TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-24 (gpt-6-sol)

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Affine-domain dimension equals transcendence degree

Statement

Let $k$ be a field, let $A$ be a finite-type $k$ -domain, and let $K = Frac (A)$ . Then $\dim A = {trdeg}_{k} K .$

Facts & Assumptions

Given: A field $k$ , a finite-type $k$ -domain $A$ , and its fraction field $K$ .

[L1]

Noether normalization provides algebraically independent elements $z_{1}, \dots, z_{d} \in A$ such that $A$ is module-finite over $k [z_{1}, \dots, z_{d}]$ (Noether normalisation yields module finiteness over a polynomial subring).

[L2]

A finite affine extension of a polynomial ring has dimension at most, and at least, the number of polynomial variables (A finite affine extension of a polynomial ring has dimension at most the number of variables, A finite affine extension of a polynomial ring has dimension at least the number of variables).

[L3]

Algebraicity is transitive in towers of fields (Algebraicity is transitive in towers of field extensions).

Proof

technique · direct

1.1L1L2givenchoose

By [L1], choose algebraically independent elements $z_{1}, \dots, z_{d} \in A$ such that $A$ is module-finite over $B = k [z_{1}, \dots, z_{d}]$ . Applying [L2] to the inclusion $B \subseteq A$ yields $\dim A = d$ .

2.1L3step 1.1

Because $A$ is integral over $B$ , every element of $K = Frac (A)$ is algebraic over the rational function field $k (z_{1}, \dots, z_{d})$ . Thus [L3] shows that $K$ is algebraic over a purely transcendental extension of degree $d$ , so ${trdeg}_{k} K = d$ .

3.1step 1.1step 2.1∎

Steps 1.1 and 2.1 give $\dim A = {trdeg}_{k} K$ .

Affine-domain dimension equals transcendence degree

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources