Lemma 101.50.2. Let \mathcal{X} be an integral algebraic stack. Then
|\mathcal{X}| is sober, irreducible, and has a unique generic point,
there exists an open substack \mathcal{U} \subset \mathcal{X} which is a gerbe over an integral scheme U.
Lemma 101.50.2. Let \mathcal{X} be an integral algebraic stack. Then
|\mathcal{X}| is sober, irreducible, and has a unique generic point,
there exists an open substack \mathcal{U} \subset \mathcal{X} which is a gerbe over an integral scheme U.
Proof. Proposition 101.49.3 tells us that |\mathcal{X}| is sober. Of course it is also irreducible and hence has a unique generic point x (by the definition of sobriety). Proposition 101.29.1 shows the existence of a dense open \mathcal{U} \subset \mathcal{X} which is a gerbe over an algebraic space U. Then U is a decent algebraic space by Lemma 101.48.5 (and the fact that \mathcal{U} is decent by Lemma 101.48.3). Since |U| = |\mathcal{U}| we see that |U| is irreducible. Finally, since \mathcal{U} is reduced the morphism \mathcal{U} \to U factors through U_{red}, see Properties of Stacks, Lemma 100.10.3. Now since \mathcal{U} \to U is flat, locally of finite presentation, and surjective (Lemma 101.28.8), this implies that U = U_{red}, i.e., U is reduced (small detail omitted). It follows that U is an integral algebraic space, see Spaces over Fields, Definition 72.4.1. Then finally, we may replace U (and correspondingly \mathcal{U}) by an open subspace and assume that U is an integral scheme, see discussion in Spaces over Fields, Section 72.4. \square
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