Math 99r - Arithmetic of Elliptic Curves

Taught by Zijian YaoNotes by Dongryul Kim

Fall 2017

This course was taught by Zijian Yao. The lectures were usually on Tuesdaysand Thursdays, but were irregular. There was no textbook, and 8 students wereenrolled.

Contents

1 September 5, 2017 41.1 Elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 September 7, 2017 62.1 Rational functions and dimension . . . . . . . . . . . . . . . . . . 62.2 Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 September 12, 2017 93.1 Group structure on an elliptic curve . . . . . . . . . . . . . . . . 93.2 Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

4 September 14, 2017 114.1 Discrete valuation ring . . . . . . . . . . . . . . . . . . . . . . . . 114.2 Differential forms . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

5 September 19, 2017 145.1 Divisor of a meromorphic form . . . . . . . . . . . . . . . . . . . 145.2 Canonical divisor and genus . . . . . . . . . . . . . . . . . . . . . 155.3 Degree-genus formula and Riemann–Roch . . . . . . . . . . . . . 15

6 September 21, 2017 176.1 Identifying the genus 1 curve . . . . . . . . . . . . . . . . . . . . 176.2 Picard group of an elliptic curve . . . . . . . . . . . . . . . . . . 186.3 Review of number theory . . . . . . . . . . . . . . . . . . . . . . 19

1 Last Update: January 15, 2018

7 September 29, 2017 207.1 Isogeny . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207.2 Invariant differential . . . . . . . . . . . . . . . . . . . . . . . . . 21

8 October 3, 2017 228.1 Group cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . 22

9 October 5, 2017 249.1 Weak Mordell–Weil theorem . . . . . . . . . . . . . . . . . . . . . 249.2 Mordell–Weil theorem . . . . . . . . . . . . . . . . . . . . . . . . 25

10 October 10, 2017 2710.1 p-adic numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2710.2 Tate modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2710.3 Endomorphism ring of an elliptic curve . . . . . . . . . . . . . . . 2910.4 Weil pairing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

11 October 12, 2017 3111.1 Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3111.2 Formal group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

12 October 17, 2017 3312.1 Formal group associated to an elliptic curve . . . . . . . . . . . . 33

13 October 19, 2017 3413.1 Weierstrass minimal model . . . . . . . . . . . . . . . . . . . . . 34

14 October 25, 2017 3614.1 Inertia actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

15 October 26, 2017 3715.1 Unramified and tamely ramified extensions of local fields . . . . . 3715.2 Good reduction and unramified Tate modules . . . . . . . . . . . 38

16 October 31, 2017 4016.1 Global review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

17 November 1, 2017 4217.1 Kummer theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

18 November 2, 2017 4318.1 Class number theorem . . . . . . . . . . . . . . . . . . . . . . . . 4318.2 Concluding Mordell–Weil . . . . . . . . . . . . . . . . . . . . . . 44

19 November 14, 2017 4519.1 Hasse’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2

20 November 29, 2017 4720.1 Modular forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

21 November 30, 2017 5021.1 Modular curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

22 December 6, 2017 5222.1 Hecke operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

23 December 7, 2017 5423.1 Eichler–Shimura construction . . . . . . . . . . . . . . . . . . . . 5423.2 Two stories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

3

Math 99r Notes 4

1 September 5, 2017

I am going to give you an actual overview of the class. Let me start witha polynomial F (x, y) ∈ Q[x, y]. The question is, what are the Q-points ofF (x, y) = 0? The projective version of this is, that for a homogeneous polyno-mial F (x, y, z) ∈ Q[x, y, z], what are equivalence classes of Q-solutions here?

We can ask, for instance, what the solutions to ax + by + c = 0, or aX +bY + cZ = 0? These are easy because they are linear. So let’s complicatethe problem and ask for quadratic solutions. For instance, are there rationalsolutions to x2 + y2 = 05 or x2 + 4y2 = 27? In the quadratic case, there isthe Hasse principle, which says that the equation has rational solutions if it hassolutions in R and in Qp.

Now let’s move to cubic equations, like

f(X,Y, Z) = X3 + Y 3 + Z3+?X?Y ?Z? + · · · .

It is easy to find the Q-points if the curve is singular; it reduces to the previouscases. So we only consider the smooth case.

Let E be this curve, and E(Q) be the set of Q-points.

Theorem 1.1. E(Q) is actually naturally an abelian group, and it is finitelygenerated.

So what is the rank of this group? This is, by definition, the algebraicrank of E.

Conjecture 1.2 (Birch–Swinnerton-Dyer). The rank r can be computed by acertain L-function.

Theorem 1.3 (Nikolav). The torsion can be computed. If f(x, y) ∈ Eaffine(Q)tor,then either y = 0 or y | ∆E. This also works for number fields.

Theorem 1.4 (Mazur). E(Q)tor is either of Z/nZ for n = 1, . . . , 10, 12 orZ/2Z× Z/mZ for m = 1, 2, 3, 4.

Given E/Q, there are Tate modules we can construct,

lim←−n

E[`n] = T`E,

which has an action of Gal(Q/Q). This T`(E) is a rank 2 Z`-module, and anthese are the source of only Galois representations. What are the traces anddeterminants of this representation? It turns out that the trace of the Frobeniusis going to be related to the size of E(F`).

1.1 Elliptic curves

Let K be any field.

Definition 1.5. An elliptic curve E/K is a smooth projective curve E ⊆ P2K

of genus 1, together with an origin O ∈ E(K).

Math 99r Notes 5

We will later prove that E(K) is naturally an abelian group with O ∈ E(K)its origin. Another remark is that if C/Q is of genus g ≥ 2, then |C(Q)| < ∞.Also, if E/K is an elliptic curve over K where charK 6= 2, 3, then E can be cutout by y2 = x3 +Ax+B.

If we are over C, a smooth curve is a Riemann surface, and they are classifiedby their genus. For instance, a genus 1 curve is a torus, and a genus 2 curve is adouble torus, etc. Note that the torus can be made into an abelian group. ButI will also give a definition of a genus that works over all fields, and it won’t beobvious that these two notions of genus agree.

1.2 Varieties

Given an algebraically closed field K, we define affine space to be

AnK

= Kn

as a set, and topologize AnK

by declaring that closed subsets are

V (f1, . . . , fr) = common solutions of fi

for fi ∈ K[x1, . . . , xn]. This is sometimes called the Zariski topology. You canverify that this is a topology. But this is too coarse to actually make cohomologywell-behaved. Instead you should look at the etale cohomology, Het. In fact,

H1et(E,Z`)∨ = T`(E),

and this was one of the first examples of etale cohomology.If V ⊆ An is an irreducible affine variety, then we define

I(V ) = f ∈ K[x1, . . . , xn] : f(V ) = 0.

When we take V = V (a) for an ideal a, we get

Lemma 1.6. I(V (a)) =√a = f : fn ∈ a for some n.

So we don’t have a way to distinguish between x = 0 or x2 = 0. We wantto distinguish them, so we make a modified definition.

Definition 1.7. An affine variety consists of the pair (V (a),K[x1, . . . , xn]/a).

This ring K[x1, . . . , xn]/a is the ring of functions defined over the variety.This is because we look at all possible functions and quotient out be all stuffthat restricts to zero.

Math 99r Notes 6

2 September 7, 2017

Usually I will not prove the technical parts in class, or it will take an absurdamount of time or I will have to go absurdly fast. The goal today is to do morealgebraic geometry and then define elliptic curves.

I’m going to work with C or K for simplicity. An affine variety is a pair(X,OX), where OX is a sheaf that associates to each open subset the functionsdefined on it. What we are interested in this class is projective varieties. Theseare zero loci of homogeneous polynomials, inside Pn. This projective space is

Pn = (x0, . . . , xn) ∈ Kn+1 \ 0/ ∼,

quotiented out by nonzero scalar multiplication. But then it is more difficult tospecify the functions on this space.

What are functions on P1? Not all homogeneous polynomials work. Forinstance, the function

f(x0, x1) = x20 + x2

1

is not well-defined. But it makes sense to ask about the vanishing locus of f .So how are we going to study projective varieties? Each projective variety

can be covered by affine varieties. The projective space P1 can be covered by

A1 = (x0 : x1) : x0 6= 0 = (1 : x1/x0) : x0 6= 0,A1 = (x0 : x1) : x0 6= 0 = (x0/x1 : 1) : x1 6= 0.

Given an affine variety like V (y2 − x3 − x) ⊆ A2, you can add some pointsto make it a projective variety in P2. Because this is a degree 3 equation, youcan introduce a third variable and make it projective:

V (Y 2Z −X3 −XZ2) ⊆ P2.

2.1 Rational functions and dimension

Given an affine variety, the regular functions on that variety is the ring of regularfunctions on it. If X = V (a) is irreducible, or a is prime, then the ring K[V ] isan integral domain. So we can define the rational functions as

K(V ) = Frac(K[V ]) =ϕφ

: ϕ, φ ∈ K[V ], φ 6= 0.

This is the analogue of meromorphic functions.What if X is projective? In this case, the rational functions on X isϕφ

: ϕ, φ are homogeneous polynomials of same degree in K[x0, . . . , xn]/a.

The dimension of a variety is a really difficult concept in algebraic geometry.But we more or less know the dimension once we see it. In the notes I gave thelaziest definition of dimension.

There is a topological dimension, which is the length of the longest chain ofirreducible closed subsets. You can also define it as the transcendental degreeof the field of rational functions. It is the theorem that these two notions agree.

Math 99r Notes 7

2.2 Curves

A curve is an irreducible 1-dimensional non-singular projective variety. Fromnow on, I want you to think of it as a discrete valuation ring.

Definition 2.1. A discrete valuation ring or a DVR is a ring with a discretevaluation

v : R− 0 → Z≥0

that looks like a logarithm of a norm.

Example 2.2. The ring C[[x]] is a DVR with the valuation

C[[x]]− 0 → Z≥0;∑n≥0

anxn 7→ minn : an 6= 0.

This is the order of the function at 0.

Theorem 2.3. Being nonsingular and dimension 1 at a point is equivalent toOX,p being a DVR, which is also equivalent to being able to talk about the orderof zeros of functions.

For instance, in the variety V (xy) ⊆ A2, the order of x3 + xy = x3 at (0, 0)is not very well defined.

Lemma 2.4. Every rational map from a curve to another curve is actuallyregular.

Lemma 2.5. Any morphism ϕ : C1 → C2 is either 0 or surjective.

Theorem 2.6. The following descriptions are equivalent over any K withchar 6= 2, 3:

(1) A curve given by y2 = x3 + ax+ b such that ∆ 6= 0

(2) A non-singular cubic curve in P2 with a point O

(3) A non-singular curve in P2 of genus 1 together with a point O, i.e., andelliptic curve.

It is quite clear that (1) implies (2), with O being the point at infinity. (2)implies (3) is the degree-genus formula, and (3) implies (1) is Riemann–Roch.

2.3 Degree

Theorem 2.7. (1) A non-singular cubic curve E ⊆ P2 with a point O ∈ E hasa canonical abelian group structure.(2) A non-singlar curve E ⊆ P2 of genus 1 with a point O has a canonical groupstructure.

We are going give these group structures differently, and these two structurescoincide.

Math 99r Notes 8

Definition 2.8. A degree of a map ϕ : C1 → C2 is the cardinality of a genericfiber.

Note that because ϕ is not injective, it induces a pullback map

ϕ∗ : K(C2)→ K(C1); f 7→ f ϕ.

Since ϕ is non-constant, this is going to be nonzero. That is, it is a fieldextension.

Lemma 2.9. The degree [K(C1) : K(C2)] is equal to degϕ.

Now because K(C2)→ K(C1) is a finite field extension, there is a norm mapK(C1)→ K(C2).

Math 99r Notes 9

3 September 12, 2017

Last time we defined a curve, which is a dimension 1 non-singular projective(irreducible) variety. We also talked about the degree of a map between curves.I was planning to talk about the genus and Riemann–Roch, but we will postponethis.

Theorem 3.1. Let K = K be a field of characteristic not equal to 2 or 3. Thenthe following are equivalent descriptions:

(a) curves defined byy2 = x3 +Ax+B

that is non-singular,

(b) A non-singular cubic in P2K together with a point O,

(c) A non-singular curve in P2K of genus 1 with a point O.

Let’s take this as a black box now.

3.1 Group structure on an elliptic curve

We declare that three points that lie on a line add up to 0. In other words, weadd two points P and Q by taking the third intersection of the curve and PQ,and reflecting it.

Why is this well-defined? By Bezout’s theorem, a degree m curve and adegree n curve intersect “generically” at mn points. This means that the inter-section of the curve and the line PQ is, generically, three points.

So write the third intersection point as #(P,Q). We are going to now define

+ : E × E → E; P +Q = #(O,#(P,Q)).

We need check that this is a group. We also want to check that + is a morphism.Well it is clear that O is a group identity, and there is the inverse −P =

#(O,P ). For associativity, you need the following lemma.

Lemma 3.2. If ` = `1 q `2 q `3 ∈ P2 and h = h1 q h2 q h3 ∈ P2 be three lines,so that ` ∩ h = P1, . . . , P9. If a (non-singular) cubic in P2 goes through 8 ofthese, then it goes through them all.

If you think hard, you will see that this proves associativity. So we almost havea canonical abelian group structure on the curve E.

3.2 Divisors

I’m going to give you an incorrect definition.

Definition 3.3. A divisor D on a curve C is a (finite) formal sum D =∑P∈C nP [P ] for nP ∈ Z.

Math 99r Notes 10

This has nothing to do with the “group structure on elliptic curve”; it is justa formal sum. For a divisor D, we define

degD =∑P∈C

nP ,

and denote the set of divisors on C by Div(C). We also call Div0(C) thesubgroup of divisors with degree 0. This fits into the short exact sequence

0→ Div0(C)→ Div(C)deg−−→ Z→ 0.

Definition 3.4. For a rational function ϕ ∈ K(C)∗, define

div(ϕ) =∑

P∈Zero(ϕ)

nP [P ]−∑

Q∈Pole(ϕ)

nQ[Q].

Lemma 3.5. The degree deg(div(ϕ)) = 0.

Proof. A rational function is ϕ : C → P1. Then div = ϕ−1(0)−ϕ−1(∞) and soit has degree 0.

Definition 3.6. A principal divisor is a divisor of the form D = div(ϕ) forsome ϕ ∈ K(C)∗.

Denote this subgroup by PDiv(C) ⊆ Div0(C). Now we can complete thepicture as the following.

0 0

0 PDiv(C) Div0(C) Pic0(C) 0

0 PDiv(C) Div(C) Pic(C) 0

Z Z

0 0

deg

We will define a map

h : E → Pic0(E); P 7→ [P ]− [O].

Proposition 3.7. h is a isomorphism of groups.

Math 99r Notes 11

4 September 14, 2017

Last time there was an exercise, claiming that if C1 and C2 are two generic cubiccurves intersecting at 9 points, and another C goes through 8 of the points, thenit passes through all 9 of them.

Proof. Let V be the vector space of homogeneous cubic polynomials inK[X0, X1, X2].This has dimension 10. If this passes through 8 given points, then this conditioncuts down the dimension to 2. Now C1 and C2 are linearly independent vectorsin this space, so C is a linear combination of these two.

Our ultimate goal is to show that the following are equivalent:

• E : y2 = x3 + ax+ b and ∆E 6= 0,

• cubic curve in P2K

with a point,

• curve of genus 1 in P2K

with genus 1.

We also had this map

h : E → Pic0(E); P 7→ [P ]− [O].

4.1 Discrete valuation ring

Now let C/K be any curve. We have the sequence

0→ Div0(C)→ Div(C) Z→ 0.

There is also a partial order on Div(C), given by∑P

nP [P ] ≥∑P

mP [P ] ⇔ nP ≥ mP for all P.

Given a rational function ϕ, we can define its divisor as

div(ϕ) =∑

P∈zero(ϕ)

eP [P ]−∑

Q∈pole(ϕ)

eQ[Q].

But how do we know the order of something? I owe you an explanation ofDVRs.

Definition 4.1. The following description are equivalent, and we call this adiscrete valuation ring or DVR:

(1) a local PID which is not a field.

(2) R = OK for a discrete valuation field with v : K Z ∪ ∞ satisfyingv(xy) = v(x)+v(y), v(x+y) ≥ min(v(x), v(y)), and v(x) =∞ if and onlyif x = 0.

Math 99r Notes 12

Example 4.2. Consider the field Qp with the p-adic valuation. The ring ofintegers here, Zp, is a DVR. Another example is the field C((x)) with the valu-ation of taking the minimal n with an 6= 0. Then the ring of integers C[[x]] is aDVR.

Now consider A1, with the affine coordinate ring C[x]. The local ring at theorigin is going to be something like

OA1,0 =fg

: g(0) 6= 0,

which has a natural valuation. On the other hand, if you look at V (xy) ⊆ A2

and look at the local ring at (0, 0), then

OZ,0 =fg

: g(0) 6= 0

is not even an integral domain. This means that being a nonsingular curve isclosely related to locally being a DVR.

Definition 4.3. A uniformizer of a DVR R is some$ ∈ R such that v($) = 1.

Then any z ∈ R is uniquely expressed as z = u$v(z), where u ∈ R×.If ϕ : C1 → C2 is a map between curves, and ϕ(P ) = Q, then we have a

pullback mapϕ∗ : OC2,Q → OC1,P .

Now consider a uniformizer in OC2,Q and OC1,P and call them tQ and tP . Wedefine the ramification index at P as

eP = ordP (ϕ∗tQ),

so that ϕ∗tQ = utePP .There is an analogous picture in algebraic number theory. If SpecOK lies

over SpecZ, then we can look at the splitting

p = Pe11 ·P

e22 · · ·Per

r .

This ej is the analogue and this analogy can be made precise.

4.2 Differential forms

Definition 4.4. For C/K a curve, the space Ω1C/K

is (the global sections of

meromorphic differentials) the vector space over K(C) generated by symbolsdf : f ∈ K(C) subject to the following relations:

(1) da = 0 if a ∈ K,

(2) d(f + g) = df + dg,

(3) d(fg) = fdg + gdf .

Math 99r Notes 13

Example 4.5. Consider a curve C : y2 = x3 + x. Let ω = dx/2y be a differen-tial. Then we have

dx

2y=

dy

3x2 + 1

and so

ω =(−9

4xy)dx+

(3

2x2 + 1

)dy ∈ Ω1

C .

Math 99r Notes 14

5 September 19, 2017

I defined differential forms and gave some examples. For C a curve over K, themeromorphic differential forms Ω1

C is a K(C)-vector space of dimension 1. Iff ∈ K(C), df ∈ Ω1

C is a basis if and only if K(C)/K(f) is a finite separableextension.

Example 5.1. Consider the curve C : y2 = (x− e1)(x− e2)(x− e3). Take thefunction x− e1. Let Pj = (ej , 0) be the intersection with the y-axis. (These arethe points of order 2.) The divisor of this is

div(x− e1) = 2[P1]− 2[∞].

Why is this? For simplicity assume e1 = 0. To find the order of zero at P1, weneed to look at the local ring and compute what power of the uniformizer it is.The local ring is

(C[x, y]/y2 − x(x− e2)(x− e3))(0,0).

In this ring, x− e2 and x− e3 are units. So this ring is

C[x, y](0,0)/(y2 − ux) ∼= C[y](0).

with the identification x = y2/u. Then y can be taken the be the uniformizerand so x has valuation 2. As an exercise, show that

div(y) = [P1] + [P2] + [P3]− 3[∞].

5.1 Divisor of a meromorphic form

Recall that a map ϕ : C1 → C2 gives a field extension K(C1)/K(C2). We canalso pullback differential forms. We can define

ϕ∗ : Ω1C2→ Ω1

C1; ϕ(df) = d(ϕ∗f).

Lemma 5.2. The map ϕ∗ is injective if and only if ϕ is separable.

If ω ∈ Ω1C is a differential, its divisor is defined to be

div(ω) =∑

P∈zero(ω)

eP [P ]−∑

Q∈pole(ω)

eQ[Q].

Here eP is defined as the order of the rational function ω/dtP = ϕP locally atP . This gives the notion of zeros and poles.

Lemma 5.3. Both zero(ω) and pole(ω) consists of finitely many points.

Example 5.4. Let’s take the curve C : y2 = x(x− 1)(x+ 1). What is div(dx)?You can show that the zeros are precisely (0, 0), (1, 0), (−1, 0) and the pole isat ∞. If you work hard, you can show that

div(dx) = [P1] + [P2] + [P3]− 3[∞].

So as a coincidence, we have

div(dxy

)= 0.

Math 99r Notes 15

5.2 Canonical divisor and genus

Given a curve, there is a canonical way of getting a divisor. Look at anydifferential form, and consider its divisor. This gives a divisor, and the classdoes not depend on the choice of the differential form. This is called canonicaldivisor.

Actually, there is a better way of looking at this. A divisor is actually anisomorphism class of line bundles. In this interpretation, we’re just looking atthe cotangent line bundle.

If D ∈ Div(C) is a divisor, we may consider

L (D) = ϕ : div(ϕ) +D ≥ 0 ∪ 0

which is a vector space over K of finite dimension. For example, L (0) = K andL (D) = 0 if degD < 0.

Define`(D) = dimK(L (D)).

If D = D′ + div(ϕ) for some ϕ, then `(D) = `(D′).

Definition 5.5. The genus of C is defined as gC = `(KC), where KC is thecanonical divisor.

Consider an curve C : y2 = x3 + ax + b. Then the form ω = dx/y hasdiv(dx/y) = 0. So the canonical divisor is 0 and `(KC) = 1 = gC .

