Lemma 100.10.1. Let \mathcal{X} be an algebraic stack. Let T \subset |\mathcal{X}| be a closed subset. There exists a unique closed substack \mathcal{Z} \subset \mathcal{X} with the following properties: (a) we have |\mathcal{Z}| = T, and (b) \mathcal{Z} is reduced.
100.10 Reduced algebraic stacks
We have already defined reduced algebraic stacks in Section 100.7.
Proof. Let U \to \mathcal{X} be a surjective smooth morphism, where U is an algebraic space. Set R = U \times _\mathcal {X} U, so that there is a presentation [U/R] \to \mathcal{X}, see Algebraic Stacks, Lemma 94.16.2. As usual we denote s, t : R \to U the two smooth projection morphisms. By Lemma 100.4.5 we see that T corresponds to a closed subset T' \subset |U| such that |s|^{-1}(T') = |t|^{-1}(T'). Let Z \subset U be the reduced induced algebraic space structure on T', see Properties of Spaces, Definition 66.12.5. The fibre products Z \times _{U, t} R and R \times _{s, U} Z are closed subspaces of R (Spaces, Lemma 65.12.3). The projections Z \times _{U, t} R \to Z and R \times _{s, U} Z \to Z are smooth by Morphisms of Spaces, Lemma 67.37.3. Thus as Z is reduced, it follows that Z \times _{U, t} R and R \times _{s, U} Z are reduced, see Remark 100.7.3. Since
|Z \times _{U, t} R| = |t|^{-1}(T') = |s|^{-1}(T') = R \times _{s, U} Z
we conclude from the uniqueness in Properties of Spaces, Lemma 66.12.3 that Z \times _{U, t} R = R \times _{s, U} Z. Hence Z is an R-invariant closed subspace of U. By the correspondence of Lemma 100.9.11 we obtain a closed substack \mathcal{Z} \subset \mathcal{X} with Z = \mathcal{Z} \times _\mathcal {X} U. Then [Z/R_ Z] \to \mathcal{Z} is a presentation (Lemma 100.9.7). Then |\mathcal{Z}| = |Z|/|R_ Z| = |T'|/\sim is the given closed subset T. We omit the proof of unicity. \square
Lemma 100.10.2. Let \mathcal{X} be an algebraic stack. If \mathcal{X}' \subset \mathcal{X} is a closed substack, \mathcal{X} is reduced and |\mathcal{X}'| = |\mathcal{X}|, then \mathcal{X}' = \mathcal{X}.
Proof. Choose a presentation [U/R] \to \mathcal{X} with U a scheme. As \mathcal{X} is reduced, we see that U is reduced (by definition of reduced algebraic stacks). By Lemma 100.9.11 \mathcal{X}' corresponds to an R-invariant closed subscheme Z \subset U. But now |Z| \subset |U| is the inverse image of |\mathcal{X}'|, and hence |Z| = |U|. Hence Z is a closed subscheme of U whose underlying sets of points agree. By Schemes, Lemma 26.12.7 the map \text{id}_ U : U \to U factors through Z \to U, and hence Z = U, i.e., \mathcal{X}' = \mathcal{X}. \square
Lemma 100.10.3. Let \mathcal{X}, \mathcal{Y} be algebraic stacks. Let \mathcal{Z} \subset \mathcal{X} be a closed substack Assume \mathcal{Y} is reduced. A morphism f : \mathcal{Y} \to \mathcal{X} factors through \mathcal{Z} if and only if f(|\mathcal{Y}|) \subset |\mathcal{Z}|.
Proof. Assume f(|\mathcal{Y}|) \subset |\mathcal{Z}|. Consider \mathcal{Y} \times _\mathcal {X} \mathcal{Z} \to \mathcal{Y}. There is an equivalence \mathcal{Y} \times _\mathcal {X} \mathcal{Z} \to \mathcal{Y}' where \mathcal{Y}' is a closed substack of \mathcal{Y}, see Lemmas 100.9.2 and 100.9.10. Using Lemmas 100.4.3, 100.8.5, and 100.9.5 we see that |\mathcal{Y}'| = |\mathcal{Y}|. Hence we have reduced the lemma to Lemma 100.10.2. \square
Definition 100.10.4. Let \mathcal{X} be an algebraic stack. Let Z \subset |\mathcal{X}| be a closed subset. An algebraic stack structure on Z is given by a closed substack \mathcal{Z} of \mathcal{X} with |\mathcal{Z}| equal to Z. The reduced induced algebraic stack structure on Z is the one constructed in Lemma 100.10.1. The reduction \mathcal{X}_{red} of \mathcal{X} is the reduced induced algebraic stack structure on |\mathcal{X}|.
In fact we can use this to define the reduced induced algebraic stack structure on a locally closed subset.
Remark 100.10.5. Let X be an algebraic stack. Let T \subset |\mathcal{X}| be a locally closed subset. Let \partial T be the boundary of T in the topological space |\mathcal{X}|. In a formula
\partial T = \overline{T} \setminus T.
Let \mathcal{U} \subset \mathcal{X} be the open substack of X with |\mathcal{U}| = |\mathcal{X}| \setminus \partial T, see Lemma 100.9.12. Let \mathcal{Z} be the reduced closed substack of \mathcal{U} with |\mathcal{Z}| = T obtained by taking the reduced induced closed subspace structure, see Definition 100.10.4. By construction \mathcal{Z} \to \mathcal{U} is a closed immersion of algebraic stacks and \mathcal{U} \to \mathcal{X} is an open immersion, hence \mathcal{Z} \to \mathcal{X} is an immersion of algebraic stacks by Lemma 100.9.3. Note that \mathcal{Z} is a reduced algebraic stack and that |\mathcal{Z}| = T as subsets of |X|. We sometimes say \mathcal{Z} is the reduced induced substack structure on T.
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