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

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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.
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Maximal ideals of an affine domain have full height

Statement

Assume the Axiom of Choice. Let $k$ be a field, let $A$ be a finite-type $k$ -domain, and let $m$ be a maximal ideal of $A$ . Then $ht (m) = \dim A .$

Facts & Assumptions

Given: The Axiom of Choice (The Axiom of Choice), a field $k$ , a finite-type $k$ -domain $A$ , and a maximal ideal $m \subset A$ .

[L1]

The residue field $A / m$ is a finite extension of $k$ (A maximal ideal of an affine algebra has finite residue field over the base field).

[L2]

Under Choice, in an affine domain $ht (m) + \dim (A / m) = \dim A$ (Height plus quotient dimension equals ambient dimension in an affine domain).

Proof

technique · direct

1.1L1given

By [L1], the quotient $A / m$ is a field. Hence $\dim (A / m) = 0$ .

2.1L2step 1.1

Applying [L2] now yields $ht (m) = \dim A$ .

3.1step 2.1∎

Thus every maximal ideal of an affine domain has full height.

Maximal ideals of an affine domain have full height

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources