LemmaStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Module finiteness is transitive along a tower of algebras

Statement

Let $φ ⁣ : A \to B$ be a homomorphism of commutative rings, so that $B$ is an $A$ -algebra (Algebras over a commutative ring, central structure maps, and algebra homomorphisms) and every $B$ -module becomes an $A$ -module through $a \cdot z : = φ (a) z$ . Suppose $B$ is generated as an $A$ -module by $b_{1}, \dots, b_{m}$ with $m \in N$ , and let $M$ be a $B$ -module generated as a $B$ -module by $u_{1}, \dots, u_{n}$ with $n \in N$ . Then the $m n$ products $b_{i} u_{j}$ generate $M$ as an $A$ -module.

In particular, if $ψ ⁣ : B \to C$ is a homomorphism of commutative rings making $C$ module-finite over $B$ , and $B$ is module-finite over $A$ , then $C$ is module-finite over $A$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras): the products $ψ (b_{i}) c_{j}$ of a finite $A$ -generating list of $B$ with a finite $B$ -generating list of $C$ generate $C$ over $A$ .

Facts & Assumptions

Given: Commutative rings $A$ and $B$ , a ring homomorphism $φ ⁣ : A \to B$ , a finite $A$ -module generating list $b_{1}, \dots, b_{m}$ of $B$ , a $B$ -module $M$ and a finite $B$ -module generating list $u_{1}, \dots, u_{n}$ of $M$ .

[L1]

An $R$ -algebra is a unital ring $A$ with a unital ring homomorphism $η_{A} ⁣ : R \to A$ of central image, and the induced scalar action $r a : = η_{A} (r) a$ makes $A$ an $R$ -module (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

[L2]

$A$ is module-finite over $R$ when it is finitely generated as an $R$ -module (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L3]

A left $R$ -module is an abelian group with an action satisfying $r (m + n) = r m + r n$ , $(r + s) m = r m + s m$ , $(r s) m = r (s m)$ and $1_{R} m = m$ (Unital left and right modules over a ring; unqualified module means left module).

[L4]

For a ring $R$ , a left $R$ -module $M$ and $S \subseteq M$ , the submodule $⟨ S ⟩_{R}$ is the set of finite sums $\sum_{i = 1}^{k} r_{i} s_{i}$ with $k \in N$ , $r_{i} \in R$ and $s_{i} \in S$ , the term with $k = 0$ being $0_{M}$ (The submodule generated by a subset consists of the finite $R$ -linear combinations of that subset).

Proof

technique · direct

1.1

The $A$ -action on $M$ is $a \cdot z = φ (a) z$ , computed in the $B$ -module $M$ ; it satisfies the module axioms because $φ$ is a ring homomorphism and $M$ is a $B$ -module. Each product $b_{i} u_{j}$ is an element of $M$ .

L1L3given

2.1

Let $z \in M$ . Since $u_{1}, \dots, u_{n}$ generate $M$ over $B$ , the finite-sum description gives $β_{1}, \dots, β_{n} \in B$ with $z = \sum_{j = 1}^{n} β_{j} u_{j}$ . Since $b_{1}, \dots, b_{m}$ generate $B$ over $A$ , the same description gives, for each $j$ , elements $α_{1 j}, \dots, α_{m j} \in A$ with $β_{j} = \sum_{i = 1}^{m} φ (α_{i j}) b_{i}$ .

L4step 1.1

3.1

Substituting and using the $B$ -module axioms, $z = \sum_{j = 1}^{n} (\sum_{i = 1}^{m} φ (α_{i j}) b_{i}) u_{j} = \sum_{i = 1}^{m} \sum_{j = 1}^{n} φ (α_{i j}) (b_{i} u_{j}) = \sum_{i, j} α_{i j} \cdot (b_{i} u_{j})$ , an $A$ -linear combination of the $m n$ products. As $z$ was arbitrary, those products generate $M$ as an $A$ -module. Taking $M = C$ with its $B$ -module structure gives the transitivity statement.

L2L3L4step 2.1algebra∎

Remarks

Both degenerate cases collapse rather than fail. If $m = 0$ then $B$ is generated over $A$ by the empty list, so $B = 0$ and $1_{B} = 0_{B}$ ; every $B$ -module then satisfies $z = 1_{B} z = 0$ , so $M = 0$ and the empty set of products generates it. If $n = 0$ then $M = 0$ directly. The count $m n$ is $0$ in both cases, which is the correct answer.
The products need not be distinct or independent. The list $b_{i} u_{j}$ may repeat entries and may be far from minimal; only the finiteness of the count is used.

Depends on

Used by

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

M. Hochster, Introduction to Commutative Algebra, Math 614, Lemma 5.4 (standard reference, not scraped)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., (16.21) (standard reference, not scraped)