Lemma 26.21.6. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

1. The morphism $f$ is quasi-separated.

2. For every pair of affine opens $U, V \subset X$ which map into a common affine open of $S$ the intersection $U \cap V$ is a finite union of affine opens of $X$.

3. There exists an affine open covering $S = \bigcup _{i \in I} U_ i$ and for each $i$ an affine open covering $f^{-1}U_ i = \bigcup _{j \in I_ i} V_ j$ such that for each $i$ and each pair $j, j' \in I_ i$ the intersection $V_ j \cap V_{j'}$ is a finite union of affine opens of $X$.

Proof. Let us prove that (3) implies (1). By Lemma 26.17.4 the covering $X \times _ S X = \bigcup _ i \bigcup _{j, j'} V_ j \times _{U_ i} V_{j'}$ is an affine open covering of $X \times _ S X$. Moreover, $\Delta _{X/S}^{-1}(V_ j \times _{U_ i} V_{j'}) = V_ j \cap V_{j'}$. Hence the implication follows from Lemma 26.19.2.

The implication (1) $\Rightarrow$ (2) follows from the fact that under the hypotheses of (2) the fibre product $U \times _ S V$ is an affine open of $X \times _ S X$. The implication (2) $\Rightarrow$ (3) is trivial. $\square$

Comment #1051 by Wessel Bindt on

In the penultimate sentence of the proof, it seems that "hypotheses of (1)" is supposed to be "hypotheses of (2)".

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