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 $X$.



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



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> 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?
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**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.



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 $p \geq 5$.


