MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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? – Martin Brandenburg Apr 1 '12 at 15:07
@Martin, why not? The double complex does compute the Tor, and in several situations it is a nicer description of it! – Mariano Suárez-Alvarez Apr 1 '12 at 18:46
@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 – Mariano Suárez-Alvarez Apr 1 '12 at 18:52
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. – Charles Staats Apr 1 '12 at 22:51
@Mariano, do people really use this double complex for Tor? See Ralph's comment below and Anton F's answer. – Yemon Choi Apr 2 '12 at 2:36
up vote 3 down vote accepted

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.

share|cite|improve this answer
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. – Mariano Suárez-Alvarez Apr 1 '12 at 18:49
@Sasha: Can you give me please a hint or a reference, why $C \otimes_R Q$ is acyclic, if $C$ is ? – Ralph Apr 1 '12 at 21:07
@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. – Sasha Apr 2 '12 at 6:45
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. – Sasha Apr 2 '12 at 9:36
Sasha, thanks, you're right. I also found a reference: Theorem I.8.6 in Brown: Cohomology of groups. – Ralph Apr 2 '12 at 9:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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