GALOIS REPRESENTATIONS and the MODULARITY THEOREM Dedicated to Dr. Mic Jackson. 1. Introduction We Are Interested in the Roots O

GALOIS REPRESENTATIONS AND THE MODULARITY THEOREM

THOMAS MORRILL

Abstract. We wish to study the algebraic closure of Q, denoted Q and its associated ring of integers Z. Often classical Galois theory is employed to examine the ﬁnite extensions of Q, but this analysis fails to apply to the inﬁnite extension Q. Following a brief review of Galois theory, we introduce the absolute Galois group over Q and examine its group representations in `-adic vector spaces. These Galois representations have fundamental connections to modular forms and elliptic curves.

Dedicated to Dr. Mic Jackson.

1. Introduction We are interested in the roots of polynomials. Restricting to polynomials with coefficients in Q yields a rich theory. The roots of polynomials in Z[x] are called algebraic numbers. It can be shown that the set of all algebraic numbers forms a field containing the set of all rational numbers Q. One method of studying algebraic numbers is to examine the finite degree extensions of Q using Galois Theory. Here we assume the reader is familiar with abstract algebra. Foundational material may be found in [DF04]. Evariste´ Galois developed his theory in the early 1800s as a tool to study the solutions of polynomial equations. Rather than study a field extension K/F directly, Galois examined its group of automorphisms, specifically those that fixed some subfield of K. So long as K is Galois over F , there is a one-to-one correspondence between the subgroups of the automorphism groups of K and fields H that lie between F and K. This fundamental theorem, along with the simplicity of the alternating group A5, was enough to demonstrate that there is no general formula to derive the roots of quintic polynomials using radicals. Algebraic number theory typically begins by applying Galois theory to finite degree extensions of Q. However, because the algebraic numbers are an infinite degree field extension of the rationals, these results do not apply to Q. In Section 2, we begin with a brief overview of algebraic number theory and some intro- ductory results. Although classical Galois theory does not apply to infinite extensions, we may attempt to generalize its methods. We are motivated to study the automorphism group

of Q, which we denote GQ. This group becomes our main object of study. In order to get a handle on the structure of GQ, we introduce the inverse limit of topological groups and rings in Section 3. We assume the reader is familiar with the basics of point- set topology, which may be found in [Mun00]. Inverse limits allow us to decompose an automorphism of Q into a sequence of automorphisms of number fields. Moreover, other objects which arise from inverse limits, such as the `-adic numbers Q`, their analog λ-adic number Kλ, and the Tate module of torsion subgroups have a natural affinity with GQ. With the automorphism group in hand, in Section 4 we compare the algebraic structure of Q to the number fields from Section 2. Often in group theory, the problem of studying an 1 abstract group may be simplified by mapping it into matrix groups, or equivalently viewing its group action on vector spaces. These tools allow us to apply linear algebra to the study of GQ. In Section 5 we introduce modular forms, another large topic in number theory, in order to showcase some results using Galois representations. First, we define the modular group Γ , also known as SL2(Z), and establish its group action on the upper half plane H of C. A modular form f(z) is a holomorphic function whose values are influenced by this group action. We assume the reader is familiar with complex analysis. Additional information may be found in [Ash71]. The space of modular forms is a complex vector space, with a basis given by newforms. The Fourier coefficients of newforms give rise to number fields, and subsequently Galois representations ρf . With this in place we prove a theorem by the French mathematician J.P. Serre regarding congruences, prime density, and the irreducibility of ρf . All figures were made in [S+11]. Elliptic curves are introduced in Section 6, and they also give rise to Galois representations. An elliptic curve is the solution set of a cubic polynomial in two variables, as viewed in projective space. As long as the curve is non-singular, it may be equipped with a geometrically powered group structure. The torsion subgroups of elliptic curves subsequently allow for the influence of GQ. Finally, in Section 7 we close the paper with some statements of the modularity theorem, a deep connection found between elliptic curves and modular forms. In plain language, the theorem states that all elliptic curves exhibit properties inherited from modular forms. It was the partial proof of this theorem in 1993 by Andrew Wiles that yielded the proof of Fermat’s Last Theorem. The proof is too large for the scope of this paper, much less the margin of a textbook, but we give a sketch of how modularity relates to xn + yn = zn. We include it here as Wiles proved his version of modularity as stated in the language of Galois representations.

2. Algebraic Number Theory We are interested in the collection of all roots of polynomials with coefficients over Q, denoted Q, called the algebraic closure of Q. Recall the Fundamental Theorem of Algebra: Theorem 1. If f(x) ∈ C[x], then f has a root in C. A proof may be found in Chapter 14 of [DF04]. By inducting on the number of roots of a polynomial f, we see that Q ⊂ C. A number α ∈ C is called algebraic if there exists an irreducible polynomial f with coefficients in Z such that f(α) = 0. We call f the minimal polynomial for α if f is irreducible and monic. Complex numbers that are not the root of some polynomial over Z are not algebraic. These are called transcendental numbers. Although transcendence theory is a very rich topic in number theory, we will not explore transcendental numbers further in this paper.Any polynomial g ∈ Z[x] that has α as a root must be divisible by f. If α is the root of a irreducible monic (lead coefficient 1) polynomial in Z, then α is called an algebraic integer. The set of√ algebraic integers is denoted by Z. For example, consider the golden ratio φ = (1 + 5)/2. As φ is a root of the irreducible polynomial f(x) = x2 − x − 1, 2 and f is a monic polynomial in Z[x], we see that φ is an algebraic integer. By factoring f(x) in C, √ ! √ ! 1 + 5 1 − 5 x2 − x − 1 = x − x − , 2 2 √ 0 we find a second algebraic integer, (1 − 5)/2, which√ we denote φ for brevity. Note that the only difference between φ and φ0 is the sign of 5; we will revisit this when dealing with automorphisms of quadratic fields. Also note that −φφ0 = 1. That is, both φ and φ0 are units in Z, as they both have a multiplicative inverse. Units will become important when we try to factor algebraic integers. More generally, if α is an algebraic number with minimal polynomial f and β is another root of f, then we say that α and β are algebraic conjugates. For example, the polynomial x2 + 1 is irreducible in Z[x]. Its complex roots are i and −i, which means that i and −i are algebraic conjugates as well as complex conjugates. In fact, complex conjugation is a special case of algebraic conjugation. To see this, let p(x) be an irreducible polynomial in Z and suppose that α is a root of p. Then p(α) = 0 = 0 = p(α) = p(α). That is, the complex conjugate of α is also one of its algebraic conjugates. However, many algebraic numbers have more than one algebraic conjugate. To clarify terminology, we will refer to the distinct roots of the minimal polynomial of α as the conjugates of α and refer to the operation a + bi = a − bi as complex conjugation. In algebraic number theory, we do not study Q directly; rather, we examine its smaller subfields. A number field F is a field Q ⊆ F ⊆ C which is a finite dimensional vector space over Q. The degree of F over Q, written [F : Q], is the dimension of F as a vector space over Q. If α1, . . . , αn ∈ C, the set Q adjoin α1, . . . , αn is given by f(α1, . . . , αn) Q(α1, . . . , αn) := f, g ∈ Q[x1, . . . , xn], g(α1, . . . , αn) 6= 0 . g(α1, . . . , αn)

We see that Q(α1, . . . , αn) is a subﬁeld of C. Likewise, Z adjoin α1, . . . , αn is given by

Z[α1, . . . , αn] := {f(α1, . . . , αn) | f ∈ Z[x1, . . . , xn]} , which forms a subring of C. Note that Q(α) contains multiplicative inverses for all its elements, whiles Z[α] does not. We state the following theorem without proof. Theorem 2. A number α ∈ C is algebraic if and only if Q(α) is a number field. Moreover, if F is a number field, then there exists α ∈ Q such that F = Q(α). A proof may be found in Chapters 1 and 2 of [ST02]. We immediately see that if β is an element of the number field F = Q(α), then Q(β) must be a finite dimensional Q vector space, as it is a subspace of Q(α). Therefore β is algebraic. We now consider our first examples of number fields. Let N be a squarefree rational 2 integer, i.e., not divisible by any n 6= 1 in Z. We√ make√ this distinction√ in order to avoid 2 cataloging√ the√ same number field twice, since if N = n M = n M, then we have that Q(√N) = Q( M). In general, if c 6= 0 is rational, then Q(α) = Q(cα). The number field Q( N) is called a quadratic field. Each quadratic field is classified as real or imaginary, depending on whether N > 0 or N < 0. 3 To clarify notation, we define the sets N and Z+ as N := {n ∈ Z | n ≥ 0} + Z := {n ∈ Z | n ≥ 1} Let N ∈ Z+ and define the Nth root of unity as 2πi ζN := e N . th Then we call the number field Q(ζN ) ⊂ C the N cylcotomic field. (The first and second cyclotomic fields are just Q.) Note that πi ζ4 = e 2 = i. That is, Q(i) is both a quadratic and a cyclotomic field. We turn our attention to field embeddings in preparation for Galois theory. An embedding is an injective field homomorphism. Suppose σ is an embedding σ : Q(α) −→ C. Then we claim σ(α) is a conjugate of α. Note that any such field homomorphism must fix 1 and therefore all of Q. If f(x) is the minimal polynomial for α, then we find that f(α) = 0 = σ(0) = σ(f(α)) = f(σ(α)). Note that if σ(α) is known, then every value of σ is determined, since every β ∈ Q(α) can be written as a rational function of α. In general, if τ is a field automorphism, then τ(f(x)) = f(τ(x)) for all f ∈ Q(x). We will frequently take advantage of this fact. We see that σ(β) = σ(f(α)) = f(σ(α)),

which is determined by σ(α). Therefore, if α = α1, . . . , αn are the distinct conjugates of α, then there are exactly n embeddings of Q(α) into C, one for each conjugate. If σ is an embedding and σ(F ) ⊆ F , then we find that σ(F ) = F , since fields have no nontrivial ideals. Therefore, σ is an automorphism of F . Conversely, as F ⊂ C, every automorphism of F is also an embedding√ of F into C. Note however, that it not always the case that σ(F ) ⊆ F . For example, 3 2 is algebraic as it is a root of the polynomial √ √ √ 3 3 3 3 2 f(x) = x − 2 = x − 2 x − 2ζ3 x − 2ζ3 . √ 3 Consider the number field Q( 2). Then the embedding defined by √ 3 σ : Q( 2) −→ C √ √ 3 3 2 7−→ 2ζ3 √ 3 contains complex numbers, where Q( 2) does not. We say that a number field F is Galois over Q if the number of distinct automorphisms τ : F → F is equal to [F : Q]. From the previous argument, this is equivalent to the condition σ(F ) ⊆ F for all embeddings σ : F → C. If F is Galois over Q then we define the Galois group of F over Q as the group of automorphisms which fix Q pointwise, with group operation given by function composition. We denote this group as Gal(F/Q). In general, the Galois group of K over F is the group of automorphisms of K that fix F pointwise. 4 In the special case of a field extension over Q, we have already seen that every embedding fixes Q pointwise. We therefore abbreviate the Galois group of F over Q as Gal(F ). Likewise, if F is a number field that is Galois over Q we say that F is a Galois number field. While Galois theory is a good tool for studying finite extensions of Q, it will not apply to Q, which is an infinite dimensional vector space over Q. However, the study of finite extensions of Q and their associated Galois groups will serve as an analog to the study of Q and its automorphisms, the absolute Galois group. If H is a subgroup of Gal(F/K), then we define the fixed field of H Fix(H) to be the collection of all α ∈ F with σ(α) = α for all σ ∈ H. We state in part the Fundamental Theorem of Galois Theory. Theorem 3. Let K be a Galois field extension of F . There is a bijection between fields F ⊆ L ⊆ K and subgroups of Gal(K/F ). A proof may be found in Chapter 14 of [DF04]. If F and K are Galois number fields, then we define the composite of F and K to be the intersection of all subfields of C containing both F and K. field is denoted as FK. It is demonstrated in Chapter 14 of [DF04] that if both F and K are number fields, then FK is√ also a Galois number field. √ √ Let N be a squarefree integer and F = Q( N). Then the conjugates of N are ± N, which both lie in F . Therefore, there are two embeddings of F into C, the trivial embedding and the map σ given by σ : F −→ C √ √ N 7−→ − N which satisfies σ(F ) ⊆ F . We conclude that every quadratic number field is Galois over Q. As an example, let N = 5. We see that √ ! √ √ √ 1 + 5 σ(1 + 5) σ(1) + σ( 5) 1 − 5 σ = = = . 2 σ(2) σ(2) 2

