Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 125 / 3

For $P=[x:y]\in {\mathbb{P}}^1(\mathbb{Q})$ represented by coprime integers, its naive height on the projective line isWrite the degree-$d$ morphism as $\Phi =[F:G]$, where $F,G\in \mathbb{Z}[X,Y]$ are homogeneous of degree $d$ with no common projective zero. Bounding their coefficients gives

For the reverse inequality, the nonvanishing of the resultant of $F$ and $G$ gives homogeneous Bézout identities expressing fixed nonzero integer multiples of powers of $X$ and $Y$ as polynomial combinations of $F$ and $G$. Evaluating at $(x,y)$, removing the common divisor of $F(x,y)$ and $G(x,y)$, and taking the larger of $|x|,|y|$ givesThus $c_{1}H(P{)}^d\le H(\Phi (P))\le c_{2}H(P{)}^d$ for constants depending only on $\Phi$.

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Let $h=\log H$ and set $h_x(P)=h(x(P))$ for $P\ne O$, with $h_x(O)=0$. The given degree-four morphism and part (a) imply that a constant $C$ exists withfor every $P\in E(\mathbb{Q})$. DefineTo check the limit, put $a_r=4^{-r}{h}_x(2^{r}P)$. ThenThe geometric series converges, so $(a_r)$ is Cauchy and the limit exists. This is the canonical height of an elliptic curve; shifting the sequence by one index immediately gives $\hat{h}(2P)=4\hat{h}(P)$.

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Because the canonical height of an elliptic curve is a quadratic form, polarization makesa symmetric bilinear form on the free part of the Mordell-Weil group. If $P_i^{\prime }={\sum }_j{U}_{ij}{P}_j$ is another integral basis, then $U\in {\mathrm{GL}}_r(\mathbb{Z})$ and the Gram matrices satisfySince $\det U=\pm 1$, their determinants agree. Thus the regulator of an elliptic curve is independent of the chosen basis.

Now let $Q_1,\dots ,Q_r$ be a basis for the free part of $E^{\prime }(\mathbb{Q})$. The images $\varphi (P_i)$ span a finite-index sublattice, so modulo torsionfor an integral matrix $A$ with nonzero determinant. The height identity givesTaking determinants in the two descriptions of this Gram matrix yieldsTherefore the required formula holds with $d=(\det A{)}^{-1}\in {\mathbb{Q}}^{\times }$.

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