How to decompose a module of $S_3$ as a direct sum of irreducible modules?

Question

Let $A=\mathbb{C}[x_1,x_2,x_3]/I$ be the polynomial algebra $\mathbb{C}[x_1,x_2,x_3]$ by the ideal generated by $x_1^2, x_2^2, x_3^2$. Then $\dim A=8$ and it has a basis $1,x_1,x_2,x_3,x_1x_2, x_1x_3, x_2x_3, x_1x_2x_3$. The symmetric group $S_3$ acts on $A$ by $s_i x_j= x_{s_i(j)}$, $i \in \{1,2\}$, $j \in \{1,2,3\}$, where $s_1, s_2$ are transposition. How to decompose $A$ as a direct sum of irreducible representations of $S_3$? Thank you very much.

Answer 1

As pointed out in the comments, it is straightforward to find the trivial submodules spanned by symmetric polynomials: $\{1\}$, $\{x_1+x_2+x_3\}$, $\{x_1x_2+x_1x_3+x_2x_3\}$ and $\{x_1x_2x_3\}$. There are also two 2-dimensional submodules spanned by $\{x_1-x_2,x_2-x_3\}$ and $\{x_1x_2-x_1x_3, x_1x_2-x_2x_3\}$.

Answer 2

I try to create an answer to summarize the comments and add some details.

As already noted, $A = \langle 1 \rangle \oplus \langle x_1, x_2, x_3 \rangle \oplus \langle x_1x_2, x_2x_3, x_3x_1 \rangle \oplus \langle x_1x_2x_3 \rangle$. The subrepresentations $\langle 1 \rangle$ and $\langle x_1x_2x_3 \rangle$ are trivial.

The subspaces $\langle x_1, x_2, x_3 \rangle$ and $\langle x_1x_2, x_2x_3, x_3x_1 \rangle$ are isomorphic to the defining representation of $S_3$, i.e., the representation on $\mathbb{C}^3$ where each $\sigma \in S_3$ acts on the canonical base as $\sigma(e_i) = e_{\sigma(i)}$, equivalently, $\sigma$ acts as its associated permutation matrix.

The subrepresentation $\langle e_1 + e_2 + e_3 \rangle < \mathbb{C}^3$ is trivial. In $\langle x_1, x_2, x_3 \rangle$ and $\langle x_1x_2, x_2x_3, x_3x_1 \rangle$, it corresponds respectively to $\langle x_1 + x_2 + x_3 \rangle$ and $\langle x_1x_2 + x_2x_3 + x_3x_1 \rangle$.

The space $\langle e_2 - e_1, e_3 - e_1 \rangle < \mathbb{C}^3$ is the subrepresentation complementary to the trivial one and corresponds to the space $$ \Big\{ \sum_{i = 1}^3 a_i e_i: \sum_{i = 1}^3 a_i = 0 \Big\}. $$

This representation is irreducible, otherwise $\mathbb{C}^3$ would decompose into a direct sum of degree $1$ representations, hence $S_3$ would act on $\mathbb{C}^3$ through a group of diagonal matrices, which would be commutative while $3 \times 3$ permutation matrices are not.

Note that $\langle e_2 - e_1, e_3 - e_1 \rangle$ corresponds to $\langle x_2 - x_1, x_3 - x_1 \rangle$ and $\langle x_2x_3 - x_1x_3 \rangle$ in $\langle x_1, x_2, x_3 \rangle$ and $\langle x_1x_2, x_2x_3, x_3x_1 \rangle$, respectively. This completes the decomposition of $A$ in irreducible representations.

A more general approach is note that the defining representation $\mathbb{C}^n$ of $S_n$ corresponds to the permutation module $M^{(n-1, 1)}$ whose irreducile decomposition is the direct sum of the Specht modules $S^{(n)}$ and $S^{(n-1, 1)}$. For more details on this topic, look at Chapter 2 of B. E. Sagan, The Symmetric Group.