The Stacks project

Lemma 99.11.5. Let $\mathcal{Z}$ be an algebraic stack. Assume $\mathcal{Z}$ satisfies the equivalent conditions of Lemma 99.11.2. Then there exists a unique strictly full subcategory $\mathcal{Z}' \subset \mathcal{Z}$ such that $\mathcal{Z}'$ is an algebraic stack which satisfies the equivalent conditions of Lemma 99.11.3. The inclusion morphism $\mathcal{Z}' \to \mathcal{Z}$ is a monomorphism of algebraic stacks.

Proof. The last part is immediate from the first part and Lemma 99.8.4. Pick a field $k$ and a morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$ which is surjective, flat, and locally of finite type. Set $U = \mathop{\mathrm{Spec}}(k)$ and $R = U \times _\mathcal {Z} U$. The projections $s, t : R \to U$ are locally of finite type. Since $U$ is the spectrum of a field, it follows that $s, t$ are flat and locally of finite presentation (by Morphisms of Spaces, Lemma 66.28.7). We see that $\mathcal{Z}' = [U/R]$ is an algebraic stack by Criteria for Representability, Theorem 96.17.2. By Algebraic Stacks, Lemma 93.16.1 we obtain a canonical morphism

\[ f : \mathcal{Z}' \longrightarrow \mathcal{Z} \]

which is fully faithful. Hence this morphism is representable by algebraic spaces, see Algebraic Stacks, Lemma 93.15.2 and a monomorphism, see Lemma 99.8.4. By Criteria for Representability, Lemma 96.17.1 the morphism $U \to \mathcal{Z}'$ is surjective, flat, and locally of finite presentation. Hence $\mathcal{Z}'$ is an algebraic stack which satisfies the equivalent conditions of Lemma 99.11.3. By Algebraic Stacks, Lemma 93.12.4 we may replace $\mathcal{Z}'$ by its essential image in $\mathcal{Z}$. Hence we have proved all the assertions of the lemma except for the uniqueness of $\mathcal{Z}' \subset \mathcal{Z}$. Suppose that $\mathcal{Z}'' \subset \mathcal{Z}$ is a second such algebraic stack. Then the projections

\[ \mathcal{Z}' \longleftarrow \mathcal{Z}' \times _\mathcal {Z} \mathcal{Z}'' \longrightarrow \mathcal{Z}'' \]

are monomorphisms. The algebraic stack in the middle is nonempty by Lemma 99.4.3. Hence the two projections are isomorphisms by Lemma 99.11.4 and we win. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 99.11: Residual gerbes

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Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 06MR. Beware of the difference between the letter 'O' and the digit '0'.