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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Finitely generated modules over a left Noetherian ring are Noetherian

Statement

Every finitely generated left module over a left Noetherian ring is Noetherian. See Left and right Noetherian rings.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A unital ring R is left Noetherian when its left regular module RR is Noetherian, and right Noetherian when the right regular module RR is Noetherian. Unqualified “Noetherian ring” means left Noetherian here; the side is stated whenever both notions occur. (Left and right Noetherian rings).

[L2]

Let M be a left R-module and SM. The submodule generated by S is SR:={NM:SN}. The family is nonempty because MM, and its intersection is a submodule by lem-submodule-criterion-sums-and-intersections. Thus SR is the smallest submodule of M containing S, just as def-generated-subgroup defines a generated subgroup. (Generated submodule, cyclic and finitely generated modules, module basis and free module).

[L3]

A finite direct sum is Noetherian if and only if every summand is Noetherian, and it is Artinian if and only if every summand is Artinian. The empty direct sum is included. (Finite direct sums preserve and reflect Noetherian and Artinian conditions).

[L4]

For every left R-module M, the free module R(M) on its underlying set admits a canonical surjection εM:R(M)M, determined by εM(em)=m. Consequently MR(M)/kerεM. (Every module is a quotient of a free module).

[L5]

In a short exact sequence 0NMQ0, the module M is Noetherian if and only if N and Q are Noetherian. (Noetherian and Artinian conditions are each exact in short exact sequences).

Proof

technique · direct
1.1

A module generated by n elements is a quotient of Rn.

L1L2L3L4givenalgebra
2.1

The left regular module is Noetherian, [L3] makes Rn Noetherian, and [L5] makes its quotient M Noetherian.

L1L3L5step 1.1given
3.1

The case n=0 is admitted: a module generated by the empty set is 0, and R0 is the zero module, so step 1.1 presents M=0 as a quotient of 0. The zero module has only the constant chain of submodules and is Noetherian, so the argument needs no separate base case. This proves the stated claim.

step 1.1step 2.1givenalgebra

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 33 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources