Lemma 34.9.4. Let $T$ be a scheme. Let $\{ f_ i : T_ i \to T\} _{i \in I}$ be a family of morphisms of schemes with target $T$. Assume that

1. each $f_ i$ is flat, and

2. the family $\{ f_ i : T_ i \to T\} _{i \in I}$ can be refined by an fpqc covering of $T$.

Then $\{ f_ i : T_ i \to T\} _{i \in I}$ is an fpqc covering of $T$.

Proof. Let $\{ g_ j : X_ j \to T\} _{j \in J}$ be an fpqc covering refining $\{ f_ i : T_ i \to T\}$. Suppose that $U \subset T$ is affine open. Choose $j_1, \ldots , j_ m \in J$ and $V_ k \subset X_{j_ k}$ affine open such that $U = \bigcup g_{j_ k}(V_ k)$. For each $j$ pick $i_ j \in I$ and a morphism $h_ j : X_ j \to T_{i_ j}$ such that $g_ j = f_{i_ j} \circ h_ j$. Since $h_{j_ k}(V_ k)$ is quasi-compact we can find a quasi-compact open $h_{j_ k}(V_ k) \subset U_ k \subset f_{i_{j_ k}}^{-1}(U)$. Then $U = \bigcup f_{i_{j_ k}}(U_ k)$. We conclude that $\{ f_ i : T_ i \to T\} _{i \in I}$ is an fpqc covering by Lemma 34.9.2. $\square$

Comment #3201 by Dario Weißmann on

Typo(s): there are two instances of 'a fpqc covering' in the statement

There are also:

• 6 comment(s) on Section 34.9: The fpqc topology

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).