2

Let $q$ be a prime and consider the finite field with $q$ elements $\mathbb{F}_{q}$. Let $k\geq 2$ and consider $a_{1},a_{2}, \ldots, a_{k}\in\mathbb{F}_{q}$. I wish to find all the linear equations (upto equivalence) such that no matter what $a_{1}, a_{2}, \ldots, a_{k}$ one chooses, $(a_{1}, a_{2}, \ldots, a_{k})$ is always solution to one of these equations.

For example, when $q = 3$ and $k = 2$, we see that $(a_{1}, a_{2})$ is always a solution to one of the equations $a_{1} = 0, a_{2} = 0, a_{1} + a_{2} = 0$ and $a_{1} + 2a_{2} = 0$. Of course, one could replace $a_{1} + a_{2} = 0$ with $2a_{1} + 2a_{2} = 0$ but we only choose one of such equations.

When $q = 3$ and $k = 3$, the equations required are $$a_{1} = 0, a_{2} = 0, a_{3} = 0, a_{1} + a_{2} + a_{3} = 0, 2a_{1} + a_{2} + a_{3} = 0, a_{1} + 2a_{2} + a_{3} = 0 \text{ and } a_{1} + a_{2} + 2a_{3} = 0.$$ In other words, no matter what $a_{1}, a_{2}, a_{3} \in\mathbb{F}_{3}$ we choose, $(a_{1}, a_{2}, a_{3})$ will always be a solution to one of the above equations. As noted in the comments, I would like all the $k$-variables to be involved non-trivially. For instance, the four equations for $k = 2$ case may suffice otherwise.

I also worked out the example for $q = 5$ and $k = 2, 3 $. However, I have not succeeded in seeing a pattern that would allow some sort of listing or identification of all of such equations for given values of $q$ and $k \geq 2 $.

Maulik
  • 396
  • The two-variables polynomial $a_1a_2(a_1+a_2)(a_1+2a_2)$ vanishes on the whole of $\Bbb{F}_3^2$ because it is $=a_2(a_1^3-a_1)-a_1(a_2^3-a_2) \bmod 3$ – reuns Jun 02 '22 at 23:48
  • One simple way is to take all nonzero vectors $(b_1,\dots,b_k)$ whose first nonzero entry equals $1$, and then consider the set of linear equations $b_1a_1+\cdots+b_ka_k=0$. This simply enumerates all possible linear equations with different forms of the same equation removed. It doesn't necessarily give a minimal set of equations for your problem (for example, when $q=k=3$ it gives $13$ equations instead of your $5$). If you want fancy words to investigate further, I've described the dual of a projective space over $\Bbb F_q$. – Greg Martin Jun 03 '22 at 00:55
  • A linear equation defines a subspace of $\Bbb{F}_q^k$ (or a coset of one if the constant term is non-zero). This is a well studied problem, and the answer is that you need at least $q+1$ subspaces to cover the whole thing. See for example this old thread and others linked to it. Several approaches outlined in the answers and links. – Jyrki Lahtonen Jun 03 '22 at 04:55
  • Oh, and the following $q+1$ equations will cover it all: $a_1=0$ and the $q$ equations $\lambda a_1+a_2=0$ where $\lambda$ varies over all the elements of $\Bbb{F}_q$. So only the two first components ever get used, but that can be changed if so desired. – Jyrki Lahtonen Jun 03 '22 at 04:59
  • 1
    If I misunderstood, and you really want a way to parametrize all the linear homoegenous examples up to equivalence, then Greg Martin's comment gives you the recipe. Otherwise please explain, how your collection with $q=3, k=2$ fails for you in the case $q=3, k=3$. – Jyrki Lahtonen Jun 03 '22 at 08:07
  • @GregMartin For $q = k = 3$ It seems that the 13 equations can be identified exactly with the 13 hyperplanes in $\mathbb{F}{3}^{3}$. When $k = 2$, there are $q+1$ hyperplanes whose equations are exactly what we need. I am interested in decreasing the number of equations, rather than just counting the number of hyperplanes in $\mathbb{F}{q}^{k}$. – Maulik Jun 03 '22 at 18:22
  • @JyrkiLahtonen I certainly desire all the $k$-variables to be involved non-trivially, rather than just using $k = 2$ case for every $k$. – Maulik Jun 03 '22 at 18:29
  • @Maulik in that case, that's important information that you should put in the question itself. – Greg Martin Jun 03 '22 at 19:02
  • @Maulik What I meant with ...that can be changed if so desired was that $q+1$ hyperplanes suffices if and only if they all contain the same $(k-2)$-dimensional subspace. That subspace need not be $a_1=a_2=0$ like it was in my example set, but I do realize that for many purposes such a selection of hyperplanes may feel "non-generic", whatever that means :-) – Jyrki Lahtonen Jun 03 '22 at 19:12
  • I think I may have seen a pattern. It appears that the equations for $k = 3$ case come from equations from $k = 2$ case. We add an equation $c = 0$ to $a = 0$ and $b = 0$. On the other hand, to the rest of the two equations, we add $ia_{3}$ for every $i \in{0, 1, \ldots, q-1}$. Therefore, we get $2+1+2(2) = 3+4 = 7$ equations for $k = 3$ case. According to this, we guess that for $k = 4$ case, we obtain $3+1+2(4) = 4+8 = 12$ equations. I guess that one could do something similar for other $q$ too. – Maulik Jun 03 '22 at 20:14
  • 1
    Like in the case $q=k=3$ you can also use the four hyperplanes $a_1=0$, $a_2+a_3=0$, $a_1+a_2+a_3=0$, $2a_1+a_2+a_3=0$. In the case $q=3, k=4$ take $a_1=0$, $a_2+a_3+a_4=0$, $a_1+a_2+a_3+a_4=0$, $2a_1+a_2+a_3+a_4=0$. If there is something wrong with these, please formulate what it is :-) – Jyrki Lahtonen Jun 04 '22 at 04:25
  • @JyrkiLahtonen oh wow! So it appears that 4 equations will suffice for every $k$ when $k = 3$. Similarly, $q+1$ equations will do it for every $q$ and $k \geq 2$. I did not see that coming. – Maulik Jun 04 '22 at 19:01

0 Answers0