Algebraic group with no $k$-rational points Let $k$ be a field, and $G$ an algebraic group. What are some nontrivial examples of an algebraic group with no $k$-rational points?
 A: Here's a simple example where there are no nontrivial $k$-rational points. Consider $\mu_n = \text{Spec } \mathbb{Z}[x]/(x^n - 1)$, the $n^{th}$ roots of unity, with the obvious multiplication, and its base change to any field. When $n = 3$ and the field is $\mathbb{Q}$ there are no nontrivial $\mathbb{Q}$-rational points. 
A: I will give an example here using $SL_n$ though this story generalizes to other reductive algebraic groups.
If we work over the algebraically closed field $k$ of characteristic $p > 0$ then $SL_n$ comes equipped with the Frobenius map $F: SL_n\to SL_n$ which raises each entry in a matrix to the $p$'th power.
If we regard $SL_n$ as a scheme, then the kernel of this map is not trivial, but it has no non-trivial points over $k$ (or in fact over any reduced ring). So why is this particular subgroup interesting?
It turns out that representations of this subgroup (seen as a group scheme) correspond to the representations of the Lie algebra of $SL_n$ (as a restricted Lie algebra).
For more about this whole story, I recommend Jantzen's Representations of Algebraic Groups. 
A: A group scheme $G$ over any base scheme $S$ satisfies (by definition!) $G(S)\neq\emptyset$. In particular, a $k$-group scheme for $k$ a field has a $k$-rational point (the identity section). But that might be the only $k$-rational point. Some examples exhibiting this behavior have already been exhibited. Another one (similar to one others but with the additive group in place of $\mathrm{SL}_n$), for $k$ of characteristic $p$, is the group scheme $\alpha_p$ of $p$-th roots of zero. Its coordinate ring is $k[X]/(X^p)$ (so this group scheme has underlying topological space a single point) and its functor of points sends a $k$-algebra $R$ to $\{r\in R:r^p=0\}$. Note that the group law is addition. When $R$ is a field, $\alpha_p(R)$ is the trivial group. 
However, there are some classes of (linear) algebraic groups for which this can't possibly happen. Namely for $k$ infinite, if $G$ is a smooth connected affine $k$-group scheme and either $k$ is perfect or $G$ is reductive, then $G(k)$ is Zariski dense in $G$.
