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Let $G$ be an affine algebraic group defined over a field of characteristic zero $K$. 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
@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
Ana, my comments prove that in the connected case there will always be more rational points (in fact for any smooth connected affine group over any infinite field if we assume non-unipotence, and without needing that constraint in char. 0). I'm not sure if you call this "yes" or "no" since you posed a negative question and hence I am getting disoriented about which word to use (but the content should be clear, as I said to Kevin earlier). I'm not sure why you ask about a "finite and reductive" group, since Speyer gave such an example without nontrivial rational points. – BCnrd Jul 15 '10 at 17:32

Question (edited here)

Let $G$ be an affine algebraic group defined over a field $k$ of characteristic zero. Is it possible for $|G(k)|=1$, even if $G$ is not trivial?

As shown by David Speyer in the comments, if $\dim G=0$ then yes. For example, let $G$ be the solutions to $z^3-1$. Then over $|G(\mathbb{Q})|=1$, but $|G(\mathbb{C})|=3$ and hence $G$ is not trivial.

On the other hand, the comments by Brian Conrad show that if $\dim G \geq 1$, then $|G(k)|\not=1$.

I think this proves it: Since the identity component of $G$ is a connected affine algebraic group over $k,$ it suffices to prove this for $G$ connected. Then, since we are in characteristic 0, $G$ is isomorphic (as a variety, but not as an algebraic group) to $(G/G_u) \times G_u$ where $G_u$ is its unipotent radical. The unipotent radical is likewise isomorphic to an affine space, and $G/G_u$ is reductive. By the Bruhat-decomposition $G/G_u$ contains an affine open subset whose $\overline{k}$-points are isomorphic to $(\overline{k}^*)^n \times \overline{k}^m$ where $\overline{k}$ is an algebraic closure of $k$.

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If $G$ is an anisotropic torus, your characterization of $k$-points is a bit off. – S. Carnahan Apr 24 at 8:51
Thank you. I edited it (I think) to fix this problem. For example, $SO(2,\mathbb{R})$ is a anisotropic torus but its $\mathbb{C}$-points are $\mathbb{C}^*$, which shows there is more than one $\mathbb{R}$-point. – Sean Lawton Apr 24 at 14:00

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