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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}

    Comments (2)

    Comment #576 by Yogesh on May 19, 2014 a 3:39 pm UTC

    Possible minor typo: In the proof, end of the first paragraph: "Then we conclude that $g'g^{-1}$ fixes $u$" ...I think it should be $g^{-1}g'$ instead of $g'g^{-1}$, but I might be being stupid and they might both be okay.

    Comment #592 by Johan (site) on May 23, 2014 a 7:55 pm UTC

    Thanks! Fixed here.

    There are also 7 comments on Section 56.14: Algebraic Spaces.

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