Let $M$ and $N$ be $R$-modules with $R$ a commutative ring with identity. When we calculate $Tor_i^R(M,N)$, usually first we choose projecive resolutions $P_.$ and $Q_.$ of $M$ and $N$, then we calculate the $i$-th homology group of the complex $P_.\otimes Q_.$. My question is: can we choose an injective resolution $I^.$ of $M$ and a projective resolution $Q_.$ of $N$, and $Tor_i^R(M,N)$ is just the $i$-th cohomology group of the cochain complex $I^.\otimes Q_.$?

The cochain groups of $I^.\otimes Q_.$ is defined by N.E.Steenrod as follows: $(I^.\otimes Q_.)^n=\sum_{i=0}^{\infty}I^{n+i}\otimes Q_i$, the coboundary operator is defined by: $d(r\otimes q)=\delta(r)\otimes q+(-1)^{n+i}r\otimes \partial(q)$.

`\bullet`

than actual periods, which give $Q^\bullet$ instead of $Q^.$: your periods look very much like dead pixels in LCD screens :P – Mariano Suárez-Alvarez♦ Apr 1 '12 at 18:52thisdouble complex for Tor? See Ralph's comment below and Anton F's answer. – Yemon Choi Apr 2 '12 at 2:36