 2. ab/N −gb). 2 eπi(bd(1−c )+c(a+d−3))/6 2. −ieπi(ac(1−d. )+d(b−c+3))/6. Eag (τ ),. (6). if c is odd, if d is odd.. Remark. Note that Property (5) implies that there are only (N − 1)/2 essentially

(9) distinct Eg , indexed over the set (Z/N Z)/ ± 1 − {0}. Hence, a product g or a sum g is understood to be running over g ∈ (Z/N Z)/ ± 1 − {0}. The functions Eg clearly have no poles or zeros in the upper half-plane. The next lemma describes the order of Eg at cusps of X1 (N ). Lemma 8 [16, Lemma 2]. The order of the function Eg at a cusp a/c of X1 (N ) with (a, c) = 1 is (c, N )B2 (ag/(c, N ))/2, where B2 (x) = {x}2 − {x} + 1/6 and {x} denotes the fractional part of a real number x. The following theorem of Yu [18] characterizes the modular units on X1 (N ) with divisors supported at the ∞-cusps in terms of Eg . Theorem B [18, Theorem 4]. Let N be a positive integer. A modular function e f on Γ1 (N ) has a divisor supported on the cusps k/N , (k, N ) = 1, if and only if f = g Eg g with the exponents eg satisfying the two conditions N if N is odd, 2 g eg ≡ 0 mod (7) 2N if N is even, g and. . eg = 0. (8). g≡±a mod N/p. for all prime factors p of N and all integers a. Again, we remark that Theorem 4 of [18] was stated in the setting of modular units with the divisor supported on the that is, the cusps lying over 0 of X0 (N ). Here we use the 0-cusps, to get Theorem B from Yu’s result. Atkin–Lehner involution N0 −1 0 3.2. Basis for Fn We now describe our basis for Fn constructed in [17]. The case of an odd prime p and the case of p = 2 are stated in Theorems C and D, respectively. Theorem C [17, Theorem 2]. Let n 0 and let N = pn+1 be an odd prime power. For a non-negative integer , we set φ = φ(p+1 )/2. Let α be a generator of the cyclic group (Z/pn+1 Z)× / ± 1 and let β be an integer such that αβ ≡ 1 mod p. Then a basis for Fn mod C×.

(10) CUSPIDAL RATIONAL TORSION SUBGROUP OF J1 (pn ). 211. is given by 2. Eαi−1 Eαβi+φn−1. fi =. 2. Eαi+φn−1 −1 Eαβi Eαp i−1. fi =. Eαp i+φn−1 −1. i = 1, . . . , φn − φn−1 − 1,. ,. i = φn − φn−1 ,. ,. (pn ). Eαi−1 (pτ ). fi =. i = φn − φn−1 + 1, . . . , φn − φn−2 ,. ,. (pn ). Eαi+φn−2 −1 (pτ ). .. .. .. . (p2 ). Eαi−1 (pn−1 τ ). fi =. (p2 ). Eαi+φ0 −1 (pn−1 τ ). , i = φn − φ1 + 1, . . . , φn − φ0 ,. (p). Eαi−1 (pn τ ). fi =. (p). Eαi (pn τ ). i = φn − φ0 + 1, . . . , φn − 1.. ,. Theorem D [17, Theorem 3]. Let n 2 and N = 2n+1 . Let α = 3 be a generator of the cyclic group (Z/2n+1 Z)× / ± 1. For 1, set φ = φ(2+1 )/2 = 2−1 . Then a basis for Fn mod C× is given by fi =. Eαi−1 Eαi+φn−1 , Eαi Eαi+φn−1 −1. i = 1, . . . , φn − φn−1 − 1,. fi =. Eα2 i−1 2 Eαi+φk−1 −1. i = φn − φn−1 ,. ,. (2n ). fi =. Eαi−1 (2τ ). (2n ). Eαi+φn−2 −1 (2τ ). .. .. , i = φn − φn−1 + 1, . . . , φn − φn−2 , .. .. (8). fi =. Eαi−1 (2n−2 τ ) (8). Eαi (2n−2 τ ). ,. i = φn − 1.. The proofs of these two theorems use the following divisor class number formula of Kubert, Lang and Yu, which will also be used in the present paper. Note that the cases p 5 were proved in [7], while the cases p = 2 and p = 3 were settled in [18]. In the same paper [18], Yu also obtained a divisor class number formula for general N , although the general result is not needed in the present paper. Theorem E [7, Theorem 3.4; 18, Theorem 5]. Let N = pn+1 be a prime power greater than 4. We have the divisor class number formula 1 B2,χ , |Cn | = pL(p) (9) 4 χ=χ0 even.

(11) 212 where. YIFAN YANG AND JENG-DAW YU. pn−1 − 2n + 2 if N = pn and p is odd, L(p) = 2n−1 − 2n + 3 if N = 2n 8,. and the product runs over all even non-principal Dirichlet characters modulo pn+1 . Remark. We should remark that the definition of generalized Bernoulli numbers used in [7, 18] is different from ours; namely, if an even Dirichlet character χ mod N has a conductor f , then their definition is given by ∞. tk 1 χf (r)tert = , B 2,χ 2 r=1 ef t − 1 k! f. k=0. where χf is the Dirichlet character modulo f that induces χ. When N is a prime power pn and χ is not principal, the two definitions differ by a 1/2 factor. Moreover, the readers are reminded that there were slight errors in the original statement of [18, Theorem 5]. See the discussion following Theorem B of [17] for details.. 4. Properties of πn and ιn Throughout this section, we follow the notation specified in Subsection 2.1. The main results in this section are Lemmas 11 and 15, which state that πn maps a principal divisor to a principal divisor, and that if ιn (D) is a principal divisor, then D itself is principal. In addition, in Lemma 12 we will prove the converse of Lemma 15, that is, if D is a principal divisor in Dn−1 , then ιn (D) is a principal divisor. The first lemma is rather trivial, but it plays a crucial role in the proof of Proposition 5. Lemma 9. We have πn ◦ ιn = [p2 ],. (10). the multiplication-by-p2 endomorphism of Dn−1 Proof. The proof is obvious. (pn+1 ). In the next lemma we compute the image of the divisor of Eg that the notation div∞ f means the Cn -part of the divisor of f .. Lemma 10. Let g be an integer. For g ≡ 0 mod pn+1 , we have ⎧ n ⎨div∞ Eg(p ) if p g, n+1 πn (div∞ Eg(p ) ) = n ⎩p2 div∞ E (p ) if p|g. g/p Proof. By Lemma 8, we have n+1. div∞ Eg(p. ). =. k. φn −1 gα pn+1 B2 Pn,k . 2 pn+1 k=0. under πn . Here we recall.

