ELEMENTARY MODULAR IWASAWA THEORY Contentshida/207b.1.18s/Lec18s.pdf · 2018-05-25 · ELEMENTARY...

ELEMENTARY MODULAR IWASAWA THEORY

HARUZO HIDA

Contents

1. Curves over a field 31.1. Plane curves 31.2. Tangent space and local rings 51.3. Projective space 81.4. Projective plane curve 91.5. Divisors 111.6. The theorem of Riemann–Roch 121.7. Regular maps from a curve into projective space 132. Elliptic curves 142.1. Abel’s theorem 142.2. Weierstrass Equations of Elliptic Curves 152.3. Moduli of Weierstrass Type 173. Modular forms and functions 203.1. Geometric modular forms 203.2. Topological Fundamental Groups 213.3. Classical Weierstrass ℘-function 233.4. Complex Modular Forms 243.5. Weierstarss ζ and σ functions 263.6. Product q-expansion 283.7. Klein forms 294. Modular units 324.1. Siegel units 334.2. Distribution on p-divisible groups 344.3. Stickelberger distribution 354.4. Rank of distribution 364.5. Cusps of X(N) 374.6. Finiteness of ClX(N) 384.7. Siegel units generate A×

pm 384.8. Fricke–Wohlfahrt theorem 424.9. Siegel units and Stickelberger’s ideal 444.10. Cuspidal class number formula 464.11. Cuspidal class number formula for X1(N). 49

Date: May 25, 2018.1

ELEMENTARY MODULAR IWASAWA THEORY 2

References 50

We discuss the first three topics of the following list:

(1) Explicit construction of modular forms and modular functions;(2) Determination of units in the elliptic modular function fields (modular units);(3) The cuspidal class group of modular curves including a proof of the cuspidal

class number formula;(4) Construction of units (called elliptic units) of the Hilbert ring class field of

imaginary quadratic fields as specialization of modular units;(5) Iwasawa theory for imaginary quadratic fields via elliptic units.

If time allows, we further go into the topics (4) and (5). A modular curve is anaffine plane curve (i.e., an open Riemann surface) classifying elliptic curves withcertain additional structure (called level structure) naturally defined over Q. As aRiemann surface, it is a quotient of the upper half complex plane by SL2(Z) (and itssubgroups). Adding finite number of points (called cusps), we can complete the curveinto a projective curve (i.e., a compact Riemann surface). Most of recent progress innumber theory and arithmetic geometry is based on the study of modular curves andmodular forms defined on them; e.g., proof of Iwasawa’s conjectures (Mazur–Wiles)and Fermat’s last theorem (Wiles).

Modular units are the units in the ring of holomorphic functions of the affinemodular curve. Divisors supported on cusps modulo principal divisors (divisors ofmodular units) give the cuspidal class group (which is the torsion subgroup of rationalpoints of the Jacobian of the modular curve). We can determine explicitly the groupof modular units via classical results of Weierstrass and Siegel (like the determinationof cyclotomic units in the field generated by roots of unity by Dirichlet–Kummer).

Striking points are that the class group is finite and that we have an explicit classnumber formula of Kubert–Lang in terms of the Dirichlet L-values at s = 2 (whilethe classical class number formula of Dirichlet/Kummer of the cyclotomic field is interms of the values at s = 1). This is done by generalizing Stickelberger’s theory ofcyclotomic class groups to the setting of modular curves. This theory gives a base ofthe proof of the Iwasawa main conjecture by Mazur–Wiles.

For each prime p > 2, sin(πa/p)/ sin(π/p)0<a<p/2 gives independent units in theinteger ring of the field of p-th root of unity for each prime p (the cyclotomic units).Analogously, the value of modular units at a point on the upper half complex planebelonging to an imaginary quadratic field K gives independent units in the (certain)Hilbert ring class field overK, and this is a base of the generalization of the cyclotomicIwasawa theory to the elliptic Iwasawa theory of imaginary quadratic fields. If timeallows, we describe these finer results (possibly including the class number formulasof the ring class field as an index of elliptic units inside the entire units).

We start with a sketch of the theory of affine plane curves and how to compactifyit in a projective space. Then we turn analytic and construct rational functions overmodular curves in analytic means. This explicit construction helps us to computecucpidal class groups inside the Jacobian of the projective modular curves.


1. Curves over a field

Any algebraic curve over an algebraically closed field can be embedded into the3-dimensional projective space P3 (e.g., [ALG, IV.3.6]) and any closed curve in P3 isbirationally isomorphic to a curve inside P2 (a plane curve; see [ALG, IV.3.10]), wegive some details of the theory of plain curve defined over a field k ⊂ C in this sectionto accustom with the theory of curves. We also write k for the algebraic closure k ofk inside C.

1.1. Plane curves. Let a be a principal ideal of the polynomial ring k[X, Y ]. Notethat polynomial rings over a field is a unique factorization domain. We thus haveprime factorization a =

∏ppe(p) with principal primes p. We call a square free if

0 ≤ e(p) ≤ 1 for all principal primes p. Fix a square-free a. The set of A-rationalpoints for any k-algebra A of a plane curve is given by the zero set

Va(A) =(x, y) ∈ A2

∣∣f(x, y) = 0 for all f(X, Y ) ∈ a.

Obviously, for a generator f(X, Y ) of a, we could have defined

Va(A) = Vf (A) =(x, y) ∈ A2

∣∣f(x) = 0,

but this does not depend on the choice of generators and depends only on the ideal a;so, it is more appropriate to write Va. As an exceptional case, we note V(0)(A) = A2.Geometrically, we think of Va(C) as a curve in C2 = V(0)(C) (the 2-dimensional“plane”). This view point is more geometric. In this sense, for any algebraicallyclosed field K over k, a point x ∈ Va(K) is called a geometric point with coefficientsin K, and V(f)(K) ⊂ V(0)(K) is called the geometric curve in V(0)(K) = K2 definedby the equation f(X, Y ) = 0.

By Hilbert’s zero theorem (Nullstellensatz; see [CRT] Theorem 5.4 and [ALG]Theorem I.1.3A), writing a the principal ideal of k[X, Y ] generated by a, we have

(1.1) a =f(X, Y ) ∈ k[X, Y ]

∣∣f(x, y) = 0 for all (x, y) ∈ Va(k).

Thus we have a bijection

square-free ideals of k[X, Y ] ↔ plane curves Va(k) ⊂ V(0)(k).The association Va : A 7→ Va(A) is a covariant functor from the category of k-algebrasto the category of sets (denoted by SETS). Indeed, for any k-algebra homomorphismσ : A → A′, Va(A) 3 (x, y) 7→ (σ(x), σ(y)) ∈ Va(A

′) as 0 = σ(0) = σ(f(x, y)) =f(σ(x), σ(y)). Thus a = a ∩ k[X, Y ] is determined uniquely by this functor, but thevalue Va(A) for an individual A may not determine a.

From number theoretic view point, studying Va(A) for a small field is important.Thus it would be better regard Va as a functor in some number theoretic setting.

If a =∏

p p for principal prime ideals p, by definition, we have

Va =⋃

p

Vp.

The plane curve Vp (for each prime p|a) is called an irreducible component of Va. Sincep is a principal prime, we cannot further have non-trivial decomposition Vp = V ∪Wwith plane curves V and W . A prime ideal p ⊂ k[X, Y ] may decompose into a


product of primes in k[X, Y ]. If p remains prime in k[X, Y ], we call Vp geometricallyirreducible.

Suppose that we have a map FA = F (φ)A : Va(A) → Vb(A) given by two poly-nomials φX(X, Y ), φY (X, Y ) ∈ k[X, Y ] (independent of A) such that FA(x, y) =(φX(x), φY (y)) for all (x, y) ∈ Va(A) and all k-algebras A. Such a map is calleda regular k-map or a k-morphism from a plane k-curve Va into Vb. Here Va and Vb

are plane curve defined over k. If A1 = Vb is the affine line, i.e., Vb(A) ∼= A for all A(taking for example b = (y)), a regular k-map Va→ A1 is called a regular k-function.Regular k-functions are just functions induced by the polynomials in k[x, y] on Va;so, Ra is the ring of regular k-functions of Va defined over k.

We write Homk-curves(Va, Vb) for the set of regular k-maps from Va into Vb. Obvi-ously, only φ? mod a can possibly be unique. We have a commutative diagram forany k-algebra homomorphism σ : A→ A′:

Va(A)FA−−−→ Vb(A)

σ

yyσ

Va(A′) −−−→

FA′

Vb(A′).

Indeed,

σ(FA((x, y))) = (σ(φX(x, y)), σ(φY (x, y)))

= (φX(σ(x), σ(y)), φY (σ(x), σ(y)) = FA′(σ(x), σ(y)).

Thus the k-morphism is a natural transformation of functors (or a morphism of func-tors) from Va into Vb. We write HomCOF (Va, Vb) for the set of natural transformations(we will see later that HomCOF (Va, Vb) is a set).

The polynomials (φX , φY ) induces a k-algebra homomorphism F : k[X, Y ] →k[X, Y ] by pull-back, that is, F (Φ(X, Y )) = Φ(φX(X, Y ), φY (X, Y )). Take a class[Φ]b = Φ + b in B = k[X, Y ]/b. Then look at F (Φ) ∈ k[X, Y ] for Φ ∈ b. Since(φX(x), φY (y)) ∈ Vb(k) for all (x, y) ∈ Va(k), Φ(φX(x, y), φY (x, y)) = 0 for all(x, y) ∈ Va(k). By Nullstellensatz, F (Φ) ∈ a ∩ k[X, Y ] = a. Thus F (b) ⊂ a, and Finduces a (reverse) k-algebra homomorphism

F : k[X, Y ]/b→ k[X, Y ]/a

making the following diagram commutative:

k[X, Y ]F−−−→ k[X, Y ]y

y

k[X, Y ]/b −−−→F

k[X, Y ]/a.

We write Ra = k[X, Y ]/a and call it the affine ring of Va. Here is a useful (buttautological) lemma which is a special case of Yoneda’s lemma (in Math 210 series):

Lemma 1.1. We have a canonical isomorphism:

HomCOF (Va, Vb) ∼= Homk-curves(Va, Vb) ∼= Homk−alg(Rb, Ra).


The first association is covariant and the second is contravariant.

Since Ra = Homk-alg(k[X] = R(Y ), Ra) = Homk-curves(Va,A) = HomCOF (Va,A), wecan recover the ring Ra as a collection of morphisms from Va into the affine line A.Here is a sketch of the proof.

Proof. First we note Va(A) ∼= HomALG/k(Ra, A) via (a, b) ↔ (Φ(X, Y ) 7→ Φ(a, b)).

Thus as functors, we have Va(?) ∼= HomALG/k(Ra, ?). We identify the two functors

A 7→ Va(A) and A 7→ Hom(Ra, A) in this way, and in this sense, we write the functorcorresponding to Va as Spec(Ra). Then the main point of the proof of the lemma isto construct from a given natural transformation F ∈ HomCOF (Va, Vb) a k-algebra

homomorphism F : Rb→ Ra giving F by Va(A) = HomALG/k(Ra, A) 3 φ FA7→ φ F ∈

HomALG/k(Rb, A) = Vb(A). Then the following exercise finishes the proof, as plainly

if we start with F , the above association gives rise to F .

Exercise 1.2. Let F = FRa(idRa) ∈ VRb(Ra) = HomALG/k

(Rb, Ra), where idRa ∈Va(Ra) = HomALG/k

(Ra, Ra) is the identity map. Then prove that F does the job.

We call Va irreducible (resp. geometrically irreducible) if a is a prime ideal (resp. a =ak[X, Y ] is a prime ideal in k[X, Y ]). For a general noetherian k-algebra A, we defineVA = Spec(A) to be the functor A 7→ Homk-alg(A, A) = Spec(A)(A), and call Spec(A)a curve associated to the ring A. In the same way as above, A ∼= HomCOF (VA,A).The curve VA is called an affine curve with affine ring A. We give many geometricnotion for plane curves in two ways, one is an intuitive definition particular to theplane curve and another a ring theoretic interpretation of the notion which is valid forgeneral VA. For example, Homk-curves(VA, VB) = HomCOF (VA, VB) = Homk-alg(B,A).

An element in the total quotient ring of Ra (resp. A) is called a rational k-functionon Va (resp. VA). If Va is irreducible, then rational k-functions form a field. This fieldis called the rational function field of Va over k.

1.2. Tangent space and local rings. Suppose a = (f(X, Y )). Write V = Va andR = Ra. Let P = (a, b) ∈ Va(K). We consider partial derivatives

∂f

∂X(P ) :=

∂f

∂X(a, b) and

∂f

∂Y(P ) :=

∂f

∂Y(a, b).

Then the line tangent to Va at (a, b) has equation

∂f

∂X(a, b)(X − a) +

∂f

∂Y(a, b)(Y − b) = 0.

We write corresponding line as TP = Vb for the principal ideal b generated by∂f∂X

(a, b)(X − a) + ∂f∂Y

(a, b)(Y − b). We call Va is non-singular or smooth at P =(a, b) ∈ Va(K) for a subfield K ⊂ C if this TP is really a line; in other word, if( ∂f

∂X(P ), ∂f

∂Y(P )) 6= (0, 0).

Example 1.1. Let a = (f) for f(X, Y ) = Y 2−X3. Then ∂f∂X

(a, b)(X−a)+ ∂f∂Y

(a, b)(Y −b) = −3a2(X − a) + 2b(Y − b) (b2 = a3). Thus this curve is singular only at (0, 0).

Example 1.2. Suppose that k has characteristic different from 2. Let a = (Y 2−g(X))for a cubic polynomial g(X) = X3 + aX + b. Then the tangent line at (x0, y0) is


given by 2y0(X − x0) − g′(x0)(Y − y0). This equation vanishes if 0 = y20 = g(x0)

and g′(x0) = 0; so, singular at only (x0, 0) for a multiple root x0 of g(X). Thus Va

is a nonsingular curve if and only if g(X) is separable if and only if its discriminant4a3 − 27b2 6= 0.

Suppose that K/k is an algebraic field extension. Then K[X, Y ]/aK[X, Y ] containsRa as a subring. The maximal ideal (X1 − a1, . . . , Xn − an) ⊂ K[X, Y ]/aK[X, Y ]induces a maximal ideal P = (X − a, Y − b) ∩Ra of Ra. The local ring OVa,P at P isthe localization

OVa ,P =ab

∣∣b ∈ Ra, b ∈ RaP,

where ab

= a′

b′if there exists s ∈ Ra \P such that s(ab′− a′b) = 0. Write the maximal

ideal of OVa ,P as mP . Then mP ∩R = P .

Lemma 1.3. The linear vector space TP (K) is the dual vector space of P/P 2 =mP /m

2P .

In general, for a maximal ideal P of A with residue field K, we define the tangentspace TP := HomK(P/P 2, K).

Proof. Write a = (f). Replacing k[X, Y ]/(f) by K[X, Y ]/(f), we may assume thatK = k. A K-derivation ∂ : OV,P → K (at P ) is a K-linear map with ∂(φϕ) =ϕ(P )∂(φ)+φ(P )∂(ϕ). WritingDV,P for the space of K-derivations at P , which is a K-vector space. Plainly for A := V(0), DA,P is a 2-dimensional vector space generated by

∂X : φ 7→ ∂φ∂X

(P ) and ∂Y : φ 7→ ∂φ∂Y

(P ). We have a natural injection i : DV,P → DA,P

given by i(∂)(φ) = ∂(φ|V ). Note that Ω(a,b) = (X − a,X − b)/(X − a,X − b)2 is a 2-dimensional vector space over K generated by X−a and Y − b. Thus DA,P and Ω(a,b)

is dual each other under then pairing (α(X−a)+β(Y −b), ∂) = ∂(α(X−a)+β(Y−b)).The projection k[X, Y ] R induces a surjection

Ω(a,b) → ΩV,P = P/P 2,

whose kernel is spanned by f mod (X−a, Y −b)2 = ∂f∂X

(a, b)(X−a)+ ∂f∂Y

(a, b)(Y −b)if a = (f), since φ(X, Y ) ≡ ∂φ

∂X(a, b)(X − a) + ∂φ

∂Y(a, b)(Y − b) mod (X − a, Y − b)2.

Thus the above duality between Ω(a,b) and DA,(a,b) induces the duality ΩV,P = P/P 2

and TP (K) given by (ω, t) = t(ω), where we regard t as a derivation OV,P → K.

We call TP the tangent space at P and ΩP = ΩV,P the cotangent space at P ofV . More generally, a k-derivation ∂ : Ra → Ra is a k-linear map satisfying theLeibniz condition ∂(φϕ) = φ∂(ϕ)+ϕ∂(φ) and ∂(k) = 0. For a k-derivation as above,f∂ : ϕ 7→ f · ∂(ϕ) for f ∈ Ra is again a k-derivation. The totality of k-derivationDerVa/k is therefore an Ra-module.

First take a = (0); so, Va = A2. By the Leibniz relation, ∂(X2) = nXn−1∂X,∂(Y m) = mY m−1∂Y and ∂(X2Y m) = nXn−1Y m∂X +mX2Y m−1∂Y for ∂ ∈ DerA2/k;

so, ∂ is determined by its value ∂(X) and ∂(Y ). Note that (∂X) ∂∂X

+ (∂Y ) ∂∂Y

inDerA2/k and the original ∂ has the same value at X and Y ; so, we have

∂ = (∂X)∂

∂X+ (∂Y )

∂

∂Y.


Thus

∂∂X, ∂

∂Y

gives a basis of DerA2/k.

Assuming Va nonsingular (including A2 = V(0)), we write the Ra-dual as ΩVa/k :=Hom(DerVa/k, Ra) (the space of k-differentials) with the duality pairing

(·, ·) : ΩVa/k ×DerVa/k → Ra.

We have a natural map d : Ra → ΩVa/k given by φ 7→ (dφ : ∂ 7→ ∂(φ)) ∈ DerVa/k.Note

(d(φϕ), ∂) = ∂(φϕ) = φ∂(ϕ) + ϕ∂(φ) = (φdϕ+ ϕdφ, ∂)

for all ∂ ∈ DerVa/k. Thus we have d(φϕ) = φdϕ+ ϕdφ, and d is a k-linear derivationwith values in ΩVa/k.

Again let us first look into ΩA2/k. Then by definition (dX, ∂) = ∂X and (dY, ∂) =

∂Y ; so, dX, dY is the dual basis of

∂∂X, ∂

∂Y

. We have dΦ = ∂Φ

∂XdX + ∂Φ

∂YdY as we

can check easily that the left hand side and right hand side as the same value on any∂ ∈ DerA2/k.

If ∂ : Ra = k[X, Y ]/(f)→ Ra is a k-derivation, we can apply it to any polynomialΦ(X, Y ) ∈ k[X, Y ] and hence regard it as ∂ : k[X, Y ]→ Ra. By the above argument,Derk(k[X, Y ], Ra) has a basis

∂

∂X, ∂

∂Y

now over Ra. Since ∂ factor through the

quotient k[X, Y ]/(f), it satisfies ∂(f(X, Y )) = (df, ∂) = 0. Thus we have

Lemma 1.4. We have an inclusion DerVa/k → (Ra∂

∂X⊕Ra

∂∂Y

) whose image is givenby ∂ ∈ Derk(k[X, Y ], Ra)|∂f = 0. This implies ΩVa/k = (RadX ⊕ RadY )/Radf for

df = ∂f∂XdX + ∂

∂YdY by duality.

Remark 1.5. If Va is an irreducible curve; so, Ra is an integral domain, for itsquotient field k(Va), k(Va)ΩVa/k = (k(Va)dX⊕ k(Va)dY )/k(Va)df is 1 dimensional, asdf 6= 0 in ΩA2/k. In particular, if we pick ψ ∈ Ra with dψ 6= 0 (i.e., a non-constant),any differential ω ∈ ΩVa/k can be uniquely written as ω = φdψ for φ ∈ k(Va).

