# The Stacks Project

## Tag 02Z0

### 56.14. Examples of algebraic spaces

In this section we construct some examples of algebraic spaces. Some of these were suggested by B. Conrad. Since we do not yet have a lot of theory at our disposal the discussion is a bit awkward in some places.

Example 56.14.1. Let $k$ be a field of characteristic $\not = 2$. Let $U = \mathbf{A}^1_k$. Set $$j : R = \Delta \amalg \Gamma \longrightarrow U \times_k U$$ where $\Delta = \{(x, x) \mid x \in \mathbf{A}^1_k\}$ and $\Gamma = \{(x, -x) \mid x \in \mathbf{A}^1_k, x \not = 0\}$. It is clear that $s, t : R \to U$ are étale, and hence $j$ is an étale equivalence relation. The quotient $X = U/R$ is an algebraic space by Theorem 56.10.5. Since $R$ is quasi-compact we see that $X$ is quasi-separated. On the other hand, $X$ is not locally separated because the morphism $j$ is not an immersion.

Example 56.14.2. Let $k$ be a field. Let $k'/k$ be a degree $2$ Galois extension with $\text{Gal}(k'/k) = \{1, \sigma\}$. Let $S = \mathop{\mathrm{Spec}}(k[x])$ and $U = \mathop{\mathrm{Spec}}(k'[x])$. Note that $$U \times_S U = \mathop{\mathrm{Spec}}((k' \otimes_k k')[x]) = \Delta(U) \amalg \Delta'(U)$$ where $\Delta' = (1, \sigma) : U \to U \times_S U$. Take $$R = \Delta(U) \amalg \Delta'(U \setminus \{0_U\})$$ where $0_U \in U$ denotes the $k'$-rational point whose $x$-coordinate is zero. It is easy to see that $R$ is an étale equivalence relation on $U$ over $S$ and hence $X = U/R$ is an algebraic space by Theorem 56.10.5. Here are some properties of $X$ (some of which will not make sense until later):

1. $X \to S$ is an isomorphism over $S \setminus \{0_S\}$,
2. the morphism $X \to S$ is étale (see Properties of Spaces, Definition 57.15.2)
3. the fibre $0_X$ of $X \to S$ over $0_S$ is isomorphic to $\mathop{\mathrm{Spec}}(k') = 0_U$,
4. $X$ is not a scheme because if it where, then $\mathcal{O}_{X, 0_X}$ would be a local domain $(\mathcal{O}, \mathfrak m, \kappa)$ with fraction field $k(x)$, with $x \in \mathfrak m$ and residue field $\kappa = k'$ which is impossible,
5. $X$ is not separated, but it is locally separated and quasi-separated,
6. there exists a surjective, finite, étale morphism $S' \to S$ such that the base change $X' = S' \times_S X$ is a scheme (namely, if we base change to $S' = \mathop{\mathrm{Spec}}(k'[x])$ then $U$ splits into two copies of $S'$ and $X'$ becomes isomorphic to the affine line with $0$ doubled, see Schemes, Example 25.14.3), and
7. if we think of $X$ as a finite type algebraic space over $\mathop{\mathrm{Spec}}(k)$, then similarly the base change $X_{k'}$ is a scheme but $X$ is not a scheme.

In particular, this gives an example of a descent datum for schemes relative to the covering $\{\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)\}$ which is not effective.

See also Examples, Lemma 100.57.1, which shows that descent data need not be effective even for a projective morphism of schemes. That example gives a smooth separated algebraic space of dimension 3 over ${\mathbf C}$ which is not a scheme.

We will use the following lemma as a convenient way to construct algebraic spaces as quotients of schemes by free group actions.

Lemma 56.14.3. Let $U \to S$ be a morphism of $\mathit{Sch}_{fppf}$. Let $G$ be an abstract group. Let $G \to \text{Aut}_S(U)$ be a group homomorphism. Assume

• $(*)$    if $u \in U$ is a point, and $g(u) = u$ for some non-identity element $g \in G$, then $g$ induces a nontrivial automorphism of $\kappa(u)$.

Then $$j : R = \coprod\nolimits_{g \in G} U \longrightarrow U \times_S U, \quad (g, x) \longmapsto (g(x), x)$$ is an étale equivalence relation and hence $$F = U/R$$ is an algebraic space by Theorem 56.10.5.

