For an elliptic curve $y^2=4x^3-g_2x-g_3$ over $\mathbb{Q}$, it is known that the local parameter at $O$ (the identity point) could be written by $-\frac{x}{y}$. Is it possible to write down the local parameter at $P\in E[\ell]$ for some prime $\ell\geq3$ with $P\neq O$ in the similar way (maybe, in terms of $x$ and $y$)?
1 Answer
Let $P=(a,b)$. Then $x-a$ is a local parameter at $P$, since it has a zero of order one at P. (It's two zeros are $P$ and $-P$.) Or if you want a local parameter that is defined over the field of definition of $E$, you can take the $\ell$-division polynomial $$\psi_\ell(x):=\prod_{P\in E[\ell]/\{\pm1\}} \bigl(x-x(P)\bigr).$$
But you may be misunderstanding what a local parameter is, because you say that $-x/y$ is the local parameter at $O$. It is just one of many local parameters. In general, a local parameter at a point $P$ on a smooth curve $C$ is simply a function $f$ in $K(C)$ with the property that $\text{ord}_P(f)=1$, or equivalently, a function that generates the maximal ideal of the local ring at $P$.
-
$\begingroup$ If $P$ is of order 2, then one should take $y$ instead (since then $x-a$ has order 2 at $P$). $\endgroup$ Commented Mar 3, 2017 at 22:01
-
$\begingroup$ @MichaelStoll Very true, and more generally if $P$ is a point of order $m$ with $m$ even, one has to be more careful. But the question specified that $\ell$ is a prime that is at least $3$. $\endgroup$ Commented Mar 3, 2017 at 22:29
-
$\begingroup$ Thank you very much. In fact, I have one more (stupid) question: I have read many literature choosing $-x/y$ as a local parameter at $O$. Is there any particular reason for that? I mean, is $-x/y$ `canonical' in some sense? $\endgroup$– User0829Commented Mar 4, 2017 at 14:24
-
$\begingroup$ Moreover, is there a canonical way of choosing local parameters $\pi_O$ and $\pi_P$ at $O$ and $P_{\neq O}\in E[\ell]$ respectively so that $\pi_O\mapsto \pi_P$ under the induced ring homomorphism $\mathcal{O}_O\to\mathcal{O}_P$ from the $\ell$-isogeny $E\xrightarrow{[\ell]}E$. $\endgroup$– User0829Commented Mar 4, 2017 at 14:28
-
1$\begingroup$ @User0829 $-x/y$ is certainly not canonical, as $x$ and $y$ themselves are not. But if you want a uniformizer of a simple form, then (a constant multiple of) $x/y$ is the simplest function that vanishes to order 1 at infinity that you can come up with. The minus sign makes the standard invariant differential look like $(1 + \dots) dt$, when $t = -x/y$. $\endgroup$ Commented Mar 4, 2017 at 18:37