Injective resolution - Imported Knowlpedia Content

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

An injective resolution of $M$ is an augmented cochain complex

0 \to M \xrightarrow{\eta} I^0 \xrightarrow{d^0} I^1 \xrightarrow{d^1} I^2 \xrightarrow{d^2} \cdots

such that:

Each $I^j$ is an injective R-module.
The sequence is exact (i.e. an exact complex).

Existence in module categories is guaranteed by injective resolutions exist.

What resolutions are for

If $F$ is a left exact functor, its right derived functors are computed by applying $F$ to an injective resolution and taking cohomology; see derived functor.

In particular, for any $R$ -module $N$ ,

\mathrm{Ext}_R^n(N,M) \cong H^n\bigl(\operatorname{Hom}_R(N, I^\bullet)\bigr),

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

Examples

Example 1: Over a field, injective resolutions are trivial

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

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

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

Example 2: An injective resolution of $\mathbb Z$

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

0\to \mathbb Z \to \mathbb Q \to \mathbb Q/\mathbb Z \to 0

is exact, hence yields an injective resolution of $\mathbb Z$ of length $1$ .

Using it, one computes

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

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

(See Ext.)

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

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

0\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 $\mathbb Z/n\mathbb Z$ .

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

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 $\operatorname{Hom}(\mathbb Z/m,\mathbb Q/\mathbb Z)\cong (\mathbb Q/\mathbb Z)[m]\cong \mathbb Z/m\mathbb Z$ . Therefore

\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 Ext.)

Example 1: Over a field, injective resolutions are trivial

Example 2: An injective resolution of Z\mathbb ZZ

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

Example 2: An injective resolution of $\mathbb Z$

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