Let $G$ be an almost simple group defined over $\mathbb{Q}$. Assume that $\Gamma$ is a subgroup of $G(\mathbb{Q})$ which is Zariski dense. Consider now the $p$adic completion $\mathbb{Q}_p$ of $\mathbb{Q}$ for some prime $p$. If we think of $\Gamma$ as a subgroup of $G(\mathbb{Q}_p)$, is it still Zariski dense?

5$\begingroup$ The answer is yes (for general reasons not specific to algebraic groups) if and only if $G(\mathbf{Q})$ is Zariskidense in $G(\mathbf{Q}_p)$. The latter is automatic if $G$ is a linear connected algebraic group, by Rosenlicht's theorem. $\endgroup$– YCorJun 25, 2022 at 12:49

$\begingroup$ @YCor Thanks for your prompt answer. Could you please give a reference or outline the proof of Rosenlicht's theorem? $\endgroup$– HiggsBosonJun 25, 2022 at 13:03

1$\begingroup$ What do you mean by "$\Gamma$ is Zariski dense"? I think YCor made an assumption about the meaning of this that is different from what I would have. (I would guess the meaning is that it is Zariski dense in $G_{\mathbb Q}$, not in $G(\mathbb Q)$.) $\endgroup$– Will SawinJun 25, 2022 at 19:50

$\begingroup$ @Will Sawin: Yes, $\Gamma$ is Zariski dense in $G_{\mathbb{Q}}$. I wonder whether this is some general statement about the Zariski density for the field extension. $\endgroup$– HiggsBosonJun 27, 2022 at 11:20

$\begingroup$ @WillSawin what do you mean by $G_\mathbf{Q}$ then? you can be Zariskidense in the set of $\mathbf{Q}$points, or in the whole variety (no mention to $\mathbf{Q}$ needed: this is "Zariskidense in $G$"). In any case for a connected linear algebraic group over $\mathbf{Q}$ this is the same. Nevertheless I indeed assumed OP means "which is Zariskidense in $G(\mathbf{Q})$. You're right that if OP assumes plain Zariskidensity, the result is trivial with no use of Rosenlicht and is not about algebraic groups. $\endgroup$– YCorJun 27, 2022 at 15:22
1 Answer
Yes.
Let $f$ be a polynomial over $\mathbb Q_p$ that vanishes on $\Gamma$. Fix a basis for $\mathbb Q_p$ over $\mathbb Q$ (or the $\mathbb Q$subspace of $\mathbb Q_p$ generated by the coefficients of $f$, which is finitedimensional). Then we can write $f = \sum_i f_i \alpha_i$ where $\alpha_i$ lie in that basis and $f_i$ have coefficients in $\mathbb Q$ by decomposing each coefficient of $f$ in that basis. By the linear independence of the basis, the $f_i$ all vanish on $\Gamma$, hence are zero on $G$, so $f$ is zero on $G$.

1$\begingroup$ This proof shows that $\Gamma$ need only be a subset of $G(\mathbb Q)$ which is ZAriski dense in $G_{\mathbb Q}$,; it need not be a subgroup. $\endgroup$ Jun 27, 2022 at 12:31

$\begingroup$ @Venkataramana Correct, and works for any algebraic variety, not necessarily an algebraic group. $\endgroup$ Jun 27, 2022 at 13:36

$\begingroup$ @WillSawin That is a nice argument. Your proof works for any algebraic variety defined over any field $K$, and any field extension $L\supset K$. $\endgroup$ Jun 27, 2022 at 19:11