Lemma 66.6.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. There exists a surjective étale morphism $U \to X$ where $U$ is a disjoint union of affine schemes. We may in addition assume each of these affines maps into an affine open of $S$.

## 66.6 Special coverings

In this section we collect some straightforward lemmas on the existence of étale surjective coverings of algebraic spaces.

**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).
$\square$

Lemma 66.6.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. There exists a Zariski covering $X = \bigcup X_ i$ such that each algebraic space $X_ i$ has a surjective étale covering by an affine scheme. We may in addition assume each $X_ i$ maps into an affine open of $S$.

**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 $S$.
$\square$

Lemma 66.6.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Then $X$ is quasi-compact if and only if there exists an étale surjective morphism $U \to X$ with $U$ an affine scheme.

**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 $X$.
$\square$

The following lemma will be obsoleted by the discussion of separated morphisms in the chapter on morphisms of algebraic spaces.

Lemma 66.6.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $U$ be a separated scheme and $U \to X$ étale. Then $U \to X$ is separated, and $R = U \times _ X U$ is a separated scheme.

**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. $\square$

Lemma 66.6.5. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. If there exists a quasi-separated scheme $U$ and a surjective étale morphism $U \to X$ such that either of the projections $U \times _ X U \to U$ is quasi-compact, then $X$ is quasi-separated.

**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. $\square$

Lemma 66.6.6. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The following are equivalent

$X$ is Zariski locally quasi-separated over $S$,

$X$ is Zariski locally quasi-separated,

there exists a Zariski open covering $X = \bigcup X_ i$ such that for each $i$ there exists an affine scheme $U_ i$ and a quasi-compact surjective étale morphism $U_ i \to X_ i$, and

there exists a Zariski open covering $X = \bigcup X_ i$ such that for each $i$ there exists an affine scheme $U_ i$ which maps into an affine open of $S$ and a quasi-compact surjective étale morphism $U_ i \to X_ i$.

**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). $\square$

The following lemma will turn out to be quite useful.

Lemma 66.6.7. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $U$ be a scheme. Let $\varphi : U \to X$ be an étale morphism such that the projections $R = U \times _ X U \to U$ are quasi-compact; for example if $\varphi $ is quasi-compact. Then the fibres of

are finite.

**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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)