Let $K$ be a field. Let $n$ be a positive integer and $P$ be a non-symmetric polynomial in $K[x_1,x_2,...,x_n]$. $S_n$ acts on $K[x_1,...,x_n]$ in an obvious way. Let $P_1,P_2,...,P_r $ be the elements of the set $\{\sigma P\}_{\sigma \in S_n}$. Can there exist a non-symmetric polynomial $f\in K[x_1,x_2,...,x_r]$ such that $f(P_1,P_2,...,P_r)$ is the zero polynomial in $K[x_1,x_2,...,x_n]$ ?

Thank you

Edit: I assume it is clear from the way my question is formulated that $|\{\sigma P\}_{\sigma \in S_n}|=r$.


1 Answer 1


Yes, that is possible. For instance, two different elements $\sigma,\tau$ of $S_n$ might act on disjoint subsets of $\{x_1,x_2,\dots,x_n\}$, potentially leading to asymmetric relations between $P, \sigma P, \tau P, \sigma\tau P$.

Example. Take $P = 2x + 2y + z + w \in k[x,y,z,w]$. The images of $P$ under the various elements of $S_4$ are $$\begin{align*} P_1 & = 2x + 2y + z + w \\ P_2 & = 2x + y + 2z + w \\ P_3 & = 2x + y + z + 2w \\ P_4 & = x + 2y + 2z + w \\ P_5 &= x + 2y + z + 2w \\ P_6 &= x + y + 2z + 2w \end{align*}$$ We're looking for polynomials $f \in k[t_1,t_2,t_3,t_4,t_5,t_6]$ such that $$\begin{align*}0 & = f(P_1,P_2,P_3,P_4,P_5,P_6).\end{align*}$$ As the $P_i$ are even linearly dependent, there are many of these polynomials. For instance $f = t_1 - t_2 - t_5 + t_6$ is one that is non-symmetric.

  • $\begingroup$ Hi. You misunderstood my question. When I said that $P_1,P_2,..,P_r$ are the elements of $\{\sigma P\}_{\sigma \in S_n}$, I mean that $P_1,...,P_r$ is the list of elements without repetition. If I allow repetition I could have just said $P_1,P_2,...,P_{n!}$ are the elements of $\{\sigma P\}_{\sigma \in S_n}$. In other words $|\{\sigma P\}_{\sigma \in S_n}|=r$ $\endgroup$
    – Amr
    Mar 31, 2014 at 12:31
  • $\begingroup$ Ok; I'll have to rethink... $\endgroup$ Mar 31, 2014 at 12:33
  • $\begingroup$ @Amr Ok, new counterexample - essentially the same idea; I'll clean up the answer later. $\endgroup$ Mar 31, 2014 at 12:59

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