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Let $C$ be a smooth, genus $g$ curve defined over a number field $K'$. Suppose we have a $K'$-valued point $P$ on $C$. We can view $C$ as a Riemann surface; then the space of holomorphic differential forms has dimension $g$ over $\mathbb{C}$.

It seems to me that we can choose a uniformizer $x$ at $P$ and a basis of differential forms $\{ \omega_i \}$ such that the expansion of each $\omega_i$ in terms of $x$ has algebraically integral coefficients; that is, there is some finite extension $K$ of $K'$, with ring of integers $A$, such that $\omega_i = \sum_{n=0}^\infty a_{i,n} x^n dx$ with all $a_{i,n} \in A$.


I would really appreciate critiques, help, or references for my reasoning. We can start with $dx$, which is a meromorphic differential form on $X$. We just need to multiply $dx$ by rational functions with zeros in the right places to get holomorphic differential forms, but we want to choose rational functions with integral expansions in $x$.

First, we make an initial choice of uniformizer $x$. If $P=(C_0:C_1: ... :C_n)$ then we know each monomial $C_iX_j-C_jX_i$ has a zero of some order at $P$; we can just choose some convenient quotient of powers of these monomials to get a zero of order one and call that expression $x$ (for now).

The poles of $dx$ will be at various points of $C$ with coordinates in some finite extension of $K'$ (as will the zeros of $dx$). It seems to me that we can build enough rational functions from monomials involving these coordinates to get what we need. (Similar to how we chose the uniformizer $x$.) We can then expand these rational functions in terms of $x$. We may get some denominators (say powers of $N$) in the expansions of our functions, but then we can replace $x$ with $x/N$ to absorb them.

What do you think? Many thanks for the assistance!

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    @Pete: He$r$e's my motivatio$n$, which I'll post as anothe$r$ questio$n$: I'$m$ thinking of modular curves and the equivalence of weight 2 modular forms and differential forms. We study the q-expansions of cusp forms f, where q=e^{2pi iz/m} is a uniformizer at i\infty on the upper half-plane. Is q a rational function on the modular curve? Or at least an integral power series in a rational function uniformizer? If we expand f in (z-a) at the point z=a, then we are using something like the shape z=ln(q+1); which I want to avoid.2011-02-07

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