Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $M$?)

share|cite|improve this question

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.