39.22 Separation conditions
This really means conditions on the morphism $j : R \to U \times _ S U$ when given a groupoid $(U, R, s, t, c)$ over $S$. As in the previous section we first formulate the corresponding diagram.
Lemma 39.22.1. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid over $S$. Let $G \to U$ be the stabilizer group scheme. The commutative diagram
\[ \xymatrix{ R \ar[d]^{\Delta _{R/U \times _ S U}} \ar[rrr]_{f \mapsto (f, s(f))} & & & R \times _{s, U} U \ar[d] \ar[r] & U \ar[d] \\ R \times _{(U \times _ S U)} R \ar[rrr]^{(f, g) \mapsto (f, f^{-1} \circ g)} & & & R \times _{s, U} G \ar[r] & G } \]
the two left horizontal arrows are isomorphisms and the right square is a fibre product square.
Proof.
Omitted. Exercise in the definitions and the functorial point of view in algebraic geometry.
$\square$
Lemma 39.22.2. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid over $S$. Let $G \to U$ be the stabilizer group scheme.
The following are equivalent
$j : R \to U \times _ S U$ is separated,
$G \to U$ is separated, and
$e : U \to G$ is a closed immersion.
The following are equivalent
$j : R \to U \times _ S U$ is quasi-separated,
$G \to U$ is quasi-separated, and
$e : U \to G$ is quasi-compact.
Proof.
The group scheme $G \to U$ is the base change of $R \to U \times _ S U$ by the diagonal morphism $U \to U \times _ S U$, see Lemma 39.17.1. Hence if $j$ is separated (resp. quasi-separated), then $G \to U$ is separated (resp. quasi-separated). (See Schemes, Lemma 26.21.12). Thus (a) $\Rightarrow $ (b) in both (1) and (2).
If $G \to U$ is separated (resp. quasi-separated), then the morphism $U \to G$, as a section of the structure morphism $G \to U$ is a closed immersion (resp. quasi-compact), see Schemes, Lemma 26.21.11. Thus (b) $\Rightarrow $ (a) in both (1) and (2).
By the result of Lemma 39.22.1 (and Schemes, Lemmas 26.18.2 and 26.19.3) we see that if $e$ is a closed immersion (resp. quasi-compact) $\Delta _{R/U \times _ S U}$ is a closed immersion (resp. quasi-compact). Thus (c) $\Rightarrow $ (a) in both (1) and (2).
$\square$
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