# Some simple algebra of rational functions by André Weil

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 $$

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?

• 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 $\not=$.) – Donu Arapura Dec 20 '19 at 3:05
• @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. – Colin McLarty Dec 20 '19 at 4:34

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 $$Q(y)$$, $$\deg Q, and the rest is $$xP(x,y)$$, where degree of $$P$$ with respect to $$x$$ and $$y$$ is at most $$k-1$$.
• 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 $x$. – Colin McLarty Dec 20 '19 at 20:03
• 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. – Alexandre Eremenko Dec 20 '19 at 21:05
• 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. – Alexandre Eremenko Dec 20 '19 at 22:46
• 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. – Colin McLarty Dec 21 '19 at 16:02