By Connell I.

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version five Jun 2009

Additional resources for Elliptic curve handbook

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The transformation and its inverse are described affinely as follows: (µ1 − µ2 )3 x = φ(x, y), y = , µ2 (x − x0 ) − (y − y0 ) x y = x0 + (µ1 − µ2 )2 (x − 1)/y = y0 + (µ1 − µ2 )2 (µ2 x − µ1 )/y , and we have bijections P = (x, y) ←→ (x , y ) = (x , (x − 1)3 /x ) ←→ x = φ(P ). The line y = ax + b meets the curve in the three points whose x -coordinates are the three roots x1 , x2 , x3 of the cubic (x −1)3 −x (ax +b) = 0. The constant term shows that x1 x2 x3 = 1. Conversely if three points Pi = (xi , yi ) ∈ Ens (K) are such that x1 x2 x3 = 1 then these xi are the roots of the above cubic where we define a = x1 +x2 +x3 −3 and b = 3−(x1 x2 +x1 x3 +x2 x3 ), and consequently the points lie on the line y = ax + b.

1, we will explain later that starting with different rational points in Nagell’s algorithm yields isomorphic Weierstrass equations), Nagell’s algorithm yields — we omit the details — E : y 2 = x3 + 44x2 + 528x. The reader may also wish to verify that the points (0, 0) and (12, 120) on E correspond to the points (1, −1, 1, 1) and (131, −259, −59, 171) on the intersection. 5 Singular points. Consider a homogeneous polynomial F = F (X0 , . . , Xn ) ∈ K[X0 , . . , Xn ] of degree d. The Taylor expansion can be written as F (X0 + λ0 , .

B) We have the polynomial identity σ2 ψ22 − τ2 φ2 = ∆ where σ2 ψ22 τ2 φ2 = 12x3 − b2 x2 − 10b4 x + b2 b4 − 27b6 , = 4x3 + b2 x2 + 2b4 x + b6 , = 48x2 + 8b2 x + 32b4 − b22 , = x4 − b4 x2 − 2b6 x − b8 . Equivalently, since x([2](x, y)) = φ2 /ψ22 and ψ2 = κ, (σ2 − x([2](x, y))τ2 )κ2 = ∆. g. K = K). Suppose that char K is either 0 or a positive prime that does not divide m. Then E(K )[m] ≈ Cm ⊕ Cm . (%) The proof will be an elegant application of basic properties of isogenies; but let us consider how a computational proof might go at this stage.