Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 125 / 4

For $x=u/v\in \mathbb{Q}$ in lowest terms, the naive height on the projective line isHomogenize the coprime numerator and denominator of $\xi$ to degree $d$. The triangle inequality gives the upper estimate $H(\xi (x))\le c_{2}H(x{)}^d$. Since the two homogenized forms have no common projective zero, their resultant is nonzero, and the Bézout identities for the resultant express fixed multiples of $u^{2d-1}$ and $v^{2d-1}$ as combinations of their values with coefficients of degree $d-1$. After cancellation this gives $H(x{)}^d\le CH(\xi (x))$, which is the lower estimate with $c_1=C^{-1}$.

Now write $x=u/v$ in lowest terms and putSince $\gcd (N,v)=1$, the equation $y^2=N/v^3$ givesClearly $M\le \gamma H(x{)}^3$ for $\gamma =1+|a|+|b|$. Homogenizing the supplied polynomial identity givesIts coefficient sum is at most ${\gamma }^2$, so $|u{|}^5\le {\gamma }^{2}H(x{)}^{2}M$. The same lower bound is immediate from $|v{|}^3\le M$ when $|v|=H(x)$. Thus

Set $h(O_E)=0$ andfor $P\ne O_E$. The canonical height of an elliptic curve isThe duplication formula is a rational function of degree four in $x$, so the rational-map height estimate in part (a) givesTherefore successive terms of $4^{-r}h([2^r]P)$ differ by at most $C_E{4}^{-r-1}$, and the limit is well defined.

Shifting the limit immediately gives $\hat{h}([2]P)=4\hat{h}(P)$. The addition formula likewise givesApply this to $[2^r]P,[2^r]Q$, divide by $2\cdot 4^r$, and pass to the limit to obtain the exact parallelogram lawTaking $Q=P$ starts an induction on $|n|$ that yieldsfor every integer $n$.

An admissible change of Weierstrass equation of an elliptic curve replaces the $x$-coordinate by a degree-one rational function, so the two logarithmic heights differ by $O(1)$. Dividing this bounded difference by $4^r$ in the defining limit shows that the canonical height is unchanged. It is intrinsic to the elliptic curve and the chosen point.

Replacing $[2^r]$ and $4^r$ by $[3^r]$ and $9^r$ gives the same canonical height. Indeed $h([3]P)=9h(P)+O_E(1)$, and the same telescoping argument constructs a quadratic height differing from the original by a bounded function. A bounded quadratic function is zero, so the two limits agree.

Part (a) givesThus replacing the $x$-coordinate height literally by the $y$-coordinate height multiplies the resulting canonical height by $3/2$. Multiplying the new naive height by $2/3$ restores the original normalization.