If N = −1, let σ be the automorphism of Q(i) given by σ(i) = −i. Then we have σ(a + bi) = σ(a) + σ(b)σ(i) = a − bi √ ∼ for a and b ∈ Q. Regardless of choice of N, we find that Gal( N) = Z/2Z. The only normal subgroups of Z/2Z are the trivial subgroup and itself. Therefore the only subfields of F are F and Q. th 2πi Recall that the N cyclotomic field is defined as Q(ζN , where ζN = e N . The conjugates of ζN are roots of the polynomial xN − 1. By the Fundamental Theorem of Algebra, this polynomial has N roots, which we parame- terize as k {ζN | 1 ≤ k ≤ N} k For the above values of k, if gcd(k, N) > 1, then ζN is also a root of N x k − 1, 2πi but ζN = e N is not. We conclude that ζN has exactly ϕ(N) conjugates, where ϕ is the k Euler totient function. each of which is a power of ζN . Each conjugate ζN is a power of ζN , and therefore is an element of Q(ζN ) We conclude that Q(ζN ), is a Galois number field. 5 Let σk be the automorphism defined by

σk : Z(ζN ) −→ Q(ζN ) k ζN 7−→ ζN

Then Gal(Q(ζN )) = {σk | k ∈ I}. For all σj, σk ∈ Gal(Q(ζN ) we have that jk σj(σk(ζN )) = ζN . N Since ζN = 1, for any m ∈ Z, we have that jk ij+mN ζN = ζN .

It follows that σj ◦σk = σik, where we reduce ij modulo N. There is a group homomorphism × ψ from Gal(Q(ζN ) to the multiplicative group (Z/NZ) given by × ψ : Gal(Q(ζN ) −→ (Z/NZ)

σk 7−→ k (mod N) with kernel given by

ker(ψ) = {σk | k ≡ 1 (mod N)}

= {σ1}. × Because |Gal(Q(ζN )| = φ(N) = |(Z/NZ) |, we conclude that ψ is an isomorphism and ∼ × Gal(Q(ζN ) = (Z/NZ) . We return our attention to the algebraic integers Z. We state the following lemma: Lemma 1. For all α and β ∈ Z, both α + β and αβ ∈ Z. A proof may be found in Chapter 2 of [ST02]. If F is a number field, we define the ring of integers of F , denoted OF , as F ∩Z. In this situation, it is necessary to distinguish between the usual integers Z and members of OF . We use the word rational to make this distinction. For example, a rational prime p is any irreducible element of Z. Note that Q ∩ Z = Z. In plain English, “rational integers are rational integers.” The containment Z ⊂ Q ∩ Z is easy. A proof of the reverse containment may be found in Chapter 2 of [Mar91]. In general, if α1, . . . , αn are algebraic integers and F = Q(α1, . . . , αn) is a number field,√ its ring of integers OF is not√ always given by Z√[α1, . . . , αn]. Consider the field F = Q( 5). 2 The√ minimal polynomial for 5 is x − 5, so 5√ is an algebraic integer. Clearly the ring Z[ 5] ⊂ OF . However, the golden ratio√ φ = (1 + 5)/2 ∈ F . We have already seen that φ is an algebraic integer. But φ∈ / Z[ 5]. Fortunately we do not need to extend√ very far to capture all of OF . In general, if N is a squarefree rational integer and F = Q( N), then √  [ N]: N ≡ 2, 3 (mod 4)  Z O = F h √ i  1+ N Z 2 : N ≡ 1 (mod 4) Note that N ≡ 0 modulo 4 is excluded by assumption. A proof of this result can be found in Chapter 2 of [Mar91]. Just as in Z, we study OF by examining how numbers factor. In order to discuss factorization, we need to make a distinction between prime numbers and irreducible numbers in OF . Recall that a unit of a ring R is any element that has a multiplicative inverse. Let x be a nonzero, non-unit element of a ring R. We say x is 6 irreducible if whenever x = αβ, one of α or β must be a unit. We say x is prime if whenever x divides a product αβ, x divides one of α or β. √ When factoring numbers, we must be careful of units. Recall the golden ratio φ = (1 + 2/2) is an algebraic integer. In the ring Z[φ], we can factor the number 2 as ! √ ! √ 1 + 5 2 = − 1 + 5 . 2 √ The second factor, (1 − 5)/2, is the conjugate of φ, which we denoted previously as φ0. Both φ and φ0 are units in Z[φ]. This factorization of 2 is actually √ 2 = −(1 + 5)(φ0) = −2φ(φ0) = 2. In general, if α, β and µ are ring elements such that α = βµ and µ is a unit, then we say that α and β are associates. Recall that a unique factorization domain is an integral domain R in which every 0 6= r ∈ R can be written as a product of irreducible elements

r = p1 . . . pn and if there exists a second factorization of r into irreducibles

r = q1 . . . qm

then m = n and the factors pi and qi may be reindexed such that for all 1 ≤ i ≤ n we have that pi and qi are associates. Consider the quadratic field Q(i) = {x + iy ∈ C | x, y, ∈ Q}. From the previous result, its ring of integers is given by Z[i] = {x + iy ∈ C | x, y ∈ Z}. This ring is also known as the Gaussian integers. As is the case with Z, the Gaussian integers can be equipped with a Euclidean algorithm. For any two Gaussian integers a, d, with d 6= 0 it is possible to write a = qd + r, where 0 ≤ |r| < |d|. It is a fact that any integral domain which admits a Euclidean algorithm is a unique factorization domain. For proof and further discussion of Euclidean domains, see Chapter 8 of [DF04]. We concude that Z[i] is a unique factorization domain. In general, unique factorization is not guaranteed for rings of integers. Many imaginary quadratic fields have rings√ of integers that are not unique factorization√ domains. To give one example, consider Q[ −5]. Its ring of integers is given by Z[ −5]. Here√ we find√ that the number 6 has both its usual factorization, 2 · 3, and a new one: (1 + −5)(1 − −5).√ By checking the moduli√ of these complex√ numbers, we find there are no non-units in Z[ −5] that divide√ 2, 3, (1+ −5), or (1− −5). We claim that each of these numbers is irreducible in Q[ −5]. To prove this, we must define the norm of an algebraic number. If α is any element of the number field F , then we define the norm of α as n Y NF (α) = σi(α) i=1

where σ1, . . . , σn are the distinct embeddings of F into C. Because each σi is an embedding, we immediately see that if α and β ∈ F are conjugates, then NF (α) = NF (β). More generally, for any α and β ∈ F we have that

NF (αβ) = NF (α)NF (β). 7 If α is an algebraic integer, then NF (α) is a rational integer. This is proven in Chapter 2 of [ST02]. If α divides β in OF then it follows that NF (α) divides NF (β) in Z. Let µ be a unit. Since the norm function is multiplicative, n −1 Y NF (µ)NF (µ ) = NF (1) = σi(1) = 1. i=1

Therefore, NF (µ) = ±1 for any unit µ. Moreover, if α and β are associates, then we have

NF (α) = NF (µ)NF (β) = ±NF (β). √ √ √ Returning to Q( −5), if a + b −5 is in OF , the norm of a + b −5 is given by √ √ √ 2 2 NF a + b −5 = a + b −5 a − b −5 = a + 5b . √ √ Then we see that NF (2) =√ 4,NF (3) = 9 and NF (1± −5) = 6. Suppose α = a+b −5 ∈ OF 2 2 divides 2. Then NF (a+b −5)√ must divide 4. That is, a +5√ b must be one of ±1 or ±2. We ﬁnd that b = 0 and NF (a + b −5) = 1. Therefore, a + b −5 is a unit and 2 is irreducible. A similar argument√ holds for 3. √ If α = a + b −5 divides 1 + −5, then either α is a unit or NF (α) = 6. There are only four such α, given by √ √ α = ±1 ± 5, α = ±1 ∓ 5. √ √ Note that −1 − −5 is an associate to 1 + −5. Furthermore, √ √ √ √ 1 + −5 1 + −5 1 + −5 −2 + −5 √ √ = √ √ = ∈/ Z[ −5]. 1 − −5 1 − −5 1 + −5 3 √ √ Because NF (2) 6= NF (1 ± −5), we conclude that Q( −5) is not a unique factorization domain. Where unique factorization fails, we can still get results by considering ideals rather than lone integers. For α1, . . . , αn ∈ OF , we denote the ideal generated by α1, . . . , αn as ( n ) X (α1, . . . , αm) := αiβi | βi ∈ OF . i=1 If A and B are ideals of a ring R, then the product of A and B is deﬁned by ( n )

X AB := aibi ai ∈ A, bi ∈ B . i=1 Recall that an ideal A 6= R is called maximal if for any ideal B with A ⊂ B ⊂ R either A = B or B = R. That is, the only ideal that contains A is the trivial ideal R.A prime ideal satisﬁes the condition αβ ∈ A ⇒ α ∈ A or β ∈ A. The following result guarantees the existence of maximal ideals. Lemma 2. If F is a number ﬁeld and

I0 ⊆ I1 ⊆ ...

is an ascending chain of ideals in OF , then there exists N ∈ N such that for all n < N we have that In = IN . 8 A proof is given in Chapter 4 of [ST02]. Therefore, if A0 is any nontrivial ideal of OF , we can ﬁnd a maximal ideal M containing A0 as follows. If A0 is not maximal, then there exists an ideal A1 ⊃ A0. If A1 is maximal, we stop. Otherwise, there exists an ideal A2 ⊃ A1 ⊃ A0, and so on. By the previous lemma, this process must eventually terminate.

Theorem 4. Every nontrivial ideal of OF may be written as a product of maximal ideals, uniquely up to the order of the factors. √ A proof may be found in Chapter 5 of [ST02]. Returning to the example in Z[ −5], we have factored the ideal (6) as √ √ (6) = (2)(3) = (1 + −5)(1 − −5). Therefore none of these ideals are maximal. We turn to non-principal ideals. The ideals (2) and (3) factor as √ (2) = (2, 1 + −5)2 √ √ (3) = (3, 1 + −5)(3, 1 − −5) √ √ √ (1 + −5) = (2, 1 + −5)(3, 1 + −5) √ √ √ (1 − −5) = (2, 1 + −5)(3, 1 − −5) which is shown in more detail in Chapter 3 of [Mar91]. Thus we have found the unique factorization of the ideal (6). It is possible to extend the norm function to encompass ideals as well as individual numbers. If A is an ideal of OF , we deﬁne the norm of an ideal A by

N(A) := |OF /A|. Lemma 3. Let R be an integral domain and let A be an ideal of R. Then A is prime if and only if R/A is an integral domain. Moreover, A is maximal if and only if R/A is a field. A proof may be found in Chapter 5 of [ST02]. Let p be a rational prime and F be a Galois number field. Suppose p be a maximal ideal containing the ideal (p). Because p is maximal, OF /p is a field. Moreover, repeated addition of 1 + p in OF /p eventually reaches 0, since p X (1 + p) = p + p = p i=1

We find that {n + p | n = 1, . . . , p} is a subfield of OF /p with exactly p elements. Therefore, OF /p is a field extension of Fp. We may construct a homomorphism φ from Gal(F ) to Aut(OF /p) by

φ : Gal(F ) −→ Aut(OF /p) σ 7−→ σˆ where σˆ(α + p) := σ(α) + p. p Because OF /p has characteristic p, we claim that the Frobenius transformation τ(x) = x is an automorphism of OF /p. With regard to multiplication, we find that ((α + p)(β + p))p = (α + p)p(β + p)p. 9 For addition, since OF /p has characteristic p, ((α + p) + (β + p))p = (α + p)p + (β + p)p, which comes easily from expanding the binomial coefficient. Lastly, if (α + p)p = (β + p)p, we see that ((α + p)p − (β + p))p = 0, which implies that (α + p) = (β + p). We define a Frobnenius element of Gal(F ) as any automorphism F robp ∈ Gal(F ) satisfying p F robp(x) ≡ x (mod p)

for all x ∈ OF . In other words, each Frobenius element is a preimage of the Frobenius automorphism of OF /p. In order to motivate the following constructions, we close the section with an example of an extension field of Q, consisting entirely of algebraic numbers, that is not a number field. Fix a prime number `. We now define ∞ [ Q(ζ`∞ ) := Q(ζ`n ). n=1

We claim that Q(ζ`∞ ) is a subfield of Q with infinite degree over Q. Let α, β ∈ Q(ζ`∞ ). Then both α and β are finite sums and products of elements α1, . . . αr and β1, . . . , βr from some Q(ζ`n ). Therefore there exists N such that Q(ζ`N ) contains all of the αi and βj. It follows that, α, β ∈ Q(ζ`N ), which shows that they are algebraic numbers. Since Q(ζ`N ) is a field, α ∞ α + β, α − β, αβ (and and β if β 6= 0) are defined as elements of Q(ζ`N ) ⊂ Q(ζ` ). To show that Q(ζ`∞ ) is an infinite degree extension of Q, it is suffient to show that Q(ζ`∞ ) is a vector space containing subspaces of arbitrarily high dimension over Q. Fix N ∈ N. M N Then there exists M ∈ N such that ϕ(` ) > N. Recall that ϕ(` ) is the dimension of Q(ζN ) over Q. Because Q(ζN ) is a subspace of Q(ζ`∞ ), we conclude that Q(ζ`∞ ) is not a number field.

3. Inverse Limits The inverse limit is a powerful structure that can be deﬁned on families of either groups or rings. We will use inverse limits to impose a topological structure on the automorphism group of Q in Section 4 and relate it to other inverse limits.

Deﬁnition 1. Let (I, ≤) be a nonempty set with a partial order. Let {Ri | i ∈ I} be a collection of rings (or groups) and {µij : Rj → Ri |i ≤ j ∈ I} be a collection of surjective ring (group) homomorphisms with the following two properties:

(1) µij ◦ µjk = µik for all i ≤ j ≤ k. (2) µii is the identity map for all i. Q The inverse limit of {Ri} is the subset of the direct product i∈I Ri consisting of all tuples (ri)i∈I that satisfy µij(rj) = ri for all i ≤ j. We denote the inverse limit of {Ri} as lim R . ←− i i∈I 10 The inverse limit is a ring (or group) with addition and multiplication deﬁned compo- nentwise. Proof and additional information about inverse limits is presented as exercises in Chapter 7 of [DF04]. Before we give an important example, it is necessary to employ some point-set topology. We assume the reader has a working knowledge of the basics: basis of a topology, subspaces, product spaces, and continuity, which may be found in [Mun00]. Let R = lim R , ←− i i∈I

and suppose that each Ri is a topological ring (i.e., addition and multiplication are continuous) and that the µij are continuous. By definition, R is a subspace of the infinite product space Y X := Ri, i∈I which we equip with the product topology. The Krull topology on R is defined as the subspace topology R inherits from X. While constructing inverse limits, we will be dealing exclusively with finite groups and rings, which we equip with the discrete topology. That is, all subsets of Ri are open sets. To find a basis element of the Krull topology, fix a finite collection of rings

Ri1 ,...,Rin . In each of these, take a singleton open set

Uij = {xij } ⊆ Ri. This gives us a basic open set in R

U := {(ri)i∈J | rij = xij , 1 ≤ j ≤ n}. Fix a rational prime `. We are now ready to construct an important example of inverse limits, `-adic integers. For every pair i ≤ j, let I = Z+ with the standard order. Let i Ri = Z/` Z for all i. Let µij be the projection map j i µij : Z/` Z −→ Z/` Z a (mod `j) 7−→ a (mod ì). Then the ring of `-adic integers is defined to be := lim /ì . Z` ←− Z Z i∈Z+ We will need the following result:

Lemma 4. If ` is a rational prime, then Z` is an integral domain. This lemma is given as an exercise in Chapter 7 of [DF04]. It is possible to embed Z into Z` as follows. Let φi be the projection map given by i φi : Z −→ Z/` Z n 7−→ n (mod `i). 11 Deﬁne φ by

φ : Z −→ Z`

n 7−→ (φi(n))i∈I .

Each φi is a ring homomorphism. Because the ring operations of Z` are deﬁned component- wise we have that φ is a ring homomorphism. Moreover, φ is injective, since i + ker(φ) = {n ∈ Z | n ≡ 0 (mod ` ) ∀i ∈ Z } = {0}.

This allows us to multiply tuples in Z` with integers by deﬁning

n(ai) := φ(n)(ai).

Recall the definition of the Krull topology. A basic open set U ⊂ Z` is the collection of tuples that agree in a fixed, finite number of components. The rings Z/ìZ used to construct + Z` were ordered with respect to Z . Therefore there is a largest component n that all of these tuples agree on. But once we have fixed the nth component, for any m < n the mth component is also determined:

am = µmn(an) th Let (ai), (bi) ∈ U. For 1 ≤ j ≤ n, the i component of (ai) − (bi) is given by ai − bj, which n must be 0 by the previous argument. For j > n because aj − bj ≡ 0 modulo ` , we conclude n that aj − bj is an integer divisible by ` . Then we have that U is actually the coset n U = (ai) + ` Z`.

We define the `-adic numbers Q` to be the field of fractions of Z`. To extend the Krull topology to Q`, define a basis of open sets B by i + B = {x + ` Z` | i ∈ Z , x ∈ Q`}, where i x + ` Z` := {x + α | α ∈ `Z`}. i Note that ` Z` is not an ideal of Q`, so these open sets are not cosets; we are just borrowing the notation. Without delving much further, we mention that Q` can also be constructed as the metric completion of Q with respect to the `-adic metric:

d`(x, y) := |x − y|`

where the `-adic norm | · |` is given by m `−k :(`, m) = (`, n) = (m, n) = 1 `k := , n ` 0 : m = 0 Frequently we will look for Galois representations in the λ-adic numbers, a ﬁeld extension of the `-adics. Let K be a number ﬁeld (not necessarily Galois over Q) with ring of integers OK . Fix a prime `. Recall that the ideal (`) factors uniquely as a product of maximal ideals

Y eλ `OK = (λ) λ|` 12 where each (λ) is a maximal ideal lying over the ideal (`). Fix such a (λ). We deﬁne the λ-adic integers as O := lim O /(λ)n. K,λ ←− K n

The λ-adic numbers Kλ are then given by the field of fractions of OK,λ. Since (λ) contains (`), equivalence modulo (`)n implies equivalence modulo (λ)n. This is sufficient to see that Z` ⊂ OK,`. Consequently Kλ is a field extension of Q`. If we are working with a field that is already labeled with a subscript, for example Kf , we write the λ-adic numbers as Kf,λ.

4. The Absolute Galois Group Rather than study number fields one at a time, we would like to investigate Q and learn about all number fields in one fell swoop. The success of algebraic number theory motivates us to study the automorphisms of Q. We therefore define (slightly abusively) the absolute Galois group of Q to be the collection of all field automorphisms of Q under the operation of function composition. We denote this group GQ. Our main method of studying the absolute Galois group is to give it an inverse limit structure.

Lemma 5. The group GQ is isomorphic to the inverse limit of the groups Gal(F ), where F ranges over all Galois number ﬁelds. Proof. Consider the family of groups

{Gal(F ) | [F : Q] < ∞,F is Galois over Q} To construct the inverse limit, we must index this set with a partial ordering and ﬁnd associated group homomorphisms. We take our indexing set to be

I = {F | [F : Q] < ∞,F is Galois over Q}, with the subset ordering. That is, F ≤ K if and only if F ⊆ K. As for homomorphisms, suppose F ⊆ K and σ ∈ Gal(K). Let ι : F,→ K be the inclusion map. We claim that the domain restriction of σ

σ|F :F −→ K α 7−→ σ(ι(α))

is an element of Gal(F ). Because σ is an automorphism of K and K ⊂ C, we ﬁnd that σ|F is an embedding of F into C. Since F is Galois, then σ|F (F ) = F , which shows that 0 σ|F ∈ Gal(F ). Suppose F ⊂ K ⊂ L with inclusion maps ι : F,→ K and ι : K,→ L. Then ι0 ◦ ι is the inclusion map from F to L. For all σ ∈ Gal(L), we have that 0 0 (σ|K )|F = σ|K ◦ ι = σ ◦ ι ◦ ι = σ ◦ (ι ◦ ι) = σ|F . To show that group operations are preserved in the restriction map, recall that σ(F ) = F . Then we have that ι ◦ (σ ◦ ι) = σ ◦ ι. Then for all σ0 ∈ Gal(K) we have that 0 0 0 0 (σ ◦ σ)|F = σ ◦ σ ◦ ι = σ ◦ ι ◦ σ ◦ ι = σ |F ◦ σ|F .

It is not diﬃcult to show that (idK )|F = idF . We conclude that |F is a homomorphism between Gal(K) and Gal(F ). Note that for σ ∈ Gal(K), we have that σ|K = σ. Then for 13 any two Galois number ﬁelds F ⊆ K the map

µF,K :Gal(K) → Gal(F )

σ → σ|F has the properties required by inverse limits. This is suﬃcient to show that the group G = lim Gal(F ) ←− F ∈I ∼ exists. To show that G = GQ, let σ ∈ GQ. From the previous argument, σ|F ∈ Gal(F ) for any F ∈ I. Moreover, if F ⊆ K ∈ I, we have that (σ|K )|F = σF . To map σ to G, we deﬁne

φ(σ) := (σ|F )F ∈I . From our results on domain restriction, we see that φ is a homomorphism. Moreover, because 1G = (idF )F ∈I , any field automorphism in the kernel of this homomorphism must fix every algebraic number pointwise. Therefore, ker(φ) = {id } and φ is injective. GQ To show that φ is surjective, let (σF )F ∈I ∈ G. We claim that (σF )F ∈I defines a field automorphism from Q to itself. Let α ∈ Q and K ∈ I containing α. We define

[(σF )F ∈I ](α) := σK (α). To show that this is well defined, suppose F is another Galois number field containing α. Recall that FK is a Galois number field containing both F and K. From the definition of G, we have that σF (α) = σFK (α) = σK (α).

Because each component σF is a ﬁeld automorphism, then (σF )F ∈I is bijective. For all α and β we have that

[(σF )F ∈I ](α + β) = [(σF )F ∈I ](α) + [(σF )F ∈I ](β) and [(σF )F ∈I ](αβ) = [(σF )F ∈I ](α)[(σF )F ∈I ](β).

