Timeline for Solving elliptic equation in rational functions
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 6, 2018 at 8:45 | answer | added | François Brunault | timeline score: 1 | |
| Aug 5, 2018 at 7:44 | comment | added | François Brunault | There is a theorem of Lang (see Silverman p. 275) which asserts that your equation has only finitely many integral solutions i.e. in $\mathbb{C}[z]$ or some localization. But this theorem or results related to the Thue equation concern integral, not rational solutions. For estimating the degree of rational solutions one needs to use heights as mentioned by Chris Wuthrich in his comment. Intuitively I would say the problem is not easy: for an elliptic curve $E/\mathbb{Q}$, the size of generators is related to the regulator of $E$, itself related to the $L$-function of $E$ via BSD conjecture. | |
| Aug 4, 2018 at 20:40 | comment | added | T. Combot | Indeed, I found in Silvermann p230 Th 6.1 the Mordell Weil Theorem and it applies when $A$ is not a square, thank you. For the degree of solutions, the only related result I found is in "THUE'S EQUATION OVER FUNCTION FIELDS" of Schmidt Thm 1, where a degree bound for Thue equation is given. I don't know if it can be used to obtain a bound for my equation, or maybe a hyperelliptic version. | |
| Aug 3, 2018 at 18:26 | comment | added | François Brunault | It depends what one means with isotrivial, but in any case, for this elliptic surface to be isomorphic to $y^2 = x^3+a'x+b'$ with $a',b'\in \mathbb{C}$ one needs to base change from $C(z)$ to $C(z)(A^{1/2})$. If $A$ is not a square then this elliptic surface does not split in the terminology of Silverman so I think the Mordell-Weil theorem applies i.e. the group of solutions is finitely generated and you can find them using standard descent. If $A$ is a square then the group of solutions obviously contains $E'(\mathbb{C})=\mathbb{C}/\Lambda$ for some elliptic curve $E'/\mathbb{C}$. | |
| Aug 3, 2018 at 14:09 | comment | added | Chris Wuthrich | This is an isotrivial elliptic surface. See chapter III in Silverman 2 ("Advanced Topics...") Yes, they form a group (unless the discriminant is $0$). For instance, if $A$ is constant then this group is not finitely generated. Most mostly it will be and the degree of the generators is closely linked to the height function on the Mordell-Weil group. | |
| Aug 3, 2018 at 13:45 | review | First posts | |||
| Aug 3, 2018 at 14:30 | |||||
| Aug 3, 2018 at 13:42 | history | asked | T. Combot | CC BY-SA 4.0 |