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Let $A$ be a commutative ring. If $M$ or $N$ aree free $A$-module then $Tor_{n}^{A}(M,N)=0$.

Since $Tor_{n}^{A}(M,N)=Tor_{n}^{A}(N,M)$ it suffices to deal with the case say when $N$ is flat right?

Take a projective resolution of $N$:

$... \rightarrow P_{n} \rightarrow P_{n-1} ... \rightarrow P_{0} \rightarrow N \rightarrow 0$

The next step is to tensor this sequence with $M$ yes?

$... \rightarrow P_{n} \otimes_{A} M \rightarrow P_{n-1} \otimes _{A} M \ .... $

But $M$ is free so by definition $M$ is isomorphic to a direct sum of copies of $A$. Hence:

$M \cong \oplus_{i \in I} A$ and since $P_{n} \otimes_{A} (\oplus_{i \in I} A) \cong \oplus_{i \in I} (P_{n} \otimes_{A} A) \cong \oplus_{i \in I} P_{n}$ we obtain:

the map $f_{n} \otimes 1: \oplus_{i \in I} P_{n} \rightarrow \oplus_{i \in I} P_{n-1}$.

Now we need to consider:

$ker(f_{n} \otimes 1)/Im(f_{n+1} \otimes 1)$ yes? Why this quotient is trivial? Can you please explain?

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"Why this quotient is trivial?" Because you chose your projection to be projective. – Fredrik Meyer Apr 10 '11 at 16:28
You should change the title to "Free module implies trivial Tor". – Zev Chonoles Apr 10 '11 at 16:34
up vote 4 down vote accepted
  • If $M$ is a projective module, then there is a very convenient projective resolution for $M$, namely $$\cdots\to0\to0\to0\to M$$ If you tensor it by any module $N$ and compute homology, then obviously $\mathrm{Tor}_p(M,N)=0$ for all $p>0$.

  • Alternatively, suppose that $N$ is any module and consider a projective resolution $$\cdots\to P_2\to P_1\to P_0$$ for $N$. Now, if $M$ is a projective module, then it is in particular flat, so if we tensor the projective resolution of $N$, which is an acyclic complex, with $M$, then the result will again be acyclic. It follows immediately that $\mathrm{Tor}_p(M,N)=0$ for all $p>0$.

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thanks! it is clear now, it was easier to consider the projective resolution you gave. Any particular books you can please recommend? I'm trying to learn this on my own using notes from the web, seems hard stuff (at least to me..) – user6495 Apr 10 '11 at 17:05
@user6495: Hilton and Stammbach's book. – Mariano Suárez-Alvarez Apr 10 '11 at 17:06

The original projective resolution $\cdots\xrightarrow{f_{n+1}} P_{n} \xrightarrow{f_n} P_{n-1} \cdots \xrightarrow{f_1} P_{0} \xrightarrow{f_0} N \rightarrow 0$ was (by definition) an exact sequence, i.e. $\ker(f_n)/\text{im}(f_{n+1})=0$ for all $n$, i.e. $\ker(f_n)=\text{im}(f_{n+1})$ for all $n$.

Now note that $\ker(f_n\otimes1)=\oplus_{i\in I}\ker(f_n)\subseteq \oplus_{i\in I}P_n$, and similarly for $\text{im}(f_{n+1}\otimes1)$.

share|cite|improve this answer
Thanks! – user6495 Apr 10 '11 at 17:05

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