Elliptic Curves and Modular Forms -...

ALGANT Master Thesis - July 2018

Elliptic Curves and Modular Forms

Candidate

Francesco Bruzzesi

Advisor

Prof. Dr. Marc N. Levine

Universita degli Studi di Milano Universitat Duisburg–Essen

Introduction

The thesis has the aim to study the Eichler-Shimura construction associatingelliptic curves to weight-2 modular forms for Γ0(N): this is the perfect topicto combine and develop further results from three courses I took in the firstsemester of the academic year 2017-18 at Universitat Duisburg-Essen (namelythe courses of modular forms, abelian varieties and complex multiplication).Chapter 1 gives a brief overview on the algebraic geometry results and toolswe will need along the whole thesis. Chapter 2 introduces and develops thetheory of elliptic curves, firstly as an algebraic curve over a generic field, andthen focusing on the fields of complex and rational numbers, in particular forthe latter case we will be able to define an L-function associated to an ellipticcurve. As we move to Chapter 3 we shift our focus to the theory of modularforms. We first treat elementary results and their consequences, then we see howit is possible to define the canonical model of the modular curve X0(N) overQ, and the integrality property of the j-invariant. Chapter 4 concerns Heckeoperators: Shimura’s book [Shi73 Chapter 3] introduces the Hecke ring and itsproperties in full generality, on the other hand the other two main referencesfor the chapter [DS06 Chapter 5] and [Kna93 Chapters XIII & IX] do notintroduce the Hecke ring at all, and its attributes (such as commutativity forthe case of interests) are proved by explicit computation. We try to take anintermediate approach to the subject and rephrase everything just in terms ofSL2(Z) and congruence subgroups. At the end of the chapter we will be able toassociate an L-function to a cusp form for Γ0(N). Lastly in Chapter 5 we areready to illustrate how to obtain an elliptic curve from a weight-2 cusp form forΓ0(N), making use of the theory of abelian varieties.

i

ii

Notes and References

Chapter 1: The organization of Sections 1.1 and 1.2 is based on [Sil09,Chapter 1 & Chapter 2]. The first statement of Proposition 1.10 is taken from[DS06, Proposition 7.2.6]; Proposition 1.14 is from [Kna93, Proposition 11.43];Proposition 1.22 comes from [Was03, Proposition C.2]. The main references forthese two sections are [Har77], [Mir95] and [Sha77]. Section 1.3 treats somestandard results on the Jacobian variety associated to a Riemann surface. Themain references are [Kir92] and [Mir95].

Chapter 2: Section 2.1 is organized as [Sil09, Chapter 3]. Proposition 2.1 istaken from [Sil09, III.1.4(i)]; Proposition 2.3 and Proposition 2.5 are based onfilling the details of [Kna93, Theorem 11.57 and Theorem 11.58] (respectively);Proposition 2.12 comes from [Sil09, Theorem III.4.10]; Lemma 2.15 is from[Kna93, Lemma 11.63]; Theorem 2.16 and Theorem 2.17 try to fill the detailsof [Kna93, Theorem 11.64 and Theorem 11.66] (respectively). The main refer-ence for Section 2.2 is [Shi73, Chapter 4]. For Section 2.3 we followed [Kna93,Chapter X].

Chapter 3: Some of the results arise as homeworks and/or are taken fromthe course of modular forms mentioned in the Introduction above and mostof the results are standards: main references are [DS06], [Miy89] and [Lan76].However Section 3.2.2 is taken from [Kna93, Chapter XI, Section 8].

Chapter 4: The structure of the Chapter is as in [Shi73, Chapter 3]. Propo-sition 4.5 is from [Shi73, Proposition 3.8]; Lemma 4.9 is from[Shi73, Lemma3.12]; Proposition 4.10 is from [Shi73, Proposition 3.14]; Lemma 4.40 and 4.41are respectively from [Shi73, Lemma 3.61 and 3.62]. Results from Theorem 4.33to Proposition 4.39 are taken from [Kna93, Chapter XI, Section 5].

Chapter 5: The organization of the whole chapter is as in [Kna93, Chap-ter XI, Sections 10 & 11]. For Section 5.1 we used as references [Lan59] and[Swi74]. Section 5.2 is from the last part of [Kna93, Chapter XI, Section 10];Proposition 5.15 is taken from [Kna93, Theorem 11.74].

iii

Notation

Z,Q,R,C integers, rationals, reals, complex numbersFp the field with p elementsH complex upper half planeRe(z), Im(z) real, imaginary part of zK algebraic closure of the field KAut(K/L) automorphism group of K fixing the subfield L ⊂ KMn(K) n× n matrices with coefficients in KGLn(K) invertible n× n matrices with coefficients in KSLn(K) n× n matrices with coefficients in K and determinant 1[A : B] index of B in A or degree of A in BW∨ dual space of the vector space WA× group of invertible elements of the ring A#S cardinality of the set S

iv

Acknowledgements

First and foremost, I would like to express my gratitude to Professor Dr.Marc N. Levine, who accepted to supervise my study in this topic and patientlyspent time to enlight me with his deep mathematical insights. He motivated meto do always better.

I want to thank all the people I met in the last two year in the Algant Master,both Professors and students. In particular Bob, John and Francesco: we sharedevery moment, all the laughs and all the dissapointments. We constantly helpedeach other, both in life and in university. We all know how frustating it feels tostudy math sometimes, and I am thankful that we overcome the difficulties ofthese two years together.

A special thanks goes to Federica, who had to bear me every single day ofthis stressful period. Even though she had to listen to my complaints, she mademe smile in every situation. Thank you for having coped with all my strugglesand problems.

Also I would like to mention my dear friend Antonio, the person who helpedme to develop a taste for number theory, I couldn’t ask for a better mentor andfriend. No matter how long we don’t see each other, we always have a great timeand connection.

I am thankful to all my hometown friends and relatives. You are the reasonwhy I keep coming back home and on all occasions it feels like time didn’t pass.

Last, but certainly most important, I want to thank my parents Lanfrancoand Laura for providing me with unfailing support and continuous encourage-ment throughout my years of study. This accomplishment would not have beenpossible without them: words cannot really describe how grateful I am.

Francesco

Contents

Introduction iNotes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiNotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiAckowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

1 Algebraic Geometry 11.1 Algebraic Varieties . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Algebraic Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2.1 Fields of positive characteristic . . . . . . . . . . . . . . . 101.2.2 Riemann-Roch theorem . . . . . . . . . . . . . . . . . . . 13

1.3 Riemann Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2 Elliptic Curves 192.1 Elliptic curves over an arbitrary field . . . . . . . . . . . . . . . . 19

2.1.1 Weierstrass form & abstract elliptic curves . . . . . . . . 192.1.2 Isogenies . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.2 Elliptic curves over C . . . . . . . . . . . . . . . . . . . . . . . . 272.2.1 The Weierstrass ℘-function . . . . . . . . . . . . . . . . . 292.2.2 Isogenies over C . . . . . . . . . . . . . . . . . . . . . . . 332.2.3 Automorphisms of an elliptic curve . . . . . . . . . . . . . 35

2.3 Elliptic curves over Q . . . . . . . . . . . . . . . . . . . . . . . . 362.3.1 L-function associated to an elliptic curve . . . . . . . . . . 372.3.2 Hasse theorem . . . . . . . . . . . . . . . . . . . . . . . . 38

3 Modular Forms 403.1 Modular forms for SL2(Z) . . . . . . . . . . . . . . . . . . . . . . 40

3.1.1 Functions of lattices . . . . . . . . . . . . . . . . . . . . . 403.1.2 The action of SL2(Z) on H . . . . . . . . . . . . . . . . . 423.1.3 Divisors of modular functions . . . . . . . . . . . . . . . . 453.1.4 The space of modular forms . . . . . . . . . . . . . . . . . 493.1.5 The modular curve X0(1) . . . . . . . . . . . . . . . . . . 52

3.2 Congruence subgroups . . . . . . . . . . . . . . . . . . . . . . . . 583.2.1 Modular functions of higher level . . . . . . . . . . . . . . 633.2.2 The canonical model of X0(N) over Q . . . . . . . . . . . 67

3.3 Integrality of the j-invariant . . . . . . . . . . . . . . . . . . . . . 693.3.1 j(z) is an algebraic number . . . . . . . . . . . . . . . . . 703.3.2 j(z) is integral . . . . . . . . . . . . . . . . . . . . . . . . 71

v

CONTENTS vi

4 Hecke Operators 744.1 The Hecke ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

4.1.1 The structure of R(Γ,∆) . . . . . . . . . . . . . . . . . . 764.2 Action on modular functions . . . . . . . . . . . . . . . . . . . . 83

4.2.1 Hecke operators on SL2(Z) . . . . . . . . . . . . . . . . . 844.2.2 Hecke operators on congruence subgroups . . . . . . . . . 86

4.3 L-function of a cusp form . . . . . . . . . . . . . . . . . . . . . . 93

5 Eichler-Shimura Theory 975.1 Complex abelian varieties and Jacobian varieties . . . . . . . . . 975.2 Technical results . . . . . . . . . . . . . . . . . . . . . . . . . . . 1015.3 Elliptic curves associated to weight-2 cusp forms . . . . . . . . . 104

5.3.1 Perspective . . . . . . . . . . . . . . . . . . . . . . . . . . 108

Bibliography 109

Chapter 1

Algebraic Geometry

Throughout this whole chapter, let K0 denote a field and K be an alge-braically closed field containing K0

1.1 Algebraic Varieties

Definition. Define the affine n-dimensional space over K as

An = An(K) = P = (x1, ..., xn) | xi ∈ K

Similarly, define the set of K0-points (or K0-rational points) of An as the set

An(K0) = P = (x1, ..., xn) ∈ An | xi ∈ K0

Remark 1.1. We have an action of the Galois group Gal(K/K0) on An: letσ ∈ Gal(K/K) and P ∈ An, then Pσ = (xσ1 , ..., x

σn). It follows that

An(K0) = P ∈ An | Pσ = P ∀σ ∈ Gal(K/K0)

Recall that by the Hilbert basis theorem any polynomial ring over a field isa Noetherian ring, thus every ideal is finitely generated.

Definition. Consider the polynomial ring in n variables K[X1, .., Xn], to anyideal J ⊂ K[X1, .., Xn] we can associate a subset of An, called affine algebraicset,

VJ = P ∈ An | f(P ) = 0 ∀f ∈ J

Viceversa, to any algebraic set V ⊂ An we can associate an ideal of polynomialsvanishing on V , called ideal of V ,

I(V ) = f ∈ K[X1, .., Xn] | f(P ) = 0 ∀P ∈ V

In particular we say that an algebraic set is defined over K0 if its ideal I(V ) canbe generated by polynomials in K0[X1, .., Xn] and we write V/K0. Then if V isdefined over K0, the set of K0-rational points of V is

V (K0) = V ∩ An(K0)

1

1.1. ALGEBRAIC VARIETIES 2

Remark 1.2. Assume that V is an algebraic set defined over K0 and let

f1, ..., fr ∈ K0[X1, .., Xn] be the generators for I(V/K0)def= I(V )∩K0[X1, .., Xn],

then

V (K0) = P = (x1, .., xn) ∈ An(K0) | f1(P ) = ... = fr(P ) = 0

Noice that Gal(K/K0) acts on K[X1, .., Xn] by acting on the coefficients of anelement f ∈ K[X1, .., Xn]. Then for f ∈ K[X1, .., Xn], P ∈ An, σ ∈ Gal(K/K0)

(f(P ))σ = fσ(Pσ)

Remark 1.3. If f ∈ K0[X1, .., Xn] and P ∈ An, then for σ ∈ Gal(K/K0) wehave f(Pσ) = f(P )σ. Therefore if V is defined over K0 we can characterizeV (K0) as

V (K0) = P ∈ V | Pσ = P ∀σ ∈ Gal(K/K0)

Definition. An affine algebraic set is irreducible if it is not union of two properaffine algebraic sets. An irreducible affine algebraic set is called affine variety

Proposition 1.4. Let V = ∅ be an affine algebraic set, then V is irreducible ifand only if I(V ) is a prime ideal.

Proof. Assume V is reducible, so that V = V1 ∪ V2 for some proper algebraicsubsets V1, V2 $ V . Since the containment is proper, there exist f ∈ I(V1)rI(V )and g ∈ I(V2)rI(V ). Therefore fg vanishes on V1∪V2 = V and thus fg ∈ I(V ).We conclude that I(V ) is not a prime ideal.Viceversa, suppose that I(V ) is not prime, then there exist f, g ∈ K0[X1, .., Xn]rI(V ) but fg ∈ I(V ). Let us define J1 = (I(V ), f) and J2 = (I(V ), g), thenV1 = V (J1), V2 = V (J2) are both strictly contained in V . On the other handV ⊂ V1∪V2 since if P ∈ V then fg(P ) = 0, so that f(P ) = 0 or g(P ) = 0 whichyields that P ∈ V1 or P ∈ V2. This shows that V is reducible.

Definition. Given an affine variety V , define its affine coordinate ring as

K[V ] = K[X1, .., Xn]/I(V )

Since I(V ) is a prime ideal, K[V ] is an integral domain, thus we can form itsquotient field K(V ) called function field of V .In case that V is defined over K0 we have K0[V ] = K0[X1, .., Xn]/I(V/K0).Since I(V ) is prime, so is I(V/K0): therefore K0[V ] is an integral domain aswell.

Remark 1.5. Recall that f is a polynomial function on V if there existsF ∈ K[X1, .., Xn] such that F (P ) = f(P ) ∀P ∈ V .Then K[V ] can be identified with the set f : V −→ K | f is a polynomial function

Definition. The dimension dim(V ) of an affine variety V is the transcendencedegree of K(V ) over K. In particular an affine curve is an affine variety ofdimension one.

Definition. Let V ⊂ An be an affine variety and let I(V ) = ⟨f1, ..., fr⟩. A pointP ∈ V is a nonsingular point if the matrix⎛⎜⎝

∂f1∂X1

(P ) . . . ∂fr∂X1

(P )...

...∂f1∂Xn

(P ) . . . ∂fr∂Xn

(P )

⎞⎟⎠ (1.1.1)


has rank n− dim(V ). V is nonsingular if each point P ∈ V is nonsingular.

Remark 1.6. These definitions do not depend upon the choice of the generatorsf1, .., fr of I(V ).

To see this, let P ∈ V and define

mP = f ∈ K[V ] | f(P ) = 0

mP is maximal in K[V ] since evaluation at P is an isomorphism from K[V ]/mP

to K and mP /m2P is a finite dimensional vector space over K.

Fact 1.7. P ∈ V is nonsingular if and only if dim(mP /m2P ) = dim(V )

Proof. See [Har77] I.5.1

Definition. Define the local ring of V at P as

K[V ]P = f/g ∈ K(V ) | g(P ) = 0

Its elements are said to be regular (or defined) at P .

K[V ]P is a local ring with maximal ideal

MP = f/g ∈ K[V ]P | f(P ) = 0 = mP ·K[V ]P

MP is maximal and consists of all non-invertible elements of K[V ]P , thereforeit is the unique maximal ideal.Let us now consider the natural map ι : mP −→MP /M

2P given by f ↦→ f +M2

P

• ker(ι) = mP ∩M2P = m2

P

• ι is surjective since for all f/g ∈MP we have f/g(P ) ∈ mP and

f

g(P )− f

g=f · (g − g(P ))g · g(P )

∈M2P

which means that ι(f/g(P )) = f/g +M2P .

We can conclude that ι induces an isomorphism mP /m2P

∼−→MP /M2P

Definition. The projective n-space over K is

Pn = Pn(K) = 0 = P = (x0, .., xn) ∈ An+1/ ∼

where (x0, .., xn) ∼ (y0, .., yn) if there exists λ ∈ K×

such that yi = λxi for alli = 0, .., n. We denote [x0, .., xn] such equivalence class, thus we can write

Pn(K) = [x0, .., xn] | xi ∈ K,∃i : xi = 0

x0, .., xn are called homogeneous coordinates of the corresponding point in Pn

and the set of K0-rational points in Pn is

Pn(K0) = [x0, .., xn] ∈ Pn | xi ∈ K0 ∀i


As for the affine case, the Galois group Gal(K/K0) acts on Pn by acting onhomogeneous coordinates:

[x0, .., xn]σ = [xσ0 , .., x

σn]

Definition. A polynomial f ∈ K[X0, .., Xn] is homogeneous of degree d if

f(λX0, .., λXn) = λdf(X0, .., Xn) for all λ ∈ K×.

An ideal J ⊂ K[X0, .., Xn] is homogeneous if it is generated by homogeneouspolynomials.

Definition. To any homogeneous ideal J ⊂ K[X0, .., Xn] we cab associate itszero set in Pn

WJ = P = [x0, .., xn] ∈ Pn | f(P ) = 0 ∀f ∈ J

called projective algebraic set.Viceversa, if W ⊂ Pn is a projective algebraic set, its homogeneous ideal ofpolynomials vanishing on it is the ideal I(W ) generated by all homogeneouspolynomials f ∈ K[X0, .., Xn] such that f(x0, .., xn) = 0 for all [x0, .., xn] ∈W .

As for the affine case, W is defined over K0 if I(W ) can be generated byelements of K0[X0, .., Xn] and we write W/K0. The set of K0-rational points ofW is W (K0) =W ∩ Pn(K0).

Definition. A projective algebraic set is irreducible if it is not union of twoproper projective algebraic sets. A irreducible projective algebraic set W iscalled projective algebraic variety.

Now notice that Pn(K) = U0 ∪ · · · ∪ Un where Ui = [x0, .., xn] ∈ Pn(K) |xi = 0.Then for each i ∈ 0, .., n the map Ui

ϕi−→ An(K) given by

[x0, .., xn] ↦→(x0xi, ..,

xi−1

xi,xi+1

xi, ..xnxi

)is a bijection (actually it is a homeomorphism with respect to the Zariski topol-ogy).It follows that any projective variety W has a standard covering

W =

n⋃i=0

Wi where Wi =W ∩ Ui

And by mean of ϕi, each Wi can be regarded as an affine variety.

Remark 1.8. The map W ↦→W0 =W ∩ U0 yields a bijectionprojective varietiesW ⊂ Pn |W * x0 = 0

←→

affine varietiesW0 ⊂ An

which inverse map is given by the projective closure. Namely if V is an affinealgebraic set with ideal I(V ), its projective closure is the projective set W = Vwhose homogeneous ideal is generated by Xd

0f(X1

X0, .., Xn

X0) | f ∈ I(V )


Definition. If W is a projective variety its function field is

K(W ) = f/g | f, g ∈ K[X0, .., Xn],homogeneous of same degree d, g /∈ I(W )/ ∼

where f/g ∼ f ′/g′ if and only if fg′ − gf ′ ∈ I(W ).The elements of K(W ) are called rational functions on W .

Lemma 1.9. Let W be a projective variety such that W * V (x0), with stan-dard covering W =W0 ∩ .. ∩Wn. We have an isomorphism

K(W )∼−→ K(W0)

Proof. We have the following maps are mutually inverse:

K(W ) −→ K(W0) given by f(X0, .., Xn)/g(X0, .., Xn) ↦→ f(1, X1, .., Xn)/g(1, X1, .., Xn)

K(W0) −→ K(W ) given by f(X1, .., Xn)/g(X1, .., Xn) ↦→ f(X1

X0, ..,

Xn

X0

)/g(X1

X0, ..,

Xn

X0

)

Definition. The dimension dim(W ) of a projective variety W is the transcen-dence degree of K(W ).

Definition. A rational function F ∈ K(W ) is regular at P ∈ W if there is arepresentation F = f/g such that g(P ) = 0. The local ring of W at P ∈W is

K[W ]P = F ∈ K(W ) | F is regular at P

with maximal ideal MP = F ∈ K[W ]P | F (P ) = 0.A point P ∈ W is a nonsingular point if dim(MP /M

2P ) = dim(W ); W is non-

singular if each point P is nonsingular.

Now let W1 ⊂ Pm and W2 ⊂ Pn be projective varieties in the respectiveprojective spaces.

Definition. A rational map F :W1 −→W2 is a tuple F = [f0, .., fn] of elementsof K(W1) such that ∀P ∈ W1 where f0, .., fn are all defined, then F (P ) =[f0(P ), .., fn(P )] belongs to W2.F is defined over K0 if W1/K0, W2/K0 and f0, .., fn can be multiplied by thesame invertible elements of K so that f0, .., fn ∈ K0(W1)

Definition. A rational map F = [f0, .., fn] : W1 −→ W2 is regular at P ∈ W1

if there exists g ∈ K(W1) such that gfi is regular at P and [gf0(P ), .., gfn(P )]is not the 0-tuple.A rational map regular at all points of W1 is called morphism.

Definition. We say that W1 and W2 are isomorphic if ∃F : W1 −→ W2 and∃G : W2 −→ W1 morphisms such that F G and G F are the respectiveidentity maps.W1/K0 and W2/K0 are isomorphic over K0 is F and G as above are definedover K0.

Definition. A rational map F :W1 −→ W2 is dominant if its image F (W1) isa Zariski dense subset of W2.

1.2. ALGEBRAIC CURVES 6

Now if F :W1 −→W2 is a rational map and F is dominant, then we have awell-defined K-algebra map

F ∗ : K(W2) −→ K(W1)

such that F ∗(g/h) = F ∗(g)/F ∗(h) = g F/h F .In particular if F is an isomorphism then F ∗ is an isomorphism of function fieldsand an isomorphism of local rings at each point.

1.2 Algebraic Curves

Proposition 1.10. Let C be a projective curve and P be a nonsingular point

on the curve. Then MP = (t) for some t ∈MP rM2P and

∞⋂h=1

MhP = 0

Proof. P ∈ C nonsingular implies that mP /m2P

∼−→ MP /M2P has dimension 1

as K-vector space, therefore mP /m2P = ⟨t⟩ for some t ∈ K[C]. We claim that

MP = t ·K[C]P .

Let Ndef= t ·K[C]P ⊆MP , and let us prove that the quotient MP /N is trivial,

so that they coincide.Since MP is a K[C]P -module, so is the quotient MP /N , and notice that

• MP · (MP /N) = (N +M2P )/N

• N +M2P = (K · t+M2

P )K[C]P

ThereforeMP · (MP /N) =MP /N

and by Nakayama’s lemma this can happen only if MP = N = (t).

Now let f ∈∞⋂h=1

MhP =

∞⋂h=1

(th) and write f = th · fh. Since fh = t · fh+1 we

obtain a chain of inclusions

(f1) ⊆ (f2) ⊆ ..

which has to stabilize due to the Noetherianity of K[C]P . Therefore there existsm > 0 such that (fm) = (fm+1), hence

fm+1 = afm = atfm+1 for some a ∈ K[C]P

Either at = 1 or fm+1 = 0. Since t ∈ MP , t is not invertible, and we canconclude that fm+1 = 0 =⇒ f = 0.

Definition. Any such t ∈MP rM2P is called uniformizer of C at P .

Define ordP (·) : K[C]P −→ Z ∪ ∞ as

ordP (f) =

+∞ if f = 0

minl ∈ Z>0 | f ∈M lP rM l+1

P if f = 0(1.2.1)

Remark 1.11. (i) If f = 0 and l = ordP (f), then f = a · tl for some a ∈K[C]×P .


(ii) We can extend ordP to the whole K(C), namely if F = f/g with f, g ∈K[C]P and g = 0 then define

ordP (F ) = ordP (f)− ordP (g)

This is well defined since if l = ordP (F ) then l is the unique power forwhich F/tl is a unit in K[C]P .

Proposition 1.12. ordP : K(C) −→ Z∪∞ satisfies the following properties:

(a) ordP (F ·G) = ordP (F ) + ordP (G)

(b) ordP (F +G) ≥ minordP (F ), ordP (G)

(c) K[C]P = F ∈ K(C) | ordP (F ) ≥ 0

(d) MP = F ∈ K(C) | ordP (F ) > 0

(e) ordP (F ) =∞⇐⇒ F = 0

In other words, K[C]P is a discrete valuation ring with valuation given byordP .

Theorem 1.13. Let C be a projective curve, P ∈ C be a nonsingular pointand let W be a projective variety. If F : C −→ W is a rational map, then F isregular at P .In particular it follows that if C is nonsingular, then F is automatically a mor-phism.

Proof. Let F = [f0, .., fn] with fj ∈ K(C) and let t be a uniformizer at P .Define m = min

0≤j≤nordP (fj), then ordP (t

−m · fj) ≥ 0 ∀j = 0, .., n and equality

holding for some j = j0. This means that t−m · fj is regular at P and (t−m ·fj0)(P ) = 0.Hence taking g = t−m exhibits that F is regular at P .

Definition. A proper subring R ⊂ K(C) containing K is called discrete valua-tion ring of K(C) over K if there exists a function v : K(C) −→ Z∪∞ calledvaluation, such that:

(a) v(FG) = v(F ) + v(G);

(b) v(F +G) ≥ minv(F ), v(G);

(c) v(F ) =∞ ⇐⇒ F = 0;

(d) R = F ∈ K(C) | v(F ) ≥ 0.

In particular if R is a discrete valuation ring of K(C) over K, then R is a

local ring with maximal ideal MRdef= F ∈ R | v(F ) > 0.

In particular, for a nonsingular projective curve C we have the following:

Proposition 1.14. Let C be a nonsingular projective curve and let R be adiscrete valuation ring of K(C) over K with valuation v. Then there exists P ∈ Csuch that v = c · ordP for some c ∈ Z>0, moreover for such P , R = K[C]P andMR =MP .


Proof. Let [x0, .., xn] be the projective coordinates of Pn, then we can regard xjas a function from C to K, mapping a point P to its j-th coordinate.For what follows, do not consider the xj ’s such that xj ∈ I(C) = f ∈K[X0, .., Xn] | f(P ) = 0 ∀P ∈ C. For the remaining indices, regard xj/x0as an element of K(C)

• If v(xj/x0) ≥ 0 ∀j pass to the affine curve C ∩ U0 = C0 (which is notempty since x0 = 0 somewhere in C).Elements of the affine coordinate ring K[C0] are polynomials in the xj/x0’s,therefore v ≥ 0 on the whole ring. It follows that K[C0] ⊂ R.

• If v(xj/x0) < 0 for some j, then let j0 be the index for which v(xj0/x0) isthe smallest possible. Then v(xj/xj0) = v(xj/x0) − v(xj0/x0) > 0 for allj.Changing j0 with 0 we can assume C0 is not empty and K[C0] ⊆ R.

Therefore Idef= MR ∩K[C0] is a proper ideal of K[C0] ∼= K[X1, .., Xn]/I(C).

Let J be the inverse image of I in K[X1, .., Xn] (so in particular I(C) ⊂ J). Thenthere are some points P ∈ An that annihilate all the elements of J , thereforethose P belongs to C and for each f ∈ I we have ordP (f) ≥ 0.We claim that ∀F ∈ K(C)

v(F ) > 0 and ordP (F ) ≥ 0 =⇒ ordP (F ) > 0 (1.2.2)

In fact since C ∩ U0 = ∅, by Lemma 1.9 we can identify K(C) with K(C0) andK[C]P with K[C0]P . Let F ∈ MR ∩ K[C0]P , F = f/g for f, g ∈ K[C0] andordP (g) = 0; g ∈ K[C0] ⊂ R, implies that f = F · g belongs to I and thereforeordP (f) > 0. It follows that ordP (F ) = ordP (f)− ordP (g) > 0.

The contrapositive of (1.2.2) is ordP (F ) = 0 =⇒ v(F ) ≤ 0; which appliedto 1/F yields

ordP (1/F ) = −ordP (F ) = 0 =⇒ 0 ≥ v(1/F ) = −v(F ) =⇒ v(F ) ≥ 0

and consequentelyordP (F ) = 0 =⇒ v(F ) = 0 (1.2.3)

Finally choose an uniformizer t ∈ K[C0] at P : since K[C0] ⊆ R we have thatv(t) ≥ 0 and if F ∈ K(C), F = tℓ · F0 with ℓ = ordP (F ) and ordP (F0) = 0 (sothat by (1.2.3) v(F0) = 0), then

v(F ) = v(tℓ) + v(F0) = ℓ · v(t) = v(t) · ordP (F )

Since R is a proper subring of K(C), v cannot be identically zero and v(t) ≥ 0:therefore v(t) > 0. We obtained that v = v(t) · ordP and c := v(t) > 0 as wewanted.

Theorem 1.15. Let L be a field of trancendental degree one and finitely gen-erated over K and let CL be the set of discrete valuation rings R of L over K.Then there exists a nonsingular projective curve C such that K(C) ∼= L and CL

can be canonically identified with the set of points of C.

Sketch of the proof.


Step 1 If B is an integral domain finitely generated as K-algebra then B is iso-morphic to a coordinate ring of an affine variety. Namely, let x1, .., xngenerate B over K, then the map Xj ↦→ xj yields a homomorphismK[X1, .., Xn] B which kernel I is prime since B is a domain. There-fore VI is the required affine variety.

Step 2 Let x ∈ L r K and consider the polynomial ring K[x]: then L is a finiteextension of the quotient field K(x). Now if B is the integral closure of K[x]in L, the construction in the previous step generate a nonsingular affinecurve (since in dimension 1 a ring is regular if and only if it is integrallyclosed)

Step 3 Let R ∈ CL, choose x ∈ R r K, then K[x] ⊆ R and since R is a discretevaluation ring, it is integrally closed in L. Hence B ⊆ R. Now let N =MR∩B, then N is maximal in B and therefore it corresponds to a uniquepoint of the affine curve constructed from B.

Step 4 Finally maps CL diagonally into the product of finitely many projectiveclosures (of affine curves). The image of CL will be the desired curve C.

Theorem 1.16. Let C1 be a projective nonsingular curve over K and let C2

be any curve over K. If F : C1 −→ C2 is a morphism, then either F (C1) is apoint (i.e. F is constant) or F (C1) = C2 (i.e. F is surjective). In the latter caseF ∗ is injective, K(C1) is a finite field extension of F ∗(K(C2)) and F is a finitemorphism.

Proof. See [Har77] II.6.8

Theorem 1.17. Let C1 and C2 be projective nonsingular curves over K and leti : K(C2) −→ K(C1) be an injective map fixing K. Then there exists a uniquenon constant map F : C1 −→ C2 such that F ∗ = i

Proof. Let C1 ⊂ Pm, and without loss of generality assume that C2 * V (x0).Then for each j = 0, ..,m, let fj := xj/x0 ∈ K(C2), and define

F = [1, i(f1), .., i(fm)] : C1 −→ C2

Such F satisfies F ∗ = i. If G = [g0, .., gm] is another map with the propertyG∗ = i, then gj/g0 = G∗(fj) = F ∗(fj) = i(fj) for all j = 0, ..,m, thereforeF = G.

Definition. Let F : C1 −→ C2 be a map between nonsingular projective curvesover K. Let us define the degree of F as

deg(F ) =

0 if F is constant

[K(C1) : F∗(K(C2))] otherwise

Moreover if F is not constant, then F is separable, inseparable, or purely insep-arable if the field extension K(C1)/F

∗(K(C2)) is respectively separable, insepa-rable or purely inseparable.

Remark 1.18. (i) If C1 and C2 are defined over K0, then

[K(C1) : F∗(K(C2))] = [K0(C1) : F

∗(K0(C2))]


(ii) Since any algebraic extension can be obtained as a separable extensionfollowed by a purely inseparable one, we denote degsF the degree of theseparable part and degiF the degree of the purely inseparable part of theextension K(C1)/F

∗(K(C2)). Then deg(F ) = degs(F ) · degi(F )

Corollary 1.19. Let F : C1 −→ C2 be a non constant map between nonsingularprojective curves over K. If deg(F ) = 1 then F is an isomorphism.

Proof. See [Sil09] II.4.1

Definition. Let F : C1 −→ C2 be a non constant map between nonsingularprojective curves. The ramification degree of F at a point P ∈ C1 is the positiveinteger

eF (P ) = ordP (F∗(t)) = ordP (t F )

where t = tF (P ) is a uniformizer at F (P ).F is unramified at P if eF (P ) = 1, and it is unramified if it is unramified ateach point of C1.

Notice that tF (P ) F = teP (F )P g for tP uniformizer at P and some g in K(C1)

such that ordP (g) = 0.

Proposition 1.20. Let F : C1 −→ C2 be a non constant map between nonsin-gular projective curves, then

(a) For all Q ∈ C2 we have∑

P∈F−1(Q)

eF (P ) = deg(F ).

(b) For all but finitely many Q ∈ C2 we have |F−1(Q)| = degs(F )

Proof. (a) See [Sha77] III.2.1

(b) See [Har77] II.6.8

Corollary 1.21. A map F : C1 −→ C2 between nonsingular projective curvesis unramified if and only if |F−1(Q)| = deg(F ) ∀Q ∈ C2.

Proof. By Proposition 1.20, we have that

|F−1(Q)| = deg(F ) ⇐⇒∑

P∈F−1(Q)

eF (P ) = |F−1(Q)|

Since eF (P ) ≥ 1, this happens if and only if eF (P ) = 1

1.2.1 Fields of positive characteristic

We now want to take a look on what happens in the case of charK = 0: theproblem in such situation is that field extensions of fields with prime character-istic are not necessarely separable.Let p be a fixed rational prime and Fp be the field of p elements with algebraicclosure Fp. Recall that for any q = pr, r ∈ Z>0, there exists a unique field Fq

of order q in Fp, namely the splitting field of Xq − X over Fp. In particularFp =

⋃r≥1

Fpr .


Definition. The q-th Frobenius map φq on Fp is φq : Fq −→ Fq , x ↦→ xq

Proposition 1.22. Let q = pr, r ∈ Z>0, then:

(a) Fq = Fp

(b) Let α ∈ Fq, then α ∈ Fqn if and only if φnq (α) = α

(c) φq is an automorphism of Fq; in particular ∀x, y ∈ Fq

φq(x+ y) = φq(x) + φq(y) and φq(xy) = φq(x)φq(y)

Proof. (a) It is a more general fact that ifM ⊂ L with L algebraic overM thenM = L.In fact if α ∈ L and L is algebraic over M , then α is algebraic over M , thusα ∈M . The other inclusion is trivial and equality follows.

(b) Let us first prove that Fq = α ∈ Fq | αq = α. Since F×q ⊂ Fq is a group of

order q − 1 we have that αq−1 = 1 for all α ∈ F×q and therefore ∀α ∈ Fq we

have αq = α. This proves the containment Fq ⊂ α ∈ Fq | αq = α.Now recall that a polynomial f(X) has multiple roots if and only if f(X) andits formal derivative f ′(X) have common roots. Since we are in characteristicp,

d

dX(Xq −X) = qXq−1 − 1 = −1

which means that Xq − X has no multiple roots and therefore we haveq = def(Xq −X) dinstict α ∈ Fq such that αq = α. Hence |Fq| = |α ∈ Fq |αq = α| and Fq ⊆ α ∈ Fq | αq = α therefore they must coincide. Finally,(b) follows by substituting q with qn.

(c) Remark that(pk

)= p!

k!(p−k)! and for all k = 1, .., p− 1 we have a factor p at

its numerator which does not cancel by the denominator, thus

∀k = 1, .., p− 1

(p

k

)= 0 mod p

It follows that φp(x+ y) = (x+ y)p = xp + yp = φp(x) + φp(y). Now sinceif q = pr then φq = φrp, we have

φq(x+ y) = φq(x) + φq(y) ∀x, y ∈ Fq

On the other hand the multiplicativity of φq follows by commutativity ofFq.Therefore φq is a homomorphism of fields, and this also gives us injectivity.Now if α ∈ Fq, then α ∈ Fqn for some n ∈ Z>0 and therefore α = φnq (α) =φq(φ

n−1q (α)). Hence α ∈ im(φq), therefore φq is surjective.

Fact 1.23. For q = pr, Fq/Fp is a Galois extension with cyclic Galois group oforder r generated by φp.

