why should i care about elliptic curves?egoins/notes/blackwell.pdfheron triangles elliptic curves...

Heron TrianglesElliptic Curves

Diophantine n-tuples

Why Should I Care About Elliptic Curves?

Edray Herber Goins

Department of MathematicsPurdue University

August 7, 2009

MAA MathFest – David Blackwell Lecture Elliptic Curves



Abstract

An elliptic curve E possessing a rational point is anarithmetic-algebraic object: It is simultaneously a nonsingularprojective curve with an affine equation y 2 = x3 + A x + B, whichallows one to perform arithmetic on its points; and a finitelygenerated abelian group E(Q) ' E(Q)tors × Zr , which allows one toapply results from abstract algebra.

In this talk, we discuss some basic properties of elliptic curves, andgive applications along the way.




David Harold Blackwell

http://www.math.buffalo.edu/mad/PEEPS/blackwell_david.html


http://www.math.buffalo.edu/mad/PEEPS/blackwell_david.html



Why should I

care about

David Blackwell?




Institute for Advanced Study, April 2001

http://modular.math.washington.edu/pics/ascent4_pics/April01/

06April01/small_dscf2855.jpg


http://modular.math.washington.edu/pics/ascent4_pics/April01/06April01/small_dscf2855.jpg

http://modular.math.washington.edu/pics/ascent4_pics/April01/06April01/small_dscf2855.jpg



From the The MacTutor History of Mathematics archive:

“At this point [after graduation in 1941] Blackwell received aprestigious one year appointment as Rosenwald Postdoctoral Fellowat the Institute for Advanced Study in Princeton. This, however, ledto problems over the fact that Blackwell was a black AfricanAmerican.

The standard practice was to have Fellows of the Institute becomehonorary Faculty members of Princeton University but, in Blackwell’scase, this caused a problem. At that time Princeton University hadnever had a black undergraduate student much less a black Facultymember and this produced opposition within the University.

The President of the University wrote to the Director of the Institutefor Advanced Study saying that the Institute were abusing thehospitality of the University with such an appointment. Fortunatelyat the time Blackwell was not fully aware of the problems hisappointment was causing.”

http://www-history.mcs.st-andrews.ac.uk/Biographies/Blackwell.html


http://www-history.mcs.st-andrews.ac.uk/Biographies/Blackwell.html



Outline of Talk

1 Heron Triangles

2 Elliptic CurvesDefinitionsChord-Tangent MethodPoincare’s Conjecture / Mordell’s Theorem

3 Diophantine n-tuplesHistoryExamples for n ≤ 5Relation with Elliptic Curves




Can you find

a right triangle

with rational sides

having area A = 6?




Motivating Question

Consider positive rational numbers a, b, and c satisfying

a2 + b2 = c2 and1

2a b = 6.

a

b

c

Recall the (a, b, c) = (3, 4, 5) triangle!




Cubic Equations

Are there more rational solutions (a, b, c) to

a2 + b2 = c2 and1

2a b = 6?

Proposition

Let x and y be rational numbers, and denote the rational numbers

a =x2 − 36

y, b =

12 x

y, and c =

x2 + 36

y.

Then

a2 + b2 = c2

1

2a b = 6

if and only if

{y2 = x3 − 36 x .

Example: (x , y) = (12, 36) corresponds to (a, b, c) = (3, 4, 5).

Can we find infinitely many rational solutions (a, b, c)?




DefinitionsChord-Tangent MethodPoincare’s Conjecture / Mordell’s Theorem

What types of properties

does this

cubic equation have?





What is an Elliptic Curve?

Definition

Let A and B be rational numbers such that 4 A3 + 27 B2 6= 0. An ellipticcurve E is the set of all (x , y) satisfying the equation

y2 = x3 + A x + B.We will also include the “point at infinity” O.

Example: y2 = x3 − 36 x is an elliptic curve.

Non-Example: y2 = x3 − 3 x + 2 is not an elliptic curve.

-12 -8 -4 0 4 8 12

-20

-10

10

20

-3 -2 -1 0 1 2 3

-5

-2.5

2.5

5





What is an Elliptic Curve?

Formally, an elliptic curve E over Q is a nonsingular projective curve ofgenus 1 possessing a Q-rational point O.

