introduction to the modular method

Introduction to the modular method

Nicolas Billerey

Laboratoire de mathématiques Blaise PascalUniversité Clermont Auvergne

BarcelonaFebruary 3, 2020

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FrameworkFermat’s Last Theorem

Extending the modular method

Table of contents

1 Framework

2 Fermat’s Last Theorem

3 Extending the modular method

Nicolas Billerey Introduction to the modular method

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Table of contents

1 Framework




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Generalized Fermat Equations (GFE)

Let A, B and C be coprime non-zero integers.We are interested in the following diophantine problem : findall sextuples (x , y , z , p, q, r) of integers such that p, q, r ≥ 2and

Axp + Byq = Cz r .

This is a widely open problem, despite lots of efforts by manymathematicians, starting with old Greeks.In this lecture : survey a tiny, yet important part of this longstory, focusing mainly on the case

A = B = C = 1 and p = q = r .


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Solutions and signatures

Given p, q, r ≥ 2, consider the Generalized Fermat Equation

Axp + Byq = Cz r . (1)

DefinitionWe call a solution any triple (x , y , z) of integers satisfying (1).A solution is said primitive if gcd(x , y , z) = 1.The triple (p, q, r) is called the signature.

In solving Eq. (1), the problem is completely different according towhether

1p

+1q

+1r> 1,= 1 or < 1.


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Fixed signature

For any integer p ≥ 4, the equation

AX p + BY p = CZp

defines a smooth, projective curve of genus (p−1)(p−2)2 ≥ 2.

By Faltings’ proof of Mordell’s conjecture, it has only finitelymany rational points.Applying Faltings’ result in a subtle way, Darmon and Granvilleobtained the following.

Theorem (Darmon-Granville)

Let p, q, r ≥ 2 be integers such that 1p + 1

q + 1r < 1. Then, there

exist only finitely many primitive solutions to the GeneralizedFermat Equation

Axp + Byq = Cz r .


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Varying signatures : general expectations

abc-Conjecture (Masser–Oesterlé)

Let ε > 0. For all non-zero coprime integers a, b, c such thata + b = c , we have

max (|a|, |b|, |c |)�ε rad (abc)1+ε .

PropositionAssume the abc-Conjecture holds. Then, there are only finitelymany triples (x , y , z) of coprime integers for which there existintegers p, q, r ≥ 2 such that

1p

+1q

+1r< 1 and Axp + Byq = Cz r .


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The special case A = B = C = 1

Fermat–Catalan ConjectureThe only primitive solutions in non-zero integers of the GeneralizedFermat Equations

xp + yq = z r , with1p

+1q

+1r< 1

correspond to the following identities :

1+23 = 32, 25+72 = 34, 73+132 = 29, 27+173 = 712, 35+114 = 1222

and177 + 762713 = 210639282, 14143 + 22134592 = 657,

92623 + 153122832 = 1137, 438 + 962223 = 300429072,

338 + 15490342 = 156133.


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Trivial solutions (I)

While solving GFE for varying signatures, it is common to keepthe notation p for a varying exponent and r for a fixed one.In these lectures, we’ll be concerned with Generalized FermatEquations of signatures (p, p, p), (p, p, r) and (r , r , p).As a consequence of abc-Conjecture, if (x , y , z) is a primitivesolution of a GFE in non-zero integers, then for largeenough p, we have

|xyz | = 1 for signature (p, p, p) ;|xy | = 1 for signature (p, p, r) ;|z | = 1 for signature (r , r , p).


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Trivial solutions (II)

DefinitionWe call such triples the trivial solutions together withsolutions (x , y , z) such that xyz = 0.

RemarkIt is worth noting that these “trivial” solutions might be a highlynon-trivial obstruction to solving the corresponding equation.


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Step 1 : The Frey curveStep 2 : ModularityStep 3 : Elliptic Galois representations and irreducibilityStep 4 : Modular Galois representations and level-loweringStep 5 : Modular forms computations and contradiction

Table of contents

1 Framework




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Statement and first reductions

Fermat’s Last TheoremFor every n ≥ 3, there is no non-trivial primitive solution to theequation

xn + yn = zn.

