Lemma 99.11.1. 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.

## 99.11 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 66.31.5 (via the discussion in Section 99.3) 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 99.11.2. 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 66.30.3). 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 71.16.1) 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 73.9.2). 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 66.23.2 and 66.37.6. It is also flat and surjective by Lemma 99.11.1. Hence (3) holds. $\square$

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

Lemma 99.11.3. 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 99.11.2 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 66.23.5). 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 73.9.3). 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 66.28.2 and 66.37.5. It is also flat and surjective by Lemma 99.11.1. Hence (2) holds. $\square$

Lemma 99.11.4. 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 99.4.3. 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 66.10.8. 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 99.8.4). Hence we win.
$\square$

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

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

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. $\square$

Example 99.11.6. Here is an example where the morphism constructed in Lemma 99.11.5 isn't an isomorphism. This example shows that imposing that residual gerbes are locally Noetherian is necessary in Definition 99.11.8. 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 64.14.4. 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 99.11.5 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 100.31).

Lemma 99.11.7. 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 99.10 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 99.11.2.

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 99.8.4. 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 93.12.4. Apply Lemma 99.11.5 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 99.4.3. Hence the two projections are isomorphisms by Lemma 99.11.4 and we win. $\square$

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

Definition 99.11.8. 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 99.11.7 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 99.11.7.

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 100.28.12 that $\mathcal{Z}_ x$ is a gerbe. Existence of residual gerbes is discussed in Morphisms of Stacks, Section 100.31.

Example 99.11.9. 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 67.13.

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

Lemma 99.11.10. 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 99.11.3). The algebraic space $T = W \times _\mathcal {Z} \mathop{\mathrm{Spec}}(k)$ is smooth over $k$ in particular regular, see Spaces over Fields, Lemma 71.16.1. Since $T \to W$ is locally of finite presentation, flat, and surjective it follows that $W$ is regular, see Descent on Spaces, Lemma 73.9.4. By definition this means that $\mathcal{Z}$ is regular.
$\square$

Lemma 99.11.11. 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 99.4.3 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 66.10.8 we see that $T = \mathop{\mathrm{Spec}}(K)$ which proves the lemma.
$\square$

Lemma 99.11.12. 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 99.11.3 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 93.12.4. Since $\mathcal{Z}_ x$ inherits all the properties of $\mathcal{Z}$ from this equivalence it is clear from the uniqueness in Lemma 99.11.7 that $\mathcal{Z}_ x$ is the residual gerbe of $\mathcal{X}$ at $x$.
$\square$

Lemma 99.11.13. 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 99.11.11. 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 99.11.4.
$\square$

Lemma 99.11.14. 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 99.11.5. 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 99.11.13 is fully faithful by Algebraic Stacks, Lemma 93.16.1. We conclude for example by Lemma 99.11.12. $\square$

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