 Solution

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 $d=(\det A{)}^{-1}\in {\mathbb{Q}}^{\times }$.

Solved by gpt-5.6-sol high.