Proof. In the statement of the lemma the symbol $\text{Aut}_S(U)$ denotes the group of automorphisms of $U$ over $S$. Assume $(*)$ holds. Let us show that $$j : R = \coprod\nolimits_{g \in G} U \longrightarrow U \times_S U, \quad (g, x) \longmapsto (g(x), x)$$ is a monomorphism. This signifies that if $T$ is a nonempty scheme, and $h : T \to U$ is a $T$-valued point such that $g \circ h = g' \circ h$ then $g = g'$. Suppose $T \not = \emptyset$, $h : T \to U$ and $g \circ h = g' \circ h$. Let $t \in T$. Consider the composition $\mathop{\mathrm{Spec}}(\kappa(t)) \to \mathop{\mathrm{Spec}}(\kappa(h(t))) \to U$. Then we conclude that $g^{-1} \circ g'$ fixes $u = h(t)$ and acts as the identity on its residue field. Hence $g = g'$ by $(*)$.

Thus if $(*)$ holds we see that $j$ is a relation (see Groupoids, Definition 38.3.1). Moreover, it is an equivalence relation since on $T$-valued points for a connected scheme $T$ we see that $R(T) = G \times U(T) \to U(T) \times U(T)$ (recall that we always work over $S$). Moreover, the morphisms $s, t : R \to U$ are étale since $R$ is a disjoint product of copies of $U$. This proves that $j : R \to U \times_S U$ is an étale equivalence relation. $\square$

Given a scheme $U$ and an action of a group $G$ on $U$ we say the action of $G$ on $U$ is free if condition $(*)$ of Lemma 56.14.3 holds. This is equivalent to the notion of a free action of the constant group scheme $G_S$ on $U$ as defined in Groupoids, Definition 38.10.2. The lemma can be interpreted as saying that quotients of schemes by free actions of groups exist in the category of algebraic spaces.

Definition 56.14.4. Notation $U \to S$, $G$, $R$ as in Lemma 56.14.3. If the action of $G$ on $U$ satisfies $(*)$ we say $G$ acts freely on the scheme $U$. In this case the algebraic space $U/R$ is denoted $U/G$ and is called the quotient of $U$ by $G$.

This notation is consistent with the notation $U/G$ introduced in Groupoids, Definition 38.20.1. We will later make sense of the quotient as an algebraic stack without any assumptions on the action whatsoever; when we do this we will use the notation $[U/G]$. Before we discuss the examples we prove some more lemmas to facilitate the discussion. Here is a lemma discussing the various separation conditions for this quotient when $G$ is finite.

Lemma 56.14.5. Notation and assumptions as in Lemma 56.14.3. Assume $G$ is finite. Then

1. if $U \to S$ is quasi-separated, then $U/G$ is quasi-separated over $S$, and
2. if $U \to S$ is separated, then $U/G$ is separated over $S$.

Proof. In the proof of Lemma 56.13.1 we saw that it suffices to prove the corresponding properties for the morphism $j : R \to U \times_S U$. If $U \to S$ is quasi-separated, then for every affine open $V \subset U$ which maps into an affine of $S$ the opens $g(V) \cap V$ are quasi-compact. It follows that $j$ is quasi-compact. If $U \to S$ is separated, the diagonal $\Delta_{U/S}$ is a closed immersion. Hence $j : R \to U \times_S U$ is a finite coproduct of closed immersions with disjoint images. Hence $j$ is a closed immersion. $\square$

Lemma 56.14.6. Notation and assumptions as in Lemma 56.14.3. If $\mathop{\mathrm{Spec}}(k) \to U/G$ is a morphism, then there exist

1. a finite Galois extension $k'/k$,
2. a finite subgroup $H \subset G$,
3. an isomorphism $H \to \text{Gal}(k'/k)$, and
4. an $H$-equivariant morphism $\mathop{\mathrm{Spec}}(k') \to U$.

Conversely, such data determine a morphism $\mathop{\mathrm{Spec}}(k) \to U/G$.

