The Stacks project

Proof. Let $T = \overline{\{ x\} } \subset |\mathcal{X}|$ be the closure of $x$. By Properties of Stacks, Lemma 100.10.1 there exists a reduced closed substack $\mathcal{X}' \subset \mathcal{X}$ such that $T = |\mathcal{X}'|$. Note that $\mathcal{I}_{\mathcal{X}'} = \mathcal{I}_\mathcal {X} \times _\mathcal {X} \mathcal{X}'$ by Lemma 101.5.6. Hence $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'$ is quasi-compact as a base change, see Lemma 101.7.3. Therefore Proposition 101.29.1 implies there exists a dense open substack $\mathcal{U} \subset \mathcal{X}'$ which is a gerbe. Note that $x \in |\mathcal{U}|$ because $\{ x\} \subset T$ is a dense subset too. Hence a residual gerbe $\mathcal{Z}_ x \subset \mathcal{U}$ of $\mathcal{U}$ at $x$ exists by Lemma 101.28.15. It is immediate from the definitions that $\mathcal{Z}_ x \to \mathcal{X}$ is a residual gerbe of $\mathcal{X}$ at $x$. $\square$

Proof. Choose a scheme $U$ and a surjective, flat, locally finite presented, and locally quasi-finite morphism $U \to \mathcal{X}$, see Theorem 101.21.3. Set $R = U \times _\mathcal {X} U$. The projections $s, t : R \to U$ are surjective, flat, locally of finite presentation, and locally quasi-finite as base changes of the morphism $U \to \mathcal{X}$. There is a canonical morphism $[U/R] \to \mathcal{X}$ (see Algebraic Stacks, Lemma 94.16.1) which is an equivalence because $U \to \mathcal{X}$ is surjective, flat, and locally of finite presentation, see Algebraic Stacks, Remark 94.16.3. Thus we may assume that $\mathcal{X} = [U/R]$ where $(U, R, s, t, c)$ is a groupoid in algebraic spaces such that $s, t : R \to U$ are surjective, flat, locally of finite presentation, and locally quasi-finite. Set

The canonical morphism $U' \to U$ is a monomorphism. Let

Because $U' \to U$ is a monomorphism we see that both projections $s', t' : R' \to U'$ factor as a monomorphism followed by a locally quasi-finite morphism. Hence, as $U'$ is a disjoint union of spectra of fields, using Spaces over Fields, Lemma 72.10.9 we conclude that the morphisms $s', t' : R' \to U'$ are locally quasi-finite. Again since $U'$ is a disjoint union of spectra of fields, the morphisms $s', t'$ are also flat. Finally, $s', t'$ locally quasi-finite implies $s', t'$ locally of finite type, hence $s', t'$ locally of finite presentation (because $U'$ is a disjoint union of spectra of fields in particular locally Noetherian, so that Morphisms of Spaces, Lemma 67.28.7 applies). Hence $\mathcal{Z} = [U'/R']$ is an algebraic stack by Criteria for Representability, Theorem 97.17.2. As $R'$ is the restriction of $R$ by $U' \to U$ we see $\mathcal{Z} \to \mathcal{X}$ is a monomorphism by Groupoids in Spaces, Lemma 78.25.1 and Properties of Stacks, Lemma 100.8.4. Since $\mathcal{Z} \to \mathcal{X}$ is a monomorphism we see that $|\mathcal{Z}| \to |\mathcal{X}|$ is injective, see Properties of Stacks, Lemma 100.8.5. By Properties of Stacks, Lemma 100.4.3 we see that

is surjective which implies (by our choice of $U'$) that $|\mathcal{Z}| \to |\mathcal{X}|$ has image $\{ x\} $. We conclude that $|\mathcal{Z}|$ is a singleton. Finally, by construction $U'$ is locally Noetherian and reduced, i.e., $\mathcal{Z}$ is reduced and locally Noetherian. This means that the essential image of $\mathcal{Z} \to \mathcal{X}$ is the residual gerbe of $\mathcal{X}$ at $x$, see Properties of Stacks, Lemma 100.11.12. $\square$

Proof. Choose an affine scheme $U$ and a smooth morphism $U \to \mathcal{X}$ such that $x$ is in the image of the open continuous map $|U| \to |\mathcal{X}|$. We may and do replace $\mathcal{X}$ with the open substack corresponding to the image of $|U| \to |\mathcal{X}|$, see Properties of Stacks, Lemma 100.9.12. Thus we may assume $\mathcal{X} = [U/R]$ for a smooth groupoid $(U, R, s, t, c)$ in algebraic spaces where $U$ is a Noetherian affine scheme, see Algebraic Stacks, Section 94.16.

Let $E \subset |U|$ be the inverse image of $\{ x\} \subset |\mathcal{X}|$. Of course $E \not= \emptyset $. Since $|U|$ is a Noetherian topological space, we can choose an element $u \in E$ such that $\overline{\{ u\} } \cap E = \{ u\} $. As usual, we think of $u = \mathop{\mathrm{Spec}}(\kappa (u))$ as the spectrum of its residue field. Let us write

Furthermore, denote $Z = \overline{\{ u\} } \subset U$ with the reduced induced scheme structure. Denote $p : F \to U$ the morphism induced by the second projection (using $s : R \to U$ in the first fibre product description of $F$). Then $E$ is the set theoretic image of $p$. The morphism $R' \to F$ is a monomorphism which factors through the inverse image $p^{-1}(Z)$ of $Z$. This inverse image $p^{-1}(Z) \subset F$ is a closed subscheme and the restriction $p|_{p^{-1}(Z)} : p^{-1}(Z) \to Z$ has image set theoretically contained in $\{ u\} \subset Z$ by our careful choice of $u \in E$ above. Since $u = \mathop{\mathrm{lim}}\nolimits W$ where the limit is over the nonempty affine open subschemes of the irreducible reduced scheme $Z$, we conclude that the morphism $p|_{p^{-1}(Z)} : p^{-1}(Z) \to Z$ factors through the morphism $u \to Z$. Clearly this implies that $R' = p^{-1}(Z)$. In particular the morphism $t' : R' \to u$ is locally of finite presentation as the composition of the closed immersion $p^{-1}(Z) \to F$ of locally Noetherian algebraic spaces with the smooth morphism $\text{pr}_1 : F \to u$; use Morphisms of Spaces, Lemmas 67.23.5, 67.28.12, and 67.28.2. Hence the restriction $(u, R', s', t', c')$ of $(U, R, s, t, c)$ by $u \to U$ is a groupoid in algebraic spaces where $s'$ and $t'$ are flat and locally of finite presentation. Therefore $\mathcal{Z} = [u/R']$ is an algebraic stack by Criteria for Representability, Theorem 97.17.2. As $R'$ is the restriction of $R$ by $u \to U$ we see $\mathcal{Z} \to \mathcal{X}$ is a monomorphism by Groupoids in Spaces, Lemma 78.25.1 and Properties of Stacks, Lemma 100.8.4. Then $\mathcal{Z}$ is (isomorphic to) the residual gerbe by the material in Properties of Stacks, Section 100.11. $\square$