Let $M$ be a connected compact Riemann surface. Let $f, g$ be two nonconstant meromorphic functions. Why is there a two-variable complex polynomial $F(x,y)$ that vanishes for $(x, y)=(f, g)$, (in other words $F(f,g)=0$)?



Two non constant meromorphic functions over a connected compact Riemann surface, could not be algebraically independent
Let $F$ be a polynomial of degree at most $n$.


For a point $x$ where $f$ has a pole of order $a$ and $g$ has a pole of order $b$, $F(f,g)$ has a pole of order at most $n\max(a,b)$. Locally near $x$, we can write $F(f,g)$ as a Laurent series $$c\_{ -N} z^{-N} + c\_{1-N} z^{1-N} + \dots + c\_{-1} z^{-1}+ c\_0 + c\_1 z + \dots $$ where $N = n \max(a,b)$ and $c\_{-N}, \dots, c\_{-1}$ are linear functions on $F$.


There are finitely many points where $f$ or $g$ has a pole. Let $d$ be the sum of $\max(a,b)$ over these points. Then we have $nd$ linear functions of the form $c\_{-k}$ at these points. If all these linear functions vanish, then $F(f,g)$ has no poles, and thus is constant.


So as long as the kernel of this set of $nd$ linear functions has dimension $>1$, there will be two linearly independent polynomials $F$ with $F(f,g)$ constant, and taking a linear combination, there will be anontrivial $F$ with $F(f,g)=0$.


Since the dimension of the space of polynomials of degree $\leq n$ is $\frac{n(n+1)}{2}$, the dimension of the kernel is at least $\frac{ n (n+1)}{2} - nd$, hence is $>1$ for n sufficiently large with respect to $d$.


