Lemma 68.13.1. Let $S$ be a scheme. Let $Z$ be an algebraic space over $S$. Let $k$ be a field and let $\mathop{\mathrm{Spec}}(k) \to Z$ be surjective and flat. Then any morphism $\mathop{\mathrm{Spec}}(k') \to Z$ where $k'$ is a field is surjective and flat.

## 68.13 Reduced singleton spaces

A *singleton* space is an algebraic space $X$ such that $|X|$ is a singleton. It turns out that these can be more interesting than just being the spectrum of a field, see Spaces, Example 65.14.7. We develop a tiny bit of machinery to be able to talk about these.

**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 $T \to Z$ is flat and surjective. It follows from Morphisms of Spaces, Lemma 67.31.5 that $\mathop{\mathrm{Spec}}(k') \to Z$ is flat. It is surjective as by assumption $|Z|$ is a singleton. $\square$

Lemma 68.13.2. Let $S$ be a scheme. Let $Z$ be an algebraic space over $S$. The following are equivalent

$Z$ is reduced and $|Z|$ is a singleton,

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

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

**Proof.**
Assume (1). Let $W$ be a scheme and let $W \to Z$ be a surjective étale 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 $\eta \to Z$ is flat (see Morphisms of Spaces, Lemma 67.30.3). It is also surjective as $|Z|$ is a singleton. In other words (2) holds.

Assume (2). Let $W$ be a scheme and let $W \to Z$ be a surjective étale morphism. Choose a field $k$ and a surjective flat morphism $\mathop{\mathrm{Spec}}(k) \to Z$. Then $W \times _ Z \mathop{\mathrm{Spec}}(k)$ is a scheme étale over $k$. Hence $W \times _ Z \mathop{\mathrm{Spec}}(k)$ is a disjoint union of spectra of fields (see Remark 68.4.1), in particular reduced. Since $W \times _ Z \mathop{\mathrm{Spec}}(k) \to W$ is surjective and flat we conclude that $W$ is reduced (Descent, Lemma 35.19.1). In other words (1) holds.

It is clear that (3) implies (2). Finally, assume (2). Pick a nonempty affine scheme $W$ and an étale morphism $W \to Z$. 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 67.39.9. It is also flat and surjective by Lemma 68.13.1. Hence (3) holds. $\square$

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

Lemma 68.13.3. Let $S$ be a scheme. Let $Z$ be an algebraic space over $S$. The following are equivalent

$Z$ is reduced, locally Noetherian, and $|Z|$ is a singleton, and

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

**Proof.**
Assume (2) holds. By Lemma 68.13.2 we see that $Z$ is reduced and $|Z|$ is a singleton. Let $W$ be a scheme and let $W \to Z$ be a surjective étale morphism. Choose a field $k$ and a locally finitely presented, surjective, flat morphism $\mathop{\mathrm{Spec}}(k) \to Z$. Then $W \times _ Z \mathop{\mathrm{Spec}}(k)$ is a scheme étale over $k$, hence a disjoint union of spectra of fields (see Remark 68.4.1), hence locally Noetherian. Since $W \times _ Z \mathop{\mathrm{Spec}}(k) \to W$ is flat, surjective, and locally of finite presentation, we see that $\{ W \times _ Z \mathop{\mathrm{Spec}}(k) \to W\} $ is an fppf covering and we conclude that $W$ is locally Noetherian (Descent, Lemma 35.16.1). In other words (1) holds.

Assume (1). Pick a nonempty affine scheme $W$ and an étale morphism $W \to Z$. 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 67.39.8. It is also flat and surjective by Lemma 68.13.1. Hence (2) holds. $\square$

Lemma 68.13.4. Let $S$ be a scheme. Let $Z' \to Z$ be a monomorphism of algebraic spaces over $S$. Assume there exists a field $k$ and a locally finitely presented, surjective, flat morphism $\mathop{\mathrm{Spec}}(k) \to Z$. Then either $Z'$ is empty or $Z' = Z$.

**Proof.**
We may assume that $Z'$ is nonempty. In this case the fibre product $T = Z' \times _ Z \mathop{\mathrm{Spec}}(k)$ is nonempty, see Properties of Spaces, Lemma 66.4.3. 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 $\mathop{\mathrm{Spec}}(k) \to Z$ factors through $Z'$. But as $\mathop{\mathrm{Spec}}(k) \to Z$ is surjective, flat and locally of finite presentation, we see that $\mathop{\mathrm{Spec}}(k) \to Z$ is surjective as a map of sheaves on $(\mathit{Sch}/S)_{fppf}$ (see Spaces, Remark 65.5.2) and we conclude that $Z' = Z$.
$\square$

The following lemma says that to each point of an algebraic space we can associate a canonical reduced, locally Noetherian singleton algebraic space.

Lemma 68.13.5. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. Then there exists a unique monomorphism $Z \to X$ of algebraic spaces over $S$ such that $Z$ is an algebraic space which satisfies the equivalent conditions of Lemma 68.13.3 and such that the image of $|Z| \to |X|$ is $\{ x\} $.

**Proof.**
Choose a scheme $U$ and a surjective étale morphism $U \to X$. Set $R = U \times _ X U$ so that $X = U/R$ is a presentation (see Spaces, Section 65.9). Set

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

Because $U' \to U$ is a monomorphism we see that the projections $s', t' : R' \to U'$ factor as a monomorphism followed by an étale morphism. Hence, as $U'$ is a disjoint union of spectra of fields, using Remark 68.4.1, and using Schemes, Lemma 26.23.11 we conclude that $R'$ is a disjoint union of spectra of fields and that the morphisms $s', t' : R' \to U'$ are étale. Hence $Z = U'/R'$ is an algebraic space by Spaces, Theorem 65.10.5. As $R'$ is the restriction of $R$ by $U' \to U$ we see $Z \to X$ is a monomorphism by Groupoids, Lemma 39.20.6. Since $Z \to X$ is a monomorphism we see that $|Z| \to |X|$ is injective, see Morphisms of Spaces, Lemma 67.10.9. By Properties of Spaces, Lemma 66.4.3 we see that

is surjective which implies (by our choice of $U'$) that $|Z| \to |X|$ has image $\{ x\} $. We conclude that $|Z|$ is a singleton. Finally, by construction $U'$ is locally Noetherian and reduced, i.e., we see that $Z$ satisfies the equivalent conditions of Lemma 68.13.3.

Let us prove uniqueness of $Z \to X$. Suppose that $Z' \to X$ is a second such monomorphism of algebraic spaces. Then the projections

are monomorphisms. The algebraic space in the middle is nonempty by Properties of Spaces, Lemma 66.4.3. Hence the two projections are isomorphisms by Lemma 68.13.4 and we win. $\square$

We introduce the following terminology which foreshadows the residual gerbes we will introduce later, see Properties of Stacks, Definition 100.11.8.

Definition 68.13.6. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. The *residual space of $X$ at $x$*^{1} is the monomorphism $Z_ x \to X$ constructed in Lemma 68.13.5.

In particular we know that $Z_ x$ is a locally Noetherian, reduced, singleton algebraic space and that there exists a field and a surjective, flat, locally finitely presented morphism

The residual space is often given by a monomorphism from the spectrum of a field.

Lemma 68.13.7. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. The residual space $Z_ x$ of $X$ at $x$ is isomorphic to the spectrum of a field if and only if $x$ can be represented by a monomorphism $\mathop{\mathrm{Spec}}(k) \to X$ where $k$ is a field. If $X$ is decent, this holds for all $x \in |X|$.

**Proof.**
Since $Z_ x \to X$ is a monomorphism, if $Z_ x = \mathop{\mathrm{Spec}}(k)$ for some field $k$, then $x$ is represented by the monomorphism $\mathop{\mathrm{Spec}}(k) = Z_ x \to X$. Conversely, if $\mathop{\mathrm{Spec}}(k) \to X$ is a monomorphism which represents $x$, then $Z_ x \times _ X \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(k)$ is a monomorphism whose source is nonempty by Properties of Spaces, Lemma 66.4.3. Hence $Z_ x \times _ X \mathop{\mathrm{Spec}}(k) = \mathop{\mathrm{Spec}}(k)$ by Morphisms of Spaces, Lemma 67.10.8. Hence we get a monomorphism $\mathop{\mathrm{Spec}}(k) \to Z_ x$. This is an isomorphism by Lemma 68.13.4. The final statement follows from Lemma 68.11.1.
$\square$

The residual space is a regular algebraic space by the following lemma.

Lemma 68.13.8. A reduced, locally Noetherian singleton algebraic space $Z$ is regular.

**Proof.**
Let $Z$ be a reduced, locally Noetherian singleton algebraic space over a scheme $S$. Let $W \to Z$ be a surjective étale morphism where $W$ is a scheme. Let $k$ be a field and let $\mathop{\mathrm{Spec}}(k) \to Z$ be surjective, flat, and locally of finite presentation (see Lemma 68.13.3). The scheme $T = W \times _ Z \mathop{\mathrm{Spec}}(k)$ is étale over $k$ in particular regular, see Remark 68.4.1. Since $T \to W$ is locally of finite presentation, flat, and surjective it follows that $W$ is regular, see Descent, Lemma 35.19.2. By definition this means that $Z$ is regular.
$\square$

Lemma 68.13.9. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Let $x \in |X|$ be a point. Assume

$|f|(|Y|)$ is contained in $\{ x\} \subset |X|$,

$Y$ is reduced, and

$X$ is locally Noetherian.

Then $f$ factors through the residual space $Z_ x$ of $X$ at $x$.

**Proof.**
Preliminary remark: since $Z_ x \to X$ is a monomorphism, it suffices to find a surjective étale morphism $Y' \to Y$ such that $Y' \to X$ factors through $Z_ x$. A remark here is that $Y'$ is reduced as well.

