## 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.

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.

Proof. Consider the fibre square

$\xymatrix{ T \ar[d] \ar[r] & \mathop{\mathrm{Spec}}(k) \ar[d] \\ \mathop{\mathrm{Spec}}(k') \ar[r] & Z }$

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

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

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

3. 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

$\mathop{\mathrm{Spec}}(k) \xrightarrow {w} W \longrightarrow Z$

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

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

2. 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

$\mathop{\mathrm{Spec}}(k) \xrightarrow {w} W \longrightarrow Z$

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

$U' = \coprod \nolimits _{u \in U\text{ lying over }x} \mathop{\mathrm{Spec}}(\kappa (u)).$

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

$R' = U' \times _ X U' = R \times _{(U \times _ S U)} (U' \times _ S U').$

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

$|U'| = |Z \times _ X U'| \to |Z| \times _{|X|} |U'|$

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

$Z' \longleftarrow Z' \times _ X Z \longrightarrow Z$

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

$\mathop{\mathrm{Spec}}(k) \longrightarrow Z_ x.$

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

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

2. $Y$ is reduced, and

3. $X$ 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

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

2. $Y$ is reduced, and

3. $x$ 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. $\square$

 This is nonstandard notation.

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