# Why are meromorphic functions on a smooth projective curve rational?

Let $$C \subset \mathbb P^n$$ be a smooth connected projective curve over $$\mathbb C$$. Then the function field $$k(C)$$ consists of all functions $$f$$ which can locally (in the Zariski topology) be written as quotients of polynomial functions on $$C$$.

On the other hand, we can also think of $$C$$ as a connected compact Riemann surface. Then the field of meromorphic functions $$\mathcal M(C)$$ consists of all $$f$$ which can locally (in the analytic topology) be written as quotients of holomorphic functions.

Some answers claim $$k(C) = \mathcal M(C)$$. Clearly, $$k(C) \subset \mathcal M(C)$$. How can I see the other inclusion?

That’s a good question! Note that it is a (very) special case of the GAGA theorems/philosophy – in ”complete enough” situations (usually properness), analytic objects and morphisms should come from algebraic geometry.

The first part of the argument is, I believe, to study the case $$C=\mathbb{P}^1$$, which is classical: $$k(C)=k(t)$$ almost by definition, and a meromorphic function on $$\mathbb{P}^1$$ has (after multiplying by a polynomial factor) poles only at infinity, so it has a power series expansion with finitely many terms (otherwise the singularity at infinity is essential) ie is a polynomial.

Next, we treat the relative cases separately: let $$f: Y \rightarrow X$$ be a nonconstant morphism of smooth projective algebraic curves (resp. compact Riemann surfaces coming from smooth projective curves*). We’ll see that the pull-back makes $$k(Y)$$ (resp. $$\mathcal{M}(Y)$$) a $$k(X)$$-algebra (resp. a $$\mathcal{M}(X)$$-algebra) of dimension exactly (resp. at most) $$\deg{f}$$.

Assume that this holds. Let $$C$$ be a smooth projective curve. We can then choose some nonconstant $$f \in k(C)^{\times}$$ which defines a nonconstant morphism $$\varphi: C \rightarrow \mathbb{P}^1$$ of algebraic curves and Riemann surfaces as well – and it has the same degree in both interpretations (because the degree is the cardinality of every fiber but finitely many).

Because pull-back commutes to analytification, it follows that $$k(C) \subset \mathcal{M}(C)$$ is an inclusion of $$\varphi^*-k(\mathbb{P}^1)$$-vector spaces with non-increasing finite dimension – so is an equality.

*in fact, Chow’s lemma and general considerations on Riemann surfaces show that every compact Riemann surface should come from algebraic geometry (as they are projective!). I would even say that the relationship comes from “earlier”: if I recall correctly, much of the theory of compact Riemann surfaces comes from the existence of nonconstant meromorphic functions – that is, their finiteness over $$\mathbb{P}^1$$.

All that remains to do is show the relative case. For algebraic curves, this is more or less the definition of degree. For Riemann surfaces, it’s trickier, and I think this argument needs quite more detail.

Let $$f \in \mathcal{M}(Y)$$. Define, for every $$x \in X$$ and $$1 \leq i \leq d$$, $$c_i(x)=\sum_{F \subset \varphi^{-1}(x), |F|=i}{\prod_{y \in F}{f(y)}}$$. $$c_i$$ is clearly holomorphic at every point above which $$\varphi$$ is not ramified and $$f$$ has no pole (so, almost every point). I believe that we can see in local charts that $$c_i$$ is meromorphic at the remaining points.

The trick is that $$\sum_{i=0}^d{(-1)^{i}c_i(\varphi(y))f^{d-i}(y)}=0$$ for almost all $$y$$, thus $$f$$ is algebraic of degree $$d$$ over $$\mathcal{M}(X)$$ (where $$d$$ is the degree of $$f$$). Thus, the pull-back $$\mathcal{M}(X) \rightarrow \mathcal{M}(Y)$$ is an integral extension of characteristic zero fields such that every monogenous subextension has degree at most $$d$$. The conclusion is then “elementary” field theory in characteristic zero.

• I find the part with the $c_i$ totally believable, actually I did the computation for $c_1$ recently. The last step should essentially be the theorem of the primitive element: Pick any $f \in \mathcal M(Y)$ of maximal degree $\leq d$. If $g \in \mathcal M(Y) \setminus \mathcal M(X)(f)$, then the extension $\mathcal M(X)(f,g)$ has strictly higher degree than $f$. Since we are in characteristic $0$, there exists a primitive element $h \in \mathcal M(X)(f,g)$, which then has higher degree than $f$, contradiction. Hence $\mathcal M(Y) = \mathcal M(X)(f)$. Jul 19, 2022 at 17:39