Reducing a standard integral lattice for S3 modulo 3 produces a reducible kS3-module

The extension $K\mathrm{/}{\mathbb{Q}}_3$ is totally ramified of degree $2$, with uniformizer $1-{\zeta }_3$, valuation ring $\mathcal{O}$, and residue field $k={\mathbb{F}}_3$. The field $K$ splits the subgroups of $S_3$: it contains the values needed for the cyclic subgroups, and the trivial, sign, and standard representations split $S_3$. If $V$ is a simple $k{S}_3$-module and $g$ generates the normal subgroup $C_3$, then $(g-1{)}^3=g^3-1=0$, so $V^{C_3}\mathrm{\ne }0$; normality and simplicity give $V^{C_3}=V$. Thus $V$ factors through $S_3\mathrm{/}{C}_3\cong C_2$ and is trivial or sign. The same calculation handles the subgroups, so $k$ also splits all of them. Hence $(K,\mathcal{O},k)$ is a splitting $3$-modular system for $S_3$.

The lattice $L$ is free of rank $2$ over $\mathcal{O}$, with basis $(1,0,-1),(0,1,-1)$, so [F1] and [L1] give the two-dimensional $k{S}_3$-module $\bar{L}=L\mathrm{/}\mathfrak{m}L$. The vector $v:=(1,1,1)\in k^3$ lies in $\bar{L}$ because it is the reduction of $(1,1,-2)\in L$, and it is fixed by every permutation in $S_3$.

The nonzero line $kv\subseteq \bar{L}$ is therefore an $S_3$-stable proper submodule of the two-dimensional module $\bar{L}$. Hence $\bar{L}$ is reducible.