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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: $d(r\otimes q)=\delta(r)\otimes q+(-1)^{n+i}r\otimes \partial(q)$.

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    $\begingroup$ Usually I just choose just a projective resolution of M (or of N), tensor it with N (or with M), and calculate the homology. Why do you deal with a double complex? $\endgroup$
    – Martin Brandenburg
    Apr 1, 2012 at 15:07
  • $\begingroup$ @Martin, why not? The double complex does compute the Tor, and in several situations it is a nicer description of it! $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 1, 2012 at 18:46
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    $\begingroup$ @Nock: By the way, if you want to draw dots on complexes, it is better to use \bullet than actual periods, which give $Q^\bullet$ instead of $Q^.$: your periods look very much like dead pixels in LCD screens :P $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 1, 2012 at 18:52
  • $\begingroup$ Nick: In my understanding, injective modules (and injective resolutions) are useful for theoretical purposes, but are rarely useful for actual computations; among other things, they are almost never finitely generated. Thus, I find myself asking whether you are using the word "calculation" figuratively (as a substitute for, say, "definition"), or whether you actually have a computation in mind for which this approach would be useful. $\endgroup$
    – Charles Staats
    Apr 1, 2012 at 22:51
  • $\begingroup$ @Mariano, do people really use this double complex for Tor? See Ralph's comment below and Anton F's answer. $\endgroup$
    – Yemon Choi
    Apr 2, 2012 at 2:36

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It is enough to replace one of the objects with its projective resolution. For example, you can replace $N$ with its projective resolution $Q$. Then the cohomology of the complex $M \otimes Q$ is equal to $Tor$'s. On the other hand, after that you can replace $M$ with ANY complex $C$ quasiisomorphic to it, for example with its injective resolution, and the cohomology of $C\otimes Q$ still will be isomorphic to $Tor$'s. The reason for this is the fact that if $C$ is acyclic then $C\otimes Q$ is also acyclic.

So, the answer is yes, you can.

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    $\begingroup$ It is probably not useless to remark that this is in fact sometimes actually done in practice. For example, tricks like this are used by Cartan-Eilenberg to construct the usual spectral sequences for changes of rings and whatnot. $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 1, 2012 at 18:49
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    $\begingroup$ @Sasha: Can you give me please a hint or a reference, why $C \otimes_R Q$ is acyclic, if $C$ is ? $\endgroup$
    – Ralph
    Apr 1, 2012 at 21:07
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    $\begingroup$ @Ralph: Since $Q^i$ is projective, the complex $C\otimes Q^i$ is acyclic. Thus the bicomplex $C\otimes Q$ has acyclic rows. Now you can either use a spectral sequence argument, or alternatively check by hand that any cocycle is a coboundary. $\endgroup$
    – Sasha
    Apr 2, 2012 at 6:45
  • $\begingroup$ OK. Let $x = (x_{ij})$ be an element in the kernel with $x_{ij} \in C_i\otimes Q_j$. Note that only finite number of $x_{ij}$ is nonzero. Use induction in $max\{j|x_{ij} \ne 0\}$. Let $x_{ij}$ be the nonzero element with maximal $j$. Then $d_C(x_{ij}) = 0$. Hence $x_{ij} = d_C(y_{i+1,j})$. Replace $x$ by $x - dy_{i+1,j}$, then the maximal $j$ will be smaller. $\endgroup$
    – Sasha
    Apr 2, 2012 at 9:36
  • $\begingroup$ Sasha, thanks, you're right. I also found a reference: Theorem I.8.6 in Brown: Cohomology of groups. $\endgroup$
    – Ralph
    Apr 2, 2012 at 9:48

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