Lemma 34.9.2 (03L7)—The Stacks project

Lemma 34.9.2. Let $T$ be a scheme. Let $\{ f_ i : T_ i \to T\} _{i \in I}$ be a family of morphisms of schemes with target $T$. The following are equivalent

$\{ f_ i : T_ i \to T\} _{i \in I}$ is an fpqc covering,
each $f_ i$ is flat and for every affine open $U \subset T$ there exist quasi-compact opens $U_ i \subset T_ i$ which are almost all empty, such that $U = \bigcup f_ i(U_ i)$,
each $f_ i$ is flat and there exists an affine open covering $T = \bigcup _{\alpha \in A} U_\alpha $ and for each $\alpha \in A$ there exist $i_{\alpha , 1}, \ldots , i_{\alpha , n(\alpha )} \in I$ and quasi-compact opens $U_{\alpha , j} \subset T_{i_{\alpha , j}}$ such that $U_\alpha = \bigcup _{j = 1, \ldots , n(\alpha )} f_{i_{\alpha , j}}(U_{\alpha , j})$.

If $T$ is quasi-separated, these are also equivalent to

each $f_ i$ is flat, and for every $t \in T$ there exist $i_1, \ldots , i_ n \in I$ and quasi-compact opens $U_ j \subset T_{i_ j}$ such that $\bigcup _{j = 1, \ldots , n} f_{i_ j}(U_ j)$ is a (not necessarily open) neighbourhood of $t$ in $T$.

Proof. We omit the proof of the equivalence of (1), (2), and (3). From now on assume $T$ is quasi-separated. We prove (4) implies (2). Let $U \subset T$ be an affine open. To prove (2) it suffices to show that for every $t \in U$ there exist finitely many quasi-compact opens $U_ j \subset T_{i_ j}$ such that $f_{i_ j}(U_ j) \subset U$ and such that $\bigcup f_{i_ j}(U_ j)$ is a neighbourhood of $t$ in $U$. By assumption there do exist finitely many quasi-compact opens $U'_ j \subset T_{i_ j}$ such that such that $\bigcup f_{i_ j}(U'_ j)$ is a neighbourhood of $t$ in $T$. Since $T$ is quasi-separated we see that $U_ j = U'_ j \cap f_{i_ j}^{-1}(U) = (f_{i_ j}|_{U'_ j})^{-1}(U)$ is quasi-compact by Schemes, Lemma 26.21.14. Thus (2) holds. Since it is clear that (2) implies (4) the proof is finished. $\square$

Comments (5)

Comment #8469 by Ryo Suzuki on June 07, 2023 at 23:05

I think Lemma 03GI is used to deduce $U_j$ is quasi-compact. It might worth to note it explicitly.

Comment #8641 by Tony Scholl on August 29, 2023 at 09:04

Here's an example to answer your question from 2021 on the blog: let $X=X_1\cup X_2$ , $X_1=X_2=\mathrm{Spec} k[t_1,t_2,\dots]$ be the non-qs scheme from Example 01KL, and $Y = X_1 \sqcup \mathrm{Spec} \mathcal{O}_{X_2,0}$ . Then the obvious morphism $f:Y\to X$ is flat and surjective, and (since $Y$ is qc) trivially satisfies (4). But it isn't an fpqc covering: $f^{-1}(X_2)= (X_1\smallsetminus \{0\}) \sqcup \mathrm{Spec} \mathcal{O}_{X_2,0}$ , so $X_2$ isn't the image of any qc open of $Y$ .

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