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?


1 Answer 1


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.

  • $\begingroup$ 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 $\mathcal M(Y) = \mathcal M(X)(f)$. $\endgroup$ Jul 19, 2022 at 17:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .