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Divisors

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

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Divisors

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  1. Divisors Presented by J.liu

  2. Outline • Elliptic curves • Definitions • Isomorphism from Div0(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 • Fp, Fpn or any subgroup of Fpn.

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

  5. Definitions • Two mappings • Deg: Div(E)→Z with Deg(Di) = 2+3+4-9 = 0 • Sum: Div(E)→E(κ) with Sum(Di) =2P+3Q+4T-9∞ • Div0 = {D: DDiv(E) and Deg(D) = 0} • Sum is a onto homomorphism form Div0(E) to E. • The kernel of Sum is the set of principle divisors. (We need a bijective Homo.!!)

  6. Definitions • Principle divisor DDiv0: 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[Pi] ai is the order of f on Pi • Note that, f is a rational function mod E

  7. Examples • Ex1: E: y2=x3-x f = x =y2/(x2-1)f has a zero on (0,0)  ord(0,0)(f) = 2, therefore, div(f) = 2[(0,0)]+… • Ex2: A line f through P • Not a tangent line of P then ord(f) = 1 • Tangent line of P, and 3P≠∞ then ord(f) = 2 • Tangent line of P, and 3P =∞ then ord(f) = 3 • ord∞(x) = -2 • ord∞(y) = -3 • 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 • f has only finitely zeros and poles • deg(div(f)) = 0 (div(f)Div0) • 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[∞]

  9. Sum is a isomorphism from Div0(E)/S→E(κ) [P]+S = T That is, [P]+{div(f)} = T ∞ S P T E(κ) Div0(E)

  10. 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 • 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[∞] • The vertical line pass through (2,4): x-2 = 0 then we have div(x-2) = [(2,4)]+[(2,-4)]-2[∞]

  11. div((y-2x)/(x-2)) = [(0,0)]+[(2,4)]-[(2,-4)]-[∞] • [(0,0)]+[(2,4)] = [(2,-4)]+[∞]+div(g) • y+x+2=0 pass through (4,5) and (6,3), then div(y+x+2) = [(4,5)]+[(6,3)]+[(2,-4)]-3[∞] • x-2 = 0 pass through (2,-4) then we have div(x-2) = [(2,4)]+[(2,-4)]-2[∞] • div((y+x+2)/(x-2)) = [(4,5)]+[(6,3)]-[(2,4)]-[∞] • [(4,5)]+[(6,3)] = [(2,4)]+[∞]+div(h) • 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. (y-2x)(y+x+2)/(x-2) • (y-2x)(y+x+2) = y2-xy-2x2+2y-4x ≡ x3-xy-2x2+2y mod (y2=x3+4x) = (x-2)(x2-y) • then D = div(x2-y) Let’s check div(x2-y) • We have simple zeros at: (0,0), (2,4), (4, 5), (6, 3) • We have ord∞(x2-y) = -4—[dominate at x2] • That is, div(x2-y) = [(0,0)]+[(2,4)]+[(4,5)]+[(6, 3)]-4[∞] = D

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