I am a little confused on the $p$-adic regulator on elliptic curves and what happens when you switch to different Weierstrass models. Restrict to ell. curves over $\mathbb Q$ for simplicity.


From my understanding, to define the p-adic height (and more precisely the $p$-adic sigma function), you need an integral Weierstrass model minimal at $p$. As long as that condition is cleared, the $p$-adic regulator should be independent of the model.


However, when I try this in practice, I get different results.
For example, (and this is the case with most examples), take the rank 1 curves on Sage:



```
E=EllipticCurve('143a1')
E.is_p_minimal(7)
E1=E.short_weierstrass_model()
E1.is_p_minimal(7)

```

It's easy to check, they're both minimal at $p=7$ and that $p=7$ is a good ordinary non-anomalous prime. (The model of $E$ is automatically The minimal Weierstrass model).


However, their regulators are completely different.



```
E.padic_regulator(7)
2*7^2 + 5*7^3 + 7^4 + 5*7^5 + 3*7^6 + 5*7^7 + 7^8 + 5*7^9 + O(7^10)
E1.padic_regulator(7)
3*7 + 3*7^2 + 4*7^3 + 3*7^4 + 5*7^6 + 7^8 + 3*7^9 + 5*7^11 + 7^12 +O(7^13)

```

Not only are they different, but their $p$-valuation as you can see is also different, which, without going through details, would imply in the first case that the Iwasawa $\lambda$ invariant would be greater than 1, whereas in the second case it would imply that $\lambda\_p(E)= 1$, which is absurd since $\lambda\_p(E)$ does not depend on the model. The correct regulator is the first one, of The minimal curve, since in this example one can check $\lambda\_p(E) >1$.


At first I was convinced it was an implementation error by Sage, so I posted pretty much the same question on [Ask Sage](https://ask.sagemath.org/question/64632/p-adic-regulator-on-different-weierstrass-models/) with the assumption that it was a bug, but for various reasons I could go through, now I have started to doubt that it was due to implementation errors, which is why I also decided to post the question here.


So in short, am I misunderstanding something? Shouldn't both the regulators be the same, since both models are minimal at $p$?



Does the $p$-adic regulator depend on Weierstrass model?
Here a long comment to settle this question.


This is really a bug in the implementation of $p$-adic heights in sagemath. I have announced it [as a bug here on the sage trac list](https://trac.sagemath.org/ticket/34790). I hope to add the code to fix it soon after this post.


The current implementation is based on [1], which in turn is a refinement of the method by Mazur-Stein-Tate mentioned before.


First of all the $p$-adic height of a point on an elliptic curve must be independent of the chosen model. The formula used looks like $\hat{h}\_p(P) = \log\_p(\sigma\_p(P)/d(P))$ up to a chosen constant factor, where $\sigma\_p$ is the canonical $p$-adic $\sigma$-function for this model and $d(P)$ is the square root of the denominator of the $x$-coordinate of $P$. However this formula only holds if $P$ has good reduction at all places for the given model and trivial reduction at $p$. Instead of figuring out the correction factors introduced when the point has bad reduction in the model, one uses the fact that $\hat{h}\_p$ is quadratic. First a certain multiple $Q=nP$ has good reduction everywhere. Then a multiple $mQ$ has trivial reduction at $p$.


The algorithm implemented took $n=\operatorname{lcm}(c\_v)$, the lowest common multiple of all Tamagawa numbers, but this is only valid if the model is minimal. Instead for primes $p$ at which the model is not minimal, one needs to correct this.


As explained in [1], David Harvey has optimised the implementation by a clever trick to avoid the next multiplication by $m$ which brings the point into the formal group. As one only needs a $p$-adic approximation of $d(mQ)$ and of $\sigma(mQ)$. This follows the idea explained in the other answer.


I would avoid to try and extend this trick to the first multiplication $nP$ which aims at getting the point to have good reduction at all primes. The reason is that there will be cancellation in the fraction obtained by the formula using the $n$-th division polynomial $f\_n$. When the point $P$ has good reduction at all primes then $d(nP) = f\_n(P)\cdot d(P)^{n^2}$ as explained in Proposition 1 [here](https://www.maths.nottingham.ac.uk/plp/pmzcw/download/pheight.pdf). Otherwise this formula does not holds as such.


[1] MR2395362 Harvey, David
[Efficient computation of p-adic heights](https://arxiv.org/abs/0708.3404).
LMS J. Comput. Math. 11 (2008), 40–59.


