Timeline for Rational functions on elliptic curves over global fields with given support
Current License: CC BY-SA 4.0
6 events
when toggle format | what | by | license | comment | |
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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 |