Lemma 30.26.1. Let $f : X \to S$ be a morphism of schemes which is locally of finite type. Let $Z \subset X$ be a closed subset. The following are equivalent

1. the morphism $Z \to S$ is proper if $Z$ is endowed with the reduced induced closed subscheme structure (Schemes, Definition 26.12.5),

2. for some closed subscheme structure on $Z$ the morphism $Z \to S$ is proper,

3. for any closed subscheme structure on $Z$ the morphism $Z \to S$ is proper.

Proof. The implications (3) $\Rightarrow$ (1) and (1) $\Rightarrow$ (2) are immediate. Thus it suffices to prove that (2) implies (3). We urge the reader to find their own proof of this fact. Let $Z'$ and $Z''$ be closed subscheme structures on $Z$ such that $Z' \to S$ is proper. We have to show that $Z'' \to S$ is proper. Let $Z''' = Z' \cup Z''$ be the scheme theoretic union, see Morphisms, Definition 29.4.4. Then $Z'''$ is another closed subscheme structure on $Z$. This follows for example from the description of scheme theoretic unions in Morphisms, Lemma 29.4.6. Since $Z'' \to Z'''$ is a closed immersion it suffices to prove that $Z''' \to S$ is proper (see Morphisms, Lemmas 29.41.6 and 29.41.4). The morphism $Z' \to Z'''$ is a bijective closed immersion and in particular surjective and universally closed. Then the fact that $Z' \to S$ is separated implies that $Z''' \to S$ is separated, see Morphisms, Lemma 29.41.11. Moreover $Z''' \to S$ is locally of finite type as $X \to S$ is locally of finite type (Morphisms, Lemmas 29.15.5 and 29.15.3). Since $Z' \to S$ is quasi-compact and $Z' \to Z'''$ is a homeomorphism we see that $Z''' \to S$ is quasi-compact. Finally, since $Z' \to S$ is universally closed, we see that the same thing is true for $Z''' \to S$ by Morphisms, Lemma 29.41.9. This finishes the proof. $\square$

Comment #7073 by Manuel Hoff on

There is a typo in the first line of the proof: '(3) $\Rightarrow$ (2)' should be replaced by '(3) $\Rightarrow$ (1)'.

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