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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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Idempotents lift uniquely in a Henselian pair

Statement

Let (A,I) be a Henselian pair. Every idempotent e‾∈A/I lifts to a unique idempotent e∈A.

Facts & Assumptions

Given: A Henselian pair (A,I) and an idempotent e‾∈A/I.

[L1]

In a Henselian pair, coprime monic factorizations lift uniquely (Henselian pairs and Henselian local rings).

Proof

technique · lift the factorization of $T^2-T$
1.1

Because e‾2=e‾, one has T2−T=(T−e‾)(T−(1−e‾)) in (A/I)[T]. The two factors are monic, and their difference is 1−2e‾, which is a unit because e‾(1−e‾)=0 forces every prime quotient to send e‾ to 0 or 1. Hence the factors are coprime.

givenalgebra
2.1

By [L1], this residue factorization lifts uniquely to T2−T=(T−e)(T−(1−e)) for some e∈A lifting e‾. Evaluating at T=e yields e2−e=0, so e is idempotent.

L1step 1.1
3.1

If e′ is another lifted idempotent, then (T−e′)(T−(1−e′)) is a second lift of the same residue factorization. By [L1], the factorization is unique, so e′=e.

L1step 2.1
4.1

Therefore idempotents lift uniquely in a Henselian pair.

step 2.1step 3.1∎

Depends on

Used by

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