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

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 $p \geq 5$.

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 $p \equiv 1 \pmod{3}$).