Introduction
In this chapter we study “local” properties of general algebraic spaces, i.e., those algebraic spaces which aren’t quasi-separated. Quasi-separated algebraic spaces are studied in [Kn]. It turns out that essentially new phenomena happen, especially regarding points and specializations of points, on more general algebraic spaces. On the other hand, for most basic results on algebraic spaces, one needn’t worry about these phenomena, which is why we have decided to have this material in a separate chapter following the standard development of the theory.
Conventions
The standing assumption is that all schemes are contained in a big fppf site \(\Sch_{fppf}\). And all rings \(A\) considered have the property that \(\Spec(A)\) is (isomorphic) to an object of this big site.
Let \(S\) be a scheme and let \(X\) be an algebraic space over \(S\). In this chapter and the following we will write \(X \times_S X\) for the product of \(X\) with itself (in the category of algebraic spaces over \(S\)), instead of \(X \times X\).
Universally bounded fibres
We briefly discuss what it means for a morphism from a scheme to an algebraic space to have universally bounded fibres. Please refer to Morphisms, Section 03J3 for similar definitions and results on morphisms of schemes.
Definition
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\), and let \(U\) be a scheme over \(S\). Let \(f : U \to X\) be a morphism over \(S\). We say the fibres of \(f\) are universally bounded1 if there exists an integer \(n\) such that for all fields \(k\) and all morphisms \(\Spec(k) \to X\) the fibre product \(\Spec(k) \times_X U\) is a finite scheme over \(k\) whose degree over \(k\) is \(\leq n\).
This definition makes sense because the fibre product \(\Spec(k) \times_Y X\) is a scheme. Moreover, if \(Y\) is a scheme we recover the notion of Morphisms, Definition 03J4 by virtue of Morphisms, Lemma 03J5.
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). Let \(V \to U\) be a morphism of schemes over \(S\), and let \(U \to X\) be a morphism from \(U\) to \(X\). If the fibres of \(V \to U\) and \(U \to X\) are universally bounded, then so are the fibres of \(V \to X\).
Proof
Let \(n\) be an integer which works for \(V \to U\), and let \(m\) be an integer which works for \(U \to X\) in Definition 03JL. Let \(\Spec(k) \to X\) be a morphism, where \(k\) is a field. Consider the morphisms \[\Spec(k) \times_X V \longrightarrow \Spec(k) \times_X U \longrightarrow \Spec(k).\] By assumption the scheme \(\Spec(k) \times_X U\) is finite of degree at most \(m\) over \(k\), and \(n\) is an integer which bounds the degree of the fibres of the first morphism. Hence by Morphisms, Lemma 03J6 we conclude that \(\Spec(k) \times_X V\) is finite over \(k\) of degree at most \(nm\).
Lemma
Let \(S\) be a scheme. Let \(Y \to X\) be a representable morphism of algebraic spaces over \(S\). Let \(U \to X\) be a morphism from a scheme to \(X\). If the fibres of \(U \to X\) are universally bounded, then the fibres of \(U \times_X Y \to Y\) are universally bounded.
Proof
This is clear from the definition, and properties of fibre products. (Note that \(U \times_X Y\) is a scheme as we assumed \(Y \to X\) representable, so the definition applies.)
Lemma
Let \(S\) be a scheme. Let \(g : Y \to X\) be a representable morphism of algebraic spaces over \(S\). Let \(f : U \to X\) be a morphism from a scheme towards \(X\). Let \(f' : U \times_X Y \to Y\) be the base change of \(f\). If \[\Im(|f| : |U| \to |X|) \subset \Im(|g| : |Y| \to |X|)\] and \(f'\) has universally bounded fibres, then \(f\) has universally bounded fibres.
Proof
Let \(n \geq 0\) be an integer bounding the degrees of the fibre products \(\Spec(k) \times_Y (U \times_X Y)\) as in Definition 03JL for the morphism \(f'\). We claim that \(n\) works for \(f\) also. Namely, suppose that \(x : \Spec(k) \to X\) is a morphism from the spectrum of a field. Then either \(\Spec(k) \times_X U\) is empty (and there is nothing to prove), or \(x\) is in the image of \(|f|\). By Properties of Spaces, Lemma 03H4 and the assumption of the lemma we see that this means there exists a field extension \(k'/k\) and a commutative diagram \[\xymatrix{ \Spec(k') \ar[r] \ar[d] & Y \ar[d] \\ \Spec(k) \ar[r] & X }\] Hence we see that \[\Spec(k') \times_Y (U \times_X Y) = \Spec(k') \times_{\Spec(k)} (\Spec(k) \times_X U)\] Since the scheme \(\Spec(k') \times_Y (U \times_X Y)\) is assumed finite of degree \(\leq n\) over \(k'\) it follows that also \(\Spec(k) \times_X U\) is finite of degree \(\leq n\) over \(k\) as desired. (Some details omitted.)
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). Consider a commutative diagram \[\xymatrix{ U \ar[rd]_g \ar[rr]_f & & V \ar[ld]^h \\ & X & }\] where \(U\) and \(V\) are schemes. If \(g\) has universally bounded fibres, and \(f\) is surjective and flat, then also \(h\) has universally bounded fibres.
Proof
Assume \(g\) has universally bounded fibres, and \(f\) is surjective and flat. Say \(n \geq 0\) is an integer which bounds the degrees of the schemes \(\Spec(k) \times_X U\) as in Definition 03JL. We claim \(n\) also works for \(h\). Let \(\Spec(k) \to X\) be a morphism from the spectrum of a field to \(X\). Consider the morphism of schemes \[\Spec(k) \times_X V \longrightarrow \Spec(k) \times_X U\] It is flat and surjective. By assumption the scheme on the left is finite of degree \(\leq n\) over \(\Spec(k)\). It follows from Morphisms, Lemma 03JB that the degree of the scheme on the right is also bounded by \(n\) as desired.
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\), and let \(U\) be a scheme over \(S\). Let \(\varphi : U \to X\) be a morphism over \(S\). If the fibres of \(\varphi\) are universally bounded, then there exists an integer \(n\) such that each fibre of \(|U| \to |X|\) has at most \(n\) elements.
Proof
The integer \(n\) of Definition 03JL works. Namely, pick \(x \in |X|\). Represent \(x\) by a morphism \(x : \Spec(k) \to X\). Then we get a commutative diagram \[\xymatrix{ \Spec(k) \times_X U \ar[r] \ar[d] & U \ar[d] \\ \Spec(k) \ar[r]^x & X }\] which shows (via Properties of Spaces, Lemma 03H4) that the inverse image of \(x\) in \(|U|\) is the image of the top horizontal arrow. Since \(\Spec(k) \times_X U\) is finite of degree \(\leq n\) over \(k\) it has at most \(n\) points.
Finiteness conditions and points
In this section we elaborate on the question of when points can be represented by monomorphisms from spectra of fields into the space.
Remark
Before we give the proof of the next lemma let us recall some facts about étale morphisms of schemes:
An étale morphism is flat and hence generalizations lift along an étale morphism (Morphisms, Lemmas 02GS and 03HV).
An étale morphism is unramified, an unramified morphism is locally quasi-finite, hence fibres are discrete (Morphisms, Lemmas 02GV, 02V5, and 01TH).
A quasi-compact étale morphism is quasi-finite and in particular has finite fibres (Morphisms, Lemmas 01TJ and 02NH).
An étale scheme over a field \(k\) is a disjoint union of spectra of finite separable field extension of \(k\) (Morphisms, Lemma 02GL).
For a general discussion of étale morphisms, please see Étale Morphisms, Section 0257.
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). Let \(x \in |X|\). The following are equivalent:
there exists a family of schemes \(U_i\) and étale morphisms \(\varphi_i : U_i \to X\) such that \(\coprod \varphi_i : \coprod U_i \to X\) is surjective, and such that for each \(i\) the fibre of \(|U_i| \to |X|\) over \(x\) is finite, and
for every affine scheme \(U\) and étale morphism \(\varphi : U \to X\) the fibre of \(|U| \to |X|\) over \(x\) is finite.
Proof
The implication (2) \(\Rightarrow\) (1) is trivial. Let \(\varphi_i : U_i \to X\) be a family of étale morphisms as in (1). Let \(\varphi : U \to X\) be an étale morphism from an affine scheme towards \(X\). Consider the fibre product diagrams \[\xymatrix{ U \times_X U_i \ar[r]_-{p_i} \ar[d]_{q_i} & U_i \ar[d]^{\varphi_i} \\ U \ar[r]^\varphi & X } \quad \quad \xymatrix{ \coprod U \times_X U_i \ar[r]_-{\coprod p_i} \ar[d]_{\coprod q_i} & \coprod U_i \ar[d]^{\coprod \varphi_i} \\ U \ar[r]^\varphi & X }\] Since \(q_i\) is étale it is open (see Remark 03II). Moreover, the morphism \(\coprod q_i\) is surjective. Hence there exist finitely many indices \(i_1, \ldots, i_n\) and a quasi-compact opens \(W_{i_j} \subset U \times_X U_{i_j}\) which surject onto \(U\). The morphism \(p_i\) is étale, hence locally quasi-finite (see remark on étale morphisms above). Thus we may apply Morphisms, Lemma 03JA to see the fibres of \(p_{i_j}|_{W_{i_j}} : W_{i_j} \to U_i\) are finite. Hence by Properties of Spaces, Lemma 03H4 and the assumption on \(\varphi_i\) we conclude that the fibre of \(\varphi\) over \(x\) is finite. In other words (2) holds.
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). Let \(x \in |X|\). The following are equivalent:
there exists a scheme \(U\), an étale morphism \(\varphi : U \to X\), and points \(u, u' \in U\) mapping to \(x\) such that setting \(R = U \times_X U\) the fibre of \[|R| \to |U| \times_{|X|} |U|\] over \((u, u')\) is finite,
for every scheme \(U\), étale morphism \(\varphi : U \to X\) and any points \(u, u' \in U\) mapping to \(x\) setting \(R = U \times_X U\) the fibre of \[|R| \to |U| \times_{|X|} |U|\] over \((u, u')\) is finite,
there exists a morphism \(\Spec(k) \to X\) with \(k\) a field in the equivalence class of \(x\) such that the projections \(\Spec(k) \times_X \Spec(k) \to \Spec(k)\) are étale and quasi-compact, and
there exists a monomorphism \(\Spec(k) \to X\) with \(k\) a field in the equivalence class of \(x\).
Proof
Assume (1), i.e., let \(\varphi : U \to X\) be an étale morphism from a scheme towards \(X\), and let \(u, u'\) be points of \(U\) lying over \(x\) such that the fibre of \(|R| \to |U| \times_{|X|} |U|\) over \((u, u')\) is a finite set. In this proof we think of a point \(u = \Spec(\kappa(u))\) as a scheme. Note that \(u \to U\), \(u' \to U\) are monomorphisms (see Schemes, Lemma 01L6), hence \(u \times_X u' \to R = U \times_X U\) is a monomorphism. In this language the assumption really means that \(u \times_X u'\) is a scheme whose underlying topological space has finitely many points. Let \(\psi : W \to X\) be an étale morphism from a scheme towards \(X\). Let \(w, w' \in W\) be points of \(W\) mapping to \(x\). We have to show that \(w \times_X w'\) is a scheme whose underlying topological space has finitely many points. Consider the fibre product diagram \[\xymatrix{ W \times_X U \ar[r]_p \ar[d]_q & U \ar[d]^\varphi \\ W \ar[r]^\psi & X }\] As \(x\) is the image of \(u\) and \(u'\) we may pick points \(\tilde w, \tilde w'\) in \(W \times_X U\) with \(q(\tilde w) = w\), \(q(\tilde w') = w'\), \(u = p(\tilde w)\) and \(u' = p(\tilde w')\), see Properties of Spaces, Lemma 03H4. As \(p\), \(q\) are étale the field extensions \(\kappa(w) \subset \kappa(\tilde w) \supset \kappa(u)\) and \(\kappa(w') \subset \kappa(\tilde w') \supset \kappa(u')\) are finite separable, see Remark 03II. Then we get a commutative diagram \[\xymatrix{ w \times_X w' \ar[d] & \tilde w \times_X \tilde w' \ar[l] \ar[d] \ar[r] & u \times_X u' \ar[d] \\ w \times_X w' & \tilde w \times_S \tilde w' \ar[l] \ar[r] & u \times_S u' }\] where the squares are fibre product squares. The lower horizontal morphisms are étale and quasi-compact, as any scheme of the form \(\Spec(k) \times_S \Spec(k')\) is affine, and by our observations about the field extensions above. Thus we see that the top horizontal arrows are étale and quasi-compact and hence have finite fibres. We have seen above that \(|u \times_X u'|\) is finite, so we conclude that \(|w \times_X w'|\) is finite. In other words, (2) holds.
Assume (2). Let \(U \to X\) be an étale morphism from a scheme \(U\) such that \(x\) is in the image of \(|U| \to |X|\). Let \(u \in U\) be a point mapping to \(x\). Then we have seen in the previous paragraph that \(u = \Spec(\kappa(u)) \to X\) has the property that \(u \times_X u\) has a finite underlying topological space. On the other hand, the projection maps \(u \times_X u \to u\) are the composition \[u \times_X u \longrightarrow u \times_X U \longrightarrow u \times_X X = u,\] i.e., the composition of a monomorphism (the base change of the monomorphism \(u \to U\)) by an étale morphism (the base change of the étale morphism \(U \to X\)). Hence \(u \times_X U\) is a disjoint union of spectra of fields finite separable over \(\kappa(u)\) (see Remark 03II). Since \(u \times_X u\) is finite the image of it in \(u \times_X U\) is a finite disjoint union of spectra of fields finite separable over \(\kappa(u)\). By Schemes, Lemma 03DP we conclude that \(u \times_X u\) is a finite disjoint union of spectra of fields finite separable over \(\kappa(u)\). In other words, we see that \(u \times_X u \to u\) is quasi-compact and étale. This means that (3) holds.
Let us prove that (3) implies (4). Let \(\Spec(k) \to X\) be a morphism from the spectrum of a field into \(X\), in the equivalence class of \(x\) such that the two projections \(t, s : R = \Spec(k) \times_X \Spec(k) \to \Spec(k)\) are quasi-compact and étale. This means in particular that \(R\) is an étale equivalence relation on \(\Spec(k)\). By Spaces, Theorem 02WW we know that the quotient sheaf \(X' = \Spec(k)/R\) is an algebraic space. By Groupoids, Lemma 02VH the map \(X' \to X\) is a monomorphism. Since \(s, t\) are quasi-compact, we see that \(R\) is quasi-compact and hence Properties of Spaces, Lemma 03DZ applies to \(X'\), and we see that \(X' = \Spec(k')\) for some field \(k'\). Hence we get a factorization \[\Spec(k) \longrightarrow \Spec(k') \longrightarrow X\] which shows that \(\Spec(k') \to X\) is a monomorphism mapping to \(x \in |X|\). In other words (4) holds.
Finally, we prove that (4) implies (1). Let \(\Spec(k) \to X\) be a monomorphism with \(k\) a field in the equivalence class of \(x\). Let \(U \to X\) be a surjective étale morphism from a scheme \(U\) to \(X\). Let \(u \in U\) be a point over \(x\). Since \(\Spec(k) \times_X u\) is nonempty, and since \(\Spec(k) \times_X u \to u\) is a monomorphism we conclude that \(\Spec(k) \times_X u = u\) (see Schemes, Lemma 03DP). Hence \(u \to U \to X\) factors through \(\Spec(k) \to X\), here is a picture \[\xymatrix{ u \ar[r] \ar[d] & U \ar[d] \\ \Spec(k) \ar[r] & X }\] Since the right vertical arrow is étale this implies that \(\kappa(u)/k\) is a finite separable extension. Hence we conclude that \[u \times_X u = u \times_{\Spec(k)} u\] is a finite scheme, and we win by the discussion of the meaning of property (1) in the first paragraph of this proof.
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). Let \(x \in |X|\). Let \(U\) be a scheme and let \(\varphi : U \to X\) be an étale morphism. The following are equivalent:
\(x\) is in the image of \(|U| \to |X|\), and setting \(R = U \times_X U\) the fibres of both \[|U| \longrightarrow |X| \quad\text{and}\quad |R| \longrightarrow |X|\] over \(x\) are finite,
there exists a monomorphism \(\Spec(k) \to X\) with \(k\) a field in the equivalence class of \(x\), and the fibre product \(\Spec(k) \times_X U\) is a finite nonempty scheme over \(k\).
Proof
Assume (1). This clearly implies the first condition of Lemma 03JU and hence we obtain a monomorphism \(\Spec(k) \to X\) in the class of \(x\). Taking the fibre product we see that \(\Spec(k) \times_X U \to \Spec(k)\) is a scheme étale over \(\Spec(k)\) with finitely many points, hence a finite nonempty scheme over \(k\), i.e., (2) holds.
Assume (2). By assumption \(x\) is in the image of \(|U| \to |X|\). The finiteness of the fibre of \(|U| \to |X|\) over \(x\) is clear since this fibre is equal to \(|\Spec(k) \times_X U|\) by Properties of Spaces, Lemma 03H4. The finiteness of the fibre of \(|R| \to |X|\) above \(x\) is also clear since it is equal to the set underlying the scheme \[(\Spec(k) \times_X U) \times_{\Spec(k)} (\Spec(k) \times_X U)\] which is finite over \(k\). Thus (1) holds.
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). Let \(x \in |X|\). The following are equivalent:
for every affine scheme \(U\), any étale morphism \(\varphi : U \to X\) setting \(R = U \times_X U\) the fibres of both \[|U| \longrightarrow |X| \quad\text{and}\quad |R| \longrightarrow |X|\] over \(x\) are finite,
there exist schemes \(U_i\) and étale morphisms \(U_i \to X\) such that \(\coprod U_i \to X\) is surjective and for each \(i\), setting \(R_i = U_i \times_X U_i\) the fibres of both \[|U_i| \longrightarrow |X| \quad\text{and}\quad |R_i| \longrightarrow |X|\] over \(x\) are finite,
there exists a monomorphism \(\Spec(k) \to X\) with \(k\) a field in the equivalence class of \(x\), and for any affine scheme \(U\) and étale morphism \(U \to X\) the fibre product \(\Spec(k) \times_X U\) is a finite scheme over \(k\),
there exists a quasi-compact monomorphism \(\Spec(k) \to X\) with \(k\) a field in the equivalence class of \(x\),
there exists a quasi-compact morphism \(\Spec(k) \to X\) with \(k\) a field in the equivalence class of \(x\), and
every morphism \(\Spec(k) \to X\) with \(k\) a field in the equivalence class of \(x\) is quasi-compact.
Proof
The equivalence of (1) and (3) follows on applying Lemma 040U to every étale morphism \(U \to X\) with \(U\) affine. It is clear that (3) implies (2). Assume \(U_i \to X\) and \(R_i\) are as in (2). We conclude from Lemma 03JS that for any affine scheme \(U\) and étale morphism \(U \to X\) the fibre of \(|U| \to |X|\) over \(x\) is finite. Say this fibre is \(\{u_1, \ldots, u_n\}\). Then, as Lemma 03JU (1) applies to \(U_i \to X\) for some \(i\) such that \(x\) is in the image of \(|U_i| \to |X|\), we see that the fibre of \(|R = U \times_X U| \to |U| \times_{|X|} |U|\) is finite over \((u_a, u_b)\), \(a, b \in \{1, \ldots, n\}\). Hence the fibre of \(|R| \to |X|\) over \(x\) is finite. In this way we see that (1) holds. At this point we know that (1), (2), and (3) are equivalent.
If (4) holds, then for any affine scheme \(U\) and étale morphism \(U \to X\) the scheme \(\Spec(k) \times_X U\) is on the one hand étale over \(k\) (hence a disjoint union of spectra of finite separable extensions of \(k\) by Remark 03II) and on the other hand quasi-compact over \(U\) (hence quasi-compact). Thus we see that (3) holds. Conversely, if \(U_i \to X\) is as in (2) and \(\Spec(k) \to X\) is a monomorphism as in (3), then \[\coprod \Spec(k) \times_X U_i \longrightarrow \coprod U_i\] is quasi-compact (because over each \(U_i\) we see that \(\Spec(k) \times_X U_i\) is a finite disjoint union spectra of fields). Thus \(\Spec(k) \to X\) is quasi-compact by Morphisms of Spaces, Lemma 03KG.
It is immediate that (4) implies (5). Conversely, let \(\Spec(k) \to X\) be a quasi-compact morphism in the equivalence class of \(x\). Let \(U \to X\) be an étale morphism with \(U\) affine. Consider the fibre product \[\xymatrix{ F \ar[r] \ar[d] & U \ar[d] \\ \Spec(k) \ar[r] & X }\] Then \(F \to U\) is quasi-compact, hence \(F\) is quasi-compact. On the other hand, \(F \to \Spec(k)\) is étale, hence \(F\) is a finite disjoint union of spectra of finite separable extensions of \(k\) (Remark 03II). Since the image of \(|F| \to |U|\) is the fibre of \(|U| \to |X|\) over \(x\) (Properties of Spaces, Lemma 03H4), we conclude that the fibre of \(|U| \to |X|\) over \(x\) is finite. The scheme \(F \times_{\Spec(k)} F\) is also a finite union of spectra of fields because it is also quasi-compact and étale over \(\Spec(k)\). There is a monomorphism \(F \times_X F \to F \times_{\Spec(k)} F\), hence \(F \times_X F\) is a finite disjoint union of spectra of fields (Schemes, Lemma 03DP). Thus the image of \(F \times_X F \to U \times_X U = R\) is finite. Since this image is the fibre of \(|R| \to |X|\) over \(x\) by Properties of Spaces, Lemma 03H4 we conclude that (1) holds. At this point we know that (1) – (5) are equivalent.
It is clear that (6) implies (5). Conversely, assume \(\Spec(k) \to X\) is as in (4) and let \(\Spec(k') \to X\) be another morphism with \(k'\) a field in the equivalence class of \(x\). By Properties of Spaces, Lemma 0H2X we have a factorization \(\Spec(k') \to \Spec(k) \to X\) of the given morphism. This is a composition of quasi-compact morphisms and hence quasi-compact (Morphisms of Spaces, Lemma 03HG) as desired.
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). The following are equivalent:
there exist schemes \(U_i\) and étale morphisms \(U_i \to X\) such that \(\coprod U_i \to X\) is surjective and each \(U_i \to X\) has universally bounded fibres, and
for every affine scheme \(U\) and étale morphism \(\varphi : U \to X\) the fibres of \(U \to X\) are universally bounded.
Proof
The implication (2) \(\Rightarrow\) (1) is trivial. Assume (1). Let \((\varphi_i : U_i \to X)_{i \in I}\) be a collection of étale morphisms from schemes towards \(X\), covering \(X\), such that each \(\varphi_i\) has universally bounded fibres. Let \(\psi : U \to X\) be an étale morphism from an affine scheme towards \(X\). For each \(i\) consider the fibre product diagram \[\xymatrix{ U \times_X U_i \ar[r]_{p_i} \ar[d]_{q_i} & U_i \ar[d]^{\varphi_i} \\ U \ar[r]^\psi & X }\] Since \(q_i\) is étale it is open (see Remark 03II). Moreover, we have \(U = \bigcup \Im(q_i)\), since the family \((\varphi_i)_{i \in I}\) is surjective. Since \(U\) is affine, hence quasi-compact we can finite finitely many \(i_1, \ldots, i_n \in I\) and quasi-compact opens \(W_j \subset U \times_X U_{i_j}\) such that \(U = \bigcup p_{i_j}(W_j)\). The morphism \(p_{i_j}\) is étale, hence locally quasi-finite (see remark on étale morphisms above). Thus we may apply Morphisms, Lemma 03JA to see the fibres of \(p_{i_j}|_{W_j} : W_j \to U_{i_j}\) are universally bounded. Hence by Lemma 03JM we see that the fibres of \(W_j \to X\) are universally bounded. Thus also \(\coprod_{j = 1, \ldots, n} W_j \to X\) has universally bounded fibres. Since \(\coprod_{j = 1, \ldots, n} W_j \to X\) factors through the surjective étale map \(\coprod q_{i_j}|_{W_j} : \coprod_{j = 1, \ldots, n} W_j \to U\) we see that the fibres of \(U \to X\) are universally bounded by Lemma 03JP. In other words (2) holds.
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). The following are equivalent:
there exists a Zariski covering \(X = \bigcup X_i\) and for each \(i\) a scheme \(U_i\) and a quasi-compact surjective étale morphism \(U_i \to X_i\), and
there exist schemes \(U_i\) and étale morphisms \(U_i \to X\) such that the projections \(U_i \times_X U_i \to U_i\) are quasi-compact and \(\coprod U_i \to X\) is surjective.
Proof
If (1) holds then the morphisms \(U_i \to X_i \to X\) are étale (combine Morphisms, Lemma 02GN and Spaces, Lemmas 02WK and 02WJ ). Moreover, as \(U_i \times_X U_i = U_i \times_{X_i} U_i\), both projections \(U_i \times_X U_i \to U_i\) are quasi-compact.
