Projective resolution of tensor product Let $M,N$ are $R$-modules and $P^\bullet, Q^\bullet$ are their projective resolutions. Can we obtain a projective resolution of $M\otimes N$ using $P^\bullet, Q^\bullet$? If I understand correctly the homology groups of $P^\bullet\otimes Q^\bullet$ are $\mathrm{Tor}(M,N)$ so it's not a projective resolution.
 A: It is not a projective resolution, see Mac Lane, Homology, Theorem 9.3 page 164.
A: There is a lot going on in your question, so I'll try and address it piece by piece. Lets work over a commutative ring so I dont need to be careful.
$Tor(M,N)$ is a bifunctor, and can be calculated as you probably know, by projective resolutions of $M$ or $N$, tensoring appropriately, and then taking the homology. However what you have described (taking a tensor product of complexes) is the proof of such a statement.
The tensor compled $P^*\otimes Q^*$ is a massive double complex. Depending on your familiarity with spectral sequences, it takes some or a lot of work, to show that since the rows of this double complex are exact, taking homology yields precisely $Tor(M,N)$. However, notice that this does not say that the tensor product complex is not projective or acyclic. Just as when you take a projective resolution, tensor and take homology, "philosophically" you are always left with at least one nonzero homology group (only the higher ones vanish in some cases), namely $Tor_0$ of what you want to calculate, in the case of a double complex, the $P^i$ and $Q^j$ dont contribute their own homology, so you dont need to worry that you're calculating stuff you dont want to.
Now to your actual question: Tensor product of flat objects remain flat. The tensor product of free things remains free (work this out!). The tensor product of projective objects, hence also remains projective!
This all bodes well for us. So now, we have a monstrous double complex, and we want to resolve the object $M\otimes N$. It doesnt quite make sense to say that "the double complex resolves $M\otimes N$", but its not far from the truth: take the total complex of this double complex. That is, direct sum up the objects of a given total degree and form a complex. To make it into a complex you have to be a bit careful: use the sign trick to correct the differentials (see here for instance) and get a complex. This complex is exact (work this out by hand, or use Kunneth). And each of the objects are projective.
