The are some $R$ for which the integers are a free $R$-module. (For example, when $R = Z$.) But there are other cases when the integers are not. (E.g. take $R = Z[G]$ with the action induced by the trivial action of $G$ on $Z$, and pick $G = Z/2$. Then if $Z$ were free then we could factor the identity map on $Z$ through the augmentation map $Z[G] \rightarrow Z$ (since free modules are projective) but this is impossible because of the action of $Z[G]$ on itself.)
Is it known for what $R$ (and what actions of $R$) equip the integers into a free $R$-module? (Or more specificially, for what groups $G$ with any action are the integers a free $Z[G]$ module?)
I stumbled on this question after doing some calculations with group cohomology, but I couldn't find the answer after some cursory internet searches.
If $\mathbb{Z}$ is a free $R$-module, then as a group, $\mathbb{Z} \cong \bigoplus_n R$. $\mathbb{Z}$ isn't a direct sum, except trivially, so $\mathbb{Z} \cong R$ as groups. So addition on $R$ has the same group structure as addition on $\mathbb{Z}$. But the integer ring is the unique ring structure on the group $\mathbb{Z}$.