The Stacks project

Lemma 26.19.7. Let $f : X \to S$ be a quasi-compact morphism of schemes. The following are equivalent

  1. $f(X) \subset S$ is closed, and

  2. $f(X) \subset S$ is stable under specialization.

Proof. We have (1) $\Rightarrow $ (2) by Topology, Lemma 5.19.2. Assume (2). Let $U \subset S$ be an affine open. It suffices to prove that $f(X) \cap U$ is closed. Since $U \cap f(X)$ is stable under specializations in $U$, we have reduced to the case where $S$ is affine. Because $f$ is quasi-compact we deduce that $X = f^{-1}(S)$ is quasi-compact as $S$ is affine. Thus we may write $X = \bigcup _{i = 1}^ n U_ i$ with $U_ i \subset X$ open affine. Say $S = \mathop{\mathrm{Spec}}(R)$ and $U_ i = \mathop{\mathrm{Spec}}(A_ i)$ for some $R$-algebra $A_ i$. Then $f(X) = \mathop{\mathrm{Im}}(\mathop{\mathrm{Spec}}(A_1 \times \ldots \times A_ n) \to \mathop{\mathrm{Spec}}(R))$. Thus the lemma follows from Algebra, Lemma 10.41.5. $\square$

Comments (2)

Comment #3070 by Dario WeiƟmann on

The property "closed under specialization" is relative to the ambient space. In the proof is closed under specialization in . This confused me for a second, maybe this could be clarified?

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  • 8 comment(s) on Section 26.19: Quasi-compact morphisms

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