The proof is by contradiction, assuming the existence of anon-trivial primitive solution (a, b, c) for some n ≥ 3.Euler proved the case n = 3 and Fermat the case n = 4.Hence we may assume n = p ≥ 5 is prime.Permuting a, b and c if necessary, we assume

aP ≡ −1 (mod 4) and bP ≡ 0 (mod 16).

We proceed in five steps. The whole argument is known as themodular method.


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The Frey curve (I)

Consider the elliptic curve Ea,b,c given by the equation

Y 2 = X (X − ap)(X + bp). (2)

We compute the standard coefficients of this model

c4 = 16(a2p + (ab)p + b2p) , ∆ = 16(abc)2p and j =

c34

∆.

The equation (2) defines a minimal model for Ea,b,c awayfrom 2.The curve Ea,b,c has bad reduction at an odd prime ` if andonly if ` | abc .Under our assumptions, the curve Ea,b,c has bad multiplicativereduction at 2.


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The Frey curve (II)

To summarize :

PropositionThe curve Ea,b,c is semi-stable and has bad reduction precisely atthe primes dividing abc . Moreover, if ` | abc is such a prime, then

v`(j) =

{−2pv`(abc) ≡ 0 (mod p) if ` odd;8− 2pv2(abc) 6≡ 0 (mod p) if ` = 2.


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Wiles’ theorem

Theorem (Wiles, Taylor–Wiles)

Let E/Q be a semi-stable elliptic curve of conductor N and let

L(E , s) =∏

p prime

11− ap(E )p−s + 1N(p)p1−2s =

∑n≥1

an(E )

ns

be its L-function. Then, fE =∑

n≥1 an(E )qn is a weight 2 newformof level N and trivial Nebentypus.

RemarkNote that the coefficients of fE are all rational integers.


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Elliptic Galois representations

Let E/Q be an elliptic curve of conductor N.For every prime number p, denote by

ρE ,p : Gal(Q/Q)→ GL2(Fp)

the Galois representation corresponding to the action ofGal(Q/Q) on the p-torsion of E .

DefinitionThe representation ρE ,p is unramified at a prime ` if ρE ,p(I`) = {1}where I` is an inertia group at ` in Gal(Q/Q).It is ramified otherwise.

RemarkThe representation ρE ,p is unramified outside Np.


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Ramification at the bad primes

Let j be the j-invariant of E .

Proposition (Tate)

Let ` 6= p be a prime such that ` ‖ N. Then, the representation ρE ,p

is unramified at ` if and only if v`(j) ≡ 0 (mod p).

Using the arithmetic properties of the Frey curve, we obtain :

CorollaryThe representation ρEa,b,c ,p is unramified away from 2 and p.


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Irreducibility

Theorem (Mazur)

The only possible torsion subgroups of E (Q) are

Z/nZ for 1 ≤ n ≤ 10 and n = 12Z/2Z⊕ Z/2nZ for 1 ≤ n ≤ 4.

As a consequence of the local description of ρEa,b,c ,p and theprevious result, we get :

TheoremThe representation ρEa,b,c ,p is (absolutely) irreducible.


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Modular Galois representations (I)

Let f =∑

n≥1 anqn be a weight 2 newform of level N and

trivial Nebentypus.For every prime p, there is a Galois representation

ρf ,p : Gal(Q/Q)→ GL2(Fp)

which is uniquely characterized (up to semi-simplification andisomorphism) by the following property : it is unramifiedoutside Np and for every prime ` - Np, the characteristicpolynomial of ρf ,p(Frob`) is the reduction of

X 2 − a`X + `.


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Modular Galois representations (II)

DefinitionA Galois representation

ρ : Gal(Q/Q)→ GL2(Fp)

is modular of level N ≥ 1 if there exists a weight 2 newform f oftrivial Nebentypus and level N such that ρ ' ρf ,p. In that case, wesay that ρ arises from f .

