Symmetric Polynomials and Automorphisms of Complex Polynomial Rings

I asked a version of this question earlier, but it was very imprecise and poorly formatted, so I decided to create a new question.

Suppose we have an ordered 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 $\mathbb{C}[x_1,x_2,...x_n]$.

I'm trying to show that $\forall P_i,P_j \in P$, there exists an automorphism $\phi$ of $\mathbb{C}[x_1,x_2,...x_n]$ such that $\phi(P_i)=P_j$. In fact, there exist $\phi$'s that form subgroup of the automorphism group of $\mathbb{C}[x_1,x_2,...x_n]$ that is isomorphic to $S_n$, where each $\phi$ corresponds to a permutation of $P$.

For example, for $n=3$, working in $\mathbb{C}[x_1,x_2,x_3]$, if we take our set $P$ to be the polynomials $(x_1+x_2),(x_2+x_3),(x_1+x_3)$, the product $(x_1+x_2)(x_2+x_3)(x_1+x_3)$ is symmetric and there exist automorphisms of $\mathbb{C}[x_1,x_2,x_3]$ that send each one to the other.

This too is vaguely phrased, but I hope I got my point across!

Here is the link to the old question:

Product of $n(n-1)/2$ polynomials of the same degree is symmetric

• You should at least link to the earlier version of this question. It is a bit different, so I don't want to rule on it being duplicate (the same goes for the answer). Anyway, if not a duplicate it is clearly either an evolved version. Often, but not always, it is preferrable to edit the first version. An important exception to that rule is if the edit makes existing answers obsolete. – Jyrki Lahtonen Jun 15 '15 at 7:24
• I did edit the previous question. However, I felt that it was all too convoluted to keep in one place. Posting the link completely slipped my mind, I will do that now. Also, if need be I will delete this one. – leibnewtz Jun 15 '15 at 7:29
• I sort of see what you mean. I need to think about this one. Don't delete! Actually, you cannot delete this on your own, but that's besides the point. – Jyrki Lahtonen Jun 15 '15 at 7:31

This conjecture is false. Choose a multiple $d=k n$ of $n$, with $k>1$, 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.
For example, if $n=3$ you can take $k=2$, and write $$x_1^2 x_2^2 x_3^2=(x_1^2)(x_2 x_3)(x_2 x_3).$$
• Thank you for your response! I am almost convinced. In your example, it is true that the product is invariant with respect to $S_3$ and that it can be divided into the appropriate number of factors, but each factor is not distinct (there are two $x_2x_3$'s). We could also (trivially) factor $x_1^2x_2^2x_3^2$ into $(x_1^2)(x_2^2)(x_3^2)$, and thereby define $6$ automorphisms of $\mathbb{C}[x_1,x_2,x_3]$ that permute the factors. Also, that is the only way to factor the monomial so that all the factors are distinct. Off the top of my head, it seems that this would hold for any $n$ and $d$. – leibnewtz Jun 15 '15 at 5:02