Timeline for Over which fields does the Mordell-Weil theorem hold?
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14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 17, 2020 at 14:39 | answer | added | Arno Fehm | timeline score: 5 | |
| Dec 20, 2019 at 17:01 | comment | added | Watson | In this paper, D. Ghioca showed that for any non-isotrivial elliptic curve $E$ over $K=\mathbb F_p(t)$, the group $E(K^{p^{-\infty}})$ is finitely generated (this does not address your Q1 though, since we need $E$ to be non-constant). | |
| Nov 14, 2018 at 18:32 | comment | added | Watson | In Elliptic Curve Handbook, Ian Connell, Mordell–Weil theorem is proved for $K$ being the fraction field of a so-called "F2-Krull domain". | |
| Oct 30, 2012 at 10:28 | answer | added | Laurent Moret-Bailly | timeline score: 11 | |
| Oct 29, 2012 at 19:37 | answer | added | Pete L. Clark | timeline score: 4 | |
| Oct 3, 2012 at 7:15 | comment | added | Qing Liu | and there are plenty of $K$-points in $A$ as $K$ is algebraically closed (use a finite surjective morphism from $A$ to $\mathbb P^d$ if you want). | |
| Oct 3, 2012 at 7:14 | comment | added | Qing Liu | @François: trivial generalizations of your examples for Q2: replace $Q_p$ by any (algebraic extension of) complete discrete valuation fields; replace $\mathbb C$ by any algebraically closed field $K$ (you don't need to use complex uniformization, just observe that $K$ is the algebraic closure of a purely transcendental extension $L$ of its prime field. Then any $A$ over $K$ is defined over a finite extension of $L$. If $A(K)$ was of finite type, then there exists a finite extension $L'$ of $L$, containing the defintion field of $A$, such that $A(K)=A(L')$. This is impossible because $K\ne L'$ | |
| Oct 1, 2012 at 20:44 | comment | added | Damian Rössler | Merci pour la référence à Kato ! Je vais regarder. | |
| Oct 1, 2012 at 17:40 | comment | added | François Brunault | @Damian : This is interesting, this means that the field $F_{p^\infty}$ will (conjecturally) answer Q1. This reminds me Kato proved the following theorem (in Astérisque 295) : if $E/\mathbf{Q}$ is an elliptic curve and $m>1$ is any integer, then $E(\mathbf{Q}(\mu_{m^\infty}))$ is finitely generated. More generally, he proves it for any abelian variety which is a quotient of $J_0(N)$, so it applies to any abelian variety over $\mathbf{Q}$ with GL_2-type. | |
| Oct 1, 2012 at 16:20 | comment | added | Damian Rössler | There is a conjecture a Mazur to the effect that for any number field F and abelian variety A over F, the group $A(F_{p^\infty})$ is finitely generated; here $p$ is a fixed prime number and $F_{p^\infty}$ is the union of the largest $p$-power subextensions of fields of the form $F(\mu_{p^n})$, for all $n\geq 1$. See for instance Lang's "Survey of Diophantine geometry", I, par. 4 (p. 29). | |
| Oct 1, 2012 at 14:35 | answer | added | Felipe Voloch | timeline score: 10 | |
| Oct 1, 2012 at 13:54 | comment | added | François Brunault | @Chandan : Right, thanks. Prompted by your observation, I realize that any intermediate extension $\mathbf{Q}_p \subset K \subset \overline{\mathbf{Q}}_p$ will work, because any $A/K$ is defined over some finite extension of $\mathbf{Q}_p$. | |
| Oct 1, 2012 at 13:29 | comment | added | Chandan Singh Dalawat | For Q2, you could take $K$ to be the maximal unramified extension of ${\bf Q}_p$ (so $K$ is not a finite extension of ${\bf Q}_p$, it is not complete for a discrete valuation, and it is not algebraically closed). | |
| Oct 1, 2012 at 13:01 | history | asked | François Brunault | CC BY-SA 3.0 |