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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.

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    $\begingroup$ You should at least be able to single out some invariant subspaces to start with. $\endgroup$ – Tobias Kildetoft Oct 1 '19 at 17:09
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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 $\{x_1x_2-x_1x_3, x_1x_2-x_2x_3\}$.

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    $\begingroup$ I think you have one too many with degree $2$ monomials and one too few with degree $1$s, since the action preserves degree. $\endgroup$ – Tobias Kildetoft Oct 2 '19 at 3:45
  • $\begingroup$ @TobiasKildetoft Yep. That was dumb. $\endgroup$ – David Hill Oct 2 '19 at 14:51
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    $\begingroup$ In degree 1, the submodule spanned by $\{x_1-x_2,x_2-x_3\}$ is the standard reflection representation of $S_3$ (in general, the reflection representation of $S_n$ acting on $F^n$ is the $(n-1)$-dim subspace spanned by $\{x_i-x_{i+1}\mid 1\leq i<n\}$, look up Coxeter groups). $\endgroup$ – David Hill Oct 4 '19 at 15:29
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    $\begingroup$ When you have an action of a group $G$ on a finite set $X$ there is a corresponding $G$-module that acts on an $|X|$-dimensional vector space $V$ with basis $\{v_x\mid x\in X\}$ given by $g.v_x=v_{g.x}$. In this basis, the subspace spanned by $\sum_{x\in X}v_x$ is a copy of the trivial $G$-module, and the complement $$W=\left\{\sum_{x\in X}a_xv_x\mid \sum_{x\in X}a_x=0\right\}$$ is also a $G$-invariant submodule (which may or may not be irreducible). In the degree 2 case, $S_3$ acts on $\{x_1x_2,x_1x_3,x_2x_3\}$. The complement of the trivial module is the degree 2 submodule in my answer. $\endgroup$ – David Hill Oct 4 '19 at 15:34
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    $\begingroup$ @Tengu see comments above. $\endgroup$ – David Hill Oct 4 '19 at 15:36

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