1. Divisors Presented by J.liu

2. Outline Elliptic curves Definitions Isomorphism from Div 0 (E)/S→E(κ) Find a function f, such that div(f) = D

3. Elliptic curves Weierstrass normal forms Over on a field F with Char(F) = p F p, F p n or any subgroup of F p n.

4. Definitions Divisor D on E is a finite linear combinations of symbols (points on E)with integer coefficients. D = Σa i [P], with P  E(  ) Ex: D i = 2[P]+3[Q]+4[T]-9[∞] Div(E) = {D: D is a divisor on E} Div is a free abelian group generated by symbols on E.

5. Definitions Two mappings 1.Deg: Div(E)→Z with Deg(Di) = 2+3+4-9 = 0 2.Sum: Div(E)→E(κ) with Sum(D i ) =2P+3Q+4T-9∞ Div 0 = {D: D  Div(E) and Deg(D) = 0} Sum is a onto homomorphism form Div 0 (E) to E. The kernel of Sum is the set of principle divisors. (We need a bijective Homo.!!)

6. Definitions Principle divisor D  Div 0 :  f  div(f) = D  Sum(D) = ∞ Divisor of a function div(f): f is a rational function defined for at least one point in E(κ), f has zero or pole on points on E then div(f) = Σa i [P i ] a i is the order of f on P i Note that, f is a rational function mod E

7. Examples Ex1: E: y 2 =x 3 -x  f = x =y 2 /(x 2 -1)  f has a zero on (0,0)  ord (0,0) (f) = 2, therefore, div(f) = 2[(0,0)]+… Ex2: A line f through P 1.Not a tangent line of P then ord(f) = 1 2.Tangent line of P, and 3P≠∞ then ord(f) = 2 3.Tangent line of P, and 3P =∞ then ord(f) = 3 4.ord ∞ (x) = -2 5.ord ∞ (y) = -3 6.ord ∞ (x+y-2) = 0 ∵ ∞ = (0,1,0) → x+y-2z≠0

8. Div(f) f is a function on E that is not identically 0 1.f has only finitely zeros and poles 2.deg(div(f)) = 0 (div(f)  Div 0 ) 3.If f has no zeros or poles then f is a constant (div(f) = 0 identity of Div 0 ) Ex: Line f(x,y) = ax+by+c pass P, Q, R on E 1. b≠0 then div(f) = [P]+[Q]+[R]-3[∞] 2. b = 0 then div(f) = [P]+[-P]-2[∞]

9. Sum is a isomorphism from Div 0 (E)/S→E(κ) ∞ Div 0 (E) E(κ) P S T [P]+S = T That is, [P]+{div(f)} = T

10. D to f, f to D Let E: y 2 =x 3 +4x over F 11 Let D = [(0,0)]+[(2,4)]+[(4, 5)]+[(6, 3)]-4[∞] it’s easy to see Sum(D) = ∞ and deg(D) = 0 Find f such that div(f) = D 1. find a line through (0,0) and (2,4): y-2x=0 which is a tangent line of (2,4) then we have div(y-2x) = [(0,0)]+2[(2,4)]-3[∞] 2.The vertical line pass through (2,4): x-2 = 0 then we have div(x-2) = [(2,4)]+[(2,-4)]-2[∞]

11. 3.div((y-2x)/(x-2)) = [(0,0)]+[(2,4)]-[(2,-4)]-[∞] 4.[(0,0)]+[(2,4)] = [(2,-4)]+[∞]+div(g) 5. y+x+2=0 pass through (4,5) and (6,3), then div(y+x+2) = [(4,5)]+[(6,3)]+[(2,-4)]-3[∞] 6.x-2 = 0 pass through (2,-4) then we have div(x-2) = [(2,4)]+[(2,-4)]-2[∞] 7.div((y+x+2)/(x-2)) = [(4,5)]+[(6,3)]-[(2,4)]-[∞] 8.[(4,5)]+[(6,3)] = [(2,4)]+[∞]+div(h) 9.D = [(0,0)]+[(2,4)]+[(4, 5)]+[(6, 3)]-4[∞] = [(2,-4)]+[∞]+div(g)+[(2,4)]+[∞]+div(h)-4[∞] = div(gh)+div(x-2)= div((y+x+2)(y-2x)/(x-2))

12. 10. (y-2x)(y+x+2)/(x-2) 11.(y-2x)(y+x+2) = y 2 -xy-2x 2 +2y-4x ≡ x 3 -xy-2x 2 +2y mod (y 2 =x 3 +4x) = (x-2)(x 2 -y) 12. then D = div(x 2 -y) Let’s check div(x 2 -y) 1.We have simple zeros at: (0,0), (2,4), (4, 5), (6, 3) 2.We have ord ∞ (x 2 -y) = -4—[dominate at x 2 ] 3.That is, div(x 2 -y) = [(0,0)]+[(2,4)]+[(4,5)]+[(6, 3)]-4[∞] = D