The Stacks project

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 $\mathop{\mathrm{Spec}}((A \otimes _ R B)/J) \to \mathop{\mathrm{Spec}}(A \otimes _ R B)$ 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. $\square$

Comments (6)

Comment #2680 by sdf on

Perhaps this is a stupid question, but how does one construct in the first line of the proof? Actually, I thought in general if is an arbitrary morphism then in general there does not exist affine with ? I know how to construct say when is a local ring, but in general not?

Comment #2681 by sdf on

Erratum: the morphism is .

Comment #2682 by on

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

Comment #2768 by sdf on

Yes I didn't read it properly, apologies.

Comment #8467 by on

Minor typo: instead of in "we see that can be identified with ," one should write .

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  • 18 comment(s) on Section 26.21: Separation axioms

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