TITLE: Given a symmetric polynomial in F_q, write it in terms as elementary symmetric polynomials. How to find out the coefficient?
QUESTION [5 upvotes]: Consider the finite field $F_q$, where $q$ is a power of an odd prime and $N$ is a power of $q$. We have a homogeneous symmetric polynomial
$$
E_{l,s}(x) = \sum_{\substack{l_1+l_2+\cdots +l_s=l \\ l_i\geqslant 1}}\ \sum_{1\leqslant i_1<i_2<\cdots <i_s\leqslant N}x_{i_1}^{l_1}x_{i_2}^{l_2}\cdots x_{i_s}^{l_s} \, ,
$$ 
where $l=q^a-1$ ($a\in \mathbb{N}$) and $l-q+2\leqslant s \leqslant l$. We can write it in terms of elementary symmetric polynomials $e_1, e_2, \ldots, e_l$, and it is easy to check the coefficient of $e_l$ is a constant in $F_q$. The question is how to find out this constant.
I guess it is $1$ for all $l-q+2\leqslant s \leqslant l$, and I verified it for $s=l,l-1,l-2$. However, I cannot find a way to prove it.

REPLY [2 votes]: We have $$E_{l,s}(x)=[\tau^st^l]\prod_{i=1}^N\left(1+\tau\cdot \frac{tx_i}{1-tx_i}\right).$$
The coefficient $M$ of $e_l$ in the symmetric polynomial $E_{l,s}$ is nothing but $E_{l,s}(\omega_1,\ldots,\omega_l,0,0,\ldots,0)$, where $\omega_i$'s are (complex) roots of unity of degree $\ell$. Indeed, for this sequence of values of the variables all other elementary symmetric polynomials $e_1,\ldots,e_{l-1}$ are equal to 0 and $e_l$ equals 1. So we get 
$$M=[\tau^st^l]\prod_{i=1}^l\left(1+\tau\cdot \frac{t\omega_i}{1-t\omega_i}\right)=[\tau^st^l]\frac{(1-t(1-\tau))^l}{1-t^l}=
[t^l]\frac{{l\choose s}(1-t)^{l-s}t^s}{1-t^l}=\\ {l\choose s}[t^{l-s}](1-t)^{l-s}=(-1)^{l-s}{l\choose s}.$$
In your situation $l=q^a-1$ and modulo $q$ we have ${l\choose s}=\prod_{i=1}^{l-s}\frac{q^a-i}i\equiv (-1)^{l-s}$, thus indeed $M\equiv 1\pmod q$.