Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06
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Positive Ext need not vanish for an injective first variable

Statement refuted

Use Q/Z as an injective abelian group in the first variable and calculate a nonzero Ext^1 against a suitable second variable, separating the correct variance from the false symmetry.

Facts & Assumptions

Given: The abelian category of abelian groups and the short exact sequence 0→Z→Q→Q/Z→0.

Counterexample

technique · direct
1.1givenalgebra

The group Q/Z is divisible, hence injective, so it is an injective object in the first variable. The displayed sequence is an extension of Q/Z by Z.

2.1step 1.1algebra∎

This extension does not split: a retraction r:Q→Z would satisfy r∣Z=1, whereas every homomorphism Q→Z is zero (if r(1)=a, then a=r(1/m)m is divisible by every m). Hence it represents a nonzero element of Ext⁡Z1(Q/Z,Z), refuting the asserted first-variable vanishing.

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