Decomposing a polynomial ring into Specht Modules Let $S_{\pi}$ where $\pi$ is an integer partition of $n$, denote the Specht module corresponding to $\pi$.
I am trying to decompose the set of all homogeneous polynomials in $x_1,x_2,...,x_n$ generated linearly (over any field of characteristic zero) by the monomials of the form $x_i^2x_jx_k$ ($i,j,k$ are distinct), into Specht modules. I managed to do it for the polynomials generated by each of the following classes of monomials with $i,j,k,l$ distinct: $x_i^3x_j,x_i^2x_j^2,x_i^4,x_ix_jx_kx_l$.
Once it is achieved for ${x_i}^2x_jx_k$ a decomposition is successfully found for the space of degree 4 homogeneous polynomials in $n$ variables where $n$ is large enough, say $n\ge20$. This is the aim.


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*First $x_i^3x_j$ $(i\ne j)$: We know that $x_i^3x_j=\displaystyle \frac{x_i^3x_j+x_ix_j^3}2+\frac{x_i^3x_j-x_ix_j^3}2$
(a) The terms of the form $\displaystyle \frac{x_i^3x_j+x_ix_j^3}2$ generate linearly a space isomorphic as $S_n$-modules (the module action is by permuting indices) to the homogeneous square-free degree 2 polynomials. This is isomorphic to $S_{(n-2,2)}\oplus S_{(n-1,1)}\oplus S_{(n)}$.
(b) The terms of the form $\displaystyle\frac{x_i^3x_j-x_ix_j^3}2$ generate linearly a space isomorphic as $S_n$-modules to the second exterior power of a vector space generated by $\{x_1,x_2,...,x_n\}$ via $x_i\wedge x_j\mapsto\displaystyle\frac{x_i^3x_j-x_ix_j^3}2$. Thus this is isomorphic to $S_{(n-2,1,1)}\oplus S_{(n-1,1)}$.
So the decomposition is
$\displaystyle S_{(n-2,2)}\oplus S_{(n-2,1,1)}\oplus 2S_{(n-1,1)}\oplus S_{(n)}$
where "$2$" indicates that we have two copies of $S_{(n-1,1)}$.

*$x_i^4$: This is simply a vector space generated by $x_i^4$, and is a direct sum of the standard and the trivial representations of $S_n$ that is $S_{(n-1)}$ and $S_{(n)}$. Thus the decomposition is $S_{(n-1,1)}\oplus S_{(n)}$.

*$x_ix_jx_kx_l$: These generate the module isomorphic to module $M_\lambda$ as in Bruce Sagan's book "The Symmetric Group" where $\lambda=(n-4,4)$ which one figures is just $S_{(n-4,4)}\oplus S_{(n-3,3)}\oplus S_{(n-2,2)}\oplus S_{(n-1,1)}\oplus S_{(n)}$.

*$x_i^2x_j^2$: These generate the module isomorphic to module $M_\lambda$ where $\lambda=(n-2,2)$ which one figures is just $S_{(n-2,2)}\oplus S_{(n-1,1)}\oplus S_{(n)}$.
The reason behind showing above is that these are alarmingly simple deductions though I can't seem to find one nearly as slick for the class $x_i^2x_jx_k$. I found that there is a submodule isomorphic to $M_{(n-3,3)}$ inside this class of polynomials. But that is the closeset I could get.
I tried dimension count as well, because the homogeneous degree-4 polynomials are of dimension ${n+3\choose 4}$.
The terms in the decompositions including the $M_{(n-3,3)}$ above sum up to
$\displaystyle S_{(n-4,4)}\oplus 2S_{(n-3,3)}\oplus 4S_{(n-2,2)}\oplus S_{(n-2,1,1)}\oplus 6S_{(n-1,1)}+5S_{(n)}$.
This has a total dimension $n^4-2n^3+35n^2+38n$ which, subtracted from ${n+3\choose4}$ is $\frac13(n^3-3n^2-4n)$ which should be the sum of the dimensions of the remaining irreducible components (that is, copies of Specht modules).
But this counting technique leads to many possible decompositions and I am kind of out of ideas on this. Could anyone help?
 A: 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.
A: 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 $$((n-1) + (n-2,1) + (n-3,2))*(1)$$ $$ = (n) + (n-1,1) + (n-2,2) + (n-1,1) + (n-2,2) + (n-2,1,1) + (n-2,2) + (n-3,3) + (n-3,2,1)$$
