Moduli Spaces of Elliptic Curves

Universidad de los Andes

Undergraduate thesis submitted for the degree of Mathematician.

Santiago Arango PinerosAdvisor: Guillermo Mantilla-Soler, Ph. D.

December 26, 2016

Abstract

This thesis is simply a record of my first steps in the world of elliptic curves. My intentionwith it is that it serves as an introductory reference on the subject for any undergraduateor graduate level student with some basic background in algebraic geometry and complexanalysis.

One of my favourite things about number theory is also one of the things that make itvery difficult to learn, and that is the fact that you need a strong background in almostall of mathematics. For this specific case, we discuss concepts from algebraic geometry,commutative algebra, group theory, topology, complex analysis and Riemann surface theory.Never the less, the paper is designed in a way that everyone can understand the main ideasregardless of the personal mathematical baggage.

In chapter one we try to convince the reader that elliptic curves are a cool thing to studyby discussing three of the most important problems in number theory, all of them closelyrelated to elliptic curves.

In chapter two we summarize fundamental results from other areas of mathematics thatare vital for proving the main theorems afterwards. We discuss Algebraic number theoryanalogues of some of the results on the algebraic geometry of curves.

Chapter three is an introduction to the basic concepts and definitions. We show that everyelliptic curves comes from a Weierstrass equation and that there is a natural group structureon them. We follow [9] very closely, although with a different structure.

In chapter four we restrict to the case of complex elliptic curves. We talk about lattices, con-gruence subgroups and their action on the upper half plane. Finally, we prove the beautifuluniformization theorem. We follow [19], developing several exercises and examples.

Chapter 5 is an introduction to moduli spaces of elliptic curves. We start by the definitions,we continue to study their topology and we end with an example of their applications, basedon the papers [15] and [26].

1

Contents

1 Motivation 41.1 The Congruent number Problem . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 The BSD Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 Fermat’s Last Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Preliminaries 92.1 Maps between Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.2.1 The action of Gal(k/k) . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3 Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.4 Riemann-Roch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.5 Complex Analysis and Riemann Surfaces . . . . . . . . . . . . . . . . . . . . 172.6 Covering Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3 What are Elliptic Curves? 213.1 Elliptic Curves and Weierstrass Equations . . . . . . . . . . . . . . . . . . . 21

3.1.1 Legendre Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.2 The group structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.2.1 Geometric addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4 Complex Elliptic Curves 294.1 Lattices and Complex Tori . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.2 Congruence Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.3 Modular Functions and Modular Forms . . . . . . . . . . . . . . . . . . . . . 334.4 Uniformization Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

5 Moduli Spaces 405.1 Modular Curves and Moduli Spaces . . . . . . . . . . . . . . . . . . . . . . . 405.2 The topology of Modular Curves . . . . . . . . . . . . . . . . . . . . . . . . 425.3 A rational elliptic curve with a point of order 11 . . . . . . . . . . . . . . . . 45

2

Notation

Unless it is stated otherwise, throughout this thesis we let k to be a perfect field, and k afixed algebraic closure of k. If R is a ring, R∗ denotes the group of units in R. As it iscustomary in number theory, the symbol # denotes cardinality.

N The set of natural numbers 1, 2, 3, ....Z The ring of integers ...,−3,−2,−1, 0, 1, 2, 3, ....Z/NZ The ring of integers modulo N ∈ N.Q The field of rational numbers.R The field of real numbers.C The field of complex numbers.H The upper half complex plane.Fp The finite field of p elements.An The affine space An(k) = kn.Pn The projective space Pn(k).k(C) the function field of the variety C over k.k(C) the function field of the variety C over k.k[C]p the local ring of C at p.O(V ) The ring of regular functions on the variety V .Op,V The local ring of germs of regular functions of the variety V near p.

3

Chapter 1

Motivation

The theory of elliptic curves is a central topic in contemporary number theory. And althoughthey are interesting on their own, the applications of the properties of this algebraic andgeometric objects are unprecedented. To back up this claim, we present three differentbeautiful and fundamental problems on which elliptic curves play a central role.

1.1 The Congruent number Problem

Which positive integers are the area of a right triangle with rational sides?

This is one of the oldest unsolved questions in number theory, and therefore in all of mathe-matics; the famous Congruent Number Problem. As many problems in number theory it isfairly easy to state.

At least since the time of Diophantus, mathematicians have been battling with this question.The problem is to classify integers n > 0 such that there exist a, b, c ∈ Q satisfying theconditions a2 + b2 = c2 and ab/2 = n. Such integers are called congruent numbers.

The first example that comes to mind is the right triangle corresponding to the Pythagoreantriple (3,4,5). Since its area is 3 · 4/2 = 6, number 6 is congruent and the correspondingright triangle has integer sides. This is not always the case, for example 5 is congruent aswell, but the corresponding rational triangle is a lot harder to find:(

3

2

)2

+

(20

3

)2

=

(41

6

)2

,

(3

2

)(20

3

)(1

2

)= 5.

The congruent triangle for number 5 was found by the Italian mathematician LeonardoPisano, commonly known as Fibonacci. He later wrote a more complete treatment of theproblem in which he stated, without a proof, that square numbers can not be congruent.The proof came 400 years later with Pierre de Fermat.

Theorem 1.1.1. (Fermat, 1640) Number 1 is not congruent.

This result is equivalent to the case n = 4 of Fermat’s last theorem. In fact, Fermat himselfused this result to show this case of his famous theorem. The connection between thecongruent number problem and elliptic curves is given by the following proposition:

4

6803298487826435051217540411340519227716149383203

2244035177043369699245575130906748631609484720418912332268928859588025535178967163570016480830

41134051922771614938320321666555693714761309610

Figure 1.1: Number 157 congruent triangle.

Theorem 1.1.2. The number n ∈ N is congruent if and only if the curve Y 2 = X3 − n2Xhas a rational point (x, y) ∈ Q×Q∗. That is, there is a one to one correspondence betweenthe sets:

En = (x, y) ∈ Q2 : y2 = x3 − n2x, y 6= 0,Cn = (a, b, c) ∈ Q3 : a2 + b2 = c2, ab/2 = n.

Proof: [25, Theorem 3.1]

The previous theorem translates the congruent number problem to finding rational pointson the curve y2 = x3 − n2x, which is our first example of an elliptic curve. The best knownresult towards the solution of the congruent number problem is due to Jerrold B. Tunnelland can be found in [7]. The full power of this theorem relies on a famous conjecture byBrian Birch and Peter Swinnerton-Dyer, which will be discussed later.

Theorem 1.1.3 (J. Tunnell, 1983). Let n be a square free positive integer. If n is odd andcongruent, then the following cardinalities are equal:

#(x, y, z) ∈ Z3 : n = 2x2+y2 + 32z2 =

1

2(#(x, y, z) ∈ Z3 : n = 2x2 + y2 + 8z2).

If n is even and congruent, the following cardinalities are equal:

#(x, y, z) ∈ Z3 :n

2= 4x2+y2 + 32z2 =

1

2(#(x, y, z) ∈ Z3 :

n

2= 4x2 + y2 + 8z2).

Conversely, if the Birch and Swinnerton-Dyer conjecture is true, these equalities imply thatn is congruent.

5

The converse part of the theorem states that if the Birch and Swinnerton-Dyer conjecture istrue, the congruent number problem is computationally solved.

Proof: (of theorem 1.1.1)Suppose by contradiction that 1 is congruent. By Tunnell’s theorem:

2 = #y ∈ Z : 1 = y2 = #(x, y, z) ∈ Z3 : 1 = 2x2 + y2 + 32z2

=1

2(#(x, y, z) ∈ Z3 : 1 = 2x2 + y2 + 8z2)

=1

2(#y ∈ Z : 1 = y2) = 1.

Therefore, 1 cannot be a congruent number.

As pointed out by Keith Conrad on his expository paper on the subject [25], this fact leads toa funny one line proof that

√2 is not rational: If

√2 where rational, the triangle (

√2,√

2, 2)would be congruent with area 1 = (

√2)2/2. Another more serious corollary is case Fermat’s

last theorem for n = 4.

Corollary 1.1.3.1 (FLT, n = 4). The equation x4 + y4 = z4 has no solutions in integersx, y, z with xyz 6= 0.

1.2 The BSD Conjecture

A polynomial with complex coefficients f(x, y) = 0 in two variables defines a curve E. Ifthe coefficients of the polynomial are rational numbers, then one can ask for rational pointson the curve. For the time being, we define an elliptic curve E to be the locus in C of anequation of the form

Y 2 = X3 + AX +B,

where the polynomial X3 +AX +B does not have repeated roots. If A,B ∈ Q we say thatE is defined over Q. As we will show later, we can define a group structure over E, in whichthe set of rational points of E, denoted by E(Q) is a subgroup of E.

Theorem 1.2.1 (Mordell-Weil). Let E be an elliptic curve definded over Q. Then E(Q) isa finitely generated abelian group.

By the classification theorem of finitely generated abelian groups, we know that

E(Q) ∼= E(Q)tor ⊕ ZRE ,

and the rank of the free part, denoted by RE is called the rank of the curve. Andrew Oggconjectured a complete classification of the torsion part, which was proven to be correct byBarry Mazur in [3, 4].

Theorem 1.2.2 (Ogg’s Conjecture, Mazur). Let E be an elliptic curve defined over Q. ThenE(Q)tor is isomorphic to one of the following groups:

Z/nZ with 1 ≤ n ≤ 10 or n = 12, or

(Z/2Z)× (Z/2mZ) with 1 ≤ m ≤ 4.

6

In LMFDB database, there are available examples of hundreds of elliptic curves for eachpossible torsion subgroup.

Concerning the group E(Q) in virtue of 1.2.2 we know everything about the torsion part. Incontrast, very little is known about the rank RE. The Birch and Swinnerton-Dyer conjecturerelates the rank of the curve, a purely algebraic property, to a purely analytic property ofthe curve: the analytic rank.

For an elliptic curve E : Y 2 = X3 + AX + B, define ∆ to be the discriminant of the cubic(which is non zero since the cubic polynomial has no repeated roots) and let

Np := #solutions of y2 ≡ x3 + Ax+B (mod p),ap := p−Np.

Definition 1.2.1. Let E be an elliptic curve defined over Q. We define the incompleteL-series for E to be:

L(E, s) :=∏p - 2∆

(1− app−s + p1−2s)−1.

It is called “incomplete” because the Euler product excludes prime numbers dividing 2∆.Viewed as a function on the complex variable s, it is known to converge for Re(s) > 3/2.In fact, it has been proven that L(E, s) has an analytic continuation to the whole complexplane.

Conjecture 1.2.3 (Birch and Swinnerton-Dyer). Let E be an elliptic curve defined over Q.The Taylor expansion of L(E, s) at s = 1 has the form

L(E, s) = c(s− 1)RE + higher order terms,

with c 6= 0.

In fact, the conjecture is a little bit stronger than this; it gives an explicit formula for c.The BSD conjecture is one of the six unsolved Millenium Prize Problems, stated by the ClayMathematics Institute in the year 2000. A correct solution to any of these problems resultsin a US $1 million prize being awarded by the institute to the discoverer(s).

1.3 Fermat’s Last Theorem

Theorem 1.3.1 (Wiles, 1995). The equation xn+yn = zn has no solutions in integers x, y, zwith xyz 6= 0, whenever n > 2 is an integer.

This is perhaps the most famous problem in the history of mathematics. Unlike the BSDconjecture, it does not have a lot of meaningful consequences. Never the less, mathemati-cians worked on this problem for approximately 350 years, and in the process, some of thefundamental theories in modern mathematics where developed.

Suppose a, b, c, n are integers such that abc 6= 0, n > 2 and

an + bn = cn.

7

Since n > 2, there are two options: n is divisible by four, in which case n = 4k for somek ∈ N and

(ak)4 + (bk)4 = (ck)4,

or n is divisible by some prime number p ≥ 3, in which case n = kp for some k ∈ N and

(ak)p + (bk)p = (ck)p.

This observation reduces the proof to the cases n = 4 and n = p for prime numbers p ≥ 3.

• Euler provided an incomplete proof of the case n = 3 in 1770. Kausler (1802), Legendre(1823), Lame (1865) and others have published proofs for this case.

• The case n = 4 was proved by Fermat himself and, as mentioned before, it is equivalentto theorem 1.1.1.

• Legendre and Dirichlet first showed the case n = 5 independently from each other,around 1825.

• The proof of the case n = 7 is due to Lame, published around 1839.

Therefore, it is enough to show it for prime numbers p ≥ 11. Elliptic curves played a centralrole in the discovery of the first proof of this famous result. The strategy was proposed byGerhard Frey [11] and Jean Pierre Serre [10]. It turns out that if there is a triple (a, b, c) ∈ Z3

with abc 6= 0 such that ap + bp = cp for p ≥ 11 a prime number, then the elliptic curve

E : Y 2 = X(X − ap)(X + bp)

would satisfy some technical properties that contradicted a famous conjecture of Taniyama,Shimura and Weil.

8

Chapter 2

Preliminaries

Problems in number theory bring together several distinct areas of mathematics and thetheory of elliptic curves is not the exception. In this chapter we give an schematic discussionof some results on the algebraic geometry of curves, complex analysis, Riemann surfaces andone important fact from Algebraic Topology. Since the focus of this work is not any of thesesubjects, we skip most of the proofs indicating a reference for the interested reader (some ofthem fairly advanced).

2.1 Maps between Curves

One often talks about an affine chart of an elliptic curve when one actually refers to the entireprojective model. In order to understand this distinction we review some basic algebraicgeometry facts. Recall that the projective space Pn is covered by n+ 1 copies of An.

Proposition 2.1.1. Let Ui ⊂ Pn be the open set defined by Xi 6= 0. The mapping ϕ : An →Ui given by

(a1, ..., an) 7→ [a1 : ... : ai−1 : 1 : ai+1 : ... : an]

is an isomorphism of varieties.

Proof: [22, Prop. 3.3, p. 18]

This proposition implies, for example, that if V is a projective variety and p ∈ V , thenthe affine variety V ′ = ϕ−1(V ∩ Ui) (where Ui is an affine chart containing p) satisfiesk[V ′]p ∼= Op,V .

Rational maps over smooth curves have the nice property of being defined at every point.

Proposition 2.1.2. Let V ⊆ Pm be a variety, p ∈ C a smooth point and φ : C → V arational map. Then φ is regular at p. In particular, if C is smooth, then φ is a morphism.

Proof: [9, Sect. 2, Prop. 2.1]

Also, every non-constant morphism of algebraic curves is necessarily surjective. As we willsee later, compact Riemann surfaces share the same property. We write C/k to denote thatthe curve C is defined over k.

9

Theorem 2.1.1. Let φ : C1 → C2 be a morphism of curves. Then φ is either constant orsurjective.

Proof: For instance [22, p. II.6.8]

If C1/k and C2/k are curves and φ : C1 → C2 is a non-constant rational map, let φ∗ :k(C2) → k(C1) be the induced k-algebra monomorphism, which is given by f 7→ f φ. Ifφ∗k(C2) denotes the image of φ∗, we have the following theorem:

Theorem 2.1.2. Let C1/k and C2/k be curves and φ : C1 → C2 a non-constant rationalmap. Then:

(i) k(C1) is a finite extension of φ∗k(C2).

(ii) If ι : k(C2) → k(C1) is a k-algebra monomorphism, then there exists a unique non-constant rational map defined over k, φ : C1 → C2 such that φ∗ = ι.

Proof: Also at [22, p. II.6.8]

Definition 2.1.1. Let φ : C1 → C2 be rational map of curves defined over k. We define thedegree of φ to be

deg(φ) =

0, if φ is constant,

[k(C1) : φ∗k(C2)], otherwise.

Corollary 2.1.2.1. Let C1, C2 be smooth curves and φ : C1 → C2 a rational map of degree1. Then φ is an isomorphism.

Proof: Since deg(φ) = 1, we have by definition that k(C1) = φ∗k(C2). So φ∗ is injectiveand surjective and hence a k-algebra isomorphism. By 2.1.2 (b), there exists a unique non-constant rational map ψ : C2 → C1 such that ψ∗ = (φ∗)−1. Since C2 is smooth, by 2.1.2 ψ isin fact a morphism. Since (φ ψ)∗ = ψ∗ φ∗ and (ψ φ)∗ = φ∗ ψ∗ are both identity maps,we have the φ and ψ are mutually inverse isomorphisms by the uniqueness assertion of 2.1.2(b).

Recall from algebraic geometry that k[C]p is a discrete valuation ring (DVR) whenever p isa smooth point (proposition 1.1 in [9]). Suppose p ∈ C is smooth. Then the valuation ofk[C]p is given by

ordp : k[C]p 7→ 0, 1, 2, 3, ... ∪ ∞, ordp(f) := supd : f ∈Mdp .

As usual, it is possible to extend this valuation to the fraction field defining

ordp : k(C) 7→ Z ∪ ∞, ordp(f/g) = ordp(f)− ordp(g).

Definition 2.1.2. A uniformizer for C at p is a function t ∈ k(C) such that ordp(t) = 1.

Since k[C]p is also a principal ideal domain, a uniformizer is simply a generator for themaximal ideal Mp.

10

Definition 2.1.3. Let φ : C1 → C2 be a non constant map between smooth curves, and letp ∈ C1. The ramification index of φ at p, denoted by eφ(p), is the quantity

eφ(p) = ordp(φ∗tφ(p)),

where tφ(p) ∈ k(C2) is a uniformizer at φ(p).

Note that φ∗tφ(p)(p) = (tφ(p) φ)(p) = tφ(p)(φ(p)) = 0, so eφ(p) ≥ 1. We say that φ isunramified at p if eφ(p) = 1. φ is unramified if it is unramified at every point.

Proposition 2.1.3. Let φ : C1 → C2 be a non constant map of smooth curves. Then, forevery q ∈ C2 ∑

p∈φ−1q

eφ(p) = deg(φ).

Proof: [22, p. II.6.8]

Example 2.1.1. Let C1 = V(y2 − x) ⊆ A2(R) and C2 = A1(R). Consider projection mapφ : C1 → C2 on the X coordinate: (a, a2) 7→ a2. Since R(C1) ∼= R(t) = R(C2), we have thatdeg(φ) = 1. On the other hand, consider the point q = 0. φ−1(0) = (0, 0) so that:∑

p∈φ−1(0)

eφ(p) = eφ((0, 0)) = ord0(t φ) = ord0(t) = 1,

so q = 0 is unramified. For q2 6= 0, we have that φ−1(q2) = (±q, q2) and:∑p∈φ−1(q2)

eφ(p) = eφ((−q, q2)) + eφ((q, q2)) = 2ordq2(t φ) = 2,

so that q2 6= 0 is ramified with ramification index 2.

Example 2.1.2 (Algebraic number theory). Consider for a moment a number field L andthe ring of integers OL, i.e the integral closure of Z in L. Since OL is a Dedekind Domain(theorem 9.5 in [2]), for every prime number p ∈ Z the extension 〈p〉OL factors as a productof prime ideals in OL, that is 〈p〉OL = be11 · · · b

egg . The exponent ei is called the ramification

index of bi.

For example, 〈2〉 = 〈1 + i〉2 in Z[i] = OQ(i). Since 1 + i is prime in Z[i], 〈2〉EZ ramifies withramification index 2.

Furthermore, the fundamental equality theorem ([16] theorem 3.34) states that

g∑i=1

eifi = [L : Q],

where fi := [OL/bi : Fp] is the degree of the residue field OL/bi. In the context of curvesk[C] plays the role of the ring of integers OL. Since every residue field over k[C] is k, theresidue class degree is always 1 and both equations agree.

11

2.2 Divisors

The fundamental theorem of arithmetic implies that the group Q∗ is the free abelian groupgenerated by the primes. Consider two integers a and b. Then

a =∏p|a

pαp and b =∏p|b

pβp .

Clearly, a divides b if and only if for every prime divisor p of a, we have αp ≤ βp. One of themotivations behind the divisor group of a curve is to define some kind of divisibility relationon it. Another motivation is to control the orders of poles and zeros of functions on a curve.

Definition 2.2.1. The divisor group of a curve C, denoted by Div(C), is the free abelianadditive group generated by the formal symbols [p], p ∈ C.

Div(C) :=

∑p∈C

mp[p] : mp ∈ Z and mp 6= 0 for finitely many p

.

Definition 2.2.2. The degree of a divisor D ∈ Div(C) is defined as the sum

deg(D) :=∑p∈C

np.

If we assume that the curve C is smooth, there is a natural way of defining a divisor for eachf ∈ k(C)∗ the group of units of the field k(C). Define

div(f) :=∑p∈C

ordp(f)[p].

To show that div(f) ∈ Div(C), it sufices to see that there are finitely many points in C atwhich f has a zero (ordp(f) > 0) or a pole (ordp(f) < 0).

Theorem 2.2.1. Let C be a smooth curve and f ∈ k(C)∗. The set of points in which f hasa zero or a pole is finite. Further, if f has no poles, then f ∈ k.

Proof: In [22, p. I.6.5] it is shown that the number of poles is finite. Given this, thenumber of zeros is automatically finite since the poles of 1/f are the zeros of f . For the laststatement, see [22, I.3.4a] .

Proposition 2.2.1. Let C be a smooth curve and let f ∈ k(C)∗.

(a) div(f) = 0 if and only if f ∈ k∗.

(b) deg(div(f)) = 0.

Proof: [9, Chapter 2, Prop. 3.1]

Part (a) of proposition 2.2.1 is analogous to the fact that the ring of regular functionsO(P1) = k.

12

Definition 2.2.3. Let D,D1, D2 ∈ Div(C).

• Div0(C) := D ∈ Div(C) : deg(D) = 0. Div0(C) is a subgroup of Div(C).

• D =∑mp[p] ∈ Div(C) is positive, denoted by D ≥ 0, if mp ≥ 0 for every p ∈ C. We

write D1 ≥ D2 to indicate that D1 −D2 is positive.

• D is principal if there exists f ∈ k(C)∗ such that D = div(f). The set of principaldivisors, denoted by Prin(C). By part (b) of proposition 2.2.1, it is a subgroup ofDiv0(C).

• The divisor class group, also called Piccard group, denoted by Pic(C), is defined tobe Pic(C) := Div(C)/Prin(C). Similarly, Pic0(C) corresponds to the quotient groupPic0(C) := Div0(C)/Prin(C).

• D1 and D2 are linearly equivalent, denoted by D1 ∼ D2, if D1 −D2 ∈ Prin(C).

Example 2.2.1. (Algebraic number theory) By definition we have the exact sequence

1→ k∗ → k(C)∗ → Div0(C)→ Pic0(C)→ 0,

which is the function field analogue of the fundamental exact sequence of algebraic numbertheory for a number field L:

1→ O∗L → L∗ → Frac(L)→ Cl(L)→ 1,

where Frac(L) is the group of fractional ideals of OL and Cl(L) := Frac(L)/Prin(L) is theideal class group of OL.

Example 2.2.2. Assume that char(k) 6= 2. Let e1, e2, e3 ∈ k be distinct and consider thecurve

C : y2 = (x− e1)(x− e2)(x− e3).

In chapter three we will show that C is a smooth curve (proposition 3.1.2) with a singlepoint at infinity, p∞. We calculate div(x− ei), for i = 1, 2, 3 and div(y2).

Recall that C is actually the projective curve in P2 given by

Y 2Z = (X − e1Z)(X − e2Z)(X − e3Z).

By symmetry, it suffices to calculate div((X − e1Z)/Z). This function has exactly one poleand one zero, namely p∞ = [0 : 1 : 0] and p1 = [e1 : 0 : 1].

• Valuation at p∞:

We restrict to the affine chart Y 6= 0. So we consider C as the affine variety given byz = (x − e1z)(x − e2z)(x − e3z) and p∞ = (0, 0). The local ring k[C]p∞ has maximal idealM = 〈x, z〉 = 〈x−e1z, z〉. Since x−eiz divides z for i = 1, 2, 3: ordp∞(z) = 3 and in particularM = 〈x− e1z〉. This implies that ordp∞((x− e1z)/z) = ordp∞(x− e1z)− ordp∞(z) = 1− 3 =−2.

13

• Valuation at p1:

We restrict to the affine chart Z 6= 0. So we consider C as the affine variety given at thebeginning and p1 = (e1, 0). The local ring k[C]p1 has maximal ideal M = 〈x− e1, y〉. Sincex− e2, x− e3 are units in k[C]p1 , we have that 〈x− e1〉 = 〈y2〉, and so M = 〈y〉. This impliesthat ordp1(x− e1) = 2.Therefore, we have by definition that

div((X − eiZ)/Z) = ordpi(x− ei)[pi] + ordp∞((x− eiz)/z)[p∞] = 2[pi]− 2[p∞].

div(Y 2) =3∑i=1

ordpi(x− ei)[pi]− ordp∞(z)[p∞] = [p1] + [p2] + [p3]− 3[p∞].

