Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 125 / 3 / a

Solution

For

E : y^{2} = x^{3} + a x + b,

the chord-and-tangent group law gives, for distinct nonopposite points

P = (x_{1}, y_{1})

and

Q = (x_{2}, y_{2})

λ = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}, x (P + Q) = λ^{2} - x_{1} - x_{2}, y (P + Q) = - y_{1} + λ (x_{1} - x (P + Q)) .

For doubling, replace the slope by

λ = \frac{3 x _{1}^{2} + a}{2 y _{1}} .

The point at infinity is the identity and

- (x, y) = (x, - y)

Let the

x

-coordinates of

P, Q, P + Q, P - Q

x_{1}, x_{2}, x_{3}, x_{4}

. Applying the addition formula with

Q

and

- Q

gives

x_{3} + x_{4} = \frac{( y _{2} - y _{1} ) ^{2} + ( y _{2} + y _{1} ) ^{2}}{( x _{2} - x _{1} ) ^{2}} - 2 (x_{1} + x_{2}) .

Substituting

y_{i}^{2} = x_{i}^{3} + a x_{i} + b

and simplifying yields

x_{3} + x_{4} = \frac{2 ( x _{1} x _{2} + a ) ( x _{1} + x _{2} ) + 4 b}{( x _{1} - x _{2} ) ^{2}} .

Multiplying the two addition formulas and eliminating

y_{1} y_{2}

in the same way gives

x_{3} x_{4} = \frac{( x _{1} x _{2} - a ) ^{2} - 4 b ( x _{1} + x _{2} )}{( x _{1} - x _{2} ) ^{2}} .

These identities are the algebraic source of two parallelogram laws. Applied to pullbacks of the pole divisor of

x

, they imply

de g (φ + ψ) + de g (φ - ψ) = 2 de g φ + 2 de g ψ

de g (n φ) = n^{2} de g φ

, this makes the degree a quadratic form. Applied to the Absolute logarithmic Weil height of the four

x

-coordinates, with bounded terms removed by passage to the limit defining the canonical height of an elliptic curve, they similarly give

h (P + Q) + h (P - Q) = 2 h (P) + 2 h (Q), h (n P) = n^{2} h (P) .

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