Fermat's Last Theorem and the Modular Method

IntroductionModular FormsElliptic Curves

Three Black BoxesProof of Fermat’s Last Theorem

Applications to Diophatine Equations

Fermat’s Last Theorem and the Modular Method

Philippe Michaud-Rodgers

University of Warwick Postgraduate Seminar

04.11.2020

Philippe Michaud-Rodgers Fermat’s Last Theorem and the Modular Method




Table of Contents

1 Introduction

2 Modular Forms

3 Elliptic Curves

4 Three Black Boxes

5 Proof of Fermat’s Last Theorem

6 Applications to Diophatine Equations





Statement of Fermat’s Last Theorem

Theorem (Wiles + many others! 1995)

The equationxn + yn = zn,

with n ≥ 3, has no non-trivial solutions for integers x , y , z .

By a non-trivial solution, we mean that xyz 6= 0.





Fermat’s Little Margin

Pierre de Fermat:

His Margin:

”I have discovered atruly remarkable proofof this result which thismargin is too small tocontain.”





First Observations

If n = p ·m and (x , y , z) satisfies xn + yn = zn, then

(xm)p + (ym)p = (zm)p.

n = 3 (Euler, 1770) and n = 4 (Fermat, 1670): elementary.

So enough to prove:

FLT

The equationxp + yp = zp,

with p ≥ 5, prime, has no non-trivial solutions for integers x , y , z .





Outline of Argument

Suppose xp + yp = zp.

Associate to (x , y , z) an “elliptic curve”, E .

Associate to E a “level Np newform”, f .

Observe that there are no newforms at level Np.

Aim: Understand this.





Table of Contents

1 Introduction

2 Modular Forms

3 Elliptic Curves

4 Three Black Boxes







Definition of a Modular Form

Let H = {τ ∈ C : im(τ) > 0}. This is the upper half plane.

Definition

A “level N” modular form (of weight 2) is a holomorphic functionf : H → C satisfying:(i) for z ∈ H,

f

(az + b

cz + d

)= (cz + d)2f (z),

whenever a, b, c , d ∈ Z, ad − bc = 1, N|c ;(ii) Another technical condition.

f has a Fourier or q-expansion:

f =∞∑n=0

cnqn, where cn ∈ C, q = e

2πiz .





Newforms: Definition

A newform at level N is a special type of level N modularform.

It has a q-expansion

f =∞∑n=1

cnqn.

A newform is rational if ci ∈ Q for all i (then in fact ci ∈ Zfor all i), and irrational otherwise.

Newforms at level 38:

f1 = q − q2 + q3 + q4 − q6 − q7 + · · ·f2 = q + q2 − q3 + q4 − 4q5 − q6 + 3q7 + · · ·





Absence of Newforms

Fact

There are no newforms at levels

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 16, 18, 22, 25, 28, 60.





Table of Contents

1 Introduction

2 Modular Forms

3 Elliptic Curves

4 Three Black Boxes







What is an Elliptic Curve?

Definition

An elliptic curve over Q is a curve given by an equation

Y 2 = X 3 + AX 2 + BX + C ,

where A,B,C ∈ Q. It is smooth.





Rational Points

E : Y 2 = X 3 + AX 2 + BX + C .A point P = (x , y) is a rational point on E if P lies on E , andx , y ∈ Q.We write E (Q) for the set of rational points.Example: Y 2 = X 3 + 17, (2, 5) ∈ E (Q)





E (Q) is a group!

Let P,Q ∈ E (Q). Then P ⊕ Q ∈ E (Q). What is P ⊕ Q?





Group Law Continued

What is the identity in the group? It is 0 (or ∞):





Group Structure

E (Q) is an abelian group (clear).

E (Q) is a finitely generated abelian group (Mordell-Weil)

Everything I have said works if you replace Q by a numberfield K , e.g. K = Q(

√2),Q(i),Q( 3

√19), . . . .





Three Important Quantities

E : Y 2 = X 3 + AX 2 + BX + C = (X − e1)(X − e2)(X − e3)

Discriminants

The Discriminant of E is the discriminant of X 3 + AX 2 + BX + C .This is ∆E = [(e1 − e2)(e2 − e3)(e1 − e3)]2.The minimal discriminant of E , ∆min,E , is the smallestdiscriminant “under transformations”.

Conductor

The conductor of E is

NE =∏

p|∆min,E

pe(p),

where ep > 0 depends on the reduction type of E at p.Philippe Michaud-Rodgers Fermat’s Last Theorem and the Modular Method




Reduction of an Elliptic Curve

E : Y 2 = X 3 + AX 2 + BX + C , p prime, A,B,C ∈ Z.

We can reduce E mod p, and write E .

View E as a curve over Fp, and write E (Fp) for its Fp points.

If p - N, then E is itself an elliptic curve over Fp (otherreduction types possible).

Set ap(E ) = p + 1−#E (Fp).

Example: Let E : Y 2 = X 3 + 5X 2 + 9X + 25 and p = 5.Then NE = 29 · 313 and E : Y 2 = X 3 − X .E (F5) = {∞, (0, 0), (1, 0), (2, 1), (2,−1), (−1, 0), (−2, 2), (−2, 2)}.So a5(E ) = 5 + 1− 8 = −2.





Elliptic Curves Recap

E : Y 2 = X 3 + AX 2 + BX + C .

E (Q) is a finitely generated abelian group.

We associate an integer N to E , called the conductor.

For each prime p - N, we have an ap(E ).





Table of Contents

1 Introduction

2 Modular Forms

3 Elliptic Curves

4 Three Black Boxes







Eichler-Shimura

Recall: a level N rational newform is a holomorphic functionf : H → C with q-expansion

f =∞∑n=1

cnqn, where ci ∈ Z, q = e

2πiz .

Eichler-Shimura

Let f be a level N rational newform. Then we can associate to fan elliptic curve Ef over Q of conductor N, so that for all primesp - N:

cp = ap(E ).





Modularity

Modularity Theorem (Wiles, Taylor + Wiles, Breuil + Conrad +Diamond + Taylor)

Let E be an elliptic curve over Q of conductor N. Then E ismodular. Meaning we can associate a level N rational newformfE =

∑cnq

n to E satisfying cp = ap(E ) for any p - N.

So we have maps (BLACK BOX 1)

{Level N rational newforms}� {Elliptic Curves /Q of Conductor N}





Arises mod p

Let E be an elliptic curve of conductor N.

Let f =∑

cnqn be a newform of level N ′.

Definition

Let p be a prime. We say E arises modulo p from f if for allprimes l - pNN ′,

cl ≡ al(E ) (mod p).

This is a natural definition given what we saw above.

Note that here N and N ′ can differ.





Level-Lowering

(BLACK BOX 2)

Level-Lowering Theorem (Ribet)

Let E be an elliptic curve over Q of conductor N and let p ≥ 5 beprime. (Suppose E has no rational p-subgroups.) Then E arisesmod p from a newform f at level Np, where

Np =N∏

q‖N,p|ordq(∆min)

q

If E satsifies hypotheses and there are no newforms at levelNp, then this gives a contradiction.





Rational p-subgroups and Mazur’s theorem

Recall: E (K ) is a group for fields K .

The ‘p’ in p-subgroup means the subgroup has order p.

‘Rational’ means the subgroup is rational, not necessarily thepoints.

e.g. {(√

2, 0), (−√

2, 0)} is rational even though its elements arenot rational.(BLACK BOX 3)

Mazur’s Theorem

Let E : Y 2 = f (X ) be an elliptic curve over Q of conductor N andlet p ≥ 5. Suppose N is squarefree and that f has 3 rational roots.Then E has no rational p-subgroups. (So can level-lower)





Table of Contents

1 Introduction

2 Modular Forms

3 Elliptic Curves

4 Three Black Boxes







The Frey Curve

Recall FLT:

FLT

The equation xp + yp = zp, with p ≥ 5, prime, has no non-trivialsolutions for integers x , y , z .

Suppose (x , y , z) is a solution and define the Frey Curve

E : Y 2 = X (X − xp)(X + yp)

This is an elliptic curve.

We want to show that E cannot exist.

First check E has no rational p-subgroups.

Then level-lower to reach a contradiction.





Frey Curve Computations

Frey Curve E : Y 2 = X (X − xp)(X + yp).

Minimal discriminant: ∆min = 2−8(xyz)2p.

Conductor:N = 2

∏p|xyz,odd

p.

Note that N is dependent on x , y , z .

N is squarefree, and X (X − xp)(X + yp) has three rationalroots.

So, by Mazur (Black Box 3), we can level-lower.





Level-Lowering the Frey Curve

We level lower (Black Box 2): E arises modulo p from anewform f at level Np.

Recall:

∆min = 2−8(xyz)2p, N = 2∏

p|xyz,odd

p, Np =N∏

q‖N,p|ordq(∆min)

q

So Np = 2.

But! There are no newforms at level 2, a contradiction.

Hoorah!





Table of Contents

1 Introduction

2 Modular Forms

3 Elliptic Curves

4 Three Black Boxes







The Modular Method

The Modular Method (level-lowering a Frey curve) can beapplied to other Diophantine equations.

Usually we do find newforms at the level Np.

There are lots of techniques to try and eliminate them.

The Modular Method can often be combined with othermethods.





Other Diophantine Examples

The Modular Method has been used extensively to solve orpartially solve:

The Generalised Fermat Equation: Axn + Byn = Czn.

Other signatures: Axp + Byp = Cz2, Axp + Byp = Cz3.

Lebesgue-Nagell equation: x2 + D = yn, D ∈ Z.

Perfect Powers in Fibonacci and Lucas sequences: yk = Fn,yk = Ln.

Fermat’s Last Theorem over totally real fields:e.g. K = Q(

√d), d ≥ 0.





Thank you for listening! :)


Fermat's Last Theorem and the Modular Method

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