Generators of the Augmentation Ideal in a Group Ring

Problems and solutions of ring theory in abstract algebra

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 $g-e$.
 
LoadingAdd 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 $g-e$.


LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Ring theory
Problems and solutions of ring theory in abstract algebra
There is Exactly One Ring Homomorphism From the Ring of Integers to Any Ring

Let $\Z$ be the ring of integers and let $R$ be a ring with unity. Determine all the ring homomorphisms...

Close