Pointless groups II

Question

This question is a sequel to Pointless groups, where I asked for a certain kind of counterexample. @DanielLitt produced an elegant and easy-to-understand counterexample, but also suggested a sense in which it might be the only counterexample.

At the suggestion of @R.vanDobbendeBruyn and @YCor, this question has yet another sequel: Pointless groups III.

Fix a field $k$ — not algebraically closed, and, in fact, this question is obviously of interest only if $k$ is finite. (EDIT: @R.vanDobbendeBruyn points out in the comments that my memory that the $k$-points are Zariski-dense in the infinite case is only guaranteed true for reductive groups, and gives a counterexample for an infinite, imperfect field.) 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?

Answer 1

One other type of example is constructed in Conrad–Gabber–Prasad's Pseudo-reductive groups, Example 11.3.2. The construction is kind of classic (and possibly predates the book), so let me recall it here:

Example. Let $k = \bar{\mathbf F}_p(t)$, and let $G \subseteq \mathbf G_a \times \mathbf G_a$ be the subgroup $$\left\{(x,y) \in \mathbf G_a \times \mathbf G_a\ \big|\ y^p = x-tx^p\right\}.$$ It is smooth and becomes isomorphic to $\mathbf G_a$ over $k(t^{1/p})$ (replace $y$ by $y+t^{1/p}x$ and eliminate $x$).

Lemma. If $p > 2$, then $G(k) = \{(0,0)\}$.

Proof. Let $(x,y) \in G(k)$. If $x = 0$, then $y = 0$ as well, so we may assume $x \neq 0$. Then $x$ cannot be a $p$-th power, for otherwise $t$ is a $p$-th power. Thus there exists $a \in \bar{\mathbf F}_p$ such that $v:=v_a(x)$ is not divisible by $p$. The equation $y^p = x-tx^p$ gives $$(y+ax)^p = x-(t-a)x^p.$$ The ultrametric triangle inequality implies that if $\alpha+\beta = \gamma$, then the set of valuations $\{v_a(\alpha),v_a(\beta),v_a(\gamma)\} \subseteq \mathbf Z \cup \{\infty\}$ has at most $2$ elements. Since $v_a((y+ax)^p)$ is divisible by $p$ but neither $v=v_a(x)$ nor $v_a((t-a)x^p)=pv+1$ is, we conclude that $v=pv+1$, i.e. $(p-1)v=-1$, which is impossible since $p > 2$ and $v \in \mathbf Z$. $\square$

For $p = 2$ there is also the point $(1/t,0)$, and again Conrad–Gabber–Prasad show that this is the only other point.

Remark. On the other hand, if $k$ is an infinite field and either $k$ is perfect or $G$ is reductive, then $G$ is unirational, hence $G(k) \subseteq G$ dense. See for instance Milne's notes, Corollaries 19.21 and 19.22. Thus, the only examples are among imperfect fields and finite fields.