LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Hensel factor lifting over a complete valued field

Statement

Let F be complete nonarchimedean, A its valuation ring, and k its residue field. Suppose $g \in A [T]$ has nonzero reduction $\overset{ˉ}{g} = h_{0} H_{0}$ , where $h_{0} \in k [T]$ is monic and $\gcd (h_{0}, H_{0}) = 1$ . Then $g = h H$ for $h, H \in A [T]$ , with h monic of degree $\deg h_{0}$ , $\overset{ˉ}{h} = h_{0}$ , $\overset{ˉ}{H} = H_{0}$ . No discreteness or monicity of g is assumed. In particular, a simple residue root of a monic polynomial lifts uniquely to a simple root in A.

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Absolute values on a field: Let $F$ be a field. An absolute value on $F$ is a function $∣ \cdot ∣ : F \to R_{\geq 0}$ such that for all $x, y \in F$ : $∣ x ∣ = 0 ⟺ x = 0, ∣ x y ∣ = ∣ x ∣ ∣ y ∣, ∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣ .$ It is nonarchimedean when it satisfies the stronger inequality $∣ x + y ∣ \leq \max {∣ x ∣, ∣ y ∣}$ for all $x, y \in F$ . It is trivial when $∣ x ∣ = 1$ for every nonzero $x \in F$ .

Proof

1.1

If the valuation is trivial, A=k=F and the original factorization suffices. If $h_{0} = 1$ , take h=1 and H=g. Otherwise put $m = \deg h_{0} > 0$ , $N = \deg g$ , lift $h_{0}$ to a monic $h_{1}$ of degree m and $H_{0}$ to $H_{1}$ of degree at most N-m. Lift a Bezout relation to polynomials r,s with $h_{1} r + H_{1} s \equiv 1$ modulo the maximal ideal. Among the finitely many nonzero coefficients of $g - h_{1} H_{1}$ and $h_{1} r + H_{1} s - 1$ , choose one of maximum absolute value, or any element with value strictly between zero and one if both errors vanish. Denote it by pi. Then both errors lie in $I [T]$ , where $I = (π) = {x : ∣ x ∣ \leq ∣ π ∣}$ and $0 < ∣ π ∣ < 1$ .

F1given

2.1

Suppose $g - h_{n} H_{n} \in I^{n} [T]$ , $h_{n}, H_{n} \equiv h_{1}, H_{1}$ modulo I, with the stated degree bounds. Put $e_{n} = (g - h_{n} H_{n}) / π^{n} \in A [T]$ . Over A/I we need $q H_{1} + Q h_{1} = e_{n}$ . Multiply the fixed Bezout relation by $e_{n}$ ; then divide $e_{n} s$ by the monic $h_{1}$ , writing $e_{n} s = h_{1} u + q$ , $\deg q < m$ . Take $Q = e_{n} r + u H_{1}$ . Modulo I, monicity of $h_{1}$ and $\deg (e_{n} - q H_{1}) \leq N$ imply $\deg Q \leq N - m$ . Delete higher coefficients of Q, which lie in I. This explicitly solves the congruence with bounded degrees.

step 1.1algebra

3.1

Set $h_{n + 1} = h_{n} + π^{n} q$ and $H_{n + 1} = H_{n} + π^{n} Q$ . Their product equals g modulo $I^{n + 1}$ , since $2 n \geq n + 1$ , and all degree bounds persist. This deterministic correction uses only the initial finite lifts and polynomial division, so no new arbitrary residue representatives are chosen at successive stages. The finitely many coefficient sequences are Cauchy because $∣ π ∣^{n} \to 0$ . Completeness gives limits h,H in A[T], with h monic of degree m, and continuity of finite multiplication gives g=hH and the prescribed reductions.

step 2.1

4.1

For a simple root $\overset{ˉ}{a}$ of monic $\overset{ˉ}{f}$ , apply the factorization to $(T - \overset{ˉ}{a}) H_{0}$ with $H_{0} (\overset{ˉ}{a}) \neq 0$ . Write $f = (T - a) H$ . Then $H (a)$ is a unit, so $f^{'} (a) \neq 0$ . If b is another root with $\overset{ˉ}{b} = \overset{ˉ}{a}$ , then $H (b)$ is a unit and $(b - a) H (b) = 0$ forces b=a. This proves both existence and uniqueness of the simple-root lift.

step 3.1algebra∎

Depends on

Absolute values on a field

Used by

Irreducible polynomial coefficients in a complete valuation ring Lemma

Dependency tree · one level

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Sources

§6, Theorem 6.5 and full proof, pp.12–13 (standard reference, not scraped)