5.3 Degree-genus formula and Riemann–Roch

I am going to black box two theorems.

Theorem 5.6 (Degree-genus formula). Consider a curve C in P2 of degree d.The genus of this curve is given by

g =(d− 1)(d− 2)

2.

Theorem 5.7 (Riemann–Roch). Let C be a smooth projective curve. For anydivisor D,

`(D)− `(KC −D) = deg(D)− gC + 1.

Implicitly, there is a Serre duality here that interchanges a certain H0 andH2. The dualizing sheaf KC plays a role in this.

For example, plug in D = 0. Then we get that this is correct. Next, if weput D = KC , we get

g − 1 = deg(KC)− g + 1

and so deg(KC) = 2g − 2.Consider a curve of genus 1 in P2. I claim that it looks like

y2 + a1xy + a3y = x3 + a2x2 + a4x+ a6,

Math 99r Notes 16

i.e., there is an isomorphism ϕ → C such that O 7→ ∞. I’ll give a sketch.Consider the divisor n[O] with n ≥ 1. By Riemann–Roch, we have

`(n[O]) = n+ `(KC − n[O]) = n.

Since `([O]) = 2, there is a non-constant element x ∈ L (2[O]). Then thereis another basis y ∈ L (3[O]). Then 1, x, x2, x3, y, xy, y2 are in L (6[O]). Thisshows that they are linearly dependent.

Math 99r Notes 17

6 September 21, 2017

For the curve y2 = (x− e1)(x− e2)(x− e3), we have

div(x− ei) = 2[Pi]− [∞], div(y) = [P1] + [P2] + [P3]− 3[∞].

Also div(dx) = div(y).We have

K(C) = Frac(K[x, y]/(y2 −

∏i(x− ei))

).

The extension K(C)/K(x) is comes from the map x : C → P1 and the degreeof this morphism is 2. This is because div(x− e1) has two points over 0 and ∞.Likewise, y : C → P1 has degree 3 and so K(C)/K(y) has degree 3.

Our first goal is to show that a genus 1 curve in P2 looks like y2+a1xy+a3y =x3 + a2x

2 + a4x + a6. Our second goal is to prove that E ∼= Pic0(E) is anisomorphism of groups.

6.1 Identifying the genus 1 curve

The tool we are going to use is Riemann–Roch: for D ∈ Div(C),

`(D)− `(KC −D) = degD − g + 1.

An application is that degKC = 2g − 2.

Corollary 6.1. Suppose g = 1. If [P ] = [Q] ∈ Pic(C) for P,Q ∈ C, thenP = Q.

Proof. Suppose div(ϕ) = P −Q. Then ϕ ∈ L ([Q]) but on the other hand,

`([Q]) = `(KC − [Q]) + deg([Q])− g + 1 = 1.

This implies that ϕ is a constant function and so P = Q.

Another observation to make is that if P ∈ C and gC = 1, then L ([P ]) = K.

Proposition 6.2. Suppose C is a genus 1 curve and let O be a point. Thenthere exist E ⊆ P2 and an isomorphism α : C

∼−→ E ⊆ P2 with α : O 7→ ∞,where E is given by

E : y2 + a1xy + a3y = x3 + a2x2 + a4x+ a6.

Proof. Consider `(n[O]) for n ≥ 1. This is

`(n[O]) = `(Kc − n[O]) + n− 1 + 1 = n.

We have `([O]) = 1 and `(2[O]) = 2. So there is a non-constant function x, suchthat x has a pole of order 2 at O. Next, `(3[O]) = 3 and so there is a functiony that has a unique pole of order 3 at y.

Math 99r Notes 18

On the other hand, `(6[O]) = 6 but

1, x, x2, x3, y, xy, y2 ∈ L (6[O]).

This implies that these are linearly dependent. This gives a relation betweenx, y ∈ K(C) satisfying

y2 + a1xy + a3y = x3 + a2x2 + a4x+ a6. (∗)

This gives a map α : C → E where E ⊆ P2 is defined by the equation (∗).We now want to show that degα = 1. This will imply that α is an isomor-

phism. Note that x has a unique pole of order 2. This shows that K(C)/K(x)is of degree 2. Likewise K(C)/K(y) is of degree 3.

K(C)

K(E)

K(x) K(y)

2 3

This shows that K(C)/K(E) has degree 1.

6.2 Picard group of an elliptic curve

Theorem 6.3. The map

h : E → Pic0(E); P 7→ [P ]− [O]

is an isomorphism of groups.

Proof. We first show that it is a bijection of sets. Let us first prove surjectivity.For any D ∈ Div0(E), we want to find a P ∈ E and ϕ such that D−([P ]−[O]) =div(ϕ). We have

`(D + [O]) = 0 + 1− 1 + 1 = 1

by Riemann–Roch. So there exists a ϕ 6= 0 such that divϕ + D + [O] ≥ 0.Because this is of degree 1, there exists some P ∈ E such that divϕ+D+[O] =[P ].

For injectivity, we need to prove that [P ] − [O] = [Q] − [O] then P = Q.This is by the first corollary today.

Now it suffices to show that h(P + Q) = h(P ) + h(Q). This can be seenfrom the geometric construction we used. If P,Q,R lie on the same line, then[P ] + [Q] + [R]− 3[O] is the divisor of the equation that cuts out that line.

Lemma 6.4. The group + : E × E → E is a morphism of varieties.

Math 99r Notes 19

6.3 Review of number theory

The big goal is to prove Mordell–Weil theorem, i.e., if E is defined over a numberfield K, E(K) is finitely generated. Next, we are going to study, for a F/Qp, anelliptic curve E over F or Qp. From this we get a Galois representation. In somesense, number theory is studying the group Gal(Q/Q), with class field theoryconcerned with only 1-dimensional representations. But the interesting thing isthat elliptic curves produce a large number of 2-dimensional Galois representa-tions of Qp/Qp. We want to understand the characters of this representationand when it is unramified.

Let us talk about extensions of number fields. You should look up Qp. InsideQ there is the ring of integers Z. For K/Q, I have the ring of integers OK ⊆ K.This is

OK = α ∈ K : α is a root of a monic integral polynomial.

These are not PIDs, e.g., 2 · 3 = (1 +√−5)(1−

√−5) ∈ Q[

√−5]. But they are

Dedekind domains and have unique factorization of prime ideals. Then (OK)pare DVRs. So they can be though of as curves. So for this extension Z → OK ,we can similarly define ramification.

If p is a prime, we can write

(p) = pe11 · · · perr .

Definition 6.5. The ei is the ramification index of pi/p in K/Q.

If K/Q is Galois, then e(pi/p) = e(pj/p).

Math 99r Notes 20

7 September 29, 2017

I have defined elliptic curves over algebraically closed fields. But this is notvery satisfactory, because we want to talk about rational solutions. In general, avariety X is defined over K if it the vanishing locus of functions with coefficientsin K. That is, when its ideal can be generated by polynomials with rationalcoefficients. For instance, (πy − πx2) ⊆ C[x, y] is defined over Q.

Let L/Q be finite. If E/L is an elliptic curve, its rational points are rationalsolutions. We could also thing of this as

E(L)GL = E(L),

where GL is the absolute Galois group.

7.1 Isogeny

Definition 7.1. Let E1, E2 be two elliptic curves (over k). An isogeny is amorphism ϕ : E1 → E2 such that ϕ(OE1

) = OE2.

We are also going to write

Hom(E1, E2) = Isog(E1, E2).

This is actually an abelian group, because if ϕ1, ϕ2 ∈ Hom(E1, E2), then we candefine.

(ϕ1 + ϕ2)(x) = ϕ1(x) +2 ϕ2(x).

Lemma 7.2. Every isogeny ϕ : E1 → E2 is a group homomorphism, i.e.,ϕ(x+ y) = ϕ(x) + ϕ(y).

Proof. We have the group isomorphism E → Pic0(E) given by P 7→ [P ] − [O].So we have

E1 Pic0(E1)

Ex Pic0(E2).

ϕ

∼=

ϕ∗

∼=

Because ϕ∗ is a group homomorphism, we also have that ϕ is a group homo-morphism.

If ϕ : E1 → E2 is non-constant, it is surjective and we can look at it.

0→ ker(ϕ)→ E1ϕ−→ E2 → 0.

If there is no ramification, the kernel is going to be have size degϕ.

Lemma 7.3. ϕ is separable if and only if ϕ∗ : ΩE2→ ΩE1

is nonzero.

Math 99r Notes 21

If ϕ is separable, then |ker(ϕ)| = degϕ.For m 6= 0 ∈ k, there is the multiplication map

[m] : E → E; P 7→ mP.

Lemma 7.4. (1) [m] 6= 0 if m is an integer that is nonzero in k.(2) [m] is separable of degree m2.

Corollary 7.5. Then ker([m]) is isomorphic to Z/mZ × Z/mZ (as a groupscheme, or over an algebraically closed field).

Proof. We know that ker([m]) is an m-torsion abelian group of order m2. Also,for n | m, there are n2 elements of order dividing n. This implies that the groupis actually Z/mZ× Z/mZ.

7.2 Invariant differential

To prove separability of [m], we need to find a differential form that does notpull back to zero.

Theorem 7.6. Let Ta be the translation by a ∈ E. There exists a differentialω such that T ∗aω = ω for all a.

The proof of this goes by giving an explicit differential in coordinates.

Corollary 7.7. Let ϕ ∈ Hom(E1, E2). Then (ϕ+ ψ)∗ω = ϕ∗ω + ψ∗ω.

Using this, we get [m]∗ω = mω. This shows the first part of (2) in thelemma, that [m] is separable.

Math 99r Notes 22

8 October 3, 2017

Our goal is to show that for E/K, that E(K) is finitely generated. This meansthat E(K) has finite rank and torsion.

Definition 8.1. The rank of E(K) is called the algebraic rank of E/K.

To show that E(K) is finitely generated, we will show that E(K)/mE(K)is finite and use height to show that E(K) is finitely generated.

8.1 Group cohomology

We have the short exact sequence

0→ E(K)[m]→ E(K)[m]−−→ E(K)→ 0,

and we are going to take the Galois invariant. Taking invariants is left exact.So it continues as

0→ E(K)[m]→ E(K)m−→ E(K)→ H1(GK , E(K)[m])→ · · · .

Our first job is to identify what Hi are.Let M be an abelian group with an action of G. Then there are abelian

groups Hi(G,M). This is created in a way such that whenever

0→M ′ϕ−→M →M ′′ → 0

is an short exact sequence of abelian groups with equivariant maps, (i.e. a shortexact sequence of G-module) then you get a long exact sequence

0→ (M ′)G →MG → (M ′′)G → H1(G,M ′)→ H1(G,M)→ H1(G,M ′′)→ · · · .

