On the generalized Fermat equation x 2l + y 2m = z p

Transcription

1 On the generalized Fermat equation x 2l + y 2m = z p Samuele Anni joint work with Samir Siksek University of Warwick University of Debrecen, 29 th Journées Arithmétiques; 6 th July 2015

2 Generalized Fermat Equation 1 Generalized Fermat Equation 2 x 2l + y 2m = z p 3 The proof

3 Generalized Fermat Equation Generalized Fermat Equation Let (p, q, r) Z 3 2. The equation x p + y q = z r is a Generalized Fermat Equation of signature (p, q, r). A solution (x, y, z) Z 3 is called non-trivial if xyz 0, primitive if gcd(x, y, z) = 1. Poonen Schaefer Stoll: (2, 3, 7). Bruin: (2, 3, 8), (2, 8, 3), (2, 3, 9), (2, 4, 5), (2, 5, 4). Many others...

4 Generalized Fermat Equation Infinite Families of Exponents: Wiles: (p, p, p). Darmon and Merel: (p, p, 2), (p, p, 3). Many other infinite families by many people... The study of infinite families uses Frey curves, modularity and level-lowering over Q (or Q-curves).

5 Generalized Fermat Equation Solve x p + y p = z l Naïve idea To solve x p + y p = z l factor over Q (ζ), where ζ is a p-th root of unity. (x + y)(x + ζy)... (x + ζ p 1 y) = z l. x + ζ j y = α j ξ l j, α j finite set. ɛ j Q (ζ) such that ɛ 0 (x + y) + ɛ 1 (x + ζy) + ɛ 2 (x + ζ 2 y) = 0. γ 0 ξ l 0 + γ 1 ξ l 1 + γ 2 ξ l 2 = 0 (γ 0, γ 1, γ 2 ) finite set. It looks like x l + y l + z l = 0 solved by Wiles. Problems Problem 1: trivial solutions (1, 0, 1) and (0, 1, 1) become non-trivial. Problem 2: modularity theorems over non-totally real fields.

6 x 2l + y 2m = z p 1 Generalized Fermat Equation 2 x 2l + y 2m = z p 3 The proof

7 x 2l + y 2m = z p Theorem (Anni-Siksek) Let p = 3, 5, 7, 11 or 13. Let l, m 5 be primes. The only primitive solutions to x 2l + y 2m = z p are (±1, 0, 1) and (0, ±1, 1). Remark: this is a bi-infinite family of equations.

8 x 2l + y 2m = z p Let l, m, p 5 be primes, l p, m p. x 2l + y 2m = z p, gcd(x, y, z) = 1. Modulo 8 we get 2 z so WLOG 2 x. Only expected solution (0, ±1, 1). { x l + y m i = (a + bi) p x l y m i = (a bi) p a, b Z gcd(a, b) = 1. x l = 1 p 1 2 ((a + bi)p + (a bi) p ( ) = a (a + bi) + (a bi)ζ j ) = a (p 1)/2 j=1 j=1 ( (θj + 2)a 2 + (θ j 2)b 2) θ j = ζ j + ζ j Q (ζ + ζ 1 ).

9 x 2l + y 2m = z p Let K := Q (ζ + ζ 1 ) then x l = a (p 1)/2 j=1 ( (θj + 2)a 2 + (θ j 2)b 2) }{{} f j (a,b) θ j = ζ j + ζ j K. p x = a = α l, f j (a, b) O K = b l j, p x = a = p l 1 α l, f j (a, b) O K = p b l j, p = (θ j 2) p. (θ 2 2)f 1 (a, b) + (2 θ 1 )f 2 (a, b) + 4(θ 1 θ 2 )a 2 = 0. }{{}}{{}}{{} u v w

10 x 2l + y 2m = z p Frey curve ( ) E : Y 2 = X (X u)(x + v), = 16u 2 v 2 w 2. Problems Problem 1: trivial solutions (0, ±1, 1) become non-trivial. Trivial solution x = 0 implies a = 0 so w = 0 = = 0. Problem 2: modularity theorems over non-totally real fields. K := Q (ζ + ζ 1 )

11 x 2l + y 2m = z p Lemma Suppose p x. Let E be the Frey curve ( ). The curve E is semistable, with multiplicative reduction at all primes above 2 and good reduction at p. It has minimal discriminant and conductor Lemma D E/K = 2 4ln 4 α 4l b 2l j b 2l k, N E/K = 2 (αb j b k ). Suppose p x. Let E be the Frey curve ( ). The curve E is semistable, with multiplicative reduction at p and at all primes above 2. It has minimal discriminant and conductor D E/K = 2 4ln 4 p 2δ α 4l b 2l j b 2l k, N E/K = 2p (αb j b k ).

