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 étale. 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).

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