Proof. Consider the fibre product $V = \mathop{\mathrm{Spec}}(k) \times_{U/G} U$. Here is a diagram $$\xymatrix{ V \ar[r] \ar[d] & U \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r] & U/G }$$ Then $V$ is a nonempty scheme étale over $\mathop{\mathrm{Spec}}(k)$ and hence is a disjoint union $V = \coprod_{i \in I} \mathop{\mathrm{Spec}}(k_i)$ of spectra of fields $k_i$ finite separable over $k$ (Morphisms, Lemma 28.34.7). We have \begin{align*} V \times_{\mathop{\mathrm{Spec}}(k)} V & = (\mathop{\mathrm{Spec}}(k) \times_{U/G} U) \times_{\mathop{\mathrm{Spec}}(k)}(\mathop{\mathrm{Spec}}(k) \times_{U/G} U) \\ & = \mathop{\mathrm{Spec}}(k) \times_{U/G} U \times_{U/G} U \\ & = \mathop{\mathrm{Spec}}(k) \times_{U/G} U \times G \\ & = V \times G \end{align*} The action of $G$ on $U$ induces an action of $a : G \times V \to V$. The displayed equality means that $G \times V \to V \times_{\mathop{\mathrm{Spec}}(k)} V$, $(g, v) \mapsto (a(g, v), v)$ is an isomorphism. In particular we see that for every $i$ we have an isomorphism $H_i \times \mathop{\mathrm{Spec}}(k_i) \to \mathop{\mathrm{Spec}}(k_i \otimes_k k_i)$ where $H_i \subset G$ is the subgroup of elements fixing $i \in I$. Thus $H_i$ is finite and is the Galois group of $k_i/k$. We omit the converse construction. $\square$

It follows from this lemma for example that if $k'/k$ is a finite Galois extension, then $\mathop{\mathrm{Spec}}(k')/\text{Gal}(k'/k) \cong \mathop{\mathrm{Spec}}(k)$. What happens if the extension is infinite? Here is an example.

Example 56.14.7. Let $S = \mathop{\mathrm{Spec}}(\mathbf{Q})$. Let $U = \mathop{\mathrm{Spec}}(\overline{\mathbf{Q}})$. Let $G = \text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ with obvious action on $U$. Then by construction property $(*)$ of Lemma 56.14.3 holds and we obtain an algebraic space $$X = \mathop{\mathrm{Spec}}(\overline{\mathbf{Q}})/G \longrightarrow S = \mathop{\mathrm{Spec}}(\mathbf{Q}).$$ Of course this is totally ridiculous as an approximation of $S$! Namely, by the Artin-Schreier theorem, see [JacobsonIII, Theorem 17, page 316], the only finite subgroups of $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ are $\{1\}$ and the conjugates of the order two group $\text{Gal}(\overline{\mathbf{Q}}/\overline{\mathbf{Q}} \cap \mathbf{R})$. Hence, if $\mathop{\mathrm{Spec}}(k) \to X$ is a morphism with $k$ algebraic over $\mathbf{Q}$, then it follows from Lemma 56.14.6 and the theorem just mentioned that either $k$ is $\overline{\mathbf{Q}}$ or isomorphic to $\overline{\mathbf{Q}} \cap \mathbf{R}$.

What is wrong with the example above is that the Galois group comes equipped with a topology, and this should somehow be part of any construction of a quotient of $\mathop{\mathrm{Spec}}(\overline{\mathbf{Q}})$. The following example is much more reasonable in my opinion and may actually occur in ''nature''.

Example 56.14.8. Let $k$ be a field of characteristic zero. Let $U = \mathbf{A}^1_k$ and let $G = \mathbf{Z}$. As action we take $n(x) = x + n$, i.e., the action of $\mathbf{Z}$ on the affine line by translation. The only fixed point is the generic point and it is clearly the case that $\mathbf{Z}$ injects into the automorphism group of the field $k(x)$. (This is where we use the characteristic zero assumption.) Consider the morphism $$\gamma : \mathop{\mathrm{Spec}}(k(x)) \longrightarrow X = \mathbf{A}^1_k/\mathbf{Z}$$ of the generic point of the affine line into the quotient. We claim that this morphism does not factor through any monomorphism $\mathop{\mathrm{Spec}}(L) \to X$ of the spectrum of a field to $X$. (Contrary to what happens for schemes, see Schemes, Section 25.13.) In fact, since $\mathbf{Z}$ does not have any nontrivial finite subgroups we see from Lemma 56.14.6 that for any such factorization $k(x) = L$. Finally, $\gamma$ is not a monomorphism since $$\mathop{\mathrm{Spec}}(k(x)) \times_{\gamma, X, \gamma} \mathop{\mathrm{Spec}}(k(x)) \cong \mathop{\mathrm{Spec}}(k(x)) \times \mathbf{Z}.$$

This example suggests that in order to define points of an algebraic space $X$ we should consider equivalence classes of morphisms from spectra of fields into $X$ and not the set of monomorphisms from spectra of fields.

We finish with a truly awful example.

Example 56.14.9. 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 10.16.10 (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).

Whatever you think is actually going on in this example, it certainly shows that some care has to be exercised when defining irreducible components, connectedness, etc of algebraic spaces.

The code snippet corresponding to this tag is a part of the file spaces.tex and is located in lines 2131–2516 (see updates for more information).

\section{Examples of algebraic spaces}
\label{section-examples}

\noindent
In this section we construct some examples of algebraic spaces.
Some of these were suggested by B.\ Conrad.
Since we do not yet have a lot of theory at our disposal the
discussion is a bit awkward in some places.

\begin{example}
\label{example-affine-line-involution}
Let $k$ be a field of characteristic $\not = 2$. Let $U = \mathbf{A}^1_k$. Set
$$j : R = \Delta \amalg \Gamma \longrightarrow U \times_k U$$
where $\Delta = \{(x, x) \mid x \in \mathbf{A}^1_k\}$ and
$\Gamma = \{(x, -x) \mid x \in \mathbf{A}^1_k, x \not = 0\}$.
It is clear that $s, t : R \to U$ are \'etale, and hence
$j$ is an \'etale equivalence relation.
The quotient $X = U/R$ is an algebraic space by
Theorem \ref{theorem-presentation}.
Since $R$ is quasi-compact we see that $X$ is quasi-separated.
On the other hand, $X$ is not locally separated because
the morphism $j$ is not an immersion.
\end{example}

\begin{example}
\label{example-non-representable-descent}
Let $k$ be a field. Let $k'/k$ be a degree $2$ Galois extension
with $\text{Gal}(k'/k) = \{1, \sigma\}$. Let $S = \Spec(k[x])$
and $U = \Spec(k'[x])$. Note that
$$U \times_S U = \Spec((k' \otimes_k k')[x]) = \Delta(U) \amalg \Delta'(U)$$
where $\Delta' = (1, \sigma) : U \to U \times_S U$. Take
$$R = \Delta(U) \amalg \Delta'(U \setminus \{0_U\})$$
where $0_U \in U$ denotes the $k'$-rational point whose $x$-coordinate is zero.
It is easy to see that $R$ is an \'etale equivalence relation on $U$ over $S$
and hence $X = U/R$ is an algebraic space by
Theorem \ref{theorem-presentation}. Here are some properties of $X$ (some
of which will not make sense until later):
\begin{enumerate}
\item $X \to S$ is an isomorphism over $S \setminus \{0_S\}$,
\item the morphism $X \to S$ is \'etale (see
Properties of Spaces,
Definition \ref{spaces-properties-definition-etale})
\item the fibre $0_X$ of $X \to S$ over $0_S$ is isomorphic to
$\Spec(k') = 0_U$,
\item $X$ is not a scheme because if it where, then $\mathcal{O}_{X, 0_X}$
would be a local domain $(\mathcal{O}, \mathfrak m, \kappa)$ with
fraction field $k(x)$, with $x \in \mathfrak m$ and residue field
$\kappa = k'$ which is impossible,
\item $X$ is not separated, but it is
locally separated and quasi-separated,
\item there exists a surjective, finite, \'etale morphism $S' \to S$
such that the base change $X' = S' \times_S X$ is a scheme (namely, if
we base change to $S' = \Spec(k'[x])$ then $U$ splits into
two copies of $S'$ and $X'$ becomes isomorphic to the affine line with
$0$ doubled, see
Schemes, Example \ref{schemes-example-affine-space-zero-doubled}), and
\item if we think of $X$ as a finite type algebraic space over
$\Spec(k)$, then similarly the base change $X_{k'}$ is a scheme
but $X$ is not a scheme.
\end{enumerate}
In particular, this gives an example of a descent datum for schemes
relative to the covering $\{\Spec(k') \to \Spec(k)\}$
which is not effective.
\end{example}

\noindent
Lemma \ref{examples-lemma-non-effective-descent-projective},
which shows that descent data need not be effective
even for a projective morphism of schemes. That example
gives a smooth separated algebraic space of dimension 3
over ${\mathbf C}$ which is not a scheme.

