# Given a symmetric polynomial in F_q, write it in terms as elementary symmetric polynomials. How to find out the coefficient?

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 $$E_{l,s}(x) = \sum_{\substack{l_1+l_2+\cdots +l_s=l \\ l_i\geqslant 1}}\ \sum_{1\leqslant i_1 where $$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.

• 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... – David E Speyer Nov 8 '19 at 16:31
• @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 $F_q$. – Max Alekseyev Nov 8 '19 at 21:47
• 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 $\prod_{i=1}^N(1-x_it)$ – zgczgczgczgczgc Nov 9 '19 at 7:44
• @zgczgczgczgczgc: Yes, you can set $x_i:=r^i$ to get the formula in my previous comment. – Max Alekseyev Nov 9 '19 at 17:09
• @MaxAlekseyev: Yes, but actually $x_i$ may not be in $F_q$ or its finite extension. – zgczgczgczgczgc Nov 9 '19 at 18:11

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 $$M\equiv 1\pmod q$$.