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.



Is it true that Sym[P]!=0 and Sym[Q]!=0 implies Sym[PQ]!=0 ?
Ok, I found the solution:


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.


Hence, in each case, we may reduce the problem to a smaller size, induction, yada yada...


