ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-01

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A fraction field is flat over its domain and may fail to be projective

Example

Assume the Axiom of Choice for the direct-summand characterization below.

Let $R = Z$ and $K = Q$ . Then $K$ is the localization $S^{- 1} Z$ with $S = Z ∖ {0}$ , so $K$ is flat over $Z$ . It is not projective over $Z$ , because otherwise it would be flat and a direct summand of a free abelian group; but no nonzero direct summand of a free abelian group is divisible, whereas $Q$ is divisible.

Facts & Assumptions

Given: The Axiom of Choice and the inclusion $Z \subset Q$ .

[L1]

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

[L3]

Under the Axiom of Choice, a projective module is a direct summand of a free module (Equivalent characterizations of projective modules).

Verification

technique · direct

1.1

Since $Q$ is the localization of $Z$ at the nonzero integers, [L1] gives that $Q$ is flat over $Z$ .

L1given

1.2

Assume the Axiom of Choice. If $Q$ were projective, [L3] would make it a direct summand of a free abelian group. Every direct summand of a free abelian group is reduced, while $Q$ is nonzero and divisible. Hence $Q$ is not projective.

L3algebra

2.1

Thus a fraction field can be flat without being projective.

algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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

Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, §9 (standard reference, not scraped)