Theorem

The Galois representation ρEa,b,c ,p is modular of level N =∏`|abc` prime

`.


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Lowering the level

Theorem (Ribet)

Let ρ : Gal(Q/Q)→ GL2(Fp) be an irreducible Galoisrepresentation. Suppose that ρ is modular of level N and let ` ‖ Nbe a prime. If ρ is finite at `, then ρ is modular of level N/`.

Applying Ribet’s theorem recursively to the representation ρEa,b,c ,p

gives the following result.

TheoremThe representation ρEa,b,c ,p is modular of level 2.


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End of the proof

There is no non-zero weight 2 newform of level 2.Hence Fermat’s Last Theorem is proved !


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Fermat curvesOther signatures

Table of contents

1 Framework




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Fermat curves

We wish to apply the strategy of the previous section to the Fermatcurves

Axp + Byp = Czp

for some fixed coprime non-zero integers A, B and C .


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Good points (I)

Let (a, b, c) be a non-trivial primitive solution.The equation

Y 2 = X (X − Aap)(X + Bbp)

still defines an elliptic curve EA,B,Ca,b,c /Q with nice arithmetic

properties.Thanks to the work of Breuil, Conrad, Diamond and Taylor,the curve E = EA,B,C

a,b,c is again modular.At least for large enough p, depending on A, B , C (but noton (a, b, c)) general results of Mazur apply to show that therepresentation

ρE ,p : Gal(Q/Q)→ GL2(Fp)

is irreducible.Nicolas Billerey Introduction to the modular method

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Good points (II)

Combining arithmetic properties of E with modularity andirreducibility for ρE ,p, Ribet’s result applies to show that ρE ,p

arises from a weight 2 newform f =∑

n≥1 an(f )qn of level Mwhich is explicit and almost independant of the solution.The isomorphism ρE ,p ' ρf ,p can be restated as follows. Thereexists a prime ideal p over p in Q such that for any prime `,the following congruences hold :{

a`(f ) ≡ a`(E ) (mod p) if ` - Npa`(f ) ≡ ±(`+ 1) (mod p) if ` - Mp and ` | N

where N denotes the conductor of E .


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Questions

Despite these good points, we are left with several questions :1 How do we compute the forms f ?2 How do we discard the previous congruences ?3 How do we deal with the “small” primes ?


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The equation xp + y p = 2zp

Theorem (Ribet, Darmon–Merel)

For every n ≥ 3, the Fermat equation xn + yn = 2zn has nonon-trivial primitive solution.

RemarkNote the existence of the trivial solution (1, 1, 1).

The case 2 < n < 31 of the above theorem is known thanks tothe work of Dénes.Let (a, b, c) be a non-trivial primitive solution for n = p ≥ 7prime.


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The Frey curve and its attached Galois representation

Proposition

The elliptic curve E = EA,B,Ca,b,c has all of its points of order 2

defined over Q and conductor

N =

{rad(abc) if abc is even25rad(abc) if abc is odd.

There is at least one odd prime of bad multiplicative reduction.

PropositionThe representation ρE ,p is absolutely irreducible.

It has conductor M =

{2 if abc is even32 if abc is odd.


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Congruences

By modularity and level-lowering, the representation ρE ,p

arises from a weight 2 newform of level M.For M = 2 (i.e. abc is even), we get a contradiction as in FLT.But for M = 32, the space Snew

2 (Γ0(32)) is 1-dimensional andspanned by a unique newform f corresponding to the (isogenyclass of the) elliptic curve F with equation Y 2 = X 3 − X .

RemarkNote that this curve is precisely the Frey curve associated with thetrivial solution (1, 1, 1).

How do we contradict the isomorphism ρE ,p ' ρF ,p ?


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A conjectural answer

Conjecture (Frey–Mazur)

There exists a constant C > 0 such that for all elliptic curves Eand F defined over Q and for all prime numbers p > C we have

ρE ,p ' ρF ,p =⇒ E and F are isogenous over Q.

At least for large enough p, this conjecture would implythat N = M = 32 and hence a contradiction, for (a, b, c) is anon-trivial solution.


