The Asymptotic Fermat Conjecture

Transcription

1 The Asymptotic Fermat Conjecture Samir Siksek (Warwick) joint work with Nuno Freitas (Warwick), and Haluk Şengün (Sheffield) 23 August 2018

2 Motivation Theorem (Wiles, Taylor Wiles 1994) Semistable elliptic curves over Q are modular. Theorem (Wiles 1994) The only solutions to the equation a p + b p + c p = 0, a, b, c Q, p 3 prime satisfy abc = 0.

3 Motivation Theorem (Jarvis and Manoharmayum 2004) Semistable elliptic curves over Q( 2) and Q( 17) are modular. Theorem (Jarvis and Meekin 2004) The only solutions to the equation a p + b p + c p = 0, a, b, c Q( 2), p 5 prime satisfy abc = the numerology required to generalise the work of Ribet and Wiles directly continues to hold for Q( 2)... however, we will explain that there are no other real quadratic fields for which this is true... Objective: To prove the Asymptotic FLT for some infinite families of number fields.

4 An Example of Serre and Mazur Let L 3 be a rational prime. Suppose (a, b, c) Z 3 is a solution to a p + b p + L r c p = 0 with abc 0 and gcd(a, b, Lc) = 1, and p 5 prime. {A, B, C} = {a p, b p, L r c p }, A 1 (mod 4), 2 B E : Y 2 = X (X A)(X + B) (Frey curve) Mazur, Wiles, Ribet = { ρ E,p ρ f,π f newform of weight 2 and level 2L. ρ E,p ρ f,π = { a q (E ) a q (f ) (mod π) if q 2Lpabc ±(q + 1) a q (f ) (mod π) if q abc, q 2Lp

5 An Example of Serre and Mazur : a p + b p + L r c p = 0 Suppose a q (f ) / Q for some q 2L. Therefore π divides C q (f ) := q(q + 1 a q (f ))(q a q (f )) a <2 q (a a q (f )) 0. But π p. p Norm(C q (f )). Conclusion: If p > B L (some explicit constant) then f is rational. Hence ρ E,p ρ E,p where E/Q is an elliptic curve of conductor 2L. Stronger conclusion: If p > B L (some bigger explicit constant) then ρ E,p ρ E,p where E/Q is an elliptic curve of conductor 2L and full 2-torsion.

6 An Example of Serre and Mazur : a p + b p + L r c p = 0 Stronger conclusion: If p > B L (some bigger explicit constant) then ρ E,p ρ E,p where E/Q is an elliptic curve of conductor 2L and full 2-torsion. Such E has model (minimal away from 2) E : Y 2 = X (X u)(x + v), = 16u 2 v 2 (u + v) 2 = ±2 m L n. u = ±2 m 1 L n 1, v = ±2 m 2 L n 2, u + v = ±2 m 3 L n 3, ±2 m 1 L n 1 ± 2 m 2 L n 2 = ±2 m 3 L n 3. Easy exercise: show that L is a Mersenne or a Fermat prime. Theorem (Serre, Mazur) If L is neither Mersenne nor Fermat, then p < B L.

7 Theoretical Pillars The proof of Fermat s Last Theorem over Q rests on three pillars: (i) Mazur s isogeny theorem. (ii) Wiles, Breuil, Conrad, Diamond, Taylor: elliptic curves /Q are modular. (iii) Ribet s level lowering theorem. Can replace (ii) and (iii) with Serre s modularity conjecture /Q (Khare and Wintenberger).

8 Theoretical Pillars over Totally Real Fields No analogue of Mazur s isogeny theorem. Can make do with Merel s uniform boundedness theorem. Modularity lifting and level optimization theorems due to Kisin, Barnet-Lamb Gee Geraghty, Breuil Diamond,.... Theorem (Calegari, Freitas Le Hung S.) Let K be a totally real field. There are at most finitely many j-invariants of elliptic curves over K that are non-modular. Theorem (Freitas Le Hung S.) Let E be an elliptic curve over a real quadratic field K. Then E is modular.

