User Eric Gregor and I were talking in chat and he mentioned this question and postulated the possibility of an approach through symmetric polynomials. After some thinking, I came to this:
Hypothesis. For any $n$, the elementary symmetric polynomial $e_n\in k[x_1,\cdots,x_n]$ can be expressed as a $k$-linear combination of $n$th powers of degree one homogeneous polynomials.$^\dagger$
$^\dagger$We assume the characteristic does not divide $n!$. We could assume it's zero for more simplicity.
Let us denote $s(a,b,\cdots,c)=(a+b+\cdots+c)^n-(a^n+b^n+\cdots+c^n).$ I have two examples:
$xy=\frac{s(x,y)}{2} \tag{$n=2$}$
$xyz=\frac{s(x,y,z)-\big(s(x,y)+s(y,z)+s(z,x)\big)}{6} \tag{$n=3$}$
My scratchwork was getting tedious so I didn't finish the $n=4$ case. Besides this I haven't really made any substantive headway, but I did derive the following equality. We denote $\mathrm{pt}\,\lambda$ the number of parts of an integer partition $\lambda\vdash n$, and $m_\lambda$ the sum of all monomials of shape $\lambda$ in $x_j,j\in J$.
$T_{J,\ell}:=\sum_{\large I\subseteq J \atop \large |I|=\ell}\left(\sum_{i\in I}x_i\right)^n=\sum_{\large \lambda\vdash n \atop \large \mathrm{pt}\lambda\le\ell}\binom{n}{\lambda}\binom{|J|-|I|}{\ell-\mathrm{pt}\,\lambda}m_\lambda.$
This can be justified as follows: expanding the inner summands of the LHS with the multinomial theorem will result in the terms $m_\lambda$ (with the appropriate multinomial coefficents), times the count of supersets $I$ ($\subseteq J$) of cardinality $\ell$ containing a particular subset $K$ of cardinality $\mathrm{pt}\,\lambda$; construct such $I$ by choosing $|K/I|$ elements out of $|J/I|$ available. In our context $J=[n]$ of course.
The reason I mention this is that the $T_{[n],\ell}$'s appear to be relevant in the computations I was going through for $n=2,3,4$ (as if inverting a linear system in the $m_\lambda$'s...). It may or may not be the correct way of thinking about the problem. I guess my question is then:
- Is the hypothesis correct?
- If so, how would we prove it?
- (Optional) If this isn't already inherently answered in the hypothetical proof, how would we explicitly compute what the combinations of powers are?