Lemma 1.6. The following four conditions are equivalent:

(1) A point P of V (k) is a smooth point.(2) OV,P is a local principal ideal domain, not a field.

(3) OV,P is a discrete valuation ring with residue field k.(4) lim←−n

OV,P/m2P∼= k[[T ]] (a formal power series ring).

Proof. Let K = k. By the above lemma, TP is a line if and only if dimTP (K) = 1if and only if dimP/P 2 = 1. Thus by Nakayama’s lemma, P is principal. Anyprime ideal of k[X, Y ] is either minimal or maximal (i.e, the ring k[X, Y ] has Krulldimension 2). Thus any prime ideal of R and OV,P is maximal. Thus (1) and (2) areequivalent. The equivalence of (2) and (3) follows from general ring theory coveredby Math 210 (see [CRT] Theorem 11.2). We leave the equivalence (3) ⇔ (4) as anexercise.

Write x, y for the image of X, Y ∈ k[X, Y ] in Ra. Any ω ∈ ΩVa/k can be written asφdx+ϕdy. Suppose that Va is nonsingular. Since OVa ,P → k[[T ]] (for P ∈ Va(k)) fora local parameter T as above, φ, ϕ, x, y have the “Taylor expansion” as an elementof k[[T ]], for example, x(T ) =

∑n≥0 an(x)T

2 with an(x) ∈ k. Thus dx, dy also have

a well define expansion, say, dx = d(∑

n≥0 an(x)T2) =

∑n≥1 an(x)T

n−1dT . Thus


we may expand ω = φdx + ϕdy =∑

n≥0 an(ω)T 2dT once we choose a parameter Tat P . This expansion is unique independent of the expression φdx + ϕdy. Indeed,if we allow meromorphic functions Φ as coefficients, as we remarked already, we canuniquely write ω = Φdx and the above expansion coincides with the Taylor expansionof Φdx. Write Max(Ra) for the set of maximal ideals of Ra. Then plainly, we havea natural inclusion Va(k) → Max(Ra) sending (a, b) to (x − a, y − b) for the imagex, y in Ra of X, Y ∈ k[X, Y ]. For P ∈ Max(Ra), we call P is smooth on Va if OV,P

is a discrete valuation ring. By the above exercise, this is consistent with the earlierdefinition (no more and no less).

For any given affine plane irreducible curve Va, we call Va is normal ifRa is integrallyclosed in its field of fractions.

Corollary 1.7. Any normal irreducible affine plane curve is smooth everywhere.

Proof. By ring theory, any localization of a normal domain is normal. Thus OV,P isa normal domain. By the exercise below, we may assume that P ∩ k[X, Y ] 6= (0).Then P is a maximal ideal, and hence K = k[X, Y ]/P is an algebraic extension of k.In this case, OV,P is a normal local domain with principal maximal ideal, which is adiscrete valuation ring (cf. [CRT] Theorem 11.1).

1.3. Projective space. Let A be a commutative ring. Write AP be the localizationat a prime ideal P of A. Thus

AP =

b

s

∣∣s ∈ A \ P/ ∼,

where bs∼ b′

s′if there exists s′′ ∈ A \ P such that s′′(s′b − sb′) = 0. An A-module M

is called locally free at P if

MP = ms|s ∈ A \ P/ ∼= AP ⊗A M

is free over AP . We callM locally free if it is free at all prime ideals of A. If rankAPMP

is constant r independent of P , we write rankAM for r.Write ALG/k for the category of k-algebras; so, HomALG/k

(A,A′) is made up ofk-algebra homomorphisms from A into A′ sending the identity 1A to the identity1A′ . Here k is a general base ring, and we write ALG for ALG/Z (as ALG is thecategory of all commutative rings with identity). We consider a covariant functorP2 = P2

/k : ALG/k → SETS given by

P2(A) =L ⊂ An+1

∣∣L and An+1/L are locally A-free with rankL = 1.

This is a covariant functor. Indeed, if σ : A → A′ is a k-algebra homomorphism,letting it act on An+1 component-wise, L 7→ σ(L) ⊗A A′ induces a map P2(A) →P2(A′). If A is a field K, then X has to be free of dimension 1 generated by anon-zero vector x = (x0, x1, . . . , xn). The vector x is unique up to multiplication bynon-zero elements of K. Thus we have proven (for a field) of the following

Lemma 1.8. Suppose that K is a field. Then we have

P2(K) ∼=x = (x0, x1, . . . , xn) ∈ Kn+1|x 6= (0, . . . , 0)

/K×.


Moreover, writing Di : ALG/k → SETS for the subfunctor Di(A) ⊂ P2(A) madeup of the classes L whose projection to the i-th coordinate is surjective onto A (i.e.,xi 6= 0), we have P2(K) =

⋃i Di(K) and Di

∼= A2 canonically for all k-algebrasA. The isomorphism: Di

∼= A2 is given by sending (x0, . . . , xn) to (x0

xi, . . . , xn

xi) ∈ A2

removing the i-th coordinate.

Proof. If L ∈ Di(A), we have the following commutative diagram

L→−−−→ An+1

‖y

yi-th proj

L∼−−−→ A

Thus L is free of rank 1 over A; so, it has a generator (x0, . . . , xn) with xi ∈ A×.Then (x0, . . . , xn) 7→ (x0

xi, . . . , xn

xi) ∈ A2 gives rise to a natural transformation of Di

onto A2 (which is an isomorphism of functors).

If K is a field, we write (x0 : x1 : · · · : xn) for the point of P2(K) represented by(x0, . . . , xn) as only the ratio matters. We assume that K is a field for a while. Whenn = 1, we see P1(K) = K× t ∞ by (x : y) 7→ x

y∈ K t ∞. Thus P1(R) is

isomorphic to a circle and P1(C) is a Riemann sphere.We now assume that n = 2. Writing L = (x : y : 0) ∈ P2(K). Then P1 ∼= L

by (x : y) 7→ (x : y : 0); so, L is isomorphic to the projective line. We haveP2(K) = D(K)tL for fields K, where D = D2. Thus geometrically (i.e., over fields),P2 is the union of the affine plane added L. We call L = L∞ (the line at ∞).

1.4. Projective plane curve. For a plane curve defined by a = (f(x, y)) for f(x, y)of degree m, F (X, Y, Z) = Zmf(X

Z, Y

Z) is a (square-free) homogeneous polynomial of

degree m in k[X, Y, Z]. If L ∈ P2(A), we can think of F (`) for ` ∈ L. We writeF (L) = 0 if F (`) = 0 for all ` ∈ L. Thus for any k-algebra A, we define the functorV a : ALG/k → SETS by

V a(A) =L ∈ P2(A)|F (L) = 0

.

If A is a field K, we sent L ∈ P2(K) to its generator (a : b : c) ∈ L when weidentified P2(K) with the (classical) projective space with homogeneous coordinate.Since F (L) = 0 if and only if F (a : b : c) = 0 in this circumstances, we have

V a(K) =(a : b : c) ∈ P2(K)|F (a, b, c) = 0

which is called a projective plane k-curve. Since D2∼= A2 canonically via (x : y :

1) 7→ (x, y) (and this coordinate is well defined even over A which is not a field), wehave V a(A) ∩D2(A) = Va(A). In this sense, we can think of V a as a completion ofVa adding the boundary V a ∩ L∞. Since in Dj

∼= A2 (j = 0, 1), V a ∩ Dj is a plane

affine curve (for example, V a ∩ D0 is defined by F (1, y, z) = 0), (L∞ ∩ V a)(k) is afinite set. Thus V a is a sort of completion/compactification of the (open) affine curveVa (we sort out this point more rigorously later). Of course, we can start with ahomogeneous polynomial F (X, Y, Z) (or a homogeneous ideal of k[X, Y, Z] generatedby F (X, Y, Z)) to define a projective plane curve. Following Lemma 1.1, we defineHomproj k-curves(V a, V b) := HomCOF (V a, V b).


Example 1.3. Suppose a = (y2 − f(x)) for a cubic f(x) = x3 + ax + b. ThenF (X, Y, Z) = Y 2Z − X3 − aXZ2 − bZ3. Since L∞ is defined by Z = 0, we findL∞ ∩ V a = (0 : 1 : 0) made of a single point (with multiplicity 3). This point wecall the origin 0 of Va.

A projective plane curve V a is non-singular (or smooth) if V a∩Dj is a non-singularplane curve for all j = 0, 1, 2. The tangent space at P ∈ V a(K) is defined as beforesince P is in one of Dj ∩ Va. By the above exercise, the tangent space (the dual ofmP /m

2P ) at P ∈ V a(K) does not depend on the choice of j with P ∈ V a ∩Dj . If a

projective plane curve C is irreducible, the rational function field over k is the fieldof fraction of OC,P for any P ∈ C(k); so, independent of C ∩Dj .

Lemma 1.9. Take a nonzero f ∈ k(C). Then there exist homogeneous polynomialsG(X, Y, Z), H(X, Y, Z) ∈ k[X, Y, Z] with deg(G) = deg(H) such that f(x : y : z) =H(x,y,z)G(x,y,z)

for all (x : y : z) ∈ C(k).

Proof. We may write on C ∩D2 f(x, y, 1) = h(x,y)g(x,y)

. If m = deg(h) = deg(g), we just

define H(X, Y, Z) = h(XZ, Y

Z)Zm and G(X, Y, Z) = g(X

Z, Y

Z)Zm. If deg(h) > deg(g),

we define H(X, Y, Z) = h(XZ, Y

Z)Zdeg(h) and G(X, Y, Z) = g(X

Z, Y

Z)Zdeg(h). If deg(h) <

deg(g), we define H(X, Y, Z) = h(XZ, Y

Z)Zdeg(g) and G(X, Y, Z) = g(X

Z, Y

Z)Zdeg(g).

Multiplying h or g by a power of Z does not change the above identity f(x, y, 1) =h(x,y)g(x,y)

, because Z = 1 on C ∩D2. Thus by adjusting in this way, we get G and H.

Example 1.4. Consider the function φ = cx + dy in k(C) for C = V a with a =(y2 − x3 − ax− b). Then C is defined by Y 2Z −X3 − aXZ2 − bZ3 = 0, and

φ(X : Y : Z) = cX

Z+ d

Y

Z=cX + dY

Z.

So φ has pole of order 3 at Z = 0 (as the infinity on C has multiplicity 3) and threezeros at the intersection of L := cx+ dy = 0 and C ∩D2 ∩ L.

Take a projective nonsingular plane k-curve C/k. Put Ci = C∩Di which is an affinenonsingular plane curve. Then we have well defined global differentialsDerCi/k. Since∂ : DerCi/k induces ∂P : OCi,P → K for any P ∈ Ci(K) by f 7→ ∂(f)(P ), we have∂P ∈ TP . If ∂i ∈ DerCi/k given for each i = 0, 1, 2 satisfies ∂i,P = ∂j,P for all (i, j) and

all P ∈ (Di ∩Dj)(k), we call ∂ = ∂ii a global tangent vector defined on C . Plainlythe totality TC/k of global tangent vectors are k-vector space. The k-dual of TC/k iscalled the space of k-differentials over k and written as ΩC/k. It is known that ΩC/k

is finite dimensional over k.

Corollary 1.10. Suppose that C is non-singular. Each φ ∈ k(C) induces φ ∈Homproj k-curves(C,P

1). Indeed, we have k(C) t ∞ ∼= Homproj k-curves(C,P1), where

∞ stands for the constant function sending all P ∈ C(A) to the image of ∞ ∈ P1(k)in P1(A).

Proof. We prove only the first assertion. Suppose k = k. Write φ(x : y : z) = h(x,y,z)g(x,y,z)

as a reduced fraction by the above lemma. For L ∈ C(A) ⊂ P2(A), we considerthe sub A-module φ(L) of A2 generated by (h(`), g(`)) ∈ A2|` ∈ L. We now show


that φ(L) ∈ P1(A); so, we will show that the map C(A) 3 L 7→ φ(L) ∈ P1(A)induces the natural transformation of C into P1. If A is local, by Lemma 1.8, L isgenerated by (a, b, c) with at least one unit coordinate. Then any ` ∈ L is of the formλ(a, b, c) and therefore φ(`) = λdeg(h)φ(a, b, c). Thus φ(L) = A ·φ(a, b, c). Since A is ak-algebra, k is naturally a subalgebra of the residue field A/m of A. Since φ(P ) for allP ∈ C(k) is either a constant in k or ∞, we may assume that (h(P ), g(P )) 6= (0, 0)for all P ∈ C(k). Since (a, b, c) 6≡ 0 mod m as (a, b, c) generates a direct summandof A3. Thus (h(a, b, c), g(a, b, c)) 6≡ (0, 0) mod m. After tensoring A/m over A,(A/m)2/(φ(L)/mφ(L)) is one dimensional. Thus by Nakayama’s lemma (e.g., [CRT]Theorem 2.2–3), A/φ(L) is generated by a single element and has to be a free moduleof rank 1 as φ(L) is a free A-module of rank 1. Thus φ(L) ∈ P1(A). If k is notalgebraically closed, replacing A by A = A⊗k k, we find φ(L)⊗k k ∈ P2(k) and henceφ(L)⊗A A/m ∈ P2(k), which implies φ(L) ∈ P2(A).

If A is not necessarily local, applying the above argument to the local ring AP forany prime ideal P of A, we find that φ(L)P = φ(LP ) and A2

P/φ(LP ) are free of rank1; so, φ(L) and A2/φ(L) are locally free of rank 1; therefore, φ(L) ∈ P2(A).

Now it is plain that L 7→ φ(L) induces a natural transformation of functors.

1.5. Divisors. The divisor group Div(C) of a non-singular projective geometricallyirreducible plane curve C is a formal free Z-module generated by points P ∈ C(k).When we consider a point P as a divisor, we write it as [P ]. For each divisor D =∑

P mP [P ], we define deg(D) =∑

P mP . Since C is nonsingular, for any point P ∈C(k), OC,P is a DVR, and the rational function field k(C) is the quotient field of OC,P

(regarding C as defined over k). Thus if we write the valuation vP : k(C) Z∪∞for the additive valuation of OC,P , we have a well defined vP (f) ∈ Z for any non-zero

rational k-function f ∈ k(C). Since mP = (tP ) and tvP (f)P ‖ f in OC,P , f has a zero of

order vp(f) at P if vP(f) > 0 and a pole of order |vp(f)| if vP (f) < 0. In other words,the Taylor expansion of f at P is given by

∑n an(f)t2P and vp(f) = min(n : an(f) 6=

0). For a global doifferential ω ∈ ΩC/k, we have its Taylor expansion∑

n an(f)t2P dtP

at each P ∈ C(k); so, we may also define vP (ω) := min(n : an(ω) 6= 0). We extendthis definition for meromorphic differentials k(C) ·ΩC/k = f ·ω|f ∈ k(C), ω ∈ ΩC/k.Here we quote Bezout’s theorem:

Theorem 1.11. Let C and C ′ be two plane projective k-curves inside P2 defined byrelatively prime homogeneous equations F (X, Y, Z) = 0 and G(X, Y, Z) = 0 of degreem and n respectively. Then counting with multiplicity, |C(k) ∩ C ′(k)| = m · n.

If C is smooth at P ∈ C ∩ C ′ in C ∩D2, φ = G(X,Y,Z)Z2 is a function vanishing at

P . The multiplicity of P in C ∩ C ′ is just vP(φ). More generally, if P = (a, b) isnot necessarily a smooth point, writing C ∩D2 = Va and C ′ ∩D2 = Vb for principalideals a, b in k[X, Y ] and regarding P as an ideal (X − a, Y − b) ⊂ k[X, Y ], themultiplicity is given by the dimension of the localization (k[x, y]/a+ b)P over k. Thesame definition works well for any points in C ∩ D0 and C ∩ D1. One can find theproof of this theorem with (possibly more sophisticated) definition of multiplicity ina text of algebraic geometry (e.g. [ALG] Theorem I.7.7).


Since there are only finitely many poles and zeros of f , we can define the divi-sors div(f) =

∑P∈C(k) vP (f)[P ], div0(f) =

∑P∈C(k),vP (f)>0 vP(f)[P ] and div∞(f) =∑

P∈C(k),vP (f)<0 vP (f)[P ] of f . Similarly, for meromorphic differential ω, we define

again div(ω) =∑

P vP (ω)[P ]. By Lemma 1.9, f(x : y : z) = h(x:y:z)g(x:y:z)

for a homogeneous

polynomial h, g in k[x, y, z] of the same degree. If the degree of equation defining C ism and C ′ is defined by h(X, Y, Z) = 0, deg0(div(f)) = |C(k) ∩ C ′(k)| = m deg(h) =m deg(g) = deg∞(div(f)). This shows deg(div(f)) = 0 as

∑P,vP (f)>0mP = m deg(h)

and −∑

P,vP (f)<0mP = m deg(g).

Lemma 1.12. Let C be a nonsingular projective plane curve. For any f ∈ k(C),deg(div(f)) = 0, and if f ∈ k(C) is regular at every P ∈ C, f is a constant in k.

Lemma 1.13. If f ∈ k(C) satisfies deg(div0(f)) = deg(div∞(f)) = 1, f : C → P1

induces an isomorphism of projective plane curve over k.

Proof. By the proof of Corollary 1.10, deg(div0(f)) is the number of points over 0(counting with multiplicity) of the regular map f : C → P1. By taking off a constantα ∈ k ⊂ P1 to f , deg(div0(f − α)) = 1 = deg(div∞(f − α)), and |f−1(α)| =deg(div0(f − α)) = 1; so, we find that f is 1-1 onto. Thus f is an isomorphism.

Write Div0(C) = D ∈ Div(C/k)| deg(D) = 0. Inside Div0(C), we have the

subgroup div(f)|f ∈ k(C)×. We call two divisors D,D′ linearly equivalent if D =div(f) + D′ for f ∈ k(C). We call that D and D′ are algebraically equivalent ifdeg(D) = deg(D′). The quotient groups J(C) = Div0(C)/div(f)|f ∈ k(C)× andPic(C) = Div(C)/div(f)|f ∈ k(C)× are called the jacobian and the Picard group ofC , respectively. Sometimes, J(C) is written as Pic0(C) (the degree 0 Picard group).

1.6. The theorem of Riemann–Roch. We write D =∑

P mP [P ] ≥ 0 (resp. D >0) for a divisor D on C if mP ≥ 0 for all P (resp. D ≥ 0 and D 6= 0). For a divisorD on Ck

L(D) = f ∈ k(C)| div(f) +D ≥ 0 ∪ 0.Plainly, L(D) is a vector space over k. It is known that `(D) = dimk L(D) <∞. Forφ ∈ k(C)×, L(D) 3 f 7→ fφ ∈ L(D − div(φ)) is an isomorphism. Thus `(D) onlydepends on the class of D in Pic(C).

Example 1.5. Let C = P1. For a positive divisor D =∑

a∈kma[a] with ma ≥ 0 and

ma > 0 for some a, regarding a ∈ k as a point [a] ∈ P1(k) = kt∞. On A1(k) = k,

forgetting about the infinity, div(f) +D ≥ 0 if f = g(x)Qa(x−a)ma for a polynomial g(x).

If deg(D) ≥ deg(g(x)), the function f does not have pole at ∞. Thus L(D) =g(x)| deg(g(x)) ≤ deg(D) and we have `(D) = 1 + deg(D) if D > 0. If C is a

plane projective curve, we can write f = h(X,Y,Z)g(X,Y,Z)

as a reduced fraction by Lemma 1.9.

Write D =∑

P mP [P ], and put |D| = P |D =∑

P mP [P ] with mP 6= 0. If |D|is inside D2 ∩ C ⊂ A2 and D > 0, we may assume that V(g(X,Y,1)) ∩ C contains |D|.Then not to have pole at C \ D2, deg(h) has to be bounded; so, `(D) < ∞. SinceL(D) ⊂ L(D+) in general, writing D = D+ +D− so that D+ ≥ 0 and −D− ≥ 0, thisshows `(D) <∞.