9 Theoretical Pillars over General Number Fields If we want to say something about the Fermat equation over a general number field K, need to assume conjectures: Serre s uniformity conjecture over K. Serre s modularity conjecture over K.

10 Over Totally Real Fields A Sample Result Lemma (Freitas S., Kraus) Let K be a totally real field in which 2 totally ramifies: 2O K = q [K:Q]. Suppose there is no elliptic curve E/K with full 2-torsion and conductor q. Then the asymptotic FLT holds for K: there is a constant B K such that if p > B K is prime and a, b, c K satisfy a p + b p + c p = 0 then abc = 0. If we have modularity for elliptic curves over K, then B K is effective. Conjecture (Kraus) Let K be a number field of narrow class number 1 with a unique prime q above 2. Then there is no elliptic curve E/K with full 2-torsion and conductor q. Evidence: LMFDB. Kraus: Do the number fields Q(ζ 2 r ) + have narrow class number 1?

11 Theorem (Kraus Conjecture, Freitas-S. 2018) Let K be a number field of narrow class number 1 with a unique prime q above 2. Then there is no elliptic curve E/K with full 2-torsion and conductor q. Proof. Suppose E exists. There is a basis for T 2 (E) such that ( ) χ2 ρ E,2 Iq = (from Tate module of Tate curve). 0 1 Claim: for all n 1, ρ E,2 n ( ) χ2 n 0 1 (mod 2 n ). Claim contradicts Serre s Open Image Theorem! So enough to prove claim. We prove claim by induction. Case n = 1 true as E has full 2-torsion. ( χ2 n + 2 Inductive step: ρ E,2 n n 1 ) φ 2 n 1 ψ n 1 (mod 2 n ), µ where φ, ψ, µ : G K Z/2Z are functions.

12 Proof. Suppose E exists. There is a basis for T 2 (E) such that ( ) χ2 ( ) ρ E,2 Iq = (from Tate module of Tate curve). 0 1 Claim: for all n 1, ρ E,2 n ( ) χ2 n 0 1 (mod 2 n ). We prove claim by induction. Case n = 1 true as E has full 2-torsion. ( χ2 n + 2 Inductive step: ρ E,2 n n 1 ) φ 2 n 1 ψ n 1 (mod 2 n ), µ where φ, ψ, µ : G K Z/2Z are functions. ( ) = φ Iq ψ Iq µ Iq 0 (mod 2). E has conductor q = { φ Im ψ Im µ Im 0 (mod 2) for all finite places m.

13 We prove claim by induction. Case n = 1 true as E has full 2-torsion. ( χ2 n + 2 Inductive step: ρ E,2 n n 1 ) φ 2 n 1 ψ n 1 (mod 2 n ) ( ) µ where φ, ψ, µ : G K Z/2Z are functions. { φ Im ψ Im µ Im 0 (mod 2) for all finite places m. But ρ E,2 n(στ) ρ E,2 n(σ) ρ E,2 n(τ) (mod 2 n ) for all σ, τ G K. 2 n 1 ψ(στ) 2 n 1 ψ(σ) χ 2 n(τ) + 2 n 1 ψ(τ) (mod 2 n ). ψ(στ) ψ(σ) χ 2 n(τ) + ψ(τ) (mod 2). ψ(στ) ψ(σ) + ψ(τ) (mod 2) (ψ : G K Z/2Z is a character).

14 ( χ2 n + 2 ρ E,2 n n 1 ) φ 2 n 1 ψ n 1 µ (mod 2 n ) ψ : G K Z/2Z is a character unramified at all finite places. But h + K = 1. Therefore ψ = 0. Similarly φ = 0 and µ = 0. Proved claim: for all n 1, ρ E,2 n ( ) χ2 n 0 1 (mod 2 n ). Theorem (Kraus Conjecture, Freitas-S. 2018) Let K be a number field of narrow class number 1 with a unique prime q above 2. Then there is no elliptic curve E/K with full 2-torsion and conductor q.

