Skip to main content
6 events
when toggle format what by license comment
Dec 11 at 21:59 vote accept yoyo
Dec 11 at 21:53 answer added Will Sawin timeline score: 4
Dec 11 at 21:52 comment added R. van Dobben de Bruyn This is true whenever $S$ is a $\mathbf Z$-linearly independent set (for elliptic curves over any base field $k$). Indeed, the group isomorphism $E \stackrel\sim\to \operatorname{Pic}^0(E)$ is given by $P \mapsto [P] - [O]$, so this means that $\{[P]-[O]\ |\ P \in S\}$ is linearly independent. If $f \in k(E)$ satisfies $\operatorname{supp}(f) \subseteq S$ then $\operatorname{div}(f) = \sum_{P\in S} n_P[P] = \sum_{P \in S} n_P([P]-[O])$ since it has degree $0$, giving a $\mathbf Z$-linear relation unless $\operatorname{div}(f) = 0$.
Dec 11 at 21:49 comment added yoyo @WillSawin Yes, you are right. I mean $div(f) = \sum_{i=0}^{r} n_i\cdot [x_i]$.
Dec 11 at 21:46 comment added Will Sawin By the support of $f$ you mean the location of the zeroes and poles of $f$?
Dec 11 at 21:17 history asked yoyo CC BY-SA 4.0