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Complex Multiplication Theory

The curve F has complex multiplication by Q(√−1).

Let G be the image of ρF ,p in GL2(Fp).

PropositionThe group G is the normalizer of a Cartan subgroup. This Cartansubgroup is split if p ≡ 1 (mod 4) and non-split if p ≡ −1(mod 4).


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The case p ≡ 1 (mod 4)

Assume ρE ,p ' ρF ,p and p ≡ 1 (mod 4).The curve E gives rise to a rational point on the modularcurve Xsplit(p).For p ≥ 17, this implies that E has potentially good reductionat all primes ` 6= 2 (Momose) and hence a contradiction.The conclusion now follows from more recent results byBilu–Parent–Rebolledo (for p ≥ 17) andBalakrishnan–Dogra–Müller–Tultman–Vonk (for p = 13).


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The case p ≡ −1 (mod 4)

A more detailed study of the representation ρF ,p gives the following.

Proposition

We have G = ρF ,p(Dp) where Dp is a decomposition group at p. Inparticular, G is the normalizer of a non-split Cartan subgroup andthe prime p does not divide abc .

Besides, Darmon and Merel proved the following integrality resultwhich contradicts the proposition above and finishes the proof oftheir theorem in the case when abc is odd.

Theorem

The j-invariant of E belongs to Z[ 1p ].


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Fermat curves with odd coefficients

Theorem (Halberstadt–Kraus)

Let A, B , C be three odd pairwise coprime integers. Then thereexists a set of primes P of positive density such that for everyprime p ∈ P , the Generalized Fermat equation Axp + Byp = Czp

has no non-trivial primitive solution.For large enough p, we have ρE ,p ' ρF ,p where F is an ellitpiccurve over Q of conductor 2rad(ABC ).Under some congruence conditions this isomorphism is bothcompatible and incompatible with the Weil pairing.

RemarkNote that this result does not require any modular formscomputation.


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Other known Frey curves over Q

Frey curves associated with the Generalized Fermat equationshave been constructed by various authors for a few signaturesincluding :

(p, p, 2), (p, p, 3) and (r , r , p)

for r = 3, 5, 7.For a non-trivial primitive solution (a, b, c) of x3 + y3 = zp

this Frey curve reads as follows

E : Y 2 = X 3 + 3abX + b3 − a3.


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The case of x3 + y 3 = zp

The following result follows from works by Euler,Darmon–Granville, Kraus, Bruin, Chen–Siksek and Freitas.

Theorem

The generalized Fermat equation x3 + y3 = zp has no non-trivialprimitive solution for p ≥ 3 in a set of primes P of density ≈ 0.844.For instance, P contains primes p such that

p < 107, or p ≡ 51, 103, 105 (mod 106), or p ≡ 2 (mod 3).

RemarkThe main obstacle here is to contradict the isomorphismρE ,p ' ρF ,p where F is the Frey curve associated with thepseudo-solution 23 + 13 = 32.


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The case of x7 + y 7 = Czp

Associated with a non-trivial primitive solution (a, b, c)of x7 + y7 = Czp are three different Frey objects :

A Frey curve over Q

Y 2 = X 3 + a2X2 + a4X + a6

where

a2 = −(a− b)2,

a4 = −2a4 + a3b − 5a2b2 + ab3 − 2b4,

a6 = a6 − 6a5b + 8a4b2 − 13a3b3 + 8a2b4 − 6ab5 + b6.


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A Frey curve over a totally real cubic field :

Y 2 = X (X − Aa,b)(X + Ba,b),

where z = −(ζ7 + ζ−17 ) and

Aa,b = (−2 + z + z2)(a + b)2

Ba,b = (4− z2)(a2 − zab + b2).

A Frey hyperelliptic curve over Q

y2 = x7 + 7abx5 + 14a2b2x3 + 7a3b3x + b7 − a7.


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This is a very rich situation !How the modular method extends to such a situation is thetopic of some of the next lectures and a great challenge forfuture research.

Thank you for your attention


introduction to the modular method

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