TITLE: A variant of the Stothers-Mason Theorem
QUESTION [6 upvotes]: 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.

REPLY [8 votes]: 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.

REPLY [4 votes]: 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.