\medskip\noindent
We will use the following lemma as a convenient way to construct
algebraic spaces as quotients of schemes by free group actions.

\begin{lemma}
\label{lemma-quotient}
Let $U \to S$ be a morphism of $\Sch_{fppf}$.
Let $G$ be an abstract group. Let $G \to \text{Aut}_S(U)$
be a group homomorphism. Assume
\begin{itemize}
\item[$(*)$] if $u \in U$ is a point, and $g(u) = u$
for some non-identity element $g \in G$, then $g$
induces a nontrivial automorphism of $\kappa(u)$.
\end{itemize}
Then
$$j : R = \coprod\nolimits_{g \in G} U \longrightarrow U \times_S U, \quad (g, x) \longmapsto (g(x), x)$$
is an \'etale equivalence relation and hence
$$F = U/R$$
is an algebraic space by Theorem \ref{theorem-presentation}.
\end{lemma}

\begin{proof}
In the statement of the lemma the symbol $\text{Aut}_S(U)$ denotes
the group of automorphisms of $U$ over $S$.
Assume $(*)$ holds. Let us show that
$$j : R = \coprod\nolimits_{g \in G} U \longrightarrow U \times_S U, \quad (g, x) \longmapsto (g(x), x)$$
is a monomorphism. This signifies that if $T$ is a nonempty
scheme, and $h : T \to U$ is a $T$-valued point such that
$g \circ h = g' \circ h$ then $g = g'$. Suppose
$T \not = \emptyset$, $h : T \to U$ and $g \circ h = g' \circ h$.
Let $t \in T$. Consider the composition
$\Spec(\kappa(t)) \to \Spec(\kappa(h(t))) \to U$.
Then we conclude that $g^{-1} \circ g'$ fixes $u = h(t)$ and
acts as the identity on its residue field. Hence $g = g'$ by $(*)$.

\medskip\noindent
Thus if $(*)$ holds we see that $j$ is a relation (see
Groupoids, Definition \ref{groupoids-definition-equivalence-relation}).
Moreover, it is an equivalence relation since on $T$-valued points
for a connected scheme $T$ we see that
$R(T) = G \times U(T) \to U(T) \times U(T)$ (recall that we always
work over $S$). Moreover, the morphisms $s, t : R \to U$ are \'etale
since $R$ is a disjoint product of copies of $U$.
This proves that $j : R \to U \times_S U$ is an \'etale equivalence relation.
\end{proof}

\noindent
Given a scheme $U$ and an action of a group $G$ on $U$ we say the action
of $G$ on $U$ is {\it free} if condition $(*)$ of Lemma \ref{lemma-quotient}
holds. This is equivalent to the notion of a free action of the constant
group scheme $G_S$ on $U$ as defined in
Groupoids, Definition \ref{groupoids-definition-free-action}.
The lemma can be interpreted as saying that quotients of schemes by
free actions of groups exist in the category of algebraic spaces.

\begin{definition}
\label{definition-quotient}
Notation $U \to S$, $G$, $R$ as in Lemma \ref{lemma-quotient}.
If the action of $G$ on $U$ satisfies $(*)$ we say $G$ {\it acts freely}
on the scheme $U$. In this case the algebraic space $U/R$ is denoted
$U/G$ and is called the {\it quotient of $U$ by $G$}.
\end{definition}

\noindent
This notation is consistent with the notation $U/G$ introduced in
Groupoids, Definition \ref{groupoids-definition-quotient-sheaf}.
We will later make sense of the quotient as an algebraic stack without
any assumptions on the action whatsoever; when we do this we will use the
notation $[U/G]$. Before we discuss the examples we prove
some more lemmas to facilitate the discussion. Here is a lemma discussing the
various separation conditions for this quotient when $G$ is finite.

