The Stacks project

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

Since $q_ i$ is étale it is open (see Remark 68.4.1). 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 29.57.9 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 66.4.3 and the assumption on $\varphi _ i$ we conclude that the fibre of $\varphi $ over $x$ is finite. In other words (2) holds. $\square$

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 $u = \mathop{\mathrm{Spec}}(\kappa (u))$ as a scheme. Note that $u \to U$, $u' \to U$ are monomorphisms (see Schemes, Lemma 26.23.7), 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

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 66.4.3. 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 68.4.1. Then we get a commutative diagram

where the squares are fibre product squares. The lower horizontal morphisms are étale and quasi-compact, as any scheme of the form $\mathop{\mathrm{Spec}}(k) \times _ S \mathop{\mathrm{Spec}}(k')$ 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 $u = \mathop{\mathrm{Spec}}(\kappa (u)) \to X$ 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

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 68.4.1). 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 26.23.11 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 $\mathop{\mathrm{Spec}}(k) \to X$ be a morphism from the spectrum of a field into $X$, in the equivalence class of $x$ such that the two projections $t, s : R = \mathop{\mathrm{Spec}}(k) \times _ X \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(k)$ are quasi-compact and étale. This means in particular that $R$ is an étale equivalence relation on $\mathop{\mathrm{Spec}}(k)$. By Spaces, Theorem 65.10.5 we know that the quotient sheaf $X' = \mathop{\mathrm{Spec}}(k)/R$ is an algebraic space. By Groupoids, Lemma 39.20.6 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 66.15.3 applies to $X'$, and we see that $X' = \mathop{\mathrm{Spec}}(k')$ for some field $k'$. Hence we get a factorization

which shows that $\mathop{\mathrm{Spec}}(k') \to X$ is a monomorphism mapping to $x \in |X|$. In other words (4) holds.

Finally, we prove that (4) implies (1). Let $\mathop{\mathrm{Spec}}(k) \to X$ 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 $\mathop{\mathrm{Spec}}(k) \times _ X u$ is nonempty, and since $\mathop{\mathrm{Spec}}(k) \times _ X u \to u$ is a monomorphism we conclude that $\mathop{\mathrm{Spec}}(k) \times _ X u = u$ (see Schemes, Lemma 26.23.11). Hence $u \to U \to X$ factors through $\mathop{\mathrm{Spec}}(k) \to X$, here is a picture

Since the right vertical arrow is étale this implies that $\kappa (u)/k$ is a finite separable extension. Hence we conclude that

is a finite scheme, and we win by the discussion of the meaning of property (1) in the first paragraph of this proof. $\square$

Proof. Assume (1). This clearly implies the first condition of Lemma 68.4.3 and hence we obtain a monomorphism $\mathop{\mathrm{Spec}}(k) \to X$ in the class of $x$. Taking the fibre product we see that $\mathop{\mathrm{Spec}}(k) \times _ X U \to \mathop{\mathrm{Spec}}(k)$ is a scheme étale over $\mathop{\mathrm{Spec}}(k)$ 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 $|\mathop{\mathrm{Spec}}(k) \times _ X U|$ by Properties of Spaces, Lemma 66.4.3. The finiteness of the fibre of $|R| \to |X|$ above $x$ is also clear since it is equal to the set underlying the scheme

which is finite over $k$. Thus (1) holds. $\square$

Proof. The equivalence of (1) and (3) follows on applying Lemma 68.4.4 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 68.4.2 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 68.4.3 (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 $\mathop{\mathrm{Spec}}(k) \times _ X U$ is on the one hand étale over $k$ (hence a disjoint union of spectra of finite separable extensions of $k$ by Remark 68.4.1) 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 $\mathop{\mathrm{Spec}}(k) \to X$ is a monomorphism as in (3), then

is quasi-compact (because over each $U_ i$ we see that $\mathop{\mathrm{Spec}}(k) \times _ X U_ i$ is a finite disjoint union spectra of fields). Thus $\mathop{\mathrm{Spec}}(k) \to X$ is quasi-compact by Morphisms of Spaces, Lemma 67.8.8.

It is immediate that (4) implies (5). Conversely, let $\mathop{\mathrm{Spec}}(k) \to X$ 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

Then $F \to U$ is quasi-compact, hence $F$ is quasi-compact. On the other hand, $F \to \mathop{\mathrm{Spec}}(k)$ is étale, hence $F$ is a finite disjoint union of spectra of finite separable extensions of $k$ (Remark 68.4.1). Since the image of $|F| \to |U|$ is the fibre of $|U| \to |X|$ over $x$ (Properties of Spaces, Lemma 66.4.3), we conclude that the fibre of $|U| \to |X|$ over $x$ is finite. The scheme $F \times _{\mathop{\mathrm{Spec}}(k)} F$ is also a finite union of spectra of fields because it is also quasi-compact and étale over $\mathop{\mathrm{Spec}}(k)$. There is a monomorphism $F \times _ X F \to F \times _{\mathop{\mathrm{Spec}}(k)} F$, hence $F \times _ X F$ is a finite disjoint union of spectra of fields (Schemes, Lemma 26.23.11). 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 66.4.3 we conclude that (1) holds. At this point we know that (1) – (5) are equivalent.

It is clear that (6) implies (5). Conversely, assume $\mathop{\mathrm{Spec}}(k) \to X$ is as in (4) and let $\mathop{\mathrm{Spec}}(k') \to X$ be another morphism with $k'$ a field in the equivalence class of $x$. By Properties of Spaces, Lemma 66.4.11 we have a factorization $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k) \to X$ of the given morphism. This is a composition of quasi-compact morphisms and hence quasi-compact (Morphisms of Spaces, Lemma 67.8.5) as desired. $\square$

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

Since $q_ i$ is étale it is open (see Remark 68.4.1). Moreover, we have $U = \bigcup \mathop{\mathrm{Im}}(q_ i)$, 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 29.57.9 to see the fibres of $p_{i_ j}|_{W_ j} : W_ j \to U_{i_ j}$ are universally bounded. Hence by Lemma 68.3.2 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 68.3.5. In other words (2) holds. $\square$

Proof. If (1) holds then the morphisms $U_ i \to X_ i \to X$ are étale (combine Morphisms, Lemma 29.36.3 and Spaces, Lemmas 65.5.4 and 65.5.3 ). 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 66.4.10. 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 65.11.4 the morphisms $U_ i \to X_ i$ are quasi-compact. $\square$