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Maximal chains in an affine domain all have the same length

Statement

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 dimA.

Facts & Assumptions

Given: A field k, a finite-type k-domain A, and a saturated prime chain (0)=p0p1pr=m.

[L1]

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

[L2]

Every maximal ideal of an affine domain has height equal to the full dimension (Maximal ideals of an affine domain have full height).

[L3]

For primes pq in an affine domain, ht(q/p)+trdegkFrac(A/q)=trdegkFrac(A/p) (Transcendence degrees along affine prime quotients add correctly).

[L4]

For every prime p of an affine domain, ht(p)+trdegkFrac(A/p)=trdegkFrac(A) (The dimension formula for affine domains).

Proof

technique · direct
1.1

By [L1], ht(p0)=0.

L1given
2.1

For each 0i<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 pipi+1 and comparing [L4] at pi and pi+1 gives ht(pi+1)=ht(pi)+1.

L3L4step 1.1given
3.1

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

L2step 2.1

Depends on

Used by

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Sources