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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 I⊴R satisfy I⊆J(R), and let M be a finitely generated left R-module. If elements x1,…,xr∈M generate M/IM, then x1,…,xr generate M.

Facts & Assumptions

Given: The Axiom of Choice (The Axiom of Choice), a commutative ring R, an ideal I⊴R with I⊆J(R), a finitely generated left R-module M, and elements x1,…,xr∈M whose images generate M/IM.

[L1]

Under AC, if a finite module Q satisfies IQ=Q and I⊆J(R), then Q=0 (Assuming the Axiom of Choice, Nakayama's lemma); applying this supplier in step 2.1 is the sole inherited use of AC here.

[L2]

The submodule IM consists of finite sums of products im with i∈I and m∈M (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
1.1L2L3given

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.

2.1step 1.1L1

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.

3.1step 2.1L3∎

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

Depends on

Used by

Dependency tree · two levels

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Sources