find a $k[G]$-module basis for a $k$ module with $G$ action Given a ring $k$, a finite group $G$ and a free $k$-module $M$ with a free action of $G$, why is $M$ a free module over the group ring $k[G]$? (how do I find a $k[G]$ basis for $M$?)
 A: Since $G$ acts freely, $M=\oplus_i \mathbb{Z}[G]e_i$ for some $\{e_i\}\subset M$, as $\mathbb{Z}[G]$ module.
Now, as $R[G]$ module, $M=R\otimes_\mathbb{Z}(\oplus_i \mathbb{Z}[G]e_i/\thicksim)=\oplus_{j\in J} R[G]\overline{e_j}$, here $\overline{e_j}$ is an equivalent class deduced by the following relation
$e_i\thicksim' e_k$ iff $\exists r\in R$, $re_i=e_k$ or $re_k=e_i$.
Deduce process: $e_i\thicksim e_s$ iff $e_i\thicksim'e_{k_1}\thicksim'\cdots \thicksim'e_{k_m}\thicksim'e_s$ for some sequence $\{e_{k_r}\}$.
$\{\overline{e_j}\}_{j\in J}$ is the basis you want.
Note. Using this way, $R$ should be PID (principal integral domain) to guarantee $\overline{e_j}$ can be $R$ generated by 1 element, because $M$ is free $R$ module.

Another way.  Exchange the order of $G$ and $R$.
Since $M$ is free $R$ module, $M=\oplus_i Re_i$ as $R$ module
Now as $R[G]$ module, $M=\mathbb{Z}[G]\otimes_\mathbb{Z}(\oplus_i Re_i/\thicksim)$
where $\thicksim$ is given by
$e_i\thicksim e_k$ iff $\exists g\in G,~ge_i=e_k$
Since $g^{-1}$ exists this time, this is an equivalent relation and need not to do the deduce process.
Let $\overline{e_j}$ be the equivalent class again, we have
$M=\oplus_jR[G]\overline{e_j}$.  $(*)$
Note. Using this way, $R$ should be an integral domain (ID) to guarantee $(*)$ is a direct sum. The second way is better.
