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




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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.