# Meromorphic functions on a Riemann surface as quotient of polynomials

Consider a compact Riemann surface $$C$$ embedded in a projective plane with coordinates $$[x:y:z]$$.
Is it possible to write any meromorphic function $$f\in M(C)$$ as a fraction $$f=\frac{P(x,y,z)}{Q(x,y,z)}$$ with $$P,Q$$ homogeneous polynomials of the same degree having no common zero on $$C$$?
Of course $$Q(x,y,z)$$ will have zeroes on $$C$$ since a (non constant) meromorphic function on $$C$$ can not be holomorphic everywhere, but I'm asking whether at such zeroes $$P$$ will be non-zero.
I am pretty sure that Riemann-Roch guarantees that the answer is yes provided the common degree of $$P$$ and $$Q$$ is large enough, but I'd be grateful for a clean write-up.

• How many zeros does a homogeneous polynomial of degree $d$ have on $zy^2=x^3+xz^2$ ? Then note that $x/z$ has only two zeros and two poles Feb 13, 2022 at 13:02
• Avoid this kind of unfriendly behavior. Bezout's theorem says that $P$ has $3\deg(P)$ zeros on the smooth projective curve $zy^2=x^3+xz^2$. So any meromorphic function not having 3n zeros and poles doesn't have any chance to be of the form you asked. Feb 19, 2022 at 18:34
• Every meromorphic function is a quotient of two homogeneous polynomials but not all are such that $P,Q$ have no common zero which is your question. Once again $P$ has $3\deg(P)$ zeros so if $P,Q$ have no common zeros (and $\deg(P)=\deg(Q)$ then $P/Q$ is a meromorphic function with $3\deg(P)$ zeros and $3\deg(P)$ poles. Feb 20, 2022 at 22:38
• I gave one in the first comment, $x/z$ on the degree 3 smooth complex projective curve (elliptic curve) $zy^2=x^3+xz^2$. It is a quotient of two homogeneous polynomials but they'll always have a common zero (here at $[0:1:0]$). I don't think that Nicolas Hemelsoet noticed the "no common zero" point. Feb 20, 2022 at 23:42

Often no, by Bezout's theorem.

Let $$E:zy^2=x^3+z^3\subset \Bbb{P^2(C)}$$

• For an homogeneous polynomial $$f\in \Bbb{C}[x,y,z]$$ of degree $$d$$ its zeros on $$E$$ are well defined, to define their multiplicities: take $$w\in \{x,y,z\}$$ not vanishing at $$P$$ and look at the order at $$P$$ of the zero of the meromorphic function $$f/w^d$$.

• $$x$$ has 3 simple zeros at $$[0:1:1],[0:-1:1],[0:1:0]$$

• $$x^d$$ has $$3d$$ zeros

• $$f/x^d$$ is a meromorphic function, with the same number of zeros and poles, so $$f$$ must have $$3d$$ zeros.

• The meromorphic function $$x/z$$ has two simple zeros at $$[0:1:1],[0:-1:1]$$ and a double pole at $$[0:1:0]$$.

It can't be that $$x/z=f/g$$ with $$f,g$$ homogeneous of same degree $$d$$ and with no common zero, as $$f/g$$ would have $$3d$$ zeros and poles whereas $$x/z$$ has $$2$$ zeros and poles.

• Congratulations on this perfect answer, which I of course "accept" (in the technical sense of this site: green check mark!) and upvote. Thanks a lot. Feb 22, 2022 at 10:38