Such a curve is birationally equivalent over Q to a cubic equation inWeierstrass form:

E : y2 = x3 + A x + B;

with rational coefficients A and B, and nonzero discriminant∆(E ) = −16

(4 A3 + 27 B2

). For any field K , define

E (K ) =

{(x1 : x2 : x0) ∈ P2(K )

∣∣∣∣ x22 x0 = x3

1 + A x1 x20 + B x3

0

};

where O = (0 : 1 : 0) is on the projective line at infinity x0 = 0.

Remark: In practice we choose either K = Q or Fp.





Chord-Tangent Method

Given two rational points on an elliptic curve E , we explain how toconstruct more.

1 Start with two rational points P and Q.

2 Draw a line through P and Q.

3 The intersection, denoted by P ∗ Q, is another rational point on E .





Example: y 2 = x3 − 36 x

Consider the two rational points

P = (6, 0) and Q = (12, 36).

-8 -4 0 4 8 12 16 20 24

-50

50

P

Q

P*Q

P ∗ Q = (18, 72)







P = (6, 0) and Q = (12, 36).

-8 -4 0 4 8 12 16 20 24

-50

50

P

P ∗ P = OMAA MathFest – David Blackwell Lecture Elliptic Curves






P = (6, 0) and Q = (12, 36).

-8 -4 0 4 8 12 16 20 24

-50

50

Q*Q

Q

Q ∗ Q = (25/4, 35/8)





Group Law

Definition

Let E be an elliptic curve defined over a field K , and denote E (K ) as theset of K -rational points on E . Define the operation ⊕ as

P ⊕ Q = (P ∗ Q) ∗ O.

-4.8 -4 -3.2 -2.4 -1.6 -0.8 0 0.8 1.6 2.4 3.2 4 4.8

3 2

-2.4

-1.6

-0.8

0.8

1.6

2.4

3.2

P

QP*Q

P+Q







P = (6, 0) and Q = (12, 36).

-8 -4 0 4 8 12 16 20 24

-50

50

P

Q

P*Q

P+Q

P ⊕ Q = (18,−72)







P = (6, 0) and Q = (12, 36).

-8 -4 0 4 8 12 16 20 24

-50

50

P

P ⊕ P = OMAA MathFest – David Blackwell Lecture Elliptic Curves






P = (6, 0) and Q = (12, 36).

-8 -4 0 4 8 12 16 20 24

-50

50

Q*Q

Q

Q+Q

Q ⊕ Q = (25/4,−35/8)





Poincare’s Theorem

Theorem (Henri Poincare, 1901)

Let E be an elliptic curve defined over a field K . Then E (K ) is anabelian group under ⊕.

Recall that to be an abelian group, the following five axioms must besatisfied:

Closure: If P, Q ∈ E (K ) then P ⊕ Q ∈ E (K ).

Associativity: (P ⊕ Q)⊕ R = P ⊕ (Q ⊕ R).

Commutativity: P ⊕ Q = Q ⊕ P.

Identity: P ⊕O = P for all P.

Inverses: [−1]P = P ∗ O satisfies P ⊕ [−1]P = O.





What types of properties

does this

abelian group have?





Poincare’s Conjecture

Conjecture (Henri Poincare, 1901)

Let E be an elliptic curve. Then E (Q) is finitely generated.

Recall that an abelian group G is said to be finitely generated if thereexists a finite generating set

{a1, a2, . . . , an}

such that, for each given g ∈ G , there are integers m1, m2, . . . , mn suchthat

g = [m1]a1 ◦ [m2]a2 ◦ · · · ◦ [mn]an.

Example: G = Z is a finitely generated abelian group because allintegers are generated by a1 = 1.





Mordell’s Theorem

Theorem (Louis Mordell, 1922)

Let E be an elliptic curve. Then E (Q) is finitely generated.

That is, there exists a finite group E (Q)tors and a nonnegative integer rsuch that

E (Q) ' E (Q)tors × Zr .

The set E (Q) is called the Mordell-Weil group of E .

The finite set E (Q)tors is called the torsion subgroup of E . Itcontains all of the points of finite order, i.e., those P ∈ E (Q) suchthat

[m]P = O for some positive integer m.

