Showing the regular representation is decomposable/indecomposable for an explicit example. Let $G=\mathbb{Z}/3\mathbb{Z}$.
a) Show that the regular representation $V=\mathbb{F}_3 G$ of $G$ over the field $\mathbb{F}_3$ is indecomposable.
b) Provide an explicit decomposition of the regular representation into irreducibles over the field $\mathbb{F}_2$.

For part a), I have found an explination a more general case here, where part of the proof claims that if $\mathbb{F}_3 G=M_1\oplus M_2$ is decomposable, then each $M_i$ contains the trivial subrepresentation. How do we know that each $M_i$ contains the trivial subrepresentation? Why can't they be irreducible?
For part b), I believe the decomposition is $\mathbb{F}_2 G=I\oplus N$, where $I=\{\alpha(g_1+g_2+g_3)\mid\alpha\in\mathbb{F}_2\}$ is the trace ideal (trivial representation) and $N=\{\alpha_1g_1+\alpha_2g_2+\alpha_3g_3\mid\alpha_i\in\mathbb{F}_2,\sum_i\alpha_i=0\}$ is the augmentation ideal. If we try the analogous decomposition for part a) as in part b) (i.e., $\mathbb{F}_3 G=$ trace ideal $\oplus$ augmentation ideal), I see that these two ideals have nontrivial intersection (since $|G|=3$ divides the characteristice of $\mathbb{F}_3$), so this is not correct. In particular, the trace ideal is a submodule of the augmentation ideal. However, it seems to me that the augmentation ideal is $G$-stable (hence irreducible) regardless of whether the $|G|$ divides the characteristic. Where is my logic going wrong in this seeming contradiction?
 A: $\DeclareMathOperator{\F}{\mathbb F_3}\DeclareMathOperator{\kk}{\mathbb k}$
Let $x$ be a generator of $C_3$. I claim there is an $\F$-algebra isomorphism $\F[t]/(t^3) \longrightarrow \F[C_3]$.
Indeed, consider the map $t\longmapsto x-1$. Since $\F$ has characteristic $3$, we compute directly that $(x-1)^3 = x^3-1= 0$ so this map is indeed well defined. Since $t+1$ maps to $x$, we see this map is surjective, and by counting dimensions we see it is injective.
One now considers the action of $t$ on $\F[t]/(t^3)$ on the basis $\{1,t,t^2\}$. The matrix you get is the usual $3\times 3$ nilpotent Jordan block, which makes this action indecomposable: a direct sum decomposition would mean that this block is nilpotent of order less than three.
For the second part you are right: that is the correct decomposition, but in the modular case it is not a direct sum decomposition. In particular, this copy of the trivial representation has no complement in the group algebra, as you noted.
But when $|G|$ and $\mathrm{char}\kk$ are coprime it is a direct sum decomposition, so this works for $G=C_3$ and $\kk=\mathbb F_2$. More precisely, you can consider the basis given by $e_1 = 1+t+t^2$, $e_2 = 1 + t^2$ and $e_3 = t + t^2$ and the subspaces $V = \langle e_1\rangle$ and $W = \langle e_2,e_3\rangle$. You can check directly that $te_2 = e_2 + e_3$ and that $te_3 = e_2$. Thus, the matrix of for this subrepresentation is
$$A = \begin{pmatrix}
0 & 1 \\
1 & 1
\end{pmatrix}$$
while $V$ is just the trivial representation. In particular, you can see directly that since $\chi_A(t) =  t^2+t+1$ has no roots in $\mathbb F_2$, the two dimensional subspace $W$ has no one dimensional subrepresentations, and thus it is irreducible. Of course $C_3$ is Abelian, but since $\mathbb F_2$ is not algebraically closed some irreducibles do not have dimension one!
Add. Let $G$ have order divisible by $p$ and $\kk$ have characteristic $p$. In $\kk[G]$, consider the sub-representation
$V$ given by the kernel of the algebra map $\varepsilon : \kk[G] \longrightarrow \kk$ such that $\varepsilon(g)=1$ for all $g\in G$. Clearly $V$ is a proper submodule of codimension $1$. Let $\eta = \sum_{g\in G} g$ and notice that $\eta\in V$.
If $W$ is a non-zero submodule of $\kk[G]$ and $w\in W$ is non-zero then $\eta w = \varepsilon(w)\eta$. If this element is zero
then $w\in V$ and $V\cap W\neq 0$, so we may assume that this
element is non-zero. If not, then since $\eta\in V$, we see that this
element is both in $W$ and in $V$, so $W\cap V\neq 0$ in any case. Thus, no pair of non-trivial submodules in $\kk[G]$ have trivial intersection.
Moreover, we have proved that if $W$ is any submodule such that $W\subsetneq V$, then $W$ must contain the trivial representation spanned by $\eta$. Indeed, if $W\subseteq V$ then picking $w\in W\smallsetminus V$ the element $\eta w = \varepsilon(w)\eta\neq 0$ is in $W$.
