= Z r T E ( Q ) rank torsion Rank < , hard to compute! - PowerPoint PPT Presentation

A F AMILY OF R ANK S IX E LLIPTIC C URVES OVER N UMBER F IELDS David Mehrle & Tomer Reiter Carnegie Mellon University August 22, 2014

E LLIPTIC C URVES E : y 2 = x 3 + ax 2 + bx + c • Elliptic curve • “ Adding” points on E makes group E ( Q ) ✘ s ✘✘✘✘✘✘✘✘✘✘ Q ✥ ✥✥✥✥✥✥✥✥✥✥ P s s s P s s P + Q E E P + P s

G ROUP S TRUCTURE M ORDELL -W EIL T HEOREM : E ( Q ) finitely generated ⇓ = Z r ⊕ T E ( Q ) ∼ “rank” “torsion” • Rank < ∞ , hard to compute! • Torsion = points of finite order

R ANK C ONJECTURE : rank is unbounded • Noam Elkies: 28 ≤ rank ( E ) ≤ 32 ← World Record! • High rank curves are hard to find! • Much interest in modern number theory • Applications to cryptography G OAL : Find family of curves of moderate rank

N UMBER F IELDS • Number field K = finite field extension of Q � √ √ • e.g. K = Q � � � − 5 = a + b − 5 | a , b ∈ Q • Many analogies with Q K Q integers Z − → O K primes 0 , 2 , 3 , 5 , . . . − → prime ideals p ⊂ O K factorization integers − → ideals � Z / � � � O K / � � norm | p | = − → N ( p ) := ( p ) � p �

E LLIPTIC S URFACES • Elliptic surface E ≈ elliptic curve / K ( T ) plug in T = t • Specialization: E − − − − − − − − − − − → E t → → − − curve/ K ( T ) curve/ K S ILVERMAN S PECIALIZATION T HEOREM : If E is an elliptic surface, then for almost all t ∈ O K , rank ( E t ) ≥ rank ( E )

I MPORTANT T HEOREM R OSEN & S ILVERMAN T HEOREM : E an elliptic surface 1 � − A E ( p ) log N ( p ) = rank ( E ) lim X X →∞ N ( p ) ≤ X � O K / � • a t ( p ) = N ( p ) + 1 − # E t p 1 � • A E ( p ) = a t ( p ) N ( p ) t ∈O K / p � count points specialize average reduce mod p � O K / E − − − − − → E t − − − − − − − − → E t − − − − − − − → a t ( p ) − − − − → A E ( p ) p

C ONSTRUCTION S TEP 1: surface E with A E ( p ) = − 6 , ∀ p − − Rosen & Silverman → S TEP 2: evaluate limit = ⇒ rank ( E ) = 6 − − Silverman − Specialization − → Family of rank 6 curves E t

S TEP 1 : E QUATIONS • Define surface E : y 2 = f ( x , T ) y 2 = f ( x , T ) = T 2 x 3 + T g ( x ) − h ( x ) g ( x ) = x 3 + ax 2 + bx + c , c � = 0 h ( x ) = Ax 3 + Bx 2 + Cx + D • Discriminant of f in T ∆ T ( x ) = g ( x ) 2 + 4 x 3 h ( x )

S TEP 1 : K EY I DEA make roots of ∆ T ( x ) distinct perfect squares K EY I DEA : • Choose roots ρ 2 i of ∆ T ( x ) 6 � x − ρ 2 � � ∆ T ( x ) = ( 4 A + 1 ) i i = 1 • Equate coefficients 6 = g ( x ) 2 + 4 x 3 h ( x ) � x − ρ 2 � � ∆ T ( x ) = ( 4 A + 1 ) i i = 1 • Solve nonlinear system for a , b , c , A , B , C , D

