Lemma 34.9.7. Let $T$ be a scheme.

1. If $T' \to T$ is an isomorphism then $\{ T' \to T\}$ is an fpqc covering of $T$.

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

3. If $\{ T_ i \to T\} _{i\in I}$ is an fpqc covering and $T' \to T$ is a morphism of schemes then $\{ T' \times _ T T_ i \to T'\} _{i\in I}$ is an fpqc covering.

Proof. Part (1) is immediate. Recall that the composition of flat morphisms is flat and that the base change of a flat morphism is flat (Morphisms, Lemmas 29.25.8 and 29.25.6). Thus we can apply Lemma 34.9.2 in each case to check that our families of morphisms are fpqc coverings.

Proof of (2). Assume $\{ T_ i \to T\} _{i\in I}$ is an fpqc covering and for each $i$ we have an fpqc covering $\{ f_{ij} : T_{ij} \to T_ i\} _{j\in J_ i}$. Let $U \subset T$ be an affine open. We can find quasi-compact opens $U_ i \subset T_ i$ for $i \in I$, almost all empty, such that $U = \bigcup f_ i(U_ i)$. Then for each $i$ we can choose quasi-compact opens $W_{ij} \subset T_{ij}$ for $j \in J_ i$, almost all empty, with $U_ i = \bigcup _ j f_{ij}(U_{ij})$. Thus $\{ T_{ij} \to T\}$ is an fpqc covering.

Proof of (3). Assume $\{ T_ i \to T\} _{i\in I}$ is an fpqc covering and $T' \to T$ is a morphism of schemes. Let $U' \subset T'$ be an affine open which maps into the affine open $U \subset T$. Choose quasi-compact opens $U_ i \subset T_ i$, almost all empty, such that $U = \bigcup f_ i(U_ i)$. Then $U' \times _ U U_ i$ is a quasi-compact open of $T' \times _ T T_ i$ and $U' = \bigcup \text{pr}_1(U' \times _ U U_ i)$. Since $T'$ can be covered by such affine opens $U' \subset T'$ we see that $\{ T' \times _ T T_ i \to T'\} _{i\in I}$ is an fpqc covering by Lemma 34.9.2. $\square$

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