LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-07

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finite local modules admit minimal free resolutions

Statement

Assume the Axiom of Choice. Every finite module $M$ over a nonzero Noetherian local ring $(R, m, k)$ has an augmented resolution $\dots \to F_{1} \to F_{0} \to M \to 0$ by finite-rank free modules, with $d_{i} (F_{i}) \subseteq m F_{i - 1}$ for $i > 0$ . Such a resolution is called minimal; it need not be bounded. This extends the bounded terminology without changing it.

Facts & Assumptions

Given: The objects and hypotheses in the statement. The Axiom of Choice (The Axiom of Choice) supplies dependent choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain) for the infinite sequence of successive finite generator/lift selections, and the choice assumptions of the cited Nakayama inputs.

[F1]

Minimal Free Resolution Over A Local Ring: A finite free resolution $F_{∙} \to N$ over local $(R, m)$ is minimal when $d_{i} (F_{i}) \subseteq m F_{i - 1}$ for every $i > 0$ .

[F2]

Assuming the Axiom of Choice, minimal generators over a local ring are exactly residue-field bases: Assume the Axiom of Choice. Let $(R, m)$ be a local ring with residue field $k = R / m$ , and let $M$ be a finitely generated left $R$ -module. A finite generating set $x_{1}, \dots, x_{r}$ of $M$ is minimal if and only if the images of $x_{1}, \dots, x_{r}$ in $M / m M$ form a $k$ -basis. In particular every minimal generating set of $M$ has the same cardinality.

[F3]

Assuming the Axiom of Choice, Nakayama's lemma: Assume the Axiom of Choice. Let $R$ be a commutative ring, let $I ⊴ R$ satisfy $I \subseteq J (R)$ , and let $M$ be a finitely generated left $R$ -module. If $I M = M$ , then $M = 0$ .

[F4]

Finite generation, ACC, and maximal-condition characterizations of Noetherian modules: For a left $R$ -module $M$ , write (1) every submodule is finitely generated, (2) ACC, and (3) every nonempty set of submodules has a maximal member. The implications $1 \Rightarrow 2$ and $3 \Rightarrow 1$ are choice-free; $2 \Rightarrow 3$ assumes DC, under which the three conditions are equivalent. See def-noetherian-module.

Proof

technique · Explicit algebraic derivation

1.1F2F3algebra

Choose a basis of $M / m M$ and lift it to $x_{1}, \dots, x_{r} \in M$ . If $N = \sum_{i} R x_{i}$ , then $M = N + m M$ , so the finite module $M / N$ satisfies $m (M / N) = M / N$ . Nakayama gives $M = N$ . Thus the corresponding map $F_{0} = R^{r} \to M$ is onto. A relation among the $x_{i}$ has all coefficients in $m$ , because their residue classes are independent. Hence $K_{0} = \ker (F_{0} \to M) \subseteq m F_{0}$ .

2.1F4F3F1step 1.1givenchoose∎

Each finite-rank free module is finitely generated, hence Noetherian by Finitely generated modules over a left Noetherian ring are Noetherian. Its kernel submodule is therefore finitely generated by the Noetherian condition in [F4]. Repeating step 1.1 on $K_{0}$ and on each successive kernel gives an exact augmented complex: the next map surjects onto precisely the preceding kernel, then includes it in the preceding free module. Its image lies in the required maximal-ideal multiple by step 1.1 at that stage. DC, supplied by AC, makes these successive finite generator/lift selections into one infinite sequence; the Nakayama inputs also use the assumed AC. If a kernel is zero, choose zero modules thereafter, making the selection relation serial; for $M = 0$ use the zero complex. The bounded case agrees with the prior definition.

Depends on

Used by

Dependency tree · two levels

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Sources

Construction before Proposition 12.27, p.120 (standard reference, not scraped)