**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.2**

Solution: Recall that the augmentation ideal of $R[G]$ is the kernel of the ring homomorphism $R[G] \rightarrow R$ given by $\sum r_ig_i \mapsto \sum r_i$; that is, it consists of all elements in $R[G]$ whose coefficients sum to 0 in $R$.

Let $S = \{ g-1 \ |\ g \in G \}$, and let $A$ denote the augmentation ideal of $R[G]$. First, note that $g-1 \in A$ for each $g \in G$, so that $(g-1) \subseteq A$. Thus $(S) \subseteq A$. Now let $\alpha = \sum_{g_i \in G} r_ig_i \in A$; then we have $\sum r_i = 0$. Consider the following.\begin{align*}\sum_{g_i \in G} r_i(g_i - 1) =&\ \sum_{g_i \in G} r_ig_i - r_i\\=&\ \left( \sum_{g_i \in G} r_ig_i \right) - (\sum_{g_i \in G} r_i)\\=&\ \alpha - 0= \alpha\end{align*}Thus $A \subseteq (S)$, and we have $A = (S)$.

Suppose further that $G = \langle \sigma \rangle$ is cyclic. We still have $(\sigma - 1) \subseteq A$. Now note that $$(\sigma - 1)\left( \sum_{t=0}^k \sigma^t \right) = \sigma^{k+1} - 1,$$ so that $\sigma^k - 1 \in (\sigma - 1)$ for all $k$. Thus $A = (\sigma - 1)$.