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This question is a sequel to Pointless groups, to which @DanielLitt produced an elegant and easy-to-understand counter-example, and Pointless groups II, where @R.vanDobbendeBruyn pointed out that my implicit claim about Zariski density of rational points of groups over infinite fields is only guaranteed for reductive groups (or perfect fields), and gave a counterexample (still elegant, though less easy for me to understand than a split torus over $\mathbb F_2$!) for an infinite but imperfect field.

At the suggestion of @R.vanDobbendeBruyn and @YCor, I am engaging in the extreme bad manners of once again modifying my search for counterexamples to rule out the counterexamples already provided. The revised question is @DanielLitt's wording.

So, 3rd time's a charm, hopefully … fix a finite field $k$. Then "group" means "smooth, connected, affine algebraic group scheme of finite type over $k$".

If $G$ is a (smooth, connected) group, and $G(k)$ is trivial, then does that imply that $k = \mathbb F_2$ and $G$ is a split torus?

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    $\begingroup$ By Corollary 1.5 here we can reduce to the case of tori. Some version of Weil's bound on number of points on a variety might give the required result. (Edit: not quite correct, I forgot the unipotent case. YCor's comment below is correct though) $\endgroup$
    – Wojowu
    Commented Jun 11, 2022 at 14:01
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    $\begingroup$ By a theorem of Grothendieck, $G$ has a maximal torus that is defined over $k$. Hence, if, for given $k$, $G$ is a counterexample of minimal dimension, then $G$ is either a torus or unipotent. In Borel's book(16.5(ii)) it is proved that quotient maps are surjective on $k$-points for finite fields. Hence one can suppose that $G$ is abelian. For perfect fields, the unipotent abelian groups are $k$-isomorphic to the standard one. So a minimal counterexample should be a torus. $\endgroup$
    – YCor
    Commented Jun 11, 2022 at 14:02
  • $\begingroup$ In turn, since a non-trivial split torus over a field $k \ne \mathbb F_2$ has non-trivial $k$-points, a minimal counterexample when $k \ne \mathbb F_2$ is an anisotropic torus. A minimal counterexample when $k = \mathbb F_2$ is not split (by the way the question was stated), so has a non-trivial anisotropic torus, which necessarily has no non-trivial $k$-points (since the full torus has none); so, even for $k = \mathbb F_2$, a minimal counterexample is anisotropic. $\endgroup$
    – LSpice
    Commented Jun 11, 2022 at 15:11
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    $\begingroup$ I think this finishes it. The number of rational points of a torus of dimension $d$ over $\mathbb{F}_q$ is computed by $(-1)^d\det(\Phi-q\cdot \text{Id})$, where $\Phi$ is the action of Frobenius on the characters. But for $q>2$, or $\Phi$ not equal to the identity, $\Phi-q\cdot \text{Id}$ has all eigenvalues of size at least $1$, and at least one eigenvalue greater than one. So $(-1)^d\det(\Phi-q\cdot \text{Id})=1$ implies $q=2$ and $\Phi=\text{Id}$, which is what we were trying to prove. $\endgroup$
    – Daniel Litt
    Commented Jun 11, 2022 at 15:16

1 Answer 1

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Yes.

Let $|k| = q$. Let $T$ be an $n$-dimensional torus defined over $k$ and let $\overline{T}$ be the base change of $T$ to $\overline{k}$. Then the character group $\text{Hom}(\overline{T}, \mathbb{G}_m)$ is isomorphic to $\mathbb{Z}^n$ and the Frobenius acts on $\text{Hom}(\overline{T}, \mathbb{G}_m)$ by a matrix $F$. If $T$ splits over $\mathbb{F}_{q^n}$, then $F$ obeys $F^n = q^n \text{Id}$. The isomorphism class of the torus $T$ is determined by the $\text{GL}_n(\mathbb{Z})$ conjugacy class of $F$, and we have $\# T(\mathbb{F}_q) = \det(F - \text{Id})$. So we are reduced to the following linear algebra problem:

If $F$ is an integer matrix obeying $F^n = q^n \text{Id}$ and $\det(F - \text{Id}) = 1$, can we deduce that $q=2$ and $F = 2 \text{Id}$?

The answer is "yes". The condition that $F^n = q^n \text{Id}$ means that the eigenvalues of $F$ are of the form $q \zeta$ for $\zeta$ a root of unity. So we have $\det(F - \text{Id}) = \prod_i (q \zeta_i - 1) = 1$ where the $\zeta_i$ are various roots of unity.

But $|q \zeta - 1| \geq 1$ for any prime power $q$ and any root of unity $\zeta$, and strictly greater unless $q=2$ and $\zeta=1$, so the only solution is $q=2$ and $F = \text{Id}$. $\square$

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    $\begingroup$ The point-counting approach was also mentioned by @DanielLitt. $\endgroup$
    – LSpice
    Commented Jun 11, 2022 at 15:58
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    $\begingroup$ The proof above implies the result for every affine group scheme $G$ (since we are counting points over a finite field $k$, we can replace $G$ by its associated reduced scheme, which is a smooth affine group scheme). There is a smooth projective variety over $k$ that parameterizes Borel subgroup schemes, and this variety is $\mathcal{O}$-acyclic, hence has a point. Thus $G$ contains a Borel $B$. This is a torsor for $R_u(B)$ over $B/R_u(B)$, and all $R_u(B)$-torsors over $k$ have $|k|^d$ points, where $d=\text{dim}R_u(B)$. So $R_u(B)$ is trivial, and $B/R_u(B)$ has a unique point . . . $\endgroup$
    – Jason Starr
    Commented Jun 11, 2022 at 17:08
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    $\begingroup$ . . . Thus $B/R_u(B)$ is a split torus and $|k|=2$ by the proof above. Since $d$ equals $0$, the semisimple part of $G$ is trivial. So $G$ equals $B$ equals $B/R_u(B)$ equals a split torus. $\endgroup$
    – Jason Starr
    Commented Jun 11, 2022 at 17:09
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    $\begingroup$ Here $\mathcal{O}$-acyclic is as in the work of Katz on the Ax-Grothendieck theorem: every smooth, projective, geometrically connected variety over a finite field such that the higher cohomology of the structure sheaf vanishes has number of rational points congruent to $1$ modulo the cardinality of the field. $\endgroup$
    – Jason Starr
    Commented Jun 11, 2022 at 21:31
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    $\begingroup$ Just a correction. That theorem is the "Ax-Katz theorem". Katz's paper does follow earlier results of Ax and Grothendieck, but the "Ax-Grothendieck theorem" is a different result (using point counts of varieties over finite fields). $\endgroup$
    – Jason Starr
    Commented Jun 11, 2022 at 22:31

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