Here, H0(G,M) = MG, and we are going to have

H1(G,M) =Z1(G,M)

B1(G,M)=(ϕ : G→M) : ϕ(g1g2) = g1ϕ(g2) + ϕ(g1)(ϕ : G→M) : ϕm(g) = gm−m for some m

.

Here are some properties:

(1) Suppose M receives a trivial G action. Then Z1(G,M) = HomGrp(G,M)and B1(G,M) = 0. So we have

H1(G,M) = HomGrp(G,M).

(2) There is something called inflation-restriction. Suppose G acts on M andH is a normal subgroup of M . Then G/H acts on MH . Then we have anexact sequence

0→ H1(G/H,MH)inf−−→ H1(G,M)

res−−→ H1(H,M).

Math 99r Notes 23

We have, by group cohomology,

0→ E(K)[m]→ E(K)m−→ E(K)

→ H1(GK , E(K)[m])→ H1(GK , E(K))m−→ H1(GK , E(K))→ · · · .

So we have an injection

0→ E(K)/mE(K) → H1(GK , E(K)[m]) H1(GK , E(K))[m]→ 0.

Lemma 8.2. Suppose L/K is finite Galois. If E(L)/mE(L) is finite, thenE(K)/mE(K) is finite.

Proof. Let us consider the map E(K)→ E(L). Let us take the quotient to get

0 ker(i) E(K)/mE(K) E(L)/mE(L)

0 H1(Gal(L/K), E(K)[m]GL) H1(GK , E(K)[m]) H1(GL, E(K)[m]).

i

But we have that Gal(L/K) is finite and E(K)[m] is finite. So this H1 isfinite and so ker(i) is finite. Because E(L)/mE(L) is finite, E(K)/mE(K) isfinite.

Let us first replace K by K([m]−1(O)). This is adjoining all the coordinatesof points in [m]−1E(O) to K. This only extends K in finite dimension. Theresult is that GK acts trivially on E(K)[m]. Then we get an inclusion

E(K)/mE(K) → HomGrp(GK , E(K)[m]).

In other words, we have a pairing

E(K)/mE(K)×GK → E(K)[m]

that is nondegenerate on the left side. What is the kernel of the right?

Proposition 8.3. The kernel on the right is GL, where L = K([m]−1E(K)).

Math 99r Notes 24

9 October 5, 2017

Let E/K be an elliptic curve, and let m ≥ 2.

Theorem 9.1. L = K([m]−1E(K)) is a finite extension over K.

Theorem 9.2 (Weak Mordell–Weil). The group E(K)/mE(K) is finite.

Theorem 9.3 (Mordell–Weil). The group E(K) is finitely generated.

Today, we are going to show that Theorem 9.1 implies Theorem 9.2, andthat this implies Theorem 9.3.

9.1 Weak Mordell–Weil theorem

Firstly, we may assume without loss of generality, that E[m] is defined over K,after taking a finite extension. This is because of the statement that K ′/K isfinite Galois then E(K ′)/mE(K ′) finite implies E(K)/mE(K) finite. The proofof this fact went by looking at the inflation-restriction sequence

0 ker(i) E(K)/mE(K) E(L)/mE(L)

0 H1(Gal(L/K), E(K)[m]GL) H1(GK , E(K)[m]) H1(GL, E(K)[m]).

i

The upshot of this is that GK acts trivially on E(K)[m], and so we get

E(K)/mE(K) → H1(GK , E[m]) = Hom(GK < E[m]).

So there is a Kummer pairing

E(K)/mE(K)×GK → E[m].

What is the kernel on the right? We need to figure out what this map is.We had applied the Galois invariants of the sequence

0→ E[m]→ E(K)×m−−→ E(K)→ 0

to get

0→ E(K)[m]→ E(K)×m−−→ E(K)

φ−→ H1(GK , E[m])→ · · · .

In general, the boundary map φ can be described. A short exact sequence0 → M ′ → M → M ′′ → 0 gives rise to a boundary map M ′′ → H1(G,M ′).This is described as, for each m ∈ M ′′ look at any preimage n ∈ M and thenlooking at the cocycle

ϕm : G→M ′; g 7→ gn− n.

Math 99r Notes 25

So let’s apply this. For P ∈ E(K), pick a Q ∈ E(K) such that mQ = P . Sothe above recipe shows that this pairing is given by

(P, σ) 7→ σQ−Q.

So the kernel of the Kummer pairing on the right is going to be elements σ ∈ GKsuch that σQ = Q for all Q ∈ [m]−1E(K). That is, it is precisely

L = K([m]−1E(K)).

In other words, the pairing actually is defined on

E(K)/mE(K) → Hom(GK/GL, E[m]) = Hom(Gal(L/K), E[m]).

If we know that Gal(L/K) is finite, then we deduce the E(K)/mE(K) is finite.

9.2 Mordell–Weil theorem

For simplicity, let’s do this over K = Q. We can’t say that that weak Mordell–Weil immediately implies Mordell–Weil. Take for instance Q. The idea is thatwe do have something like the Euclidean algorithm on Z but not on Q. BecauseE(Q)/mE(Q) is finite, we can label representatives as P1, . . . , Pr. Then for allP , we can write P = mQ1 +Pi1 for some Pi1 , then Q1 = mQ2 +Pi2 and so on.If we eventually hit a fixed finite set R1, . . . , Rs that we understand well, weare in good shape.

Lemma 9.4. Suppose A is an abelian group such that there exists an m ≥ 2such that A/mA is finite. Suppose there exists a height function h : A → Rin the following sense:

(1) there exists a constant β such that h(mx) ≥ m2h(x)− β,

(2) for each y ∈ A, there exists a constant γy such that h(x+ y) ≤ 2h(x) + γyfor all x ∈ A,

(3) for every δ, the set z ∈ A : h(z) < δ is finite.

Then A is finitely generated.

Proof. Take a representative P1, . . . , Pr ⊆ A of A/mA. Write

P = mQ1 + Pi1 , Q1 = mQ2 + Pi2 , Q2 = mQ3 + Pi3 , . . . .

Then the condition (1) implies

h(Qn+1) ≤ 1

m2(β + h(mQn)) =

1

m2(β + h(Qn − Pin+1)).

Let γ = maxγ(−P1), . . . , γ(−Pr). Then we can continue

h(Qn+1) ≤ 1

m2(β + h(Qn − Pin+1

)) ≤ 1

m2(β + γ + 2h(Qn)) ≤ 2

m2h(Qn) + c.

So if we iterate this process, we are going to get h(Qn) bounded for n sufficientlylarge. Then (3) shows that there can be only finitely many such possible Qn.Now A is generated by Pi and the finitely many possible such Qn.

Math 99r Notes 26

Theorem 9.5. Consider an elliptic curve E/Q given by y2 = x3 + Ax + B.Define the function

h : E(Q)→ R;(axbx, y)7→ log(max|ax|, |bx|).

This is a height function.

Proof. I will only prove (2) and (3). You know how to compute the group lawin coordinates, and here it is. If P1 = (x1, y1) and P2 = (x2, y2), then

x(P1 + P2) =( y1 − y2

x1 − x2

)2

− (x1 + x2).

Now let P0 = (x0, y0) = (a0d20, b0d30

), and let P = ( ad2 ,bd3 ). Plug it in, and you

get

x(P0 + P ) =(aa0 +Ad2d2

0)(ad20 + a0d

2) + 2Bd4d40 − 2bdb0d0

(ad20 − a0d2)2

.

Thenh(x(P0 + P )) ≤ Cx0,y0 + log max|a|2, |d|4, |bd|.

So we are almost done. P is a point on the elliptic curve, so

d2b2 = d2a3 +Aad6 +Bd8 ≤ CA,B max|a|4, |d|8.

This shows that

h(x(P0 + P )) ≤ Cx0,y0,A,B + log max|a|, |d|2.

You can check (1) similarly.The third condition is clear, because there are just finitely many rationals

with bounded height.

Corollary 9.6. E(Q) is finitely generated.

Next time we will start looking at Tate modules.

Math 99r Notes 27

10 October 10, 2017

Today we are going to talk about Tate modules. We will first define p-adicnumbers.

10.1 p-adic numbers

If p is a prime, we can consider

· · · → Z/p3Z→ Z/p2Z→ Z/pZ→ Z.

Then we can define Zp to be its limit, Zp = lim←−n Z/pnZ. This consists of

elements

(x1, x2, . . .) ∈ Z/pZ× Z/p2Z× · · · , xn mod pn−1 = xn−1.

The ring Zp has characteristic zero, because if N < pk, then N 6= 0 ∈ Z/pkZand hence N 6= 0 ∈ Zp.

Every Z/pkZ has a discrete topology, and we can use this to give Zp aprofinite topology. The topology on Zp is totally disconnected, i.e., for anyx 6= y, they are in different connected components.

Algebraically, Zp is a discrete valuation ring with maximal ideal pZp andresidue field Fp. We also write Qp for the fractional field of Zp. Because Qp isa local field, there is a short exact sequence

0→ IQp→ GQp

→ GFp→ 0,

where I is the inertia group. Here, we know GFp∼= Z.

10.2 Tate modules

Let E/k be an elliptic curve, and ` 6= char k be a prime. We have

E[`n] ∼= Z/`nZ× Z/`nZ.

We can take the inverse limit, and define the Tate module as

T`E = lim←−n

E[`n] ∼= Z` × Z`.

Now the Galois group Gk acts on E[`n], and so we get a continuous actionof Gk on T`E, with respect to the profinite topologies on each side. Then weget a group homomorphism

ρE : Gk → AutZ`T`E = GL2(Z`).

That is, there is a `-adic representation attached to E. This is also related toetale cohomology with Z` and Z` coefficients. More precisely, H1

et(E,Z`)∨ ∼=T`E.

Math 99r Notes 28

We can use Tate modules to study isogenies between elliptic curves. Supposewe have some isogeny ϕ:E1 → E2. Clearly this induces a map E1[`n]→ E2[`n],and if we take the inverse limit, we get

ϕ` : T`E1 → T`E2.

That is, we get a map

Hom(E1, E2)→ HomZ`(T`E1, T`E2).

Proposition 10.1. This map is injective.

Proof. Suppose ϕ : E1 → E2 such that ϕ` = 0. Then ϕ : E1[`n]→ E2[`n] is zerofor all n, so kerϕ has infinite cardinality. This is impossible unless ϕ = 0.

Proposition 10.2. Actually, Hom(E1, E2) ⊗ Z` → HomZ`(T`E1, T`E2) is an

injection.

Here, note that Hom(E1, E2) is torsion-free.

Proof. Let ϕ ∈ Hom(E1, E2)⊗ Z` such that ϕ` = 0. We can write

ϕ = αψ1 + · · ·+ αsψs,

where αi ∈ Z`.Fix n ≥ 1 and choose integers a1, . . . , an such that ai ≡ αi (mod `n). Now

ψ = a1ψ1 + · · ·+ asψs

is a true isogeny, and we have ψ(E1[`n]) = 0. Then we can write ψ = [`n] λfor some λ ∈ Hom(E1, E2). But we can’t yet conclude that αi ≡ 0 (mod `n),which is what we need.

So let us go back to representing ϕ. Let

M = Zψ1 + Zψ2 + · · ·+ Zψs ⊆ Hom(E1, E2)

be a Z-submodule. Then let

Mdiv = ψ′ ∈ Hom(E1, E2) : [m] ψ′ ∈M for some m.

Note that M ⊆ Mdiv ⊆ 1dM , where d = lcmi degψi. This shows that Mdiv

is still a free abelian group of finite rank. (Everything is torsion-free becauseHom(E1, E2) is torsion-free.)

Now replace ψ1, . . . , ψs by a basis of Mdiv. We had ψ = [`n] λ for some λ,and by divisibility, we also have λ ∈Mdiv. So we can write

λ = b1ψ1 + · · ·+ bsψs,

and by linear independence of ψi, we get ai = `nbi. That is, αi ≡ 0 (mod `n).

Math 99r Notes 29

Corollary 10.3. rankZ Hom(E1, E2) ≤ 4.

Theorem 10.4 (Isogeny theorem). We have an isomorphism Homk(E1, E2)⊗Z`∼=−→ HomZ`

(T`E1, T`E2)Gk , when

(1) k is a finite field (Tate 1966, Endomorphisms of abelian varieties overfinite fields)

(2) k is a number field (Faltings 1983, Finiteness theorems for abelian varietiesover number fields).

10.3 Endomorphism ring of an elliptic curve

Let ϕ : E1 → E2 be an isogeny. This map induces a map

ϕ∗ : Pic0E2 → Pic0E1;∑

miPi 7→∑

miϕ−1(Pi).

Recall that we also have a map λE : E∼=−→ Pic0(E). So we can define the dual

ϕ∨ : E2

λE2−−→ Pic0(E2)ϕ∗−−→ Pic0(E1)

λ−1E1−−→ E1.

This is called the dual isogeny.

Proposition 10.5. Let ϕ : E1 → E2 be a degree m isogeny.

(1) ϕ∨ ϕ = [m] on E1 and ϕ ϕ∨ = [m] on E2.

(2) (ϕ+ ψ)∨ = ϕ∨ + ψ∨.

(3) (ϕ ψ)∨ = ψ∨ ϕ∨.

(4) (ϕ∨)∨ = ϕ.

(5) [m]∨ = [m].

This is what we will use the understand End(E). We have already notedthat rankZ End(E) ≤ 4, and we also have this anti-involution ∨ satisfying (3)and (4).

Proposition 10.6. Suppose R is a ring and ι : R → R is a anti-involutionsatisfying rkZR ≤ 4. Also assume ι satisfies

• ι(ϕ+ ψ) = ι(ϕ) + ι(ψ),

• ι(ϕψ) = ι(ψ)ι(ϕ),

• ι(m) = m,

• ϕι(ϕ) ∈ Z≥0.

Then R is one of the following:

(1) R = Z

Math 99r Notes 30

(2) R is a subring of an imaginary quadratic field of finite Z-rank,

(3) R is a subring of the quaternion algebra of finite Z-rank.

Definition 10.7. If E/k is an elliptic curve of End(E) ) Z, then we say thatE has complex multiplication.

If k has characteristic zero, only (1) and (2) can happen. If k has character-istic p, then only (2) and (3) can happen. This is because there is the Frobeniusthat acts on k.

10.4 Weil pairing

Suppose E/k is an elliptic curve.

Proposition 10.8. For a positive integer n > 0, relatively prime to char k,there exists a non-degenerate, Galois equivariant, bilinear pairing

en : E[n]× E[n]→ µn ⊆ k

that is alternating, i.e., en(x, x) = 1, and is compatible with different n. Thisis called the Weil pairing.

Hence this induces

e` : T`E × T`E → µ`∞ = Z`(1).

Another way to think about this is that we have the cup product on H1, whichtakes them to H2, and then some Poincare duality gives the map.

Here is another way to think about it. Note that∧

2T`E ∼= Z`(1). Thiscorresponds to the determinant map

det ρE : Gk → Z×`

of the character.

Math 99r Notes 31

11 October 12, 2017

For an elliptic curve, we can look at its rational points E(Q), which is finitelygenerated. On the other hand, we now have a way of looking at torsion pointsE[m] or T`(E), which also receives an action from Gal(Qp/Qp). Here, we canalso look at local models, like E/Qp, and consider T`(Qp) having an action ofGal(Qp/Qp).

11.1 Reduction

Recall the proof of weak Mordell–Weil. WE had the embedding

E(K)/mE(K) → H1(GK/K , E(K)[m]),

and then we had an embedding

Gal(L/K) → Hom(E(K)/mE(K), E(K)[m]).

This shows that L/K is an abelian, and all elements are killed after multiplyingm.

Now there is a theorem that says that if L/K is abelian and there is someN such that σN = 1 for all σ ∈ Gal(L/K), and L/K is unramified over all butfinitely many primes, then it is finite.

Here is the philosophy. We have two ways of mapping Zp to a field:

Fp Zp Qp.

Now the curve xy = p in A2 is nonsingular if we change to Qp but it is singularif we change to Fp.

Now Zp can be considered as a line, and the two fibers are going to be theone defined over Qp and the one over Fp. We can ask, for a curve E definedover Qp, whether there is a curve E such that it has good reduction over Fp.

This is our strategy for studying E(K)[m], for K/Qp. There is some myste-rious “Neron model” E so that we have E(K) ∼= E(OK . Then we can actuallyreduce

E(K)[m]

ker = E E(K) Es(ks)

E(OK)

∼=

reduction

reduction

The key theorem is now that E(K)[m] → Es(ks) is a injective map. If youchange the diagram, this is equivalent to that ×m : E → E is injective.

Let us try to think what E looks like. We have

pZp → Zp → Fp, 1 + pZp → Z×p → F×p .

Here, E really is going to be a formal group.

Math 99r Notes 32

11.2 Formal group

The group law for pZ can be written as (x, y) 7→ x + y, and the group law for1 + pz can be written as (x, y) 7→ x+ y + xy.

Definition 11.1. A commutative formal group law is a power series F (x, y) ∈K[[x, y]] such that

(1) F (x, y) = x+ y + (higher order terms),

(2) F (F (x, y), z)) = F (x, F (y, z)),

(3) F (x, y) = F (y, x).

Example 11.2. There are F (x, y) = x+ y, F (x, y) = x+ y+xy, and there areF (x, y) for elliptic curves.

Lemma 11.3. (1) F (x, 0) = x and F (y, 0) = y.

(2) There exists a unique i(x) such that F (x, i(x)) = F (i(x), x) = 0.

Proof. (1) is an exercise. (2) follows from Hensel’s lemma.

A morphism between two formal group laws from F (x, y) to G(x, y) is anf ∈ K[[t]] such that

f(F (x, y)) = G(f(x), f(y)).

Definition 11.4. Let me give a naıve definition. Let K/Qp. A formal groupover K is (mK , F ) where F is a formal group law and mK ⊆ OK is the maximalideal.

Here, if x, y ∈ mK , then F (x, y) actually converges to an element in mK ,and it turns out that this forms a commutative group on mK .

Proposition 11.5. Let (mK , F ) be a formal group.

(1) [m]T = mT + higher order terms.

(2) If a ∈ OK is invertible and f(T ) = aT + higher order terms, then thereexists a g such that f(g(T )) = f(g(T )) = T .

Proof. (1) We can do this by induction.(2) This is by Hensel’s lemma. I claim that there exist gn(T ) ∈ OK [[T ]]

for all n such that f(gn(T )) ≡ T (mod Tn+1) and gn+1(T ) ≡ gn(T ) (mod Tn).Take g1(T ) = a−1T . Now take gn+1 = gn + λTn. Then we will have

f(gn+1(T )) ≡ f(gn(T )) + aλTn = T + aλTn + bTn (mod Tn+1).

Then set λ = −b/a.

Math 99r Notes 33

12 October 17, 2017

We were talking about formal groups. This is a power series F (x, y) ∈ R[[x, y]]satisfying

(1) F (x, y) = x+ y + higher order terms,

(2) F (x, F (y, z)) = F (F (x, y), z),

(3) F (x, y) = F (y, x),

(4) there exists a i(x) such that F (x, i(x)) = F (i(x), x) = 0,

(5) F (x, 0) = x, F (0, y) = y.

Here, (4) and (5) are implied by (1), (2), (3).

12.1 Formal group associated to an elliptic curve

If I can associate a power series to each group, you get a better feeling of whatthe group looks like. Let us consider the elliptic curve that looks like

y2 + a1xy + a3y = x3 + a2x2 + a4x+ a6

and choose z = −xy and w = − 1y . Then we get

w = z3 + a1zw + a2z2w + · · · = f(z, w),

and we can regard w = w(z) as a function of z to parametrize. We would thenhave

w(z) = f(z, w) = f(z, f(z, w)) = f(z, f(z, f(z, w))) = · · ·= z3(1 +A1z +A2z

2 + · · · )

which is a power series in z. Then we get a parametrization

x(z) =1

z2(1 +A1z + · · · ), y(z) = − 1

z3(1 +A1z + · · · ).

This is how you get the formal group. If you have z1, z2 ∈ E, you have agroup law z3 = z3(z1, z2) on the elliptic curve, and you can check that

F (z1, z2) = z1 + z2 + · · ·

and it is a formal group law because addition on an elliptic curve is commutativeand associative.

Definition 12.1. A morphism f : F → G is a power series f ∈ R[[T ]] suchthat f(0) = 0 and f(F (x, y)) = G(f(x), f(y)).

Proposition 12.2. Let f = aT + higher terms with a ∈ R×. Then there existsa unique g such that f(g(T )) = g(f(T )) = T .

Corollary 12.3. If m ∈ R×, then [m] : F → F is an isomorphism.

Math 99r Notes 34

13 October 19, 2017

13.1 Weierstrass minimal model

Recall that the curve E : y2 = x3 + Ax + B is nonsingular if and only if∆E = 4A3 + 27B2 6= 0. Let E/Kv be a curve over a local field. (For thislecture, take Kv = Qp.) If the coefficients are A,B ∈ Qp, then you can changevariables to get

Y 2 = X3 + (Au4)X + (Bu6).

Now we want to choose the minimal u such that Au4, Bu6 ∈ Zp.

Definition 13.1. The Weierstrass minimal model for E/Qp is the Y 2 =X3 +A′X +B′ such that A′, B′ ∈ Zp and v(∆′) is minimal among all possiblechoices. (This is unique up to units.)

Let E be a minimal model. Define the special fiber Es as

Es = E ⊗Zp Fp : y2 = x3 +Ax+B.

Then this is a curve over Fp. This has no reason to be an elliptic curve. Forinstance, when

y2 = x3 +1

p5,

we would get Es : y2 = x3 + p = x3, so it is singular. This is a particular caseof Neron models.

