Why Should I Care About Elliptic Curves?

Transcription

1 Why Should I Care About? Edray Herber Goins Department of Mathematics Purdue University August 7, 2009

2 Abstract An elliptic curve E possessing a rational point is an arithmetic-algebraic object: It is simultaneously a nonsingular projective curve with an affine equation y 2 = x 3 + A x + B, which allows one to perform arithmetic on its points; and a finitely generated abelian group E(Q) E(Q) tors Z r, which allows one to apply results from abstract algebra. In this talk, we discuss some basic properties of elliptic curves, and give applications along the way.

3 David Harold Blackwell

4 Why should I care about David Blackwell?

5 Institute for Advanced Study, April April01/small_dscf2855.jpg

6 From the The MacTutor History of Mathematics archive: At this point [after graduation in 1941] Blackwell received a prestigious one year appointment as Rosenwald Postdoctoral Fellow at the Institute for Advanced Study in Princeton. This, however, led to problems over the fact that Blackwell was a black African American. The standard practice was to have Fellows of the Institute become honorary Faculty members of Princeton University but, in Blackwell s case, this caused a problem. At that time Princeton University had never had a black undergraduate student much less a black Faculty member and this produced opposition within the University. The President of the University wrote to the Director of the Institute for Advanced Study saying that the Institute were abusing the hospitality of the University with such an appointment. Fortunately at the time Blackwell was not fully aware of the problems his appointment was causing.

7 Outline of Talk Heron Triangles 1 Heron Triangles 2 3 History Examples for n 5 Relation with

8 Can you find a right triangle with rational sides having area A = 6?

9 Motivating Question Consider positive rational numbers a, b, and c satisfying a 2 + b 2 = c 2 and 1 2 a b = 6. b c a Recall the (a, b, c) = (3, 4, 5) triangle!

10 Cubic Equations Heron Triangles Are there more rational solutions (a, b, c) to Proposition a 2 + b 2 = c 2 and 1 2 a b = 6? Let x and y be rational numbers, and denote the rational numbers Then a = x 2 36, b = 12 x y y, and c = x y a 2 + b 2 = c a b = 6 if and only if { y 2 = x 3 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)?

11 What types of properties does this cubic equation have?

12 What is an Elliptic Curve? Definition Let A and B be rational numbers such that 4 A B 2 0. An elliptic curve E is the set of all (x, y) satisfying the equation y 2 = x 3 + A x + B. We will also include the point at infinity O. Example: y 2 = x 3 36 x is an elliptic curve. Non-Example: y 2 = x 3 3 x + 2 is not an elliptic curve

13 What is an Elliptic Curve? Formally, an elliptic curve E over Q is a nonsingular projective curve of genus 1 possessing a Q-rational point O. Such a curve is birationally equivalent over Q to a cubic equation in Weierstrass form: E : y 2 = x 3 + A x + B; with rational coefficients A and B, and nonzero discriminant (E) = 16 ( 4 A B 2). For any field K, define E(K) = { (x 1 : x 2 : x 0 ) P 2 (K) x 2 2 x 0 = x1 3 + A x 1 x0 2 + B x0 3 where O = (0 : 1 : 0) is on the projective line at infinity x 0 = 0. Remark: In practice we choose either K = Q or F p. } ;

14 Given two rational points on an elliptic curve E, we explain how to construct 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.

15 Example: y 2 = x 3 36 x Consider the two rational points P = (6, 0) and Q = (12, 36). P*Q 50 Q P P Q = (18, 72)

16 Example: y 2 = x 3 36 x Consider the two rational points P = (6, 0) and Q = (12, 36). 50 P P P = O

17 Example: y 2 = x 3 36 x Consider the two rational points P = (6, 0) and Q = (12, 36). 50 Q Q*Q Q Q = (25/4, 35/8)

18 Group Law Heron Triangles Definition Let E be an elliptic curve defined over a field K, and denote E(K) as the set of K-rational points on E. Define the operation as P Q = (P Q) O P Q 1.6 P*Q P+Q

19 Example: y 2 = x 3 36 x Consider the two rational points P = (6, 0) and Q = (12, 36). P*Q 50 Q P P+Q P Q = (18, 72)

20 Example: y 2 = x 3 36 x Consider the two rational points P = (6, 0) and Q = (12, 36). 50 P P P = O

21 Example: y 2 = x 3 36 x Consider the two rational points P = (6, 0) and Q = (12, 36). 50 Q Q*Q Q+Q -50 Q Q = (25/4, 35/8)

22 Poincaré s Theorem Heron Triangles Theorem (Henri Poincaré, 1901) Let E be an elliptic curve defined over a field K. Then E(K) is an abelian group under. Recall that to be an abelian group, the following five axioms must be satisfied: 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.