2.2.1 The action of Gal(k/k)

Gal(k/k) acts on An in a natural way: if σ ∈ Gal(k/k) and p = (x1, ..., xn) ∈ An, definepσ := (σ(x1), ..., σ(xn)). It is clear that the set on k-rational points An(k) is the fixed subsetof An by Gal(k/k). Similarly, Gal(k/k) acts on Pn by acting on homogeneous coordinates.

Now, if C/k, the curve inherits the action of Gal(k/k) and there is a natural way of definingthe action of Gal(k/k) over Div(C):

Dσ :=∑p∈C

np[pσ], σ ∈ Gal(k/k).

If e ∈ Gal(k/k) is the identity automorphism, by definition we have De = D. Also, ifσ, τ ∈ Gal(k/k) then Dστ =

∑np[p

στ ] =∑np[(p

σ)τ ] = (Dσ)τ . So it is an action.

Definition 2.2.4. We say that D ∈ Div(C) is defined over k if Dσ = D for every σ ∈Gal(k/k). The group of divisors defined over k is denoted by Divk(C). Pick(C),Div0

k(C),and Pic0

k(C) are defined analogously.

2.3 Differentials

In differential geometry, one defines the tangent space of a point a in some manifold M asthe vector space of point derivations at a with point-wise addition and multiplication withscalars, and it is denoted by TaM . The cotangent space at a is then defined as the dual spaceof TaM and differentials are the elements of these vector spaces. These constructions havealgebro-geometric analogues, but for the time being, we take a more syntactic approach:

Definition 2.3.1. Let C be a curve. The space of meromorphic differential forms on C,denoted by ΩC , is the k(C)-vector space generated by the dx for x ∈ k(C), subject to therelations:

(a) d(ax+ y) = adx+ dy for all x, y ∈ k(C) and a ∈ k. (Linearity)

14

(b) d(xy) = xdy + ydx for all x, y ∈ k(C). (Leibniz’s Rule)

Proposition 2.3.1. Let C be a curve.

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

(b) Let x ∈ k(C). Then dx is a k(C) basis for ΩC if and only if k(C)/k(x) is a finiteseparable extension.

Proof: See [5] 27, A,B.

The following proposition follows easily from the fact that the space of meromorphic differ-ential forms on C is a one dimensional k(C)-vector space.

Proposition 2.3.2. Let p ∈ C and t ∈ k(C) be a uniformizer at p.

(a) For every ω ∈ ΩC there exists a unique function g ∈ k(C), depending on ω and t, suchthat ω = gdt. We denote g by ω/dt.

(b) Let f ∈ k(C) be regular at p. Then df/dt is regular at p.

(c) The quantity ordp(ω/dt) is independent of the choice of uniformizer t. It is denoted byordp(ω).

(d) For all but finitely many p in C, ordp(ω) = 0.

Proof: See [9] chapter 2, proposition 4.3.

Definition 2.3.2. Let ω ∈ ΩC . The divisor associated to ω is

div(ω) =∑p∈C

ordp(ω)[p] ∈ Div(c).

We say that ω is regular, or holomorphic, if ordp(ω) ≥ 0 for all p ∈ C. It is non-vanishing ifordp(ω) ≤ 0 for all p ∈ C.

Since ΩC is a one dimensional k(C)-vector space, if ω1, ω2 ∈ ΩC are non zero differentialsthen there is a unique function f ∈ k(C)∗ such that ω1 = fω2. Therefore div(ω1) =div(f) + div(ω2). This observation motivates the following definition.

Definition 2.3.3. The the canonical divisor class on C is the image in Pic(C) of div(ω)where ω ∈ ΩC is any non zero differential. Any divisor in this class is called a canonicaldivisor.

2.4 Riemann-Roch

Theorem 2.4.1 (Mordell’s conjecture, Faltings). Let L be a number field and C a curvedefined over L. If C has genus greater than 2, then #C(L) is finite.

15

It is known that the genus of Fermat curves Y n + Xn = 1 is greater than two for n ≥ 4.Therefore, when Faltings proved Mordell’s conjecture in 1984, we all ready knew that therecould only be finitely many counterexamples of Fermat’s last theorem for each n.

A proper definition of the genus of a curve requires some more background in algebraicgeometry. For the purpose of this work, it will be enough to define it in terms of theRiemann-Roch theorem

Definition 2.4.1. Let D ∈ Div(C). Define

L(D) := f ∈ k(C)∗ : div(f) ≥ −D ∪ 0.

L(D) is a finite dimensional k-vector space, and its dimension is denoted by `(D).

Theorem 2.4.2. Let D ∈ Div(C). Then L(D) is a finite dimensional k-vector space.

Proof: See [22, p. II.5.19]

Theorem 2.4.3 (Riemann-Roch). Let C be a smooth curve and let ω be a canonical divisoron C. There is an integer g ≥ 0, called the genus of C, such that for every divisor D ∈Div(C):

`(D)− `(ω −D) = deg(D) + g − 1.

Proof: See [21, Chapter 1]

Definition 2.4.2. Let C be a smooth curve. We define the genus of C to be the integer ggiven by the Riemann-Roch theorem.

The following corollary will be very important in the near future, and the proof is straight-forward given Riemann-Roch.

Corollary 2.4.3.1. Let C, ω and g be as in 2.4.3. Then

(a) `(ω) = g.

(b) deg(ω) = 2g − 2.

(c) If deg(D) > 2g − 2 then `(D) = deg(D) + g − 1.

Proposition 2.4.1. Let C/k be a smooth curve and let D ∈ Divk(C). Then L(D) has abasis consisting of functions in k(C).

Proof: See [9, Section 2, Prop. 5.8]

The proof for corollary 2.4.3.1 follows directly from the theorem choosing appropriate divi-sors. For a complete proof of corollary 2.4.3.1 and proposition 2.4.1 see [9, Chapter 2].

16

2.5 Complex Analysis and Riemann Surfaces

All of the definitions and propositions previously stated have complex analytic analogues.When k = C, complex analysis and the theory of Riemann surfaces give additional structureto the discussion. We summarise the fundamental results needed for the rest of this thesis.

Theorem 2.5.1. Let∑

n>0 an = A be an absolutely convergent series of complex numbers.Then every rearrangement of

∑n>0 an converges to A.

Proof: See [1, Theorem 3.55].

Theorem 2.5.2. If fn(z) is a sequence of analytic functions on a domain D that convergesuniformly to f(z) on D, then f(z) is analytic on D.

Proof: See [18, Chapter V, Sect. 2]

Theorem 2.5.3 (Weierstrass M -test). Suppose Mk is a sequence of positive real numberssuch that

∑Mk converges. If fk(z) are complex-valued functions on a set E such that

|fk(z)| ≤Mk for all z ∈ E, then∑fk(z) converges uniformly on E.

Proof: See [18, Chapter V, Sect. 2].

Recall that an entire function is a function that is analytic at every point in the complexplane.

Theorem 2.5.4 (Liouville’s Theorem). Every entire and bounded function is constant.

Proof: [18, Chapter IV, Sect. 5]

Theorem 2.5.5 (Morera’s Theorem). Let D ⊆ C be a connected not empty open subset andf : D → C a continuous function. If

∫∂Rf(z)dz = 0 for every closed rectangle R ⊆ D, with

sides parallel to the coordinate axes, then f is analytic on D.

Proof: [18, Chapter IV, Sect. 6] .

Theorem 2.5.6 (Open Mapping Theorem). Let D ⊆ C be a connected not empty opensubset and f : D → C an analytic non-constant map. Then f is open.

Proof: See [8, p. 214] .

A Riemann surface is a one dimensional complex manifold, that is essentially a topologicalspace for which every point has a neighborhood homeomorphic to an open subset of C. Weinclude the following definition for the readers that are not familiar with differential geometryor Riemann surface theory:

Definition 2.5.1. Let X be a connected Hausdorff topological space.

• A chart around x ∈ X is a pair (U, ζ) such that U is a neighborhood of x and ζ : U →ζ(U) ⊆ C is an homeomorphism.

17

• Given two charts (U1, ζ1), (U2, ζ2), we get two transition maps

ζ2 ζ−11 : ζ1(U1 ∩ U2)→ ζ2(U1 ∩ U2),

ζ1 ζ−12 : ζ2(U1 ∩ U2)→ ζ1(U1 ∩ U2).

• Two charts are called compatible if their transition maps are analytic.

• A compatible family of charts covering X defines a differentiable structure on X.

• A Riemann surface is a connected Hausdorff topological space together with a complexstructure.

• Let V be an open subset of a Riemann surface X. A function f : V → C is said to beanalytic if for all charts (U, ζ) in the complex structure ofX, the map fζ−1 : ζ(U)→ Cis analytic.

Example 2.5.1. Clearly the whole complex plane is a Riemann surface with complex struc-ture (C, x 7→ x). Moreover, every open connected subspace of C is also a Riemannsurface. In particular the open unit disc D := z ∈ C : |z| < 1 and the upper half planeH := z ∈ C : Im(z) > 0 are Riemann surfaces.

The next theorem is one of the fundamental results of the theory of Riemann surfaces. Itconnects the worlds of classical algebraic geometry and compact Riemann surfaces.

Theorem 2.5.7. Every compact Riemann surface is analytically isomorphic to a projectivenon singular curve.

Proof: [14, Chapter VII] . The previous theorem comes from a more general result from differential geometry on com-plex manifolds, due to Chow:

Theorem 2.5.8 (Chow). Let X be a compact complex manifold analytically embedded in inPn(C). Then:

(i) X is a non-singular projective variety.

(ii) Every meromorphic function is a rational function.

(iii) Every meromorphic differential is a rational differential.

(iv) Every analytic map between varieties is algebraic.

Proof: [23, p. 166]

Chow’s theorem says even more, it follows that there is a complete equivalence of the categoryof compact Riemann surfaces and the category of projective non-singular curves, includingmaps, divisors, differentials, Riemann-Roch and so on. An example of this phenomena is thefollowing lemma, which is the Riemann surface analogue of 2.1.1:

18

Lemma 2.5.9. Let X, Y compact Riemann surfaces and f : X → Y analytic and nonconstant. Then f is surjective.

Proof: X is compact and connected, so f(X) is compact by continuity and since Y iscompact, f(X) is closed. Since f is non constant and analytic, the open mapping theorem(2.5.6) implies that f(X) is open. In conclusion, Y is connected and f(X) is open and closed,therefore f(X) = Y .

Theorem 2.5.10. Every simply connected Riemann surface is isomorphic to exactly one ofthe following three:

(i) The Riemann sphere P1(C).

(ii) The complex plane.

(iii) The open unit disc D.

Proof: See [13, pp. VI, 4.7].

This is a famous theorem proved independently by Poincare and Koebe in 1907. It impliesthat every compact Riemann surface of genus zero is isomorphic to P1(C).

2.6 Covering Spaces

Definition 2.6.1. A covering space for a topological space X is a space X together with acontinuous function p : X → X called covering projection, satisfying the following condition:There exists an open cover Uα of X such that for each α, p−1(Uα) is a disjoint union of

open sets in X, each of which is mapped by p homeomorphically to Uα.

Definition 2.6.2. Let X be a topological space and let X be a covering space for X withcovering projection p. (X, p) is called a universal covering ox X if X is simply connected.

We present a weak version of the lifting theorem that will be enough for the purpose of thiswork.

Theorem 2.6.1 (Lifting Theorem). Let p : X → X be a universal covering. Let f : Y → Xbe another covering. If we choose a point x0 ∈ X and a pair of points y0 ∈ p−1(x0), z0 ∈f−1(x0) there is exactly one map f : X → Y with p = f f and z0 = f(y0).

X

Y X

pf

f

Proof: See [20].

It follows from this theorem that a universal covering is unique, up to isomorphism.

19

Example 2.6.1 (Helix over S1). Consider the circle X = S1 := z ∈ C : |z| = 1. X = Rwith p : R → S1 given by t 7→ eit is a covering space for the circle since any two open arcswhose union is S1 have the desired property.For instance, take

U1 := eit : t ∈ (−4π/3, π/3), U2 := eit : t ∈ (−π/3, 4π/3).

Then p−1(U1) =⋃k∈Z(−4π/3 + 2πk, π/3 + 2πk), successive neighborhoods are disjoint since

they are translations of (−4π/3, π/3)∩ (2π/3, 7π/3) = ∅, and each (−4π/3+2πk, π/3+2πk)is mapped homeomorphically to U1 since p is 2π-periodic. The argument for U2 is exactlythe same. Since R is simply connected, it is the covering space of the one dimensional sphere.

20

Chapter 3

What are Elliptic Curves?

3.1 Elliptic Curves and Weierstrass Equations

Algebraic curves are classified by their genus. All algebraic curves of genus zero can bedescribed as conics in P2(k). The next step is to classify algebraic curves of genus 1. Incontrast to most authors, we start by stating the most general definition of an elliptic curve.

Definition 3.1.1. An elliptic curve is a pair (E,O) where E is a non-singular projectivecurve of genus 1 and O ∈ E is a distinguished point called the origin, base point or point atinfinity. The elliptic curve E is said to be defined over k if E/k as a curve and O ∈ E(k).

Example 3.1.1 (Fermat’s Cubcic). Consider the curve E : X3 +Y 3 +Z3 = 0 ⊆ P2(C) withbase point O = [1 : −1 : 0]. Let f(X, Y, Z) = X3 +Y 3 +Z3. Then ∇(f) = (3X2, 3Y 3, 3Z2) =(0, 0, 0) if and only if (X, Y, Z) = (0, 0, 0) so that E is smooth. Clearly E/Q since the curvehas rational coefficients and O ∈ E(Q). The pair (E,O) is an elliptic curve since E hasgenus 1.

Example 3.1.2 (Selmer’s Cubic). Consider the curve E : 3X3 + 4Y 3 + 5Z3 = 0. This cubicis the standard counterexample of Hasse’s local-global principle for rational curves of degreehigher than 2.

Proposition 3.1.1. The curve E has only the trivial solution over Q but there exist a nontrivial solution over every completion Qν.

Proof: [24].

This proposition implies that E can not be an elliptic curve defined over Q, since E(Q) = ∅.

Since the origin is easily deduced from the context, it is common to omit the ordered pairnotation and refer only to the curve when defining an elliptic curve.

Definition 3.1.2. A Weierstrass Equation is an homogeneous polynomial p(X, Y, Z) ∈k[X, Y, Z] of the form:

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

2Z + a4XZ2 + a6Z

3.

21

To ease notation, one usually writes a Weierstrass Equation in non-homogeneous coordinatesx = X/Z and y = Y/Z:

Y 2 + a1XY + a3Y = X3 + a2X2 + a4X + a6.

Although our definition for an elliptic curve seems abstract, every elliptic curve is the locusof a Weierstrass Equation. This fact is a direct consequence of the Riemann-Roch theorem.

Theorem 3.1.1. Let E be an elliptic curve defined over k.

a) There exist functions X, Y ∈ k(E) such that the map

φ : E 7→ P2, φ = [X : Y : 1],

gives an isomorphism of E onto a curve given by

C : Y 2 + a1XY + a3Y = X3 + a2X2 + a4X + a6,

with coefficients a1, a2, a3, a4, a6 ∈ k and satisfying φ(O) = [0 : 1 : 0].

b) Conversely, every smooth curve C as defined above is an elliptic curve defined over kwith base point O = [0 : 1 : 0].

The functions x, y are called Weierstrass coordinate functions.

Proof:

(a) For n = 1, 2, 3, ... consider the vector spaces L(n[O]), i.e the space of functions definedeverywhere on E, except at O, with a pole of order at most n at O. Since deg(n[O]) = n >0 = 2g − 2, Riemann-Roch implies that dim(L(n[O])) = n.

Therefore, you can choose functions x, y ∈ k(E) such that 1, x is a basis for L(2[O]) and1, x, y is a basis for L(3[O]). By proposition 2.4.1 you may choose x, y ∈ k(E). Observethat x must have a pole of order 2 and y a pole of order 3 at O, since other wise 1, xwouldn’t generate L(2[O]) and 1, x, y wouldn’t generate L(3[O]).

Notice that 1, x, y, x2, xy, y2, x3 ∈ L(6[O]) are seven different elements of a vector space ofdimension 6. Therefore, there exist (again by proposition 2.4.1) A1, A2, ..., A7 ∈ k, not allzero, such that

A1 + A2x+ A3y + A4x2 + A5xy + A6y

2 + A7x3 = 0.

Note that A6 and A7 are not both zero, since otherwise every element would have a differentorder pole at O, and so all Ai’s would be zero, which is impossible. Applying the linearchange of variables x 7→ −A6A7X, y 7→ A6A

27Y and dividing by A3

6A47 yields

Y 2 − A5

A6A7

XY +A3

A26A

27

Y = X3 − A4

A6A27

X2 +A2

A26A

37

X − A1

A36A

47

.

22

Defining a1 = −A5/A6A7, a3 = A3/A26A

27, a2 = −A4/A6A

27, a4 = A2/A

26A

37 and a6 = −A1/A

36A

47 ∈

k, we have that the image of φ = [X : Y : 1] is precisely C.

φ : E → C is a rational function by construction. Since E is smooth, we have that φ is amorphism by 2.1.2. Also, since it isn’t constant, φ is onto by 2.1.1.

To prove that φ is an isomorphism, we know by 2.1.2.1 that it suffices to show that deg(φ) =1. We have that k∗(C) = k(x, y). Consider the map x = [x : 1] defined by

x(p) =

[x(p) : 1], if x is regular at p,

[1 : 0], otherwise.

By proposition 2.1.3,

deg(x) =∑

p∈x−1([1:0])

eX(p) = eX(O) = ordO(x) = 2.

Defining y = [y : 1] as x and a similar calculation yields that deg(y) = 3. Therefore,[k(E) : k(x, y)] divides 2 and 3, so it must be equal to 1. This shows that φ is an isomor-phism.

(b) Let E be given by a non-singular Weierstrass equation. The invariant differential asso-ciated to the Weierstrass equation for E

ω =dx

2y + a1x+ a3

∈ ΩE

has no poles or zeros, that is div(ω) = 0 (see [9] chapter 3, proposition 1.5). Therfore,Riemann-Roch theorem (2.4.3 b.) implies that 2genus(E)− 2 = deg(div(ω)) = 0, so E hasgenus 1. Taking [0 : 1 : 0] as the base point, E is an elliptic curve.

Example 3.1.3. (Fermat’s Cubic Weierstrass Equation) Let E be the curve defined in

example 3.1.1. Consider the curve E : zy2 = x3 − 432z3 and the change of variables

φ : E → E, φ([X : Y : Z]) = [−12Z : 36(Y −X) : X + Y ] = [x : y : z].

The reader may verify that φ is an isomorphism sending O 7→ [0 : 1 : 0] with inverse

φ−1([x : y : z]) =

[−36z + y

−72:

36z + y

72: − x

12

].

Assuming char(k) 6= 2, we can simplify the equation by completing square. The substitutiony 7→ 1

2(y − a1x− a3) gives an equation of the form

E : Y 2 = 4X3 + b2X2 + 2b4X + b6,

withb2 = a2

1 + 4a2, b4 = 2a4 + a1a3, b6 = a23 + 4a6.

23

Furthermore, assuming char(k) 6= 2, 3, the substitution (x, y) 7→(x−3b2

36, y

108

)eliminates the

x2 term:E : Y 2 = X3 − 27c4x− 54c6,

withc4 = b2

2 − 24b4, c6 = −b32 + 36b2b4 − 216b6.

In any case, we have that if the characteristic of k is not two, an elliptic curve is given byan equation of the form y2 = f(x), where deg(f) = 3. There is a straightforward algebraiccriterion for determining when such a curve is smooth:

Proposition 3.1.2. Let C : y2 = f(x) with deg(f) = 3. Then C is smooth if and only iff(x) has no repeated roots.

Proof: Let g(x, y) = y2 − f(x). Since C = V (g(x, y)) is a curve, we have that a pointp = (a, b) ∈ C is singular if and only if(

∂g(a, b)

∂x,∂g(a, b)

∂y

)= (−f ′(a), 2b) = (0, 0)

therefore p = (a, 0) and since p ∈ C, we have that 0 = f ′(a) = f(a), so that a is a repeatedroot for f(x). Hence p ∈ C is singular if and only if there exists a ∈ k a repeated root forf(x).

We continue to assume char(k) 6= 2, 3. The Weierstrass equations of our elliptic curvessimplify to

Y 2 = X3 + AX +B. (3.1)

Note that by proposition 3.1.2, equation 3.1 defines an elliptic curve given that the poly-nomial discriminant disc(X3 + AX + B) 6= 0. Associated to this equation there are twofundamental quantities.

Definition 3.1.3. 1. The Discriminant of the Weierstrass equation, denoted by ∆, de-fined by

∆ := 16∆(f) = −16(4A3 + 27B2).

2. The j-invariant of the elliptic curve E : Y 2 = X3 + AX +B, ∆ 6= 0, defined as

j = −1728(4A)3

∆.

The j-invariant defined above gives a correspondence from the set of isomorphism classes ofelliptic curves defined over k to the field k.

You may have noticed number 1728 in the definition of the j-invariant. It turns out that1728 is a very interesting number, since 1728 + 1 = 1729. As Srinivasa Ramanujan onceexplained to Godfried Hardy, 1729 is the smallest number that can be written as the sum oftwo different cubes, namely:

1729 = 13 + 123 = 103 + 93.

24

Proposition 3.1.3. Assume char(k) 6= 2, 3.

(i) Two elliptic curves are isomorphic over k if and only if they have the same j-invariant.

(ii) Let j0 ∈ k. There exists an elliptic curve defined over k(j0) with j-invariant j0.

The proof consists on a series of calculations that rely on the fact that the only change ofvariables that preserve a Weierstrass equation of the form (3.1) is

X = uX ′, Y = u3Y ′ for some u ∈ k∗.

For example, the case AB 6= 0 (j 6= 0, 1728) of 3.1.3 part (ii) follows by considering the curve

E : Y 2 +XY = X3 − 36

j0 − 1720x− 1

j0 − 1728,

and calculating j(E). For the complete proof, look at [9] chapter 3 proposition 1.4

3.1.1 Legendre Form

The Legendre form of an elliptic curve is just an other kind of Weierstrass that is sometimesconvenient to use.

Proposition 3.1.4. Assume char(k) 6= 2.

(i) Every elliptic curve E/k is isomorphic over k to a curve in Legendre Form

Eλ : Y 2 = X(X − 1)(X − λ)

for some λ ∈ k − 0, 1.

(ii) The associationk − 0, 1 → k, λ 7→ j(Eλ),

is surjective and exactly six-to-one, except above j = 0 and j = 1728 where it istwo-to-one and three-two-one respectively.

Proof: [9] chapter 3, proposition 1.7.

3.2 The group structure

There is a completely geometric way of defining a binary operation on an elliptic curve tomake it a group (even more, a Lie group). Alternatively, we show that for every ellipticcurve E there is a one to one correspondence from Pic0(E) onto E. Therefore, E inheritsthe group structure from Pic0(E).

Recall that two divisors are said to be equivalent if their difference is a principal divisor,that is: D1 ∼ D2 if and only if they D1 = D2 in Pic0(E).

25

Lemma 3.2.1. Let C be a curve of genus one, and let p, q ∈ C. Then [p] ∼ [q]⇔ p = q.

Proof:(⇐) Clear.(⇒) Let f ∈ k(C)∗ be such that div(f) = [p] − [q]. Then div(f) ≥ −[q], that is, div(f) ∈L([q]). By Riemann-Roch (2.4.3.1 c.) we have that `([q]) = deg([q]) = 1. Since k ⊆ L([q]),f ∈ k and p = q.

Theorem 3.2.2. Let (E,O) be an elliptic curve.

(a) For every D ∈ Div0(E) there exists a unique point p ∈ E such that D ∼ [p]− [O].

(b) Let σ : Div0(E)→ E be the map defined by part (a). Then σ is surjective.

(c) The map σ : Pic0(E)→ E defined by D + Prin(E) 7→ σ(D) is a bijection.

(d) The map κ : E → Pic0(E) defined by p 7→ [p]− [O] + Prin(E) is an inverse for σ.

Proof:

(a) Consider the vectorspace L(D + [O]) = f ∈ k(E)∗ : div(f) ≥ −D − [O]. SinceD ∈ Div0(E), deg(D + [O]) = 1 > 2g − 2 = 0. Therefore, corolary 2.4.3.1 implies that`(D + [O]) = deg(D + [O]) = 1. Let f ∈ k(E) be a generator of L(D + [O]). Since

div(f) ≥ −D − [O] and deg(div(f)) = 0,

it follows that there exists some p ∈ E such that div(f) = −D − [O] + [p]. ThereforeD ∼ [p]− [O] and the existence is proven. For the uniqueness part, suppose p′ ∈ E has thesame property. Then

[p]− [O] ∼ [p′]− [O]⇐⇒ [p] ∼ [p′]

and by lemma 3.2.1 we have p = p′.

(b) Let p ∈ E. The [p] − [O] ∈ Div0(E). Since ∼ is reflexive, we have by the uniquenesspart of (a) that σ([p]− [O]) = p.

(c) We show next that σ is well defined and injective. Let D1, D2 ∈ Div0(e) and letpi = σ(Di). Suppose D1 ∼ D2. Since Di ∼ [pi]− [O], [p1] ∼ [p2] and once more lemma 3.2.1implies that p1 = p2 and we have that σ is well defined. Suppose now that p1 = p2 = p.Then D1 −D2 ∼ [p]− [p] ∼ 0 which implies D1 ∼ D2, i.e σ is one to one.

(d) For simplicity, call σ = σ. Computing the compositions:

(σ κ)(p) = σ([p]− [O] + Prin(E)) = p.

(κ σ)([p]− [O] + Prin(E)) = κ(σ([p]− [O] + Prin(E))) = κ(p) = [p]− [O] + Prin(E).

Now that we have a one to one correspondence between E and Pic0(E), E inherits the groupstructure from Pic0(E). The addition on E is the only possible one that makes κ a groupisomorphism, that is:

p1 +E p2 := σ(κ(p1) + κ(p2)).

26

3.2.1 Geometric addition

Start by considering an elliptic curve of the form E : y2 = x3 + Ax + B where A,B ∈ k.Consider the homogeneous model of the curve by setting x = X/Z and y = Y/Z, whichyields the equation

Y 2Z = X3 + AXZ2 +BZ3. (3.2)

The intersection of the cubic (3.2) with the line at infinity is obtained by substituting Z = 0into the equation, which gives X3 = 0. So the cubic has exactly one point at infinity withmultiplicity 3, namely O = [0 : 1 : 0]. We consider O to be the zero element. Consider twopoints p, q ∈ E. Since every line intersects the curve in three points (counting multiplicity),the line through p and q intersects E at a third point which we call p ∗ q. Take p + q to bethe third point of intersection between the line that goes through p∗q and O. Look at figure3.1 for a visual representation of this paragraph.

To see that both definitions agree, let f = αX +βY +γZ be the line that intersects p, q andp ∗ q and g = α′X + β′Y + γ′Z the line that intersects O, p ∗ q and p+ q. Then:

div(f/Z) = [p] + [q] + [p ∗ q]− 3[0]

div(g/Z) = [O] + [p ∗ q] + [p+ q]− 3[O]

= [p ∗ q] + [p+ q]− 2[O]

Then,

div(f/g) = [p] + [q] + [p ∗ q]− 3[0]− ([p ∗ q] + [p+ q]− 2[O])

= [p] + [q]− [p+ q]− [O].

Since [p] + [q]− [p+ q]− [O] is principal, we have that [p] + [q]− [p+ q] + Prin(E) = Prin(E).

Example 3.2.1. Consider the curve E : y2 = x3 + 17. Consider the two points

p = (−1, 4), q = (2, 5) ∈ E.

First, we find the point p∗q = (x3, y3). The line through these points is given by the equationy = 1

3x+ 13

3. Substituting into the cubic:(

1

3x+

13

3

)2

= x3 + 17 ⇐⇒ x3 − 1

9x2 − 26

9x− 16

9= 0.

Since x3 − 19x2 − 26

9x− 16

9factors as (x + 1)(x− 2)(x− x3) and the sum of the roots is the

negative of the x2 coefficient:

−1 + 2 + x3 =1

9=⇒ x3 = −8

9.

Therefore y3 = 13(−8

9) + 13

3= 109

27and p ∗ q = (−8/9, 109/27). The line through p ∗ q and O

is precisely the vertical line through p ∗ q, so that p+ q = (x3,−y3) = (−8/9,−109/27.)

27

x

y

p•

q•

•p ∗ q

•p+ q

Figure 3.1: Geometric addition on the real locus of an Elliptic Curve.

28

Chapter 4

Complex Elliptic Curves

4.1 Lattices and Complex Tori

Definition 4.1.1. A lattice Λ is a discrete subgroup of C which contains an R-basis. Thatis, Λ = ω1Z⊕ ω2Z where ω1, ω2 is a basis of C over R. We write Λ = 〈ω1, ω2〉.

There is a normalizing convention to take ω1/ω2 in the upper half plane H := τ ∈ C :Im(τ) > 0.

Lemma 4.1.1. Consider two lattices Λ = 〈ω1, ω2〉 and Λ′ = 〈ω′1, ω′2〉. Then Λ = Λ′ if andonly if [

ω′1ω′2

]= γ

[ω1

ω2

], for some γ ∈ SL2(Z).

If γ was a matrix in GL2(Z) the result would be clear, since it would be stating that thereexists an invertible Z-change of basis. The condition that γ has determinant 1 comes fromthe normalizing condition ω1/ω2, ω

′1/ω

′2 ∈ H.

Definition 4.1.2. A fundamental domain for a lattice Λ is a set of the form

F(Λ) := t1ω1 + t2ω2 : 0 ≤ t1, t2 ≤ 1.

Example 4.1.1. Take the root of x2 + x + 1 with positive imaginary part ω = −12

+ i√

32

and consider the lattice Λω = 〈ω, 1〉. In figure 4.1 you can observe the lattice points, thegenerators and the fundamental domain F(Λω).

Definition 4.1.3. A complex torus is the quotient abelian group of the complex plane by alattice,

C/Λ = z + Λ : z ∈ C.

Since the fundamental domain F(Λ) is a complete set of representatives for C/Λ, a complextorus is the parallelogram spanned by the generators of lambda with it’s opposite sidesidentified. Moreover, C/Λ is a compact Riemann Surface.

29

ω

1

Figure 4.1: Ring of Eisenstein Integers.

Example 4.1.2 (Complex structure of C/Λτ ). Let τ ∈ H and Λτ = 〈τ, 1〉. Observe that forevery a, b ∈ Z we have

|aτ + b| ≥

|Im(τ)|, if a 6= 0,

1, if a = 0.

Let r = 12

min1, |Im(τ)| and consider the open disc Dr(z) := w ∈ C : |z − w| < r. Letz1, z2 ∈ Dr(z) be distinct with z1 = z2 + ω for some ω ∈ Λτ . Then

min1, |Im(τ)| ≤ |aτ + b| = |ω| ≤ |z1 − z2| ≤ |z1 − z|+ |z − z2| < 2r = min1, |Im(τ)|,

which is a contradiction. This implies that the projection λ : C → C/Λτ restricted to theopen set Uz := Dr(z) is a bijection. Moreover, since C/Λτ has the quotient topology, λ Uz isa homeomorphism. Take ψz := λ −1

Uz, then the collection (Λ(Uz)), ψz : λ(Uz)→ Uzz∈F(Λτ )

forms an atlas for C/Λ, making it into a Riemann surface.

Theorem 4.1.2. Let ϕ : C/Λ → C/Λ′ be an analytic map between complex tori such thatϕ(Λ) = Λ′. Then, there is some α ∈ C with αΛ ⊆ Λ′ such that ϕ(z + Λ) = αz + Λ′.

Proof: C is the universal covering space for C/Λ and C/Λ′. By the lifting theorem, thereexists a continuous map ϕ : C→ C such that ϕ(0) = 0 and the following diagram commutes:

C C

C/Λ C/Λ′λ

ϕ

λ′

ϕ

30

Since the vertical arrows are local isomorphisms, ϕ is analytic. Let ω ∈ Λ and consider themap fω : C→ C given by

z 7→ ϕ(z + ω)− ϕ(z).

Note that by commutativity of the diagram:

λ′(fω(z)) = ϕ(λ(ω)) = Λ′,

so that fω takes values in Λ′. Since fω is continuous and C is connected and it maps intoa discrete space, fω must be constant. Since ω ∈ Λ is arbitrary, the function d

dzϕ is Λ-

periodic. Consequently, it defines an analytic function C/Λ → C. Since C/Λ is compact,ddzϕ is bounded and analytic in the whole complex plane, and thus it must be constant by

Liouville’s theorem, say ϕ′(z) = α. Then ϕ(z) = αz + β. Since ϕ(0) = 0, we have thatβ = 0, and the result follows.

Corollary 4.1.2.1. Every analytic map between complex tori ϕ : C/Λ → C/Λ′ such thatϕ(Λ) = Λ′ is a group homomorphism.

Proof: Clearly z + Λ 7→ αz + Λ′ is an homomorphism for every α ∈ C.

Corollary 4.1.2.2. The complex tori C/Λ and C/Λ′ are isomorphic as Riemann surfacesif and only if αΛ = Λ′ for some α ∈ C∗.

Example 4.1.3. Let Λ = 〈ω1, ω2〉 be a lattice with τ = ω1/ω2 ∈ H. Define Λτ = 〈τ, 1〉. Invirtue of corollary 4.1.2.2 there is an isomorphism of complex tori given by

ϕτ : C/Λ→ C/Λτ , z + Λ 7→ z/ω2 + Λτ .

From now on we will focus on complex tori given by lattices of the form Λτ = 〈τ, 1〉 whereτ ∈ H. Denote Eτ := C/Λτ .

Example 4.1.4 (N -torsion subgroup). Consider the isomorphism [N ] : Eτ → Eτ given bymultiplication by N . Clearly ker([N ]) =

kτ+lN

+ Λτ : 0 ≤ k, l ≤ N − 1

is the N -torsionsubgroup of Eτ , which we denote by Eτ [N ].

Note that Eτ ∼= (R/Z)× (R/Z) as groups. Since the N -torsion subgroup of (R/Z)× (R/Z)is precisely (Z/NZ)× (Z/NZ), we have that Eτ [N ] ∼= (Z/NZ)× (Z/NZ).

4.2 Congruence Subgroups

In this section we consider a certain family of subgroups of the 2 × 2 special linear groupwith integer entries SL2(Z), also known as the modular group.

Lemma 4.2.1. SL2(Z) is generated by S =

[0 −11 0

]and T =

[1 10 1

].

31

τ

0

1

τ + 1

Figure 4.2: 5-torsion points of Eτ .

Proof:First of all, note that

S2 = −I, T k =

[1 k0 1

], S

[a bc d

]=

[−c −da b

], T k

[a bc d

]=

[a+ kc b+ kdc d

].

Clearly G = 〈S, T 〉 ≤ SL2(Z). Let γ = [a, b; c, d] ∈ SL2(Z). Since det(γ) = ad − bc = 1, aand c are relatively prime. If |c| > |a|, take γ to be Sγ, so we may suppose |a| > |c|. By thedivision algorithm, there exist q, r ∈ Z such that a = cq + r with 0 ≤ |r| < |c|. Multiplyingγ by T−q ∈ G:

T−q[a bc d

]=

[r b− qdc d

].

Next, multiply this matrix by S and continue in this fashion. By the division algorithm, oneeventually gets a matrix gγ with left lower entry equal to zero. Since this matrix belongs toSL2(Z), it must be of the form ±T k for some k ∈ Z. Thus, we obtain that γ = g−1T k org−1S2T k for some g ∈ G and the proof is complete.

Consider the group homomorphism that reduces modulo N each coordinate in a matrix:

redN : SL2(Z)→ SL2(Z/NZ),

[a bc d

]7→[a bc d

].

Definition 4.2.1. The principal congruence subgroup of level N of SL2(Z) is defined asΓ(N) := ker(redN).

It is clear from the definition that Γ(N) is a normal subgroup of SL2(Z). In addition, sinceredN is surjective, we have that

SL2(Z)/Γ(N) ∼= SL2(Z/NZ),

which implies that Γ(N) is of finite index in SL2(Z).

32

Definition 4.2.2. A subgroup Γ of SL2(Z) is called a congruence subgroup if there existsN ∈ N such that Γ(N) ⊆ Γ.

We have from the previous observation that every congruence subgroup has finite index.There are two congruence subgroups of special interest for us, Γ0(N) and Γ1(N):

Γ0(N) :=

[a bc d

]∈ SL2(Z) : c ≡ 0 (mod N)

,

Γ1(N) :=

[a bc d

]∈ SL2(Z) : c ≡ 0 (mod N), a ≡ d ≡ 1 (mod N)

.

It is clear from the definitions that Γ(N) ( Γ1(N) ( Γ0(N) ⊆ SL2(Z), so they are indeedcongruence subgroups, in particular they have finite index.

4.3 Modular Functions and Modular Forms

The Lie group SL2(R) acts on the upper half plane as follows:[a bc d

](τ) =

aτ + b

cτ + d, τ ∈ H, ad− cd = 1.

A straightforward calculation shows that

Im

(aτ + b

cτ + d

)=

Im(τ)

|cτ + d|2> 0,

To see that SL2(R) × H → H defined above is an action, let γ1, γ2 ∈ SL2(R) and I be the2× 2 identity matrix. Then:

I · τ =

[1 00 1

]· τ =

τ + 0

0 · τ + 1= τ.

γ2 · (γ1 · τ) =

[e fg h

]·([a bc d

]· τ)

=

[e fg h

]·(aτ + b

cτ + d

)=e(aτ+bcτ+d

)+ f

g(aτ+bcτ+d

)+ h

=eaτ + eb+ f(cτ + d)

gaτ + gb+ h(cτ + d)=

(ea+ fc)τ + eb+ fd

(ga+ hc)τ + gb+ hd

=

[ea+ fc eb+ fdga+ hc gb+ hd

]· τ = γ2γ1 · τ.

Note that −I · τ = τ , so that the action is not faithfull. Nevertheless, restricting the actionto PSL2(R) = SL2(R)/I,−I makes it faithfull. Since SL2(Z) ≤ SL2(R), we have that themodular group acts on H. In particular, every congruence subgroup acts on the upper halfplane in this way.

33

Definition 4.3.1. Let f : H→ C be a meromorphic function, Γ a congruence subgroup andk an integer. We say f is weakly modular of weight k for Γ if

f(γ · τ) = (cτ + d)kf(τ), for γ =

[a bc d

]∈ Γ.

Note that weak modularity of weight 0 for SL2(Z) is just SL2(Z)-invariance. Recall from4.2.1 that SL2(Z) = 〈S, T 〉, so f : H→ C is weakly modular of weight k for SL2(Z) and onlyif

f(τ + 1) = f(T · τ) = f(τ) and f(−1/τ) = f(S · τ) = tkf(τ).

Lemma 4.2.1 implies that the space of orbits SL2(Z)\H (written on the left to signify thatit is a left action) can be identified with the set D := τ ∈ H : |Re(τ)| ≥ 1/2, |Im(τ)| ≥ 1displayed in figure 4.3.

Figure 4.3: Fundamental domain for SL2(Z)\H.

Suppose f is weakly modular of weight k for Γ some congruence subrgroup. Then

f(τ) = f(−I · τ) = (−1)kf(τ).

This implies that the only weakly modular function of odd weight is the zero function.

Definition 4.3.2. Let k be an integer. A function f : H → C is a modular form of weightk if:

(i) f is analytic in H.

(ii) f is weakly modular of weight k for SL2(Z).

(iii) f is analytic at ∞.

Perhaps a more intuitive way of thinking of modular forms is as analytic functions on thequotient SL2(Z)\H. The condition of being analytic at infinity will become clear soon.

Our first non trivial example of a modular form is a two-dimensional generalization of theRiemann ζ-function. For a complete exposition of this topic read the paper [6] by Zagier.

34

Example 4.3.1 (Eisenstein Series). Consider a lattice Λ and let k > 2 be an even integer.We define the k-Eisenstein series with respect to Λ by

Gk(Λ) :=∑

06=ω∈Λ

1

ωk.

If we consider lattices of the form Λτ for τ ∈ H, Gk(τ) := Gk(Λτ ) may be considered as afunction Gk : H → C. We show that Gk(τ) is a modular form of weight k, and we start byshowing two important lemmas:

Lemma 4.3.1. Let k ≥ 3 and L = Z2−(0, 0). The series∑

(a,b)∈L

max|a|, |b|−k converges.

Proof:Let sn := #(x, y) ∈ L : max|x|, |y| = n, i.e the number of integer points on the edge ofa square of radius n centered at the origin. Since each side of such a square contains 2n+ 1integer points, sn ≤ 4(2n+ 1) = 8n+ 4. Then:∑

(a,b)∈L

max|a|, |b|−k ≤∞∑n=1

snnk≤

∞∑n=1

8n+ 4

nk= 8

∞∑n=1

1

nk−1+ 4

∞∑n=1

1

nk<∞.

Lemma 4.3.2. Let a, b > 0 be fixed. Define Ωa,b := τ ∈ H : |Re(τ)| ≤ a, |Im(τ)| ≥ b.There exists C > 0 such that

|τ + δ| > max1, |δ|, ∀ τ ∈ Ωa,b, δ ∈ R.

Proof: Let τ ∈ Ωa,b and δ ∈ R. We consider the following four cases:

a) If |δ| < 1: |τ + δ| ≥ |Im(τ + δ)| = |Im(τ)| ≥ b = bmax1, |δ|.b) If 1 ≤ |δ| ≤ 3a and |Im(τ)| > a: |τ+δ| ≥ |Im(τ+δ)| = |Im(τ)| ≥ a ≥ 1

3|δ| = 1

3max1, |δ|.

c) If 1 ≤ |δ| ≤ 3a and b ≤ |Im(τ)| ≤ a: |τ + δ|/|δ| is a continuous function defined on acompact set, and thus takes a non-zero minimum m. So |τ + δ| ≥ m|δ| = mmax1, |δ|.d) If |δ| ≥ 1 and |δ| > 3a: |τ + δ| ≥ |Re(τ + δ)| = |Re(τ) + δ| ≥ ||δ| − |Re(τ)|| ≥ ||δ| − |a|| ≥|δ| − 1

3|δ| = 2

3|δ| = 2

3max1, |δ|.

Therefore, taking C := minb, 1/3,m we have the result.

1. The function Gk : H→ C is analytic for k ≥ 3.

Proof: Let τ ∈ H. Choose a, b > 0 such that τ ∈ Ωa,b. Recall that the Riemann ζ-functionis defined by ζ(s) =

∑∞n=1 n

−s, for Re(s) > 1.

|Gk(τ)| ≤∑

(c,d)∈L

1

|cτ + d|k=∑

(0,d)∈L

1

|d|k+∑

(c,d)∈Lc 6=0

1

|cτ + d|k= 2ζ(k) +

∑c 6=0δ=d/c

1

|c|k|τ + δ|k

35

By lemma 4.3.2, there exists C > 0 such that |τ + δ| ≥ C max1, |δ|. Also, using lemma4.3.1:

≤ 2ζ(k) +1

Ck

∑c 6=0δ=d/c

1

|c|k max1, |δ|k=≤ 2ζ(k) +

1

Ck

∑(c,d)∈Lc 6=0

1

max|c|, |d|k<∞ .

Therefore, Gk(τ) converges absolutely. By theorem 2.5.1, every rearrangement converges tothe same value. Consider the following rearrangement:

Gk(τ) = 2ζ(k) +∞∑c=1

(∞∑d=0

2

(cτ + d)k

),

here we use the fact that k is even since Gk is weakly modular of weight k. Let Mc :=∑∞d=0 2(cτ + d)−k. By the Weierstrass M -test 2.5.3, Gk(τ) converges uniformly as well. In

virtue of theorem 2.5.2, Gk(τ) is analytic on Ωa,b, for every a, b > 0. Now let R be a closedrectangle contained in H. Take a, b > 0 such that R ⊆ Ωa,b. Since Gk is analytic in Ωa,b,∫∂RGk(z)dz = 0, so Morera’s theorem 2.5.5 implies that Gk is analytic in H.

2. Gk is weakly modular of weight k.

Proof: Since by lemma 4.2.1 SL2(Z) = 〈S, T 〉, it suffices to very weak modularity for thesetwo matrices. Since S · τ = −1/τ and T · τ = τ + 1, that is Gk(−1/τ) = τ kGk(τ) andGk(τ + 1) = Gk(τ).

Gk(−1/τ) =∑

(a,b)∈L

1

(−a/τ + b)k=∑

(a,b)∈L

τ k

(−a+ bτ)k= τ kGk(τ),

Gk(τ + 1) =∑

(a,b)∈L

1

(a(τ + 1) + b)k=

∑(a,a+b)∈L

1

(aτ + (a+ b))k= Gk(τ).

3. Gk is analytic at ∞.

Proof: It suffices to show that Gk(τ) is bounded when Im(τ)→∞. Let a, b > 0 be sch thatτ ∈ Ω. Since Re(τ) + yi ∈ Ωa,b for every y > Im(τ), Gk(τ) is bounded when Im(τ)→∞.

We continue our study of modular functions by defining the analogues of the discriminantand j-invariant functions of elliptic curves discussed in chapter 3.1.

Definition 4.3.3. Let g2(τ) := 60G4(τ) and g3(τ) := 140G6(τ).

1. The discriminant function ∆ : H→ C is defined by

∆(τ) := g2(τ)3 − 27g3(τ)2.

2. The modular invariant is defined by

j(τ) := 1728g2(τ)3

∆(τ).

36

Notice that by example 4.3.1, ∆ is weakly modular of weight twelve and analytic on H. Also,we will show in the following section (corollary 4.4.1.1) that ∆(τ) 6= 0 for every τ ∈ H. Thisimplies j is analytic on H and SL2(Z) invariant since the numerator and denominator havethe same weight.

Definition 4.3.4. A cusp form of weight k is a modular form of weight k whose Fourierexpansion has leading coefficient a0 = 0, i.e.

f(τ) =∞∑n=1

anqn, q = e2πiτ .

Since q = e2πiτ = e2πiRe(τ)e−2πIm(τ), f is a cusp form when limIm(τ)→∞ = 0. ∆(τ) is a cuspform of weight twelve. For a proof, look at [19] exercise 1.1.7(d).

Proposition 4.3.1. The modular invariant j : H→ C is surjective.

For the proof of this proposition, we rely on the fact that the Fourier expansion of j is

j(τ) =1

q+ 744 + 196884q + ... ,

(see [12] chapter 15, remark 7.4) which implies that limIm(τ)→∞ j(τ) =∞.

Proof: Since j is analytic and non constant on H, the open mapping theorem 2.5.6 impliesthat j(H) is open. Since C is connected and j is continuous, it suffices to show that j(H) isalso closed.Let j(τn) be a sequence in j(H) that converges to some z ∈ C. Consider the sequenceτn ⊂ H. Since j is SL2(Z)-invariant, j(τ) = j(τ+k) = j(−1/τ) for every k ∈ Z. Therefore,

without loss of generality, we may take τn ⊆ R := τ ∈ H : |Re(τ)| ≤ 12, |Im(τ)| ≥

√3

2.

If Im(τn) where unbounded, limn→∞ Im(τn) =∞ and in turn limn→∞ j(τn) =∞, which isa contradiction. Therefore Im(τn) is bounded by some M ≥

√3/2 and the sequence τn

lies in the rectangle RM = τ ∈ R : |Im(τ)| ≤ M. Since RM is compact, there is somesubsequence of τn that converges to some τ ′ ∈ RM ⊆ H. The continuity of j implies thatj(τ ′) = z, which concludes the proof.

4.4 Uniformization Theorem

In this section we prove one of the most beautiful results of this thesis. The assertion isthat there is a bijective correspondence between isomorphism classes of complex lattices andisomorphism classes of complex elliptic curves.

Definition 4.4.1. Let Λ be a lattice. The Weierstrass ℘-function with respect to Λ is definedby

℘(Λ, z) :=1

z2+∑

0 6=ω∈Λ

(1

(z − ω)2− 1

ω2

).

Theorem 4.4.1. Let Λ be a lattice and let ℘ be the Weierstrass ℘-function with respect toΛ. Then:

37

(i) The Laurent expansion of ℘ is

℘(z) =1

z2+

∞∑n=2n even

(n+ 1)Gn+2(Λ)zn,

for all z such that 0 < |z| < inf|ω| : ω ∈ Λ− 0.

(ii) The functions ℘ and ℘′ satisfy:

(℘′(z))2 = 4℘(z)3 − g2(Λ)℘(z)− g3(Λ),

where g2(Λ) := 60G4(Λ) and g3(Λ) := 140G6(Λ).

(iii) Let Λ = 〈ω1, ω2〉 and ω3 = ω1 + ω2. Then the cubic equation satisfied by ℘ and ℘′,y2 = 4x3− g2(Λ)x− g3(Λ), factors as y2 = 4(x−℘(ω1/2))(x−℘(ω2/2))(x−℘(ω3/2)).This equation is non-singular.

Proof:(i) Recall that the series expansion for 1/(1 − z)2 around 0 is

∑∞n=0(n + 1)zn for |z| < 1.

Note that if 0 < |z| < |ω|, we have

1

(z − ω)2− 1

ω2=

1

ω2

(1

(1− z/ω)2− 1

)= − 1

ω2+∞∑n=0

(n+ 1)zn

ωn+2=∞∑n=1

(n+ 1)zn

ωn+2.

Let δ = inf|ω| : ω ∈ Λ− 0 and 0 < |z| < δ. Take some ε > 0 such that ε < |z| < δ − ε.Then z lies on the annulus A := w ∈ C : ε ≤ |w| ≤ δ − ε, which is a compact subset ofC− Λ, and thus ℘(z) converges absolutely and uniformly on A and we may apply theorem2.5.1 to ℘(z):

℘(z) =1

z2+∑

06=ω∈Λ

(1

(z − ω)2− 1

ω2

)=

1

z2+∑

06=ω∈Λ

(∞∑n=1

(n+ 1)zn

ωn+2

)

=1

z2+∞∑n=1

( ∑06=ω∈Λ

(n+ 1)zn

ωn+2

)=

1

z2+∞∑n=1

( ∑06=ω∈Λ

1

ωn+2

)(n+ 1)zn

=1

z2+∞∑n=1

Gn+2(Λ)(n+ 1)zn =1

z2+

∞∑n=2n even

Gn+2(Λ)(n+ 1)zn .