(12) CUSPIDAL RATIONAL TORSION SUBGROUP OF J1 (pn ). 213. Recall that πn (Pn,k ) = πn (Pn,h ) if and only if h ≡ k mod φn−1 . Thus, we have k+hφn−1. φn−1 −1 p−1 n+1 pn+1 gα Pn−1,k B2 . πn (div∞ Eg(p ) ) = 2 pn+1 k=0. h=0. Now assume that p does not divide g; then as h goes through 0 to p − 1, the residue classes of gαk+hφn−1 mod pn+1 go through gαk , gαk + pn , . . . , gαk + (p − 1)pn . Hence, by equation (3), we find that k. φn−1 −1 n gα pn ∞ (pn+1 ) )= B2 πn (div Eg Pn−1,k = div∞ Eg(p ) . n 2 p k=0. When p | g, all ga. k+hφn−1. are congruent to gak mod pn+1 . Therefore, we have. n φn−1 −1 (g/p)αk (pn ) ∞ (pn+1 ) 2 p πn (div Eg )=p · B2 Pn−1,k = p2 div∞ Eg/p . 2 pn k=0. This proves the lemma. Lemma 11. Assume that n 1. If D is a principal divisor in Dn , then πn (D) is a principal divisor in Dn−1 . More precisely, if fi , with i = 1, . . . , φn − 1, is the basis for Fn given in Theorem C, then for p 3 we have ⎧ 0, i = 1, . . . , φn − φn−1 , ⎪ ⎪ ⎪ ⎪ ⎪ n 2 ⎪ (p ) ⎪ Eαi−1 (τ )p ⎪ ⎪ ⎪ div , i = φn − φn−1 + 1, . . . , φn − φn−2 , ⎪ (pn ) ⎪ p2 ⎪ E i+φn−2 −1 (τ ) ⎪ ⎪ α ⎪ ⎪ ⎪ ⎪ .. .. ⎨ . . πn (div fi ) = ⎪ 2 ⎪ 2 (p ) ⎪ ⎪ Eαi−1 (pn−2 τ )p ⎪ ⎪ , i = φn − φ1 + 1, . . . , φn − φ0 , div ⎪ ⎪ (p2 ) ⎪ Eαi+φ0 −1 (pn−2 τ )p2 ⎪ ⎪ ⎪ ⎪ ⎪ (p) ⎪ n−1 p2 ⎪ τ) ⎪ ⎪div Eαi−1 (p ⎪ , i = φn − φ0 + 1, . . . , φn − 1. ⎩ (p) n−1 p2 Eαi (p τ) A similar result also holds for p = 2. Proof. Here we prove that the case p is an odd prime; the proof of the case p = 2 is similar, and is omitted. We first show that πn (div fi ) = 0 for i = 1, . . . , φn − φn−1 . By Lemma 10, we have (pn+1 ). (pn ). πn (div∞ Eαi−1 ) = div∞ Eαi−1 ,. (pn+1 ). (pn ). πn (div∞ Eαi+φn−1 −1 ) = div∞ Eαi+φn−1 −1 . (pn ). (pn ). However, since αφn−1 ≡ 1 mod pn , we have Eαi−1 = ±Eαi+φn−1 −1 by Lemma 7. It follows that (pn+1 ). (pn+1 ). πn (div fi ) = πn (div∞ fi ) = πn (div∞ Eαi−1 /Eαi+φn−1 −1 ) = 0 for i = 1, . . . , φn − φn−1 . For i = φn − φn−1 + 1, . . . , φn − φn−2 , by (2) we have (pn ). (pn ). (pn+1 ). (pn+1 ). fi = Eαi−1 (pτ )/Eαi+φn−2 −1 (pτ ) = Epαi−1 (τ )/Epαi+φn−2 −1 (τ )..

(13) 214. YIFAN YANG AND JENG-DAW YU. By Lemma 10, we have πn (div fi ) = πn (div. ∞. fi ) = div. ∞. (pn ). 2. Eαi−1 (τ )p. (pn ). Eαi+φn−2 −1 (τ )p2. .. Using the criteria given in Theorem B, we find the last function is in Fn−1 and (pn ). πn (div fi ) = div. 2. Eαi−1 (τ )p. (pn ). Eαi+φn−2 −1 (τ )p2. .. This proves the case i = φn − φn−1 + 1, . . . , φn − φn−2 . The remaining cases i = φn − φn−2 + 1, . . . , φn − 1 can be proved in the same way. This gives us the lemma. In the next few lemmas, we will establish the fact that D ∈ Dn−1 is principal if and only if ιn (D) ∈ Dn is principal. Lemma 12. If D is a principal divisor in Dn−1 , then ιn (D) is a principal divisor in Dn . Proof. Let f ∗ be one of the functions in the basis of Fn−1 given in Theorem C (or Theorem D if p = 2). Define f (τ ) = f ∗ (pτ ). From the explicit description of the basis, we see that f (τ ) is either one or a product of the functions appearing in our basis for Fn . We now show that div f = ιn (f ∗ ). (pn ) eg ∗ k n+1 ∈ Cn , we choose a matrix σ = that f (τ ) = g Eg (τ ) . For a cusp α /p Assume αk b pn+1 d. in Γ0 (pn+1 ). Then we have n. n. Eg(p ) (pστ ) = Eg(p. ). αk pn. pb (pτ ) . d. Using Lemma 7, we find (pn ). n. Eg(p ) (pστ ) = Eαk g (pτ ) (pn ). for some root of unity , and consequently the order of Eg k. α g pn B2 p· , 2 pn (pn ). which is the same as p times the order of Eg div f = ιn (div f ∗ ). This proves the lemma.. (pτ ) at αk /pn+1 is given by. (τ ) at αk /pn . From this, we conclude that. The proof of the converse statement is more difficult. It relies on the next two lemmas. Lemma 13. Let N 4 be a prime power, let m = φ(N )/2 and let ai , with 1 i m, be the integers in the range 1 ai N/2 such that (ai , N ) = 1. Let M be the m × m matrix −1 denotes the multiplicative which has an (i, j)th entry that is N B2 (ai a−1 j /N )/2, where aj inverse of aj mod N . Then we have 1 B2,χ = 0, det M = 4 χ where χ runs over all even characters modulo N ..