Theorem 1.14 (Riemann-Roch). Let C = V a be a non-singular projective curvedefined over a field k. Then for g = dimk ΩC/k and a divisor K of degree 2g − 2 of

the form div(ω) for a meromorphic differential ω on C such that `(D) = 1 − g +deg(D) + `(K − D) for all divisor D on C(k) and the equality holds for sufficientlypositive divisor D. If g = 1, K = 0.

The divisor K is called a canonical divisor K (whose linear equivalence class isunique). Note that

L(K) = f ∈ k(C)| div(fω) = div(f) + div(ω) ≥ 0 ∼= ΩC/k

by f 7→ fω ∈ ΩC/k. Then by the above theorem,

g(C) = dimΩC/k = `(K) = 1− g + deg(K) + `(0) = 2 + deg(K)− g(C),

and from this, we conclude deg(K) = 2g(C)−2. One can find a proof of this theoremin any algebraic geometry book (e.g., [ALG] IV.1 or [GME] Theorem 2.1.3).

Corollary 1.15. If g(C) = 1, then `(D) = deg(D) if deg(D) > 0.

Proof. For a non-constant f ∈ k(E), deg(div(f)) = 0 implies that f has a polesomewhere. If D > 0, f ∈ L(−D) does not have pole; so, constant. Since D > 0,f vanishes at P ⊂ D. Thus f = 0. More generally, if deg(D) > 0 and φ ∈ L(−D),then 0 > deg(−D) = deg(φ) − deg(D) ≥ 0; so, φ = 0. Thus if deg(D) > 0, then`(−D) = 0. Since K = 0, we have by the Riemann-Roch theorem that `(D) =deg(D) + `(0−D) = deg(D) if deg(D) > 0.

Because of deg(div(f)) = 0, if D 0, `(−D) = 0. Thus in particular `(K−D) = 0if D 0. Thus the above theorem implies what Riemann originally proved:

Corollary 1.16 (Riemann). Let C = V a be a non-singular projective curve definedover a field k. Then there exists a non-negative integer g = g(C) such that `(D) ≥1−g+deg(D) for all divisor D on C(k) and the equality holds for sufficiently positivedivisor D.

By the above example, we conclude g(P1) = 0 from the corollary.

Exercise 1.17. Prove ΩP1/k = 0.

1.7. Regular maps from a curve into projective space. Tak a divisor D ona nonsingular projective plane curve C . Suppose `(D) = n > 0. Take a basis

(f1, f2, . . . , fn) of L(D). Thus we can write fj =hj

gjwith homogeneous polynomials

gj , hj having deg(gj) = deg(hj). Replacing (gj, hj) by (g′0 := g1g2 · · · gn, h′j := hjg

(j))

for g(j) =∏

i6=j gi, we may assume deg(g′j) = deg(h′j) for all j, and further dividing

them by the GCD of (h′1, . . . , h′n, g

′0), we may assume that fj =

hj

g0with deg(hj) =

deg(g0) for all j and (g0, h1, . . . , hn) do not have nontrivial common divisor.

Lemma 1.18. Let the assumptions on (g0, h1, . . . , hn) be as above. Suppose that(g0(P ), h1(P ), . . . , hn(P )) 6= (0, 0, . . . , 0) for all P ∈ C(k). Define L ∈ C(A) ⊂P2(A), φA(L) for an A-submodule of A3 generated by φ(`) = (g0(`), h1(`), . . . , hn(`)) ∈An+1 for all ` ∈ L. Then φ = φAA : C → P2 is a k-morphism of the projectiveplane k-curve C into P2

/k.


The proof of the above lemma is the same as that of Corollary 1.10; so, we leave itto the reader.

2. Elliptic curves

An elliptic curve E/k is a non-singular projective geometrically irreducible planecurve with point 0E specified having g(E) = 1. Here we define g(E), regarding E isdefined over k. We study elliptic curves in more details.

2.1. Abel’s theorem. When we regard P ∈ E(k) as a divisor, we just write [P ]. So3[P ] is a divisor supported on P with multiplicity 3. We prove

Theorem 2.1 (Abel). Let E/k be an elliptic curve with origin 0E. The correspondence

P 7→ [P ]− [0E ] induces a bijection E(k) ∼= J(E). In particular, E(k) is an ableiangroup.

Proof. Injectivity: if [P ] − [Q] = [P ] − [0E] − ([Q] − [0E]) = div(f) with P 6= Qin E(k), by Lemma 1.13, f is an isomorphism. This is wrong as g(P1) = 0 whileg(E) = 1. Thus P = Q.

Surjectivity: Pick D ∈ Div0(E). Then D + [0E] has degree 1; so, `(D + [0E]) = 1by Corollary 1.15, and we have φ ∈ L(D+[0E]). Then div(φ)+D+[0E ] ≥ 0 and hasdegree 1. Any non-negative divisor with degree 1 is a single point [P ]. Thus D+ [0E]is linearly equivalent to [P ]; so, the map is surjective.

Corollary 2.2. If 0 6= ω ∈ ΩE/k, then div(ω) = 0.

Proof. Since E(k) is a group, for each P ∈ E(k), TP : Q 7→ Q + P gives an auto-morphism of E. Thus ω TP is another element in ΩE/k. Since dimΩE/k = 1, we

find ω TP = λ(P )ω for λ ∈ k. Since ω 6= 0, at some point P ∈ E(k), vP (ω) = 0.Since vQ(ω TP ) = vP+Q(ω) and we can bring any point to P by translation, we havevP (ω) = 0 everywhere. Thus div(ω) = 0.

We can show easily λ(P ) = 1 for all P (see [GME] §2.2.3). The nonzero differentialsω in ΩE/k are called nowhere vanishing differentials as div(ω) = 0. They are uniqueup to constant multiple.

Exercise 2.3. Take a line L defined by aX+bY +cZ on P2 and suppose its intersec-tion with an elliptic curve E ⊂ P2 to be P,Q,R. Prove that [P ]+[Q]+[R]∼ 3[0E ].

A field k is called a perfect field if any finite field extension of k is separable (i.e.,generated by θ over k whose minimal equation over k does not have multiple roots).Fields of characteristic 0 and finite fields are perfect.

Remark 2.4. If k is perfect, k/k is possibly an infinite Galois extension; so, by Galoistheory, we have a bijection between open subgroups G of Gal(k/k) and finite extensions

K/k inside k by G 7→ kG

= x ∈ k|σ(x) = x for all σ ∈ G and K 7→ Gal(k/K).Since the isomorphicm E(k) ∼= J(C) is Galois equivariant, we have

E(K) ∼= J(E)Gal(k/K) = D ∈ J(E)|σ(D) = D for all σ ∈ G,


where σ ∈ Gal(k/k) acts on D =∑

P mP [P ] by σ(D) =∑

P mP [σ(P )]. Basically bydefinition, we have

J(E)(K) := J(E)Gal(k/K) =D ∈ Pic0(E)|σ(D) = Ddiv(f)|f ∈ K(E)× .

Since any subfield K ⊂ k is a union of finite extensions, the identity E(K) ∼= J(E)(K)is also true for an infinite extension K/k inside K . Actually we have a good defini-tion of Pic(E)(A) for any k-algebra A, and we can generalize the identity E(K) ∼=J(E)(K) to all k-algebras A in place of fields K inside k.

2.2. Weierstrass Equations of Elliptic Curves. We now embed E/k into thetwo-dimensional projective space P2

/k using a basis of L(3[0]) and determine the

equation of the image in P2/k. Choose a parameter T = t0 at the origin 0 = 0E.

We first consider L(n[0]) which has dimension n if n > 0. We have L([0]) = k andL(2[0]) = k1 + kx. Since x has to have a pole of order 2 at 0, we may normalizex so that x = T−2(1 + higher terms) in k[[T ]]. Here x is unique up to translation:x 7→ x + a with a ∈ k. Then L(3[0]) = k1 + kx + ky. We may then normalize y sothat y = −T−3(1 + higher terms) (following the tradition, we later rewrite y for 2y;thus, the normalization will be y = −2T−3(1 + higher terms) at the end). Then y isunique up to the affine transformation: y 7→ y + ax+ b (a, b ∈ k).Proposition 2.5. Suppose that the characteristic of the base field k is different from2 and 3. Then for a given pair (E, ω) of an elliptic curve E and a nowhere-vanishingdifferential ω both defined over k, we can find a unique base (1, x, y) of L(3[0]) suchthat E is embedded into P2

/k by (1, x, y) whose image is defined by the affine equation

(2.1) y2 = 4x3 − g2x− g3 with g2, g3 ∈ k,and ω on the image is given by dx

y. Conversely, a projective algebraic curve defined by

the above equation is an elliptic curve with a specific nowhere-vanishing differentialdxy

if and only if the discriminant ∆(E, ω) = g32 − 27g2

3 of 4X3 − g2X − g3 does notvanish.

An equation of an elliptic curve E as in (2.1) is called a Weierstrass equation of E,which is determined by the pair (E, ω).

Proof. By the dimension formulas, counting the order of poles at 0 of monomials ofx and y, we have

L(4[0]) = k + kx+ ky + kx2,

L(5[0]) = k + kx+ ky + kx2 + kxy and

L(6[0]) = k + kx+ ky + kx2 + kxy + kx3

= k + kx+ ky + kx2 + kxy + ky2,

from which the following relation results,

(2.2) y2 + a1xy + a3y = x3 + a2x2 + a4x+ a6 with aj ∈ k,

because the poles of order 6 of y2 and x3 have to be canceled. We homogenize theequation (2.2) by putting x = X

Zand y = Y

Z(and multiplying by Z3). Write C for


the projective plane k-curve in P2 defined by the (homogenized) equation. Thus wehave a k-regular map: φ : E → C ⊂ P2 given by P 7→ (x(P ) : y(P ) : 1). Thusthe function field k(E) contains the function field k(C) by the pull back of φ. Bydefinition, k(C) = k(x, y). Since div∞(x) = 2[0E] for x = X

Z: E → P1, this gives

a covering of degree 2; so, [k(E) : k(x)] = 2. Similarly [k(E) : k(y)] = 3. Since[k(E) : k(C)] is a common factor of [k(E) : k(x)] = 2 and [k(E) : k(y)] = 3, we getk(E) = k(C). Thus if C is smooth, E ∼= C by φ as a smooth geometrically irreduciblecurve is determined by its function field. Therefore, assuming C is smooth, E/k canbe embedded into P2

/k via P 7→ (x(P ), y(P )). The image is defined by the equation

(2.2).Let T be a local parameter at 0E normalized so that

ω = (1 + higher degree terms)dT.

Anyway ω = (a + higher degree terms)dT for a ∈ k×, and by replacing T by aT , weachieve this normalization. The parameter T normalized as above is called a param-eter adapted to ω. Then we may normalize x so that x = T−2 +higher degree terms.We now suppose that 2 is invertible in k. Then we may further normalize y so thaty = −2T−3 + higher degree terms (which we will do soon but not yet; so, for themoment, we still assume y = T−3 + higher degree terms).

The above normalization is not affected by variable change of the form y 7→ y+ax+band x 7→ x+ a′. Now we make a variable change y 7→ y + ax+ b in order to removethe terms of xy and y (i.e., we are going to make a1 = a3 = 0):

(y + ax+ b)2 + a1x(y + ax+ b) + a3(y + ax+ b)

= y2 + (2a + a1)xy + (2b+ a3)y + polynomial in x.

Assuming that 2 is invertible in k, we take a = −a1

2and b = −a3

2. The resulting

equation is of the form y2 = x3 +b2x2 +b4x+b6. We now make the change of variable

x 7→ x+ a′ to make b2 = 0:

y2 = (x+ a′)3 + b2(x+ a′)2 + b4(x+ a′) + b6 = x3 + (3a′ + b2)x2 + · · · .

Assuming that 3 is invertible in k, we take a′ = − b23. We can rewrite the equation as

in (2.1) (making a variable change −2y 7→ y). By the variable change as above, wehave y = −2T−3(1+higher terms), and from this, we conclude ω = dx

y. The numbers

g2 and g3 are determined by T adapted to a given nowhere-vanishing differential formω.

If the discriminant ∆(E, ω) of g(x) = 4x3−g2x−g3 vanishes, C has only singularityat (x0 : 0 : 1) for a multiple root x0 of g(x) = 0. If g(x) has a double zero, C isisomorphic over k to the curve defined by y2 = x2(x− a) for a 6= 0. Let t = x

y. Then

for P ∈ E(k) mapping to (0, 0), vP (y) = vP (x); so, P is neither a zero nor a pole oft. The function t never vanish outside 0E (having a pole at (a, 0)). It has a simplezero at 0E by the normalization of x and y. Thus deg(div0(t)) = 1, and k(C) = k(t),which is impossible as k(C) = k(E) and g(E) = 1. The case of triple zero can beexcluded similarly. Thus we conclude ∆(E, ω) 6= 0 (⇔ C is smooth: Example 1.3),and we have E ∼= C by φ.


Conversely, we have seen that any curve defined by equation (2.1) is smooth inExample 1.3 if the cubic polynomial F (X) = 4X3− g2X − g3 has three distinct rootsin k. In other words, if the discriminant ∆(E, ω) of F (X) does not vanish, E issmooth.

For a given equation, Y 2 = F (X), the algebraic curve E defined by the homoge-neous equation Y 2Z = 4X3−g2XZ

2−g3Z3 in P2

/k has a rational point 0 = (0, 1, 0) ∈E(k), which is ∞ in P2. Thus E is smooth over k if and only if ∆(E, ω) 6= 0 (anexercise following this proof).

We show that there is a canonical nowhere-vanishing differential ω ∈ ΩE/k if E isdefined by (2.1). If such an ω exists, all other holomorphic differentials ω′ are of theform fω with div(f) ≥ 0, which implies f ∈ k; so, g = dimk ΩE/k = 1, and E/k isan elliptic curve. It is an easy exercise to show that y−1dx does not vanish on E (anexercise following this proof).

We summarize what we have seen. Returning to the starting elliptic curve E/k, forthe parameter T at the origin, we see by definition

x = T−2(1 + higher degree terms) and y = −2T−3(1 + higher degree terms).

This shows

dx

y=−2T−3(1 + · · · )−2T−3(1 + · · · )dT = (1 + higher degree terms)dT = ω.

Thus the nowhere-vanishing differential form ω to which T is adapted is given by dxy

.

Conversely, if ∆ 6= 0, the curve defined by y2 = 4x3 − g2x − g3 is an elliptic curveover k with origin 0 =∞ and a standard nowhere-vanishing differential form ω = dx

y.

This finishes the proof.

Exercise 2.6. (1) If C is defined by y2 = x3, prove k(C) = k(t) for t = xy.

(2) Compute vP (dx/y) explicitly at any point P on E(k).(3) Show that if ∆ 6= 0, the curve defined by y2 = 4x3 − g2x− g3 is also smooth

at 0 =∞.

2.3. Moduli of Weierstrass Type. We continue to assume that the characteristicof k is different from 2 and 3. Suppose that we are given two elliptic curves (E, ω)/k

and (E ′, ω′)/k with nowhere-vanishing differential forms ω and ω′. We call two pairs(E, ω) and (E ′, ω′) isomorphic if we have an isomorphism ϕ : E → E ′ with ϕ∗ω′ = ω.Here for ω′ = fdg, ϕ∗ω′ = (f ϕ)d(g ϕ); in other words, if σ : k(E ′) → k(E) isthe isomorphism of the function fields associated with ϕ, ϕ∗ω′ = σ(f)d(σ(g)). LetT ′ be the parameter at the origin 0 of E ′ adapted to ω′. If ϕ : (E, ω) ∼= (E ′, ω′),then the parameter T = ϕ∗T ′ mod T 2 is adapted to ω (because ϕ∗ω′ = ω). Wechoose coordinates (x, y) for E and (x′, y′) for E ′ relative to T and T ′ as above. Bythe uniqueness of the choice of (x, y) and (x′, y′), we know ϕ∗x′ = x and ϕ∗y′ = y.Thus the Weierstrass equations of (E, ω) and (E ′, ω′) coincide. We write g2(E, ω) andg3(E, ω) for the g2 and g3 of the coefficients of the Weierstrass equation of (E, ω). Ifa field K has characteristic different from 2 and 3, we have

(2.3) P(K) :=[(E, ω)/K

] ∼=(g2, g3) ∈ K2

∣∣∆(E, ω) 6= 0 ∼= HomALG(R, K),


where R := Q[g2, g3,1

g32−27g2

3

] (the polynomial ring of variables gj with g32 − 27g2

3

inverted) and [·] indicates the set of isomorphism classes of the objects inside thebracket and Spec(R)(K) for a ringR is the set of all algebra homomorphisms: R→ K.The last isomorphism sends (g2, g3) to the algebra homomorphism φ with φ(X) = g2

and φ(Y ) = g3.There is an elliptic curve E defined by Y 2Z = 4X3 − g2XZ

2 − g3Z3 over R.

This is a universal curve in the sense that for any pair (E, ω)/A defined by Y 2Z =

4X3 − a2XZ2 − a3Z

3 with ω = dXY

over A, we have a unique morphism R ϕ−→ A suchthat ϕ(gj) = aj induces the pair (E, ω). In other words, (2.3) means that for eachφ ∈ HomALG(R, K), there is a unique object (E, ω)/K defined by the equation Y 2Z =

4X3 − φ(g2)XZ2 − φ(g3)Z

3 with ω = dXY

(not just an isomorphism class of (E, ω)/K;so, for such representability, it is absolutely necessary that Aut(E, ω)/K) = id asotherwise, we would have several choices in the isomorphism class of (E, ω)/K .

We now classify elliptic curves E eliminating the contribution of the differentialfrom the pair (E, ω). If ϕ : E ∼= E ′ for (E, ω) and (E ′, ω′), we have ϕ∗ω′ = λωwith λ ∈ K×, because ϕ∗ω′ is another nowhere-vanishing differential. Therefore westudy K×-orbit: (E, ω) mod K× under the action of λ ∈ K× given by (E, ω)/K 7−→(E, λω)/K , computing the dependence of gj(E, λω) (j = 2, 3) on λ for a given pair(E, ω)/K . Let T be the parameter adapted to ω. Then λT is adapted to λω. We see

x(E, ω) =(1 + Tφ(T ))

T 2⇒x(E, λω) =

(1 + higher terms)

(λT )2= λ−2x(E, ω),

y(E, ω) =(−2 + Tψ(T ))

T 3⇒y(E, λω) =

(−2 + higher terms)

(λT )3= λ−3y(E, ω).

Since y2 = 4x3 − g2(E, ω)x− g3(E, ω), we have

(λ−3y)2 = 4λ−6x3 − g2(E, ω)λ−6x− λ−6g3(E, ω)

= 4(λ−2x)3 − λ−4g2(E, ω)(λ−2x)− λ−6g3(E, ω).

This shows

(2.4) g2(E, λω) = λ−4g2(E, ω) and g3(E, λω) = λ−6g3(E, ω).

Thus we have

Theorem 2.7. If two elliptic curves E/K and E ′/K are isomorphic, then choosing

nowhere-vanishing differentials ω/E and ω′/E′, we have gj(E

′, ω′) = λ−2jgj(E, ω) for

λ ∈ K×. The constant λ is given by ϕ∗ω′ = λω.

We define the J -invariant of E by J(E) = (12g2(E,ω))3

∆(E,ω). Then J only depends on E

(not the chosen differential ω). If J(E) = J(E ′), then we have

(12g2(E, ω))3

∆(E, ω)=

(12g2(E′, ω′))3

∆(E ′, ω′)⇐⇒ gj(E

′, ω′) = λ−2jgj(E, ω)

for a twelfth root λ of ∆(E, ω)/∆(E ′, ω′). Note that the twelfth root λ may not bein K if K is not algebraically closed.


