False statementConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedverified 2026-09-26 (gpt-6-sol)

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Kunneth over a PID is not always a tensor-product isomorphism

Statement

The assertion that Kunneth over a PID has no Tor correction is false.

Refutation

Given: two copies $C, D$ of the two-term free complex $0 \to Z \overset{2}{\to} Z \to 0$ .

1.1givenalgebra

Both $C$ and $D$ have $H_{0} ≅ Z / 2$ and $H_{1} = 0$ . The total tensor complex has groups $Z$ in degree two, $Z^{2}$ in degree one, and $Z$ in degree zero, with differentials $d_{2} (k) = (- 2 k, 2 k)$ and $d_{1} (a, b) = 2 (a + b)$ .

2.1step 1.1algebra∎

Thus $\ker d_{1} = {(k, - k) : k \in Z}$ and $H_{1} (C \otimes_{Z} D) = \ker d_{1} / im d_{2} ≅ Z / 2$ . But $⨁_{p + q = 1} H_{p} C \otimes_{Z} H_{q} D = 0$ , so the cross product cannot be a tensor-product isomorphism in degree one.

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Weibel, An Introduction to Homological Algebra (standard reference, not scraped)