Homological Algebra - Tor

Question

I am trying to prove the following:

If A and B are abelian groups with mA = 0 = nB, where (m, n) = 1 , Then $Tor_{1}^{\mathbb{Z}}\left( A,B \right)=0$. Conclude that, in this case, exactness of $0\to D\to C\to B\to 0$ implies exactness of $0\to A\otimes D\to A\otimes C\to A\otimes B\to 0$

Can someone teach me how to prove this? Thanks

It's generally considered polite to phrase your question as a request rather than just a statement to prove. What have you tried so far? — tharris, Dec 17 '15 at 22:03
What have you tried? For the first part you need to compute that Tor group, and that can be done very easily: pick a projective resolution of $A$, tensor it with $B$ and compute the homology. — Mariano Suárez-Álvarez, Dec 17 '15 at 22:10

score 2 · Answer 1 · edited Dec 18 '15 at 00:59

2

Hint:

Since multiplication by $m$ is the zero-morphism on $A$, it is also the zero-morphism on $\operatorname{Tor}^{\mathbf Z}_1(A,B)$.

For similar reasons, multiplication by $n$ is the zero-morphism on $\operatorname{Tor}^{\mathbf Z}_1(A,B)$.

Thus $\operatorname{Tor}^{\mathbf Z}_1(A,B)$ is killed by $m$ and by $n$, hence by any element in the ideal $(m,n)$ – which contains $1$.

edited Dec 18 '15 at 00:59

Dustan Levenstein

12,941

answered Dec 18 '15 at 00:52

Bernard

175,478

Homological Algebra - Tor

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