LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-30

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Finite morphisms survive base change and composition

Statement

Assume the Axiom of Choice (AC) for the base-change assertion. For every finite morphism $f : X \to S$ and every morphism $T \to S$ , the projection $X \times_{S} T \to T$ is finite. Finite morphisms are closed under composition, and that composition assertion uses no AC.

Facts & Assumptions

Given: Schemes and morphisms as in the statement; AC is assumed only for the base-change assertion.

[F1]

A morphism $f : X \to S$ is finite when, for every affine open $U = Spec A \subseteq S$ , its inverse image is affine $f^{- 1} (U) = Spec B$ and $B$ is a finite $A$ -module (Finite morphisms of schemes).

[F2]

Assuming AC, a morphism is finite if there is an affine open cover of its target on which inverse images are affine and the coordinate ring maps are module-finite (Finite is affine and local on its target).

[F3]

Every point of a scheme has an affine open neighbourhood; this also applies to an open subscheme (Schemes).

[F4]

Every diagram of schemes $X \to S \leftarrow T$ has a fibre product (Existence of all scheme fibre products).

[F5]

If $A \to B$ and $A \to A^{'}$ are ring maps, then $Spec B \times_{Spec A} Spec A^{'}$ is $Spec (B \otimes_{A} A^{'})$ , with the projection induced by $A^{'} \to B \otimes_{A} A^{'}$ (Affine fibre products are spectra of tensor products).

[F6]

The inverse image $f^{- 1} (U)$ represents $X \times_{S} U$ ; restricting a fibre product to opens mapping into a common open base gives the corresponding fibre product over that open (Restricting fibre products to open subschemes).

[F7]

Module-finite means finitely generated as a module over the stated ring (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[A1]

AC says every family of nonempty sets has a choice function. In this item it is used only to apply the converse local criterion [F2] (The Axiom of Choice).

Proof

technique · direct

1.1F3F4

Let $f : X \to S$ be finite and $g : T \to S$ any morphism; the fibre product $P = X \times_{S} T$ exists by [F4]. Index a family by all pairs $(U, W)$ where $U = Spec A \subseteq S$ is affine, $W \subseteq T$ is affine open, and $g (W) \subseteq U$ . These opens cover $T$ : for each $t \in T$ , an affine neighbourhood $U$ of $g (t)$ exists, and the open subscheme $g^{- 1} (U)$ has an affine neighbourhood $W$ of $t$ by [F3]. Taking all such pairs makes no simultaneous pointwise choice.

1.2F1F7

Let $X \overset{f}{\to} Y \overset{g}{\to} S$ be finite, without assuming AC, and fix an arbitrary affine open $U = Spec A \subseteq S$ . By [F1], write $g^{- 1} (U) = Spec B$ with $B$ finite over $A$ . This is an affine open of $Y$ , so [F1] applied to $f$ gives $f^{- 1} (g^{- 1} (U)) = Spec C$ with $C$ finite over $B$ . Fix finite module generating lists $b_{1}, \dots, b_{m}$ for $B / A$ and $c_{1}, \dots, c_{n}$ for $C / B$ . The products $b_{i} c_{j}$ generate $C$ over $A$ : for $c = \sum_{j} d_{j} c_{j}$ , write each $d_{j} = \sum_{i} a_{i j} b_{i}$ to obtain $c = \sum_{i, j} a_{i j} b_{i} c_{j}$ . Thus the inverse image of every affine $U$ is affine and finite over $A$ , so [F1] proves $g \circ f$ finite. Only two finite generating lists are fixed, with no choice from an arbitrary family.

2.1F1F5F6F7step 1.1

Fix one pair from step 1.1, with $W = Spec A^{'}$ and $U = Spec A$ . By [F1], write $f^{- 1} (U) = Spec B$ with $B$ finite over $A$ . The inverse image $P \times_{T} W$ identifies with $f^{- 1} (U) \times_{U} W$ by [F6], and hence with $Spec (B \otimes_{A} A^{'})$ by [F5]. If $b_{1}, \dots, b_{n}$ generate $B$ as an $A$ -module, then $b_{1} \otimes 1, \dots, b_{n} \otimes 1$ generate $B \otimes_{A} A^{'}$ as an $A^{'}$ -module, since if $b = \sum_{i} a_{i} b_{i}$ with $a_{i} \in A$ , then $b \otimes a^{'} = \sum_{i} (b_{i} \otimes 1) (1 \otimes ρ (a_{i}) a^{'})$ , where $ρ : A \to A^{'}$ is the base-change map, and every tensor is a finite sum of pure tensors. Thus the pullback over each $W$ is affine and finite.

3.1A1F2step 1.1step 2.1

The affine opens from step 1.1 cover $T$ , and step 2.1 verifies the affine and module-finite conditions on every member. Assuming AC, the converse in [F2] therefore shows that $P \to T$ is finite. This is the only choice-dependent step in the base-change assertion.

4.1

If $T = \emptyset$ , then $P = \emptyset$ and the base-changed map is finite. On an empty affine chart or a chart whose pullback is empty, the coordinate ring is $0$ , generated as a module by the empty set; tensoring with the zero ring again gives zero. Identity maps are finite because their affine chart module is generated by $1$ , and the composition argument includes identity factors. There is no endpoint parameter or iff claim. The base-change conclusion carries AC exactly as stated, while the composition proof is choice-free. [F1, F5, step 1.2, step 2.1, step 3.1] $□$

Finite morphisms survive base change and composition

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources