## 99.10 Reduced algebraic stacks

We have already defined reduced algebraic stacks in Section 99.7.

Lemma 99.10.1. 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.

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 93.16.2. As usual we denote $s, t : R \to U$ the two smooth projection morphisms. By Lemma 99.4.5 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 65.12.5. The fibre products $Z \times _{U, t} R$ and $R \times _{s, U} Z$ are closed subspaces of $R$ (Spaces, Lemma 64.12.3). The projections $Z \times _{U, t} R \to Z$ and $R \times _{s, U} Z \to Z$ are smooth by Morphisms of Spaces, Lemma 66.37.3. Thus as $Z$ is reduced, it follows that $Z \times _{U, t} R$ and $R \times _{s, U} Z$ are reduced, see Remark 99.7.3. 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 65.12.3 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 99.9.10 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 99.9.6). Then $|\mathcal{Z}| = |Z|/|R_ Z| = |T'|/\sim$ is the given closed subset $T$. We omit the proof of unicity. $\square$

Lemma 99.10.2. 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 99.9.10 $\mathcal{X}'$ 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 99.10.3. 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 99.9.2 and 99.9.9. Using Lemmas 99.4.3, 99.8.5, and 99.9.5 we see that $|\mathcal{Y}'| = |\mathcal{Y}|$. Hence we have reduced the lemma to Lemma 99.10.2. $\square$

Definition 99.10.4. 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 99.10.1. 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 99.10.5. 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 99.9.11. 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 99.10.4. 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 99.9.3. 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 $T$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).