Tor functor

Remarks

Up to a canonical isomorphism, $\operatorname{Tor}_n^R(M, N)$ does not depend on the resolution, and resolving $N$ instead of $M$ gives the same modules. One has $\operatorname{Tor}_0^R(M, N) \cong M \otimes_R N$ and $\operatorname{Tor}_n^R(M, N) \cong \operatorname{Tor}_n^R(N, M)$. Example: for integers $m, n \ge 1$, tensoring $0 \to \Z \xrightarrow{\cdot m} \Z \to \Z/m\Z \to 0$ with $\Z/n\Z$ gives $\Z/n\Z \xrightarrow{\cdot m} \Z/n\Z$, whose kernel is cyclic of order $\gcd(m, n)$; so $\operatorname{Tor}_1^\Z(\Z/m\Z, \Z/n\Z) \cong \Z/\gcd(m,n)\Z$. Non-example: the exact sequence $0 \to \Z/m\Z \xrightarrow{\id} \Z/m\Z \to 0$ is not a projective resolution over $\Z$ for $m \ge 2$, and using it would wrongly give $\operatorname{Tor}_1 = 0$. A module $M$ is flat if and only if $\operatorname{Tor}_1^R(M, N) = 0$ for all $R$-modules $N$. For a noncommutative ring $R$, the same construction works for a right module $M$ and a left module $N$.