Injective Cogenerator

Definition.

Let $R$ be a commutative ring. An $R$-module $C$ is a cogenerator for the category of $R$-modules if for any nonzero $R$-module $M$ and any nonzero element $m\in M$, there exists an $R$-module homomorphism $f:M\to C$ such that $f(m)\ne 0$. An injective cogenerator is a module that is both injective and a cogenerator.

Theorem.

Let $R$ be a commutative ring, $I$ an injective cogenerator for the category of left $R$-modules, and $M$ an $R$-module. Then $M$ can be embedded into a direct product of copies of $I$. In particular, $M=0$ iff ${\mathrm{Hom}}_R(M,I)=0$.

Proof.

Let $M$ be a left $R$-module and let $I$ be an injective cogenerator. Let the index set be $\mathcal{F}={\mathrm{Hom}}_R(M,I)$, the set of all $R$-module homomorphisms from $M$ to $I$.

Consider the direct product $P={\prod }_{f\in \mathcal{F}}{I}_f$, where each $I_f$ is a copy of $I$. As a direct product of injective modules, $P$ is an injective $R$-module.

Define the evaluation homomorphism $\psi :M\to P$ by $\psi (m)=(f(m){)}_{f\in \mathcal{F}}$. This is an $R$-module homomorphism.

To show $\psi$ is a monomorphism, its kernel must be shown to be trivial. Let $m\in M$ be a nonzero element. Since $I$ is a cogenerator, there exists a homomorphism $g\in \mathcal{F}$ such that $g(m)\ne 0$. The $g$-th component of $\psi (m)$ is $g(m)$, which is nonzero. Therefore, $\psi (m)\ne 0$. This implies that $\ker (\psi )=\{0\}$.

Thus, $\psi$ is a monomorphism from $M$ into the injective module $P$.

$\square$

Theorem.

For any ring $R$, the category of left $R$-modules possesses an injective cogenerator.

Proof.

The proof proceeds in two main steps. First, an injective cogenerator for the category of abelian groups ($\mathbb{Z}$-modules) is established. Second, this is used to construct an injective cogenerator for any category of $R$-modules.

The module $\mathbb{Q}/\mathbb{Z}$ is an injective cogenerator for the category of $\mathbb{Z}$-modules. As a divisible group, it is injective. To show it is a cogenerator, let $A$ be a nonzero abelian group and $a\in A$ be a nonzero element. Consider the cyclic subgroup $⟨a⟩$. If the order of $a$ is infinite, then $⟨a⟩\cong \mathbb{Z}$, and the map $f:⟨a⟩\to \mathbb{Q}/\mathbb{Z}$ defined by $f(ka)=k/2+\mathbb{Z}$ for $k\in \mathbb{Z}$ is a nonzero homomorphism. If the order of $a$ is a finite integer $n>1$, then $⟨a⟩\cong \mathbb{Z}/n\mathbb{Z}$, and the map $f:⟨a⟩\to \mathbb{Q}/\mathbb{Z}$ defined by $f(ka)=k/n+\mathbb{Z}$ for $k\in \mathbb{Z}$ is a well-defined, nonzero homomorphism. By the injectivity of $\mathbb{Q}/\mathbb{Z}$, this map $f$ extends to a homomorphism $g:A\to \mathbb{Q}/\mathbb{Z}$ such that $g(a)=f(a)\ne 0$.

Let $C$ be an injective cogenerator for $\mathbb{Z}$-modules, such as $C=\mathbb{Q}/\mathbb{Z}$. Let $I={\mathrm{Hom}}_{\mathbb{Z}}(R,C)$. The module $I$ is a left $R$-module via the action $(r\cdot \varphi )(s)=\varphi (sr)$ for $r,s\in R$ and $\varphi \in I$.

Injectivity of $I$: The functor ${\mathrm{Hom}}_R(-,I)$ is exact because ${\mathrm{Hom}}_R(-,{\mathrm{Hom}}_{\mathbb{Z}}(R,C))\cong {\mathrm{Hom}}_{\mathbb{Z}}(-{\otimes }_{R}R,C)\cong {\mathrm{Hom}}_{\mathbb{Z}}(-,C)$, and the latter is an exact functor as $C$ is $\mathbb{Z}$-injective. Therefore, $I$ is an injective left $R$-module.

Cogenerator Property of $I$: Let $M$ be a nonzero left $R$-module and $m\in M$ be a nonzero element. There exists a nonzero $\mathbb{Z}$-homomorphism $g:M\to C$. By Hom-tensor adjunction

$${\mathrm{Hom}}_R(M,{\mathrm{Hom}}_{\mathbb{Z}}(R,C))\cong {\mathrm{Hom}}_{\mathbb{Z}}(M{\otimes }_{R}R,C)$$

this corresponds to a nonzero $R$-homomorphism $f:M\to {\mathrm{Hom}}_{\mathbb{Z}}(R,C)=I$. Thus, $I$ is a cogenerator for left $R$-modules.

$\square$