# Algebraically integral expansions of differential forms?

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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The result you want is true. It seems to me that you may be making too much of it: by definition, a smooth projective genus $g$ curve over a number field $K$ has $g$ $K$-linearly independent global regular differential forms. These will look like $\sum_i f_i dg_i$. If you have any denominators in the coefficients of the rational functions appearing in your differential, multiplying by a suitable positive integer $N$ will get rid of them. Aren't we done? – Pete L. Clark Feb 7 '11 at 17:11
@Pete: I think it depends crucially on the chosen uniformizer at P. We can't just multiply the differential form by N: Consider 1/(1-x/N), which has unbounded denominators when expanded in x, but we could just use x/N instead of x. Also, the expansion may be integral for a uniformizer x, but not integral for y=ln(x+1), which is also a local uniformizer at P. It seems like there is some "natural" family of uniformizers that makes this work (probably rational functions), and I'm trying to nail it down precisely. – Hypocycloid Feb 7 '11 at 18:20
for one thing, you seem to be identifying a differential with its power series expansion, which is valid but certainly not necessary. (And I don't think it's helping you out here.) Second: $y = \ln x+1$ is not a rational function on any compact Riemann surface I know... – Pete L. Clark Feb 7 '11 at 18:24
@Pete: Thanks for the comments. You're right; it is clear to me now that rational functions are what I want to work with. By uniformizer at P, I mean any analytic function on a neighborhood of P with a simple zero at P. – Hypocycloid Feb 7 '11 at 19:21
@Pete: Here's my motivation, which I'll post as another question: 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. – Hypocycloid Feb 7 '11 at 19:26