Definition. The Frobenius map on Fn

p is φp : Fn

p −→ Fn

p , (x1, .., xn) ↦→(xp1, .., x

pn)


Remark 1.24. φp : Fn

p −→ Fn

p is a bijection with fixed points Fnp , which induces

a bijection at level of projective classes

φp : Pn(Fp) −→ Pn(Fp) [x0, .., xn] ↦→ [xp0, .., xpn]

For brevity write X = (X0, .., Xn), e = (e0, .., en) and consider a homoge-neous polynomial f(X) =

∑eaeX

e ∈ Fp[X0, .., Xn], then define

f (q)(X) =∑e

φq(ae)Xe ∈ Fp[X0, .., Xn]

and since we are in characteristic p if follows that

f (q)(Xq) = (f(X))q (1.2.4)

Remark 1.25. Let q = pr, and let C be a projective curve over Fp with idealof polynomials I(C) = ⟨f1, .., fr⟩. Then we can define a new curve C(q) over Fp

which ideal of polynomial I(C(q)) is the ideal generated by f (q) | f ∈ I(C).We have that the Frobenius map on Pn induces a q-th Frobenius morphism

φq : C −→ C(q) given by [x0, .., xn] ↦→ [xq0, .., xqn]

In fact ∀[x0, .., xn] ∈ C

f (q)(φq(x0, .., xn)) = f (q)(xq0, .., xqn)

(1.2.4)=

(f(x0, .., xn)

=0

)qIn particular notice that if C/Fp then C(p) = C

Theorem 1.26. Notation as above, we have:

(a) φ∗q(K(C(q))) = K(C)qdef= fq | f ∈ K(C)

(b) φq is purely inseparable;

(c) deg(φq) = q

Proof. (a) Remark that every algebraically closed field is perfect, therefore eachelement of K is a q-th power. Hence (K[X0, .., Xn])

q = K[Xq0 , .., X

qn].

Now let f/g ∈ K(C) with f, g homogeneous polynomial of the same degree.Then for any φ∗q(f/g) ∈ φ∗q(K(C(q))) we have

φ∗q(f/g) = f(Xq0 , .., X

qn)/g(X

q0 , .., X

qn) = fq0 (X

q0 , .., X

qn)/g

q0(X

q0 , .., X

qn) ∈ K(C)q

where f0 and g0 are polynomials which coefficients to the q-th power arethe coefficients of f and g respectively.

On the other hand K(C)q is the subfield of K(C) of elements

f(X0, .., Xn)q/g(X0, .., Xn)

q

Therefore they coincide.

(b) Follows from (a).


(c) deg(φq) = ordP (tφq(P ) ·φq) = q where the first equality follows from Propo-sition 1.20 together with the fact that φq is a injective.

Corollary 1.27. Let C1, C2 be nonsingular projective curves, defined over afield of prime characteristic p. Then any map F : C1 −→ C2 factor as F = Fsφqwhere q = degi(F ), φq : C1 −→ C

(q)1 is the q-th Frobenius morphism and

Fs : C(q) −→ C2 is a separable morphism.

Proof. Let L = F ∗(K(C2))sep be the maximal separable extension of F ∗(K(C2))in K(C1). Then K(C1)/L is purely inseparable of degree q = degi(F ), and inparticular we have that K(C1)

q ⊂ L.From Theorem 1.26 K(C1)

q = φ∗q(K(C(q)1 )) and [K(C1) : φ

∗q(K(C

(q)1 ))] = q.

By comparing the degrees, it follows that L = φ∗q(K(C(q)1 )). Thus we have the

inclusionsF ∗(K(C2)) ⊂ φ∗q(K(C

(q)1 )) ⊂ K(C1)

which correspond to

C1φq−→ C

(q)1

Fs−→ C2

1.2.2 Riemann-Roch theorem

Definition. The abelian group of divisors of a nonsingular curve C, Div(C), isthe free abelian group generated by the points of C. In other words D ∈ Div(C)is a formal sum

D =∑P∈C

nP (P ) with nP ∈ Z, nP = 0 for almost all P

The degree of a divisor D is the (actual) sum deg(D) =∑

P∈C

nP ∈ Z.

The subgroup of degree-0 divisors is

Div0(C) = D ∈ Div(C) | deg(D) = 0

Remark that if C is defined over K0, then Aut(K/K0) acts on Div(C) by

Dσ =∑P∈C

nP (Pσ)

We say that D is defined over K0 if Dσ = D for all σ ∈ Aut(K/K0) The divisorof f ∈ K(C)× is

div(f) =∑P∈C

ordP (f) · (P ) =∑

P∈f−1(0)

ordP (f) · (P )−∑

P∈f−1(∞)

ordP (f) · (P )

Divisors of the form div(f) are called principal divisors and its set is denoted byDivl(C). Notice that principal divisors have degree 0. Define the Picard groupof C as the quotient

Pic(C) = Div(C)/Divl(C)


Remark 1.28. If F : C1 −→ C2 is a dominant morphism between nonsingularprojective curves, then F induces maps on divisors in both directions:

F ∗ :Div(C2) −→ Div(C1) F∗ : Div(C1) −→ Div(C2)

(Q) ↦→∑

P∈F−1(Q)

eF (P ) · (P ) (P ) ↦→ (F (P ))

Definition. Let C be a nonsingular projective curve, the space of meromorphicdifferential 1-forms on C, denoted ΩC , is the K-vector space generated by thesymbols dx for x ∈ K(C) such that ∀x, y ∈ K(C), ∀u ∈ K:

1. d(x+ y) = dx+ dy

2. d(xy) = xdy + ydx

3. du = 0

If F : C1 −→ C2 is a nonconstant map between nonsingular projectivecurves, then the map between function fields F ∗ : K(C2) −→ K(C1) induces amap on differentials F ∗ : ΩC2

−→ ΩC1given by∑

fidxi ↦→∑

F ∗(fi)d(F∗xi)

Proposition 1.29. Let C be a curve, then

(a) ΩC is a 1-dimensional K(C)-vector space.

(b) Let x ∈ K(C), then dx is a basis for ΩC over K(C) if and only if K(C)/K(x)is a finite separable extension.

(c) Let F : C1 −→ C2 be a non constant map between curves, then F isseparable if and only if F ∗ : ΩC2

−→ ΩC1is injective (equivalently, non

zero).

(d) Let P ∈ C and let t = tP be a uniformizer at P . Then for every ω ∈ ΩC

there exists a unique function f ∈ K(C) (depending on ω and t), such that

ω = fdt

Proof. See [Sil09] II.4.2, II.4.3

Definition. Let C be a curve, let ω ∈ ΩC , let P ∈ C and t be a uniformizer atP . The order of ω at P is ordP (ω) = ordP (f) if ω = fdt. Then we can definethe divisor associated to ω as

div(ω) =∑P∈C

ordP (ω) · (P ) ∈ Div(C)

We say that the differential ω is regular (or holomorphic) if ordP (ω) ≥ 0 for allP ∈ C.

Notice that if ω1, ω2 ∈ ΩC are nonzero, then there exists f ∈ K(C)× suchthat ω2 = fω1 and consequently div(ω2) = div(f) + div(ω1).


Definition. Therefore define the canonical class of C as the image in Pic(C)of div(ω) for any non zero differential ω ∈ ΩC .

Remark that for a curve C, there is a partial order in Div(C) as follows:we say that a divisor D =

∑P∈C

nP (P ) is positive (and write D ≥ 0) if nP ≥

0 ∀P ∈ C. Then for D1, D2 ∈ Div(C) the partial order is given by D1 ≥ D2 ifD1 −D2 ≥ 0

Definition. Let D be a divisor on a curve C, the linear system associated toD is the (finite-dimensional) K-vector space

L(D) = f ∈ K | div(f) +D ≥ 0

Denote its dimension over K by ℓ(D)

Definition. Let C be a curve defined over K, define the genus of C, denotedby g = g(C) as the dimension of L(κ) for any κ canonical divisor on C

Let κ be a canonical divisor, say κ = div(ω), then L(κ) = f ∈ K(C) |div(f) + κ ≥ 0, or in other words f ∈ L(κ), hence 0 ≤ div(f) + κ =div(f) + div(ω) = div(fω). This means that fω is holomorphic. Viceversa,if fω is holomorphic, then f ∈ L(κ). It follows that

L(κ) ∼= ω ∈ ΩC | ω is holomorphic =: Ω1(C)

Proposition 1.30. Let D be a divisor on a curve C, then

(a) If deg(D) < 0, then L(D) = 0 and ℓ(D) = 0.

(b) If D′ ∈ Div(C) is linearly equivalent to D i.e. there exists f ∈ K(C)× suchthat D′ = D + div(f), then L(D) ∼= L(D′) and consequently ℓ(D′) = ℓ(D).

Proof. (a) Let f ∈ L(D) r 0 then div(f) + D ≥ 0 then 0 ≤ deg(div(f)) +deg(D) = deg(D) ≤ 0 so we get a contradiction. Thus L(D) = 0.

(b) D′ = D + div(f) then the map L(D′) −→ L(D) such that g ↦→ g · f is anisomorphism of K-vector spaces.

Theorem 1.31. (Riemann-Roch) Let C be a nonsingular projective curveand let κ be a canonical divisor on C. Then for all divisors D ∈ Div(D)

ℓ(D)− ℓ(κ−D) = deg(D)− g + 1

Proof. See [Har77] IV.1

Corollary 1.32. Let κ be a canonical divisor on C and let D be a divisor onC, then:

(a) deg(κ) = 2g − 2

(b) If deg(D) > 2g − 2 then ℓ(D) = deg(D)− g + 1

Proof. (a) Apply Riemann-Roch theorem with D = κ, so that

g − 1 = ℓ(κ)− ℓ(0) = deg(κ)− g + 1

thus deg(κ) = 2g − 2.

1.3. RIEMANN SURFACES 16

(b) By (a), deg(κ−D) < 0, hence by Proposition 1.30 ℓ(κ−D) = 0, and thenthe result follows by Riemann-Roch.

Proposition 1.33. Let C/K0 be a nonsingular projective curve and let D be adivisor defined over K0. Then the K-vector space L(D) has a basis defined overK0 i.e. consisting of elements of K0(C).


Theorem 1.34. (Hurwitz) Let F : C1 −→ C2 be a nonconstant separablemap between nonsingular projective curves. Let g1 = g(C1) and g2 = g(C2).Then

2g1 − 2 ≥ (deg(F ))(2g2 − 2) +∑P∈C1

(eF (P )− 1)

Moreover, we have equality if and only if either char(K) = 0 or char(K) = p > 0and p does not divide eF (P ) for all P ∈ C1.


1.3 Riemann Surfaces

From now untill the end of the chapter, suppose that K = C. Then affineand projective spaces come with the complex topology, in addition to the Zariskitopology. Then one can naturally give the points of a variety over C a topol-ogy inherited from the subspace topology. A little extra work (with the inversefunction theorem and other analytic arguments) shows you that, if the variety isnonsingular, you have a nonsingular complex manifold. Nonetheless, in generalthe converse is false: there are many complex manifolds that do not arise fromnonsingular algebraic varieties in this manner.However, in dimension 1, a miracle happens, and the converse is true: all com-pact one dimensional complex manifolds (i.e. compact Riemann Surfaces) areanalytically isomorphic to the complex points of a nonsingular projective onedimensional variety (i.e. Curves), endowed with the complex topology insteadof the Zariski topology. (See [Gun66] Chapter 10)

It follows that all the terminology and results we developed about curves overan arbitrary algebraically closed field hold for compact Riemann surfaces. Inparticular the genus g of a compact Riemann surface has the costumary topo-logical meaning and one can show that the two definition coincide.

Let X be a Riemann surface of genus g ≥ 1, since X is a closed orientablesurface of genus g, the ordinary homology group H1(X,Z) is free abelian on2g generators. Let α1, .., αg, β1, .., βg be the standard basis over Z i.e. with theproperty that

αiβj = δij and αiαj = 0 = βiβj ∀i = j

Now consider the map∫: H1(X,Z) −→ Ω1(X)∨

def= HomC(Ω

1(X),C) given by

[γ] ↦→[ω ↦→

∫γ

ω

](1.3.1)


Since∫

: H1(X,Z) −→ Ω1(X)∨ is injective, we can view the elements of thefirst homology groups as mapping holomorphic differentials on X to C, moreexplicitely:

Proposition 1.35. Let X be a compact Riemann surface of genus g ≥ 1 andlet ω1, .., ωg be a basis over C for Ω1(X) (which has dimension g since it isisomorphic to L(κ) for a canonical divisor κ). Then the 2g vectors

λ1 =

⎛⎜⎝∫α1ω1

...∫α1ωg

⎞⎟⎠ , . . . , λ2g =

⎛⎜⎝∫βgω1

...∫βgωg

⎞⎟⎠ ∈ Cg

are linearly independent over R and therefore λ1, .., λ2g are a Z basis for a latticeΛ(X) in Cg.

Proof. See [Swi74] I.1.7

In particular since dimC(Ω1(X)) = g, that is also the complex dimension

of its dual space. It follows that H1(X,Z) free abelian subgroup of rank 2g,generated by 2g indipendent elements over R, hence it is a lattice in Ω1(X)∨.

Definition. The Jacobian variety of a compact Riemann surface X is

J(X) = Ω1(X)∨/H1(X,Z)

which is realized as the g-dimensional complex torus Cg/Λ(X)

Now fix a point x0 ∈ X and define the Abel-Jacobi map Φ : X −→ J(X) by

x ↦→(∫ x

x0ω1, . . . ,

∫ x

x0ωg

)Remark 1.36. Since J(X) is a group, we can extend, by Z-linearity, Φ to amap Φ : Div(X) −→ J(X) , D =

∑x∈X

ordx(D) · (x) ↦→∑x∈X

ordx(D) · (Φ(x))

For a generic divisor D, this definition depends on the base point x0, howeverfor a 0-degree divisor we have:

Lemma 1.37. The Abel-Jacobi map Φ : Div(X) −→ J(X) restricted toDiv0(X) is independent of the base point x0.

Proof. Let y0 be another base point and let γ be a path from x0 to y0. Then inthe formula of Φ(x), if we change x0 to y0, we see that the image changes by

λ =(∫

γω1, . . . ,

∫γωg

)mod Λ(X)

Such element λ ∈ J(X) is indipendent of x, hence if∑nx = 0, then Φ

(∑nx ·

(x))changes by

∑nx · (λ) = (λ) ·

∑nx = 0.

Theorem 1.38. (Abel) Let X be a compact Riemann surface of genus g ≥ 1and let D be a divisor on X. Then

D is principal ⇐⇒ deg(D) = 0 and Φ(D) = 0 ∈ J(X)

Or in other words the map from Pic0(X) to J(X) given by[∑x

nx · x

]↦→∑x

nx

∫ x

x0

is an isomorphism.


Proof. See [Mir95] VIII.2.2

Corollary 1.39. Let X be a compact Riemann surfact of genus g ≥ 1. Thenthe Abel-Jacobi map is injective

Proof. Assume is not, then let Φ(x) = Φ(y) for some x = y ∈ X. Since Φis additive, Φ(x − y) = 0. Therefore by Abel’s theorem, x − y is a principaldivisor, which means that there exists some meromorphic function f on X witha simple pole at y and no other poles. This implies that f is an isomorphismbetween X, of genus greater than 1, and P1(C), which has genus 0, hence acontradiction.

Proposition 1.40. Let X be a compact Riemann surface of genus 1, then J(X)is isomorphic to X.

Proof. Since X has genus one, J(X) is a one-dimensional complex torus, henceit is also a compact Riemann surface of genus one. Therefore for X of genusone, Φ is an injective holomorphic map between compact Riemann surfaces,therefore an isomorphism.

Chapter 2

Elliptic Curves

2.1 Elliptic curves over an arbitrary field

2.1.1 Weierstrass form & abstract elliptic curves

Definition. A Weierstrass equation over a field K is a cubic equation (in ho-mogeneous coordinates) of the form

E : Y 2Z + a1XY Z + a3Y Z2 = X3 + a2X

2Z + a4XZ2 + a6Z

3, a1, .., a6 ∈ K0

(2.1.1)If a1, .., a6 ∈ K0 then we say that E is defined over K0 .

We identify the Weierstrass equation E with the projective curve in P2 given

by the points that annihilate F (X,Y, Z)def= Y 2Z + a1XY Z + a3Y Z

2 − X3 −a2X

2Z − a4XZ2 − a6Z3 i.e.

E = [X,Y, Z] ∈ P2 | Y 2Z+a1XY Z+a3Y Z2−X3−a2X2Z−a4XZ2−a6Z3 = 0

Remark that the only K-rational point on the line at infinity Z = 0 is O =

[0, 1, 0], which is nonsingular since ∂F (X,Y,Z)∂Z (O) = 1 = 0.

Thus we can study the curve by working with the non-homogeneous coordinatesx=X/Z, y=Y/Z:

E : y2 + a1xy + a3y = x3 + a2x2 + a4x+ a6 (2.1.2)

with associated curve

E = (x, y) ∈ A2 | y2 + a1xy+ a3y−x3− a2x2− a4x− a6 = 0∪ ∞ = [0, 1, 0]

Let us now simplify the equation in (2.1.2) under some assumption regardingthe characteristic of K.First assume that Char(K) = 2, then the variable change

y ↦→ (y − a1x− a3)2

completes the square and yields

y2 = 4x3 + b2x2 + 2b4x+ b6 (2.1.3)

19

2.1. ELLIPTIC CURVES OVER AN ARBITRARY FIELD 20

where b2 = a21 + 4a2, b4 = 2a4 + a1a3 and b6 = a23 + 4a6.Moreover, if char(K) = 3, then

(x, y) ↦→(x− 3b2

36,y

108

)produces the equation

y2 = x3 − 27c4x− 54c6 (2.1.4)

where c4 = b22 − 24b4 and c6 = −b32 + 36b2b4 − 216b6

Definition. Let b8 = a21a6 + 4a2a6 − a1a3a4 + a2a23 − a24, then define the dis-

criminant ∆ of the curve given by the equation in (2.1.2) by

∆ = ∆(E) = −b22b8 − 8b34 − 27b26 + 9b2b4b6 (2.1.5)

If ∆ = 0 it makes sense to define the so-called j-invariant of E

j = jE = c34/∆ (2.1.6)

Lastly, define the invariant differential associated to E.

ω =dx

2y + a1x+ a3=

dy

3x2 + 2a2x+ a4 − a1y(2.1.7)

Remark that for char(K) = 2, 3 we have

∆ =c34 − c261728

(2.1.8)

Definition. An admissible change of variable in a Weierstrass equation is oneof the form

X = u2X ′ + r ; Y = u3Y ′ + su2X ′ + t ; Z = Z ′ for u, r, s, t ∈ K, u = 0(2.1.9)

This change of variable fixes the point at infinity [0, 1, 0] and carries the lineZ = 0 to itself. Moreover one can compute all the new coefficients and seethat

u12∆′ = ∆ ; j′ = j and ω′ = uω (2.1.10)

Proposition 2.1. The curve E given by a Weierstrass equation is nonsingularif and only if ∆ = 0

Proof. Let E = (x, y) ∈ A2 | f(x, y) = y2+a1xy+a3y−x3−a2x2−a4x−a6 =0 ∪ ∞.We already saw that the point at infinity is never singular.Now suppose that E is singular at a some point P = (x0, y0) and consider theadmissable change of variable given by x′ = x−x0 and y′ = y− y0 which leaves∆ invariant. Therefore we can assume that E is singular at the point (0, 0) ∈ A2.Singularity means that

0 = f(0, 0) = −a6 ; 0 =∂f(x, y)

∂x(0, 0) = −a4 ; 0 =

∂f(x, y)

∂y(0, 0) = a3


It follows that f(x, y) = y2 + a1xy − a2x2 − x3 and consequently

b2 = a21 + 4a2 and b4 = b6 = b8 = 0 =⇒ ∆ = −(a2 + 4a2)2 · 0 + 0 = 0

where ∆ is calculated using the formula in (2.1.5).For the other implication assume that char(K) = 2 to simplify the computation.As in (2.1.3),

f(x, y) = y2 − 4x3 − b2x2 + 2b4x+ b6

Then E is singular if and only if there is a point (x0, y0) such that y20 = 4x30 −b2x

20 + 2b4x0 + b6 and

0 =∂f

∂x(x0, y0) = 12x20 + 2b2x0 + 2b4 ; 0 =

∂f

∂y(x0, y0) = 2y0

The last equality forces y0 = 0 and thus (x0, 0) is singular if and only if it is adouble root of the polynomial q(x) = 4x3 − b2x2 + 2b4x + b6 if and only if thediscriminant of q(x) is 0. Recall that for such polynomial its discriminant is

∆q = b22 · (2b4)2 − 16(2b4)3 − 4b32b6 − 27 · 16b26 + 18 · 4b2(2b4)b6 =

= 4(b22b24 − 32b34 − b32b6 − 108b26 + 36b2b4b6) =

(∗)= 4(−4b22b8 − 32b34 − 108b26 + 36b2b4b6 = 16∆(E)

where the equality in (∗) follows from the fact that b24 − b2b6 = −4b8. Thiscompletes the proof for char(K) = 2. The case of characteristic 2 is treated in[Sil09] A.1.2

Proposition 2.2. Let E be a singular curve given by a Weierstrass equation.Then E is birational to P1.

Proof. As in the proof of Proposition 2.1, after an admissible linear change ofvariables we assume that E is singular at the point (0, 0) and thus E = (x, y) ∈A2 | y2 + a1xy = x3 + a2x

2.Then the rational maps

E →→ P1 P1 →→ E

(x, y) →→ [x, y] [x, y] →→(

y2+a1xy−a2xx2 , y

3+a1xy2−a2x

2yx3

)are mutually inverse.

Definition. An elliptic curve is a pair (E,O), where E us a nonsingular curveof genus 1 and O belongs to E. The elliptic curve is defined over K0 if E isdefined over K0 and O ∈ E(K0).

Proposition 2.3. Let (E,O) be an elliptic curve defined over K0. Then thereexist f, g ∈ K0(E) such that the morphism

F : E −→ P2, F = [f, g, 1]

induces an isomorphism of E to a nonsingular curve given by a Weierstrassequation

Y 2Z + a1XY Z + a3Y Z2 = X3 + a2X

2Z + a4XZ2 + a6Z

3

for a1, .., a6 ∈ K0 such that F (0) = [0, 1, 0].


Proof. • Claim 1: There exists f, g ∈ K0(E) that satisfy a Weierstrass equa-tion.

By Riemann-Roch theorem we have that, since g = g(E) = 1, the dimensionof the linear system L(n(O)) is n for any n ≥ 1. By Proposition 1.33 we canchoose a basis of L(n(O)) defined over K0. Therefore let f ∈ K0(E) be such that1, f is a basis over K0 for L(2(O)) and then choose any g so that 1, f, g isa basis over K0 for L(3(O)). Notice that f must have a pole of order exactly 2at O and g a pole of order exactly 3 at O. Therefore we have that

Space L(2(0)) L(3(0)) L(4(0)) L(5(0)) L(6(0))Dimension 2 3 4 5 6Generators 1, f 1, f, g 1, f, g, f2 1, f, g, f2, fg 1, f, g, f2, fg, f3, g2

Since seven functions are in L(6(O)), they must be linearly dependent. In par-ticular the coefficients of f3 and g2 cannot be 0, otherwise we would have a poleof order 6 which doesn’t cancel. After scaling coefficients we will have a linearrelation

f3 + a2f2 + a4f + a6 = g2 + a1gf + a3g

Therefore we have a map F : E −→ P2 given by F = [f, g, 1] whose image is inthe curve X3 + a2X

2Z + a4XZ2 + a6Z

3 = Y 2Z + a1XY Z + a3Y Z2 =: C.

By Theorem 1.13 and Theorem 1.16 F is a surjective morphism. Moreover re-mark that since f (respectively g) has a pole of order 2 (resp. 3) at O, locallywe have f = 1

z2 (1 + ...) (resp. g = 1z3 (1 + ...)), therefore F (O) = [f, g, 1]|O =

[z3f, z3g, z3]|O = [0, 1, 0].

• Claim 2: F has degree 1, so that it is an isomorphism by Corollary 1.19.

In fact by Proposition 1.20 the morphisms [f, 1] : E −→ P1 and [g, 1] : E −→ P1

have degree 2 and 3 respectively. Hence [K0(E) : K0(f)] = 2 and [K0(E) :K0(g)] = 3, and this implies that [K0(E) : K0(f, g)] divides 2 and 3 therefore[K0(E) : K0(f, g)] = 1.

• Claim 3: C is a nonsingular curve.

Suppose not, then by Proposition 2.2 there is a degree 1 rational map h : C −→P1 and therefore hF is a degree 1 morphism between nonsingular curves, hencean isomorphism by Corollary 1.19. This is a contradiction since E has genus 1and P1 has genus 0 but the genus of a curve is a topological invariant. We canconclude that C is nonsingular and F is an isomorphism.

Also the converse holds true:

Proposition 2.4. If a curve E is given by a nonsingular Weierstrass equation(2.1.2) over K and O is taken as the usual point at infinity, then (E,O) is anelliptic curve.

Proof. To see this we only need to compute the genus of E. One first provesthat for the invariant differential ω, div(ω) is a canonical divisor of degree 0(See [Sil09] III.1.5). Then Riemann-Roch theorem with D = div(ω) yields

g = dimL(div(ω)) = deg(div(ω)) + dimL(0)− g + 1 = 0 + 1− g + 1


Hence g = −g + 2 ⇒ g = 1.

Proposition 2.5. Let (E,O) and (E′, O′) be two elliptic curves defined overK0 given by Weierstrass equations over K0 with O and O′ corresponding to∞. If there exists an isomorphism F : E −→ E′ defines over K0 such thatF (∞) = ∞, then E and E′ are related by an admissible change of variable(2.1.9) with coefficients in K0.

Proof. Suppose that E, respectively E′, corresponds to the set of coordinates(f, g), resp. (x, y), as in Proposition 2.3. Then 1, f and 1, x are bases ofL(2(O)) and L(2(O′)) respectively, while 1, f, g and 1, x, y are bases ofL(3(O)) and L(3(O′)). Since F is an isomorphism of curves, it induces an iso-

morphism between function fields. Hence f ′def= F ∗(x) = x F ∈ L(2(O)) and

g′ = F ∗(y) ∈ L(3(O)). In other words, there exist u1, u2, r, s1, t ∈ K0, u1, u2 = 0such that

f ′ = u1f + r and g′ = u2g + s1f + t

Since (f, g), (f ′, g′) satisfy the Weierstrass equation in E, we obtain that u31 =u22. Letting u = u2/u1 and s = s1/u

2 gives a change of variable as in (2.1.9).

Let us now introduce a group operation on the elliptic curve.

Definition. Let (E,O) be an elliptic curve, regarded as a nonsingular curve inP2. Let P,Q ∈ E and consider a line ℓ through these two points (we understandthat the line is tangent to the curve at P if P = Q). By Bezout’s theorem thereis a third point R in ℓ ∩E, thus let ℓ′ be the line through R and O. Define thethird point in ℓ′ ∩ E as P +Q.

Proposition 2.6. This operation makes E into an abelian group with O asidentity element. Moreover the inclusion E(K0) ⊂ E(K) is a group homomor-phism.

Proof. The core of the proof is to show associativity (See[Kna93] Theorem3.8). The fact that the operation is commutative and that O is the identityelement follows immediately from the above definition. Since by definition andassociativity, three points P,Q,R on a line sum to O means that Q+R = −Pi.e. we have an additive inverse.

Fact 2.7. One can prove that for an elliptic curve over K0 given by a Weierstrassequation over K0, addition and negative are morphisms defined over K0.

Proof. See [Sil09] III.3.6

We can now reformulate Proposition 1.40 and obtain a description of theaddition on E by mean of divisors.

Theorem 2.8. Let (E,O) be an elliptic curve over K, then the map φ :Div(E) −→ E given by

∑nP (P ) ↦→

∑[nP ]P induces an isomorphism

Pic0(E) = Div0(E)/Divl(E)∼−→ E

Therefore the principal divisors of E are characterized by∑nP (P ) ∈ Divl(E) ⇐⇒

∑nP = 0 and

∑[nP ]P = O (2.1.11)


2.1.2 Isogenies

From now on we write E for an elliptic curve, and drop the reference to thegroup identity, which will be denoted O and for Weierstrass equations will bethe point at ∞.

Definition. Let E and E′ be elliptic curves over K. An isogeny between themis a non-constant morphism F : E −→ E′ such that F (O) = O.

Notice that by Theorem 1.16 all isogenies are surjective morphisms.

Proposition 2.9. Let F : E −→ E′ be a morphism such that F (0) = 0, thenF is a group homomorphism.

Proof. If F = 0 there is nothing to prove. Assume F = 0, then F is a finitemorphism and by Remark 1.28 we have a homomorphism F∗ : Div(E) −→Div(E′) which induces a homomorphism, denoted again by F∗, Pic

0(E) −→Pic0(E′). Therefore by Theorem 2.10, we can write F as the composition ofthree homomorphisms

F : E ∼= Pic0(E)F∗−→ Pic0(E′) ∼= E′

Example 2.10. Let E be an elliptic curve then

1. The multiplication by an integer, denoted by [m], is an isogeny from E toE.

2. If E is defined over Z/pZ then the Frobenius map is an isogeny.

Let E and E′ be elliptic curves over K, denoted by Hom(E,E′) the set ofall isogenies from E to E′.

Remark 2.11. Hom(E,E′) is an abelian group with pointwise sum.

Proof. Let F,G ∈ Hom(E,E′), then F +G is the composition of

E −→ E × E such that P ↦→ (P, P )

F ×G : E × E −→ E′ × E′ such that (P,Q) ↦→ (F (P ), G(Q))

+ : E′ × E′ −→ E′ such that (P,Q) ↦→ P +Q

Moreover, if E = E′, Hom(E,E) = End(E) is a ring, with multiplicationgiven by composition of morphisms.For P ∈ E define a translation map TP : E −→ E by Q ↦→ Q+ P . Then T ∗

P isan automorphism of K(E).

Proposition 2.12. Let F : E −→ E′ be a nonzero isogeny between ellipticcurves over K. Then ker(F ) is finite and the map P ↦→ T ∗

P induces an isomor-phism

ker(F ) ∼= Aut(K(E)/F ∗(K(E′))

)Furthermore, if F is separable then F is unramified, deg(F ) = #ker(F ) andK(E)/F ∗(K(E′)) is a Galois extension.


Proof. Since F is non constant, it surjects E to E′ and it is a finite morphism,thus is particular has finite fibers, hence finite kernel.Now if P ∈ ker(F ) and f ∈ F ∗(K(E′)) then f = g F for some g ∈ K(E′).Hence T ∗

P (g F ) = g F TP = g F = f since F TP = F ∀P ∈ ker(F ).Therefore T ∗

P fixes F ∗(K(E′)) for any P ∈ ker(F ). Moreover

T ∗P T ∗

Q(f) = f TQ TP = f TP+Q = T ∗P+Q(f) = T ∗

Q+P (f)

Hence P ↦→ T ∗P is a homomorphism.

By Corollary 1.20 (b), #F−1(Q) = degs(F ) for all but finitely many Q ∈ E′.Now if F is an isogeny, the equality holds for all Q ∈ E′, in fact ∀Q1, Q2 ∈ E′

there exists (by surjectivity of F ) R ∈ E such that F (R) = Q1−Q2 and since Fis a homomorphism, the map TR : F−1(Q1) −→ F−1(Q2) sending P ∈ F−1(Q1)to P + R ∈ F−1(Q2) is a bijection. It follows that #F−1(Q) = degs(F ) for allQ ∈ E′. In particular #ker(F ) = #F−1(O) = degs(F ) and by Galois theory

#Aut(K(E)/F ∗(K(E′))

)≤ degs(F ) = #ker(F )

It follows that it is enough to prove that the map P ↦→ T ∗P is injective. But this

is obvious since if T ∗P = T ∗

Q then TP = TQ and therefore

∀ R ∈ E, R+ P = TP (R) = TQ(R) = Q+R

Thus in particular P = O + P = O +Q = Q.Finally assume that F is separable, then

#F−1(Q) = degs(F ) = deg(F ) ∀Q ∈ E′

hence F is unramified by Corollary 1.21 and setting Q = O produces

#ker(F ) = #F−1(O) = deg(F )

By the isomorphism P ↦→ T ∗P it follows

[K(E) : F ∗(K(E′))] = deg(F ) = #ker(F ) = #Aut(K(E)/F ∗(K(E′))

)

Lemma 2.13. 1− φq is a separable isogeny for any q = pr, r ≥ 1.

Proof. Assume not, then 1− φq factors as

1− φrp = 1− φq = f φsp

for some f : E −→ E separable morphism and s ≥ 1. Hence

1 = f φsp + φrq =

(f φs−r

p + 1) φr−1p φp if s ≥ r

(f + φr−sp ) φs−1

p φp if r > s

which contradicts the fact that 1 is an isomorphism, while the right hand sidecannot be since φp is an inseparable map of degree p.


Remark 2.14. Let E be an elliptic curve defined over Fp, then Proposition2.12 gives us a way of computing #E(Fp). We saw that the Frobenius map φfixes exactly those points of E that are in E(Fp), or in other words

Q ∈ E(Fp) ⇐⇒ φp(Q) = Q ⇐⇒ ([1]−φp)(Q) = O ⇐⇒ Q ∈ ker([1]−φp)

hence #E(Fp) = #ker([1]− φp).Since 1− φp is a separable isogeny, by Proposition 2.12

#E(Fp) = #ker([1]− φp) = deg([1]− φp) (2.1.12)

Lemma 2.15. Let F : E1 −→ E2 and G : E1 −→ E3 be nonzero isogeniesbetween elliptic curves over K. Let F be separable and such that ker(F ) ⊆ker(G). Then there exists a unique isogeny H : E2 −→ E3 such that G = H F .Moreover, if F and G are both defined over K0, so is H.

Proof. By Proposition 2.12 we have

Aut(K(E1)/F

∗(K(E2)))∼= ker(F ) ⊂ ker(G) ∼= Aut

(K(E1)/G

∗(K(E3)))

Hence is σ fixes F ∗(K(E2)), then it also fixes G∗(K(E3)). Moreover since Fis separable K(E1)/F

∗(K(E2)) is a Galois extension, therefore G∗(K(E3)) ⊂F ∗(K(E2)) is an inclusion of fields. Theorem 1.17 yields a nonconstant morphismH : E2 −→ E3 such that F ∗ H∗ = G∗. By uniqueness follows that G = H F ,and this implies that H is an isogeny since O = G(O) = H(F (O)) = H(O) asrequired.If F and G are defined over K0 we can apply Theorem 1.17 to G∗(K0(E3)) ⊂F ∗(K0(E2)) to produce H defined over K0 aswell.

Theorem 2.16. Let F : E −→ E′ be a nonzero isogeny of degree m, betweenelliptic curves over K. Then there exists a unique isogeny F : E′ −→ E suchthat F F = [m]. Such F is called dual isogeny of F .

Proof. Factor F as F = Fs φq for Fs a separable morphism and the q-thFrobenius map φq. Then it is enough to prove existence of dual isogeny for each

factor. In fact, assume there exist Fs and φq, then

(φq Fs) (Fs φq) = φq [degs(F )] φq = [degs(F )] φq φq = [m]

Hence φq Fs = (Fs φ1).

• Let Fs be a separable morphism of degree m, then #ker(Fs) = m andtherefore [m] : E −→ E annhilates ker(F ). It follows that ker(F ) ⊂ker([m]) and we found ourselves in the assumption of Lemma 2.15 withFs : E −→ E′ and [m] : E −→ E. Therefore there exists a uniqueH : E′ −→ E such that [m] = H Fs. Thus H = Fs.

• Let φq be the q-th Frobenius map for q = pr, then φq = φrp. Thus it

it enough to prove existence of φp. Recall that deg(φp) = p, and sincechar(K) = p, multiplication by p is not separable1. Since [p] is not sepa-

rable, it has a factorization [p] = Gs φhp , thus take φp = Gs φh−1p .

1 It follows from Proposition 1.29 (c), since [p]∗ : ω ↦→ 0, where ω is the invariant differential

2.2. ELLIPTIC CURVES OVER C 27

Finally let us prove uniqueness: let F , F ′ be two dual isogeny of F , then

(F − F ′) F = [m]− [m] = 0

But by assumption F is nonzero, therefore F = F ′.

Definition. We say that E is isogenous to E′ if there exists a non zero isogenyfrom E to E′. The Theorem implies that ”being isogenous” is an equivalencerelation.

The next result is a sort of converse of Proposition 2.12

Theorem 2.17. Let E be an elliptic curve over K and let S be a finite subgroupof E. Then there exist an elliptic curve E′ unique up to isomorphism and aseparable isogeny F : E −→ E′ such that ker(F ) = S. Moreover if E is definedover K0 and S is stable under Aut(K/K0), then E

′ is also defined over K0. Suchelliptic curve is often denoted E/S.

Proof. We give a proof in the case char(K) = 0, for a proof in positive charac-teristic see [DS06 Section 7.8].To any P ∈ S associate the automorphism T ∗

P : f ↦→ f TP , ∀f ∈ K(E). SinceT ∗P+P ′ = T ∗

P T ∗P ′ , S can be identified with a finite subgroup of Aut(K(E)),

namelyS ∼= T ∗

P | P ∈ S

Let K(E)S be the subfield of K(E) fixed by the elements in S, then K(E)/K(E)S

is a Galois extension with Galois group Gal(K(E)/K(E)S) ∼= S. Moreover,K(E)S is a function field of dimension one over K, thus there exist a uniquenonsingular projective curve E′ over K and a K-isomorphism F ∗ : K(E′)

∼−→K(E)S → K(E). The fields-curves correspondence in Theorem 1.17 yields asurjective morphism F : E −→ E′.We are left to show that such F is unramified and the genus of E′ is 1. For anyX ∈ E, let Y = F (X) ∈ E′, then ∀P ∈ S, ∀f ∈ K(E)S we have

f(X + P ) = f TP (X) = T ∗P f(X) = f(X) =⇒ F (X + P ) = F (X) = Y

=⇒ X + S = X + P | P ∈ S ⊆ F−1(Y )

On the other hand we have

#F−1(Y ) ≤∑

X∈F−1(Y )

eX(F ) = deg(F ) = [K(E) : F ∗(K(E′)] = [K(E) : K(E)S ] = #S

where the last equality follows by Galois theory. Hence we can conclude thatF−1(Y ) = X + S and therefore F is unramified and deg(F ) = #S.Finally Theorem 1.34 in characteristic 0 yields that the genus of E′ is 1.

2.2 Elliptic curves over CThe theory of elliptic curves over C becomes much easier, since it is equiva-

lent (in a categorial sense) to the theory of lattices in C or, equivalently again,to the one of complex tori.


Definition. (i) A lattice Λ in C is a subgroup of (C,+) of the form Λ =Zw1 ⊕ Zw2 such that Λ⊗Z R = Rw1 +Rw2 = C i.e. w1, w2 is a R-basisfor C.

(ii) An ordered basis

(w1

w2

)of a lattice Λ is normalized if w1/w2 ∈ H

where H := z ∈ C | Im(z) > 0.

(iii) Two lattices Λ and Λ′ are homothetic if there exists α ∈ C× such thatΛ′ = αΛ.

Remark 2.18. (a) (i) is equivalent to saying that w1, w2 ∈ C× and w1/w2 ∈Cr R.

(b) Λ is a lattice if and only if Λ is a free rank 2 discrete subgroup of (C,+).

Let us denote by L the set of all lattices and by B the set of normalizedbases.

Then we have a surjective map π : B → L given by

(w1

w2

)↦→ Zw1 + Zw2.

Lemma 2.19. The fibers of π are all of the form SL2(Z) · b for b ∈ B. Hence πinduces a bijection π : SL2(Z)\B −→ L

Proof. Notice that b =

(w1

w2

)and b′ =

(w′

1

w′2

)are the same basis for a lattice Λ

if and only if there exists a matrix γ =

(a bc d

)∈ GL2(Z) such that

(w′

1

w′2

)=

(a bc d

)(w1

w2

)=

(aw1 + bw2

cw1 + dw2

)Setting τ = w1/w2 and τ ′ = w′

1/w′2 gives that τ, τ ′ belong to C r R and τ ′ =

aτ+bcτ+d . Noticing that Im(τ ′) = det(γ)

|cτ+d|2 Im(τ) we deduce that b, b′ are normalized

basis if and only if det(γ) > 0, therefore γ ∈ SL2(Z).

Notice that there is an action of C× both on L and B. Namely, let α ∈ C×,

Λ ∈ L and

(w1

w2

)∈ B, then αΛ ∈ L and

(αw1

αw2

)∈ B.

In particular we have an identification B/C× −→ H given by

[w1

w2

]↦→ w1/w2.

The actions of C× and SL2(Z) on B commute: therefore the action of SL2(Z)on B induces an action of SL2(Z) on H by fractional linear transformation

SL2(Z)× H −→ H((a bc d

), τ)↦→ aτ + b

cτ + d

(2.2.1)

Corollary 2.20. The bijection π induces a natural bijection

π : SL2(Z)\H −→ L/C×

In particular any lattice Λ is homothetic to a lattice of the form Λτ = Zτ+Zfor some τ ∈ H.


Remark 2.21. We will see that the set L/C× = [Λ] = homothety class of the lattice Λis in bijection with the set of isomorphic classes of complex tori [C/Λ] =isomorphism class of C/Λ where isomorphism means holomorphic isomorphism.Moreover, the set of isomorphism classes of 1-dimensional complex tori is in bi-jection with the set of isomorphism classes of elliptic curves over C (with respectto algebraic isomorphism).

2.2.1 The Weierstrass ℘-function

Definition. Let Λ = Zw1 + Zw2 be a lattice in C. A meromorphic functionf : C −→ C is an elliptic function of period Λ if f(z +w) = f(w) for all w ∈ Λ.This amount to saying that f(z + wi) = f(z) for i = 1, 2.

Definition. Let Λ = Zw1 + Zw2 be a lattice in C, the function ℘ : C −→ P1

given by

℘Λ(z) = ℘(z) =1

z2+∑w∈Λw =0

( 1

(z − w)2− 1

w2

)is called Weierstrass ℘-function for Λ.

Proposition 2.22. Let Λ = Zw1 +Zw2 be a lattice in C, and ℘ be the Weier-strass ℘-function for Λ, then:

(a) ℘(z) is an even function whose singularities are double poles at lattice points;

(b) ℘ is an elliptic function of period Λ;

(c) deg(℘) = 2;

(d) ℘ is surjective.

Proof. (a) One first proves that∑

w∈Λw =0

(1

(z−w)2 −1w2

)converges absolutely and

uniformely on compact subsets of Cr Λ.Then to see that it has a double pole at lattice points, notice that thefunction hv(z) = ℘(z)− 1

(z−v)2 for some v ∈ Λ converges uniformely on (CrΛ)∪v. Hence in a neighbourhood of v, ℘(z) = 1

(z−v)2 +holomorphic part.

Finally absolute and uniform convergence on compact subsets allow us toreorder the summands and therefore since Λ = −Λ

℘(−z) = 1

(−z)2+∑w∈Λw =0

( 1

(−z − w)2− 1

w2

)=

1

z2+∑w∈Λw =0

( 1

(z − (−w)2− 1

(−w)2)= ℘(z)

(b) Since ℘ is meromorphic on C, its derivative ℘′ is also meromorphic. Moreover

℘′(z) = − 2z3 +

∑w∈Λw =0

(−2

(z−w)3

)= −2

∑w∈Λw =0

(1

(z−w)3

), hence ℘′(z + w) =

℘′(z) for all w ∈ Λ since w + Λ = Λ. Therefore ℘′ is an elliptic function,hence ℘(z +wi) = ℘(z) + ci for some ci ∈ C. Taking z = −wi/2 yields that℘(wi

2 ) = ℘(−wi

2 ) + ci. But by (a), ℘ is even, thus ci = 0 for both i = 1, 2.We conclude that ℘ is an elliptic function.

(c) Follows from Proposition 1.20 together with the fact that ℘ has poles oforder 2 at any lattice point.


(d) Since ℘ is an elliptic function of period Λ, it factors through the complecttorus C/Λ, which is compact. Now ℘ is non constant therefore it must surjectto P1(C).

Lemma 2.23. ℘′ has degree 3 and it has three distinct zeroes in the funda-

mental domain∏ def

= t1w1 + t2w2 ∈ C | 0 ≤ t1, t2 < 1, at w1

2 , w2

2 and w3

2 ,where w3 = w1 + w2, each of multiplicity one.

Proof. ℘′ has only poles of order 3 in w ∈ Λ, thus deg(℘′) = 3. Since ℘ is even,℘′ is odd, and we saw that ℘′ is elliptic with period Λ. Therefore

℘′(wi

2

)= −℘′(− wi

2

)= −℘′(− wi

2+ wi

)= −℘′(wi

2

)Hence ℘′(wi

2

)= 0 for i = 1, 2, 3 and as deg(℘′) = 3 these must be all the zeroes

of ℘′ in∏

and therefore have multiplicity one.

Definition. For n ≥ 3, we call Eisenstein series of weight n the quantity

En(Λ) =∑w∈Λw =0

1

wn

Theorem 2.24. Define g2 = g2(Λ) = 60E4(Λ) and g3 = g3(Λ) = 140E6(Λ),then for all z ∈ C:

(℘′(z))2 = 4℘(z)3 − g2℘(z)− g3 (2.2.2)

Proof. Recall the equality 1 + t+ t2 + .. = 11−t for |t| < 1. Deriving we obtain

1

(1− t)2= 1 + 2t+ 3t2 + ... =

∞∑k=0

(k + 1)tk

Hence for z near 0 we have

1

(z − w)2− 1

w2=

1

w2

( 1

(z/w − 1)2−1)=

1

w2

(1+2(z/w)+...−1

)=

∞∑k=1

k + 1

wk+2zk

Therefore we can write ℘(z) as

℘(z) =1

z2+

∞∑k=1

(k+1)(∑

w∈Λw =0

1

wk+2

)zk =

1

z2+

∞∑k=1

Ek+2(Λ)zk =

1

z2+3E4(Λ)z2+5E6(Λ)z4+..

and

℘′(z) = − 2

z3+ 6E4(Λ)z + 20E6(Λ)z3 + .. =⇒ (℘′(z))2 =

4

z6− 24E4(Λ)

1

z2+ ...

Therefore setting h(z) = (℘′(z))2 − 4℘(z)3 + g2℘(z) + g3 we see that h has nopoles in 0 ∈

∏since all terms with zn for n < 0 cancel out, and h(0) = 0. As

℘ and ℘′ are holomorphic in∏

r0, h is holomorphic in∏

and Λ-periodic.Therefore h is constant and h(0) = 0 implies that h ≡ 0 as we wanted toshow.


Definition. For a lattice Λ ⊂ C, define its discriminant as

∆(Λ) = g2(Λ)3 − 27g3(Λ)

2

Remark 2.25. Recall that for a cubic polynomial of the form P (x) = 4x3 −g2x− g3 its discriminant is ∆P = g32 − 27g23 = 16(e1 − e2)2(e1 − e3)2(e2 − e3)2where e1, e2, e3 are the roots of P (x). It follows that for any lattice Λ in C,

∆P = ∆(Λ) = 0

Proof. This is equivalent to prove that the roots of the polynomial are distinct.Since ℘ is even, ℘′ is odd. Now let v ∈ w1

2 ,w2

2 ,w3

2 where Λ = Zw1 +Zw2 andw3 = w1+w2; then by Lemma 2.23 ℘′(v) = 0 and they are simple zeroes. Henceby the functional equation

0 = ℘′(v)2 = 4℘(v)3 − g2℘(v)− g3

Therefore ℘(wi

2 ) is a root of P (x). We are reduced to show that ℘(wi) = ℘(wj)for i = j so that we have 3 distinct roots and nonzero discriminant.Consider the function ℘(z + wi

2 ) − ℘(wi

2 ) which is even since ℘ is and it is 0at z = 0. Therefore it has a zero of order at least 2 at z = 0, but since it hasdegree 2, it has order exactly 2 at z = 0 and no other zeroes. Thus in particularfor z =

wi+wj

2 with i = j, we obtain 0 = ℘(wj

2 )− ℘(wi

2 ) and this concludes theproof.

Therefore E = [x, y, z] ∈ P2 | y2z = 4x3− g2(Λ)xz2− g3(Λ)z3 is a nonsin-gular projective curve for any lattice Λ.Moreover it makes sense to define the quantity

j(Λ) = 1728g2(Λ)

3

∆(Λ)

for any lattice Λ in C, called j-invariant.

Theorem 2.26. Let π : C −→ T = C/Λ be the canonical surjection on thequotient. Consider Φ : T −→ P2 given by

t = π(z) ↦→

[℘(z), ℘′(z), 1] if z /∈ Λ

[0, 1, 0] if z ∈ Λ

Then Φ is holomorphic (as a map of complex manifolds) with image

Φ(T ) = Edef= [x, y, z] ∈ P2 | y2z = 4x3 − g2xz2 − g3z3

Moreover, Φ is injective and has maximal rank (one) on T . Thus since T iscompact, Φ : T −→ E is a biholomorphism, hence T ∼= E.

Proof. Φ is holomorphic on T r 0: in fact let t ∈ T , t = 0, then Φ(t) ∈ U2 =[x0, x1, x2] ∈ P2 | x2 = 0 ⊂ P2. We must check that for V ⊆ T r 0, the mapF := ϕ2 Φ (π−1

V )−1 is holomorphic on V = π−1V ⊆ C, where ϕ2[x0, x1, x2] ↦→(

x0

x2, x1

x2

)∈ C2. We have

F (z) = ϕ2(Φ(πV (z))) = ϕ2([℘(z), ℘′(z), 1]) = (℘(z), ℘′(z))


which are holomorphic in T r 0.Moreover we can immidiatly see that Φ has maximal rank in T r 0: in factthis is the case if and only if for all z ∈ Cr Λ,

JC(F ) =

(℘′(z)℘′′(z)

)=(00

)Suppose not, then ℘′(z) = 0, thus z = wi

2 for i = 1, 2, 3, but we saw that theseare simple zeroes, therefore ℘′′(wi

2

)= 0.

Let us now show that Φ is holomorphic at t = 0: we have Φ(0) = [0, 1, 0] ∈ U1.Let V be a neighbourhood of 0 ∈ C and let z = 0 in V , then

F0(z) = (ϕ1 Φ π)(z) =( ℘(z)℘′(z)

,1

℘′(z)

)(We can divide by ℘′(z) since we are near 0). Recall that

℘(z) =1

z2(1 + a1z + ...) ℘′(z) = − 2

z3(1 + b1z + ..)

=⇒ F0(z) =(− z

2(1 + c1z + ...),−z

3

2(1 + d1z + ..)

)Therefore F0(z) is holomorphic at 0.Using the same expasion near 0 we see that

JC(F0) =

((℘/℘′)′(z)(1/℘′)′(z)

)|z=0

=

(− 1

2 + z(...)− 3

2z2 + z(...)

)|z=0

=

(− 1

20

)=(00

)

So we can conclude that Φ has maximal rank everywhere on T .To see that Φ(T ) = E , let P = [x, y, z] ∈ E: if z = 0 then x = 0, thusP = [0, 1, 0] = Φ(0) ∈ Φ(T ). Thus assume z = 0 and write P = [x, y, 1], since ℘is surjective, let x ∈ C be such that x = ℘(x). As P ∈ E we get

y2 = 4x3 − g2x− g3 = 4℘(x)3 − g2℘(x)− g3 = (℘′(x))2

Hence y = ℘′(x) or y = −℘′(x). For the former case we have P = Φ(π(x)), forthe latter notice that y = −℘′(x) = ℘′(−x) and x = ℘(x) = ℘(−x), thereforeP = Φ(π(−x)).The other inclusion Φ(T ) ⊂ E is trivial by Theorem 2.27.We are left to prove that Φ is injective: let z1, z2 ∈

∏be such that Φ(π(z1)) =

Φ(π(z2)). If z1 = 0 then Φ(π(0)) = [0, 1, 0] = Φ(π(z2)), which yields that z2 = 0.If z1, z2 = 0, then

℘(z1) = ℘(z2) (i)

℘′(z1) = ℘′(z2) (ii)

(i) implies that either z1 = z2 ( and we are done) or z1 = −z2 + w for somew ∈ Λ. Then

℘′(z2) = ℘′(z1) = ℘′(−z2 + w) = −℘′(z2)

hence ℘′(z2) = 0 so that z2 = wi

2 and z1 =wj

2 . Since ℘(z1) = ℘(z2), i = j andz1 = z2 as we wanted.


Notice that the cubic equation y2 = 4x3 − g2x− g3 is in the form of (2.1.3)therefore E defined by such cubic is an elliptic curve over C.The Theorem shows how to associate to a lattice Λ in C an elliptic curve andimplements this association by the biholomorphism Φ. However also the converseis true, for any elliptic curve E there exists a lattice Λ such that E is realizedas C/Λ.

2.2.2 Isogenies over CLet E and E′ be elliptic curves isomorphic to C/Λ = T and C/Λ′ = T ′

respectively, for Λ,Λ′ lattices in C. Then every isogeny of E into E′ correspondsto a holomorphic homomorphism of T onto T ′, and viceversa.

Proposition 2.27. With the same notation as above, we have that

Hom(E,E′) ∼= Hom(T, T ′) = α ∈ C | αΛ ⊂ Λ′

Proof. Assume that f is non-constant and consider the following diagram:

C

π1

↓↓

C

π2

↓↓T

f →→ T ′

Now by the Hurwitz theorem, f has no ramification points, therefore it is acovering of T ′ (as well as π2) and π1 is a covering of T . Since we have twouniversal covering of T ′ given by π2 and f π1, there exists F : C −→ Cisomorphism of covering such that π2 F = f π1, and in particular F iscontinuous. Fixing a point O = π1(0) ∈ T , we have that f(O) = O ∈ T ′

and ∀b ∈ π−12 (O) there exists a unique F isomorphism of coverings such that

F (0) = b. In particular since we require that F is a group homomorphism, thereis a unique F so that F (0) = 0.Let us now show that F (z) = αz for some α ∈ C such that αΛ ⊂ Λ′: the factthat F passes to quotients means that for all w ∈ Λ there exists w′ ∈ Λ′ suchthat F (z+w) = F (z)+w′. Now notice that w′ = F (z+w)−F (z) is indipendentfrom z since the map F (∗ + w) − F (∗) is continuous, C is connected and Λ′ isdiscrete. Hence taking derivatives yields the equality F ′(z + w) = F ′(z) , or inother words F ′ is an elliptic function of period Λ. Therefore it defines a mapg : T −→ C holomorphic such that F ′ = gπ1. On the other hand T is compact,hence g is constant, and so F ′ is : say F ′(z) = α = 0 ∀z ∈ C and consequentlyF (z) = αz since F (0) = 0.In particular for any w ∈ Λ, F (z +w) = F (z) +w′ implies aw = w′ ∈ Λ′ henceαΛ ⊂ Λ′.If f is constant, then F is constant, and in particular it is 0 (since F (0) = 0).Viceversa, let T = C/Λ, T ′ = C/Λ′ be complex tori and F (z) = αz with αΛ ⊂Λ′. Then there exists f : T −→ T ′ holomorphic homomorphism, in fact such Fpasses to the quotient since π2 F (Λ) = π2(αΛ) = 0 = π1(Λ). Therefore thereexists f continuous such that f π1 = π2 F . Finally f is holomorphic sincelocally f|U = π2|

V ′ F π−1

1|Vis composition of holomorphic maps.

In particular we have that

End(E) ∼= α ∈ C | αΛ ⊂ Λ


and this always contains Z, since multiplication by integers maps Λ into Λ.

Definition. We say that an elliptic curve E has complex multiplication ifEnd(E) % Z.

Proposition 2.28. Let Λ = Zw1 + Zw2 and Λ′ = Zw′1 + Zw′

2 be normalizedlattices in C. Then E ∼= T = C/Λ and E′ ∼= T ′ = C/Λ′ are isogenous (resp.isomorphic) if and only if there exists an element γ ∈ GL+

2 (Q) ∩M2(Z) withGL+

2 (Q) = γ ∈ GL2(Q) | det(γ) > 0 (resp. SL2(Z)) such that γ · τ ′ = τ ,where τ = w1/w2, τ

′ = w′1/w

′2 and the action of γ on H is by fractional linear

transformation i.e.

(a bc d

)· τ ′ = aτ ′+b

cτ ′+d

Proof. If 0 = α ∈ Hom(E,E′) then αΛ ⊂ Λ′, therefore there exist a, b, c, d ∈ Zsuch that

αw1 = aw′1 + bw′

2

αw2 = cw′1 + dw′

2

Thus we obtain γ =

(a bc d

)∈M2(Z) ∩GL2(Q) such that τ = γ · τ ′. Moreover

det(γ) > 0 since both bases are normalized.

Conversely, if γ · τ ′ = τ for γ =

(a bc d

)∈ M2 ∩ GL+

2 (Q) then for λ = cτ ′ + d

we have

λ

(τ1

)=

(a bc d

)(τ ′

1

)=⇒ (λw′

2/w2)

(w1

w2

)=

(a bc d

)(w′

1

w′2

)Hence αΛ ⊂ Λ′ for α = λ

w′2

w2. In particular αΛ = Λ′ if and only if γ ∈ SL2(Z).

Proposition 2.29. Let Λ = Zw1 + Zw2 with τ = w1/w2 ∈ H. Then C/Λhas complex multiplication if and only if there exists a nonscalar element ξ ∈GL+

2 (Q) such that ξ · τ = τ .

Proof. Let α ∈ End(E), then since αΛ ⊂ Λ, there exist a, b, c, d ∈ Z such thatατ = aτ + b

α = cτ + d=⇒ α

(τ1

)=

(a bc d

)(τ1

)=: ξ

(τ1

)for ξ ∈M2(Z)∩GL+

2 (Q)

(2.2.3)Now notice that α is an eigenvalue for ξ for how ξ is defined, on the other handα is an eigenvalue for ξ since

α

(τ1

)= ξ

(τ1

)thus, in a compact way, we have

(τ τ1 1

)(α 00 α

)= ξ

(τ τ1 1

)Therefore α ∈ Z ⇐⇒ b = 0 = c which yields a = d ⇐⇒ ξ is scalar.

Proposition 2.30. Let Λ = Zw1 + Zw2 with τ = w1/w2 ∈ H. Then C/Λ hascomplex multiplication if and only if Q(τ) is a quadratic imaginary field andEnd(E) is an order in Q(τ).


Proof. C/Λ has complex multiplication if and only if α as in Proposition 2.32is not in Z hence it is not real (since τ ∈ H) and ξ is not scalar. Thereforeeliminating τ from (2.2.2) we obtain the quadratic equation

α2 − (a+ d)α+ ad− bc = 0 (∗)

Since α = cτ + d, it follows that Q(τ) = Q(α) is a quadratic imaginary field.Moreover, (∗) tells us that α is integral over Z, thus contained in the ring ofintegers OQ(τ). Therefore End(E) ⊂ OQ(τ) and it is free of rank 2 because OQ(τ)

is torsion free and End(E) contains 1 and an imaginary element: we concludethat End(E) is an order in Q(τ).

From now on suppose that the elliptic curve E ∼= T = C/Λ has complexmultiplication by the maximal order OK , for a quadratic imaginary field Kand that Λ is of the form Zτ + Z. Under these assumptions, the conditionEnd(E) = OK is equivalent to saying that OKΛ ⊂ Λ. Therefore Λ is a fractionalOK-ideal.

Fact 2.31. If I is a fractional OK-ideal then there exists λ ∈ K× such thatλI ⊂ OK and therefore it is an OK-ideal.

In particular, we have m ∈ Z>0 such that mΛ ⊂ OK and consequently mΛis a nonzero ideal of OK .Conversely, if a is a fractional OK-ideal, then the torus C/a satisfies End(C/a) =OK . In fact OKa ⊂ a, thus End(C/a) ⊃ OK and equality follows by maximalityof OK in K.Recall that the class group of K is

Pic(OK) =(multiplicative group of fractional ideals)

(subgroup of principal fractional ideals)=

=(multiplicative semigroup of nonzero ideals of OK)

∼where I1 ∼ I2 if and only if there exist α1, α2 ∈ OK nonzero, such that α1I1 =α2I2.Therefore it follows that:

Proposition 2.32. For a complex torus T , denote by [T ] its isomorphism classin the category of complex tori. After fixing one of the two possible embeddingsι : OK −→ C, the map

Pic(OK) −→ [T ] | End(T ) = OK defined by [ι(a)] ↦→ [C/ι(a)]

is a bijection.

Corollary 2.33. [T ] | End(T ) = OK is finite since Pic(OK) is.

2.2.3 Automorphisms of an elliptic curve

Let Aut(E) denote the group of all automorphisms of a given elliptic curveE defined over C. If E has no complex multiplication, Aut(E) = End(E)× =Z× = ±1.Therefore assume that E has complex multiplication and let End(E) ∼= O ⊆ OK

be an order in an imaginary quadratic field K, so that Aut(E) = O× ⊆ O×K .

Recall that O×K = µ ∩K where µ = z ∈ C | zn = 1 for some n ∈ N.

2.3. ELLIPTIC CURVES OVER Q 36

Proposition 2.34. O×K = ±1 except for the following cases:

(a) if K = Q(i) then O×K = ±1,±i ∼= Z/4Z;

(b) if K = Q(ρ) with ρ = e2πi/3, then O×K = ±1,±ρ,±ρ2.

Proof. u ∈ O×K if and only if u ∈ OK and Norm(u) = |u| = 1. Recall that

for a quadratic imaginary field K, Gal(K/Q) = ⟨c⟩ where c is the restriction ofcomplex conjugation toK. Since u ∈ OK , u = a+

√db for a, b ∈ Z or a, b ∈ 1

2+Zand d is a negative square-free integer such that K = Q(

√d).

Now one has that

Norm(a+√db) = (a+

√db)(a−

√db) = a2 + |d|b2

Therefore for |d| > 5, Norm(u) ≥ |d|b2 ≥ |d|4 > 1, for all b = 0. It follows that

u ∈ O×K if and only if a2 = 1, b = 0 i.e. u = ±1.

We are left with 4 cases:

• d = −1 ≡ 3 mod 4 =⇒ OK = Zi+ Z. Hence Norm(u) = a2 + b2 = 1 ifand only if a2 = 1, b = 0 or a = 0, b2 = 1, if and only if u = ±1 or u = ±i.

• d = −2 ≡ 2 mod 4 =⇒ OK = Zi√2+Z and then Norm(u) = a2+2b2 = 1

if and only if a2 = 1, b = 0 if and only if u = ±1.

• d = −3 ≡ 1 mod 4 =⇒ OK = Z 1+√3

2 +Z and Norm(u) = a2 +3b2 = 1with a, b ∈ 1

2Z if and only if a2 = 1, b = 0 or a2 = 14 = b2 if and only if

u = ±1 or u = ± 12 ±

√32 .

• d = −4 is not square-free.

In these 2 cases, O is the maximal order of K and the class number is 1,therefore by Proposition 2.35, there is exactly one elliptic curve up to isomor-phism over C such that End(E) ∼= OK .Now notice that the elliptic curve E : y2 = 4x3 − g3 has the 4 automorphisms

(x, y) ↦→ (x,±y), (x, y) ↦→ (−x,±iy)

And the elliptic curve E : y2 = 4x3 − g2x has the 6 automorphisms

(x, y) ↦→ (ρjx,±y) for j = 0, 1, 2

2.3 Elliptic curves over QLet E be an elliptic curve defined over Q and we may assume that its

Weierstrass equation has integer coefficients: in fact the change of variable(x, y) = (u2x′, u3y′) gives a Weierstrass equation with coefficients a′i = ai/u

i.Therefore a suitable choice of u makes this an integral equation. Then its dis-criminant ∆ will be an integer as well and which p-adic norm satisfies

|∆|p ≤ 1 with equality if and only if p - ∆


Definition. A Weierstrass equation (2.1.2) is minimal for the prime p if forany admissible change of variable over Q, such that the new coefficients arep-integral, the power of p dividing ∆ cannot be decreased or, equivalently |∆|pcannot be increased.A Weierstrass equation (2.1.2) is a global minimal Weierstrass equation if it isminimal for all primes and its coefficients are integers.

From now on we will assume that the elliptic curve E is given by a globalminimal Weierstrass equation. This is in fact not a restriction, as one sees bythe following results:

Proposition 2.35. Let p be a fixed prime and let E be an elliptic curve overQ, then

(a) There exists an admissible change of variable over Q such that the resultingequation is minimal for p.

(b) If E has p-integral coefficients, then the change of variable in (a) is suchthat u, r, s, t are p-integral.

(c) Two equations that are minimal for p are related by an admissible changeof varibale such that |u|p = 1 and r, s, t are p-integral.

Proof. See [Kna93] Proposition 10.2

Theorem 2.36. (Neron) Let E be an elliptic curve over Q, then there existsan admissible change of variable over Q such that the resulting equation is aglobal minimal Weierstrass equation. Moreover any two such global minimalWeierstrass equations are related by a change of variable with u = ±1 andr, s, t ∈ Z.

Proof. See [Kna93] Theorem 10.3

2.3.1 L-function associated to an elliptic curve

Let E be an elliptic curve defined over Q and assume that E is given by aglobal minimal Weierstrass equation. Now for each prime p, write Ep for thereduction of E modulo p: such curve is defined over Fp and it is singular if andonly if p | ∆ (For a complete discussion of singular Weierstrass equation see[Kna93] III.5 or [Sil09] III.1,III.2). For both singular and nonsingular casesdefine

ap = p+ 1−#Ep(Fp) (2.3.1)

Definition. The L-funtion associated to an elliptic curve E is

L(s, E) =∏p|∆

(1− app−s

)−1 ·∏p-∆

(1− app−s + p1−2s

)−1(2.3.2)

Before giving a convergence result regarding L(s, E), we first need to provethe Hasse bound for elliptic curves over finite fields.


2.3.2 Hasse theorem

Notice that if E is defined over Fq with q = pr, we have the obvious boundon 1 ≤ #E(Fp) ≤ 2q+1 since for any value of x there can be at most two valuesof y, and the point at ∞. In particular for q = p, it follows that |ap| ≤ p.

Definition. Let A be an abelian group, a map d : A −→ R is a quadratic formif:

(i) d(a) = d(−a) ∀a ∈ A;

(ii) B : A×A −→ R given by (a, b) ↦→ d(a+ b)− d(a)− d(b) is bilinear.

If additionally d(a) ≥ 0 ∀a ∈ A and d(a) = 0 ⇐⇒ a = 0 then d is said to bepositive definite.

Example 2.37. If E ∼= C/Λ is an elliptic curve over K0 then deg : End(E) −→Z is a positive definite quadratic form. In fact for α ∈ End(E), deg(α) = N(α) =α · α (see [Cox97 Theorem 10.14]) and therefore it is immidiate that N(α) =N(−α) and N(α) = 0 if and only if α = 0. Since B is symmetric it is enoughto check bilinearity on the first variable:

B(aα+ bβ, γ) = (aα+ bβ + γ)(aα+ bβ + γ)− (aα+ bβ)(aα+ bβ)− γγ =

= (aα+ bβ)γ + γ(aα+ bβ) = aαγ + aαγ + bβγ + bβγ =

= (aα+ γ)(aα+ γ)− a2αα− γγ + (bβ + γ)(bβ + γ)− b2ββ − γγ =

= B(aα, γ) +B(bβ, γ) ∀a, b ∈ R, α, β, γ ∈ End(E)

Lemma 2.38. Let d : A −→ Z be a positive definite quadratic form, then

|d(a− b)− d(a)− d(b)| ≤ 2√d(a)d(b) (2.3.3)

Proof. Let B : A× A −→ Z given by (a, b) ↦→ d(a− b)− d(a)− d(b): such B isbilinear. Notice that for all m,n ∈ Z, we have d(ma) = m2d(a) and

0 ≤ d(ma− nb) = m2d(a) +mnB(a, b) + n2d(b)

Setting m = −B(a, b) and n = 2d(a) to obtain

0 ≤ d(a)(4d(a)d(b)−B(a, b)2

)If d(a) = 0, then a = 0 and the both sides of (2.3.3) are zero and there is nothingto prove. Otherwise d(a) > 0 and dividing we obtain

4d(a)d(b) ≥ B(a, b)2 =⇒ |d(a− b)− d(a)− d(b)| ≤ 2√d(a)d(b)

Theorem 2.39. (Hasse) Let E be an elliptic curve defined over Fq, q = pr,then

q + 1− 2√q ≤ #E(Fq) ≤ q + 1 + 2

√q (2.3.4)


Proof. Let φ = φq be the q-th Frobenius map and recall that Fq is the fixedfield of φq, so that E(Fq) = ker(1− φ) and by Lemma 2.13 follows that 1− φqis a separable isogeny. Therefore #ker(1− φ) = deg(1− φ) and by the previouslemma

|deg(1− φ)− deg(1)− deg(φ)| = |#E(Fq)− 1− q| ≤ 2√deg(1)deg(φ) = 2

√q

Corollary 2.40. For an elliptic curve E defined over Q, the Euler productL(s, E) as in (2.3.2) converges for Re(s) > 3

2 and is given there by an absolutelyconvergent Dirichlet series.

Proof. For a proof that L(s, E) is given by a Dirichlet series see [Kna93] Corol-lary 10.6.We only need to prove convergence for the infinite product

∏p-∆

(1 − app

−s +

p1−2s)−1

since L(s, E) is obtain by multiplicating this factor with a finite prod-uct. L(s, E) converges absolutely if and only is∑

p-∆

| − app−s + p1−2s| ≤ ∞

By Theorem 2.39 with q = p we get |ap| ≤ 2√p, thus∑

p-∆

|app−s + p1−2s| ≤∑p-∆

2|p−s+ 12 |+ p1−2s| ≤

≤∑p-∆

pRe(−s+ 12 )(2 + pRe(−s+ 1

2 ))≤∑p

pRe(−s+ 12 ) ≤

∑n

nRe(−s+ 12 )

Which converges for Re(s) ≥ 32 .

Chapter 3

Modular Forms

3.1 Modular forms for SL2(Z)3.1.1 Functions of lattices

As in 2.2 denote by L the set of all lattices and by B the set of normalizedbasis.

Definition. A map F : L −→ P1(C) = C ∪ ∞ is a function of lattices ofweight k ∈ Z if F (αΛ) = α−kF (Λ) for all α ∈ C and all Λ ∈ L.

Example 3.1. For k ≥ 3, consider the Eisestein series Ek : L −→ P1(C) definedas Λ ↦→

∑0=w∈Λ

1wk . One can prove that such series is absolute convergent, so

that Ek is a function of lattices of weight k. Note that Ek ≡ 0 for all k ≥ 3 odd.

Let F : L −→ P1(C) be a function of lattices of weight k. Then we can viewF as

F = F π : SL2(Z)\B −→ P1(C)

Then the map σ : H −→ SL2(Z)\B such that τ ↦→[τ1

]yields, by further

composingf = F π σ : H −→ P1(C)

Since the map H −→ SL2(Z)\B/C× is surjective, f determines F . As F hasweight k, we have

F([w1

w2

])= F

(w2

[w1/w2

1

])= w−k

2 f(w1

w2

)We now wonder which functions f : H −→ P1(C) arise from F : L −→ P1(C).

