TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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The dimension formula for affine domains

Statement

Let $k$ be a field, let $A$ be a finite-type $k$ -domain, and let $p \in Spec (A)$ . Then

ht (p) + {trdeg}_{k} Frac (A / p) = {trdeg}_{k} Frac (A) .

Facts & Assumptions

Given: A field $k$ , a finite-type $k$ -domain $A$ , and a prime ideal $p \subset A$ .

[F1]

For a finite-type $k$ -domain $S$ and a prime ideal $q \subset S$ , the local dimension formula gives $\dim S = \dim (S_{q}) + {trdeg}_{k} Frac (S / q) .$

[L1]

Because $p$ is prime, the quotient $A / p$ is a domain, so its residue field at the generic point is $Frac (A / p)$ ( $R / P$ is an integral domain if and only if $P$ is a prime ideal).

[L2]

The height of $p$ is the dimension of the local ring $A_{p}$ (Height equals local dimension).

[L3]

For affine domains, dimension equals transcendence degree of the fraction field (Affine-domain dimension equals transcendence degree).

Proof

technique · direct

1.1

Applying [F1] to $S = A$ and $q = p$ gives $\dim A = \dim (A_{p}) + {trdeg}_{k} Frac (A / p)$ , where [L1] identifies the residue field term with $Frac (A / p)$ .

F1L1given

2.1

By [L2] and [L3], one has $\dim (A_{p}) = ht (p)$ and $\dim A = {trdeg}_{k} Frac (A)$ . Substituting these into step 1.1 yields $ht (p) + {trdeg}_{k} Frac (A / p) = {trdeg}_{k} Frac (A)$ .

L2L3step 1.1∎

Depends on

Used by

Dependency tree · two levels

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Sources

J. S. Milne, A Primer of Commutative Algebra, v4.03, §§18, 21 (standard reference, not scraped)
The Stacks Project, Section 10.116: Dimension of finite type algebras over fields, reprise (standard reference, not scraped)
Melvin Hochster, Dimension theory and systems of parameters (standard reference, not scraped)