The Stacks project

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

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

  2. 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.

  3. 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 34.9.10 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. $\square$


Comments (0)

There are also:

  • 1 comment(s) on Section 73.9: Fpqc topology

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Your email address will not be published.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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03MR. Beware of the difference between the letter 'O' and the digit '0'.