Projective resolution

Let $R$ be a ring and $M$ an R-module.

A projective resolution of $M$ is an augmented chain complex

\cdots \xrightarrow{d_2} P_1 \xrightarrow{d_1} P_0 \xrightarrow{\varepsilon} M \to 0

such that:

Each $P_i$ is a projective R-module.
The sequence is exact (equivalently, the augmented complex is an exact complex).

Equivalent characterizations

Equivalently, if $P_\bullet$ denotes the chain complex $\cdots\to P_1\to P_0\to 0$ (forgetting the augmentation), then

H_0(P_\bullet)\cong M,\qquad H_i(P_\bullet)=0 \text{ for all } i>0,

where $H_i$ is the homology of the underlying chain complex.

Remarks

A projective resolution is free if each $P_i$ is free.

Existence in module categories is guaranteed by projective resolutions exist.

Applications

For any $R$ -module $N$ , the homology of $P_\bullet\otimes_R N$ computes Tor: $H_i(P_\bullet\otimes_R N)\cong \mathrm{Tor}_i^R(M,N).$ (Tensor is right exact, so one needs derived functors.)
Applying $\operatorname{Hom}_R(P_\bullet,N)$ produces a cochain complex whose cohomology computes Ext: $H^i(\operatorname{Hom}_R(P_\bullet,N))\cong \mathrm{Ext}_R^i(M,N).$

Examples

Over $R=\mathbb Z$ , the sequence

0\to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathbb Z/n\mathbb Z \to 0

is exact, and $\mathbb Z$ is free (hence projective). Thus it is a projective resolution of $\mathbb Z/n\mathbb Z$ of length $1$ .

Tensor the resolution in Example 1 with $\mathbb Z/m\mathbb Z$ :

0\to \mathbb Z/m \xrightarrow{\times n} \mathbb Z/m \to 0.

Then

\mathrm{Tor}_1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m) \cong H_1 \cong \ker(\times n:\mathbb Z/m\to\mathbb Z/m) \cong \mathbb Z/\gcd(m,n)\mathbb Z.

(See Tor.)

Let $R=k[x]$ and $M=R/(x)\cong k$ . Then

0\to R \xrightarrow{\times x} R \to k \to 0

is a free (hence projective) resolution of $k$ of length $1$ . For instance,

\mathrm{Tor}_1^{k[x]}(k,k) \cong H_1\bigl((0\to R\xrightarrow{x}R\to 0)\otimes_{k[x]}k\bigr) \cong H_1(0\to k\xrightarrow{0}k\to 0)\cong k.

(See Tor.)