CorollaryStatement: Literature-sourcedProof: Literature-sourcedverified 2026-09-24 (gpt-6-sol)

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A Noetherian local domain has dimension zero exactly when it is a field

Statement

Assume the Axiom of Choice (The Axiom of Choice).

Let $(R, m)$ be a Noetherian local domain. Then $\dim R = 0$ if and only if $R$ is a field.

Facts & Assumptions

Given: The Axiom of Choice and a Noetherian local domain $(R, m)$ .

[L1]

A local ring has a unique maximal ideal $m$ (A local ring is a nonzero commutative ring with a unique maximal ideal).

[L2]

In a domain the zero ideal is prime (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors, Prime ideals and maximal ideals in a commutative ring).

[L3]

Krull dimension is the supremum of lengths of strict prime chains (Krull dimension of a nonzero ring).

[L4]

Under the assumed Choice, an element outside the unique maximal ideal of a local ring is a unit (Assuming the Axiom of Choice, a nonzero commutative ring is local exactly when its nonunits form an ideal, exactly when one of $x$ and $1 - x$ is a unit for every $x$ ).

Proof

technique · direct

1.1L1L2L3L4given

Suppose $\dim R = 0$ . By [L2], the chain $(0) \subseteq m$ is a prime chain. If it were strict, [L3] would give $\dim R \geq 1$ , contradiction. Hence $m = (0)$ , so every nonzero element is outside the maximal ideal and is a unit by the AC-qualified [L4]. Thus $R$ is a field.

1.2L1L3given

Conversely, if $R$ is a field, its unique maximal ideal is $(0)$ . Therefore the only prime ideal is $(0)$ , and [L3] gives $\dim R = 0$ .

A Noetherian local domain has dimension zero exactly when it is a field

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources