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Let $E$ be an elliptic curve over a global field $k$. Let $x_1, \dots, x_r$ be a set of generators of $E(k) / E(k)_{tor}$ (or more generally, a $\mathbb Q$-basis of $E(k)_{\mathbb Q}$), and let $x_0$ be the zero point of $E(k)$.

Let $f\neq 0$ be a rational function on $E$ with support $\text{supp}(f) \subseteq S = \{x_0, ..., x_r\}$. Then I am wondering: must we have $f\in k^\times$?

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  • $\begingroup$ By the support of $f$ you mean the location of the zeroes and poles of $f$? $\endgroup$
    – Will Sawin
    Commented Dec 11 at 21:46
  • $\begingroup$ @WillSawin Yes, you are right. I mean $div(f) = \sum_{i=0}^{r} n_i\cdot [x_i]$. $\endgroup$
    – yoyo
    Commented Dec 11 at 21:49
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    $\begingroup$ 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 $\operatorname{div}(f) = 0$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Dec 11 at 21:52

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Since $f$ is a rational function, $\operatorname{div}(f)$ must be zero in the divisor class group. For an elliptic curve, the divisor class group is isomorphic to $E(k) + \mathbb Z$ where the projection to $E(k)$ sends $[x_i]$ to $x_i$, so $\sum_i n_i x_i=0$ in $E(k)$.

You have assumed that the $[x_i]$ are linearly independent in $E(k)\otimes \mathbb Q$, so the $n_i$ must be zero, which means $f$ is constant.

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