Conversely, for a given j 6∈ 0, 1, the elliptic curve defined by y2 = 4x3 − gx − gfor g = 27j

j−1has J -invariant 123j. If j = 0 or 1, we can take the following elliptic curve

with J = 0 or 123. If J = 0, then y2 = 4x3 − 1 and if J = 123, then y2 = 4x3 − 4x.Thus we have

Corollary 2.8. If K is algebraically closed, then J(E) = J(E ′) ⇔ E ∼= E ′ for twoelliptic curves over K. Moreover, for any field K, there exists an elliptic curve Ewith a given J(E) ∈ K.

Exercise 2.9. (1) Prove that gj(E′, ω′) = λ−2jgj(E, ω) for suitable ω and ω′ and

a suitable twelfth root λ of ∆(E, ω)/∆(E ′, ω′) if J(E) = J(E ′).(2) Explain what happens if J(E) = J(E ′) but E 6∼= E ′ over a field K not neces-

sarily algebraically closed.

Note that R is a graded ring such that g2 is of degree 4 and g3 is of degree 6. Thenthe degree 0 subring R0 = Q[J ]. Note that Spec(Q[J ]) is the 1 dimensional affinespace A1, as Spec(Q[J ])(A) = Homalg(Q[J ], A) ∼= A by φ 7→ φ(J) ∈ A.

Consider functors

P1,N(A) :=[(E, ω, φ)/A|φ : N−1Z/Z → E[N ] := Ker(E

N−→ E)],

E1,N (A) :=[(E, φ)/A|φ : N−1Z/Z → E[N ] := Ker(E

N−→ E)](2.5)

for a positive integer N . The functor has natural transformation P1,N → P sending(E, ω, φ)/A to (E, ω)/A. This is represented by Y1(N) := E[N ]−⋃

0<d|N,D 6=N E[d] which

is affine in the sense that Y1(N) = Spec(R1,N) for the ringR1,N := HomCOF (Y1(N),A),and R1,N is finite locally free over R. By the action ω 7→ λω on P1,N, the multiplica-tive group Gm given by A 7→ A× acts on R1,N . The subring A1,N := H0(Gm,R1,N)fixed by this action represents E1,N , i.e.,

E1,N(A) = HomQ-alg(A1,N, A).

The corresponding plane curve Y1(N)/Q := Spec(A1,N) is the modular curve of levelΓ1(N). In other words, for a triple (E, ω, φ : Z/NZ → E[N ])/A, the value φ(1) giverise to a unique point of Y1(N)(A) over the point corresponding (E, ω)/A ∈ P1,N(A).

Similarly,

P1,N(A) :=[(E, ω, φ)/A|φ : (N−1Z/Z)2 ∼= E[N ] := Ker(E

N−→ E)],

E1,N(A) :=[(E, φ)/A|φ : (N−1Z/Z)2 ∼= E[N ] := Ker(E

N−→ E)](2.6)

is represented by

Y(N) = (x, y) ∈ E[N ]×Y E[N ]|x ∧ y ∈ E[N ] ∧ E[N ]−⋃

0<d|N,D 6=N

E[d] ∧ E[d]

and Y (N) = Spec(AN ) with AN = H0(Gm,RN ), respectively, where Y(N) =Spec(RN ) for RN := HomCOF (Y(N),A)


Remark 2.10. We will see later that the curve Y (N) is irreducible over Q but be-comes reducible over Q[µN ] (the cyclotomic field of N-th root of unity), looking intothe q-expansion of Weierstrass ℘-functions. The function field Q(Y (N)) thereforecontains Q[µN ] as algebraic closure of Q in it.

3. Modular forms and functions

We give an algebraic definition of modular forms and then relate it to classicaltheory (due to Weierstarss, Klein, Flicke).

3.1. Geometric modular forms. Let A be an algebra over Q. We restrict the func-tor P to ALG/A and write the restriction P/A. Then by (2.3), for RA := A[g2, g3,

1∆

],

P/A(?) = HomALG/A(RA, ?).

A morphism of functors φ : P/A → A1/A is by definition given by maps φR : P/A(R)→

A1(R) = R indexed by R ∈ ALG/A such that for any σ : R→ R′ in HomALG/A(R,R′),

φR′((E, ω) ⊗R R′) = σ(f((E, ω)/R)). Note that A1/A(?) = HomALG/A

(A[X], ?) by

R 3 a↔ (ϕ : A[X]→ R) ∈ HomALG/A(A[X], ?) with ϕ(X) = a. Thus in particular,

φRA: P(RA) = HomALG/A

(RA,RA)→ A1(A[X],RA) = RA.

Thus φRA(idRA

) ∈ RA; so, write φRA(idRA

) = Φ(g2, g3) for a two variable rationalfunction Φ(x, y) ∈ A[x, y, 1

x3−27y2 ]. Let E/RAbe the universal elliptic curve over RA

defined by Y 2Z = 4X3 − g2XZ2 − g3Z

3 with the universal differential ω = dXY

. Ifwe have (E, ω)/R, we have a unique A-algebra homomorphism σ : RA → R given byσ(gj) = gj(E, ω); in other words, (E, ω)/R

∼= (E, ω)RA⊗RA

R. Thus

φR(E, ω) = φR((E, ω)⊗RAR) = σ(φRA

(E, ω))

= σ(φRA(idRA

)) = Φ(σ(g2), σ(g3)) = Φ(g2(E, ω), g3(E, ω)).

Theorem 3.1. Any functor morphism φ : P/A → A1/A is given by a rational function

Φ ∈ RA of g2 and g3 so that φ(E, ω) = Φ(g2(E, ω), g2(E, ω)) for every eliiptic curve(E, ω) over an A-algebra.

Define a weight function w : A[g2, g3] → Z by w(ga2g

b3) = 4a + 6b, and for general

polynomials Φ =∑

a,b ca,bga2g

b3, we put w(Φ) = max(w(ga

2gb3)|ca,b 6= 0). A polynomial

Φ =∑

a,b ca,bga2g

b3 of g2 and g3 is called isobaric if ca,b 6= 0⇒ 4a + 6b = w.

A weight w modular form defined over A is a morphism of functors P/A → A1/A

given by an isobaric polynomial of g2 and g3 of weight w with coefficients in A.Write Gw(A) for the A-module of modular forms of weight w. Then f ∈ Gw(A) isa functorial rule assigning each isomorphism class of (E, ω)/R for an A-algebra R anelement f(E, ω) ∈ R satisfying the following properties:

(G0) f ∈ A[g2, g3],(G1) If (E, ω) is defined over an A-algebra R, we have f(E, ω) ∈ R, which depends

only on the isomorphism class of (E, ω) over R,(G2) f((E, ω) ⊗R R

′) = σ(f(E, ω)) for A-algebra homomorphism σ : R→ R′,(G3) f((E, λω)/R) = λ−wf(E, ω) for any λ ∈ R×.


Exercise 3.2. For a field K ⊃ Q, prove for 0 < w ∈ 2Z,

dimK Gw(K) =

[w12

]if w ≡ 2 mod 12,[

w12

]+ 1 otherwise.

We can define modular form of level N rational over A replacing by the functor P inthe previous section by P1,N or PN . We lst the functorial properties:

(GN0) f is integral over ∈ A[g2, g3],(GN1) If (E, ω, φ) is defined over an A-algebra R, we have f(E, ω, ω) ∈ R, which

depends only on the isomorphism class of (E, ω) over R,(GN2) f((E, ω, φ)⊗R R

′) = σ(f(E, ω, φ)) for A-algebra homomorphism σ : R→ R′,(GN3) f((E, λω, φ)/R) = λ−wf(E, ω, φ) for any λ ∈ R×.

Here φ is the level structure depending on our choice of P1,N and PN .We then define Gk(Γ1(N);A) (resp. Gk(Γ(N);A) for the space of modular forms

of weight k defined over A for P1,N (resp. PN).

3.2. Topological Fundamental Groups. In the following three sections, we wouldlike to give a sketch of Weierstrass’ theory of elliptic curves defined over the complexfield C. By means of Weierstrass P–functions, we can identify E(C) (for each ellipticcurve E/C) with a quotient of C by a lattice L. In this way, we can identify [(E, ω)/C]with the space of lattices in C. This method is analytic.

We can deduce from the analytic parameterization (combining with geometric tech-nique of Weil-Shimura) many results on the moduli space of elliptic curves, like, theexact field of definition of the moduli, determination of the field of moduli (of eachmember), and so on (e.g., [IAT] Chapter 6). We have come here in a reverse way:starting algebraically, mainly by the Riemann-Roch theorem, we have determined aunique Weierstrass equation over A for a given pair (E, ω)/A, and therefore, we knowthe exact shape of the moduli space before setting out in studying analytic method.After studying analytic theory over C, combining these techniques, we start studyingmodular units.

Let (E, ω)/C be an elliptic curve over C. Then

E(C) = E(g2, g3)(C) = (x : y : z) ∈ P2(C)|y2z − 4x3 + g2z2x+ g3z

3 = 0,

and E(C) is a compact Riemann surface of genus 1. A path γ : y → x on E(C)is a piecewise smooth continuous map γ from the interval [0, 1] into E(C) (underthe Euclidean topology on E(C)) such that γ(0) = y and γ(1) = x. Two pathsγ, γ′ : x → x are homotopy equivalent (for which we write γ ≈ γ′) if there is a bi-continuous map ϕ : [0, 1]× [0, 1]→ E(C) such that ϕ(0, t) = γ(t) and ϕ(1, t) = γ′(t).Let Z be the set of all equivalence classes of paths emanating from 0.

More generally, for each complex manifold M , we can think of the space Z =Z(M) of homotopy classes of paths emanating from a fixed point x ∈ M . An openneighborhood U of x is called simply connected if Z(U) ∼= U by projecting (γ : x→ y)down to y. For example, if U is diffeomorphic to an open disk with center x, it issimply connected (that is, every loop is equivalent to x). If γ : x→ y and γ′ : y → z


are two paths, we define their product path γγ′ : x→ z by

γγ′(t) =

γ(2t) if 0 ≤ t ≤ 1/2

γ′(2t− 1) if 1/2 ≤ t ≤ 1.

By this multiplication, πM = π(M,x) = γ ∈ Z(M)|γ : x → x/≈ becomes agroup called the topological fundamental group of M . Taking a fundamental systemof neighborhoods Uy of y ∈ M made of simply connected open neighborhoods of y,we define a topology on Z(M) so that a fundamental system of neighborhoods ofγ : x → y is given by γU |U ∈ Ux. Then πM acts on Z(M) freely without fixedpoints. By definition, we have a continuous map π : πM\Z(M) → M given by π(γ :x→ y) = y, which is a local isomorphism. Since π−1(x) = x, π : πM\Z(M) ∼= M isa homeomorphism. Since π : Z(M)→ M is local isomorphism, we can regard Z(M)as a complex manifold. This space Z(M) is called a universal covering space of M .

We now return to the original setting: Z = Z(E(C)), and write Π = π(E, 0). SinceE(C) is a commutative group, writing its group multiplication additively, we definethe sum γ + γ′ on Z by, noting that γ and γ′ originate at the origin 0,

(γ + γ′)(t) =

γ(2t) if 0 ≤ t ≤ 1/2

γ(1) + γ′(2t− 1) if 1/2 ≤ t ≤ 1.

Then (γ + γ′)(1) = γ(1) + γ′(1), and we claim that γ + γ′ ≈ γ′ + γ. In fact, on thesquare [0, 1] × [0, 1], we consider the path α on the boundary connecting the origin(0, 0) and (1, 1) passing (0, 1), and write β the opposite path from (0, 0) to (1, 1)passing (1, 0). They are visibly homotopy equivalent. Thus we have a continuousmap φ : [0, 1] × [0, 1] → [0, 1] × [0, 1] such that φ(0, t) = α(t) and φ(1, t) = β(t).Define

f : [0, 1]× [0, 1]→ E(C) by f(t, t′) = γ(t) + γ′(t′).

Then it is easy to see f φ(0, t) = (γ′ + γ)(t) and f φ(1, t) = (γ + γ′)(t).By the above addition, Z is an additive complex Lie group. Since γ + γ′ = γγ′ if

γ ∈ Π and γ′ ∈ Z by definition, Π is an additive subgroup of Z and Π\Z ∼= E(C),where the quotient is made through the group action.

Now we define, choosing a C∞–path [γ] in each class of γ ∈ Z and a nowherevanishing differential form ω on E, a map I : Z → C by γ 7→

∫[γ]ω ∈ C. Since ω is

holomorphic on Z, the value of I is independent of the choice of the representative[γ] by Cauchy’s integration theorem. Since ω is translation invariant on E(C), itis translation invariant on Z and I(γ + γ′) = I(γ) + I(γ′). In particular, I is alocal homeomorphism because E(C) is one dimensional and for simply connected U ,Z(U) ∼= I(U). The pair (E(C), ω) is isomorphic locally to the pair of the additivegroup C and du for the coordinate u on C, because du is the unique translationinvariant differential (up to constant multiple). Since I−1([0]) = 0, I is a linearisomorphism into C. For an open neighborhood U of 0 with U ∼= Z(U) 3 γ 7→I(γ) =

∫γω ∈ C giving an isomorphism onto a small open disk D in C centered at 0,

we have two γ1, γ2 ∈ U giving rise to a two R-linearly independent I(γj) (j = 1, 2).Then I(mγ1 + nγ2) = mI(γ1) + nI(γ2) for all m,n ∈ Z. Replacing γj by 1

aγj ∈ Z(U)


such that I( 1aγj) =

I(γj)

afor any positive integer a, by the same argument, we find

I(mγ1 + nγ2) = mI(γ1) + nI(γ2) for all m,n ∈ Q; so, I is a surjective isomorphism.This also shows that if α : E → E is an endomorphism of E with α(0E) = 0E, α

lifts an endomorphism of Z sending a path γ from 0E to z ∈ C to a path α(γ) fromα(0E) = 0E to α(z). In particular, α(γ + γ′) = α(γ)+α(γ′). Thus α induces a linearmap from C = Z to C. Since α is holomorphic (as it is a polynomial map of thecoordinates of P2

/C), α is a C-linear map. We thus get a natural inclusion:

(3.1) End(E/C) → C.

Writing L = LE for I(Π), we can find a base w1, w2 of L over Z. Thus we have amap

P(C) 3 (E, ω) 7−→ LE ∈ L|L : lattice in C = Lat,

and we have (E(C), ω) ∼= (C/LE , du). Therefore the map: P(C) → Lat is injective.We show its surjectivity in the next subsection.

By the above fact combined with (3.1), we get

Proposition 3.3. We have a ring embedding End(E/C) → u ∈ C|u · LE ⊂ LE,and hence End(E/C) is either Z or an order of an imaginary quadratic field.

Proof. The first assertion follows from (3.1). Pick α ∈ End(E/C) corresponding u ∈ C

as above. Note that LE = Zw1+Zw2. Then uw1 = aw1+bw2 and uw2 = cw1+dw2 forintegers a, b, c, d. In short, writing w = ( w1

w2) and ρ(α) = ( a b

c d ), we get uw = ρ(α)w;so, ρ : End(E/C)→ M2(Z) is a ring homomorphism. By the first assertion, the imagehas to be an order of imaginary quadratic field or just Z.

When End(E/C) 6= Z, E is said to have complex multiplication.

3.3. Classical Weierstrass ℘-function. For a given L ∈ Lat, we define the Weier-strass ℘–functions by

xL(u) = ℘(u) =1

u2+

∑

`∈L−0

1

(u− `)2− 1

`2

=

1

u2+g2

20u2 +

g3

28u4 + · · ·

yL(u) = ℘′(u) = − 2

u3− 2

∑

`∈L−0

1

(u− `)3= −2u−3 + · · · ,

where

g2 = g2(L) = 60∑

`∈L−0

1

`4and g3 = g3(L) = 140

∑

`∈L−0

1

`6.

Then ϕ = y2L−4x3

L+g2xL+g3 is holomorphic everywhere. Since these functions factorsthrough the compact space C/L, ϕ has to be constant, because any non-constantholomorphic function is an open map (the existence of power series expansion andthe implicit function theorem). Since xL and yL do not have constant terms, weconclude ϕ = 0.

We have obtained a holomorphic map (xL, yL) : C/L−0 → A2/C

. Looking at theorder of poles at 0, we know the above map is of degree 1, that is, an isomorphismonto its image and extends to

Φ = (xL : yL : 1) = (u3xL : u3yL : u3) : C/L→ P2/C.


Thus we have an elliptic curve EL = Φ(C/L) = E(g2(L), g3(L)). We then have

ωL =dxL

yL= du.

This shows

Theorem 3.4. (Weierstrass) We have [(E, ω)/C] ∼= Lat.

We would like to make the space Lat a little more explicit. We see easily thatw1, w2 ∈ (C×)2 span a lattice if and only if Im(w1/w2) 6= 0. Let H = z ∈ C| Im(z) >0. By changing the order of w1 and w2 without affecting their lattice, we may assumethat Im(w1/w2) > 0. Thus we have a natural isomorphism of complex manifolds:

B =v = ( w1

w2) ∈ (C×)2

∣∣∣ Im(w1/w2) > 0∼= C× × H via ( w1

w2) 7→ (w2, w1/w2).

Since v and v′ span the same lattice L if and only if v′ = αv for α ∈ SL2(Z),

Lat ∼= SL2(Z)\B.This action of α = ( a b

c d ) ∈ SL2(Z) on B can be interpreted on C× × H as follows:

α(u, z) = (cu+ d, α(z)) for α(z) =az + b

cz + d.

By definition, ℘(u) is an even function. Let L = Zw1 +Zw2 and put w3 = w1 +w2.Put ej := ℘(

wj

2). Then ℘(u)− ej has zero at

wj

2. Since ℘ is even, the order of zero is

even; so, ℘′(u) has zero atwj

2. Therefore, ej are roots of 4x3 − g2x− g3. Comparing

the zeros and poles of ℘′2 and 4(℘− e1)(℘− e2)(℘− e3), we get

(3.2) ℘′2 = 4(℘− e1)(℘− e2)(℘− e3) and ∆ = 16[(e1 − e2)(e2 − e3)(e3 − e1)]2.

Since E(C) is smooth ⇔ ∆ 6= 0, ∆ never vanishes over H.

3.4. Complex Modular Forms. We want to write down definitions of modularforms over C. We consider f ∈ Gw(C). Writing L(v) = L(w1, w2) for the latticespanned by v ∈ B, we can regard f as a holomorphic function on B by f(v) =f(EL(v), ωL(v)). Then the conditions (G0–3) can be interpreted as

(G0) f ∈ C[g2(v), g3(v)];(G1) f(αv) = f(v) for all α ∈ SL2(Z);(G2) f ∈ C[g2(v), g3(v),∆(v)−1];(G3) f(λv) = λ−wf(v) (λ ∈ C×).

We may also regard f ∈ Gw(C) as a function on H by f(z) = f(v(z)) for v(z) =2πi ( z

1 ) (z ∈ H). Here multiplying ( z1 ) by 2πi is to adjust the rationality coming from

q-expansion to the rationality coming from the universal ring Q[g2, g3], as we will seelater (2πi)−jg(( z

1 )) has Fourier expansion in Q[[q]] for q = exp(2πiz). Then we havethe following interpretation:

(G0) f ∈ C[g2(z), g3(z)];(G1,3) f(α(z)) = f(z)(cz + d)w for all α = ( a b

c d ) ∈ SL2(Z);(G2) f ∈ C[g2(z), g3(z),∆(z)−1].


Since ( 1 10 1 ) (z) = z + 1, any f ∈ C[g2(z), g3(z),∆

−1(z)] is translation invariant.Defining e(z) = exp(2πiz) for i =

√−1, the function e : C→ C× induces an analytic

isomorphism: C/Z ∼= C×. Let q = e(z) be the variable on C×. Since f is translationinvariant, f can be considered as a function of q. Thus it has a Laurent expansionf(q) =

∑n−∞ a(n, f)q2. We have the following examples (see the following section

and [LFE] Chapter 5):

12g2 = 1 + 240∞∑

n=1

∑

0<d|nd3

q2 ∈ Z[[q]]×,

−63g3 = 1− 504∞∑

n=1

∑

0<d|nd5

q2 ∈ Z[[q]]×,(3.3)

∆ = q

∞∏

n=1

(1− qn)24 ∈ q(Z[[q]]×).

This shows that

J =(12g2)

3

∆= q−1 + · · · ∈ q−1(1 + Z[[q]]).

In particular, we may regard g2 and g3 as elements of Q[[q]].We consider a projective plane curve E∞/Z[[q]] called the Tate curve defined over the

power series ring Z[[q]] by the equation Y 2Z = 4X3− g2(q)XZ2− g3(q)Z

3 and defineω∞ = dX

Y. Since ∆ is a unit in Z[1/6]((q)) := Z[1/6][[q]][ 1

q], we see that (E∞, ω∞)

gives an elliptic curve over Z[1/6]((q)) with nowhere vanishing differential ω∞. Forany f ∈ Gw(A), f(q) = f((E∞, ω∞)⊗Q((q)) A((q))) ∈ A[[q]] is called the q–expansionof f . In particular, if f ∈ Gw(C), the q–expansion f(q) coincides with the analyticFourier expansion via q = e(z), because f is an isobaric polynomial in g2 and g3 andby definition g2(q) and g3(q) are their analytic expansions.

Write P1(J)/Q for the projective line over Z[ 16] whose coordinate is given by J

(in other words, P1(J) = D0 ∪ D1 over local rings with D1 = A1 defined by theaffine ring Z[ 1

6][J ]). Since the coordinate at ∞ of P1(J) can be given by J−1 (J−1 ∈

q(1 + qZ[[q]])), we know that Z[[q]] = Z[[J−1]] and

(3.4) OP1(J),∞ ∼= Z[1/6][[q]] via q–expansion,

where OP1(J),∞ is the (q)–adic completion of the local ring OP1(J),∞ at ∞.Since we have

M1(C) = Lat/C× = H× C×/(SL2(Z)× C×) ∼= SL2(Z)\H,which is isomorphic to P1(J)−∞ by J . Thus we see that (G0) over C is equivalentto

(G0′) f is a holomorphic function on H satisfying the automorphic property (G1,3),and its analytic q–expansion f(q) is contained in C[[q]].

More generally, for modular forms f ∈ Gw(A), we can interpret (G0) as

(G0′′) f : P/A → A1/A is a morphism of functors satisfying the automorphic property

(G3) in §3.1, and its algebraic q–expansion f(E∞, ω∞) is contained in A[[q]].


3.5. Weierstarss ζ and σ functions. Pick a lattice L of C; so, for E(C) = C/L,L = π1(E(C)), and put L′ := ` ∈ L|` 6= 0. We define Weierstrass σ-function by thefollowing infinite product:

(σ) σ(u) = σ(u;L) = u∏

`∈L′

(1− u

`

)exp((u/`) +

1

2(u/`)2).

Plainly σ(λu;λL) = λσ(u;L) (homogeneous of degree 1). Taking the logarithmicderivative of σ formally, we get Weierstrass ζ-function given by the following infinitesum:

(ζ) ζ(u) = ζW (u) =1

u+

∑

`∈L′

[1

u− ` +1

`+u

`2

],

which converges absolutely and locally uniformly outside L as the denominator of theterm is of degree 2 in `. Therefore σ also converges. By definition,

ζ ′(u) =dζ

du(u) = −℘(u) and ζ(λu;λl) =

1

λζ(u;L) (homogeneous of degree −1).

Since ℘-function is periodic, we find ζ(u+ `) = ζ(u) + η(`) for a linear map η = ηW :L → C. If L = Zw1 + Zw2 with Im(w1/w2) > 0, we define ηj := η(wj) and extendη to a R-linear map from C into C by η(a1w1 + a2w2) := a1η1 + a2η2. Also we caneasily check

σ(−u) = −σ(u) and ζ(−u) = −ζ(u).

Proposition 3.5 (Legendre relation). We have η2w1 − η1w2 = 2πi.

Proof. Take the fundamental parallelogram P with 4 vertices α, α+wj and α+w1+w2

so that L∩P = 0. Write the path connecting α, α+w1 as γ. Then γ+w2 connectα + w2 and α + w1 + w2. Similarly, writing δ for the path connecting α to α + w2,then δ + w1 connects α+ w1 and α+ w1 + w2. On the one hand, we have

∫

∂P

ζ(u)du =

∫

γ

ζ(u)du−∫

γ

ζ(u+ w2)du+

∫

δ

ζ(u+ w1)du−∫

δ

ζ(u)du

=

∫

γ

ζ(u)du−∫

γ

ζ(u)du+ η2

∫

γ

du+

∫

δ

ζ(u)du+ η1

∫

δ

du−∫

δ

ζ(u)du

= η2w1 − η1w2

On the other havd, ζ(u) has pole of residue 1 at 0 and no other pole in P ; so,∫

∂P

ζ(u)du = 2πi · Resu=0ζ = 2πi

as desired.

Theorem 3.6. For a ∈ C not in L, we have

℘(u)− ℘(a) = −σ(u+ a)σ(u− a)σ2(u)σ2(a)

.


Proof. We may assume that a in the parallelogram P . The function ℘(u)− ℘(a) haszeros at a and −a (as ℘ is an even function) and has a double pole at 0. The product

expansion of σ tells us that the same holds for σ(u+a)σ(u−a)σ2(u)

; so,

℘(u)− ℘(a) = Cσ(u+ a)σ(u− a)

σ2(u)

for a constant C . We see easily

limu→0

σ2(u)/u2 = 1 and limu→0

u2℘(u) = 1.

So C = −1/σ2(a).

A theta function for E is an entire function θ satisfying the following functionalequation”

(θ) θ(u+ `) = θ(z) exp(2πi[l(u, `) + c(`)]),

where l is C-linear in u and R-linear in ` and a function c : L→ C. There is a trivialtheta function φ(u) = exp(au2 + bu) as

φ(u+ `) = exp(a(u+ `)2 + b(u+ `)) = φ(u) exp(2au`+ c(`))

for c(`) := a`2 + b`.

Theorem 3.7. The σ-function is a theta function; i.e.,

σ(u+ `) = ψ(`) exp(η(`)(u+ `/2))σ(u),

where ψ(`) =

1 if ` ∈ 2L,

−1 if ` 6∈ 2L.

Proof. We haved

dulog

σ(u+ `)

σ(u)= η(`)

by definition. By the fundamental theorem of Calculus, we get

logσ(u+ `)

σ(u)= η(`)u+ c(`).

Exponentiating, we get

σ(u+ `) = σ(u) exp(η(`)u+ c(`)).

Here c mod (2πi) is well defined.Define ψ(`) := exp(η(`)u+ c(`))/ exp(η(`)(u+ `/2)), and we compute ψ. Suppose

` 6∈ 2L. Set u = −`/2. Then σ(`/2) = ψ(`)σ(−`/2); so, ψ(`) = −1 as σ is an oddfunction and σ(`/2) 6= 0 by ` 6∈ 2L.

We see

ψ(2`) exp(η(2`)(u+ `)) =σ(u+ 2`)

σ(u)=σ(u+ 2`)

σ(u+ `)

σ(u+ `)

σ(u)

= ψ(`)2 exp(η(`)(u+3

2`) + η(`)(u+

1

2`)).


In other words, we have ψ(`) = ψ(`/2)2. Iterating this to reach `/2n 6∈ L first time,we get ψ(`) = (−1)2n = 1.

3.6. Product q-expansion. Hereafter we take L = Z + Zz for z ∈ H. We the write

(u;L) as (u, z). Define ϕ(z) = ϕ(u, z) := exp(−12η2u

2)q1/2u σ(u; z), where η2 = η(1) for

η = ηW : L→ C given by η(`) = ζ(u+ `)− ζ(u) and qu = exp(2πiu). Then we see

(3.5) ϕ(u+ 1) = ϕ(u) and ϕ(u+ z) = − 1

quϕ(u).

By Theorem 3.7, we find

ϕ(u+ 1) = exp(−1

2η2(u+ 1)2 + πi(u+ 1))σ(u+ 1)

= − exp(−1

2η2(u+ 1)2 + πi(u+ 1) + η2(u+

1

2))σ(u) = ϕ(u).

Similarly for η1 = η(z)

ϕ(u+ z) = exp(−1

2η2(u+ z)2) + πi(u+ z))σ(u+ τ )

= − exp(−1

2η2(u+ z)2) + πi(u+ z) + η1(u+ z/2))σ(u).

By Legendre’s relation: η2z − η1 = 2πi, we can eliminate η1 and get (3.5). Indeed,we have

− 1

2η2(u+ z)2) + πi(u+ z) + η1(u+ z/2)

= −1

2η2u

2 − η2uz −1

2η2z

2 + πi(u+ z) + (η2z − 2πi)u+(η2z − 2πi)z

2

= −1

2η2u

2 + πi(u+ z)− 2πiu− πiz = −1

2η2u

2 + πiu− 2πiu

as desired.

Theorem 3.8. Let qw = exp(2πiw) for w = u, z. Then we have

σ(u, z) = (2πi)−1 exp(1

2η2z

2)(q1/2u − q−1/2

u )

∞∏

n=1

(1− qnz qu)(1− qn

z /qu)

(1− qnz )2

.

Proof. We prove the equivalent form of the product formula of ϕ(u, z) = g(u):

g(u) := (2πi)−1(qu − 1)

∞∏

n=1

(1− qnz qu)(1− qn

z /qu)

(1− qnz )2

.

By definition, it is easy to see g(u + 1) = g(u) and g(u + z) = − 1qug(u) and that g

has exactly the same zeros of order 1 at u = 0 as σ (and hence as ϕ). Thus ϕ/g isan entire function on E(C) = C/L, and hence ϕ = C · g for a constant C . The wecompute C by C = limu→0 ϕ(u)/g(u) = 1.


Corollary 3.9. We have

∆(z) = (2πi)12qz

∞∏

n=1

(1− qnz )24.

This can be proven by the above theorem because ∆ = 16[(e1−e2)(e2−e3)(e3−e1)]2

by (3.2) with ej = ℘(wj

2) and invoking Theorem 3.6

ei − ej = ℘(wi/2) − ℘(wj/2) = −σ((wi + wj)/2)σ((wi − wj)/2)

σ2(wi/2)σ2(wj/2).

Define the Dedekind η-function (as a 24-th root of ∆ by

(η) η(z) = ηD(z) = q1/24z

∞∏

n=1

(1− qnz ).

Theorem 3.10 (R. Dedekind). We have ηD(z+1) = ηD(z) and ηD(−1/z) =√−izηD(z),

where the square root√−iz has positive value on the imaginary axis in H.

Proof. The first formula follows from the definition. Since ∆ is on SL2(Z) of weight

12, ηD(−1/z)√zηD(z)

is holomorphic on H with∣∣∣ηD(−1/z)√

zηD(z)

∣∣∣ = 1. By maximum principle, it must

be a constant C . Putting z = i, we have 1 = C√i; so, C =

√−i.

3.7. Klein forms. We modify the Weierstrass σ-function into the so-called Kleinform which is a modular form of weight 1. Let L = L(v) = Zw1 + Zw2 for v =

(w1

w2

)

and η = ηW be Weierstrass eta function given by ηW (`;L) = ζ(u+ `;L) − ζ(`;L) for` ∈ L ∈ Lat (not the Dedekind eta function). Extend ηW : L→ C linearly to the Q

span Q · L.

Definition 3.1. For a = (a1, a2) ∈ Q2−Z2 (a row vector), define ka(v) = ka(L(v)) :=exp(−ηW (a · v;L(v))a · v/2)σ(a · v;L(v)), where a · v = a1w1 +a2w2 (matrix product).

Since σ(λu, λL) = λσ(u, L) (homogeneous of weight−1) and ζ(λu, λL) = λ−1ζ(u, L)(homogeneous of weight 1), we have

(3.6) ka(λv) = λka(v).

Theorem 3.11. Let a ∈ 1N

Z2 but a 6∈ Z2. Then ka is a meromorphic modular formof weight −1 on Γ(2N2) (whose poles concentrated at cusps), and k2N

a is on Γ(N) andif N is odd, kN

a is on Γ(N).

Proof. Since ka(v) = ka(L(v)) := exp(−ηW (a · v;L(v))a · v/2)σ(a · v;L(v)), for α ∈SL2(Z), we have

kaα(v) = exp(−ηW (aα · v;L(v))aα · v/2)σ(aα · v;L(v))

= exp(−ηW (a · αv;L(v))a · αv/2)σ(a · αv;L(v)).

In short

(3.7) kaα(v) = ka(αv).

We are going to show for b = (b1, b2) ∈ Z2

(3.8) ka+b(v) = ε(a, b)ka(v) for ε(a, b) = (−1)b1b2+b1+b2e((a1b2 − a2b1)/2),


where e(z) = exp(2πiz). By (3.7) combined with (3.8), as long as ε(a,mZ2)j = 1for 0 < m ∈ Z, we find that kj

a is on Γ(mN) as Γ(mN) induces an identity on(N−1Z/mZ)2.

By Theorem 3.7, we know

σ(u+ `) = ψ(`) exp(η(`)(u+ `/2))σ(u)

with ψ(`) =

1 if ` ∈ 2L,

−1 if ` 6∈ 2L.We can then write ψ(b1w1 + b2w2) = (−1)b1b2+b1+b2 for

b = (b1, b2) ∈ Z2. Take ` = b · v ∈ L(v). Then

ka+`(v) = exp(−η((a+ b) · v)(a+ b) · v/2)σ((a+ `) · v)= ψ(`) exp(−η((a + b) · v)(a+ b) · v/2) exp(η(b · v)(a · v + b · v/2))σ(a · v)

= ψ(`) exp(−η((a+ b) · v)(a+ b) · v/2 + η(b · v)(2a+ b) · v/2 + η(a · v)a · v/2)ka(v)

The inside of the exponential function is

− η((a + b) · v)(a+ b) · v/2 + η(b · v)(2a+ b) · v/2 + η(a · v)a · v/2

=1

2(−a2b1(η2w1 − η1w2) + b2a1(η2w1 − η1w2))

(∗)= πi(a1b2 − a2b1).

Here the identity at (∗) is by Legendre’s relation η2w1 − η1w2 = 2πi. Thus (3.8)follows.

Thus if m = N2, we have ka+`(v) = ψ(`)ka(v) and hence, if m = 2N2, ka dependsonly on a ∈ N−1Z2/2NZ2 as desired. Since ε(a, b) is a 2N -th root of unity, k2N

a onlydepends on a ∈ (N−1Z/Z)2; so, it is on Γ(N) in the stabilizer of a ∈ (N−1Z/Z)2.

To deal with kNa′ for odd N , we need to be more careful. Let α = ( a b

c d ) ∈ Γ(N).Write a′ = (r, s)/N . Then a′α = ( r

N+ (a−1

Nr + c

Ns), s

N+ ( b

Nb+ d−1

Ns)). Therefore

(3.9) ka′(αv) = ka′α(v) = ka′+b′(v) = εa′(α)ka′(v),

where from (3.8), for Na′ = (r, s),

εa′(α) = −(−1)( a−1

Nr+ c

Ns+1)( b

Nr+d−1

Ns+1)e((br2 + (d − a)rs− cs2)/2N2)

because b′ = (a−1Nr + c

Ns, b

Nb+ d−1

Ns).

We now need to show

(3.10) εa′(α)N = 1 if N is odd.

Equivalently, as N ≡ 1 mod 2,

(3.11) ((a− 1)r + cs+ 1)(br + (d− 1)s+ 1) + (br2 + (d− a)rs− cs2)) ≡ 1 mod 2.

If r, s ∈ 2Z, it is plain. Suppose (r, s) ≡ (1, 0) mod 2. Then (3.11) is equivalent to

a(b+ 1) + b ≡ 1 mod 2.

If b is odd, this is plain. By ad − bc = 1, if b is even, then a is odd; so, the resultfollows. In the same way, we can settle the case where (r, s) ≡ (0, 1) mod 2.

Now suppose (r, s) ≡ (1, 1) mod 2. Then (3.11) is equivalent to

(a+ c)(b+ d) + a+ b+ c+ d ≡ 1 mod 2.


Again if b is even, then a and d are odd, and hence

(a + c)(b+ d) + a+ b+ c+ d ≡ (1 + c) + (1 + c) + 1 ≡ 1 mod 2.

We can settle the case where c even in the same way. If bc ≡ 1 mod 2, then a or dis even. Supposing that a is even, again we have

(a+ c)(b+ d) + a + b+ c+ d ≡ 1 + d + 1 + 1 + d ≡ 1 mod 2.

We settle the case where d even in the same way. This finishes the proof.

Let p ≥ 5 be a prime and put N = pm for m > 0. Form : (N−1Z/Z)2−(0, 0 → Z,we say m satisfies (QN) if

(QN)∑

a 6=0

m(a)a21 ≡

∑

a 6=0

m(a)a22 ≡

∑

a 6=0

m(a)a1a2 ≡ 0 mod N−1Z,

which is equivalent to, writing a1 = rN

, a2 = sN

and m(a) = m(r, s),

(Q′N)

∑

(r,s) 6=0

m(r, s)r2 ≡∑

(r,s) 6=0

m(r, s)s2 ≡∑

(r,s) 6=0

m(r, s)rs ≡ 0 mod N,

Remark 3.12. If m(a) = 1 for all a, then

∑

a

m(a)a21 = N−2

N−1∑

r=0

N−1∑

s=0

r2 = N−1N−1∑

r=0

r2 =(N − 1)(2N − 1)

6∈ Z.

Similarly

∑

a

m(a)a1a2 = N−2

N−1∑

r=0

N−1∑

s=0

rs = N−2(N−1∑

r=0

r)2 = N−2N2(N − 1)2

4∈ Z.

Thus a constant function m satisfies (QN).

We like to prove

Theorem 3.13 (D. Kubert). Let m is as above. Then km =∏

a 6=0 km(a)a is on Γ(N)

if and only if m satisfies (QN ).

We give a sketch of the proof done by D. Kubert.

Proof. Let e(z) = exp(2πiz) and define εa(α) by kaα(v) = εa(α)ka for α = ( a bc d ) ∈

Γ(N). By (3.8), we have ε(a, b) = (−1)b1b2+b1+b2 exp(πi(a1b2− a2b1)). Replacing b by(a−1

Nr + c

Ns, b

Nb+ d−1

Ns) as in (3.9) and writing a = ( r

N, s

N),

εa(α) = −(−1)( a−1

Nr+ c

Ns+1)( b

Nr+d−1

Ns+1)e((br2 + (d− a)rs− cs2)/2N2).

Then we need to compute εm(α) =∏

a εa(α)m(a). Define, writing m(r, s) = m( rN, s

N),

(3.12) E(r, s; a, b, c, d) := m(r, s)[ abN2

r2 +c(d− 1)− c

N2s2 +

(b

N+a− 1

N

)r

+

(d− 1

N+

c

N

)s+

(bc

N2+

(a− 1)(d − 1)

N2

)rs+

d− aN2

rs].


In the definition of this formula in [MUN, page 69], the term d−aN2 rs is wrongly written

as d−aNrs. Then we have, for Z := (r, s) ∈ Z2 ∩ [0, N)2|(r, s) 6= (0, 0),

εm(α) = exp(πi∑

(r,s)∈Z

E(r, s; a, b, c, d)).

From this, taking α = ( 1 N0 1 ), we have E(r, s; 1, N, 0, 1) = m(r, s)( 1

Nr2 + r) and hence

∑

(r,s)

m(r, s)(rN + r2) ≡ 0 mod 2N.

Similarly, taking α = ( 1 0N 1 ), we conclude∑

(r,s)

m(r, s)(sN + s2) ≡ 0 mod 2N.

However from (3.10), we can replace the modulus 2N by N . This implies the firsttwo identities of (QN).

