Lemma 26.21.7. Let $f : X \to S$ be a morphism of schemes.

1. If $f$ is separated then for every pair of affine opens $(U, V)$ of $X$ which map into a common affine open of $S$ we have

1. the intersection $U \cap V$ is affine.

2. the ring map $\mathcal{O}_ X(U) \otimes _{\mathbf{Z}} \mathcal{O}_ X(V) \to \mathcal{O}_ X(U \cap V)$ is surjective.

2. If any pair of points $x_1, x_2 \in X$ lying over a common point $s \in S$ are contained in affine opens $x_1 \in U$, $x_2 \in V$ which map into a common affine open of $S$ such that (a), (b) hold, then $f$ is separated.

Proof. Assume $f$ separated. Suppose $(U, V)$ is a pair as in (1). Let $W = \mathop{\mathrm{Spec}}(R)$ be an affine open of $S$ containing both $f(U)$ and $f(V)$. Write $U = \mathop{\mathrm{Spec}}(A)$ and $V = \mathop{\mathrm{Spec}}(B)$ for $R$-algebras $A$ and $B$. By Lemma 26.17.3 we see that $U \times _ S V = U \times _ W V = \mathop{\mathrm{Spec}}(A \otimes _ R B)$ is an affine open of $X \times _ S X$. Hence, by Lemma 26.10.1 we see that $\Delta ^{-1}(U \times _ S V) \to U \times _ S V$ can be identified with $\mathop{\mathrm{Spec}}(A \otimes _ R B/J)$ for some ideal $J \subset A \otimes _ R B$. Thus $U \cap V = \Delta ^{-1}(U \times _ S V)$ is affine. Assertion (1)(b) holds because $A \otimes _{\mathbf{Z}} B \to (A \otimes _ R B)/J$ is surjective.

Assume the hypothesis formulated in (2) holds. Clearly the collection of affine opens $U \times _ S V$ for pairs $(U, V)$ as in (2) form an affine open covering of $X \times _ S X$ (see e.g. Lemma 26.17.4). Hence it suffices to show that each morphism $U \cap V = \Delta _{X/S}^{-1}(U \times _ S V) \to U \times _ S V$ is a closed immersion, see Lemma 26.4.2. By assumption (a) we have $U \cap V = \mathop{\mathrm{Spec}}(C)$ for some ring $C$. After choosing an affine open $W = \mathop{\mathrm{Spec}}(R)$ of $S$ into which both $U$ and $V$ map and writing $U = \mathop{\mathrm{Spec}}(A)$, $V = \mathop{\mathrm{Spec}}(B)$ we see that the assumption (b) means that the composition

$A \otimes _{\mathbf{Z}} B \to A \otimes _ R B \to C$

is surjective. Hence $A \otimes _ R B \to C$ is surjective and we conclude that $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A \otimes _ R B)$ is a closed immersion. $\square$

Comment #2680 by sdf on

Perhaps this is a stupid question, but how does one construct $W$ in the first line of the proof? Actually, I thought in general if $f\colon X\operatorname{Spec} R \rightarrow Y$ is an arbitrary morphism then in general there does not exist affine $U\subset Y$ with $f(X)\subset U$? I know how to construct $U$ say when $R$ is a local ring, but in general not?

Comment #2681 by sdf on

Erratum: the morphism is $f\colon X=\operatorname{Spec} R \rightarrow Y$.

Comment #2682 by on

In condition (1) we assume $U$ and $V$ map into a common affine open of $S$. Thus the existence of $W$ is simply assumed to be tryue. OK?

Comment #2768 by sdf on

Yes I didn't read it properly, apologies.

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