Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a compact Riemann surface, and denote by $m_X$ the following field: $ m_X := \{ f:X \to \mathbb{P}_\mathbb{C} : f- \text{meromorphic} \} - \{\infty \} $

What is the natural injection of the field of rational functions $\mathbb{C}(z)$ into $m_X$ ?

p.s- $\mathbb{P}_\mathbb{C}$ denotes the Riemann sphere.

Thanks in advance !!!

share|cite|improve this question
If you have a distinguished function $f_0\in m_X$, then the map $g\mapsto g\circ f_0$ from $\mathbb C(z)$ to $m_X$ appears to be natural enough. Otherwise I don't know what the natural injection would be. – user31373 Jun 11 '12 at 16:26
Great, Thanks ! – joshua Jun 11 '12 at 17:13
up vote 1 down vote accepted

Let me say that emphatically:

There is no canonical injection of $\mathbb C(z)$ into $\mathcal M(X)$

To give an embedding $\mathbb C(z) \hookrightarrow\mathcal M(X)$ exactly amounts to choosing a non-constant morphism $m:X\to \mathbb P^1(\mathbb C)$.
If such a choice is made, the deduced field embedding $\mathbb C(z) \hookrightarrow\mathcal M(X)$ will send $z\mapsto m$, where $m$ is now seen as a meromorphic function on $X$ .

share|cite|improve this answer
Thanks a lot !!! – joshua Jun 11 '12 at 17:16
You are welcome, joshua. – Georges Elencwajg Jun 11 '12 at 17:32

There is no natural map. $m_X$ is a field and $\mathbb C \subset m_X$ is a field extension. Obviously every $f \in m_X$ that is transcendental over $\mathbb C$ defines such an injection.

The really hard part is to show that there exists a non-constant meromorphic function at all!

share|cite|improve this answer
Thanks a Lot !!! – joshua Jun 11 '12 at 17:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.