Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06
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Injective dimension at most n iff higher Ext vanishes

Statement

Assume enough injectives. For an object N and n≥0, id⁡(N)≤n if and only if Ext⁡k(M,N)=0 for every object M and every k>n.

Facts & Assumptions

Given: An object N, an integer n≥0, and enough injectives.

Proof

technique · direct
1.1givenconstruct

If id⁡(N)≤n, the nth cosyzygy is injective. Repeated second-variable shifting from Ext dimension shifting in the second variable reduces every Ext⁡k(M,N) with k>n to positive Ext into that injective cosyzygy, which vanishes by Positive injective-resolution Ext vanishes on an injective second variable.

2.1step 1.1algebra∎

Conversely, repeated shifting makes Ext⁡1(M,ΣnN) vanish for every M. The defining injective-dimension convention in Injective dimension of an object and the splitting criterion for the final cosyzygy show it is injective, so the chosen resolution has length at most n.

Depends on

Used by

Dependency tree · two levels

9 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