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?