Let $R$ be a commutative ring with identity and $G=<\sigma>$ is a cyclic group.
Prove that the augmentation ideal in the group ring $RG$ is generated by $\sigma-1$.
My attempt:
Since $R$ is commutative, we know ideal generated by $\sigma-1$ is $\{r(\sigma-1):r\in R\}$. And note that any element in Augmentation ideal can be represented by $$ \sum_{i=0}^{n-1}r_i\sigma^i $$
where $r_i\in R$ and the order of $\sigma$ is $n$. So I want to show that
$$\sum_{i=0}^{n-1}r_i\sigma^i =r(\sigma-1) $$ for some $r\in R$. But I cannot proceed anymore than this because of powers of $\sigma$. Am I in wrong direction? I will be happy with any help. Thank you in advance.
The augmentation ideal is generated as an $R$-module by the elements $\newcommand{\si}{\sigma}\si^k-1$. When $k\ge1$, $\si^k-1=(\si-1) (\si^{k-1}+\cdots+\si+1)$ lies in $\left<\si-1\right>$. When $k=0$, $\si^k-1=0$. When $k<0$, $\si^k-1=-\si^k(\si^{|k|}-1)$ which is also in $\left<\si-1\right>$, since $\si^{|k|}-1$ is.