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 $f : A \to B$ be an integral ring map, and let $q_{1} \subseteq q_{2}$ be prime ideals of $B$ with $f^{- 1} (q_{1}) = f^{- 1} (q_{2}) = p$ . Then $q_{1} = q_{2}$ .

Facts & Assumptions

Given: An integral ring map $f : A \to B$ and prime ideals $q_{1} \subseteq q_{2}$ of $B$ with common contraction $p \subseteq A$ .

[L1]

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

[L2]

The localisation $A_{p}$ is local with maximal ideal $p A_{p}$ ( $R_{p}$ is local with unique maximal ideal $p R_{p}$ ).

[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 $q_{1}$ and $q_{2}$ correspond to primes $S^{- 1} q_{1} \subseteq S^{- 1} q_{2}$ of $S^{- 1} B$ , and by [L1] the localized map $A_{p} \to S^{- 1} B$ remains integral.

2.1L2L4step 1.1

By [L2], the contraction of each $S^{- 1} q_{i}$ to $A_{p}$ is the maximal ideal $p A_{p}$ . Therefore [L4] makes both $S^{- 1} q_{1}$ and $S^{- 1} q_{2}$ maximal ideals of $S^{- 1} B$ . 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 $q_{1} = q_{2}$ .

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources