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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 $M$?)

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1 Answer 1

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.

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