23 What types of properties does this abelian group have?

24 Poincaré s Conjecture Heron Triangles Conjecture (Henri Poincaré, 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 there exists a finite generating set {a 1, a 2,..., a n } such that, for each given g G, there are integers m 1, m 2,..., m n such that g = [m 1 ]a 1 [m 2 ]a 2 [m n ]a n. Example: G = Z is a finitely generated abelian group because all integers are generated by a 1 = 1.

25 Mordell s Theorem Heron Triangles 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 r such that E(Q) E(Q) tors Z r. 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. It contains all of the points of finite order, i.e., those P E(Q) such that [m]p = O for some positive integer m. The nonnegative integer r is called the Mordell-Weil rank of E.

26 Example: y 2 = x 3 36 x Consider the three rational points P 1 = (0, 0), P 2 = (6, 0), and P 3 = (12, 36). [2]P 1 = [2]P 2 = O, i.e., both P 1 and P 2 have order 2. They are torsion. 50 (-6,0) (6,0) (0,0) E(Q) tors = P 1, P 2 Z 2 Z 2

27 Example: y 2 = x 3 36 x Consider the three rational points P 1 = (0, 0), P 2 = (6, 0), and P 3 = (12, 36). [2]P 3 = (25/4, 35/8) and [3]P 3 = (16428/529, /12167) P [2]P [3]P E(Q) = P 1, P 2, P 3 Z 2 Z 2 Z

28 Classification of Torsion Subgroups Theorem (Barry Mazur, 1977) Let E is an elliptic curve, then { Z n where 1 n 10 or n = 12; E(Q) tors Z 2 Z 2m where 1 m 4. Remark: Z n denotes the cyclic group of order n. Example: The elliptic curve y 2 = x 3 36 x has torsion subgroup E(Q) tors Z 2 Z 2 generated by P 1 = (0, 0) and P 2 = (6, 0). Mordell s Theorem states that E(Q) E(Q) tors Z r. What can we say about the Mordell-Weil rank r?

29 Records for Prescribed Torsion and Rank E(Q)tors Highest Rank r Found By year Discovered Trivial 28 Elkies 2006 Z 2 19 Elkies 2009 Z 3 13 Eroshkin 2007, 2008, 2009 Z 4 12 Elkies 2006 Z 5 6 Dujella, Lecacheux 2001 Z 6 8 Z 7 5 Eroshkin Dujella, Eroshkin Elkies, Dujella Dujella, Kulesz Elkies Z 8 6 Elkies 2006 Z 9 4 Fisher 2009 Z 10 4 Dujella 2005, 2008 Elkies 2006 Z 12 4 Fisher 2008 Z 2 Z 2 15 Elkies 2009 Z 2 Z 4 8 Elkies Eroshkin Dujella, Eroshkin Z 2 Z 6 6 Elkies 2006 Z 2 Z 8 3 Connell Dujella Campbell, Goins Rathbun Flores, Jones, Rollick, Weigandt Fisher , 2001, 2006, ,

30 How does this help answer the motivating questions?

31 Rational Triangles Revisited Can we find infinitely many right triangles (a, b, c) having rational sides and area A = 6? Proposition Let x and y be rational numbers, and denote the rational numbers Then a = x 2 36, b = 12 x y y, and c = x y a 2 + b 2 = c a b = 6 if and only if { y 2 = x 3 36 x. Example: (x, y) = (12, 36) corresponds to (a, b, c) = (3, 4, 5).

