Let $X$ be a smooth (integral) projective curve over a field $k$ and let $K$ be the function field of $X$.

I'd like to prove that there is a bijective correspondence between the closed points of $X$ and the set of places (i.e. equivalent absolute values on $K$) which are trivial on $k$.

Clearly for every closed point $p$ I have the local ring $\mathcal O_{X,p}$ which is a DVR, therefore I get a desired valuation on $K$, hence a metric.

But what about the inverse of this map? From an absolute value of $K$ I want to get a closed point of $X$ inverting the above construction.

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    $\begingroup$ Do you know the valuative criteria for properness? This gives you, essentially, what you want. $\endgroup$ – Alex Youcis Nov 30 '16 at 1:47
  • $\begingroup$ I don't know it, I'll check this notion. Thank you. $\endgroup$ – manifold Nov 30 '16 at 1:51
  • $\begingroup$ How do I apply the criteria of properness to my problem? $\endgroup$ – manifold Nov 30 '16 at 13:31
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    $\begingroup$ You have a valuation $v$ on $K$. This corresponds to a valuation subring $\mathcal{O}_v\subseteq K$. Note then that, in particular you know that we have the diagram $$\begin{matrix}\text{Spec}(K) & \xrightarrow{i} & X\\ \downarrow & &\downarrow\\ \text{Spec}(\mathcal{O}) & \to & \text{Spec}(k)\end{matrix}$$ Where the map $i$ is the inclusion of the spectrum of the function field into $X$ (i.e. it's the inclusion of the generic point). Note that we are using that $v(k^\times)=0$ to say that this diagram commutes (why?). So, by the valuative criteria for properness this $\endgroup$ – Alex Youcis Nov 30 '16 at 13:45
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    $\begingroup$ gives us an extension $\text{Spec}(\mathcal{O})\to X$. Suppose that the closed point of $\mathcal{O}$ maps to the closed point $x$ of $X$. I claim then that $\mathcal{O}_{X,x}=\mathcal{O}$ showing that $v=v_x$. But, this follows since the map $\mathcal{O}_{X,x}\to\mathcal{O}$ is local (why?) and DVRs are maximal by domination amongst subrings of $K$. Thus, this implies that $\mathcal{O}=\mathcal{O}_{X,x}$. $\endgroup$ – Alex Youcis Nov 30 '16 at 13:46

Valuations are the same as places, which are just maps from a valuation ring into its quotient field. The center of a place is the prime ideal of all functions on which the place is zero. In the case of a curve this will be a maximal ideal, and if the field is algebraically closed you get a point of the curve. This theory is exposited in Zariski and Samuel: Commutative Algebra Vol.II. In general many places will correspond to a given point, but if the point is simple there will be only one place for that point.

  • $\begingroup$ Well, as the OP mentioned, you need to assume that the kernel of the valuation contains the field of constants. $\endgroup$ – Alex Youcis Nov 30 '16 at 1:46
  • $\begingroup$ I am assuming this. $\endgroup$ – Rene Schipperus Nov 30 '16 at 1:48
  • $\begingroup$ So, I took a look to Zariski&Samuel book. The chapter on valuations is huge! By the way I have two comments/questions: 1) By the term "place" I refer to the definition you find in Neukirch ANT (ch.3). But I suppose that the two notions are the same. 2) Is the result true if I don't assume $k$ to be algebraically closed? $\endgroup$ – manifold Nov 30 '16 at 1:49
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    $\begingroup$ If $k$ is not closed, cant you just take the ideal for an algebraic point and use that to define a valuation ? $\endgroup$ – Rene Schipperus Nov 30 '16 at 1:52

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