# Proof of $X(\mathcal{O}_K)\simeq X_K(K)$

I have problems to understand a proof of the following theorem (Algebraic Geometry and Arithmetic Curves, Qing Liu, Theo 3.3.25, page 107).

Theorem: let $\mathcal{O}_K$ be a valuation ring over $K$, $X$ a proper $\mathcal{O}_K$-scheme. Then the canonical map $X(\mathcal{O}_K)\to X_K(K)$ is bijective.

If I have well understand the canonical map associate to $\varphi:\mathrm{Spec}\mathcal({O}_K)\to K$ the map $\varphi_K:\mathrm{Spec}(K)\to X_K$ base change of $\varphi$ under $\mathrm{Spec}(K)\to\mathrm{Spec}(\mathcal{O}_K)$.

The proof is going so:

Proof: injectivity: ok

surjectivity: let $\pi:\mathrm{Spec}(K)\to X_K\in X_K(K)$. Let $x=\pi((0))\in X_K$.

Let $Z=\overline{\{x\}}\subseteq X$: problem 1, why can we suppose that $x\in X$? Should we reasonning on $y$ image of $x$ by $X_K\to X$?

We endowed $Z$ with the structure of reduced closed subscheme, so $Z$ is integral: ok

The $\mathcal{O}_K$-scheme $Z$ is proper: ok

The point $x$ is closed in $X_K$: ok

The point $x$ is dense in $Z_K$: problem 2, why? I don't have the beaginning of a explanation.

Then $Z_K=\{x\}$: ok

Image of $Z\to\mathrm{Spec}(\mathcal{O}_K)$ is $\mathrm{Spec}(\mathcal{O}_K)$: ok

Let $t\in Z_\mathfrak{m}$ ($\mathfrak{m}\subseteq\mathcal{O}_K$ is the maximal ideal). $\mathcal{O}_{Z,t}$ is dominating $\mathcal{O}_K$: ok

The field of fraction of $\mathcal{O}_{Z,t}$ is $\mathcal{O}_{Z,x}$ and $\mathcal{O}_{Z,x}=K$: problem 3, why $\mathcal{O}_{Z,x}=K$? It should be linked with $\{x\}=Z_K$ but how?

Then $\mathcal{O}_{Z,t}=\mathcal{O}_K$ and so we have $\mathrm{Spec}(\mathcal{O}_K)\to X$: ok

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Problem 1: $X_K/K$ is the generic fiber of $X\to Spec(\mathcal{O}_K)$, so just include $x$ into the whole space: $X_K\hookrightarrow X$. – Matt Jan 1 '13 at 16:50
Ok I understand: problem 1 solved. Thanks – Macadam Jan 1 '13 at 20:14

Problem 2: $x$ is dense in $Z$, so is a fortiori dense in $Z_K$.
Problem 3: As $x$ is the generic point of $Z$, $O_{Z,x}$ is equal to the residue field of $Z$ at $x$. As $x$ is a rational point of $X_K/K$, this residue field is $K$. Another way to see this is $Z\to \mathrm{Spec}(O_K)$ is birational.
Problem 2: ok I understand. Details: we can consider $Z_K\subseteq X_K\subset X$ because $X_K\hookrightarrow X$ (Matt comment) and $Z_K\hookrightarrow Z$ (same argument) and $Z_K\hookrightarrow X_K$ (stability of closed immersions under base change). Other point, $x\in Z_K$ because $Z_K$ is the fiber of $Z$ over $(0)\in\mathcal{O}_K$ and that $x$ is above $(0)$ by definition. – Macadam Jan 2 '13 at 15:08
Problem 3: ok I understand. Details: As $X_K\hookrightarrow X$ open immersion then $\mathcal{O}_{X_K,x}=\mathcal{O}_{X,x}$ so $\mathrm{Frac}(\mathcal{O}_{X,x})=\mathrm{Frac}(\mathcal{O}_{X_K,x})=K$ (because $x$ is a rationnal point of $X_K/K$). But $Z\hookrightarrow X$ is a closed immersion so $\mathcal{O}_{X,x}\twoheadrightarrow\mathcal{O}_{Z,x}$ and then $\mathcal{O}_{Z,x}\simeq\mathrm{Frac}(\mathcal{O}_{X,x})\simeq K$. – Macadam Jan 2 '13 at 15:19
@GabrielSoranzo: for Problem 2, you don't need to bother with immersions. Any fiber $X_y$ is canonically a topological subspace of $X$. – user18119 Jan 2 '13 at 17:02