Given the usual action of $S_{3}$ on $X:=\{1,2,3\}$, consider the linearization to obtain a representation of $S_{3}$ on $\mathbb{F}_{2}[X]$ (where $\mathbb{F}_{2}$ is the field with 1 element). Determine whether this representation is decomposable as a direct sum of irreducible representations (if not, explain why, if so, give the explicit decomposition).
We can encode each vector as, $(\alpha_1,\alpha_2,\alpha_3)$ where $\alpha_i \in \mathbb{F}_{2}$, and for example: $(1,2) ~ \cdot ~ (\alpha_1,\alpha_2,\alpha_3) = (\alpha_2,\alpha_1,\alpha_3)$. We know that the augmentation submodule always exist and is irreducible of degree $2$, where: $$I = \{(\alpha_1,\alpha_2,\alpha_3) \in \mathbb{F}_{2}^{3}: \alpha_1 + \alpha_2 + \alpha_3 = 0 \}.$$ We then have that: $$V = \mathbb{F}_{2}\{ (1,1,1)\},$$ is a $S_{3}$ invariant subspace, and irreducible since it is of degree $1$ (spanned by one vector). So then we have that,
$$\mathbb{F}_{2}S_{3} = V \oplus I.$$
Is my proof correct? Any issues. Thanks!