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Let $\mathbb{F}_q^k$ be a $k$-dimensional vectorspace over a finite field, and let $l \ge 0$ be an integer. The question is how to construct a (maximally) large set $A \subset \mathbb{F}_q^k$ such that if you pick any $k+l$ distinct elements from $A$, then these points span $\mathbb{F}_q^k$.

This question is a more general version of my previous question, which handled the case $l=0$.

In more geometric terms the question sounds like this: Construct a large set of points in $\mathbb{P}^{k-1}(\mathbb{F}_q)$ such that no $k+l$ lie on the same hyperplane.

In the answer to my previous question it was pointed out that for the case $l=0$ these objects are called $\mathcal{K}$-arcs and they are well known.

I believe for $l>0$ these objects are called $(\mathcal{K},k+l)$-arcs, but I could not find any construction for them.

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    Double check this for typos. The letters seem a bit weird to me. What is $n$ ever used for?. Where does $l$ come from? Is $q=p^l$ the case? If not, your need to specify the exponent that gives rise to $q$ in some way because the answer depends on that exponent. – 2017-01-18
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    Oops you are right. I changed an $n$ in a $k$. l can be any non negative number – 2017-01-18
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    I did not specify the exponent $e$ such that $q = p^e$ because I didn't need it in the question. – 2017-01-18
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    Can you add a bit more context, especially about what values of $l$ you're looking at? For example, for $l=q^k-k$ then obviously the entire space works... – 2017-01-18
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    Im particularly interested in small $l$, that is 1,2,3 or $l = k$. – 2017-01-18

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As you've mentioned, this is a natural generalization of $\mathcal{K}$-arcs. Apart from few particular cases, not much is known about these objects. Most of the results/constructions in the literature are still dealing with the $\mathbb{P}^2(\mathbb{F}_q)$-case. It seems that, even in this case, general sharp bounds are somewhat lacking.

A simple procedure to construct a "large" set of $n$ points in $\mathbb{P}^2(\mathbb{F}_q)$, with no $r+1$ in the same line is:

Pick an irreducible projective curve $\mathcal{C}: f(x,y,z)=0$ of degree $r>1$ with a "large" number $n$ of $\mathbb{F}_q$-points. Then by Bezout's theorem, this set has at most $r$ collinear points.

To illustrate the case $(k,l)=(3,1)$ in your notation, just take an elliptic curve $\mathcal{C}$ attaining the Hasse-Weil-Serre bound: $n=1+q+[2q]$.

For additional details (bounds/constructions), you can look at the paper (and references therein): http://www.sciencedirect.com/science/article/pii/S1071579705000250

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    The same thing would work in higher dimensions, right? If you have an irreducible 1-dimensional component $\mathcal{C}$ of degree $r>1$ then by bezout's theorem every hyperplane contains at most $r$ points. I believe the irreducibility constraint is only there to assure that there are no lines in $\mathcal{C}$? Then my question becomes, what are some 1d components with lots of $\mathbb{F}_q$ points? – 2017-01-18