Existence of projective resolutions

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

Statement

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

such that each $P_i$ is a projective module and the complex

is exact in all positive degrees and has $H_0 \cong M$ via the augmentation.

Theorem (existence). Every $R$-module $M$ admits a projective resolution. In fact, one can choose each $P_i$ to be a free module (a free resolution).

Equivalently, the category of $R$-modules has enough projectives: every module is a quotient of a projective module.

Construction (standard free resolution)

Choose a surjection $F_0 \twoheadrightarrow M$ with $F_0$ free (e.g. take $F_0 = R^{(M)}$, the free module on the underlying set of $M$). Let

Then choose a surjection $F_1 \twoheadrightarrow K_1$ with $F_1$ free, set $K_2 := \ker(F_1 \twoheadrightarrow K_1)$, and iterate. Splicing these short exact sequences produces an exact complex

with all $F_i$ free, hence projective.

Cross-links: exact sequences , chain complexes , projective resolutions .

Examples

Example 1: A length-1 resolution of $\mathbb Z/n\mathbb Z$ over $\mathbb Z$

As a $\mathbb Z$-module,

is exact, and the two copies of $\mathbb Z$ are free (hence projective). Thus it is a projective resolution of $\mathbb Z/n\mathbb Z$.

Example 2: The principal-ideal case $R/(f)$

For any ring $R$ and element $f\in R$, the sequence

is exact if and only if multiplication by $f$ is injective (e.g. if $f$ is not a zero-divisor). When exact, it gives a projective (free) resolution of $R/(f)$ of length $1$.

Example 3: $k$ as a $k[x]$-module

Let $R=k[x]$ and $k \cong R/(x)$ with $x$ acting by $0$. Then

is a free resolution of $k$ as an $R$-module.