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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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2 Answers2

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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
  • Does this also hold for abelian varieties ? – Poitou-Tate Jul 02 '25 at 05:20
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    I imagine you might get trouble if you define the heights using noncompatible line bundles. But see Theorem B.5.6 (d) in Hindry Silverman 'Diophantine Geometry' – leoli1 Jul 24 '25 at 16:39
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I believe the statement does not hold in general.

For a counterexample see ExampleIII.4.5 in the second edition of Silverman's The Arithmetic of Elliptic Curves. In the example, two isogenies of degree $2$ were computed, where one takes, say, $E$ to $E'$ and the other takes $E'$ to $E$. The existence of these isogenies implies that $rk(E) \le rk(E')$ and $rk(E') \le rk(E)$, respectively, which implies that $rk(E) = rk(E') = r$.

Let us say your claim holds. Then, for $R_{E}$ representing the regulator of $E$ and a similar notation used for $E'$, we have that $R_{E} = 2^r R_{E'}$ and $R_{E'} = 2^r R_{E}$, which cannot be true unless $r=0$ which is not nessecarily the case. (This is because one can easily prove that in this case, if $P_1, ...,P_r$ represent the free generators of $E$, then, $f(P_1), ...,f(P_r)$ represent the free generators of $E'$. Additionally, note that for $n \times n$ matrices $A$ and $B = kA$, where $k$ is a scalar, we have that $det(B) = k^n det(A)$.

I mostly agree with the statement given by leoli1, however, I think the reason we can provide the above counterexample is that although $h_g(f(Q))=h_{g \circ f}(Q)$, as in Tate's formula, we have not shown that $[2^n]f(P)$ is equal to $f([2^n]P)$.

Please let me know if I am wrong or if anything is unclear!

Helia
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    $[2^n]f(P)$ is equal to $f([2^n]P)$ since $f$ is an isogeny (hence group homomorphism). Note OP's question was whether $\langle f(P), f(Q)\rangle =(\deg f)\langle P,Q\rangle$. This is true as proved in my answer. However it does not imply $R_{E'}=(\deg f)^rR_E$ since the image of a basis for $E(K)$ under $f$ need not be a basis for $E(K')$ (you need to track cokernel, torsion etc., see the formula here) – leoli1 Jul 24 '25 at 16:35
  • Thank you very much for your helpful comment! I have now realised the difference working over integers makes, as I had mistakenly applied the intuition from linear algebra over real numbers. – Helia Jul 26 '25 at 19:19