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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.

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    $\begingroup$ I had an approach which hit a snag, but it still seems worth leaving the following comments. Let $F$ be the field of size $q^a$. If we evaluate the elementary symmetric polynomial $e_k$ on the set $F^{\times}$, we get $0$ for $k<\ell$ and $-1$ at $k=\ell$, so it is equivalent to evaluate $E_{\ell,s}$ on the set $F^{\times}$. (2) We can evaluate on $F$ instead, because all the terms which use the $0$ element will contribute $0$. I have more ideas after that, but I'm not sure if they are good after all... $\endgroup$ – David E Speyer Nov 8 '19 at 16:31
  • $\begingroup$ @DavidESpeyer: Your observation implies that the coefficient of $e_l$ equals the coefficient of $t^l$ in $$- e_s\left(\frac{rt}{1-rt},\frac{r^2t}{1-r^2t},\dots,\frac{r^Nt}{1-r^Nt}\right),$$ where $r$ is a primitive element of $F_q$. $\endgroup$ – Max Alekseyev Nov 8 '19 at 21:47
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    $\begingroup$ Thank you for the comments. I have figured it out, so no need to think about it anymore... The way is to set a function $$ G(y,t)=\prod_{i=1}^N (y+x_it+x_i^2t^2+\cdots) $$ and multiple by $\prod_{i=1}^N(1-x_it)$ $\endgroup$ – zgczgczgczgczgc Nov 9 '19 at 7:44
  • $\begingroup$ @zgczgczgczgczgc: Yes, you can set $x_i:=r^i$ to get the formula in my previous comment. $\endgroup$ – Max Alekseyev Nov 9 '19 at 17:09
  • $\begingroup$ @MaxAlekseyev: Yes, but actually $x_i$ may not be in $F_q$ or its finite extension. $\endgroup$ – zgczgczgczgczgc Nov 9 '19 at 18:11
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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 $M\equiv 1\pmod q$.

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