FALSE: reduction mod p of an ordinary irreducible is always irreducible

The space $K{\otimes }_{\mathcal{O}}L$ is the two-dimensional sum-zero subspace of $K^3$. The cycle $(123)$ has distinct eigenlines there with eigenvalues ${\zeta }_3$ and ${\zeta }_3^2$ (for example, they are spanned by $(1,{\zeta }_3^2,{\zeta }_3)$ and $(1,{\zeta }_3,{\zeta }_3^2)$). A transposition conjugates the cycle to its inverse and swaps these eigenlines. Thus neither line is stable under $S_3$; a proper nonzero submodule of a two-dimensional representation would be a stable line, so this ordinary representation is irreducible.

By [F1], reducing modulo the maximal ideal gives the $k{S}_3$-module $\bar{L}=L\mathrm{/}(1-{\zeta }_3)L\mathrm{.}$ Coordinate reduction identifies $\bar{L}$ with the two-dimensional sum-zero subspace of $k^3$: if all coordinates of a vector in $L$ are divisible by $1-{\zeta }_3$, dividing them by this element still gives a vector in $L$. The image of $(1,1,-2)\in L$ is the nonzero vector $(1,1,1)$ because $k={\mathbb{F}}_3$. It is fixed by every permutation matrix, so its line is a nonzero proper invariant subspace of $\bar{L}$. Thus the reduction is reducible.

Thus an ordinary irreducible representation can have reducible reduction modulo $p$. In the decomposition process recorded by [F2], this reduction is therefore not irreducible. The statement is false.