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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.

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    $\begingroup$ 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 $\endgroup$
    – reuns
    Feb 13, 2022 at 13:02
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    $\begingroup$ 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. $\endgroup$
    – reuns
    Feb 19, 2022 at 18:34
  • $\begingroup$ 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. $\endgroup$
    – reuns
    Feb 20, 2022 at 22:38
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    $\begingroup$ 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. $\endgroup$
    – reuns
    Feb 20, 2022 at 23:42
  • $\begingroup$ @reuns: Sorry, I had forgotten about your example. I'll think about it and meanwhile I have deleted my unfair comments. $\endgroup$
    – lefuneste
    Feb 21, 2022 at 12:32

1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – lefuneste
    Feb 22, 2022 at 10:38

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