I am having trouble understanding the Tor functor as presented in Dummit and Foote.

Given $\dotsb\to P_n\to P_{n-1}\to\dotsb\to P_0\to B\to 0$ as a projective resolution with homomorphisms $d_i:P_n\to P_{n-1}$.

The book defines projective resolution as an exact sequence as above with each $P_i$ being a projective module.

An exact sequence is defined as a sequence such that for each $P_i$ in our sequence, $\ker(d_n)={\rm im}(d_{n+1})$.

We then tensor each module with the module $A$, giving $\dotsb\to A\otimes P_n\to\dotsb\to A\otimes P_0\to A\otimes B\to 0$ and related homomorphisms, $1\otimes d_n:A\otimes P_n\to A\otimes P_{n-1}$.

The definition in Dummit and Foote says $\operatorname{Tor}_n^R(A,B)=\ker(1\otimes d_n)/{\rm im}(1\otimes d_{n+1})$.

Is $\dotsb\to A\otimes P_n\to\dotsb\to A\otimes P_0\to A\otimes B\to 0$ always exact? It feels like it is not, since if it was exact, it would be fairly uninteresting.

Assuming it is not exact, what does $\operatorname{Tor}_n^R(A,B)$ look like? In other words, how are $\ker(1\otimes d_n)$ and ${\rm im}(1\otimes d_{n+1})$ different?

Specifically I am trying to figure out this problem (Given a torsion $R$-module $A$ where $R$ is an integral domain, $\mathrm{Tor}_n^R(A,B)$ is also torsion.), but I clearly am having trouble understanding the definitions, so recalling them is not really helping me.

Edit: Let's also assume that I will want to solve other problems involving the $\operatorname{Tor}$ functor, so I want to understand it in general and not just well enough to answer the linked question.

  • $\begingroup$ No, it is not always exact. If it were, this procedure would not calculate anything interesting. But none of this really matters to your previous question: that one really does just follow from the definitions. $\endgroup$ – Zhen Lin Mar 12 '14 at 18:49
  • $\begingroup$ The definitions are not clear to me. It may follow from the definitions, but that is kind of irrelevant until the definitions actually make sense to me. $\endgroup$ – Cyllindra Mar 12 '14 at 19:02

"Fairly uninteresting" is an understatement: if the tensored complex were always exact, then every $\text{Tor}$ would be identically zero! The question "what does $\text{Tor}$ look like?" is rather vague, but let me give a proposition which may shed some light, and also gives a different approach to your linked question: $\newcommand{\Tor}{\text{Tor}}$

Proposition: Let $R$ be a Noetherian reduced ring. Then every $\Tor^R_i(\_, \_)$ for $i \ge 1$ is a torsion $R$-module.

Proof: Let $A, B$ be any $R$-modules. To show that $\Tor^R_i(A,B)$ is torsion, it suffices to show that $\Tor^R_i(A,B) \otimes_R Q(R) = 0$, where $Q(R)$ is the total ring of quotients of $R$ (i.e. $Q(R) = S^{-1}R$, where $S$ is the set of nonzerodivisors in $R$). But $Q(R)$ is flat over $R$ (being a localization), so $\Tor^R_i(A,B) \otimes_R Q(R) \cong \Tor_i^{Q(R)}(A \otimes_R Q(R), B \otimes_R Q(R))$, which vanishes for every $i \ge 1$, since $R$ reduced $\implies Q(R)$ is a finite direct product of fields $\implies Q(R)$ has global dimension zero.

In particular, one can deduce that over Noetherian reduced rings, the torsion modules are precisely the modules of the form $\Tor_i(\_, \_)$ for $i \ge 1$ (one can even take $i = 1$).

  • $\begingroup$ I appreciate the help, but my issue at the moment is definition overload -- And here you have introduced several new terms that I have minimal exposure to. I haven't seen reduced ring (or Noetherian reduced), total ring of quotients (although your definition is very clear), localization (although my text has a section on it that our class skipped), or global dimension (not mentioned in my text). $\endgroup$ – Cyllindra Mar 12 '14 at 20:44
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    $\begingroup$ I see. Well in that case, might I advise you to try and work out some explicit examples? Resolutions are concrete objects, and writing out a few can be quite enlightening $\endgroup$ – zcn Mar 13 '14 at 3:52

The tensor product is not an exact functor. So the only thing you have when you tensor an exact sequence $f:A \to B$, $g: B \to C$, say by $M$, is $$(1_M \otimes g) \circ (1_M \otimes f) = 1_M \otimes (g \circ f) = 1_M \otimes 0 = 0$$

then all you can be sure is that $Im(1_M\otimes f) \subset Ker(1_M \otimes g)$, and you are aloud to form the quotient.

The functors $Tor_n^R(M,-)$ measures how far is the module $M$ from being flat. For example, $M$ is flat if and only if $Tor_1^R(M,N)=0$ for all module $N$. And you say that $M$ has $Tor$-dimension $2$ if $Tor_2^R(M,N)=0$ for all module $N$, and so on.

You can also calculate the $Tor$ groups by flat resolutions, and remember that:

$$free \implies projective \implies flat$$


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