This means that f factors as f = σF , or equivalently such that ∀γ =

(a bc d

)∈

SL2(Z)

f(γτ) = f(aτ + b

cτ + d

)= F

([aτ+bcτ+d

1

])=

(cτ + d)kF([aτ + bcτ + d

])= (cτ + d)kF

([τ1

])= (cτ + d)kf(τ)

40

3.1. MODULAR FORMS FOR SL2(Z) 41

Proposition 3.2. There is a one-to-one correspondence between

F : L −→ P1(C) of weight k

←→

⎧⎪⎪⎨⎪⎪⎩f : H −→ P1(C)f(aτ + b

cτ + d

)= (cτ + d)kf(τ)

∀ γ =

(a bc d

)∈ SL2(Z)

⎫⎪⎪⎬⎪⎪⎭Definition. We say that f : H −→ P1(C) is a weakly modular function of weightk ∈ Z for SL2(Z) if

(i) f(γτ) = (cτ + d)kf(τ) for all γ =

(a bc d

)∈ SL2(Z) and all τ ∈ H;

(ii) f is meromorphic on H.

Remark 3.3. (a) Since f is meromorphic, f−1(∞) is a discrete subset of H.

(b) If k is odd, then f ≡ 0 as one sees by taking γ =

(−1 00 −1

)∈ SL2(Z)

which acts on H as the identity, and by (1) we get f(τ) = −f(τ), thusf(τ) = 0.

(c) If f = 0 is a weakly modular function of weight k, then 1/f is a weaklymodular function of weight −k.

To get to the definition of modular functions and modular forms we willrequire a growth condition at infinity.Consider the exponential function H −→ C defined by τ ↦→ qτ := e2πiτ .If we write τ = x + iy ∈ H for x ∈ R, y ∈ R>0, then qτ = e2πixe−2πy and inparticular e2πix = cos(2πx) + isin(2πx) ∈ S1 = z ∈ C : |z| = 1.The mapping x ↦→ e2πix induces a real analytic isomorphism R/Z −→ S1 ; onthe other hand the strictly decreasing real analytic map y ↦→ e−2πy yields a realanalytic isomorphism R>0 −→]0, 1[.It follows that τ ↦→ qτ gives a holomorphic isomorphism

H/Z −→ z ∈ C | 0 < |z| < 1 =: D∗

where the action of Z on H is the translation (n, τ) ↦→ τ + n, ∀n ∈ Z, τ ∈ H.Now if f : H −→ P1(C) is a weakly modular function of weight k, it inducesa map f : H/Z −→ P1(C) since the translation τ ↦→ τ + n, for any n ∈ Z, is

described by the action of γ =

(1 n0 1

)∈ SL2(Z) and f(τ +n) = f(γτ) = f(τ).

Therefore f induces a function f∗ : D∗ −→ P1(C) by f∗(qτ ) = f(τ) and therelation |qτ | = e−2πy = e−2πIm(τ) shows that q → 0 as Im(τ)→∞.Consequently we say that f is meromorphic (resp. holomorphic) at ∞ is f∗ ismeromorphic (resp. holomorphic) at the center of D∗.

Definition. (i) A modular function of weight k ∈ Z for SL2(Z) is a weaklymodular function of weight k for SL2(Z), which is meromorphic on H andat infinity.

(ii) A modular form of weight k ∈ Z for SL2(Z) is a weakly modular functionof weight k for SL2(Z), which is holomorphic on H and at infinity.


(iii) A cusp form of weight k ∈ Z for SL2(Z) is a modular form of weight k forSL2(Z), that vanishes at infinity.

Remark 3.4. (a) Note that the set of modular functions (resp. modular forms,resp. cusp forms) is equipped with a natural structure of C-vector space;

(b) Modular functions of weight 0 form a field;

(c) If f is meromorphic at infinity, then in a neighbourhood of qτ = 0 (orequivalently for Im(τ)≫ 0 ) we have a Laurant series expansion

f(τ) = f∗(qτ ) =

∞∑n=−N

anqnτ =

∞∑n=−N

ane2πiτ with an = an(f) ∈ C

(3.1.1)This Fourier expansion is called the q-expansion of f .In particular, if f is a modular function, that it is also meromorphic on H,then its q-expansion converges on H.

3.1.2 The action of SL2(Z) on H

We now want to determine a set of representatives for the quotient SL2(Z)\Hwith respect the left action of SL2(Z) on H given by fractional linear transfor-mation .Define Γ = SL2(Z)/±1: since −1 ∈ SL2(Z) acts trivially on H, then Γ actson H. In particular the action of Γ on H is faithful, which means that

γτ = τ ∀τ ∈ H implies γ = 1 ∈ Γ

Let S ∈ Γ be an element represented by S =

(0 −11 0

)∈ SL2(Z), then Sτ =

− 1τ ; and let T ∈ Γ be represented by T =

(1 10 1

)∈ SL2(Z), then Tτ = τ + 1.

Remark that S2 = 1 and (ST )3 = 1 = (TS)3 in Γ; moreover since Tnτ = τ + nfor all n ∈ Z, the subgroup of Γ generated by T is infinite.Also define define D = τ ∈ H | |τ | ≥ 1, |Re(τ)| ≤ 1/2, then:

ρ −ρi

−1 1− 12

12

0

D


Theorem 3.5. (1) For all τ ∈ H there exists γ ∈ Γ such that γτ ∈ D.

(2) If τ = τ ′ ∈ D and γτ = τ ′ for some γ ∈ Γ then either Re(τ) = ± 12 (and

then τ ′ = τ ∓ 1) or |τ | = 1 (and then τ ′ = Sτ).

(3) For τ ∈ D, StabΓ(τ) = γ ∈ Γ | γτ = τ is trivial except for the followingthree cases:

(a) τ = i, then StabΓ(i) = ⟨S⟩ ∼= Z/2Z;(b) τ = ρ = e2πi/3, then StabΓ(ρ) = ⟨ST ⟩ ∼= Z/3Z;(c) τ = −ρ = e2πi/6, then StabΓ(−ρ) = ⟨TS⟩ ∼= Z/3Z

Proof. (1): Define G = ⟨S, T ⟩ ⊆ Γ. We will prove that for any τ ∈ H there existsg ∈ G such that gτ ∈ D.

If g ∈ G is represented by g =

(a bc d

)∈ SL2(Z) then Im(gτ) = Im(τ)

|cτ+d|2 . Hence

there exists g ∈ G such that Im(gτ) is maximal since ∀k, the set (c, d) ∈ Z2 :|cτ + d|2 < k2 is finite.Fix g ∈ G for which Im(gτ) is maximal, then ∃n ∈ Z such that |Re(Tngτ)| ≤ 1

2 .We claim that Tngτ ∈ D, or in other words that |Tngτ | ≥ 1.Suppose it is not, then |Tngτ | < 1, therefore

Im(STngτ) =Im(Tngτ)

|Tngτ |> Im(Tngτ) = Im(gτ)

which contradicts the maximality property with which we selected g.(2) & (3): If τ ′ = γτ with τ, τ ′ ∈ D, γ ∈ Γ, we can assume that Im(τ ′ =

Im(γτ) ≥ Im(τ) i.e. |cτ + d| ≤ 1 (∗).It follows that Im(cτ + d) = |c|Im(τ) ≤ |cτ + d| ≤ 1. Since τ ∈ D, Im(τ) ≥Im(ρ) =

√32 , hence |c|

√32 ≤ 1 which forces c ∈ 0,±1. Also remark that the

case c = −1 reduces to the case c = 1 by replacing γ with −γ.

• c = 0 yields γ =

(±1 b0 ±1

)which means that γ is a translation. Since

|Re(τ)|, |Re(γτ | ≤ 12 the only possibilities are b = 0 if τ belongs to the

interior of D or b = ±1 and consequently |Re(τ)| = 12 , Re(γτ) = −Re(τ).

• If c = 1, (∗) becomes |τ + d| ≤ 1.

If d = 0 then |τ | ≤ 1 hence |τ | = 1 and γ is represented by γ =

(a −11 0

)hence τ ′ = a − 1

τ = T aSτ . If |Re(τ)| = 1/2 then a = 0 hence γ = S andSi = i, which also yields S ∈ StabΓ(i).If |Reτ | = 1/2 then τ = ρ,−ρ. For τ = ρ, γρ = −ρ hence a = 0 or a = −1,for τ = −ρ we have a = 0 or a = 1.If d = 0 write τ = x+iy, then τ+d = (x+d)+iy and an easy computationfrom (∗) shows that d = ±1. For such values of d follows that τ = ρ ifd = 1 and τ = −ρ for d = −1. The former yields that a − b = 1 andγρ = a − 1

1+ρ = a + ρ hence a = 0 or a = 1. Similarly for the latter casewe get a = 0 or a = −1.

To conclude, it is enough to show that G = Γ: let τ belong to the interior ofD and let γ ∈ Γ, then γτ ∈ H. The proof of part (1) shows that there existsg ∈ G = ⟨S, T ⟩ such that gγτ ∈ D. By parts (2) & (3), since τ is in the interior ofD, follows that gγτ = τ , hence gγ ∈ StabΓ(τ) = 1, therefore γ = g−1 ∈ G.


Definition. A subset A of H is a fundamental domain for Γ if :

(i) A is open and connected;

(ii) The canonical map A −→ Γ\H is injective;

(iii) The canonical map A −→ Γ\H is surjective.

Corollary 3.6. The interior of D is a fundamental domain for Γ.

Notice that if A is a fundamental domain, then γA is also a fundamentaldomain for any γ ∈ Γ (Since γ is a holomorphic isomorphism of H).

Remark 3.7. The theorem shows that Γ\H equipped with the quotient topol-ogy is homeomorphic to the complex plane A1(C).

Corollary 3.8. There is a bijection between function of lattices of weight k ∈ Zand functions on H such that

f(Tτ) = f(τ + 1) = f(τ) (3.1.2a)

f(Sτ) = f(−1

τ) = τkf(τ) (3.1.2b)

Proposition 3.9. For τ ∈ H, k ≥ 3, the so-called Eisestain series of weight k

Ek(τ) =∑

(m,n)∈Z2

(m,n)=(0,0)

1

(mτ + n)k(3.1.3)

defines a modular form of weight k.In particular ∀k ≥ 3 odd, Ek ≡ 0; for k ≥ 4 even, Ek does not vanish at infinity

and Ek(∞) = 2ζ(k), where ζ(s) =∞∑

n=1

1ns is the Riemann Zeta function.

Proof. For k ≥ 3 and for any lattice Λ ∈ L, the series Ek =∑

0=w∈Λ

1wk converges

absolutely and Ek(τ) = Ek(Λτ ) for Λτ = Zτ + Z i.e. Ek is a function on Hcorresponding to Ek by the bijection in Proposition 3.2.In particular Ek ≡ 0 for all k odd and Ek satisfies the functional equation

Ek(γτ) = (cτ + d)kEk(τ)

We are left to show that Ek is holomorphic on H and at infinity, and that itsatisfies

Ek(∞) = E∗k(0) = lim

q→0E∗

k(q) = 2ζ(k)

Assume at first that τ ∈ D, then

|mτ + n|2 = m2|τ |2 + 2mnRe(τ) + n2 ≥ m2 −mn+ n2 = |mρ+ n|2

where, as before, ρ = e2πi/3. Therefore∑(m,n) =(0,0)

1

|mτ + n|2≤

∑(m,n)=(0,0)

1

|mρ+ n|2= Ek(ρ) ≤ ∞


Hence for τ ∈ D we have absolute and uniform convergence on compact subsets,therefore since 1

(mτ+n)kis holomorphic on D, Ek(τ) is.

Now write H =⋃γ∈Γ

γD and let τ ∈ γD, then γ−1τ ∈ D and therefore for

γ =

(a bc d

)Ek(τ) = Ek(γγ

−1τ) = (c(γ−1τ + d)kEk(γ−1τ)

The right hand side has both factors holomorphic, thus we conclude that Ek(τ)is holomorphic on each γD, hence on H.Due to uniform convergence we may compute lim

q→0E∗

k(q) term by term and

obtain:1

(mτ + n)kIm(τ)→∞−−−−−−→

q→0

0 if m = 01nk if m = 0

This calculation shows that Ek(τ) is holomorphic at infinity i.e. E∗k(q) extends

to the point q = 0 to a holomorphic function by defining E∗k(0) = lim

q→0E∗

k(q)

3.1.3 Divisors of modular functions

Let f : H −→ P1(C) be a modular function of weight k ∈ Z, f = 0. Sincef is meromorphic, it makes sense to define for any τ ∈ H the order of f at τ

ordτ (f) ∈ Z. Remark that for all γ =

(a bc d

)∈ SL2(Z) we have ordγτ (f) =

ordτ (f) since f(γτ) = (cτ +d)kf(τ) and τ = −d/c since τ ∈ H while −d/c ∈ R.

Definition. Let f be as above and let π : H −→ Γ\H be the canonical projec-tion. Then define the divisor of f to be

div(f) =∑

x∈Γ\H

( 1

#StabΓ(x)ordx(f)

)· x+ (ord∞(f)) · ∞ =

=1

2ordi(f) · i+

1

3ordρ(f) · ρ+ ord∞(f) · ∞+

∑x=i,ρ

ordx(f) · x(3.1.4)

where ord∞(f)def= ord0(f

∗) and for x = π(τ), ordx(f)def= ordτ (f).

Given a divisor div(f) =∑

x∈Γ\H

(1

#StabΓ(x)ordx(f)

)·x+(ord∞(f)) ·∞, its degree

is

deg(div(f)) =1

2ordi(f) +

1

3ordρ(f) + ord∞(f) +

∑x=i,ρ

ordx(f) (3.1.5)

Theorem 3.10. Let f be a modular function of weight k ∈ Z, f = 0. Then

deg(div(f)) =k

12

Proof. First of all notice that te sum in (3.1.5) if finite. In fact since in a neigh-bourhood of ∞ there are no zeroes nor poles other than ∞ itself, we reduce to


consider singularities in a compact set of the form KM = D ∩ τ ∈ H | Im(τ) ≤M: now since singularities form a discrete set and belong to a compact set,they are finite.We now want to deduce the theorem using the argument principle by perform-

ing contour integration of dlog(f) = f ′

f dτ on the curve given by the boundaryof KM . There is a problem on having singularities on such boundary, thus letus assume at first for simplicity that the only possible zeroes or poles on theboundary are at ρ,−ρ, i and consider the contour integration given by

A E

B

B′

C C ′D′

D

C

ρ −ρ

i

so that we avoid ρ,−ρ, i. Choose a parametrization for the curve C which ispiecewise differentiable, makes it positively oriented and such that the windingnumber around any point in the interior is 1. The argument principle states that

1

2πi

∫C

f ′(τ)

f(τ)dτ =

∑x∈Γ\Hx=i,ρ

ordx(f)

provided that the arcsBB′ ,

CC ′ ,

DD′ and the segment EA are chosen

such that all the singularities of Γ\H r ρ, i lie inside C. Since under theseconditions ∑

x∈Γ\Hx =i,ρ

ordx(f)

is constant, it must be equal to the quantity we obtain by taking the limits ofthe radius of those arcs going to zero (provided that the limits exist).Choosing symmetrical arcs, notice that the invariance of f under T : τ ↦→ τ +1implies

(I)(∫ B

A

dlog(f) +

∫ E

D′dlog(f)

)= 0

Consider∫ A

Edlog(f) : the map q : τ ↦→ e2πiτ transforms EA into a circle α

around 0 of radius depending on M = Im(A) and which is negatively oriented.By change of variable we get

(II)1

2πi

∫ A

E

dlog(f) =1

2πi

∫α

f∗ = −ord0(f∗) = −ord∞(f)

Consider∫ B′

Bdlog(f): we are integrating along an arc of radius ε > 0. Call Cε(ρ)

the circle of radius ε centered at ρ negatively oriented. Then by the argument


principle1

2πi

∫Cε(ρ)

dlog(f) = −ordρ(f)

In a neighbourhood of ρ we have f ′

f =ordρ(f)τ−ρ + (holom. part). Parametrize the

arcBB′ as t ↦→ ρ+ ε−it for a ≤ t ≤ b, then

1

2πi

∫ B′

B

dlog(f) =ordρ(f)

2πi

∫ B′

B

dτ

τ − ρ+

1

2πi

∫ B′

B

(holom)dτ =

=ordρ(f)

2πi

∫ b

a

−−ie−it

e−itdτ +

∫ b

a

(holom)dτ = −ordρ(f)

2π(b− a) +

∫ b

a

(holom)dτ

As ε → 0, B → B′ and∫ B′

B(holom)dτ → 0. Moreover b − a → (angle of the

arc), therefore

b− a2π

−→2π6

2π=

1

6=⇒ −ordρ(f)

2π(b− a)→ −ordρ(f)

6(III)

The same computation shows that

(III)1

2πi

∫ D′

D

dlog(f) −→ 1

6ord−ρ(f) = −

1

6ordρ(f)

Also similarly we obtain

(III)1

2πi

∫ C′

C

dlog(f) −→ 1

2ordi(f)

We are left to compute( ∫ C

B′ dlog(f) +∫D

C′ dlog(f)).

Notice that S : τ ↦→ − 1τ takes

B′C to

C ′D but with the opposite orientation

(call itDC ′ ). We relate

∫D

C′ to∫ C

B′ by the change of variable τ ↦→ Sτ . Now

since f(Sτ) = τkf(τ) it follows that df(Sτ)dτ = kτk−1f(τ) + τkf ′(τ). Therefore

df(Sτ)

f(Sτ)=k

τdτ +

f ′τ

f(τ)dτ which implies

(IV )1

2πi

(∫ C

B′dlog(f) +

∫ D

C′dlog(f)

)=

1

2πi

∫ C

B′

(df(τ)f(τ)

− df(Sτ)

f(Sτ)

)=

=1

2πi

∫ C

B′−kτdτ =

−k2πi

∫ C

B′

dτ

τ

ε→0−→ − k

2πi

(− π

6i)=

k

12

Combining together the results in (I), (II), (III) and (IV ) proves the theoremunder our initial assumptions that the only singularities on C are on ρ,−ρ, i.For the general situation, if τ0 is a zero or a pole in C then |Re(τ0)| = 1

2 and/or|τ0| = 1, then also Tτ0 (or T−1τ0) and/or Sτ0 is a singularity and we can takearcs around such that the two contributions cancel out: in other words one arcinclude the point in the interior of C while the arc for the other point leaves itoutside.


τ0 Tτ0

A E

B

B′

C C ′D′

D

C

ρ −ρ

i

We already defined the quatities g2 and g3 for lattices in Theorem 2.27, andby mean of the one-to-one correspondence given by Proposition 3.2, we haveg2(τ) = 60E4(τ) and g3(τ) = 140E6(τ). Also define, in this new setting, themodular discriminant as ∆(τ) = g2(τ)

3 − 27g3(τ)2 and the modular j-invariant

j(τ) = 1728 g2(τ)∆(τ) .

Proposition 3.11. (a) ∆(τ) is a cusp form of weight 12 which is nowherevanishing on H and it has a simple zero at ∞.

(b) j(τ) is a modular function of weight 0 with a simple pole at ∞.

Proof. Even though we could use the correspondence, we rather take a differentapproach.

(a) We already know that g2(τ) and g3(τ) are modular forms of weight 4 and6 respectively, such that

g2(∞) = g∗2(0) = 60 · 2ζ(4) = 120ζ(4) = 0 g3(∞) = 280ζ(6) = 0

Moreover by Theorem 3.10 we have

div(deg(g2)) =4

12=

1

3=⇒ div(g2) = 1 · ρ

div(deg(g3)) =6

12=

1

2=⇒ div(g3) = 1 · i

where both implications follow by the fact that ordx(f) ∈ Z for all x ∈ H.

Now since ζ(4) = π4

90 and ζ(6) = π6

945 it follows that g2(∞) = 4π4

3 and

g3(∞) = 8π6

27 .After these computation follows that ∆(τ) is a modular form of weight 12since both g2(τ)

3 and g3(τ)2 are modular forms of weight 12 and ∆ = 0 on

H since ∆(i),∆(ρ) = 0. Finally

∆(∞) = π12((4

3

)3 − 27( 8

27

)2)= 0

which means that ∆(τ) is a cusp form. By Theorem 3.10, deg(div(∆)) = 1,therefore we have a simple zero at infinity.


(b) j(τ) is a modular function of weight 0 since both ∆(τ) and g2(τ)3 are of

weight 12. Moreover j(τ) is holomorphic on H since ∆(τ) = 0 for all τ ∈ H.More precisely deg(div(j)) = 0, in fact from the computation above followsthat div(j) = 1 · ρ− 1 · ∞.

Theorem 3.12. j induces a bijection j : Γ\H −→ C (denoted again by j).

Proof. Since j has weight 0, it is Γ-invariant i.e. j(γτ) = j(τ) for all γ ∈ Γ,τ ∈ H. Hence it descends to a holomorphic map j : Γ\H −→ C.We must check that ∀λ ∈ C there exists a unique τ = τλ ∈ H such that j(τ) = C.This is equivalent to saying that fλ(τ) = 1728g2(τ)

3− λ∆(τ) has a unique zeroin Γ\H −→ C. Notice that such fλ(τ) is a modular form of weight 12, therefore

1 = deg(div(fλ)) = ord∞(fλ) +1

2ordi(fλ) +

1

3ordρ(fλ) +

∑x =i,ρ

ordx(fλ)

Since ∆(∞) = 0 but g2(∞) = 0, we have ord∞(fλ) = 0. Since fλ is a modularform, it is holomorphic on H, thus has no poles. Therefore we have to solve

1 =a

2+b

3+ c for a, b, c ∈ N

which yields (a, b, c) ∈ (2, 0, 0), (0, 3, 0), (0, 0, 1) and in each case fλ has onlyone zero as we wanted.

3.1.4 The space of modular forms

Let us denote byMk (resp. Sk) the C-vector space of modular (resp. cusp)forms of weight k ∈ Z.

Proposition 3.13. (a) dimC(Mk/Sk

)≤ 1 and equality holds if k ≥ 4.

(b) Mk = 0 for k < 0 and k = 2.

(c) Multiplication by ∆ induces an isomorphism

Mk−12·∆−→ Sk

andMk = Sk ⊕ CEk for any k ≥ 12.

(d) M0 = C,M4 = CE4,M6 = CE6,M8 = CE8 andM10 = CE10

Proof. (a) Since Sk is the kernel of the map Mk −→ C given by f ↦→ f(∞),we have dimC

(Mk/Sk

)≤ 1. Now if k ≥ 4, then Ek ∈ Mk r Sk therefore

Mk = S ⊕ CEk.

(b) If f ∈Mk, f = 0 then ordx(f) ≥ 0 for all x ∈ Γ\H∪∞. On the other handdeg(div(f)) = k

12 therefore there are no nonzero modular form of negativeweight, and for k = 2 we get deg(div(f)) = 1

6 but

1

6=a

2+b

3+ c+ d has no solutions in N4.


(c) We know that div(∆) = 1 · ∞ and ∆(τ) = 0 for all τ ∈ H. If f ∈ Sk thenord∞(f) ≥ 1 therefore f/∆ ∈ Mk−12. Since ord∞(∆) = 1, it follows thatord∞(f/∆) ≥ 0 . Therefore the C-linear maps Mk−12 −→ Sk, f ↦→ f · ∆and Sk −→Mk−12, f ↦→ f/∆ are mutually inverse.

(d) We need to examine the cases k = 0, 4, 6, 8, 10. By (b) and (c), for theseweights Sk = 0 (otherwise by (c) we would have nonzero modular forms ofnegative weight, contradicting (b) ). On the other hand by (a), dimCMk ≤ 1and just notice that 0 = 1 ∈M0 and for k ∈ 4, 6, 8, 10, 0 = Ek ∈Mk.

Consider the C-algebra structure on M∞def=

∑k≥0

Mk ⊂ f : H −→ C |

f is holomorphic.

Lemma 3.14. M∞ is actually a direct sum:M∞ =⨁k≥0

Mk.

Proof. Consider a relation of the form∑i∈I

λifi = 0 for some finite set of indices

I and fi ∈Mi. We check that λi = 0 ∀i ∈ I: choose τ0 ∈ H such that fi(τ0) = 0for each i ∈ I. However

∑i∈I

λifi = 0 implies∑i∈I

λifi(γτ0) = 0 ∀γ ∈ Γ. Let

γ =

(a bc d

)then

0 =∑i∈I

λifi(γτ0) =∑i∈I

λi(cτ0 + d)ifi(τ0)

which means that the polynomial∑i∈I

λifi(τ0)Xi has the (infinitely many) roots

cτ0+d, thus it is the zero polynomial. Therefore λifi(τ0) = 0∀i ∈ I =⇒ λi = 0∀isince by our initial choice fi(τ0) = 0 ∀i ∈ I.

Theorem 3.15. The map C[X,Y ] −→ M∞ given by P (X,Y ) ↦→ P (g2, g3) isan isomorphism of C-algebras.

Proof. This map is clearly a C-algebra homomorphism.

• Surjectivity: it is enough to check that the monomials in gα2 gβ3 | 4α+6β =

k, α, β ≥ 0 generate Mk. We already know this for k = 0, 2, 4, 6 byProposition 3.13. Therefore assume that k ≥ 8 and proceed by inductionon k.Any k ≥ 8 even can be written as 4α+ 6β with α, β ≥ 0; we have gα2 g

β3 ∈

Mk and gα2 gβ3 (∞) = 0. Now let f belong toMk, then there exists λ ∈ C

such that (f−λgα2 gβ3 )(∞) = 0, so that (f−λgα2 g

β3 ) ∈ Sk and consequently

by Proposition 3.13 (c), f − λgα2 gβ3 = ∆ · g for some g ∈ Mk−12. By

inductive hypothesis g = P (g2, g3) for some P (X,Y ) ∈ C[X,Y ] and ∆ =g32 − 27g23 , hence

f = P (g2, g3)(g32 − 27g23) + λgα2 g

β3 =: P (g2, g3)

• Injectivity: let us prove that P (g2, g3) = 0 implies P (X,Y ) = 0 or equiv-alently that there are no non-trivial relations between g2 and g3.


By contradiction assume that there is a nontrivial algebraic relation amongg2 and g3. We can assume that such relation is homogeneous in the sensethat P (g2, g3) ∈Mk for some k and P (X,Y ) is sum of monomials λXαY β

with 4α+6β = k. Without loss of generality we can (and do) assume thatk is minimal, then the relation is of the form

(I) gm2 + g3P1(g2, g3) = 0 or (II) gn3 + g2P2(g2, g3) = 0

Recall that div(g2) =13 ·ρ and div(g3) =

12 · i, then evaluating (I) at i and

(II) at ρ we get a contradiction.

Remark 3.16. (i) The isomorphism C[X,Y ]∼−→M∞ is actually an isomor-

phism of graded C-algebras, where the grading in C[X,Y ] is defined by

deg(XαY β) = 4α+ 6β

(ii) The isomorphism Mk−12·∆−→ Sk implies that S∞ =

⨁k≥0

Sk is a principal

ideal ofM∞ generated by ∆.

We are now ready to describe all modular functions of weight 0: this amountsto characterize all meromorphic functions on Γ\H which are also meromorphicat infinity.

Theorem 3.17. Let FSL2(Z) be the set of a modular functions of weight 0.Then FSL2(Z)

∼= C(j), or in other words f is a modular function of weight 0 ifand only if there exists P (t) ∈ C(t) such that f = P (j).

Proof. If f = P (j) for some P (t) ∈ C(t) then f is a modular function of weight0 since j is. Conversely, let f be a modular function of weight 0. Recall thatj induces a bijection Γ\H −→ C and div(j) = 1 · ρ − 1 · ∞. Notice that forany τ0 ∈ H, j(τ) − j(τ0) is also a modular function of weight 0, with divisordiv(j−j(τ0)) = 1 ·τ0−1 ·∞. Therefore without loss of generality we can assumethat f is holomorphic on H (otherwise multiply f by a function of the form∏i

(j − j(τi)

)ki, with ki ≥ 0). Now by multiplying f for ∆n, n ≥ 0 we can also

assume that g = f∆n is a modular form inM12n. Therefore by Theorem 3.15we can write

g =∑α,β

4α+6β=12n

λαβgα2 g

β3

Setting p = α3 and q = β

2 , follows that p, q ≥ 0 are integers and p+ q = n, hence

f =g

∆n=

∑α,β

4α+6β=12n

λαβ(g32)

p(g23)q

∆p ·∆q

Thus we are left with weight 0 modular functionsg32

∆ = 11728j and (since ∆ =

g32 − 27g23)g23

∆ =g32−∆27∆ = 1

27·1728j −127 which is linear is j, and this concludes

the proof.

Corollary 3.18. The modular function of weight 0 for Γ form a field isomorphicto C(t).


3.1.5 The modular curve X0(1)

As sets write Y0(1)def= Γ\H and X0(1)

def= Y0(1) ∪ ∞.

Definition. The extended upper half plane is

H∗ = H ∪Q ∪ ∞ = H ∪ P1(Q)

Remark that the action of Γ on H extends to H∗. Identifying P1(Q) with

(Q2 r 0

)/Q× =

[xy

]| (x, y) ∈ Q2 r 0modulo linearity

then the action of SL2(Z) on Q2 r 0 given by(

xy

)↦→(a bc d

)(xy

)yields an action of Γ on P1(Q). In fact P1(Q) is identified with Q ∪ ∞ by[

xy

]↦→ x

y∈

Q if y = 0

∞ if y = 0

And with such identification, Γ acts on Q ∪ ∞ by

γ · xy=axy + b

cxy + dand γ · ∞ =

a

c(3.1.6)

for any γ =

(a bc d

)∈ Γ.

Definition. The elements of Γ\P1(Q) are called cusps of X0(1).

Lemma 3.19. (a) Γ\P1(Q) contains a single element, which we will denote by∞.

(b) StabΓ(∞) = ⟨T ⟩.

Proof. (a) Given any rational number xy we want to show that there exists γ ∈ Γ

such that γ xy =∞. If γ =

(a bc d

)mod ± 1, then

γ · xy=axy + b

cxy + d=ax+ by

cx+ dy

Without loss of generality assume that x, y ∈ Z, (x, y) = 1, then by theBezout identity there exist (a, b) ∈ Z2 such that ax + by = 1 therefore(a b−y x

)∈ SL2(Z) and

(a b−y x

)· xy=∞.


(b) If γ · ∞ = ∞ then by (3.1.6) γ corresponds to some

(a bc d

)with c = 0.

This implies that ad = 1, thus a = d = ±1 and γ =

(±1 b0 ±1

)= T±b. We

conclude that StabΓ(∞) = ⟨T ⟩.

Let us now define a topology on H∗:

• If τ ∈ H an open neighbourhood of τ will be an open neighbourhood of τin C contained in H (i.e. equip H with the induced topology from C).

• If τ = ∞ ∈ P1(Q) we define a base of open neighbourhoob to be thecollection of sets of the form

τ ∈ H | Im(τ) > A, A ∈ R>0 ∪ ∞ (3.1.7)

• If a ∈ Q we define a base of open neighbourhood of a to be the collectionof sets of the form

a ∪ open disc in H tangent to R at a (3.1.8)

Remark 3.20. (i) Given a ∈ Q, let γ ∈ Γ be such that γ∞ = a (which existsby Lemma 3.19), then γ sends elements in the base of open neighbourhoodof ∞ to elements of open neighbourhood of a.

(ii) Γ acts on the topological space H∗ via homeomorphisms.

(iii) H∗ with the topology we just introduced is Hausdorff and connected.

Definition. Define the topology on X0(1) as the quotient topology induced bythe canonical projection π : H∗ −→ X0(1). In other words U ⊂ X0(1) is open ifand only if π−1(U) is open in H∗.

Lemma 3.21. π is an open map.

Proof. If V ⊂ H∗ is open, then π−1(π(V)

)=⋃γ∈Γ

γV is open since union of open

sets in H∗. Therefore π(V) is open in X0(1) by definition of its topology.

In the following we want to show that X0(1) is a compact Riemann surface.Before proving that we need a preparation lemma. Let us denote I(τ1, τ2) = γ ∈Γ | γτ1 = τ2 for any given τ1, τ2 ∈ H∗ and let I(V1,V2) = γ ∈ Γ | γV1∩V2 = ∅for any given open V1,V2 ⊂ H∗. Then:

Lemma 3.22. For any pair τ1, τ2 ∈ H∗ there exist open neighbourhoods V1of τ1 and V2 of τ2 such that I(V1,V2) = I(τ1, τ2) (and such an action is calledproperly discontinuous).

Proof. Due to the relations

I(γ1τ1, γ2τ2) = γ2I(τ1, τ2)γ−11 and I(γ1V1, γ2V2) = γ2I(V1,V2)γ−1

1

we may assume that τ1, τ2 ∈ D ∪ ∞ and we have 3 cases:


• Let τ1, τ2 ∈ D, and notice that I(D,D) is finite and contains the elements

1Γ, T, S, T−1, ST, TS, (ST )2, (TS)2

Define F =⋃

γ∈I(D,D)

γD ⊃ D and remark that also I(F ,F) is finite since

F ⊆⋃

γ1,γ2∈I(D,D)

I(γ1D, γ2D)

Consider the elements γ ∈ I(F ,F) r I(τ1, τ2) =: I. For any such γ ∈ Iholds that γτ1 = τ2, thus we may fix Uγ ∋ γτ1 and Wγ ∋ τ2 open setssuch that Uγ ∩Wγ = ∅. Then define

V1 = F ∩ (⋂γ∈I

γ−1Uγ) ∋ τ1 and V2 = F ∩ (⋂γ∈IWγ) ∋ τ2

V1 and V2 are open and I(V1,V2) = I(τ1, τ2): in fact if γ ∈ I(V1,V2) rI(τ1, τ2) then γ ∈ I and by definition of V1 and V2 we have γ ∈ I(γ−1Uγ ,Wγ) =I(Uγ ,Wγ)γ. Therefore 1Γ ∈ I(Uγ ,Wγ) which yields Uγ ∩ Wγ = ∅ a con-tradiction.

• Let τ1 = τ ∈ D and τ2 = ∞. Recall that for γ ∈ Γ given by

(a bc d

)we have Im(γτ) = Im(τ)

|cτ1+d|2 . Now consider an open disc Dτ ∋ τ , then

by the relation above we have Im(γτ) < M = M(Dτ ) for some M ∈R>0 indipendent of τ . Let D∞ = z ∈ H | Im(z) > M ∪ ∞, thenI(Dτ , D∞) = ∅ = I(τ,∞).

• Let τ1 = τ2 = ∞, then I(∞,∞) = StabΓ(∞) = ⟨T ⟩. Set V = τ ∈ H |Im(τ) > 1 ∪ ∞ and let γ ∈ I(V,V) so that γV ∩ V = ∅. Thereforeγτ ′1 = τ ′2 for τ ′1, τ

′2 ∈ V r ∞. Write τ ′1 = Tnz1 and τ ′2 = Tmz2 for some

z1, z2 ∈ D. Then we have

γTnz1 = Tmz2 =⇒ z2 = T−mγTnz1

Since Im(zi) > 1, by Theorem 3.5 follows that T−mγTn ∈ T, T−1, 1Γand consequently, γ ∈ ⟨T ⟩.

Theorem 3.23. X0(1) is a connected, compact and Hausdorff topologicalspace.

Proof. H∗ is connected and π is continuous, hence X0(1) is connected. Now letUii∈I be an open covering of X0(1), then π−1(Ui)i∈I is an open coveringof H∗. Say ∞ ∈ π−1(Ui0) for some i0 ∈ I, then some set of the form 3.1.7 iscontained in π−1(Ui0). Since Dr (π−1(Ui0)∩D) is compact because closed andbounded, there is a finite set of indeces i1, .., ik such that

D r (π−1(Ui0) ∩ D) ⊂k⋃

j=1

π−1(Uij )


Therefore X0(1) ⊂ Ui0 ∪ .. ∪ Uik , thus it is compact.We are left to prove that X0(1) is Hausdorff: let x1, x2 ∈ X0(1), x1 = x2 andchoose τi ∈ H∗ such that π(τi) = xi, i = 1, 2. Since x1 = x2 we have I(τ1, τ2) = ∅therefore for i = 1, 2 by Lemma 3.22 we can select open neighbourhood Vi of τiin H∗ such that I(V1,V2) = ∅ and this implies that π(V1) ∩ π(V2) = ∅. Since πis an open map, π(Vi) is an open neighbourhood of xi in X0(1). We concludethat X0(1) is an Hausdorff topological space.

We are now left to define a complex structure on X0(1) such that makes itinto a compact Riemann surface.For each x ∈ X0(1) fix τ ∈ H∗ such that π(τ) = x and fix a open neighbourhoodVx ⊂ H∗ of τ such that I(Vx,Vx) = I(τ, τ) = StabΓ(τ). Then notice that

π(Vx) = StabΓ(τ)\Vx (3.1.9)

and π(Vx) is a open neighbourhood of x. Let us now define a homeomorphismψx : π(Vx) −→ Ux ⊆ C.

• Let x /∈ π(ρ), π(i), π(∞) and as above let τ ∈ H∗ be such that π(τ) =x. Then StabΓ(τ) = 1Γ and by (3.1.9) π(Vx) is homeomorphic to Vx.Therefore choose Ux = Vx and ψx = idVx

.

• Let x = π(∞) and choose τ = ∞. Recall that by Lemma 3.22, for V∞we have I(V∞,V∞) = StabΓ(∞) = ⟨T ⟩. Hence π(V∞) = TZ\V∞ and thelocal parameter τ ↦→ qτ = e2πiτ defines a homeomorphism

ψ∞ : TZ\V∞ −→ z ∈ C | |z| < e−2π ⊂ C such that [∞] ↦→ 0

• Let x = π(ρ), π(i), set τ = ρ, i respectively, write γ be a generator forStabΓ(τ) (as we found in Theorem 3.5) and let k be the order of γ (whichis 3 and 2 respectively).Recall that we have a holomorphic isomorphism

gx : H −→ D = z ∈ C | |z| < 1 given by w ↦→ w − τw − τ

(3.1.10)

Thus in particular τ ↦→ 0. Since γ is a holomorphic isomorphism of H offinite order k, then

γ∗ = gx γ g−1x : D −→ D

is a holomorphic isomorphism of order k in D such that γ∗(0) = 0.By complex analysis one can prove that γ∗ is of the form z ↦→ ξz for someξ ∈ S1 = u ∈ C | |u| = 1 and the order of γ∗ is k. Therefore such ξ is aprimitive k-th root of unity. Then define

ψx : StabΓ\Vx = ⟨γ⟩\Vx −→ D by [z] ↦→ gx(z)k

Notice that ψx is well-defined in fact by definition of γ∗ we have

gx(γz) = γ∗(gx(z)) =⇒ gx(γz)k =

(γ∗(gx(z))

)k=(ξgx(z)

)k= gx(z)

k


Moreover ψx is continuous since induced by the continuous map z ↦→gx(z)

k. Let us now prove that ψx is also injective: let gx(z)k = gx(z

′)k,then

gx(z) = ξrgx(z′) = (γ∗)r(gx(z

′)) = gx(γrz′) for some r ≥ 1

By injectivity of gx z = γrz′ and therefore z ≡ z′ mod ⟨γ⟩.Finally, since gx is a homeomorphism and w ↦→ wk is an open map on Dwe get that ψx is an open map, hence a homeomorphism onto ψx(⟨γ⟩\Vx).

A (boring) computation proves that the transition maps ψy ψ−1x are holomor-

phic for all x, y ∈ X0(1). Therefore we can conclude that X0(1) is a compactRiemann surface, called modular curve of level 1.

Proposition 3.24. The modular j-function j : X0(1) −→ P1(C) is a holomor-phic isomorphism of compact Riemann surfaces, mapping [∞] to ∞.

Proof. Let us proceed in multiple steps.

Claim 1: Let k be a positive integer, let νk : D −→ D be the holomorphicmap given by z ↦→ zk and let ξk = e2π/k. Now if f : D −→ C is a holomorphicfunction on D such that f(ξkz) = f(z) for all z ∈ D then f induces a mapg : D −→ C such that f = g νk, ord0(f − f(0)) ≥ k and g is holomorphic.

In fact let g(z)def= f(w) for w ∈ k

√z, then since w ∈ k

√z, also ξikw ∈ k

√z for all

i = 0, .., k − 1 and f(ξkz) = f(z) implies that g : D −→ C is well-defined andf = g νk. Now computing deriatives of f (which exists since f is holomorphic)one gets

f ′(z) = ξkf′(ξkz) ; f

′′(z) = ξ2kf′′(ξz) ; and so on..

Therefore f (m)(0) = 0 for all m such that n - m and consequently near z = 0we have

f(z) = f(0) +

∞∑i=1

anizni (∗)

which in particular implies ord0(f − f(0)) ≥ n.Finally using (∗) and the definition of g we get near z = 0

g(zn) = f(z) = f(0) +

∞∑i=1

anizni

and changing variable w = zn yields

g(w) = g(0) +

∞∑i=1

aniwni

Therefore g, having a power series expansion around 0 and radius of convergenceat least 1 1 is holomorphic in D.Claim 2: j induces a holomorphic function on Y0(1).

1In fact we have that composition of two holomorphic functions H = F G only makessense whenever G is analytic in a neighbourhood of z0 and F is analytic in a neighbourhoodof G(z0). In our case we know that f is holomorphic on D and νk is analytic in all C, thus ghas to be holomorphic in D.


Recall that j is holomorphic on H and is Γ-invariant therefore defines a map(which is a bijection by Theorem 3.12) : Y0(1) −→ C. Now consider thecommutative diagram

D

F

g−1x

→→

νk

↓↓

H

π

↓↓

j

→→ C

Dπ →→

G

Y0(1)

→→

where νk is defined as above with k = #StabΓ(x), g−1x is the inverse of the

biholomorphism in (3.1.10) centered at x, π is the canonical projection on thequotient, π is such that π = π exp with exp : H∗ −→ D, and finally F = j g−1

x

and G = π. Due to commutativity we have F = G νk and F is holomorphicsince g−1

x is a biholomorphism and j is holomorphic on H (by its definition). Toshow that j induces a holomorphic function on Y0(1) (as Riemann surfaces) isequivalent to show that G is holomorphic:

• For x = i, ρ, ν1 : z ↦→ z therefore F = G and G is holomorphic.

• For x = i we have k = 2 and if we prove that F (z) = F (−z) we canconclude by Claim 1. By the commutativity we in fact we have

F (z) = G(ν2(z)) = G(z2) = G(ν2(−z)) = F (−z)

• Similarly for x = ρ we get

F (z) = G(ν3(z)) = G(z3) = G(ν3(ξ3z)) = F (ξ3z)

Therefore is holomorphic.Claim 3: extends to a meromorphic function on X0(1) with a simple pole atthe cusp.Recall that from the proof of Proposition 3.11 we have

div(j) = 1 · ρ− 1 · ∞

And since π(P1(Q)) = [∞] ∈ X0(1), the claim follows.Claim 4 j defines an isomorphism of Riemann surfaces from X0(1) to P1(C).From Claim 3 we have that = j (with some abuse of notation) is a meromorphicfunction from X0(1) to C, now from the theory of Riemann surfaces we havethat Mer(X) = Hol(X,P1(C)) r c = ∞. Therefore this is equivalent to saythat j is a holomorphic function between the compact Riemann surfaces X0(1)and P1(C). Finally, since j has a single simple pole at ∞, deg(j) = 1, hence itis a biholomorphism.

Corollary 3.25. X0(1) has genus 0.

Remark that in view of Theorem 3.17, Proposition 3.24 and the curves-fieldscorrespondence, we have that the function field C(X0(1)) coincides with C(j).Moreover, since P1 can be defined over Q, the modular function j allows us todefine the structure of non-singular projective curve over Q for X0(1).

3.2. CONGRUENCE SUBGROUPS 58

3.2 Congruence subgroups

Definition. Let N ∈ N, then define the principal congruence subgroup of levelN to be

Γ(N) =

(a bc d

)∈ SL2(Z)

(a bc d

)≡(1 00 1

)mod N

A subgroup Γ of SL2(Z) is a congruence subgroup of level N if Γ(N) ⊂ Γ forsome N ∈ N. In particular define

Γ0(N) =

(a bc d

)∈ SL2(Z)

(a bc d

)≡(∗ ∗0 ∗

)mod N

Γ1(N) =

(a bc d

)∈ SL2(Z)

(a bc d

)≡(1 ∗0 1

)mod N

Remark 3.26. (i) Since Γ(N) is the kernel of the natural homomorphism

given by reduction modulo N , SL2(Z) −→ SL2(Z/NZ), thus Γ(N) isnormal in SL2(Z). Moreover the homomorphism is actually surjective,hence it induces the isomorphism

SL2(Z)/Γ(N)∼−→ SL2(Z/NZ) (3.2.1)

It follows that the index of Γ(N) in SL2(Z) is finite.

(ii) Γ(N) ⊂ Γ1(N) ⊂ Γ0(N) ⊂ SL2(Z): therefore also Γ1(N) and Γ0(N) havefinite index in SL2(Z), as well as any congruence subgroup.

(iii) Notice that for N > 2, −1 /∈ Γ(N),Γ1(N).

Proof. The only non-trivial statement is the surjectivity of SL2(Z) −→ SL2(Z/NZ),so that

1 −→ Γ(N) −→ SL2(Z)mod (N)−→ SL2(Z/NZ) −→ 1

is an exact sequence.

Let γ =

(a b

c d

)∈ SL2(Z/NZ) i.e. ad− bc = 1.

Write a, b, c, d for the representatives of a, b, c, d in Z ∩ 0, .., N − 1, so in par-

ticular ad − bc = 1 +Nk for some k ∈ Z. We want to lift the 1st column

(ac

)in a ”smart” way:

• If (a, c) = 1 then by the Bezout identity there exist β, δ ∈ Z such thataδ − cβ = 1 and we can choose β = b + Nr and δ = d + Ns for suitabler, s ∈ Z. In fact

a(d+Ns)− c(b+Nr) = ad− bc+N(as− cr) = 1 +Nk +N(as− cr)

By coprimality of a and c, there exist r, s ∈ Z such that as− cr = −k.

• If (a, c) = 1, then notice that necessarely (a, c,N) = 1, otherwise f =(a, c,N) > 1 is not invertible mod (N) but f(a′d − bc′) = 1 and we geta contradiction.


Established that, we are now looking for a lifting a′ of a such that a′ ≡ amod (N) and (a′, c) = 1.Define l =

∏p primep|c,p-a

p and let a′ = a+ lN then we claim that (a′, c) = 1.

In fact, ∀p | c we have a ≡ 0 mod (p) since

– If p | a then p - l, N hence lN ≡ 0 mod (p) ⇒ a′ ≡ lN ≡ 0 mod (p)

– If p - a then p | l hence a′ ≡ a+ 0 ≡ 0 mod (p)

This implies that a′, c are coprime, and select them as lift for the 1stcolumn. Using the same argument, there exist β = b+Nr, δ = d+Ns ∈ Zsuch that a′δ − cβ = 1 for suitable r, s ∈ Z. In fact

a′(d+Ns)−c(b+Nr) = (a+lN)d−bc+N(a′s−rc) = 1+Nk+Nld+N(a′s−rc)

By the Bezout identity, we can choose r, s such that a′s− rc = −k − ld.

For a modular subgroup Γ define the sets YΓ = Γ\H and XΓ = Γ\H∗ =YΓ ⊔ CΓ where CΓ = Γ\P1(Q) is the set of cusps of Γ.

Lemma 3.27. CΓ is a finite set.

Proof. Consider the finite set decomposition SL2(Z) = Γγ1 ⊔ .. ⊔ Γγr, γi ∈SL2(Z) and r = [SL2(Z) : Γ]. We know that for all x ∈ P1(Q) there existsγ ∈ SL2(Z) such that γ · ∞ = x. Write γ = γ′ · γi for some γ′ ∈ Γ and γi asabove. Then

x = γ · ∞ = γ′ · γi · ∞ =⇒ [x] = [γi∞] in CΓ

Hence CΓ = [γi · ∞], i = 1, .., r which is finite.

Remark that from the proof we deduced more, namely that #CΓ ≤ [SL2(Z) :Γ]. In general such inequality may be strict. For example we have that [SL2(Z) :Γ0(p)] = p+ 1 and CΓ0(p) = [∞], [0] for any rational prime p.

Theorem 3.28. The natural projection π = πΓ : H∗ −→ XΓ equips XΓ witha natural structure of compact Riemann surface, XΓ is called modular curve oflevel Γ.

Proof. Let us prove the theorem for X0(N), X1(N) and X(N) which are thequotients of H∗ by Γ0(N),Γ1(N) and Γ(N) respectively.

Recall that D∗ def= D ∪ ∞ surjects onto X0(1) and let us prove that D∗ is

compact in H∗. Let Uii∈I be any open covering of D∗, then there exist someindex i0 ∈ I such that∞ ∈ Ui0 . Therefore there is a neighboorhood of∞ in H∗,say VM = z ∈ H : |z| > M ∪ ∞ for some M ∈ R>0, such that Ui0 containsVM ∩D∗. It follows that D∗∩Vc

M+1 is closed and bounded in H, hence compact.Since, trivially, D∗ ∩ Vc

M+1, ⊂ C ⊂⋃i∈I

Ui, we get that Uii∈I is a covering of

D∗ ∩ VcM+1 and by compactness, there exists a finite subcovering Ui1 , ..,Uik

of it such that D∗ ∩ VcM+1 ⊂

k⋃j=1

Uij . We can conclude that the set Ui0 , ..,Uik


is a finite subcovering the initial covering, therefore D∗ is compact in H∗.Step 1 X0(N), X1(N) and X(N) are compact.Since Γ(N) ⊂ Γ1(N) ⊂ Γ0(N) we have canonical (continuous) surjections

X(N)π1−→ X1(N)

π0−→ X0(N). Therefore it is enough to show that X(N) iscompact, since if so, then the continuous image of a compact is compact.Let [SL2(Z) : Γ(N)] = r <∞ and consider the right coset decomposition

SL2(Z) = Γ(N)γ1 ⊔ · · · ⊔ Γ(N)γr

for γ1, .., γr ∈ SL2(Z) cosets representatives. Define K =n⋃

i=1

γi(D∗), then K is

compact in H∗ since finite union of compact subsets (since the action of SL2(Z)is continuous, γiD∗ is compact for all i’s), and such such K surjects to X(N).In fact let [τ ] ∈ X(N) and let us prove that there exists ξ ∈ K such thatπ(ξ) = [τ ], or equivalently that there exists γ ∈ Γ(N) for which γξ = τ . Wealready know that for every τ ∈ H∗ there exist γ ∈ SL2(Z) and η ∈ D∗ such thatγη = τ . Now using the coset decomposition we have that SL2(Z) ∋ γ = γ′ · γifor some γ′ ∈ Γ(N) and γi ∈ SL2(Z) representative. Letting ξ := γiη, it followsthat ξ ∈ K and π(ξ) = [τ ].Step 2 For any τ ∈ H the stabilizer of τ in Γ(N) is either ±1 or trivial.In fact remark that StabΓ(N)(τ) ⊆ StabSL2(Z)(τ) for all τ ∈ H and StabΓ(N)(τ) =StabSL2(Z)(τ) ∩ Γ(N). Let us first analize the case for τ ∈ D:

• If τ = i, ρ,−ρ, by Theorem 3.5 we already have StabSL2(Z)(τ) = ±1 andjust notice that ±1 ∈ Γ(2) but −1 /∈ Γ(N) if N ≥ 3, hence StabΓ(2)(τ) =±1 and for N ≥ 3, StabΓ(N)(τ) = ±1.

• If τ = i we have StabSL2(Z)(i) = ±1,±Swhere S =

(0 1−1 0

). However

±S /∈ Γ(N), hence the stabilizer of i is the same of the generic τ above.

• If τ = ρ,−ρ, similarly we obtain StabSL2(Z)(τ) = ±1,±ST,±(ST )2;however ±ST,±(ST )2 /∈ Γ(N) for any N . It follows that also the stabilizerof ρ and −ρ is as in the general case.

For any other point ξ ∈ H there exist γ ∈ SL2(Z) and τ ∈ D such thatγ(τ) = ξ, and

StabSL2(Z)(ξ) = γStabSL2(Z)(τ)γ−1

In fact if η ∈ StabSL2(Z)(τ) then γηγ−1(ξ) = γη(τ) = γ(τ) = ξ.

Viceversa, if ν ∈ StabSL2(Z)(ξ) then γ−1νγ(τ) = γ−1ν(ξ) = τ .

Thus setting σ = γ−1νγ we obtain ν = γσγ−1. Therefore since the stabilizersare conjugated in SL2(Z) they have the same group structure (conjugation isan automorphism).It follows that we always are in one of the two cases:

∀τ ∈ H : StabΓ(2)(τ) = ±1; or if N ≥ 3 StabΓ(N)(τ) = 1

Step 3 Riemann surface structure on Y (N).Now recall that the action of SL2(Z)/±1 is properly discontinuous, thereforealso the actions of Γ(2)/±1 and of Γ(N), for N ≥ 3, are. Let [τ ] ∈ Y (N), andtake Uγiτ a neighbourhood of γiτ in H (where γi is a coset representatives as


above) such that ∀γ ∈ Γ(2) r ±1 or Γ(N) r 1 it holds γUγiτ ∩ Uγiτ = ∅.Then we obtain that π(Uγiτ ) = StabΓ(N)(τ)\Uγiτ =: U[τ ] is a neighbourhood of

[τ ] in Y (N). Taking as local coordinates for [τ ], (U[τ ], π−1|Uγiτ

) yields a Riemann

surface structure on Y (N).Step 4 Riemann surface structure on Y0(N) and Y1(N).Let us prove that for Γ = Γ0(N),Γ1(N) and for any τ ∈ H, ±StabΓ(τ)/ ± 1 isa cyclic group of order at most three. As above

StabΓ(τ) = StabSL2(Z)(τ) ∩ Γ ∼= γ−1StabSL2(Z)(ξ)γ ∩ Γ = StabΓ(ξ)

where ξ ∈ D and γ ∈ SL2(Z) are such that γ(ξ) = τ .Therefore it is enough to prove for elements in D, by Theorem 3.5 every pointhas cyclic stabilizer of order at most 6, and by quotienting for ±1 we get thatsuch quotient is cyclic of order at most 3.Therefore define the charts as follows :

• If τ ∈ H has trivial stabilizer in then we can consider Uτ ⊂ H ⊂ C and thecanonical projection π|Uτ

: Uτ −→ π(Uτ ) is already a homeomorphism, so

take (π(Uτ ), π−1|Uτ

) as local chart around π(τ).

• If τ has non-trivial stabilizer, say of order k, then fix a biholomorphismg−1τ : D −→ H with the property that g−1

τ (0) = τ . Now since the stabilizercan be viewed as a cyclic subgroup of order k of the automorphisms of Dsuch that they fix 0, then they operate on D as multiplication by powersof τ . This yields a commutative diagram

Dg−1τ →→

νk

↓↓

H

π

↓↓D

π →→ Y

where Y is either Y0(N) or Y1(N) and νk : w ↦→ wk. Then there is a neigh-bourhood V of 0 in D such that π(V) = π(U) = StabΓ(τ)\U . Therefore acomplex chart around τ is given by (π(V), π−1

|V ).

This construction makes Y0(N) and Y1(N) into a Riemann surface.Step 5 Stabilizers of the cusps.

We are left to study the cusps: let us show that for z ∈ P1(Q) we have

StabΓ(z) = γ−1StabSL2(Z)(∞)γ ∩ Γ

where γ ∈ SL2(Z) is such that γ(z) =∞.Since StabΓ(z) = StabSL2(Z)(z) ∩ Γ it is enough to prove the equality

StabSL2(Z)(z) = γ−1StabSL2(Z)(∞)γ

We already know that StabSL2(Z)(∞) = ⟨±T ⟩ where T =

(1 10 1

). Now let

η ∈ StabSL2(Z)(∞), then

γ−1ηγ(z) = γ−1η(∞) = γ−1(∞) = z


hence γ−1ηγ ∈ StabSL2(Z)(z) for any η ∈ StabSL2(Z)(∞). Viceversa, let ν ∈StabSL2(Z)(z), then

γνγ−1(∞) = γν(z) = γ(z) =∞

hence γνγ−1 =: η belongs to StabSL2(Z)(∞), thus ν = γ−1ηγ.We conclude that StabΓ(z) = γ−1⟨±T ⟩γ ∩ Γ. Now consider two cases:

• If z =∞, we have

StabΓ(∞) = T−n⟨±T ⟩Tn∩Γ = ⟨±T ⟩∩Γ =

⎧⎪⎪⎪⎨⎪⎪⎪⎩⟨±T ⟩ if Γ = Γ0(N),Γ1(2)

⟨T ⟩ if Γ = Γ1(N), N ≥ 3

⟨TN =

(1 N

0 1

)⟩ if Γ = Γ(N)

• If z = ∞, then z ∈ Q so we may assume that z = αβ for some α, β ∈ Z

coprime and β = 0.

First of all let us characterize γ =

(a bc d

)∈ SL2(Z) such that γ(z) =∞:

γ(z) =∞ =⇒ aα+ bβ

cα+ dβ=∞

This forces cα+ dβ = 0 (1) and aα+ bβ = 0 (2).

(1) =⇒ c = −kβ and d = kα for some k ∈ Z

The condition on the determinant ad− bc = 1 implies that k = ±1 (oth-erwise we get a contradiction), so without loss of generality assume k = 1, and also that a = β, b = −α (otherwise γ would be singular). Since α, βare comprime, by the Bezout identity, there exist suitable a, b ∈ Z suchthat aα+ bβ = 1, thus choosing such pair we have

γ =

(a b−β α

)and γ−1 =

(α −bβ a

)Let us show ∀N ∈ N and ∀z ∈ Q we have StabΓ(N)(z) = ∅ (and conse-quentely for all other Γ’s stabilizers are not empty). Consider the following:(

α −bβ a

)(1 N0 1

)(a b−β α

)=

(α −bβ a

)(a−Nβ b+Nα−β α

)=

=

(aα+ bβ −Nαβ bα+Nα2 − bαaβ −Nβ2 − aβ aα+ bβ +Nαβ

)=

(1−Nαβ Nα2

Nβ2 1 +Nαβ

)≡(1 00 1

)mod (N)

Furthermore, each stabilizer is subgroup of a cyclic group, thus it is cyclic,

and it contains as a subgroup ⟨γ−1

(1 N0 1

)γ⟩ which is infinite, hence it

is itself infinite.

Step 6 Local parameter at the cusps.

Let [s] ∈ CΓ be a cusp, s ∈ P1(Q). Consider the open subset Vs = H∪s of H∗,let γ ∈ SL2(Z) be such that γ(s) =∞. Then we have the exponential map to D


given by τ ↦→ e2πiγ(τ)/n. Since the action of SL2(Z) is continuous, γ(Vs) = V∞and we just computed the stabilizer of∞ for different Γ’s: StabΓ(∞) is generated

by

(1 b0 1

)for a suitable b ≥ 1. Choosing n = b is enough to have a bijection

StabΓ(s)\Vs−→D given by τ ↦→ e2πiγ(τ)/n

Therefore consider the composition of maps

Vs(left) action of γ−→ V∞

exp−→ Dπ−→ Γ\H∗

So s ↦→ 0 ∈ D and there exists U neighbourhood of 0 in D such that π|U is ahomeomorphism. Therefore take (StabΓ(s)\Vs, π|U ) as a local chart around thecusp [s].

3.2.1 Modular functions of higher level

Definition. Let Γ be a congruence subgroup of SL2(Z),

(i) A weakly modular function of weight k ∈ Z for Γ is a meromorphic function

f : H −→ P1(C) such that ∀γ =

(a bc d

)∈ Γ,∀τ ∈ H it satisfies the

functional equation f(γτ) = (cτ + d)kf(τ).

(ii) A modular function of weight k for Γ is a weakly modular function for Γwhich is also meromorphic at all the cusps in CΓ.

(iii) A modular form of weight k for Γ is a modular function for Γ which isholomorphic at all the cusps in CΓ.

(iv) A cusp form of weight k for Γ is a modular form for Γ which vanishes atall the cusps in CΓ.

Let us explain more in details what this means. We use the local parameterat [s] ∈ CΓ, that we introduced in Step 6 of the proof of Theorem 3.28, to definewhat it means to be meromorphic, holomorphic, vanishing at [s].Let f : H −→ P1(C) be a weakly modular function of weight k for Γ. For any

σ =

(a bc d

)define j(σ, τ) = (cτ + d) and consider the meromorphic function

on H given byf |[σ]k(τ) = f(σ(τ))j(σ, τ)−k (3.2.2)

Hence for σ ∈ Γ we have f |[σ]k(τ) = f(τ) due to the modularity property for Γ.Moreover a simple computation shows that f |[σ]k is a weakly modular functionof weight k for σ−1Γσ. Now let γ ∈ SL2(Z) be such that γs =∞, then f |[γ−1]kis invariant for τ ↦→ τ +n, where n is as in Step 6. Define a function f∗s on someopen disc of C centered at 0 with origin remuved, by

f∗s (e2πiτ/n) = f |[γ−1](τ) (3.2.3)

Then we say that f is meromorphic (resp. holomorphic, resp vanishes) at thecusp [s] if f∗s is meromorphic (resp. holomorphic, resp. vanishes) at 0. In otherwords

f |[γ−1]k(τ) =∑i≥n0

aiqi/n = Φ(q) for q = e2πiτ (3.2.4)


called Puiseaux expansion.We can define divisors of modular functions for congruence subgroups in a sim-ilar way as we did for SL2(Z).Let f be a modular function of weight k for Γ: let us define ordx(f) for allx ∈ XΓ.

• If x corresponds to a point z0 ∈ H, then fix a holomorphic isomorphismg = gz0 : H −→ D such that g(z0) = 0. Let e = #StabΓ(z0), then t = g(z)e

is the local normal form at x, therefore define ordx(f) =ord(z−z0)(f)

e .

• If x corresponds to a cusp s, let γ ∈ SL2(Z) be such that γ(s) =∞. Then

we saw that ±γStabΓ(s)γ =

±(1 h0 1

)m

m ∈ Z

for some h ∈ Z>0.

Let q = e2πiz/h, since f |[γ−1]k(z) is invariant under z ↦→ z+h there existsa meromorphic function Φ such that Φ(q) = f |[γ−1]k(z), therefore defineordx(f) = ord0(Φ).

Finally define div(f) =∑

x∈XΓ

ordx(f) · x ∈ DivQ(XΓ) = Div(XΓ)⊗Z Q.

We saw at the end of 3.1 that the analytic isomorphism given by the modularfunction j of weight 0, j : X0(1)

∼−→ P1(C) allows us to define the structureof projective nonsingular curve over Q. Untill the end of these section and thenext one, we want to adress the following question:Question: For a congruence subgroup Γ of SL2(Z), can we similarly define onXΓ the structure of nonsingular curve over Q (or at least over a finite extensionof Q)?We will focus on Γ = Γ0(N) and imitating the case of X0(1) we start by describ-ing the field of modular functions FΓ0(N) = C(X0(N)) of weight 0 for Γ0(N).

Remark 3.29. (i) Notice that j ∈ FΓ0(N) = C(X0(N)) since j ∈ FSL2(Z) =C(X0(1)).

(ii) Define jN by jN (τ) = j(Nτ) for any τ ∈ H, then jN ∈ C(X0(N)). In fact

let γ =

(a bNc d

)∈ Γ0(N), then

(a Nbc d

)∈ SL2(Z) and

jN (γτ) = j(N

aτ + b

Ncτ + d

)= j(a(Nτ) +Nb

c(Nτ) + d

)= j((

a Nbc d

)·Nτ

)= j(Nτ) = jN (τ)

Therefore jN is Γ0(N)-invariant and holomorphic on H since j is. Moreoverj is meromorphic at [∞], hence jN is meromorphic at the cusps of Γ0(N).

We first need to compute the q-expansion of j at ∞:

Fact 3.30. The modular function j of weight 0 has a q-expansion of the form

j(τ) =1

q+∑n≥0

a(n)qn

with a(n) ∈ Z and q = e2πiτ

Proof. See [Kna93] Corollary 8.2


Theorem 3.31. (The modular equation)

(a) C(X0(N)) ∼= C(j, jN );

(b) jN satisfies a non-zero polynomial in C(j)[Y ];

(c) If FN (j, Y ) ∈ C(j)[Y ] is the monic polynomial of minimal degree such thatFN (j, jN ) = 0, then FN (X,Y ) ∈ Z[X,Y ].

Proof. Claim 1: Any f ∈ C(X0(N)) satisfies a polynomial in C(j)[Y ] of degreeequal to r = [SL2(Z) : Γ0(N)].

Consider the coset decomposition SL2(Z) =r⨆

i=0

Γ0(N)γi for γi ∈ SL2(Z); for

any f ∈ C(X0(N)) define the polynomial Ff (Y ) =r∏

i=1

(Y − f(γiτ)). By its

definition, degFf (Y ) = r and f(τ) is a root since there is an index for whichγi ∈ Γ0(N). Notice that f(γiτ) only depends upon the right coset Γ0(N)γisince f(γiτ) = f(γγiτ) for any γ ∈ Γ0(N) as f is Γ0(N)-invariant. Moreoverthe coefficients of Ff (Y ) are symmetric polynomials in f(γiτ)’s and right mul-tiplication of the above cosets by γ ∈ SL2(Z) induces a permutation σ = σγ ofthe cosets: Γ0(N)γiγ = Γ0(N)γσ(i). It follows that the coefficients of Ff (Y ) areSL2(Z)-invariant : if ci(τ) is a coefficient, then ci(τ) = S(f(γ1τ), .., f(γrτ)) =S(f(γσ(1)τ), .., f(γσ(r)τ)) = ci(γτ) for S = S(z1, .., zr) a symmetric polynomial.We can conclude that ci(τ) ∈ C(j), so that Ff (Y ) ∈ C(j)[Y ].Claim 2 C(X0(N)) is a finite extension of C(j) of degree less or equal to r.Choose f ∈ C(X0(N)) such that its minimal polynomial Pf (Y ) ∈ C(j)[Y ] hasmaximal degree, then

(∗) C(X0(N)) = C(j, f)

In fact, suppose there exists g ∈ C(X0(N))rC(j, f), then [C(j)(f, g) : C(j)] >[C(j)(f) : C(j)] and by the Primitive element theorem C(j)(f, g) = C(j)(h) andtherefore degPh(Y ) > degPf (Y ) contradicting maximality for f . Therefore (∗)holds and by Claim 1 follows that [C(X0(N)) : C(j)] ≤ r.Claim 3 C(X0(N)) = C(j, jN ) and [C(X0(N)) : C(j)] = r.By Claim 2, it is enough to prove that PjN (Y ) the minimal polynomial for jNover C(j) has exactly degree r. We actually reduce to check that

FN (j, Y ) = FjN (Y ) =

r∏i=1

(Y − jN (γiτ)) (3.2.5)

is the minimal polynomial of jN . Since FN (j, Y ) is satisfied by jN , it is enoughto show that

(i) The roots jN (γiτ) of FN (j, Y ) are all conjugate over C(j);

(ii) These roots are pairwise distinct.

Now if PjN (Y ) = PjN (j, Y ) is the minimal polynomial of jN over C(j), thenPjN (j(τ), jN (τ)) = 0 for all τ ∈ H. Therefore

0 = PjN (j(γiτ), jN (γiτ)) = PjN (j(τ), jN (γiτ)) = PjN (jN (γiτ))


where the second equality follows by the SL2(Z)-invariance of j. Hence jN (γiτ)are conugate.Suppose that jN (γiτ) = jN (γkτ) for all τ ∈ H, or more explicitly j(Nγiτ) =j(Nγkτ). This means that the holomorphic isomorphism

η : H −→ H τ ↦→(N 00 1

)γkγ

−1i

(N−1 00 1

)τ

is such that j(τ) = j(ητ) for all τ ∈ H. Since j is a holomorphic isomorphism

from SL2(Z)\H to C, η is induced by a matrix γ =

(a bc d

)∈ SL2(Z), i.e. for

all τ ∈ H

γτ =

(N 00 1

)γkγ

−1i

(N−1 00 1

)τ

Therefore

γkγ−1i =

(N−1 00 1

)γ

(N 00 1

)=

(N−1 00 1

)(a bc d

)(N 00 1

)=

(a N−1bNc d

)Notice that the same computation from the second equality for a generic elementof SL2(Z) yields(

N−1 00 1

)SL2(Z)

(N 00 1

)∩ SL2(Z) ⊆ Γ0(N)

We can conclude that γkγ−1i ∈ Γ0(N) and therefore Γ0(N)γi = Γ0(N)γk.

Claim 4 FN = FN (j, Y ) ∈ C(j, Y ) actually belongs to Z[j, Y ].The first step is to check that FN (j, Y ) ∈ C[j, Y ]: since the coefficients ci(τ)’s ofFN are symmetric polynomials in holomorphic functions on H, they are them-selves holomorphic on H and belong to C(j). Therefore for any i, they are of

the form λiPi(j)Qi(j)

for some λi ∈ C and Pi(X), Qi(X) monic polynomials with

distinct roots. This acutally implies that Qi(X) = 1 for all indeces i, because ifX − βi is a linear factor, then ci(τ) would have a pole at each τ ∈ H such thatj(τ) = β (which exists by surjectivity of j) and this contradicts holomorphicityof ci(τ). It follows that FN (j, Y ) ∈ C[j, Y ].

Now consider jN (γτ) = j((N 0

0 1

)γkτ), then by linear algebra there exists

γ ∈ SL such that γ

(N 00 1

)γk =

(a b0 d

)with ad = N, 0 ≤ b < d. Therefore

jN (γkτ) = j((a b

0 d

))= j(aτ + b

d

)(3.2.6)

and recall that j(τ) = 1q +

∑n≥0

a(n)qn with a(n) ∈ Z, q = e2πiτ .

Notice that e2πiaτ+b

d = e2πib/de2πiaτ/d, thus since d | N , jN (γkτ) has a q-expansion in q1/N in the ring Z[e2πi/N ] for any k ∈ 1, .., r. Therefore the sameholds for the coefficients ck(τ) of FN (j, Y ) which are symmetric polynomials injN (γkτ). We previously proved that ck(τ) = Pk(j) for some P (X) ∈ C[X], andwe now claim that the coefficients of P (X) are algebraic integers in Z[e2πi/N ].Write

P (j) =

m∑n=0

anj(τ)n =

∑k≥k0

bkqk


and notice that an ∈ Z[bk | k ≥ k0]. By comparison we have

am = b−m ; am−1 = b1−m −mama(1)

am−2 = b2−m − (m+ 1)am−1a(1)−m(m− 1)

2ama(2); and so on..

Therefore FN (X,Y ) =∑m,n

am,nXmY n with am,n algebraic integers and a0,r = 1

since FN (j, Y ) is monic. Substituting the q-expansions of j and jN we ob-tain a system of linear equations in the indeterminates am,n corresponding toFN (j, jN ) = 0. Such system has a unique solution since FN arises from theunique monic minimal polynomial of jN over C(j). Moreover the coefficients ofsuch system are rational, therefore by the theory of linear systems the uniquesolution (am,n)m,n has rational entries. Since they are also algebraic integers,they belong to Z.

Remark that Theorem 3.31 (b) tells us that jN is algebraic over C(j).

3.2.2 The canonical model of X0(N) over QLet us state but not prove a technical lemma:

Lemma 3.32. Let C(x, y) be a field such that x is transcendental over C and yis algebraic over C(x). Then for a subfield K0 ⊂ C, the following are equivalent:

(a) y is algebraic over K0(x) and the minimal polynomial of y over K0(x) re-mains irreducible over C(x);

(b) [K0(x, y) : K0(x)] = [C(x, y) : C(x)];

(c) C ∩K0(x, y) = K0;

(d) K0 ∩K0(x, y) = K0

Proof. See [Kna93] Theorem 11.36

Proposition 3.33. Let C(f, g) be a function field of dimension 1 over C, letK0 ⊂ C be a subfield such that K0(f, g) ∩ C = K0. For x ∈ K0(f, g)rK0 let Bbe the integral closure of C[x] in C(f, g) and write V for the nonsingular affinecurve associated (as in Step 1 of Theorem 1.15). Then V is defined over K0 andK0(V ) ∼= K0(f, g), C(V ) ∼= C(f, g).

Proof. Suppose f is transcendental over C and g algebraic over C(f). By Lemma3.32, f is transcendental over K0 and g algebraic over K0(f), therefore K0(f, g)has transcendence degree one over K0. Since K0(f, g) ∩ C = K0, such x ∈K0(f, g)rK0 cannot belong to C and therefore x is transcendental over K0. Itfollows that x is a transcendental basis for K0(f, g) over K0 and in particularf, g are algebraic over K0(x). This means that K0(f, g)/K0(x) is a finite algebraicextension, thus by the Primitive element theorem there exists some y ∈ K0(f, g)such that K0(f, g) = K0(x, y) and consequently C(f, g) = C(x, y). Lemma 3.32is indipendent of the generators, thus

n := [K0(f, g) : K0(x)] = [C(f, g) : C(x)]


Since C[x] is a principal ideal domain, there exists a basis x1, .., xn of C(f, g)over C(x) consisting of elements of B, such that

B =

n∑i=1

C[x]xi (3.2.7)

Thus we take x1, .., xn, x as generators of B over C so that the map

ϑ : C[X1, .., Xn+1] −→ B

such that for i = 1, .., n, Xi ↦→ xi and Xn+1 ↦→ x shows that B is the affinecoordinate ring of the curve V defined by I = kerϑ.We need to show that V is canonically defined over K0 or in other words thatI is (finitely) generated by elements in K0[X1, .., Xn+1]. To accomplish that,write B0 for the integral closure of K0[x] in K0(f, g), then there exists a basisy1, .., yn of K0(f, g) over K0(x) with yi ∈ B0 such that

B0 =

n∑i=1

K0[x]yi (3.2.8)

and consider the map

ϑ0 : K0[X1, .., Xn+1] −→ B0

such that Xi ↦→ yi for i = 1, .., n and Xn+1 ↦→ x, with kernel I0 = kerϑ0.Since K0[X1, .., Xn+1] is a Noetherian ring, I0 is finitely generated. So let I0 =⟨P1, .., Pℓ | Pj ∈ K0[X1, .., Xn+1]⟩. Now applying the exact functor · ⊗K0

C tothe exact sequence

0 −→ I0 −→ K0[X1, .., Xn+1] −→ B0 −→ 0

it yields the short exact sequence

0 −→ I0 ⊗K0C −→ C[X1, .., Xn+1]

ϑC0−→ B0 ⊗K0

C −→ 0

where ϑC0 = ϑ0 ⊗ idC. Now I0 ⊗K0 C is generated by P1, .., Pℓ over C and

B0 ⊗ C =n∑

i=1

C[x]yi. By Lemma 3.32 (a), yi’s are linearly indipendent over

C(x) hence we have two bases for C(f, g) over C(x), namely xi and yi.Therefore we have a C(x)-linear map ϕ : C(f, g) −→ C(f, g) such that xi ↦→ yifor all i = 1, .., n. In particular ϕ is injective since sends a basis to a basis andϕ ϑ = ϑC0 as C-linear maps. We can conclude that

I = kerϑ = kerϕ ϑ = kerϑC0 = I0 ⊗ C = ⟨P1, .., Pℓ⟩

where the second equality follows by injectiveness of ϕ.Moreover by construction, the affine coordinate ring of V is B and the functionfield is the quotient field of B, which by (3.2.7) is C(f, g). Finally if V is definedover K0, we have

I(V/K0) = I(V ) ∩K0[X1, .., Xn+1] = I ∩K0[X1, .., Xn+1] =

= (I0 ⊗ C) ∩K0[X1, .., Xn+1] = I0Thus by definition K0[V ] = K0[X1, .., Xn+1]/I0 = B0 and by (3.2.8) its quotientfield is K0(f, g).

3.3. INTEGRALITY OF THE J-INVARIANT 69

We can conclude that if K0 ⊂ C is a subfield and C is a nonsingular projectivecurve defined over K0, then there exists f, g ∈ K0(C) such that:

K0(C) = K0(f, g); C(C) = C(f, g) and C ∩K0(C) = K0

Viceversa, if C(f, g) is a function field of dimension 1 over C and K0 is a subfieldof C such that K0(f, g)∩C = K0 then there exists a nonsingular projective curveC defined over K0 such that

K0(C) ∼= K0(f, g) and C(C) ∼= C(f, g)

Furthermore, such C is unique due to the field-curves correspondence we saw inChapter 1. We can finally apply everything with f = j, g = jN and K0 = Q toobtain

Theorem 3.34. There exist a nonsingular projective curve C defined over Qand a biholomorphic map ϕ : X0(N) −→ C(C) such that

ϕ∗(C(C)) = C(X0(N)) = C(j, jN ) and ϕ∗(Q(C)) = Q(j, jN )

Such curve is unique up to isomorphism defined over Q and ϕ is uniquely de-termined by the isomorphism of Q(C) with Q(j, jN ).

Definition. The pair (ϕ,C) is called canonical model for X0(N) over Q.

Remark 3.35. 1. Moreover one can prove that an element of C(j, jN ) is inQ(j, jN ) if and only if its q-expansion at ∞ has coefficients in Q. ( See[Kna93] Corollary 11.50 or [Shi73] Proposition 6.9)

2. An alternative way to construct the canonical model of X0(N) over Q isvia

Y0(N) −→ V (FN ) = (x, y) ∈ A2 | FN (x, y) = 0 ; [τ ] ↦→ (j(τ), jN (τ))

The equation FN (X,Y ) = 0 defines a curve C over Q and by remuvingsingular points one obtains an affine nonsingular curve Cns over Q. ThenCns can be embedded in a projective nonsingular curve C such that (j, jN )extends to an isomorphism X0(N) −→ C(C).

3. A similar discussion applies to FΓ for Γ = Γ1(N),Γ(N) and one obtainsmodels for XΓ over Q or over Q(e2πi/N ) respectively. However the methodis less explicit and requires to study the automorphisms of C on FΓ (See[Shi73] Chapter 6).

3.3 Integrality of the j-invariant

This whole section will be dedicated to prove the following:

Theorem 3.36. If z ∈ H belongs to an imaginary quadratic field then j(z) isan algebraic integer.

The two steps of the proof will be to show that j(z) is an algebraic integer,and then its integrality.


3.3.1 j(z) is an algebraic number

Claim 1: For any z ∈ C r Q there exists a transcendental basis Sz of C over Qcontaining z.

Let z ∈ C r Q and consider the inclusions Q ⊂ Q(z) ⊂ C. Obviously z is atrnscendental basis of Q(z) over Q. Now if z is a transcendental basis of Cover Q then set Sz = z. Otherwise there exists a transcendental basis S′ of Cover Q(z). So define Sz = S′ ∪ z and this is a transcendental basis of C overQ, in fact :

• C is algebraic over Q(Sz) by ”transitivity of being algebraic” (it is a moregeneral fact that if F ⊂ E ⊂ K are such that E is algebraic over F andK is algebraic over E then K is algebraic over F ).

• The map Q[Xs : s ∈ Sz] −→ C such that Xs ↦−→ s is injective because wecan view it as composition of two injective maps, namely

Q[Xz][Xs : s ∈ S′] →→ Q(z)[Xs : s ∈ S′] →→ C

Xz →→ z →→ z

Xs →→ Xs

→→ s

Step 2: Sz is uncountable.

If Sz was a countable transcendental basis z1, z2, .. of C over Q then C wouldbe algebraic over Q(z1, z2, ..). Now since the polynomials over Q can be identifiedwith finite sequences of rational numbers and countable product of a countableset is still countable, we would have |C| = |Q| contradicting uncountability ofC.

Step 3: The orbit Aut(C) · z is uncountable.

Fix z ∈ C. Since Sz is uncountable, hence Sym(Sz) the set of permutationsof Sz is an uncountable set. Since C is algebraic over Q(Sz) we can extend bylinearity any such permutation to an automorphism of C. In particular

Aut(C) · z ⊃ Sym(Sz) · z = σz | σ ∈ Sym(Sz) = s | s ∈ Sz

which is uncountable by Step 2.

Step 4: There is only a countable amount of isomorphism classes of complex toriwith complex multiplication.

To begin with recall that every complex torus T is isomorphic to one of theform Tτ = C/(Zτ + Z) for some τ ∈ H and for a torus to have complex multi-plication means that End(T ) = O for some order O in a quadratic imaginaryfield K. Also recall the bijection of Proposition 2.35

Pic(OK)←→ [T ] : End(T ) = OK

given by [a] ↦−→ [C/a], which implies that the latter set is finite since the first is.Moreover since any orderO is of the formO = Z+cOK for some c ∈ Z>0, so that


there are countably many orders in any fixed imaginary quadratic field extensionK. Finally every quadratic imaginary field extension is of the form Q(

√−d) for

some square-free positive integer d, in other words there is a countable number ofquadratic imaginary field K/Q. Putting everything together, the claim follows.

Step 5: Let Λ = Zτ + Z, so that for E ∼= T = C/Λ we have j(τ) = j(E) = j(Λ).If E is given by y2 = 4x3 − g2(Λ)x − g3(Λ), then for any σ ∈ Aut(C) letEσ : y2 = 4x3 − σ−1g2(Λ))x− σ−1(g3(Λ)) and we have j(E) = σ(j(Eσ)).Therefore if E has complex multiplication, then Aut(C) · j(E) ⊂ C is atmost countable.

Simply by computation we have

σ(j(Eσ)) = σ

(1728

σ−1(g2(Λ))3

σ−1(g2(Λ))3 − 27σ−1(g3(Λ))2

)=

= 1728(g2(Λ))

3

g2(Λ)3 − 27g3(Λ)2= j(E)

Let ψσ : (x, y) ↦→ (σ(x), σ(y)) be the isomorphism from Eσ to E, thenψ−1σ End(E)ψσ = End(Eσ). It follows that if E has complex multiplication, then

also EσΛ has complex multiplication (in fact conjugating is a group isomorphism).

Now assume that E has complex multiplication and consider

Aut(C) · j(Λ) = σ−1j(Λ) | σ−1 ∈ Aut(C) = j(Eσ) | σ−1 ∈ Aut(C)

Since j is an invariant for the isomorphism class of a torus and by Step 4 weonly have a countable number of classes with complex multiplication. Hence theorbit Aut(C) · j(Λ) is at most countable.

Step 6: If C/Λ has complex multiplication, then j(Λ) ∈ Q.

Contropositive of Step 3 is that Aut(C) · z is countable implies that z ∈ Q.And this is what we just proved about j(Λ).

3.3.2 j(z) is integral

Fix n ∈ Z>1 and let Adef=

α =

(a b0 d

)ad = n, 0 ≤ b < d, (a, b, d) = 1

then we have a coset decomposition

SL2(Z)(n 00 1

)SL2(Z) =

⨆α∈A

SL2(Z)α (3.3.1)

and consider the polynomial Fn(X, j) =∏

α∈A

(X − j α). Recall the q-expansion

of j, j(z) = 1q +

∑n≥0

a(n)qn with a(n) ∈ Z, q = e2πiz. Hence for α ∈ A

j(αz) = j((a b

0 d

)· z)= j(

az + b

d) = ξ−b

d q−a/d +∑n≥0

a(n)ξnbd qna/d (3.3.2)


where ξd = e2πi/d. Therefore j(αz) has a q-expansion in q1/d with coefficientspowers of ξd. Since d | n, these are algebraic integers of Q(ξn). In other words wehave that Fn(X, j) ∈ Z[ξn]((q1/n))[X]. And we claim that actually Fn(X, j) ∈Z[X, j].

Step 1: The coefficients of Fn(X, j) are invariant under the action of Gal(Q(ξn)/Q) ∼=(Z/nZ)∗ =: G, hence they belong to Z((q1/n)).

Let σ ∈ G, then σ(ξn) = ξtn for some t such that (n, t) = 1. Trasforming the

coefficients of jα by σ in (3.3.2) we obtain jβ for some β =

(a b′

0 d

)∈ A with

b′ ≡ bt mod d. Since α ↦→ β gives a permutation of the set A, the q-expansionin (3.3.2) has coefficients in Z.

Step 2: Let f be a meromorphic function of SL2(Z)\H∗ holomorphic on SL2(Z)\Hwhose q-expansion has integral coefficients. Then f ∈ Z[j].

Proceed by induction on n = −ord∞(f) = −ord0(f∗).If n ≤ 0 then f is holomorphic at ∞ thus holomorphic on SL2(Z)\H∗ which iscompact. Therefore f is constant and its q-expansion has integral coefficients,hence f ∈ Z.Assume true for n: let f = c−(n+1)q

−(n+1)+∑k≥n

ckqk. Then ord∞(f−c−(n+1)j

n+1) ≥

n, thus by inductive step g = f−c−(n+1)jn+1 ∈ Z[j]. Hence f = g+c−(n+1)j

n+1 ∈Z[j].

We can now conclude, in fact the coefficients of Fn(X, j) are meromorphicfunctions on SL2(Z)\H∗ since they are SL2(Z)-invariant and holomorphic onSL2(Z)\H. By Step 1 every coefficient has q-expansion in Z, thus they lie inZ[j] by Step 2.

Lemma 3.37. If n is not a square, then the leading coefficients of Hn(X)def=

Fn(X,X) ∈ Z[X] is ±1.

Proof. n not a square implies that in (3.3.2) a/d = 1 (as ad = n). Therefore theleading coefficient of j − j α is a root of unity, and so is the leading coefficientof Hn(j). But since it is also rational, it has to be ±1.

Finally, let z ∈ H be such that Q(z) = K is a quadratic imaginay field, sothat E ∼= C/(Zz + Z) has complex multiplication by some order O ⊂ OK.Let us first assume thatO = OK. Then there exists µ ∈ OK such that NormK/Q(µ) =n > 0 is square-free. In fact if K = Q(i), take µ = 1+i, to get Norm(µ) = µ ·µ =(1 + i)(1 − i) = 2; if K = Q(

√−m) for some square-free m > 1, let µ = i

√m,

then Norm(µ) = µ · µ = m. In both cases, Norm(µ) is a square-free integer andclearly µ ∈ OK since it is a root of the monic polynomial X4+4 and X2+NK/Qrespectively. Moreover, since µ is imaginary, it actually belongs to OK r Z.Since µ ∈ OK r Z, µ ∈ End(E), and the multiplication by µ sends the latticeΛ = Zz + Z to a sublattice. Namely there are a, b, c, d ∈ Z such that

µz = az + b and µ = cz + d (∗)

and we can define ξ ∈M2(Z) by the relation

µ

(z1

)=

(a bc d

)(z1

)=: ξ

(z1

)


Now notice that µ and µ are eigenvalues for ξ: in fact µ is an eigenvalue forhow ξ is defined; µ is an eigenvalue since by (∗)

µ

(z1

)= ξ

(z1

)Therefore

det(ξ) = det

(µ 00 µ

)= µµ = n = Norm(µ)

which, by the choice of µ, is square-free.

Since detξ = n, ξ ∈ SL2(Z)(n 00 1

)SL2(Z) Thus there exists α ∈ A such that

j(αz) = j(ξz) = j(z)

and the second equality follows by the fact that ξ · z = z. So that

0 = Fn(j(z), j(αz)) = Fn(j(z), j(ξz)) = Fn(j(z), j(z)) = Hn(j(z)) (3.3.3)

Since n is square-free, by Lemma 3.37, the leading coefficient of Hn is ±1 andwe conclude that j(z) is an algebraic integer.

Now if O $ OK, then by Proposition 2.31 there exists β ∈ GL+2 (Q)∩M2(Z)

such that for z′ = β(z) we have End(C/(Zz′ + Z)) = OK. Therefore j(z) isintegral over Z[j(z′)] and by the previous argument j(z′) is integral. Thereforej(z) is.

Chapter 4

Hecke Operators

In this chapter, we will introduce the action of Hecke operators on modularforms, modular curves and integral homology, and at the end of the chapter wewill see how to associate a L-function to a cusp form.

4.1 The Hecke ring

Let Γ1 and Γ2 be congruence subgroups of SL2(Z). Then for any α ∈GL+

2 (Q) consider the double coset Γ1αΓ2 and denote R1,2 the Z-module gener-ated by these double cosets.

Lemma 4.1. If Γ is a congruence subgroup of SL2(Z) and α ∈ GL+2 (Q), then

the intersection αΓα−1 ∩ SL2(Z) is a congruence subgroup of SL2(Z).

Proof. Since αΓ(N)α−1 = (bα)Γ(N)(bα)−1 for all b ∈ Q∗, we can assume thatα has integer coefficients, i.e. belongs to M2(Z). Now if d = det(α), then weclaim that

Γ(Nd) ⊆ αΓ(N)α−1 ⊆ αΓα−1

In fact, define α′ = dα−1 ∈ M2(Z) and let γ ≡ 1 mod Nd (i.e. γ ∈ Γ(Nd)),then

α′γα ≡ α′α =

(d 00 d

)mod (Nd)

Hence α−1γα ≡ 1 mod N , which in particular implies that α−1γα ∈ M2(Z).Moreover det(α−1γα) = 1, so that α−1γα ∈ Γ(N) and we conclude that γ ∈αΓα−1.

Proposition 4.2. If Γ1 and Γ2 are congruence subgroups of SL2(Z), thenΓ1\Γ1αΓ2 and Γ1αΓ2/Γ2 are finite for all α ∈ GL+

2 (Q).

Proof. Let Γ3 = α−1Γ1α ∩ Γ2 ⊂ G2. Then, the map given by [αγ] ↦→ [γ] is abijection between Γ1\Γ1αΓ2 and Γ3\Γ2. Since Γ2 (by assumption) and Γ3 (byLemma 4.1) are congruence subgroups, the quotient Γ3\Γ2 is finite.The ”dual” argument holds for Γ1αΓ2/Γ2.

It follows that we can write

Γ1αΓ2 =

n⨆i=1

Γ1αi =

m⨆j=1

α′jΓ2 (4.1.1)

74

4.1. THE HECKE RING 75

Now let us consider the special case Γ1 = Γ2

Definition. Let Γ be a congruence subgroup of SL2(Z) and let ∆ be a semi-group such that Γ ⊆ ∆ ⊆M+

2 (Z) = γ ∈M2(Z) | det(γ) > 0. Then define theHecke ring with respect to Γ and ∆ to be

R(Γ,∆) =

∑k

ck · ΓγkΓ | ck ∈ Z, γk ∈ ∆

For every ΓγΓ ∈ R(Γ,∆) define its degree deg(ΓγΓ) as the number of cosetsΓξ contained in ΓγΓ, and extend the definition to the whole ring by Z-linearity.Let us define a composition law on R(Γ,∆) as follows: consider ΓαΓ =

n⨆i=1

Γαi

and ΓβΓ =m⨆j=1

Γβj in R(Γ,∆), then

(ΓαΓ)(ΓβΓ) =

n∑i=1

m∑j=1

Γαiβj =∑

ckΓγkΓ (4.1.2)

where the second sum is over all distinct double cosets R(Γ,∆) ∋ ΓγkΓ ⊆ΓαΓβΓ and

ck = #(i, j) | Γγk = Γαiβj =#(i, j) | ΓγkΓ = ΓαiβjΓ

#(Γ\ΓγkΓ)(4.1.3)

(For the second equality in (4.1.3) see [Shi73] Proposition 3.2)

Proposition 4.3. The composition law defined in (4.1.2) is well-defined andassociative.

Proof. To see that such law is well-defined we need to show that each ck dependsonly upon the double cosets ΓαΓ, ΓβΓ and ΓγkΓ and not on the choice of therepresentatives αi, βj and γk. Choosen γk, we have Γαiβj = Γγk if and onlyif Γαi = Γγkβ

−1j and for a given index j, the last equality holds exactly for one

i. It follows that

#(i, j) | Γγk = Γαiβj = #j | γkβ−1j ∈ ΓαΓ =

= #j | βj ∈ Γα−1Γγk = #j | Γβj ⊂ Γα−1Γγk

Since this last quantity coincides with the number of cosets Γξ in ΓβΓ∩Γα−1Γγk,it is indipendent of the representatives αi and βj. Now if ΓγkΓ = Γγ′kΓ,then γk = ϑγ′kδ for some ϑ, δ ∈ Γ, hence ΓβjΓ∩Γα−1Γγk = (ΓβjΓ∩Γα−1Γγ′k)δ,so that it is also indipendent of γk.For associativity see [Shi73] Proposition 3.4.

Lemma 4.4. Let Γ be a congruence subgroup of SL2(Z). Let α ∈ GL+2 (Q) be

such that #(Γ\ΓαΓ) = #(ΓαΓ/Γ), then there exists a common set of represen-tatives for the two quotients, in other words there exists αi such that

ΓαΓ =

n⨆i=1

Γαi =

n⨆i=1

αiΓ


Proof. Let βi and δi be such that ΓαΓ =n⨆

i=1

Γβi =n⨆

i=1

δiΓ. It is enough to show

that Γβi ∩ δiΓ = ∅ for all indeces i, in fact if that is the case, then we can pickαi ∈ Γβi ∩ δiΓ for every i and obtain the result since Γβi = Γαi and δiΓ = αiΓ.Now assume that Γβi ∩ δiΓ = ∅ for some i, then Γβi ⊂

⨆j =i

δjΓ and therefore

ΓαΓ =⨆j =i

δjΓ which is a contradiction.

Proposition 4.5. Let Γ and ∆ be as above. If there exists an anti-involutionϕ : ∆ −→ ∆ such that ϕ(Γ) = Γ and ΓαΓ = Γϕ(α)Γ for all α ∈ ∆, then R(Γ,∆)is a commutative ring with unit Γ1Γ = Γ.

Proof. Let α ∈ ∆ and ΓαΓ =n⨆

i=1

Γβi. Since

ΓαΓ = Γϕ(α)Γ = ϕ(ΓαΓ) = ϕ(

n⨆i=1

Γβi) =

n⨆i=1

ϕ(βi)Γ

it follows that #(Γ\ΓαΓ) = #(ΓαΓ/Γ), so that we can apply Lemma 4.4.Now let α, β ∈ ∆, then we can write

ΓαΓ =

n⨆i=1

Γαi =

n⨆i=1

αiΓ and ΓβΓ =

m⨆i=1

Γβj =

m⨆i=1

βjΓ

Therefore

(ΓαΓ)(ΓβΓ) =∑

ckΓγkΓ and (ΓαΓ)(ΓβΓ) =∑

c′kΓγkΓ

with the same components ΓγkΓ. Now using the characterization on the righthand side of (4.1.3) we see that ck = c′k.

4.1.1 The structure of R(Γ,∆)

In this section our goal will be to determine the structure of R(Γ,∆).Assume at first that Γ = SL2(Z) and ∆ =M+

2 (Z). By the theory of elementary

divisors, the diagonal matrices

(a1 00 a2

)= diag(a1, a2), with ai ∈ Z>0 such

that a1 divides a2, form a set of representatives for Γ\∆/Γ. In particular trans-position on matrices is an anti-automorphism which fixes every double cosetΓαΓ since we can assume α to be diagonal. Therefore, in view of Proposition4.5, R(Γ,∆) is a commutative ring and generated by the elements of the formT (a1, a2) = ΓαΓ with α = diag(a1, a2), ai ∈ Z>0, a1 | a2.To obtain informations regarding how elements multiply in the ring, our strat-egy will be to assign a lattice to each coset Γα and then to count the numberof lattices instead of the number of cosets.Consider Q2 as the space of row-vectors, then GL2(Q) acts on the right by mul-tiplication. A submodule Λ of Q2 is a lattice in Q2 if finitely generated over Zand Λ⊗Z Q = Q2.We can restate the fundamental theorem of elementary divisors in terms oflattices:


Fact 4.6. Let Λ1,Λ2 be lattices in Q2. Then there exist vectors u1, u2 ∈ Q2

and scalars b1, b2 ∈ Q such that Λ1 = Zu1 + Zu2, Λ2 = Zb1u1 + Zb2u2, withb2 ∈ b1Z.

The set b1, b2 is called set of elementary divisors of Λ2 with respect to Λ1

and we writeΛ1 : Λ2 = b1, b2

In particular, Λ2 ⊆ Λ1 if and only if b1, b2 ∈ Z, and then [Λ1 : Λ2] = b1b2;

moreover if α =

(b1 00 b2

)then Λ1 : Λ1α = b1, b2.

Let us write Λ = Z2, then by Lemma 2.19,

Γ = SL2(Z) = γ ∈ GL2(Q) | Λγ = Λ,det(γ) > 0

Remark 4.7. Notice for α, β ∈ ∆,

Λα = Λβ ⇐⇒ Λ = Λβα−1 ⇐⇒ βα−1 ∈ Γ (since det(βα−1) > 0)

⇐⇒ β ∈ Γα ⇐⇒ Γβ = Γα

Lemma 4.8. Let Λ1 and Λ2 be lattices in Q2. Then Λ : Λ1 = Λ : Λ2 if andonly if there exists α ∈ Γ such that Λ1α = Λ2.

Proof. Let Λ : Λ1 = Λ : Λ2 = a1, a2, then there exist u1, u2, v1, v2 ∈ Q2

such that Λ = Zu1 +Zu2 = Zv1 +Zv2, Λ1 = Za1u1 +Za2u2 and Λ2 = Za1v1 +Za2v2. Then define α ∈ GL2(Q) by uiα = vi for i = 1, 2. Then Λα = Λ andΛ1α = Λ2 however det(α) = ±1. If det(α) = 1 we are done, otherwise replacev1 with −v1.For the converse we have that

Λ : Λ2 = Λ : Λ1α = Λα : Λ1α = Λ : Λ1

Lemma 4.9. Let ΓαΓ = T (a1, a2), then Γξ ↦→ Λξ is a one-to-one correspon-dence between cosets Γξ in T (a1, a2) and lattices Λ′ such that Λ : Λ′ =a1, a2

Proof. We can assume α = diag(a1, a2). Γξ ⊂ ΓαΓ, thus there exists some δ ∈ Γsuch that Γξ = Γαδ, and consequentely:

Λ : Λξ = Λ : Λαδ = Λ : Λα = a1, a2

Viceversa, if Λ : Λ′ = a1, a2 then by Lemma 4.8 there exists γ ∈ Γ suchthat Λ′ = Λαγ and Γαγ ⊂ ΓαΓ. The correspondence is one-to-one by Remark4.7.

In particular it follows that degT (a1, a2) = #Λ′ | Λ : Λ′ = a1, a2.

Proposition 4.10. Let (ΓαΓ)(ΓβΓ) =∑ck(ΓγkΓ) with ck ∈ Z. Then ck

corresponds to the number of lattices Λ′ such that

Λ : Λ′ = Λ : Λβ and Λ′ : Λγk = Λ : Λα


Proof. Let ΓαΓ =⨆i

Γαi and ΓβΓ =⨆j

Γβj . Then by (4.1.3) and Lemma 4.9

ck = #(i, j) | Γαiβj = Γγk = (i, j) | Λαiβj = Λγk

Assume Λαiβj = Λγk and define Λ′ = Λβj , then Λ : Λ′ = Λ : Λβ andΛ′ : Λγk = Λβj : Λαiβj = Λ : Λαi = Λ : Λα. Viceversa, let Λ′ be alattice such that Λ : Λ′ = Λ : Λβ and Λ′ : Λγk = Λ : Λα, then Λ′ = Λβjfor one and only one index j, so that Λ : Λγkβ

−1j = Λ : Λα. By Lemma

4.9 Λγkβ−1j = Λαi for some i. Therefore each Λ′ determines a pair (i, j) and

viceversa.

Remark 4.11. Notice that from the definition of the multiplicative law inR(Γ,∆), it immidiatly follows that T (n, n)T (a1, a2) = T (na1, na2) for any pos-itive integer n. In particular T (n, n) is not a zero-divisor in R(Γ,Λ).

Definition. The elements T (n, n) ∈ R(Γ,∆) are called diamond operators.

Proposition 4.12. For any T (a1, a2), T (b1, b2) ∈ R(Γ,∆) such that (a2, b2) = 1we have

T (a1, a2)T (b1, b2) = T (a1b1, a2b2)

Proof. Since T (a1, a2) = T (a1, a1)T (1, a) for a = a2/a1 and T (b1, b2) = T (b1, b1)T (1, b)for b = b2/b1 and the ring R(Γ,∆) is commutative, we reduce to check the equal-ity T (1, a)T (1, b) = T (1, ab) for a, b coprime.

Let α =

(1 00 a

)and β =

(1 00 b

)and suppose that T (1, a)T (1, b) =

∑ck(ΓγkΓ).

Claim 1: ck = 1 for any γk.Apply Proposition 4.10 and consider Λ1 and Λ2 such that

Λ : Λ1 = Λ : Λ2 = Λ : Λβ and Λ1 : Λγk = Λ2 : Λγk = Λ : Λα

Then [Λ1 + Λ2 : Λ1] = [Λ2 : Λ1 ∩ Λ2]. However the left hand side divides[Λ : Λ1] = b since Λ1 + Λ2 ⊂ Λ, and the right hand side divides [Λ : Λα] = asince Λγk ⊂ Λ1 ∩ Λ2. But a and b are coprime, therefore Λ1 + Λ2 = Λ1 andΛ1 ∩ Λ2 = Λ2, hence Λ1 = Λ2 and we conclude that ck = 1.Claim 2: The only double coset contained in ΓαΓβΓ is ΓαβΓ.

Write ΓαΓβΓ =s⨆

i=1

ΓγiΓ. We want to show that s = 1, which is equivalent to

say that for any ξ1, ξ2 ∈ ΓαΓβΓ we have Γξ1Γ = Γξ2Γ. Let ξ1, ξ2 ∈ ΓαΓβΓ,then Λξi ⊂ Λi ⊂ Λ for some Λi as in Claim 1 and

Λ/Λξ1 ∼= Λ/Λ1 ⊕ Λ1/Λξ1 ∼= Λ/Λα⊕ Λ/Λβ ∼= Λ/Λξ2

where the second isomorphism follows from coprimality of a and b. Therefore ξ1and ξ2 have the same set of elementary divisors. Hence the equality Γξ1Γ = Γξ2Γand s = 1. Since ΓαβΓ ⊆ ΓαΓβΓ, they have to coincide.

As a consequence every element of R(Γ,∆) is the product of elements ofthe form T (1, pr) and T (p, p) with p rational primes and r ≥ 1. Thereforeunderstanding the structure of R(Γ,∆) reduces to understand the one of Rp:the subring generated by T (1, pr), T (p, p), r ≥ 1.

Lemma 4.13. T (1, p) and T (p, p) are algebraically indipendent.


Proof. Assume notand let Q(X,Y ) ∈ Z[X,Y ] be a non zero polynomial suchthat Q(T (1, p), T (p, p)) = 0. Factoring the lowest power of T (p, p) we can write

Q(T (1, p), T (p, p)) = T (p, p)mQ(T (1, p), T (p, p))

Which implies Q(T (1, p), T (p, p)) = 0 since T (p, p) is not a zero divisors. Nowwrite

0 = Q(T (1, p), T (p, p)) = a0Q0(T (1, p))+a1Q1(T (1, p))T (p, p)+..+anQn(T (1, p))T (p, p)n

Therefore aiQi(T (1, p)) = 0 for all i = 0, .., n. Since T (1, p) is a transcendentalelement this implies that aiQi(X) = 0, and consequentely Q = 0 contradictingour initial assumption.

Theorem 4.14. The ring Rp is the polynomial ring Z[T (1, p), T (p, p)].

Proof. Since Rp is generated by T (p, p) and T (1, pr), for any r ≥ 1, it is enoughto show that T (1, pr) is a polynomial in T (p, p), T (1, p) for any r. Proceed byinduction:

• T (1, p)2 = (Γ

(1 00 p

)Γ)2 =

∑ck(ΓγkΓ). Now notice that we are looking

for γk’s diagonal, with a1 | a2 and necessarely det(γk) = p2. Therefore theonly possibilities are γk ∈

diag(1, p2),diag(p, p)

.

By (4.1.3) let’s compute multiplicities and show that for γk =

(1 00 p2

),

ck = 1:

ck = #

Λ′ | Λ : Λ′ = Λ : Λ · diag(1, p)& Λ′ : Λ · diag(1, p2) = Λ : Λ · diag(1, p)

= 1

This is enough to prove that T (1, p2) is a polynomial in T (1, p) and T (p, p).

• Assume that T (1, pk) is a polynomial is T (1, p) and T (p, p) for any k ≤ rand let us prove it for r+1. Similarly as above, we have that T (1, pr)T (1, p) =∑ck(ΓγkΓ) with γk diagonal and necessarely det(γk) = pr+1. In particu-

lar γ = diag(1, pr+1) appears in the sum and its multiplicity is 1 by (4.1.3).All the remaining factors ΓγkΓ are of the form T (pi, pj) with 1 ≤ i ≤ j,i + j = r + 1 and therefore we can factor them as T (p, p)T (pi, pj) with0 ≤ i ≤ j, i+ j = r−1. Since by inductive hypotesis they are polynomialsin T (p, p), T (1, p) as well as T (1, pr)T (1, p), the result follows.

Corollary 4.15. We can conclude that

R(Γ,∆) = Z[T (1, p), T (p, p) | p rational prime]

In particular R(Γ,∆) is an integral domain.

Definition. For any integer n ≥ 1, let ∆(n) = α ∈ ∆ | det(α) = n and define

T (n) =∑

α∈∆(n)

ΓαΓ

called Hecke operator.


Lemma 4.16. T (n) belongs to R(Γ,∆).

Proof. We need to see that the sum is finite. We claim that there is a one-to-

one correspondence between the set C(n) =

(a b0 d

)∈ ∆(n) 0 ≤ b < d

and

Γ\∆(n).

• Surjectivity: Let A =

(a bc d

)∈ ∆(n), we want to show that there exists

γ =

(α βθ δ

)∈ Γ such that γA ∈ C(n). Consider two different cases:

– If c = 0 we have ad = n and in particular a, d = 0.(α βθ δ

)(a b0 d

)=

(aα bα+ dβaθ bθ + dδ

)To belong to C(n), take θ = 0 hence αδ = 1 and dδ > 0, thus letδ = sgn(d) = α. We are left to show that there exists β for which0 ≤ sgn(d) · b + dβ < sgn(d) · d. Consider the Euclidian division ofsgn(d) · b by sgn(d) · d with remainder 0 ≤ r < |d| so that

sgn(d) · b = sgn(d) · d · q + r

β = −sgn(d) · q satisfies 0 ≤ sgn(d) · b+ dβ = r < sgn(d) · d = |d| asrequired

– If c = 0, let a′

c′ = ac with (a′, c′) = 1. Then there exist α, β ∈ Z such

that αa′ + βc′ = 1: therefore(α β−c′ a′

)(a bc d

)=

(aα+ cβ bα+ dβ

0 a′d− bc′)

=: T ∈ ∆(n)

Let us now show that there exist γ′ =

(z uv w

)∈ Γ such that γ′T ∈

C(n). Write T =

(A B0 D

)to simplify notation, then

γ′T =

(zA zB + uDvA vB + wD

)Choosing v = 0 and z = w = sgn(D), gives γ′T =

(|A| B + uD0 |D|

).

We are left to show that there exists u ∈ Z such that 0 ≤ B + uD <|D|. Again by mean of the Euclidean division we have B = QD +R

with 0 ≤ R < |D|. Thus take u = −Q and this yields that γ′γ

(a bc d

)belongs to C(n).

This proves that C(n) surjects to Γ\M∗(n).

• Injectivity: Suppose that A,A′ ∈ C(n) are in the same class, so that there

exists γ =

(α βθ δ

)for which

A =

(a b0 d

)=

(α βθ δ

)(a′ b′

0 d′

)=

(a′α b′α+ d′βa′θ b′θ + d′δ

)= γA′


This equality forces θ = 0 hence αδ = 1. Since d, d′ > 0 from d = d′δfollows that δ has positive sign, thus δ = 1 = α. Therefore a = a′, d = d′.Now b = b′ + d′β and by hypothesis 0 ≤ b, b′ < d = d′ hence

|d′β| = |b− b′| < d′ =⇒ β = 0 (sinceβ ∈ Z)

i.e. b = b′ which gives A = A′.

Proposition 4.17. (a) For any p prime we have T (p) = T (1, p).

(b) If n,m are coprime then T (n)T (m) = T (nm).

(c) For any n ≥ 1 we have

T (n) =∑ad=na|d

T (a, d)

(d) For any r ≥ 2 we have T (1, pr) = T (pr)− T (p, p)T (pr−2).

(e) For any r ≤ s we have TprTps =r∑

i=0

piT (pi, pi)Tpr+s−2i , and in particular

T (p)T (pr) = T (pr+1) + pT (p, p)T (pr−1) (4.1.4)

Proof. (a) It is immidiate from the definition of T (p).

(b) Follows from the definition and Proposition 4.12.

(c)

T (n) =∑

α∈∆(n)

ΓαΓ =∑ad=na|d

Γ

(a 00 d

)Γ =

∑ad=na|d

T (a, d)

where the second equality follows from the theory of elementary divisors.

(d) We have the three equalities: T (1, pr) = Γ

(1 00 pr

)Γ;

T (pr) =∑

α∈∆(pr)

ΓαΓ =∑

i+j=r0≤i≤j

Γ

(pi 00 pj

)Γ = Γ

(1 00 pr

)Γ+Γ

(p 00 pr−1

)Γ+..+Γ

(ph 00 pr−h

)Γ

where h ∈ Z>0 is such that h ≤ k − h and k − h− 1 < h+ 1.

T (p, p)T (pr−2) =(Γ

(p 00 p

)Γ)·( ∑

i+j=r−20≤i≤j

Γ

(pi 00 pj

)Γ)=∑

i+j=r1≤i≤j

Γ

(pi 00 pj

)Γ =

Γ

(p 00 pr−1

)Γ + Γ

(p 00 pr−1

)Γ + ..+ Γ

(ph 00 pr−h

)Γ

where the second equality follows from Remark 4.11.Putting these three formula together we obtain (b).


(e) See [Shi73] Theorem 3.24 (4).

Now let ∆′ = ∆Ndef=

α ∈M2(Z) | det(α) > 0, α ≡

(∗ ∗0 ∗

)mod N

and

Γ′ = Γ0(N). Let T ′(n, n) = Γ′(n 00 n

)Γ′ and define the Hecke operators:

Definition. For any integer n ≥ 1 define

T ′(n) =∑

α∈∆′(n)

Γ′αΓ′ ∈ R(Γ′,∆′)

Lemma 4.18. The CN (n) =

(a b0 d

)ad = n, 0 ≤ b < d, (a,N) = 1

is a

complete set of coset representatives for the right coset Γ0(N) on ∆N (n) =α ∈ ∆N | det(α) = n

Proof. An analogous argument as in Lemma 4.16 gives the result.

With some more work, one can obtain a structure theorem for R(Γ′,∆′):namely

Theorem 4.19. R(Γ′,∆′) is the polynomial ring generated over Z by the ele-ments

T ′(1, p), T ′(p, p) ∀p - N and T ′(p) = T ′(1, p) ∀p | N

Moreover, these elements are algebraically indipendent.In other words R(Γ′,∆′) is the homomorphic image of R(Γ,∆) via the map⎧⎪⎨⎪⎩

T (n) ↦→ T ′(n) ∀nT (p, p) ↦→ T ′(p, p) ∀p - NT (p, p) ↦→ 0 ∀p | N

Proof. See [Shi73] Theorem 3.34.

Corollary 4.20. (a) For any prime p - N , and r ≥ 1:

T (pr)T (p) = T (pr+1) + pT (p, p)T (pr−1)

(b) For any prime p | N , and r ≥ 1:

T (pr) = T (p)r

(c) T (n)T (m) = T (nm) if (n,m) = 1.

Proof. All the relations immidiatly follows from Proposition 4.17 and Theorem4.19.

4.2. ACTION ON MODULAR FUNCTIONS 83

4.2 Action on modular functions

Let k ∈ Z and recall the action of GL+2 (C) on the space of modular functions

for some congruence subgroup, as in (3.2.2), given by

(σ, f) ↦→ f |[σ]k(τ) = det(σ)k−1f(σ(τ))j(σ, τ)−k (4.2.1)

Let Γ1 and Γ2 be congruence subgroups of SL2(Z). Then we have a map

Mk(Γ1) −→Mk(Γ2) given by f ↦→ fρ

for each ρ ∈ R1,2.Explicitly if ρ = Γ1αΓ2 for α ∈ GL+

2 (Q) such that Γ1αΓ2 =⨆Γ1αj , we have

f ↦→ f |[Γ1αΓ2] = det(α)k/2−1n∑

j=1

f |[αj ]k (4.2.2)

Firstly, let us show that f |[Γ1αΓ2]k is a weakly modular form of weight k with

respect to Γ2. Let γ ∈ Γ2, then det(γ) = 1 and Γ1αΓ2 =n⨆

j=1

αjγ, therefore

f |[Γ1αΓ2]k[γ]k = det(α)k/2−1n∑

j=1

f |[αj ]k[γ]k = det(α)k/2−1n∑

j=1

f |[αjγ]k = f |[Γ1αΓ2]k

We are left to see that f |[Γ1αΓ2] is holomorphic at the cusps of Γ2. For γ ∈SL2(Z) we have f |[Γ1αΓ2]k[γ]k = det(α)k/2−1

n∑j=1

f |[αjγ]k, but since f is holo-

morphic at the cusps, each f |[αjγ]k is holomorphic at ∞ therefore the wholesum is.Moreover by the last formula also follows that if f ∈ Sk(Γ1) then f |[Γ1αΓ2]k ∈Sk(Γ2).Then (4.2.2) extends Z-linearly to R1,2.

In what follows we will focus on the case Γ1 = Γ2, so that we have anaction of R(Γ,∆) onMk(Γ), and in particular on Sk(Γ). Before doing so, let usintroduce an inner product in the space of cusp forms:

Definition. Let Γ be a congruence subgroup of SL2(Z), define the Peterssoninner product

⟨ · , · ⟩ : Sk(Γ)× Sk(Γ) −→ C

(f, g) ↦→ 1

V (Γ)

∫XΓ

f(τ)g(τ)Im(τ)kdµ(τ)

where dµ is the measure on XΓ induced by the SL2(Z)-invariant hyperbolicmeasure dxdy

y2 on H (see [Kna93] 8.12) and V (Γ) is the (finite) volume of XΓ,

Vol(XΓ) (see [Miy89] 1.9.1).

Notice that even though f and g are not necessarely Γ-invariant, the integralis well-defined since f(τ)g(τ)Im(τ)k and dµ(τ) are. It follows from its definitionthat the Petersson inner product is Hermitian and positive definite.


4.2.1 Hecke operators on SL2(Z)Let us consider the previous setting with Γ = SL2(Z) = Γ1 = Γ2, and focus

on the action of the Hecke operators as in (4.2.2). Let αiµi=1 be a completeset of representatives for the right cosets Γα of Γ on ∆(n) and let f ∈ Mk(Γ)then

Tk(n)f = nk/2−1

µ∑i=1

f |[αi]k (4.2.3)

We saw in Lemma 4.16 that C(n) =

(a b0 d

)ad = n, 0 ≤ b < d

is a complete

set of coset representatives. Therefore explicitly, for every αi ∈ C(n), we have

f |[αi](τ) = nk/2f(aτ + b

d

)d−k (4.2.4)

Hence Tk(n)f(τ) = nk−1∑ad=n0≤b<d

f(aτ + b

d

)d−k (4.2.5)

Lemma 4.21. Let n, d be positive integers, then

d−1∑b=0

e2πinb/d =

d if d | n0 if d - n

(4.2.6)

Proof. If d | n then let a = n/d ∈ Z and therefore e2πia = 1. Since we have dsummands the result follows.If d - n then e2πin/d = 1 and

d−1∑b=0

e2πinb/d =

d−1∑b=0

(e2πin/d

)b=

1−(e2πin/d

)d1− e2πin/d

= 0

Proposition 4.22. Let f ∈ Mk(Γ) have q-expasion f(τ) =∞∑

m=0cmq

m. Then

the q-expasion of Tk(n)f is

Tk(n)f(τ) =

∞∑ℓ=0

bℓqℓ

where b0 = c0∑

0<d|ndk−1; b1 = cn and bℓ =

∑a|(n,ℓ)

ak−1cnℓ/a2 if ℓ > 1.

Proof. Using the q-expasion of f , (4.2.5) and (4.2.6) we obtain

Tk(n)f(τ) = nk−1∑ad=n0≤b<d

f(aτ + b

d

)d−k = nk−1

∞∑m=0

∑ad=n0≤b<d

cme2πin(aτ+b)/dd−k =

= nk−1∞∑

m=0

∑ad=nd>0

cmdqmad−k+1 =

∞∑m=0

∑0<a|n

cmn/aak−1qma =

∞∑ℓ=0

bℓqℓ


Let us compute the coefficient of powers of q on the left hand side: b0 is∑0<a|n

c0ak−1; the coefficient of q1 comes form a = m = 1 so that b1 = cn;

for powers greater than 1, we are looking at triples (m, a, d) such that ma = ℓ(hence m = ℓ/a) and a | n. Therefore cnm/a = cℓm/a2 with a | ℓ,m. Hence the

coefficient of qℓ is bℓ =∑

a|(ℓ,n)ak−1cnℓ/a2 .

In particular for f(τ) =∑m≥0

cmqm and p prime, Tk(p)f(τ) has q-expansion

with coefficients

bm =

cpm if p - ncpm + pk−1cm/p if p | n

Remark that for f ∈Mk(Γ), and T (n, n), (4.2.2) gives:

f |T (p, p) = pk−2f (4.2.7)

Therefore by (4.1.4), as an operator onMk(Γ)

Tk(pr)Tk(p) = Tk(p

r+1) + pk−1Tk(pr−1) ∀p prime,∀r ≥ 1

Tk(m)Tk(n) = Tk(mn) if (m,n) = 1(4.2.8)

Theorem 4.23. (Petersson) The Hecke operators Tk(n) on Sk(Γ) are normalwith respect to the Petersson inner product.

Proof. One first proves that for all α ∈ GL+2 (Q) holds [ΓαΓ]∗k = [Γ(detα)α−1Γ]k

where ∗ denotes the adjoint operator. Then for every α in ∆(n)

ΓαΓ = Γ(detα)α−1Γ

since α and (detα)α−1 have the same set of elementary divisors. Therefore theHecke operators are self-adjoint. Since they also commute with each other, theyare normal.

Theorem 4.24. There exists a basis of Sk(Γ) of simultaneous eigenvectors forthe operators T (n).

Proof. It follows immediately from the complex spectral theorem.

Proposition 4.25. Let f ∈ Sk(Γ) be a simultaneous eigenvector for Tk(n) suchthat Tk(n)f = λ(n)f . If f has q-expansion at infinity f(τ) =

∑n≥1

cnqn, then

cn = λ(n)c1

Proof. From Proposition 4.22 b1 = cn, on the other hand from being an eigen-vector for Tk(n) follows that b1 = λ(n)c1. Since f is an eigenvector for everyHecke operator, the equality cn = λ(n)c1 holds for every n.

Remark 4.26. As consequences we can see that

(i) f = 0 implies c1 = 0.

(ii) The eigenvalues λ(n)’s determines f up to a scalar factor.


(iii) The previous point allows us to normalize f so that its q-expansion hasc1 = 1, and therefore cn is the eigenvalue for Tk(n).

(iv) In particular if f is normalized, from (4.2.8) follows that:

• For any p prime, cprcp = cpr+1 + pk−1cpr−1 .

• If (m,n) = 1 then cncm = cnm.

4.2.2 Hecke operators on congruence subgroups

Fix a positive integer N and let Γ′ = Γ0(N),∆′ = ∆N (n). Then for f ∈Mk(Γ

′), (4.2.2) gives :

T ′k(n)f = nk/2−1

∑α∈CN (n)

f |[α]k (4.2.9)

and recall that T ′k(n) carriesMk(Γ

′) (resp. Sk(Γ′)) to itself.

Proposition 4.27. Let f ∈Mk(Γ′) have q-expansion at infinity f(τ) =

∞∑m=0

cnqn,

then T ′k(n)f has q-expansion

T ′k(n)f(τ) =

∞∑ℓ=0

bℓqℓ

where b0 = c0∑

0<a|n(a,N)=1

ak−1; b1 = cn, and bℓ =∑

0<a|(m,ℓ)(a,N)=1

ak−1cnℓ/a2 if ℓ > 1.

Proof. The proof is essentialy the same as the one of Proposition 4.22; howeverreplacing C(n) with CN (n) produces the comprimality condition of a with Nalong all the step of the argument.

From Theorem 4.19 and equations in (4.2.8) we obtain:

T ′k(p

r)T ′k(p) = T ′

k(pr+1) + pk−1T ′

k(pr−1) ∀p - N, ∀r ≥ 1

T ′(pr) = T ′(p)r ∀p | N, ∀r ≥ 1 (4.2.10)

T ′k(m)T ′

k(n) = T ′k(mn) if (m,n) = 1

We would now want to obtain a similar result as in Theorem 4.23, however:

Theorem 4.28. (Petersson) For every positive integer n such that (n,N) = 1,the Hecke operators T ′(n) are self adjoint with respect to the Petersson innerproduct in Sk(Γ′).

In fact the same argument used to prove Theorem 4.23 only holds true if nis coprime with N , otherwise not necessarely T ′

k(n) is self-adjoint. Distinguish-ing forms coming from the lower level would allow to remove the restriction(n,N) = 1 (see DS06 5.6) but this exceeds the purpuse of the thesis.

Nontheless since Hecke operators commute, Sk(Γ′) splits into orthogonalsum of simultaneous eigenspaces for the operators T ′

k(n) with (n,N) = 1.


Definition. A cusp form f ∈ Sk(Γ) which is a eigenvector for all T ′k(n),

(n,N) = 1, is called eigenform.

Proposition 4.29. Let f ∈ Sk(Γ′) be an eigenform such that T ′k(n)f = λ(n)f

for all n coprime with N . If f has q-expansion at infinity f(τ) =∑n≥1

cnqn, then

cn = λ(n)c1

Proof. From Proposition 4.27 b1 = cn, on the other hand from being an eigen-vector for T ′

k(n) follows that b1 = λ(n)c1. Since f is an eigenform, the equalitycn = λ(n)c1 holds for every n, (n,N) = 1.

Remark 4.30. As consequences we can see that

(i) f = 0 implies c1 = 0.

(ii) The eigenvalues λ(n)’s determines f up to a scalar factor.

(iii) The previous point allows us to normalize f so that its q-expansion hasc1 = 1, and therefore cn is the eigenvalue for T ′

k(n) with (n,N) = 1.

(iv) In particular if f is normalized, from (4.2.8) follows that:

• For any p prime, p - N , cprcp = cpr+1 + pk−1cpr−1 .

• For any p prime, p | N , cpr = crp.

• If (m,n) = 1 then cncm = cnm.

Our next task is to deduce that T2(n)’s act on S2(Γ0(N)) as matrices withinteger coefficients, so that their eigenvalues will be algebraic integers.

Proposition 4.31. Let x1, .., xr ∈ X0(N) correspond to all the elliptic point ofΓ0(N) (i.e. the points in H∗ with nontrivial stabilizer) of order e1, .., er respec-tively. Let s1, .., sm be the inequivalent cusps for Γ0(N). Let f be a modularfunction of weight k for Γ0(N) and for k even define ω = f(z)dzk/2. Then

(a) div(f) = div(ω) + k2

( r∑i=1

( ei−1ei

) · xi +m∑j=1

sj

)if k is even.

(b) deg(div(f)) = k2

(2g − 2 +m+

r∑i=1

( ei−1ei

))for any k.

Proof. (a) Assume that k is even.

• Recall that if x corresponds to a point z0 ∈ H then the local nor-mal form is given by t = g(z)e where g : H −→ D is a holomorphicisomorphism such that g(z0) = 0 and e = #StabΓ(z0). Notice thatdtdz = eg(z)i−1g′(z) therefore ordt=0

(dtdz

)= e−1

e , thus we obtain

(∗) ordx(ω) = ordx(fdzk/2) = ordx(f)+

k

2ordx(dz) = ordx(f)+

k

2(e−1−1)


• Let s be a cusp corresponding to s ∈ H∗, let γ ∈ SL2(Z) be such that

γ(s) =∞ and let q = e2πiz/h if

(1 h0 1

)generates ±γStabΓ(s)γ−1.

The local parameter is z = γ(t) and dqdz = 2πi

h q, hence f(t)dtk/2 =

f |[γ−1]k(z)dzk/2 = Φ(q)

(dzdq

)k/2(dq)k/2. We obtain

(∗∗) ords(ω) = ord0(Φ)−k

2= ords(f)−

k

2

Putting together (∗) and (∗∗) we obtain (a).

(b) By Riemann-Roch theorem we have deg(ω) = k2 (2g − 2) and (b) for k

even immidiately follows. On the other hand if k is odd, apply the samecomputation in (a) with f2 and notice that div(f) = 1

2div(f2).

Corollary 4.32. S2(Γ0(N)) is isomorphic to the space Ω1(X0(N)) of holomor-phic 1-forms onX0(N) via the map f ↦→ f(z)dz. In particular dimCS2(Γ0(N)) =g.

It follows that the T (n)’s act on Ω1(X0(N)), and by composing on the rightwe obtain an action of the Hecke operators T (n)’s on Ω1(X0(N))∨ and in par-ticular an action on H1(X0(N),Z). If we prove that Hecke operators act as anendomorphism of H1(X0(N),Z) we will be able to conclude that T (n)’s act asmatrices with integer coefficients.

Definition. We say that γ ∈ Γ0(N) is elliptic (resp. parabolic) if |Trγ| < 2(resp. |Trγ| = 2).

Theorem 4.33. Let Γep be the normal subgroup of Γ0(N) generated by all theelliptic and parabolic elements. After fixing a point τ0 ∈ H∗, there is a canonicalisomorphism

H1(X0(N),Z) ∼= Γ0(N)ab/Γabep

Proof. See [Kna93] Proposition 11.22

The correspondences Ω1(X0(N)) ∼= S2(Γ0(N)) andH1(X0(N),Z) ∼= Γ0(N)ab/Γabep

respect integration, i.e. if ω ↔ f and c↔ [γ] then

⟨c, ω⟩ def=∫c

ω =

∫ γτ0

τ0

f(τ)dτ (4.2.11)

Let us show that the right hand side of (4.2.11) is indipendent of τ0 and is 0 forelliptic and parabolic elements.

• Let τ1 ∈ H∗ then∫ γτ1

τ1

f(τ)dτ =

∫ τ0

τ1

f(τ)dτ +

∫ γτ0

τ0

f(τ)dτ +

∫ γτ1

γτ0

f(τ)dτ =

=

∫ τ0

τ1

f(τ)dτ +

∫ γτ0

τ0

f(τ)dτ +

∫ τ1

τ0

f(τ)dτ =

∫ γτ0

τ0

f(τ)dτ

where the secont equality follows from the fact that f is Γ0(N)-invariant.


• Remark that given a cusp form f and fixed a point τ0 ∈ H∗, the associationγ ↦→

∫ γτ0τ0

f(τ)dτ is a homomorphism from Γ0(N) to (C,+).

Therefore if γ is an elliptic element, then it has finite order in Γ0(N), sodoes its image in (C,+). We conclude that

∫ γτ0τ0

f(τ)dτ = 0.

• Now let γ =

(a bc d

)∈ Γ0(N) be a parabolic element and assume a+d = 2

. Then

γz =az + b

cz + d= z ⇐⇒ cz2 + (d− a)z − b = 0 ⇐⇒ z =

a− 1

c

(for the case a+ d = −2 one gets a+1c ).

This means that γz = z has a double root at z0 = a−1c . Now let ρ ∈ SL2(Z)

be such that ρ(z0) =∞, then γ′ = ργρ−1 ∈ StabSL2(Z)(∞). In particular

γ′ = ±(1 h0 1

)for some h ∈ Z such that the width of the cusp z0 divides

h. Therefore∫ γτ0

τ0

f(τ)dτ =

∫ ρ−1γ′ρτ0

τ0

f(τ)dτ =

=

∫ γ′ρτ0

ρτ0

f(ρ−1τ)j(ρ−1, τ)−2dτ =

∫ ρτ0+h

ρτ0

f |[ρ−1]2(τ)dτ = 0

where the last equality follows from the fact that f is a cusp form, thereforeit has zero residue.

Going back to the action of T (n) in H1(X0(N),Z): let ∆N (n) =µ⨆

i=1

Γ0(N)αi

where #CN (n) = µ and αi ∈ CN (n). Then notice that for every element γ ∈

Γ0(N) we have ∆N (n)γ ⊆ ∆N (n) =µ⨆

i=1

Γ0(N)αi. Therefore

αiγ = γiαj(i) (4.2.12)

for some γi ∈ Γ0(N) and i ↦→ j(i) a permutation of 1, .., µ.Then the action of T (n) on H1(X0(N),Z) is

T (n)[γ] =

µ∑i=1

[γi] (4.2.13)

with γi ∈ Γ0(N) are as in (4.2.12).

Proposition 4.34. The action of T (n) on H1(X0(N),Z) in (4.2.13) is well-defined, Z-linear and indipendent from the coset representatives αi ∈ CN (n).

Proof. To verify that T (n) is well-defined and Z-linear we need to show thatT (n)[γ1γ2] = T (n)[γ1] + T (n)[γ2] for every γ1, γ2 ∈ Γ0(N) and it is zero on Γep.

• Let αiγ1 = γiαj(i) and αjγ2 = δjαk(j). Then

αi(γ1γ2) = (γiαj(i))γ2 = γiδj(i)αk(j(i))

Hence since i ↦→ j(i) is a permutation of 1, .., µ we get

T (n)[γ1γ2] =

µ∑i=1

[γiδj(i)] =

µ∑i=1

[γi] +

µ∑i=1

[δj(i)] = T (n)[γ1] + T (n)[γ2]


• Let γ ∈ Γ0(N) be elliptic so that there exists r ≥ 1 such that γr = 1, then

rT (n)[γ] = T (n)[γr] = T (n)[1] =

µ∑i=1

[1] = 0

since [1] = 0 (1 is parabolic). Now H1(X0(N),Z) is torsion free, henceT (n)[γ] = 0.

• Let γ be parabolic and let αiγ = γiαj(i). Then there exists some r ≥ 1 forwhich the r-th iteration of j is the identity, we obtain

αiγr = γiαj(i)γ

r−1 = γiγj(i)αj2(i)γr−2 = ... = γiγj(i) · .. · γjr−1(i)αi

So we have that αiγrα−1

i = γiγj(i) · .. · γjr−1(i) ∈ Γ0(N) and it is parabolicsince parabolic elements form a subgroup and conjugating leaves themstable. Therefore in (4.2.12) for γr we can use αiγ

rα−1i since they belong

to Γ0(N) for each index i and αiγr = (αiγ

rα−1i )αi. Hence

rT (n)[γ] = T (n)[γr] =

µ∑i=1

[αiγrα−1

i ] = 0

because each αiγrα−1

i is parabolic. As before the torsion free property ofH1(X0(N),Z) yields that T (n)[γ] = 0.

We are left to show that the action is indipendent from coset representatives: letβi be another set of representatives, then βi = δiαi for some δi ∈ Γ0(N). Nowif αiγ = γiαj(i), then βiγ = δi(γiαj(i)) = δiγiδ

−1j(i)βj(i). Therefore we conclude

that

µ∑i=1

[δiγiδ−1j(i)] =

µ∑i=1

[δi] +

µ∑i=1

[γi]−µ∑

i=1

[δj(i)] =

µ∑i=1

[γi] = T (n)[γ]

Now we claim that if c ∈ H1(X0(N),Z) and ω ∈ Ω1(X0(N)), with ω corre-sponding to f ∈ S2(Γ0(N)), then

⟨T (n)c, ω⟩ =∫T (n)c

ω =

∫c

T (n)ω = ⟨c, T (n)ω⟩ (4.2.14)

where T (n)ω is the holomorphic 1-form corresponding to T2(n)f .

Let γ correspond to c, if T (n)[γ] =µ∑

i=1

[γi], then

∫T (n)c

ω =

µ∑i=1

∫ γiτ0

τ0

f(τ)dτ

where τ0 ∈ H is such that the path [τ0, γτ0] in H∗ has projection in X0(N)homologous to c.On the other hand∫

c

T (n)ω =

∫ γτ0

τ0

T2(n)f(τ)dτ =

µ∑i=1

∫ γτ0

τ0

f |[αi]2(τ)dτ =


=

µ∑i=1

∫ γτ0

τ0

f(αiτ)j(αi, τ)−2dτ

(∗)=

µ∑i=1

∫ αiγτ0

αiτ0

f(τ ′)dτ ′ =

µ∑i=1

∫ γiαj(i)τ0

αiτ0

f(τ)dτ =

µ∑i=1

∫ αj(i)τ0

αiτ0

f(τ)dτ =0

+

µ∑i=1

∫ γiαj(i)τ0

αj(i)τ0

f(τ)dτ =

µ∑i=1

∫ γiτi

τi

f(τ)dτ

where the first 3 equalities are tautological; (∗) follows by the change of vari-

able τ ′ = αiτ in each summand; andµ∑

i=1

∫ αj(i)τ0αiτ0

f(τ)dτ = 0 because i ↦→ j(i)

is a permutation, thus the whole sum is a loop and f being a cusp form has0-residue. Therefore the last equality follows by setting τi = αj(i)τ0, and thissum is equal to

∫T (n)c

ω since each integral is indipendent from the base point

(as we proved in (4.2.11)).

Now extending by linearity ⟨c, ω⟩ we have it defined for c ∈ H1(X0(N),R)and H1(X0(N),C).

Remark 4.35. If ⟨c, ω⟩ = 0 for all ω ∈ H1(X0(N),R), then c = 0.

Proof. Let c1, .., c2g be a Z-basis for H1(X0(N),R), then c =2g∑i=1

rici for some

ri ∈ R. Let ω1, .., ωg be a basis of Ω1(X0(N)) over C, then any ω ∈ Ω1(X0(N)) is

of the form ω =g∑

j=1

sjωj . Since ⟨c, ω⟩ = 0 for all ω we must have2g∑i=1

ri⟨ci, ωj⟩ = 0

for all j = 1, .., g. By Proposition 1.35 this can only occur if ri = 0 for all i,hence c = 0.

Consider the involution H −→ H given by τ ↦→ τ∗ := −τ . This map in-

duces an involution on X0(N): let γ =

(a bc d

)∈ Γ0(N) and let τ2 = γτ1,

then τ∗2 = γ∗τ∗1 with γ∗ =

(a −b−c d

)∈ Γ0(N). Therefore by Theorem 4.33 we

have an involution on H1(X0(N),Z) which extends linearly to H1(X0(N),R)and H1(X0(N),C). Since the eigenvalues of an involution are ±1, write H+

1

and H−1 for the eigenspace of +1 and −1 respectively. Then H1(X0(N),R) =

H+1 (X0(N),R)⊕H−

1 (X0(N),R) and, similarly,H1(X0(N),C) = H+1 (X0(N),C)⊕

H−1 (X0(N),C).

Notice that τ ↦→ τ∗ induces an involution also in the space S2(Γ0(N)), namelyf ↦→ f∗ given by f∗(τ) = −f(τ∗). Moreover if c ↦→ c∗ is the involution on the1-cycle corresponding to γ ↦→ γ∗, then

⟨c∗, f⟩ =∫c∗f(τ)dτ =

∫c

−f(τ∗)dτ =

∫c

−f(τ∗)dτ = ⟨c, f∗⟩

Remark 4.36. If f = f∗ then (if)∗ = −if and viceversa, therefore if we letS+2 = f ∈ S2(Γ0(N)) | f = f∗ and S−2 = f ∈ S2(Γ0(N)) | f∗ = −f,we obtain they have equal dimension g over R (since S2(Γ0(N)) has complexdimension g).


.Now if c ∈ H+

1 (X0(N),R) and f ∈ S+2 , then

⟨c, f⟩ = ⟨c∗, f⟩ = ⟨c, f∗⟩ = ⟨c, f⟩ =⇒ ⟨c, f⟩ ∈ R

This implies that dimRH+1 (X0(N),R) = g = dimRH

−1 (X0(N),R). In fact ar-

guing by contradiction, assume dimRH+1 (X0(N),R) > g, then by rank-nullity

theorem∅ = ker

(H+

1 (X0(N),R) −→ Hom(S+2 ,R))

where the map is c ↦→ ⟨c, · ⟩. This means that there exists some 0 = c0 ∈H+

1 (X0(N),R) such that ⟨c0, f⟩ = 0 for all f ∈ S+2 and by bilinearity also⟨c0, if⟩ = 0. By Remark 4.36 ⟨c0,S2(Γ0(N))⟩ = 0 but by Remark 4.35 this canonly occour if c0 = 0. Therefore dimH+

1 (X0(N),R) ≤ g. The ”dual” argumentshows that dimH−

1 (X0(N),R) ≤ g and consequentely they are both equalitysince their sum is necessarely 2g.

Proposition 4.37. The linear extension of T (n) toH1(X0(N),C) mapsH+1 (X0(N),C)

to itself and the restriction of T (n) to H+1 (X0(N),C) is the transpose of T2(n)

on Ω1(X0(N)).

Proof. If c ∈ H1(X0(N),C) corresponds to the path [τ0, γτ0] and due to theindependence of the base point we can freely select τ0 to be imaginary. With suchchoice of τ0, we obtain that c∗ corresponds to [τ0, γ

∗τ0], therefore [γ]∗ = [γ∗].Now if αiγ = γiαj(i) and extend the involution ∗ to every matrix by(

a bc d

)∗

=

(a −b−c d

)then such map is multiplicative and therefore α∗

i γ∗ = γ∗i α

∗j(i). Since α

∗i are coset

representatives and the action is indipendent from those, it follows that

T (n)[γ∗] =

µ∑i=1

[γ∗i ] =(T (n)[γ]

)∗Hence T (n)c∗ = (T (n)c)∗ on H1(X0(N),C). Now if c ∈ H+

1 (X0(N),C), then

T (n)c = T (n)c∗ = (T (n)c)∗

i.e. T (n)c ∈ H+1 (X0(N),C) and we conclude that T (n) maps H+

1 (X0(N),C) toitself.The argument before the proposition prove that H1(X0(N),C) and Ω1(X0(N))are dual to one another, with perfect pairing ⟨c, ω⟩. Therefore

⟨c, T (n)tf⟩ = ⟨T (n)c, f⟩ = ⟨c, T2(n)f⟩

where the second equality follows from (4.2.14).

Lemma 4.38. H+1 (X0(N),R) ∩ H1(X0(N),Z) is a lattice in H+

1 (X0(N),R)and a Z-basis for the intersection is a basis for H+

1 (X0(N),C).

Proof. It is immidiate that H+1 (X0(N),R)∩H1(X0(N),Z) is discrete and spans

H+1 (X0(N),R).

4.3. L-FUNCTION OF A CUSP FORM 93

Proposition 4.39. There exists a basis of S2(Γ0(N)) for which the operatorsT2(n) act as matrices with integer coefficients.

Proof. Let c1, .., cg ∈ H+1 (X0(N),R)∩H1(X0(N),Z) be a C-basis forH+

1 (X0(N),C)and let f1, .., fg be the dual basis of S2(Γ0(N)). Since T (n) maps H+

1 (X0(N),C)to itself, it also maps H+

1 (X0(N),R)∩H1(X0(N),Z) to itself and therefore act-ing as an endomorphism of H1(X0(N),Z), its associated matrix has integercoefficients. Since by Proposition 4.37 T2(n) is its transpose, also T2(n) is actingas a matrix with integer coefficient.

We have finally reached our goal: since T2(n) is given by an integer matrix,its characteristic polynomial is monic with integer coefficients. Therefore weconclude that the eigenvalues of T2(n) are algebraic integers.

4.3 L-function of a cusp form

Definition. Let f ∈Mk(Γ0(N)) have q-expansion at ∞ f(τ) =∑n≥0

cnqn. The

L-function associated to f is the Dirichlet series

L(s, f) =∑n≥1

cnns

where s is a complex variable.

Our goal will be to determine convergence properties and a functional equa-tion for the L-function associated to a cusp form.

Lemma 4.40. Let f ∈ Sk(Γ0(N)), then for all τ = x + iy ∈ H there exists apositive constant M indipendent from x such that |f(x+ iy| ≤My−k/2.Conversely, if f is a modular form for Γ0(N) of weight k such that |f(x+ iy| ≤My−k/2 for some positive constantM indipendent from x, then f ∈ Sk(Γ0(N)).

Proof. For any modular function f for Γ0(N) of weight k define a functionhf = h : H −→ R by setting

h(τ) = h(x+ iy) = |f(τ)|yk/2

Recall the equality for all γ ∈ SL2(Z), Im(γτ) = Im(τ)|j(γ, τ)|−2, which impliesthat h is Γ0(N) invariant, and therefore factors through X0(N).Now suppose that f ∈ Sk(Γ0(N)), then in a neighbourhood of any cusp s we

have f |[γ−1]k(τ) = Φ(q) for some γ ∈ SL2(Z) such that γs =∞ and Φ(q)q→0−→ 0.

Hence h(γ−1(τ)) = |Φ(q)|Im(τ)k/2 −→ 0. This implies that h is continuous onX0(N). Since X0(N) is compact, h is bounded and the result follows.On the other hand if h(τ) is bounded, then Φ must be holomorphic at q = 0 and

|Φ(q)| = MIm(τ)k/2

Im(τ)→∞−→ 0. Therefore Φ(0) = 0, and thus f ∈ Sk(Γ0(N)).

Lemma 4.41. Let f ∈ Sk(Γ0(N)) and f(τ) =∑n≥1

cnqn with q = e2πiτ/h where

h is the width of the cusp [∞]. Then there exists a constant C indipendent of nfor which |cn| ≤ Cnk/2 for all n ≥ 1.


Proof. Consider the function in the q variable F (q) =∑n≥1

cnqn; by Cauchy

formula follows that

cn =1

2πi

∫|q|=r

F (q)q−n−1dq

for some small r > 0.Now let τ have Im(τ) = y = h

2πn , by Lemma 4.40, |F (q)| ≤ My−k/2 =

M(

2πnh

)k/2and |q| = |e(2πi(x+iy))/h| = e−2πy/h = e−1/n. Therefore taking

r = e−1/n in the integral, yields

|cn| ≤1

2π

∫r=e−1/n

|F (q)||q−n−1|dq =[q = e−1/ne2πiϑ ⇒dq = 2πie−1/ne2πiϑdϑ

]=

=2π

2π

∫ 1

0

M(2πnh

)k/2e(n+1)/ne−1/ndϑ =Me

(2πnh

)k/2= Cnk/2

As a consequence we obtain a result about absolute convergence:

Proposition 4.42. Let L(s, f) =∑n≥1

cnns be the L-function associated to the

cusp form f =∑n≥1

cnqn ∈ Sk(Γ0(N)), then L(s, f) converges absolutely for

Re(s) > k2 + 1

Proof. ∑n≥1

|cnn−s| =∑n≥1

|cn|n−Re(s) ≤∑n≥1

Bnk/2−Re(s)

the inequality follows from Lemma 4.41 and the last sum converges for Re(s)−k2 > 1.

Let L(s, f) be the L-function associated to some cusp form f for Γ0(N) ofweight k, then defineR(s, f) = Ns/2(2π)−sΓ(s)L(s, f) where Γ(s) =

∫∞0e−xxs−1dx

is the Gamma function.

Theorem 4.43. Let f(τ) =∑n≥1

cnqn ∈ Sk(Γ0(N)). Then L(s, f) is absolutely

convergent for Re(s) > k2 + 1 can be holomorphically continued to the whole

complex plane and it satisfies the functional equation

R(s, f) = ikR(k − s, f |[ρ]k)

with ρ =

(0 −1N 0

).

Proof. Define the Mellin transformation of f h(s) =∫∞0f(iy)ys−1dy for y ∈

R>0 and proceed formally:

h(s) =

∫ ∞

0

f(iy)ys−1dy =

∫ ∞

0

∑n≥1

cne−2πnyys−1dy =

∑n≥1

cn

∫ ∞

0

e−2πnyys−1dy =


=

[σ = 2πny ⇒dσ = 2πndy

(†)]=∑n≥1

cn

∫ ∞

0

e−σ( σ

2πn

)s−1 dσ

2πn=

= (2π)−s∑n≥1

cnn−s

=L(s,f)

·∫ ∞

0

e−σσs−1dσ =Γ(s)

= (2π)−sL(s, f)Γ(s) = R(s, f)N−s/2

where the change of variable (†) is done for each n.Claim: For Re(s) > k

2 + 1 all the previous formal steps are actually rigorous.In fact:∫ ε

0

f(iy)ys−1dy ≤∫ ε

0

|f(iy)ys−1|dy ≤M∫ ε

0

y−k/2yk/2dyε→0−→ 0 for Re(s) >

k

2+1

where the second inequality follows from Lemma 4.40.∫ ∞

E

f(iy)ys−1dy ≤∫ ∞

E

|f(iy)|yRe(s)−1dy =

∫ ∞

E

|∑n≥1

cne−2πny|yRe(s)−1dy ≤

≤∫ ∞

E

(∑n≥1

|cn|)e−2πyyRe(s)−1dy ≤ A

∫ ∞

E

e2πyyRe(s)−1dyE→∞−→ 0 ∀s ∈ C

Now since f(iy) =∑n≥1

cne−2πny is uniformely convergent for y ≥ ε

∫ E

ε

f(iy)ys−1dy =∑n≥1

cn

∫ E

ε

e−2πnyys−1dy

Then for any δ > 0 small there exists some B big enough such that

∑n≥B

cn

∫ E

ε

e−2πnyys−1dy ≤∑n≥B

|cn|∫ ∞

0

e−2πnyyRe(s)−1dy(†)=

= Γ(Re(s))(2π)−Re(s)∑n≥B

|cn|n−Re(s) < δ

Therefore we can conclude that∫ ∞

0

f(iy)ys−1dy −B∑

n≥1

cn

∫ ∞

0

e−2πnyys−1dy =

= limε→0E→∞

∫ E

ε

f(iy)ys−1dy −B∑

n≥1

cn

∫ E

ε

e−2πnyys−1dy ≤ δ

Which proves the claim.Similarly if g = f |[ρ]k we obtain 1∫ ∞

0

g(iy)ys−1dy = Γ(s)(2π)−sL(s, g) = R(s, g)N−s/2

1One can easily prove that g is a cusp form for Γ0(N) of the same weight k of f .


Let us now prove the functional equation and deduce from that that L(s, f) canbe holomorphically extended. Define A = N−1/2 and split the integral as∫ ∞

0

f(iy)ys−1dy =

∫ A

0

f(iy)ys−1dy +

∫ ∞

A

f(iy)ys−1dy

By the computation we have done above, the first term converges for Re(s) >k2 + 1 and the second converges for any s ∈ C. Now we have

g(iy) = f |[ρ]k(iy) = det(ρ)k/2j(ρ, iy)−kf(ρ(iy)) = Nk/2(Niy)−kf(i/Ny)

Hence f(i/Ny) = g(iy)(iy)kNk/2. Therefore changing variable in the integraly ↦→ 1

Ny , dy ↦→1N

(− dy

y2

)yields:∫ A

0

f(iy)ys−1dy =

∫ A

∞f(i/Ny)

( 1

Ny

)s−1 1

N

(−dyy2)=

∫ ∞

A

f(i/Ny)( 1

Ny

)s−1 1

Ny2dy =

=

∫ ∞

A

g(iy)(iy)kNk/2N−sy−s−1dy = ikNk/2−s

∫ ∞

A

g(iy)yk−s−1dy

which is convergent for any s ∈ C. Similarly∫ ∞

A

f(iy)ys−1dy = ikNk/2−s

∫ A

0

g(iy)yk−s−1dy

which is convergent for Re(s) > k2 + 1. Therefore

R(s, f)N−s/2 = ikNk/2−sR(k − s, g)N−(k−s)/2 =⇒ R(s, f) = ikR(k − s, g)

as we wanted to prove. and R(s, f)N−s/2 = Γ(s)(2π)−sL(s, f) can be holo-morphically continued to the whole complex plane. Since Γ(s)−1 is an entirefunction, also L(s, f) can be extended holomorphically for any s ∈ C.

Chapter 5

Eichler-Shimura Theory

5.1 Complex abelian varieties and Jacobian va-rieties

In this section we want to introduce the Jacobian variety of a nonsingularprojective curve defined over Q. We will first give some results regarding abelianvarieties over the complex numbers.

Definition. A lattice Λ in Cg is a discrete subgroup of maximal rank, or equiv-alently a free abelian group in Cg containing a R-basis of Cg.Let Λ be a lattice in Cg, the quotient Cg/Λ is called complex torus of dimensiong.For a complex torus Cg/Λ, a positive definite Hermitian form H = S + iE onCg such that H(Λ× Λ) ⊂ Z is called a Riemann form.

Definition. A (complex) abelian variety is a pair (A,O) where A is a non-singular projective variety over C, O ∈ A and A has an abelian group structurewith identity O and such that the operations of addition and negative are mor-phisms.An abelian variety (A,O) is defined over Q if A is defined over Q, O ∈ A(Q)and addition and negative morphisms are defined over Q.The dimension of an abelian variety is the dimension as non-singular projectivevariety.

Definition. Let A be an abelian variety and C be a subvariety. If C has a groupstructure compatible with that of A, then C is an abelian subvariety of A.

Remark 5.1. As for the case of elliptic curves over C, the theory of complexabelian varieties reduces to the one of complex tori admitting a Riemann form.(See [Swi74])

In what follows, if A is an abelian variety of dimension n, let us denoteby V the n-dimensional complex vector space overlying A, in other words letA = V/Λ for a lattice Λ in V .

97

5.1. COMPLEX ABELIAN VARIETIES AND JACOBIANVARIETIES 98

Theorem 5.2. Let (A1 = V1/Λ1, O1), (A2 = V2/Λ2, O2) be abelian varieties,write πi : Vi −→ Ai for the canonical projections, and let f : A1 −→ A2 bea holomorphic map such that f(O1) = O2. Then f is a group homomorphismand it is induced by a C-linear map F : V1 −→ V2 such that F (Λ1) ⊂ Λ2 andπ2 F = f π1.

Proof. See [Swi74] III.7 Theorem 32

Definition. A map f : A1 −→ A2 is a homomorphism between abelian varietiesif it is a morphism of varieties and a homomorphism of groups. A homomorphismf is an isogeny if the map F : V1 −→ V2 as in Theorem 5.2 is an isomorphism.

Remark 5.3. Let f : A1 −→ A2 be a homomorphism of abelian varieties. Thefollowing are equivalent:

(a) f is an isogeny;

(b) dim(A1) = dim(A2) and f has finite kernel;

(c) dim(A1) = dim(A2) and f is surjective.

As for the case of elliptic curves, given an isogeny f : A1 −→ A2, there existan integer N and an isogeny g : A2 −→ A1 such that f g and gf are respectivemultiplication by N , therefore being isogenous is an equivalence relation.We denote by Hom(A1, A2) the additive group (with respect to pointwise sum)of isogenies between the abelian varieties A1 and A2.One of the key results in the theory of complex abelian varieties is the so-calledPoincare’s complete reducibility theorem:

Theorem 5.4. (Poincare complete reducibility theorem) Let A1 = V1/Λ1

and A2 = V2/Λ2 be two abelian varieties and let f ∈ Hom(A1, A2). Then

A3def= im(f) is an abelian variety and there exist B1, B2 abelian varieties and

α, β such that the following diagram is commutative:

A1f →→

α

↓↓

A2

A3 ×B1p →→ A3

i →→ A3 ×B2

β

↑↑

where p and i are the obvious projection and inclusion (respectively).

Proof. Step 1: A3 is an abelian variety.By Theorem 5.2 f : A1 −→ A2 is induced by F : V1 −→ V2 which is C-linearand F (Λ1) ⊂ Λ2. Hence F (Λ1) is discrete. Moreover RΛ1 = V1 thus RF (Λ1) =F (V1), and we can conclude that F (Λ1) is a lattice in F (V1). Therefore A3 =F (V1)/F (Λ1) =: V3/Λ3 is a complex torus. We are left to show that there existsa Riemann form on A3 with respect to Λ3. If H2 = S2 + iE2 is a Riemann formon A2 with respect to Λ2, then H2|F (V1)

is a Riemann form with respect to Λ3.Hence A3 is an abelian variety.Step 2: Construction of B1 and α.Define W = kerF , we claim that W ∩ Λ1 is a lattice in W . It is discrete sincecontained in Λ1 and consider F|Λ1

: Λ1 −→ Λ2. We can choose bases λ1, .., λ2n


and µ1, .., µ2n1 for Λ1 and Λ2 respectively, such that the matrix associated tothe linear operator is of the form⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

d1. . .

dm0

. . .

0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠Therefore ker(F|Λ1

) = λm+1, .., λ2n and kerF = R⟨λn+1, .., λ2n⟩ = W . HenceW ∩ Λ1 is a lattice in W as we wanted.Now if H1 = S1 + iE1 one can prove (with some linear algebra) that

(i) W⊥H1 =W⊥E1 =:W⊥;

(ii) Λ1 ∩W⊥ is a lattice in W⊥;

(iii) (Λ1 ∩W )⊕ (Λ1 ∩W⊥) has finite index in Λ1.

Now consider the projection p1 : V =W⊕W⊥ −→W such that v = w+w′ ↦→ wand let Λ = p1(Λ1). Since Λ1 ∩W ⊂ Λ ⊂ W we have RΛ = W . Moreover Λ isdiscrete: in fact by (iii) we have that (Λ1 ∩W )⊕ (Λ1 ∩W⊥) has finite index inΛ1, hence

Λ1 ∩W = p1((Λ1 ∩W )⊕ (Λ1 ∩W⊥)) ⊂ p1(Λ1) = Λ

has finite index in Λ. We conclude that Λ is a lattice inW and define B1 =W/Λ.Let us prove that B1 is an abelian variety: let N = [Λ : Λ1 ∩W ], then N Λ ⊂Λ1 ∩W ⊂ Λ1. Define H = N2H1, which is positive definite since N > 0 and H1

is positive definite by definition. Moreover H(Λ× Λ) ⊂ Z because

H(λ, λ′) = N2H1(λ, λ′) = H1(Nλ

∈Λ1

, Nλ′∈Λ1

) ∈ Z

Therefore B1 is an abelian variety.Now consider the map V1 =W⊥ ⊕W α−→ V3 ⊕W given by

α =

(f|

W⊥ 0

0 idW

)and f|

W⊥ is an isomorphism of C-vector spaces since injective and dim(W⊥) =dim(V3). Therefore dim(V1) = dim(V3 ⊕W ) and thus α induces the isogeny α

V1 =W⊥ ⊕W α →→

↓↓

V3 ⊕W

↓↓A1

α →→ A3 ×B1

Step 3: Construction of B2 and β.

We have V3 ⊂ V2, let W ′ = V⊥H23 . Similarly as before V3 ∩ Λ2 is a lattice in V3

1Same n since f is an isogeny, therefore dim(V1) = dim(V2)


and W ′ ∩ Λ2 is a lattice in W ′. Define B2 = W ′/(W ′ ∩ Λ2). Then we have anisomorphism of C-vector spaces β :W ′⊕V3 −→ V2 which induces an isogeny β:

W ′ ⊕ V3β →→

↓↓

V2

↓↓B2 ×A3

β →→ A2

Remark 5.5. If A1,A2 and f : A1 −→ A2 are defined over Q in Theorem 5.4,then also A3 can be defined over Q.

Corollary 5.6. Let A1 be an abelian variety, C an abelian subvariety of A,then the quotient A1/C has an abelian variety structure in the sense that thereexist a unique (up to isomorphism) pair (A2, f) where A2 is an abelian variety,f : A1 −→ A2 is a surjective homomorphism such that ker(f) = C and satisfyingthe following universal property: if g : A1 −→ B is a homomorphism of abelianvarieties such that ker(g) ⊃ C, then there exists f ′ : A2 −→ B homomorphismof abelian varieties such that g = f ′ f .Moreover is if A1 and C are defined over Q then A2 and f can be defined over Q;if that is the case, and also B and g are defined over Q, then f ′ can be definedover Q.

Definition. An abelian variety A is simple if it is not isogenous to the productof two non-trivial abelian subvarieties. This is equivalent to saying that A hasno non-trivial subvarieties or no non-trivial quotient variety.

Corollary 5.7. Any abelian variety A is isogenous to the product of simplevarieties, and factors are unique up to isogeny.

Remark 5.8. Let A1 and A2 be abelian subvarieties of an abelian variety Bdefined over Q. Then A1 +A2, which is the image of A1 ×A2 → B ×B −→ Bby (a1, a2) ↦→ a1 + a2, is defined over Q. In fact A1 ×A2 and + : B ×B −→ Bare both defined over Q. Moreover

dimA1 + dimA2 ≥ dim(A1 +A2) ≥ dimAi i = 1, 2

As in 1.3, if C is a nonsingular projective curve over C, then C has a com-plex manifold structure, and therefore it is a compact Riemann surface. Conse-quentely, we can define the Jacobian variety of C.

Theorem 5.9. Let C be a nonsingular projective curve over C, let J(C) theJacobian variety of its underlying compact Riemann surface and let Φ : C −→J(C) be the Abel-Jacobi map with base point x0 ∈ C. Then J(C) is a nonsigularprojective variety such that:

(a) it has a group structure which makes it into an abelian variety;

(b) Φ is a morphism;

5.2. TECHNICAL RESULTS 101

(c) if F : C −→ A is a morphism into an abelian variety A, then F factorsthrough J(C), i.e. there exists f : J(C) −→ A homomorphism of abelianvarieties such that

F = f Φ+ F (x0)

Moreover, if C is defined over Q, then J(C) can be defined over Q in such away that (a),(b) and (c) holds true with structures defined over Q.

Proof. See [Lan59] II.2 Theorem 8 & Theorem 9.

As a consequence notice that J(X0(N)) is defined over Q since, by Theorem3.34, X0(N) is defined over Q.

5.2 Technical results

In this section we want to see how to transform Hecke operators into elementsof End(J(X0(N))) by applying Theorem 5.9 to X0(N) and further that thesenew elements are defined over Q.In order to do that, we first need a result due to Chow:

Theorem 5.10. (Chow) If V1, V2 are non-singular projective varieties over Cand F : V1 −→ V2 is a holomorphic map between their underlying complexmanifolds, then F is a morphism defined over C.

Proof. See [Cho49] Theorem VII.

As a consequence the ring of holomorphic homomorphisms from J(X0(N)) intoitself corresponds to End(J(X0(N))) the ring of homomorphisms of abelianvarieties over C.Suppose that X0(N) has genus g ≥ 1 and fix a basis ω1, .., ωg of the spaceΩ1(X0(N)) and as in 1.3 define

J = J(X0(N)) = Ω1(X0(N))∨/H1(X0(N),Z)

Write π : H∗ −→ X0(N) for the canonical projection and, after fixing a pointx0 ∈ X0(N), Φ : X0(N) −→ J for the Abel-Jacobi map. If we define fj(τ)dτ =π∗(ωj) for j = 1, .., g and π∗ the pullback of π, then f1, .., fg is a basis of

S2(Γ0(N)) and the composition map Φdef= Φ π : H∗ −→ J is given by

Φ(τ) =(∫ τ

τ0

f1(ξ)dξ, ..,

∫ τ

τ0

fg(ξ)dξ)

for some τ0 ∈ π−1(x0) (5.2.1)

Definition. Let V ⊂ An(K) be an affine variety, the tangent space to V atP ∈ V is

TP (V ) =

(x1, .., xn) ∈ An

n∑i=1

∂f

∂Xi(P )(xi − Pi

)= 0 ∀f ∈ I(V )

where Pi is the i-th component of P .


One can prove (See [Hul03] Theorem 3.14) that TPV ∼= m∨P /(m

2P )

∨, there-fore if P is a nonsingular point dimTP (V ) = dimV .Now if f : J −→ J is a holomorphic homomorphism, then by Theorem 5.2 it isinduced by some F : Ω1(X0(N))∨ −→ Ω1(X0(N))∨. Consequentely we have an

C-linear map between the tangent spaces: df =(d(pr)F d(pr)−1

): T0(J) −→

T0(J)(∼= Cg), where pr : Ω1(X0(N))∨ −→ J and d(pr) is the isomorphism fromT0(Ω

1(X0(N))∨) to T0(J).Since any dinstict f yields a different df ,we proved the existence of an injectivering homomorphism

End(J) −→Mg(C) , f ↦→ [matrix associated to df ] (5.2.2)

Let z1, .., zg be coordinates on Cg, then dz1, .., dzg form a basis for Ω1(J),and let e1, .., eg be the dual basis for T0(J), in other words, such that:

Ω1(J)× T0(J) −→ C,(dzi, ej

)↦→ dzi(ej) = ⟨dzi, ej⟩ = δij (5.2.3)

If we regard T0(J) as the space of invariant vector fields on J , we can extendand consider valid this formula at any point of J .Now let f ∈ End(J), we saw that f induces a map df : T0(J) −→ T0(J),therefore we have a map between dual spaces, namely δf : Ω1(J) −→ Ω1(J)such that((

δf)(u))(v) = u

((df)(v))∀u ∈ Ω1(J),∀v ∈ T0(J) (5.2.4)

Remark 5.11. We have that Φ∗(dzi) = fi(τ)dτ .

Proof. Since Φ∗ = π∗ Φ∗, it is enough to show that Φ∗(dzi) = ωi for all

i = 1, .., g, and this follows by its definition: Φ(x) =(∫ x

x0ω1, ..,

∫ x

x0ωg

)∈ J .

As a consequence we see that Φ∗ is a C-vector spaces isomorphism sincemaps a basis to a basis. Hence we can define µ : S2(Γ0(N)) −→ Ω1(J) as

µ(f) =(Φ∗)−1(

f(τ)dτ)

(5.2.5)

Now recall the coset decomposition from Chapter 4: ∆N (n) =µ⨆

i=1

Γ0(N)αi

and write T (n) in place of T ′(n). T (n) acts on X0(N) by T (n) : π(τ) ↦→µ∑

i=1

(π(αiτ)) ∈ Div(X0(N)). Also recall that since J is an additive group, we

can extend Φ : X0(N) −→ J to the surjective map2 Φ# : Div(X0(N)) −→ J .Therefore composing we obtain

T#(n) : π(τ)T (n)↦→

µ∑i=1

(π(αiτ))Φ#

↦→

⎛⎜⎜⎜⎝∑i

∫ αiτ

τ0f1(ξ)dξ

...∑i

∫ αiτ

τ0fg(ξ)dξ

⎞⎟⎟⎟⎠ (5.2.6)

2Surjectivity follows from Theorem 1.38


which shows that T#(n) is holomorphic, and consequentely by Theorem 5.10 itis a morphism of varieties over C.Let us now apply Theorem 5.9 (c) with A = J to obtain that

T#(n)(π(τ)) = t(n)(Φ(τ)

)+ T#(n)(π(τ0)) ∀τ ∈ H∗ (5.2.7)

for some t(n) ∈ End(J).Let us be more explicit in order to compute dt(n); we have:

t(n)

⎛⎜⎝∫ τ1τ0f1(ξ)dξ...∫ τ1

τ0fg(ξ)dξ

⎞⎟⎠ = t(n) Φ(τ1) =

= T#(n)(π(τ1))− T#(n)(π(τ0) =0

=

⎛⎜⎜⎜⎝∑i

∫ αiτ1τ0

f1(ξ)dξ

...∑i

∫ αiτ1τ0

fg(ξ)dξ

⎞⎟⎟⎟⎠(5.2.8)

where T#(n)(π(τ0) = 0 since we integrate on a loop and each fi has zero-residuebeing a cusp form.Now since t(n) is additive, compute t(n) Φ(τ2) for some τ2 ∈ H∗, subtract anduse (5.2.8) to obtain the equality

t(n)

⎛⎜⎝∫ τ1τ2f1(ξ)dξ...∫ τ1

τ2fg(ξ)dξ

⎞⎟⎠ =

⎛⎜⎜⎜⎝∑i

∫ αiτ1τ2

f1(ξ)dξ

...∑i

∫ αiτ1τ2

fg(ξ)dξ

⎞⎟⎟⎟⎠ (5.2.9)

Finally, by differentiating (5.2.9) for τ2 → τ1 in every direction follows that for

all τ1 ∈ H∗, ∀j = 1, .., g, dt(n)dτ1

∫ τ1τ2fj(ξ)dξ = dt(n)fj(τ1) and on the other hand

d

dτ1

∫ αiτ1

τ2

fj(ξ)dξ =d(αiτ1)

dτ1

d

d(αiτ1)

∫ αiτ1

τ2

fj(ξ)dξ =d(αiτ1)

dτ1f(αiτ1)

Therefore

dt(n)

⎛⎜⎝f1(τ1)...fg(τ1)

⎞⎟⎠ =

⎛⎜⎜⎜⎝∑i

f1(αiτ1)dαiτ1dτ1

...∑i

fg(αiτ1)αiτ1dτ1

⎞⎟⎟⎟⎠ =

⎛⎜⎜⎜⎝∑i

f1|[αi]2(τ1)

...∑i

fg|[αi]2(τ1)

⎞⎟⎟⎟⎠ =

⎛⎜⎝T2(n)f1(τ1)...T2(n)fg(τ1)

⎞⎟⎠(5.2.10)

By (5.2.2), let A = A(T2(n)) =( a11 .. a1g

. .

. .ag1 .. agg

)be the matrix associated to dt(n),

so that

A

⎛⎜⎝f1...fg

⎞⎟⎠ =

⎛⎜⎝T2(n)f1...T2(n)fg

⎞⎟⎠Theorem 5.12. (Shimura-Tanyama) For any f ∈ S2(Γ0(N))

(δt(n))(µ(f)) = µ(T2(n)f)

5.3. ELLIPTIC CURVES ASSOCIATED TO WEIGHT-2 CUSPFORMS 104

Proof. Write aj forj-th column of A(T2(n)). Let e1, .., eg be the basis of T0(J)

as above and let f =g∑

k=1

ckfk. Then for all j = 1, .., g we have

⟨(δt(n))(µ(f), ej⟩ = ⟨(µ(f), dt(n)ej⟩ =∑k

ck⟨µ(fk), dt(n)ej⟩ =

=∑k

ck⟨µ(fk), A(T2(n))ej⟩ =∑k

ck⟨dzk, aj⟩ =∑k

ckakj

On the other hand:

⟨µ(T2(n)f), ej⟩ =∑k

ck⟨µ(T2(n)fk), ej⟩ =∑k

ck⟨µ(ak1f1 + ..+ akgfg), ej⟩ =

=∑k

g∑i=1

ckaki⟨µ(fi), ej⟩ =∑k,i

ckaki ⟨dzi, ej⟩ =0 if i =j

=∑k

ckakj

Lemma 5.13. The elements t(n) : J −→ J are defined over Q.

Proof. See [Kna93] Lemma 11.76.

The last result we need before moving onto the next section is a structuretheorem due to Wedderburn:

Theorem 5.14. (Wedderburn) Let T be a finite-dimensional associative andcommutative algebra with identity over a field K and letR be its nilradical. Thenthere exists a semisimple algebra S such that as a K-vector space T = S ⊕R.Moreover S is direct sum of ideals, each of which is a simple algebra isomorphicto a finite algebraic field extension of K.

Proof. See [Jac43] Chapter 5, Theorem 37.

5.3 Elliptic curves associated to weight-2 cuspforms

We now have all the tools in order to prove the following:

Theorem 5.15. (Eichler-Shimura) Let f ∈ S2(Γ0(N)) be a normalized eigen-

form with q-expansion at ∞ f(τ) =∞∑

n=1cnq

n, q = e2πiτ . If cn ∈ Z for all n ≥ 1,

then there exists a pair (E, ν) such that:

(a) (E, ν) is the quotient of J by an abelian subvariety A ⊂ J defined over Qand E is an elliptic curve defined over Q.

(b) t(n) ∈ End(J) leaves A stable and acts on E as multiplication by cn.

(c) µ(f) is a nonzero multiple of ν∗(ω) where ω is the invariant differential ofE.


(d) Λfdef=∫ γτ0

τ0f(ξ)dξ | γ ∈ Γ0(N)

is a lattice in C and E ∼= C/Λf over C.

Proof. Step 1: Construction of A.Let T be the commutative Q-subalgebra of EndQ(J) = End(J)⊗Z Q generatedby all the t(n) in End(J). By (5.2.2), t(n) can be identified with its differentialdt(n), which is represented by the g × g matrix A. By Proposition 4.39, A hasinteger coefficients, therefore T is isomorphic to a subalgebra of Mg(Q), andin particular it is finite-dimensional over Q. We are under the hypothesis ofTheorem 5.14, hence

T = S ⊕R = (K1 ⊕ ..⊕Kr)⊕R

where R is the Nilradical of T and each Ki is an ideal of S isomorphic to a finitealgebraic extension of Q. By Theorem 5.12 we have

(δt(n))(µ(f)) = µ(T2(n)f) = µ(cnf) = cnµ(f)

where the second equality follows from the fact that f is a normalized eigenform.It follows that we have a well-defined Q-algebra homomorphism

ρ : T −→ Q given by t(n) ↦→ cn

Notice thatR ⊂ kerρ since Q is torsion free. Since t(1) = idT = (e1, .., er, 0) ↦→ 1,ρ is surjective. Moreover each ei is idempotent i.e. e2i = ei, therefore its imageis either 1 or 0. Due to surjectivity there exists an index for which it is 1, andwithout loss of generality assume it is e1. Hence ρ(K1) = Q and since K1 is afield, it is isomorphic to Q.Now let ρ : Q −→ K1 be the inverse map of ρ to K1, define U = (K2 ⊕ .. ⊕Kr)⊕R ⊂ T and consider α ∈ U ∩End(J). Since α ∈ U ⊂ T, it is a polynomialin t(n)’s with rational coefficients, thus there exists M ∈ Z such that Mα is apolynomial in t(n)’s with integer coefficients. By Lemma 5.13 t(n)’s are definedover Q, hence Mα is. Now since im(Mα) = im(α) ⊂ J , im(α) is an abeliansubvariety of J defined over Q. By Remark 5.8

Adef=

∑α∈U∩End(J)

im(α)

is an abelian subvariety of J defined over Q. Consequentely, J/A is an abelianvariety in the sense of Corollary 5.6: there exists (E, ν) such that E is an abelianvariety and ν : J −→ E is surjective with kerν = A, satisfying the followinguniversal property: whenever ϕ : J −→ C is a homomorphism of abelian va-rieties such that A = kerν ⊆ kerϕ, then ϕ factors through ν, i.e. there existsψ : E −→ C such that ϕ = ψ ν.(Notice that to prove (a) we are left to show that dim(E) = 1).Step 2: Action of t(n) on E.Let β ∈ T∩End(J) and a =

∑αk(xk) ∈ A, where αk ∈ U ∩End(J) and xk ∈ J .

Since U is an ideal of T, βαk ∈ U ∩ End(J) and therefore

β(a) =∑

βαk(xk) ∈∑

(βαk)(J) ⊂ A

In other words we proved that for all β ∈ T ∩ End(J), β(A) ⊂ A. In particularsince t(n) belongs to T∩End(J), it leaves A stable. But further, since t(n)(A) ⊂


A, we have ker(ν t(n)) ⊃ kerν = A: by the universal property of (E, ν) thereexists t(n) ∈ End(E) such that

t(n) ν = ν t(n) (5.3.1)

Hence t(n) acts on E as t(n). By construction t(n)− ρ(cn) = t(n)− ρ(ρ(t(n))) ∈U , and if [cn] denotes the multiplication by cn, then ρ(cn)− [cn] ∈ U . Thereforet(n)− [cn] ∈ U ∩End(J), and, by the construction of A follows that t(n) = [cn].This proves (b).Step 3: dim(E) ≥ 1.This is equivalent to show that A $ J : since R is the Nilradical of T, there existssome m ≥ 0 such that K1Rm = 0 but K1Rm+1 = 0. Therefore let β ∈ K1Rm

be a nonzero element, then, as above, there exists some integerM ≥ 1 for whichMβ ∈ End(J), and notice that Mβ ∈ K1Rm,Mβ = 0, hence ker(Mβ) $ J .On the other hand, for all α ∈ U we have (Mβ)α = 0 since by constructionK1 ·Kj = 0 for all j = 2, .., r and RRm = 0. Therefore if a =

∑αk(xk) ∈ A, we

get (Mβ)(a) = 0. In other words we have a nonzero element of End(J) whichis identically zero on A. Hence A ⊂ ker(Mβ) $ J .Step 4: dim(E) ≤ 1.

Since dim(E) ≥ 1, there exists 0 = ω′ ∈ Ω1(E). Let ν∗ : Ω1(E) −→ Ω1(J) bethe pullback map induced by ν. Since ν is a morphism of complex tori over Cit is separable and therefore by Proposition 1.29 ν∗ is injective. Now taking thepullback of (5.3.1) yeilds

ν∗ δt(n) = δt(n) ν∗ (5.3.2)

Since t(n) = [cn], δt(n) = cn and therefore

δt(n)(ν∗(ω′) = ν∗(cnω

′) = cnν∗(ω′) (5.3.3)

Define f ′ = µ−1(ν∗(ω′)), then

µ(T2(n)f′) = (δt(n))(µ(f ′)) = (δt(n))(ν∗(ω′)) = cnν

∗(ω′) = cnµ(f′) (5.3.4)

Since µ is an isomirphism we obtain that T2(n)f′ = cnf

′ for all n. If dim(E) > 1,then there exist ω′, ω′′ = 0 in Ω1(E) linearly independent. Since ν∗ is injective,also ν∗(ω′) and ν∗(ω′′) are linearly independent. However if f ′′ = µ−1(ν∗(ω′′))then by the above argument, also T2(n)f

′′ = cnf′′ for all n. Since µ is an

isomorphism, f ′ and f ′′ are linearly independent, contradicting Remark 4.30(ii). We conclude that dim(E) = 1 and we proved (a).In particular, taking ω′′ = ω the invariant differential of E, yields that f and f ′

are linearly dependent and (c) follows.Step 5: Λf is a lattice in C.Recall from Section 1.3 that J is realized as the complex torus Cg/Λ0 with Λ0 =

Λ(X0(N)). Write f =g∑

k=1

rkfk for f1, .., fk basis of S2(Γ0(N)) and rk ∈ C, let

λj =

⎛⎜⎝∫cjω1

...∫cjωg

⎞⎟⎠ with c1, .., c2g basis of H1(X0(N),Z) and ω1, .., ωg basis of


Ω1(X0(N)). Let us compute µ(f) on Λ0:

µ(f)(λj) = ⟨µ(f), λj⟩ =∑k

rk⟨µ(fk), λj⟩ =∑k

rk⟨dzk, λj⟩ =∑k

rk

∫cj

fk(ξ)dξ =

∫cj

f(ξ)dξ

=⇒ µ(f)(Λ0

)=∑j

Z∫cj

f(ξ)dξ = Λf (5.3.5)

Let TO(A) be the tangent space to A at O: Cg−1 ∼= TO(A) ⊂ TO(J) ∼= Cg. Sinceµ(f) is a nonzero multiple of ν∗(ω) we have

kerµ(f) = u ∈ TO(J) | ⟨ν∗(ω), u⟩ = 0 = u ∈ TO(J) | ⟨ω, (dν)(u)⟩ = 0 == u ∈ TO(J) | (dν)(u) = 0 = ker(dν) = TO(A)

(5.3.6)

where the third equality follows from the fact that ω is a basis of Ω1(E). There-fore ker(Φ|TO(A)

: TO(A) −→ A) = TO(A) ∩ Λ0 and this is a lattice in TO(A) of

rank 2g−2 since A is compact as it is an abelian variety over C (thus a complextorus).Let x1, .., x2g−2 be a Z-basis for TO(A) ∩ Λ0 and let x2g−1, x2g ∈ Λ0 be such

that Λdef= Zx1 + ..+ Zx2g has rank 2g. Then it follows that Λ has finite index

m in Λ0, so that mΛ0 ⊂ Λ. Therefore we have

C = µ(f)(TO(J)) = µ(f)(Rx1 + ..+ Rx2g

=Λ⊗ZR

)= µ(f)

(Rx2g−1 + Rx2g

)

Hence µ(f)(x2g−1) and µ(f)(x2g) are linearly independent over R. Moreover:

µ(f)(Zx2g−1 + Zx2g

)= µ(f)

( 2g∑j=1

Zxj)= µ(f)

(Λ)⊂ µ(f)

(Λ0

) =Λf

⊂

⊂ µ(f)(m−1Λ

)= µ(f)

(m−1Zx2g−1 +m−1Zx2g

)Since the first and last terms of the chain have rank 2, we deduce that Λf if afree abelian subgroup of (C,+) of rank 2 and spans C over R. Therefore it is alattice.Step 6: C/Λf is isomoprhic to E over C.Let E′ = C/Λf be the elliptic curve associated to Λf . Now if π′ : C −→ E′ isthe canonical projection, then π′ µ(f) : Cg −→ E′ has kernel (µ(f)−1)

(Λf

)=

Λ0 + kerµ(f) = Λ0 + TO(A). Since Φ : Cg −→ J is a universal coverining withkernel Λ0 ⊂ ker(π′ µ(f)), π′ µ(f) factors through Φ, namely there existsη : J −→ E′ such that π′ µ(f) = η Φ. Since η is a holomorphic map, byTheorem 5.10 it is a morphism over C and its kernel is Φ(TO(A) + Λ0) = A.Therefore ker(η) = ker(ν) and we can apply the universal property of ν to obtainψ : E −→ E′ defined over C such that η = ψ ν. Finally since ker(ψ) = 0,we have deg(ψ) = #ker(ψ) = 1 and by Corollary 1.19 follows that ψ is anisomorphism.

Remark 5.16. Properties (a) and (b) of Theorem 5.15 characterize A uniquelyand consequentely (E, ν) up to isomorphism over Q.


5.3.1 Perspective

• Work of Igusa shows that the L-functions of E and f in Theorem 5.15coincide as Euler products except for possibly finitely many primes (thosedividing N).

• With additional assumptions on f , the two L-functions L(s, E) and L(s, f)match exactly (See [Shi73, Chapter 7]).

• The modularity theorem states that any elliptic curve over Q can be ob-tained via a rational map with integer coefficients from the classical mod-ular curve X0(N) for some integer N (See [DS06, Chapters 8 & 9]).

• Andrew Wiles proved the modularity theorem for some class of ellipticcurves. This was enough (together with Ribet’s Theorem in [Rib90]) toimply Fermat’s last theorem (See [Wil95]).

Bibliography

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[Cox97] David Cox, Primes of the Form x2+ny2: Fermat, Class Field Theory, and ComplexMultiplication, Wiley, second edition, 1997.

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[Miy89] Toshitsune Miyake, Modular forms, Translated by Y. Maeda, Springer Monographsin Mathematics, Springer-Verlag Berlin Heidelberg, 1989.

[Rib90] Kenneth A. Ribet, On modular representations of Gal(Q/Q) arising from modularforms, Invent. Math., 100:431-476, 1990.

[Sha77] I.R. Shafarevich, Basic Algebraic Geometry, Translated by K.A. Hirsch, Study Edi-tion, Springer-Verlag, Berlin 1977.

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