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Let $k$ be an algebraically closed field and let $X$ be a non-singular algebraic projective curve over $k$. Very often, when most books present the function field $k(X)$, they say that "it is the analogue of the field of meromorphic functions on a Riemann surface". I don't agree with this statement, and I'm going to explain the reason:


If $p\in X$, then the local ring $\mathcal O_{X,p}$ is a discrete valuation field (remember that $X$ is non-singular) with respect to a (normalized) valuation $\text{ord}_p$. Then $k(X)$ is the field of fractions of $\mathcal O_{X,p}$ and it is in natural way a discrete valuation field with respect to $\mathcal O_{X,p}$. At this point we can take the completion $\widehat {k(X)}$ and moreover we notice that $k\subseteq k(X)$ is a set of representatives for the residue field $\mathcal O_{X,p}/\mathfrak m_p$. Thanks to the theory of valuations we have the following equality:

$$\widehat {k(X)}\cong k((\pi))=\left\{\sum_{i\ge-k}a_i\pi^i\,: a_i\in k\right\}$$

where $\pi\in\mathcal O_{X,p}$ is a uniformizer parameter. Moreover we can choose $\pi=(t_1-p_1,\ldots,t_n-p_n)$ where $p=(p_1,\ldots,p_n)$ and $t_i:= I(X)+T_i\in \mathcal O_X(X)$.

Edit: Note that when $k$ is not algebraically closed we have the Laurent expansion only for the $k$-rational points.


By summarizing, for every element of $f\in \widehat {k(X)}$ we find the Laurent expansion of $f$ around $p$, so I would say that:

$\widehat {k(X)}$ is "the algebraic equivalent" of the field of meromorphic functions, and not $k(X)$.

Is my reasoning right? Why is this process of completion not always mentioned?

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  • $\begingroup$ Every closed point of $X$ gives rise to a discrete valuation of the function field. How can you choose one which you want to call the algebraic equivalent of the field of meromorphic functions on a Riemann surface? Also, when you say $\mathscr{O}_{X,p}$ is a "discrete valuation field," you mean "discrete valuation ring." $\endgroup$ – Keenan Kidwell Jan 27 '16 at 23:33
  • $\begingroup$ Yes of course I mean d.v. ring. $\endgroup$ – Dubious Jan 27 '16 at 23:34
  • $\begingroup$ Moreover: you are right! For every (closed) point I have a discrete valuation so for every Laurent expansion I have to chance valuation. On the other hand, note that $k(X)=\text{Frac}(\mathcal O_X(X)) $ , and on compact Rieman surfaces the field of meromorphic functions is not the field of fractions of holomorphic functions. So, I continue to believe that the analogy is wrong. $\endgroup$ – Dubious Jan 27 '16 at 23:42
  • $\begingroup$ Dear @Dubious, I'm not sure what you're looking for with this question. It doesn't really have an answer. You are correct that you can complete $k(X)$ with respect to the discrete valuation corresponding to a closed point, and you get Laurent series expansions, etc.. Perhaps people regard $k(X)$ as "more algebraic" than something obtained via a completion process (particularly one which depends on a choice of closed point). $\endgroup$ – Keenan Kidwell Jan 27 '16 at 23:51
  • $\begingroup$ Yes, there is not a complete answer for this question. Anyone is free to define his own meromorphic functions. I'm only trying to see which are the most natural analogies between the world of manifolds and the world of algebraic varieties. This site contains plenty of people that generally are able to give very inspiring hints, so that's the reason of my "question". $\endgroup$ – Dubious Jan 27 '16 at 23:55
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I don't follow your reasoning. The classical sense in which $k(X)$ is analogous to meromorphic functions on a Riemann surface is that when $k = \mathbb{C}$ it is literally meromorphic functions on a Riemann surface. The completion you describe is completely different: it's a completion of germs of meromorphic functions at a particular point, whereas meromorphic functions should have Laurent series expansions everywhere, not just at a particular point.

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  • $\begingroup$ +1, I think this is a fairly definitive answer to the semi-question posed in the OP. $\endgroup$ – Keenan Kidwell Jan 28 '16 at 19:25
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Locally a meromorphic function is a quotient of holomorphic functions. I guess that's where the analogy comes from.

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  • $\begingroup$ That's true! but we lose the power series expansion $\endgroup$ – Dubious Jan 27 '16 at 23:48

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