In my course of associative algebra we covered modules and an excercise involved showing that every ring $R$ can also be viewed as an $R$-module. This was straightforward enough. Is it also true that all rings are $\mathbb{Z}$-modules? It fits the definition. Should it not bother me that a ring can be a module over two rings at the same time? Is this true for any other pair of rings?

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    $\begingroup$ All abelian groups are $\mathbf Z$-modules. $\endgroup$
    – Bernard
    Aug 27, 2015 at 13:33

1 Answer 1


$\mathbb{Z}$-modules are precisely abelian groups. As every ring is an abelian group, it is a $\mathbb{Z}$-module.

It is entirely possible to be a module over more than one ring. For example, if $M$ is an $R$-module then it is also an $S$-module for any subring $S$ of $R$ (you seem to be interested in the case where $M = R$). Another example is given by $\mathbb{Z}/m\mathbb{Z}$ which is a module over $\mathbb{Z}$ (as it is a ring), but it is also a module over $\mathbb{Z}/m\mathbb{Z}$. It is worth noting however that the properties of the module depend on which ring you are using. For example, $\mathbb{Z}/m\mathbb{Z}$ is a free module over $\mathbb{Z}/m\mathbb{Z}$, but not over $\mathbb{Z}$.


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