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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.

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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 $p \geq 5$.

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    $\begingroup$ @GilKalai Yes. "A resolution of the sylvester-gallai problem of J.-P. serre", Kelly, Discrete & Computational Geometry, 1986. See also arxiv.org/abs/math/0403023 for a simpler proof and for results over the quaternions. $\endgroup$ – David E Speyer Dec 31 '16 at 17:33
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    $\begingroup$ This configuration looks a lot like the "torsion points" of the "elliptic curve" given by the cubic equation $xyz=0$. The smooth locus is the disjoint union of three copies of $\mathbb{G}_m$, we take their $n$-torsion points. In that sense the example resembles the configuration of the $3$-torsion of an elliptic curve mentioned by the OP. $\endgroup$ – Piotr Achinger Dec 31 '16 at 17:59
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    $\begingroup$ @PiotrAchinger Indeed, and when I read OP's question, my immediate reaction was "well, we can certainly take the $n$-torsion of an elliptic curve and it will give a S-G configuration over some field", but this does not work because sometimes you get a line of tangency meeting the curve in only $2$ points; so then when I read David Speyer's answer, my first reaction was, this can't work because the same problem will happen; but it does work. $\endgroup$ – Gro-Tsen Dec 31 '16 at 18:06
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    $\begingroup$ @GilKalai Well, it only applies over fields of characteristic zero, of course: If $\mathbb{F}_p$ is in your field than the points of $\mathbb{P}^n(\mathbb{F}_p)$ form an SG configuration in $\mathbb{P}^n$. But yes, proving it over the complex numbers implies all characteristic zero fields. $\endgroup$ – David E Speyer Dec 31 '16 at 19:19
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    $\begingroup$ The dual configuration appears in this paper by Dolgachev (§5) as well as in the book Geradenkonfigurationen und Algebraische Flächen by Barthel &al (Vieweg 1987) (§2.3) under the name "Ceva configuration". So I guess a logical name for this one would be "Menelaus configuration" since the Ceva and Menelaus theorems are dual. $\endgroup$ – Gro-Tsen Jan 13 '17 at 22:33
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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 $p \equiv 1 \pmod{3}$).

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  • $\begingroup$ Thanks for your answer and the reference! This also shows that the SG theorem fails over $\mathbf{Q}_9$, the unramified quadratic extension of $\mathbf{Q}_3$. $\endgroup$ – François Brunault Dec 31 '16 at 17:05
  • $\begingroup$ I have to say, David Speyer's answer is better than mine in every respect (except perhaps that we would like to know who first described the S-G configuration he describes). $\endgroup$ – Gro-Tsen Dec 31 '16 at 17:21
  • $\begingroup$ I was reminded of it by Theorem 3.11 in your reference, but I'm pretty sure I've seen it before. $\endgroup$ – David E Speyer Dec 31 '16 at 17:22
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    $\begingroup$ Ah, so my answer wasn't so useless after all! But I should have read it more carefully. I also seem to (retrospectively) remember having seen this configuration somewhere, and I'd like to remember where… $\endgroup$ – Gro-Tsen Dec 31 '16 at 17:27

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