A question on symmetric functions

Let $0 \leq m \leq n$ be integers. The group $S_n$ of permutations acts on the ring $\mathbb{Z}[X_1,\dots,X_n]$ by permuting the coordinates, with fixed subring $\mathbb{Z}[\sigma_1,\dots,\sigma_n]$, where the $\sigma_j$'s are the elementary symmetric functions.

Let $H$ be the stabilizer of $P = \prod_{j=m+1}^n X_j^m$, which is the subgroup of permutations leaving $[m+1,n]$ globally invariant, and consider the expansion $$\prod_{\sigma \in S_n/H} (T - {}^{\sigma} P) = T^{\binom{n}{m}} + a_1 T^{\binom{n}{m} - 1} + a_2 T^{\binom{n}{m}-2} + \dots +a_{\binom{n}{m}}.$$ All $a_j$'s are fixed by $S_n$, hence can be written as $a_j = A_j(\sigma_1,\dots,\sigma_n)$, for a unique polynomial $A_j$.

Is it always true that the degree of $A_j$ is at most $jm$ ?

Note that if we give weight $j$ to $\sigma_j$, then $A_j$ is easily seen to have degree $jm(n-m)$, but here I'm asking for the degree of $A_j$ where we give weight $1$ to $\sigma_j$.

The answer is "yes" (and easy) when $m=0,1,n-1,n$. I also checked the case $m=2,n=4$, so that one can assume $n \geq 5$. I have strong reasons to believe, or at least to expect, that the result is true in general.

• If I understand you correctly, then $a_j$ is $\pm$ the sum of all possible products of $j$ distinct $n-m$-element subsets of $X_1, X_2, \ldots, X_n$. This is a symmetric polynomial of degree at most $j\left(n-m\right)$, so the degree of $A_j$ is also at most $j\left(n-m\right)$ (by a standard claim in the fundamental theorem on elementary symmetric polynomials). – darij grinberg Apr 8 '16 at 18:49
• But in $P$, the $X_j$'s are raised to the power $m$... – js21 Apr 8 '16 at 22:42
• Oh, I see. OK, let me first try to generalize your question. I let $e_0, e_1, e_2, \ldots$ be the elementary symmetric functions in countably many indeterminates $x_1, x_2, x_3, \ldots$. (These are symmetric functions -- thus, power series, not polynomials -- but I can of course evaluate them at finitely many values.) I claim that $e_j\left(\left(x_{i_1}^d x_{i_2}^d \cdots x_{i_k}^d\right)_{1 \leq i_1 < i_2 < \cdots < i_k}\right)$ can be written as a polynomial of degree $\leq jd$ in $e_1, e_2, e_3, \ldots$ for any $j$, $d$ and $k$. (Here, ... – darij grinberg Apr 9 '16 at 2:59
• ... I use the notation $e_j\left(\mathfrak{f}\right)$ for the result of evaluating the symmetric function $e_j$ at the entries of a finite family $\mathfrak{f}$ of elements of a commutative ring.) Your question follows from my claim when we set $d = m$ and $k = n-m$ and set all $x_i$ with $i > n$ to zero. So it remains to prove my claim. The first step is to reduce it to the case when the ground ring is $\mathbb{Q}$: Indeed, it is clear that $e_j\left(\left(x_{i_1}^d x_{i_2}^d \cdots x_{i_k}^d\right)_{1 \leq i_1 < i_2 < \cdots < i_k}\right)$ can be written as a ... – darij grinberg Apr 9 '16 at 3:01
• ... polynomial in $e_1, e_2, e_3, \ldots$; the degree of this polynomial surely does not change when we pass from $\mathbb{Z}$ to $\mathbb{Q}$. Now that we are over $\mathbb{Q}$, we can use Newton's formulas to reduce the problem to the following one: Show that $p_j\left(\left(x_{i_1}^d x_{i_2}^d \cdots x_{i_k}^d\right)_{1 \leq i_1 < i_2 < \cdots < i_k}\right)$ can be written as a polynomial of degree $\leq jd$ in $e_1, e_2, e_3, \ldots$, where $p_j$ is the $j$-th power-sum symmetric function. This is a lot simpler, ... – darij grinberg Apr 9 '16 at 3:03