Computing Tor using flat resolutions I am reading Matsumura's book Commutative Algebra, second edition. Doing a proof(3.E) in the chapter about flatness the author appears to assume that the reader knows that one can use flat projections of an $A$-module $M$ to compute $\mathrm{Tor}_i^A(M,N)$.
Is this legitimate?
Where can I find a reference of this?
 A: One can in general prove that if $F:A\to B$ is a functor between abelian categories with enough projectives, the following holds. For every $F$-acyclic resolution $C_* \to M$ of an object $M$ of $A$, it is true that
$$L_iF(M) \simeq H_i(FC_*)$$
To do this, one uses the horseshoe lemma to construct a "resolution" $R_{**}$ of the augmented complex $C_* \to M$: $R_{**}$ is a bicomplex of projective modules with exact rows and such that each column is a resolution of $C_i$ ($C_0=M$). In particular its first column is a projective resolution $P_*\to M$ of $M$. Now consider the bicomplex $F_{**}$ obtained by deleting the $P_*$ from $R_{**}$. There are maps of bicomplexes ($FP_*$ and $FC_*$ being one column bicomplexes) obtained from the edge arrows of the bigger double complex (draw a picture)
$$\eta : F F_{**} \to FP_*$$
$$\nu : FF_{**} \to FC_*$$
and I claim $\operatorname{Tot}(\eta)$ and $\operatorname{Tot}(\nu)$ are quasi-isomorphisms. To see this, one notes the cone of such morphisms are acyclic. Now the cone of $\operatorname{Tot}(\eta)$ is the total complex of $FR_{**}$ minus $FC_*$, and the rows of this complex are acyclic because the $P_*$, being projective, are $F$-acyclic. Similarly, the cone of $\operatorname{Tot}(\nu)$ is $FR_{**}$ minus $FP_*$, and the columns of this complex are acyclic since the $C_*$ are $F$-acyclic. It follows both maps are quasi-isomorphisms, so 
$$L_i F(M) = H_i(FP_*) \simeq H_i(\operatorname{Tot}(FF_{**})) \simeq H_i(FC_*)$$ 
as desired. 
A: This is pretty important, so I guess it can be found in some form in most of the introductory homological algebra books. One reference is Rotman's Introduction to Homological algebra, Theorem 7.5 (on the first glance, the proof given there seems somewhat computational, but that also means elementary, so why not).