If (2) holds then let \(X_i \subset X\) be the open subspace corresponding to the image of the open map \(|U_i| \to |X|\), see Properties of Spaces, Lemma 06NF. The morphisms \(U_i \to X_i\) are surjective. Hence \(U_i \to X_i\) is surjective étale, and the projections \(U_i \times_{X_i} U_i \to U_i\) are quasi-compact, because \(U_i \times_{X_i} U_i = U_i \times_X U_i\). Thus by Spaces, Lemma 02WZ the morphisms \(U_i \to X_i\) are quasi-compact.
Conditions on algebraic spaces
In this section we discuss the relationship between various natural conditions on algebraic spaces we have seen above. Please read Section 03I7 to get a feeling for the meaning of these conditions.
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). Consider the following conditions on \(X\):
\((\alpha)\) For every \(x \in |X|\), the equivalent conditions of Lemma 03JS hold.
\((\beta)\) For every \(x \in |X|\), the equivalent conditions of Lemma 03JU hold.
\((\gamma)\) For every \(x \in |X|\), the equivalent conditions of Lemma 03JV hold.
\((\delta)\) The equivalent conditions of Lemma 03JT hold.
\((\epsilon)\) The equivalent conditions of Lemma 03IH hold.
\((\zeta)\) The space \(X\) is Zariski locally quasi-separated.
\((\eta)\) The space \(X\) is quasi-separated
\((\theta)\) The space \(X\) is representable, i.e., \(X\) is a scheme.
\((\iota)\) The space \(X\) is a quasi-separated scheme.
We have \[\xymatrix{ & (\theta) \ar@{=>}[rd] & & & & \\ (\iota) \ar@{=>}[ru] \ar@{=>}[rd] & & (\zeta) \ar@{=>}[r] & (\epsilon) \ar@{=>}[r] & (\delta) \ar@{=>}[r] & (\gamma) \ar@{<=>}[r] & (\alpha) + (\beta) \\ & (\eta) \ar@{=>}[ru] & & & & }\]
Proof
The implication \((\gamma) \Leftrightarrow (\alpha) + (\beta)\) is immediate. The implications in the diamond on the left are clear from the definitions.
Assume \((\zeta)\), i.e., that \(X\) is Zariski locally quasi-separated. Then \((\epsilon)\) holds by Properties of Spaces, Lemma 03W7.
Assume \((\epsilon)\). By Lemma 03IH there exists a Zariski open covering \(X = \bigcup X_i\) such that for each \(i\) there exists a scheme \(U_i\) and a quasi-compact surjective étale morphism \(U_i \to X_i\). Choose an \(i\) and an affine open subscheme \(W \subset U_i\). It suffices to show that \(W \to X\) has universally bounded fibres, since then the family of all these morphisms \(W \to X\) covers \(X\). To do this we consider the diagram \[\xymatrix{ W \times_X U_i \ar[r]_-p \ar[d]_q & U_i \ar[d] \\ W \ar[r] & X }\] Since \(W \to X\) factors through \(X_i\) we see that \(W \times_X U_i = W \times_{X_i} U_i\), and hence \(q\) is quasi-compact. Since \(W\) is affine this implies that the scheme \(W \times_X U_i\) is quasi-compact. Thus we may apply Morphisms, Lemma 03JA and we conclude that \(p\) has universally bounded fibres. From Lemma 03JO we conclude that \(W \to X\) has universally bounded fibres as well.
Assume \((\delta)\). Let \(U\) be an affine scheme, and let \(U \to X\) be an étale morphism. By assumption the fibres of the morphism \(U \to X\) are universally bounded. Thus also the fibres of both projections \(R = U \times_X U \to U\) are universally bounded, see Lemma 03JN. And by Lemma 03JM also the fibres of \(R \to X\) are universally bounded. Hence for any \(x \in X\) the fibres of \(|U| \to |X|\) and \(|R| \to |X|\) over \(x\) are finite, see Lemma 03JQ. In other words, the equivalent conditions of Lemma 03JV hold. This proves that \((\delta) \Rightarrow (\gamma)\).
Lemma
Let \(S\) be a scheme. Let \(\mathcal{P}\) be one of the properties \((\alpha)\), \((\beta)\), \((\gamma)\), \((\delta)\), \((\epsilon)\), \((\zeta)\), or \((\theta)\) of algebraic spaces listed in Lemma 03JX. Then if \(X\) is an algebraic space over \(S\), and \(X = \bigcup X_i\) is a Zariski open covering such that each \(X_i\) has \(\mathcal{P}\), then \(X\) has \(\mathcal{P}\).
Proof
Let \(X\) be an algebraic space over \(S\), and let \(X = \bigcup X_i\) is a Zariski open covering such that each \(X_i\) has \(\mathcal{P}\).
The case \(\mathcal{P} = (\alpha)\). The condition \((\alpha)\) for \(X_i\) means that for every \(x \in |X_i|\) and every affine scheme \(U\), and étale morphism \(\varphi : U \to X_i\) the fibre of \(\varphi : |U| \to |X_i|\) over \(x\) is finite. Consider \(x \in X\), an affine scheme \(U\) and an étale morphism \(U \to X\). Since \(X = \bigcup X_i\) is a Zariski open covering there exits a finite affine open covering \(U = U_1 \cup \ldots \cup U_n\) such that each \(U_j \to X\) factors through some \(X_{i_j}\). By assumption the fibres of \(|U_j | \to |X_{i_j}|\) over \(x\) are finite for \(j = 1, \ldots, n\). Clearly this means that the fibre of \(|U| \to |X|\) over \(x\) is finite. This proves the result for \((\alpha)\).
The case \(\mathcal{P} = (\beta)\). The condition \((\beta)\) for \(X_i\) means that every \(x \in |X_i|\) is represented by a monomorphism from the spectrum of a field towards \(X_i\). Hence the same follows for \(X\) as \(X_i \to X\) is a monomorphism and \(X = \bigcup X_i\).
The case \(\mathcal{P} = (\gamma)\). Note that \((\gamma) = (\alpha) + (\beta)\) by Lemma 03JX hence the lemma for \((\gamma)\) follows from the cases treated above.
The case \(\mathcal{P} = (\delta)\). The condition \((\delta)\) for \(X_i\) means there exist schemes \(U_{ij}\) and étale morphisms \(U_{ij} \to X_i\) with universally bounded fibres which cover \(X_i\). These schemes also give an étale surjective morphism \(\coprod U_{ij} \to X\) and \(U_{ij} \to X\) still has universally bounded fibres.
The case \(\mathcal{P} = (\epsilon)\). The condition \((\epsilon)\) for \(X_i\) means we can find a set \(J_i\) and morphisms \(\varphi_{ij} : U_{ij} \to X_i\) such that each \(\varphi_{ij}\) is étale, both projections \(U_{ij} \times_{X_i} U_{ij} \to U_{ij}\) are quasi-compact, and \(\coprod_{j \in J_i} U_{ij} \to X_i\) is surjective. In this case the compositions \(U_{ij} \to X_i \to X\) are étale (combine Morphisms, Lemmas 02GN and 02GP and Spaces, Lemmas 02WK and 02WJ ). Since \(X_i \subset X\) is a subspace we see that \(U_{ij} \times_{X_i} U_{ij} = U_{ij} \times_X U_{ij}\), and hence the condition on fibre products is preserved. And clearly \(\coprod_{i, j} U_{ij} \to X\) is surjective. Hence \(X\) satisfies \((\epsilon)\).
The case \(\mathcal{P} = (\zeta)\). The condition \((\zeta)\) for \(X_i\) means that \(X_i\) is Zariski locally quasi-separated. It is immediately clear that this means \(X\) is Zariski locally quasi-separated.
For \((\theta)\), see Properties of Spaces, Lemma 03JH.
Lemma
Let \(S\) be a scheme. Let \(\mathcal{P}\) be one of the properties \((\beta)\), \((\gamma)\), \((\delta)\), \((\epsilon)\), or \((\theta)\) of algebraic spaces listed in Lemma 03JX. Let \(X\), \(Y\) be algebraic spaces over \(S\). Let \(X \to Y\) be a representable morphism. If \(Y\) has property \(\mathcal{P}\), so does \(X\).
Proof
Assume \(f : X \to Y\) is a representable morphism of algebraic spaces, and assume that \(Y\) has \(\mathcal{P}\). Let \(x \in |X|\), and set \(y = f(x) \in |Y|\).
The case \(\mathcal{P} = (\beta)\). Condition \((\beta)\) for \(Y\) means there exists a monomorphism \(\Spec(k) \to Y\) representing \(y\). The fibre product \(X_y = \Spec(k) \times_Y X\) is a scheme, and \(x\) corresponds to a point of \(X_y\), i.e., to a monomorphism \(\Spec(k') \to X_y\). As \(X_y \to X\) is a monomorphism also we see that \(x\) is represented by the monomorphism \(\Spec(k') \to X_y \to X\). In other words \((\beta)\) holds for \(X\).
The case \(\mathcal{P} = (\gamma)\). Since \((\gamma) \Rightarrow (\beta)\) we have seen in the preceding paragraph that \(y\) and \(x\) can be represented by monomorphisms as in the following diagram \[\xymatrix{ \Spec(k') \ar[r]_-x \ar[d] & X \ar[d] \\ \Spec(k) \ar[r]^-y & Y }\] Also, by definition of property \((\gamma)\) via Lemma 03JV (2) there exist schemes \(V_i\) and étale morphisms \(V_i \to Y\) such that \(\coprod V_i \to Y\) is surjective and for each \(i\), setting \(R_i = V_i \times_Y V_i\) the fibres of both \[|V_i| \longrightarrow |Y| \quad\text{and}\quad |R_i| \longrightarrow |Y|\] over \(y\) are finite. This means that the schemes \((V_i)_y\) and \((R_i)_y\) are finite schemes over \(y = \Spec(k)\). As \(X \to Y\) is representable, the fibre products \(U_i = V_i \times_Y X\) are schemes. The morphisms \(U_i \to X\) are étale, and \(\coprod U_i \to X\) is surjective. Finally, for each \(i\) we have \[(U_i)_x = (V_i \times_Y X)_x = (V_i)_y \times_{\Spec(k)} \Spec(k')\] and \[(U_i \times_X U_i)_x = \left((V_i \times_Y X) \times_X (V_i \times_Y X)\right)_x = (R_i)_y \times_{\Spec(k)} \Spec(k')\] hence these are finite over \(k'\) as base changes of the finite schemes \((V_i)_y\) and \((R_i)_y\). This implies that \((\gamma)\) holds for \(X\), again via the second condition of Lemma 03JV.
The case \(\mathcal{P} = (\delta)\). Let \(V \to Y\) be an étale morphism with \(V\) an affine scheme. Since \(Y\) has property \((\delta)\) this morphism has universally bounded fibres. By Lemma 03JN the base change \(V \times_Y X \to X\) also has universally bounded fibres. Hence the first part of Lemma 03JT applies and we see that \(Y\) also has property \((\delta)\).
The case \(\mathcal{P} = (\epsilon)\). We will repeatedly use Spaces, Lemma 02WL. Let \(V_i \to Y\) be as in Lemma 03IH (2). Set \(U_i = X \times_Y V_i\). The morphisms \(U_i \to X\) are étale, and \(\coprod U_i \to X\) is surjective. Because \(U_i \times_X U_i = X \times_Y (V_i \times_Y V_i)\) we see that the projections \(U_i \times_Y U_i \to U_i\) are base changes of the projections \(V_i \times_Y V_i \to V_i\), and so quasi-compact as well. Hence \(X\) satisfies Lemma 03IH (2).
The case \(\mathcal{P} = (\theta)\). In this case the result is Categories, Lemma 03KC.
Reasonable and decent algebraic spaces
In Lemma 03JX we have seen a number of conditions on algebraic spaces related to the behaviour of étale morphisms from affine schemes into \(X\) and related to the existence of special étale coverings of \(X\) by schemes. We tabulate the different types of conditions here: \[\boxed{ \begin{matrix} (\alpha) & \text{fibres of \'etale morphisms from affines are finite} \\ (\beta) & \text{points come from monomorphisms of spectra of fields} \\ (\gamma) & \text{points come from quasi-compact monomorphisms of spectra of fields} \\ (\delta) & \text{fibres of \'etale morphisms from affines are universally bounded} \\ (\epsilon) & \text{cover by \'etale morphisms from schemes quasi-compact onto their image} \end{matrix} }\]
The conditions in the following definition are not exactly conditions on the diagonal of \(X\), but they are in some sense separation conditions on \(X\).
Definition
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\).
We say \(X\) is decent if for every point \(x \in X\) the equivalent conditions of Lemma 03JV hold, in other words property \((\gamma)\) of Lemma 03JX holds.
We say \(X\) is reasonable if the equivalent conditions of Lemma 03JT hold, in other words property \((\delta)\) of Lemma 03JX holds.
We say \(X\) is very reasonable if the equivalent conditions of Lemma 03IH hold, i.e., property \((\epsilon)\) of Lemma 03JX holds.
We have the following implications among these conditions on algebraic spaces: \[\xymatrix{ \text{representable} \ar@{=>}[rd] & & & \\ & \text{very reasonable} \ar@{=>}[r] & \text{reasonable} \ar@{=>}[r] & \text{decent} \\ \text{quasi-separated} \ar@{=>}[ru] & & & }\] The notion of a very reasonable algebraic space is obsolete. It was introduced because the assumption was needed to prove some results which are now proven for the class of decent spaces. The class of decent spaces is the largest class of spaces \(X\) where one has a good relationship between the topology of \(|X|\) and properties of \(X\) itself.
Example
The algebraic space \(\mathbf{A}^1_{\mathbf{Q}}/\mathbf{Z}\) constructed in Spaces, Example 02Z7 is not decent as its “generic point” cannot be represented by a monomorphism from the spectrum of a field.
Remark
Reasonable algebraic spaces are technically easier to work with than very reasonable algebraic spaces. For example, if \(X \to Y\) is a quasi-compact étale surjective morphism of algebraic spaces and \(X\) is reasonable, then so is \(Y\), see Lemma 03L2 but we don’t know if this is true for the property “very reasonable”. Below we give another technical property enjoyed by reasonable algebraic spaces.
Lemma
Let \(S\) be a scheme. Let \(X\) be a quasi-compact reasonable algebraic space. Then there exists a directed system of quasi-compact and quasi-separated algebraic spaces \(X_i\) such that \(X = \colim_i X_i\) (colimit in the category of sheaves). Moreover we can arrange it such that
for every quasi-compact scheme \(T\) over \(S\) we have \(\colim X_i(T) = X(T)\),
the transition morphisms \(X_i \to X_{i'}\) of the system and the coprojections \(X_i \to X\) are surjective and étale, and
if \(X\) is a scheme, then the algebraic spaces \(X_i\) are schemes and the transition morphisms \(X_i \to X_{i'}\) and the coprojections \(X_i \to X\) are local isomorphisms.
Proof
We sketch the proof. By Properties of Spaces, Lemma 03H6 we have \(X = U/R\) with \(U\) affine. In this case, reasonable means \(U \to X\) is universally bounded. Hence there exists an integer \(N\) such that the “fibres” of \(U \to X\) have degree at most \(N\), see Definition 03JL. Denote \(s, t : R \to U\) and \(c : R \times_{s, U, t} R \to R\) the groupoid structural maps.
Claim: for every quasi-compact open \(A \subset R\) there exists an open \(R' \subset R\) such that
\(A \subset R'\),
\(R'\) is quasi-compact, and
\((U, R', s|_{R'}, t|_{R'}, c|_{R' \times_{s, U, t} R'})\) is a groupoid scheme.
Note that \(e : U \to R\) is open as it is a section of the étale morphism \(s : R \to U\), see Étale Morphisms, Proposition 024T. Moreover \(U\) is affine hence quasi-compact. Hence we may replace \(A\) by \(A \cup e(U) \subset R\), and assume that \(A\) contains \(e(U)\). Next, we define inductively \(A^1 = A\), and \[A^n = c(A^{n - 1} \times_{s, U, t} A) \subset R\] for \(n \geq 2\). Arguing inductively, we see that \(A^n\) is quasi-compact for all \(n \geq 2\), as the image of the quasi-compact fibre product \(A^{n - 1} \times_{s, U, t} A\). If \(k\) is an algebraically closed field over \(S\), and we consider \(k\)-points then \[A^n(k) = \left\{(u, u') \in U(k) : \begin{matrix} \text{there exist } u = u_1, u_2, \ldots, u_n \in U(k)\text{ with} \\ (u_i , u_{i + 1}) \in A \text{ for all }i = 1, \ldots, n - 1. \end{matrix} \right\}\] But as the fibres of \(U(k) \to X(k)\) have size at most \(N\) we see that if \(n > N\) then we get a repeat in the sequence above, and we can shorten it proving \(A^N = A^n\) for all \(n \geq N\). This implies that \(R' = A^N\) gives a groupoid scheme \((U, R', s|_{R'}, t|_{R'}, c|_{R' \times_{s, U, t} R'})\), proving the claim above.
Consider the map of sheaves on \((\Sch/S)_{fppf}\) \[\colim_{R' \subset R} U/R' \longrightarrow U/R\] where \(R' \subset R\) runs over the quasi-compact open subschemes of \(R\) which give étale equivalence relations as above. Each of the quotients \(U/R'\) is an algebraic space (see Spaces, Theorem 02WW). Since \(R'\) is quasi-compact, and \(U\) affine the morphism \(R' \to U \times_{\Spec(\mathbf{Z})} U\) is quasi-compact, and hence \(U/R'\) is quasi-separated. Finally, if \(T\) is a quasi-compact scheme, then \[\colim_{R' \subset R} U(T)/R'(T) \longrightarrow U(T)/R(T)\] is a bijection, since every morphism from \(T\) into \(R\) ends up in one of the open subrelations \(R'\) by the claim above. This clearly implies that the colimit of the sheaves \(U/R'\) is \(U/R\). In other words the algebraic space \(X = U/R\) is the colimit of the quasi-separated algebraic spaces \(U/R'\).
Properties (1) and (2) follow from the discussion above. If \(X\) is a scheme, then if we choose \(U\) to be a finite disjoint union of affine opens of \(X\) we will obtain (3). Details omitted.
Lemma
Let \(S\) be a scheme. Let \(X\), \(Y\) be algebraic spaces over \(S\). Let \(X \to Y\) be a representable morphism. If \(Y\) is decent (resp. reasonable), then so is \(X\).
Proof
Translation of Lemma 03KF.
Lemma
Let \(S\) be a scheme. Let \(X \to Y\) be an étale morphism of algebraic spaces over \(S\). If \(Y\) is decent, resp. reasonable, then so is \(X\).
Proof
Let \(U\) be an affine scheme and \(U \to X\) an étale morphism. Set \(R = U \times_X U\) and \(R' = U \times_Y U\). Note that \(R \to R'\) is a monomorphism.
Let \(x \in |X|\). To show that \(X\) is decent, we have to show that the fibres of \(|U| \to |X|\) and \(|R| \to |X|\) over \(x\) are finite. But if \(Y\) is decent, then the fibres of \(|U| \to |Y|\) and \(|R'| \to |Y|\) are finite. Hence the result for “decent”.
To show that \(X\) is reasonable, we have to show that the fibres of \(U \to X\) are universally bounded. However, if \(Y\) is reasonable, then the fibres of \(U \to Y\) are universally bounded, which immediately implies the same thing for the fibres of \(U \to X\). Hence the result for “reasonable”.
Points and specializations
There exists an étale morphism of algebraic spaces \(f : X \to Y\) and a nontrivial specialization between points in a fibre of \(|f| : |X| \to |Y|\), see Examples, Lemma 06UK. If the source of the morphism is a scheme we can avoid this by imposing condition (\(\alpha\)) on \(Y\).
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). Let \(U \to X\) be an étale morphism from a scheme to \(X\). Assume \(u, u' \in |U|\) map to the same point \(x\) of \(|X|\), and \(u' \leadsto u\). If the pair \((X, x)\) satisfies the equivalent conditions of Lemma 03JS then \(u = u'\).
Proof
Assume the pair \((X, x)\) satisfies the equivalent conditions for Lemma 03JS. Let \(U\) be a scheme, \(U \to X\) étale, and let \(u, u' \in |U|\) map to \(x\) of \(|X|\), and \(u' \leadsto u\). We may and do replace \(U\) by an affine neighbourhood of \(u\). Let \(t, s : R = U \times_X U \to U\) be the étale projection maps.
Pick a point \(r \in R\) with \(t(r) = u\) and \(s(r) = u'\). This is possible by Properties of Spaces, Lemma 03BW. Because generalizations lift along the étale morphism \(t\) (Remark 03II) we can find a specialization \(r' \leadsto r\) with \(t(r') = u'\). Set \(u'' = s(r')\). Then \(u'' \leadsto u'\). Thus we may repeat and find \(r'' \leadsto r'\) with \(t(r'') = u''\). Set \(u''' = s(r'')\), and so on. Here is a picture: \[\xymatrix{ & r'' \ar[rd]^s \ar[ld]_t \ar@{~>}[d] & \\ u'' \ar@{~>}[d] & r' \ar[rd]^s \ar[ld]_t \ar@{~>}[d] & u''' \ar@{~>}[d] \\ u' \ar@{~>}[d] & r \ar[rd]^s \ar[ld]_t & u'' \ar@{~>}[d] \\ u & & u' }\] In Remark 03II we have seen that there are no specializations among points in the fibres of the étale morphism \(s\). Hence if \(u^{(n + 1)} = u^{(n)}\) for some \(n\), then also \(r^{(n)} = r^{(n - 1)}\) and hence also (by taking \(t\)) \(u^{(n)} = u^{(n - 1)}\). This then forces the whole tower to collapse, in particular \(u = u'\). Thus we see that if \(u \not = u'\), then all the specializations are strict and \(\{u, u', u'', \ldots\}\) is an infinite set of points in \(U\) which map to the point \(x\) in \(|X|\). As we chose \(U\) affine this contradicts the second part of Lemma 03JS, as desired.
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). Let \(U \to X\) be an étale morphism from a scheme to \(X\). Assume \(u, u' \in |U|\) map to the same point \(x\) of \(|X|\), and \(u' \leadsto u\). If \(X\) is locally Noetherian, then \(u = u'\).
Proof
The discussion in Schemes, Section 01J5 shows that \(\mathcal{O}_{U, u'}\) is a localization of the Noetherian local ring \(\mathcal{O}_{U, u}\). By Properties of Spaces, Lemma 0BAM we have \(\dim(\mathcal{O}_{U, u}) = \dim(\mathcal{O}_{U, u'})\). By dimension theory for Noetherian local rings we conclude \(u = u'\).
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). Let \(x, x' \in |X|\) and assume \(x' \leadsto x\), i.e., \(x\) is a specialization of \(x'\). Assume the pair \((X, x')\) satisfies the equivalent conditions of Lemma 03JV. Then for every étale morphism \(\varphi : U \to X\) from a scheme \(U\) and any \(u \in U\) with \(\varphi(u) = x\), exists a point \(u'\in U\), \(u' \leadsto u\) with \(\varphi(u') = x'\).
Proof
We may replace \(U\) by an affine open neighbourhood of \(u\). Hence we may assume that \(U\) is affine. As \(x\) is in the image of the open map \(|U| \to |X|\), so is \(x'\). Thus we may replace \(X\) by the Zariski open subspace corresponding to the image of \(|U| \to |X|\), see Properties of Spaces, Lemma 06NF. In other words we may assume that \(U \to X\) is surjective and étale. Let \(s, t : R = U \times_X U \to U\) be the projections. By our assumption that \((X, x')\) satisfies the equivalent conditions of Lemma 03JV we see that the fibres of \(|U| \to |X|\) and \(|R| \to |X|\) over \(x'\) are finite. Say \(\{u'_1, \ldots, u'_n\} \subset U\) and \(\{r'_1, \ldots, r'_m\} \subset R\) form the complete inverse image of \(\{x'\}\). Consider the closed sets \[T = \overline{\{u'_1\}} \cup \ldots \cup \overline{\{u'_n\}} \subset |U|, \quad T' = \overline{\{r'_1\}} \cup \ldots \cup \overline{\{r'_m\}} \subset |R|.\] Trivially we have \(s(T') \subset T\). Because \(R\) is an equivalence relation we also have \(t(T') = s(T')\) as the set \(\{r_j'\}\) is invariant under the inverse of \(R\) by construction. Let \(w \in T\) be any point. Then \(u'_i \leadsto w\) for some \(i\). Choose \(r \in R\) with \(s(r) = w\). Since generalizations lift along \(s : R \to U\), see Remark 03II, we can find \(r' \leadsto r\) with \(s(r') = u_i'\). Then \(r' = r'_j\) for some \(j\) and we conclude that \(w \in s(T')\). Hence \(T = s(T') = t(T')\) is an \(|R|\)-invariant closed set in \(|U|\). This means \(T\) is the inverse image of a closed (!) subset \(T'' = \varphi(T)\) of \(|X|\), see Properties of Spaces, Lemmas 03BW and 03BX. Hence \(T'' = \overline{\{x'\}}\). Thus \(T\) contains some point \(u_1\) mapping to \(x\) as \(x \in T''\). I.e., we see that for some \(i\) there exists a specialization \(u'_i \leadsto u_1\) which maps to the given specialization \(x' \leadsto x\).
