Lemma 100.46.2. Let $\mathcal{X}$ be an algebraic stack satisfying the equivalent conditions of Lemma 100.46.1. Then there exists an integral morphism of algebraic stacks

$\mathcal{X}^\nu \longrightarrow \mathcal{X}$

such that for every scheme $U$ and smooth morphism $U \to \mathcal{X}$ the fibre product $\mathcal{X}^\nu \times _\mathcal {X} U$ is the normalization of $U$.

Proof. Let $U \to \mathcal{X}$ be a surjective smooth morphism where $U$ is a scheme. Set $R = U \times _\mathcal {X} U$. Recall that we obtain a smooth groupoid $(U, R, s, t, c)$ in algebraic spaces and a presentation $\mathcal{X} = [U/R]$ of $\mathcal{X}$, see Algebraic Stacks, Lemmas 93.16.1 and 93.16.2 and Definition 93.16.5. The assumption on $\mathcal{X}$ means that the normalization $U^\nu$ of $U$ is defined, see Morphisms, Definition 29.54.1. By Morphisms of Spaces, Lemma 66.49.5 taking normalization commutes with smooth morphisms of algebraic spaces. Thus we see that the normalization $R^\nu$ of $R$ is isomorphic to both $R \times _{s, U} U^\nu$ and $U^\nu \times _{U, t} R$. Thus we obtain two smooth morphisms $s^\nu : R^\nu \to U^\nu$ and $t^\nu : R^\nu \to U^\nu$ of algebraic spaces. A formal computation with fibre products shows that $R^\nu \times _{s^\nu , U^\nu , t^\nu } R^\nu$ is the normalization of $R \times _{s, U, t} R$. Hence the smooth morphism $c : R \times _{s, U, t} R \to R$ extends to $c^\nu$ as well. Similarly, the inverse $i : R \to R$ (an isomorphism) induces an isomorphism $i^\nu : R^\nu \to R^\nu$. Finally, the identity $e : U \to R$ lifts to $e^\nu : U^\nu \to R^\nu$ for example because $e$ is a section of $s$ and $R^\nu = R \times _{U, s} U^\nu$. We claim that $(U^\nu , R^\nu , s^\nu , t^\nu , c^\nu )$ is a smooth groupoid in algebraic spaces. To see this involves checking the axioms (1), (2)(a), (2)(b), (3)(a), and (3)(b) of Groupoids, Section 39.13 for $(U^\nu , R^\nu , s^\nu , t^\nu , c^\nu , e^\nu , i^\nu )$. For example, for (1) we have to see that the two morphisms $a, b : R^\nu \times _{s^\nu , U^\nu , t^\nu } R^\nu \times _{s^\nu , U^\nu , t^\nu } R^\nu \to R^\nu$ we obtain are the same. This holds because we know that the corresponding pair of morphisms $R \times _{s, U, t} R \times _{s, U, t} R \to R$ are the same and the morphisms $a$ and $b$ are the unique extensions of this morphism to the normalizations. Similarly for the other axioms.

Consider the algebraic stack $\mathcal{X}^\nu = [U^\nu /R^\nu ]$ (Algebraic Stacks, Theorem 93.17.3). Since we have a morphism $(U^\nu , R^\nu , s^\nu , t^\nu , c^\nu ) \to (U, R, s, t, c)$ of groupoids in algebraic spaces, we obtain a morphism $\nu : \mathcal{X}^\nu \to \mathcal{X}$ of algebraic stacks. Since $R^\nu = R \times _{s, U} U^\nu$ we see that $U^\nu = \mathcal{X}^\nu \times _\mathcal {X} U$ by Groupoids in Spaces, Lemma 77.25.3. In particular, as $U^\nu \to U$ is integral, we see that $\nu$ is integral. We omit the verification that the base change property stated in the lemma holds for every smooth morphism from a scheme to $\mathcal{X}$. $\square$

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