32 Rational Triangles Revisited The elliptic curve E : y 2 = x 3 36 x has Mordell-Weil group E(Q) = P 1, P 2, P 3 Z 2 Z 2 Z as generated by the rational points P 1 = (0, 0), P 2 = (6, 0), and P 3 = (12, 36). P 3 is not a torsion element, so we find triangles for each [m]p 3 : [1]P 3 = (12, 36) = (a, b, c) = (3, 4, 5) «25 [ 2]P 3 = 4, 35 8 «16428 [ 3]P 3 = 529, «49 = (a, b, c) = 70, , « = (a, b, c) = , , There are infinitely many rational right triangles with area A = 6!

33 Can elliptic curves help answer similar questions?

34 History Heron Triangles History Examples for n 5 Relation with Around 250 B.C., Diophantus of Alexandria posed the following: Find four numbers such that the product of any two of them increased by unity is a perfect square. He presented the set { 1 16, 33 16, 17 4, 105 } as a solution, because = = = «2 « « = = = «2 «2 «2 Around 1650, Pierre de Fermat found the solution {1, 3, 8, 120}.

35 History Heron Triangles History Examples for n 5 Relation with Cover of the 1621 translation of Diophantus Arithmetica

36 Motivating Question Heron Triangles History Examples for n 5 Relation with A set {m 1, m 2,..., m n } of n rational numbers is called a rational Diophantine n-tuple if m i m j + 1 is the square of a rational number for i j. More information can be found at the website Remark: When the word rational is omitted, it is assumed that each m i is an integer. We will focus on distinct, nonzero rational numbers. Existence Problem: Given a positive integer n, does there exist a rational Diophantine n-tuple? Extension Problem: For which N can a given a rational Diophantine n-tuple {m 1, m 2,..., m n } be extended to a rational Diophantine N-tuple {m 1, m 2,..., m n,..., m N }?

37 Rational Diophantine 2-tuples History Examples for n 5 Relation with Theorem (Campbell-G, 2007) There exist infinitely many rational Diophantine 2-tuples. Proof: We show existence by classifying all 2-tuples {m 1, m 2 }. Consider the substitution m 1 t 1 = 1 + m 1 m m 1 = 2 t 1 1 t m 1 t 2 2 t 2 = 1 + m m 2 m = 2 t 2 1 t 1 t 2 Hence each {m 1, m 2 } corresponds to (t 1, t 2 ) A 2 (Q).

38 Rational Diophantine 3-tuples History Examples for n 5 Relation with Theorem (Campbell-G, 2007) There exist infinitely many rational Diophantine 3-tuples. Proof: We show existence by classifying all 3-tuples {m 1, m 2, m 3 }. Consider the substitution m 1 t 1 = 1 + m 1 = 2 t 1 (1 + t 1 t 2 + t 1 t2 2 t 3) m 1 m t1 2 m 2 t2 2 t2 3 t 2 = 1 + m m 2 m = 2 t 2 (1 + t 2 t 3 + t 2 t3 2 t 1) 1 t1 2 m t2 2 t2 3 3 t 3 = 1 + m 3 m m 3 = 2 t 3 (1 + t 3 t 1 + t 3 t1 2 t 2) 1 t1 2 t2 2 t2 3 Hence each {m 1, m 2, m 3 } corresponds to (t 1, t 2, t 3 ) A 3 (Q).

39 Rational Diophantine 5-tuples History Examples for n 5 Relation with Theorem (Leonhard Euler, 1783) There exist infinitely many rational Diophantine 5-tuples. Proof: Extend {m 1, m 2 } to {m 1, m 2, m 3, m 4 } by choosing m 1 = 2 t 1 1 t 1 t 2 m 2 = 2 t 2 1 t 1 t 2 m 3 = 2 (1 + t 1) (1 + t 2 ) 1 t 1 t 2 m 4 = 4 (1 + t 1 t 2 ) (1 + 2 t 1 + t 1 t 2 ) (1 + 2 t 2 + t 1 t 2 ) (1 t 1 t 2 ) 3 Remark: These 4-tuples correspond to (t 1, t 2 ) A 2 (Q). In particular, not every rational Diophantine 4-tuple is in this form!

