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

2$\begingroup$ Because the field of meromorphic functions on $M$ has transcendance degree $1$. You'll find the proof in any book on Riemann surfaces. $\endgroup$ – abx Jan 18 at 14:52

6$\begingroup$ This is a fact that lots of people need to know, and it isn't something every mathematician should know, nor is it easy to find a proof of it if you aren't in the area. I think this is the sort of question I want to be welcome on MO. $\endgroup$ – David E Speyer Jan 18 at 17:36

1$\begingroup$ @DavidESpeyer There is a funny point. Before you left this comment I got a downvote, and $1$ or $2$ (or maybe $3$) upvotes. After your comment, the number of these upvotes has reached to a total of $9$. $\endgroup$ – NeoTheComputer Jan 20 at 14:59

2$\begingroup$ NeoTheComputer, I think it is a reasonable discussion to have as to what is appropriate on MO, and @DavidESpeyer making this argument as to why you asked a good question convinced people (including me). So I'm personally glad you asked. $\endgroup$ – Ravi Vakil Feb 1 at 0:49
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 $$c_{ N} z^{N} + c_{1N} z^{1N} + \dots + c_{1} z^{1}+ c_0 + c_1 z + \dots $$ 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 $d$.