Definition 73.9.1. Let $S$ be a scheme, and let $X$ be an algebraic space over $S$. An *fpqc covering of $X$* is a family of morphisms $\{ f_ i : X_ i \to X\} _{i \in I}$ of algebraic spaces such that each $f_ i$ is flat and such that for every affine scheme $Z$ and morphism $h : Z \to X$ there exists a standard fpqc covering $\{ g_ j : Z_ j \to Z\} _{j = 1, \ldots , m}$ which refines the family $\{ X_ i \times _ X Z \to Z\} _{i \in I}$.

## 73.9 Fpqc topology

We briefly discuss the notion of an fpqc covering of algebraic spaces. Please compare with Topologies, Section 34.9. We will show in Descent on Spaces, Proposition 74.4.1 that quasi-coherent sheaves descent along these.

In other words, there exists indices $i_1, \ldots , i_ m \in I$ and morphisms $h_ j : Z_ j \to X_{i_ j}$ such that $f_{i_ j} \circ h_ j = h \circ g_ j$. Note that if $X$ and all $X_ i$ are representable, this is the same as a fpqc covering of schemes by Topologies, Lemma 34.9.12.

Lemma 73.9.2. Any fppf covering is an fpqc covering, and a fortiori, any syntomic, smooth, étale or Zariski covering is an fpqc covering.

**Proof.**
We will show that an fppf covering is an fpqc covering, and then the rest follows from Lemma 73.7.2. Let $\{ f_ i : U_ i \to U\} _{i \in I}$ be an fppf covering of algebraic spaces over $S$. By definition this means that the $f_ i$ are flat which checks the first condition of Definition 73.9.1. To check the second, let $V \to U$ be a morphism with $V$ affine. We may choose an étale covering $\{ V_{ij} \to V \times _ U U_ i\} $ with $V_{ij}$ affine. Then the compositions $f_{ij} : V_{ij} \to V \times _ U U_ i \to V$ are flat and locally of finite presentation as compositions of such (Morphisms of Spaces, Lemmas 67.28.2, 67.30.3, 67.39.7, and 67.39.8). Hence these morphisms are open (Morphisms of Spaces, Lemma 67.30.6) and we see that $|V| = \bigcup _{i \in I} \bigcup _{j \in J_ i} f_{ij}(|V_{ij}|)$ is an open covering of $|V|$. Since $|V|$ is quasi-compact, this covering has a finite refinement. Say $V_{i_1j_1}, \ldots , V_{i_ Nj_ N}$ do the job. Then $\{ V_{i_ kj_ k} \to V\} _{k = 1, \ldots , N}$ is a standard fpqc covering of $V$ refinining the family $\{ U_ i \times _ U V \to V\} $. This finishes the proof.
$\square$

Lemma 73.9.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.

If $X' \to X$ is an isomorphism then $\{ X' \to X\} $ is an fpqc covering of $X$.

If $\{ X_ i \to X\} _{i\in I}$ is an fpqc covering and for each $i$ we have an fpqc covering $\{ X_{ij} \to X_ i\} _{j\in J_ i}$, then $\{ X_{ij} \to X\} _{i \in I, j\in J_ i}$ is an fpqc covering.

If $\{ X_ i \to X\} _{i\in I}$ is an fpqc covering and $X' \to X$ is a morphism of algebraic spaces then $\{ X' \times _ X X_ i \to X'\} _{i\in I}$ is an fpqc covering.

**Proof.**
Part (1) is clear. Consider $g : X' \to X$ and $\{ X_ i \to X\} _{i\in I}$ an fpqc covering as in (3). By Morphisms of Spaces, Lemma 67.30.4 the morphisms $X' \times _ X X_ i \to X'$ are flat. If $h' : Z \to X'$ is a morphism from an affine scheme towards $X'$, then set $h = g \circ h' : Z \to X$. The assumption on $\{ X_ i \to X\} _{i\in I}$ means there exists a standard fpqc covering $\{ Z_ j \to Z\} _{j = 1, \ldots , n}$ and morphisms $Z_ j \to X_{i(j)}$ covering $h$ for certain $i(j) \in I$. By the universal property of the fibre product we obtain morphisms $Z_ j \to X' \times _ X X_{i(j)}$ over $h'$ also. Hence $\{ X' \times _ X X_ i \to X'\} _{i\in I}$ is an fpqc covering. This proves (3).