The nonnegative integer r is called the Mordell-Weil rank of E .






Consider the three rational points

P1 = (0, 0), P2 = (6, 0), and P3 = (12, 36).

[2]P1 = [2]P2 = O, i.e., both P1 and P2 have order 2. They are torsion.

-8 -4 0 4 8 12 16 20 24

-50

50

(6,0)(-6,0)

(0,0)

E (Q)tors = 〈P1, P2〉 ' Z2 × Z2






Consider the three rational points

P1 = (0, 0), P2 = (6, 0), and P3 = (12, 36).

[2]P3 = (25/4,−35/8) and [3]P3 = (16428/529,−2065932/12167).

-8 0 8 16 24 32

-160

-80

80

160

P

[2]P

[3]P

E (Q) = 〈P1, P2, P3〉 ' Z2 × Z2 × ZMAA MathFest – David Blackwell Lecture Elliptic Curves




Classification of Torsion Subgroups

Theorem (Barry Mazur, 1977)

Let E is an elliptic curve, then

E (Q)tors '

{Zn where 1 ≤ n ≤ 10 or n = 12;

Z2 × Z2m where 1 ≤ m ≤ 4.

Remark: Zn denotes the cyclic group of order n.

Example: The elliptic curve y2 = x3 − 36 x has torsion subgroupE (Q)tors ' Z2 × Z2 generated by P1 = (0, 0) and P2 = (6, 0).

Mordell’s Theorem states that

E (Q) ' E (Q)tors × Zr .

What can we say about the Mordell-Weil rank r?





Records for Prescribed Torsion and Rank

E(Q)tors Highest Rank r Found By year Discovered

Trivial 28 Elkies 2006

Z2 19 Elkies 2009

Z3 13 Eroshkin 2007, 2008, 2009

Z4 12 Elkies 2006

Z5 6 Dujella, Lecacheux 2001

Z6 8Eroshkin

Dujella, EroshkinElkies, Dujella

200820082008

Z7 5Dujella, Kulesz

Elkies20012006

Z8 6 Elkies 2006

Z9 4 Fisher 2009

Z10 4DujellaElkies

2005, 20082006

Z12 4 Fisher 2008

Z2 × Z2 15 Elkies 2009

Z2 × Z4 8Elkies

EroshkinDujella, Eroshkin

200520082008

Z2 × Z6 6 Elkies 2006

Z2 × Z8 3

ConnellDujella

Campbell, GoinsRathbun

Flores, Jones, Rollick, WeigandtFisher

20002000, 2001, 2006, 2008

20032003, 2006

20072009

http://web.math.hr/~duje/tors/tors.html


http://web.math.hr/~duje/tors/tors.html




How does this

help answer

the motivating questions?





Rational Triangles Revisited

Can we find infinitely many right triangles (a, b, c) having rational sidesand area A = 6?

Proposition

Let x and y be rational numbers, and denote the rational numbers

a =x2 − 36

y, b =

12 x

y, and c =

x2 + 36

y.

Then

a2 + b2 = c2

1

2a b = 6

if and only if

{y2 = x3 − 36 x .

Example: (x , y) = (12, 36) corresponds to (a, b, c) = (3, 4, 5).





Rational Triangles Revisited

The elliptic curve E : y2 = x3 − 36 x has Mordell-Weil group

E (Q) = 〈P1, P2, P3〉 ' Z2 × Z2 × Z

as generated by the rational points

P1 = (0, 0), P2 = (6, 0), and P3 = (12, 36).

P3 is not a torsion element, so we find triangles for each [m]P3:

[1]P3 = (12, 36) =⇒ (a, b, c) = (3, 4, 5)

[−2]P3 =

„25

4,

35

8

«=⇒ (a, b, c) =

„49

70,

1200

70,

1201

70

«[−3]P3 =

„16428

529,

2065932

12167

«=⇒ (a, b, c) =

„7216803

1319901,

2896804

1319901,

7776485

1319901

«

There are infinitely many rational right triangles with area A = 6!





Can elliptic curves

help answer

similar questions?




HistoryExamples for n ≤ 5Relation with Elliptic Curves

History

Around 250 B.C., Diophantus of Alexandria posed the following:

“Find four numbers such that the product of any two of themincreased by unity is a perfect square.”

