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Let $R$ be any commutative ring, let $V^\bullet$ and $W^\bullet$ be (co)chain complexes of $R$-modules, indexed cohomologically. We can also assume that they have both cohomology in nonpositive degrees. Using K-flat resolutions we can define the derived tensor product:

$$ V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet. $$ I'm looking for assumptions on $V^\bullet$ or $W^\bullet$ that ensure that the following "Künneth formula" holds: $$ H^*(V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet) \cong H^*(V^\bullet) \otimes_R H^*(W^\bullet), $$ where the right hand side is the tensor product of graded $R$-modules. Searching the literature, it seems that there should be some spectral sequence involving Tor of the cohomologies, as mentioned for instance in the nLab entry.

  • Is there a more precise reference for such a result?
  • Looking at these Künneth formulas and the spectral sequence, I suspect that the following is true: $H^*(V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet) \cong H^*(V^\bullet) \otimes_R H^*(W^\bullet)$ holds if $V^\bullet$ or $W^\bullet$ has flat cohomologies, namely $H^k(V^\bullet)$ (say) is a flat $R$-module for all $k$. Is it correct?

A little more wildly, could one expect that without assumptions one has $$ H^*(V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet) \cong H^*(V^\bullet) \overset{\mathbb L}{\otimes}_R H^*(W^\bullet), $$ so taking K-flat resolutions of the cohomology graded $R$-module if necessary?

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This isn't a complete answer, but here are some thoughts. I'm sceptical that it is often true.

If $R$ is semisimple then the result holds, because every module over a semisimple ring is projective.

We may as well take $V$ and $W$ to be complexes of projectives, and then ask when is $H(V\otimes W)\cong H(V)\otimes H(W)$? If $R$ is a field this is the usual Kuenneth theorem.

When $V$ has flat cohomology the spectral sequence is not hard to obtain: let $V' \to V$ and $W' \to W$ be flat resolutions; because both $V$ and $W$ are cohomologically bounded above we can take $V'$ and $W'$ to be bounded above. The direct sum total complex of the double complex $V' \otimes W'$ computes the Tor groups. Taking cohomology in the vertical direction and using flatness of $W'$ tells us that the $E_1$ page of the associated spectral sequence is $H(V)\otimes W'$. Now taking cohomology in the horizontal direction and using flatness of $H(V)$ tells us that the $E_2$ page is $H(V)\otimes H(W)\Rightarrow \mathrm{Tor}(V,W)$, as desired.

The same argument when $V$ does not necessarily have flat cohomology gives a spectral sequence with $E_2$ page $\mathrm{Tor}(HV,W)\Rightarrow \mathrm{Tor}(V,W)$. In general, Tor spectral sequences tend to look like this (as in e.g. https://stacks.math.columbia.edu/tag/061Y) - a spectral sequence of the form $H(V)\otimes H(W)\Rightarrow \mathrm{Tor}(V,W)$ will not exist without this flatness assumption on cohomology (just think about when $V$ and $W$ are genuine $R$-modules!). Of course, when $V$ has flat cohomology then the tensor product $H(V)\otimes H(W)$ is already the derived tensor product.

However, this doesn't answer your main question in the flat cohomology case: you have a spectral sequence, and you are interested in knowing when it collapses (or more generally when the $E_2$ page equals the $E_\infty$ page).

As for your wild question, when $V$ and $W$ are genuine modules, you are asking that $V\otimes^\mathbb{L}W$ be a formal complex. This is true in certain situations: for example the HKR theorem is true on the level of cochains and gives a quasi-isomorphism between the Hochschild complex and the graded module of polyvector fields. Or if $R$ is a ring, $r_1,\ldots, r_n$ is a (finite) regular sequence, $K$ the Koszul complex, and $M$ an $R$-module annihilated by all of the $r_i$ then you can easily check that $R/(r_1,\ldots, r_n)\otimes^\mathbb{L}_R M \simeq K\otimes_R M$ is formal. In general it won't be true, but I can't think of an example off the top of my head.

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