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. $\square$

Comment #3070 by Dario Weißmann on

The property "closed under specialization" is relative to the ambient space. In the proof $U\cap f(X)$ is closed under specialization in $U$. This confused me for a second, maybe this could be clarified?

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