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.

## 99.10 Reduced algebraic stacks

We have already defined reduced algebraic stacks in Section 99.7.

**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 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

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

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 $T$.

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)