# High-dimensional pointless varieties over finite fields

Let $$\mathbb{F}_q$$ be a finite field. Do there exist smooth projective varieties over $$\mathbb{F}_q$$ of arbitrarily high dimension that have no $$\mathbb{F}_q$$-points and no non-constant maps to lower-dimensional varieties?

Two lines of attack that don't work: Brauer-Severi varieties (the Brauer group of $$\mathbb{F}_q$$ is trivial) and varieties that become simple abelian over $$\mathbb{F}_q$$ (they have a rational point to begin with as mentioned by Wojowu).

• What do you mean by "smooth" if there are no points? Maybe geometrically smooth (namely, smooth over $\bar{\mathbb{F}}_q$)? Jul 30 at 10:19
• @FrancescoPolizzi Why is lack of points a problem for smoothness? Jul 30 at 10:39
• RE the suggestion in the edit - no, this is not possible. It is a theorem due to Lang that if a variety over a finite field which becomes an abelian variety after a base change, then it has a rational point. Jul 30 at 11:45
• @FrancescoPolizzi I believe you can just require regularity of all algebras in some cover by affines, no? Jul 30 at 14:11
• @FrancescoPolizzi: As $\mathbb{F}_q$ is perfect, smooth over $\mathbb{F}_q$ just means that the scheme is regular, which is defined in terms of the local rings at all scheme-theoretic points (in fact it just suffices to check the closed points, which can be thought of as Galois orbits of $\bar{\mathbb{F}}_q$ points.). Also by fpqc descent, smooth over $\bar{\mathbb{F}}_q$ implies smooth over $\mathbb{F}_q$ (and naturally the latter implies the former by base change). Jul 30 at 15:12

Sure.

By Poonen's Bertini theorem, for every $$q$$ and $$n$$, the limit as $$d$$ goes to $$\infty$$ of the proportion of degree $$d$$ hypersurfaces in $$\mathbb P^n$$ that are smooth and have no $$\mathbb F_q$$-points is positive.

Being smooth hypersurfaces in $$\mathbb P^n$$ for (let's say) $$n>3$$, they have Picard rank $$1$$, and thus no maps to lower-dimensional varieties.

For an explicit smooth hypersurface of dimension $$n-2$$, take $$\alpha \in \mathbb F_{q^n}$$ with $$\operatorname{tr}(\alpha)\neq 0$$. Choose a basis for $$\mathbb F_{q^n}$$, so that elements of $$\mathbb F_{q^n}$$ can be written as vectors over $$\mathbb F_q$$, with multiplication given by an $$n$$-tuple of quadratic polynomials.

Consider the function $$\operatorname{tr} ( \alpha x^{ \frac{q^n-1}{q-1}})$$ from $$\mathbb F_{q^n}$$ to $$\mathbb F_q$$. In coordinates, this is a homogeneous polynomial of degree $$\frac{q^n-1}{q-1}$$ in $$n$$ variables, hence defines a projective hypersurface, and sends every nonzero element of $$\mathbb F_{q^n}$$ to a nonzero element of $$\mathbb F_q$$, hence has no $$\mathbb F_q$$-points.

Over $$\overline{\mathbb F_q}$$, we can choose our coordinate system to be the $$n$$ embeddings of $$\mathbb F_{q^n}$$ into $$\overline{\mathbb F_q}$$, and in this coordinate system the equation is $$\sum_i c_i x_i^{ \frac{q^n-1}{q-1}}$$ where $$c_i \neq 0$$ is the image of $$\alpha_i$$ under the $$n$$th embedding, so this hypersurface is also smooth.

• Is there some simple explicit family of hypersurfaces that comes to mind?
– vope
Jul 30 at 12:29
• @vope After thinking a bit I found one (inspired by Leech Lattice's example). Jul 30 at 12:54

Just take the Weil restriction with respect to $$\mathbb{F}_{q^r} / \mathbb{F}_q$$ of a pointless curve. This will typically admit no $$\mathbb{F}_q$$-morphism to a variety of lower dimension.

