# Two non constant meromorphic functions over a connected compact Riemann surface, could not be algebraically independent

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 $$F(f,g)=0$$)?

• Because the field of meromorphic functions on $M$ has transcendance degree $1$. You'll find the proof in any book on Riemann surfaces.
– abx
Jan 18, 2021 at 14:52
• 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. Jan 18, 2021 at 17:36
• @DavidESpeyer There is a funny point. Before you left this comment I got a down-vote, and $1$ or $2$ (or maybe $3$) up-votes. After your comment, the number of these up-votes has reached to a total of $9$. Jan 20, 2021 at 14:59
• 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. Feb 1, 2021 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 $$c_{ -N} z^{-N} + c_{1-N} z^{1-N} + \dots + c_{-1} z^{-1}+ c_0 + c_1 z + \dots$$ 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 $$d$$.