To finish the proof, choose a point \(r \in R\) such that \(s(r) = u\) and \(t(r) = u_1\) (using Properties of Spaces, Lemma 03H4). As generalizations lift along \(t\), and \(u'_i \leadsto u_1\) we can find a specialization \(r' \leadsto r\) such that \(t(r') = u'_i\). Set \(u' = s(r')\). Then \(u' \leadsto u\) and \(\varphi(u') = x'\) as desired.
Lemma
Let \(S\) be a scheme. Let \(f : Y \to X\) be a flat morphism of algebraic spaces over \(S\). Let \(x, x' \in |X|\) and assume \(x' \leadsto x\), i.e., \(x\) is a specialization of \(x'\). Assume the pair \((X, x')\) satisfies the equivalent conditions of Lemma 03JV (for example if \(X\) is decent, \(X\) is quasi-separated, or \(X\) is representable). Then for every \(y \in |Y|\) with \(f(y) = x\), there exists a point \(y' \in |Y|\), \(y' \leadsto y\) with \(f(y') = x'\).
Proof
(The parenthetical statement holds by the definition of decent spaces and the implications between the different separation conditions mentioned in Section 03I7.) Choose a scheme \(V\) and a surjective étale morphism \(V \to Y\). Choose \(v \in V\) mapping to \(y\). Then we see that it suffices to prove the lemma for \(V \to X\). Thus we may assume \(Y\) is a scheme. Choose a scheme \(U\) and a surjective étale morphism \(U \to X\). Choose \(u \in U\) mapping to \(x\). By Lemma 03K2 we may choose \(u' \leadsto u\) mapping to \(x'\). By Properties of Spaces, Lemma 03H4 we may choose \(z \in U \times_X Y\) mapping to \(y\) and \(u\). Thus we reduce to the case of the flat morphism of schemes \(U \times_X Y \to U\) which is Morphisms, Lemma 03HV.
Stratifying algebraic spaces by schemes
In this section we prove that a quasi-compact and quasi-separated algebraic space has a finite stratification by locally closed subspaces each of which is a scheme and such that the glueing of the parts is by elementary distinguished squares. We first prove a slightly weaker result for reasonable algebraic spaces.
Lemma
Let \(S\) be a scheme. Let \(W \to X\) be a morphism of a scheme \(W\) to an algebraic space \(X\) which is flat, locally of finite presentation, separated, locally quasi-finite with universally bounded fibres. There exist reduced closed subspaces \[\emptyset = Z_{-1} \subset Z_0 \subset Z_1 \subset Z_2 \subset \ldots \subset Z_n = X\] such that with \(X_r = Z_r \setminus Z_{r - 1}\) the stratification \(X = \coprod_{r = 0, \ldots, n} X_r\) is characterized by the following universal property: Given \(g : T \to X\) the projection \(W \times_X T \to T\) is finite locally free of degree \(r\) if and only if \(g(|T|) \subset |X_r|\).
Proof
Let \(n\) be an integer bounding the degrees of the fibres of \(W \to X\). Choose a scheme \(U\) and a surjective étale morphism \(U \to X\). Apply More on Morphisms, Lemma 07RY to \(W \times_X U \to U\). We obtain closed subsets \[\emptyset = Y_{-1} \subset Y_0 \subset Y_1 \subset Y_2 \subset \ldots \subset Y_n = U\] characterized by the property stated in the lemma for the morphism \(W \times_X U \to U\). Clearly, the formation of these closed subsets commutes with base change. Setting \(R = U \times_X U\) with projection maps \(s, t : R \to U\) we conclude that \[s^{-1}(Y_r) = t^{-1}(Y_r)\] as closed subsets of \(R\). In other words the closed subsets \(Y_r \subset U\) are \(R\)-invariant. This means that \(|Y_r|\) is the inverse image of a closed subset \(Z_r \subset |X|\). Denote \(Z_r \subset X\) also the reduced induced algebraic space structure, see Properties of Spaces, Definition 047X.
Let \(g : T \to X\) be a morphism of algebraic spaces. Choose a scheme \(V\) and a surjective étale morphism \(V \to T\). To prove the final assertion of the lemma it suffices to prove the assertion for the composition \(V \to X\) (by our definition of finite locally free morphisms, see Morphisms of Spaces, Section 03ZT). Similarly, the morphism of schemes \(W \times_X V \to V\) is finite locally free of degree \(r\) if and only if the morphism of schemes \[W \times_X (U \times_X V) \longrightarrow U \times_X V\] is finite locally free of degree \(r\) (see Descent, Lemma 02VO). By construction this happens if and only if \(|U \times_X V| \to |U|\) maps into \(|Y_r|\), which is true if and only if \(|V| \to |X|\) maps into \(|Z_r|\).
Lemma
Let \(S\) be a scheme. Let \(W \to X\) be a morphism of a scheme \(W\) to an algebraic space \(X\) which is flat, locally of finite presentation, separated, and locally quasi-finite. Then there exist open subspaces \[X = X_0 \supset X_1 \supset X_2 \supset \ldots\] such that a morphism \(\Spec(k) \to X\) where \(k\) is a field factors through \(X_d\) if and only if \(W \times_X \Spec(k)\) has degree \(\geq d\) over \(k\).
Proof
Choose a scheme \(U\) and a surjective étale morphism \(U \to X\). Apply More on Morphisms, Lemma 086R to \(W \times_X U \to U\). We obtain open subschemes \[U = U_0 \supset U_1 \supset U_2 \supset \ldots\] characterized by the property stated in the lemma for the morphism \(W \times_X U \to U\). Clearly, the formation of these closed subsets commutes with base change. Setting \(R = U \times_X U\) with projection maps \(s, t : R \to U\) we conclude that \[s^{-1}(U_d) = t^{-1}(U_d)\] as open subschemes of \(R\). In other words the open subschemes \(U_d \subset U\) are \(R\)-invariant. This means that \(U_d\) is the inverse image of an open subspace \(X_d \subset X\) (Properties of Spaces, Lemma 07TW).
Lemma
Let \(S\) be a scheme. Let \(X\) be a quasi-compact algebraic space over \(S\). There exist open subspaces \[\ldots \subset U_4 \subset U_3 \subset U_2 \subset U_1 = X\] with the following properties:
setting \(T_p = U_p \setminus U_{p + 1}\) (with reduced induced subspace structure) there exists a separated scheme \(V_p\) and a surjective étale morphism \(f_p : V_p \to U_p\) such that \(f_p^{-1}(T_p) \to T_p\) is an isomorphism,
if \(x \in |X|\) can be represented by a quasi-compact morphism \(\Spec(k) \to X\) from a field, then \(x \in T_p\) for some \(p\).
Proof
By Properties of Spaces, Lemma 03H6 we can choose an affine scheme \(U\) and a surjective étale morphism \(U \to X\). For \(p \geq 0\) set \[W_p = U \times_X \ldots \times_X U \setminus \text{all diagonals}\] where the fibre product has \(p\) factors. Since \(U\) is separated, the morphism \(U \to X\) is separated and all fibre products \(U \times_X \ldots \times_X U\) are separated schemes. Since \(U \to X\) is separated the diagonal \(U \to U \times_X U\) is a closed immersion. Since \(U \to X\) is étale the diagonal \(U \to U \times_X U\) is an open immersion, see Morphisms of Spaces, Lemmas 06CR and 05W1. Similarly, all the diagonal morphisms are open and closed immersions and \(W_p\) is an open and closed subscheme of \(U \times_X \ldots \times_X U\). Moreover, the morphism \[U \times_X \ldots \times_X U \longrightarrow U \times_{\Spec(\mathbf{Z})} \ldots \times_{\Spec(\mathbf{Z})} U\] is locally quasi-finite and separated (Morphisms of Spaces, Lemma 03KN) and its target is an affine scheme. Hence every finite set of points of \(U \times_X \ldots \times_X U\) is contained in an affine open, see More on Morphisms, Lemma 07S0. Therefore, the same is true for \(W_p\). There is a free action of the symmetric group \(S_p\) on \(W_p\) over \(X\) (because we threw out the fix point locus from \(U \times_X \ldots \times_X U\)). By the above and Properties of Spaces, Proposition 07S6 the quotient \(V_p = W_p/S_p\) is a scheme. Since the action of \(S_p\) on \(W_p\) was over \(X\), there is a morphism \(V_p \to X\). Since \(W_p \to X\) is étale and since \(W_p \to V_p\) is surjective étale, it follows that also \(V_p \to X\) is étale, see Properties of Spaces, Lemma 03FS. Observe that \(V_p\) is a separated scheme by Properties of Spaces, Lemma 0BBM.
We let \(U_p \subset X\) be the open subspace which is the image of \(V_p \to X\). By construction a morphism \(\Spec(k) \to X\) with \(k\) algebraically closed, factors through \(U_p\) if and only if \(U \times_X \Spec(k)\) has \(\geq p\) points; as usual observe that \(U \times_X \Spec(k)\) is scheme theoretically a disjoint union of (possibly infinitely many) copies of \(\Spec(k)\), see Remark 03II. It follows that the \(U_p\) give a filtration of \(X\) as stated in the lemma. Moreover, our morphism \(\Spec(k) \to X\) factors through \(T_p\) if and only if \(U \times_X \Spec(k)\) has exactly \(p\) points. In this case we see that \(V_p \times_X \Spec(k)\) has exactly one point. Set \(Z_p = f_p^{-1}(T_p) \subset V_p\). This is a closed subscheme of \(V_p\). Then \(Z_p \to T_p\) is an étale morphism between algebraic spaces which induces a bijection on \(k\)-valued points for any algebraically closed field \(k\). To be sure this implies that \(Z_p \to T_p\) is universally injective, whence an open immersion by Morphisms of Spaces, Lemma 05W5 hence an isomorphism and (1) has been proved.
Let \(x : \Spec(k) \to X\) be a quasi-compact morphism where \(k\) is a field. Then the composition \(\Spec(\overline{k}) \to \Spec(k) \to X\) is quasi-compact as well (Morphisms of Spaces, Lemma 03HG). In this case the scheme \(U \times_X \Spec(\overline{k})\) is quasi-compact. In view of the fact (seen above) that it is a disjoint union of copies of \(\Spec(\overline{k})\) we find that it has finitely many points. If the number of points is \(p\), then we see that indeed \(x \in T_p\) and the proof is finished.
Lemma
Let \(S\) be a scheme. Let \(X\) be a quasi-compact, reasonable algebraic space over \(S\). There exist an integer \(n\) and open subspaces \[\emptyset = U_{n + 1} \subset U_n \subset U_{n - 1} \subset \ldots \subset U_1 = X\] with the following property: setting \(T_p = U_p \setminus U_{p + 1}\) (with reduced induced subspace structure) there exists a separated scheme \(V_p\) and a surjective étale morphism \(f_p : V_p \to U_p\) such that \(f_p^{-1}(T_p) \to T_p\) is an isomorphism.
Proof
The proof of this lemma is identical to the proof of Lemma 0BBN. Let \(n\) be an integer bounding the degrees of the fibres of \(U \to X\) which exists as \(X\) is reasonable, see Definition 03I8. Then we see that \(U_{n + 1} = \emptyset\) and the proof is complete.
Lemma
Let \(S\) be a scheme. Let \(X\) be a quasi-compact, reasonable algebraic space over \(S\). There exist an integer \(n\) and open subspaces \[\emptyset = U_{n + 1} \subset U_n \subset U_{n - 1} \subset \ldots \subset U_1 = X\] such that each \(T_p = U_p \setminus U_{p + 1}\) (with reduced induced subspace structure) is a scheme.
Proof
Immediate consequence of Lemma 07S9.
The following result is almost identical to [GruRay, Proposition 5.7.8].
Lemma
Let \(X\) be a quasi-compact and quasi-separated algebraic space over \(\Spec(\mathbf{Z})\). There exist an integer \(n\) and open subspaces \[\emptyset = U_{n + 1} \subset U_n \subset U_{n - 1} \subset \ldots \subset U_1 = X\] with the following property: setting \(T_p = U_p \setminus U_{p + 1}\) (with reduced induced subspace structure) there exists a quasi-compact separated scheme \(V_p\) and a surjective étale morphism \(f_p : V_p \to U_p\) such that \(f_p^{-1}(T_p) \to T_p\) is an isomorphism.
Proof
The proof of this lemma is identical to the proof of Lemma 0BBN. Observe that a quasi-separated space is reasonable, see Lemma 03JX and Definition 03I8. Hence we find that \(U_{n + 1} = \emptyset\) as in Lemma 07S9. At the end of the argument we add that since \(X\) is quasi-separated the schemes \(U \times_X \ldots \times_X U\) are all quasi-compact. Hence the schemes \(W_p\) are quasi-compact. Hence the quotients \(V_p = W_p/S_p\) by the symmetric group \(S_p\) are quasi-compact schemes.
The following lemma probably belongs somewhere else.
Lemma
Let \(S\) be a scheme. Let \(X\) be a quasi-separated algebraic space over \(S\). Let \(E \subset |X|\) be a subset. Then \(E\) is étale locally constructible (Properties of Spaces, Definition 0ECU) if and only if \(E\) is a locally constructible subset of the topological space \(|X|\) (Topology, Definition 005G).
Proof
Assume \(E \subset |X|\) is a locally constructible subset of the topological space \(|X|\). Let \(f : U \to X\) be an étale morphism where \(U\) is a scheme. We have to show that \(f^{-1}(E)\) is locally constructible in \(U\). The question is local on \(U\) and \(X\), hence we may assume that \(X\) is quasi-compact, \(E \subset |X|\) is constructible, and \(U\) is affine. In this case \(U \to X\) is quasi-compact, hence \(f : |U| \to |X|\) is quasi-compact. Observe that retrocompact opens of \(|X|\), resp. \(U\) are the same thing as quasi-compact opens of \(|X|\), resp. \(U\), see Topology, Lemma 0069. Thus \(f^{-1}(E)\) is constructible by Topology, Lemma 005I.
Conversely, assume \(E\) is étale locally constructible. We want to show that \(E\) is locally constructible in the topological space \(|X|\). The question is local on \(X\), hence we may assume that \(X\) is quasi-compact as well as quasi-separated. We will show that in this case \(E\) is constructible in \(|X|\). Choose open subspaces \[\emptyset = U_{n + 1} \subset U_n \subset U_{n - 1} \subset \ldots \subset U_1 = X\] and surjective étale morphisms \(f_p : V_p \to U_p\) inducing isomorphisms \(f_p^{-1}(T_p) \to T_p = U_p \setminus U_{p + 1}\) where \(V_p\) is a quasi-compact separated scheme as in Lemma 07ST. By definition the inverse image \(E_p \subset V_p\) of \(E\) is locally constructible in \(V_p\). Then \(E_p\) is constructible in \(V_p\) by Properties, Lemma 054E. Thus \(E_p \cap |f_p^{-1}(T_p)| = E \cap |T_p|\) is constructible in \(|T_p|\) by Topology, Lemma 09YE (observe that \(V_p \setminus f_p^{-1}(T_p)\) is quasi-compact as it is the inverse image of the quasi-compact space \(U_{p + 1}\) by the quasi-compact morphism \(f_p\)). Thus \[E = (|T_n| \cap E) \cup (|T_{n - 1}| \cap E) \cup \ldots \cup (|T_1| \cap E)\] is constructible by Topology, Lemma 09YK. Here we use that \(|T_p|\) is constructible in \(|X|\) which is clear from what was said above.
Integral cover by a scheme
Here we prove that given any quasi-compact and quasi-separated algebraic space \(X\), there is a scheme \(Y\) and a surjective, integral morphism \(Y \to X\). After we develop some theory about limits of algebraic spaces, we will prove that one can do this with a finite morphism, see Limits of Spaces, Section 0ACX.
Lemma
Let \(S\) be a scheme. Let \(j : V \to Y\) be a quasi-compact open immersion of algebraic spaces over \(S\). Let \(\pi : Z \to V\) be an integral morphism. Then there exists an integral morphism \(\nu : Y' \to Y\) such that \(Z\) is \(V\)-isomorphic to the inverse image of \(V\) in \(Y'\).
Proof
Since both \(j\) and \(\pi\) are quasi-compact and separated, so is \(j \circ \pi\). Let \(\nu : Y' \to Y\) be the normalization of \(Y\) in \(Z\), see Morphisms of Spaces, Section 0BAZ. Of course \(\nu\) is integral, see Morphisms of Spaces, Lemma 0823. The final statement follows formally from Morphisms of Spaces, Lemmas 0ABP and 0825.
Lemma
Let \(S\) be a scheme. Let \(X\) be a quasi-compact and quasi-separated algebraic space over \(S\).
There exists a surjective integral morphism \(Y \to X\) where \(Y\) is a scheme,
given a surjective étale morphism \(U \to X\) we may choose \(Y \to X\) such that for every \(y \in Y\) there is an open neighbourhood \(V \subset Y\) such that \(V \to X\) factors through \(U\).
Proof
Part (1) is the special case of part (2) where \(U = X\). Choose a surjective étale morphism \(U' \to U\) where \(U'\) is a scheme. It is clear that we may replace \(U\) by \(U'\) and hence we may assume \(U\) is a scheme. Since \(X\) is quasi-compact, there exist finitely many affine opens \(U_i \subset U\) such that \(U' = \coprod U_i \to X\) is surjective. After replacing \(U\) by \(U'\) again, we see that we may assume \(U\) is affine. Since \(X\) is quasi-separated, hence reasonable, there exists an integer \(d\) bounding the degree of the geometric fibres of \(U \to X\) (see Lemma 03JX). We will prove the lemma by induction on \(d\) for all quasi-compact and separated schemes \(U\) mapping surjective and étale onto \(X\). If \(d = 1\), then \(U = X\) and the result holds with \(Y = U\). Assume \(d > 1\).
We apply Morphisms of Spaces, Lemma 0ABS and we obtain a factorization \[\xymatrix{ U \ar[rr]_j \ar[rd] & & Y \ar[ld]^\pi \\ & X }\] with \(\pi\) integral and \(j\) a quasi-compact open immersion. We may and do assume that \(j(U)\) is scheme theoretically dense in \(Y\). Then \(U \times_X Y\) is a quasi-compact, separated scheme (being integral over \(U\)) and we have \[U \times_X Y = U \amalg W\] Here the first summand is the image of \(U \to U \times_X Y\) (which is closed by Morphisms of Spaces, Lemma 03KO and open because it is étale as a morphism between algebraic spaces étale over \(Y\)) and the second summand is the (open and closed) complement. The image \(V \subset Y\) of \(W\) is an open subspace containing \(Y \setminus U\).
The étale morphism \(W \to Y\) has geometric fibres of cardinality \(< d\). Namely, this is clear for geometric points of \(U \subset Y\) by inspection. Since \(|U| \subset |Y|\) is dense, it holds for all geometric points of \(Y\) by Lemma 07S8 (the degree of the fibres of a quasi-compact étale morphism does not go up under specialization). Thus we may apply the induction hypothesis to \(W \to V\) and find a surjective integral morphism \(Z \to V\) with \(Z\) a scheme, which Zariski locally factors through \(W\). Choose a factorization \(Z \to Z' \to Y\) with \(Z' \to Y\) integral and \(Z \to Z'\) open immersion (Lemma 0G2D). After replacing \(Z'\) by the scheme theoretic closure of \(Z\) in \(Z'\) we may assume that \(Z\) is scheme theoretically dense in \(Z'\). After doing this we have \(Z' \times_Y V = Z\). Finally, let \(T \subset Y\) be the induced closed subspace structure on \(Y \setminus V\). Consider the morphism \[Z' \amalg T \longrightarrow X\] This is a surjective integral morphism by construction. Since \(T \subset U\) it is clear that the morphism \(T \to X\) factors through \(U\). On the other hand, let \(z \in Z'\) be a point. If \(z \not \in Z\), then \(z\) maps to a point of \(Y \setminus V \subset U\) and we find a neighbourhood of \(z\) on which the morphism factors through \(U\). If \(z \in Z\), then we have an open neighbourhood of \(z\) in \(Z\) (which is also an open neighbourhood of \(z\) in \(Z'\)) which factors through \(W \subset U \times_X Y\) and hence through \(U\).
Lemma
Let \(S\) be a scheme. Let \(X\) be a quasi-compact and quasi-separated algebraic space over \(S\) such that \(|X|\) has finitely many irreducible components.
There exists a surjective integral morphism \(Y \to X\) where \(Y\) is a scheme such that \(f\) is finite étale over a quasi-compact dense open \(U \subset X\),
given a surjective étale morphism \(V \to X\) we may choose \(Y \to X\) such that for every \(y \in Y\) there is an open neighbourhood \(W \subset Y\) such that \(W \to X\) factors through \(V\).
Proof
The proof is the (roughly) same as the proof of Lemma 09YB with additional technical comments to obtain the dense quasi-compact open \(U\) (and unfortunately changes in notation to keep track of \(U\)).
Part (1) is the special case of part (2) where \(V = X\).
Proof of (2). Choose a surjective étale morphism \(V' \to V\) where \(V'\) is a scheme. It is clear that we may replace \(V\) by \(V'\) and hence we may assume \(V\) is a scheme. Since \(X\) is quasi-compact, there exist finitely many affine opens \(V_i \subset V\) such that \(V' = \coprod V_i \to X\) is surjective. After replacing \(V\) by \(V'\) again, we see that we may assume \(V\) is affine. Since \(X\) is quasi-separated, hence reasonable, there exists an integer \(d\) bounding the degree of the geometric fibres of \(V \to X\) (see Lemma 03JX).
By induction on \(d \geq 1\) we will prove the following induction hypothesis \((H_d)\):
for any quasi-compact and quasi-separated algebraic space \(X\) with finitely many irreducible components, for any \(m \geq 0\), for any quasi-compact and separated schemes \(V_j\), \(j = 1, \ldots, m\), for any étale morphisms \(\varphi_j : V_j \to X\), \(j = 1, \ldots, m\) such that \(d\) bounds the degree of the geometric fibres of \(\varphi_j : V_j\to X\) and \(\varphi = \coprod \varphi_j : V = \coprod V_j \to X\) is surjective, the statement of the lemma holds for \(\varphi : V \to X\).
If \(d = 1\), then each \(\varphi_j\) is an open immersion. Hence \(X\) is a scheme and the result holds with \(Y = V\). Assume \(d > 1\), assume \((H_{d - 1})\) and let \(m\), \(\varphi : V_j \to X\), \(j = 1, \ldots, m\) be as in \((H_d)\).
Let \(\eta_1, \ldots, \eta_n \in |X|\) be the generic points of the irreducible components of \(|X|\). By Properties of Spaces, Proposition 06NH there is an open subscheme \(U \subset X\) with \(\eta_1, \ldots, \eta_n \in U\). By shrinking \(U\) we may assume \(U\) affine and by Morphisms, Lemma 02NW we may assume each \(\varphi_j : V_j \to X\) is finite étale over \(U\). Of course, we see that \(U\) is quasi-compact and dense in \(X\) and that \(\varphi_j^{-1}(U)\) is dense in \(V_j\). In particular each \(V_j\) has finitely many irreducible components.
Fix \(j \in \{1, \ldots, m\}\). As in Morphisms of Spaces, Lemma 0ABS we let \(Y_j\) be the normalization of \(X\) in \(V_j\). We obtain a factorization \[\xymatrix{ V_j \ar[rr] \ar[rd]_{\varphi_j} & & Y_j \ar[ld]^{\pi_j} \\ & X }\] with \(\pi_j\) integral and \(V_j \to Y_j\) a quasi-compact open immersion. Since \(Y_j\) is the normalization of \(X\) in \(V_j\), we see from Morphisms of Spaces, Lemmas 0ABP and 0825 that \(\varphi_j^{-1}(U) \to \pi_j^{-1}(U)\) is an isomorphism. Thus \(\pi_j\) is finite étale over \(U\). Observe that \(V_j\) is scheme theoretically dense in \(Y_j\) because \(Y_j\) is the normalization of \(X\) in \(V_j\) (follows from the characterization of relative normalization in Morphisms of Spaces, Lemma 0823). Since \(V_j\) is quasi-compact we see that \(|V_j| \subset |Y_j|\) is dense, see Morphisms of Spaces, Section 0831 (and especially Morphisms of Spaces, Lemma 088G). It follows that \(|Y_j|\) has finitely many irreducible components. Then \(V_j \times_X Y_j\) is a quasi-compact, separated scheme (being finite over \(V_j\)) and \[V_j \times_X Y_j = V_j \amalg W_j\] Here the first summand is the image of \(V_j \to V_j \times_X Y_j\) (which is closed by Morphisms of Spaces, Lemma 03KO and open because it is étale as a morphism between algebraic spaces étale over \(Y\)) and the second summand is the (open and closed) complement.
The étale morphism \(W_j \to Y_j\) has geometric fibres of cardinality \(< d\). Namely, this is clear for geometric points of \(V_j \subset Y_j\) by inspection. Since \(|V_j| \subset |Y_j|\) is dense, it holds for all geometric points of \(Y_j\) by Lemma 07S8 (the degree of the fibres of a quasi-compact étale morphism does not go up under specialization). By \((H_{d - 1})\) applied to \(V_j \amalg W_j \to Y_j\) we find a surjective integral morphism \(Y_j' \to Y_j\) with \(Y_j'\) a scheme, which Zariski locally factors through \(V_j \amalg W_j\), and which is finite étale over a quasi-compact dense open \(U_j \subset Y_j\). After shrinking \(U\) we may and do assume that \(\pi_j^{-1}(U) \subset U_j\) (we may and do choose the same \(U\) for all \(j\); some details omitted).
We claim that \[Y = \coprod\nolimits_{j = 1, \ldots, m} Y'_j \longrightarrow X\] is the solution to our problem. First, this morphism is integral as on each summand we have the composition \(Y'_j \to Y \to X\) of integral morphisms (Morphisms of Spaces, Lemma 03ZR). Second, this morphism Zariski locally factors through \(V = \coprod V_j\) because we saw above that each \(Y'_j \to Y_j\) factors Zariski locally through \(V_j \amalg W_j = V_j \times_X Y_j\). Finally, since both \(Y'_j \to Y_j\) and \(Y_j \to X\) are finite étale over \(U\), so is the composition. This finishes the proof.
Schematic locus
In this section we prove that a decent algebraic space has a dense open subspace which is a scheme. We first prove this for reasonable algebraic spaces.
Proposition
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). If \(X\) is reasonable, then there exists a dense open subspace of \(X\) which is a scheme.
Proof
By Properties of Spaces, Lemma 03JH the question is local on \(X\). Hence we may assume there exists an affine scheme \(U\) and a surjective étale morphism \(U \to X\) (Properties of Spaces, Lemma 03FX). Let \(n\) be an integer bounding the degrees of the fibres of \(U \to X\) which exists as \(X\) is reasonable, see Definition 03I8. We will argue by induction on \(n\) that whenever
\(U \to X\) is a surjective étale morphism whose fibres have degree \(\leq n\), and
\(U\) is isomorphic to a locally closed subscheme of an affine scheme
then the schematic locus is dense in \(X\).
Let \(X_n \subset X\) be the open subspace which is the complement of the closed subspace \(Z_{n - 1} \subset X\) constructed in Lemma 07S8 using the morphism \(U \to X\). Let \(U_n \subset U\) be the inverse image of \(X_n\). Then \(U_n \to X_n\) is finite locally free of degree \(n\). Hence \(X_n\) is a scheme by Properties of Spaces, Proposition 07S6 (and the fact that any finite set of points of \(U_n\) is contained in an affine open of \(U_n\), see Properties, Lemma 01ZY).
Let \(X' \subset X\) be the open subspace such that \(|X'|\) is the interior of \(|Z_{n - 1}|\) in \(|X|\) (see Topology, Definition 03HN). Let \(U' \subset U\) be the inverse image. Then \(U' \to X'\) is surjective étale and has degrees of fibres bounded by \(n - 1\). By induction we see that the schematic locus of \(X'\) is an open dense \(X'' \subset X'\). By elementary topology we see that \(X'' \cup X_n \subset X\) is open and dense and we win.
Theorem
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). If \(X\) is decent, then there exists a dense open subspace of \(X\) which is a scheme.
Proof
Assume \(X\) is a decent algebraic space for which the theorem is false. By Properties of Spaces, Lemma 03JH there exists a largest open subspace \(X' \subset X\) which is a scheme. Since \(X'\) is not dense in \(X\), there exists an open subspace \(X'' \subset X\) such that \(|X''| \cap |X'| = \emptyset\). Replacing \(X\) by \(X''\) we get a nonempty decent algebraic space \(X\) which does not contain any open subspace which is a scheme.
Choose a nonempty affine scheme \(U\) and an étale morphism \(U \to X\). We may and do replace \(X\) by the open subscheme corresponding to the image of \(|U| \to |X|\). Consider the sequence of open subspaces \[X = X_0 \supset X_1 \supset X_2 \ldots\] constructed in Lemma 086T for the morphism \(U \to X\). Note that \(X_0 = X_1\) as \(U \to X\) is surjective. Let \(U = U_0 = U_1 \supset U_2 \ldots\) be the induced sequence of open subschemes of \(U\).
Choose a nonempty open affine \(V_1 \subset U_1\) (for example \(V_1 = U_1\)). By induction we will construct a sequence of nonempty affine opens \(V_1 \supset V_2 \supset \ldots\) with \(V_n \subset U_n\). Namely, having constructed \(V_1, \ldots, V_{n - 1}\) we can always choose \(V_n\) unless \(V_{n - 1} \cap U_n = \emptyset\). But if \(V_{n - 1} \cap U_n = \emptyset\), then the open subspace \(X' \subset X\) with \(|X'| = \Im(|V_{n - 1}| \to |X|)\) is contained in \(|X| \setminus |X_n|\). Hence \(V_{n - 1} \to X'\) is an étale morphism whose fibres have degree bounded by \(n - 1\). In other words, \(X'\) is reasonable (by definition), hence \(X'\) contains a nonempty open subscheme by Proposition 03JI. This is a contradiction which shows that we can pick \(V_n\).
By Limits, Lemma 01Z2 the limit \(V_\infty = \lim V_n\) is a nonempty scheme. Pick a morphism \(\Spec(k) \to V_\infty\). The composition \(\Spec(k) \to V_\infty \to U \to X\) has image contained in all \(X_d\) by construction. In other words, the fibred \(U \times_X \Spec(k)\) has infinite degree which contradicts the definition of a decent space. This contradiction finishes the proof of the theorem.
Lemma
Let \(S\) be a scheme. Let \(X \to Y\) be a surjective finite locally free morphism of algebraic spaces over \(S\). For \(y \in |Y|\) the following are equivalent
\(y\) is in the schematic locus of \(Y\), and
there exists an affine open \(U \subset X\) containing the preimage of \(y\).
Proof
If \(y \in Y\) is in the schematic locus, then it has an affine open neighbourhood \(V \subset Y\) and the inverse image \(U\) of \(V\) in \(X\) is an open finite over \(V\), hence affine. Thus (1) implies (2).
Conversely, assume that \(U \subset X\) as in (2) is given. Set \(R = X \times_Y X\) and denote the projections \(s, t : R \to X\). Consider \(Z = R \setminus s^{-1}(U) \cap t^{-1}(U)\). This is a closed subset of \(R\). The image \(t(Z)\) is a closed subset of \(X\) which can loosely be described as the set of points of \(X\) which are \(R\)-equivalent to a point of \(X \setminus U\). Hence \(U' = X \setminus t(Z)\) is an \(R\)-invariant, open subspace of \(X\) contained in \(U\) which contains the fibre of \(X \to Y\) over \(y\). Since \(X \to Y\) is open (Morphisms of Spaces, Lemma 042S) the image of \(U'\) is an open subspace \(V' \subset Y\). Since \(U'\) is \(R\)-invariant and \(R = X \times_Y X\), we see that \(U'\) is the inverse image of \(V'\) (use Properties of Spaces, Lemma 03H4). After replacing \(Y\) by \(V'\) and \(X\) by \(U'\) we see that we may assume \(X\) is a scheme isomorphic to an open subscheme of an affine scheme.
Assume \(X\) is a scheme isomorphic to an open subscheme of an affine scheme. In this case the fppf quotient sheaf \(X/R\) is a scheme, see Properties of Spaces, Proposition 07S6. Since \(Y\) is a sheaf in the fppf topology, obtain a canonical map \(X/R \to Y\) factoring \(X \to Y\). Since \(X \to Y\) is surjective finite locally free, it is surjective as a map of sheaves (Spaces, Lemma 05VM). We conclude that \(X/R \to Y\) is surjective as a map of sheaves. On the other hand, since \(R = X \times_Y X\) as sheaves we conclude that \(X/R \to Y\) is injective as a map of sheaves. Hence \(X/R \to Y\) is an isomorphism and we see that \(Y\) is representable.
At this point we have several different ways for proving the following lemma.
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). If there exists a finite, étale, surjective morphism \(U \to X\) where \(U\) is a scheme, then there exists a dense open subspace of \(X\) which is a scheme.
Proof
The morphism \(U \to X\) is finite locally free. Hence there is a decomposition of \(X\) into open and closed subspaces \(X_d \subset X\) such that \(U \times_X X_d \to X_d\) is finite locally free of degree \(d\). Thus we may assume \(U \to X\) is finite locally free of degree \(d\). In this case, let \(U_i \subset U\), \(i \in I\) be the set of affine opens. For each \(i\) the morphism \(U_i \to X\) is étale and has universally bounded fibres (namely, bounded by \(d\)). In other words, \(X\) is reasonable and the result follows from Proposition 03JI.
Proof
The question is local on \(X\) (Properties of Spaces, Lemma 03JH), hence may assume \(X\) is quasi-compact. Then \(U\) is quasi-compact. Then there exists a dense open subscheme \(W \subset U\) which is separated (Properties, Lemma 03J1). Set \(Z = U \setminus W\). Let \(R = U \times_X U\) and \(s, t : R \to U\) the projections. Then \(t^{-1}(Z)\) is nowhere dense in \(R\) (Topology, Lemma 03HR) and hence \(\Delta = s(t^{-1}(Z))\) is an \(R\)-invariant closed nowhere dense subset of \(U\) (Morphisms, Lemma 03HX). Let \(u \in U \setminus \Delta\) be a generic point of an irreducible component. Since these points are dense in \(U \setminus \Delta\) and since \(\Delta\) is nowhere dense, it suffices to show that the image \(x \in X\) of \(u\) is in the schematic locus of \(X\). Observe that \(t(s^{-1}(\{u\})) \subset W\) is a finite set of generic points of irreducible components of \(W\) (compare with Properties of Spaces, Lemma 0BAQ). By Properties, Lemma 01ZV we can find an affine open \(V \subset W\) such that \(t(s^{-1}(\{u\})) \subset V\). Since \(t(s^{-1}(\{u\}))\) is the fibre of \(|U| \to |X|\) over \(x\), we conclude by Lemma 0BA1.
Proof
(This proof is essentially the same as the second proof, but uses fewer references.) Assume \(X\) is an algebraic space, \(U\) a scheme, and \(U \to X\) is a finite étale surjective morphism. Write \(R = U \times_X U\) and denote \(s, t : R \to U\) the projections as usual. Note that \(s, t\) are surjective, finite and étale. Claim: The union of the \(R\)-invariant affine opens of \(U\) is topologically dense in \(U\).
Proof of the claim. Let \(W \subset U\) be an affine open. Set \(W' = t(s^{-1}(W)) \subset U\). Since \(s^{-1}(W)\) is affine (hence quasi-compact) we see that \(W' \subset U\) is a quasi-compact open. By Properties, Lemma 03J1 there exists a dense open \(W'' \subset W'\) which is a separated scheme. Set \(\Delta' = W' \setminus W''\). This is a nowhere dense closed subset of \(W''\). Since \(t|_{s^{-1}(W)} : s^{-1}(W) \to W'\) is open (because it is étale) we see that the inverse image \((t|_{s^{-1}(W)})^{-1}(\Delta') \subset s^{-1}(W)\) is a nowhere dense closed subset (see Topology, Lemma 03HR). Hence, by Morphisms, Lemma 03HX we see that \[\Delta = s\left((t|_{s^{-1}(W)})^{-1}(\Delta')\right)\] is a nowhere dense closed subset of \(W\). Pick any point \(\eta \in W\), \(\eta \not \in \Delta\) which is a generic point of an irreducible component of \(W\) (and hence of \(U\)). By our choices above the finite set \(t(s^{-1}(\{\eta\})) = \{\eta_1, \ldots, \eta_n\}\) is contained in the separated scheme \(W''\). Note that the fibres of \(s\) is are finite discrete spaces, and that generalizations lift along the étale morphism \(t\), see Morphisms, Lemmas 02GS and 03HV. In this way we see that each \(\eta_i\) is a generic point of an irreducible component of \(W''\). Thus, by Properties, Lemma 01ZV we can find an affine open \(V \subset W''\) such that \(\{\eta_1, \ldots, \eta_n\} \subset V\). By Groupoids, Lemma 03JE this implies that \(\eta\) is contained in an \(R\)-invariant affine open subscheme of \(U\). The claim follows as \(W\) was chosen as an arbitrary affine open of \(U\) and because the set of generic points of irreducible components of \(W \setminus \Delta\) is dense in \(W\).
Using the claim we can finish the proof. Namely, if \(W \subset U\) is an \(R\)-invariant affine open, then the restriction \(R_W\) of \(R\) to \(W\) equals \(R_W = s^{-1}(W) = t^{-1}(W)\) (see Groupoids, Definition 03BC and discussion following it). In particular the maps \(R_W \to W\) are finite étale also. It follows in particular that \(R_W\) is affine. Thus we see that \(W/R_W\) is a scheme, by Groupoids, Proposition 03BM. On the other hand, \(W/R_W\) is an open subspace of \(X\) by Spaces, Lemma 02WU. Hence having a dense collection of points contained in \(R\)-invariant affine open of \(U\) certainly implies that the schematic locus of \(X\) (see Properties of Spaces, Lemma 03JH) is open dense in \(X\).
Residue fields and henselian local rings
For a decent algebraic space we can define the residue field and the henselian local ring at a point. For example, the following lemma tells us the residue field of a point on a decent space is defined.
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). Consider the map \[\{\Spec(k) \to X \text{ monomorphism where }k\text{ is a field}\} \longrightarrow |X|\] This map is always injective. If \(X\) is decent then this map is a bijection.
Proof
We have seen in Properties of Spaces, Lemma 03E1 that the map is an injection in general. By Lemma 03JX it is surjective when \(X\) is decent (actually one can say this is part of the definition of being decent).
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). If a point \(x \in |X|\) can be represented by a monomorphism \(\Spec(k) \to X\), then the field \(k\) is unique up to unique isomorphism. For a decent algebraic space such a monomorphism exists for every point by Lemma 03K4 and hence the following definition makes sense.
Definition
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Let \(x \in |X|\). The residue field of \(X\) at \(x\) is the unique field \(\kappa(x)\) which comes equipped with a monomorphism \(\Spec(\kappa(x)) \to X\) representing \(x\).
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of decent algebraic spaces over \(S\). Let \(x \in |X|\) be a point. Set \(y = f(x) \in |Y|\). Then the composition \(\Spec(\kappa(x)) \to Y\) is in the equivalence class defining \(y\) and hence factors through \(\Spec(\kappa(y)) \to Y\). In other words we get a commutative diagram \[\xymatrix{ \Spec(\kappa(x)) \ar[r]_-x \ar[d] & X \ar[d]^f \\ \Spec(\kappa(y)) \ar[r]^-y & Y }\] The left vertical morphism corresponds to a homomorphism \(\kappa(y) \to \kappa(x)\) of fields. We will often simply call this the homomorphism induced by \(f\).
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of decent algebraic spaces over \(S\). Let \(x \in |X|\) be a point with image \(y = f(x) \in |Y|\). The following are equivalent
\(f\) induces an isomorphism \(\kappa(y) \to \kappa(x)\), and
the induced morphism \(\Spec(\kappa(x)) \to Y\) is a monomorphism.
Proof
Immediate from the discussion above.
The following lemma tells us that the henselian local ring of a point on a decent algebraic space is defined.
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). For every point \(x \in |X|\) there exists an étale morphism \[(U, u) \longrightarrow (X, x)\] where \(U\) is an affine scheme, \(u\) is the only point of \(U\) lying over \(x\), and the induced homomorphism \(\kappa(x) \to \kappa(u)\) is an isomorphism.
Proof
We may assume that \(X\) is quasi-compact by replacing \(X\) with a quasi-compact open containing \(x\). Recall that \(x\) can be represented by a quasi-compact (mono)morphism from the spectrum a field (by definition of decent spaces). Thus the lemma follows from Lemma 0BBN.
Definition
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). Let \(x \in X\) be a point. An elementary étale neighbourhood is an étale morphism \((U, u) \to (X, x)\) where \(U\) is a scheme, \(u \in U\) is a point mapping to \(x\), and the morphism \(u = \Spec(\kappa(u)) \to X\) is a monomorphism. A morphism of elementary étale neighbourhoods \((U, u) \to (U', u')\) is defined as a morphism \(U \to U'\) over \(X\) mapping \(u\) to \(u'\).
If \(X\) is not decent then the category of elementary étale neighbourhoods may be empty.
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Let \(x\) be a point of \(X\). The category of elementary étale neighborhoods of \((X, x)\) is cofiltered (see Categories, Definition 04AZ).
Proof
The category is nonempty by Lemma 0BBP. Suppose that we have two elementary étale neighbourhoods \((U_i, u_i) \to (X, x)\). Then consider \(U = U_1 \times_X U_2\). Since \(\Spec(\kappa(u_i)) \to X\), \(i = 1, 2\) are both monomorphisms in the class of \(x\) (Lemma 0EMX) , we see that \[u = \Spec(\kappa(u_1)) \times_X \Spec(\kappa(u_2))\] is the spectrum of a field \(\kappa(u)\) such that the induced maps \(\kappa(u_i) \to \kappa(u)\) are isomorphisms. Then \(u \to U\) is a point of \(U\) and we see that \((U, u) \to (X, x)\) is an elementary étale neighbourhood dominating \((U_i, u_i)\). If \(a, b : (U_1, u_1) \to (U_2, u_2)\) are two morphisms between our elementary étale neighbourhoods, then we consider the scheme \[U = U_1 \times_{(a, b), (U_2 \times_X U_2), \Delta} U_2\] Using Properties of Spaces, Lemma 03FV we see that \(U \to X\) is étale. Moreover, in exactly the same manner as before we see that \(U\) has a point \(u\) such that \((U, u) \to (X, x)\) is an elementary étale neighbourhood. Finally, \(U \to U_1\) equalizes \(a\) and \(b\) and the proof is finished.
Definition
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Let \(x \in |X|\). The henselian local ring of \(X\) at \(x\), is \[\mathcal{O}_{X, x}^h = \colim \Gamma(U, \mathcal{O}_U)\] where the colimit is over the elementary étale neighbourhoods \((U, u) \to (X, x)\).
Here is the analogue of Properties of Spaces, Lemma 04KF.
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Let \(x \in |X|\). Let \((U, u) \to (X, x)\) be an elementary étale neighbourhood. Then \[\mathcal{O}_{X, x}^h = \mathcal{O}_{U, u}^h\] In words: the henselian local ring of \(X\) at \(x\) is equal to the henselization \(\mathcal{O}_{U, u}^h\) of the local ring \(\mathcal{O}_{U, u}\) of \(U\) at \(u\).
Proof
Since the category of elementary étale neighbourhood of \((X, x)\) is cofiltered (Lemma 0BGV) we see that the category of elementary étale neighbourhoods of \((U, u)\) is initial in the category of elementary étale neighbourhood of \((X, x)\). Then the equality follows from More on Morphisms, Lemma 05KS and Categories, Lemma 04E7 (initial is turned into cofinal because the colimit defining henselian local rings is over the opposite of the category of elementary étale neighbourhoods).
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Let \(\overline{x}\) be a geometric point of \(X\) lying over \(x \in |X|\). The étale local ring \(\mathcal{O}_{X, \overline{x}}\) of \(X\) at \(\overline{x}\) (Properties of Spaces, Definition 04KG) is the strict henselization of the henselian local ring \(\mathcal{O}_{X, x}^h\) of \(X\) at \(x\).
Proof
Follows from Lemma 0EMY, Properties of Spaces, Lemma 04KF and the fact that \((R^h)^{sh} = R^{sh}\) for a local ring \((R, \mathfrak m, \kappa)\) and a given separable algebraic closure \(\kappa^{sep}\) of \(\kappa\). This equality follows from Algebra, Lemma 08HT.
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Let \(x \in |X|\). The residue field of the henselian local ring of \(X\) at \(x\) (Definition 0BGW) is the residue field of \(X\) at \(x\) (Definition 0EMW).
Proof
Choose an elementary étale neighbourhood \((U, u) \to (X, x)\). Then \(\kappa(u) = \kappa(x)\) and \(\mathcal{O}_{X, x}^h = \mathcal{O}_{U, u}^h\) (Lemma 0EMY). The residue field of \(\mathcal{O}_{U, u}^h\) is \(\kappa(u)\) by Algebra, Lemma 04GN (the output of this lemma is the construction/definition of the henselization of a local ring, see Algebra, Definition 04GQ).
Remark
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of decent algebraic spaces over \(S\). Let \(x \in |X|\) with image \(y \in |Y|\). Choose an elementary étale neighbourhood \((V, v) \to (Y, y)\) (possible by Lemma 0BBP). Then \(V \times_Y X\) is an algebraic space étale over \(X\) which has a unique point \(x'\) mapping to \(x\) in \(X\) and to \(v\) in \(V\). (Details omitted; use that all points can be represented by monomorphisms from spectra of fields.) Choose an elementary étale neighbourhood \((U, u) \to (V \times_Y X, x')\). Then we obtain the following commutative diagram \[\xymatrix{ \Spec(\mathcal{O}_{X, \overline{x}}) \ar[r] \ar[d] & \Spec(\mathcal{O}_{X, x}^h) \ar[r] \ar[d] & \Spec(\mathcal{O}_{U, u}) \ar[r] \ar[d] & U \ar[r] \ar[d] & X \ar[d] \\ \Spec(\mathcal{O}_{Y, \overline{y}}) \ar[r] & \Spec(\mathcal{O}_{Y, y}^h) \ar[r] & \Spec(\mathcal{O}_{V, v}) \ar[r] & V \ar[r] & Y }\] This comes from the identifications \(\mathcal{O}_{X, \overline{x}} = \mathcal{O}_{U, u}^{sh}\), \(\mathcal{O}_{X, x}^h = \mathcal{O}_{U, u}^h\), \(\mathcal{O}_{Y, \overline{y}} = \mathcal{O}_{V, v}^{sh}\), \(\mathcal{O}_{Y, y}^h = \mathcal{O}_{V, v}^h\) see in Lemma 0EMY and Properties of Spaces, Lemma 04KF and the functoriality of the (strict) henselization discussed in Algebra, Sections 0BSG and 0BSK.
Points on decent spaces
In this section we prove some properties of points on decent algebraic spaces. The following lemma shows that specialization of points behaves well on decent algebraic spaces. Spaces, Example 02Z8 shows that this is not true in general.
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Let \(U \to X\) be an étale morphism from a scheme to \(X\). If \(u, u' \in |U|\) map to the same point of \(|X|\), and \(u' \leadsto u\), then \(u = u'\).
Proof
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Let \(x, x' \in |X|\) and assume \(x' \leadsto x\), i.e., \(x\) is a specialization of \(x'\). Then for every étale morphism \(\varphi : U \to X\) from a scheme \(U\) and any \(u \in U\) with \(\varphi(u) = x\), exists a point \(u'\in U\), \(u' \leadsto u\) with \(\varphi(u') = x'\).
Proof
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Then \(|X|\) is Kolmogorov (see Topology, Definition 004X).
Proof
Let \(x_1, x_2 \in |X|\) with \(x_1 \leadsto x_2\) and \(x_2 \leadsto x_1\). We have to show that \(x_1 = x_2\). Pick a scheme \(U\) and an étale morphism \(U \to X\) such that \(x_1, x_2\) are both in the image of \(|U| \to |X|\). By Lemma 03IL we can find a specialization \(u_1 \leadsto u_2\) in \(U\) mapping to \(x_1 \leadsto x_2\). By Lemma 03IL we can find \(u_2' \leadsto u_1\) mapping to \(x_2 \leadsto x_1\). This means that \(u_2' \leadsto u_2\) is a specialization between points of \(U\) mapping to the same point of \(X\), namely \(x_2\). This is not possible, unless \(u_2' = u_2\), see Lemma 03K5. Hence also \(u_1 = u_2\) as desired.
Proposition
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Then the topological space \(|X|\) is sober (see Topology, Definition 004X).
Proof
We have seen in Lemma 03K3 that \(|X|\) is Kolmogorov. Hence it remains to show that every irreducible closed subset \(T \subset |X|\) has a generic point. By Properties of Spaces, Lemma 03IQ there exists a closed subspace \(Z \subset X\) with \(|Z| = |T|\). By definition this means that \(Z \to X\) is a representable morphism of algebraic spaces. Hence \(Z\) is a decent algebraic space by Lemma 03KF. By Theorem 086U we see that there exists an open dense subspace \(Z' \subset Z\) which is a scheme. This means that \(|Z'| \subset T\) is open dense. Hence the topological space \(|Z'|\) is irreducible, which means that \(Z'\) is an irreducible scheme. By Schemes, Lemma 01IS we conclude that \(|Z'|\) is the closure of a single point \(\eta \in T\) and hence also \(T = \overline{\{\eta\}}\), and we win.