Therefore, (σF )F ∈I deﬁnes an element σ in G . We ﬁnd that φ(σ) = (σF )F ∈I and conclude ∼ Q that GQ = G.

The benefit of this construction is that we may now impose the Krull topology on GQ. The open and closed sets of this topology carry information about the subfields of Q. Recall our overview of the Krull topology on Z`. Fix a finite number of Galois number fields F1,...,Fn. Each has an associated finite Galois group Gal(Fi). In each, take the open singleton set {σi}, where σi ∈ Gal(Fi). This gives us a basic open set in the Krull topology on GQ,

U = {σ ∈ GQ | σ|Fi = σi, 1 ≤ i ≤ n}.

If each σi is the identity automorphism on Fi, the U is the collection of all automorphisms σ that ﬁx Fi pointwise. Let F be a Galois number ﬁeld. Then by using the argument above, we see that the mapping φ

φ : GQ −→ Gal(F )

σ 7−→ σ|F is a surjective homomorphism. Then the kernel, ker(σ) = τ ∈ GQ | σ|F = 1Gal(F ) , 14 is an open normal subgroup of GQ. Recall, when looking at finite extensions of Q, we defined Frobenius elements. We wish to do the same with the infinite extension Q. Let p be a rational prime and let p be any maximal ideal in Z containing p. Just as we saw in Section 2, the field Z/p is a field extension of Fp. The decomposition group of p is given by

Dp = {σ ∈ GQ : σ(p) = p}. For all σ ∈ Dp, we see that σ is a preimage of the Frobenius automorphism on Fp. An absolute Frobenius element over p is any of these preimages, which we denote F robp. If we map D → G , then Frob is only deﬁned up to the kernel of this mapping, as it is a p Fp p preimage. We call the kernel of this map the inertia group of p, σ Ip = {σ ∈ Dp : x ≡ x (mod p) for all x ∈ Z}.

Absolute Frobenius elements are essential to characterizing GQ: Theorem 5. Let P be a ﬁnite set of rational primes. For each maximal ideal p lying over a rational prime p∈ / P, choose an absolute Frobenius element Frobp. The set of such elements is dense in GQ

A proof may be found in Chapter 9 of [DS05]. Even with the Krull topology, GQ is a large and unwieldy group. Any given automorphism σ permutes the elements of infinitely many number fields simultaneously. We use Galois representations to simplify our problems. Definition 2. Let d be a positive integer, let F be a field, and let d ∈ Z+.A d-dimensional Galois representation is a continuous homomorphism

ρ : GQ −→ GLd(F ),

where GLd(F ) is the group of invertible d × d matrices with entries in F . Mapping an abstract group into a matrix group is a common technique in group theory. See [FH91] for examples of representations of ﬁnite groups. Galois representations allow us to apply linear algebra to help analyze GQ. In particular, we can assign automorphisms matrix invariants, such as determinant and trace. This is essential to our results in Section 5.

We require Galois representations to be continuous because the topology on GQ carries information about the subfields of Q that we want to preserve. Recall from linear algebra that any given matrix only defines a linear transformation with respect to an established basis. We can create a new transformation by changing the basis without changing the information carried in the matrix. We say that two Galois representations ρ and ρ0 are equivalent if there exists an invertible matrix m ∈ GL(F ) such that ρ0(σ) = m−1ρ(σ)m 0 for all σ ∈ GQ. We denote equivalent representations by ρ ∼ ρ . We will mostly be interested in Galois representations into vector spaces over Q`. Because Frobp is only defined up to Ip, ρ(Frobp) is only well defined if Ip ⊂ ker ρ. We say that ρ is unramified at p if Ip ⊂ ker ρ for any p containing (p). Otherwise, ρ is ramified at p. Let N ∈ Z+. Let φ be a group homomorphism × × φ :(Z/NZ) −→ C 15 A Dirichlet character modulo N is a function + × χ : Z −→ C ( φ(n) : gcd(n, N) = 1 n 7−→ 0 : otherwise. Because φ is a homomorphism, the only values χ can take are ϕ(N)th roots of unity and 0, as these are the only complex numbers with multiplicative order dividing ϕ(N). The trivial character is given by the constant function χ(n) ≡ 1. Lemma 6. The set of homomorphisms × × φ :(Z/NZ) 7−→ C form a group with multiplication defined by (φψ)(n) = φ(n)ψ(n). This group is isomorphic to (Z/NZ)×. This is presented as an exercise in Chapter 5 of [DF04]. This group is called the dual group of (Z/NZ)× and denoted (Z\/NZ)×. We now construct an example of a Galois representation. Recall that for all N, ∼ × Gal(Q(ζN )) = (Z/NZ) .

Let πN be the projection map

πN : GQ −→ Gal(Q(ζN )

σ 7−→ σ|Q(ζN . Let χ be homomorphism × × χ :(Z/NZ) −→ C and φ be an isomorphism × ψ : Gal(Q(ζN ) −→ (Z/NZ) .

Then we have a group homomorphism ρχ = ψ ◦ φ ◦ πN × ρχ : GQ −→ C = GL1(C).

Because the image of ρχ consists of ﬁnitely many points, to see that ρχ is continuous, we −1 only need to consider the inverse image of a single point. When ρχ (z) is nonempty, we have that −1 −1 ρχ (z) = πN ({σ1, . . . , σn | σi ∈ Gal(Q(ζN )}) n [ = {(σi) ∈ GQ | σi|j = σj} j=1 which is open in GQ. Therefore, ρχ is a Galois representation. 16 For another example, recall from Section 2: ∞ [ Q(ζ`∞ ) = Q(ζ`n ). n=1

Let GQ,` = Aut(Q(ζ`∞ )). In the same way we decomposed GQ, we can write GQ,` as an inverse limit: i G ∼ lim Gal( (ζ i )) ∼ lim /` = . Q,` = ←− Q ` = ←− Z Z Z` i∈Z i∈Z

Thus we can deﬁne the `-adic cyclotomic character of GQ by × χ` : GQ −→ Q`

σ 7−→ (mi)i∈Z

mi where ζ`i = σ(ζ`i ). The following sections demonstrate applications of Galois representations in the studies of modular forms and elliptic curves.

5. Modular Forms In order to discuss modular forms, first we must define the modular group Γ and its group action on the upper half plane of C. For our purposes, Γ is defined to be the group of matrices a b a b a, b, c, d ∈ Z, det = 1 , c d c d under the operation of matrix multiplication. Another name for this group is SL2(Z), the special linear group of degree 2 over Z. Be aware that other texts define the modular group differently. For example, [Leh69] defines the modular group as the quotient space of SL2(Z) −1 0 with the normal subgroup generated by 0 −1 . a b The complex upper half plane is given by H = {z ∈ C | Im(z) > 0}. For γ = c d ∈ Γ and z ∈ H, we define the linear fractional transformation of γ on z by a b az + b z := . c d cz + d Note that −1 0 −z 1 0 z = = z = z, 0 −1 −1 0 1 which motivates the alternative definition of the modular group. There is the possibility for division by 0 that we must be careful about. If cz + d = 0, because z 6= 0, then c and d must both not be zero. If one was zero, then cz + d = 0 implies that the other would be zero, which contradicts det(γ) = 1. We can conclude that z = −d/c, a rational number, which is not in H. Allowing linear fractional transformation on the rationals requires us to formally define the point at infinity, which we write as ∞. For rational numbers x, we define linear fractional transformation as above, so long as cx + d 6= 0. Otherwise, we define γx = ∞. Linear fractional transformation won’t be a group action if we get stuck at ∞, so we also define a b a ∞ := , c d c 17 where again, if c = 0, we take a/0 to mean ∞. For our purposes, it is convenient to think of ∞ as lying infinitely far along the positive i axis above 0. Strictly speaking we are using the one point compactification of C. We will prove the following result. Lemma 7. The group Γ acts on both H and Q ∪ {∞} by linear fractional transformation. Proof. To show that linear fractional transformation is a group action on H, we first show a b that Im(γz) > 0 for all γ ∈ Γ and z ∈ H. Let γ = c d ∈ Γ and z = x + iy with x, y ∈ R and y > 0. Then a b az + b (ax + b) + iay (ax + b) + iay (cx + d) − icy z = = = . c d cz + d (cx + d) + icy (cx + d) + icy (cx + d) − icy Simplifying yields [(ax + b)(cx + d) + acy2] + i[ay(cx + d) − cy(ax + b)] (cx + d)2 + (cy)2 The imaginary part of this complex number is given by ay(cx + d) − cy(ax + b) acxy + ady − acxy − bcy (ad − bc)y = = . (cx + d)2 + (cy)2 (cx + d)2 + (cy)2 (cx + d)2 + (cy)2 We have already shown that cz + d 6= 0. A similar argument holds for cz + d 6= 0. Since the denominator is the sum of squares of real numbers and nonzero, it must be positive. As for the numerator, note that ad − bc is the determinant of γ. By definition of Γ this is just 1. Because y > 0 and the denominator is positive, Im(γz) > 0. We’ve already seen that the identity matrix satisfies 1 0 z = z 0 1

0 0 a b 0 e f for all z ∈ H. All that remains is to show that γ (γz) = (γ γ)z. Let γ = c d and γ = g h . Then for any z ∈ H, e f a b e f az + b e( az+b ) + f γ(γ0z) = z = = cz+d . g h c d g h cz + d az+b g( cz+d ) + h cz+d Multiplying by cz+d yields aez + be + cfz + df (ae + cf)z + (be + df) ae + cf be + df = = z = (γγ0)z. agz + bg + chz + dh (ag + ch)z + (bg + hf) ag + ch bg + dh Therefore, Γ acts on H. As for x ∈ Q ∪ {∞}, either γx is a rational number or γx = ∞, so we land in the correct a b set. The previous calculations will hold so long as x and c d x are both not ∞. If x = ∞, then either c = 0 or c 6= 0. If the former, then e f e ae ae + 0f be + df ∞ = = = ∞ g h g ag ag + 0g bg + dh Otherwise, e f a e( a ) + f ae + cf ae + cf be + df = c = = ∞. g h a ag + ch bg + dh c g( c ) + h ga + ch 18 m If γx = ∞ and x = n , then we must have cm + dn = 0. Then, ae + cf be + df m (ae + cf)( m ) + (be + df) aem + cfm + ben + dfn = n = . ag + ch bg + dh m n (ag + ch)( n ) + (bg + dh) agm + chm + bgn + dhn Collecting the coefficients of e, f, g, and h yields e(am + bn) + f(cm + dn) e(am + bn) e e f = = = ∞, g(am + bn) + h(cm + dn) g(am + bn) g g h which is what we want. Note that am + bn cannot be zero. Otherwise, the vector (m, n)T would be a nontrivial element of ker(γ), which contradicts det(γ) 6= 0. We say that two points z and w are Γ-equivalent if they belong to the same orbit under the Γ-action. One consequence of this result is that all points of Q ∪ {∞} are Γ-equivalent. This is because we can write any rational number in lowest terms as m/n. Since m and n are relatively prime, the Euclidean Algorithm grants the existence of integers a and b such that am − bn = 1. Then the matrix m b ∈ Γ n a and we have m b m ∞ = . n a n Since each rational number is Γ-equivalent to ∞ and Γ-equivalence is transitive, there is exactly one Γ-equivalence class of Q ∪ {∞}. Note that if Γ0 is a subgroup of Γ, then points that are Γ-equivalent may not be Γ0-equivalent. For example, consider the principal congruence subgroup of level N, a b a b 1 0 Γ(N) := ∈ Γ ≡ (mod N) , c d c d 0 1 where matrix equivalence modulo N means a ≡ b ≡ 1 and b ≡ c ≡ 0 modulo N. Suppose ∞ is Γ(3) equivalent to 2. Then for some a = 3m + 1 and c = 3n, with m and n ∈ Z, we find that m b a 1 + 3m ∞ = 2 = = , n a c 3n which implies that 6n = 1 + 3m 3(2n − m) = 1 which is impossible. We conclude that Γ(3) has more orbits than Γ. We call the Γ0-orbits of Q ∪ {∞} the cusps of Γ0. If Γ0 is a subgroup of the modular group with finite index in Γ that also contains Γ(N), we say that Γ0 is a congruence subgroup of level N. If N is understood, we may also say that Γ0 is a congruence subgroup. If we have integers M and N satisfying N|M, then 19 equivalence modulo M implies equivalence modulo N. Therefore Γ(M) ⊂ Γ(N). The following congruence subgroups are valuable to study. a b Γ (N) := ∈ Γ: c ≡ 0 (mod N) 0 c d a b Γ (N) := ∈ Γ: a, b ≡ 1 mod N; c ≡ 0 (mod N) . 1 c d

