PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Homology of the derived tensor product is tor

Statement

For a right $R$ -module $N$ , a left $R$ -module $M$ , and $n \geq 0$ , with supplied projective resolutions, $H^{- n} (N [0] \otimes_{R}^{L} M [0]) ≅ {Tor}_{n}^{R} (N, M)$ naturally in both modules.

Facts & Assumptions

Given: For a right $R$ -module $N$ , a left $R$ -module $M$ , and $n \geq 0$ , with supplied projective resolutions, $H^{- n} (N [0] \otimes_{R}^{L} M [0]) ≅ {Tor}_{n}^{R} (N, M)$ naturally in both modules.

[F1]

The derived tensor is represented using either projective replacement with the two-replacement balancing zigzag (Derived tensor product in the bounded above setting).

[F2]

Balanced Tor is resolution tensor homology, with maps induced by comparison maps (The balanced Tor bifunctor).

Proof

1.1

Represent the derived tensor by $N \otimes_{R} P_{M}$ with the projective resolution reindexed by $P_{M}^{- i} = (P_{M})_{i}$ . Its degree $- n$ cohomology is exactly $H_{n} (N \otimes_{R} (P_{M})_{∙})$ , including $n = 0$ and zero modules.

2.1

This homology is the definition of balanced Tor on the left-resolution side. The common two-resolution tensor complex gives the same balance on the other side, and comparison maps on resolutions induce exactly the maps used in that definition. Thus the identification is natural and compatible with either supplied resolution.

F2step 1.1∎

Depends on

Used by

ex-derived-tensor-of-two-cyclic-abelian-groups.md Example

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

10.6.1–10.6.4 and Exercise 10.6.1, p. 395; elementary finite-diagonal replacement for spectral sequence proof (standard reference, not scraped)