Let $U$ be an affine scheme and let $U \to X$ be an étale morphism such that $x$ is in the image of $|U| \to |X|$. Since $X$ is locally Noetherian, $U$ is a Noetherian affine scheme. By assumption (1) we see that $Y' = U \times _ X Y \to Y$ is surjective as well as étale. Denote $E \subset |U|$ the set of points mapping to $x$. There are no nontrivial specializations between the elements of $E$, see Lemma 68.7.2. The morphism $Y' \to U$ maps $|Y'|$ into $E$. By our construction of $Z_ x$ in the proof of Lemma 68.13.5 we know that $\coprod _{u \in E} u \to X$ factors through $Z_ x$. Hence it suffices to prove that $Y' \to U$ factors through $\coprod _{u \in E} u \to X$. After replacing $Y'$ by an étale covering by a scheme (which we are allowed by our preliminary remark), this follows from Morphisms, Lemma 29.58.2. $\square$

Lemma 68.13.10. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Let $x \in |X|$ be a point. Assume

$|f|(|Y|)$ is contained in $\{ x\} \subset |X|$,

$Y$ is reduced, and

$x$ can be represented by a quasi-compact monomorphism $x : \mathop{\mathrm{Spec}}(k) \to X$ where $k$ is a field (for example if $X$ is decent).

Then $f$ factors through the residual space $Z_ x = \mathop{\mathrm{Spec}}(k)$ of $X$ at $x$.

**Proof.**
By Lemma 68.13.7 we have $Z_ x = \mathop{\mathrm{Spec}}(k)$.

Preliminary remark: since $\mathop{\mathrm{Spec}}(k) \to X$ is a monomorphism, it suffices to find a surjective étale morphism $Y' \to Y$ such that $Y' \to X$ factors through $Z_ x$. A remark here is that $Y'$ is reduced as well.

After replacing $X$ by a quasi-compact open neighbourhood of $x$, we may assume $X$ quasi-compact. By Lemma 68.8.3, $x$ is a point of $T \subset U \subset X$ where $T \to U$ (resp. $U \to X$) is a closed (resp. open) immersion, and $T$ is a scheme. By Properties of Spaces, Lemma 66.4.9, $f$ factors through $U$, so we may assume $U = X$. Then $f$ factors through $T$ because $Y$ is reduced, see Properties of Spaces, Lemma 66.12.4. So we may assume that $X = T$ is a scheme. By our preliminary remark we may assume $Y$ is a scheme too. This reduces us to Morphisms, Lemma 29.58.1. $\square$

Example 68.13.11. Here is a counter example to Lemmas 68.13.9 and 68.13.10 in case $X$ is neither locally Noetherian nor decent. Let $k$ be a field. Let $G$ be an infinite profinite group. Let $Y$ be $G$ viewed as a zero-dimensional affine $k$-group scheme, i.e., $Y = \mathop{\mathrm{Spec}}(\text{locally constant maps } G \to k)$. Let $\Gamma $ be $G$ viewed as a discrete $k$-group scheme, acting on $X$ by translations. Put $X = Y/\Gamma $. This is a one-point algebraic space, with projection $q : Y \to X$. Let $e \in G$ be the origin (any element would do), and view it as a $k$-point of $Y$. We get a $k$-point $x :\mathop{\mathrm{Spec}}(k) \to X$ which is a monomorphism since it is a section of $X \to \mathop{\mathrm{Spec}}(k)$. We claim that (although $Y$ is affine and reduced and $|X| = \{ x\} $), the morphism $q$ does not factor through any morphism $\mathop{\mathrm{Spec}}(K) \to X$, where $K$ is a field. Otherwise it would factor through $x$ by Properties of Spaces, Lemma 66.4.11. Now the pullback of $q$ by $x$ is $\Gamma \to \mathop{\mathrm{Spec}}(k)$, with the projection $\Gamma \to Y$ being the orbit map $g \mapsto g \cdot e$. The latter has no section, whence the claim.

Lemma 68.13.12. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$ with residual space $Z_ x \subset X$. Assume $X$ is locally Noetherian. Then $x$ is a closed point of $|X|$ if and only if the morphism $Z_ x \to X$ is a closed immersion.

**Proof.**
If $Z_ x \to X$ is a closed immersion, then $x$ is a closed point of $|X|$, see Morphisms of Spaces, Lemma 67.12.3. Conversely, assume $x$ is a closed point of $|X|$. Let $Z \subset X$ be the reduced closed subspace with $|Z| = \{ x\} $ (Properties of Spaces, Lemma 66.12.3). Then $Z$ is locally Noetherian by Morphisms of Spaces, Lemmas 67.23.7 and 67.23.5. Since also $Z$ is reduced and $|Z| = \{ x\} $ it $Z = Z_ x$ is the residual space by definition.
$\square$

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