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I was this past year working with a bright high-schooler on algebraic geometry following Reid's book Undergraduate Algebraic Geometry, and we got all the way to proving that there is at least one line on a non-singular cubic surface over an algebraically closed field that has a point whose tangent space intersects the original surface in a nodal plane cubic of type $y^2=x^3$ (I believe in char=3 there is another type of nodal plane cubic, which we didn't examine in detail). However, one fun thing is of course looking at examples over arbitrary fields where the lines are all defined over that base field.

There are well-known examples over the reals, but I was wondering about what happens over finite fields in low characteristic. In particular what is the "smallest", in the sense of the smallest field and the simplest surface, and "best", in the sense of the most symmetric while having all lines distinct over the base field, example. Ideally all the points of intersection are in a single affine chart, so one could make a toy model of the affine part of the (rational points on the) surface together with the (rational points on the) lines, in some combinatorially-pleasing embedding.

The stretch goal is to know what is the automorphism group of the configuration of 27 lines, in this particular small example.

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    $\begingroup$ In case it isn't obvious: I want a specific finite field, an explicit equation for the cubic surface, and an explicit description (in whatever way) for the lines. $\endgroup$
    – David Roberts ♦
    Commented Dec 12 at 13:36
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    $\begingroup$ A cubic surface where all $27$ lines are defined over a finite field $\mathbb F_q$ has $q^2+7q +1$ points over $\mathbb F_q$. The embedding in $\mathbb P^3$ gives the inequality $q^2+7q +1 \leq q^3 + q^2+q+1$ which requires $q\geq \sqrt{6}$, so we must have $q$ at least $3$. $\endgroup$
    – Will Sawin
    Commented Dec 12 at 13:48
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    $\begingroup$ For the Fermat hypersurfaces in characteristic different from $2$ and $3,$ all twenty-seven lines are rational if the base field contains a primitive sixth root of unity. If the base field is $\mathbb{F}_q$, this is equivalent to the congruence $q\equiv\ 1\ (\text{mod}\ 6)$. So $q=7$ gives one example. $\endgroup$
    – Jason Starr
    Commented Dec 12 at 14:41
  • $\begingroup$ See also (for $q=4$): Hirschfeld, James W. P., The double-six of lines over PG(3,4). J. Austral. Math. Soc. 4 (1964), 83–89. $\endgroup$
    – Felipe Voloch
    Commented Dec 14 at 2:52
  • $\begingroup$ Just to clarify Jason Starr's comment above. Actually if you have the third roots of unity then you already have the sixth roots of unity, again away from characteristic 2 and 3. So this is actually equivalent to $q \equiv 1 \bmod 3$. $\endgroup$
    – Daniel Loughran
    Commented Dec 14 at 20:07

1 Answer 1

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It is the Fermat cubic surface over $\mathbb{F}_4$ or (if you prefer $\mathbb{F}_p$) over $\mathbb{F}_7$.

There are quite a few papers on this topic. Firstly in [1] Swinnerton-Dyer showed (amongst other things) there is a smooth split cubic surface over $\mathbb{F}_q$ if and only if $q \neq 2,3,5$ (we say a cubic surface is split if all 27 lines are defined over the ground field).

The existence a smooth split cubic surface over $\mathbb{F}_4$, but not $\mathbb{F}_5$, is some funny combinatorial coincidence about the geometry of the projective plane over finite fields of even characteristic.

In characteristic 2 all kinds of funny things happen. For example over $\mathbb{F}_4$ the Fermat cubic surface has automorphism group $\mathrm{PSU}_4(\mathbb{F}_2)$, which is the largest possible automorphism group, see [2]. This surface is split over $\mathbb{F}_4$ hence is the required surface.

Over $\mathbb{F}_7$ it is also split and is known to have the largest possible automorphism group of all smooth cubic surfaces over an algebraically closed field of characteristic not equal to $3$ (again see [2]). But since there are 6th roots of unity in $\mathbb{F}_7$, all automorphisms are actually defined over $\mathbb{F}_7$.

[1] Swinnerton-Dyer, Peter, Cubic surfaces over finite fields. Math. Proc. Cambridge Philos. Soc. 149 (2010), no. 3, 385–388.

[2] Vikulova, Anastasia V. The most symmetric smooth cubic surface over a finite field of characteristic 2. Finite Fields Appl. 98 (2024), Paper No. 102470, 25 pp.

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    $\begingroup$ The thesis of Raymond Cheng studies generalizations of cubic Fermat surfaces in characteristic $2$ to “$q$-bic varieties.” These each have a large (unitary) group of automorphisms. $\endgroup$
    – Jason Starr
    Commented Dec 12 at 14:56
  • $\begingroup$ I don't understand your comment about the surface over F_7, do you mean the base change to the alg closure has the largest possible automorphism group? Also, can you address the matter of getting all the intersections in the affine chart, and the (sets of rational points of the) lines being pairwise distinct? $\endgroup$
    – David Roberts ♦
    Commented Dec 12 at 23:09
  • $\begingroup$ The Fermat cubic is the surface with the largest automorphism group over $\bar{\mathbb{F}}_7$. However since every automorphism over $\bar{\mathbb{F}}_7$ is actually defined over $\mathbb{F}_7$, it also has the same property over $\mathbb{F}_7$. (This is not true over $\mathbb{F}_2$, say, where the surface $x^2t + y^2z + z^2y + t^2x = 0$ has the largest automorphism group over $\mathbb{F}_2$, despite the Fermat cubic having a larger automorphism group over $\mathbb{F}_4$). $\endgroup$
    – Daniel Loughran
    Commented Dec 13 at 9:19
  • $\begingroup$ Regarding getting the intersections of the lines in the same affine chart, I'm not quite sure. You can try looking at the Fermat cubic surface over $\mathbb{F}_{13}$ then looking for a plane which doesn't contain any intersection points, then make a change of variables to make this plane the hyperplane at infinity. But doing this may lose the nice simple symmetric form of the equation. $\endgroup$
    – Daniel Loughran
    Commented Dec 13 at 9:24
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    $\begingroup$ I deleted my previous comment about intersection points in an affine chart; what it said was correct but misleading. I now claim you need $q$ to be at least 19 to get all intersections in one affine chart. This follows essentially from the Hasse bound. I don't know if this is sharp. $\endgroup$
    – Lazzaro Campeotti
    Commented Dec 13 at 10:11

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