Lecture 2 Elliptic curves over finite fields The Group structure - PowerPoint PPT Presentation

Elliptic curves over F q F. Pappalardi Lecture 2 Elliptic curves over finite fields The Group structure Reminder from Monday the j -invariant Research School: Algebraic curves over finite fields Points of finite order CIMPA-ICTP-UNESCO-MESR-MINECO-PHILIPPINES Points of order 2 Points of order 3 University of the Phillipines Diliman, July 24, 2013 Points of finite order The group structure sketch of proof Important Results Hasse’s Theorem Waterhouse’s Theorem Rück’s Theorem Further reading Francesco Pappalardi Dipartimento di Matematica e Fisica Università Roma Tre 2.1

Elliptic curves over F q Elliptic curves over F q F. Pappalardi Definition (Elliptic curve) An elliptic curve over a field K is the data of a non singular Reminder from Weierstraß equation Monday E : y 2 + a 1 xy + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6 , a i ∈ K the j -invariant Points of finite order Points of order 2 If p = char K > 3, Points of order 3 Points of finite order The group structure sketch of proof ∆ E := 1 − a 5 1 a 3 a 4 − 8 a 3 1 a 2 a 3 a 4 − 16 a 1 a 2 2 a 3 a 4 + 36 a 2 1 a 2 � 3 a 4 Important Results 2 4 Hasse’s Theorem − a 4 1 a 2 4 − 8 a 2 1 a 2 a 2 4 − 16 a 2 2 a 2 4 + 96 a 1 a 3 a 2 4 + 64 a 3 Waterhouse’s Theorem 4 + Rück’s Theorem Further reading a 6 1 a 6 + 12 a 4 1 a 2 a 6 + 48 a 2 1 a 2 2 a 6 + 64 a 3 2 a 6 − 36 a 3 1 a 3 a 6 − 144 a 1 a 2 a 3 a 6 − 72 a 2 1 a 4 a 6 − 288 a 2 a 4 a 6 + 432 a 2 � 6 2.2

Elliptic curves over F q Elliptic curves over K F. Pappalardi After applying a suitable affine transformation we can always assume that E / K has a Weierstraß equation of the following form Example (Classification ( p = char K )) Reminder from Monday E p ∆ E the j -invariant Points of finite order y 2 = x 3 + Ax + B 4 A 3 + 27 B 2 ≥ 5 Points of order 2 Points of order 3 Points of finite order y 2 + xy = x 3 + a 2 x 2 + a 6 a 2 The group structure 2 6 sketch of proof y 2 + a 3 y = x 3 + a 4 x + a 6 Important Results a 4 2 3 Hasse’s Theorem Waterhouse’s Theorem 4 A 3 C − A 2 B 2 − 18 ABC Rück’s Theorem y 2 = x 3 + Ax 2 + Bx + C 3 Further reading + 4 B 3 + 27 C 2 Let E / F q elliptic curve, ∞ := [ 0 , 1 , 0 ] . Set q : y 2 = x 3 + Ax + B } ∪ {∞} E ( F q ) = { ( x , y ) ∈ F 2 2.3

� if P � = Q line through P and Q Elliptic curves over F q If P , Q ∈ E ( F q ) , r P , Q : F. Pappalardi tangent line to E at P if P = Q , r P , ∞ : vertical line through P � x y � y 2 � y � x 3 � 3 x 2 � x � 1 � x y � y 2 � y � x 3 � 3 x 2 � x � 1 3 3 � R Reminder from Monday 2 2 the j -invariant Points of finite order 1 1 Q Points of order 2 P Points of order 3 Points of finite order 0 � 0 The group structure P+ Q sketch of proof P -P � 1 � 1 Important Results Hasse’s Theorem Waterhouse’s Theorem � 2 � 2 Rück’s Theorem Further reading � 3 � 3 � 2 � 1 0 1 2 3 4 � 2 � 1 0 1 2 3 4 − P := P ′ r P , ∞ ∩ E ( F q ) = { P , ∞ , P ′ } � P + E Q := − R r P , Q ∩ E ( F q ) = { P , Q , R } � 2.4