As for the third term involving rs, again we only need to compute modulo N inplace of 2N . We take α =

(1−N N−N 1+N

). Since we know the first two identities of (QN),

we can ignore the square terms in r, s and also terms only having denominator N of(3.12). Then

E(r, s; 1−N,N,−N, 1 +N) ≡ m(r, s)

(bc

N2+

(a− 1)(d− 1)

N2+d− aN2

)rs

≡ m(r, s)2

Nrs mod Z.

Thus the last identity of (QN) holds if km is on Γ(N).Conversely, we already know from (3.10) that ka is on Γ(N2) if N = pm with p ≥ 5.

Writing γ ∈ Γ(N) as γ = 1 + Nγ1 and sending γ1 to (γ1 mod N) ∈ M2(Z/NZ), wecan easily verify that i : Γ(N)/Γ(N2) ∼= sl2(Z/NZ) = γ1 ∈M2(Z/NZ)|Tr(γ1) = 0.This is an isomorphism of groups regarding sl2(Z/NZ) as an additive group. Thenit is easy to see that i ( 1 N

0 1 ), i ( 1 0N 1 ) and i

(1−N N−N 1+N

)generate the right-hand-side.

Therefore the relation (QN) is sufficient for km is on Γ(N).

4. Modular units

The units A×N and A×

1,N of affine ring of the modular curves Y1(N) and Y (N) arecalled modular units. They are holomorphic and does not vanishes on H, and indeedoften they have product expansion similar to the ∆-function. The compactificationP1(J) of A1 = Q[J ] has one cusp ∞. Since A = A1,N and AN are finite flat overY (1) = Spec(A1), the normalization of P1(J) in A gives a projective curve X finiteflat over P1(J). The curve X for A1,N (resp. AN ) is denoted by X1(N) (resp. X(N)).Points above ∞ is called cusps of X. The cuspidal divisor group of X is defined tobe

ClX :=D ∈⊕

P :cusps Z[P ]| deg(D) = 0div(u)|u: modular units .


We study this cuspidal divisor group and prove its finiteness in this section and aformula for its order in the following section. Moreover we see it is a cyclic moduleover a suitable group algebra.

4.1. Siegel units. We now introduce a typical modular unit called Siegel units whoseorder at the infinity cusp is essentially given by the rational values of the secondBernoulli polynomial B2(x) = x2− x+ 1

6(so, the values of Hurwitz zeta functions at

s = −1).

Definition 4.1. We define the Siegel unit ga(z) for a = (a1, a2) ∈ Q2 ∩ [0, 1)2 by

ga(z) = ka(z)∆(z)1/12,

where ∆1/12(z) = 2πiq1/12∏∞

n=1(1 − qn)2 for q = exp(2πiz). For a general a ∈ Q2,taking the fraction part 〈a〉 := (〈a1〉, 〈a2〉) we define ga := g〈a〉. Here x−〈x〉 ∈ Z with〈x〉 ∈ [0, 1) for x ∈ R.

Note that ∆1/12(z) is the square of the Dedekind η-function ηD up to constant.

Proposition 4.1. We have

(4.1) ga(z) = −qB2(a1)/2 exp(2πia2(a1 − 1)/2)(1 − exp(2πia2)qa1)

×∞∏

n=1

[(1− exp(2πia2)qn+a1)(1− exp(−2πia2)q

n−a1)],

where q = exp(2πiz).

Proof. This is just a computation out of the product expansion of the σ-function(combined with definition of ka) and ∆1/12. An important point later is the leadingterm qB2(a1)/2; so, we describe the computation for the leading term. Here is thecontribution of each of σ, ka and ∆−1/12 to the leading term:

(σ) exp(12η2(a1z + a2)

2);(k) exp(−(η1a1 + η2a2)(a1z + a2)/2);

(∆) q1/12 = exp(2πiz12

).

The product of the terms (σ), (k) and (∆) has inside exp the following:

1

2η2(a1z + a2)

2 − (η1a1 + η2a2)(a1z + a2)/2 +2πiz

12

=1

2(η1a

21z − η1a1a2 − η2a2a1z − η2a

22 + η2a

21z

2 + 2η2a1a2z + η2a22) +

2πiz

12

=1

2a2

1(−η1 + η2z) +a1a2

2(−η1 + η2z) +

2πiz

12

(∗)= 2πi(

a21

2+a1a2

2) +

2πiz

12= 2πi(B2(a1)/2 + a2(a1 − 1)/2).

The identity (∗) follows from Legendre’s relation: η2z − η1 = 2πi.

By the analysis of Klein forms and the fact that ∆ ∈ S12(SL2(Z)), we get

Theorem 4.2. Suppose that 0 6= a ∈ (N−1Z/Z)2. Then g12Na is a modular function

on Γ(N), and it does not have poles and zeros on H. Further for α ∈ SL2(Z), wehave g12N

aα (z) = g12Na (α(z)); therefore, gaα(z) = ζa,αga(α(z)) for ζa,α ∈ µ12N(Q).


Note that gaα/g α is holomorphic everywhere over X(2N2)(C) by Theorem 3.11;so, it is a non-zero constant. Thus the ratio is a 12N -th root of unity.

Remark 4.3. We see that g12Na generate the function field Q(Y (N)) over Q as it

is generated by ∆−1/12℘(a · v;L(v)) for a running over (N−1Z/Z)2 (essentially loga-rithmic derivative of ga) basically by definition. Since its q-expansion involves N-throots of unity, Q(Y (N)) becomes reducible over Q[µN ] (see [EFN, Chapter 7] to seeShimura’s proof of the fact that each geometrically irreducible component of Y (N) isdefined over Q[µN ]).

Proposition 4.4. We have∏

Na=0,a 6=0 g12Na ∈ Q×, where a runs over N−1Z2∩ [0, 1)2.

Proof. Since the product ε :=∏

Na=0,a 6=0 g12Na is invariant under Gal(X(N)/P1(J)) =

Gal(Q(Y (N))/Q(J)), we find that ε ∈ Q[J ]×; so, ε is a constant in Q×.

Remark 4.5. It is known that∏

Na=0,a 6=0 g12Na = N12N (see [MUN, Chapter 2, §4]).

Theorem 4.6. Let ZN := a ∈ N−1Z2∩ [0, 1)2|a 6= (0, 0) for N = pm (with a prime

p ≥ 5), and put gm :=∏

a∈ZNg

m(a)a for a function m : ZN → Z. Then gm ∈ A×

N ifand only if

∑a∈ZN

m(a) ≡ 0 mod 12 and (QN) is satisfied.

Proof. By Theorem 3.13, km is on Γ(N) if and only if (QN) is satisfied. Let M :=∑a∈ZN

m(a). Thus we need to show ∆M/12 is on Γ(N) if and only if M ≡ 0 mod 12.

Since p ≥ 5, we need to prove that f := ∆1/2 is on Γ(6) but not on SL2(Z). Forα = ( a b

c d ) ∈ SL2(Z), we have f |α(z) = f(α(z))(cz + d)−6 = χ(α)f(z). Indeed, byq=expansion, we have χ ( 1 1

0 1 ) = −1 and by Theorem 3.10, χ ( 0 1−1 0 ) = −1. Since ∆ is

on SL2(Z) and the above two elements generate SL2(Z), χ is a character of SL2(Z)of order 2. By the strong approximation theorem (e.g., [LFE, §6.1]), the abelianquotient of SL2(Z) is isomorphic to the abelian quotient of SL2(Z/6Z). Therefore fis on Γ(6) as desired.

Actually, for η = ηD, we know η2 = ∆1/12 is on Γ(12), η4 is on Γ(3), ∆1/2 is onΓ(2) (see [MUN, §3.5]).

4.2. Distribution on p-divisible groups. We seek to prove that A×N for the affine

ring AN of X = Y (N) is generated by Siegel units. This is to show

ClX :=D ∈

⊕P :cusps ZP | deg(D) = 0

〈div(ga)|a ∈ (N−1Z/Z)2〉 ,

where the denominator is the Z-span of div(ga). Hereafter we assume that N = pm

for a prime p ≥ 5 for simplicity. See [MUN] for a general theory. To show linearindependence of div(ga)a∈(N−1Z/Z)2, we recall a theory of distributions.

Let W be the unique discrete valuation ring unramified over Zp of rank 2 (theWitt vector ring with coefficients in Fp2). Then the field K of fractions of W is theunique unramified quadratic extension of Qp generated by (p2 − 1)-st roots of unity.Let V = Zp or W and put Q = V ⊗Zp Qp (the field of fractions of V ). We sayµ : Q/V =

⋃n p

−nV/V → A for an abelian group A with identity is a distribution of


weight k ≥ 0 if µ : Q/V → A is a function satisfying

pk(m−n)∑

pm−ny=x

µ(y) = µ(x)

for all m > n ≥ 0. If we modify µ into µpn(a) = pknµ( apn ) for a ∈ V/pnV , this

is equivalent to the usual distribution relation∑

a≡b mod pn µpm(a) = µpn(b) for alla ∈ V/pmV and b ∈ V/pnV .

We restrict µ to p−mV/V and write it as µ|m if we need to indicate its order pm. Fora distribution µ, rank(µ|m) = dimQ Im(µ|m)⊗Z Q. A distributionM : p−mV/V →Ais called universal if for any distribution µ : p−mV/V → A, there exists a uniquehomomorphism φ : A → A such that µ = φ M. Universal distribution is uniqueup to isomorphism if exists. Indeed A =

⊕a∈p−mV/V Z(a)/R (the free abelian group

generated by symbol (a) indexed by a ∈ p−nV/V and R is submodule generated by(x) − pk(m−n)

∑pm−ny=x(y) for all possible x ∈ p−nV/V ). PlainlyM(x) = (x) is the

universal distribution.

4.3. Stickelberger distribution. Let Cm := (V/pmV )× and consider its group ringA[Cm] (assuming A is a ring). For a ∈ Cm, we write σa for the group element inQ[Cm] corresponding to a ∈ Cm. We define Stµ = Stµ|m : p−mV/V → A[Cm] byStµ(x) :=

∑a∈Cm

µ(ax)σ−1a , which is obviously a distribution. Let Bn(x) be n-th

Bernoulli polynomial; so,tetx

et − 1=

∞∑

n=1

Bn(x)xn

n!.

By definition, Bn(1 − x) = (−1)nBn(x). For example, B1(x) = x − 12

and B2(x) =x2 − x + 1

6, . . . . Fix 0 < n ∈ Z. Let V = Zp and define β(x) = pn−1Bn(〈x〉) for

x ∈ p−mZ/Z with representative 0 ≤ 〈x〉 < 1. Here is an obvious way of creating adistribution:

Lemma 4.7. Let V = Zp, and fix 0 < n ∈ Z. Then β is a distribution with valuesin Q.

Proof. Define Hurwitz zeta function by

ζ(s, x) =

∞∑

n=0

1

(x+ n)s(Re(s) > 1, 0 < x ≤ 1).

This function can be analytically continued to s ∈ C and holomorphic outside s = 1and known that ζ(1−n, x) = −Bn(x)

n(0 < n ∈ Z) for the Bernoulli polynomial Bn(x)

(cf. [LFE, §2.3]). By definition,

f−sζ(s,a

f) =

∞∑

n≡a mod f,n>0

1

ns.

Thus ∑

a∈(p−m′Z/Z)×,pm′−ma=b

p−m′sζ(s,a

pm′) = p−msζ(s,

b

pm).

Thus β is a distribution of weight n− 1.


For a function f : Z/pmZ→ C, we define

ζ(s, f) =∞∑

n=1

f(n)n−s =∑

a∈p−mZ/Z

f(pma)p−msζ(s, 〈a〉).

We have the following functional equation of Hurwitz zeta function (e.g. [LFE, §2.3]):

(4.2) ζ(s, f) =1

2πi

(2π

pm

)s

Γ(1− s)[ζ(1− s, f−)eπis/2 − ζ(1− s, f)e−πis/2,

where f(x) = f(−x) and f is the Fourier transform over Z/pmZ given by

f(a) =∑

b∈Z/pmZ

f(b) exp(−2πi〈p−mab〉).

We consider the trace map Tr : W → Zp, which induces a surjection Tr : K/W →Qp/Zp, where K = W ⊗Zp Qp. Then β Tr is a distribution defined on K/W . Wethen have Stickelberger distribution Stβ|m and StβTr|m with values in Q[Cm].

4.4. Rank of distribution. For a distribution µ|m : Cm → A, we write 〈µ|m〉 bethe submodule of A generated by the values of µ|m. Then we define rankµ|m =dimQ〈µ|m〉 ⊗Z Q. By distribution relation, the value at an element x of order pn 6= 0is the sum of values at y with py = x. Thus we can modify the value at 0 to be zerotaking off µ(0)

∑a σa from Stµ. This does not affect distribution relation and the

new modified distribution will be denoted as St0µ. This modified distribution satisfiesdeg(St0µ|m) = 0 for the degree map of the group algebra A[Cm]. Since By definition,

Bn(1− x) = (−1)nBn(x); so, Bn(〈x〉) = (−1)nBn(〈x〉). Thus St0βTr factors throughCm/±1 if n is even.

Theorem 4.8. Let k be the weight of β. For a prime p, if pm > 3, we have

rank St0βTr|m =

|(W/pmW )×/±1| − 1 if (−1)k+1 = 1,

|(W/pmW )×/±1| if (−1)k+1 = −1.

Proof. Write Stm = StβTr|m simply. Since

Stm(x) =∑

a∈Cm

β(Tr(ax))σ−1a ,

we have St(x) · σb =∑

a∈Cmβ(Tr(ax))σ−1

ab−1 =∑

a∈Cmβ(Tr(abx))σ−1

a = Stm(bx).Thus the module 〈Stm〉 is stable under the multiplication by σb; so, to prove thatrankStm = |Cm| − 1 or |Cm|, we need to show the χ-eigenspace of 〈Stm〉⊗Z Q is non-trivial for all characters χ 6= 1 of Cm with χ(−1) = (−1)n. We therefore compute

∑

a∈p−mW/W

β(〈Tr(a)〉)χ(pma) =∑

a∈Cm

β(〈Tr(a/pm)〉)χ(a).

By distribution relation, 〈Stm〉 ⊃ 〈Stm′〉 for all m′ < m, we only need to do this forprimitive characters modulo pm. Then∑

a∈Cm

β(〈Tr(a/pm)〉)χ(Tr(a)) =∑

a∈Cm

p−msχ(Tr(a))ζ(s, 〈Tr(a/pm)〉)|s=1−n = ζ(1−n, fχ),


where fχ(a) =∑

x∈Cm,Tr(x)=a χ(x) for a ∈ Z/pmZ. We compute the Fourier transformof fχ:

fχ(a) =

N∑

b=1

fχ(b)e−2πib〈a/pm〉 =

N∑

b=1

∑

x∈Cm,Tr(x)=b

χ(x)e−2πi〈ba/pm〉

=∑

x∈Cm

χ(x)e−2πi〈Tr(xa/pm)〉 = G(χ)χ(−a)

for the Gauss sum G(χ) =∑

x∈Cmχ(x)e2πi〈Tr(xp−m)〉 6= 0 (as χ is primitive). Thus by

the functional equation (4.2), we get from −nNn−1ζ(1− n, 〈a〉) = Bn(〈a〉)

(4.3)∑

a∈Cm

β(〈Tr(a/pm)〉)χ(Tr(a))

=−nNn−1

2πi

(2π

pm

)s

Γ(1− s)[ζ(1− s, f−χ )eπis/2 − ζ(1− s, fχ)e

−πis/2|s=1−n

=−nNn−1

2πi

(2π

pm

)1−n

Γ(n)[eπi(1−n)/2 − χ(−1)e−πi(1−n)/2]G(χ)L(n, χ|Z)

=−nNn−1

2πi

(2π

pm

)1−n

Γ(n)[i1−n − χ(−1)(−i)1−n]G(χ)L(n, χ|Z)

=−nNn−1

2πi

(2π

pm

)1−n

Γ(n)in−1[(−1)n−1 − χ(−1)]G(χ)L(n, χ|Z)

which does not vanish if and only if χ(−1) = (−1)n. This finishes the proof.

4.5. Cusps of X(N). We prove

Theorem 4.9. Let Cm = (W/pmW )× and embed Cm into GL2(Z/pmZ) by its action

on W/pmW ∼= (Z/pmZ)2. Then Cm/±1 acts on the cusp of X(pm) freely andtransitively.

Proof. Let N = pm. The set SN of cusps is one to one onto correspondence withΓ(N)\P1(Q)/±1. Since Y (N) classifies elliptic curves E with level structure φ :(Z/NZ)2 ∼= E[N ], GL2(Z/NZ) acts on Y (N) and X(N) and X(N)/GL2(Z/NZ) =P1(J). Since ±1 ⊂ Aut(E), the action factors through GL2(Z/NZ)/±1. Theaction sends the Siegel unit g12N

a to g12Naα as in Theorem 4.2, and hence as is shown

by Shimura, α ∈ GL2(Z/NZ) in the image α of α ∈ SL2(Z) acts on Q(X(N)) by

f 7→ f α and on Q(X(N)) ∩ Q = Q(µN ), g ∈ GL2(Z/NZ) acts by ζN 7→ ζdet(g)N .

Also ( 1 00 d ) acts on q-expansion of Q(X(N)) through its coefficients by σa : ζN 7→ ζd

N .This follows from the fact that Q(X(N)) = Q(g12N

a ) for ga in Theorem 4.2.Writing one geometrically irreducble component of X(N) as X(N), we there-

fore have Q(X(N)) = Q(µN )(X(N)) anbd X(N) is geometrically irreducible oevrQ(µN ). Thus Gal(X(N)/P1(J)) ∼= GL2(Z/NZ)/±1 canonically. Since P1(J) hasone cusp ∞, the cusp of Γ(N)\H is one to one onto SL2(Z)(∞) ∼= SL2(Z)\P1(Q) ∼=Γ(N)\SL2(Z)/Γ∞ for

Γ∞ =± ( 1 m

0 1 )∣∣m ∈ Z

.


Since Γ(N)\H gives rise to a geometrically irreducible component ofX(N) indexed byGal(Q(µN )/Q). Thus the stabilizer of ∞ in Gal(X(N)/P1(J)) = GL2(Z/NZ)/±1is given by the image M(Z/NZ) of

( 1 u

0 d )∣∣u ∈ Z/NZ, d ∈ (Z/NZ)×

.

We have

SN = GL2(Z/NZ)/M(Z/NZ) ∼=v :=

(a

c

)∈ (Z/NZ)2

∣∣order of v = N

/±1.

Taking a basis τ, 1 of W over Zp, we have for w = a + bτ with a, b ∈ Zp,

w(1, τ ) = (w,wτ ) = (1, τ ) ( a ∗b ∗ ) .

This shows that CmM(Z/NZ) = GL2(Z/NZ)/±1 (the Iwasawa decomposition)and SN

∼= Cm/±1, and the action of Cm on the left is transitive, as desired.

4.6. Finiteness of ClX(N). Note that the parameter at the cusp ∞ ∈ X(N) is q1/N

(as the parameter at the cusp of P1(J) is q). We take a basis of W of the form (12, τ )

with Tr(τ ) = 0. Then Tr(a112

+ a2τ ) = a1. In this way, we identify W ∼= Z2p and

p−mW/W with (p−mZ/Z)2. Then we consider Siegel units ga for a ∈ p−mW/W withTr(a) = a1. Identify Spm with Cm/±1. Thus by Proposition 4.1, noting that theparameter at ∞ of X(N) is given by q1/N , we find

(4.4) div(ga) = N∑

b∈Cm/±1

1

2B2(〈Tr(ab)〉)σ−1

b (∞),

Though this formula is defined for 0 6= a ∈ p−mW/W , we can put g0 := 1 and thenthe a 7→ div(ga) ∈ Div0(X(N)) is a distribution proportional to St := St0βTr|m, where

Div0(X(N)) is the degree 0 divisor group of X(N). Write Div0(SN ) ⊂ Div0(X(N))for the subgroup generated by cusps of X(N). Thus ClX(N) is a surjective image ofDiv0(SN )/〈div(g12pm

a 〉a∈p−mW/W . By Theorem 4.8, Div0(SN )/〈div(g12pm

a 〉a∈p−mW/W isa finite group. Therefore we obtain

Theorem 4.10. If N = pm > 3 for a prime p, we have ClX(N) is finite, andA×

N/Q[µN ]× is non-trivial.

