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I recently read a thread on this site that solved the following problem: let $K:=\mathbb{F}_q$ be a finite field of $q$ elements and $i$ an integer. Then $\sum\limits_{\alpha \in K} \alpha^i = 0$ unless $i$ is divisible by $q-1$.

I'm interested in a generalisation of this problem. What happens if instead of summing over $K$, we sum over an additive subgroup $G$ of $K$? For instance, at one extreme we still have $\sum\limits_{\alpha \in G} \alpha^i \neq 0$ if $i$ is divisible by $q-1$, since we get a sum of $|G|-1$ ones and a single zero. On the other hand, if $|G|=p$ then $G = \beta \times \mathbb{F}_p$ where $\beta \in K$ and $\mathbb{F}_p$ is the prime subgroup of $K$, and we get that $\sum\limits_{\alpha \in G} \alpha^i \neq 0$ if and only if $i$ is divisible by $p-1$. Can we say anything more general?

Qwerty
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Jon Elmer
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2 Answers2

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A beginning.

Any additive subgroup $G$ of $\Bbb{F}_q$ is a vector space over the prime field $\Bbb{F}_p$. Consequently the polynomial $$ T_G(x)=\prod_{a\in G}(x-a) $$ is a linearized polynomial (see for example Lidl & Niederreiter). This implies that all the non-zero terms of $T_G(x)$ have degrees that are powers of $p$. In other words $$ T_G(x)=x^{p^m}+\sum_{j=0}^{m-1}a_jx^{p^j}, $$ where $a_j\in \Bbb{F}_q$, and $m=\dim G$.

The highest degree non-leading term has thus degree $p^{m-1}$. This means that the elementary symmetric polynomials of $G$ of degree $<p^{m-1}(p-1)$ all vanish. Consequently the power sum $$ S_G(j):=\sum_{a\in G}a^j $$ vanishes whenever $j<p^{m-1}(p-1)$.

Probably more can be said. Some vanishing power sums are consequences of the above facts about the elementary symmetric polynomials. It often happens that when you write the power sum $S_G(j)$ in terms of the elementary symmetric polynomials (= the coefficients of $T_G(x)$) - use the Newton (-Girard) identities to achieve that (or Waring's formula) - then all the terms will have low degree terms as factors, and thus the entire power sum will vanish.

Observe that $$ T_{\Bbb{F}_q}(x)=x^q-x. $$ In that case we have more information, namely that $a_j=0$ for all $j>0$. Therefore we can say more!

Jyrki Lahtonen
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    Thanks for this, it's exactly the kind of thing I was looking for. In fact, you can go a little further with this. There's a convenient way to express the Newton-Girard identities in matrix form (I think I need to click "answer" to type this..) – Jon Elmer May 16 '16 at 10:07
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    May ask you a question about that problem please? We consider the sum $\sum_{\alpha\in \mathbb{F}_q}\alpha^i$, where $i\in \mathbb{Z}$. How do you define $\alpha^i$ if $\alpha=0$ and $i\leq 0$? – RFZ Jan 16 '23 at 18:21
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S_G(n) = \begin{pmatrix} e_1 & 1 & 0 & \ldots & 0 \\ 2e_2 & e_1 & 1 & \ldots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ ne_n & e_{n-1} & \ldots & e_2 & e_1\end{pmatrix} where $e_i$ is the elementary symmetric polynomial in $G$ of degree $i$, and $n<p^m$. Now Jyrki's argument showed that $e_i = 0$ unless $i=p^m-1, p^m-p, \ldots, p^m-p^{m-1}$. Looking at the left hand column of the matrix we see straight away that $S_G(n) = 0$ for $n<p^m-1$, since for all $i<p^m-1$ with $e_i \neq 0$ we have $i=0$ in $\mathbb{F}_q$. The same argument shows that $S_G(p^m-1) = -e_{p^m-1}$. I'd be very interested to know if there's any particular reason $e_{p^m-1}$ should be non-zero in general.

Jon Elmer
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    So $S_G(n)$ is the determinant of that matrix? Why didn't I know about this :-) !!!! Anyway, $e_{p^m-1}$ is the coefficient $a_0$ of the first degree term of the polynomial $T_G(x)$. But $a_0$ is the product of the non-zero elements of $G$, and hence always $a_0\neq0$. If we had $a_0=0$, then $T_G(x)$ itself would be a $p$th power, and all its zeros would have multiplicity $p$... – Jyrki Lahtonen May 16 '16 at 11:05
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    Apparently I was asleep not reaching the conclusion that $S_G(n)=0$ for all $n<p^m-1$. This is a common argument in proving that the BCH-bound (from coding theory) is tight for some values of a designed distance - exactly because the power sums over a subspace vanish. Sorry about that – Jyrki Lahtonen May 16 '16 at 11:07
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    Yes, I did mean the determinant :). Thanks again for your help. – Jon Elmer May 16 '16 at 11:47