Künneth formula and tensor products

For chain complexes over a commutative ring, the total tensor complex has degree-n module $\bigoplus_{p+q=n}C_p\otimes D_q$. Its boundary sends a tensor a⊗b to $\partial a\otimes b+(-1)^pa\otimes\partial b$ when a has degree p. Applying the boundary twice cancels the mixed terms because their signs differ. Omitting the degree sign would generally violate boundary squared zero even if both original complexes were perfectly valid.

The singular cross product connects this algebraic tensor complex to chains on the topological product. The Eilenberg–Zilber theorem supplies a chain-homotopy equivalence, rather than asserting that every singular simplex in the product is literally one product simplex. This bridge is the topological input to the Künneth calculation. Tensor products of cycles yield genuine product classes, while the algebra of torsion measures what can remain beyond these classes in homology of the total complex.

For singular homology with coefficients in a PID R, the chain modules are free and the Künneth theorem gives a short exact sequence. The left term sums tensor products of factor homology in total degree n. The right term sums $\operatorname{Tor}_1^R$ in total degree n−1. The latter indexing means torsion in degrees p and q can contribute to product degree p+q+1. The sequence splits as R-modules, but generally not naturally; writing an abstract direct sum does not make its summands canonical under every map.

Over a field, Tor vanishes and the cross-product map gives the usual direct sum of tensor products in the actual total degree. Over Z, a free factor in either slot also has zero Tor contribution. For positive integers m and n, tensoring Z/m with Z/n gives a cyclic group of order their greatest common divisor, and their first Tor group has the same abstract order. Equal isomorphism types do not erase their different degree positions or the maps by which they arise.

A short free resolution makes both cyclic-group formulas concrete. Tensoring the resolution of Z/m with an abelian group G gives kernel of multiplication by m for Tor and cokernel of that multiplication for the tensor product. For G=Z/n these groups have the stated greatest-common-divisor order. This calculation also shows why Tor disappears when the relevant module is free: multiplication by a nonzero integer on a free abelian group has no nonzero kernel.

Homological Künneth uses direct sums and does not require finite-dimensional factor groups in this singular PID statement. Cohomology involves Hom and may require additional finiteness hypotheses for a simple tensor description; the dual of an infinite-dimensional tensor product is not automatically the tensor product of the duals. A graded ring description also needs product sign conventions beyond the additive formula. Report exactly which version is being used, rather than transferring a convenient field calculation into integral cohomology or into an infinite-type setting without checking it.