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Let $K$ be a function field over $\mathbb{C}$, i.e. a finitely generated extension of $\mathbb{C}$ of transcendence degree 1. Suppose that $x, y \in K^\ast$ are such that $x + y = 1$. Then the Stothers-Mason Theorem states that if $x \not \in \mathbb{C}$ we have $$ H_K(x) \leq |S| + 2g_K - 2, $$ where $g_K$ is the genus of $K$ and $S := \{v \in M_K : v(x) \neq 0 \text{ or } v(y) \neq 0\}$ are the places of $K$ such that $v(x) \neq 0$ or $v(y) \neq 0$.

The Stothers-Mason Theorem can be used to count the number of solutions to unit equations in function fields, but it leads to the appearance of $g$ in the resulting upper bounds. Therefore I am wondering if the upper bound in the Stothers-Mason Theorem can be replaced by $$ H_K(x) \leq c \gamma |S|, $$ where $c$ is some absolute constant and $\gamma$ is the gonality, that is $$ \gamma := \min_{t \in K} [K : \mathbb{C}(t)]. $$ Note that this upper bound is weaker than the Stothers-Mason Theorem if $|S|$ is of size at least $2g_K - 2$, but a lot stronger if $|S|$ is very small.

So far I have been able to find the following: Brownawell and Masser in their paper ``vanishing sums in function fields'' find examples for every value of $g$ such that equality holds in the Stothers-Mason Theorem for infinitely many values of $|S|$. Unfortunately $|S|$ is of size at least $g$ in their examples, so it does not say anything about the truth of $H_K(x) \leq c \gamma |S|$.

Edit: changed Mason's Theorem to Stothers-Mason Theorem.

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    $\begingroup$ Hope you don't mind: I corrected the spelling of Stothers's name (Stothers, not Stother). $\endgroup$ Commented Jun 7, 2017 at 20:56

3 Answers 3

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No.

Take the curve with equation $z^n = x (x-1)$ for $n$ a large odd number. This is a degree $n$ covering of $\mathbb P^1$, totally ramified over $x=0,x=1$, and $x=\infty$. Those $3$ points are the only places where the valuation of $x$ or $y$ is nonzero. Hence $|S|=3$

Furthermore, the map $z$ is a degree $2$ map to $\mathbb P^1$, which shows that $\gamma=2$

The height of $x$ is the degree of the map $x$ (I presume), and so is equal to $n$. This is unbounded, but $c \gamma |S|= 6 c$ is bounded.

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First, to give credit where it is due, this should be called Stother's theorem, or maybe the Stother-Mason theorem, since Stother actually published the result first, at least over $\mathbb C[t]$, and (quoting Wikipedia) Mason "rediscovered it shortly thereafter." (As did I, independently, with a somewhat different proof, but that's another story.)

Second, and more to the point of your question, you might look at the very short proof of Stother's inequality in

The S-unit equation over function fields, Proc. Camb. Philos. Soc. 95 (1984), 3-4.

The proof is an elementary application of the Riemann-Hurwitz formula. It's possible that it can be adapted to give the sort of result that you want.

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  • $\begingroup$ Thank you! I will take a look at the paper and have edited the question accordingly. $\endgroup$
    – P. Koymans
    Commented Jun 7, 2017 at 14:18
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If your intention is to bound the number of solutions to the unit equation $x+y=1, x,y \in G$, Beukers and Schlikewei (Acta Arithmetica 78(1996), 189-199) proved that it is bounded by $2^{8r + 8}$ where $r$ is the rank of the finitely generated group $G$. Now the $S$-units have rank bounded by $|S|-1$ only modulo constants. I am not sure how to deal with the constants.

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  • $\begingroup$ My main intention was to reprove the Beukers and Schlickewei result using the height bound as stated in the question. Such a proof would have the additional advantage that it works in characteristic $p$ if one makes the obvious restriction that $x, y \not \in G^p$. $\endgroup$
    – P. Koymans
    Commented Jun 8, 2017 at 0:17

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