Let E be an elliptic curve over Q, and let Emin be its minimal model. Letus assume that Es is an elliptic curve.

Lemma 13.2. E(Qp) = Emin(Zp).

Lemma 13.3. Consider

0→ ker(r)→ Emin(Zp)r−→ Es(Fp)→ 0

and M(E) the formal group (m = (p), FE). There is an isomorphism M(E)→ker(r).

Corollary 13.4. The map [m] : E(Qp) → E(Qp) with p - m induces an iso-morphism on ker(r).

Proof. Because m ∈ Zp×

, the map [m] on the formal group is invertible.

So we get a diagram, and applying the snake lemma gives the following

Math 99r Notes 35

corollary.

0 0

0 E(Qp)[m] Es(Fp)[m]

0 ker(r) E(Qp) Es(Fp) 0

0 ker(r) E(Qp) Es(Fp) 0

[m]∼= [m] [m]

Corollary 13.5. If Es is nonsingular, i.e., E has good reduction, then

ψ : E(Qp)[m] → Es(Fp)

is injective for p - m.

Then consider Gal(Qp/Qp) acting on E[m] = E(Qp)[m]. Also note thatthere is

0→ Ip → Gal(Qp/Qp) Gal(Fp/Fp)→ 0.

Lemma 13.6. The map ψ is Gal(Qp/Qp)-equivariant.

This shows that Ip trivially on E[m]. That is E[m] is unramified. Take` 6= p. Then Ip acta trivially on E[`n] for all n, so Ip acts trivially on the Tatemodule T`E = lim←−nE[`n].

Math 99r Notes 36

14 October 25, 2017

We had the minimal model E/Zp. We need to show that E(Qp) ∼= E(Zp) and

0→M(E)→ E(Zp)→ Es(Fp)→ 0.

Corollary 14.1. If E has good reduction, then GQpacting on E[m] is unram-

ified.

14.1 Inertia actions

Take Qp as an example. Consider the absolute Galois group GQp= Gal(Qp/Qp).

There is a mapGQp Z ∼= Gal(Fp/Fp)→ 0.

How is this defined? We can look at Zp sitting inside Qp. Another strategy is

to find a subextension L/Qp so that Gal(L/Qp) is Z.We use

L = Qunrp = maximal unramified subextension L ⊆ Qp of Qp.

Lemma 14.2. There exists a short exact sequence

0→ IK → GK → Gal(Kunr/K)→ 0,

where IK is defined as

IL/K = σ ∈ Gal(L/K) : σ(x) ≡ x mod P for all x ∈ OL, IK = lim−→L

IL/K .

Math 99r Notes 37

15 October 26, 2017

15.1 Unramified and tamely ramified extensions of localfields

We have the short exact sequence

0→ I(K/Qp)→ Gal(K/Qp) Gal(k/Fp)→ 0.

Here, I(K/Qp) = σ ∈ Gal : σx ≡ x (mod $). As an aside, there also is awild inertia

P (K/Qp) = σ : σx ≡ x (mod $2).

Then P is the unique p-Sylow subgroup of I.

Definition 15.1. Let G act on a set S. This is unramified of I acts trivially.It is also called tamely ramified if P acts trivially.

Definition 15.2. An extension of local fields L/K is unramified if e = 1. Itis tamely ramified if p - e and wildly ramified if p | e.

What this all means is that we have a tower

L

L1 = LP

L0 = LI

K

pvp(e), wild

e/pvp(e), tame

f, unramified

You can also do this for infinite extensions. The field LP is the maximaltamely ramified subextension. So we will have

Qp

Qtamp = QPQp

p

Qunrp = QIQp

p

Qp

Math 99r Notes 38

Lemma 15.3. Let p - m and consider a root of unity ζm. Then K = Qp(ζm)is unramified over Qp.

Proof. Compute the discriminant. Or show that the surjection G(K/Qp) →G(Fp(ζm)/Fp) bijective by counting elements.

Proposition 15.4. We have Qunrp =

⋃p-mQp(ζm) and Qtam

p =⋃p-nQ(p1/n).

Then Gal(Qunrp /QP ) ∼= Gal(Fp/Fp) and we get

0→ IQp→ GQp

→ Gal(Fp/Fp)→ 0.

Here, the tame part of the inertia group is

IQp/PQp

∼=∏r 6=p

Zr.

15.2 Good reduction and unramified Tate modules

Let us go back to elliptic curves. Fix any ` 6= p and E/Qp. We have the action

Gal(Qp/Qp) over T`(E).

Theorem 15.5. (1) The Tate module is unramified if and only if E has goodreduction.(2) The Tate module is tamely unramified if and only if E has semistable re-duction.

Recall that if E/K with K/Qp finite, a minimal Weierstrass model for E isgiven by

y2 = x3 + ax+ b

where a, b ∈ OK and ∆ is minimal as possible. Also recall that E is definedto have good reduction if and only if ∆min ∈ O×K . This is equivalent to∆(Es) 6= 0. What can go wrong with elliptic curves? If it is singular, then iteither has a node or a cusp.

Definition 15.6. A bad reduction can be either semistable, which means thatthere is a node, or additive, which means that there is a cusp.

In the case when E has a good reduction, there is a map

E(K)→ Es(k)→ 0; (x : y : z) 7→ (x : y : z).

Here we need to scale so that x, y, z ∈ OK and one of them is a unit. This mapis surjective because we have Hensel’s lemma.

Lemma 15.7. We have a map

ME

∼=−→ ker(r : E(K)→ Es(k)).

Math 99r Notes 39

Let us recall ME = (mK , FE). Here, we have used z = −xy and w = − 1y .

Then there is a map

z 7→( z

w(z),− 1

w(z)

)∈ E(K).

This maps to the origin, i.e., infinity point, because

vK

( z

w(z)

)= vK

( z

z3(1 + · · · )

)= −2vK(z) ≤ −2

and likewise vK(−1/w(z)) < 0.It is clear that it is a group homomorphism because the formal group law

came from addition. To show that this is a bijection, we create an inverse map

ker(r)→ ME ; (x, y) 7→ −xy∈ mK .

This really lies in mk because y2 = x3+ax+b with a, b ∈ OK implies vK(y) = 3vand vK(x) = 2v for some v < 0.

Corollary 15.8. If E has good reduction and p - m, then E(K)[m] → Es(k).

Corollary 15.9. If E has good reduction then T`(E) is unramified.

Proof. We need to show that IK = I(K/K) acts on E[`n] trivially. Take K ′/Ksuch that E[`n] ⊂ E(K ′). So

E(K ′)[`n] → Es(K′)

is Galois-equivariant and so IK fixes E[`n].

Math 99r Notes 40

16 October 31, 2017

We have seen that E/Kv has good reduction if and only if for all ` 6= p, T`(E)is unramified. I will show you that if T`(E) is unramified for some ` 6= p thenE has good reduction.

16.1 Global review

So far we’ve been looking at E(K) and E[m], but a priori there aren’t anyrelations between them. Now to prove Mordell–Weil, we need to show thatK([m]−1(E(K))) is finite over K. This is because we have

E(K)/mE(K) → H1(GK , E[m]).

We may assume without loss of generality that E(K) contains E[m], so thenGK acts trivially on E[m]. Then

E(K)/mE(K) → Hom(GK , E[m]).

The GK part kernel is going to be GL where K([m]−1(E(K))). So if we get thatthis is finite, then E(K)/mE(K) embeds into Hom(Gal(L/K), E[m]), which isfinite.

But let me use some facts about Gal(L/K). The Weil pairing gives that

Gal(L/K) → Hom(E(K)/mE(K), E[m]).

This shows that L/K is an abelian extension, and also it is Galois group ism-torsion. Now it is a fact in number theory that an abelian extension that isunramified at almost all primes and has finite exponent is finite.

Proposition 16.1. If L/K is abelian of exponent m, and it is unramified overalmost all primes, then it is finite.

If p is not infinite, not a divisor of m, and has good reduction, then we willshow that L/K is unramified at p.

Theorem 16.2. Let E/K and L = K([m]−1E(K)), and let S be the primeswith bad reduction, or divisors of m, or ∞. Then L/K is unramified at every pnot in S.

Now L/K being unramified at p means that LP /Kp for all P dividing p. Toprove this theorem, it suffices to show that for all Q ∈ [m]−1E(K),

K(Q) = K ′/K

is unramified at all p not in S.But recall that E/Kp has good reduction if and only if E[`k] is unramified

for all k and ` not dividing p. Then E[m] is unramified at p, because m is made

Math 99r Notes 41

out of primes not dividing p. We want to show that for any σ ∈ IP , σQ−Q = 0.But

[m](σQ−Q) = σ([m]Q)− [m]Q = 0.

Also because p has good reduction, we can look at the image under reductionE(K ′P )→ Es(k

′). But because σ is inertial, it is sent to 0 in Es(k′). So

0→ ker→ E(K ′P ) Es(k′)→ 0

shows that σQ−Q is in the kernel. But [m] acts isomorphically on the kernel.This shows that σQ−Q = 0.

Math 99r Notes 42

17 November 1, 2017

We want to prove the following:

Proposition 17.1. If L/K is abelian of exponent m and L/K is unramified atalmost all primes, then L/K is finite.

Proposition 17.2. Let E/K be an elliptic curve and let L = K([m]−1E(K)).If p is a finite prime with good reduction, such that E has good reduction, thenL/K is unramified at p.

Proof. We did this last time. We want to show that for any Q ∈ [m]−1(K),K ′ = K(Q)/K is unramified at p. That is, for any P over p, we want to showthat K(Q)P = K ′P /Kp is unramified.

We need to show that for all σ ∈ IP , σQ−Q = 0. To show this, recall that

0→ ker→ E(K ′P )→ Es(k′P )→ 0

is exact. But σQ − Q ∈ E(K ′P ) maps to 0 so lies in the kernel. On theother hand, the kernel is the associated formal group. Also [m](σQ − Q) =σ([m]Q)− [m]Q = 0. But m is not divisible by p and so σQ = Q.

17.1 Kummer theory

To prove Proposition 17.1, we can assume that ζm ∈ K without loss of generality.This is because we can apply the theorem to L(ζm)/K(ζm).

What I claim is that every abelian extension of exponent m is contained in acompositum of fields K( n

√a)/K. If you think about it, finding an finite abelian

extension E/K of exponent m is finding a quotient

GK → Gal(E/K).

This maps to Z/mZ, so this is sort of looking at H1(GK ,Z/mZ).So consider the sequence

0→ µm → K×m−→ K× → H1(GK , µm)→ H1(GK ,K

×)→ · · · .

But Hilbert 90 gives H1(GK ,K×

) = 0. So we get the isomorphism

K×/(K×)m ∼= Hom(GK , µm)

because µm ⊆ K×.

