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Fibres of proper morphisms are proper

Statement

Assume the Axiom of Choice. Let $f : X \to S$ be a proper morphism of schemes and let $s \in S$ be a point, not necessarily closed. Then the scheme-theoretic fibre $X_{s} = X \times_{S} Spec κ (s)$ is proper over $Spec κ (s)$ . In particular every fibre of a proper morphism, including a generic fibre and the empty fibre over a point not in the image, is a proper $κ (s)$ -scheme.

Facts & Assumptions

Given: A proper morphism $f : X \to S$ , a point $s \in S$ , and the canonical morphism $Spec κ (s) \to S$ .

[F1]

The scheme-theoretic fibre is $X_{s} = X \times_{S} Spec κ (s)$ , viewed as a $κ (s)$ -scheme; empty fibres are allowed and $s$ need not be closed. (Scheme-theoretic fibre)

[F2]

Assume AC. For every proper morphism $f : X \to S$ and every morphism $S^{'} \to S$ , the base-changed morphism $f_{S^{'}} : X \times_{S} S^{'} \to S^{'}$ is proper. (Properness survives arbitrary base change)

[F3]

The Axiom of Choice: every family of nonempty sets has a choice function. (The Axiom of Choice)

Proof

technique · direct: read the fibre as a base change

1.1F1

By [F1] the scheme-theoretic fibre is the fibre product $X \times_{S} Spec κ (s)$ , with structure morphism the second projection, and this is exactly the base change of $f$ along the canonical morphism $Spec κ (s) \to S$ . Its source is $X_{s}$ and its target is $Spec κ (s)$ .

2.1F2step 1.1

The base-changed morphism $f_{Spec κ (s)}$ is proper by [F2], applied to the proper morphism $f$ and the morphism $Spec κ (s) \to S$ . Combining with the identification of step 1.1, the fibre $X_{s} \to Spec κ (s)$ is proper.

3.1F1F2F3∎

The argument uses the Axiom of Choice exactly through [F2], which assumes it; nothing else in the proof selects from a family of nonempty sets. If $s \notin f (X)$ , then $X_{s}$ is empty, which is the empty affine scheme $Spec 0$ and is proper over $Spec κ (s)$ by the base-change statement applied to the empty source; if $κ (s)$ -fibres are taken over a generic point, the same computation applies, since [F1] does not require $s$ to be closed. The statement has no endpoint or infinite-length cases.

Depends on

Used by

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Dependency tree · two levels

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Sources

The Stacks Project, Morphisms of Schemes, Lemma 29.42.5 and the fibre definition of Section 29.20 (standard reference, not scraped)
Vakil, The Rising Sea, §11.3.4 (standard reference, not scraped)