(14) CUSPIDAL RATIONAL TORSION SUBGROUP OF J1 (pn ). Proof.. The proof of det M =. 1 χ. 4. 215. B2,χ. can be found in [17, Lemma 7], and will not be repeated here. To see why the determinant is non-zero, by a straightforward computation we observe that B2,χ0 =. 1 (1 − p) = 0, 6. (11). where χ0 is the principal character. Also, Theorem E in particular implies that B2,χ = 0. χ=χ0 even. Therefore, we conclude that det M = 0. (pn+1 ) eg Lemma 14. Assume that pn+1 5. Assume that f (τ ) = g Eg (τ ) is a modular unit in Fn , where g ∈ (Z/pn+1 Z)× / ± 1. Suppose that, for each integer k, the orders of f (τ ) at αk+hφn−1 /pn+1 take the same values for all h = 0, . . . , p − 1. Then we have eg = 0 for all g satisfying p g. Proof. By Lemma 8, if p | g, then the orders of Eg at αk+hφn−1 /pn+1 , with h = 0, . . . , p − 1, (pn+1 ) eg (τ ) has the same order at are all pn+1 B2 (αk (g/p)/pn )/2. Therefore, if f (τ ) = g Eg. e k+hφn−1 n+1 /p for all h = 0, . . . , p − 1 for any fixed k, then the partial product p g Eg g also α. eg has the same property. Now given k, let us assume that the order of p g Eg at αk+hφn−1 /pn+1 is A. Then, by Lemma 8, we have k+hφn−1. p−1 gα pn+1 B2 eg pA = . 2 pn+1 n+1 h=0 pg,gp. /2. Then, by (3) in Lemma 6, we have pA =. pg. pn eg B2 2. gαk pn. =. gpn /2,pg. pn B2 2. gαk pn. p−1. eg+hpn .. h=0.

(15) p−1 Now since f (τ ) is assumed to be in Fn , by Theorem B, we have h=0 eg+hpn = 0 for all g. Therefore, we have A = 0. This is true for all αk /pn+1 . In other words, we have k. gα eg B2 =0 n+1 p n+1 gp. /2,p g. for all non-negative integers k. Now write g = αj and consider the square matrix which has an (j, k)-entry that is B2 (αj+k−2 /pn+1 ). By Lemma 13, the determinant of this matrix is non-zero. Therefore, all eg , p g, are equal to 0. This completes the proof. With the above lemmas, we are now ready to prove the converse to Lemma 12. Lemma 15. Assume that p is a prime and n 1 is an integer such that pn 5. Let D be a divisor in Dn−1 . If ιn (D) ∈ Dn is principal, then D is a principal divisor in Dn−1 ..

(16) 216. YIFAN YANG AND JENG-DAW YU. Proof. Let φn−1 −1. . D=. nk Pn−1,k ∈ Dn−1 .. k=0. Assume that ιn (D) is principal; that is, assume that there exists a function f (τ ) = (pn+1 ) eg (τ ) ∈ Fn such that g Eg φn−1 −1 p−1. div f = p. . k=0. In other words, we have. nk Pn,k+hφn−1 .. h=0. k+hφn−1. pn+1 gα eg B2 = pnk 2 pn+1 g. for all h for a given k. Since ιn (D) has the same order at αk+hφn−1 /pn+1 for all h = 0, . . . , p − 1 for a fixed k, we have eg = 0 whenever p g by Lemma 14. Thus, we have. (g/p)αk pn eg B2 = nk , 2 pn p|g. which in turn implies that the function f ∗ (τ ) =. . (pn ). Eg/p (τ )eg. p|g ∗. satisfies div f = D. It remains to show that f ∗ is a modular unit contained in Fn−1 , that is, that f ∗ satisfies conditions (7) and (8) of Theorem B. Since f ∈ Fn , by Theorem B, the exponents eg satisfy eg = 0 g≡±ap mod pn. for all a. The same exponents eg then satisfy . eg = 0,. g: g/p≡±a mod pn−1. which is condition (8) for the level N = pn . It remains to consider condition (7). Observe that ιn (D) is a multiple of p, whence we have k. gα pn+1 B2 eg p 2 pn+1 . (12). p|g. for all k. We first consider the cases p 3. Setting k = 0 in (12), we have eg (g 2 − gpn+1 ) ≡ 0 mod pn+2 , p|g. or equivalently. . eg (g/p)2 ≡ 0 mod pn .. p|g ∗. In other words, f satisfies the quadratic condition (7) of Theorem B. This settles the cases p 3. For p = 2, we see that equation (12) with k = 0 yields eg (g 2 − 2n+1 g) ≡ 0 mod 2n+3 , 2|g.

(17) CUSPIDAL RATIONAL TORSION SUBGROUP OF J1 (pn ). that is,. . Partition the sum 4|g, we clearly have. eg (g/2)2 − 2n (g/2) ≡ 0 mod 2n+1 .. 2|g.

(18) g. 217. eg (g/2) into two parts .

(19) g≡0 mod 4. and.

(20). g≡2 mod 4 .. For the terms with. eg (g/2) ≡ 0 mod 2.. g≡0 mod 4. For the terms with g ≡ 2 mod 4, we have eg (g/2) ≡ g≡2 mod 4. . eg mod 2.. g≡2 mod 4. Since eg satisfy condition (8) for N = 2n+1 , we must have eg = 0. g≡2 mod 4. Therefore, we have. . eg (g/2) ≡ 0 mod 2.. 2|g. It follows that. . eg (g/2)2 ≡. 2|g. . eg (g/2)2 − 2n (g/2) ≡ 0 mod 2n+1 ,. 2|g n. which is (7) for N = 2 . This proves the case p = 2, and the proof of the lemma is complete. From Lemmas 12 and 15, we immediately get the following corollary. Corollary 16. = [ιn (D)].. ι∗n ([D]). The homomorphism ιn induces an embedding ι∗n : Cn−1 → Cn given by. 5. Proof of Propositions 5.1. Proof of Proposition 1 Lemma 17. Let p 3 be an odd prime. Let ω be a generator of the group of Dirichlet characters modulo p. Then we have the congruence ⎧ (p−1)/2−2 ⎪ B2i+2 ⎪ (p−1)/2−1 ⎨ mod p if p 5, − i+1 p B2,ω2i ≡ i=1 ⎪ ⎪ i=1 ⎩ −1 mod 3 if p = 3, where B2,ω2i are the generalized Bernoulli numbers and B2i+2 are Bernoulli numbers. Proof. The case p = 3 can be verified directly. We now assume that p 5. Since the product is a rational number, we may regard ω as the Teichm¨ uller character × ω : Z× p → μp−1 from Zp to the group of (p − 1)st roots of unity in Zp characterized by ω(a) ≡ a.

