# Showing algebraic dependence of meromorphic functions on a compact Riemann surface

I have been given the following question to do: Let $$f,g$$ be meromorphic functions on a compact Riemann Surface $$R$$. Show that there is some polynomial such that $$P(f,g) = 0$$ (i.e. show that any two meromorphic functions on $$R$$ are algebraically dependent). I have seen this result over the torus which follows from looking at the Weierstrass $$\wp$$ function, however I have no idea how to generalise that to every compact Riemann Surface.

There is a hint which says I should let $$d = m+n$$ where $$m,n$$ are the valencies of $$f,g$$ respectively and consider $$P(f,g) = \sum\limits_{j = 0}^d\sum\limits_{k = 0}^d a_{jk}f(z)^jg(z)^k$$ and show that this has at most $$d^2$$ poles and that I can choose the $$a_{jk}$$ so that $$P(f,g)$$ has at least $$d^2+2d$$ roots and so is constant by the valency theorem.

Showing that there are at most $$d^2$$ poles is easy but I don't know how to select the $$a_{jk}$$ to get $$d^2+2d$$ roots. I don't see whether I should try and find them explicitly (seems hard) or use some indirect argument (but I can't see where to start). Any help is much appreciated.

You can prove that there exists two polynomial $$p,q\in \mathbb{C}[x,y]$$ such that

$$ord_a(\frac{p(f,g)}{q(f,g)})\geq 0$$ for every $$a\in X$$

so it is an olomorphic function on $$X$$ that is a Compact R.S. so there exist a costant $$c$$ such that

$$\frac{p(f,g)}{q(f,g)})=c$$

Then

$$p(f,g)-cq(f,g)=0$$

But I don’t know how built these two polynomials. We can indicate $$\{a_1,\dots , a_n\}$$ the set of the poles of $$f$$ and $$\{b_1,\dots b_m\}$$ the set of the poles of $$g$$.

If $$l:=ord_b(g)<0$$ and $$ord_b(f)\geq 0$$ then you have that

$$ord_b((f-f(b))^l g)\geq 0$$

So you can resolve the problem in this particular case.

• Why I’m wrong? I don’t understand – Federico Fallucca Jan 11 at 12:39