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Height plus quotient dimension equals ambient dimension in an affine domain

Statement

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

ht (p) + \dim (A / p) = \dim A .

Facts & Assumptions

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

[L1]

The affine-domain dimension formula identifies $ht (p) + {trdeg}_{k} Frac (A / p) = {trdeg}_{k} Frac (A) .$ (The dimension formula for affine domains).

[L2]

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

Proof

technique · direct

1.1

Replace the two transcendence degrees in [L1] using [L2], once for $A$ and once for the quotient domain $A / p$ .

L1L2given

2.1

The result is exactly $ht (p) + \dim (A / p) = \dim A$ .

step 1.1∎

Depends on

Used by

Maximal ideals of an affine domain have full height Corollary
Coordinate ideals show the height bound is sharp Example
Localisation can strictly lower dimension Example
Relative height in a quotient of k[x,y,z] Example
The affine dimension formula on a plane curve domain Example

Dependency tree · two levels

9 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, §§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)