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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29
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Comparable primes with the same contraction are equal under an integral map

Statement

Let f:A→B be an integral ring map, and let q1⊆q2 be prime ideals of B with f−1(q1)=f−1(q2)=p. Then q1=q2.

Facts & Assumptions

Given: An integral ring map f:A→B and prime ideals q1⊆q2 of B with common contraction p⊆A.

[L1]

Integrality is preserved by localisation (Integrality and integral closure commute with localisation).

[L2]

The localisation Ap is local with maximal ideal pAp (Rp is local with unique maximal ideal pRp).

[L3]

Prime ideals of a localisation correspond exactly to primes disjoint from the denominator set, with strict inclusions preserved (Prime ideals of a localization are exactly the primes disjoint from the denominator set).

[L4]

In an integral extension, a prime upstairs is maximal if and only if its contraction is maximal (Under an integral extension, a prime is maximal if and only if its contraction is maximal).

Proof

technique · direct
1.1L1L3given

Let S:=A∖p. By [L3], the primes q1 and q2 correspond to primes S−1q1⊆S−1q2 of S−1B, and by [L1] the localized map Ap→S−1B remains integral.

2.1L2L4step 1.1

By [L2], the contraction of each S−1qi to Ap is the maximal ideal pAp. Therefore [L4] makes both S−1q1 and S−1q2 maximal ideals of S−1B. Since one is contained in the other, they are equal.

3.1L3step 2.1∎

Applying the inverse bijection of [L3] to the equality of step 2.1 gives q1=q2.

Depends on

Used by

Dependency tree · two levels

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Sources