Sylvester–Gallai theorem for small sets in a finite field The well-known Sylvester–Gallai Theorem states that a set of $n>2$ points in $R^2$ not all on a line contains two points such that the line passing through these two points does not contain a third point in the set.
One of the reasons this is somewhat tricky to prove is that it is false in some geometries, including finite field geometry. This can be seen by considering, for example, all $p^2$ points in $F_p^2$ a plane over a finite field.
My question is the following:

Let $n$ be a set of (say) $n=p^{1/100}$ points in $F_p^2$ a large finite field of prime order. Then does the conclusion of the Sylvester-Gallai theorem hold?

I am restricting the question to prime order fields to (1) avoid issues with subfields, and to (2) try to prevent the analog of the Hesse Configuration, a counterexample to the corresponding statement over the complex numbers.
Many of the proofs of the real case use the ordering of the field, so will run into trouble in a finite field. On the other hand, once the sets are really really small (much smaller than what is being considered here) there might be hope of formally translating the problem to the complexes and using the characterizations of possible counter-examples [see: https://arxiv.org/pdf/math/0403023.pdf].
I assume this question would been studied before, but I can't find a reference. I'm mostly interested in a positive result, so if there is a counter-example, I'd be interested to know if there is an understanding of counterexamples, which there is in the complex case.
 A: The 9-point "Hesse configuration" requires only a cube root of unity $\omega$,
so it works for any $p \equiv 1 \bmod 3$ with $n=9$.  In the projective plane
these points are the cyclic permutations of $(1:-\omega^i:0)$
$(i=0,1,2)$.  To put this configuration in the affine plane,
use a projective linear transformation that makes the line at infinity
miss all nine points.  That's certainly possible once $p$ is large enough that
$n = 9$ is smaller than $p^{1/100}$.
A: The answer is no, basically thanks to Menelaus's theorem.  First observe as a routine application of the Bombieri-Vinogradov theorem that there are infinitely many primes $p$ such that $p-1$ contains a factor $d \sim p^{1/100}$, so in particular the multiplicative group ${\bf F}_p^\times$ contains a subgroup $H$ of order $d$.  Now let $ABC$ be any triangle in the plane ${\bf P F}_p^2$ and consider the set of all points $E$ on the line $AC$ such that the ratio $CE/EA$ (defined in the obvious fashion) lies in $H$, together with the set of all points $D$ on the line $CB$ such that $BD/DC$ lies in $H$, and the set of all points $F$ on the line $AB$ such that
$-AF/FB$ lies in $H$.  Menelaus's theorem tells us that this is a Sylvester-Gallai configuration of order $3d \sim p^{1/100}$ which is not all contained in one line.  It contains some points at infinity, but one can apply a generic projective transformation to create an affine configuration (or alternatively one can scale the three copies of $H$ here appropriately).
This "Menelaus configuration" also basically appears (in the context of the p-adics) in this answer of David Speyer to a related MathOverflow question; Ben Green and I also encountered similar configurations in our paper on this topic.  In the spirit of that paper (which conveys the moral that sets with few ordinary lines tend to arise from cubic curves, although we could only make this heuristic rigorous over the reals), one could imagine that these Menelaus configurations are essentially the only Sylvester-Gallai configurations in this regime (the other cubic curves beyond triangles not giving good examples due to the presence of tangent lines, or (in the case of three concurrent lines) because the underlying group is of prime order), though that would seem rather beyond the reach of our current technology to prove.  (In high dimensions, there are however some results constraining the rank of Sylvester-Gallai configurations over finite fields; see Section 5.3 of this survey of Dvir.)
