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Let us consider a hyperelliptic equation $$Y^2=A_nX^n+A_{n-1}X^{n-1}+\dots+A_0$$ where $A_i\in\mathbb{C}[z]$. I am interested in rational solutions $X,Y\in\mathbb{C}(z)$ when genus is $\geq 2$ and equation is not isotrivial. In "Diophantine geometry on curves over function fields" Theorem 5.12 proves that such equation has always finitely many solutions, and in the conclusion it is claimed that a bound on the degree is obtainable. In Parshin "Algebraic curves over function field I", p1168, a bound is given whose meaning is not clear for me, and which seems too big in practice.

I wonder if a reasonnable bound for the degree of the solutions depending on the degree of the $A_i$ and $n$ exists?

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There are such bounds, although I don't have a reference handy. I'll try to find one later, unless someone else posts one first. Let $f(X)$ be the polynomial on the RHS, and let $D_f\in\mathbb C[z]$ be its discriminant. Then one can obtain bounds that look like $$\deg(Y^{2(n-1)}/D_f) \le(\text{polynomial in $n$})\cdot(\text{# of roots of $D_f(z)$}).$$ For a similar result that's for polynomial solutions $X,Y\in\mathbb C[z]$, but which applies for $n\ge3$, so for genus $g\ge1$, see Proposition 8 in [1] for a proof of $$ \deg(Y^{2(n-1)}/D_f) \le 4n(n-1)\cdot(\text{# of roots of $D_f(z)$}). $$

[1] The canonical height and integral points on elliptic curves, Invent.math. 93, 419-450 (1988)

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  • $\begingroup$ Thank you very much for your answer. Yes if you have a reference for the hyperelliptic result, such a bound seems very nice. By the way, for g=1, is there some hope for a similar bound on a Mordell Weil generators set? $\endgroup$
    – T. Combot
    Commented Oct 1, 2020 at 18:16
  • $\begingroup$ > By the way, for g=1, is there some hope for a similar bound on a Mordell Weil generators set? Unfortunately, the answer is a resounding no. There's a conjecture (I think) that there exists a bound that's roughly exponentially worse than this. But I don't think that there are any theorems. (A good analogy is the Pell equation $x^2-Dy^2=1$, some of them have small'ish solutions, but presumably many of them have smallest solution that's exponential in $D$.) $\endgroup$ Commented Oct 1, 2020 at 18:37

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