(21) 218. YIFAN YANG AND JENG-DAW YU. mod p for all a ∈ Z× p . For 2i = p − 3 it is well known that B2,ω 2i is contained in Zp and satisfies B2,ω2i ≡. B2i+2 mod p. i+1. (For a proof, follow the argument in [15, Corollary 5.15].) Also, for 2i = p − 3, we have pB2,ωp−3 =. p−1 . ω. −2. (a)(a − pa + p /6) ≡ 2. 2. a=1. p−1 . ω. −2. (a)a ≡ 2. a=1. p−1 . 1 ≡ −1 mod p.. a=1. Then the lemma follows. Proof of Proposition 1. The cases p = 2 and p = 3 can be easily seen from the fact that the modular curves X1 (8) and X1 (9) have genus zero. Now assume that p 5. By Theorem E, the order of the divisor group C0 is given by . (p−3)/2. |C0 | = p. i=1. 1 B 2i . 4 2,ω. Using Lemma 17, we obtain |C0 | ≡ −. . (p−5)/2. 1 4(p−3)/2. i=1. B2i+2 mod p. i+1. By the assumption that p is a regular prime, none of B4 , . . . , Bp−3 is divisible by p. Therefore, p does not divide |C0 |. 5.2. Proof of Proposition 2 Among the five propositions, this proposition is perhaps the most complicated to prove. Recall that, given a free Z-module Λ

(22) of finite rank r with basis {a1 , . . . , ar } and a submodule Λ generated by b1 , . . . , bs with bi = rij aj , the standard method to determine the group structure of Λ/Λ is to compute the Smith normal form of the matrix (rij ). Then the p-rank of the group Λ/Λ is simply the number of diagonals in the Smith normal form that are divisible by p. Thus, in order to prove Proposition 2, we need to know very precisely the linear dependence over Fp among the divisors of modular units generating Fn . In the first two lemmas, we will show that the divisors of the first φn − φn−1 functions in the basis for Fn are linearly independent over Fp . Lemma 18. Let p be a prime and let n 1 be an integer such that pn+1 5. Let α be a generator of (Z/pn+1 Z)× / ± 1. Let fi , with i = 1, . . . , φn − 1, be the basis for Fn given in Theorem C or Theorem D. Let M = (mij ) be the square matrix of size φn − φn−1 such that mij is the order of fi at the cusp αj−1 /pn+1 . Then we have det M = p. χ even primitive. 1 B2,χ , 4. where χ runs over all even primitive Dirichlet characters modulo pn+1 and is either 1 or −1..

(23) CUSPIDAL RATIONAL TORSION SUBGROUP OF J1 (pn ). 219. Proof. Let A = (aij ) be the φn × φn matrix with aij = pn+1 B2 (αi+j−2 /pn+1 )/2, which is the order of Eαi−1 at Pn,j−1 = αj−1 /pn+1 . Define ⎞ ⎛ I −I 0 ··· ··· ··· ⎜ 0 I −I · · · · · · · · · ⎟ ⎟ ⎜ ⎜ .. .. .. .. ⎟ ⎜ . . . . ⎟ V1 = ⎜ ⎟, ⎜· · · · · · · · · I −I 0 ⎟ ⎟ ⎜ ⎝· · · · · · · · · 0 I −I ⎠ I I ··· ··· I I where the matrix consists of p2 blocks, each of which is of size φn−1 × φn−1 , and I is the identity matrix of dimension φn−1 . Let β be an integer such that αβ ≡ 1 mod p. Set also ⎛ ⎞ 1 −β 2 0 ··· ··· ··· ⎜ 0 1 −β 2 · · · · · · · · ·⎟ ⎜ ⎟ ⎜ .. .. .. .. ⎟ ⎜ . . . . ⎟ V2 = ⎜ ⎟, 2 ⎜· · · · · · ⎟ 0 · · · 1 −β ⎜ ⎟ ⎝· · · · · · ··· 0 p 0⎠ 0 0 ··· ··· 0 I where the identity matrix at the lower right corner has dimension φn−1 . Then, for i = 1, . . . , φn − φn−1 , the (i, j)-entry of the matrix V2 V1 A is the order of fi at Pn,j−1 , while for i = φn − φn−1 + 1, . . . , φn , the (i, j)-entry of V2 V1 A is i+j+hφn−1 −2. p−1 α pn+1 B2 . 2 pn+1 h=0. By equation (3) in Lemma 6, this is equal to i+j−2. α pn B2 . 2 pn. (13). Observe that B2 (αi+j−2 /pn ) = B2 (αi+j+kφn−1 −2 /pn ) for all integers k; that is, V2 V1 A takes the form ⎞ ⎛ order of fi at αj−1 /pn+1 V2 V1 A = ⎝ for i = 1, . . . , φn − φn−1 ⎠, A A · · · · · · A A where A is a square matrix of size φn−1 Now let ⎛ I ⎜0 ⎜ ⎜ .. ⎜. U1 = ⎜ ⎜. ⎜ .. ⎜ ⎝0 0. which has an (i, j)-entry that is given by (13). 0 I .. . .. . 0 0. ··· ···. ··· ···. ··· ···. ··· ···. 0 0 .. . .. . I 0. ⎞ I I⎟ ⎟ .. ⎟ .⎟ ⎟, .. ⎟ .⎟ ⎟ I⎠ I. and consider V2 V1 AU1 . For i = 1, . . . , φn − φn−1 and j = φn − φn−1 + 1, . . . , φn , the (i, j)-entry of V2 V1 AU1 is given by p−1 . (order of fi at Pn,j+hφn−1 −1 ).. h=0.

