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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

1.1L1givenchoose

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 $π^{n} = u^{- 1} x \in I$ .

2.1L1F1step 1.1

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 $m^{n}$ .

3.1step 2.1algebra∎

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.

Ideals in a DVR are powers of the maximal ideal

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources