# Are $\mathbb{Z}$ and $\mathbb{Z}_n$ the only rings (with identity) whose modules are equivalent to abelian groups?

Let $$R$$ be a ring with identity. Let $$M$$ and $$N$$ be $$R$$-modules. Let $$f$$ be an (arbitrary) group homomorphism from $$M$$ to $$N$$. Under what conditions on $$R$$,$$M$$, and $$N$$ is $$f$$ also a $$R$$-module homomorphism?

Is $$R=\mathbb{Z}$$ or $$R=\mathbb{Z}_n$$ necessary? If so, is there a general method to construct group homomorphisms which are not module homomorphisms?

Thanks!

• $R=\mathbb{Q}$ seems to work as well. – Mindlack Feb 7 at 16:17
• If $M=N=R$, then any module homomorphism $f\colon M\to N$ is determined by $f(1)$ as we must have $f(x)=f(x\cdot1)=x\cdot f(1)$. Thus if $R$ is a ring such that its abelian group admits an automorphism that is not of the form $f\colon x\mapsto c\cdot x$ for some $c\in R$, then $R$ does not have the desired property. Thus we can restrict our focus to rings $R$ such that every automorphism (as additive group) is a multiplication by a constant. Among these are $\Bbb Z$, $\Bbb Z/n\Bbb Z$, and $\Bbb Q$, but right now I don't see a) that these are all or b) that this implies the property – Hagen von Eitzen Feb 7 at 17:15
• It seems like we can also any subring of $\mathbb Q,$ such as $\mathbb Z[1/2]$ has this property? I could be wrong. – Thomas Andrews Feb 7 at 17:30