Knowlpedia

Injective resolution

An exact cochain complex starting at M and continuing with injective modules, used to compute Ext and right derived functors.

Let RR be a and MM an .

An injective resolution of MM is an augmented

0→M→ηI0→d0I1→d1I2→d2⋯0 \to M \xrightarrow{\eta} I^0 \xrightarrow{d^0} I^1 \xrightarrow{d^1} I^2 \xrightarrow{d^2} \cdots

such that:

  1. Each IjI^j is an .
  2. The sequence is exact (i.e. an ).

Existence in module categories is guaranteed by .

What resolutions are for

If FF is a functor, its right derived functors are computed by applying FF to an injective resolution and taking ; see .

In particular, for any RR-module NN,

ExtRn(N,M)≅Hn(Hom⁡R(N,I∙)),\mathrm{Ext}_R^n(N,M) \cong H^n\bigl(\operatorname{Hom}_R(N, I^\bullet)\bigr),

where Hom⁡R(N,I∙)\operatorname{Hom}_R(N,I^\bullet) is a cochain complex (see ).

Examples

Example 1: Over a field, injective resolutions are trivial

If R=kR=k is a field, every kk-vector space is injective. Thus an injective resolution of a kk-vector space VV can be taken as

0→V→=V→0→0→⋯ ,0\to V \xrightarrow{=} V \to 0 \to 0 \to \cdots,

and therefore Extkn(W,V)=0\mathrm{Ext}_k^n(W,V)=0 for all n>0n>0 (see ).

Example 2: An injective resolution of Z\mathbb Z

In the category of abelian groups (R=ZR=\mathbb Z), both Q\mathbb Q and Q/Z\mathbb Q/\mathbb Z are injective (they are divisible groups). The sequence

0→Z→Q→Q/Z→00\to \mathbb Z \to \mathbb Q \to \mathbb Q/\mathbb Z \to 0

is exact, hence yields an injective resolution of Z\mathbb Z of length 11.

Using it, one computes

ExtZ1(Z/nZ,Z)≅H1(Hom⁡(Z/n, [0→Q→Q/Z→0])).\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n\mathbb Z,\mathbb Z) \cong H^1\bigl(\operatorname{Hom}(\mathbb Z/n,\, [0\to \mathbb Q\to \mathbb Q/\mathbb Z\to 0])\bigr).

Since Hom⁡(Z/n,Q)=0\operatorname{Hom}(\mathbb Z/n,\mathbb Q)=0 and Hom⁡(Z/n,Q/Z)≅(Q/Z)[n]≅Z/nZ\operatorname{Hom}(\mathbb Z/n,\mathbb Q/\mathbb Z)\cong (\mathbb Q/\mathbb Z)[n]\cong \mathbb Z/n\mathbb Z, this gives

ExtZ1(Z/n,Z)≅Z/nZ.\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,\mathbb Z)\cong \mathbb Z/n\mathbb Z.

(See .)

Example 3: An injective resolution of Z/nZ\mathbb Z/n\mathbb Z and Ext1\mathrm{Ext}^1 between cyclic groups

Embed Z/nZ\mathbb Z/n\mathbb Z into Q/Z\mathbb Q/\mathbb Z as the nn-torsion subgroup (Q/Z)[n](\mathbb Q/\mathbb Z)[n]. Multiplication by nn on Q/Z\mathbb Q/\mathbb Z has kernel (Q/Z)[n](\mathbb Q/\mathbb Z)[n], so

0→Z/nZ→Q/Z→×nQ/Z→00\to \mathbb Z/n\mathbb Z \to \mathbb Q/\mathbb Z \xrightarrow{\times n} \mathbb Q/\mathbb Z \to 0

is exact, with injective terms, hence an injective resolution of Z/nZ\mathbb Z/n\mathbb Z.

Applying Hom⁡(Z/m,−)\operatorname{Hom}(\mathbb Z/m,-) yields

0→Hom⁡(Z/m,Q/Z)→×nHom⁡(Z/m,Q/Z)→0,0\to \operatorname{Hom}(\mathbb Z/m,\mathbb Q/\mathbb Z)\xrightarrow{\times n} \operatorname{Hom}(\mathbb Z/m,\mathbb Q/\mathbb Z)\to 0,

and Hom⁡(Z/m,Q/Z)≅(Q/Z)[m]≅Z/mZ\operatorname{Hom}(\mathbb Z/m,\mathbb Q/\mathbb Z)\cong (\mathbb Q/\mathbb Z)[m]\cong \mathbb Z/m\mathbb Z. Therefore

ExtZ1(Z/m,Z/n)≅coker⁡(×n:Z/m→Z/m)≅Z/gcd⁡(m,n)Z.\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/m,\mathbb Z/n) \cong \operatorname{coker}(\times n:\mathbb Z/m\to\mathbb Z/m) \cong \mathbb Z/\gcd(m,n)\mathbb Z.

(See .)