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Let us write the irreducible equation relating $x$ and $y$ as
$$P_k(x)y^k+P_{k-1}(x)y^{k-1}+\ldots+P_0(x)=0.$$
Consider the Newton polygon (the graph of the smallest concave function $\phi$ with
$\phi(j)\geq \deg P_j,\; 0\leq j\leq k$.
Condition on the poles of $x$ and $y$ tells us that
$P_k=\mathrm{const}$, and all
slopes of this graph are $\geq -1.$ This ...

answered Dec 20 '19 at 5:09

Alexandre Eremenko

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3

Edit. Thanks to nfdc23 for pointing out a couple of corrections. As nfdc23 points out, these kinds of things are important in passing from a birational group law to a (regular) group law, so they may go back to Weil. Also, I vaguely remember something about some of this in SGA 3, so that might also be a reference.
As requested, I am posting my comments as ...

2

The second half of the statment is not true, but if we replace "meromorphic" by holomorphic, it becomes a true proposition:
$f(z)dz$ is holomorphic if and only if $f(z)$ is holomorphic on $\mathbb{C}$ and $z^2f(z)$ tends to a finite limit as $z \rightarrow \infty$.
This is easy to prove because $$f(z_1)dz_1=-f(1/z_2)/z_2^2 dz_2$$ when changing local ...

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There is an amazing connection of this question to discrete geometry. Suppose $f$ is of the form
$$
f(x_1,\ldots,x_k,z_1,\ldots,z_m) = \frac{z^v}{\prod_{i=1}^k (1-x_i z^{b_i})}
$$
for $v, b_1, \ldots, b_i \in \mathbb{Z}^m$. Clearly the general case, at least when $f$ is rational reduces to this one. Then the integral is simply the sum
$$
\sum_{a\in \mathbb{Z}...

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