Elliptic curves over F q F. Pappalardi Theorem The addition law on E / K (K field) has the following properties: (a) P + E Q ∈ E ∀ P , Q ∈ E Reminder from (b) P + E ∞ = ∞ + E P = P ∀ P ∈ E Monday the j -invariant (c) P + E ( − P ) = ∞ ∀ P ∈ E Points of finite order (d) P + E ( Q + E R ) = ( P + E Q ) + E R ∀ P , Q , R ∈ E Points of order 2 Points of order 3 Points of finite order (e) P + E Q = Q + E P ∀ P , Q ∈ E The group structure So ( E ( ¯ K ) , + E ) is an abelian group. sketch of proof Important Results Hasse’s Theorem Remark: Waterhouse’s Theorem Rück’s Theorem If E / K ⇒ ∀ L , K ⊆ L ⊆ ¯ K , E ( L ) is an abelian group. Further reading − P = − ( x 1 , y 1 ) = ( x 1 , − a 1 x 1 − a 3 − y 1 ) 2.5

Elliptic curves over F q Formulas for Addition on E (Summary) F. Pappalardi E : y 2 + a 1 xy + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6 P 1 = ( x 1 , y 1 ) , P 2 = ( x 2 , y 2 ) ∈ E ( K ) \ {∞} , Addition Laws for the sum of affine points Reminder from Monday • If P 1 � = P 2 the j -invariant Points of finite order ⇒ P 1 + E P 2 = ∞ Points of order 2 • x 1 = x 2 Points of order 3 • x 1 � = x 2 Points of finite order λ = y 2 − y 1 ν = y 1 x 2 − y 2 x 1 The group structure x 2 − x 1 x 2 − x 1 sketch of proof • If P 1 = P 2 Important Results Hasse’s Theorem ⇒ P 1 + E P 2 = 2 P 1 = ∞ Waterhouse’s Theorem • 2 y 1 + a 1 x + a 3 = 0 Rück’s Theorem • 2 y 1 + a 1 x + a 3 � = 0 Further reading 3 x 2 a 3 y 1 + x 3 1 + 2 a 2 x 1 + a 4 − a 1 y 1 1 − a 4 x 1 − 2 a 6 λ = , ν = − 2 y 1 + a 1 x + a 3 2 y 1 + a 1 x 1 + a 3 Then P 1 + E P 2 = ( λ 2 − a 1 λ − a 2 − x 1 − x 2 , − λ 3 − a 2 1 λ + ( λ + a 1 )( a 2 + x 1 + x 2 ) − a 3 − ν ) 2.6

Elliptic curves over F q Formulas for Addition on E (Summary for special equation) F. Pappalardi E : y 2 = x 3 + Ax + B P 1 = ( x 1 , y 1 ) , P 2 = ( x 2 , y 2 ) ∈ E ( K ) \ {∞} , Addition Laws for the sum of affine points Reminder from Monday • If P 1 � = P 2 the j -invariant Points of finite order P 1 + E P 2 = ∞ • x 1 = x 2 ⇒ Points of order 2 Points of order 3 • x 1 � = x 2 Points of finite order λ = y 2 − y 1 ν = y 1 x 2 − y 2 x 1 The group structure x 2 − x 1 x 2 − x 1 sketch of proof • If P 1 = P 2 Important Results Hasse’s Theorem P 1 + E P 2 = 2 P 1 = ∞ Waterhouse’s Theorem • y 1 = 0 ⇒ Rück’s Theorem • y 1 � = 0 Further reading 3 x 2 x 3 1 + A 1 − Ax 1 − 2 B λ = 2 y 1 , ν = − 2 y 1 Then P 1 + E P 2 = ( λ 2 − x 1 − x 2 , − λ 3 + λ ( x 1 + x 2 ) − ν ) 2.7

Elliptic curves over F q Notations F. Pappalardi Finite fields 1 F p = { 0 , 1 , . . . , p − 1 } is the prime field; 2 F q is a finite field with q = p n elements; Reminder from 3 F q = F p [ ξ ] , f ( ξ ) = 0, f ∈ F p [ X ] irreducible, ∂ f = n ; Monday 4 F 4 = F 2 [ ξ ] , ξ 2 = 1 + ξ ; the j -invariant Points of finite order 5 F 8 = F 2 [ α ] , α 3 = α + 1 but also F 8 = F 2 [ β ] , β 3 = β 2 + 1, Points of order 2 ( β = α 2 + 1); Points of order 3 Points of finite order 6 F 101 101 = F 101 [ ω ] , ω 101 = ω + 1 The group structure sketch of proof Important Results Hasse’s Theorem Algebraic Closure of F q Waterhouse’s Theorem Rück’s Theorem Further reading 1 ∀ n ∈ N , we fix an F q n 2 We also require that F q n ⊆ F q m if n | m 3 We let F q = � n ∈ N F q n 4 F q is algebraically closed 2 If F ( x , y ) ∈ F q [ x , y ] a point of the curve F = 0, means ( x 0 , y 0 ) ∈ F q s.t. F ( x 0 , y 0 ) = 0. 2.8

