# Rational points

Let G be an affine algebraic group defined over k, with k a field of characteristic zero. Suppose G has only one single k-point, can we conclude that G does not have more points?

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I can't see what you are asking here. You say $G$ has just one $k$-point and then ask if it has any more. Are you sure this is what you wanted to ask? –  Robin Chapman Jul 14 '10 at 17:18
Here is a simple counterexample: $k=\mathbb{Q}$, and $G = \mu_3$, the group of third roots of unity. The underling scheme of $G$ has only one $\mathbb{Q}$ point, but three $\mathbb{C}$ points. –  David Speyer Jul 14 '10 at 17:38
Do you intend to insist that $G$ is connected (which would rule out David Speyer's example), or equivalently geometrically connected? Anyway, the answer is still negative (assuming $G \ne 1$!): over fields of characteristic 0, every smooth connected affine group is unirational and hence has a Zariski-dense locus of rational points. This relies crucially on char. 0, as well as structural facts from the theory of connected reductive groups. So if $G$ is of positive dimension, the answer is "no" (by consideration of its identity component). –  BCnrd Jul 14 '10 at 18:10
Maybe I'm missing something. Doesn't that mean the answer to the OP's question is yes in the connected case? –  Kevin Ventullo Jul 14 '10 at 20:36
@Kevin: the question is asked as a negative ("can we conclude...does not..."), so it may be confusing, but I think the answer is "no"; anyway, the content of the answer is clear (there are more $k$-pts!). Here is a proof valid for non-unipotent smooth affine groups of positive dimension over any infinite field $k$: by Grothendieck, such groups always have a non-trivial $k$-torus, and those are unirational, QED. A variant works over any infinite perfect field in the unipotent case. But over imperfect fields it can fail: over $k(t)$ for $k$ of char. $p > 2$, take $G = {y^p = x - t x^p}$. –  BCnrd Jul 14 '10 at 23:05