If N ≤ 3, X(N) has genus 0; so, AN = Q[µN ][λ] for a modular function λ. Thisimplies AN = Q[µN ]× and ClX(N) with 1 ≤ N ≤ 3 is trivial.

4.7. Siegel units generate A×pm. We start with a theorem due to Shimura but we

have given a sketch of a proof by using a solution of the moduli problem for thefunctor ℘:

Theorem 4.11. If f ∈ Q(X) for X = X0(N), X1(N) and X(N), for every σ ∈GL2(Z/NZ) = Gal(X(N)/P1(J)), the q-expansion of fσ has bounded denominator;in other word, for some 0 < M ∈ Z, Mfσ(q) has integral q-expansion coefficients.

For a function m : Q2/Z2 → Z with finite support, we define g(m) :=∏

a gm(a)〈a〉 .

Here 〈a〉 = (〈a1〉, 〈a2〉). Since ga+b = cga with c ∈ µ2N for b ∈ Z2 as long asNa ∈ Z2 (see (3.8)), up to constant, for any choice of representatives [a] of a ∈ Q2/Z2,


g(m)/∏

a gm(a)[a] ∈ µ2N(C). Note that End(Q2/Z2) = M2(Z) for Z = lim←−N

Z/NZ =∏l Zl. By Theorems 3.11 and 4.2, ga mod scalars only depends on a mod Z2, we can

think of mσ(a) := m(aσ−1) (modulo scalars). Then σ ∈ GL2(Z) = lim←−NGL2(Z/NZ)

acts (modulo scalars) on g(m) by g(m) 7→ σg(m) := g(mσ).We write g(m) =

∑∞n=n0

anqn/N with an0

6= 0 and put g∗(m) := a−1n0q−n0g(m) which

is determined uniquely for any scalar multiples of g(m). Since

ga = q(1/2)B2(〈a1〉)(1− exp(2πia2)qa1)

∞∏

n=1

(1− exp(2πia2)qn+a1)(1− exp(−2πia2)q

n−a1),

we see

(4.5) g∗a =

(1− e2πia2qa1)

∏∞n=1(1− e2πia2qn+a1)(1− e−2πia2qn−a1) if 〈a1〉 6= 0,∏∞

n=1(1− e2πia2qn+a1)(1− e−2πia2qn−a1) if 〈a1〉 = 0.

An immediate feature from this definition is

Lemma 4.12. We have g∗a α = g∗aα for α ∈ GL2(Z) and g∗a ∈ Z[µN ][[qN]].

Since ka only depends on a ∈ N−1Z2/2NZ2 ∼= Z/2N2Z, ga only depends on a ∈N−1Z/12NZ2 (basically by (3.8)), GL2(Z) (Z =

∏l Zl) acts on ga via its quotient

GL2(Z/12N2) as Aut(N−1Z/12NZ2) = GL2(Z/12N

2Z).

Proof. By (3.8), ka+b = εka and ∆1/12 α = ε∆1/12 for roots of unity ε, ε. Thereforega = ka∆

1/12 also satisfies gaα = ζ(α)ga for a root of unity ζ(α) (dependent on α).Thus (ga α)∗ = (ζ(α)gaα)∗ = g∗aα. The integrality of g∗a is obvious from the productexpansion.

Here by (3.8), ga α = εga for a root of unity ε; so, Therefore, for qN = q1/N ifNa ∈ Z2, we have g∗a = 1+ qNf(qN ) for f(qN ) ∈ Q[µN)[[qN]]. Thus as a formal powerseries, for any 1 < M ∈ Z, we have a well defined

(g∗a)1/M =

∞∑

n=0

(1/M

n

)qnNf(qN)n ∈ Q[µN ][[qN]]

with(

1/M

n

)=

(1/M)((1/M) − 1)((1/M) − 2) · · · ((1/M) − n + 1)

n!

=(1−M)(1− 2M) · · · (1− (n− 1)M)

Mnn!.

However the coefficients of g∗a1/M and hence of g∗(m)1/M would have growing de-

nominator caused by the denominator of 1/M ; in other words, if g∗(m)1/M remainsintegral, we expect that m is divisible by M . We explore this expectation. By theabove formula, we remark that for a prime l,

(4.6) g∗(m)1/l remains in Q[µN ][[q1/N]] if g∗(m) ∈ Q[µN ][[q1/N]].

Fix a prime p and consider a primitive root of unity ζpn := exp(2πipn ). We need the

following lemma:


Lemma 4.13. The ring Z[ζpn] is the integer ring of Q[ζpn] and the roots of Φn(X)(i.e., all primitive pn-th roots gives rise to a basis of Z[ζpn] over Z, p fully ramifies inZ[ζpn], and ζpn − 1 generates a unique prime ideal of Z[ζpn] over p. Moreover, taking

a decomposition µpn =⊔pn−1

j=1 ζjµp, S = µpn − ζjj is a basis of Z[ζpn] over Z.

Proof. Note that ζp satisfies the equation Φ1(X) = Xp−1X−1

= 1+X+X2 + · · ·+Xp−1 =∏p−1

j=1(X−ζjp). Since Φ1(X+1) = (X+1)p−1

X=

∑pj=1

(pj

)Xj−1 = Xp−1 +pXp−2 + · · ·+p

is an Eisenstein polynomial, Φ1(X +1) is irreducible, and hence Φ1(X) is irreducible.

In the same manner, ζpn is a root of Φn(X) = Xpn−1

Xpn−1−1= Φ1(X

pn−1

) is irreducible.

Therefore Q(ζpn)/Q is a field extension of degree equal to deg(Φn) = pn−1(p− 1). Inparticular, in Q(ζpn)/Q, p fully ramifies with ζpn − 1 giving the unique prime ideal

(ζpn − 1) over p; so, (ζpn − 1)pn−1(p−1) = (p) in Q[ζpn], and Zp[ζpn − 1] is the p-adicinteger ring of Qp[ζpn]. For all other prime l 6= p, taking a prime l of Q[ζpn] above l,

ζpn := (ζpn mod l) is a primitive pn-th root in a finite field of characteristic l; so, ζj

pn

are all distinct for j = 1, . . . , pn−1(p− 1). Thus Ql[ζpn] is unramified at l, and Zl[ζpn]is an unramified valuation ring over Zl; so, it is the l-adic integer ring of Ql[ζpn]. Thisshows that Z[ζpn] ∼= Z[X]/(Φn(X)) is the integer ring of Q[µpn] (cf. [CRT, §9]).

Since∑

ζ∈µpζ = 0, therefore

∑ζ∈µp

ζjζ = 0. Since Z[µpn ] =∑

ζ∈µpnZζ, by∑

ζ∈µpζjζ = 0, we find Z[µpn] =

∑ζ∈S Zζ. Since |S| = deg Φn(X) = [Q[µpn] : Q], the

set S must be a basis of Z[µpn].

Corollary 4.14. For a prime l, if l|∑ζ∈µpnaζζ in Z[µpn], for any pair ζ, ζ ′ with

ζ ′ζ−1 ∈ µp, l|(aζ − aζ′).

Proof. Decompose µpn =⊔pn−1

j=1 ζjµp. By Lemma 4.13,∑pn−1

j=1

∑ζ∈µpζj−ζj bζζ is di-

visible by l if and only if l|bζ for all ζ ∈ µpn − ζjj . Since −ζj =∑

ζ∈µpζj−ζj ζ, we

have∑

ζ∈µpnaζζ =

∑j

∑µpζj−ζj(aζ − aζj)ζ, which shows the result.

We say

• an element m : Q2/Z2 → Z has prime period if for any a with m(a) 6= 0, thereexists a prime p such that pa ∈ Z2,• 0 < n ∈ Z occurs as a denominator of m if there exists a ∈ Q2/Z2 with ordern such that m(a) 6= 0.

Note that g∗(mσ/l) has l-integral coefficients for all σ ∈ GL2(Z) by Lemma 4.12

Lemma 4.15. Let l be a prime number. Assume that m has prime period. Thenthere exists an expression g∗(m) = cg∗(m′) with a constant c such that m′ has valuesin lZ and every denominator occuring in m′ also occur in m.

Proof. For each prime p, we put g∗p(m/l) =∏

pa=0(g∗a)

m(a)/l. Then by (4.6), g∗p(m/l)

lives in Q[µp][[q1/p]] with the identity g∗p(m/l)(0) = 1 of its leading term. There-

fore, the coefficient in q1/p of g∗p(m/l) and g∗(m/l) are equal. In the same way, the

coefficient in q1/p of g∗p(mσ/l) and g∗(mσ/l) are equal for any σ ∈ GL2(Z).We claim

If a, b in Q2/Z2 has order p, then l|m(a)−m(b).


Writing 〈a〉 for a subgroup generated by a, to prove the claim we may assume 〈a〉 ∩〈b〉 = 0 as we can write m(a)−m(b) = m(a)−m(a′) + m(a′) −m(b) for a choicea′ ∈ (p−1Z/Z)2 with 〈a′〉 ∩ 〈a〉 = 〈a′〉 ∩ 〈b〉 = 0. Thus a, b is a basis over Fp of(p−1Z/Z)2, and we can find σ ∈ GL2(Fp) so that aσ = (1

p, 0) and bσ = (1

p, 1

p). Then

the coefficient in q1/p of g∗p(mσ/l) is equal to the coefficient in q1/p of the followingproduct:

∏

c

[(1− exp(2πic2)q

c1)∞∏

n=1

(1− exp(2πic2)qc1qn)(1− exp(−2πic2)q

−c1qn)

]m(c)/l

,

where c runs over elements of the following form:

c = aσ + j(bσ − aσ) (j = 0, . . . , p− 1).

This is because the coefficient of q1/p comes from a with a1 = 1p. Therefore the

coefficient is equal to

−1

l

∑

c

m(c)ζc for ζc = exp(2πij

p).

Thus by Corollary 4.14, l-integrality of this sum implies l|m(c) −m(c′) for any twoc 6= c′. We take j = 0, 1 (i.e., c = aσ and c′ = aσ + bσ − aσ = bσ), and we getl|m(aσ)−m(bσ) as claimed.

We now finish the proof. Take a0 of order p. Then we have∏

pa=0,a 6=0

=∏

pa=0,a 6=0

gm(a)−m(a0)a

∏

pa=0,a 6=0

gm(a0)a .

By Proposition 4.4, we know c :=∏

pa=0,a 6=0 gm(a0)a ∈ Q×, and by the above claim

m′ = m−m(a0) has values in lZ.

We now prove

Theorem 4.16. Let l, p be prime numbers. Assume that g∗(mσ/l) :=∏

a(g∗aσ−1)m(a)/l

has l-integral coefficients for all σ ∈ GL2(Z) and that m is supported on (p−rZ/Z)2

for some r > 0. Then there exists an even function m′ : Q2/Z2 → lZ supported on(p−rZ/Z)2 for some r′ > 0 such that

∏

a

gm(a)a = c

∏

a

gm′(a)a

for a constant c ∈ Q[µpr ]×.

The left-hand-side resides inA×N and the right-hand-side is stable under GL2(Z/NZ);

so, c ∈ Q[µpr ]×.

Proof. The case of r = 1 follows from Lemma 4.15. We proceed by induction on r;so, we may assume that m(a) 6= 0 for some a of order pr . We claim

If pra = prb = 0 in Q2/Z2 with p(a− b) = 0, then l|m(a)−m(b).

If 〈a〉 = 〈b〉, choose a′ of order pr so that 〈a′〉 6= 〈a〉. Then again m(a) − m(b) =m(a) − m(a′) + m(a′) − m(b), and hence, we may assume that 〈a〉 6= 〈b〉 (so, inparticular, a − b 6= 0). Choose c so that pr−1c = b − a. As p(b − a) = 0 and (we


may assume) b− a 6= 0, c has order pr. Since a and a− b form a basis of (p−1Z/Z)2,by Nakayama’s lemma, a and c form a basis of (p−rZ/Z)2. Through base-change

via multiplication by σ, we may assume that a =(

1pr , 0

)and c =

(0, 1

pr

). Define

g∗pr(m/l) :=∏

pra=0,pr−1a 6=0(g∗a)

m(a)/l. As in the proof of Lemma 4.15, we look into the

coefficient in q1/prof g∗pr(m/l) and g∗(m/l) which are equal. It is given by

−1

l

pr−1∑

j=0

m

(1

pr,j

pr

)ζjpr (ζpr = exp(

2πi

pr)).

Again by Corollary 4.14, we find l|m(a)−m(b) as claimed.Given a coset C of (p−1Z/Z)2 in (p−rZ/Z)2, we see p · C = c for a single c ∈

(p1−rZ/Z)2, and the distribution relation Proposition 4.4 tells us∏

x∈C

gx = λgc

for a constant λ ∈ Q[µ12pr]×. Taking c0 ∈ C , we see∏

x∈C

gm(x)x =

∏

x∈C

gm(x)−m(c0)x

∏

x∈C

gm(c0)x = λm(c0)gm(c0)

c

∏

x∈C

gm(x)−m(c0)x .

Since c has order ≤ pr−1, for∏

c gm(c0)c , induction assumption applies. As for the

remaining∏

x∈C gm(x)−m(c0)x , we have proven l|m(x)−m(c0). This finishes the proof.

Since we already know that Siegel unit g12Na a∈p−mW/W (for N = pm) generate

subgroup of finite index in A×N . So for any g ∈ A×

N , we find 0 < M ∈ Z such thatgM = g(m) for some m : (p−mZ/Z)2 − (0, 0) → Z. Then by Theorem 4.16, we canremove prime-factor by prime-factor from M replacing the exponent m; so, we obtain

Corollary 4.17. Suppose p ≥ 5 and that N is a p-power. Then g ∈ A×N , there exists

m : (N−1Z/Z)2 − (0, 0) → Z such that g = g(m) up to constants.

Thus the next step is to make explicit the set

m : (N−1Z/Z)2 − (0, 0) → Z|g(m) ∈ A×N.

4.8. Fricke–Wohlfahrt theorem. We need a theorem due to Fricke–Klein (1980)and its generalization by Wohlfahrt (1963). Let ∆ ⊂ SL2(Z) be a subgroup of finiteindex. Such a group ∆ is called a congruence subgroup if ∆ contains a principalcongruence subgroup

Γ(N) = α ∈ SL2(Z)|α ≡ 1 mod N ·M2(Z)for a positive integer N . If ∆ is a congruence subgroup, the smallest N such that∆ ⊃ Γ(N) is called the arithmetic level of ∆. For each cusp s of general ∆, let∆s = α ∈ ∆|α(s) = s. Taking γ = γs ∈ SL2(Z) with γ(s) =∞, we find

γ±1∆sγ−1 ⊂

± ( 1 m

0 1 )∣∣m ∈ Z

=: Γ∞

as a subgroup of finite index Ns. Since the coset Γ∞γs is determined uniquely by s,Ns depends only on s. We define the geometric level of ∆ to be the least commonmultiple of Ns for s running over all cusps of ∆.


Theorem 4.18. If ∆ is a congruence subgroup of SL2(Z), its geometric level andarithmetic level coincide. In other words, writing N for the geomeric level of ∆, if∆ ⊃ Γ(N ′) and E(N) is the minimal normal subgroup of SL2(Z) generated by ( 1 N

0 1 ),then Γ(N) = Γ(N ′)E(N).

We give a proof due to Wohlhart. Let Γ := Γ(N ′)E(N). We define M to be theorder of T = ( 1 1

0 1 ) in SL2(Z)/Γ. Since U = jT j−1 = ( 1 01 1 ) also has order M . Plainly

M |N . Thus we need to prove that Γ(M) ⊂ Γ(N), which shows M = N .

Proof. We will define subsets E1, E2, and E3 of Γ(M) whose union equals Γ(M)and show that each Ei is a subset of Γ(N). To this end, let A ∈ Γ(M) and writeA = ( a b

c d ) . First, let E1 be the collection of elements whose upper right and lowerleft entries are divisible by N and suppose A ∈ E1. If we let

A0 =(

a ad−11−ad d(2−ad)

)

then det(A0) = 1, and since ad − bc = 1, we have ad ≡ 1 mod N and so A ≡ A0

mod N . Equivalently, AA−10 − 1 ∈ Γ(N). It suffices to show A0 ∈ Γ(N). This follows

from writing V = TU−1T 3U−1T and computing A0 = (ST d−1S)−1(V T a−1V −1)T d−1

with S = TU−1T is an element of E(N) since a ≡ d ≡ 1 mod M .Now let E2 be the set of matrices who diagonal entries are relatively prime to N .

If A ∈ E2 then gcd(a,N) = 1 so a is a unit modulo N say with inverse a′. We canwrite c = Mc and let k = −a′c. Then

(STmS)−kA = ( 1 0mk 1 ) ( a b

c d ) =(

a bc+amk d+bmk

)

and c + amk = m(c − a′ac) is congruent to 0 modulo N . Let d = d + bmk. Taking

the determinant of (STmS)−kA, we see that d is relatively prime to N . Therefore we

can also choose ` so that b+ dm` ≡ 0 mod N . In this case,

Tm`(STmS)−kA = ( 1 m`0 1 )

(a b

c+amk ed)

=(

∗ b+edm`

c+amk ed

)

and this is an element of E1. This says A is an element of E(M)E1 and is thereforein Γ(N). Finally, define E3 = Γ(M)\(E1 ∪ E2). If A ∈ E3, then gcd(a, n) 6= 1 orgcd(d, n) 6= 1. Suppose first that gcd(a, n) 6= 1. Since ad−bc = 1, we know gcd(a, b) =1. Then since a ≡ 1 mod M , we also have gcd(a, bM) = 1. Recall Dirichlet’s theoremon primes in arithmetic progressions: Suppose r and s are relatively prime integers.Then r + st|t ∈ Z contains infinitely many prime numbers. In particular, takingr = a and s = bm, we can choose an integer k so that a + bmk is a prime numberlarger than N . Then

A(STmS)−k = ( a bc d ) ( 1 0

mk 1 ) =(

a+bmk bc+dmk d

)

has upper left entry relatively prime to N . If gcd(d, n) = 1, then A(STmS)−k ∈ E2

and we are done. Otherwise, let a = a+bmk, let c = c+dmk and note that ad−bc = 1so gcd(c, d) = 1. As gcd(d,m) = 1, just as above we can choose ` so that d + cm` isa prime number larger than N . Then

A(STmS)−kTm` = ( ea bec d ) ( 1 m`

0 1 ) =( ea ∗

ec d+ecmk

)


and A(STmS)−kTm` must be in E2. Thus A is an element of E2E(M) and the theoremfollows.

4.9. Siegel units and Stickelberger’s ideal. We put

D = DN =⊕

P :cusp of X(N)

Z[P ],

D0 = D0N = D ∈

⊕

P :cusp of X(N)

Z[P ]| deg(D) = 0,

F = FN = div(u) ∈⊕

P :cusp of X(N)

Z[P ]|u ∈ A×N,

SN = div(gm)|gm =∏

a∈N−1Z2/Z2−0gm(a)

a ∈ AN,

S = div(gm)|gm =∏

a∈Q2/Z2−0gm(a)

a ∈ AN

(4.7)

Then we have D0 ⊃ F ⊃ S and ClX(N) = D0/F, which is a finite abelian group byTheorem 4.10.

Theorem 4.19 (Kubert–Lang). We have S = SN = FN if N = pr for a primep ≥ 5.

When N is not a prime power as in the theorem, then [F : S] is a 2-power (see[MUN, Theorem 5.1.1]).

