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I have a question about Qing Liu's proof of $X(\mathcal O_K)\rightarrow X_K(K)$. It is theorem 3.25 in his book.

Theorem 3.25. Let $X$ be a proper scheme over a valuation ring $\mathcal O_K$ and let $K = Frac(\mathcal O_K)$. Then the canonical map $X(\mathcal O_K)\rightarrow X_K(K)$ is bijective.

I struggle to understand every steps of his proof. After complicated constructions, he obtains a morphism $Spec(\mathcal O_K)\rightarrow X$ and complete the proof. But why this morphism satisfies the property we want? The construction is too strange for me and I think I'm lost somewhere.

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Hydrogen
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1 Answers1

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Maybe it is better to see the construction this way: a map from $\operatorname{Spec}(K)$ to $X_k$ gives a closed point $x$ of $X_K$. The underlying topological space of $\operatorname{Spec}(O_K)$ has two important points: a generic point coming from $\operatorname{Spec} K$ and a closed point $s$. So if we want to extend the map from $\operatorname{Spec}(K)\to X_K$ to a map $\operatorname{Spec}(O_K)\to X$ we have to send the generic point of $\operatorname{Spec}(O_K)$ to $x$ and because we want the continuity on topological spaces we have to send $s$ to a point in the closure of $x$ for example $z$. so topologically the image of our map must be in $Spec(O_{\bar{\{x\}},z})\subset Spec X$

Now that we know map on topological space for knowing the map between schemes we only have to give a map between $O_z\to O_K$ extending the morphism between $O_x\to K$. The proof in Liu's book uses the definition of properness to give such a map: in fact both ring are isomorphic:we have the map $Spec(O_z)\to Spec(O_K)$ we know that the generic point is in the image of this map and by properness the image is closed so this map is subjective on topological space and hence coming from a dominating map between local ring but both of this ring have the same field of fraction(because we know we have a map $$O_z\to K$$) and by general properties of valuation ring the map must be an isomorphism so its inverse is the extension we want. note that we never used openness of generic point in the proof.

ali
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  • Thank you for your answer. I know that DVR has only 2 points. Is this still true for a general valuation ring? I'm also not sure if $Spec(K)\rightarrow Spec(\mathcal O_K)$ is an open immersion. I don't think so but Liu seems to use this in the part of injectivity. Sorry for asking such basic questions. – Hydrogen Aug 30 '20 at 22:54
  • Yes a valuation ring has only two prime ideal zero and its mmaximal ideal i think the proof in the case of dvr works more or less for any valuation ring. Again beacuse we only jave to point the generic point spec K is open in spec O – ali Aug 31 '20 at 05:48
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    Liu is a good book but it is very algebraic i think it is good to sometimes look at other sources to get the geometric picture – ali Aug 31 '20 at 05:55
  • Thank you for your advice. In fact, I want to throw Liu's book these days. It drives me crazy. Could you recommend some extra materials, rich in geometric pictures. – Hydrogen Aug 31 '20 at 16:13
  • It seems that some counterexample can be given in general valuation rings. https://math.stackexchange.com/questions/3415695/valuation-rings-only-have-two-prime-ideals – Hydrogen Aug 31 '20 at 16:14
  • @Hydrogen you are right I was thinking about $\mathbb{R}$ valued valuation ring. but the geometric picture is similar in general/the generic point must go to $x$ the closed point must go to a point in the closure of $x$ in the fiber over s for example z and the liu's proof shows that you have a map from $Spec(O_K)\to spec(O_Z)$ – ali Aug 31 '20 at 17:28
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    https://math.stackexchange.com/questions/998/best-algebraic-geometry-text-book-other-than-hartshorne there are some good suggestions here – ali Aug 31 '20 at 17:31
  • Thank you. I think I understand the construction of underlying topogical map after reading your answer. But I'm still confused about the ring map. Why Qing constructs a ring map $Spec(\mathcal O_{Z,t})\rightarrow X$? Could you explain it with more details? I'm sorry this might be a basic question, but I'm not really familar with this. – Hydrogen Sep 01 '20 at 21:51
  • As $Spec(K)$ may not be an open subscheme of $Spec(\mathcal O_K)$, can we still regard this construction as a morphism extension in some sense? – Hydrogen Sep 01 '20 at 21:52
  • I updated my answer – ali Sep 02 '20 at 09:24
  • Your answer helps a lot. But I think I need more time to think about it. Anyway, thank you so much! – Hydrogen Sep 02 '20 at 17:49