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](https://stacks.math.columbia.edu/tag/06XY):


$$
V^\bullet \overset{\mathbb L}{\otimes}\_R W^\bullet.
$$
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](https://ncatlab.org/nlab/show/K%C3%BCnneth+theorem#OverGeneralRings).


* 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?



Cohomology of derived tensor product of complexes and Künneth spectral sequence
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.