Proof. We know that the q-expansion coefficients of gm lies in Q[µN ] if m is supportedon N−1Z2 − Z2. Since ga α = gaα up to non-zero constant by Theorem 4.2, if aninteger which is not a factor of N occurs as a divisor of m, gmα 6∈ C[[q1/N]] by lookinginto the leading term of gm. Therefore to have gm ∈ AN , it is necessary to have msupported on N−1Z2 − Z2. Thus S ⊂ XN := div(gm)|m : N−1Z2/Z2 − 0 → Z.Since XN ⊂ Q[µN ]][[qN2]], we need to show that XN ∩ C(X(N)) = S. This followsfrom Theorem 4.18.

By Theorem 4.9, we identify R = RN = Z[Cm/±1] with D by sending α ∈Cm/±1 to [α(∞)]. Recall (4.4):

div(ga) = N∑

b∈Cm/±1

1

2B2(〈Tr(ab)〉)σ−1

b .

We then put

(4.8) θ = N∑

b∈Cm/±1

1

2B2(〈Tr(b/N)〉)σ−1

b ∈ Q · RN = Q[Cm/±1].

Taking a = Tr(1/N), we find div(ga) = θ; so,

(4.9) deg(θ) = deg(div(ga)) = 0.

Since div(gaα) = div(gaα) for α ∈ GL2(Z), we find S = R∩Rθ by Fricke–Wohlfahrttheorem, where Rθ is the fractional ideal inside Q[Cm/±1] generated by θ. As in


§4.6, we identify Wm = W/pmW with (Z/pmZ)2 by sending 12a1 + a2τ to a = (a1, a2).

Note that Cm = Wm − pWm; so, if a, b ∈ Wm are generators of Z/pmZ-module Wm,a, b, a+b and a−b belong to Cm, where the sum a+b and difference a−b is computedin the additive group Wm. Let I be the ideal of RN generated by parallelograms:

π(a, b) = (a+ b) + (a− b)− 2(a)− 2(b) ∈ RN ,

where we regard a, b ∈ Cm ⊂ (Z/pmZ)2 and (a + b), (a− b), (a) and (b) is the imageof a+ b, a− b, a and b in Cm/±1.Lemma 4.20. Let N = pm with m ≥ 1 and BN be the set of pairs (a, b) in Z2/NZ2

such that a, b, a + b and a − b all have order N . Define I by the ideal of RN

generated by π(a, b)(a,b)∈BN. Then I contains

∑a∈Cm

m(a/N)[a] if and only if m :N−1Z2/Z2 − (0, 0) → Z satisfies (QN).

If m(a) 6= 0 for a of order d less than N with N = dM , by the distribution relation:ga = c

∏Mb=a gb for b’s of order N , replacing m by m′ such that m′(b) = m(a) for all

b with Mb = a and m′(a) = 0 without changing any other values, we may assumethat m(a) 6= 0 implies that a has order N (i.e., m is supported on Cm).

Proof. Multiplying N , we prove the equivalence between(Q′

N)∑

(r,s) 6=0

m(r/N, s/N)r2 ≡∑

(r,s) 6=0

m(r/N, s/N)s2 ≡∑

(r,s) 6=0

m(r/N, s/N)rs ≡ 0 mod N,

and∑

(r,s)m( rN, s

N)[(r, s)] ∈ I . Without taking congruence modulo N of r and s, it

is a plain computation to check that π(a, b) for a = (r, s) and b = (r′, s′) satisfies thefollowing condition:

(Q)∑

(r,s) 6=0

m(r/N, s/N)r2 =∑

(r,s) 6=0

m(r/N, s/N)s2 =∑

(r,s) 6=0

m(r/N, s/N)rs = 0.

For example,

Left-hand-side of (Q) for π(a, b) = (r + r′)2 + (r − r′)2 − 2r2 − 2r′2

= 0.

Since r and s are actually classes modulo N , the equality in (Q) becomes the identity(Q′

N) modulo N .Let deg(m) =

∑(r,s)m(r, s). To forget about congruence classes, we argue in the

group ring Z[Z2] regarding m =∑

(r,s)m(r/N, s/N)(r, s) ∈ Z[Z2]. In other words,

assuming (Q) for m ∈ Z[Z2] and prove that m is a linear combination of π(a, b) forfinitely many (a, b) as long as deg(m) is even. Since we take modulo N at the end andN is odd, by multiplying 2 (which is a unit in Z/NZ), we may assume that deg(m)is even. Let I be the Z-linear span in Z[Z2] generated by π(a, b) with a+ b, a− b, a, ball having order N in (Z/NZ)2. Since the image of I in RN and NRN span the idealof elements satisfying (Q′

N ), we are going to prove that any m : Z2 → Z satisfying(Q) is in I.

Put h(r, s) = |r| + |s| and define h(m) = maxm(r,s) 6=0 h(r, s). Suppose m(r, s) 6= 0with maximal h(r, s) ≥ 3. As we remarked, a := (r, s) has order N in (Z/NZ)2. Sincethe argument is the same for each quadrant by rotation of ±90 and 180 degrees wemay assume r ≥ 0 and s > 0 with r + s ≥ 3. We now show that we are able to


choose (0, 0) 6= (i, j) ∈ [0, r/2] × [0, s/2] such that x = (r − i, s − j) and y = (i, j)have both order N in (Z/NZ)2. Then π(x, y) = (x + y) + (x − y) − 2(x) − 2(y) =(a) + (r − 2i, r − 2j) − 2(x) − 2(y), and m′ := m − m(a)π(x, y) removes a from itssupport and h(m′) ≤ h(m), and repeating this process, eventually we can bring mmodulo I to another function supported in P := (c, d) : | ± c± d| ≤ 3.

We separate our argument in the following four cases:

(1) (r, s)mod p ∈ (Z/pZ)2 − 0, 12,(2) (r, s) ≡ (0, 1) mod p,(3) (r, s) ≡ (1, 0) mod p,(4) (r, s) ≡ (1, 1) mod p.

First suppose r ≥ 2 and s ≥ 2, we may choose in Case (1) and (2), y = (1, 0);Case (3) y = (0, 1) and Case (4) y = (2, 0) (as r > 2 and s > 2 in Case (4) byN ≥ 5). If r = 1 (then s ≥ 2), we are in Case (3), and y = (0, 1) still works. Inthe same way, if s = 1, y = (1, 0) works. If r = 0, s ≥ 3, we can choose y = (0, j)with j ∈ 1, 2 so that s − 2j 6≡ 0 mod p. This is possible as p ≥ 5. If s = 0,we take y = (j, 0) with j ∈ 1, 2 so that r − 2j 6≡ 0 mod p. Thus we are able tochoose x, y as desired, and by induction, we may now assume that m is supportedon P0 := (±2, 0), (0,±2), (±1, 0), (0,±1), (±1,±1). We can explicitly check thatm : P0 → Z satisfying (Q′

N ) implies that m is a linear combination of π(a, b)’s.To pin down, we follow [MUN, §3.1]. Note that (1, 1) + (1,−1) ≡ 2(1, 0) + 2(0, 1)

mod I and (2, 0) + (0, 2) ≡ 2(1, 1) + 2(1,−1) mod I. By this, we are reduced to alinear combination of P1 := (1, 0), (0, 1), (1, 1), (2, 0). Note that 8(1, 0) − 2(2, 0) =α − β − 2γ ∈ I with α = π((2, 0), (3, 0)) − π((1, 0), (4, 0)), β = π((1, 0), (2, 0)) andγ = π((1, 0), (3, 0)). Suppose m : P1 → Z satisfies (Q′

N), and write x = m(1, 0),y = m(0, 1), z = m(2, 0) and u = m(1, 1). Then, we have

x+ 4z + u = 0, y + u = 0, u = 0⇒ y = u = 0

Since deg(m) = x + z is even, z is even. Writing z = 2a and x = −8a. Since8(1, 0) − 2(2, 0) ∈ I, we are done.

This lemma shows

Theorem 4.21. Let I12 := α ∈ I | deg(α) ≡ 0 mod 12. Then we have SN =RNθ ∩RN = I12θ.

4.10. Cuspidal class number formula. Let R0 = R0N = D ∈ RN | deg(D) = 0.

Then we have

Theorem 4.22. Suppose N = pm (m > 0) with a prime p ≥ 5. Then we have

|ClX(N)| = [R0N : SN ] =

6N3

|Cm|∏

χ 6=1

N

2B2,χ

for the generalized Bernoulli number B2,χ =∑

a∈Cm/±1B2(〈Tr(a)N〉), where χ runs

over all even character of Cm.

The above B2,χ is a non-zero multiple of L(−1, χ) (see (4.3)).We prepares a series of lemmas to prove the formula. Let s :=

∑σ∈Cm/±1 σ ∈ RN

and put θ′ := θ − N12s. Then for D ∈ RN , D · s = deg(D)s, and div(∆) = N · s.


Lemma 4.23. We have R0N ∩ (Iθ′ +R div(∆)) = I12θ.

Proof. Since ξs = deg(ξ)s and div(∆) = Ns, we find R div(∆) = Zdiv(∆). Forsimplicity, write G := Cm/±1. Recall deg(θ) = 0 by (4.9); so,

deg(θ′) = −N12|G| = −N

12deg(s).

Thus, for ξ ∈ I , ξθ = ξθ′ + 112

deg(ξ)Ns = ξθ′ + 112

deg(ξ) div(∆) ∈ R0. This implies

R0N ∩ (Iθ′ +R div(∆)) = R0

N ∩ (Iθ′ + Zdiv(∆)) ⊂ I12θ.

The reverse inclusion follows from Theorem 4.21.

Lemma 4.24. Let Rd := D ∈ R| deg(D) ∈ dZ. Then we have

deg(R0 + Iθ′ +RNs) = deg(Iθ′ +RNs) =N |G|

6Z;

so, Rd = R0 + Iθ′ +RNs for d = N |G|6

.

Proof. Since θ has degree 0 and Q ·R0 is the augmentation ideal of Q[G], deg(Iθ) = 0.Thus we conclude

deg(Iθ′) = deg(I(N

12s)) = 2

N |G|12

Z =N |G|

6Z.

Here we used the fact that deg(π(a, b)) = −2 and hence deg(I) = 2Z. Since |G| =(p2 − 1)/2 ≡ 0 mod 6, deg(Iθ′) ⊂ Z. Therefore

deg(deg(Iθ′ +RNs)) =N |G|

6Z +N |G|Z =

N |G|6

Z

as p ≥ 5.

Then for d = N |G|6

, we have inclusions:

R ⊃ Rd ⊃ Iθ′ +RN · s ⊃ Iθ′.

Lemma 4.25. We have [R : Rd] = d and [Iθ′ +RN · s : Iθ′] = |G|12

.

Proof. By the surjectivity of deg : R→ Z, we see

[R : Rd] = [deg(R) : deg(Rd)] = [Z : dZ] = d.

By the isomorphism theorem, we have [Iθ′+RN ·s : Iθ′] = [RN ·s : RN ·s∩Iθ′]. Weargue via modular forms. The module Iθ′ is the module generated by divisors of km

for m : N−1Z2/Z2−0 → Z satisfying (QN). ThusD ∈ Iθ′∩RN ·s = Iθ′∩R(div(∆))means that D = div(km) = ν div(∆) with ν ∈ Z. Thus km = C ·∆ν with 0 6= C ∈ C.Comparing the weight, we get

ν = − 1

12

∑

a

m(a).

Since ga = ka∆1/12, we find gm = C . This implies that m is a constant function

as gm α = gmα for α = GL2(Z/NZ)/±1 = Gal(X(N)/P1(J)) (and ga for a ∈N−1Z2/Z2 − 0 are independent; see Proposition 4.4). Since a constant function msatisfies (QN) (see Remark 3.12), m is arbitrary, and ξNs 7→ ξ div(∆) = Nsdeg(ξ) ∈


ZNs sends RNs (resp. RNs ∩ Iθ′) isomorphic to ZNs (resp. |G|12

ZNs). This we

conclude [RN · s : RN · s ∩ Iθ′] = |G|12

.

For any two lattices L and L′ of a finite dimensional Q-vector space, we define

[L : L′] = [L:L∩L′][L′:L∩L′]

∈ Q, which behaves just like index. For example, we have

[L : L′][L′ : L′′] = [L : L′′]

and the value equals the index [L : L′] if L ⊃ L′. Moreover for a linear transfor-mation T taking L onto L′, we have [L : L′] = | det(T )|. Since [R0 : R0θ] 6= 0 (by

Theorem 4.10) and deg(θ′) = −N |G|126= 0, we find [R : Rθ′] 6= 0.

Lemma 4.26. We have [RN : RNθ′] = deg(θ′)

∏χ 6=1 χ(θ).

Proof. The index is just the determinant of the multiplication by θ′ on R. Thedeterminant can be calculated in C[G] which is the product of the eigenvalues oneach eigenspaces. The χ eigenspace is of diemsion 1, and hence we get χ(θ) as itseigenvalue if χ 6= 1 (since θ ∈ R0). For the trivial eigenspace, R/R0, the eigenvalue

is deg(θ′) = N |G|12

.

Lemma 4.27. We have [R : I ] = [Rθ′ : Iθ′] = N3.

Proof. We may identify Cm with P := (r, s) ∈ (Z/NZ)2|(r) + (s) = Z/NZ (havingorder N). Then we claim that R/I is free of rank 3 with basis (0, 1), (1, 0), (1, 1).Indeed, in the proof of Lemma 4.20, we have shown that R/I is generated by P1 :=(1, 0), (0, 1), (1, 1), (2, 0) and 8(1, 1) ≡ 2(2, 0) mod I ; so, we can remove (2, 0).

We need to show that for any given (r, s) ∈ P , we have a unique triple (x, y, z) ∈(Z/NZ)3 such that (r, s)+x(1, 0)+y(0, 1)+ z(1, 1) such that this sum satisfies (Q′

N).The condition (Q′

N ) is equivalent to

r2 + x+ z = 0, s2 + y + z = 0 and rs+ z = 0,

which can be solved:

x = −rs− r2, y = −rs− s2 and z = −rs.

Proof of Theorem 4.22: Recall d = N |G|6

= [R : Rd]. We have

|ClX(N)| = [R0 : I12θ]Lemma 4.23

= [R0 : R0 ∩ (Iθ′ +RN · s)]Lemma 4.24

= [Rd : (Iθ′ +RN · s)] =[R : Iθ′]

[(Iθ′ +RN · s) : Iθ′][R : Rd]

Lemma 4.25=

12 · 6N |G|2 [R : Rθ′][Rθ′ : Iθ′]

Lemma 4.27=

12 · 6N2

|G|2 [R : Rθ′]

Lemma 4.26=

12 · 6N2

|G|2 deg(θ′)∏

χ 6=1

χ(θ) =6N3

|G|∏

χ 6=1

χ(θ),

where the last identity follows from deg(θ′) = N |G|12

. This finishes the proof as χ(θ) =N2B2,χ by definition.


4.11. Cuspidal class number formula for X1(N). We give a description of thecuspidal class number formula for X1(N) without proof. Let N = pr for a prime p.Let X0(p) = Γ0(p)\(H tP1(Q)), where

Γ0(N) :=( a b

c d )∣∣c ≡ 0 mod N

.

It is easy to see

SL2(Z) = Γ0(p) tp⊔

j=1

Γ0(p)δ(

1 j0 1

)

for δ = ( 0 −11 0 ). Thus the set of cusps S0(N) of X0(N) can be explicitly given by

S0(p) = Γ0(p)\SL2(Z)/Γ∞ = 0,∞ = δ(∞),∞,where

Γ∞ := γ ∈ SL2(Z)|γ(∞) =∞) =± ( 1 m

0 1 )∣∣m ∈ Z

.

Thus one can classify the set S1(pr) cusps of X1(p

r) into three classes:

S1(pr) = S∞ t Sm t S0.

The classification goes as follows: Since(

x by d

)( 1 m

0 1 ) =(

x xm+by ym+d

)for

(x by d

)∈ SL2(Z),

Γ1(pr)\SL2(Z)/Γ∞ ∼=

x := ( x

y ) ∈ (Z/NZ)2∣∣x has order N

/ ∼,

where ( xy ) ∼ ( x+my

y ) for m ∈ Z. Thus

S0 ∼=(

0y

) ∣∣y ∈ (Z/NZ)×/±1

and S∞ ∼=( x

0 )∣∣x ∈ (Z/NZ)×/±1

.

The set S∞ is made up of cusps unramified over ∞ ∈ X0(p). The action of W :=(0 −1pr 0

)induces S0 ∼= S∞. On S?, the group Gr = (Z/prZ)×/± acts transitively

and freely, and hence, writing D? :=⊕

s∈S? Z[s] and D?0 := D ∈ D?| deg(D) = 0,

we have Z[Gr] ∼= D? by sending∑

g agg to∑

g agg(?) for ? = 0,∞.

Take a model X1(pr) of Γ1(p

r)\(HtP1(Q)) classifying a pair (E, µpr → E), and itsdual model X∗

1 (pr) classifying (E,Z/pr → E). Write X1(pr)−cusps = Spec(A1,N)

and X∗1 (pr)− cusps = Spec(A∗

1,N). Define

F∞ := div(u)|u ∈ A?1,N with A∞

1,N := u ∈ A×1,N | Supp(div(u)) ⊂ S∞

F0 := div(u)|u ∈ A∗,01,N with A∗,0

1,N := u ∈ (A∗1,N)×| Supp(div(u)) ⊂ S0.

(4.10)

The partial cuspidal group is defined as follows:

Cl∞X1(N) := D∞0 /F

? and Cl0X∗

1(N) := D0

0/F0.

By the action of the involution W , we have Cl∞X1(N)∼= Cl0X∗

1(N). In this case, the

Stickelberger element is the traditional one:

θ = N∑

b∈G

1

2B2(〈b/N〉)σ−1

b ,

where σb ∈ Gal(X1(pr)/X0(p

r)) associated to b ∈ G. Let I be the ideal of R = Z[G]generated by σc − c2 and I0 := ξ ∈ I | deg(ξ) = 0. Then the following lemma iseasier than Theorem 4.21 (see [MUN, Lemma 6.1.2]:


Lemma 4.28. For θ′ = θ − pr

12s with s =

∑σ∈G σ, Rθ′ ∩ R = Iθ′ and R0θ

′ ∩ R =I0θ

′ = I0θ.

Consider functions m : (p−rZ/Z→ p1−rZ/Z)/±1 → Z satisfying

(1) For every coset C of p−1Z/Z,∑

a∈C m(a) = 0,(2)

∑am(a)(pra)2 ≡ 0 mod pr.

It is shown in [MUN, Theorem 3.1] that the units satisfying the above two conditionsgenerates A∗,0

1,N up to scalars. This enable us to get

Theorem 4.29 (Kubert–Lang). We have

|Cl∞X1(N)| = |Cl0X∗

1(N)| =

ppr−1

p2r−2

∏

χ:G→Q×

,χ 6=1

1

2B2,χ.

But we do not know the cyclicity of ClX1(N) over Z[G] except for the case whereN = p.

References

[ALG] R. Hartshorne, Algebraic Geometry, Graduate Texts in Mathematics 52, Springer, NewYork, 1977.

[CRT] H. Matsumura, Commutative Ring Theory, Cambridge Studies in Advanced Mathematics8, Cambridge Univ. Press, New York, 1986.

[EEK] A. Weil, Elliptic Functions according to Eisenstein and Kronecker, Springer, 1976[EFN] S. Lang, Elliptic functions, second edition, GTM 112, 1987, Springer.[GME] H. Hida, Geometric Modular Forms and Elliptic Curves, World Scientific, Singapore, 2000.[IAT] G. Shimura, Introduction to the Arithmetic Theory of Automorphic Functions, Princeton

University Press, Princeton, NJ, and Iwanami Shoten, Tokyo, 1971.[LFE] H. Hida, Elementary Theory of L-Functions and Eisenstein Series, LMSST 26, Cambridge

University Press, Cambridge, England, 1993.[MUN] D. Kubert and S. Lang, Modular units, Grundlehren der Mathematischen Wissenschaften

[Fundamental Principles of Mathematical Science], 244, 1981, Berlin, New York: Springer.

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