= Solution
Because the <canonical height of an elliptic curve> is a quadratic form, polarization makes
$$
B(P,Q)=\widehat h(P+Q)-\widehat h(P)-\widehat h(Q)
$$
a symmetric bilinear form on the free part of the <Mordell-Weil group>. If $P'_i=\sum_jU_{ij}P_j$ is another integral basis, then $U\in\operatorname{GL}_r(\mathbb Z)$ and the Gram matrices satisfy
$$
M'=UMU^T.
$$
Since $\det U=\pm1$, their determinants agree. Thus the <regulator of an elliptic curve> is independent of the chosen basis.
Now let $Q_1,\ldots,Q_r$ be a basis for the free part of $E'(\mathbb Q)$. The images $\phi(P_i)$ span a finite-index sublattice, so modulo torsion
$$
\phi(P_i)=\sum_jA_{ij}Q_j
$$
for an integral matrix $A$ with nonzero determinant. The height identity gives
$$
B'(\phi P_i,\phi P_j)=\deg(\phi)B(P_i,P_j).
$$
Taking determinants in the two descriptions of this Gram matrix yields
$$
(\det A)^2\operatorname{Reg}(E'/\mathbb Q)
=\deg(\phi)^r\operatorname{Reg}(E/\mathbb Q).
$$
Therefore the required formula holds with $d=(\det A)^{-1}\in\mathbb Q^\times$.
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