Torsion points on elliptic curves and gonalities of modular curves

Transcription

1 Torsion points on elliptic curves and gonalities of modular curves with a focus on gonalities of modular curves. Mathematisch Instituut Universiteit Leiden Graduation talk

2 Outline 1 2 3

3 What is known S(d) := {p prime K /Q: [K : Q] d, E/K : E(K ) [p] 0} Primes(n) := {p prime p n} S(d) is finite (Merel) S(d) Primes((3 d/2 + 1) 2 ) (Oesterlé) S(1) = Primes(7) (Mazur) S(2) = Primes(13) (Kamienny, Kenku, Momose) S(3) = Primes(13) (Parent) S(4) = Primes(17) (Kamienny, Stein, Stoll) to be published.

4 New results in my thesis S(d) := {p prime K /Q: [K : Q] d, E/K : E(K ) [p] 0} Primes(n) := {p prime p n} S(5) Primes(19) {29, 31, 41} S(6) Primes(41) {73} S(7) Primes(43) {59, 61, 67, 71, 73, 113, 127} This is in the "Torsion Points" part of my thesis. Today I will not talk about this, but about how to show S(5) = Primes(19). This joint work with Michael Stoll and will be published together with the S(4) result.

5 j-invariant Over C the j-invariant gives a 1-1 correspondence: j : {E/C}/ C Now C = H/SL 2 (Z) where SL 2 (Z) acts on H by: [ ] a b τ = aτ + b c d cτ + d Analitic description E = C/(τZ + Z): j(e) = τ mod SL 2 (Z) Algebraic description E = Z (y 2 x 3 ax b) j(e) = a3 4a b 2

6 Analitic description of the modular curve Y 1 (N) {[ ] [ ] a b a b Γ 1 (N) := SL c d 2 (Z) c d [ ] } mod N Y 1 (N)(C) := H/Γ 1 (N) There is again a 1-1 correspondence: ψ : {(E, P) E/C, P E of order N}/ 1:1 Y 1 (N)(C) Analitic description (E, P) = (C/(τZ + Z), 1/N mod τz + Z) ψ(e, P) = τ mod SL 2 (Z)

7 Algebraic description of the modular curve Y 1 (N) Proposition Let K be a field, E/K and P E(K ) of order N 4. Then there are unique b, c K such that E = Z (Y 2 + cxy + by X 3 bx 2 ) and P = (0, 0) R := Z[b, c, 1 ] with := b 3 (16b 2 + (8c 2 36c + 27)b + (c 1)c 3 ) E/R elliptic curve given by Y 2 + cxy + by = X 3 + bx 2 P := (0 : 0 : 1) Let Φ N, Ψ N, Ω N R be s.t. (Φ N Ψ N : Ω N : Ψ 3 N ) = NP The equation Ψ N = 0 means P has order dividing N. Define F N by removing form Ψ N all factors coming from some Ψ d with d N. Y 1 (N) Z[1/N] := Spec(R[1/N]/F N )

8 Algebraic description of the modular curve Y 1 (N) R := Z[b, c, 1 ] E/R elliptic curve given by Y 2 + cxy + by = X 3 + bx 2 P := (0 : 0 : 1) Let Φ N, Ψ N, Ω N R be s.t. (Φ N Ψ N : Ω N : Ψ 3 N ) = NP Define F N by removing form Ψ N all factors coming from some Ψ d with d N. Y 1 (N) Z[1/N] := Spec(R[1/N]/F N ) Let N 4 and let K be a field with char(k ) N then ψ : {(E, P) E/K, P E(K ) of order N}/ 1:1 Y 1 (N)(K ) Let (E, P) = (Z (y 2 cxy by x 3 bx 2 ), (0, 0)) then ψ(e, P) = (b, c)

9 Relation between Y 1 (N) and S(d) The 1-1 correspondence gives ψ : {(E, P) E/K, P E(K ) of order N}/ 1:1 Y 1 (N)(K ) S(d) := {p prime K /Q: [K : Q] d, E/K : E(K ) [p] 0} = = {p prime K /Q: [K : Q] d, Y 1 (p)(k ) } So we want to know whether Y 1 (29), Y 1 (31) and Y 1 (41) contain points of degree 5 over Q.

10 X 1 (N) and cusps Let N 5. Then Y 1 (N) can be embedded in a projective Z[1/N]-scheme X 1 (N). Let K = K and N prime. Then #(X 1 (N)(K )\Y 1 (N)(K )) = N 1. These N 1 elements are called the cusps. Over Q we have #(X 1 (N)(Q)\Y 1 (N)(Q)) = (N 1)/2. i.e. only half of the cusps are defined over Q.