It is apparent that Γ(N) ⊂ Γ1(N) ⊂ Γ0(N). To study the orbits of this group action, we need to establish a good collection of orbit representatives. For any subgroup Γ0 of the 0 modular group, a fundamental region for Γ is an open set RΓ0 ⊂ H ∪ {∞} such that no two 0 points in RΓ0 are Γ equivalent and the closure of RΓ0 contains at least one point from each orbit under Γ0. We claim that every matrix in Γ may be written as a product in terms of the matrices

1 1 0 −1 S = ,T = . 0 1 1 0 This result is given as an exercise in Chapter 1 of [DS05]. Notice that Sz = z + 1 and T z = −1/z. This allows us to choose representatives orbit under Γ. If z1 and z2 satisfy the conditions 1 1 − < Re(z ) < , 1 < |z | 2 i 2 i then z1 and z2 cannot be Γ equivalent. We obtain a fundamental region of Γ, −1 1 RΓ = z ∈ H < Re(z) < , 1 < |z| , 2 2 which is shown in Figure 1. Recal from Section 4 the deﬁnition of a Dirichlet character. We may now deﬁne a modular form.

Deﬁnition 3. Let k ∈ Z. Let χ be a Dirichlet character. Let Γ0 be a congruence subgroup of level N. A function f : H ∪ {∞} → C is called a holomorphic modular form of weight k, level N, and Nebentypus χ if (1) f is holomorphic on H, k a b 0 (2) f(γz) = χ(d)(cz + d) f(z) for all γ = c d ∈ Γ (3) f is holomorphic at the cusps of Γ0

0 The set of holomorphic modular forms is denoted Mk(Γ , χ). Our ﬁrst example of a 0 modular form is the constant function f(z) ≡ 0. If f and g ∈ Mk(Γ , χ) and w ∈ C, then for all γ ∈ Γ we have that wf(γz) = wχ(d)(cz + d)kf(z) = χ(d)(cz + d)kwf(z) and f(γz) + g(γz) = χ(d)(cz + d)kf(z) + χ(d)(cz + d)kg(z) = χ(d)(cz + d)k(f(z) + g(z))

0 That is, Mk(Γ , χ) is a complex vector space. 20 Figure 1. A fundamental region for Γ.

Let k > 2 be an even integer. The Eisenstien series of weight k is given by X 1 (1) G (z) := . k (mz + n)k (0,0)6=(m,n)∈Z2 We omit the proof that this sum converges to a holomorphic function, which is presented as a b 0 an exercise in Chaper 1 of [DS05]. For all γ = c d ∈ Γ we have that X 1 G (γz) = k a b k 2 (m c d z + n) (0,0)6=(m,n)∈Z X 1 = (m az+b + n)k (0,0)6=(m,n)∈Z2 cz+d X 1 = (maz + mb + cnz + nd)k (0,0)6=(m,n)∈Z2 X 1 (2) = . ((ma + cn)z + (mb + nd))k (0,0)6=(m,n)∈Z2 21 Because det(γ) = det(γT )=1, we find that the linear transformation A : Z2 → Z2 given by m a c m A = n b d n is invertible. That is, the sum in (2) is just a permutation of the summands from (1). If f is holomorphic on H ∪ {∞}, it is convenient to make the coordinate change q = e2πiz. Because z = x + iy ∈ H, we see that q = e−ye2πix is in the open unit disk {w ∈ C | |w| < 1}. z The function e is 2πi periodic, so this coordinate change is bijective when we restrict to RΓ. We take q = 0 to correspond with z = ∞. Because f is holomorphic at ∞, we can write f as a power series in q, X n f(z) = a(n)q ; a : N → C, z ∈ RΓ n≥0 Note that qn = (e−ye2πix)n = e−nyen2πix = e−ny(cos(n2πx) + i sin(n2πx)). We call the values of a(n) the Fourier coefficients of f. By taking a Fourier expansion of f at ∞, we can treat f as a formal power series in q. Let K be a number field with ring of integers OK . The set of formal power series in q with coefficients in OK is denoted by ( ∞ ) X n OK [[q]] = a(n)q | a(n) ∈ OK for all n . n=0 We consider these power series abstractly, without any regard to convergance or domain. Let m be an ideal of OK [[q]]. For any f ∈ OK [[q]], we define the order of f with respect to m as ordm(f) := min{n ∈ N | a(n) ∈/ m}, We also define ordm(f) := ∞ if a(n) ∈ m for all n. We say that two power series f and g are congruent modulo m if ordm(f − g) = ∞, or equivalently, af (n) ≡ ag(n) (mod m) for all n, where af (n) and ag(n) denote the Fourier coefficients of f and g respectively. We denote this equivalence f ≡ g (mod m). P n If f = a(n)q ∈ Mk(Γ0(N), χ) is a modular form whose Fourier coefficients lie in some number field K, then we may consider f as an element of OK [[q]]. In this case, we can rule out certain values of ordm(f), depending on the index of Γ0(N) in Γ and k, the weight of f.

Lemma 8. Let K be a number field with ring of integers OK . Let N > 0 and k ∈ Z. Let χ P n be a Dirichlet character. Let f(z) = a(n)q ∈ Mk(Γ0(N), χ) ∩ OK [[q]]. Suppose m is an ideal of OK . If k ord (f) > [Γ : Γ (N)] m 12 0 then ordm(f) = ∞. 22 A proof of this lemma may be found in [Stu87]. Another way to phrase this result is that there exists an integer M depending only on N and k such that if the first M Fourier coefficients of f are elements of m, then all of the Fourier coefficients of f are in m. We will revisit this lemma at the end of the section. Different spaces of modular forms arise from replacing property (3) of holomorphic modular forms with other conditions. A weakly holomorphic modular form is allowed to have poles at the cusps of Γ0. The set of weakly holomorphic modular forms of weight k for Γ0 is denoted ! 0 0 Mk(Γ , χ). A cusp form must vanish at each cusp of Γ . The set of cusp forms of weight k 0 0 for Γ is denoted Sk(Γ , χ). When the character χ is the trivial character, we abbreviate the 0 ! 0 0 notation as Mk(Γ ), Mk(Γ ), and Sk(Γ ), respectively. We have that 0 0 ! 0 Sk(Γ ) ⊂ Mk(Γ ) ⊂ Mk(Γ ). Recall that if N and M are integers and N|M, then Γ(M) ⊆ Γ(N). When we consider the spaces of modular forms associated to these groups, the inclusion reverses. As Γ(N) contains more elements than Γ(M), any function that is modular of level N must meet more restrictions. Therefore, Mk(Γ(N)) ⊆ Mk(Γ(M)). For a given modular form f, we are interested in finding the least N such that f is modular of level N. This question can be answered by using a special class of linear operators on 0 Mk(Γ ).

Deﬁnition 4. Let f(z) ∈ Mk(Γ0(N), χ) with Fourier coeﬃcients a(n), n ≥ 0. Let p be a rational prime. The Hecke operator Tk,N,χ(p) acts on f via

X n f(z)|T (p) := a(pn) + χ(p)pk−1a qn k,N,χ p n where we define a( p ) = 0 if n is not divisible by p. This is extended to Tk,N,χ(n) for composite n as follows: ν+1 ν k−1 ν−1 (1) For all ν ≥ 1, Tk,N,χ(p ) = Tk,N,χ(p )Tk,N,χ(p) − χ(p)p Tk,N,χ(p ) (2) If (m, n) = 1, then Tk,N,χ(mn) = Tk,N,χ(m)Tk,N,χ(n). The theory of Hecke operators is covered in great detail by [DS05]. If there exists λ ∈ C such that f(z)|Tk,N,χ(n) = λf(z), 0 then we say that f is an eigenform of Tk,N,χ(n). A newform is a cusp form f ∈ Sk(Γ , χ) that is an eigenform ofl Tk,N,χ(n) for all n and has a(1) = 1. The latter requirement allows us to select representatives of newforms. The set of newforms of weight k, level N, and new Nebentypus χ is denoted Sk (Γ0(N), χ). It can be shown that Sk(Γ0(N), χ) always has a 0 finite basis of newforms {f1, f2, . . . fd}. Specifically, every cusp form f ∈ Sk(Γ , χ) may be written as a finite sum X f(z) = α(i, δ)fi(δ), where δ ranges over the set of integral divisors of N. This is demonstrated in chapter 5 of [DS05]. In order to provide interesting results involving Galois representations, we next cite theorems whose proofs are beyond the scope of this paper.

23 P n new Theorem 6. Let f = a(n)q ∈ S2 (Γ1(N)) be a newform. (1) The Fourier coefficients of f are algebraic integers. (2) Q({a(n) | n ≥ 0}) is a number field. We denote this field Kf . (3) For all n,

f|Tk,N,χ(n) = a(n)f(z). The proof comes from manipulating Hecke operators, which is given in [DS05]. These are used to construct a Galois representation associated to a newform.

Theorem 7. Let f ∈ S2(N, χ) be a newform with number ﬁeld Kf . Let ` be prime. Then for each maximal ideal (λ) ⊃ (`) of OKF , there is a two dimensional Galois representation

ρf,λ : GQ −→ GL2(Kf,λ) The proof of this theorem may be found in Chapter 9 of [DS05], which goes beyond the scope of this paper.

new Theorem 8. Let f(z) ∈ Sk (Γ0(N), χ) be a newform with ∞ X f(z) = a(n)qn. n=1 Let F be a number ﬁeld containing the Fourier coeﬃcients a(n) and the values of χ. If (m) is an ideal of OF with norm M, then there is a Galois representation

ρ : GQ → GL2(OK /mOK ) such that (1) ρ is unramiﬁed at all primes p - MN, and (2) for every prime p - MN,

Tr(ρ(Frobp)) ≡ a(p) (mod m). That is, the coeﬃcients of newforms are determined modulo m by the trace of the Frobenius elements Frobp. A proof of this may be found in [Ser76]. We will also need a result regarding prime numbers. Let P be the set of all prime numbers. Suppose P is a subset of P. Then we say that P contains a positive proportion of primes if |{p ∈ P | p < n}| lim > 0. n→∞ |{p ∈ P | p < n}| This implies that P contains inﬁnitely many primes.

Theorem 9. (The Tchebotarev Density Theorem) Let σ ∈ GQ. Then the set −1 {p ∈ P | τFrobpτ = σ for some τ in GQ} contains a positive proportion of primes. Proof of this theorem is given in Chapter 8 of [Mar91]. With these ingredients, we are able to close the section by proving a theorem first demonstated by J.P. Serre, adapting the proof from [Tre06]. 24 Theorem 10. Let k ∈ Z and N ∈ Z+, and let χ be a Dirichlet character modulo N. Suppose that F is a number field with ring of integers OF , and let m be an ideal of OF with norm M. Then a positive proportion of the rational primes p satisfying p ≡ −1 modulo MN also satisfy f(z)|Tk,N,χ(p) ≡ 0 (mod m) for all cuspforms f ∈ Sk(Γ0(N), χ) with Fourier coefficients in OF .