He presented the set

{1

16,

33

16,

17

4,

105

16

}as a solution, because

1

16· 33

16+ 1 =

„17

16

«2

1

16· 17

4+ 1 =

„9

8

«2

1

16· 105

16+ 1 =

„19

16

«2

33

16· 17

4+ 1 =

„25

8

«2

33

16· 105

16+ 1 =

„61

16

«2

17

4· 105

16+ 1 =

„43

8

«2

Around 1650, Pierre de Fermat found the solution {1, 3, 8, 120}.





History

Cover of the 1621 translation of Diophantus’ Arithmetica

http://en.wikipedia.org/wiki/Diophantus


http://en.wikipedia.org/wiki/Diophantus




Motivating Question

A set {m1, m2, . . . , mn} of n rational numbers is called a rationalDiophantine n-tuple if mi mj + 1 is the square of a rational number fori 6= j . More information can be found at the website

http://web.math.hr/~duje/dtuples.html

Remark: When the word “rational” is omitted, it is assumed that eachmi is an integer. We will focus on distinct, nonzero rational numbers.

Existence Problem: Given a positive integer n, does there exist arational Diophantine n-tuple?

Extension Problem: For which N can a given a rationalDiophantine n-tuple {m1, m2, . . . , mn} be extended to a rationalDiophantine N-tuple {m1, m2, . . . , mn, . . . , mN}?


http://web.math.hr/~duje/dtuples.html




Rational Diophantine 2-tuples

Theorem (Campbell-G, 2007)

There exist infinitely many rational Diophantine 2-tuples.

Proof: We show existence by classifying all 2-tuples {m1, m2}. Considerthe substitution

t1 =m1

1 +√

m1 m2 + 1

t2 =m2

1 +√

m2 m1 + 1

←→

m1 =

2 t11− t1 t2

m2 =2 t2

1− t1 t2

Hence each {m1, m2} corresponds to (t1, t2) ∈ A2(Q).






Theorem (Campbell-G, 2007)


Proof: We show existence by classifying all 3-tuples {m1, m2, m3}.Consider the substitution

t1 =m1

1 +√

m1 m2 + 1

t2 =m2

1 +√

m2 m3 + 1

t3 =m3

1 +√

m3 m1 + 1

←→

m1 =2 t1 (1 + t1 t2 + t1 t2

2 t3)

1− t21 t2

2 t23

m2 =2 t2 (1 + t2 t3 + t2 t2

3 t1)

1− t21 t2

2 t23

m3 =2 t3 (1 + t3 t1 + t3 t2

1 t2)

1− t21 t2

2 t23

Hence each {m1, m2, m3} corresponds to (t1, t2, t3) ∈ A3(Q).






Theorem (Leonhard Euler, 1783)


Proof: Extend {m1, m2} to {m1, m2, m3, m4} by choosing

m1 =2 t1

1− t1 t2

m2 =2 t2

1− t1 t2

m3 =2 (1 + t1) (1 + t2)

1− t1 t2

m4 =4 (1 + t1 t2) (1 + 2 t1 + t1 t2) (1 + 2 t2 + t1 t2)

(1− t1 t2)3

Remark: These 4-tuples correspond to (t1, t2) ∈ A2(Q). In particular,not every rational Diophantine 4-tuple is in this form!






Proof (cont’d): Finally extend to {m1, m2, m3, m4, m5} by choosing

m5 = 16 m12 m13 m23 m14 m24 m34

(m142 m24+m14 m24

2−m142 m34−4 m14 m24 m34−m24

2 m34+m14 m342+m24 m34

2)2

where we have set

m12 =p

m1 m2 + 1 =1 + t1 t2

1 − t1 t2

m13 =p

m1 m3 + 1 =1 + 2 t1 + t1 t2

1 − t1 t2

m23 =p

m2 m3 + 1 =1 + 2 t2 + t1 t2

1 − t1 t2

m14 =p

m1 m4 + 1 =1 + 4 t1 + 6 t1 t2 + 4 t21 t2 + t21 t22

(1 − t1 t2)2

m24 =p

m2 m4 + 1 =1 + 4 t2 + 6 t1 t2 + 4 t1 t22 + t21 t22

(1 − t1 t2)2

m34 =p

m3 m4 + 1 =3 + 4 (t1 + t2) + 10 t1 t2 + 4 (t1 + t2) t1 t2 + 3 t21 t22

(1 − t1 t2)2

One checks that this forms a rational Diophantine 5-tuple.