11 A useful proposition of Michael Stoll Proposition Let C/Q be a smooth proj. geom. irred. curve with Jacobian J, d 1 and l a prime of good reduction for C. Let P C(Q) and ι : C (d) J the canonical map normalized by ι(dp) = 0. Suppose that: 1 there is no non-constant f Q(C) of degree d. 2 J(Q) is finite. 3 l > 2 or J(Q)[2] J(F l ). 4 C(Q) C(F l ) 5 The intersection of ι(c (d) (F l )) J(F l ) with the image of J(Q) under reduction mod l is contained in the image of C d (F l ). Then C(Q) is the set of points of degree d on C.

12 Verifying the hypotheses Mazurs result on S(1) implies that if p > 7 then the only rational points on X 1 (p)(q) are the rational cusps. So if hypotheses 1 5 are satisfied for X 1 (p) and d with p > 7 and some l then p / S(d). Stoll has shown hypotheses 2 5 are satisfied for l = 2, d = 5 and C = X 1 (29), X 1 (31) or X 1 (41). What remains for proving that S(5) = Primes(19) is: For p = 29, 31 and 41 there is no non constant f Q(X 1 (p)) of degree 5. For p = 41 this was already known. For p = 29, 31 this is proved in the "gonalities" part of my thesis.

13 Definition of gonality Definition Let K be a field and C/K be a smooth proj. geom. irred. curve then the K -gonality of C is: gon K (C) := min f K (C)\K [K (C) : K (f )] = min f K (C)\K deg f Theorem (Abramovich) Let N be a prime then: gon C (X 1 (N)) (N2 1). If Selberg s eigenvalue conjecture holds then: gon C (X 1 (N)) (N2 1). So gon Q (X 1 (41)) gon C (X 1 (41)) 7/1600(41 2 1) > 7. But, even with the conjecture, this doesn t give a good enough bound for gon Q (X 1 (29)) and gon Q (X 1 (31))

14 The F l gonality is smaller than the Q-gonality Proposition Let C/Q be a smooth proj. geom. irred. curve and l be a prime of good reduction of C then: gon Q (C) gon Fl (C Fl ) To use this we need to know how compute the F l gonality of C.Let div + d C F l div + C Fl be the set of effective divisors of degree d. Then #(div + d C F l ) <.The following algorithm computes the F l -gonality: Step 1 set d = 1 Step 2 While for all D div + d C F l : dim H 0 (C, D) = 1 set d = d + 1 Step 3 Output d. This is too slow to compute gon F2 (X 1 (29)) and gon F2 (X 1 (31))

15 Divisors dominating all functions of degree d C/F l a smooth proj. geom. irr. curve. View f F l (C) as a map f : C P 1 F l. For g Aut C, h Aut P 1 F l : deg f = deg h f g Definition A set of divisors S div C dominates all functions of degree d if for all dominant f : C P 1 F l of degree d there are D S, g Aut C and h Aut P 1 F l such that div h f g D Proposition If S div C dominates all functions of degree d then gon Fl C min(d + 1, inf D S, f H 0 (C,D), degf 0 deg f ). Example: div + d C dominates all functions of degree d.

16 A smaller set of divisors dominating functions of degree d Proposition Define n := #C(F l )/(l + 1) and D := p C(F l ) p. Then div + d n C + D := { s + D s div + d n C} dominates all functions of degree d. Proof. There is a g Aut P 1 F q such that g f has poles at at least n distinct points in C(F q ). If f has degree d then there is an element s + d nc such that div g f s D.

17 An even smaller set of divisors dominating functions of degree d Proposition If S div C dominates all functions of degree d and S div C is such that for all s S there are s S and g Aut C such that g(s ) s. Then S also dominates all functions of degree d. This means that only 1 representative of each Aut C orbit of S is needed. This will be usefull in the cases C = X 1 (p) with p = 29, 31. In these case we have an automorphism of C for each d (Z/pZ) / {±1} given by (E, P) (E, dp). This gives 14 and 15 automorphisms respectively.

18 Computing the F 2 -gonality of X 1 (29) and X 1 (31) Proposition gon F2 (X 1 (29)) = 11 and gon F2 (X 1 (31)) = 12 Proof. For a "smart" choice of S div X 1 (p) dominating all function of degree d with d = 10 (respectively 11) I computed: gon Fl (X 1 (p)) min(d + 1, inf deg f ). D S, f H 0 (X 1 (p),d), degf 0 using Magma. This gives lower bounds 11 (resp. 12). During this computation I found functions of deg 11 (resp. 12).

19 The Q-gonality of X 1 (29) and X 1 (31) Over Q there are known functions of degree 11 (respectively 13) on X 1 (29) (respectively X 1 (31)). Corollary gon Q (X 1 (29)) = 11 and gon Q (X 1 (31)) {12, 13} Actually,gon Q (X 1 (31)) = 12 because recently Mark van Hoeij found a function of degree 12 on X 1 (31) defined over Q.

20 Summary S(5) = Primes(19) (was Primes(271)) S(6) Primes(41) {73} (was Primes(773)) S(7) Primes(127) (was Primes(2281)) Work in progress: Using Michael Stoll s ideas I am close to proving: Primes(19) {37} S(6) Primes(19) {37, 73}