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(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 $U$ U_0$ is a direct summand of $U$.

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

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(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 ${V_i}$ \{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 $U$ is a direct summand of $U$.

Referring to your examples of $R$ above: When $R = \mathbb{Z}$ then $U_0 = \mathbb{Q}/\mathbb{Z}$. When $R = \mathbb{K}$ then $U_0 = \mathbb{K}$ (in particular, every nonzero $\mathbb{K}$-vector space is a cogenerator).
show/hide this revision's text 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 $U$ is a direct summand of $U$.

Referring to your examples of $R$ above: When $R = \mathbb{Z}$ then $U_0 = \mathbb{Q}/\mathbb{Z}$. When $R = \mathbb{K}$ then $U_0 = \mathbb{K}$ (in particular, every nonzero $\mathbb{K}$-vector space is a cogenerator).