Tor

Choose a projective resolution $P_\bullet \to M$ by projective right $R$-modules:

Tensor with $N$ to obtain a chain complex $P_\bullet \otimes_R N$, and define

where $H_n(-)$ denotes homology.

This is well-defined up to canonical isomorphism and is functorial in both variables.

• $\mathrm{Tor}^R_0(M,N)\cong M\otimes_R N$.
• Because tensor is right exact, the derived functors $\mathrm{Tor}^R_n$ measure precisely the failure of tensor to be left exact. In particular, a module $N$ is flat iff $\mathrm{Tor}^R_1(-,N)=0$ (equivalently, iff tensoring with $N$ preserves injections).
• Any short exact sequence gives rise to a long exact sequence in Tor, a special case of the long exact sequence for derived functors.

Example 1: Vector spaces over a field

If $k$ is a field and $V,W$ are $k$-vector spaces, every $k$-module is free (hence projective), so

Example 2: $\mathrm{Tor}^{\mathbb Z}_1(\mathbb Z/n, A)\cong A[n]$

Use the standard projective resolution of $\mathbb Z/n$:

Tensor with $A$ to get

The homology at the left term is

Also $(\mathbb Z/n)\otimes_{\mathbb Z}A\cong A/nA$, so $\mathrm{Tor}^{\mathbb Z}_0(\mathbb Z/n,A)\cong A/nA$.

Example 3: $\mathrm{Tor}^{\mathbb Z}_1(\mathbb Z/n,\mathbb Z/m)\cong \mathbb Z/\gcd(n,m)$

From Example 2 with $A=\mathbb Z/m$, the $n$-torsion subgroup has order $\gcd(n,m)$ and is cyclic, hence

(As in the Ext computations over $\mathbb Z$, these cyclic modules have projective dimension $1$, so $\mathrm{Tor}^{\mathbb Z}_i(\mathbb Z/n,-)=0$ for $i\ge 2$.)