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 Zariski-dense in $G(\mathbf{Q}_p)$. The latter is automatic if $G$ is a linear connected algebraic group, by Rosenlicht's theorem. $\endgroup$
    – YCor
    Jun 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? $\endgroup$ Jun 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 Sawin
    Jun 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. $\endgroup$ Jun 27, 2022 at 11:20
  • $\begingroup$ @WillSawin what do you mean by $G_\mathbf{Q}$ then? you can be Zariski-dense in the set of $\mathbf{Q}$-points, or in the whole variety (no mention to $\mathbf{Q}$ needed: this is "Zariski-dense 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 Zariski-dense in $G(\mathbf{Q})$. You're right that if OP assumes plain Zariski-density, the result is trivial with no use of Rosenlicht and is not about algebraic groups. $\endgroup$
    – YCor
    Jun 27, 2022 at 15:22

1 Answer 1



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 finite-dimensional). 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. $\endgroup$
    – Will Sawin
    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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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