## Tag `02Z2`

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

Lemma 56.14.3. Let $U \to S$ be a morphism of $\textit{Sch}_{fppf}$. 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 $\mathop{\rm Spec}(\kappa(t)) \to \mathop{\rm Spec}(\kappa(h(t))) \to U$. 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. $\square$

The code snippet corresponding to this tag is a part of the file `spaces.tex` and is located in lines 2216–2240 (see updates for more information).

```
\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}
```

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