TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Every flat ring map satisfies going down

Statement

Assume the Axiom of Choice for the prime-ideal lifting step.

Let $f : R \to S$ be a flat homomorphism of commutative rings. Then $f$ satisfies going down: whenever $p_{1} \subseteq p_{2} \subset R, q_{2} \subset S, q_{2} \cap R = p_{2},$ there exists a prime ideal $q_{1} \subseteq q_{2}$ with $q_{1} \cap R = p_{1} .$

Facts & Assumptions

Given: A flat ring map $f : R \to S$ and primes $p_{1} \subseteq p_{2} \subset R$ , $q_{2} \subset S$ with $q_{2} \cap R = p_{2}$ .

[L1]

Localizations of flat maps are flat (Every localization is flat, and localizing a flat module preserves flatness).

[L2]

A flat local map is faithfully flat on the localized spectra criterion (A flat ring map is faithfully flat exactly when it detects proper ideals and is surjective on spectra).

Proof

technique · direct

1.1L1given

Localize at $p_{2}$ and then at $q_{2}$ . We obtain a flat local homomorphism $R_{p_{2}} \to S_{q_{2}}$ by [L1].

1.2L2algebra

Because this localized map is local and flat, [L2] makes it faithfully flat. Applying the spectral characterization of faithful flatness to the prime ideal $p_{1} R_{p_{2}}$ of the source produces a prime $q_{1} S_{q_{2}} \subseteq S_{q_{2}}$ lying over it. Contracting back to $S$ gives a prime $q_{1} \subseteq q_{2}$ with $q_{1} \cap R = p_{1} .$

Every flat ring map satisfies going down

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources