LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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Strict prime chains contract strictly under integral extensions

Statement

Let $f : A \to B$ be an integral ring map, and let

q_{0} ⊊ q_{1} ⊊ \dots ⊊ q_{n}

be a strict chain of prime ideals of $B$ . Then

f^{- 1} (q_{0}) ⊊ f^{- 1} (q_{1}) ⊊ \dots ⊊ f^{- 1} (q_{n})

is a strict chain of prime ideals of $A$ .

Facts & Assumptions

Given: An integral ring map $f : A \to B$ and a strict prime chain $q_{0} ⊊ \dots ⊊ q_{n}$ in $B$ .

[L1]

Under an integral map, comparable primes with the same contraction are equal (Comparable primes with the same contraction are equal under an integral map).

Proof

technique · direct

1.1

Contraction is inclusion-preserving, so $f^{- 1} (q_{0}) \subseteq \dots \subseteq f^{- 1} (q_{n})$ .

given

2.1

Suppose two adjacent contractions were equal: $f^{- 1} (q_{i}) = f^{- 1} (q_{i + 1})$ for some $i < n$ . Then the comparable primes $q_{i} ⊊ q_{i + 1}$ would have the same contraction, contradicting [L1]. Therefore every adjacent contraction is strict.

L1step 1.1given

3.1

Since every adjacent inclusion is strict, the whole contracted chain is strict.

step 2.1∎

Depends on

Comparable primes with the same contraction are equal under an integral map

Used by

Injective integral extensions preserve Krull dimension Corollary

Dependency tree · two levels

6 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, Corollary 7.4 (standard reference, not scraped)
Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., Theorem (14.3)(2) (standard reference, not scraped)