Let $\mathbb{F}$ be a finite field of characteristic $p$ and let $G$ be a cyclic group of order $p^n$. I read in a paper that all ideals of the group ring $\mathbb{F}[G]$ are of the form $I_n^j$ where $I_n$ is the augmentation ideal of $\mathbb{F}[G]$. (The authors did not expand on this). Why is this true?
By definition, the augmentation ideal is the kernel of the map $\mathbb{F}[G]\twoheadrightarrow \mathbb{F}$ given by $\sum_{\sigma\in G}c_\sigma \sigma\mapsto \sum_{\sigma\in G}c_\sigma$. So $I_n$ is clearly maximal, and it is furthermore not hard to see that it's generated by the set $\{\sigma-1\mid \sigma\in G\}$. That said, I'm having a hard time convincing myself that all ideals in $\mathbb{F}[G]$ are just powers of the augmentation ideal. It is clear that $\mathbb{F}[G]$ is isomoprhic to $\mathbb{F}[x]/(x^{p^n}-1)$, so ideals in $\mathbb{F}[G]$ biject with those in $\mathbb{F}[x]$ containing $(x^{p^n}-1)$, but I can't seem to get much further at this point.
This becomes easy once you realize the group ring as a quotient ring of the polynomial algebra: $$ \Bbb{F}[G]\simeq \Bbb{F}[x]/(x^{p^n}-1). $$ The key is that in characteristic $p$ we have, as a consequence of Freshman's Dream, the factorization $$ x^{p^n}-1=(x-1)^{p^n} $$ in $\Bbb{F}[x]$.
Let $I\subseteq \Bbb{F}[G]$ be an ideal. Pull it back all the way to the polynomial algebra via the projection map $\pi:\Bbb{F}[x]\to \Bbb{F}[G]$. The preimage $\pi^{-1}(I)$ is an ideal of $\Bbb{F}[x]$. Because that ring is a PID, the ideal is generated by the minimal degree non-zero polynomial $m(x)$ in it. but $x^{p^n}-1\in\pi^{-1}(I)$, so $m(x)\mid x^{p^n}-1$. The polynomial algebra is a UFD. Done.