Math 99r Notes 43

18 November 2, 2017

We need to prove the following:

Proposition 18.1. Let L/K be an abelian extension of exponent m and un-ramified outside a finite set. Then L/K is finite.

This follows from the following two facts:

Theorem 18.2. (a) (Class number theorem) If K/Q is finite, then ClK isfinite.

(b) (Unit theorem) O×K is finitely generated.

Note that the Picard group of SpecOK is just the class group. This isbecause the divisors are just fractional ideals and principal divisors are principalfractional ideals.

18.1 Class number theorem

I am going to define the adele as

A∞Q =∏′

pQp =

(xp) ∈

∏p

Qp : xp ∈ Zp for almost every p

, AQ = R×A∞Q .

Then Q naturally includes in AQ. I am going to give this a topology such thatthe open sets are

∏p Up with Up = Zp for almost all p and Up open in Qp. It is

a fact that AQ becomes a topological ring with this structure.Now we want to give a topology on A×Q so that it is a topological group. But

if I just give the subspace topology, it might be that the inversion on A×Q is notcontinuous. So we embed

A×Q → AQ × AQ; x 7→ (x, x−1)

and give a subspace topology here. So we get a embedding Q× → A×Q .We can similarly define

AK =∏′

v finiteKv ×

∏v|∞

Kv.

ThenK× → A×K CK = A×K/K

×.

This idelic class group CK actually surjects onto Cl(K). The map is given by(xp) 7→ (p)vp(x).

Because CK ClK , it would be nice if CK is finite. But this is not truebecause we have infinite stuff. So we define

A′K =

x = (xv) :

∏v

|x|v = 1

⊆ A×K .

It is a fact that K× ⊆ A1K . Also, A′K/K× ClK .

Now, A′K is compact. Also, K× is an open subgroup. This shows thatA′K/K× is finite and discrete. Thus it is finite, and it follows that ClK is finite.

Math 99r Notes 44

18.2 Concluding Mordell–Weil

For a finite set of prime S containing all infinite primes, we consider

OK,S = x ∈ K : v(x) ≥ 0 for all v /∈ S.

Then O×K,S is finitely generated. This is because it is finitely generated over

O×K , and O×K is finitely generated by the unit theorem.Now let us go back to the proposition. We know that

L ⊆ K( m√a : a ∈ K×/(K×)m).

But when isK( m√a)/K unramified outside S? We can compute the discriminant

and it is going to be

am−1 ·Disc(K(ζn)/K) = ±am−1mm.

So if it is going to be unramified at v /∈ S, we would need v(a) ≡ 0 (mod m).On the other hand, we note that

O×K,S Σ = a ∈ K×/(K×)n : m | v(a) for all v /∈ S.

This is because we have elements xi ∈ K such that vj(xi) = δij for all vj ∈ S.Also L ⊆ K( m

√a : a ∈ Σ). Then we get O×K,S/(O

×K,S)m Σ and O×K,S is

finitely generated so O×K,S/(O×K,S)m is finite. Then Σ is finite, and then the

extension L/K is finite.

Math 99r Notes 45

19 November 14, 2017

There is a modular form

f(q) = q∏n

(1− qn)2(1− q11n)2

= q − 2q2 − q3 + 2q4 + q5 + 2q6 − 2q7 − 2q9 − 2q10 + q11 − 2q12 + 4q13 + · · ·

of weight 2.

19.1 Hasse’s theorem

We are now going to consider elliptic curves over a finite field. To get a feeling,consider an elliptic curve

E : y2 + y = x3 − x2

over Fp for all p such that p - ∆.

Question. What is the size of E(Fp)?

Over Q, we’ve been showing that this is finitely generated. Over Fp, it isgoing to finite anyways. Actually there is a trivial bound

#E(Fp) ≤ 2p+ 1.

This is because for each x there is at most 2 solutions y and there is alsothe infinity point. But heuristically, half of the time you will be able to solvefor x and half of the time you will not be able to solve it. So you expect#E(Fp) ∼ p+ 1.

Theorem 19.1 (Hasse). |#E(Fp)− (p+ 1)| ≤ 2√p.

I would like to also consider αp = (p+1)−#E(Fp) and see what this means.Let us compute this number.

p 2 3 5 7 13 17#E(Fp) 5 5 5 10 10 20αp −2 −1 1 −2 4 −2

Table 1: Number of rational points of E over Fp

Note that this is exactly the coefficients of the weight 2 modular form I havewritten above.

Definition 19.2. An elliptic curve is modular if there exists a modular formf such that its q-expansion

f(q) =∑

anqn

satisfies ap = αp.

Math 99r Notes 46

It is a big theorem of Wiles and other people such that a lot of elliptic curvesare modular.

Let me give another heuristic for Hasse’s theorem. What is going to happenis there are going to be two complex numbers x1, x2 of absolute value

√p such

that αp = x1 +x2. Or maybe we can find a linear map T so that αp is the traceof T and p is the determinant.

Suppose p is a good prime, so that E/Q gives E/Fp. Now there is a Galoisrepresentation

Gal(Qp/Qp) on T`(E)

and because it is unramified, this gives an action of Gal(Fp/Fp) on T`(E). Nowthere is a Frobenius σp that generates Gal(Fp/Fp). The claim that the action ofσp has trace αp and determinant p. Then you need to show that the eigenvalueshave absolute value

√p.

Now I am going to give you a cheat proof that doesn’t work in general, butworks for elliptic curves.

Proof. Recall a bunch of things from before. If φ : E1 → E2 is an isogeny, wecan define its dual isogeny φ : E2 → E1 such that

φ φ = [deg φ], φ φ = [deg φ].

Also recall that φ : E1 → E2 is separable if and only if φ∗ : Ω1E2→ Ω1

E1is

nonzero. For example, the map

ϕ : E → E; (x, y) 7→ (xp, yp)

is not separable. This is because

ϕ∗( dx

2y + a1x+ a3

)=d(xp)

· · ·=pxp−1dx

· · ·= 0.

Because (ϕ + ψ)∗ω = ϕ∗ω + ψ∗ω, we have (1 − ϕ)∗ω = ω and so 1 − ϕ isseparable. This implies that the size of the fiber of 1−ϕ is the degree of 1−ϕ.

What is E(Fp)? It is the fixed points of E(Fp) under the Frobenius. So it isthe kernel of 1− ϕ, which has size the degree of 1− ϕ. Then we need to show

2√p ≥ |deg(1− ϕ)− p− 1| = |deg(1− ϕ)− deg(1)− deg(ϕ)|.

Because 2√

deg(1) deg(ϕ) = 2√p, it suffices to show that deg : End(E) → Z is

a quadratic form. But recall that [deg(ϕ)] = ϕ ϕ. Then

〈f, g〉 = deg(f + g)− deg(f)− deg(g)

is a bilinear form.

Math 99r Notes 47

20 November 29, 2017

We’ve been proving the Weil conjectures for elliptic curves.

Theorem 20.1 (Weil conjectures). For smooth and proper V/Fp of dimensiond, we define

Z(V, t) = exp

(∑n≥0

#(V (Fpn))

ntn).

Then

(i) Z(V, t) ∈ Q(t),

(ii) it satisfies a functional equation,

(iii) the Riemann Hypothesis holds. That is, if

Z(V, t) =P1(t) · · ·P2d−1(t)

P0(t)P2(t) · · ·P2d(t)

with

Pi =

degPi∏j=1

(1− βij(t)) ∈ Z[t],

each βij is “pure of weight j”, i.e., |i(α)| = pj/2 for all i : Q → C.

If we defineζ(V, s) = Z(V, p−s),

this means that all the zeros of ζ(V, s) have imaginary parts half-integers, andpoles have imaginary parts integers.

Last time we showed that

Z(E, t) =(1− αt)(1− βt)(1− t)(1− pt)

.

So what we need to show that α and β have pure weight of 1.

20.1 Modular forms

From my point of view, the motivation is to construct Galois representations.Basically given a modular form, you get a geometric object. This is going to bewhat we are going to do shortly. As I have said, getting a module form from ageometric object is the modularity conjecture. From a geometric object, we geta Galois representation of GQ. The converse direction is called the Fontaine–Mazur conjecture and is also hard. The Langlands is between modular formsand Galois representations.

Consider the upper half planeH = =(z) > 0 and look at the SL2(Z)-actionon H, given by (

a bc d

): z 7→ az + b

cz + d.

Math 99r Notes 48

In SL2(Z), given N there exist subgroups

Γ0(N) =

(∗ ∗0 ∗

)mod N

, Γ1(N) =

(1 ∗0 1

)mod N

,

Γ(N) =

(1 00 1

)mod N

.

These all acts properly discontinuously, so we can define Γ\H = YΓ as a Riemannsurface.

Here is the philosophy. The moduli space of elliptic curves is SL2(Z) \ H.Then Γ \ H is going to parametrize elliptic curves with some level structure.

Definition 20.2. Note that Γ0(N)/Γ1(N) ∼= (Z/NZ)×, with the identificationgiven by the entry d. Consider Γ1(N) ⊆ Γ ⊆ Γ0(N). A modular form ofweight k and level Γ is a holomorphic function f : H → C such that

(1) for any γ = ( a bc d ) ∈ Γ,

f(γτ) = (cτ + d)kf(τ),

(2) it “grows mildly” at ∞.

This is an odd definition. A more natural definition can be made in termsof sections of line bundles. The first condition means that f is a section ωk onYΓ = H/Γ. If ωk is the differentials, we get

d(γz) =1

(cz + d)2dz

and so weight is −2.The reason I want Γ1 in Γ is that ( 1 1

0 1 ) ∈ Γ. Then f(τ + 1) = f(τ). So wehave a Fourier expansion

f(τ) =∑

ane2πiτn.

The mild growth condition is then going to mean that this is holomorphic at 0,so that we are adding over n ≥ 0.

Definition 20.3. We say that f is a cusp form if a0 = 0. Mk(Γ) are themodular forms, and Sk(Γ) ⊆Mk(Γ) are the cusp forms.

Next time we are going to define a diamond operator 〈d〉 and also I amgoing to define the Hecke operator. For every p - N , we can define operatorsTp on S2(Γ). Let T be the algebra generated by Tp. Also, they commute witheach other, so we can simultaneously diagonalize. A simultaneous eigenvectoris called an eigenform. Let us write

Tpf = apf.

Math 99r Notes 49

It turns out that ap = ap(f) is the p-th coefficient.There is a concrete formula for Tp. It is given by

Tp(f) =1

p

p−1∑i=0

f(τ + i

p

)+ p〈d〉f(pτ).

If p | N , there are operators Up on S2(Γ) and then we can consider T+ thealgebra generated by Up.

Math 99r Notes 50

21 November 30, 2017

The Langlands conjecture is roughly trying to establishρ : GQ → GLn(Qp)

“geometric”

←→

automorphic representations

of GLn(AQ)

.

