Lemma 97.11.7. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$ be a point. The following are equivalent

there exists an algebraic stack $\mathcal{Z}$ and a monomorphism $\mathcal{Z} \to \mathcal{X}$ such that $|\mathcal{Z}|$ is a singleton and such that the image of $|\mathcal{Z}|$ in $|\mathcal{X}|$ is $x$,

there exists a reduced algebraic stack $\mathcal{Z}$ and a monomorphism $\mathcal{Z} \to \mathcal{X}$ such that $|\mathcal{Z}|$ is a singleton and such that the image of $|\mathcal{Z}|$ in $|\mathcal{X}|$ is $x$,

there exists an algebraic stack $\mathcal{Z}$, a monomorphism $f : \mathcal{Z} \to \mathcal{X}$, and a surjective flat morphism $z : \mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$ where $k$ is a field such that $x = f(z)$.

Moreover, if these conditions hold, then there exists a unique strictly full subcategory $\mathcal{Z}_ x \subset \mathcal{X}$ such that $\mathcal{Z}_ x$ is a reduced, locally Noetherian algebraic stack and $|\mathcal{Z}_ x|$ is a singleton which maps to $x$ via the map $|\mathcal{Z}_ x| \to |\mathcal{X}|$.

**Proof.**
If $\mathcal{Z} \to \mathcal{X}$ is as in (1), then $\mathcal{Z}_{red} \to \mathcal{X}$ is as in (2). (See Section 97.10 for the notion of the reduction of an algebraic stack.) Hence (1) implies (2). It is immediate that (2) implies (1). The equivalence of (2) and (3) is immediate from Lemma 97.11.2.

At this point we've seen the equivalence of (1) – (3). Pick a monomorphism $f : \mathcal{Z} \to \mathcal{X}$ as in (2). Note that this implies that $f$ is fully faithful, see Lemma 97.8.4. Denote $\mathcal{Z}' \subset \mathcal{X}$ the essential image of the functor $f$. Then $f : \mathcal{Z} \to \mathcal{Z}'$ is an equivalence and hence $\mathcal{Z}'$ is an algebraic stack, see Algebraic Stacks, Lemma 91.12.4. Apply Lemma 97.11.5 to get a strictly full subcategory $\mathcal{Z}_ x \subset \mathcal{Z}'$ as in the statement of the lemma. This proves all the statements of the lemma except for uniqueness.

In order to prove the uniqueness suppose that $\mathcal{Z}_ x \subset \mathcal{X}$ and $\mathcal{Z}'_ x \subset \mathcal{X}$ are two strictly full subcategories as in the statement of the lemma. Then the projections

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

are monomorphisms. The algebraic stack in the middle is nonempty by Lemma 97.4.3. Hence the two projections are isomorphisms by Lemma 97.11.4 and we win.
$\square$

## Comments (2)

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