Existence of injective resolutions

Every module embeds into an injective module, hence admits an injective resolution.

Existence of injective resolutions

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

Statement

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

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

such that each $I^n$ is an injective module .

Theorem (existence). Every $R$ -module $M$ admits an injective resolution.

Equivalently, the category of $R$ -modules has enough injectives: every module embeds into an injective module.

A standard route to this theorem uses either injective envelopes or explicit “large” injective modules constructed via character modules; Baer’s criterion is a key tool in many proofs and examples.

Construction (iterating embeddings)

Choose a monomorphism $M \hookrightarrow I^0$ with $I^0$ injective.
Let $C^0 := \operatorname{coker}(M \hookrightarrow I^0)$ .
Choose a monomorphism $C^0 \hookrightarrow I^1$ with $I^1$ injective, and iterate.

Splicing the resulting short exact sequences yields an exact cochain complex

0 \to M \to I^0 \to I^1 \to I^2 \to \cdots

as required.

Cross-links: injective resolutions , left exactness of Hom , derived functors .

Examples

Example 1: An injective resolution of $\mathbb Z$ as a $\mathbb Z$ -module

Injective $\mathbb Z$ -modules are the divisible abelian groups. The inclusion $\mathbb Z \hookrightarrow \mathbb Q$ has cokernel $\mathbb Q/\mathbb Z$ , and both $\mathbb Q$ and $\mathbb Q/\mathbb Z$ are divisible, hence injective. Thus

0 \longrightarrow \mathbb Z \longrightarrow \mathbb Q \longrightarrow \mathbb Q/\mathbb Z \longrightarrow 0

is an injective resolution of $\mathbb Z$ of length $1$ .

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

The subgroup of $\mathbb Q/\mathbb Z$ consisting of elements of order dividing $n$ is isomorphic to $\mathbb Z/n\mathbb Z$ , giving an injection $\mathbb Z/n\mathbb Z \hookrightarrow \mathbb Q/\mathbb Z$ . Moreover, multiplication by $n$ on $\mathbb Q/\mathbb Z$ is surjective with kernel $\mathbb Z/n\mathbb Z$ . Hence

0 \longrightarrow \mathbb Z/n\mathbb Z \longrightarrow \mathbb Q/\mathbb Z \xrightarrow{\;\cdot n\;} \mathbb Q/\mathbb Z \longrightarrow 0

is an injective resolution of $\mathbb Z/n\mathbb Z$ of length $1$ .

Example 3: Over a field, resolutions are trivial

If $R=k$ is a field, every $k$ -vector space is injective (and projective). So for any $k$ -module $V$ ,

0 \to V \xrightarrow{\mathrm{id}} V \to 0 \to \cdots

is an injective resolution.