# Is it true that Sym[P]!=0 and Sym[Q]!=0 implies Sym[PQ]!=0 ?

Let $P,Q$ be homogenous polynomials in variables $x=x_1,\dots,x_n$ resp. $y=y_1,\dots,y_m.$

We know that $Sym_x[P]$ and $Sym_y[Q]$ are not identically zero. Does it follow that $Sym_{x \cup y}[PQ]$ is also not identically zero?

Here, $Sym_x$ denotes symmetrization, that is, sum over all permutations of $x.$

This is how far I got: Let $x^\alpha$ and $y^\beta$ be the first monomial in the symmetrization of $P$ resp. $Q,$ in the RevLex ordering.

Then the first monomial in $Sym_{x \cup y}[PQ]$ must be $w^\gamma$ where $w$ denotes a mixture of $x,y$ and $\gamma$ is is the sorted version of $\alpha \cup \beta.$ For example, $\alpha = 5,3,3,1$ and $\beta = 4,3,2,1$ yields $\gamma = 5,4,3,3,3,2,1,1.$

If one cannot find $\alpha'>\alpha$ and $\beta'>\beta$ such that $Sort[\alpha'\cup \beta'] = Sort[\alpha\cup \beta]$ then this means that the first monomial in the symmetrization of $PQ$ is determined ONLY by the first monomials in $P$ and $Q,$ which we know are non-zero.

-
Case 1: $\alpha_1 > \beta_1,$ then $\alpha'$ must start with $\alpha,$ since if $\beta'$ does, then $\beta'<\beta,$ a contradiction. Thus, we may reduce the problem to one with less degree of $P.$
Case 2: $\alpha_1 = \beta_1,$ then both $\alpha'$ and $\beta'$ must start with $\alpha_1$ since otherwise, one of $\alpha'$ and $\beta'$ will have too many of $\alpha_1,$ and contradicting the requirement of being greater.