TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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Every nonzero fraction is a unit times a power of a uniformiser

Statement

Let $V$ be a discrete valuation ring with fraction field $K$ , let $v$ be its discrete valuation, and let $π \in V$ be a uniformiser. Then every nonzero element $x \in K^{\times}$ has a unique expression $x = u π^{n}$ with $u \in V^{\times}$ and $n \in Z$ .

Facts & Assumptions

Given: A discrete valuation ring $V \subseteq K$ , its discrete valuation $v$ , and a uniformiser $π \in V$ .

[F1]

A uniformiser is an element of value $1$ in a discrete valuation ring (Uniformising parameters).

[F2]

A discrete valuation ring is the nonnegative locus of a surjective valuation $v : K \to Z \cup {\infty}$ (Discrete valuation rings).

Proof

technique · direct

1.1F1F2given

Let $x \in K^{\times}$ and put $n : = v (x) \in Z$ . Since $v (π) = 1$ by [F1], one has $v (x π^{- n}) = v (x) - n v (π) = 0$ . Setting $u : = x π^{- n}$ , [F2] gives $u \in V$ and $u^{- 1} \in V$ , hence $u \in V^{\times}$ . Therefore $x = u π^{n}$ .

2.1step 1.1F2algebra∎

Suppose also that $x = u^{'} π^{m}$ with $u^{'} \in V^{\times}$ and $m \in Z$ . Applying $v$ gives $n = v (x) = v (u) + n = v (u^{'}) + m = m$ because units have valuation $0$ . Then $u = x π^{- n} = x π^{- m} = u^{'}$ . So the expression is unique.

Every nonzero fraction is a unit times a power of a uniformiser

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources