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Maximal chains in an affine domain all have the same length

Statement

Assume the Axiom of Choice. Let k be a field and let A be a finite-type k-domain. Every saturated prime chain from (0) to a maximal ideal of A has length dim⁡A.

Facts & Assumptions

Given: The Axiom of Choice (The Axiom of Choice), a field k, a finite-type k-domain A, and a saturated prime chain (0)=p0⊊p1⊊⋯⊊pr=m.

[L1]

In a domain, the minimal prime (0) has height zero (Minimal primes are exactly the primes of height zero).

[L2]

Under Choice, every maximal ideal of an affine domain has height equal to the full dimension (Maximal ideals of an affine domain have full height).

[L3]

Under Choice, for primes p⊆q in an affine domain, ht⁡(q/p)+trdeg⁡kFrac⁡(A/q)=trdeg⁡kFrac⁡(A/p) (Transcendence degrees along affine prime quotients add correctly).

[L4]

Under Choice, for every prime p of an affine domain, ht⁡(p)+trdeg⁡kFrac⁡(A/p)=trdeg⁡kFrac⁡(A) (The dimension formula for affine domains).

Proof

technique · direct
1.1L1given

By [L1], ht⁡(p0)=0.

2.1L3L4step 1.1given

For each 0≤i<r, the quotient A/pi is a domain and the chain is saturated, so there is no prime strictly between 0 and pi+1/pi in A/pi. Hence ht⁡(pi+1/pi)=1. Applying [L3] to pi⊆pi+1 and comparing [L4] at pi and pi+1 gives ht⁡(pi+1)=ht⁡(pi)+1.

3.1L2step 2.1∎

Starting from step 1.1 and iterating step 2.1, we obtain ht⁡(m)=r. By [L2], ht⁡(m)=dim⁡A. Therefore the saturated chain has length dim⁡A, and every such chain has the same length.

Depends on

Used by

Dependency tree · two levels

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Sources