Proof. Let f1, f2, . . . , fd be a basis of newforms for Sk(Γ0(N), χ), with Fourier expansions at ∞ given by ∞ X n fi(z) = ai(n)q . n=1 Suppose that K is a ﬁnite ﬁeld extension of F containing the set + [ + {ai(n) | 1 ≤ i ≤ d, n ∈ Z } {χ(n) | n ∈ Z }.

Let Sk(Γ0(N), χ)OF /m be the set of equivalence classes of cusp forms Sk(Γ0(N), χ)∩OF [[q]] re-

duced modulo m. For each g(z) ∈ Sk(Γ0(N), χ)OF /m, we may choose an hg(z) ∈ Sk(Γ0(N), χ) with coeﬃcients in OF such that

hg(z) ≡ g(z) (mod m). Because K is a number ﬁeld, m has ﬁnitely many cosets. Lemma 8 implies that for g(z) not congruent to 0 modulo m, k ord (h ) ≤ [Γ : Γ (N)]. m g 12 0

That is, there are only finitely many equivalence classes in Sk(Γ0(N), χ)OF /m and only finitely many hg. Write each hg in terms of the basis {f1, . . . , fd} d X hg(z) = α(i, δ)fi(δz). i=1 Using Theorem 6, the coefficients α(i, δ) are algebraic numbers. Since there are finitely many g(z), there are only finitely many α(i, δ). Then we may extend K to contain all the α(i, δ). Because every algebraic number α is the ratio of two algebraic integers, there exists an algebraic integer D such that αD is also an algebraic integer. Moreover, we may assume that D is a rational integer by replacing D with NK (D). Applying this processes to each Q α(i, δ) there exists a nonzero integer C = C(i, δ) such that Cα(i, δ) ∈ OK for all α(i, δ). 0 0 Let m = (C)m and M = CM. Since the fi(z) are newforms, if p - N then

fi(δz)|Tk,N,χ(p) = ai(p)fi(δz) for each i and δ. Then d X X hg(z)|Tk,N,χ(p) = α(i, δ)ai(p)fi(δiz). i=1 δ|N

Since each f(z) ∈ Sk(Γ0(N), χ) is congruent modulo m to one of the hg(z), it is sufficient to show that a positive proportion of primes p ≡ −1 modulo MN also satisfy ai(p) ≡ 0 modulo m0 for each i. For each i, Theorem 8 implies the existence of a Galois representation 0 ρi : GQ → GL2(OK /m OK ), 25 unramified outside M 0N, such that 0 Tr(ρ(Frobp) ≡ ai(p) mod m 0 for each p - M N. Consider ζMN . Define a homomorphism × φ : GQ −→ (Z/MNZ) σ 7−→ a a where σ(ζMN ) = ζMN . Now take the direct sum d d M × M 0 ρ := φ ⊕ ρi : GQ → (Z/MNZ) ⊕ GL2(OK /m OK ). i=1 i=1 Let H be the kernel of ρ and let E be the fixed field of H. Since H is normal and closed ∼ in GQ, the extension E/Q is Galois and Gal(E) = GQ/H is isomorphic to the image of ρ, which is finite. Therefore E is a number field and ρ factors through Gal(E). The restriction 0 ρ|E must also be unramified outside of M N. Let c ∈ Gal(E) be the automorphism c(α) = α. Then ρi(c) is conjugate to the matrix 1 0 0 −1 for each i, so Tr(ρi(c)) = 0. The Tchebotarev density theorem implies that there is a positive 0 proportion of primes p - M N such that Frobp is conjugate to c in GQ. For each such p, 0 ai(p) ≡ Tr(ρi(Frobp) ≡ Tr(ρi(c)) ≡ 0 (mod m ) p −1 for each i. Now Frobp|Q[ζ] : ζ → ζ , so φ(Frobp) = p. But φ(c) = ζ , so (c) = −1. Since Frobp is conjugate to c, we have p ≡ −1 mod MN, which is what we want. 6. Elliptic Curves In order to discuss elliptic curves, we must first define the projective plane. Let F be any field. We will most often use C, but thinking of the field as R can be more geometrically intuitive. For more detail, see Appendix C of [LR11] or Appendix A of [ST92]. We make the convention that F 3 consists of points (X,Y,Z) while F 2 consists of points (x, y). Define a relation on F 3\{(0, 0, 0)},

(X,Y,Z) ∼ (λX, λY, λZ); λ ∈ F \{0}.

Intuitively, (X1,Y1,Z1) is related to (X2,Y2,Z2) if and only if the two points lie on a line through the point (0, 0, 0). Showing that ∼ is an equivalence relation amounts to showing that F \{0} contains 1 and is closed under multiplication and division. Because F is a field, this is trivial. The set of such equivalence classes is called the projective plane over F and is denoted by P2(F ). If the field F is fixed, then we abbreviate this either as the projective plane or symbolically as P2. The equivalence classes in P2 are denoted by [X : Y : Z]. In order to minimize jargon, we continue to refer to these classes as points in homogeneous coordinates, or just points for short. Note that any point can be represented with different coordinates. For example, 1 2 [1 : 2 : 3] = : : 1 . 3 3 26 Note that [0 : 0 : 0] is not a point on the projective plane. Any ordered pair (x, y) ∈ F 2 can be viewed uniquely as the point [x : y : 1] in the projective plane. If [x : y : 1] = [X : Y : 1], because the Z coordinates of these points are both 1, we can conclude that (X,Y ) = (x, y). On the other hand, if Z 6= 0, the points [X : Y : Z] and [X/Z : Y/Z : 1] are identical. Points in the set {[X : Y : Z] | Z 6= 0} may be uniquely represented as ordered pairs (X/Z, Y/Z), as scaling X,Y, and Z has no effect on the new coordinates. Therefore, this set is a copy of F 2 sitting in the projective plane. We call the set {[X : Y : 1] ∈ P2} affine space. The remaining points have the form [X : Y : 0]. These are called points at infinity. In Euclidean geometry, we are interested in finding rational points on curves, i.e., points 2 (x, y) ∈ Q . Identifying rational points in homogeneous coordinates takes an√ extra√ step, though. For example, a point can have irrational coordinates when written as [ 2 : 8 : 0] and rational coordinates as [1 : 2 : 0]. We say that a point [X : Y : Z] ∈ P2(C) is rational if there exists a nonzero λ such that λX, λY, and λZ are all rational numbers. More generally, if F is a subfield of C, a point [X : Y : Z] is in the set P2(F ) if there exists nonzero λ such that λX, λY, and λZ ∈ F . To study the geometry of the projective plane, we can lift structures up from the Euclidean plane F 2. If p(x, y) is a polynomial with coefficients in F , then we say p gives rise to the curve C = {(x, y) ∈ F 2 | p(x, y) = 0}. This is denoted by C : p(x, y) = 0. A set of the form p−1(0) is also called a zero locus. We can associate the algebraic properties of the polynomial p to C. We claim that F [x, y] is a unique factorization domain, which is proven in Chapter 9 of [DF04]. Therefore any polynomial factors uniquely (up to multiplication by a constant) as n Y p(x, y) = pi(x, y), i=1

where pi is an irreducible polynomial for all i. Then if C is the curve associated to p, we say that the irreducible components of C are the curves Ci associated to the pi. Because a point (x, y) ∈ C satisfies p(x, y) = 0 and F is a field, it follows that pi(x, y) = 0 for some i. Then we have that n C = ∪i=1Ci. If C has one irreducible component, or equivalently if p is irreducible, then we say that C is irreducible. If C1 and C2 are curves with distinct irreducible components we say that they have no common components. For polynomials in multiple variables, the total degree of a term is the sum of the degrees for each variable separately. For example, in two variables, the degree of xmyn is m + n. The degree of p is just the maximum of the total degrees of its terms. If deg(p) = 1 , then call C : p(x, y) = 0 a line. If deg(p) = 2, then C is a quadratic. If deg(p) = 3 then C is a cubic and the pattern follows as in elementary geometry. We must establish how to evaluate polynomials on projective points before we can study the curves they generate. For example, let p(X,Y,Z) = Y 2 − X3. Then for the point [1 : 1 : 1] = [2 : 2 : 2], we find that p([X : Y : Z]) is not well defined. It could be 0, -2, or 27 one of infinitely many other choices. To avoid this, we restrict our attention to homogeneous polynomials. A homogeneous polynomial is any polynomial where each term has the same total degree. A homogeneous polynomial of degree N has the form

X nX nY nZ p(X,Y,Z) = αI X Y Z . I∈I

In the summation above, we multi-index the terms to conserve space. That is, the index I is actually the ordered triple (nX , nY , nZ ) coming from the ﬁnite set

I = {I = (nX , nY , nZ ) ∈ N | nX + nY + nZ = N}.

Then for any nonzero λ ∈ F , we have that

X nX nY nZ p(λX, λY, λZ) = αI (λX) (λY ) (λZ) I∈I

X (nX +nY +nZ ) nX nY nZ = λ αI X Y Z I∈I

X N nX nY nZ = λ αI X Y Z I∈I

N X nX nY nZ = λ αI X Y Z I∈I = λN p(X,Y,Z).

That is, if P = [X : Y : Z] is a point on the projective plane and p(X,Y,Z) = 0, we can say that p(P ) = 0 regardless of the choice of X,Y and Z. (Hence the name homogeneous coordinates.) So the set C = {[X : Y : Z] | p(X,Y,Z) = 0} is well deﬁned. Note that this does not work if we try to solve p(X,Y,Z) = α 6= 0, but we only care about the zeros of p anyway. We call the curve C ⊂ P2 a projective curve. Again, we write this as C : p(X,Y,Z) = 0. Just as in the Euclidean case, we label C according to the degree of p. Consider the polynomial

p(X,Y,Z) = Z.

Note that any polynomial with a single term is homogeneous by default. This defines the curve C = {[X,Y,Z] | Z = 0}, which is the collection of points at infinity. Because deg(p) = 1, the points at infinity all lie on a projective line, called the line at infinity. If C is a curve given by the polynomial p(x, y) in F 2, we can find a corresponding curve in the projective plane by homogenizing p. First, replace x and y in the expression of p with X/Z and Y/Z respectively. Then multiply by ZN , where N is the degree of p. This process yieldsp ˆ(X,Y,Z), which is a homogeneous polynomial in X,Y, and Z of degree N. More explicitly, if p is a degree N polynomial, 28 X i j p(x, y) = aijx y i j X Y X X Y p , = a Z Z ij Z Z i j X Y X X Y ZN p , = ZN a Z Z ij Z Z X i j N−i−j = aijX Y Z =:p ˆ(X,Y,Z)

Note that for [X : Y : 1] in affine space,p ˆ(X,Y, 1) = p(X,Y ), so the curve Cb defined byp ˆ in the projective plane passes through the affine points that correspond to C in F 2. We take this as grounds to relax the notation, writing points [X : Y : 1] in affine space as (X,Y ). In affine space, the line between two points (a1, b1) and (a2, b2) is given by

C : −(b2 − b1)x + (a2 − a1)y + [a1(b2 − b1) − b1(a2 − a1)], which comes from manipulating the point-slope formula. For any two points P1 and P2 in P2, with Pi = [Ai : Bi : Ci] the projective line passing between P1 and P2 is given by the curve

C :[B1(C2 − C1) − C1(B2 − B1)]X

+[C1(A2 − A1) − A1(C2 − C1)]Y

+[A1(B2 − B1) − B1(A2 − A1)]Z = 0.