40 Rational Diophantine 5-tuples History Examples for n 5 Relation with Proof (cont d): Finally extend to {m 1, m 2, m 3, m 4, m 5 } by choosing m 5 = 16 m 12 m 13 m 23 m 14 m 24 m 34 (m 142 m 24+m 14 m 242 m 142 m 34 4 m 14 m 24 m 34 m 242 m 34+m 14 m 342 +m 24 m 342 ) 2 where we have set m 12 = p m 1 m = 1 + t 1 t 2 1 t 1 t 2 m 13 = p m 1 m = t 1 + t 1 t 2 1 t 1 t 2 m 23 = p m 2 m = t 2 + t 1 t 2 1 t 1 t 2 m 14 = p m 1 m = t t 1 t t2 1 t 2 + t2 1 t2 2 (1 t 1 t 2 ) 2 m 24 = p m 2 m = t t 1 t t 1 t2 2 + t2 1 t2 2 (1 t 1 t 2 ) 2 m 34 = p m 3 m = (t 1 + t 2 ) + 10 t 1 t (t 1 + t 2 ) t 1 t t2 1 t2 2 (1 t 1 t 2 ) 2 One checks that this forms a rational Diophantine 5-tuple.

41 Example Heron Triangles History Examples for n 5 Relation with Euler realized that if he chose t 1 = 1/3 and t 2 = 1 then he could expand Fermat s solution to Diophantus problem to five numbers: { } , 3, 8, 120, Indeed, it is easy to verify that = = = = = = = = = = «2 «2 «2 «2

42 History Examples for n 5 Relation with Where does Euler s construction come from?

43 History Examples for n 5 Relation with Extending rational Diophantine 3-tuples to 5-tuples Theorem (Andrej Dujella, 2001) Let {m 1, m 2, m 3 } be a nontrivial rational Diophantine 3-tuple, and denote E : y 2 = (m 1 x + 1) (m 2 x + 1) (m 3 x + 1) Define m ij = m i m j + 1 for the two rational points Q 1 = (0, 1), ( m Q 2 = 12 m 13+m 12 m 23+m 13 m 23+1 m 1 m 2 m 3, ) (m 12+m 13) (m 12+m 23) (m 13+m 23) m 1 m 2 m 3. {m 1, m 2, m 3, m 4 } is a rational Diophantine 4-tuple if and only if m 4 is the x-coordinate of Q 1 [2]P for some P E(Q). Choose a point P E(Q). Let m 4 and m 5 be the x-coordinates of Q 1 [2]P and (Q 1 [2]Q 2 ) [2]P. Then {m 1, m 2, m 3, m 4, m 5 } is a rational Diophantine 5-tuple.

44 Euler s Construction Revisited History Examples for n 5 Relation with Start with a rational Diophantine 2-tuple {m 1, m 2 }, and choose m 3 = m 1 + m m 12 in terms of m 12 = m 1 m This defines a rational Diophantine 3-tuple {m 1, m 2, m 3 }. Form the curve E : y 2 = (m 1 x + 1) (m 2 x + 1) (m 3 x + 1). Choose the rational point P = [ 2]Q 1 so that the x-coordinate of Q 1 [2]P = [ 3] (0, 1) is m 4 = 4 m 12 (m 1 + m 12 ) (m 2 + m 12 ). ( ) One checks that Q 1 Q 2 = 1 m 3, 0 with this choice of m 3, so that the x-coordinate of (Q 1 [2]Q 2 ) [2]P = [ 5] (0, 1) is m 5 = 16 m 12 m 13 m 23 m 14 m 24 m 34 (m 142 m 24+m 14 m 242 m 142 m 34 4 m 14 m 24 m 34 m 242 m 34+m 14 m m 24 m 34 2 ) 2. These are Euler s choices for the 5-tuple {m 1, m 2, m 3, m 4, m 5 }!

45 with Large Rank History Examples for n 5 Relation with For each nontrivial rational Diophantine 3-tuple {m 1, m 2, m 3 }, the elliptic curve E : y 2 = (m 1 x + 1) (m 2 x + 1) (m 3 x + 1) has torsion subgroup containing Z 2 Z 2. It appears that Q 1 and Q 2 are in general two independent 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 { , , } Z 2 Z (Elkies, 2009) { , , } Z 2 Z (Elkies, 2005) { , , } Z 2 Z (Elkies, 2006) { , , } Z 2 Z (Campbell-G, 2003)

46 History Examples for n 5 Relation with Questions?