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Let $(X,\mathcal O_X)$ be an integral regular curve over $\mathbb C$. It follows that $X$ is also a connected Riemann surface.

From the algebraic point of view we can define the function field of $X$ as $K(X)=\mathcal O_{X,\eta}$ (where $\eta$ is the generic point). On the other hand from the analytic point of view we have the sheaf of meromorphic functions $\mathcal M$ on $X$ (here we have the usual definition of meromorphic functions by holomorphic charts).

Now, as far as I understand the vague statement of Serre's GAGA, we should have that meromorphic functions are the same as elements of $K(X)$. Does it mean that $\mathcal M$ is the constant sheaf: $$U\to K(X)\;?$$ Honestly this seems very strange to me and probably it is incredibly wrong. So, can you please explain the relation between $K(X)$ and the sheaf $\mathcal M$?

Important remark (this originated my question): I'm confused by the definition of the so called sheaf of stalks of meromorphic functions on $(X,\mathcal O_X)$, which is denoted by $\mathcal K_X$. You may find the definition of $\mathcal K_X$ for example on Liu's book (chap. 7). Since $X$ is integral one can show that $\mathcal K_X$ is the constant sheaf $K(X)$! So it seems that, at least in the algebraic setting, the sheaf of meromorphic functions is constant...

Dubious
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    The sheaf of meromorphic functions and $K(X)$ coincide only if $X$ is compact. For example, if $X=\mathbb{C}$, $K(X)$ is just the set of rational functions on $X$. But, $e^z$ is a holomorphic function (and thus trivially meromorphic) on $X$, which is not a rational function. – Mohan Sep 15 '16 at 01:36
  • Dear @Mohan, the sheaf of meromorphic functions does not coincide with the sheaf of rational functions even on a compact Riemann surface: see my answer. – Georges Elencwajg Sep 16 '16 at 07:42
  • @GeorgesElencwajg, Of course you are right. – Mohan Sep 16 '16 at 13:04

1 Answers1

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a) For every Riemann surface $X$ (compact or not) the sheaf $\mathcal {Rat}$ of rational functions is different from the sheaf $\mathcal {M}$ of meromorphic functions.
More precisely for every strict open subset $U\subsetneq X$, we have $\mathcal {Rat}(U)\subsetneq \mathcal {M}(U)$.
This is already crystal clear on the simplest example in the world: for $U=\mathbb C\subsetneq \mathbb P^1(\mathbb C)=X$ obviously $e^z\in \mathcal {M}(U)\setminus \mathcal {Rat}(U)$.

b) The sheaf $\mathcal {Rat}$ is constant for every Riemann surface $X$ (compact or not) while the sheaf $\mathcal {M}$ is non-constant for every such $X$.
What GAGA says (but was known in the special case of Riemann surfaces before Serre's article) is that the set of global sections of these sheaves are equal in the case of a compact Riemann surface $X$: although $\mathcal {Rat}\subsetneq \mathcal {M}$, we nevertheless have $\mathcal {Rat}(X)= \mathcal {M}(X)$.

Georges Elencwajg
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    Hi Georges, I was wondering if I can clarify the definition of $\mathcal{R}at$ with you? I have also seen you use the notation $\text{Frac}(U)$. If $U$ is an open subset of $\mathbb{C}$, or a Riemann surface, or a complex manifold, am I correct in defining $\text{Frac}(U)$ as follows? That is, $\text{Frac}(U)$ is the set of all functions $\varphi$ such that there exists $f,g \in \mathcal{O}(U)$ (holomorphic on $U$) and we may write $\varphi = f/g$. –  Jun 21 '18 at 23:13
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    I have never used the notation $Frac (U)$. What makes sense is the notation $Frac (\mathcal O(U))$. It consists of quotients $f/g$, where $f,g \in \mathcal O(U)$ such that $g$ is not identically zero on any connected component of $U$. The ring $Frac (\mathcal O(U))$ is a field if and only if $U$ is connected. It is a deep theorem that for $U$ non-compact $\mathcal M(U)=Frac (\mathcal O(U))$. – Georges Elencwajg Jun 22 '18 at 06:54