15 The proof gives a stronger theorem! Theorem ( Strong Kraus Conjecture, Freitas-S. 2018) Let K be a number field of odd narrow class number with a unique prime q above 2. Then there is no elliptic curve E/K with non-trivial 2-torsion and conductor q. Theorem (Freitas S.) Let K be a totally real field with odd narrow class number in which 2 totally ramifies. Then the asymptotic FLT holds for K: there is a constant B K such that if p > B K is prime and a, b, c K satisfy a p + b p + c p = 0 then abc = 0.

16 Asymptotic Fermat for Q(ζ 2 r) + Iwasawa Growth Formula = Q(ζ 2 r ), Q(ζ 2 r ) + have odd class numbers. Theorem (Freitas S.) Then the effective asymptotic FLT holds for Q(ζ 2 r ) +. Proof. By Thorne, we have modularity for elliptic curves over Q(ζ 2 r ) +. Suppose there is a non-trivial solution to the Fermat equation with p large. Then there is an elliptic curve E/Q(ζ 2 r ) + with full 2-torsion and conductor q 2. View E as an elliptic curve over Q(ζ 2 r ) (which has odd narrow class number). The conductor is q, the unique prime above 2. Contradiction.

17 Fermat over General Number Fields Let K be a number field. Assume two conjectures: (i) Serre s modularity conjecture over K. (ii) An Eichler Shimura conjecture : if f is a newform over K of parallel weight 2 and level N and rational Hecke eigenvalues, then it corresponds either to an elliptic curve of conductor N, or a fake elliptic curve of conductor N 2. Theorem (Şengün S.) Let K be a number field for which (i) and (ii) hold. Let S := {q 2}, T := {q 2 : F q = F 2 }. Suppose T. Suppose that for every solution (λ, µ) of λ + µ = 1, λ, µ O S there is some q T such that max( ord q (λ), ord q (µ) ) ord q (16). Then the asymptotic FLT holds over K.

18 Theorem (Şengün S.) Let K be a number field for which (i) and (ii) hold. Let S := {q 2}, T := {q 2 : F q = F 2 }. Suppose T. Suppose that for every solution (λ, µ) of λ + µ = 1, λ, µ O S there is some q T such that max( ord q (λ), ord q (µ) ) ord q (16). Then the asymptotic FLT holds over K. Corollary Let K = Q( d) where d > 0 and d 2 or 3 (mod 4). Assume (i), (ii). Then asymptotic FLT holds for K. Proof. Easy exercise: show that the only solutions to the S-unit equation are (1/2, 1/2), ( 2, 1), ( 1, 2).

19 Sketch of the proof: Absolute Irreducibility Suppose there is a non-trivial solution (a, b, c) to the Fermat equation with p large. Let E be the Frey curve. Let q T (i.e. F q = F 2 ). As a p + b p + c p = 0, one of the the three terms has larger q-valuation than the other two. From explicit formulae ord q (j(e )) < 0, p ord q (j(e )). Theory of Tate curve: p #ρ(i q ). p #ρ(g K ). Almost semistability+merel+element of order p in image = ρ E,p is absolutely irreducible.

20 Sketch of the proof: Serre s Modularity Conjecture Apply Serre s modularity conjecture to ρ E,p. Obtain mod p eigenform of weight 2 and some small level N ; i.e. an element of H i (Y 0 (N ), F p ). 0 H i (Y 0 (N ), Z) F p H i (Y 0 (N ), F p ) H i+1 (Y 0 (N ), Z)[p] 0. Key fact: H i+1 (Y 0 (N ), Z) is finitely generated. For p large, H i (Y 0 (N ), Z)[p] = 0. Can lift the mod p eigenform to a complex eigenform f. p is large. f has rational Hecke eigenvalues.

21 Proof Sketch: dealing with fake elliptic curve Can f correspond to a fake elliptic curve A? B. Jordan: A has potentially everywhere good reduction. ρ f,p (I q ) < C. Contradicts p ρ E,p(I q ). f corresponds to an elliptic curve E/K....

22 Real Quadratic Fields Theorem (Freitas S.) The asymptotic FLT holds for 5/6 of real quadratic fields. If we assume the Eichler Shimura conjecture, the asymptotic FLT holds for almost all real quadratic fields.

23 Thank You!