For decent algebraic spaces dimension works as expected.
Lemma
Let \(S\) be a scheme. Dimension as defined in Properties of Spaces, Section 04N3 behaves well on decent algebraic spaces \(X\) over \(S\).
If \(x \in |X|\), then \(\dim_x(|X|) = \dim_x(X)\), and
\(\dim(|X|) = \dim(X)\).
Proof
Proof of (1). Choose a scheme \(U\) with a point \(u \in U\) and an étale morphism \(h : U \to X\) mapping \(u\) to \(x\). By definition the dimension of \(X\) at \(x\) is \(\dim_u(|U|)\). Thus we may pick \(U\) such that \(\dim_x(X) = \dim(|U|)\). Let \(d\) be an integer. If \(\dim(U) \geq d\), then there exists a sequence of nontrivial specializations \(u_d \leadsto \ldots \leadsto u_0\) in \(U\). Taking the image we find a corresponding sequence \(h(u_d) \leadsto \ldots \leadsto h(u_0)\) each of which is nontrivial by Lemma 03K5. Hence we see that the image of \(|U|\) in \(|X|\) has dimension at least \(d\). Conversely, suppose that \(x_d \leadsto \ldots \leadsto x_0\) is a sequence of specializations in \(|X|\) with \(x_0\) in the image of \(|U| \to |X|\). Then we can lift this to a sequence of specializations in \(U\) by Lemma 03IL.
Part (2) is an immediate consequence of part (1), Topology, Lemma 0B7I, and Properties of Spaces, Section 04N3.
Lemma
Let \(S\) be a scheme. Let \(X \to Y\) be a locally quasi-finite morphism of algebraic spaces over \(S\). Let \(x \in |X|\) with image \(y \in |Y|\). Then the dimension of the local ring of \(Y\) at \(y\) is \(\geq\) to the dimension of the local ring of \(X\) at \(x\).
Proof
The definition of the dimension of the local ring of a point on an algebraic space is given in Properties of Spaces, Definition 04NA. Choose an étale morphism \((V, v) \to (Y, y)\) where \(V\) is a scheme. Choose an étale morphism \(U \to V \times_Y X\) and a point \(u \in U\) mapping to \(x \in |X|\) and \(v \in V\). Then \(U \to V\) is locally quasi-finite and we have to prove that \[\dim(\mathcal{O}_{V, v}) \geq \dim(\mathcal{O}_{U, u})\] This is Algebra, Lemma 00QF.
Lemma
Let \(S\) be a scheme. Let \(X \to Y\) be a locally quasi-finite morphism of algebraic spaces over \(S\). Then \(\dim(X) \leq \dim(Y)\).
Proof
This follows from Lemma 0ABW and Properties of Spaces, Lemma 0BAN.
The following lemma is a tiny bit stronger than Properties of Spaces, Lemma 03DZ. We will improve this lemma in Lemma 047Z.
Lemma
Let \(S\) be a scheme. Let \(k\) be a field. Let \(X\) be an algebraic space over \(S\) and assume that there exists a surjective étale morphism \(\Spec(k) \to X\). If \(X\) is decent, then \(X \cong \Spec(k')\) where \(k/k'\) is a finite separable extension.
Proof
The assumption implies that \(|X| = \{x\}\) is a singleton. Since \(X\) is decent we can find a quasi-compact monomorphism \(\Spec(k') \to X\) whose image is \(x\). Then the projection \(U = \Spec(k') \times_X \Spec(k) \to \Spec(k)\) is a monomorphism, whence \(U = \Spec(k)\), see Schemes, Lemma 03DP. Hence the projection \(\Spec(k) = U \to \Spec(k')\) is étale and we win.
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 02Z6. We develop a tiny bit of machinery to be able to talk about these.
Lemma
Let \(S\) be a scheme. Let \(Z\) be an algebraic space over \(S\). Let \(k\) be a field and let \(\Spec(k) \to Z\) be surjective and flat. Then any morphism \(\Spec(k') \to Z\) where \(k'\) is a field is surjective and flat.
Proof
Consider the fibre square \[\xymatrix{ T \ar[d] \ar[r] & \Spec(k) \ar[d] \\ \Spec(k') \ar[r] & Z }\] Note that \(T \to \Spec(k')\) is flat and surjective hence \(T\) is not empty. On the other hand \(T \to \Spec(k)\) is flat as \(k\) is a field. Hence \(T \to Z\) is flat and surjective. It follows from Morphisms of Spaces, Lemma 05VY that \(\Spec(k') \to Z\) is flat. It is surjective as by assumption \(|Z|\) is a singleton.
Lemma
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 \(\Spec(k) \to Z\) where \(k\) is a field, and
there exists a locally of finite type, surjective, flat morphism \(\Spec(k) \to Z\) 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 \(\eta = \Spec(\kappa(\eta)) \to W\) is flat. We see that the composition \(\eta \to Z\) is flat (see Morphisms of Spaces, Lemma 03MN). 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 \(\Spec(k) \to Z\). Then \(W \times_Z \Spec(k)\) is a scheme étale over \(k\). Hence \(W \times_Z \Spec(k)\) is a disjoint union of spectra of fields (see Remark 03II), in particular reduced. Since \(W \times_Z \Spec(k) \to W\) is surjective and flat we conclude that \(W\) is reduced (Descent, Lemma 06QM). 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 \[\Spec(k) \xrightarrow{w} W \longrightarrow Z\] is locally of finite type by Morphisms of Spaces, Lemmas 03XG and 06LT. It is also flat and surjective by Lemma 06QV. Hence (3) holds.
The following lemma singles out a slightly better class of singleton algebraic spaces than the preceding lemma.
Lemma
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 \(\Spec(k) \to Z\) where \(k\) is a field.
Proof
Assume (2) holds. By Lemma 06QW 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 \(\Spec(k) \to Z\). Then \(W \times_Z \Spec(k)\) is a scheme étale over \(k\), hence a disjoint union of spectra of fields (see Remark 03II), hence locally Noetherian. Since \(W \times_Z \Spec(k) \to W\) is flat, surjective, and locally of finite presentation, we see that \(\{W \times_Z \Spec(k) \to W\}\) is an fppf covering and we conclude that \(W\) is locally Noetherian (Descent, Lemma 034C). 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 \(w : \Spec(k) \to W\) is of finite presentation, see Morphisms, Lemma 01TV. Hence the composition \[\Spec(k) \xrightarrow{w} W \longrightarrow Z\] is locally of finite presentation by Morphisms of Spaces, Lemmas 03XQ and 0468. It is also flat and surjective by Lemma 06QV. Hence (2) holds.
Lemma
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 \(\Spec(k) \to Z\). Then either \(Z'\) is empty or \(Z' = Z\).
Proof
We may assume that \(Z'\) is nonempty. In this case the fibre product \(T = Z' \times_Z \Spec(k)\) is nonempty, see Properties of Spaces, Lemma 03H4. Now \(T\) is an algebraic space and the projection \(T \to \Spec(k)\) is a monomorphism. Hence \(T = \Spec(k)\), see Morphisms of Spaces, Lemma 06MG. We conclude that \(\Spec(k) \to Z\) factors through \(Z'\). But as \(\Spec(k) \to Z\) is surjective, flat and locally of finite presentation, we see that \(\Spec(k) \to Z\) is surjective as a map of sheaves on \((\Sch/S)_{fppf}\) (see Spaces, Remark 02YN) and we conclude that \(Z' = Z\).
The following lemma says that to each point of an algebraic space we can associate a canonical reduced, locally Noetherian singleton algebraic space.
Lemma
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 06QX 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 0261). Set \[U' = \coprod\nolimits_{u \in U\text{ lying over }x} \Spec(\kappa(u)).\] 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 03II, and using Schemes, Lemma 03DP 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 02WW. As \(R'\) is the restriction of \(R\) by \(U' \to U\) we see \(Z \to X\) is a monomorphism by Groupoids, Lemma 02VH. Since \(Z \to X\) is a monomorphism we see that \(|Z| \to |X|\) is injective, see Morphisms of Spaces, Lemma 06RV. By Properties of Spaces, Lemma 03H4 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 06QX.
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 03H4. Hence the two projections are isomorphisms by Lemma 06QY and we win.
We introduce the following terminology which foreshadows the residual gerbes we will introduce later, see Properties of Stacks, Definition 06MU.
Definition
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). Let \(x \in |X|\). The residual space of \(X\) at \(x\)2 is the monomorphism \(Z_x \to X\) constructed in Lemma 06QZ.
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 \[\Spec(k) \longrightarrow Z_x.\] The residual space is often given by a monomorphism from the spectrum of a field.
Lemma
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 \(\Spec(k) \to X\) 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 \(Z_x = \Spec(k)\) for some field \(k\), then \(x\) is represented by the monomorphism \(\Spec(k) = Z_x \to X\). Conversely, if \(\Spec(k) \to X\) is a monomorphism which represents \(x\), then \(Z_x \times_X \Spec(k) \to \Spec(k)\) is a monomorphism whose source is nonempty by Properties of Spaces, Lemma 03H4. Hence \(Z_x \times_X \Spec(k) = \Spec(k)\) by Morphisms of Spaces, Lemma 06MG. Hence we get a monomorphism \(\Spec(k) \to Z_x\). This is an isomorphism by Lemma 06QY. The final statement follows from Lemma 03K4.
The residual space is a regular algebraic space by the following lemma.
Lemma
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 \(\Spec(k) \to Z\) be surjective, flat, and locally of finite presentation (see Lemma 06QX). The scheme \(T = W \times_Z \Spec(k)\) is étale over \(k\) in particular regular, see Remark 03II. Since \(T \to W\) is locally of finite presentation, flat, and surjective it follows that \(W\) is regular, see Descent, Lemma 06QN. By definition this means that \(Z\) is regular.
Lemma
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 0H1Q. The morphism \(Y' \to U\) maps \(|Y'|\) into \(E\). By our construction of \(Z_x\) in the proof of Lemma 06QZ 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 0H1N.
Lemma
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 \(x : \Spec(k) \to X\) where \(k\) is a field (for example if \(X\) is decent).
Then \(f\) factors through the residual space \(Z_x = \Spec(k)\) of \(X\) at \(x\).
Proof
By Lemma 0H1R we have \(Z_x = \Spec(k)\).
Preliminary remark: since \(\Spec(k) \to X\) 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 0BBN, \(x\) 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 03IE, \(f\) factors through \(U\), so we may assume \(U = X\). Then \(f\) factors through \(T\) because \(Y\) is reduced, see Properties of Spaces, Lemma 03JJ. 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 0H1M.
Example
Here is a counter example to Lemmas 0H1S and 0H1T 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., \(Y = \Spec(\text{locally constant maps } G \to k)\). Let \(\Gamma\) be \(G\) viewed as a discrete \(k\)-group scheme, acting on \(Y\) 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 \(x : \Spec(k) \to X\) which is a monomorphism since it is a section of \(X \to \Spec(k)\). We claim that (although \(Y\) is affine and reduced and \(|X| = \{x\}\)), the morphism \(q\) does not factor through any morphism \(\Spec(K) \to X\), where \(K\) is a field. Otherwise it would factor through \(x\) by Properties of Spaces, Lemma 0H2X. Now the pullback of \(q\) by \(x\) is \(\Gamma \to \Spec(k)\), with the projection \(\Gamma \to Y\) being the orbit map \(g \mapsto g \cdot e\). The latter has no section, whence the claim.
Lemma
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 04CD. 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 03IQ). Then \(Z\) is locally Noetherian by Morphisms of Spaces, Lemmas 06ED and 04ZK. Since also \(Z\) is reduced and \(|Z| = \{x\}\) it \(Z = Z_x\) is the residual space by definition.
Decent spaces
In this section we collect some useful facts on decent spaces.
Lemma
Any locally Noetherian decent algebraic space is quasi-separated.
Proof
Namely, let \(X\) be an algebraic space (over some base scheme, for example over \(\mathbf{Z}\)) which is decent and locally Noetherian. Let \(U \to X\) and \(V \to X\) be étale morphisms with \(U\) and \(V\) affine schemes. We have to show that \(W = U \times_X V\) is quasi-compact (Properties of Spaces, Lemma 0AHR). Since \(X\) is locally Noetherian, the schemes \(U\), \(V\) are Noetherian and \(W\) is locally Noetherian. Since \(X\) is decent, the fibres of the morphism \(W \to U\) are finite. Namely, we can represent any \(x \in |X|\) by a quasi-compact monomorphism \(\Spec(k) \to X\). Then \(U_k\) and \(V_k\) are finite disjoint unions of spectra of finite separable extensions of \(k\) (Remark 03II) and we see that \(W_k = U_k \times_{\Spec(k)} V_k\) is finite. Let \(n\) be the maximum degree of a fibre of \(W \to U\) at a generic point of an irreducible component of \(U\). Consider the stratification \[U = U_0 \supset U_1 \supset U_2 \supset \ldots\] associated to \(W \to U\) in More on Morphisms, Lemma 086R. By our choice of \(n\) above we conclude that \(U_{n + 1}\) is empty. Hence we see that the fibres of \(W \to U\) are universally bounded. Then we can apply More on Morphisms, Lemma 07RY to find a stratification \[\emptyset = Z_{-1} \subset Z_0 \subset Z_1 \subset Z_2 \subset \ldots \subset Z_n = U\] by closed subsets such that with \(S_r = Z_r \setminus Z_{r - 1}\) the morphism \(W \times_U S_r \to S_r\) is finite locally free. Since \(U\) is Noetherian, the schemes \(S_r\) are Noetherian, whence the schemes \(W \times_U S_r\) are Noetherian, whence \(W = \coprod W \times_U S_r\) is quasi-compact as desired.
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\).
If \(|X|\) is a singleton then \(X\) is a scheme.
If \(|X|\) is a singleton and \(X\) is reduced, then \(X \cong \Spec(k)\) for some field \(k\).
Proof
Assume \(|X|\) is a singleton. It follows immediately from Theorem 086U that \(X\) is a scheme, but we can also argue directly as follows. Choose an affine scheme \(U\) and a surjective étale morphism \(U \to X\). Set \(R = U \times_X U\). Then \(U\) and \(R\) have finitely many points by Lemma 03JV (and the definition of a decent space). All of these points are closed in \(U\) and \(R\) by Lemma 03K5. It follows that \(U\) and \(R\) are affine schemes. We may shrink \(U\) to a singleton space. Then \(U\) is the spectrum of a henselian local ring, see Algebra, Lemma 06RS. The projections \(R \to U\) are étale, hence finite étale because \(U\) is the spectrum of a \(0\)-dimensional henselian local ring, see Algebra, Lemma 04GG. It follows that \(X\) is a scheme by Groupoids, Proposition 03BM.
Part (2) follows from (1) and the fact that a reduced singleton scheme is the spectrum of a field.
Remark
We will see in Limits of Spaces, Lemma 07VU that an algebraic space whose reduction is a scheme is a scheme.
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Consider a commutative diagram \[\xymatrix{ \Spec(k) \ar[rr] \ar[rd] & & X \ar[ld] \\ & S }\] Assume that the image point \(s \in S\) of \(\Spec(k) \to S\) is a closed point and that \(\kappa(s) \subset k\) is algebraic. Then the image \(x\) of \(\Spec(k) \to X\) is a closed point of \(|X|\).
Proof
Suppose that \(x \leadsto x'\) for some \(x' \in |X|\). Choose an étale morphism \(U \to X\) where \(U\) is a scheme and a point \(u' \in U'\) mapping to \(x'\). Choose a specialization \(u \leadsto u'\) in \(U\) with \(u\) mapping to \(x\) in \(X\), see Lemma 03IL. Then \(u\) is the image of a point \(w\) of the scheme \(W = \Spec(k) \times_X U\). Since the projection \(W \to \Spec(k)\) is étale we see that \(\kappa(w) \supset k\) is finite. Hence \(\kappa(w) \supset \kappa(s)\) is algebraic. Hence \(\kappa(u) \supset \kappa(s)\) is algebraic. Thus \(u\) is a closed point of \(U\) by Morphisms, Lemma 01TE. Thus \(u = u'\), whence \(x = x'\).
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Consider a commutative diagram \[\xymatrix{ \Spec(k) \ar[rr] \ar[rd] & & X \ar[ld] \\ & S }\] Assume that the image point \(s \in S\) of \(\Spec(k) \to S\) is a closed point and that the field extension \(k/\kappa(s)\) is finite. Then \(\Spec(k) \to X\) is a finite morphism. If \(\kappa(s) = k\) then \(\Spec(k) \to X\) is a closed immersion.
Proof
By Lemma 07U5 the image point \(x \in |X|\) is closed. Let \(Z \subset X\) be the reduced closed subspace with \(|Z| = \{x\}\) (Properties of Spaces, Lemma 03IQ). Note that \(Z\) is a decent algebraic space by Lemma 0ABT. By Lemma 047Z we see that \(Z = \Spec(k')\) for some field \(k'\). Of course \(k \supset k' \supset \kappa(s)\). Then \(\Spec(k) \to Z\) is a finite morphism of schemes and \(Z \to X\) is a finite morphism as it is a closed immersion. Hence \(\Spec(k) \to X\) is finite (Morphisms of Spaces, Lemma 03ZR). If \(k = \kappa(s)\), then \(\Spec(k) = Z\) and \(\Spec(k) \to X\) is a closed immersion.
Lemma
Let \(S\) be a scheme. Suppose \(X\) is a decent algebraic space over \(S\). Let \(x \in |X|\) be a closed point. Then \(x\) can be represented by a closed immersion \(i : \Spec(k) \to X\) from the spectrum of a field.
Proof
We know that \(x\) can be represented by a quasi-compact monomorphism \(i : \Spec(k) \to X\) where \(k\) is a field (Definition 03I8). Let \(U \to X\) be an étale morphism where \(U\) is an affine scheme. As \(x\) is closed and \(X\) decent, the fibre \(F\) of \(|U| \to |X|\) over \(x\) consists of closed points (Lemma 03K5). As \(i\) is a monomorphism, so is \(U_k = U \times_X \Spec(k) \to U\). In particular, the map \(|U_k| \to F\) is injective. Since \(U_k\) is quasi-compact and étale over a field, we see that \(U_k\) is a finite disjoint union of spectra of fields (Remark 03II). Say \(U_k = \Spec(k_1) \amalg \ldots \amalg \Spec(k_r)\). Since \(\Spec(k_i) \to U\) is a monomorphism, we see that its image \(u_i\) has residue field \(\kappa(u_i) = k_i\). Since \(u_i \in F\) is a closed point we conclude the morphism \(\Spec(k_i) \to U\) is a closed immersion. As the \(u_i\) are pairwise distinct, \(U_k \to U\) is a closed immersion. Hence \(i\) is a closed immersion (Morphisms of Spaces, Lemma 03M4). This finishes the proof.
Locally separated spaces
It turns out that a locally separated algebraic space is decent.
Lemma
Let \(A\) be a ring. Let \(k\) be a field. Let \(\mathfrak p_n\), \(n \geq 1\) be a sequence of pairwise distinct primes of \(A\). Moreover, for each \(n\) let \(k \to \kappa(\mathfrak p_n)\) be an embedding. Then the closure of the image of \[\coprod\nolimits_{n \not = m} \Spec(\kappa(\mathfrak p_n) \otimes_k \kappa(\mathfrak p_m)) \longrightarrow \Spec(A \otimes A)\] meets the diagonal.
Proof
Set \(k_n = \kappa(\mathfrak p_n)\). We may assume that \(A = \prod k_n\). Denote \(x_n = \Spec(k_n)\) the open and closed point corresponding to \(A \to k_n\). Then \(\Spec(A) = Z \amalg \{x_n\}\) where \(Z\) is a nonempty closed subset. Namely, \(Z = V(e_n; n \geq 1)\) where \(e_n\) is the idempotent of \(A\) corresponding to the factor \(k_n\) and \(Z\) is nonempty as the ideal generated by the \(e_n\) is not equal to \(A\). We will show that the closure of the image contains \(\Delta(Z)\). The kernel of the map \[(\prod k_n) \otimes_k (\prod k_m) \longrightarrow \prod\nolimits_{n \not = m} k_n \otimes_k k_m\] is the ideal generated by \(e_n \otimes e_n\), \(n \geq 1\). Hence the closure of the image of the map on spectra is \(V(e_n \otimes e_n; n \geq 1)\) whose intersection with \(\Delta(\Spec(A))\) is \(\Delta(Z)\). Thus it suffices to show that \[\coprod\nolimits_{n \not = m} \Spec(k_n \otimes_k k_m) \longrightarrow \Spec(\prod\nolimits_{n \not = m} k_n \otimes_k k_m)\] has dense image. This follows as the family of ring maps \(\prod_{n \not = m} k_n \otimes_k k_m \to k_n \otimes_k k_m\) is jointly injective.
Lemma
A locally separated algebraic space is decent.
Proof
Let \(S\) be a scheme and let \(X\) be a locally separated algebraic space over \(S\). We may assume \(S = \Spec(\mathbf{Z})\), see Properties of Spaces, Definition 03BS. Unadorned fibre products will be over \(\mathbf{Z}\). Let \(x \in |X|\). Choose a scheme \(U\), an étale morphism \(U \to X\), and a point \(u \in U\) mapping to \(x\) in \(|X|\). As usual we identify \(u = \Spec(\kappa(u))\). As \(X\) is locally separated the morphism \[u \times_X u \to u \times u\] is an immersion (Morphisms of Spaces, Lemma 03KN). Hence More on Groupoids, Lemma 04Q9 tells us that it is a closed immersion (use Schemes, Lemma 01IQ). As \(u \times_X u \to u \times_X U\) is a monomorphism (base change of \(u \to U\)) and as \(u \times_X U \to u\) is étale we conclude that \(u \times_X u\) is a disjoint union of spectra of fields (see Remark 03II and Schemes, Lemma 03DP). Since it is also closed in the affine scheme \(u \times u\) we conclude \(u \times_X u\) is a finite disjoint union of spectra of fields. Thus \(x\) can be represented by a monomorphism \(\Spec(k) \to X\) where \(k\) is a field, see Lemma 03JU.
Next, let \(U = \Spec(A)\) be an affine scheme and let \(U \to X\) be an étale morphism. To finish the proof it suffices to show that \(F = U \times_X \Spec(k)\) is finite. Write \(F = \coprod_{i \in I} \Spec(k_i)\) as the disjoint union of finite separable extensions of \(k\). We have to show that \(I\) is finite. Set \(R = U \times_X U\). As \(X\) is locally separated, the morphism \(j : R \to U \times U\) is an immersion. Let \(U' \subset U \times U\) be an open such that \(j\) factors through a closed immersion \(j' : R \to U'\). Let \(e : U \to R\) be the diagonal map. Using that \(e\) is a morphism between schemes étale over \(U\) such that \(\Delta = j \circ e\) is a closed immersion, we conclude that \(R = e(U) \amalg W\) for some open and closed subscheme \(W \subset R\). Since \(j'\) is a closed immersion we conclude that \(j'(W) \subset U'\) is closed and disjoint from \(j'(e(U))\). Therefore \(\overline{j(W)} \cap \Delta(U) = \emptyset\) in \(U \times U\). Note that \(W\) contains \(\Spec(k_i \otimes_k k_{i'})\) for all \(i \not = i'\), \(i, i' \in I\). By Lemma 088I we conclude that \(I\) is finite as desired.
Valuative criterion
For a quasi-compact morphism from a decent space the valuative criterion is necessary in order for the morphism to be universally closed.
Proposition
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\). Assume \(f\) is quasi-compact, and \(X\) is decent. Then \(f\) is universally closed if and only if the existence part of the valuative criterion holds.
Proof
In Morphisms of Spaces, Lemma 03KA we have seen one of the implications. To prove the other, assume that \(f\) is universally closed. Let \[\xymatrix{ \Spec(K) \ar[r] \ar[d] & X \ar[d] \\ \Spec(A) \ar[r] & Y }\] be a diagram as in Morphisms of Spaces, Definition 03IX. Let \(X_A = \Spec(A) \times_Y X\), so that we have \[\xymatrix{ \Spec(K) \ar[r] \ar[rd] & X_A \ar[d] \\ & \Spec(A) }\] By Morphisms of Spaces, Lemma 03HF we see that \(X_A \to \Spec(A)\) is quasi-compact. Since \(X_A \to X\) is representable, we see that \(X_A\) is decent also, see Lemma 03KF. Moreover, as \(f\) is universally closed, we see that \(X_A \to \Spec(A)\) is universally closed. Hence we may and do replace \(X\) by \(X_A\) and \(Y\) by \(\Spec(A)\).
Let \(x' \in |X|\) be the equivalence class of \(\Spec(K) \to X\). Let \(y \in |Y| = |\Spec(A)|\) be the closed point. Set \(y' = f(x')\); it is the generic point of \(\Spec(A)\). Since \(f\) is universally closed we see that \(f(\overline{\{x'\}})\) contains \(\overline{\{y'\}}\), and hence contains \(y\). Let \(x \in \overline{\{x'\}}\) be a point such that \(f(x) = y\). Let \(U\) be a scheme, and \(\varphi : U \to X\) an étale morphism such that there exists a \(u \in U\) with \(\varphi(u) = x\). By Lemma 03K2 and our assumption that \(X\) is decent there exists a specialization \(u' \leadsto u\) on \(U\) with \(\varphi(u') = x'\). This means that there exists a common field extension \(K \subset K' \supset \kappa(u')\) such that \[\xymatrix{ \Spec(K') \ar[r] \ar[d] & U \ar[d] \\ \Spec(K) \ar[r] \ar[rd] & X \ar[d] \\ & \Spec(A) }\] is commutative. This gives the following commutative diagram of rings \[\xymatrix{ K' & \mathcal{O}_{U, u} \ar[l] \\ K \ar[u] & \\ & A \ar[lu] \ar[uu] }\] By Algebra, Lemma 00IA we can find a valuation ring \(A' \subset K'\) dominating the image of \(\mathcal{O}_{U, u}\) in \(K'\). Since by construction \(\mathcal{O}_{U, u}\) dominates \(A\) we see that \(A'\) dominates \(A\) also. Hence we obtain a diagram resembling the second diagram of Morphisms of Spaces, Definition 03IX and the proposition is proved.
Relative conditions
This is a (yet another) technical section dealing with conditions on algebraic spaces having to do with points. It is probably a good idea to skip this section.
Definition
Let \(S\) be a scheme. We say an algebraic space \(X\) over \(S\) has property \((\beta)\) if \(X\) has the corresponding property of Lemma 03JX. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\).
We say \(f\) has property \((\beta)\) if for any scheme \(T\) and morphism \(T \to Y\) the fibre product \(T \times_Y X\) has property \((\beta)\).
We say \(f\) is decent if for any scheme \(T\) and morphism \(T \to Y\) the fibre product \(T \times_Y X\) is a decent algebraic space.
We say \(f\) is reasonable if for any scheme \(T\) and morphism \(T \to Y\) the fibre product \(T \times_Y X\) is a reasonable algebraic space.
We say \(f\) is very reasonable if for any scheme \(T\) and morphism \(T \to Y\) the fibre product \(T \times_Y X\) is a very reasonable algebraic space.
We refer to Remark 03L4 for an informal discussion. It will turn out that the class of very reasonable morphisms is not so useful, but that the classes of decent and reasonable morphisms are useful.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\). We have the following implications among the conditions on \(f\): \[\xymatrix{ \text{representable} \ar@{=>}[rd] & & & & \\ & \text{very reasonable} \ar@{=>}[r] & \text{reasonable} \ar@{=>}[r] & \text{decent} \ar@{=>}[r] & (\beta) \\ \text{quasi-separated} \ar@{=>}[ru] & & & & }\]
Proof
This is clear from the definitions, Lemma 03JX and Morphisms of Spaces, Lemma 03KM.
Here is another sanity check.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\). If \(X\) is decent (resp. is reasonable, resp. has property \((\beta)\) of Lemma 03JX), then \(f\) is decent (resp. reasonable, resp. has property \((\beta)\)).
Proof
Let \(T\) be a scheme and let \(T \to Y\) be a morphism. Then \(T \to Y\) is representable, hence the base change \(T \times_Y X \to X\) is representable. Hence if \(X\) is decent (or reasonable), then so is \(T \times_Y X\), see Lemma 0ABT. Similarly, for property \((\beta)\), see Lemma 03KF.
Lemma
Having property \((\beta)\), being decent, or being reasonable is preserved under arbitrary base change.
Proof
This is immediate from the definition.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\). Let \(\omega \in \{\beta, decent, reasonable\}\). Suppose that \(Y\) has property \((\omega)\) and \(f : X \to Y\) has \((\omega)\). Then \(X\) has \((\omega)\).
Proof
Let us prove the lemma in case \(\omega = \beta\). In this case we have to show that any \(x \in |X|\) is represented by a monomorphism from the spectrum of a field into \(X\). Let \(y = f(x) \in |Y|\). By assumption there exists a field \(k\) and a monomorphism \(\Spec(k) \to Y\) representing \(y\). Then \(x\) corresponds to a point \(x'\) of \(\Spec(k) \times_Y X\). By assumption \(x'\) is represented by a monomorphism \(\Spec(k') \to \Spec(k) \times_Y X\). Clearly the composition \(\Spec(k') \to X\) is a monomorphism representing \(x\).
Let us prove the lemma in case \(\omega = decent\). Let \(x \in |X|\) and \(y = f(x) \in |Y|\). By the result of the preceding paragraph we can choose a diagram \[\xymatrix{ \Spec(k') \ar[r]_x \ar[d] & X \ar[d]^f \\ \Spec(k) \ar[r]^y & Y }\] whose horizontal arrows monomorphisms. As \(Y\) is decent the morphism \(y\) is quasi-compact. As \(f\) is decent the algebraic space \(\Spec(k) \times_Y X\) is decent. Hence the monomorphism \(\Spec(k') \to \Spec(k) \times_Y X\) is quasi-compact. Then the monomorphism \(x : \Spec(k') \to X\) is quasi-compact as a composition of quasi-compact morphisms (use Morphisms of Spaces, Lemmas 03HF and 03HG). As the point \(x\) was arbitrary this implies \(X\) is decent.
Let us prove the lemma in case \(\omega = reasonable\). Choose \(V \to Y\) étale with \(V\) an affine scheme. Choose \(U \to V \times_Y X\) étale with \(U\) an affine scheme. By assumption \(V \to Y\) has universally bounded fibres. By Lemma 03JN the morphism \(V \times_Y X \to X\) has universally bounded fibres. By assumption on \(f\) we see that \(U \to V \times_Y X\) has universally bounded fibres. By Lemma 03JM the composition \(U \to X\) has universally bounded fibres. Hence there exists sufficiently many étale morphisms \(U \to X\) from schemes with universally bounded fibres, and we conclude that \(X\) is reasonable.
Lemma
Having property \((\beta)\), being decent, or being reasonable is preserved under compositions.
Proof
Let \(\omega \in \{\beta, decent, reasonable\}\). Let \(f : X \to Y\) and \(g : Y \to Z\) be morphisms of algebraic spaces over the scheme \(S\). Assume \(f\) and \(g\) both have property \((\omega)\). Then we have to show that for any scheme \(T\) and morphism \(T \to Z\) the space \(T \times_Z X\) has \((\omega)\). By Lemma 03L0 this reduces us to the following claim: Suppose that \(Y\) is an algebraic space having property \((\omega)\), and that \(f : X \to Y\) is a morphism with \((\omega)\). Then \(X\) has \((\omega)\). This is the content of Lemma 0ABY.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\), \(g : Z \to Y\) be morphisms of algebraic spaces over \(S\). If \(X\) and \(Z\) are decent (resp. reasonable, resp. have property \((\beta)\) of Lemma 03JX), then so does \(X \times_Y Z\).
Proof
Namely, by Lemma 0ABX the morphism \(X \to Y\) has the property. Then the base change \(X \times_Y Z \to Z\) has the property by Lemma 03L0. And finally this implies \(X \times_Y Z\) has the property by Lemma 0ABY.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\). Let \(\mathcal{P} \in \{(\beta), decent, reasonable\}\). Assume
\(f\) is quasi-compact,
\(f\) is étale,
\(|f| : |X| \to |Y|\) is surjective, and
the algebraic space \(X\) has property \(\mathcal{P}\).
Then \(Y\) has property \(\mathcal{P}\).
Proof
Let us prove this in case \(\mathcal{P} = (\beta)\). Let \(y \in |Y|\) be a point. We have to show that \(y\) can be represented by a monomorphism from a field. Choose a point \(x \in |X|\) with \(f(x) = y\). By assumption we may represent \(x\) by a monomorphism \(\Spec(k) \to X\), with \(k\) a field. By Lemma 03JU it suffices to show that the projections \(\Spec(k) \times_Y \Spec(k) \to \Spec(k)\) are étale and quasi-compact. We can factor the first projection as \[\Spec(k) \times_Y \Spec(k) \longrightarrow \Spec(k) \times_Y X \longrightarrow \Spec(k)\] The first morphism is a monomorphism, and the second is étale and quasi-compact. By Properties of Spaces, Lemma 03KX we see that \(\Spec(k) \times_Y X\) is a scheme. Hence it is a finite disjoint union of spectra of finite separable field extensions of \(k\). By Schemes, Lemma 03DP we see that the first arrow identifies \(\Spec(k) \times_Y \Spec(k)\) with a finite disjoint union of spectra of finite separable field extensions of \(k\). Hence the projection morphism is étale and quasi-compact.
Let us prove this in case \(\mathcal{P} = decent\). We have already seen in the first paragraph of the proof that this implies that every \(y \in |Y|\) can be represented by a monomorphism \(y : \Spec(k) \to Y\). Pick such a \(y\). Pick an affine scheme \(U\) and an étale morphism \(U \to X\) such that the image of \(|U| \to |Y|\) contains \(y\). By Lemma 03JV it suffices to show that \(U_y\) is a finite scheme over \(k\). The fibre product \(X_y = \Spec(k) \times_Y X\) is a quasi-compact étale algebraic space over \(k\). Hence by Properties of Spaces, Lemma 03KX it is a scheme. So it is a finite disjoint union of spectra of finite separable extensions of \(k\). Say \(X_y = \{x_1, \ldots, x_n\}\) so \(x_i\) is given by \(x_i : \Spec(k_i) \to X\) with \([k_i : k] < \infty\). By assumption \(X\) is decent, so the schemes \(U_{x_i} = \Spec(k_i) \times_X U\) are finite over \(k_i\). Finally, we note that \(U_y = \coprod U_{x_i}\) as a scheme and we conclude that \(U_y\) is finite over \(k\) as desired.
Let us prove this in case \(\mathcal{P} = reasonable\). Pick an affine scheme \(V\) and an étale morphism \(V \to Y\). We have the show the fibres of \(V \to Y\) are universally bounded. The algebraic space \(V \times_Y X\) is quasi-compact. Thus we can find an affine scheme \(W\) and a surjective étale morphism \(W \to V \times_Y X\), see Properties of Spaces, Lemma 03H6. Here is a picture (solid diagram) \[\xymatrix{ W \ar[r] \ar[rd] & V \times_Y X \ar[r] \ar[d] & X \ar[d]_f & \Spec(k) \ar@{..>}[l]^x \ar@{..>}[ld]^y \\ & V \ar[r] & Y }\] The morphism \(W \to X\) is universally bounded by our assumption that the space \(X\) is reasonable. Let \(n\) be an integer bounding the degrees of the fibres of \(W \to X\). We claim that the same integer works for bounding the fibres of \(V \to Y\). Namely, suppose \(y \in |Y|\) is a point. Then there exists a \(x \in |X|\) with \(f(x) = y\) (see above). This means we can find a field \(k\) and morphisms \(x, y\) given as dotted arrows in the diagram above. In particular we get a surjective étale morphism \[\Spec(k) \times_{x, X} W \to \Spec(k) \times_{x, X} (V \times_Y X) = \Spec(k) \times_{y, Y} V\] which shows that the degree of \(\Spec(k) \times_{y, Y} V\) over \(k\) is less than or equal to the degree of \(\Spec(k) \times_{x, X} W\) over \(k\), i.e., \(\leq n\), and we win. (This last part of the argument is the same as the argument in the proof of Lemma 03JO. Unfortunately that lemma is not general enough because it only applies to representable morphisms.)
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\). Let \(\mathcal{P} \in \{(\beta), decent, reasonable, very\ reasonable\}\). The following are equivalent
\(f\) is \(\mathcal{P}\),
for every affine scheme \(Z\) and every morphism \(Z \to Y\) the base change \(Z \times_Y X \to Z\) of \(f\) is \(\mathcal{P}\),
for every affine scheme \(Z\) and every morphism \(Z \to Y\) the algebraic space \(Z \times_Y X\) is \(\mathcal{P}\), and
there exists a Zariski covering \(Y = \bigcup Y_i\) such that each morphism \(f^{-1}(Y_i) \to Y_i\) has \(\mathcal{P}\).
If \(\mathcal{P} \in \{(\beta), decent, reasonable\}\), then this is also equivalent to
there exists a scheme \(V\) and a surjective étale morphism \(V \to Y\) such that the base change \(V \times_Y X \to V\) has \(\mathcal{P}\).
Proof
The implications (1) \(\Rightarrow\) (2) \(\Rightarrow\) (3) \(\Rightarrow\) (4) are trivial. The implication (3) \(\Rightarrow\) (1) can be seen as follows. Let \(Z \to Y\) be a morphism whose source is a scheme over \(S\). Consider the algebraic space \(Z \times_Y X\). If we assume (3), then for any affine open \(W \subset Z\), the open subspace \(W \times_Y X\) of \(Z \times_Y X\) has property \(\mathcal{P}\). Hence by Lemma 03KE the space \(Z \times_Y X\) has property \(\mathcal{P}\), i.e., (1) holds. A similar argument (omitted) shows that (4) implies (1).
The implication (1) \(\Rightarrow\) (5) is trivial. Let \(V \to Y\) be an étale morphism from a scheme as in (5). Let \(Z\) be an affine scheme, and let \(Z \to Y\) be a morphism. Consider the diagram \[\xymatrix{ Z \times_Y V \ar[r]_q \ar[d]_p & V \ar[d] \\ Z \ar[r] & Y }\] Since \(p\) is étale, and hence open, we can choose finitely many affine open subschemes \(W_i \subset Z \times_Y V\) such that \(Z = \bigcup p(W_i)\). Consider the commutative diagram \[\xymatrix{ V \times_Y X \ar[d] & (\coprod W_i) \times_Y X \ar[l] \ar[d] \ar[r] & Z \times_Y X \ar[d] \\ V & \coprod W_i \ar[l] \ar[r] & Z }\] We know \(V \times_Y X\) has property \(\mathcal{P}\). By Lemma 03KF we see that \((\coprod W_i) \times_Y X\) has property \(\mathcal{P}\). Note that the morphism \((\coprod W_i) \times_Y X \to Z \times_Y X\) is étale and quasi-compact as the base change of \(\coprod W_i \to Z\). Hence by Lemma 03L2 we conclude that \(Z \times_Y X\) has property \(\mathcal{P}\).
Remark
An informal description of the properties \((\beta)\), decent, reasonable, very reasonable was given in Section 03I7. A morphism has one of these properties if (very) loosely speaking the fibres of the morphism have the corresponding properties. Being decent is useful to prove things about specializations of points on \(|X|\). Being reasonable is a bit stronger and technically quite easy to work with.
Here is a lemma we promised earlier which uses decent morphisms.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\). Assume \(f\) is quasi-compact and decent. (For example if \(f\) is representable, or quasi-separated, see Lemma 03M5.) Then \(f\) is universally closed if and only if the existence part of the valuative criterion holds.
Proof
In Morphisms of Spaces, Lemma 03KA we proved that any quasi-compact morphism which satisfies the existence part of the valuative criterion is universally closed. To prove the other, assume that \(f\) is universally closed. In the proof of Proposition 03KJ we have seen that it suffices to show, for any valuation ring \(A\), and any morphism \(\Spec(A) \to Y\), that the base change \(f_A : X_A \to \Spec(A)\) satisfies the existence part of the valuative criterion. By definition the algebraic space \(X_A\) has property \((\gamma)\) and hence Proposition 03KJ applies to the morphism \(f_A\) and we win.
Points of fibres
Let \(S\) be a scheme. Consider a cartesian diagram [0AC1]\[\begin{equation} \xymatrix{ W \ar[r]_q \ar[d]_p & Z \ar[d]^g \\ X \ar[r]^f & Y } \end{equation}\] of algebraic spaces over \(S\). Let \(x \in |X|\) and \(z \in |Z|\) be points mapping to the same point \(y \in |Y|\). We may ask: When is the set [0AC2]\[\begin{equation} F_{x, z} = \{ w \in |W| \text{ such that }p(w) = x\text{ and }q(w) = z\} \end{equation}\] finite?
Example
If \(X, Y, Z\) are schemes, then the set \(F_{x, z}\) is equal to the spectrum of \(\kappa(x) \otimes_{\kappa(y)} \kappa(z)\) (Schemes, Lemma 01JT). Thus we obtain a finite set if either \(\kappa(y) \subset \kappa(x)\) is finite or if \(\kappa(y) \subset \kappa(z)\) is finite. In particular, this is always the case if \(g\) is quasi-finite at \(z\) (Morphisms, Lemma 01TG).
Example
Let \(K\) be a characteristic \(0\) field endowed with an automorphism \(\sigma\) of infinite order. Set \(Y = \Spec(K)/\mathbf{Z}\) and \(X = \mathbf{A}^1_K/\mathbf{Z}\) where \(\mathbf{Z}\) acts on \(K\) via \(\sigma\) and on \(\mathbf{A}^1_K = \Spec(K[t])\) via \(t \mapsto t + 1\). Let \(Z = \Spec(K)\). Then \(W = \mathbf{A}^1_K\). Picture \[\xymatrix{ \mathbf{A}^1_K \ar[r]_q \ar[d]_p & \Spec(K) \ar[d]^g \\ \mathbf{A}^1_K/\mathbf{Z} \ar[r]^f & \Spec(K)/\mathbf{Z} }\] Take \(x\) corresponding to \(t = 0\) and \(z\) the unique point of \(\Spec(K)\). Then we see that \(F_{x, z} = \mathbf{Z}\) as a set.
Lemma
In the situation of (0AC1) if \(Z' \to Z\) is a morphism and \(z' \in |Z'|\) maps to \(z\), then the induced map \(F_{x, z'} \to F_{x, z}\) is surjective.
Proof
Set \(W' = X \times_Y Z' = W \times_Z Z'\). Then \(|W'| \to |W| \times_{|Z|} |Z'|\) is surjective by Properties of Spaces, Lemma 03H4. Hence the surjectivity of \(F_{x, z'} \to F_{x, z}\).
Lemma
In diagram (0AC1) the set (0AC2) is finite if \(f\) is of finite type and \(f\) is quasi-finite at \(x\).
Proof
The morphism \(q\) is quasi-finite at every \(w \in F_{x, z}\), see Morphisms of Spaces, Lemma 0ABM. Hence the lemma follows from Morphisms of Spaces, Lemma 0ABN.
Lemma
In diagram (0AC1) the set (0AC2) is finite if \(y\) can be represented by a monomorphism \(\Spec(k) \to Y\) where \(k\) is a field and \(g\) is quasi-finite at \(z\). (Special case: \(Y\) is decent and \(g\) is étale.)
Proof
By Lemma 0AC5 applied twice we may replace \(Z\) by \(Z_k = \Spec(k) \times_Y Z\) and \(X\) by \(X_k = \Spec(k) \times_Y X\). We may and do replace \(Y\) by \(\Spec(k)\) as well. Note that \(Z_k \to \Spec(k)\) is quasi-finite at \(z\) by Morphisms of Spaces, Lemma 0ABM. Choose a scheme \(V\), a point \(v \in V\), and an étale morphism \(V \to Z_k\) mapping \(v\) to \(z\). Choose a scheme \(U\), a point \(u \in U\), and an étale morphism \(U \to X_k\) mapping \(u\) to \(x\). Again by Lemma 0AC5 it suffices to show \(F_{u, v}\) is finite for the diagram \[\xymatrix{ U \times_{\Spec(k)} V \ar[r] \ar[d] & V \ar[d] \\ U \ar[r] & \Spec(k) }\] The morphism \(V \to \Spec(k)\) is quasi-finite at \(v\) (follows from the general discussion in Morphisms of Spaces, Section 03MI and the definition of being quasi-finite at a point). At this point the finiteness follows from Example 0AC3. The parenthetical remark of the statement of the lemma follows from the fact that on decent spaces points are represented by monomorphisms from fields and from the fact that an étale morphism of algebraic spaces is locally quasi-finite.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\). Let \(y \in |Y|\) and assume that \(y\) is represented by a quasi-compact monomorphism \(\Spec(k) \to Y\). Then \(|X_k| \to |X|\) is a homeomorphism onto \(f^{-1}(\{y\}) \subset |X|\) with induced topology.
Proof
We will use Properties of Spaces, Lemma 03IR and Morphisms of Spaces, Lemma 06RV without further mention. Let \(V \to Y\) be an étale morphism with \(V\) affine such that there exists a \(v \in V\) mapping to \(y\). Since \(\Spec(k) \to Y\) is quasi-compact there are a finite number of points of \(V\) mapping to \(y\) (Lemma 03JV). After shrinking \(V\) we may assume \(v\) is the only one. Choose a scheme \(U\) and a surjective étale morphism \(U \to X\). Consider the commutative diagram \[\xymatrix{ U \ar[d] & U_V \ar[l] \ar[d] & U_v \ar[l] \ar[d] \\ X \ar[d] & X_V \ar[l] \ar[d] & X_v \ar[l] \ar[d] \\ Y & V \ar[l] & v \ar[l] }\] Since \(U_v \to U_V\) identifies \(U_v\) with a subset of \(U_V\) with the induced topology (Schemes, Lemma 01K1), and since \(|U_V| \to |X_V|\) and \(|U_v| \to |X_v|\) are surjective and open, we see that \(|X_v| \to |X_V|\) is a homeomorphism onto its image (with induced topology). On the other hand, the inverse image of \(f^{-1}(\{y\})\) under the open map \(|X_V| \to |X|\) is equal to \(|X_v|\). We conclude that \(|X_v| \to f^{-1}(\{y\})\) is open. The morphism \(X_v \to X\) factors through \(X_k\) and \(|X_k| \to |X|\) is injective with image \(f^{-1}(\{y\})\) by Properties of Spaces, Lemma 03H4. Using \(|X_v| \to |X_k| \to f^{-1}(\{y\})\) the lemma follows because \(X_v \to X_k\) is surjective.
Lemma
Let \(X\) be an algebraic space locally of finite type over a field \(k\). Let \(x \in |X|\). Consider the conditions
\(\dim_x(|X|) = 0\),
\(x\) is closed in \(|X|\) and if \(x' \leadsto x\) in \(|X|\) then \(x' = x\),
\(x\) is an isolated point of \(|X|\),
\(\dim_x(X) = 0\),
\(X \to \Spec(k)\) is quasi-finite at \(x\).
Then (2), (3), (4), and (5) are equivalent. If \(X\) is decent, then (1) is equivalent to the others.
Proof
Parts (4) and (5) are equivalent for example by Morphisms of Spaces, Lemmas 04NW and 06LS.
Let \(U \to X\) be an étale morphism where \(U\) is an affine scheme and let \(u \in U\) be a point mapping to \(x\). Moreover, if \(x\) is a closed point, e.g., in case (2) or (3), then we may and do assume that \(u\) is a closed point. Observe that \(\dim_u(U) = \dim_x(X)\) by definition and that this is equal to \(\dim(\mathcal{O}_{U, u})\) if \(u\) is a closed point, see Algebra, Lemma 00OU.
If \(\dim_x(X) > 0\) and \(u\) is closed, by the arguments above we can choose a nontrivial specialization \(u' \leadsto u\) in \(U\). Then the transcendence degree of \(\kappa(u')\) over \(k\) exceeds the transcendence degree of \(\kappa(u)\) over \(k\). It follows that the images \(x\) and \(x'\) in \(X\) are distinct, because the transcendence degree of \(x/k\) and \(x'/k\) are well defined, see Morphisms of Spaces, Definition 04NM. This applies in particular in cases (2) and (3) and we conclude that (2) and (3) imply (4).
Conversely, if \(X \to \Spec(k)\) is locally quasi-finite at \(x\), then \(U \to \Spec(k)\) is locally quasi-finite at \(u\), hence \(u\) is an isolated point of \(U\) (Morphisms, Lemma 01TH). It follows that (5) implies (2) and (3) as \(|U| \to |X|\) is continuous and open.
Assume \(X\) is decent and (1) holds. Then \(\dim_x(X) = \dim_x(|X|)\) by Lemma 0A4J and the proof is complete.
Lemma
Let \(X\) be an algebraic space locally of finite type over a field \(k\). Consider the conditions
\(|X|\) is a finite set,
\(|X|\) is a discrete space,
\(\dim(|X|) = 0\),
\(\dim(X) = 0\),
\(X \to \Spec(k)\) is locally quasi-finite,
Then (2), (3), (4), and (5) are equivalent. If \(X\) is decent, then (1) implies the others.
Proof
Parts (4) and (5) are equivalent for example by Morphisms of Spaces, Lemma 04NW.
Let \(U \to X\) be a surjective étale morphism where \(U\) is a scheme.
If \(\dim(U) > 0\), then choose a nontrivial specialization \(u \leadsto u'\) in \(U\) and the transcendence degree of \(\kappa(u)\) over \(k\) exceeds the transcendence degree of \(\kappa(u')\) over \(k\). It follows that the images \(x\) and \(x'\) in \(X\) are distinct, because the transcendence degree of \(x/k\) and \(x'/k\) is well defined, see Morphisms of Spaces, Definition 04NM. We conclude that (2) and (3) imply (4).
Conversely, if \(X \to \Spec(k)\) is locally quasi-finite, then \(U\) is locally Noetherian (Morphisms, Lemma 01T6) of dimension \(0\) (Morphisms, Lemma 0397) and hence is a disjoint union of spectra of Artinian local rings (Properties, Lemma 0AAX). Hence \(U\) is a discrete topological space, and since \(|U| \to |X|\) is continuous and open, the same is true for \(|X|\). In other words, (4) implies (2) and (3).
Assume \(X\) is decent and (1) holds. Then we may choose \(U\) above to be affine. The fibres of \(|U| \to |X|\) are finite (this is a part of the defining property of decent spaces). Hence \(U\) is a finite type scheme over \(k\) with finitely many points. Hence \(U\) is quasi-finite over \(k\) (Morphisms, Lemma 02NG) which by definition means that \(X \to \Spec(k)\) is locally quasi-finite.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\) which is locally of finite type. Let \(x \in |X|\) with image \(y \in |Y|\). Let \(F = f^{-1}(\{y\})\) with induced topology from \(|X|\). Let \(k\) be a field and let \(\Spec(k) \to Y\) be in the equivalence class defining \(y\). Set \(X_k = \Spec(k) \times_Y X\). Let \(\tilde x \in |X_k|\) map to \(x \in |X|\). Consider the following conditions
\(\dim_x(F) = 0\),
\(x\) is isolated in \(F\),
\(x\) is closed in \(F\) and if \(x' \leadsto x\) in \(F\), then \(x = x'\),
\(\dim_{\tilde x}(|X_k|) = 0\),
\(\tilde x\) is isolated in \(|X_k|\),
\(\tilde x\) is closed in \(|X_k|\) and if \(\tilde x' \leadsto \tilde x\) in \(|X_k|\), then \(\tilde x = \tilde x'\),
\(\dim_{\tilde x}(X_k) = 0\),
\(f\) is quasi-finite at \(x\).
Then we have \[\xymatrix{ (\href{decent-spaces.html#decent-spaces-item-dimension-top-k-fibre}{0ACF}) \ar@{=>}[r]_{f\text{ decent}} & (\href{decent-spaces.html#decent-spaces-item-isolated-in-k-fibre}{0ACG}) \ar@{<=>}[r] & (\href{decent-spaces.html#decent-spaces-item-no-specializations-in-k-fibre}{0ACH}) \ar@{<=>}[r] & (\href{decent-spaces.html#decent-spaces-item-k-fibre-at-x-dim-0}{0ACI}) \ar@{<=>}[r] & (\href{decent-spaces.html#decent-spaces-item-quasi-finite-at-x}{0ACJ}) }\] If \(Y\) is decent, then conditions (0ACD) and (0ACE) are equivalent to each other and to conditions (0ACG), (0ACH), (0ACI), and (0ACJ). If \(Y\) and \(X\) are decent, then all conditions are equivalent.
Proof
By Lemma 0AC9 conditions (0ACG), (0ACH), and (0ACI) are equivalent to each other and to the condition that \(X_k \to \Spec(k)\) is quasi-finite at \(\tilde x\). Thus by Morphisms of Spaces, Lemma 0ABM they are also equivalent to (0ACJ). If \(f\) is decent, then \(X_k\) is a decent algebraic space and Lemma 0AC9 shows that (0ACF) implies (0ACG).
If \(Y\) is decent, then we can pick a quasi-compact monomorphism \(\Spec(k') \to Y\) in the equivalence class of \(y\). In this case Lemma 0AC8 tells us that \(|X_{k'}| \to F\) is a homeomorphism. Combined with the arguments given above this implies the remaining statements of the lemma; details omitted.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\) which is locally of finite type. Let \(y \in |Y|\). Let \(k\) be a field and let \(\Spec(k) \to Y\) be in the equivalence class defining \(y\). Set \(X_k = \Spec(k) \times_Y X\) and let \(F = f^{-1}(\{y\})\) with the induced topology from \(|X|\). Consider the following conditions
\(F\) is finite,
\(F\) is a discrete topological space,
\(\dim(F) = 0\),
\(|X_k|\) is a finite set,
\(|X_k|\) is a discrete space,
\(\dim(|X_k|) = 0\),
\(\dim(X_k) = 0\),
\(f\) is quasi-finite at all points of \(|X|\) lying over \(y\).
Then we have \[\xymatrix{ (\href{decent-spaces.html#decent-spaces-item-fibre-finite}{0ACL}) & (\href{decent-spaces.html#decent-spaces-item-k-fibre-finite}{0ACP}) \ar@{=>}[l] \ar@{=>}[r]_{f\text{ decent}} & (\href{decent-spaces.html#decent-spaces-item-k-fibre-discrete}{0ACQ}) \ar@{<=>}[r] & (\href{decent-spaces.html#decent-spaces-item-k-fibre-no-specializations}{0ACR}) \ar@{<=>}[r] & (\href{decent-spaces.html#decent-spaces-item-k-fibre-dim-0}{0ACS}) \ar@{<=>}[r] & (\href{decent-spaces.html#decent-spaces-item-quasi-finite-at-points-fibre}{0ACT}) }\] If \(Y\) is decent, then conditions (0ACM) and (0ACN) are equivalent to each other and to conditions (0ACQ), (0ACR), (0ACS), and (0ACT). If \(Y\) and \(X\) are decent, then (0ACL) implies all the other conditions.
Proof
By Lemma 0ACA conditions (0ACQ), (0ACR), and (0ACS) are equivalent to each other and to the condition that \(X_k \to \Spec(k)\) is locally quasi-finite. Thus by Morphisms of Spaces, Lemma 0ABM they are also equivalent to (0ACT). If \(f\) is decent, then \(X_k\) is a decent algebraic space and Lemma 0ACA shows that (0ACP) implies (0ACQ).
The map \(|X_k| \to F\) is surjective by Properties of Spaces, Lemma 03H4 and we see (0ACP) \(\Rightarrow\) (0ACL).
If \(Y\) is decent, then we can pick a quasi-compact monomorphism \(\Spec(k') \to Y\) in the equivalence class of \(y\). In this case Lemma 0AC8 tells us that \(|X_{k'}| \to F\) is a homeomorphism. Combined with the arguments given above this implies the remaining statements of the lemma; details omitted.
Monomorphisms
Here is another case where monomorphisms are representable. Please see More on Morphisms of Spaces, Section 0B89 for more information.
Lemma
Let \(S\) be a scheme. Let \(Y\) be a disjoint union of spectra of zero dimensional local rings over \(S\). Let \(f : X \to Y\) be a monomorphism of algebraic spaces over \(S\). Then \(f\) is representable, i.e., \(X\) is a scheme.
Proof
This immediately reduces to the case \(Y = \Spec(A)\) where \(A\) is a zero dimensional local ring, i.e., \(\Spec(A) = \{\mathfrak m_A\}\) is a singleton. If \(X = \emptyset\), then there is nothing to prove. If not, choose a nonempty affine scheme \(U = \Spec(B)\) and an étale morphism \(U \to X\). As \(|X|\) is a singleton (as a subset of \(|Y|\), see Morphisms of Spaces, Lemma 06RV) we see that \(U \to X\) is surjective. Note that \(U \times_X U = U \times_Y U = \Spec(B \otimes_A B)\). Thus we see that the ring maps \(B \to B \otimes_A B\) are étale. Since \[(B \otimes_A B)/\mathfrak m_A(B \otimes_A B) = (B/\mathfrak m_AB) \otimes_{A/\mathfrak m_A} (B/\mathfrak m_AB)\] we see that \(B/\mathfrak m_AB \to (B \otimes_A B)/\mathfrak m_A(B \otimes_A B)\) is flat and in fact free of rank equal to the dimension of \(B/\mathfrak m_AB\) as a \(A/\mathfrak m_A\)-vector space. Since \(B \to B \otimes_A B\) is étale, this can only happen if this dimension is finite (see for example Morphisms, Lemmas 03WU and 03JA). Every prime of \(B\) lies over \(\mathfrak m_A\) (the unique prime of \(A\)). Hence \(\Spec(B) = \Spec(B/\mathfrak m_A)\) as a topological space, and this space is a finite discrete set as \(B/\mathfrak m_A B\) is an Artinian ring, see Algebra, Lemmas 00J6 and 00JB. Hence all prime ideals of \(B\) are maximal and \(B = B_1 \times \ldots \times B_n\) is a product of finitely many local rings of dimension zero, see Algebra, Lemma 00JA. Thus \(B \to B \otimes_A B\) is finite étale as all the local rings \(B_i\) are henselian by Algebra, Lemma 06RS. Thus \(X\) is an affine scheme by Groupoids, Proposition 03BM.
Generic points
This section is a continuation of Properties of Spaces, Section 0BAP.
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Let \(x \in |X|\). The following are equivalent
\(x\) is a generic point of an irreducible component of \(|X|\),
for any étale morphism \((Y, y) \to (X, x)\) of pointed algebraic spaces, \(y\) is a generic point of an irreducible component of \(|Y|\),
for some étale morphism \((Y, y) \to (X, x)\) of pointed algebraic spaces, \(y\) is a generic point of an irreducible component of \(|Y|\),
the dimension of the local ring of \(X\) at \(x\) is zero, and
\(x\) is a point of codimension \(0\) on \(X\)
Proof
Conditions (4) and (5) are equivalent for any algebraic space by definition, see Properties of Spaces, Definition 04NA. Observe that any \(Y\) as in (2) and (3) is decent by Lemma 0ABU. Thus it suffices to prove the equivalence of (1) and (4) as then the equivalence with (2) and (3) follows since the dimension of the local ring of \(Y\) at \(y\) is equal to the dimension of the local ring of \(X\) at \(x\). Let \(f : U \to X\) be an étale morphism from an affine scheme and let \(u \in U\) be a point mapping to \(x\).
Assume (1). Let \(u' \leadsto u\) be a specialization in \(U\). Then \(f(u') = f(u) = x\). By Lemma 03K5 we see that \(u' = u\). Hence \(u\) is a generic point of an irreducible component of \(U\). Thus \(\dim(\mathcal{O}_{U, u}) = 0\) and we see that (4) holds.
Assume (4). The point \(x\) is contained in an irreducible component \(T \subset |X|\). Since \(|X|\) is sober (Proposition 03K6) we \(T\) has a generic point \(x'\). Of course \(x' \leadsto x\). Then we can lift this specialization to \(u' \leadsto u\) in \(U\) (Lemma 03IL). This contradicts the assumption that \(\dim(\mathcal{O}_{U, u}) = 0\) unless \(u' = u\), i.e., \(x' = x\).
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Let \(T \subset |X|\) be an irreducible closed subset. Let \(\xi \in T\) be the generic point (Proposition 03K6). Then \(\text{codim}(T, |X|)\) (Topology, Definition 02I3) is the dimension of the local ring of \(X\) at \(\xi\) (Properties of Spaces, Definition 04NA).
Proof
Choose a scheme \(U\), a point \(u \in U\), and an étale morphism \(U \to X\) sending \(u\) to \(\xi\). Then any sequence of nontrivial specializations \(\xi_e \leadsto \ldots \leadsto \xi_0 = \xi\) can be lifted to a sequence \(u_e \leadsto \ldots \leadsto u_0 = u\) in \(U\) by Lemma 03IL. Conversely, any sequence of nontrivial specializations \(u_e \leadsto \ldots \leadsto u_0 = u\) in \(U\) maps to a sequence of nontrivial specializations \(\xi_e \leadsto \ldots \leadsto \xi_0 = \xi\) by Lemma 03K5. Because \(|X|\) and \(U\) are sober topological spaces we conclude that the codimension of \(T\) in \(|X|\) and of \(\overline{\{u\}}\) in \(U\) are the same. In this way the lemma reduces to the schemes case which is Properties, Lemma 02IZ.
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). Assume
every quasi-compact scheme étale over \(X\) has finitely many irreducible components, and
every \(x \in |X|\) of codimension \(0\) on \(X\) can be represented by a monomorphism \(\Spec(k) \to X\).
Then \(X\) is a reasonable algebraic space.
Proof
Let \(U\) be an affine scheme and let \(a : U \to X\) be an étale morphism. We have to show that the fibres of \(a\) are universally bounded. By assumption (1) the scheme \(U\) has finitely many irreducible components. Let \(u_1, \ldots, u_n \in U\) be the generic points of these irreducible components. Let \(\{x_1, \ldots, x_m\} \subset |X|\) be the image of \(\{u_1, \ldots, u_n\}\). Each \(x_j\) is a point of codimension \(0\). By assumption (2) we may choose a monomorphism \(\Spec(k_j) \to X\) representing \(x_j\). By Properties of Spaces, Lemma 0BAQ we have \[U \times_X \Spec(k_j) = \coprod\nolimits_{a(u_i) = x_j} \Spec(\kappa(u_i))\] This is a scheme finite over \(\Spec(k_j)\) of degree \(d_j = \sum_{a(u_i) = x_j} [\kappa(u_i) : k_j]\). Set \(n = \max d_j\).
Observe that \(a\) is separated (Properties of Spaces, Lemma 03FZ). Consider the stratification \[X = X_0 \supset X_1 \supset X_2 \supset \ldots\] associated to \(U \to X\) in Lemma 086T. By our choice of \(n\) above we conclude that \(X_{n + 1}\) is empty. Namely, if not, then \(a^{-1}(X_{n + 1})\) is a nonempty open of \(U\) and hence would contain one of the \(x_i\). This would mean that \(X_{n + 1}\) contains \(x_j = a(u_i)\) which is impossible. Hence we see that the fibres of \(U \to X\) are universally bounded (in fact by the integer \(n\)).
Lemma
Let \(S\) be a scheme. Let \(X\) be an algebraic space over \(S\). The following are equivalent
\(X\) is decent and \(|X|\) has finitely many irreducible components,
every quasi-compact scheme étale over \(X\) has finitely many irreducible components, there are finitely many \(x \in |X|\) of codimension \(0\) on \(X\), and each of these can be represented by a monomorphism \(\Spec(k) \to X\),
there exists a dense open \(X' \subset X\) which is a scheme, \(X'\) has finitely many irreducible components with generic points \(\{x'_1, \ldots, x'_m\}\), and the morphism \(x'_j \to X\) is quasi-compact for \(j = 1, \ldots, m\).
Moreover, if these conditions hold, then \(X\) is reasonable and the points \(x'_j \in |X|\) are the generic points of the irreducible components of \(|X|\).
Proof
In the proof we use Properties of Spaces, Lemma 0BAQ without further mention. Assume (1). Then \(X\) has a dense open subscheme \(X'\) by Theorem 086U. Since the closure of an irreducible component of \(|X'|\) is an irreducible component of \(|X|\), we see that \(|X'|\) has finitely many irreducible components. Thus (3) holds.
Assume \(X' \subset X\) is as in (3). Let \(\{x'_1, \ldots, x'_m\}\) be the generic points of the irreducible components of \(X'\). Let \(a : U \to X\) be an étale morphism with \(U\) a quasi-compact scheme. To prove (2) it suffices to show that \(U\) has finitely many irreducible components whose generic points lie over \(\{x'_1, \ldots, x'_m\}\). It suffices to prove this for the members of a finite affine open cover of \(U\), hence we may and do assume \(U\) is affine. Note that \(U' = a^{-1}(X') \subset U\) is a dense open. Since \(U' \to X'\) is an étale morphism of schemes, we see the generic points of irreducible components of \(U'\) are the points lying over \(\{x'_1, \ldots, x'_m\}\). Since \(x'_j \to X\) is quasi-compact there are finitely many points of \(U\) lying over \(x'_j\) (Lemma 03JV). Hence \(U'\) has finitely many irreducible components, which implies that the closures of these irreducible components are the irreducible components of \(U\). Thus (2) holds.
Assume (2). This implies (1) and the final statement by Lemma 0BB8. (We also use that a reasonable algebraic space is decent, see discussion following Definition 03I8.)
Generically finite morphisms
This section discusses for morphisms of algebraic spaces the material discussed in Morphisms, Section 02NV and Varieties, Section 0AB5 for morphisms of schemes.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\). Assume that \(f\) is quasi-separated of finite type. Let \(y \in |Y|\) be a point of codimension \(0\) on \(Y\). The following are equivalent:
the space \(|X_k|\) is finite where \(\Spec(k) \to Y\) represents \(y\),
\(X \to Y\) is quasi-finite at all points of \(|X|\) over \(y\),
there exists an open subspace \(Y' \subset Y\) with \(y \in |Y'|\) such that \(Y' \times_Y X \to Y'\) is finite.
If \(Y\) is decent these are also equivalent to
the set \(f^{-1}(\{y\})\) is finite.
Proof
The equivalence of (1) and (2) follows from Lemma 0ACK (and the fact that a quasi-separated morphism is decent by Lemma 03M5).
Assume the equivalent conditions of (1) and (2). Choose an affine scheme \(V\) and an étale morphism \(V \to Y\) mapping a point \(v \in V\) to \(y\). Then \(v\) is a generic point of an irreducible component of \(V\) by Properties of Spaces, Lemma 0BAQ. Choose an affine scheme \(U\) and a surjective étale morphism \(U \to V \times_Y X\). Then \(U \to V\) is of finite type. The morphism \(U \to V\) is quasi-finite at every point lying over \(v\) by (2). It follows that the fibre of \(U \to V\) over \(v\) is finite (Morphisms, Lemma 0AAY). By Morphisms, Lemma 02NW after shrinking \(V\) we may assume that \(U \to V\) is finite. Let \[R = U \times_{V \times_Y X} U\] Since \(f\) is quasi-separated, we see that \(V \times_Y X\) is quasi-separated and hence \(R\) is a quasi-compact scheme. Moreover the morphisms \(R \to V\) is quasi-finite as the composition of an étale morphism \(R \to U\) and a finite morphism \(U \to V\). Hence we may apply Morphisms, Lemma 02NW once more and after shrinking \(V\) we may assume that \(R \to V\) is finite as well. This of course implies that the two projections \(R \to V\) are finite étale. It follows that \(V/R = V \times_Y X\) is an affine scheme, see Groupoids, Proposition 03BM. By Morphisms, Lemma 03GN we conclude that \(V \times_Y X \to V\) is proper and by Morphisms, Lemma 01WN we conclude that \(V \times_Y X \to V\) is finite. Finally, we let \(Y' \subset Y\) be the open subspace of \(Y\) corresponding to the image of \(|V| \to |Y|\). By Morphisms of Spaces, Lemma 03ZQ we conclude that \(Y' \times_Y X \to Y'\) is finite as the base change to \(V\) is finite and as \(V \to Y'\) is a surjective étale morphism.
If \(Y\) is decent and \(f\) is quasi-separated, then we see that \(X\) is decent too; use Lemmas 03M5 and 0ABY. Hence Lemma 0ACK applies to show that (4) implies (1) and (2). On the other hand, we see that (2) implies (4) by Morphisms of Spaces, Lemma 0ABN.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\). Assume that \(f\) is quasi-separated and locally of finite type and \(Y\) quasi-separated. Let \(y \in |Y|\) be a point of codimension \(0\) on \(Y\). The following are equivalent:
the set \(f^{-1}(\{y\})\) is finite,
the space \(|X_k|\) is finite where \(\Spec(k) \to Y\) represents \(y\),
there exist open subspaces \(X' \subset X\) and \(Y' \subset Y\) with \(f(X') \subset Y'\), \(y \in |Y'|\), and \(f^{-1}(\{y\}) \subset |X'|\) such that \(f|_{X'} : X' \to Y'\) is finite.
Proof
Since quasi-separated algebraic spaces are decent, the equivalence of (1) and (2) follows from Lemma 0ACK. To prove that (1) and (2) imply (3) we may and do replace \(Y\) by a quasi-compact open containing \(y\). Since \(f^{-1}(\{y\})\) is finite, we can find a quasi-compact open subspace of \(X' \subset X\) containing the fibre. The restriction \(f|_{X'} : X' \to Y\) is quasi-compact and quasi-separated by Morphisms of Spaces, Lemma 073B (this is where we use that \(Y\) is quasi-separated). Applying Lemma 0ACZ to \(f|_{X'} : X' \to Y\) we see that (3) holds. We omit the proof that (3) implies (2).
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\). Assume \(f\) is locally of finite type. Let \(X^0 \subset |X|\), resp. \(Y^0 \subset |Y|\) denote the set of codimension \(0\) points of \(X\), resp. \(Y\). Let \(y \in Y^0\). The following are equivalent
\(f^{-1}(\{y\}) \subset X^0\),
\(f\) is quasi-finite at all points lying over \(y\),
\(f\) is quasi-finite at all \(x \in X^0\) lying over \(y\).
Proof
Let \(V\) be a scheme and let \(V \to Y\) be a surjective étale morphism. Let \(U\) be a scheme and let \(U \to V \times_Y X\) be a surjective étale morphism. Then \(f\) is quasi-finite at the image \(x\) of a point \(u \in U\) if and only if \(U \to V\) is quasi-finite at \(u\). Moreover, \(x \in X^0\) if and only if \(u\) is the generic point of an irreducible component of \(U\) (Properties of Spaces, Lemma 0BAQ). Thus the lemma reduces to the case of the morphism \(U \to V\), i.e., to Morphisms, Lemma 0BAH.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\). Assume \(f\) is locally of finite type. Let \(X^0 \subset |X|\), resp. \(Y^0 \subset |Y|\) denote the set of codimension \(0\) points of \(X\), resp. \(Y\). Assume
\(Y\) is decent,
\(X^0\) and \(Y^0\) are finite and \(f^{-1}(Y^0) = X^0\),
either \(f\) is quasi-compact or \(f\) is separated.
Then there exists a dense open \(V \subset Y\) such that \(f^{-1}(V) \to V\) is finite.
Proof
By Lemmas 0BB9 and 0ABV we may assume \(Y\) is a scheme with finitely many irreducible components. Shrinking further we may assume \(Y\) is an irreducible affine scheme with generic point \(y\). Then the fibre of \(f\) over \(y\) is finite.
Assume \(f\) is quasi-compact and \(Y\) affine irreducible. Then \(X\) is quasi-compact and we may choose an affine scheme \(U\) and a surjective étale morphism \(U \to X\). Then \(U \to Y\) is of finite type and the fibre of \(U \to Y\) over \(y\) is the set \(U^0\) of generic points of irreducible components of \(U\) (Properties of Spaces, Lemma 0BAQ). Hence \(U^0\) is finite (Morphisms, Lemma 0AAY) and after shrinking \(Y\) we may assume that \(U \to Y\) is finite (Morphisms, Lemma 02NW). Next, consider \(R = U \times_X U\). Since the projection \(s : R \to U\) is étale we see that \(R^0 = s^{-1}(U^0)\) lies over \(y\). Since \(R \to U \times_Y U\) is a monomorphism, we conclude that \(R^0\) is finite as \(U \times_Y U \to Y\) is finite. And \(R\) is separated (Properties of Spaces, Lemma 03FZ). Thus we may shrink \(Y\) once more to reach the situation where \(R\) is finite over \(Y\) (Morphisms, Lemma 0BAI). In this case it follows that \(X = U/R\) is finite over \(Y\) by exactly the same arguments as given in the proof of Lemma 0ACZ (or we can simply apply that lemma because it follows immediately that \(X\) is quasi-separated as well).
Assume \(f\) is separated and \(Y\) affine irreducible. Choose \(V \subset Y\) and \(U \subset X\) as in Lemma 0AD0. Since \(f|_U : U \to V\) is finite, we see that \(U \subset f^{-1}(V)\) is closed as well as open (Morphisms of Spaces, Lemmas 04NX and 04NZ). Thus \(f^{-1}(V) = U \amalg W\) for some open subspace \(W\) of \(X\). However, since \(U\) contains all the codimension \(0\) points of \(X\) we conclude that \(W = \emptyset\) (Properties of Spaces, Lemma 0BAR) as desired.
Birational morphisms
The following definition of a birational morphism of algebraic spaces seems to be the closest to our definition (Morphisms, Definition 01RO) of a birational morphism of schemes.
Definition
Let \(S\) be a scheme. Let \(X\) and \(Y\) algebraic spaces over \(S\). Assume \(X\) and \(Y\) are decent and that \(|X|\) and \(|Y|\) have finitely many irreducible components. We say a morphism \(f : X \to Y\) is birational if
\(|f|\) induces a bijection between the set of generic points of irreducible components of \(|X|\) and the set of generic points of the irreducible components of \(|Y|\), and
for every generic point \(x \in |X|\) of an irreducible component the local ring map \(\mathcal{O}_{Y, f(x)} \to \mathcal{O}_{X, x}\) is an isomorphism (see clarification below).
Clarification: Since \(X\) and \(Y\) are decent the topological spaces \(|X|\) and \(|Y|\) are sober (Proposition 03K6). Hence condition (1) makes sense. Moreover, because we have assumed that \(|X|\) and \(|Y|\) have finitely many irreducible components, we see that the generic points \(x_1, \ldots, x_n \in |X|\), resp. \(y_1, \ldots, y_n \in |Y|\) are contained in any dense open of \(|X|\), resp. \(|Y|\). In particular, they are contained in the schematic locus of \(X\), resp. \(Y\) by Theorem 086U. Thus we can define \(\mathcal{O}_{X, x_i}\), resp. \(\mathcal{O}_{Y, y_i}\) to be the local ring of this scheme at \(x_i\), resp. \(y_i\).
We conclude that if the morphism \(f : X \to Y\) is birational, then there exist dense open subspaces \(X' \subset X\) and \(Y' \subset Y\) such that
\(f(X') \subset Y'\),
\(X'\) and \(Y'\) are representable, and
\(f|_{X'} : X' \to Y'\) is birational in the sense of Morphisms, Definition 01RO.
However, we do insist that \(X\) and \(Y\) are decent with finitely many irreducible components. Other ways to characterize decent algebraic spaces with finitely many irreducible components are given in Lemma 0BB9. In most cases birational morphisms are isomorphisms over dense opens.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\) which are decent and have finitely many irreducible components. If \(f\) is birational then \(f\) is dominant.
Proof
Follows immediately from the definitions. See Morphisms of Spaces, Definition 0ABL.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a birational morphism of algebraic spaces over \(S\) which are decent and have finitely many irreducible components. If \(y \in Y\) is the generic point of an irreducible component, then the base change \(X \times_Y \Spec(\mathcal{O}_{Y, y}) \to \Spec(\mathcal{O}_{Y, y})\) is an isomorphism.
Proof
Let \(X' \subset X\) and \(Y' \subset Y\) be the maximal open subspaces which are representable, see Lemma 0BB9. By Lemma 0BBB the fibre of \(f\) over \(y\) is consists of points of codimension \(0\) of \(X\) and is therefore contained in \(X'\). Hence \(X \times_Y \Spec(\mathcal{O}_{Y, y}) = X' \times_{Y'} \Spec(\mathcal{O}_{Y', y})\) and the result follows from Morphisms, Lemma 0BAB.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a birational morphism of algebraic spaces over \(S\) which are decent and have finitely many irreducible components. Assume one of the following conditions is satisfied
\(f\) is locally of finite type and \(Y\) reduced (i.e., integral),
\(f\) is locally of finite presentation.
Then there exist dense opens \(U \subset X\) and \(V \subset Y\) such that \(f(U) \subset V\) and \(f|_U : U \to V\) is an isomorphism.
Proof
By Lemma 0BB9 we may assume that \(X\) and \(Y\) are schemes. In this case the result is Morphisms, Lemma 0BAC.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a birational morphism of algebraic spaces over \(S\) which are decent and have finitely many irreducible components. Assume
either \(f\) is quasi-compact or \(f\) is separated, and
either \(f\) is locally of finite type and \(Y\) is reduced or \(f\) is locally of finite presentation.
Then there exists a dense open \(V \subset Y\) such that \(f^{-1}(V) \to V\) is an isomorphism.
Proof
By Lemma 0BB9 we may assume \(Y\) is a scheme. By Lemma 0BBC we may assume that \(f\) is finite. Then \(X\) is a scheme too and the result follows from Morphisms, Lemma 0BAJ.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\) which are decent and have finitely many irreducible components. If \(f\) is birational and \(V \to Y\) is an étale morphism with \(V\) affine, then \(X \times_Y V\) is decent with finitely many irreducible components and \(X \times_Y V \to V\) is birational.
Proof
The algebraic space \(U = X \times_Y V\) is decent (Lemma 0ABU). The generic points of \(V\) and \(U\) are the elements of \(|V|\) and \(|U|\) which lie over generic points of \(|Y|\) and \(|X|\) (Lemma 0ABV). Since \(Y\) is decent we conclude there are finitely many generic points on \(V\). Let \(\xi \in |X|\) be a generic point of an irreducible component. By the discussion following Definition 0ACV we have a cartesian square \[\xymatrix{ \Spec(\mathcal{O}_{X, \xi}) \ar[d] \ar[r] & X \ar[d] \\ \Spec(\mathcal{O}_{Y, f(\xi)}) \ar[r] & Y }\] whose horizontal morphisms are monomorphisms identifying local rings and where the left vertical arrow is an isomorphism. It follows that in the diagram \[\xymatrix{ \Spec(\mathcal{O}_{X, \xi}) \times_X U \ar[d] \ar[r] & U \ar[d] \\ \Spec(\mathcal{O}_{Y, f(\xi)}) \times_Y V \ar[r] & V }\] the vertical arrow on the left is an isomorphism. The horizontal arrows have image contained in the schematic locus of \(U\) and \(V\) and identify local rings (some details omitted). Since the image of the horizontal arrows are the points of \(|U|\), resp. \(|V|\) lying over \(\xi\), resp. \(f(\xi)\) we conclude.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a birational morphism between algebraic spaces over \(S\) which are decent and have finitely many irreducible components. Then the normalizations \(X^\nu \to X\) and \(Y^\nu \to Y\) exist and there is a commutative diagram \[\xymatrix{ X^\nu \ar[r] \ar[d] & Y^\nu \ar[d] \\ X \ar[r] & Y }\] of algebraic spaces over \(S\). The morphism \(X^\nu \to Y^\nu\) is birational.
Proof
By Lemma 0BB9 we see that \(X\) and \(Y\) satisfy the equivalent conditions of Morphisms of Spaces, Lemma 0BB1 and the normalizations are defined. By Morphisms of Spaces, Lemma 0BB4 the algebraic space \(X^\nu\) is normal and maps codimension \(0\) points to codimension \(0\) points. Since \(f\) maps codimension \(0\) points to codimension \(0\) points (this is the same as generic points on decent spaces by Lemma 0ABV) we obtain from Morphisms of Spaces, Lemma 0BB4 a factorization of the composition \(X^\nu \to X \to Y\) through \(Y^\nu\).
Observe that \(X^\nu\) and \(Y^\nu\) are decent for example by Lemma 0ABT. Moreover the maps \(X^\nu \to X\) and \(Y^\nu \to Y\) induce bijections on irreducible components (see references above) hence \(X^\nu\) and \(Y^\nu\) both have a finite number of irreducible components and the map \(X^\nu \to Y^\nu\) induces a bijection between their generic points. To prove that \(X^\nu \to Y^\nu\) is birational, it therefore suffices to show it induces an isomorphism on local rings at these points. To do this we may replace \(X\) and \(Y\) by open neighbourhoods of their generic points, hence we may assume \(X\) and \(Y\) are affine irreducible schemes with generic points \(x\) and \(y\). Since \(f\) is birational the map \(\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}\) is an isomorphism. Let \(x^\nu \in X^\nu\) and \(y^\nu \in Y^\nu\) be the points lying over \(x\) and \(y\). By construction of the normalization we see that \(\mathcal{O}_{X^\nu, x^\nu} = \mathcal{O}_{X, x}/\mathfrak m_x\) and similarly on \(Y\). Thus the map \(\mathcal{O}_{X^\nu, x^\nu} \to \mathcal{O}_{Y^\nu, y^\nu}\) is an isomorphism as well.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of algebraic spaces over \(S\). Assume
\(X\) and \(Y\) are decent and have finitely many irreducible components,
\(f\) is integral and birational,
\(Y\) is normal, and
\(X\) is reduced.
Then \(f\) is an isomorphism.
Proof
Let \(V \to Y\) be an étale morphism with \(V\) affine. It suffices to show that \(U = X \times_Y V \to V\) is an isomorphism. By Lemma 0B4D and its proof we see that \(U\) and \(V\) are decent and have finitely many irreducible components and that \(U \to V\) is birational. By Properties, Lemma 0357 \(V\) is a finite disjoint union of integral schemes. Thus we may assume \(V\) is integral. As \(f\) is birational, we see that \(U\) is irreducible and reduced, i.e., integral (note that \(U\) is a scheme as \(f\) is integral, hence representable). Thus we may assume that \(X\) and \(Y\) are integral schemes and the result follows from the case of schemes, see Morphisms, Lemma 0AB1.
Lemma
Let \(S\) be a scheme. Let \(f : X \to Y\) be an integral birational morphism of decent algebraic spaces over \(S\) which have finitely many irreducible components. Then there exists a factorization \(Y^\nu \to X \to Y\) and \(Y^\nu \to X\) is the normalization of \(X\).
Proof
Consider the map \(X^\nu \to Y^\nu\) of Lemma 0BBG. This map is integral by Morphisms of Spaces, Lemma 081Z. Hence it is an isomorphism by Lemma 0B4E.
Jacobson spaces
We have defined the Jacobson property for algebraic spaces in Properties of Spaces, Remark 03E7. For representable algebraic spaces it agrees with the property discussed in Properties, Section 01P1. The relationship between the Jacobson property and the behaviour of the topological space \(|X|\) is not evident for general algebraic spaces \(|X|\). However, a decent (for example quasi-separated or locally separated) algebraic space \(X\) is Jacobson if and only if \(|X|\) is Jacobson (see Lemma 0BA6).
Lemma
Let \(S\) be a scheme. Let \(X\) be a Jacobson algebraic space over \(S\). Any algebraic space locally of finite type over \(X\) is Jacobson.
Proof
Let \(U \to X\) be a surjective étale morphism where \(U\) is a scheme. Then \(U\) is Jacobson (by definition) and for a morphism of schemes \(V \to U\) which is locally of finite type we see that \(V\) is Jacobson by the corresponding result for schemes (Morphisms, Lemma 02J5). Thus if \(Y \to X\) is a morphism of algebraic spaces which is locally of finite type, then setting \(V = U \times_X Y\) we see that \(Y\) is Jacobson by definition.
Lemma
Let \(S\) be a scheme. Let \(X\) be a Jacobson algebraic space over \(S\). For \(x \in X_{\text{ft-pts}}\) and \(g : W \to X\) locally of finite type with \(W\) a scheme, if \(x \in \Im(|g|)\), then there exists a closed point of \(W\) mapping to \(x\).
Proof
Let \(U \to X\) be an étale morphism with \(U\) a scheme and with \(u \in U\) closed mapping to \(x\), see Morphisms of Spaces, Lemma 06EH. Observe that \(W\), \(W \times_X U\), and \(U\) are Jacobson schemes by Lemma 0BA3. Hence finite type points on these schemes are the same thing as closed points by Morphisms, Lemma 01TB. The inverse image \(T \subset W \times_X U\) of \(u\) is a nonempty (as \(x\) in the image of \(W \to X\)) closed subset. By Morphisms, Lemma 02J4 there is a closed point \(t\) of \(W \times_X U\) which maps to \(u\). As \(W \times_X U \to W\) is locally of finite type the image of \(t\) in \(W\) is closed by Morphisms, Lemma 01TB.
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent Jacobson algebraic space over \(S\). Then \(X_{\text{ft-pts}} \subset |X|\) is the set of closed points.
Proof
If \(x \in |X|\) is closed, then we can represent \(x\) by a closed immersion \(\Spec(k) \to X\), see Lemma 0AHB. Hence \(x\) is certainly a finite type point.
Conversely, let \(x \in |X|\) be a finite type point. We know that \(x\) can be represented by a quasi-compact monomorphism \(\Spec(k) \to X\) where \(k\) is a field (Definition 03I8). On the other hand, by definition, there exists a morphism \(\Spec(k') \to X\) which is locally of finite type and represents \(x\) (Morphisms, Definition 02J1). We obtain a factorization \(\Spec(k') \to \Spec(k) \to X\). Let \(U \to X\) be any étale morphism with \(U\) affine and consider the morphisms \[\Spec(k') \times_X U \to \Spec(k) \times_X U \to U\] The quasi-compact scheme \(\Spec(k) \times_X U\) is étale over \(\Spec(k)\) hence is a finite disjoint union of spectra of fields (Remark 03II). Moreover, the first morphism is surjective and locally of finite type (Morphisms, Lemma 01T8) hence surjective on finite type points (Morphisms, Lemma 06EB) and the composition (which is locally of finite type) sends finite type points to closed points as \(U\) is Jacobson (Morphisms, Lemma 01TB). Thus the image of \(\Spec(k) \times_X U \to U\) is a finite set of closed points hence closed. Since this is true for every affine \(U\) and étale morphism \(U \to X\), we conclude that \(x \in |X|\) is closed.
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Then \(X\) is Jacobson if and only if \(|X|\) is Jacobson.
Proof
Assume \(X\) is Jacobson and that \(T \subset |X|\) is a closed subset. By Morphisms of Spaces, Lemma 06EK we see that \(T \cap X_{\text{ft-pts}}\) is dense in \(T\). By Lemma 0BA5 we see that \(X_{\text{ft-pts}}\) are the closed points of \(|X|\). Thus \(|X|\) is indeed Jacobson.
Assume \(|X|\) is Jacobson. Let \(f : U \to X\) be an étale morphism with \(U\) an affine scheme. We have to show that \(U\) is Jacobson. If \(x \in |X|\) is closed, then the fibre \(F = f^{-1}(\{x\})\) is a finite (by definition of decent) closed (by construction of the topology on \(|X|\)) subset of \(U\). Since there are no specializations between points of \(F\) (Lemma 03K5) we conclude that every point of \(F\) is closed in \(U\). If \(U\) is not Jacobson, then there exists a non-closed point \(u \in U\) such that \(\{u\}\) is locally closed (Topology, Lemma 02I7). We will show that \(f(u) \in |X|\) is closed; by the above \(u\) is closed in \(U\) which is a contradiction and finishes the proof. To prove this we may replace \(U\) by an affine open neighbourhood of \(u\). Thus we may assume that \(\{u\}\) is closed in \(U\). Let \(R = U \times_X U\) with projections \(s, t : R \to U\). Then \(s^{-1}(\{u\}) = \{r_1, \ldots, r_m\}\) is finite (by definition of decent spaces). After replacing \(U\) by a smaller affine open neighbourhood of \(u\) we may assume that \(t(r_j) = u\) for \(j = 1, \ldots, m\). It follows that \(\{u\}\) is an \(R\)-invariant closed subset of \(U\). Hence \(\{f(u)\}\) is a locally closed subset of \(X\) as it is closed in the open \(|f|(|U|)\) of \(|X|\). Since \(|X|\) is Jacobson we conclude that \(f(u)\) is closed in \(|X|\) as desired.
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent locally Noetherian algebraic space over \(S\). Let \(x \in |X|\). Then \[W = \{x' \in |X| : x' \leadsto x,\ x' \not = x\}\] is a Noetherian, spectral, sober, Jacobson topological space.
Proof
We may replace by any open subspace containing \(x\). Thus we may assume that \(X\) is quasi-compact. Then \(|X|\) is a Noetherian topological space (Properties of Spaces, Lemma 04ZF). Thus \(W\) is a Noetherian topological space (Topology, Lemma 0052).
Combining Lemma 0BB6 with Properties of Spaces, Lemma 0A4G we see that \(|X|\) is a spectral topological space. By Topology, Lemma 0A31 we see that \(W \cup \{x\}\) is a spectral topological space. Now \(W\) is a quasi-compact open of \(W \cup \{x\}\) and hence \(W\) is spectral by Topology, Lemma 0902.
Let \(E \subset W\) be an irreducible closed subset. Then if \(Z \subset |X|\) is the closure of \(E\) we see that \(x \in Z\). There is a unique generic point \(\eta \in Z\) by Proposition 03K6. Of course \(\eta \in W\) and hence \(\eta \in E\). We conclude that \(E\) has a unique generic point, i.e., \(W\) is sober.
Let \(x' \in W\) be a point such that \(\{x'\}\) is locally closed in \(W\). To finish the proof we have to show that \(x'\) is a closed point of \(W\). If not, then there exists a nontrivial specialization \(x' \leadsto x'_1\) in \(W\). Let \(U\) be an affine scheme, \(u \in U\) a point, and let \(U \to X\) be an étale morphism mapping \(u\) to \(x\). By Lemma 03IL we can choose specializations \(u' \leadsto u'_1 \leadsto u\) mapping to \(x' \leadsto x'_1 \leadsto x\). Let \(\mathfrak p' \subset \mathcal{O}_{U, u}\) be the prime ideal corresponding to \(u'\). The existence of the specializations implies that \(\dim(\mathcal{O}_{U, u}/\mathfrak p') \geq 2\). Hence every nonempty open of \(\Spec(\mathcal{O}_{U, u}/\mathfrak p')\) is infinite by Algebra, Lemma 02IG. By Lemma 03K5 we obtain a continuous map \[\Spec(\mathcal{O}_{U, u}/\mathfrak p') \setminus \{\mathfrak m_u/\mathfrak p'\} \longrightarrow W\] Since the generic point of the LHS maps to \(x'\) the image is contained in \(\overline{\{x'\}}\). We conclude the inverse image of \(\{x'\}\) under the displayed arrow is nonempty open hence infinite. However, the fibres of \(U \to X\) are finite as \(X\) is decent and we conclude that \(\{x'\}\) is infinite. This contradiction finishes the proof.
Local irreducibility
We have already defined the geometric number of branches of an algebraic space at a point in Properties of Spaces, Section 06DJ. The number of branches of an algebraic space at a point can only be defined for decent algebraic spaces.
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Let \(x \in |X|\) be a point. The following are equivalent
for any elementary étale neighbourhood \((U, u) \to (X, x)\) the local ring \(\mathcal{O}_{U, u}\) has a unique minimal prime,
for any elementary étale neighbourhood \((U, u) \to (X, x)\) there is a unique irreducible component of \(U\) through \(u\),
for any elementary étale neighbourhood \((U, u) \to (X, x)\) the local ring \(\mathcal{O}_{U, u}\) is unibranch,
the henselian local ring \(\mathcal{O}_{X, x}^h\) has a unique minimal prime.
Proof
The equivalence of (1) and (2) follows from the fact that irreducible components of \(U\) passing through \(u\) are in \(1\)-\(1\) correspondence with minimal primes of the local ring of \(U\) at \(u\). The ring \(\mathcal{O}_{X, x}^h\) is the henselization of \(\mathcal{O}_{U, u}\), see discussion following Definition 0BGW. In particular (3) and (4) are equivalent by More on Algebra, Lemma 0BQ0. The equivalence of (2) and (3) follows from More on Morphisms, Lemma 0CB4.
Definition
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Let \(x \in |X|\). We say that \(X\) is unibranch at \(x\) if the equivalent conditions of Lemma 0DQ6 hold. We say that \(X\) is unibranch if \(X\) is unibranch at every \(x \in |X|\).
This is consistent with the definition for schemes (Properties, Definition 0BQ2).
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Let \(x \in |X|\) be a point. Let \(n \in \{1, 2, \ldots\}\) be an integer. The following are equivalent
for any elementary étale neighbourhood \((U, u) \to (X, x)\) the number of minimal primes of the local ring \(\mathcal{O}_{U, u}\) is \(\leq n\) and for at least one choice of \((U, u)\) it is \(n\),
for any elementary étale neighbourhood \((U, u) \to (X, x)\) the number irreducible components of \(U\) passing through \(u\) is \(\leq n\) and for at least one choice of \((U, u)\) it is \(n\),
for any elementary étale neighbourhood \((U, u) \to (X, x)\) the number of branches of \(U\) at \(u\) is \(\leq n\) and for at least one choice of \((U, u)\) it is \(n\),
the number of minimal prime ideals of \(\mathcal{O}_{X, x}^h\) is \(n\).
Proof
The equivalence of (1) and (2) follows from the fact that irreducible components of \(U\) passing through \(u\) are in \(1\)-\(1\) correspondence with minimal primes of the local ring of \(U\) at \(u\). The ring \(\mathcal{O}_{X, x}\) is the henselization of \(\mathcal{O}_{U, u}\), see discussion following Definition 0BGW. In particular (3) and (4) are equivalent by More on Algebra, Lemma 0BQ0. The equivalence of (2) and (3) follows from More on Morphisms, Lemma 0CB4.
Definition
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). Let \(x \in |X|\). The number of branches of \(X\) at \(x\) is either \(n \in \mathbf{N}\) if the equivalent conditions of Lemma 0DQ8 hold, or else \(\infty\).
Catenary algebraic spaces
This section extends the material in Properties, Section 02IV and Morphisms, Section 02J7 to algebraic spaces.
Definition
Let \(S\) be a scheme. Let \(X\) be a decent algebraic space over \(S\). We say \(X\) is catenary if \(|X|\) is catenary (Topology, Definition 02I1).
If \(X\) is representable, then this is equivalent to the corresponding notion for the scheme representing \(X\).
Lemma
Let \(S\) be a locally Noetherian and universally catenary scheme. Let \(\delta : S \to \mathbf{Z}\) be a dimension function. Let \(X\) be a decent algebraic space over \(S\) such that the structure morphism \(X \to S\) is locally of finite type. Let \(\delta_X : |X| \to \mathbf{Z}\) be the map sending \(x\) to \(\delta(f(x))\) plus the transcendence degree of \(x/f(x)\). Then \(\delta_X\) is a dimension function on \(|X|\).
Proof
Let \(\varphi : U \to X\) be a surjective étale morphism where \(U\) is a scheme. Then the similarly defined function \(\delta_U\) is a dimension function on \(U\) by Morphisms, Lemma 02JW. On the other hand, by the definition of relative transcendence degree in (Morphisms of Spaces, Definition 04NM) we see that \(\delta_U(u) = \delta_X(\varphi(u))\).
Let \(x \leadsto x'\) be a specialization of points in \(|X|\). by Lemma 03IL we can find a specialization \(u \leadsto u'\) of points of \(U\) with \(\varphi(u) = x\) and \(\varphi(u') = x'\). Moreover, we see that \(x = x'\) if and only if \(u = u'\), see Lemma 03K5. Thus the fact that \(\delta_U\) is a dimension function implies that \(\delta_X\) is a dimension function, see Topology, Definition 02I9.
Lemma
Let \(S\) be a locally Noetherian and universally catenary scheme. Let \(X\) be an algebraic space over \(S\) such that \(X\) is decent and such that the structure morphism \(X \to S\) is locally of finite type. Then \(X\) is catenary.
Proof
The question is local on \(S\) (use Topology, Lemma 02I2). Thus we may assume that \(S\) has a dimension function, see Topology, Lemma 02IC. Then we conclude that \(|X|\) has a dimension function by Lemma 0ED5. Since \(|X|\) is sober (Proposition 03K6) we conclude that \(|X|\) is catenary by Topology, Lemma 02IA.
By Lemma 0ED6 the following definition is compatible with the already existing notion for representable algebraic spaces.
Definition
Let \(S\) be a scheme. Let \(X\) be a decent and locally Noetherian algebraic space over \(S\). We say \(X\) is universally catenary if for every morphism \(Y \to X\) of algebraic spaces which is locally of finite type and with \(Y\) decent, the algebraic space \(Y\) is catenary.
If \(X\) is an algebraic space, then the condition “\(X\) is decent and locally Noetherian” is equivalent to “\(X\) is quasi-separated and locally Noetherian”. This is Lemma 0BB6. Thus another way to understand the definition above is that \(X\) is universally catenary if and only if \(Y\) is catenary for all morphisms \(Y \to X\) which are quasi-separated and locally of finite type.
Lemma
Let \(S\) be a scheme. Let \(X\) be a decent, locally Noetherian, and universally catenary algebraic space over \(S\). Then any decent algebraic space locally of finite type over \(X\) is universally catenary.
Proof
This is formal from the definitions and the fact that compositions of morphisms locally of finite type are locally of finite type (Morphisms of Spaces, Lemma 03XG).
Lemma
Let \(S\) be a scheme. Let \(f : Y \to X\) be a surjective finite morphism of decent and locally Noetherian algebraic spaces. Let \(\delta : |X| \to \mathbf{Z}\) be a function. If \(\delta \circ |f|\) is a dimension function, then \(\delta\) is a dimension function.
Proof
Let \(x \mapsto x'\), \(x \not = x'\) be a specialization in \(|X|\). Choose \(y \in |Y|\) with \(|f|(y) = x\). Since \(|f|\) is closed (Morphisms of Spaces, Lemma 04NZ) we find a specialization \(y \leadsto y'\) with \(|f|(y') = x'\). Thus we conclude that \(\delta(x) = \delta(|f|(y)) > \delta(|f|(y')) = \delta(x')\) (see Topology, Definition 02I9). If \(x \leadsto x'\) is an immediate specialization, then \(y \leadsto y'\) is an immediate specialization too: namely if \(y \leadsto y'' \leadsto y'\), then \(|f|(y'')\) must be either \(x\) or \(x'\) and there are no nontrivial specializations between points of fibres of \(|f|\) by Lemma 0ACK.
The discussion will be continued in More on Morphisms of Spaces, Section 0EDL.