Home » Homological Algebra

Projective resolution

An exact chain complex of projective modules ending in a given module M, used to compute Tor and Ext.
Projective resolution

Let RR be a and MM an .

A projective resolution of MM is an augmented

⋯→d2P1→d1P0→εM→0 \cdots \xrightarrow{d_2} P_1 \xrightarrow{d_1} P_0 \xrightarrow{\varepsilon} M \to 0

such that:

  1. Each PiP_i is a .
  2. The sequence is exact (equivalently, the augmented complex is an ).

Equivalently, if P∙P_\bullet denotes the chain complex ⋯→P1→P0→0\cdots\to P_1\to P_0\to 0 (forgetting the augmentation), then

H0(P∙)≅M,Hi(P∙)=0 for all i>0, H_0(P_\bullet)\cong M,\qquad H_i(P_\bullet)=0 \text{ for all } i>0,

where HiH_i is the of the underlying chain complex.

A projective resolution is free if each PiP_i is .

Existence in module categories is guaranteed by .

What resolutions are for

  • For any RR-module NN, the homology of P∙⊗RNP_\bullet\otimes_R N computes : Hi(P∙⊗RN)≅ToriR(M,N). H_i(P_\bullet\otimes_R N)\cong \mathrm{Tor}_i^R(M,N). (Tensor is , so one needs derived functors.)
  • Applying Hom⁡R(P∙,N)\operatorname{Hom}_R(P_\bullet,N) produces a whose cohomology computes : Hi(Hom⁡R(P∙,N))≅ExtRi(M,N). H^i(\operatorname{Hom}_R(P_\bullet,N))\cong \mathrm{Ext}_R^i(M,N).

Examples

Example 1: A projective resolution of Z/nZ\mathbb Z/n\mathbb Z

Over R=ZR=\mathbb Z, the sequence

0→Z→×nZ→Z/nZ→0 0\to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathbb Z/n\mathbb Z \to 0

is exact, and Z\mathbb Z is free (hence projective). Thus it is a projective resolution of Z/nZ\mathbb Z/n\mathbb Z of length 11.

Example 2: Computing Tor1Z(Z/n,Z/m)\mathrm{Tor}_1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m)

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

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

Then

Tor1Z(Z/n,Z/m)≅H1≅ker⁡(×n:Z/m→Z/m)≅Z/gcd⁡(m,n)Z. \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 .)

Example 3: A resolution of kk as a k[x]k[x]-module

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

0→R→×xR→k→0 0\to R \xrightarrow{\times x} R \to k \to 0

is a free (hence projective) resolution of kk of length 11. For instance,

Tor1k[x](k,k)≅H1((0→R→xR→0)⊗k[x]k)≅H1(0→k→0k→0)≅k. \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 .)