The Sylvester-Gallai theorem over $p$-adic fields

Question

The famous Sylvester-Gallai theorem states that for any finite set $X$ of points in the plane $\mathbf{R}^2$, not all on a line, there is a line passing through exactly two points of $X$.

What happens if we replace $\mathbf{R}$ by $\mathbf{Q}_p$?

It is well-known that the theorem fails if we replace $\mathbf{R}$ by $\mathbf{C}$: the set $X$ of flexes of a non-singular complex cubic curve has the property that every line passing through two points of $X$ also passes through a third.

For example, the flexes of the Fermat curve $C:X^3+Y^3+Z^3=0$ are given by the equation $XYZ=0$ and are all defined over the field $\mathbf{Q}(\zeta_3)$ generated by a cube root of unity $\zeta_3$. As a consequence, if a field $K$ contains $\mathbf{Q}(\zeta_3)$ then the set of flexes of $C$ gives a counterexample to the Sylvester-Gallai theorem over $K$. For example, for any prime $p \equiv 1 \pmod{3}$, the field $\mathbf{Q}_p$ contains $\mathbf{Q}(\zeta_3)$ so that Sylvester-Gallai fails over $\mathbf{Q}_p$.

I don't know what happens in the case $p=3$ or $p \equiv 2 \pmod{3}$. Note that the set of flexes of $C$ is not defined over $\mathbf{Q}_p$ anymore, but nothing prevents more complicated configurations of points giving counterexamples to the theorem over $\mathbf{Q}_p$.

More generally, what happens over an arbitrary field $K$? Is it true that the Sylvester-Gallai theorem holds over $K$ if and only if $K$ does not contain the cube roots of unity?

EDIT. David Speyer's beautiful example shows that the Sylvester-Gallai theorem fails over $\mathbf{Q}_p$ for any prime $p \geq 5$. Furthermore, regarding the problem of deciding whether SG holds over a given field $K$ (which looks like a difficult question, at least to me), this and Gro-Tsen's example show that the condition that $K$ does not contain the cube roots of unity clearly needs to be refined. In order for SG to hold over a characteristic $0$ field $K$, it is necessary that $K$ does not contain any root of unity of order $\geq 3$. I don't know whether this is also a sufficient condition.

Answer 1

If $n \geq 3$ and $K$ is a field of characteristic not dividing $n$, containing a primitive $n$-th root of unity $\zeta$, then the $3n$ points of the form $(1:-\zeta^a:0)$, $(0:1:-\zeta^b)$, $(-\zeta^c:0:1)$ are a Sylvester-Gallai configuration. In particular, taking $n=p-1$, this gives an SG configuration over $\mathbb{Q}_p$ for $p \geq 5$.

Answer 2

Over any field $K$ of characteristic $\neq 2$ and containing $i := \sqrt{-1}$, there exists in $\mathbb{P}^2(K)$ a Sylvester-Gallai configuration with $12$ points given in affine coordinates by $(0,0)$, $(0,1)$, $(1,0)$, $(1,1)$, $(a,a)$, $(a,b)$, $(b,a)$, $(b,b)$ where $a := \frac{1+i}{2}$ and $b := \frac{1-i}{2}$, $\infty\cdot(0,1)$, $\infty\cdot(1,0)$, $\infty\cdot(1,i)$ and $\infty\cdot(1,-i)$ (I have used affine coordinates rather than projective ones because I think it makes it easier to check: here obviously, $\infty\cdot(x,y)$ refers to the point at infinity on the line connecting $(0,0)$ and $(x,y)$). This is taken from Kelly & Nwankpa, "Affine Embeddings of Sylverter-Gallai [sic] Designs", J. Combinatorial Theory (A) 14 (1973) 422–438, "design B" in theorem 3.10 (note that the paper incorrectly writes $b = \frac{-1+i}{2}$: this is just a typo). Of course, for an infinite field, you can always realize this in the affine plane.

In particular, the Sylvester-Gallai theorem fails for $\mathbb{Q}_p$ when $p \equiv 1\pmod{4}$ (and not just for $p \equiv 1 \pmod{3}$).