Elliptic curves over F q The j -invariant F. Pappalardi Let E / K : y 2 = x 3 + Ax + B , p ≥ 5 and ∆ E := 4 A 3 + 27 B 2 . � − u − 2 x x ← u ∈ K ∗ ⇒ E − → E u : y 2 = x 3 + u 4 Ax + u 6 B − u − 3 y y ← Reminder from Monday Definition the j -invariant 4 A 3 Points of finite order The j –invariant of E is j = j ( E ) = 1728 Points of order 2 4 A 3 + 27 B 2 Points of order 3 Points of finite order Properties of j –invariants The group structure sketch of proof 1 j ( E ) = j ( E u ) , ∀ u ∈ K ∗ Important Results 2 j ( E ′ / K ) = j ( E ′′ / K ) ⇒ ∃ u ∈ ¯ K ∗ s.t. E ′′ = E ′ Hasse’s Theorem u Waterhouse’s Theorem Rück’s Theorem if K = F q can take u ∈ F q 12 Further reading 3 j � = 0 , 1728 ⇒ E : y 2 = x 3 + 3 j 2 j 1728 − j x + 1728 − j , j ( E ) = j 4 j = 0 ⇒ E : y 2 = x 3 + B , j = 1728 ⇒ E : y 2 = x 3 + Ax → { ¯ 5 j : K ← K –affinely equivalent classes of E / K } . 6 p = 2 , 3 different definition 2.9

Elliptic curves over F q Examples of j invariants F. Pappalardi From monday E 1 : y 2 = x 3 + 1 and E 2 : y 2 = x 3 + 2 # E 1 ( F 5 ) = # E 2 ( F 5 ) = 6 and j ( E 1 ) = j ( E 2 ) = 0 � x ← − 2 x E 1 and E 2 affinely equivalent √ √ Reminder from over F 5 [ 3 ] = F 25 ( twists ) y ← − 3 y Monday the j -invariant Points of finite order Definition (twisted curve) Points of order 2 Points of order 3 Let E / F q : y 2 = x 3 + Ax + B , µ ∈ F ∗ q \ ( F ∗ q ) 2 . Points of finite order The group structure sketch of proof E µ : y 2 = x 3 + µ 2 Ax + µ 3 B Important Results Hasse’s Theorem Waterhouse’s Theorem is called twisted curve. Rück’s Theorem Further reading Exercise: prove that • j ( E ) = j ( E µ ) • E and E µ are F q [ √ µ ] –affinely equivalent • # E ( F q 2 ) = # E µ ( F q 2 ) • usually # E ( F q ) � = # E µ ( F q ) 2.10

Elliptic curves over F q Determining points of order 2 F. Pappalardi Let P = ( x 1 , y 1 ) ∈ E ( F q ) \ {∞} , P has order 2 ⇐ ⇒ 2 P = ∞ ⇐ ⇒ P = − P So Reminder from − P = ( x 1 , − a 1 x 1 − a 3 − y 1 ) = ( x 1 , y 1 ) = P = ⇒ 2 y 1 = − a 1 x 1 − a 3 Monday the j -invariant If p � = 2, can assume E : y 2 = x 3 + Ax 2 + Bx + C Points of finite order Points of order 2 Points of order 3 ⇒ y 1 = 0 , x 3 1 + Ax 2 − P = ( x 1 , − y 1 ) = ( x 1 , y 1 ) = P = 1 + Bx 1 + C = 0 Points of finite order The group structure sketch of proof Important Results Note Hasse’s Theorem Waterhouse’s Theorem • the number of points of order 2 in E ( F q ) equals the Rück’s Theorem number of roots of X 3 + Ax 2 + Bx + C in F q Further reading • roots are distinct since discriminant ∆ E � = 0 • E ( F q 6 ) has always 3 points of order 2 if E / F q • E [ 2 ] := { P ∈ E (¯ F q ) : 2 P = ∞} ∼ = C 2 ⊕ C 2 2.11