The Stacks project

Example 63.14.9. Let $k$ be a field. Let $A = \prod _{n \in \mathbf{N}} k$ be the infinite product. Set $U = \mathop{\mathrm{Spec}}(A)$ seen as a scheme over $S = \mathop{\mathrm{Spec}}(k)$. Note that the projection maps $\text{pr}_ n : A \to k$ define open and closed immersions $f_ n : S \to U$. Set

\[ R = U \amalg \coprod \nolimits _{(n, m) \in \mathbf{N}^2, \ n \not= m} S \]

with morphism $j$ equal to $\Delta _{U/S}$ on the component $U$ and $j = (f_ n, f_ m)$ on the component $S$ corresponding to $(n, m)$. It is clear from the remark above that $s, t$ are ├ętale. It is also clear that $j$ is an equivalence relation. Hence we obtain an algebraic space

\[ X = U/R. \]

To see what this means we specialize to the case where the field $k$ is finite with $q$ elements. Let us first discuss the topological space $|U|$ associated to the scheme $U$ a little bit. All elements of $A$ satisfy $x^ q = x$. Hence every residue field of $A$ is isomorphic to $k$, and all points of $U$ are closed. But the topology on $U$ isn't the discrete topology. Let $u_ n \in |U|$ be the point corresponding to $f_ n$. As mentioned above the points $u_ n$ are the open points (and hence isolated). This implies there have to be other points since we know $U$ is quasi-compact, see Algebra, Lemma 10.17.10 (hence not equal to an infinite discrete set). Another way to see this is because the (proper) ideal

\[ I = \{ x = (x_ n) \in A \mid \text{all but a finite number of }x_ n\text{ are zero}\} \]

is contained in a maximal ideal. Note also that every element $x$ of $A$ is of the form $x = ue$ where $u$ is a unit and $e$ is an idempotent. Hence a basis for the topology of $A$ consists of open and closed subsets (see Algebra, Lemma 10.21.1.) So the topology on $|U|$ is totally disconnected, but nontrivial. Finally, note that $\{ u_ n\} $ is dense in $|U|$.

We will later define a topological space $|X|$ associated to $X$, see Properties of Spaces, Section 64.4. What can we say about $|X|$? It turns out that the map $|U| \to |X|$ is surjective and continuous. All the points $u_ n$ map to the same point $x_0$ of $|X|$, and none of the other points get identified. Since $\{ u_ n\} $ is dense in $|U|$ we conclude that the closure of $x_0$ in $|X|$ is $|X|$. In other words $|X|$ is irreducible and $x_0$ is a generic point of $|X|$. This seems bizarre since also $x_0$ is the image of a section $S \to X$ of the structure morphism $X \to S$ (and in the case of schemes this would imply it was a closed point, see Morphisms, Lemma 29.20.2).


Comments (0)

There are also:

  • 7 comment(s) on Section 63.14: Examples of algebraic spaces

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02Z8. Beware of the difference between the letter 'O' and the digit '0'.