TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30

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Proper morphisms are closed

Statement

Let $f : X \to S$ be a proper morphism of schemes. Then $f$ is a closed map of topological spaces: for every closed subset $Z \subseteq ∣ X ∣$ its image $f (Z)$ is closed in $∣ S ∣$ . In particular $f (X)$ is closed. Moreover, for every $S$ -scheme $T$ the base-changed projection $f_{T} : X \times_{S} T ⟶ T$ is closed, so the image of $X \times_{S} T$ , and of any closed subset of it, is closed in $∣ T ∣$ .

Facts & Assumptions

Given: A proper morphism $f : X \to S$ and an arbitrary morphism $T \to S$ .

[F1]

A morphism of schemes $f : X \to S$ is proper if and only if it is separated, of finite type, and universally closed. (Proper morphisms)

[F2]

A morphism $f : X \to S$ is universally closed if for every $S$ -scheme $T$ the base-changed projection $f_{T} : X \times_{S} T \to T$ is a closed map, and explicitly, for every closed subset $Z \subseteq ∣ X_{T} ∣$ , its image $f_{T} (∣ Z ∣)$ is closed in $∣ T ∣$ . (Universally closed morphisms)

[F3]

For an $S$ -scheme $f : X \to S$ and a morphism $h : S^{'} \to S$ , the base change is $X_{S^{'}} = X \times_{S} S^{'}$ , with structure map the second projection. (Base change of objects, morphisms and properties)

[F4]

A fibre product of $X \to S \leftarrow Y$ is a scheme $P$ with projections $p : P \to X$ and $q : P \to Y$ such that for every morphism $a : T \to X$ and $b : T \to Y$ with $f a = g b$ there is exactly one $h : T \to P$ satisfying $p h = a$ and $q h = b$ . (Fibre product of schemes)

Proof

technique · direct: closedness is universal closedness read at the identity base change

1.1F1

Since $f$ is proper, [F1] makes it universally closed, and in particular of finite type; only universal closedness is used below.

1.2F2F3F4

Take the $S$ -scheme $T = S$ , with structure morphism the identity, so that [F3] identifies the base-changed projection with $q : X \times_{S} S \to S$ . By [F2] this $q$ is a closed map. The universal property [F4] applies to $f : X \to S$ and the identity $g = {id}_{S}$ with test morphisms $a = {id}_{X}$ and $b = f$ : the required identity $f \circ a = {id}_{S} \circ f$ holds, so there is a unique $u : X \to X \times_{S} S$ with $p \circ u = {id}_{X}$ and $q \circ u = f$ . The same universal property applied to $p$ and $q$ in place of $a$ and $b$ shows that $u \circ p$ and ${id}_{X \times_{S} S}$ have the same composites with $p$ and $q$ , hence $u \circ p$ is the identity and $u$ is an isomorphism. Therefore $f = q \circ u$ is closed, so the image of every closed $Z \subseteq ∣ X ∣$ is closed in $∣ S ∣$ ; taking $Z = ∣ X ∣$ gives that $f (X)$ is closed.

1.3F2F3

Now let $T \to S$ be any $S$ -scheme. By [F2] the projection $f_{T} : X \times_{S} T \to T$ is closed, which by [F3] is exactly the assertion that the base change of $f$ along $T \to S$ maps closed subsets of $X \times_{S} T$ to closed subsets of $∣ T ∣$ . In particular the image of the closed subset $X \times_{S} T$ itself is closed in $∣ T ∣$ , and the same holds for every closed subset.

If $X$ is empty, then every image is empty and closed, and if $S$ is empty the identity base-change case of step 1.2 is the empty morphism; both are covered by the same argument. No step uses a choice principle, a Noetherian hypothesis, or a separatedness hypothesis beyond the one hidden in properness. ∎

Proper morphisms are closed

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources