Decomposing a polynomial ring into Specht Modules

The span of the monomials of the form $x_i^2x_jx_k$ is the Young permutation module $M^{(n-3,2,1)}$. (Proof. Observe that $x_1^2x_2x_3$ has stabiliser $\langle (2,3)\rangle \times S_{\{4,\ldots,n\}}$, so the relevant Young subgroup is $S_{n-3} \times S_2 \times S_1$.) Using Kostka numbers (equivalently, multiplicities of Schur functions in complete symmetric funtions) this decomposes as

$$M^{(n-3,2,1)} \cong S^{(n-3,2,1)} \oplus S^{(n-3,3)} \oplus S^{(n-2,1,1)} \oplus 2S^{(n-2,2)} \oplus 2S^{(n-1,1)} \oplus S^{(n)}$$

provided that $n \ge 6$. (I'm using superscripts for Specht modules since this is the notation I'm used to.)

To make this explicit, Specht's original construction of Specht modules shows that $S^{(n-3,2,1)}$ is generated by the product of Vandermonde determinants

$$\left| \begin{matrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ x_1^2 & x_2^2 & x_3^2 \end{matrix} \right| \left| \begin{matrix} 1 & 1 \\ x_4 & x_5 \end{matrix}\right| $$

and it's clear that the unique trivial submodule is spanned by

$$x_1^2x_2x_3 + x_1^2x_2x_4 + \cdots + x_{n-2}x_{n-1}x_n^2,$$

i.e. the sum of all monomials whose exponents are $2$, $1$, $1$, $0, \ldots, 0$ in some order.

For the other factors it is harder to make them explicit, but this can be done by using semistandard homomorphisms (see for instance James' lecture notes). Since we're working in characteristic zero, any non-zero homomorphism from a Specht module must be injective. I'll give some details here. By this theory,

$$\mathrm{Hom}_{\mathbb{C}S_n}(S^{(n-1,1)}, M^{(n-3,2,1)})$$

is spanned by the semistandard homomorphisms for the two semistandard tableaux of shape $(n-1,1)$ and content $(n-3,2,1)$: these have rows $1\ldots122,3$ and $1\ldots123$, $2$, respectively. Each such homomorphism extends to $M^{(n-1,1)}$. Taking as a model for $M^{(n-1,1)}$ the natural permutation module $\langle e_1,\ldots, e_n\rangle$, the corresponding extended homomorphisms are

$$e_n \mapsto (x_1x_2^2 + x_1x_3^2 + \cdots + x_{n-2}x_{n-1}^2)x_n$$

and

$$e_n \mapsto (x_1x_2+x_1x_3+\cdots + x_{n-2}x_{n-1})x_n^2,$$

respectively. One then has to restrict these homomorphisms to $S^{(n-1,1)} \subseteq M^{(n-1,1)}$ (for instance it is generated by $e_n-e_1$) to get two submodules of $M^{(n-3,2,1)}$ isomorphic to $S^{(n-1,1)}$, as in the claimed decomposition.

In general, these decompositions can be computed using the Pieri rule. (This is essentially the same as Mark's answer)

Fix an integer partition of $d$ as $\lambda = 1^{e_1} \dots r^{e_r}$. Then consider the $S_n$-set of ordered partitions into $r+1$ blocks $[n] = b_0 \sqcup \dots \sqcup b_r$ such that the $i$th block has size $e_i$ for $i > 0$ and the $0$th block has size $n - \sum_{i} e_i$. This set has a canonical bijection with a set of monomials, given by $$b_0 \sqcup \dots \sqcup b_r \mapsto (\prod_k (\prod_{i \in b_k} x_i)^k.$$

The case you are asking about is $\lambda = 2^1 1^2.$

Let us write $V_\lambda$ for the vector space with basis the $S_n$ set associated to $\lambda$. Then we can write $V_\lambda$ as an induced representation as follows $$V_\lambda = {\rm Ind}^{S_n}_{S_{n - \sum_{i} e_i} \times \prod_{i = 1}^r S_{e_i}} \left(\bigotimes_{i= 0}^r {\rm triv} \right).$$

The effect of this induction product can be computed via the Pieri rule, starting with the integer partition $(n- \sum_i e_i)$ corresponding to the trivial representation of $S_{n - \sum_i e_i}$ and adding horizontal strips of length $e_1, \dots, e_r$. This should translate into a combinatorial rule for multiplicities involving skew tableaux with content specified by the $e_i$, but I have not thought this through carefully.

In your case, we begin with the partition $(n-3)$. Multiplying by the horizontal strip $(e_1) = (2)$ we get $(n-1) + (n-2,1) + (n-3,2)$. Multiplying this by $(e_2) = (1)$ we get $$((n-1) + (n-2,1) + (n-3,2))*(1)$$ $$ = (n) + (n-1,1) + (n-2,2) + (n-1,1) + (n-2,2) + (n-2,1,1) + (n-2,2) + (n-3,3) + (n-3,2,1)$$