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.

  • $\begingroup$ why can't $P_{ij}$ be an arbitrary symmetric function of $x_i$ and $x_j$? $\endgroup$ – Tad Jun 14 '15 at 22:42
  • $\begingroup$ I hadn't considered that. Do you mean something like $x_1^2+x_1+x_2+x_2^2$? I guess that means my question as stated is incorrect. My main motivation though was to show that each factor must be of the same form. For example, if we have the polynomial above, then another factor would have to be $x_3^2+x_1+x_3+x_1^2$. Would this be true? $\endgroup$ – leibnewtz Jun 14 '15 at 22:54
  • $\begingroup$ Also, it seems that there could only be two variables per factor $\endgroup$ – leibnewtz Jun 14 '15 at 22:55
  • $\begingroup$ Yes, that's what I had in mind. Also, it's not true that there can be only two variables per factor; you could have $P_{ij}=f(\sigma(x_1,\ldots,x_n),g(x_i,x_j),h(x_1,\ldots,\hat{x_i},\ldots,\hat{x_j},\ldots,x_n))$ where $\sigma$ is $S_n$-invariant, $g$ is $S_2$-invariant, $h$ is $S_{n-2}$-invariant and depends on all variables except $x_i$ and $x_j$, and $f$ is arbitrary. $\endgroup$ – Tad Jun 15 '15 at 2:23

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.

  • $\begingroup$ Thank you for your response! I am almost convinced. In your example, it is true that the product is invariant with respect to S3 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 $d$, as well. $\endgroup$ – leibnewtz Jun 15 '15 at 5:03
  • $\begingroup$ I am far from certain though $\endgroup$ – leibnewtz Jun 15 '15 at 5:06
  • $\begingroup$ You could also factor it as $(x_1x_2)(x_2x_3)(x_3x_1)$. $\endgroup$ – Tad Jun 15 '15 at 11:30
  • $\begingroup$ In that case you can find automorphisms of the polynomial ring in three variables that permute the factors. For example, define $\phi_1:\mathbb{C} [x_1,x_2,x_3] \to \mathbb{C}[x_1,x_2,x_3]$ by $\phi_1(x_1)=x_2$, $\phi_1(x_2)=x_1$, and $\phi_1(x_3)=x_3$. Then $\phi_1(x_1x_2)=\phi_1(x_1)\phi_1(x_2)=x_1x_2$, $\phi_1(x_2x_3)=...=(x_3x_1)$ and $\phi_1(x_3x_1)=x_2x_3$. You can similarly define five more $\phi_i$'s so that the set $\{\phi_1,\phi_2,...,\phi_6\}$ is isomorphic to $S_3$. So you're right you can factor it in multiple ways, but the only permissible ways to factor it support the conjecture $\endgroup$ – leibnewtz Jun 15 '15 at 18:55
  • $\begingroup$ A better example: $x_1^4x_2^4x_3^4=(x_1^3x_2)(x_1x_2x_3^2)(x_2^2x_3^2)$; none of the factors are equivalent under any automorphism. Incidentally it's interesting to point out that the standard $3\times3$ magic square $\begin{matrix}6&1&8\\7&5&3\\2&9&4\end{matrix}$ yields a factorization of $x_1^{15}x_2^{15}x_3^{15}$ into three factors, where all the individual exponents which occur are different. $\endgroup$ – Tad Jun 15 '15 at 21:14

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