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

  • 1
    $\begingroup$ 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? $\endgroup$ 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! $\endgroup$ Apr 1, 2012 at 18:46
  • 2
    $\begingroup$ @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 $\endgroup$ 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. $\endgroup$ 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

1 Answer 1


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.

  • 1
    $\begingroup$ 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. $\endgroup$ Apr 1, 2012 at 18:49
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.