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Recently, I have come across following problem. The origin is the following: we study a set of $q^2$ points in $AG(3,q)$ with some special properties. Using the polynomial method (the full derivation is quite long, so I won't copy it here), we find two sets of multisets of elements of $GF(q)$ (coming from their coordinates) satisfying certain equations. In order to extract more information from the main polynomial under consideration, we need more relations on the coefficients, which are combinations of elements of these two sets. For the details, see below.

The two (multi)sets of elements of $GF(q)$ consist both of $q^2$ elements, say $\{a_i\}_{i=1}^{q^2}$ and $\{b_i\}_{i=1}^{q^2}$. For these two sets it is given that (which comes from the fact that some coefficients of the main polynomial are zero):

\begin{align*} \sum_{i=1}^{q^2} a_i^rb_i^s = 0 \end{align*}

for all $r,s$ such that $0 \leq r,s \leq q-1$ and $r + s \leq q$. Is there anything we can infer for higher degree, i.e., when the sum $r + s > q$ ? This would imply that other terms also vanish in the main polynomial, from which we can hopefully derive information on the set of $q^2$ points.

For example, I'd like to know if also $\sum_{i=1}^{q^2} a_i^{q-2}b_i^3$ equals zero.

Sam Mattheus
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  • This is an interesting question. Need to find suitable nested subspaces spanned by the power vectors ... I have a hunch that a coding theoretical construction/argument can be used here. Alas, I need to get some shut-eye time now. – Jyrki Lahtonen Dec 31 '16 at 00:15
  • If it helps, it would already be really helpful to have some information in the special case that $q$ is prime. – Sam Mattheus Jan 01 '17 at 17:29
  • I forgot about this. But I can make the following trivial observation. Some extra relations will hold unless all the $q$ elements of $GF(q)$ occur in both sequences $(a_i){i=1}^{q^2}$ and $(b_i){i=1}^{q^2}$. This is because otherwise the sequence $(a_i^{q-1})_i$ will be a linear combination of the sequences $(a_i^r)_i$, $r=0,1,2,\ldots,q-2$. Similarly for the $b$-sequences. I hazard a guess that it is possible to find sequences such that no such extra relations will hold. I'm not gonna bet a family fortune on it. A beer may be :-) – Jyrki Lahtonen Jan 01 '17 at 23:18
  • Hi Jyrki, thanks for the suggestion. I hope the context is now a bit clearer. I have tried to see what you meant in your previous comments. I thought about it in the following way: we can view the set of sequences of $q^2$ elements over $GF(q)$ as a vector space over $GF(q)$ (addition is element-wise). Then, looking at an appropiate subspace, one could hope to find that $(a_i)_i, (a_i^2)_i, \dots, (a_i^{q-1})_i$ are linearly dependent and thus we can write that last one as a combination of the former. Unfortunately, I can't quite fill in the details. – Sam Mattheus Jan 03 '17 at 12:36
  • Sam, that is the idea. If there are $q$ distinct values among the $a_i$s, then there will be a non-zero Vandermonde determinant saying that those $q$ sequence (with exponent taking values from $0$ to $q-1$) will be linearly independent in $GF(q)^{q^2}$. So your conditions say that the vector of $r$th powers of the $a_i$s is orthogonal to the sequence of the $s$th powers of the $b_i$s whenever $r+s\le q$. This reminds me of the situation with Reed-Solomon codes and their duals (except that those are shorter, which is ok, because it gives us more elbow room here). – Jyrki Lahtonen Jan 03 '17 at 14:33
  • But, with the little time I have invested on this, I couldn't quite make it work. I like to keep this open, so that I can think about it more when I find the time. – Jyrki Lahtonen Jan 03 '17 at 14:33

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