# Maximal set such that any $k$ elements form a basis

Consider ${\mathbb{F}_q}^2$, a two-dimensional vector space over a finite field. It is clear that we can pick $\frac{q^2-1}{q-1}$ vectors of ${\mathbb{F}_q}^2$ such that no two points lie in the same $1$-dimensional subspace.

The question I have is a bit more general: In a $k$-dimensional vectorspace over a finite field, how many vectors can we choose such that no $k$ vectors lie in a subspace of dimension $k-1$?

In other words, what is the maximal size of a set $A\subset {\mathbb{F}_q}^k$ such that if you pick $k$ distinct elements from $A$ they form a basis?

Maybe giving the exact number is too difficult, can you give some lower bounds?

Thanks a lot!

A lower bound is $q+1$.

For instance, for $k=2$, the set $$A=\{[1,t]: t \in \mathbb{F}_q\} \cup \{[0,1]\}$$ has $q+1$ elements, with no two of them being linearly dependent. One can recognize the set $A$ as being the projective space $\mathbb{P}^1(\mathbb{F}_q)$.

It is not hard to see that the question is equivalent to asking for the largest number of points in $\mathbb{P}^n(\mathbb{F}_q)$, with no $n+1$ of them belonging to the same hyperplane. These are well known objects objects in Finite Geometry called $\mathcal{K}$-arcs.

So for the general case, the rational normal curve in $\mathbb{P}^{k-1}(\mathbb{F}_q)$

$$A=\{[1,t,\ldots,t^{k-1}]: t \in \mathbb{F}_q\} \cup \{[0,\ldots,0,1]\}$$

gives rise to a set of $q+1$ points in $(\mathbb{F}_q)^{k}$ with the desired property.

• Thanks for your excellent answer. For completeness, let me add that if $q>k$ the example you give is conectured to be optimal, except in the case that $q$ is even and $k=1$ or $k = q-1$, according to this source : encyclopediaofmath.org/index.php/Arc_(projective_geometry) Jan 12, 2017 at 18:26