I am trying to prove a simple fact about polynomials in the multivariate polynomial ring $\mathbb{C}[x_1,x_2,...x_n]$, for $n \gt 3$ but I've been getting stuck.
EDIT: After a comment by Tad I realized that as stated the question is incorrect. A slightly different, and I think more correct, version of the question is this: Suppose we have a set of $n(n-1)/2$ distinct polynomials $P=\{P_1,P_2,...,P_{n(n-1)/2}\}$, each of the same degree, such that the product $\prod_{i=1}^{n(n-1)/2}P_i$ is symmetric with respect to the usual action of $S_n$ on $\mathbb{C}[x_1,x_2,...x_n]$.Then $\forall P_i,P_j \in P$, there exists an automorphism $\phi$ of $\mathbb{C}[x_1,x_2,...x_n]$ such that $\phi(P_i)=P_j$. Below is the question in its original formulation.
Suppose we have a set of $n(n-1)/2$ distinct polynomials $P=\{P_1,P_2,...,P_{n(n-1)/2}\}$, each of the same degree, such that the product $\prod_{i=1}^{n(n-1)/2}P_i$ is symmetric with respect to the usual action of $S_n$ on $\mathbb{C}[x_1,x_2,...x_n]$. Then $\prod_{i=1}^{n(n-1)/2}P_i$ is either of the form $c(x_1+x_2)(x_1+x_3)....(x_{n-1}+x_n)$ or $c(x_1x_2)(x_1x_3)...(x_{n-1}x_n)$, where $c \in \mathbb{C}$; i.e. the set $P$ contains the factors of the first form or the factors of the second form.
I guess another way to put it is that the symmetric polynomials $c\prod_{i\lt j}(x_i+x_j)$ and $c\prod_{i \lt j}(x_ix_j)$ both have $n(n-1)/2$ factors (not counting $c$) and are the only symmetric polynomials in $n$ variables that have $n(n-1)/2$ factors and where the degrees of the factors are pairwise the same.
Is this true? It makes sense to me but there may be something missing. I've been trying to do it by contradiction (contradict the fact that it is a symmetric polynomial), but so far I haven't gotten anywhere. Another thought I had was to try to show that every factor must have 2 distinct variables. Any help or a counterexample would be greatly appreciated.
This conjecture is false. Choose a multiple $d=k n$ of $n$, such that $n(n-1)/2$ divides $d$ (that is, $d/\binom{n}{2} = 2k/(n-1)$ is an integer $r$.) The monomial $x_1^k\cdots x_n^k$, which has total degree $d$, can be divided up into $n(n-1)/2$ degree-$r$ factors in any number of arbitrary ways, having no invariance properties whatsoever.