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Let $\cdots\rightarrow F_1 \rightarrow F_0 \rightarrow M \rightarrow 0$ be a free resolution of the $A$-module $M$. Let $N$ be an $A$-module. I saw in some notes that we have an exact sequence $0 \rightarrow \operatorname{Tor}(M,N) \rightarrow F_1 \otimes N \rightarrow F_0 \otimes N \rightarrow M \otimes N \rightarrow 0$. Why is that true?


Let me explain the source of my confusion. In my study, $\operatorname{Tor}_n(M,N)$ was defined as the homology of dimension $n$ of the double complex $K_{p,q}=F_p \otimes Q_q$, where $F_{\cdot}, Q_{\cdot}$ are projective resolutions of $M,N$ respectively. It was shown that $\operatorname{Tor}_n(M,N) = \operatorname{Tor}_n(F_{\cdot} \otimes N)=H_n(F_{\cdot} \otimes N) = \frac{Ker(F_n \otimes N \rightarrow F_{n-1} \otimes N)}{Im(F_{n+1} \otimes N \rightarrow F_{n} \otimes N)}$. Now if the sequence $0 \rightarrow \operatorname{Tor}(M,N) \rightarrow F_1 \otimes N \rightarrow F_0 \otimes N \rightarrow M \otimes N \rightarrow 0$ is exact, then $\operatorname{Tor}_1(M,N) = \operatorname{Ker}(F_1 \otimes N \rightarrow F_{0} \otimes N)$, which means that $\operatorname{Im}(F_{2} \otimes N \rightarrow F_{1} \otimes N)=0$, which does not make sense. What am I missing?

PS: The reference that i am using is Matsumura's Commutative Ring Theory Appendix B. That's as far as i have gone so far with homological algebra.

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Because ${\rm Tor}(-,N)$ is the derived functor of $\otimes N$ and ${\rm Tor}(F,N)=0$ if $F$ is flat ($\Leftarrow$ free). – Berci Feb 25 '13 at 0:51
I am looking for a simpler argument. I am not familiar with derived functors. – Manos Feb 25 '13 at 0:55
Aha.. Then how ${\rm Tor}$ is defined? Probably the mentioned exact sequence itself defines it, no? – Berci Feb 25 '13 at 0:57
You have a point. Let me think. – Manos Feb 25 '13 at 1:06
@Berci: I thought about it and i identified what confuses me. Please take a look at my edit :) – Manos Feb 25 '13 at 21:52
up vote 2 down vote accepted

You are right to be confused because the statement isn't true as written, for the reason you say. Example: take $A=k[x]/x^2$ and $M=k$ and $N=A$. Then $\operatorname{Tor}_A(M,N)=0$ because $N$ is free therefore flat. A free resolution for $M=k$ begins $$ \cdots A \stackrel{x}\to A \stackrel{x}\to A \to k \to 0 $$ Tensoring with $N=A$ gives $$ \cdots \to A \otimes_A A \to A \otimes_A A \to k \otimes_A A \to 0$$ Your sequence is $$0 \to 0 \to A \to A \to k \to 0 $$ (because the Tor group vanished and $A\otimes _A A \cong A$ and $A \otimes_A k\cong k$) which can't possibly be exact just by considering dimensions.

The correct statement is that $$ 0 \to \operatorname{Tor}_1^A(M,N) \to \ker d_0 \otimes_A N \to F_0 \otimes_A N \to M\otimes_A N \to 0 $$ is exact, where $d_0$ is the map $F_0 \to N$ in the free resolution. This is the long exact sequence for associated to $0 \to \ker d_0 \to F_0 \to N \to 0$; the 0 on the left is $\operatorname{Tor}_1^A(M,F_0)$ which vanishes by flatness of $F_0$.

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Indeed. But maybe $A$ is one of those rings where free resolutions terminate after 2 steps, say, a principal ideal domain. – Zhen Lin Feb 26 '13 at 17:22
Perhaps I should have said "isn't true without extra hypotheses", but the OP's "$\cdots \to F_1$" indicates their resolution doesn't necessarily stop after $F_1$. – m_t_ Feb 26 '13 at 17:30
mt_ is correct. I am assuming nothing more than stated. mt_, let me study your answer and i will get back to you. – Manos Feb 26 '13 at 17:32
Excellent answer, thanks. – Manos Mar 3 '13 at 17:00

The tensor functor is right-exact. We have an exact sequence

$F_1\to F_0\to M\to 0$,

implying that the complex

$F_2\otimes N\to F_1\otimes N\to F_0\otimes N\to M\otimes N\to 0$,

is exact except possibly at $F_1\otimes N$. Now how can you calculate $\mathrm{Tor}_1(M,N)$? You should be able to figure it out from here.

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Maybe it should be pointed out to the OP that we have a zero on the left because $\text{Tor}_1(F,N) = 0$ if $F$ is a free - module. – user38268 Feb 25 '13 at 3:42
@JasonPolak: Could you please take a look at my edit? – Manos Feb 25 '13 at 21:52

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