Lemma 29.13.2 (01SL)—The Stacks project

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Table of contents
Part 2: Schemes
Chapter 29: Morphisms of Schemes
Section 29.13: Quasi-affine morphisms
Lemma 29.13.2 (cite)

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Lemma 29.13.2. A quasi-affine morphism is separated and quasi-compact.

Proof. Let $f : X \to S$ be quasi-affine. Quasi-compactness is immediate from Schemes, Lemma 26.19.2. Let $U \subset S$ be an affine open. If we can show that $f^{-1}(U)$ is a separated scheme, then $f$ is separated (Schemes, Lemma 26.21.7 shows that being separated is local on the base). By assumption $f^{-1}(U)$ is isomorphic to an open subscheme of an affine scheme. An affine scheme is separated and hence every open subscheme of an affine scheme is separated as desired. $\square$

Comments (2)

Comment #4294 by Xuande Liu on July 05, 2019 at 05:43

Without loss of generality we can assume that $S=\mathit{Spec}(B)$ is affine. Let $A=\Gamma(X,\mathcal O_X)$ . And hence we can factorize $f$ as $X\to \mathit{Spec}(A)\to\mathit{Spec(B)}$ . The former morphism is an open immersion and the latter is morphism between affine schemes. Therefore $f$ is separated.

Comment #4459 by Johan on September 01, 2019 at 16:19

Much better indeed. THanks very much. See fix here.

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