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

Statement

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

Facts & Assumptions

Given: 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).

Proof

technique · direct

1.1

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 therefore is a unit. Thus $R$ is a field.

L1L2L3given

1.2

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

L1L3given

2.1

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

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

11 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

J. S. Milne, A Primer of Commutative Algebra, v4.03, §21 (standard reference, not scraped)
Melvin Hochster, Dimension theory and systems of parameters (standard reference, not scraped)