Please explain to me about small $\text{Tor}$ functor problem.

I use $\text{Tor(A,B)}$ define at http://en.wikipedia.org/wiki/Tor_functor.

we take a projective resolution:

$\cdots\rightarrow P_2 \rightarrow P_1 \rightarrow P_0 \rightarrow A\rightarrow 0$ (1)

then remove the A term and tensor the projective resolution with B to get the complex:

$\cdots \rightarrow P_2\otimes_R B \rightarrow P_1\otimes_R B \rightarrow P_0\otimes_R B \rightarrow 0$ (2)

and take the homology of this complex.

Clearly, since right exactness of $\otimes$-functor, so from (1) we have (2) is right exact sequence. It's mean, homology $H_n(x)=0$. But it's impossible!

I'm really misunderstand! Thanks for regarding!


1 Answer 1


Tensoring with $B$ doesn't take you from (1) to (2). It takes you from (1) to

$$\cdots P_2 \otimes_R B \to P_1\otimes_R B \rightarrow P_0 \otimes_R B \rightarrow A\otimes_R B \rightarrow 0$$

which is indeed exact at the last few terms.

  • $\begingroup$ it's mean (2) isn't right exact? $\endgroup$
    – Rachel
    Oct 8, 2013 at 3:56
  • $\begingroup$ The sequence below is indeed not exact, assuming that's what you mean by calling a chain complex "right exact". $$P_2\otimes_R B \rightarrow P_1\otimes_R B \rightarrow P_0\otimes_R B \rightarrow 0$$ However, the following sequence is exact: $$P_1\otimes_R B \rightarrow P_0\otimes_R B \to A \otimes_R B\rightarrow 0 $$ $\endgroup$
    – user14972
    Oct 8, 2013 at 3:58
  • $\begingroup$ However, honnomophic: $P_1\otimes_R B \rightarrow P_0\otimes_R B$ is epic? right? $\endgroup$
    – Rachel
    Oct 8, 2013 at 4:03
  • $\begingroup$ @Hoang: Usually no, because its cokernel is $A\otimes_R B$ which is usually not zero. The exactness of the original chain complex and right exactness of $\otimes_R$, however, do tell you that $P_0 \otimes_R B \to A \otimes_R B$ is epic. $\endgroup$
    – user14972
    Oct 8, 2013 at 4:06
  • $\begingroup$ OK, thanks Hurkyl, I think that I understand! :) $\endgroup$
    – Rachel
    Oct 8, 2013 at 4:12

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