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Is there some relationship between the $p$-adic regulators of isogenous curves over $\mathbb{Q}$? I've done some computations and their ratio seems to be related (equivalent in all calculations so far) to the ratio of their modular degree.

There doesn't seem to be too much literature on this. Or perhaps it's a just a corollary of some known theorem and thus not very interesting. If I'm looking at the wrong place, could somebody please advise on this?

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Whether you take the $p$-adic or the real regulator won't make a difference for this question.

Let $\varphi :E\to E'$ be an isogeny defined over a number field $K$. Consider the map $\varphi_K : E(K)\to E'(K)$. Then $$ \frac{\operatorname{Reg}(E')}{(\# E'(K)_{\mathrm{tors}} )^2 } \cdot \frac{\# \operatorname{coker} \varphi_K}{\#\ker \varphi_K} = \frac{\operatorname{Reg}(E)}{(\# E(K)_{\mathrm{tors}} )^2 } \cdot \frac{\# \operatorname{coker} \hat\varphi_K}{\#\ker \hat\varphi_K}$$ where $\hat\varphi$ is the dual isogeny. This sort of formulae are used when proving that BSD is invariant under isogeny, see for instance Milne's Arithmetic Duality I.7. Papers by Dokchitsers contain variations of this formula.

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  • $\begingroup$ Maybe I just should have said that both the p-adic and the real BSD formula are invariant under isogenies. $\endgroup$
    – Chris Wuthrich
    Commented Aug 7, 2014 at 8:01
  • $\begingroup$ ... and as a consequence: I don't think there is a simple relation to the quotient of the minimal modular degree. Take non-semistable examples and examples with non-trivial Sha to find counter-examples. $\endgroup$
    – Chris Wuthrich
    Commented Aug 7, 2014 at 8:04
  • $\begingroup$ I should have specified that I am looking at good and ordinary primes for now, but thanks for the lead. I will look at those papers. $\endgroup$
    – Haikal Yeo
    Commented Aug 7, 2014 at 8:06
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    $\begingroup$ What $p$ does not matter at all. The quotient is independent of $p$. In fact any quadratic form $h$ on $E(K)$ with $h(\varphi(P)) = \deg(\varphi) \cdot h(P)$ will give the same result. $\endgroup$
    – Chris Wuthrich
    Commented Aug 7, 2014 at 8:08

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