The last equality follows from the fact that there are no non-zero modular functions of oddweight.

(ii) Using the result from (i), we have that

℘(z) =1

z2+G4(Λ)3z2 +G6(Λ)5z4 +O(z6) (4.1)

℘′(z) = − 2

z3+G4(Λ)6z +G6(Λ)20z3 +O(z5) . (4.2)

38

Calculating, both ℘′(z)2 and 4℘(z)3− g2℘(z)− g3 reduce to 4/z6− 24G4/z2− 80G6 +O(z2).

Their difference is analytic and Λ-periodic, therefore bounded and by Liouville’s theorem2.5.4 it is constant. Since O(z2)→ 0 when z → 0, we have the result.

(iii) Note that ℘′ is odd:

−℘′(z) = 2∑ω∈Λ

1

(z − ω)3= 2

∑ω∈Λ

1

(z + ω)3= −2

∑ω∈Λ

1

(−z − ω)3= ℘′(−z).

Suppose z has order two in C/Λ. Then z ≡ −z (mod Λ) implies that ℘′(z) = ℘′(−z) =−℘′(z) and therefore ℘′(z) = 0. Thus, ℘′ has zeros at the order two points in C/Λ, whichare precisely ωi/2 for i = 1, 2, 3. By (ii), the polynomial factors as claimed. Since the ωi/2are clearly distinct, the curve is smooth.

Corollary 4.4.1.1. The function ∆ is non-vanishing in H.

Proof: Let τ ∈ H and consider lattice Λτ = 〈τ, 1〉. By theorem 4.4.1 part (iii), the cubicpτ (x) = 4x3 − g2(τ)x− g3(τ) has distinct roots. Then

0 6= disc(pτ ) = −16(−g2(τ))3 − 432g3(τ)2 = 16(g2(τ)3 − 27g3(τ)2) = 16∆(τ).

Since τ was arbitrary, we have the result.

Theorem 4.4.2 (Uniformization). Given an elliptic curve defined over C

E : Y 2 = 4X3 − a2X − a3, a32 − 27a2

3 6= 0,

there exists a lattice Λ such that a2 = g2(Λ) and a3 = g3(Λ).

Proof: First, consider the case a2a3 6= 0. Since j : H → C surjects (proposition 4.3.1),there exists τ ∈ H such that j(τ) = 1728a3

2/(a32 − 27a2

3). Calculating:

1728g2(τ)3

g2(τ)3 − 27g3(τ)2=

1728a32

a32 − 27a2

3

=⇒ (a32 − 27a2

3)g2(τ)3 = (g2(τ)3 − 27g3(τ)2)a32

=⇒ a32g2(τ)3 − 27a2

3g2(τ)3 = g2(τ)3a32 − 27g3(τ)2a3

2

=⇒ −27a2

3g2(τ)3 = −27g3(τ)2a3

2

=⇒ a23g2(τ)3 = g3(τ)2a3

2 . (4.3)

Let ω2 ∈ C. Define ω1 := τω2 and consider the lattice Λ = 〈ω1, ω2〉. Then

g2(Λ) = 60G4(Λ) = 60∑

0 6=ω∈Λ

1

ω4= 60

∑(a,b)∈L

1

(aω1 + bω2)4=

60

ω42

∑(a,b)∈L

1

(aτ + b)4= ω−4

2 g2(τ),

similarly g3(Λ) = ω−62 g3(τ). Choosing ω2 ∈ C such that ω−4

2 = a2/g2(τ) we have thatg2(Λ) = ω−4

2 g2(τ) = (a2/g2(τ))g2(τ) = a2. Further more, equality (4.3) implies that ω−122 =

a32/g2(τ)3 = a2

3/g2(τ)2, thus ω−62 = ±a3/g3(τ), so that g3(Λ) = ω−6

2 g3(τ) = ±a3. Replacingω2 by iω2 if necessary completes the proof.

39

Chapter 5

Moduli Spaces

We have seen in example 4.1.3 that the upper half plane H parametrizes the space of complexlattices. This parametrization is not unique since Z-linear changes of basis don’t modify thegenerated lattice, as seen in lemma 4.1.1. Therefore, there is a one to one correspondencebetween the set of lattices and the set of orbits by the left group action SL2(Z)\H. Inaddition, theorems 4.4.1 and 4.4.2 tell us that there is a one two one correspondence betweencomplex lattices and isomorphism classes of complex elliptic curves, thus:

SL2(Z)\H ! Ell(C)

SL2(Z) · τ −→ C/Λτ

SL2(Z) ·(ω1

ω2

)←− C/〈ω1, ω2〉

Loosely speaking, a moduli space is a geometric space whose points represent algebro-geometric objects of some kind. This said, Y (1) := SL2(Z)\H is a moduli space of complexelliptic curves. In this chapter we will study Y (1) and other moduli spaces that encode ad-ditional information. We will show that these spaces are in fact algebraic curves themselves,some of which turn out to be elliptic curves as well.

5.1 Modular Curves and Moduli Spaces

Definition 5.1.1. Let N ∈ N.

• An enhanced elliptic curve for Γ0(N) is an ordered pair (E,C) where E is a complexelliptic curve and CEE is a cyclic subgroup of order N . Define the equivalence relation:

(E,C) ∼ (E ′, C ′)⇐⇒ There exists an isomorphism ϕ : E → E ′,with ϕ(C) = C ′.

Equivalent classes are denoted by square brackets [E,C] and the set of equivalenceclasses is denoted by:

S0(N) := Enhanced elliptic curves for Γ0(N)/ ∼ .

40

• An enhanced elliptic curve for Γ1(N) is an ordered pair (E, p) where E is a complexelliptic curve and p ∈ E is a point of order N . Define the equivalence relation:

(E, p) ∼ (E ′, p′)⇐⇒ There exists an isomorphism ϕ : E → E ′,with ϕ(p) = p′.

Equivalent classes are denoted by square brackets [E, p] and the set of equivalenceclasses is denoted by:

S1(N) := Enhanced elliptic curves for Γ1(N)/ ∼ .

• An enhanced elliptic curve for Γ(N) is an ordered pair (E, (p, q)) where E is a complexelliptic curve and (p, q) ∈ E × E is a pair of points that generate E[N ]. Define theequivalence relation:

(E, (p, q)) ∼ (E ′, (p′, q′))⇐⇒ ∃ϕ : E → E ′ isomorphism, with (ϕ(p), ϕ(q)) = (p′, q′).

Equivalent classes are denoted by square brackets [E, (p, q)] and the set of equivalenceclasses is denoted by:

S(N) := Enhanced elliptic curves for Γ(N)/ ∼ .

We show that S1(N) is well defined, i.e. that ∼ is in fact an equivalence relation. Theverification for S0(N) and S(N) are analogous.

Proof: ∼ is clearly reflexive since the identity map is an isomorphism. The symmetryfollows by taking the inverse map. For the transitivity, suppose that (E1, p1) ∼ (E2, p2) and(E2, p2) ∼ (E3, p3). By the uniformization theorem (4.4.2) Ei = C/Λi and pi = zi + Λi fori = 1, 2, 3. By corollary 4.1.2.2, there exist α1 and α2 such that the isomorphisms are givenby ϕi : z + Λi 7→ αiz + Λi+1 with αiΛi = Λi+1 for i = 1, 2. Since m2m1Λ1 = m2Λ2 = Λ3

the map ϕ2 ϕ1 : z + Λ1 7→ m2m1z + Λ3 is an isomorphism and ϕ2 ϕ1(p1) = m2m1p1 =m2m1z1 + Λ3 = m2z2 + Λ3 = z3 + Λ3 = p3.

S0(N), S1(N) and S(N) are moduli spaces of complex elliptic curves and N -torsion data.When N = 1, S0(1) ∼= S1(1) ∼= S(1) ∼= Complex elliptic curves/ ∼.

Definition 5.1.2. Let Γ be any congruence subgroup of SL2(Z) acting on H from the left.The modular curve Y (Γ) is defined as the quotient space of orbits under Γ:

Y (Γ) := Γ\H = Γτ : τ ∈ H.

The modular curves for Γ0(N),Γ1(N) and Γ(N) are denoted by Y0(N), Y1(N) and Y (N)respectively.

Theorem 5.1.1. Let N ∈ N. And denote Eτ := C/Λτ .

(i) S0(N) = [Eτ , 〈1/N + Λτ 〉] : τ ∈ H, and the map ψ0 : S0(N) → Y0(N) given by[Eτ , 〈1/N + Λτ 〉] 7→ Γo(N)τ is a bijection.

41

(ii) S1(N) = [Eτ , 1/N + Λτ ] : τ ∈ H, and the map ψ1 : S1(N) → Y1(N) given by[Eτ , 1/N + Λτ ] 7→ Γ1(N)τ is a bijection.

(iii) S(N) = [Eτ , (τ/N + Λτ , 1/N + Λτ )] : τ ∈ H, and the map ψ0 : S0(N)→ Y0(N) givenby [Eτ , (τ/N + Λτ , 1/N + Λτ )] 7→ Γ(N)τ is a bijection.

We give a proof for (ii):

Proof: Take a point [E, p] ∈ S1(N). By the uniformization theorem (4.4.2) and example4.1.3, there exists some τ ′ ∈ H and c, d ∈ Z such that E = C/Λτ ′ and p = (cτ ′+ d)/N + Λτ ′ .Since p has order N , m.c.d.(c, d,N) = 1, so there exist a, b, k ∈ Z such that ad − bc −kN = 1. Therefore, the matrix γ = [a, b ; c, d] reduces modulo N into SL2(Z/NZ). SinceredN : SL2(Z) → SL2(Z/NZ) is surjective, we may take γ ∈ SL2(Z). Let τ = γ · τ ′ andα = cτ ′ + d. Then:

αΛτ = α(τZ⊕ Z) = ατZ⊕ αZ = (aτ ′ + b)Z⊕ (cτ ′ + d)Z = τ ′Z⊕ Z = Λ′τ ,

where the penultimate equality follows from lemma 4.1.1. By corollary 4.1.2.2, there is ananalytic group isomorphism ϕ : Eτ → E given by z + Λτ → αz + Λ′τ . Since ϕ(1/N + Λτ ) =(cτ ′ + d)/N + Λ′τ = p, we have that [E, p] = [Eτ , 1/N + Λτ ].

To show that ψ1 is a bijection, suppose that Γ1(N)τ = Γ1(N)τ ′, for some τ, τ ′ ∈ H. Thenτ = γ · τ ′ for some γ = [a, b ; c, d] ∈ Γ1(N). Once again, defining α = cτ ′ + d we have

αΛτ = Λ′τ , and α

(1

N+ Λτ

)=cτ ′ + d

N+ Λ′τ .

Since (c, d) ≡ (0, 1) (mod N), c = kN and d = 1 + eN for some k, e ∈ Z, so that

cτ ′ + d

N+ Λ′τ =

1

N+ kτ ′ + e+ Λ′τ =

1

N+ Λ′τ .

Therefore, [Eτ , 1/N + Λτ ] = [E ′τ , 1/N + Λ′τ ] and ψ1 is one to one. Since it is surjective bydefinition, we have the result.

5.2 The topology of Modular Curves

Given a congruence subgroup Γ ≤ SL2(Z), the corresponding modular curve has been definedas the set of orbits Y (Γ) = Γ\H. The projection map π : H→ Y (Γ) which maps each pointτ to its corresponding orbit Γτ is surjective. Therefore, the natural topology over Y (Γ) isthe quotient topology. That is, we declare a subset U ⊆ Y (Γ) open if π−1(U) is open in H.Since H is connected, Y (Γ) is connected.

Lemma 5.2.1. π : H→ Y (Γ) is an open map.

Proof: Let U ∈ H be an open set. π(U) is open if and only if π−1(π(U)) is open in H.Note that Tγ : τ 7→ γ · τ is a continuous map with continuous inverse Tγ−1 . So Tγ is ahomeomorphism for every γ ∈ Γ. Since

π−1(π(U)) =⋃γ∈Γ

γU =⋃γ∈Γ

Tγ(U),

we have the result.

42

Proposition 5.2.1. Let U1, U2 be open subsets of H. Then

π(U1) ∩ π(U2) = ∅ inY (Γ) ⇐⇒ Γ(U1) ∩ U2 = ∅ inH.

Proof: The proof follows by definition:

π(U1) ∩ π(U2) 6= ∅ ⇐⇒ ∃ τ1 ∈ U1, τ2 ∈ U2 : π(τ1) = π(τ2)

⇐⇒ ∃ τ1 ∈ U1, τ2 ∈ U2, γ ∈ Γ : τ1 = γτ2 ⇐⇒ Γ(U1) ∩ U2 6= ∅.

Proposition 5.2.2. Let τ1, τ2 ∈ H be given. There exist neighborhoods U1, U2 of τ1, τ2

respectively with the following property:

∀ γ ∈ SL2(Z) : γ(U1) ∩ U2 6= ∅ =⇒ γ · τ1 = τ2.

Proof: [19, Chapter 2, Prop. 2.1.1]

Corollary 5.2.1.1. The modular curve Y (Γ) is Hausdorff for any congruence subgroupΓ ≤ SL2(Z).

Proof: Let π(τ1), π(τ2) be distinct points in Y (Γ). Take U1, U2 neighborhoods of τ1, τ2

respectively as in proposition 5.2.2. Since γ · τ1 6= τ2 for all γ ∈ Γ, the proposition impliesthat Γ(U1) ∩ U2 = ∅. Therefore, we have by proposition 5.2.1 that π(U1) ∩ π(U2) = ∅. Sinceπ is open (lemma 5.2.1), Y (Γ) is Hausdorff.

Until now, we know that Y (Γ) is a connected Hausdorff space. Never the less, Y (Γ) is notcompact. We would like modular spaces to be compact Riemann surfaces. One way to solvethe problem is to compactify H, since quotient spaces of compact spaces are compact.

Theorem 5.2.2. Let Y be a topological space. Then Y is locally compact Hausdorff if andonly if there exists a space X satisfying the following conditions:

(i) Y is a subspace of X.

(ii) The set X − Y consists of a single point.

(iii) X is a compact Hausdorff space.

If X and X ′ are spaces satisfying these conditions, then there a homeomorphism of X withX ′ that equals the identity on Y .

43

Proof: [17] chapter 3, section 29, theorem 29.1.

So if we take the one point compactification H = H ∪ ∞ of H, we need to extend theaction of Γ to include the point ∞. Intuitively, thinking of ∞ as limy→0

iy

for y ∈ R:[a bc d

]· z =∞ ⇐⇒ cz + d = 0 ⇐⇒ z = −d/c ∈ Q,

limy→0

[a bc d

]· iy

= limy→0

(ai/y + b

ci/y + d

)= lim

y→0

(ai+ yb

ci+ dy

)=a

c∈ Q.

Therefore, to extend the action to H we need to include the rational numbers as well. DefineH∗ := H ∪ ∞ ∪ Q. To show that Γ acts on ∞ ∪ Q, identify this set with P1(Q)(m/n 7→ [m : n] and ∞ 7→ [1 : 0]) and define,

γ · [m : n] := [am+ bn : cm+ dn], γ = [a, b ; c, d] ∈ Γ.

it is easy to see that it is an action and that

γ · −d/c = γ · [−d : c] = [1 : 0] =∞,γ · ∞ = γ · [1 : 0] = [a : c] = a/c.

Definition 5.2.1. Let Γ be any congruence subgroup of SL2(Z) acting on H∗ from the left.The modular curve X(Γ) is defined as the quotient space of orbits under Γ:

X(Γ) := Y (Γ) ∪ Γ\(Q ∪ ∞).

The modular curves for Γ0(N),Γ1(N) and Γ(N) are denoted by X0(N), X1(N) and X(N)respectively. The points Γr ∈ Γ\(Q ∪ ∞) are called the cusps of X(Γ).

The natural topology of H∗ consisting on the open sets of H and sets of the form H∗ − Cwhere C is some closed ball in H, contains too many points of Q∪∞ to make the quotientX(Γ) Hausdorff. We define the topology on X(Γ) as follows:

For every M > 0, define the set neighborhoods HM := τ ∈ H : Im(τ) > M, and foreach γ ∈ SL2(Z) consider the set γ(HM ∪ ∞) with γ = [a, b ; c, d]. Then we have that 0 <Im(γ ·τ) = Im(τ)/|cτ+d|2 < M . Also, since γ ·∞ = a/c and fractional linear transformationsare conformal, γ(HM ∪ ∞) is a disc tangent to the real axis at a/c containing this point(see figure 5.1).

The topology on H∗ is the one generated by open sets in H and the sets γ(HM ∪ ∞), forall γ ∈ SL2(Z). Finally, we give X(Γ) the quotient topology by the extended projectionπ : H∗ → X(N).

Proposition 5.2.3. The modular curve X(Γ) is Hausdorff, connected and compact.

Proof: We already know this for Y (Γ) ⊆ X(Γ). For the rest of the proof look at [19,Chapter 2, Prop. 2.4.2].

It turns out that the topology just defined endows X(Γ) with the structure of a compactRiemann surface. In fact, the bijection X(1)→ P 1(C) given by the j function is an isomor-phism of Riemann surfaces. Therefore, theorem 2.5.7 implies that the curves X(Γ) are infact projective non-singular algebraic curves.

44

iM

R

iR

Figure 5.1: HM and some neighborhoods of points in Q ∪ ∞.

5.3 A rational elliptic curve with a point of order 11

So what comes next? Until now, we have seen that the modular curves X(Γ) are compactRiemann surfaces for every congruence subgroup Γ ≤ SL2(Z). By theorem 2.5.7, the modularcurves X(Γ) are complex algebraic curves. One would like to know, for example, when arethese modular curves defined over the rational numbers and if this gives any additionalinformation about the elliptic curves they parametrize.

One of the most important results concerning modular spaces of elliptic curves is the mod-ularity theorem. The complex version (the one we have the tools to understand) states thatevery elliptic curve defined over C with rational j-invariant is modular:

Theorem 5.3.1 (Modularity Theorem, version XC). Let E be a complex elliptic curve withj(E) ∈ Q. Then, for some integer N there exists a surjective analytic function of compactRiemann surfaces from the modular curve X0(N) to the elliptic curve E.

To illustrate the kind of things we can do with the theory we have developed, we show thatthere exist an elliptic curve defined over Q with a point of order 11. This seems to be avery difficult problem to solve in terms of explicit polynomials. It turns out that there arebeautiful solutions in terms of modular curves.

Proposition 5.3.1. The modular curve X0(11) is an elliptic curve defined over Q.

First, we show that the curve X0(11) is defined over Q. Consider X0(11) as a complexalgebraic curve X0(11)C and let L = C(X0(11)C) be its function field. If we show that thereis a field extension F/Q of transcendence degree 1 such that L = FC, we may define X0(11)Qas the algebraic curve corresponding to this field extension.

Consider the elliptic curve

E : Y 2 = 4X3 − 27t

t− 1728X − 27

t− 1728,

45

defined over the field k = Q(t), for some indeterminate variable t. Similar to he j-invariantof the curve right after proposition 3.1.3, this curve has the property that j(E) = t. LetE[11] be 11-torsion subgroup of E. Just as in the complex case, E[11] ∼= (Z/11Z)× (Z/11Z)as groups, so we may think of it as the two dimensional vector space over F11. Let p, q ∈ Ebe a basis for E[11] and consider the field extension k(p, q) of k. By k(p, q) we actually meanthe extension of k obtained by adjoining fixed coordinates of p, q ∈ P2(k).

It turns out that the torsion elements of a curve are always algebraic over k, so the extensionk(p, q)/k is algebraic. It is a non trivial fact that the extension k(p, q)/Q is Galois.

Gal(k(p, q)/q) acts trivially on E[11] since it acts on its generators. So there is a naturalmap Gal(k(p, q)/Q) → Aut(E[11]), which sends each Galois automorphism σ to the auto-morphism induced by the action of σ on the generators. Since Aut(E[11]) ∼= GL2(F11), wemay identify the image of σ with a matrix with columns σ(p) and σ(q). It is clear that thismap is injective. A much deeper result is that in fact it is a group isomorphism.

Let G ≤ GL2(F11) be the subgroup that maps the cyclic subgroup generated by q to itself.Then

G =

[a 0c b

]: a, b ∈ F11, a 6= 0

.

Via the isomorphism Gal(k(p, q)/Q) ∼= Aut(E[11]), G corresponds to a subgroup of theGalois group Gal(k(p, q)/Q. Define F to be the fixed field of this group in k(p, q). It turnsout that FC = L. Furthermore, X0(11) is of genus 1, so it is an elliptic curve defined overQ. In fact, one can calculate the Weierstrass equation for it using the Fourier expansion ofsome modular forms in Γ0(11):

X0(11)Q : Y 2 + Y = X3 −X2 − 10X − 20

Using sage, one easily calculates the rank and torsion for this elliptic curve using the com-mands:

E = EllipticCurve([0,-1,1,-10,-20])

E.rank()

E.torsion subgroup()

It turns out that X0(11) has rank zero and torsion subgroup the group of order 5 withgenerator (5, 5), therefore:

X0(11)(Q) = O, (5, 5), (16,−61), (16, 60), (5,−6).

The curve X0(11) has two cusps that don’t correspond to points in S0(11). Never the less,there are three rational points in Y0(11) ⊆ X0(11) that correspond to pairs (E,C) where Eis an elliptic curve defined over C and C is a cyclic subgroup of order 11. Therefore, thereexist elliptic curves with points of order 11.

46

Bibliography

[1] Walter Rudin. Principles of mathematical analysis. Vol. 3. McGraw-Hill New York,1964.

[2] Michael Francis Atiyah and Ian Grant Macdonald. Introduction to commutative alge-bra. Vol. 2. Addison-Wesley Reading, 1969.

[3] Barry Mazur. “Modular curves and the Eisenstein ideal”. In: Publications Mathematiquesde l’Institut des Hautes Etudes Scientifiques 47.1 (1977), pp. 33–186.

[4] Barry Mazur and D Goldfeld. “Rational isogenies of prime degree”. In: Inventionesmathematicae 44.2 (1978), pp. 129–162.

[5] Hideyuki Matsumura. Commutative algebra. 1980.

[6] Don Zagier. “Eisenstein series and the Riemann zeta-function”. In: Automorphic forms,representation theory and arithmetic. Springer, 1981, pp. 275–301.

[7] Jerrold B Tunnell. “A classical Diophantine problem and modular forms of weight3/2”. In: Inventiones mathematicae 72.2 (1983), pp. 323–334.

[8] Walter Rudin. Real and complex analysis (3rd). 1986.

[9] Joseph H. Silverman. The arithmetic of elliptic curves. Graduate texts in mathematics.Springer, 1986.

[10] Jean-Pierre Serre et al. “Sur les representations modulaires de degre 2 de Gal(Q/Q)”.In: Duke Mathematical Journal 54.1 (1987), pp. 179–230.

[11] Gerhard Frey. “Links between solutions of A- B= C and elliptic curves”. In: Numbertheory. Springer, 1989, pp. 31–62.

[12] JH Silvermann. “Advanced topics in the arithmetic of elliptic curves”. In: Graduatetexts in math 151 (1994).

[13] Henri Cartan. Elementary theory of analytic functions of one or several complex vari-ables. Courier Corporation, 1995.

[14] Rick Miranda. Algebraic curves and Riemann surfaces. Vol. 5. American MathematicalSoc., 1995.

[15] David E Rohrlich and George F Rohrlich. “Modular curves, Hecke correspondences,and L-functions”. In: Modular forms and Fermat’s last theorem. Springer, 1997, pp. 41–100.

[16] James S Milne. “Algebraic number theory”. In: (1998).

47

[17] James R Munkres. Topology. Prentice Hall, 2000.

[18] Theodore Gamelin. Complex analysis. Springer Science & Business Media, 2003.

[19] Fred Diamond and Jerry Shurman. A first course in modular forms. Vol. 228. SpringerScience & Business Media, 2006.

[20] Martin Schlichenmaier. An introduction to Riemann surfaces, algebraic curves andmoduli spaces. Springer Science & Business Media, 2010.

[21] Serge Lang. Introduction to algebraic and abelian functions. Vol. 89. Springer Science& Business Media, 2012.

[22] Robin Hartshorne. Algebraic geometry. Vol. 52. Springer Science & Business Media,2013.

[23] Phillip Griffiths and Joseph Harris. Principles of algebraic geometry. John Wiley &Sons, 2014.

[24] Keith Conrad. Selmer’s Example. Expository Papers.

[25] Keith Conrad. The Congruent Number Problem. Expository Papers.

[26] Tom Weston. The modular curves X0(11) and X1(11). Expository Papers.

48