Suppose G is an algebraic group defined over F, the algebraic closure of F is K. Consider the Zariski topology on G(K), is G(F) Zariski dense in G(K)?
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4$\begingroup$ Obviously the answer is no if the field F is finite, so your question needs more context and a more precise formulation. Are you dealing just with linear algebraic groups? $\endgroup$– Jim HumphreysCommented Feb 21, 2011 at 18:04
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$\begingroup$ Other necessary condition when $F$ is imperfect: if $G(F)$ is dense in $G$, then $G_{\mathrm{red}}$ is a smooth algebraic group over $F$. Proof: the regular locus of $G_{\mathrm{red}}$ is open and non-empty, so contains a rational point. This point is then smooth. By translation, $G_{\mathrm{red}}$ is smooth at origine, hence smooth everywhere. This implies that it is an algebraic group because it is geometrically reduced. $\endgroup$– Qing LiuCommented Feb 22, 2011 at 17:46
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$\begingroup$ Unfortunately my arguments are not correct as pointed out by Brian Conrad by email. He propose to repair in the following way: consider the $F$-scheme of finite type $X=G_{\mathrm{red}$. The fact that $X(F)$ is dense implies (as in above comment) that the smooth locus $U$ of $X$ is dense (any non-empty open subset of $X$ contains a rational and regular hence smooth point). Denote by $i : U\to X$ the inclusion. Then $O_X\to i_*O_U$ is injective because $X$ is reduced. As $U$ is geometrically reduced, this implies that $X$ too. One can also so that $X_{\bar{F}}$ is $(S_1)$ $\endgroup$– Qing LiuCommented Feb 23, 2011 at 20:21
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$\begingroup$ (continued) and reduced at generic points. Now using the group structure on $G$ we see that $G_{\mathrm{red}}$ is an algebraic group, smooth because it is geometrically reduced. $\endgroup$– Qing LiuCommented Feb 23, 2011 at 20:22
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1$\begingroup$ @QingLiu's comment, with the missing brace restored: "Unfortunately my arguments are not correct as pointed out by @BCnrd by email. He proposes to repair in the following way: consider the $F$-scheme of finite type $X=G_\text{red}$. The fact that $X(F)$ is dense implies (as in the above comment) that the smooth locus $U$ of $X$ is dense (any non-empty open subset of $X$ contains … mathoverflow.net/questions/56192#comment950605_56192 $\endgroup$– LSpiceCommented Oct 27, 2020 at 18:49
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3 Answers
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If $G$ is connected and reductive over an arbitrary field $k$, then $G$ is $k$-unirational. Thus if $k$ is moreover infinite, $G(k)$ is Zariski-dense in $G(\overline{k})$. This is due to Rosenlicht (Ann. Mat. Pura Appl. (4) 43 (1957), 25--50; MR0090101) in the case of perfect $k$ and Grothendieck (SGAIII) in the arbitrary case.
Certainly one wants to restrict to linear groups, since an abelian variety over a global field can have a finite number of rational points.
Perhaps someone else can speak to the linear, non-reductive case.
answered Feb 21, 2011 at 19:34
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2$\begingroup$ Connectedness is essential: suppose that $G=\mu_n, n>1$ and $k$ contains no non-trivial $n$th root of unity (for example, $k=\mathbb{Q}$ and $n=3$), then $G(k)$ has order 1, whereas $G(\bar{k})$ has order $n.$ $\endgroup$ Commented Feb 23, 2011 at 2:47
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To amplify Pete's answer, there is a reasonable discussion in Section 18 of the second edition of Borel's Linear Algebraic Groups (Springer GTM 126). In particular, his Corollary 18.3 following a discussion of unirationality in linear algebraic groups defined over a field $k$ states:
Let $G$ be connected, $k$ infinite. If either $k$ is perfect, or $G$ is reductive, $G(k)$ is Zariski-dense in $G$.
He goes on to note an example given by Rosenlicht in the paper cited by Pete, giving a one-dimensional unipotent group over an infinite but imperfect field $k$ for which $G(k)$ fails to be Zariski-dense. I'm not sure what further results are out there in the literature, but one certainly has to be cautious.
answered Feb 21, 2011 at 20:02
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4$\begingroup$ For those who don't have access to Rosenlischt's paper: Let $k$ be an imperfect field of characteristic $p$, and let $t \in k \setminus k^p$. Then the variety $x^p \equiv t y^p$ is a subgroup of $\mathbb{G}_a^2$ and, other then the origin, contains no $k$ points. $\endgroup$ Commented Feb 21, 2011 at 21:41
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$\begingroup$ I don't know whether or not this is Rosenlicht's example, but it certainly seems like the most obvious one. $\endgroup$ Commented Feb 21, 2011 at 21:41
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$\begingroup$ @David: I saw Rosenlicht's paper long ago but don't have a copy. Your example is almost certainly the one he used. It appears also in some related counterexamples discussed in 12.1 of T.A. Springer's book Linear Algebraic Groups. $\endgroup$ Commented Feb 21, 2011 at 22:30
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7$\begingroup$ BCnrd e-mails me to point out that this example is not smooth (check the Jacobian condition). This is one of those bizarre examples of a non-smooth group scheme in characteristic p. (It is also an example of a scheme which is reduced, but not geometrically reduced.) One might think that these features are crucial but they are not. BCnrd points out that Rosenlicht's example, $y^p = x + t x^p$, for $p>2$, has none of these peculiar features, but still does not have a Zariski dense set of k points. $\endgroup$ Commented Feb 22, 2011 at 2:35
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$\begingroup$ The indirect comments from BCnrd are very helpful. (I did say "almost certainly" to hedge my bets.) Actually I just discovered I'm able to bring up that 1957 paper on my screen (via the library). On page 46 Rosenlicht writes his example in the form
$y^p−y=tx^p$with$t$indeterminate over a field of odd characteristic; this defines a form of the affine line over the function field, etc. $\endgroup$ Commented Feb 22, 2011 at 13:51
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A (smooth, connected) unipotent group $U$ is said to be $k$-split if there is a filtration by $k$-subgroups for which the successive quotients are isomorphic to $\mathbf{G}_{a/k}$. The examples mentioned in comments (e.g. the subgroup of $\mathbf{G}_a^2$ defined by $y^p - y = tx^p$) are non-split unipotent groups. Any $k$-split unipotent group $U$ is even a rational variety (in fact, $k$-isomorphic as a variety to $\mathbf{A}^n$) so it is clear that $U(k)$ is Zariski dense in $U(k_{alg})$ when $k$ is infinite.
More generally, let $G$ be a (smooth) linear algebraic group over $k$ and assume that the unipotent radical of $G$ is defined and split over $k$ (both of these conditions can fail). Then as a $k$-variety, $G$ is just the product of its reductive quotient $G_{red}$ and its unipotent radical (result of Rosenlicht). In particular, $G$ is unirational and if $k$ is infinite, $G(k)$ is dense in $G(k_{alg})$.
Of course, this observation isn't that interesting -- in some sense, it just "identifies" the problem.
answered Feb 22, 2011 at 20:58
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$\begingroup$ Just a tiny comment, four years later as I read through Milne's notes jmilne.org/math/CourseNotes/ala.html which reference this question, that your $G_{\text{red}}$ (the quotient of $G$ by its unipotent radical) is not the same as $G_{\text{red}}$ in, for example, Qing Liu's comment (the underlying reduced scheme). $\endgroup$– LSpiceCommented Aug 1, 2015 at 18:56