Lemma 100.11.1. Let $\mathcal{Z}$ be an algebraic stack. Let $k$ be a field and let $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$ be surjective and flat. Then any morphism $\mathop{\mathrm{Spec}}(k') \to \mathcal{Z}$ where $k'$ is a field is surjective and flat.

## 100.11 Residual gerbes

In the Stacks project we would like to define the *residual gerbe* of an algebraic stack $\mathcal{X}$ at a point $x \in |\mathcal{X}|$ to be a monomorphism of algebraic stacks $m_ x : \mathcal{Z}_ x \to \mathcal{X}$ where $\mathcal{Z}_ x$ is a reduced algebraic stack having a unique point which is mapped by $m_ x$ to $x$. It turns out that there are many issues with this notion; existence is not clear in general and neither is uniqueness. We resolve the uniqueness issue by imposing a slightly stronger condition on the algebraic stacks $\mathcal{Z}_ x$. We discuss this in more detail by working through a few simple lemmas regarding reduced algebraic stacks having a unique point.

**Proof.**
Consider the fibre square

Note that $T \to \mathop{\mathrm{Spec}}(k')$ is flat and surjective hence $T$ is not empty. On the other hand $T \to \mathop{\mathrm{Spec}}(k)$ is flat as $k$ is a field. Hence $T \to \mathcal{Z}$ is flat and surjective. It follows from Morphisms of Spaces, Lemma 67.31.5 (via the discussion in Section 100.3) that $\mathop{\mathrm{Spec}}(k') \to \mathcal{Z}$ is flat. It is clear that it is surjective as by assumption $|\mathcal{Z}|$ is a singleton. $\square$

Lemma 100.11.2. Let $\mathcal{Z}$ be an algebraic stack. The following are equivalent

$\mathcal{Z}$ is reduced and $|\mathcal{Z}|$ is a singleton,

there exists a surjective flat morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$ where $k$ is a field, and

there exists a locally of finite type, surjective, flat morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$ where $k$ is a field.

**Proof.**
Assume (1). Let $W$ be a scheme and let $W \to \mathcal{Z}$ be a surjective smooth morphism. Then $W$ is a reduced scheme. Let $\eta \in W$ be a generic point of an irreducible component of $W$. Since $W$ is reduced we have $\mathcal{O}_{W, \eta } = \kappa (\eta )$. It follows that the canonical morphism $\eta = \mathop{\mathrm{Spec}}(\kappa (\eta )) \to W$ is flat. We see that the composition $\eta \to \mathcal{Z}$ is flat (see Morphisms of Spaces, Lemma 67.30.3). It is also surjective as $|\mathcal{Z}|$ is a singleton. In other words (2) holds.

Assume (2). Let $W$ be a scheme and let $W \to \mathcal{Z}$ be a surjective smooth morphism. Choose a field $k$ and a surjective flat morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$. Then $W \times _\mathcal {Z} \mathop{\mathrm{Spec}}(k)$ is an algebraic space smooth over $k$, hence regular (see Spaces over Fields, Lemma 72.16.1) and in particular reduced. Since $W \times _\mathcal {Z} \mathop{\mathrm{Spec}}(k) \to W$ is surjective and flat we conclude that $W$ is reduced (Descent on Spaces, Lemma 74.9.2). In other words (1) holds.

It is clear that (3) implies (2). Finally, assume (2). Pick a nonempty affine scheme $W$ and a smooth morphism $W \to \mathcal{Z}$. Pick a closed point $w \in W$ and set $k = \kappa (w)$. The composition

is locally of finite type by Morphisms of Spaces, Lemmas 67.23.2 and 67.37.6. It is also flat and surjective by Lemma 100.11.1. Hence (3) holds. $\square$

The following lemma singles out a slightly better class of singleton algebraic stacks than the preceding lemma.

Lemma 100.11.3. Let $\mathcal{Z}$ be an algebraic stack. The following are equivalent

$\mathcal{Z}$ is reduced, locally Noetherian, and $|\mathcal{Z}|$ is a singleton, and

there exists a locally finitely presented, surjective, flat morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$ where $k$ is a field.

