The Stacks project

Example 26.11.9. There exists a scheme without closed points. Namely, let $R$ be a local domain whose spectrum looks like $(0) = \mathfrak p_0 \subset \mathfrak p_1 \subset \mathfrak p_2 \subset \ldots \subset \mathfrak m$. Then the open subscheme $\mathop{\mathrm{Spec}}(R) \setminus \{ \mathfrak m\}$ does not have a closed point. To see that such a ring $R$ exists, we use that given any totally ordered group $(\Gamma , \geq )$ there exists a valuation ring $A$ with valuation group $(\Gamma , \geq )$, see [Krull]. See Algebra, Section 10.50 for notation. We take $\Gamma = \mathbf{Z}x_1 \oplus \mathbf{Z}x_2 \oplus \mathbf{Z}x_3 \oplus \ldots$ and we define $\sum _ i a_ i x_ i \geq 0$ if and only if the first nonzero $a_ i$ is $> 0$, or all $a_ i = 0$. So $x_1 \geq x_2 \geq x_3 \geq \ldots \geq 0$. The subsets $x_ i + \Gamma _{\geq 0}$ are prime ideals of $(\Gamma , \geq )$, see Algebra, notation above Lemma 10.50.17. These together with $\emptyset$ and $\Gamma _{\geq 0}$ are the only prime ideals. Hence $A$ is an example of a ring with the given structure of its spectrum, by Algebra, Lemma 10.50.17.