(24) 220. YIFAN YANG AND JENG-DAW YU. By Lemma 11, this sum is equal to 0. In other words, ⎛ ⎜ ⎜ ⎜ ⎜ V2 V1 AU1 = ⎜ ⎜ ⎜ ⎝ A. 0 0 .. .. M ···. ···. A. ···. ⎞. ⎟ ⎟ ⎟ ⎟ ⎟, 0 ⎟ ⎟ 0 ⎠ pA. where M is the (φn − φn−1 ) × (φn − φn−1 ) matrix specified in the lemma. This shows that det(V2 V1 AU1 ) = pφn−1 (det M )(det A ). On the other hand, by Lemma 13, we have 1 B2,χ , det A = ± 4 n+1. . det A = ±. χ mod pn. χ mod p. 1 B2,χ . 4. Also, det V1 = pφn−1 ,. det V2 = p,. Combining everything, we conclude that 1 B2,χ det M = ±p 4 n+1 χ mod p. χ mod pn. det U1 = 1.. 1 B2,χ = ±p 4. χ even primitive mod pn+1. 1 B2,χ , 4. as claimed in the lemma. Here we give an example to exemplify the above argument. Example. Consider the case p = 3 notation as above, we have ⎛ 191 143 59 ⎜ 59 −61 ⎜ 143 ⎜ ⎜ 59 −61 −109 ⎜ ⎜ 23 ⎜ −61 −109 1 ⎜ ⎜ −109 23 −97 A= 108 ⎜ ⎜ ⎜ 23 −97 −37 ⎜ ⎜ −97 −37 −121 ⎜ ⎜ 191 ⎝ −37 −121 −121 191 143 where the ⎛ 1 ⎜0 ⎜ ⎜0 ⎜ ⎜0 ⎜ V2 = ⎜ ⎜0 ⎜0 ⎜ ⎜0 ⎜ ⎝0 0. and n = 2. We choose α = 2 and β = −1. With the −61 −109 −109 23 23 −97 −97 −37 −37 −121 −121 191. 191 143. 143 59. 59 −61. (i, j)-entry is 27B2 (2i+j−2 /27)/2, ⎞ −1 0 0 0 0 0 0 0 1 −1 0 0 0 0 0 0⎟ ⎟ 0 1 −1 0 0 0 0 0⎟ ⎟ 0 0 1 −1 0 0 0 0⎟ ⎟ 0 0 0 1 −1 0 0 0⎟ ⎟, 0 0 0 0 3 0 0 0⎟ ⎟ 0 0 0 0 0 1 0 0⎟ ⎟ 0 0 0 0 0 0 1 0⎠ 0 0 0 0 0 0 0 1. ⎛. ⎞ 23 −97 −37 −121 ⎟ −97 −37 −121 191 ⎟ ⎟ −37 −121 191 143 ⎟ ⎟ ⎟ −121 191 143 59 ⎟ ⎟ 191 143 59 −61 ⎟ ⎟, ⎟ 143 59 −61 −109 ⎟ ⎟ 59 −61 −109 23 ⎟ ⎟ ⎟ −61 −109 23 −97 ⎠ −109 23 −97 −37. 1 ⎜0 ⎜ ⎜0 ⎜ ⎜0 ⎜ V1 = ⎜ ⎜0 ⎜0 ⎜ ⎜1 ⎜ ⎝0 0. 0 1 0 0 0 0 0 1 0. ⎞ 0 −1 0 0 0 0 0 0 0 −1 0 0 0 0⎟ ⎟ 1 0 0 −1 0 0 0⎟ ⎟ 0 1 0 0 −1 0 0⎟ ⎟ 0 0 1 0 0 −1 0 ⎟ ⎟. 0 0 0 1 0 0 −1⎟ ⎟ 0 1 0 0 1 0 0⎟ ⎟ 0 0 1 0 0 1 0⎠ 1 0 0 1 0 0 1.

(25) CUSPIDAL RATIONAL TORSION SUBGROUP OF J1 (pn ). Then. ⎞ 0 2 0 1 −2 4 −1 0 −4 ⎜ 2 0 1 −2 4 −1 0 −4 0 ⎟ ⎟ ⎜ ⎜ 0 1 −2 4 −1 0 −4 0 2 ⎟ ⎟ ⎜ ⎜ 1 −2 4 −1 0 −4 0 2 0 ⎟ ⎟ ⎜ 4 −1 0 −4 0 2 0 1 ⎟ V2 V1 A = ⎜ ⎟, ⎜−2 ⎟ ⎜ 4 −8 −5 −5 7 7 1 1 −2 ⎟ ⎜ ⎟ ⎜ a b c a b c a b c ⎟ ⎜ ⎝ b c a b c a b c a ⎠ c a b c a b c a b. 221. ⎛. a = 11/36 = 9B2 (1/9)/2, b = −1/36 = 9B2 (2/9)/2, c = −13/36 = 9B2 (4/9)/2.. Here the first six rows are the orders of E1 E11 , E2 E8 at the cusps 2j−1 /27. Then ⎛ 1 0 0 0 0 0 1 ⎜0 1 0 0 0 0 0 ⎜ ⎜0 0 1 0 0 0 0 ⎜ ⎜0 0 0 1 0 0 1 ⎜ U1 = ⎜ ⎜0 0 0 0 1 0 0 ⎜0 0 0 0 0 1 0 ⎜ ⎜0 0 0 0 0 0 1 ⎜ ⎝0 0 0 0 0 0 0 0 0 0 0 0 0 0 We find. E2 E5 , E4 E11. E4 E10 , E8 E5. E 8 E7 , E11 E10. E11 E13 , E5 E7. E53 3 E13. the matrices U1 and V2 V1 AU1 are ⎞ ⎛ 0 0 0 2 0 1 −2 4 0 0 ⎜ 2 1 0⎟ 0 1 −2 4 −1 0 0 ⎟ ⎜ ⎜ 0 0 1⎟ 1 −2 4 −1 0 0 0 ⎟ ⎜ ⎜ 1 −2 0 0⎟ 4 −1 0 −4 0 0 ⎟ ⎜ ⎜−2 1 0⎟ 4 −1 0 −4 0 0 0 , V V AU = 2 1 1 ⎟ ⎜ ⎜ 4 −8 −5 −5 0 1⎟ 7 7 0 0 ⎟ ⎜ ⎟ ⎜ a 0 0⎟ b c a b c 3a 3b ⎜ ⎝ b 1 0⎠ c a b c a 3b 3c 0 1 c a b c a b 3c 3a. ⎛. ⎞ 0 2 0 1 −2 4 ⎜ 2 0 1 −2 4 −1 ⎟ ⎜ ⎟ ⎜ 0 1 −2 4 −1 0 ⎟ ⎜ ⎟ = −5833 = −3 det M = det ⎜ 4 −1 0 −4 ⎟ ⎜ 1 −2 ⎟ χeven ⎝−2 4 −1 0 −4 0 ⎠ 4 −8 −5 −5 7 7. primitive mod27. Lemma 19. Let p be a regular prime and let n 1 be an integer. Then we have 1 p B2,χ ≡ 1 mod p, 4 χ even primitive where the product runs over all even primitive Dirichlet characters modulo pn+1 . Proof.. ⎞ 0 0⎟ ⎟ 0⎟ ⎟ 0⎟ ⎟ 0⎟ ⎟. 0⎟ ⎟ 3c ⎟ ⎟ 3a⎠ 3b. First of all, for any non-trivial even Dirichlet character χ we have n+1 p. χ(a) = 0. a=1. and n+1 p. a=1. n+1. n+1. p pn+1 p 1 aχ(a) + (pn+1 − a)χ(pn+1 − a) = aχ(a) = χ(a) = 0. 2 a=1 2 a=1. 1 B2,χ . 4.