Not an answer, this is just a (cw) out-of-curiosity example of the recipe from Will Sawin's answer. The following polynomial defines a smooth projective surface without any $$\mathbb F_3$$-points: $$\scriptstyle -t^{40}-zt^{39}-x^3t^{37}-x^4t^{36}+y^4t^{36}-xy^3t^{36}-yz^3t^{36}-x^3yt^{36}-x^9t^{31}+y^9t^{31}-z^9t^{31}+x^{10}t^{30}-y^{10}t^{30}-xz^9t^{30}-yz^9t^{30}-x^9yt^{30}-y^{12}t^{28}-z^{12}t^{28}-x^3y^9t^{28}+x^3z^9t^{28}+x^9y^3t^{28}+x^9z^3t^{28}+x^{13}t^{27}-z^{13}t^{27}-xy^{12}t^{27}+yz^{12}t^{27}-x^3y^{10}t^{27}+x^4y^9t^{27}+y^4z^9t^{27}+xy^3z^9t^{27}-x^3yz^9t^{27}+x^9y^4t^{27}-x^9z^4t^{27}+y^9z^4t^{27}+x^{10}z^3t^{27}+y^{10}z^3t^{27}-xy^9z^3t^{27}+x^9yz^3t^{27}+x^{12}yt^{27}-x^{12}zt^{27}-y^{12}zt^{27}-x^9y^3zt^{27}-x^{27}t^{13}-y^{27}t^{13}-z^{27}t^{13}-xy^{27}t^{12}+xz^{27}t^{12}-yz^{27}t^{12}+y^{30}t^{10}+z^{30}t^{10}-x^3y^{27}t^{10}+y^3z^{27}t^{10}-y^{27}z^3t^{10}-x^{31}t^9-y^{31}t^9-z^{31}t^9-xy^{30}t^9+xz^{30}t^9+yz^{30}t^9+x^3y^{28}t^9-x^3z^{28}t^9-y^3z^{28}t^9+x^4z^{27}t^9+y^4z^{27}t^9+x^3yz^{27}t^9-x^{27}y^4t^9-x^{27}z^4t^9-y^{27}z^4t^9+xy^{27}z^3t^9-x^{30}yt^9-x^{30}zt^9+x^3y^{27}zt^9+x^{27}y^3zt^9-y^{36}t^4-x^9y^{27}t^4-x^9z^{27}t^4+y^9z^{27}t^4-y^{27}z^9t^4-x^{37}t^3+y^{37}t^3-z^{37}t^3-xy^{36}t^3+xz^{36}t^3+yz^{36}t^3-x^{10}z^{27}t^3-xy^9z^{27}t^3-x^{27}z^{10}t^3+y^{27}z^{10}t^3+x^{28}y^9t^3-x^{28}z^9t^3+y^{28}z^9t^3-x^{27}yz^9t^3+x^{36}yt^3-x^{36}zt^3-y^{36}zt^3-x^{27}y^9zt^3-x^{39}t-y^{39}t+z^{39}t-x^3y^{36}t+x^3z^{36}t-x^9z^{30}t-y^9z^{30}t-x^{12}z^{27}t+y^{12}z^{27}t-x^3y^9z^{27}t+x^9y^3z^{27}t-x^{27}y^{12}t-x^{27}z^{12}t+y^{30}z^9t-x^3y^{27}z^9t-x^{36}y^3t-x^{36}z^3t-y^{36}z^3t+x^{27}y^9z^3t+x^{40}-y^{40}+z^{40}-xy^{39}+yz^{39}-x^4z^{36}+xy^3z^{36}-x^9y^{31}-x^9z^{31}+y^9z^{31}+x^{10}z^{30}+xy^9z^{30}-x^{12}y^{28}+x^{12}z^{28}+y^{12}z^{28}-x^{13}y^{27}+y^{13}z^{27}-xy^{12}z^{27}-x^4y^9z^{27}-x^9y^4z^{27}-x^{10}y^3z^{27}+x^{12}yz^{27}-x^{27}y^{13}+x^{27}z^{13}+y^{27}z^{13}+xy^{27}z^{12}+x^{30}y^{10}-x^{30}z^{10}-y^{30}z^{10}-x^3y^{27}z^{10}+x^{27}y^3z^{10}-x^{31}y^9+x^{31}z^9+y^{31}z^9-xy^{30}z^9+x^3y^{28}z^9-x^4y^{27}z^9+x^{27}y^4z^9+x^{36}z^4-x^9y^{27}z^4+x^{27}y^9z^4-x^{37}y^3-x^{37}z^3+y^{37}z^3-xy^{36}z^3-x^{27}y^{10}z^3-x^{28}y^9z^3+x^{36}yz^3+x^{39}y+x^{39}z-y^{39}z+x^9y^{30}z-x^{12}y^{27}z-x^{30}y^9z+x^{36}y^3z$$ (this is the trace of $$(x+y\alpha+z\alpha^2+t\alpha^3)^{40}\alpha$$ where $$\alpha$$ is the generator of the multiplicative group of $$\mathbb F_{81}$$ and $$x,y,z,t\in\mathbb F_3$$).

In fact, using that $$x=x^3=x^5=...$$ and $$x^2=x^4=x^6=...$$ for any $$x\in\mathbb F_3$$, one can turn the above into a much shorter example: $$-t^4-t^3 x+t^3 y+t^2 x z-t^2 y^2-t x^3-t x^2 y+t x^2 z+t x y^2+t y^3-t y z^2+t z^3+x^4+x^3 y-x^3 z+x^2 z^2+x y^2 z-x z^3-y^4-y^3 z+z^4$$ However the latter might be non-smooth, as Will Sawin points out in the comment.

• The shorter example may not be smooth, though. I suspect it's equal to the norm form, which has singularities larger fields (but you still have some freedom in how to write it, for example in the $x^3z$ and $xz^3$ terms, and it's possible the way you did is smooth). Jul 31 at 3:26
• @WillSawin Have not thought about it! Will add the warning. Jul 31 at 5:00