# Tag Info

7

$\mathbb{Z}$-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 ...

2

Let $A = \mathbb Z$ and $M = \mathbb Z/2\mathbb Z$.

1

I think the following should work, but take it with a grain of salt. Let $\mathbb Z(p^\infty)$ denote the $p$-Prüfer group, that is, the injective envelope of $\mathbb Z(p)$. The module $I=\prod_p \mathbb Z(p^\infty)$ is an injective module containing your module $M=\prod_p \mathbb Z(p)$ as a submodule. Now consider the module  E = M + \bigoplus_p \mathbb ...

Only top voted, non community-wiki answers of a minimum length are eligible