(26) 222. YIFAN YANG AND JENG-DAW YU. Thus, we have B2,χ = p. n+1. n+1 p. a2 p2n+2. a=1. −. a pn+1. 1 + 6. χ(a) =. n+1 p. 1 pn+1. χ(a)a2 .. (14). a=1. Now we consider that the case p is an odd regular prime first. Fix a generator α of the multiplicative group (Z/pn+1 Z)× . For a non-negative integer m, write r(m) =

(27) αm /pn+1 and s(m) = αm /pn+1 − r(m). We have pn+1 s(m)2 =. α2m − 2αm r(m) + pn+1 r(m)2 pn+1. Therefore, if a is the integer in the range 0 < a < pn+1 such that αm ≡ a mod pn+1 , then a2. −. pn+1. α2m = −2αm r(m) + pn+1 r(m)2 pn+1. (15). is an integer. Denote this integer by δ(m). Then, by (14), we may write B2,χ = =. 1 pn+1 1. n+1 p. χ(a)a2. a=1 pn (p−1)−1. . pn+1. pn (p−1)−1 m. χ(α )α. +. m=0. . χ(αm )δ(m). m=0. 2pn (p−1). =. 2m. 1−α + pn+1 (1 − χ(α)α2 ). pn (p−1)−1. . χ(αm )δ(m).. (16). m=0. n. Note that the number (1 − α2p (p−1) )/pn+1 is an integer. Therefore, (1 − χ(α)α2 )B2,χ is an algebraic integer. Let ω and θ denote the Dirichlet characters satisfying ω(α) = ζp−1 ,. θ(α) = ζpn ,. respectively, where ζm = e2πi/m . Set χij = ω 2i θj . Then the set of even primitive Dirichlet characters modulo pn+1 is precisely {χij = ω 2i θj : 0 i < (p − 1)/2, 0 j < pn , p j}. From (16), for all j not divisible by p, we have (1 − ω 2i (α)α2 )B2,ω2i − (1 − χij (α)α2 )B2,χij pn (p−1)−1. . = (1 − ω (α)α ) 2i. 2. pn (p−1)−1. ω (α )δ(m) − (1 − χij (α)α ) 2i. m. 2. m=0 pn (p−1)−1. . = (1 − ω 2i (α)α2 ). . χij (αm )δ(m). m=0. ω 2i (αm )δ(m)(1 − θj (αm )). m=0 pn (p−1)−1. − ω (α)α (1 − θ (α)) 2i. 2. j. . χij (αm )δ(m). m=0. ≡0. mod 1 − ζpn .. (Note that when i = 0, we find that ω 0 = χ0 is principal, and (14) does not hold in this case. However, the difference is pn times a p-unit, and the above congruence still holds.) In other.

(28) CUSPIDAL RATIONAL TORSION SUBGROUP OF J1 (pn ). 223. words, n. p . pn−1 (p−1) (1 − χij (α)α2 )B2,χij ≡ (1 − ω 2i (α)α2 )B2,ω2i mod 1 − ζpn .. j=1, pj. It follows that . n. . p . i=0. j=1, pj. (p−1)/2−1. (1 − χ(α)α )B2,χ = 2. χ even primitive. ⎛ ≡⎝. . (1 − χij (α)α2 )B2,χij. (p−1)/2−1. ⎞pn−1 (p−1). (1 − ω 2i (α)α2 )B2,ω2i ⎠. mod 1 − ζpn .. i=0. Now consider the product in the last expression. We have . (p−1)/2−1. (1 − ω 2i (α)α2 ) = 1 − αp−1 .. i=0. Since α is a generator for (Z/pn+1 Z)× , we have 1 − αp−1 = pu for some integer u relatively prime to p. Also, according to (11) and Lemma 17, we have ⎧ (p−1)/2−2 ⎪ ⎪ (p−1)/2−1 B2i+2 ⎨ 1 − mod p if p 5, B2,ω2i ≡ p 6 i+1 i=1 ⎪ ⎪ i=0 ⎩−1 if p = 3. By the assumption that p is a regular prime, this product is relatively prime to p. Therefore, we have ⎛ ⎞p−1 (p−1)/2−1 ⎝ (1 − ω 2i (α)α2 )B2,ω2i ⎠ ≡ 1 mod p, i=0. and consequently. . (1 − χ(α)α2 )B2,χ ≡ 1 mod 1 − ζpn .. χ even primitive. Since the product is a rational integer, the congruence actually holds modulo p. Finally, because α is a generator of (Z/pn+1 Z)× , there exists an integer u relatively prime to p such that k αp (p−1) ≡ 1 − upk+1 mod pk+2 for all k 0. Thus, . n. (1 − χ(α)α2 ) =. χ even primitive. From this we conclude that p. 1 − αp (p−1) upn+1 + · · · ≡ p mod p2 . = upn + · · · 1 − αpn−1 (p−1). χ even primitive. (17). 1 B2,χ ≡ 1 mod p. 4. This completes the proof of the case where p is an odd regular prime. Now consider the case p = 2 with n 2. Choose α = 3 to be a generator of (Z/2n+1 Z)× / ± 1. n−1 Set ζ = e2πi/2 , and let θ be the Dirichlet character satisfying θ(−1) = 1 and θ(3) = ζ. Then the set of even primitive Dirichlet characters modulo 2n+1 is given by {θj : 1 j 2n−1 , 2 j}..

