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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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Factor lifting implies simple-root lifting

Statement

Let (A,m) be a Henselian local ring with residue field k=A/m. Let fA[T] be monic, and let ak be a simple root of f. Then there exists a unique aA lifting a such that f(a)=0.

Facts & Assumptions

Given: A Henselian local ring (A,m), a monic polynomial fA[T], and a simple residue root a of f.

[L1]

A simple residue root gives a coprime factorization f=(Ta)h in k[T] (A simple residue root determines a coprime residue factorisation).

[L2]

A Henselian pair lifts coprime monic factorizations uniquely (Henselian pairs and Henselian local rings, Lifted coprime factorisations are unique).

Proof

technique · lift the linear factor
1.1

By [L1], write f=(Ta)h with Ta and h coprime. Since (A,m) is Henselian, [L2] gives a lifted factorization f=(Ta)h with aA reducing to a. Evaluating at T=a gives f(a)=0.

L1L2given
2.1

If a is another lift of a with f(a)=0, then f=(Ta)h for some monic h, and this is another lift of the same residue factorization. By [L2], the lifted factorization is unique, so Ta=Ta and hence a=a.

L2step 1.1given
3.1

Therefore factor lifting implies unique lifting of every simple residue root.

step 1.1step 2.1

Depends on

Used by

Dependency tree · two levels

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Sources