Find a generating set for the augmentation ideal of a group ring

Let R be a commutative ring with 1 \neq 0 and G a finite group. Prove that the augmentation ideal in the group ring R[G] is generated by \{ g-1 \ |\ g \in G \}. Prove that if G = \langle \sigma \rangle is cyclic then the augmentation ideal is generated by \sigma - 1.

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.

\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

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).

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