I found some absurd observation which I could not fix by myself. For an elliptic curve $E$ over $\mathbb{Q}$, let $\overline{E}=E\otimes\overline{\mathbb{Q}}$. Every multiplication-by-$n$ map $\overline{E}\to \overline{E}:P\mapsto nP$ defines an étale covering of $E$. It means that for any $n$ and $n$-torsion point $P$, the morphism $\mathcal{O}_{\overline{E},O}\to\mathcal{O}_{\overline{E},P}$ sends a local parameter $\varpi_O$ at $O$ to $u_P\cdot\varpi_P$ where $u_P$ is a unit in $\mathcal{O}_{\overline{E},P}$ and $\varpi_P$ is a local parameter at $P$. Note that, basically, for any $Q\in\overline{E}(\overline{\mathbb{Q}})$, a local parameter at $Q$ is any $f\in K(\overline{E})$ such that $v_Q(f)=1$, which menas $\varpi_O$ is a function on $\overline{E}$ which has order 1 zero at $O$. However, as we have seen in the above, $\varpi_O\mapsto u_P\cdot\varpi_P$ so it also has order 1 zero at $P$. Since we chose $n$ arbitrary, $\varpi_O$ becomes a function having zeros (of order 1) at every torsion points which is definitely absurd. Could anyone point out what I am confusing?

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    $\begingroup$ Firstly, $\varpi_O$ is not a global function on $\bar E$, but only a local one. In fact, there is no global function on an elliptic curve with exactly one zero (this might be fixable, but really you should be thinking about local functions). Secondly, it is not $\varpi_O$ that has a zero at every $n$-torsion point, but rather $[n]^* \varpi_O$. This depends on $n$, so you do not get a single function that works for all torsion points. (If this is still confusing, try giving the two copies of $\bar E$ different names, and see on which space $\varpi_O$ and $u_P \varpi_P$ live.) $\endgroup$ Mar 10, 2017 at 2:34
  • $\begingroup$ @R.vanDobbendeBruyn Thank you very much. I am wondering whether this is known or not: if we take $\varpi_O=x/y$ and $\varpi_P=x-x(P)$ where $x,y$ are variables defining $E$ and $x(P)$ is the $x$-coordinates of $P\in E[n]$. Then can we explicitly write down $\frac{[n]^*\varpi_O}{\varpi_P}$? $\endgroup$ Mar 10, 2017 at 12:19
  • $\begingroup$ Do you assume a special choice for the local parameter at $\Bbb O$ and at $P$? Surely any relationship between $[n]^*\varpi_{\Bbb O}$ and $\varpi_P$ would depend strongly on such choice, isn’t that so? And do you have an answer to your question when $\Bbb O=P$? $\endgroup$
    – Lubin
    Mar 10, 2017 at 20:56
  • $\begingroup$ @Lubin I am specifically interested in the case $\varpi_O=x/y$ and $\varpi_P=x-x(P)$. When $P=O$, I think that $\frac{[n]^*\varpi_O}{\varpi_P}$ should be of the form $n+\text{higher terms in }\varpi_P$. $\endgroup$ Mar 11, 2017 at 22:06

1 Answer 1


I can only speak about the case that $p=\Bbb O$, when there’s one uniformizer involved. I’ll follow you and call $x/y=\varpi$ — now we need no subscripts. If you’re satisfied with a formal response to your question, then all is answered by the formal group of the elliptic curve in question. Since you’re working over $\Bbb Q$, you’re in luck here, due to the existence of the logarithm of the formal group.

You may express the regular differential $\psi$ of your elliptic curve in the form $g(\varpi)d\varpi$ — if you don’t see how to do this, I can explain either with an edit, or by e-mail. But once you have $\psi$, you can integrate it and multiply by a suitable element of $\Bbb Q$ to get it in the form $L(\varpi)=\varpi+\sum_ia_i\varpi^i\in\Bbb Q[[\varpi]]$. This is the logarithm of the formal group, and formally, the addition on the elliptic curve is given by a power series $F(x,y)$ which fits in with the logarithm thus: $L\bigl(F(x,y)\bigr)=L(x)+L(y)$. The $[n]$-endomorphism of the elliptic curve has $L([n](x))=nL(x)$, and in particular, $[n]'(0)=n$, which verifies the conjecture expressed in your last comment. If, for a particular elliptic curve, you should need the higher terms of $[n](x)$, you can get them by an easy degree-by-degree computation, or notice that $[n](x)$ is a root of the power-series $L(X)-nL(x)\in R[[X]]$, where $R$ is the complete ring $\Bbb Q[[x]]$.


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