## Tag `02Z8`

Chapter 56: Algebraic Spaces > Section 56.14: Examples of algebraic spaces

Example 56.14.9. Let $k$ be a field. Let $A = \prod_{n \in \mathbf{N}} k$ be the infinite product. Set $U = \mathop{\rm Spec}(A)$ seen as a scheme over $S = \mathop{\rm Spec}(k)$. 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 10.16.10 (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 10.20.1.) 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 57.4. 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 28.19.2).

The code snippet corresponding to this tag is a part of the file `spaces.tex` and is located in lines 2449–2505 (see updates for more information).

```
\begin{example}
\label{example-infinite-product}
Let $k$ be a field.
Let $A = \prod_{n \in \mathbf{N}} k$ be the infinite product.
Set $U = \Spec(A)$ seen as a scheme over $S = \Spec(k)$.
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 \'etale.
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 \ref{algebra-lemma-quasi-compact}
(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 \ref{algebra-lemma-idempotent-spec}.)
So the topology on $|U|$ is totally disconnected, but nontrivial.
Finally, note that $\{u_n\}$ is dense in $|U|$.
\medskip\noindent
We will later define a topological space $|X|$ associated to $X$, see
Properties of Spaces, Section \ref{spaces-properties-section-points}.
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
\ref{morphisms-lemma-algebraic-residue-field-extension-closed-point-fibre}).
\end{example}
```

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