Divisors
description
Transcript of Divisors
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