TITLE: Decomposing a polynomial ring into Specht Modules
QUESTION [9 upvotes]: 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.

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?

REPLY [4 votes]: 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 $$((n-1) + (n-2,1) + (n-3,2))*(1)$$ $$ = (n) + (n-1,1) + (n-2,2) + (n-1,1) + (n-2,2) + (n-2,1,1) + (n-2,2) + (n-3,3) + (n-3,2,1)$$