Let $G$ be a permutation group acting doubly transitive on $\{1, \dots, n\}$, and let $\mathbb{k}$ be any field such that $\mathrm{char}(\mathbb{k})$ does not divide $n$. Then $G$ operates on $V = \mathbb{k}^n$ by permuting coordinates, and $V$ decomposes into a direct sum of the invariant submodule $\mathbb{k} \cdot (1, \dots, 1)^t$ and the augmentation module $W = \{ x \in V : \sum_i x_i = 0\}$.
If $\mathbb{k}$ is of characteristic zero, it is well known that $W$ is irreducible. An elementary proof, I can think of, generalizes to the case, where $\mathrm{char}(\mathbb{k})$ does not divide the order of a stabilizer of $G$ with respect to two elements.
Is $W$ irreducible in general?
The statement is false in general, as the following example shows:
Let $G$ be the subgroup of $S_7$ generated by $(1,2,3,4,5,6,7)$ and $(2,6)(3,7,4,5)$. $G$ is isomorphic to $\mathrm{PGL}(3,2)$ with its natural action, so $G$ is doubly transitive. Now $W$ is reducible for $\mathbb{k} = \mathbb{F_2}$, as the vector $(1,1,1,0,1,0,0)^t$ spans a $3$-dimensional submodule of $W$.