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Let $R$ be a commutative ring and $M$ and $N$ be $R$-modules (I am not sure if one really needs commutativity in the following). It is well-known that $Ext_{R}^n(M,N)$ for $n>1$ parametrizes $n$-extension of $N$ by $M$, i.e. exact sequences $$ 0\rightarrow M\rightarrow C_{1}\rightarrow \dots \rightarrow C_{n}\rightarrow N\rightarrow 0 $$ mod certain equivalent relations. Another way to see $Ext_{R}^n(M,N)$ is via derived category; it can be seen as a hom space in $D(R-mod)$ $$ Ext_{R}^n(M,N)=Hom_{D(R-mod)}(M,N[n]). $$

I now want to understand what $Tor_{R}^n(M,N)$ represent. How should one understand $Tor_{R}^n(M,N)$ intuitively?

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It describes tensor product in the derived category, doesn't it? –  Qiaochu Yuan Sep 16 '12 at 3:48
    
Yes, that is one way to understand the higher tensors. Are there meaning more than that? –  M. K. Sep 16 '12 at 6:23

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up vote 4 down vote accepted

Similar to $\rm{Ext}^n$, $\rm{Tor}^R_n(A,B)$ for any $n$ can be viewed as an abelian group described explicitly as a so-called torsion product, which as a set consists of equivalence classes of triples $(f,L,g)$ where $L$ is a length $n$-complex of finitely generated projective modules, $f:L\to A$ and $g:\rm{Hom}(L,R)\to B$ are chain maps with $A,B$ interpreted as chain complexes concentrated in degree zero. This is classic, but you probably haven't heard of it because it is ignored in modern treatments of homological algebra. The equivalence relation and how to add these torsion elements is described for instance in Mac Lane's book "Homology" (p.154) and this is where I learned about these.

In general, you should always expect at least some kind of explicit description for the derived functors of a given functor because there are usually "standard" resolutions lurking in the background coming from (co)triples, although I am not sure how this description of $\rm{Tor}$ was thought of.

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Thank you for the information. I have never heard of torsion product. I will take a look at Mac Lane's book. –  M. K. Oct 1 '12 at 7:31

Let $R=\mathbb{Z}$ for simplicity and $n=1$. The $\mathrm{Tor}^1(-,-)$ is characterized by the following properties:

  1. $\mathrm{Tor}^1(F,M)$ for all free $\mathbb{Z}$-module $F$.
  2. $\mathrm{Tor}^1(\mathbb{Z}/n\mathbb{Z},M)=Ker(M\rightarrow M, m\mapsto nm)$
  3. $\mathrm{Tor}^1(L\oplus M, N)=\mathrm{Tor}^1(L,N)\oplus \mathrm{Tor}^1(M,N)$
  4. $\mathrm{Tor}^1(M,N)=\mathrm{Tor}(N,M)$

$\mathrm{Tor}^1(-,-)$ is essentially the common torsion of its arguments.

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While this would appear incomplete, I've never heard that there's a meaningful interpretation up in the module category for the higher Tors than the first. –  Kevin Carlson Sep 19 '12 at 4:14

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