(29) 224. YIFAN YANG AND JENG-DAW YU. Since θ is even, we have 2n+1 1 B2,θj = 4 2. . θj (a)B2. a∈(Z/2n+1 Z)× /±1. a . 2n+1. By a similar calculation as before, we find that if θj is not principal, then 1 1 B2,θj = n+2 θj (a)a2 . 4 2 n+1 × a∈(Z/2. Z) /±1. Now for a non-negative integer m, define δ(m) = −. . 32m + 2n+1 2n+1. 3m 2n+1. 2. as in (15). Following the computation in (16), we get n. 1 1 1 − 32 B2,θj = n+2 + 4 2 (1 − 9ζ j ) 2 n. φ(2n+1 )/2−1. . θj (3m )δ(m).. m=0. n−1. Now we have 32 = (1 + 8)2 ≡ 1 + 2n+2 mod 2n+3 . Also, from (15), we see that δ(m) is always even. Thus, (1 − 9ζ j )B2,θj /4 is an algebraic integer. By the same argument as before, we find 1−9 1 − 9ζ j B2,χ0 − B2,θj ≡ 0 mod 1 − ζ, 4 4 for all odd j, and thus 1 − 9χ(3) B2,χ ≡ 1 mod 2. 4 χ even primitive. Finally, from (17), we have. . (1 − 9χ(3)) ≡ 2 mod 4.. χ even primitive. This proves the case p = 2. Proof of Proposition 2. Let α be a generator of (Z/pn+1 Z)× / ± 1. Specifically, for p = 2, we set α = 3 and, for an odd prime p, we let α be an integer such that α generates (Z/pZ)× , but αp−1 ≡ 1 mod p2 . Let fi , with i = 1, . . . , φn − 1, be the generators of Fn given in Theorem C or Theorem D. Let M be the (φn − 1) × φn matrix which has an (i, j)-entry that is the order of fi at αj−1 /pn+1 . Let U and V be the unimodular matrices such that M = U M V is in the Smith normal form; that is, if M = (mij ), then we have the following: (1) m11 |m22 | . . .; (2) mij = 0 if i = j. Also (mii = 0 for all i since the rank of M is φn − 1.) Then the p-rank of Cn is equal to the number of mii that are divisible by p. In other words, if we consider M as a matrix over Fp , then our p-rank is actually equal to φn − 1 − (the rank of M over Fp ). We now determine the rank of M over Fp . From Lemmas 18 and 19, we know that the first φn − φn−1 rows of M are linearly independent over Fp . Thus, the rank of M over Fp is at least φn − φn−1 = pn−1 (p − 1)2 /2. It remains to prove that the remaining rows are all linearly dependent of the first φn − φn−1 rows modulo p..

(30) CUSPIDAL RATIONAL TORSION SUBGROUP OF J1 (pn ). 225. We first consider row φn − φn−1 + 1 to row φn − φ0 . (For p = 2, consider row φn − φn−1 + 1 to row φn − φ2 .) Let be an integer between 1 and n − 1. (For p = 2, let 1 n − 3.) By Theorems C and D, for i from φn − φn− + 1 to φn − φn−−1 , the ith row of M is the divisor of the function (pn−+1 ). fi = Eαi−1. (pn−+1 ). (p τ )/Eαi+φn−−1 −1 (p τ ),. which by Lemma 8 is given by i+k−1. i+φn−−1 +k−1. φn −1 α α pn−+1 p · B2 − B2 Pn,k . n−+1 2 p pn−+1 . (18). k=0. Now αφn−−1 ≡ −(1 + upn− ) mod pn−+1 for some integer u not divisible by p. (For p = 2, we have αφn−−1 ≡ 1 + 2n− mod 2n−+1 instead when n − 3.) Then a straightforward calculation gives i+φn−−1 +k−1. i+k−1. α α pn−+1 uα2(i+k−1) mod 1. − B ≡ − B2 2 2 pn−+1 pn−+1 p This shows that if 2, then the divisor of fi for i from φn − φn− + 1 to φn − φn−−1 is divisible by p. For such , the rows do not contribute anything to the rank of M over Fp . When = 1, the above computation shows that the ith row of M for i from φn − φn−1 + 1 to φn − φn−2 is congruent to −uα2(i−1) (1, α2 , α4 , . . . , α2φn −2 ) mod p. On the other hand, the (φn − φn−1 )th row of M is the divisor of Eαp φn −φn−1 −1 /Eαp φn −1 . By a similar computation, we find that it is congruent to −uα2(φn −φn−1 −1) (1, α2 , α4 , . . . , α2φn −2 ). From this we see that row φn − φn−1 + 1 to row φn − φ0 of M are all multiples of the (φn − φn−1 )th row of M mod p. Finally, for i = φn − φ0 + 1, . . . , φn − 1, we find that the ith row is congruent to the 0 vector. Therefore, the rank of M over Fp is precisely φn − φn−1 . We conclude that the p-rank of Cn is given by φn − 1 − (φn − φn−1 ) = φn−1 − 1 = pn−1 (p − 1)/2 − 1. This completes the proof of the proposition. 5.3. Proof of Proposition 3 Let fi , with i = 1, . . . , φn − 1, denote the basis for Fn given in Theorem C or Theorem D and let fi , with i = 1, . . . , φn−1 − 1, be the basis for Fn−1 . By Lemma 11, we have πn (div fi ) = 0 for i = 1, . . . , φn − φn−1 , and ⎛ div f1 ⎜ .. ⎝ .. div fφ n−1 −1. ⎞ 1 ⎟ ⎠= 2 p. R 0. ⎛. ⎞ πn (div fφn −φn−1 +1 ) 0 ⎜ ⎟ .. ⎝ ⎠, . I πn (div fφn −1 ). (19).

(31) 226. YIFAN YANG AND JENG-DAW YU. where I is the identity matrix of size φn−2 − 1 and ⎛ 1 −β 2 0 ··· 2 ⎜ 0 ··· 1 −β ⎜ ⎜ .. ⎜ R=⎜ . ⎜· · · · · · ··· 1 ⎜ ⎝· · · · · · ··· 0 ··· ··· ··· 0. ··· ···. ··· ··· .. .. ⎞. ⎟ ⎟ ⎟ ⎟ ⎟ 0 ⎟ ⎟ −β 2 ⎠ p. −β 2 1 0. is a square matrix of size φn−1 − φn−2 which has superdiagonals that are all −β 2 and which has diagonals that are all 1, except for the last one, which is p. Therefore, the index of πn (Pn ) in Pn−1 is given by n−1. p2(φn−1 −1)−1 = pp. (p−1)−3. .. The structure of the factor group Pn−1 /πn (Pn ) can be easily seen from the matrix above. This completes the proof of the proposition. 5.4. Proof of Proposition 4 Consider the group homomorphism π : Dn → πn (Dn )/πn (Pn ) = Dn−1 /πn (Pn ) sending D ∈ Dn to the coset πn (D) + πn (Pn ). The homomorphism is clearly onto, and the kernel is the group ker π = Pn + ker πn . Thus, we have Dn /(Pn + ker πn ) Dn−1 /πn (Pn ). Now the group on the left-hand side is isomorphic to Dn /(Pn + ker πn ) (Dn /Pn ) ((Pn + ker πn )/Pn ). Therefore, to prove that the p-part of Cn = Dn /Pn is isomorphic to that of Dn−1 /πn (Pn ), it suffices to show that the order of (Pn + ker πn )/Pn is not divisible by p. From the definition of πn , it is easy to see that the kernel of πn is generated by divisors of the form D = Pn,k − Pn,k+φn−1 . Let fi , with i = 1, . . . , φn − 1, be the basis for Fn given in Theorem C or Theorem D. If we write D as a linear combination D=. φ n −1. ri div fi ,. ri ∈ Q,. i=1. of div fi , then the order of D + Pn in the divisor class group Cn divides the least common multiple of the denominators of ri . We need to show that this number is not divisible by p. We first prove that ri = 0 for i = φn − φn−1 + 1, . . . , φn − 1. By Lemma 11, we have 0 = πn (D) =. φ n −1. ri πn (div fi ).. (20). i=φn −φn−1 +1. Let A = ( R0 I0 ) be the square matrix of size φn−1 − 1 in (19). Then ⎛ ⎞ ⎛ div f1 πn (div fφn −φn−1 +1 ) ⎜ ⎟ ⎜ .. .. 2 −1 ⎠=p A ⎝ ⎝ . . πn (div fφn −1 ). we have ⎞. div fφ n−1 −1. ⎟ ⎠,.

(32) CUSPIDAL RATIONAL TORSION SUBGROUP OF J1 (pn ). 227. where fi , with i = 1, . . . , φn−1 − 1, is the basis for Fn−1 given in Theorem C or Theorem D, and (20) can be written as ⎛ ⎞ div f1 ⎜ ⎟ .. 0 = p2 (rφn −φn−1 +1 , . . . , rφn −1 )A−1 ⎝ ⎠. . div fφ n−1 −1. Since div fi are linearly independent over Q, we must have (rφn −φn−1 +1 , . . . , rφn −1 )A−1 = (0, . . . , 0). It follows that ri = 0 for all i = φn − φn−1 + 1, . . . , φn − 1, and φn −φn−1. D=. . ri div fi .. i=1. Now, without loss of generality, we may assume that the integer k in D = Pn,k − Pn,k+φn−1 satisfies 0 < k < φn − 2φn−1 . (Let b and d be integers such that αd − bpn+1 = 1. Note that if a modular unit f (τ ) ∈ Fn has a divisor mD for some integer m, then the function f ((ατ + b)/(pn+1 τ + d)) has a divisor m(Pn,k−1 − Pn,k+φn−1 −1 ). Thus Pn,k − Pn,k+φn−1 and Pn,k−1 − Pn,k+φn−1 −1 have the same order in the divisor class group Cn .) Let M be the square matrix of size φn − φn−1 which has an (i, j)-entry that is the order of fi at Pn,j−1 . Then the order of D in the divisor class group Cn will divide the determinant of the matrix M . By Lemmas 18 and 19 and the assumption that p is a regular prime, the determinant of M is not divisible by p. This shows that the order of D + Pn in Cn is not divisible by p, and therefore |(Pn + ker πn )/Pn | is not divisible by p. This proves the proposition. 5.5. Proof of Proposition 5 By Proposition 3, we see that Pn−1 /πn (Pn ) is clearly contained in ker[p2 ]. Now suppose that D + πn (Pn ) ∈ Dn−1 /πn (Pn ) is in the kernel of [p2 ]. We have p2 D ∈ πn (Pn ). With (10), this can be written as πn (ιn (D)) ∈ πn (Pn ), or equivalently ιn (D) ∈ Pn + ker πn . Let fi , with i = 1, . . . , φn − 1, be the basis for Fn given in Theorem C or Theorem D. By Lemma 11, we have div fi ∈ ker πn for i = 1, . . . , φn − φn−1 . Hence, ιn (D) =. φ n −1. mi div fi + D. i=φn −φn−1 +1. for some integers mi and some divisor D in ker πn . Now note that if we define an inner product ·, · on Dn by c0 Pn,0 + c1 Pn,1 + . . . , d0 Pn,0 + d1 Pn,1 + . . . = c0 d0 + c1 d1 + . . . , then, for i = φn − φn−1 + 1, . . . , φn − 1, div fi is in the orthogonal complement of ker πn . The same thing is also true for ιn (D) for any D ∈ Dn−1 . It follows that the divisor D above is actually 0 and we have ιn (D) ∈ Pn . Finally, by Lemma 15, the fact that ιn (D) is principal implies that D itself is principal. This completes the proof of the proposition. Acknowledgements. The authors would like to thank Professor Jing Yu for his interest in this work. The authors are also very thankful to the referee for a thorough and careful reading of the manuscript. Part of the work was done while the first author was visiting the MaxPlanck-Institut f¨ ur Mathematik at Bonn. He would like to thank the institute for providing a stimulating research environment..

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