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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Ideals in a DVR are powers of the maximal ideal

Statement

Let $V$ be a discrete valuation ring with maximal ideal $m = (π)$ , where $π$ is a uniformiser. Then every nonzero ideal $I \subseteq V$ is of the form

I = (π^{n}) = m^{n}

for a unique integer $n \geq 0$ .

Facts & Assumptions

Given: A discrete valuation ring $V$ with maximal ideal $m = (π)$ , where $π$ is a uniformiser.

[L1]

Every nonzero element of the fraction field of $V$ is uniquely $u π^{n}$ with $u$ a unit and $n \in Z$ (Every nonzero fraction is a unit times a power of a uniformiser).

[F1]

A uniformiser generates the maximal ideal of a DVR (Uniformising parameters).

Proof

technique · direct

1.1

Let $I \neq 0$ be an ideal of $V$ . Because $I \subseteq V$ , every nonzero element of $I$ has valuation in $Z_{\geq 0}$ . Choose $x \in I ∖ {0}$ with minimal valuation $n$ . By [L1], $x = u π^{n}$ for a unit $u$ , so $π^{n} = u^{- 1} x \in I$ .

L1givenchoose

2.1

If $y \in I$ is nonzero, then [L1] gives $y = u^{'} π^{m}$ for some unit $u^{'}$ and $m \geq 0$ . Minimality of $n$ yields $m \geq n$ , so $y = u^{'} π^{m - n} π^{n} \in (π^{n})$ . Thus $I \subseteq (π^{n})$ , while step 1.1 gave $(π^{n}) \subseteq I$ . Hence $I = (π^{n})$ . Since [F1] gives $m = (π)$ , this is also $m^{n}$ .

L1F1step 1.1

3.1

If $(π^{n}) = (π^{m})$ , then $π^{n} \in (π^{m})$ and $π^{m} \in (π^{n})$ , so $n \geq m$ and $m \geq n$ . Therefore $n = m$ , and the exponent is unique.

step 2.1algebra∎

Depends on

Used by

Every DVR is a PID Corollary
Prime ideals and dimension of a DVR Corollary
The p-adic valuation ring Example
Uniformisers and ideal arithmetic in a DVR Example
Equivalent characterizations of a DVR Theorem
Length and valuation in a DVR Theorem

Dependency tree · two levels

4 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

Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., (23.1) (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, Discrete valuation rings after Example 20.1 (standard reference, not scraped)