$Tor(A,B) = 0$ if $A,B$ is torsion free

Question

I don't understand the proof given in Hatcher p.265 of $Tor(A,B) = 0$ if $A,B$ is torsion free.

The proof is the following :

The line I don't get is "This means [...] can be reduced to $0$ by a finite number of application of the definining relations for tensor product".

Why this should follow from the definition of tensor product, and why only $B$ is involved in this process ? Feels that something is hidden under the radar.

Any help would be appreciated.

Answer 1

The tensor product of two modules $A$ and $B$ over a commutative ring $R$ is defined : $$ A \otimes_{R} B:=F(A \times B) / G $$ where $F(A \times B)$ is the free $R$ -module generated by the cartesian product and $G$ is the $R$ -module generated by the relations below.

1. Distributivity $(v, w) + (v', w) \sim (v + v', w)$ and $(v, w) + (v, w') \sim (v, w + w')$.
2. Scalar multiples $c(v, w) \sim (cv, w)$ and $c(v, w) \sim (v, cw)$.

An example to what hatcher means is the following. $$2\otimes 1+(-1)\otimes 2=(2-2)\otimes 1=0 \in \mathbb{Z} \bigotimes \mathbb{Z}$$ We used the relations a finite number of times to reduce the sum on the left to $0$. In this case we had to use the elements of $2, 1 \in B=\mathbb{Z}$. In this example $1,2$ generates $B_0$ that in this particuler example is also $\mathbb{Z}$.