1
$\begingroup$

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?

$\endgroup$
9
  • 6
    $\begingroup$ 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? $\endgroup$ Jul 14, 2010 at 17:18
  • 5
    $\begingroup$ 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. $\endgroup$ Jul 14, 2010 at 17:38
  • 3
    $\begingroup$ 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). $\endgroup$
    – BCnrd
    Jul 14, 2010 at 18:10
  • 3
    $\begingroup$ @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 $G = {y^p = x - t x^p}$. $\endgroup$
    – BCnrd
    Jul 14, 2010 at 23:05
  • 1
    $\begingroup$ 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. $\endgroup$
    – BCnrd
    Jul 15, 2010 at 17:32

1 Answer 1

3
$\begingroup$

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 $k$.

$\endgroup$
2
  • 1
    $\begingroup$ If $G$ is an anisotropic torus, your characterization of $k$-points is a bit off. $\endgroup$
    – S. Carnahan
    Apr 24, 2016 at 8:51
  • $\begingroup$ 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. $\endgroup$ Apr 24, 2016 at 14:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.