Symmetric function $p_k({x_1},{x_2},...,{x_n})={x_1}^k+{x_2}^k+...+{x_n}^k$ on $GF(q)=\{0,a_1,a_2,...,a_{q-1}\}$

Asked Jan 11 '19 at 07:48

Active Jan 11 '19 at 07:48

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This is the first time for me on symmetric functions. Let us consider the symmetric function $p_k({x_1},{x_2},...,{x_n})={x_1}^k+{x_2}^k+...+{x_n}^k$ on $GF(q)=\{0,a_1,a_2,...,a_{q-1}\}$.

I want to show that:

a) $p_{q-1}(a_1,...,a_{q-1})=-1$

b) $p_{k}(a_1,...,a_{q-1})= 0$ for all $0 < k \leq q-2$.

Some hints or procedure? $GF(q)$ is cyclic so for the first point there is some recursivity, but how can I handle this problem?

asked Jan 11 '19 at 07:48

Alessar

2

Don't forget that the nonzero elements of $GF(q)$ form a cyclic group. – Angina Seng Jan 11 '19 at 07:56
This has been done on the site already. Did you search? – Jyrki Lahtonen Jan 11 '19 at 19:42
I seem to have answered it myself here as well as here. Not voting to close as a dupe though. No reason to think that would have been the first occurrence. Also, it is surely bad form to use my dupehammer to direct attention to my own answer. – Jyrki Lahtonen Jan 11 '19 at 19:46
Hmm. I may be feeding the search engine with wrong buzzwords... – Jyrki Lahtonen Jan 11 '19 at 20:10
Matt Samuel came up with the same solution proving that there is nothing original with mine. – Jyrki Lahtonen Jan 11 '19 at 20:13
@JyrkiLahtonen a little comment, thanks to you and to stack, in march I passed my Algebra exams. Thank you so much (ps you can delete this comment after read for cleanup) – Alessar Jul 05 '19 at 13:54

Symmetric function $p_k({x_1},{x_2},...,{x_n})={x_1}^k+{x_2}^k+...+{x_n}^k$ on $GF(q)=\{0,a_1,a_2,...,a_{q-1}\}$

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