\begin{lemma}
\label{lemma-quotient-finite-separated}
Notation and assumptions as in Lemma \ref{lemma-quotient}.
Assume $G$ is finite. Then
\begin{enumerate}
\item if $U \to S$ is quasi-separated, then $U/G$ is quasi-separated
over $S$, and
\item if $U \to S$ is separated, then $U/G$ is separated over $S$.
\end{enumerate}
\end{lemma}

\begin{proof}
In the proof of Lemma \ref{lemma-properties-diagonal}
we saw that it suffices to prove the
corresponding properties for the morphism $j : R \to U \times_S U$.
If $U \to S$ is quasi-separated, then for every affine open $V \subset U$
which maps into an affine of $S$
the opens $g(V) \cap V$ are quasi-compact. It follows that $j$ is
quasi-compact.
If $U \to S$ is separated, the diagonal $\Delta_{U/S}$ is a closed
immersion. Hence $j : R \to U \times_S U$ is a finite coproduct
of closed immersions with disjoint images. Hence $j$ is a closed immersion.
\end{proof}

\begin{lemma}
\label{lemma-quotient-field-map}
Notation and assumptions as in Lemma \ref{lemma-quotient}.
If $\Spec(k) \to U/G$ is a morphism, then there exist
\begin{enumerate}
\item a finite Galois extension $k'/k$,
\item a finite subgroup $H \subset G$,
\item an isomorphism $H \to \text{Gal}(k'/k)$, and
\item an $H$-equivariant morphism $\Spec(k') \to U$.
\end{enumerate}
Conversely, such data determine a morphism $\Spec(k) \to U/G$.
\end{lemma}

\begin{proof}
Consider the fibre product $V = \Spec(k) \times_{U/G} U$.
Here is a diagram
$$\xymatrix{ V \ar[r] \ar[d] & U \ar[d] \\ \Spec(k) \ar[r] & U/G }$$
Then $V$ is a nonempty scheme \'etale over $\Spec(k)$ and hence is a
disjoint union $V = \coprod_{i \in I} \Spec(k_i)$
of spectra of fields $k_i$ finite separable over $k$
(Morphisms, Lemma \ref{morphisms-lemma-etale-over-field}).
We have
\begin{align*}
V \times_{\Spec(k)} V
& =
(\Spec(k) \times_{U/G} U) \times_{\Spec(k)}(\Spec(k) \times_{U/G} U) \\
& =
\Spec(k) \times_{U/G} U \times_{U/G} U \\
& =
\Spec(k) \times_{U/G} U \times G \\
& =
V \times G
\end{align*}
The action of $G$ on $U$ induces an action of $a : G \times V \to V$.
The displayed equality means that
$G \times V \to V \times_{\Spec(k)} V$, $(g, v) \mapsto (a(g, v), v)$
is an isomorphism. In particular we see that for every $i$ we have
an isomorphism $H_i \times \Spec(k_i) \to \Spec(k_i \otimes_k k_i)$
where $H_i \subset G$ is the subgroup of elements fixing $i \in I$.
Thus $H_i$ is finite and is the Galois group of $k_i/k$.
We omit the converse construction.
\end{proof}

\noindent
It follows from this lemma for example that
if $k'/k$ is a finite Galois extension, then
$\Spec(k')/\text{Gal}(k'/k) \cong \Spec(k)$.
What happens if the extension is infinite? Here is an example.

\begin{example}
\label{example-Qbar}
Let $S = \Spec(\mathbf{Q})$.
Let $U = \Spec(\overline{\mathbf{Q}})$.
Let $G = \text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ with obvious
action on $U$. Then by construction property $(*)$ of
Lemma \ref{lemma-quotient} holds and we obtain an algebraic space
$$X = \Spec(\overline{\mathbf{Q}})/G \longrightarrow S = \Spec(\mathbf{Q}).$$
Of course this is totally ridiculous as an approximation of $S$!
Namely, by the Artin-Schreier theorem,
see \cite[Theorem 17, page 316]{JacobsonIII},
the only finite subgroups of $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$
are $\{1\}$ and the conjugates of the order two group
$\text{Gal}(\overline{\mathbf{Q}}/\overline{\mathbf{Q}} \cap \mathbf{R})$.
Hence, if
$\Spec(k) \to X$ is a morphism with $k$ algebraic over $\mathbf{Q}$,
then it follows from Lemma \ref{lemma-quotient-field-map} and the theorem
just mentioned that either $k$ is $\overline{\mathbf{Q}}$ or isomorphic to
$\overline{\mathbf{Q}} \cap \mathbf{R}$.
\end{example}

\noindent
What is wrong with the example above is that
the Galois group comes equipped with a topology,
and this should somehow be part of any construction
of a quotient of $\Spec(\overline{\mathbf{Q}})$.
The following example is much more reasonable in my opinion
and may actually occur in nature''.

\begin{example}
\label{example-affine-line-translation}
Let $k$ be a field of characteristic zero.
Let $U = \mathbf{A}^1_k$ and let $G = \mathbf{Z}$.
As action we take $n(x) = x + n$, i.e., the action of
$\mathbf{Z}$ on the affine line by translation.
The only fixed point is the generic point and it
is clearly the case that $\mathbf{Z}$ injects into
the automorphism group of the field $k(x)$. (This is
where we use the characteristic zero assumption.)
Consider the morphism
$$\gamma : \Spec(k(x)) \longrightarrow X = \mathbf{A}^1_k/\mathbf{Z}$$
of the generic point of the affine line into the quotient.
We claim that this morphism does not factor through any
monomorphism $\Spec(L) \to X$ of the spectrum of
a field to $X$. (Contrary to what happens for schemes, see
Schemes, Section \ref{schemes-section-points}.) In fact, since
$\mathbf{Z}$ does not have any nontrivial finite subgroups we see from
Lemma \ref{lemma-quotient-field-map} that for any such
factorization $k(x) = L$. Finally, $\gamma$ is not a monomorphism
since
$$\Spec(k(x)) \times_{\gamma, X, \gamma} \Spec(k(x)) \cong \Spec(k(x)) \times \mathbf{Z}.$$
\end{example}

\noindent
This example suggests that in order to define points of an algebraic space
$X$ we should consider equivalence classes of morphisms from spectra
of fields into $X$ and not the set of monomorphisms from spectra of fields.

\medskip\noindent
We finish with a truly awful example.

\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}

\noindent
Whatever you think is actually going on in this example, it certainly
shows that some care has to be exercised when defining irreducible
components, connectedness, etc of algebraic spaces.

Comment #1408 by yogesh more on April 15, 2015 a 5:52 pm UTC

In Lemma 48.14.3, does G have to be finite? I ask because in the proof it says $s,t:R \to U$ are etale, where $R=G \times U$. But maybe I'm just missing something.

Comment #1416 by Johan (site) on April 15, 2015 a 7:01 pm UTC

@#1408: Where are we assuming that $G$ is finite? I do not think we are. So what is your question?

Comment #1420 by yogesh more on April 16, 2015 a 9:21 pm UTC

I guess I'm being dumb, sorry. I thought etale map had to have quasifinite fibers, and the map t has fibers the cardinality of G because it is the projection $G \times U \to U$ (maybe this is my mistake?), and the lemma asserts t is etale.

Comment #1434 by Johan (site) on April 22, 2015 a 1:12 pm UTC

Nope, being etale is local on the source, so infinite disjoint unions of a scheme are etale over that scheme. Cheers!

Comment #1447 by yogesh more on May 5, 2015 a 6:42 am UTC

Ok, thanks, I see my mistake. The map on rings, \emph{after} localizing on the source scheme, is what is quasifinite; the original map between schemes need not be quasi-finite. Thanks!

Comment #2476 by Jiayu Zhao on April 7, 2017 a 2:55 am UTC

In Lemma 14.6, why the extension $k_i/k$ become Galois? Also in example 14.8, I'm not sure where you can use Lemma 14.6, I mean if the morphism $\gamma:Spec(k(x))\to X$ functor through some $Spec(L)\to X$, then $L$ should be a subfield of $k(x)$, not an extension.

Comment #2508 by Johan (site) on April 14, 2017 a 12:28 am UTC

OK, I tried to clarify. Changes are here. Can you look again?

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