S TEP 1 : L EGENDRE S YMBOL L EMMA : − A E ( p ) = # { perfect-square roots of ∆ T ( x ) } • Legendre Symbol:  + 1 a is a square mod p  � a �  = − 1 a not a square mod p p  0 a ∈ p  � f ( x , t ) � � • a t ( p ) = − p x ∈O K / p 1 − 1 � f ( x , t ) � � � � • A E ( p ) = a t ( p ) = N ( p ) N ( p ) p t ∈O K / p t ∈O K / p x ∈O K / p

S TEP 1 : L EGENDRE S UMS L EMMA : − A E ( p ) = # { perfect-square roots of ∆ T ( x ) } • Evaluate Legendre sum � f ( x , t ) � � − N ( p ) A E ( p ) = p x , t ∈O K / p • Quadratic Legendre sum in t � �  x ( N ( p ) − 1 ) x root of ∆ T ( x ) � f ( x , t ) �  p � = � � p x − else  t ∈O K / p p

S TEP 1 : C OMPUTING A E ( p ) L EMMA : − A E ( p ) = # { perfect-square roots of ∆ T ( x ) } • Evaluate Legendre sum � f ( x , t ) � � − N ( p ) A E ( p ) = p x , t ∈O K / p � f ( x , t ) � � f ( x , t ) � � � = + p p x nonroot x root of ∆ T ( x ) t ∈O K / p t ∈O K / p � # perfect-square � = N ( p ) = 6 N ( p ) roots of ∆ T ( x )

C ONSTRUCTION � S TEP 1: surface E with A E ( p ) = − 6 , ∀ p − − Rosen & Silverman → S TEP 2: evaluate limit = ⇒ rank ( E ) = 6 − − Silverman − Specialization − → Family of rank 6 curves E t

S TEP 2 : U SE P REVIOUS S TEP R OSEN & S ILVERMAN T HEOREM : 1 � lim − A E ( p ) log N ( p ) = rank ( E ) X X →∞ N ( p ) ≤ X • Step 1: A E ( p ) = − 6 1 � � � 1 / rank ( E ) = lim log N ( p ) 6 X X →∞ N ( p ) ≤ X • Hope lim X →∞ ( . . . ) = 1

S TEP 2 : E VALUATE L IMIT L ANDAU P RIME I DEAL T HEOREM : � log N ( p ) ≈ X N ( p ) ≤ X 1 � � 1 / � rank ( E ) = lim log N ( p ) = 1 6 X X →∞ N ( p ) ≤ X ⇓ rank ( E ) = 6 �

E XAMPLE • K = Q • E : y 2 = f ( x , T ) = T 2 x 3 + T g ( x ) + h ( x ) f ( x , T ) = x 3 + ax 2 + bx + c g ( x ) = Ax 3 + Bx 2 + Cx + D h ( x ) 6 � • Choose roots 1 2 , . . . , 6 2 , ( x − i 2 ) ∆ T ( x ) = ( 4 A + 1 ) i = 1 8 . 9161 × 10 18 a = 16660111104 A ≈ − 8 . 1137 × 10 20 = − 1603174809600 ≈ b B 2 . 6497 × 10 22 c = 2149908480000 C ≈ − 3 . 4311 × 10 23 D ≈

T HE NON -G ALOIS CASE L E ( L ) ⊆ • K / Q not Galois E ( K ) K • L / Q Galois ⊆ E ( Q ) Q T HEOREM : rank E ( L ) ≥ rank E ( K ) ≥ rank E ( Q ) If E / K has coefficients in Q , then rank ( E ) = 6 C OROLLARY :

C REDITS P RESENTED BY : David Mehrle dmehrle@cmu.edu Tomer Reiter tomer.reiter@gmail.com J OINT WORK WITH : Joseph Stahl josephmichaelstahl@gmail.com Dylan Yott dtyott@gmail.com A DVISED BY : Steven J. Miller sjm1@williams.edu Alvaro Lozano-Robledo alozano@math.uconn.edu S PECIAL THANKS TO : The PROMYS Program Boston University The SMALL REU Williams College F UNDED BY : NSF Grants DMS1347804, DMS1265673, the PROMYS Program, and Williams College