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In André Weil's dissertation, he considers two meromorphic functions $x,y$ on a complex curve.  He assumes every pole of $y$ is a pole of $x$, and its multiplicity as a pole of $y$  is no greater than its multiplicity as a pole of $x$.  Then he says there is some natural number $k$ and some complex $a\neq 0$ such that  $$ay^k+xP(x,y)+Q(y)=0$$ where $P(x,y)$ and $Q(y)$ are polynomials in $x,y$ with degree $<k$. 

I see the proof for genus 0 curves. There the field of meromorphic functions is $\mathbb{C}(z)$ and one-variable polynomial algebra suffices (unless I've made a mistake). But I do not see it for other curves. Can someone tell me how it is done?

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    $\begingroup$ I assume that the curve is compact, then the field of meromorphic functions on it is finitely generated over $\mathbb{C}$ with transcendence degree $1$. Therefore any $x,y$ would satisfy a polynomial relation. It can be rewritten in the above form. (I'm guessing the symbol between $a$ and $0$ is $\not=$.) $\endgroup$
    – Donu Arapura
    Dec 20, 2019 at 3:05
  • $\begingroup$ @DonuArapura Yes he surely means compact, though writing in 1927 the closest he comes to saying so is to say he is using the "birational viewpoint." I expect you are right about the strategy he had in mind, and i will try to work through what he says using that idea. $\endgroup$
    – Colin McLarty
    Dec 20, 2019 at 4:34

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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 implies that $$\deg P_j\leq m-j,\quad 0\leq j\leq k-1.$$ Now unite all constant terms of $P_j$ times powers of $y$ into the polynomial $Q(y)$, $\deg Q<k$, and the rest is $xP(x,y)$, where degree of $P$ with respect to $x$ and $y$ is at most $k-1$.

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  • $\begingroup$ This is likely the right route. But I do not see how the condition on poles shows $P_k$ has no poles. The condition allows shared poles of $x$ and $y$ to have arbitrarily higher multiplicity in $x$ than $y$. I suspect the proof may need a step of multiplying through by some high enough power of $y$ to make all multiplicities of poles in $y^k$ surpass their multiplicity in $x$. $\endgroup$
    – Colin McLarty
    Dec 20, 2019 at 20:03
  • $\begingroup$ The proof is complete. If all poles of $y$ are poles of $x$ then $P_k$ is constant. Look at the properties of the Newton polygon. $\endgroup$
    – Alexandre Eremenko
    Dec 20, 2019 at 21:05
  • $\begingroup$ Can you say a little more about verifying that property of the Newton polygon? I am not familiar with calculating Newton polygons. $\endgroup$
    – Colin McLarty
    Dec 20, 2019 at 21:40
  • $\begingroup$ Algebraic equation wrt $y$ has $k$ solutions for generic $x$. If $P_k$ is not constant, it has a root at $x_0$ and as $x\to x_0$ some of these roots tend to $\infty$. Which means that at some point of the Riemann surface $x$ equals $x_0$ while $y$ has a pole. See any book on algebraic functions for the details. $\endgroup$
    – Alexandre Eremenko
    Dec 20, 2019 at 22:46
  • $\begingroup$ I see that argument. But the the conclusion on degrees in $x$ and $y$ seems to follow just by counting the multiplicity of any one pole. I do not see the role of the Newton polygon. $\endgroup$
    – Colin McLarty
    Dec 21, 2019 at 16:02

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