Ideals of non semi-simple group rings. I worked for a long time on complex group rings and complex twisted group rings. In those cases the algebra is semi-simple and its structure is well understood from the decomposition to irreducible (projective in the twisted case) representations. For example
$$\mathbb{C}S_3\cong \mathbb{C}\oplus \mathbb{C}\oplus M_2(\mathbb{C}).$$
Now I am trying to deal with a non-simple case in which the group is non-commutative (in the commutative case it is much easier).
Now, I am stuck in the following example.
Let 
$$G=C_7\rtimes C_3,$$
where the action of $C_3$ on $C_7$ is by sending its generator $\sigma$ to $\sigma ^4$.
Describe (as best as you can) the ring structure of the group ring
$$\mathbb{F}_3G.$$
Here the group ring is not semi-simple. However, I am trying to find a maximal (length) chain of ideals $I_0,I_1,\ldots ,I_k$ such that
$$\{0\}=I_0\subseteq I_1 \subseteq I_2 \subseteq \ldots \subseteq I_k=\mathbb{F}_3G.$$
So far I made no progress.
Thanks in advance for any help.
 A: To understand the explicit structure of group algebras like this, a useful idea is that of a skew group ring, which is a ring theoretic analogue of a semidirect product.
Let $G$ be a group acting by ring automorphisms on a ring $A$. Then the elements of the skew group ring $A\ast G$ are formal finite sums $\sum_{g\in G}a_gg$ with $a_g\in A$, with multiplication following from the rule that $gag^{-1}$ is the result of acting by $g$ on $a$, for $g\in G$ and $a\in A$. So, for example, if a group $G$ acts on another group $H$, inducing an action on the group algebra $kH$ for any field $k$, then the group algebra $k[H\rtimes G]$ of the semidirect product is the same as the skew group algebra $(kH)*G$.
Two special cases suffice to understand the group algebra of the question.
First, if $G$ acts trivially on a field $k$, then clearly $k\ast G$ is just the normal group algebra $kG$.
Second, if a finite group $G$ of order $n$ acts faithfully on a field $K$ with fixed field $k$, so $[K:k]=n$, then $K\ast G$ acts by $k$-linear endomorphisms on $K$ by 
$$\left(\sum_{g\in G}\lambda_gg\right)(\mu)=\sum_{g\in G}\lambda_gg(\mu),$$
inducing a map of $k$ algebras 
$$K\ast G\to\operatorname{End}_k(K)\cong M_n(k)$$
which is injective since the set of field automorphisms of a field $K$ is linearly independent over $K$, and therefore an isomorphism by considering dimensions.
Going back to the example in the question, $\mathbb{F}_3C_7\cong\mathbb{F}_3\times\mathbb{F}_{3^6}$ as $\mathbb{F}_3$-algebras (this follows from the fact that $\mathbb{F}_{3^6}$ is the smallest extension of $\mathbb{F}_3$ containing a primitive seventh root of $1$), and $C_3$ acts trivially on the first factor and faithfully on the second (with fixed field $\mathbb{F}_9$), so
$$\mathbb{F}_3[C_7\rtimes C_3]\cong (\mathbb{F}_3\ast C_3)\times(\mathbb{F}_{3^6}\ast C_3)
\cong\mathbb{F}_3C_3\times M_3(\mathbb{F}_9).$$
A: I don't know much about this topic, but I did a quick calculation in Magma on $A = {\mathbb F}_3G$. It appears to be a direct sum of a simple algebra of dimension $18$ and a uniserial algebra of dimension $3$ with $3$ trivial composition factors. If you do this over the field of order $9$, then the $18$-dimensional algebra splits into a direct sum of two $9$-dimensionals.
I am sure that an expert in modular representation theroy could explain all of this! The projective indecomposable modules consist of a $3$-dimensional module with three trivial consitutents and an irreducible $6$-dimensional modal that splits in half over ${\mathbb F}_9$.
