Group ring ($R[G]$)

The group ring $R[G]$ consists of finite formal sums ${\sum }_{g\in G}{r}_{g}g$ with coefficients in a ring $R$. Addition is coefficientwise and multiplication extends the group operation and multiplication in $R$ distributively. A group algebra is the case in which $R$ is a field.

If $A$ is a left Noetherian ring and a right Noetherian ring, and $G$ is a poly-(cyclic or finite) group, then $A[G]$ satisfies both chain conditions. Induct along the subnormal series. A finite factor gives a finite free extension on each side, handled by finite-module extension preserves Noetherianity. An infinite cyclic factor gives a skew Laurent polynomial ring, with the twisting automorphism induced by conjugation by a lift of the cyclic generator. The noncommutative Hilbert basis theorem and clearing negative powers prove the required chain conditions at that step. In particular $\mathbb{Z}[G]$ is Noetherian.

The augmentation homomorphism $\epsilon :R[G]\to R$ sends ${\sum }_g{r}_{g}g$ to ${\sum }_g{r}_g$. Its kernel is the augmentation ideal.

In the ordinary group algebra of the additive p-adic integers, let $I$ be the augmentation ideal. The kernel of the map to the group algebra of ${\mathbb{Z}}_p/p^r{\mathbb{Z}}_p$ is generated by $[p^{r}h]-1=([h]-1{)}^{p^r}$, so it is contained in $I^{p^r}$. The quotient is the cyclic-group algebra ${\mathbb{F}}_p[T]/(T^{p^r})$, so its augmentation ideal has zero $p^r$th power and the reverse inclusion holds. Every finite degree of the associated graded ring can be read from such a quotient, giving ${\mathrm{gr}}_I{\mathbb{F}}_p[{\mathbb{Z}}_p]\cong {\mathbb{F}}_p[X]$. The completion is ${\mathbb{F}}_p[[T]]$, but completion is not needed for the graded calculation.

In characteristic $p$, the augmentation ideal $I$ of the group algebra of a finite $p$-group is nilpotent. Choose a central element $z$ of order $p$ and put $t=z-1$, so $t^p=0$. The quotient by $t$ is the group algebra of the smaller quotient group. Induction gives $I^m\subseteq tkP$ for some $m$, hence $I^{mp}=0$. For a normal $p$-subgroup $P$ of $G$, conjugation stability gives $(IkG{)}^m=I^{m}kG$, proving nilpotence of the kernel of $kG\to k(G/P)$.

The group ring of an infinite cyclic group is the Laurent polynomial ring $R=\mathbb{Z}[t,t^{-1}]$, with augmentation $\epsilon (t)=1$. Multiply any element in the kernel by a power of $t$ to obtain a polynomial vanishing at one. The factor theorem makes it divisible by $t-1$. Thus the augmentation ideal is $(t-1)R$. Since $R$ is an integral domain, multiplication by $t-1$ is an injective module map, so this ideal is a free $R$-module of rank one, even though its additive group has infinite rank.