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?


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    $\begingroup$ $R=\mathbb{Q}$ seems to work as well. $\endgroup$ – Mindlack Feb 7 at 16:17
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    $\begingroup$ 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 $\endgroup$ – Hagen von Eitzen Feb 7 at 17:15
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    $\begingroup$ It seems like we can also any subring of $\mathbb Q,$ such as $\mathbb Z[1/2]$ has this property? I could be wrong. $\endgroup$ – Thomas Andrews Feb 7 at 17:30

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