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

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

May 08, 2023 520 likes •726 views

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

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SLIDE 1

A FAMILY OF RANK SIX ELLIPTIC CURVES

OVER NUMBER FIELDS

David Mehrle & Tomer Reiter Carnegie Mellon University August 22, 2014

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SLIDE 2

ELLIPTIC CURVES

  • Elliptic curve

E : y2 = x3 + ax2 + bx + c

  • “

Adding” points on E makes group E(Q)

✥✥✥✥✥✥✥✥✥✥ ✥ s

P

s

Q

s s P+Q

E

✘✘✘✘✘✘✘✘✘✘ ✘ s

P

s s

P+P

E

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SLIDE 3

GROUP STRUCTURE

MORDELL-WEIL THEOREM: E(Q) finitely generated

⇓

E(Q) ∼ = Z r ⊕ T

“rank” “torsion”

  • Rank < ∞,

hard to compute!

  • Torsion = points of finite order
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SLIDE 4

RANK

CONJECTURE: 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

GOAL: Find family of curves of moderate rank

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SLIDE 5

NUMBER FIELDS

  • Number field K = finite field extension of Q
  • e.g. K = Q

√ −5

  • =
  • a + b

√ −5 | a, b ∈ Q

  • Many analogies with Q

Q K integers Z − → OK primes 0, 2, 3, 5, . . . − → prime ideals p ⊂ OK factorization integers − → ideals norm |p| =

  • Z/

(p)

  • −

→ N(p) :=

  • OK/

p

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SLIDE 6

ELLIPTIC SURFACES

  • Elliptic surface E ≈ elliptic curve / K(T)
  • Specialization:

E

plug in T = t

− − − − − − − − − − − → Et − → − →

curve/K(T) curve/K

SILVERMAN SPECIALIZATION THEOREM: If E is an elliptic surface, then for almost all t ∈ OK, rank (Et) ≥ rank (E)

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SLIDE 7

IMPORTANT THEOREM

ROSEN & SILVERMAN THEOREM: E an elliptic surface lim

X→∞

1 X

  • N(p)≤X

−AE(p) log N(p) = rank (E)

  • at(p) = N(p) + 1 − #Et

OK/

p

  • AE(p) =

1 N(p)

  • t∈OK/p

at(p) E

specialize

− − − − − → Et

reduce mod p

− − − − − − − − → Et OK/

p

count points − − − − − − − → at(p)

average

− − − − → AE(p)

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SLIDE 8

CONSTRUCTION

STEP 1: surface E with AE(p) = −6, ∀ p Rosen & Silverman

− − →

STEP 2: evaluate limit = ⇒ rank (E) = 6 Silverman Specialization

− − − − →

Family of rank 6 curves Et

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SLIDE 9

STEP 1 : EQUATIONS

  • Define surface E : y2 = f(x, T)

y2 = f(x, T) = T2x3 + T g(x) − h(x) g(x) = x3 + ax2 + bx + c, c = 0 h(x) = Ax3 + Bx2 + Cx + D

  • Discriminant of f in T

∆T(x) = g(x)2 + 4 x3h(x)

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SLIDE 10

STEP 1 : KEY IDEA

KEY IDEA: make roots of ∆T(x) distinct perfect squares

  • Choose roots ρ2

i of ∆T(x)

∆T(x) = (4A + 1)

6

  • i=1
  • x − ρ2

i

  • Equate coefficients

∆T(x) = (4A + 1)

6

  • i=1
  • x − ρ2

i

  • = g(x)2 + 4 x3h(x)
  • Solve nonlinear system for a, b, c, A, B, C, D
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SLIDE 11

STEP 1 : LEGENDRE SYMBOL

LEMMA: −AE(p) = # {perfect-square roots of ∆T(x)}

  • Legendre Symbol:

a p

  • =

     +1 a is a square mod p −1 a not a square mod p a ∈ p

  • at(p) = −
  • x∈OK/p

f(x, t) p

  • AE(p) =

1 N(p)

  • t∈OK/p

at(p) = −1 N(p)

  • t∈OK/p
  • x∈OK/p

f(x, t) p

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SLIDE 12

STEP 1 : LEGENDRE SUMS

LEMMA: −AE(p) = # {perfect-square roots of ∆T(x)}

  • Evaluate Legendre sum

−N(p)AE(p) =

  • x,t ∈OK/p

f(x, t) p

  • Quadratic Legendre sum in t
  • t ∈OK/p

f(x, t) p

  • =

   (N(p) − 1)

  • x

p

  • x root of ∆T(x)

−

  • x

p

  • else
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SLIDE 13

STEP 1 : COMPUTING AE(p)

LEMMA: −AE(p) = # {perfect-square roots of ∆T(x)}

  • Evaluate Legendre sum

−N(p)AE(p) =

  • x,t ∈OK/p

f(x, t) p

  • =
  • x root of ∆T(x)

t∈OK/p

f(x, t) p

  • +
  • x nonroot

t∈OK/p

f(x, t) p

  • =

N(p) #perfect-square

roots of ∆T(x)

  • = 6N(p)
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SLIDE 14

CONSTRUCTION

STEP 1: surface E with AE(p) = −6, ∀ p

Rosen & Silverman

− − →

STEP 2: evaluate limit = ⇒ rank (E) = 6 Silverman Specialization

− − − − →

Family of rank 6 curves Et

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SLIDE 15

STEP 2 : USE PREVIOUS STEP

ROSEN & SILVERMAN THEOREM: lim

X→∞

1 X

  • N(p)≤X

−AE(p) log N(p) = rank (E)

  • Step 1: AE(p) = −6
  • 1/

6

  • rank (E) =

lim

X→∞

1 X

  • N(p)≤X

log N(p)

  • Hope lim

X→∞(. . .) = 1

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SLIDE 16

STEP 2 : EVALUATE LIMIT

LANDAU PRIME IDEAL THEOREM:

  • N(p)≤X

log N(p) ≈ X

  • 1/

6

  • rank (E) =

lim

X→∞

1 X

  • N(p)≤X

log N(p) = 1

⇓

rank (E) = 6

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SLIDE 17

EXAMPLE

  • K = Q
  • E : y2 = f(x, T)

f(x, T) = T2x3 + T g(x) + h(x) g(x) = x3 + ax2 + bx + c h(x) = Ax3 + Bx2 + Cx + D

  • Choose roots 12, . . . , 62,

∆T(x) = (4A + 1)

6

  • i=1

(x − i2) a = 16660111104 A ≈ 8.9161 × 1018 b = −1603174809600 B ≈ −8.1137 × 1020 c = 2149908480000 C ≈ 2.6497 × 1022 D ≈ −3.4311 × 1023

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SLIDE 18

THE NON-GALOIS CASE L E(L) ⊆

  • K/Q not Galois

K E(K)

  • L/Q Galois

⊆ Q E(Q)

THEOREM: rank E(L) ≥ rank E(K) ≥ rank E(Q) COROLLARY: If E/K has coefficients in Q, then rank (E) = 6

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SLIDE 19

CREDITS

PRESENTED BY:

David Mehrle

dmehrle@cmu.edu

Tomer Reiter

tomer.reiter@gmail.com JOINT WORK WITH:

Joseph Stahl

josephmichaelstahl@gmail.com

Dylan Yott

dtyott@gmail.com ADVISED BY:

Steven J. Miller

sjm1@williams.edu Alvaro Lozano-Robledo alozano@math.uconn.edu SPECIAL THANKS TO: The PROMYS Program Boston University The SMALL REU Williams College FUNDED BY: NSF Grants DMS1347804, DMS1265673, the PROMYS Program, and Williams College