Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

Thank you for your help

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.