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 $n \geq 0$ , $id (N) \leq n$ 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 $n \geq 0$ , and enough injectives.

Proof

technique · direct

1.1givenconstruct

If $id (N) \leq n$ , the $n$ th 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 ${Ext}^{1} (M, Σ^{n} N)$ 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

Dependency tree · 2 levels

Sources

Charles A. Weibel, An Introduction to Homological Algebra, Chapters 3–4