We verify this formula by evaluating the coordinates of P1 and P2. Note that we have not proven that C is the only line passing through P1 and P2. So even in cases where our intuition fails, such as P2(C) or P2(Z/`Z) we can still work from a purely computational point of view. Let p1(x, y) = ax + by + c1 and p2(x, y) = ax + by + c2, where c1 6= c2 and not both of 2 a and b are zero. Then their associated projective lines C1 and C2 do not intersect in F . However, when we move to projective space, we can find a point of intersection. Let Z = 0. Then we have aX + bY + cZ = aX + bY + dZ = 0 which implies that aX + bY = 0. Either X = (−b/a)Y , or Y = (−a/b)X, depending on whether a or b is zero. In either case, we have a solution that is a point at infinity, [−b/a : 1 : 0] or [1 : −a/b : 0]. Note that if both a and b are both nonzero, this is the same point. Therefore, any two parallel projective lines meet at a point at infinity. Moreover, the point of intersection is unique. When solving p1 − p2 = 0, it may be the case that p1 − p2 has repeated roots. Consider a 2 tangent intersection. Let p1(x, y) = y − x and p2(x, y) = y − 2x − 1. Then we see that 2 2 2 p1(x, y) − p2(x, y) = y − x − y + 2x + 1 = −x + 2x + 1 = −(x − 1) 29 If P = (x, y) is a root of p1 − p2, then the curves generated by p1 and p2 intersect at P . Note that if these curves are tangent to one another at P , then P will be a double root of p1 − p2. In general, if P occurs as a root p1 − p2 with n times, then we say that C1 and C2 intersect at P with multiplicity n. In the example above, C1 and C2 intersect at (1, 1) with multiplicity 2. It is no accident that the intersection a line with another line (defined by polynomials of degree 1) is exactly one (1 × 1) point. Note that this only holds as long as the lines are not identical, i.e., they do not share a component. More generally, if C1 and C2 have a common component, then their intersection is much larger than if they did not have common components. We state without proof the following theorem.

Theorem 11. (Bézout’sTheorem) Let C1 : p1(x, y) = 0 and C2 : p2(x, y) = 0 be projective curves over an algebraically closed field F with no common components, where p1 and p2 have degree d1 and d2 respectively. Then C1 and C2 intersect at d1d2 points counting multiplicity. A proof is outlined in Appendix A of [ST92], along with more advanced intersection results. As a corollary, the line L1 passing between two points in projective space is unique, since any other line L2 passing through the same points must share a component with L1. Because polynomials of degree 1 have only one irreducible factor, we find that L1 = L2. Note that the intersection of a cubic and a line that have no common components consists of exactly three points in the projective plane, counting multiplicity. We are ready to define elliptic curves. Let p(x, y) be a cubic polynomial in Q[x, y]. An elliptic curve E over Q is the projective curve C in P2(Q) given by the homogenization of p. The expression E/Q is shorthand for “E is an elliptic curve over Q.” An elliptic curve is smooth if the tangent vector does not vanish, i.e., ∂p ∂p , 6= (0, 0). ∂x ∂y If E is smooth and there exists a rational point O on E, we can then take advantage of Bézout’stheorem to define a group structure on the points of E. For two points P and Q, the line PQ must intersect E at a third point, which we call P ∗ Q. We can repeat this process with the points O and P ∗Q. Then P +Q is defined to be O∗(P ∗Q). If P = Q, then we take PP to be the line tangent to E at P . This is the reason we are restricted to smooth elliptic curves. Note that because any line intersects E in only three points, O ∗(P ∗O) = P . That is, O is the identity element. We call it the origin. A proof that + is associative and additional details are available in Chapter 1 of [ST92]. We say that two elliptic curves E : p(X,Y,Z) = 0 and E0 : p0(X,Y,Z) = 0 with ori- gins O and O0 respectively are birationally equivalent if there exists an invertible change of coordinates φ : P2 → P2 [X0 : Y 0 : Z0] = P 0 = f(P ) 0 X = fX (X,Y,Z) 0 Y = fY (X,Y,Z) 0 Z = fZ (X,Y,Z) 30 Figure 2. Addition on an elliptic curve viewed in R2.

0 where fX , fY , and fZ are rational functions in X,Y, and Z such that O = f(O) and p0(f(P )) = 0 if and only if p(P ) = 0. Thus, we can simplify a polynomial using a change of variables without disturbing its geometric structure. If F is a field with characteristic different from 2 and 3, the expression of an elliptic curve E/F can be simplified to the form

E : y2 = x3 + ax2 + bx + c. The speciﬁcs are explained in [ST92]. An equation in this form is called a Weierstrauss equation. In this form, E being smooth is equivalent to p(x) having no repeated roots. The homogenized form of E is given by

E : Y 2Z = X3 + AX2Z + BXZ2 + CZ3 We can see that a curve given in Weierstrauss form has a unique point at infinity: [0 : 1 : 0]. Since this is a rational point, any elliptic curve in Weierstrauss form has the group structure outlined above, where we take O to be [0 : 1 : 0]. Recall that [0 : 1 : 0] = [0 : λ : 0] for all nonzero λ ∈ F . In affine space, we can think of O as sitting infinitely far from (0, 0) in the y direction. Further, for any P 6= O, we can assume P has the form [X : Y : 1]. Therefore P can be expressed in affine coordinates as (X,Y ). We claim that a point P = (X,Y ) = [X : Y : 1] in affine coordinates is in E(F ) if and only if X and Y ∈ F . Clearly if X and Y are in F then P ∈ E(F ). As for the converse, suppose there exists nonzero t such that tX, tY , and tZ ∈ F . As Z = 1, we have that t ∈ F . Then both X and Y ∈ F . When an elliptic curve is given by a Weierstrauss equation, we can find explicit formulas for the group law. If P = (x1, y1) and Q = (x2, y2) are points on E with P 6= Q then the 31 line passing between P and Q is given by y = λx + ν, where y2 − y1 λ := , ν := y1 − λx1. x2 − x1

Note that if x2 = x1, then either P = −Q or P = Q. In the former case, we know that P − P = O and forgo any further derivation. The latter is excluded by assumption. Substituting the right hand side of this equation for y in the Weierstrauss equation yields y2 = (λx + ν)2 = x3 + ax2 + bx + c, which reduces to 3 2 2 2 0 = x + (a − λ )x + (b − 2λν)x + (c − ν ) = (x − x1)(x − x2)(x − x3).

The third root of this cubic, x3 is the x coordinate of P + Q in affine coordinates. Equating the coefficients of the x2 term gives 2 λ − a = x1 + x2 + x3 Therefore we conclude that 2 P + Q = (λ − a − x1 − x2, λx3 + ν). To find the affine coordinates of P + P , we must use implicit differentiation. Replacing λ above with ∂x 3x2 + 2ax + b = ∂y 2y yields the formula for the x coordinate of P + P , x4 − 2bx2 − 8cx + b2 − 4ac x = . 3 4x3 + 4ax2 + 4bx + 4c Similar formulas may be derived for the y coordinates, which may be found in [ST92] For any field K ⊂ C we define the set E(K) = {[X : Y : Z] ∈ P2(K) | p(X,Y,Z) = 0]}. Because the point addition formulas are rational functions of x and y, we find that E(K) is a of E(C). When F is a Galois number field, we have the following result. Lemma 9. If F is a Galois number field and E/Q is an elliptic curve in Weierstrauss form, then Gal(F ) acts on E(F ). Moreover, for all points P and Q ∈ E(F ), we have that σ(P + Q) = σ(P ) + σ(Q). If P has order N, then σ(P ) has order N. Proof. For σ ∈ Gal(F ) and P ∈ E(F ) define ( (σ(x), σ(y)) : P = (x, y) σ(P ) := O : P = O. To show σ(P ) ∈ E(F ), it suffices to show that σ(P ) satisfies the Weierstrauss equation. This holds trivially for O. For P 6= O, we find that σ(y2) = σ(x3 + ax + c) (σ(y))2 = (σ(x))3 + aσ(x) + c 32 Figure 3. Repeated roots in f(x) give rise to singular curves y2 = f(x). Viewed in R2.

Because F is Galois, σ(x) and σ(y) ∈ F . Observe that the identity automorphism ﬁxes x and y. Suppose σ and τ ∈ Gal(F ). Then for P 6= O, τ(σ(P )) = τ((σ(x), σ(y))) = (τ(σ(x)), τ(σ(y))) = ((τ ◦ σ)(x), (τ ◦ σ)(y)) = (τ ◦ σ)(P )

Recall that the point addition formula is a rational function of x1, x2, y1 and y2, whether P 6= Q or P = Q. Writing x3(x1, x2, y1, y2) and y3(x1, x2, y1, y2) as rational functions of these coordinates, we see that

σ(P + Q) = (σ(x3(x1, x2, y1, y2)), σ(y3(x1, x2, y1, y2)))

= (x3(σ(x1), σ(x2), σ(y1), σ(y2)), y3(σ(x1), σ(x2), σ(y1), σ(y2))) = σ(P ) + σ(Q)

Note that if x1 = x2 then σ(x1) = σ(x2). That is, σ(P + Q) and σ(P ) + σ(Q) are derived using the same formula. Again, cases involving O are trivial.

We must introduce additional structure in order to study the torsion points of E. Let ω1 and ω2 be complex numbers that are linearly independent over R with ω1/ω2 ∈ H.A lattice in C is the additive group M Λ := ω1Z ω2Z = {n1ω1 + n2ω2 | n1, n2 ∈ Z}, which is a subgroup of (C, +). The quotient group C/Λ is called a complex torus. For any z ∈ C, the Weierstrauss ℘-function is given by 33 Figure 4. If f(x) has distinct roots, then E : y2 = f(x) is a smooth elliptic curve. Viewed in R2.

1 X 1 1 ℘(z) := + − . z2 (z − ω)2 ω2 06=ω∈Λ Clearly ℘ has poles of order 2 at all z ∈ Λ. We omit the proof that this sum converges uniformly to a holomorphic function on compact subsets of C/Λ , which is outlined in Chapter 1 of [DS05]. We can therefore ﬁnd its derivative is found by diﬀerentiating termwise: X 1 ℘0(z) = −2 (z − ω)3 ω∈Λ Note for any ω0 ∈ Λ, we may rewrite ((z + ω0) − ω) = (z − (ω − ω0)), which only permutes the summands. Therefore, ℘(z + ω) = ℘(z) and ℘0(z + ω) = ℘0(z) for any z ∈ C\Λ. Recall the Eisenstein series introduced in Section 5: X 1 G (z) = ; k > 2 even, z ∈ . k (cz + d)k H (0,0)6=(c,d)∈Z

Generalize Gk as a function of lattices, X 1 G (Λ) = ; k > 2 even. k ωk 06=ω∈Λ L We see that Gk(z) = Gk(Λ) for Λ = zZ 1Z. For z∈ / Λ, 0 2 3 (℘ (z)) ) = 4(℘(z)) − g2(Λ)℘(z) − g3(Λ), 34 Figure 5. Points with order dividing 2 on a complex torus. where g2(Λ) := 60G4(Λ), and g3(Λ) := 140G6(Λ). This is proved in [DS05] by analyzing the asymptotic behavior of ℘ and ℘0. We conclude 0 2 3 that (℘ (z), ℘(z)) is a point on the elliptic curve E : y = 4x − g2(Λ)x − g3(Λ). Conversely, suppose that E : y2 = 4x3 − ax = c is a smooth elliptic curve. We claim that the modular function j : H → C given by 3 1728(g2(z)) j(z) := 3 2 , (g2(z)) − 27(g3(z)) is surjective. Then there exists a z ∈ H such that 3 3 1728(g2(z)) a 3 2 = 3 2 . (g2(z)) − 27(g3(z)) a − 27c Careful manipulation of this equation yields the lattice we are looking for. Additional details are available in Chapter 1 of [DS05]. Thus, complex tori give rise to smooth elliptic curves, and every smooth elliptic curve gives rise to a complex torus. Note that each of these constructions is a group. Again, [DS05] explains why mapping these groups to each other is an isomorphism. As E is an abelian group, we may view it as a Z module. For any point P ∈ E(C) and any positive N ∈ Z, define the operation NP via repeated addition, i.e., 3P := P + P + P . We extend to nonpositive integers by defining −NP := N(−P ), where −P denotes the additive inverse of P with respect to the group action, and 0P := O. The N-torsion subgroup of E is given by E[N] := {P ∈ E(C): NP = O}. Because E is abelian it is easy to verify that E[N] is a subgroup of E. By moving from E to its associated complex torus, we can find that E[N] is isomorphic to (Z/NZ)2. Consider the parallelogram defined by ω1 and ω2 in C. We can view addition in C/Λ as addition in C followed by the quotient map φ : C → C/Λ. 35 Lemma 10. If (x, y) ∈ E[N], then both x and y are algebraic numbers. Moreover, if E[N] = {O, (x1, y1),..., (xm, ym)} then the field extension Q(E[N]), defined as

Q(E[N]) := Q(x1, . . . , xm, y1, . . . , ym) is a Galois number field. A proof is given in Chapter 6 of [ST92]. Fix ` a prime. Then for all i ≤ j ∈ Z+, we have j i a natural mapping µij : E[` ] → E[` ] given by j−i µij(P ) := ` P. j−i Here we view multiplication by ` in the group E. It is apparent that µijµjk = µjk for all i ≤ j ≤ k. We also have µii = id as multiplying by 1 fixes E pointwise. Then the groups E[`n] indexed over n meet the requirements for defining an inverse limit. Let E be an elliptic curve over Q and let ` be prime. Then the `-adic Tate module of E is the inverse limit Ta (E) := lim E[`n]. ` ←− n n ∼ n 2 For each n, we know that E[` ] = Z/` Z . Therefore, we can choose an ordered basis n (Pn,Qn) for E[` ]. We make the restriction that for all n,

`Pn = Pn−1, `Qn = Qn−1.

Each choice of basis (Pn,Qn) gives an isomorphism n n 2 φn : E[` ] −→ (Z/` Z)

anPn + bnQn 7−→ (an, bn). Therefore, we ﬁnd that the mapping 2 φ : Ta`(E) −→ Z`

(anPn + bnQn)n∈Z+ 7−→ ((an, bn))n∈Z+ is an isomorphism. For each n, we have that Q(E[`n]) is a Galois number field. Moreover, we have seen that Q acts on E[`n]. Specifically, for P ∈ E[`n], we have that σ(`P ) = `σ(P ) ∈ E[`n−1] for all σ ∈ GQ. Therefore we may define the action of σ on tuples

σ((an)n∈Z+ ) := (σ(an))n∈Z+ , which is an automorphism of Ta`(E). The choice of ordered basis (Pn,Qn) determines an isomorphism n n Aut(E[` ])) −→ GL2(Z/` Z), which gives us the isomorphism

Aut(Ta`(E)) −→ GL2(Z`) ⊂ GL2(Q`). Taken together, we have constructed a map

ρE,` : GQ −→ GL2(Q`).

We claim that ρE,` is continuous, which is given in Chapter 9 of [DS05]. Therefore, ρE,` is a Galois representation. With this representation and ρf,λ from the previous section, we may now state the Modularity Theorem. 36 7. The Modularity Theorem We began the paper by outlining the basics of algebraic number theory. In order to study the algebraic numbers, we examined finite degree field extensions of Q. The elements of such a field F share their minimal polynomials with their conjugates, which allowed us to define a norm on F and study divisibility of elements and ideals in the ring of integers OF . As a finite degree field extension, F is the perfect candidate for applying Galois theory, which determines the subfields of F . We moved on to study Q as an infinite degree field extension of Q. In order to build up tools analogous to the previous setting, we started with the automorphism group of Q. Using the inverse limit construction, we can decompose these automorphisms as elements of the Galois groups of Galois number fields. This allowed us to topologize GQ such that the open and closed sets were related to the field structure of Q. To simplify our study of this automorphism group, we constructed Galois representations, which are just continuous homomorphisms of GQ into matrix groups. We developed everything after this point in order to demonstrate the usefulness of Galois representations as a tool. In the study of modular forms, every newform has a Fourier expansion ∞ X f(z) = a(n)qn, i=1 where the coefficients a(n) are algebraic numbers. When we examined the values of a(n) with Galois representations, we found extraordinary congruence results. In elliptic curves, we started by defining a geometric group structure on the zero locus of

y2 = x3ax2 + bx + c.

For any field F ⊂ C, the set of ordered pairs in F 2 on the elliptic curve formed a subgroup that is preserved by field automorphisms on the coordinates. This provided another opportunity to construct Galois representations. It is no accident that both modular forms and elliptic curves give rise to Galois representations. One of the great feats of mathematics in the 20th century was the proof of the Modularity Theorem. Generally speaking, the theorem states that all elliptic curves over Q inherit properties from specific modular forms. Although it is well beyond the scope of this already ambitious paper to give a proof of the Modularity Theorem, we would like to conclude with an overview. What is now called the Modularity Theorem began in the 1950s as the Taniyama-Shimura- Weil conjecture. It involves many fields, including algebraic geometry, analysis and number theory. There are many equivalent statements of the Modularity Theorem, the difference between statements being the type of structure linking the modular forms to the elliptic curves. Our labels for the statements of the Modularity Theorem are taken from [DS05]. The famed proof of Fermat’s Last Theorem by Andrew Wiles is actually a corollary to his 1995 proof of a special case of the Modularity Theorem. (The general theorem was proved in 2001 by Breuil, Conrad, Diamond, and Taylor working in collaboration.) According to [ST02], Wiles was only able to make the bulk of his progress after working in isolation for seven years. 37 The version that Wiles proved is phrased in the language of Galois representations. To associate arbitrary Galois representations to those arising from modular forms, we define modular Galois representations.

Deﬁnition 5. An irreducible Galois representation ρ : GQ → GL(Q`) with det ρ = χ` is called modular if there exists a newform f ∈ S2(Γ0(Mf )) such that Kf,λ = Q` for some maximal ideal λ of OK lying over ` and that ρf,λ ∼ ρ.

−1 Recall that ρ and ρf,λ are equivalent if for a fixed matrix m we have that, m ρf,λm = ρ. 0 In other words, ρ(GQ) and ρ (GQ) are conjugate subgroups. Theorem 12. (Modularity Theorem, version R) Let E be an elliptic curve over Q. Then ρE,` is modular for some `. According to [DS05], Andrew Wiles was able to prove the preceding theorem in the special case where E is semistable. An elliptic curve E : y2 = p(x) is semistable if whenever an odd prime p divides the discriminant of E, only two of the roots of p(x) are congruent modulo p. The adventurous reader can find his result spread between [Wil95] and [TW95]. Theorem 13. (Modularity Theorem, strong Version R) Let E be an elliptic curve over Q with conductor N. Then there exists a newform f ∈ S2(Γ0(N)) with rational Fourier coefficients such that

ρf,` ∼ ρE,` for all `. Consider the equation xn + yn = zn. If n = st is composite, and a, b and c are a nontrivial solution, then as, bs and cs is a solution of the reduced equation xt + yt = zt. Therefore, it is sufficient to prove Fermat’s Last Theorem for odd prime powers. The proof of Fermat’s Last Theorem is a contradiction proof. We adapt the summaries as presented in [DS05] and [ST02]. Let p be an odd prime and suppose there exist nonzero integers a, b and c with no common factor such that ap + bp = cp. Either c ≡ 1 or c ≡ 0 modulo 2. If c ≡ 1, then a and b cannot both be odd. In this case we may assume b is even. Otherwise, if c ≡ 0, then because p is odd we may rearrange the equation ap + (−c)p = (−b)p and relabel b and c. Therefore, we may assume without loss of generality that b is even. We define the Frey elliptic curve associated to this solution F : y2 = x(x − ap)(x + bp) We wish to show that F has impossible properties. Following from the Modularity Theorem, F is modular for some `. Therefore, we can find a newform f ∈ S2(Γ0(2)), where N is the conductor of F. However, there are no such newforms. See [DS05] for more details. That is, the existence of F implies the existence of a newform that does not exist. Because F was obtained from the solution ap + bp = cp, such a solution cannot exist. 38 References [Ash71] Robert B. Ash. Complex variables. Academic Press, New York, 1971. [DF04] David S. Dummit and Richard M Foote. Abstract Algebra. John Wiley and Sons, Inc, Hoboken, 2004. Third Edition. [DS05] Fred Diamond and Jerry Shurman. A first course in modular forms, volume 228 of Graduate Texts in Mathematics. Springer-Verlag, New York, 2005. [FH91] William Fulton and Joe Harris. Representation theory, volume 129 of Graduate Texts in Mathe- matics. Springer-Verlag, New York, 1991. A first course, Readings in Mathematics. [Leh69] Joseph Lehner. Lectures on modular forms, volume 61 of National Bureau of Standards, Applied Mathematics Series. Superintendent of Documents, U.S. Government Printing Office, Washington, D.C., 1969. [LR11] Alvaro´ Lozano-Robledo. Elliptic curves, modular forms, and their L-functions, volume 58 of Stu- dent Mathematical Library. American Mathematical Society, Providence, RI, 2011. IAS/Park City Mathematical Subseries. [Mar91] Daniel A. Marcus. Number Fields. Springer, New York, 1991. [Mun00] James R. Munkres. Topology: Second Edition. Prentice-Hall Inc., Upper Saddle River, N.J., 2000. [S+11] W. A. Stein et al. Sage Mathematics Software (Version 4.7.1). The Sage Development Team, 2011. http://www.sagemath.org. [Ser73] Jean-Pierre Serre. Congruences et formes modulaires [d’après H. P. F. Swinnerton-Dyer]. In Séminaire Bourbaki, 24e année(1971/1972), Exp. No. 416, pages 319–338. Lecture Notes in Math., Vol. 317. Springer, Berlin, 1973. [Ser75] Jean-Pierre Serre. Valeurs propres des opérateursde Hecke modulo l. pages 109–117. Astérisque, Nos. 24–25, 1975. [Ser76] Jean-Pierre Serre. Divisibilitéde certaines fonctions arithmétiques. Enseignement Math. (2), 22(3- 4):227–260, 1976. [ST92] Joseph H. Silverman and John Tate. Rational points on elliptic curves. Undergraduate Texts in Mathematics. Springer-Verlag, New York, 1992. [ST02] Ian Stewart and David Tall. Algebraic number theory and Fermat’s last theorem. A K Peters Ltd., Natick, MA, third edition, 2002. [Stu87] Jacob Sturm. On the congruence of modular forms. In Number theory (New York, 1984–1985), volume 1240 of Lecture Notes in Math., pages 275–280. Springer, Berlin, 1987. [Tay02] R. Taylor. Galois representations. pages 449–474, 2002. [Tre06] S. Treneer. Congruences for coefficients of weakly holomorphic modular forms. Ph.D. Thesis (Ad- visor: S. Ahlgren), 2006. [TW95] Richard Taylor and Andrew Wiles. Ring-theoretic properties of certain Hecke algebras. Ann. of Math. (2), 141(3):553–572, 1995. [Wil95] Andrew Wiles. Modular elliptic curves and Fermat’s last theorem. Ann. of Math. (2), 141(3):443– 551, 1995.