12 x 2l + y 2m = z p Lemma Suppose p x. Let E be the Frey curve ( ). The curve E is semistable, with multiplicative reduction at all primes above 2 and good reduction at p. It has minimal discriminant and conductor Lemma D E/K = 2 4ln 4 α 4l b 2l j b 2l k, N E/K = 2 (αb j b k ). Suppose p x. Let E be the Frey curve ( ). The curve E is semistable, with multiplicative reduction at p and at all primes above 2. It has minimal discriminant and conductor D E/K = 2 4ln 4 p 2δ α 4l b 2l j b 2l k, N E/K = 2p (αb j b k ).

13 The proof 1 Generalized Fermat Equation 2 x 2l + y 2m = z p 3 The proof

14 The proof Let l be a prime, and E elliptic curve over totally real field K. The mod l Galois Representation attached to E is given by ρ E,l : G K Aut(E[l]) = GL 2 (F l ) G K = Gal(K/K). The l-adic Galois Representation attached to E is given by ρ E,l : G K Aut(T l (E)) = GL 2 (Z l ), where T l (E) = lim E[l n ] is the l-adic Tate module. Definition E is modular if there exists a cuspidal Hilbert modular eigenform f such that ρ E,l ρ f,l.

15 The proof Proof of Fermat s Last Theorem uses three big theorems: 1 Mazur: irreducibility of mod l representations of elliptic curves over Q for l > 163 (i.e. absence of l-isogenies). 2 Wiles (and others): modularity of elliptic curves over Q. 3 Ribet: level lowering for mod l representations this requires irreducibility and modularity. Over totally real fields we have 1 Merel s uniform boundedness theorem for torsion. No corresponding result for isogenies. 2 Partial modularity results, no clean statements. 3 Level lowering for mod l representations works exactly as for Q : theorems of Fujiwara, Jarvis and Rajaei. Requires irreducibility and modularity.

16 The proof Reducible representations Let E be a Frey curve as in ( ). Lemma Suppose ρ E,l is reducible. Then either E/K has non-trivial l-torsion, or is l-isogenous to an an elliptic curve over K that has non-trivial l-torsion. Lemma For p = 5, 7, 11, 13, and l 5, with l p, the mod l representation ρ E,l is irreducible.

17 The proof Modularity Theorem (Anni-Siksek) Let K be a real abelian number field. Write S 5 = {q 5}. Suppose (a) 5 is unramified in K; (b) the class number of K is odd; (c) for each non-empty proper subset S of S 5, there is some totally positive unit u of O K such that Norm Fq/F 5 (u mod q) 1. q S Then every semistable elliptic curve E over K is modular. This theorem buids over results of Thorne and Skinner & Wiles.

18 The proof Corollary For p = 5, 7, 11, 13, the Frey curve E is modular. Proof. For p = 7, 11, 13 apply the above. For p = 5 we have K = Q ( 5). Modularity of elliptic curves over quadratic fields was proved by Freitas, Le Hung & Siksek.

19 The proof Let E/K be the Frey curve ( ), then ρ E,l is modular and irreducible. Then ρ E,l ρ f,λ for some Hilbert cuspidal eigenform f over K of parallel weight 2 that is new at level N l, where { 2O K if p x N l = 2p if p x. Here λ l is a prime of Q f, the field generated over Q by the eigenvalues of f. For p = 3 the modular forms to consider are classical newform of weight 2 and level 6: there is no such newform and so we conclude.

20 The proof If p 1 (mod 4), the field K = Q (ζ + ζ 1 ) has a unique subfield K of degree (p 1)/4. In the case p x the Frey curve E is defined over K. This not true in the case p x, but we can take a twist of the Frey curve some that it is defined over K. This way we can work with Hilbert modular cuspforms over a totally real field of lower degree. The conductor of this Frey curve can be computed similarly to the previous case.

21 The proof p Case Level Eigenforms f [Q f : Q ] 5 5 x 2O K 5 x 2p 7 7 x 2O K 7 x 2p f x 2O K f x 2p f 3, f x 2B 2 f 10,..., f 21 3 f 13 22, f f 24,..., f 27 f 5,..., f x 2B f 28, f 29 1 f 30, f 31 3 In each case we deduce a contradiction using the q-expansions of the Hilber modular forms in the table coefficients and the study of the Frey curve described before.

22 The proof Final remarks In order to solve x 2l + y 2m = z p for p 17 we need to be able to compute Hilbert modular cuspforms over totally real field of high degree. Anyway the following theorem hold: Theorem (Anni-Siksek) Let p be an odd prime and let K = Q (ζ + ζ 1 ). Let O K be the ring of integers of K and p be the unique prime ideal above p. Suppose that there are no elliptic curves E/K with full 2-torsion and conductors 2O K, 2p. Then there is an ineffective constant C p (depending only on p) such that for all primes l, m C p, the only primitive solutions to x 2l + y 2m = z p are (x, y, z) = (±1, 0, 1) and (0, ±1, 1).

23 The proof On the generalized Fermat equation x 2l + y 2m = z p Samuele Anni joint work with Samir Siksek University of Warwick University of Debrecen, 29 th Journées Arithmétiques; 6 th July 2015 Thanks!