Divisors

Presented by J.liu

Outline

• Elliptic curves

• Definitions

• Isomorphism from Div0(E)/S→E(κ)

• Find a function f, such that div(f) = D

Elliptic curves

• Weierstrass normal forms

642

23

312 axaxaxyaxyay

BAxxy 32

equations sWeierstras

• Over on a field F with Char(F) = p

• Fp, Fpn or any subgroup of Fpn.

Definitions

• Divisor D on E is a finite linear combinations of symbols (points on E)with integer coefficients. D = Σai[P], with PE()

• Ex: Di = 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.

Definitions

• Two mappings1. Deg: Div(E)→Z with Deg(Di) = 2+3+4-9 = 0

2. Sum: Div(E)→E(κ) with Sum(Di) =2P+3Q+4T-9∞

• Div0 = {D: DDiv(E) and Deg(D) = 0}• Sum is a onto homomorphism form Div0

(E) to E.• The kernel of Sum is the set of principle di

visors. (We need a bijective Homo.!!)

Definitions• Principle divisor DDiv0: 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) = Σai[P

i] ai is the order of f on Pi

• Note that, f is a rational function mod E

Examples • Ex1: E: y2=x3-x f = x =y2/(x2-1)f has a ze

ro 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) = 12. Tangent line of P, and 3P≠∞ then ord(f) = 23. 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∵

Div(f)

• f is a function on E that is not identically 01. f has only finitely zeros and poles

2. deg(div(f)) = 0 (div(f)Div0)

3. If f has no zeros or poles then f is a constant (div(f) = 0 identity of Div0)

• 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[∞]

Sum is a isomorphism from Div0(E)/S→E(κ)

∞

Div0(E) E(κ)

P

S

T

[P]+S = T

That is, [P]+{div(f)} = T

D to f, f to D

• Let E: y2=x3+4x over F11 • 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[∞]

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 di

v(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))

10. (y-2x)(y+x+2)/(x-2) 11.(y-2x)(y+x+2) = y2-xy-2x2+2y-4x

≡ x3-xy-2x2+2y mod (y2=x3+4x) = (x-2)(x2-y)

12. then D = div(x2-y)Let’s check div(x2-y)1. We have simple zeros at: (0,0), (2,4), (4, 5), (6, 3)

2. We have ord∞(x2-y) = -4—[dominate at x2]3. That is, div(x2-y) = [(0,0)]+[(2,4)]+[(4,5)]+

[(6, 3)]-4[∞] = D