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CorollaryStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26
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Assuming the Axiom of Choice, generators modulo an ideal in the Jacobson radical lift to generators

Statement

Assume the Axiom of Choice.

Let R be a commutative ring, let IR satisfy IJ(R), and let M be a finitely generated left R-module. If elements x1,,xrM generate M/IM, then x1,,xr generate M.

Facts & Assumptions

Given: A commutative ring R, an ideal IR with IJ(R), a finitely generated left R-module M, and elements x1,,xrM whose images generate M/IM.

[L1]

If a finite module Q satisfies IQ=Q and IJ(R), then Q=0 (Assuming the Axiom of Choice, Nakayama's lemma).

[L2]

The submodule IM consists of finite sums of products im with iI and mM (The submodule IM generated by products of elements of an ideal I with elements of a module M).

[L3]

A finite list of elements generates the submodule it spans (Generated submodule, cyclic and finitely generated modules, module basis and free module).

Proof

technique · direct
1.1

Let N be the submodule of M generated by x1,,xr. The hypothesis on M/IM means every element of M is congruent modulo IM to an element of N, so M=N+IM.

L2L3given
2.1

Passing to the quotient Q=M/N, step 1.1 gives IQ=Q. Since Q is a quotient of the finite module M, it is finite, so [L1] gives Q=0.

step 1.1L1
3.1

The equality Q=0 means M=N, so x1,,xr generate M.

step 2.1L3

Depends on

Used by

Dependency tree · two levels

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