**Proof.**
Assume (2) holds. By Lemma 100.11.2 we see that $\mathcal{Z}$ is reduced and $|\mathcal{Z}|$ is a singleton. Let $W$ be a scheme and let $W \to \mathcal{Z}$ be a surjective smooth morphism. Choose a field $k$ and a locally finitely presented, surjective, flat morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$. Then $W \times _\mathcal {Z} \mathop{\mathrm{Spec}}(k)$ is an algebraic space smooth over $k$, hence locally Noetherian (see Morphisms of Spaces, Lemma 67.23.5). Since $W \times _\mathcal {Z} \mathop{\mathrm{Spec}}(k) \to W$ is flat, surjective, and locally of finite presentation, we see that $\{ W \times _\mathcal {Z} \mathop{\mathrm{Spec}}(k) \to W\} $ is an fppf covering and we conclude that $W$ is locally Noetherian (Descent on Spaces, Lemma 74.9.3). In other words (1) holds.

Assume (1). Pick a nonempty affine scheme $W$ and a smooth morphism $W \to \mathcal{Z}$. Pick a closed point $w \in W$ and set $k = \kappa (w)$. Because $W$ is locally Noetherian the morphism $w : \mathop{\mathrm{Spec}}(k) \to W$ is of finite presentation, see Morphisms, Lemma 29.21.7. Hence the composition

is locally of finite presentation by Morphisms of Spaces, Lemmas 67.28.2 and 67.37.5. It is also flat and surjective by Lemma 100.11.1. Hence (2) holds. $\square$

Lemma 100.11.4. Let $\mathcal{Z}' \to \mathcal{Z}$ be a monomorphism of algebraic stacks. Assume there exists a field $k$ and a locally finitely presented, surjective, flat morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$. Then either $\mathcal{Z}'$ is empty or $\mathcal{Z}' \to \mathcal{Z}$ is an equivalence.

**Proof.**
We may assume that $\mathcal{Z}'$ is nonempty. In this case the fibre product $T = \mathcal{Z}' \times _\mathcal {Z} \mathop{\mathrm{Spec}}(k)$ is nonempty, see Lemma 100.4.3. Now $T$ is an algebraic space and the projection $T \to \mathop{\mathrm{Spec}}(k)$ is a monomorphism. Hence $T = \mathop{\mathrm{Spec}}(k)$, see Morphisms of Spaces, Lemma 67.10.8. We conclude that $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$ factors through $\mathcal{Z}'$. Suppose the morphism $z : \mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$ is given by the object $\xi $ over $\mathop{\mathrm{Spec}}(k)$. We have just seen that $\xi $ is isomorphic to an object $\xi '$ of $\mathcal{Z}'$ over $\mathop{\mathrm{Spec}}(k)$. Since $z$ is surjective, flat, and locally of finite presentation we see that every object of $\mathcal{Z}$ over any scheme is fppf locally isomorphic to a pullback of $\xi $, hence also to a pullback of $\xi '$. By descent of objects for stacks in groupoids this implies that $\mathcal{Z}' \to \mathcal{Z}$ is essentially surjective (as well as fully faithful, see Lemma 100.8.4). Hence we win.
$\square$

Lemma 100.11.5. Let $\mathcal{Z}$ be an algebraic stack. Assume $\mathcal{Z}$ satisfies the equivalent conditions of Lemma 100.11.2. 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 100.11.3. 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 100.8.4. 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 67.28.7). We see that $\mathcal{Z}' = [U/R]$ is an algebraic stack by Criteria for Representability, Theorem 97.17.2. By Algebraic Stacks, Lemma 94.16.1 we obtain a canonical morphism

which is fully faithful. Hence this morphism is representable by algebraic spaces, see Algebraic Stacks, Lemma 94.15.2 and a monomorphism, see Lemma 100.8.4. By Criteria for Representability, Lemma 97.17.1 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 100.11.3. By Algebraic Stacks, Lemma 94.12.4 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

are monomorphisms. The algebraic stack in the middle is nonempty by Lemma 100.4.3. Hence the two projections are isomorphisms by Lemma 100.11.4 and we win. $\square$

Example 100.11.6. Here is an example where the morphism constructed in Lemma 100.11.5 isn't an isomorphism. This example shows that imposing that residual gerbes are locally Noetherian is necessary in Definition 100.11.8. In fact, the example is even an algebraic space! Let $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ be the absolute Galois group of $\mathbf{Q}$ with the pro-finite topology. Let

(we omit a precise explanation of the meaning of the last equal sign). Let $G$ denote the absolute Galois group $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ with the discrete topology viewed as a constant group scheme over $\mathop{\mathrm{Spec}}(\overline{\mathbf{Q}})$, see Groupoids, Example 39.5.6. Then $G$ acts freely and transitively on $U$. Let $X = U/G$, see Spaces, Definition 65.14.4. Then $X$ is a non-noetherian reduced algebraic space with exactly one point. Furthermore, $X$ has a (locally) finite type point:

Indeed, every point of $U$ is actually closed! As $X$ is an algebraic space over $\overline{\mathbf{Q}}$ it follows that $x$ is a monomorphism. So $x$ is the morphism constructed in Lemma 100.11.5 but $x$ is not an isomorphism. In fact $\mathop{\mathrm{Spec}}(\overline{\mathbf{Q}}) \to X$ is the residual gerbe of $X$ at $x$.

It will turn out later that under mild assumptions on the algebraic stack $\mathcal{X}$ the equivalent conditions of the following lemma are satisfied for every point $x \in |\mathcal{X}|$ (see Morphisms of Stacks, Section 101.31).

Lemma 100.11.7. 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 100.10 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 100.11.2.

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 100.8.4. 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 94.12.4. Apply Lemma 100.11.5 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

are monomorphisms. The algebraic stack in the middle is nonempty by Lemma 100.4.3. Hence the two projections are isomorphisms by Lemma 100.11.4 and we win. $\square$

Having explained the above we can now make the following definition.

Definition 100.11.8. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$.

We say the

*residual gerbe of $\mathcal{X}$ at $x$ exists*if the equivalent conditions (1), (2), and (3) of Lemma 100.11.7 hold.If the residual gerbe of $\mathcal{X}$ at $x$ exists, then the

*residual gerbe of $\mathcal{X}$ at $x$*^{1}is the strictly full subcategory $\mathcal{Z}_ x \subset \mathcal{X}$ constructed in Lemma 100.11.7.

In particular we know that $\mathcal{Z}_ x$ (if it exists) is a locally Noetherian, reduced algebraic stack and that there exists a field and a surjective, flat, locally finitely presented morphism

We will see in Morphisms of Stacks, Lemma 101.28.12 that $\mathcal{Z}_ x$ is a gerbe. Existence of residual gerbes is discussed in Morphisms of Stacks, Section 101.31.

Example 100.11.9. Let $X$ be a scheme and let $x \in X$ be a point. Then the monomorphism $x \to X$ is the residual gerbe of $X$ at $x$ where we, as usual, identify $x$ with the scheme $x = \mathop{\mathrm{Spec}}(\kappa (x))$. If $X$ is an algebraic space and $x \in |X|$, then the residual gerbe at $x$ (which is called the residual space) always exists, see Decent Spaces, Section 68.13.

The residual gerbe, if it exists, is a regular algebraic stack by the following lemma.

Lemma 100.11.10. A reduced, locally Noetherian algebraic stack $\mathcal{Z}$ such that $|\mathcal{Z}|$ is a singleton is regular.

**Proof.**
Let $W \to \mathcal{Z}$ be a surjective smooth morphism where $W$ is a scheme. Let $k$ be a field and let $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$ be surjective, flat, and locally of finite presentation (see Lemma 100.11.3). The algebraic space $T = W \times _\mathcal {Z} \mathop{\mathrm{Spec}}(k)$ is smooth over $k$ in particular regular, see Spaces over Fields, Lemma 72.16.1. Since $T \to W$ is locally of finite presentation, flat, and surjective it follows that $W$ is regular, see Descent on Spaces, Lemma 74.9.4. By definition this means that $\mathcal{Z}$ is regular.
$\square$

Lemma 100.11.11. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. Assume that the residual gerbe $\mathcal{Z}_ x$ of $\mathcal{X}$ exists. Let $f : \mathop{\mathrm{Spec}}(K) \to \mathcal{X}$ be a morphism where $K$ is a field in the equivalence class of $x$. Then $f$ factors through the inclusion morphism $\mathcal{Z}_ x \to \mathcal{X}$.

**Proof.**
Choose a field $k$ and a surjective flat locally finite presentation morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}_ x$. Set $T = \mathop{\mathrm{Spec}}(K) \times _\mathcal {X} \mathcal{Z}_ x$. By Lemma 100.4.3 we see that $T$ is nonempty. As $\mathcal{Z}_ x \to \mathcal{X}$ is a monomorphism we see that $T \to \mathop{\mathrm{Spec}}(K)$ is a monomorphism. Hence by Morphisms of Spaces, Lemma 67.10.8 we see that $T = \mathop{\mathrm{Spec}}(K)$ which proves the lemma.
$\square$

Lemma 100.11.12. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. Let $\mathcal{Z}$ be an algebraic stack satisfying the equivalent conditions of Lemma 100.11.3 and let $\mathcal{Z} \to \mathcal{X}$ be a monomorphism such that the image of $|\mathcal{Z}| \to |\mathcal{X}|$ is $x$. Then the residual gerbe $\mathcal{Z}_ x$ of $\mathcal{X}$ at $x$ exists and $\mathcal{Z} \to \mathcal{X}$ factors as $\mathcal{Z} \to \mathcal{Z}_ x \to \mathcal{X}$ where the first arrow is an equivalence.

**Proof.**
Let $\mathcal{Z}_ x \subset \mathcal{X}$ be the full subcategory corresponding to the essential image of the functor $\mathcal{Z} \to \mathcal{X}$. Then $\mathcal{Z} \to \mathcal{Z}_ x$ is an equivalence, hence $\mathcal{Z}_ x$ is an algebraic stack, see Algebraic Stacks, Lemma 94.12.4. Since $\mathcal{Z}_ x$ inherits all the properties of $\mathcal{Z}$ from this equivalence it is clear from the uniqueness in Lemma 100.11.7 that $\mathcal{Z}_ x$ is the residual gerbe of $\mathcal{X}$ at $x$.
$\square$

Lemma 100.11.13. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $x \in |\mathcal{X}|$ with image $y \in |\mathcal{Y}|$. If the residual gerbes $\mathcal{Z}_ x \subset \mathcal{X}$ and $\mathcal{Z}_ y \subset \mathcal{Y}$ of $x$ and $y$ exist, then $f$ induces a commutative diagram

**Proof.**
Choose a field $k$ and a surjective, flat, locally finitely presented morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}_ x$. The morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Y}$ factors through $\mathcal{Z}_ y$ by Lemma 100.11.11. Thus $\mathcal{Z}_ x \times _\mathcal {Y} \mathcal{Z}_ y$ is a nonempty substack of $\mathcal{Z}_ x$ hence equal to $\mathcal{Z}_ x$ by Lemma 100.11.4.
$\square$

Lemma 100.11.14. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $x \in |\mathcal{X}|$ with image $y \in |\mathcal{Y}|$. Assume the residual gerbes $\mathcal{Z}_ x \subset \mathcal{X}$ and $\mathcal{Z}_ y \subset \mathcal{Y}$ of $x$ and $y$ exist and that there exists a morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ in the equivalence class of $x$ such that

is an isomorphism. Then $\mathcal{Z}_ x \to \mathcal{Z}_ y$ is an isomorphism.

**Proof.**
Let $k'/k$ be an extension of fields. Then

is the base change of the morphism in the lemma by the faithfully flat morphism $\mathop{\mathrm{Spec}}(k' \otimes k') \to \mathop{\mathrm{Spec}}(k \otimes k)$. Thus the property described in the lemma is independent of the choice of the morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ in the equivalence class of $x$. Thus we may assume that $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}_ x$ is surjective, flat, and locally of finite presentation. In this situation we have

with $R = \mathop{\mathrm{Spec}}(k) \times _\mathcal {X} \mathop{\mathrm{Spec}}(k)$. See proof of Lemma 100.11.5. Since also $R = \mathop{\mathrm{Spec}}(k) \times _\mathcal {Y} \mathop{\mathrm{Spec}}(k)$ we conclude that the morphism $\mathcal{Z}_ x \to \mathcal{Z}_ y$ of Lemma 100.11.13 is fully faithful by Algebraic Stacks, Lemma 94.16.1. We conclude for example by Lemma 100.11.12. $\square$

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