What are examples of cogenerators in R-mod?

Question

Fill in the blank, please :)

Let $\mathcal C$ be a complete and cocomplete abelian category. A generator in $\mathcal C$ is an object $X \in \mathcal C$ such that every object $Y \in \mathcal C$ is a colimit of a (small) diagram made entirely of $X$s; in this way, $X$ knows everything there is to know about $\mathcal C$. When $\mathcal C$ is the category of all modules of some ring $R$, then an example of a generator is $R$, thought of as an $R$-module. A cogenerator in $\mathcal C$ is an object $X \in \mathcal C$ such that every object $Y \in \mathcal C$ is a limit of a (small) diagram made entirely of $X$s; in this way, $X$ knows everything there is to know about $\mathcal C$. When $\mathcal C$ is the category of all modules of some ring $R$, then an example of a cogenerator is ________.

The only examples I know are: when $R = \mathbb Z$, an example of a cogenerator is the rational circle $\mathbb Q / \mathbb Z$; when $R = \mathbb K$ is a field, an example of a cogenerator is $\mathbb K$. But by some version of the Law of Small Numbers, these examples are not enough for me to see how to (or, in fact, whether it is possible to) generalize.

Answer 1

(This is closely related to Hailong's comment above.)

You can say (albeit rather abstractly) what any cogenerator must look like. The following can be found in T.Y. Lam's Lectures on Modules and Rings, Theorem 19.10. Let $\{V_i\}$ be a complete set of simple right $R$-modules, with injective hulls $E(V_i)$. Then $U_0 = \bigoplus E(V_i)$ is a cogenerator, called the canonical cogenerator of $\mathrm{Mod}_R$, and any module $U_R$ is a cogenerator for $\mathrm{Mod}_R$ if and only if $U_0$ can be embedded in $U$.

Note that if $R$ is right noetherian, then the direct sum of injective right $R$-modules is again injective. Hence $U_0$ is injective, and in this case $U_R$ is a cogenerator if and only if $U_0$ embeds in $U$, if and only if $U_0$ is a direct summand of $U$.

Referring to your examples of $R$ above: If $R = \mathbb{Z}$ then $U_0 = \mathbb{Q}/\mathbb{Z}$. If $R = \mathbb{K}$ then $U_0 = \mathbb{K}$. Both of these rings are noetherian, so the previous paragraph applies. (In particular, every nonzero $\mathbb{K}$-vector space is a cogenerator).

Answer 2

For any ring $R$, the $R$-module $Hom_{\mathbb Z}(R,\mathbb Q/\mathbb Z)$ is an injective cogenerator of the category of $R$-modules. Here $\mathbb Q/\mathbb Z$ can be replaced with $\mathbb R/\mathbb Z$ or any other injective cogenerator of the category of abelian groups. When $R$ is an algebra over a field $\mathbb K$, another example of an injective cogenerator of the category of $R$-modules is $Hom_{\mathbb K}(R,\mathbb K)$.