Let $\{ X_ i \to X\} _{i\in I}$ and $\{ X_{ij} \to X_ i\} _{j\in J_ i}$ be as in (2). Let $h : Z \to X$ be a morphism from an affine scheme towards $X$. By assumption there exists a standard fpqc covering $\{ Z_ j \to Z\} _{j = 1, \ldots , n}$ and morphisms $h_ j : Z_ j \to X_{i(j)}$ covering $h$ for some indices $i(j) \in I$. By assumption there exist standard fpqc coverings $\{ Z_{j, l} \to Z_ j\} _{l = 1, \ldots , n(j)}$ and morphisms $Z_{j, l} \to X_{i(j)j(l)}$ covering $h_ j$ for some indices $j(l) \in J_{i(j)}$. By Topologies, Lemma 34.9.11 the family $\{ Z_{j, l} \to Z\} $ is a standard fpqc covering. Hence we conclude that $\{ X_{ij} \to X\} _{i \in I, j\in J_ i}$ is an fpqc covering. $\square$

Lemma 73.9.4. Let $S$ be a scheme, and let $X$ be an algebraic space over $S$. Suppose that $\{ f_ i : X_ i \to X\} _{i \in I}$ is a family of morphisms of algebraic spaces with target $X$. Let $U \to X$ be a surjective étale morphism from a scheme towards $X$. Then $\{ f_ i : X_ i \to X\} _{i \in I}$ is an fpqc covering of $X$ if and only if $\{ U \times _ X X_ i \to U\} _{i \in I}$ is an fpqc covering of $U$.

**Proof.**
If $\{ X_ i \to X\} _{i \in I}$ is an fpqc covering, then so is $\{ U \times _ X X_ i \to U\} _{i \in I}$ by Lemma 73.9.3. Assume that $\{ U \times _ X X_ i \to U\} _{i \in I}$ is an fpqc covering. Let $h : Z \to X$ be a morphism from an affine scheme towards $X$. Then we see that $U \times _ X Z \to Z$ is a surjective étale morphism of schemes, in particular open. Hence we can find finitely many affine opens $W_1, \ldots , W_ t$ of $U \times _ X Z$ whose images cover $Z$. For each $j$ we may apply the condition that $\{ U \times _ X X_ i \to U\} _{i \in I}$ is an fpqc covering to the morphism $W_ j \to U$, and obtain a standard fpqc covering $\{ W_{jl} \to W_ j\} $ which refines $\{ W_ j \times _ X X_ i \to W_ j\} _{i \in I}$. Hence $\{ W_{jl} \to Z\} $ is a standard fpqc covering of $Z$ (see Topologies, Lemma 34.9.11) which refines $\{ Z \times _ X X_ i \to X\} $ and we win.
$\square$

Lemma 73.9.5. Let $S$ be a scheme, and let $X$ be an algebraic space over $S$. Suppose that $\mathcal{U} = \{ f_ i : X_ i \to X\} _{i \in I}$ is an fpqc covering of $X$. Then there exists a refinement $\mathcal{V} = \{ g_ i : T_ i \to X\} $ of $\mathcal{U}$ which is an fpqc covering such that each $T_ i$ is a scheme.

**Proof.**
Omitted. Hint: For each $i$ choose a scheme $T_ i$ and a surjective étale morphism $T_ i \to X_ i$. Then check that $\{ T_ i \to X\} $ is an fpqc covering.
$\square$

To be continued...

## 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 (2)

Comment #8699 by Torsten on

Comment #9373 by Stacks project on