Example

Euler realized that if he chose t1 = 1/3 and t2 = 1 then he could expandFermat’s solution to Diophantus’ problem to five numbers:{

1, 3, 8, 120,777480

8288641

}.

Indeed, it is easy to verify that

1 · 3 + 1 = 22

1 · 8 + 1 = 32

3 · 8 + 1 = 52

1 · 120 + 1 = 112

3 · 120 + 1 = 192

8 · 120 + 1 = 312

1 · 777480

8288641+ 1 =

„3011

2879

«2

3 · 777480

8288641+ 1 =

„3259

2879

«2

8 · 777480

8288641+ 1 =

„3809

2879

«2

120 · 777480

8288641+ 1 =

„10079

2879

«2





Where does

Euler’s construction

come from?





Extending rational Diophantine 3-tuples to 5-tuples

Theorem (Andrej Dujella, 2001)

Let {m1, m2, m3} be a nontrivial rational Diophantine 3-tuple, and denote

E : y2 = (m1 x + 1) (m2 x + 1) (m3 x + 1)

Define mij =√

mi mj + 1 for the two rational points

Q1 = (0, 1),

Q2 =(

m12 m13+m12 m23+m13 m23+1m1 m2 m3

, (m12+m13) (m12+m23) (m13+m23)m1 m2 m3

).

{m1, m2, m3, m4} is a rational Diophantine 4-tuple if and only if m4

is the x-coordinate of Q1 ⊕ [2]P for some P ∈ E (Q).

Choose a point P ∈ E (Q). Let m4 and m5 be the x-coordinates ofQ1 ⊕ [2]P and (Q1 ⊕ [2]Q2)⊕ [2]P. Then {m1, m2, m3, m4, m5} is arational Diophantine 5-tuple.





Euler’s Construction Revisited

Start with a rational Diophantine 2-tuple {m1, m2}, and choose

m3 = m1 + m2 + 2 m12 in terms of m12 =√

m1 m2 + 1.

This defines a rational Diophantine 3-tuple {m1, m2, m3}.

Form the curve E : y2 = (m1 x + 1) (m2 x + 1) (m3 x + 1). Choose therational point P = [−2]Q1 so that the x-coordinate ofQ1 ⊕ [2]P = [−3] (0, 1) is

m4 = 4 m12 (m1 + m12) (m2 + m12).

One checks that Q1 ⊕ Q2 =(− 1

m3, 0)

with this choice of m3, so that

the x-coordinate of (Q1 ⊕ [2]Q2)⊕ [2]P = [−5] (0, 1) is

m5 = 16 m12 m13 m23 m14 m24 m34

(m142 m24+m14 m24

2−m142 m34−4 m14 m24 m34−m24

2 m34+m14 m342+m24 m34

2)2 .

These are Euler’s choices for the 5-tuple {m1, m2, m3, m4, m5}!MAA MathFest – David Blackwell Lecture Elliptic Curves




Elliptic Curves with Large Rank

For each nontrivial rational Diophantine 3-tuple {m1, m2, m3}, the ellipticcurve E : y2 = (m1 x + 1) (m2 x + 1) (m3 x + 1) has torsion subgroupcontaining Z2 × Z2. It appears that Q1 and Q2 are in general twoindependent points of infinite order.

This suggests that certain 3-tuples will yield curves with large rank:

Rat’l Diophantine 3-tuple Torsion Rank Largest Known{12702323 , 5916

2323 , 66459386132412535672267

}Z2 × Z2 9 15 (Elkies, 2009){

− 225525129 , 5129

22552 , − 5246319014458651

}Z2 × Z4 7 8 (Elkies, 2005){

3912396976 , 12947200

418209 , 424271104

}Z2 × Z6 4 6 (Elkies, 2006){

145408 , − 408

145 ,− 14543959160

}Z2 × Z8 3 3 (Campbell-G, 2003)





Questions?


why should i care about elliptic curves?egoins/notes/blackwell.pdfheron triangles elliptic curves...

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