Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 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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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⁡kK.

Facts & Assumptions

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

[L1]

Noether normalization provides algebraically independent elements z1,…,zd∈A such that A is module-finite over k[z1,…,zd] (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 z1,…,zd∈A such that A is module-finite over B=k[z1,…,zd]. Applying [L2] to the inclusion B⊆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(z1,…,zd). Thus [L3] shows that K is algebraic over a purely transcendental extension of degree d, so trdeg⁡kK=d.

3.1step 1.1step 2.1∎

Steps 1.1 and 2.1 give dim⁡A=trdeg⁡kK.

Depends on

Used by

…and 2 more results.

Dependency tree · two levels

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Sources