The Stacks project

Proof. Let $V \to X$ be a surjective étale morphism. Let $V = \bigcup _{i \in I} V_ i$ be a Zariski open covering such that each $V_ i$ maps into an affine open of $S$. Then set $U = \coprod _{i \in I} V_ i$ with induced morphism $U \to V \to X$. This is étale and surjective as a composition of étale and surjective representable transformations of functors (via the general principle Spaces, Lemma 65.5.4 and Morphisms, Lemmas 29.9.2 and 29.36.3). $\square$

Proof. By Lemma 66.6.1 we can find a surjective étale morphism $U = \coprod U_ i \to X$, with $U_ i$ affine and mapping into an affine open of $S$. Let $X_ i \subset X$ be the open subspace of $X$ such that $U_ i \to X$ factors through an étale surjective morphism $U_ i \to X_ i$, see Lemma 66.4.10. Since $U = \bigcup U_ i$ we see that $X = \bigcup X_ i$. As $U_ i \to X_ i$ is surjective it follows that $X_ i \to S$ maps into an affine open of $S$. $\square$

Proof. If there exists an étale surjective morphism $U \to X$ with $U$ affine then $X$ is quasi-compact by Definition 66.5.1. Conversely, if $X$ is quasi-compact, then $|X|$ is quasi-compact. Let $U = \coprod _{i \in I} U_ i$ be a disjoint union of affine schemes with an étale and surjective map $\varphi : U \to X$ (Lemma 66.6.1). Then $|X| = \bigcup \varphi (|U_ i|)$ and by quasi-compactness there is a finite subset $i_1, \ldots , i_ n$ such that $|X| = \bigcup \varphi (|U_{i_ j}|)$. Hence $U_{i_1} \cup \ldots \cup U_{i_ n}$ is an affine scheme with a finite surjective morphism towards $X$. $\square$

Proof. Let $X' \subset X$ be the open subscheme such that $U \to X$ factors through an étale surjection $U \to X'$, see Lemma 66.4.10. If $U \to X'$ is separated, then so is $U \to X$, see Spaces, Lemma 65.5.4 (as the open immersion $X' \to X$ is separated by Spaces, Lemma 65.5.8 and Schemes, Lemma 26.23.8). Moreover, since $U \times _{X'} U = U \times _ X U$ it suffices to prove the result after replacing $X$ by $X'$, i.e., we may assume $U \to X$ surjective. Consider the commutative diagram

In the proof of Spaces, Lemma 65.13.1 we have seen that $j : R \to U \times _ S U$ is separated. The morphism of schemes $U \to S$ is separated as $U$ is a separated scheme, see Schemes, Lemma 26.21.13. Hence $U \times _ S U \to U$ is separated as a base change, see Schemes, Lemma 26.21.12. Hence the scheme $U \times _ S U$ is separated (by the same lemma). Since $j$ is separated we see in the same way that $R$ is separated. Hence $R \to U$ is a separated morphism (by Schemes, Lemma 26.21.13 again). Thus by Spaces, Lemma 65.11.4 and the diagram above we conclude that $U \to X$ is separated. $\square$

Proof. We may think of $X$ as an algebraic space over $\mathbf{Z}$. Consider the cartesian diagram

Since $U$ is quasi-separated the projection $U \times U \to U$ is quasi-separated (as a base change of a quasi-separated morphism of schemes, see Schemes, Lemma 26.21.12). Hence the assumption in the lemma implies $j$ is quasi-compact by Schemes, Lemma 26.21.14. By Spaces, Lemma 65.11.4 we see that $\Delta $ is quasi-compact as desired. $\square$

Proof. Assume $U_ i \to X_ i \subset X$ are as in (3). To prove (4) choose for each $i$ a finite affine open covering $U_ i = U_{i1} \cup \ldots \cup U_{in_ i}$ such that each $U_{ij}$ maps into an affine open of $S$. The compositions $U_{ij} \to U_ i \to X_ i$ are étale and quasi-compact (see Spaces, Lemma 65.5.4). Let $X_{ij} \subset X_ i$ be the open subspace corresponding to the image of $|U_{ij}| \to |X_ i|$, see Lemma 66.4.10. Note that $U_{ij} \to X_{ij}$ is quasi-compact as $X_{ij} \subset X_ i$ is a monomorphism and as $U_{ij} \to X$ is quasi-compact. Then $X = \bigcup X_{ij}$ is a covering as in (4). The implication (4) $\Rightarrow $ (3) is immediate.

Assume (4). To show that $X$ is Zariski locally quasi-separated over $S$ it suffices to show that $X_ i$ is quasi-separated over $S$. Hence we may assume there exists an affine scheme $U$ mapping into an affine open of $S$ and a quasi-compact surjective étale morphism $U \to X$. Consider the fibre product square

The right vertical arrow is surjective étale (see Spaces, Lemma 65.5.7) and $U \times _ S U$ is affine (as $U$ maps into an affine open of $S$, see Schemes, Section 26.17), and $U \times _ X U$ is quasi-compact because the projection $U \times _ X U \to U$ is quasi-compact as a base change of $U \to X$. It follows from Spaces, Lemma 65.11.4 that $\Delta _{X/S}$ is quasi-compact as desired.

Assume (1). To prove (3) there is an immediate reduction to the case where $X$ is quasi-separated over $S$. By Lemma 66.6.2 we can find a Zariski open covering $X = \bigcup X_ i$ such that each $X_ i$ maps into an affine open of $S$, and such that there exist affine schemes $U_ i$ and surjective étale morphisms $U_ i \to X_ i$. Since $U_ i \to S$ maps into an affine open of $S$ we see that $U_ i \times _ S U_ i$ is affine, see Schemes, Section 26.17. As $X$ is quasi-separated over $S$, the morphisms

as base changes of $\Delta _{X/S}$ are quasi-compact. Hence we conclude that $R_ i$ is a quasi-compact scheme. This in turn implies that each projection $R_ i \to U_ i$ is quasi-compact. Hence, applying Spaces, Lemma 65.11.4 to the covering $U_ i \to X_ i$ and the morphism $U_ i \to X_ i$ we conclude that the morphisms $U_ i \to X_ i$ are quasi-compact as desired.

At this point we see that (1), (3), and (4) are equivalent. Since (3) does not refer to the base scheme we conclude that these are also equivalent with (2). $\square$

Proof. Denote $R = U \times _ X U$, and $s, t : R \to U$ the projections. Let $u \in U$ be a point, and let $x \in |X|$ be its image. The fibre of $|U| \to |X|$ over $x$ is equal to $s(t^{-1}(\{ u\} ))$ by Lemma 66.4.3, and the fibre of $|R| \to |X|$ over $x$ is $t^{-1}(s(t^{-1}(\{ u\} )))$. Since $t : R \to U$ is étale and quasi-compact, it has finite fibres (as its fibres are disjoint unions of spectra of fields by Morphisms, Lemma 29.36.7 and quasi-compact). Hence we win. $\square$