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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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  • 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.
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Maximal residue-injective subfields exist

Statement

Assume the Axiom of Choice.

Let (A,m) be an equicharacteristic local ring. Then there exists a subfield KA that is maximal, under inclusion, among subfields whose residue map to A/m is injective.

Facts & Assumptions

Given: An equicharacteristic local ring (A,m) and the Axiom of Choice.

[L1]

The residue field's prime field embeds in A, so the family of residue-injective subfields is nonempty (The prime field lifts in the equicharacteristic case).

[L2]

Assuming the Axiom of Choice, every nonempty poset in which every chain has an upper bound has a maximal element (Zorn's lemma).

Proof

technique · apply Zorn's lemma to the poset of residue-injective subfields
1.1

Let S be the set of subfields KA for which the residue map KA/m is injective. By [L1], S is nonempty. Order S by inclusion.

L1givenconstruct
2.1

If CS is a chain, then KCK is again a subfield of A: closure under the field operations is inherited from some chain member containing the finitely many elements involved. Its residue map is still injective, because a nonzero element of the union already lies in one chain member where injectivity holds. Thus every chain in S has an upper bound in S.

step 1.1givenalgebra
3.1

By [L2], the poset S has a maximal element. That is exactly a maximal residue-injective subfield of A.

L2step 2.1

Depends on

Used by

Dependency tree · two levels

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