Lemma 64.14.3. 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 64.10.5.

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 39.3.1). 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. $\square$

Comment #576 by Yogesh on

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.

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