An injectivity criterion for $\mathbb Z/m\mathbb Z$ as $\mathbb Z/m\mathbb Z$ module

Question

Let $R=\mathbb Z/m\mathbb Z$ where $m>1$. Show that $R$ is injective as an $R$-module iff for any abelian group $A$ such that $mA=0$ and for any subgroup generated by an element $a\in A$ of order $m$ is a direct summand in $A$.

It seems to me the problem must somehow be reduced to a case of $m=p^k$ where $p$ is prime but I don't know how. I don't really understand how to chose complement to $\langle a\rangle$, too.

Could you please help me?

Answer 1

Let's first just uncode what the given statement has to do with $R$-modules for $R=\mathbb{Z}/m\mathbb{Z}$. An "abelian group $A$ such that $mA=0$" is just the same thing as an $R$-module. A "subgroup generated by an element $a\in A$ of order $m$" is then just a submodule of $A$ that is isomorphic to $R$. So the given condition just says "for all $A$ and $B$, if $A$ is an $R$-module and $B$ is a submodule of $A$ isomorphic to $R$, then $B$ is a direct summand of $A$".

This condition "for all $A$ and $B$, if $A$ is an $R$-module and $B$ is a submodule of $A$ isomorphic to $I$, then $B$ is a direct summand of $A$" is one of the several commonly used equivalent definitions of what it means for an $R$-module $I$ to be injective. So, if this is your definition, you're done. If you use some other definition, you just have to prove this definition is equivalent to your definition. (Depending on what exactly your definition is, one direction of this equivalence may be a bit nontrivial. You may find it helpful to use fact that every module is a submodule of an injective module, and so the condition above then would imply your module is a direct summand of an injective module.)

Answer 2

Actually, $\mathbb{Z}/m\mathbb{Z}$ is an injective $\mathbb{Z}/m\mathbb{Z}$-module. More generally, Baer's criterion states that an $R$-module $A$ is injective if and only if for every ideal $I \subseteq R$ any morphism $I \to A$ can be extended to a morphism $R \to A$. For $R=\mathbb{Z}/m\mathbb{Z}$ the category of $R$-modules is isomorphic to the category of abelian groups with $mA=0$, and ideals have the form $n \mathbb{Z}/m\mathbb{Z} \cong \mathbb{Z}/(m/n)\mathbb{Z}$ for some $n \mid m$. A morphism $I \to A$ corresponds to some element $a \in A$ with $(m/n) a = 0$, and it extends to $R \to A$ iff $a \in nA$. So we see that $A$ is injective iff $$A \xrightarrow{~~n~~} A \xrightarrow{m/n} A $$ is exact for all $n \mid m$. This easily checked for $A=\mathbb{Z}/m\mathbb{Z}$, so $R$ is injective as an $R$-mdodule.

From this the implication $\Leftarrow$ follows, and $\Rightarrow$ follows by applying injectivity to the inclusion $\mathbb{Z}/m\mathbb{Z} \cong \langle a \rangle \hookrightarrow A$ and the identity of $\mathbb{Z}/m\mathbb{Z}$.

Maybe the proof of the equivalence was intended to be different, but it works.