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Let $f: E \to E'$ be an isogeny of degree $n$. Is the following correct? Regarding the height pairing of elliptic curves that appears when defining the regulator: $\langle f(a), f(b) \rangle = \deg f \langle a, b \rangle$ holds.

When $f$ is multiplication by $m$($m\in \Bbb{Z}$), $\langle f(a), f(b) \rangle = m^2 \langle a, b \rangle$ holds by induction , but I wonder this can be generalized to arbitrary isogeny $\phi$.

Poitou-Tate
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This follows from Tate's formula for the canonical height: Let $g$ be any even nonconstant function on $E'$. Then $\hat h(P) = \frac1{\deg g}\lim_{n\to\infty}4^{-n}h_g(2^nP)$ for $P\in E'$, where $h_g(Q) = \log H(g(Q))$, $H$ the height on the projective line. Then note that we compute the height on $E$ using $g\circ f$ and $h_g(f(Q)) = h_{g\circ f}(Q)$. Finally, since the formula holds for $\hat h$, it follows for the pairing as well.

leoli1
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  • Is Tate's formula the part that states $h_g(f(Q)) = h_{g\circ f}(Q)$? Could you please tell me a reference where the proof is written? – Poitou-Tate Jun 15 '25 at 06:57
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    No, by Tate's formula I mean $\hat h(P) = \frac1{\deg g}\lim_{n\to\infty}4^{-n}h_g(2^nP)$ (for a reference see e.g. Silverman's AEC VIII, or Hindry Silverman Diophantine Geometry). The equality $h_g(f(Q)) = h_{g\circ f}(Q)$ is immediate from $h_g(P) = \log H(g(P))$. – leoli1 Jun 15 '25 at 19:03
  • Thank you very much. For non-isomorphic isogenies, height pairings used to define regulators never coincide... – Poitou-Tate Jun 15 '25 at 20:59