Class field theory is the n = 1 trivial case of Langlands. Thef left side issomething like 1-dimensional representations of Gab

Q and the right hand side is

characters Q× \ A×Q → C×.But this is hard, and people look at the local version instead. Let ` 6= p.

Then ρ` : GQ`

→ GLn(Qp)ϕ-s.s.

←→

automorphic representations

of GLn(Q`)

.

But we should have something for ` = ∞. But there is no Galois group whoserepresentation corresponds to automorphic representations of GLn(R).

Grothendieck actually looked at something that might work. Recall that Zlies in GFp

and so we can look at the inverse image.

0 I GQ`GF`

0

0 I W` Z 0

Here W` is topologized so that I is open. Then the left hand side wouldcorrespond to representations W` → GLn(Qp). In the ` = ∞ case, we haveW∞ = C× o ±1 that behaves well.

Lemma 21.1. Let α, β be the eigenvalues of the map φ acting on T`(E). Then|α| = |β| = p1/2.

Proof. The characteristic polynomial is f(X) = X2 − tr(φ)X + det(φ) = (X −α)(X−β). Our goal is to show that f(X) = (X−α)(X−β) ≥ 0 for all X ∈ R,or maybe X ∈ Q. But note that

0 ≤ det(n−mφ) ·m2 = det(n−mφ|T`(E)) = det( nm − φT`(E))m2 = f( nm )m2.

So we are done.

21.1 Modular curves

We were trying to get an elliptic curve over Q form a modular form of weight 2,with Q coefficients. The reason we are working over Q is that if we work witha number field we would get an abelian variety.

What was an elliptic curve? We defined it as a genus 1 curve with a markedpoint. But over C, this just looks like C/Λ where Λ is a lattice. Because we can

Math 99r Notes 51

scale Λ, we can fix one point to be 1. Then we can determine an elliptic curveby this one complex number. But there are different ways of writing down thesame lattice, so the moduli space of elliptic curves is

Ell = GL2(Z) \ H±1 = SL2(Z) \ H.

So if Γ1(N) ⊆ Γ ⊆ Γ0(N), then there is a map YΓ SL2(Z) \ H. BecauseSL2(Z)\H is the moduli space of elliptic curves, YΓ should be the moduli spaceof elliptic curves with some “level Γ” structure. You should think about whatthis structure really means. It is related to the p-torsion points.

We can define

Γ0(N) \ H = Y0(N), Γ1(N) \ H = Y1(N).

These are not necessarily compact spaces, but you make a one-point compacti-fication and make it into a curve X0(N) and X1(N). These are called modularcurves. It can be proved that they have nice integral models. Generally, YΓ

can be compactified to XΓ.

Math 99r Notes 52

22 December 6, 2017

We had a moduli interpretation of SL2(Z)\H as elliptic curves. The exercise wasto figure out what Γ0(p)\H corresponds to. Then answer is that it correspondsto a pair of lattices Λ ⊆ Λ′ of index p, and this corresponds to an elliptic curvewith a order p subgroup. The reason is that in Γ0(p) you are only allowed tosend e1 to xe1 + (py)e2 and so you have to fix the sublattice generated by e1

and pe2. To see how this is an elliptic curve with a order p subgroup, you lookat 1

pΛ′ inside Λ.

What does Γ1(p)\H classify? These are elliptic curves with a point of orderp. Again, this sends e1 to something that is e1 modulo p. So the choice of e1

modulo p determines everything. So looking at 1pe1 gives a order p point. Then

the mapΓ1(p) \ H → Γ0(p) \ H

is given by mapping (E,Q) to (E, 〈Q〉).Likewise, Γ(p)\H corresponds to elliptic curves E with a choice of p-torsion

points α, β that are linearly independent over Fp.

22.1 Hecke operators

Recall that Mk(Γ) are the modular forms, and Sk(Γ) are the cusp forms. Sup-pose that p - N . These are maps

Tp : Mk(Γ)→Mk(Γ).

Here is a description. Suppose f : H → C satisfies f(γz) = j(γz)−kf(z), wherej(z) = 1

cz+d .Now consider the two maps

Γ0(pN) \ H

Γ0(N) \ H Γ0(N) \ H

f2f1

where the left map is given by (E,H × H ′) 7→ (E,H) and the right map isgiven by (E,H ×H ′) 7→ (E/H ′, H), where H and H ′ are subgroups of order Nand p. Now we get a map from Γ0(N) \ H to Γ0(N) \ H on divisors, given bypull-back and push-forward, i.e., looking at the preimages on the left map andthen projecting down on the right map.

Now in algebraic geometry, you can also pull back and push forward a sectionand then get another section. Here, what you’re doing is looking at what a pointis mapped to, and then add them up.

Let me summarize the properties of Hecke operators.

• Tp acts on Sk(Γ).

• Tp commute with each other.

Math 99r Notes 53

• Each Tp is self-adjoint with respect to an inner product on Sk(Γ). So wehave a simultaneous diagaonalization

Sk(Γ) =⊕v

C〈v〉.

So if T is the C-algebras generated by Hecke operators, then each v cor-responds to a character θ : T→ C. Such is called a Hecke eigensystem.

• If f =∑anq

n and a1 = 1, then θf (Tp) = ap.

So the crucial thing is to study the space of modular forms, rather thanindividual modular forms. This was a big shift of the theory, due to Hecke.

Math 99r Notes 54

23 December 7, 2017

Last time we were talking about the modular curve Y0(N) = Γ0(N) \ H andY1(N) = Γ1(N) \ H with compactifications Y0(N) → X0(N) and Y1(N) →X1(N). For N ≥ 5, the curves X1(N) are going to be smooth schemes overZ[N−1]. Then X0(N) is a smooth Deligne–Mumford stack over Z[N−1].

Last time we described the Hecke correspondence. Fix N and a prime `. Forp - N`, we have two projections

X0(pN)

X0(N) X0(N)

π1 π2

and then we can pull-back push-forward. The map is actually

J0(N) = Pic0(X0(N))→ J0(N).

In general, Pic0 is going to be represented by a scheme. So the Jacobian J0(N)is also going to be a scheme and the Hecke correspondence is a map on schemes.

Let me tell you a bit more about abelian varieties. These are projectivegroup schemes, and over C they are always going to look like Cg/Λ where Λ isa rank 2g lattice. Here, you need to specify more data, namely polarization, tomake this into a scheme. We can also look at Tate modules. If you have anabelian variety A/Q, then

T`(A) = lim←−A[`n]

has a GQ-action, and so gives a Galois representation

ρA,` : GQ → GL2g(Z`).

23.1 Eichler–Shimura construction

Let f be a cusp eigenform of level Γ0(N), normalized so that a1 = 1. Thismeans that it gives a character θ : T → C given by Tp 7→ ap. It turns out thatap all lie in some number field, so let K = Q(ap). Let [K : Q] = g. Considerthe ideal

a = ker(θ) ⊆ Tacting on J0(N). (This is defined over Z[N−1].) The Jacobian J0(N) has goodreduction away from N , and we can define

Af = J0(N)/aJ0(N).

Lemma 23.1. Af is an abelian variety of dimension g.

There is a general construction in algebraic geometry, and this gives you nelliptic curve. Or for simplicity assume that g = 1. Then we get a Tate moduleV = T`(Af ). Here we want to show that

ap = tr(ϕp|V ).

Math 99r Notes 55

We first look at a integral model J0(N)Zp , with the Hecke operators Tpacting on it. We want to show that this action is integral. This is done by basechanging to Fp, i.e., looking at the special fiber.

J0(N)ZpJ0(N)Fp

SpecZp SpecFp

Theorem 23.2 (Eichler–Shimura). On J0(N)Fp , we have Tp = F + V .

Here, F is the Frobenius, and V is the Vershiebung, the dual of the Frobeniusmap. But because we have good reduction, the trace of Tp can be computed asthe sum of the traces of F and V .

Corollary 23.3. ap is indeed tr(ϕp|V ).

This is a complicated construction, but it is the start of everything.

23.2 Two stories

We have studied elliptic curves along with Riemann–Roch, and we studied thePicard group and the group law. We looked at the Tate module, and provedMordell–Weil. Then we studied elliptic curves over finite fields, we can actuallycount points here. This led to the more general study of Weil conjectures andmodular forms.

Fermat’s last theorem says that there is no solution to xp+yp = zp. If thereis such a solution, we can consider the elliptic curve

E : y2 = x(x− αp)(x− βp)

where α and β are some solutions. This is called the Frey curve. SupposeE comes from a modular form f . Then by the Eichler–Shimura construction,we get a Galois representation ρf of level N . (Here f is of weight 2 and levelN .) Ribet made this conjecture that to this we can associate an f of weight 2and level 2. Then f ∈ S2(Γ0(2)). But the dimension is the genus of S2(Γ0(2)),which is 0. This gives a contradiction, and so there cannot be a solution. So allyou need to prove Fermat’s last theorem is the modularity conjecture.

Theorem 23.4 (Taylor–Weils, modularity lifting). Consider the representationρ : GQ → GL2(Zp) → GL2(Fp). If ρ is “modular” and ρ satisfies some niceconditions, then ρ is modular.

Here “modular” means that it is a reduction of some modular form. Thisis called the “R = T” theorem. Here is how you use this. It can be shownthat every GQ → GL2(F2) that is “nice” is going to be modular. So if ρE has“nice” reduction ρE , then we can use modularity lifting to conclude that ρE ismodular. But what if it is not nice? You can show that there exists a E′ such

Math 99r Notes 56

that E′[3] is nice and E′[5] ∼= E[5]. But E′[3] nice implies that ρE′ modular,and so E′[5] is nice. Then you can use modularity lifting again to get that ρEis modular.

Conjecture 23.5 (Birch–Swinnerton-Dyer). Let E/Q be an elliptic curve. Thenwe can define the algebraic rank ralg = rkE(Q). This is equal to the analyticrank ran, defined as the order of zero of L(E, s).

Here, L(E, s) was not even defined meromorphically on the complex planebefore the modularity conjecture. How they did analytic continuation was byusing the fact that L(f, s) is equal to L(E, s).

Index

additive reduction, 38adele, 43affine space, 5algebraic rank, 4, 22

canonical divisor, 15complex multiplication, 30curve, 7cusp form, 48

degree, 8discrete valuation ring, 7, 11divisor, 9dual isogeny, 29

elliptic curve, 4

formal group, 32formal group law, 32Frey curve, 55

genus, 15

Hasse’s theorem, 45Hecke operator, 52height function, 25

isogeny, 20

Kummer pairing, 24

modular curves, 51modular elliptic curve, 45modular form, 48Mordell–Weil, 24

p-adic numbers, 27principal divisor, 10

ramification index, 12, 19rational function, 6Riemann–Roch, 15

semistable reduction, 38

tamely ramified, 37Tate module, 27

uniformizer, 12unramified, 37

Weierstrass minimal model, 34Weil cojectures, 47Weil pairing, 30wildly ramified, 37

Zariski topology, 5

57