Good afternoon, I'm trying to solve an elliptic equation of the form $$AY^2=4X^3+aX+b$$ where $A\in\mathbb{C}[z]$, $a,b\in\mathbb{C}$ and the unknowns $X,Y\in\mathbb{C}(z)$. In Mason ``Diophantine equation over function fields'', such equations are solved for $X,Y\in\mathbb{C}[z]$, but not in the rational functions. Do the solutions admit a group structure? Is it possible to bound the degree of generators? Only in function of the degree of A?
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$\begingroup$ 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. $\endgroup$– Chris WuthrichCommented Aug 3, 2018 at 14:09
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$\begingroup$ 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 $E'/\mathbb{C}$. $\endgroup$– François BrunaultCommented Aug 3, 2018 at 18:26
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$\begingroup$ 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. $\endgroup$– T. CombotCommented Aug 4, 2018 at 20:40
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$\begingroup$ 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. $\endgroup$– François BrunaultCommented Aug 5, 2018 at 7:44
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Fabien Pazuki pointed out to me that the analogue of the Lang-Silverman conjecture for elliptic curves over function fields has been proved by Hindry and Silverman in their article The canonical height and integral points on elliptic curves (Inventiones 1988).
This should imply that (in the case $A$ is not a square) the degree of a non-torsion point $(X(z),Y(z))$ on your elliptic curve is bounded from below by a constant times the degree of $A$.
answered Aug 6, 2018 at 8:45
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$\begingroup$ Do you mean that any non torsion point should be of degree at least C.deg A where C is some universal constant? I seems to me that, assuming $A$ square free, for any solution $X,Y$, the max of their numerators/denominators degree is $\geq 1/4 \deg A$: Either the number of roots of $AY^2$ is at $\geq 3/4 \deg A$ or $Y$ has more that $1/4\deg A$ poles (and then is of degree $\geq \deg A/4$). As the left hanside is cubic in $X$, I get $X$ of degree $\geq \deg A/4$. $\endgroup$ Commented Aug 6, 2018 at 20:51
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$\begingroup$ @T. Combot. Yes I meant an absolute constant $C>0$, but it is very small (see Hindry-Silverman, Thm 6.1). Anyway, you're absolutely right, your equation is just a quadratic twist and it is easier to find a lower bound for the degree of a non-torsion point, and the constant is better (your computation is correct). There is a similar result for quadratic twists of an elliptic curve over $\mathbb{Q}$, see Le Boudec's article arXiv 1404.7738. (...) $\endgroup$ Commented Aug 9, 2018 at 9:20
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$\begingroup$ (...) Le Boudec's conjecture suggests that there should be an even better lower bound (exponential in $\deg A$ if I'm not mistaken) for almost all quadratic twists in the case of $\mathbb{C}(z)$. As for upper bounds, there is a conjecture of Lang stating that for any $E/\mathbb{Q}$ there exists a basis $(P_1,\ldots,P_r)$ of $E(\mathbb{Q})$ with the height of each $P_i$ bounded, see Conjectured Diophantine estimates on elliptic curves, Progress in Math. 35. This bound is polynomial (not logarithmic) in the discriminant of $E$, even for quadratic twists. (...) $\endgroup$ Commented Aug 9, 2018 at 9:31
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$\begingroup$ (...) This relies on the truth of BSD conjecture + analytic estimates on the central derivatives of the $L$-function, while over $\mathbb{C}(z)$ there is no $L$-function available so other methods would be needed. $\endgroup$ Commented Aug 9, 2018 at 9:31