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For a prime $q$ and an integer $n<q$, consider working over the finite field of $q^n$ elements. Denote by $s_n^k$ the $k$-th elementary symmetric polynomial in $n$ variables. That is, $s_n^k(x_1,\ldots,x_n) = \sum_{i_1<\ldots<i_k}x_{i_1}\cdot\ldots\cdot x_{i_k}$.

I have a situation where for specific $c_1,\ldots,c_n \in \mathbb{F}_{q^n}$, $s_n^k(c_1,\ldots,c_n)=0$ for every $1 \le k \le n$. My question is, what does it imply on $c_1,\ldots,c_n$? Are they all necessarily zero? Due to Newton's identities, it also implies that the power sums is zero, but I could not continue further.

Thank you very much.

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Considering polynomial $(x-c_1)\dots(x-c_n)=x^n-s^1_nx^{n-1}+\dots+(-1)^{n-1}s^{n-1}_nx+(-1)^ns_n^n=0$ we see that $c_1=\dots=c_n=0$(that doesn't depend on field we working over)

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  • $\begingroup$ Thank you both very much, a shame that I missed that... $\endgroup$ – DanDan Oct 8 '14 at 19:46
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Hint: For all $c_1,c_2,\ldots,c_n$ (over any field) we have $$ (x-c_1)(x-c_2)\cdots(x-c_n)=x^n-s_n^1x^{n-1}+s_n^2x^{n-2}-\cdots+(-1)^{n-1}s_n^{n-1}x+(-1)^ns_n^n $$ simply by expanding the product on the left hand side.

Therefore the elements $c_1,c_2,\ldots,c_n$ are all zeros of the polynomial $$ P(x)=x^n-s_n^1x^{n-1}+s_n^2x^{n-2}-\cdots+(-1)^{n-1}s_n^{n-1}x+(-1)^ns_n^n. $$

What zeros does that polynomial have in your case?

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