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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 !!!

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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

2 Answers 2

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 $X$ .

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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!

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Thanks a Lot !!! –  joshua Jun 11 '12 at 17:16

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