# Generators of the Augmentation Ideal in a Group Ring ## Problem 302

Let $R$ be a commutative ring with $1$ and let $G$ be a finite group with identity element $e$. Let $RG$ be the group ring. Then the map $\epsilon: RG \to R$ defined by
$\epsilon(\sum_{i=1}^na_i g_i)=\sum_{i=1}^na_i,$ where $a_i\in R$ and $G=\{g_i\}_{i=1}^n$, is a ring homomorphism, called the augmentation map and the kernel of $\epsilon$ is called the augmentation ideal.

(a) Prove that the augmentation ideal in the group ring $RG$ is generated by $\{g-e \mid g\in G\}$.

(b) Prove that if $G=\langle g\rangle$ is a finite cyclic group generated by $g$, then the augmentation ideal is generated by $g-e$. Add to solve later

## Proof.

### (a) The augmentation ideal in $RG$ is generated by $\{g-e \mid g\in G\}$.

Let $I=\ker(\epsilon)$ be the augmentation ideal and let $J$ be the ideal generated by elements of the form $g-e$, $g\in G$.
Since $\epsilon(g-e)=1-1=0$, the generator $g-e\in I$. Hence $J \subset I$.

On the other hand, to show that $I \subset J$ let $\sum_{i=1}^na_i g_i$ be an arbitrary element in the augmentation ideal $I$.
Then we have
$\epsilon(\sum_{i=1}^na_i g_i)=\sum_{i=1}^na_i=0. \tag{*}$ Then we have
\begin{align*}
\sum_{i=1}^na_i g_i&=\sum_{i=1}^na_i (g_i-e)+\sum_{i=1}^na_ie\\
&=\sum_{i=1}^na_i (g_i-e)+(\sum_{i=1}^na_i)e\\
&\stackrel{(*)}{=} \sum_{i=1}^na_i (g_i-e).
\end{align*}
Therefore, the element $\sum_{i=1}^na_i g_i$ is in the ideal $J$.
Putting the two inclusions together give $I=J$, which completes the proof of (a).

### (b) The augmentation ideal is generated by $g-e$ if $G=\langle g\rangle$ is cyclic.

Now suppose $G=\langle g\rangle$ is a finite cyclic group of order $n$.
By part (a), the augmentation ideal is generated by
$\{g^i-e\mid i=0, 1,\dots, n-1\}.$

Note that we have
$g^k-e=(g-e)(g^{k-1}+g^{k-2}+\cdots+g+e)$ for $k \geq 2$.
This implies that $g^k-e$ is contained in the ideal generated by $g-e$ for $k\geq 2$.
Hence the augmentation ideal of the cyclic group $G$ is generated by $g-e$. Add to solve later

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