Lemma 30.2.5. Let X be a scheme. The following are equivalent
X has affine diagonal \Delta : X \to X \times X,
for U, V \subset X affine open, the intersection U \cap V is affine, and
there exists an open covering \mathcal{U} : X = \bigcup _{i \in I} U_ i such that U_{i_0 \ldots i_ p} is affine open for all p \ge 0 and all i_0, \ldots , i_ p \in I.
In particular this holds if X is separated.
Proof. Assume X has affine diagonal. Let U, V \subset X be affine opens. Then U \cap V = \Delta ^{-1}(U \times V) is affine. Thus (2) holds. It is immediate that (2) implies (3). Conversely, if there is a covering of X as in (3), then X \times X = \bigcup U_ i \times U_{i'} is an affine open covering, and we see that \Delta ^{-1}(U_ i \times U_{i'}) = U_ i \cap U_{i'} is affine. Then \Delta is an affine morphism by Morphisms, Lemma 29.11.3. The final assertion follows from Schemes, Lemma 26.21.7. \square
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