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

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Irreducible polynomial coefficients in a complete valuation ring

Statement

Let F be complete nonarchimedean. If $f \in F [T]$ is monic irreducible of positive degree and $∣ f (0) ∣ \leq 1$ , then every coefficient of f has absolute value at most one.

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Hensel factor lifting over a complete valued field: 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.

Proof

1.1givenalgebra

Suppose some coefficient has value greater than one. Choose one of maximal value, say $a_{j}$ , and put $g = f / a_{j}$ . All coefficients of g lie in A, its constant and leading coefficients reduce to zero, and some intermediate coefficient reduces to a nonzero element. Consequently $\overset{ˉ}{g} = T^{r} H_{0}$ with $1 \leq r \leq \deg \overset{ˉ}{g} < \deg f$ and $H_{0} (0) \neq 0$ .

2.1F1step 1.1∎

The two residue factors are coprime, so nonmonic Hensel lifting gives $g = h H$ with h monic of degree r. Both factors have positive degree, since $0 < r < \deg g$ . Multiplying back by $a_{j}$ contradicts irreducibility. Thus no coefficient exceeds one. Degree one and f=T are included without contradiction to the zero constant-term case.

Depends on

Hensel factor lifting over a complete valued field

Used by

Unique extension of a nonarchimedean absolute value Theorem

Dependency tree · two levels

2 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

§6, proof of Theorem 6.4, p.11 (standard reference, not scraped)