Reducibility of Representations over Finite fields So, there are several standard ways of proving irreducibility/reducibility for representations over fields where the characteristic doesn't divide $|G|$ such as Maschke's theorem, jordan normal form, using eigenvectors, etc. However, I was wondering, how does this change over a field $F$ of characteristic that does divide $|G|$? Is there any simple example of how one would approach this? As an example, how would one show that the permutation representation of the symmetry group $S_{3}$ is not completely reducible over a field $F$ of characteristic 3?
 A: Let $V = k^3$ be the permutation representation of $S_3$ and consider $M = k\left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right]$.  This is a one dimensional invariant subspace.  If $V$ is completely reducible then there is some two dimensional invariant subspace $N$ such that $V = N \oplus M$.
Assume such an $N$ exists and assume $\left[\begin{smallmatrix} a \\ b \\ c \end{smallmatrix}\right]$ is a vector in $N$.  As $N$ is invariant I can apply the permutation $(1 \ 2 \ 3)$ to this and get that $\left[\begin{smallmatrix} c \\ a \\ b \end{smallmatrix}\right]$ is in $N$.  Similarly $\left[\begin{smallmatrix} b \\ c \\ a \end{smallmatrix}\right]$ is in $N$.  Add them all together and I get that $(a + b + c)\left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right]$ is in $N$.  But $N \cap M = 0$ so it must be that $a + b + c = 0$.
The set of vectors satisfying $a + b + c = 0$ is a two dimensional subspace of $V$ and it contains $N$, so it must equal $N$.  But it also contains $M$, and this contradicts the assumption that $N \cap M = 0$.
