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?



Local parameters and etale coverings of of elliptic curves
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 $\Bbb Q[[x]]$.


