# Zariski density preserved under $p$-adic completion?

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?

• 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.
– YCor
Jun 25, 2022 at 12:49
• @YCor Thanks for your prompt answer. Could you please give a reference or outline the proof of Rosenlicht's theorem? Jun 25, 2022 at 13:03
• 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 $G(\mathbb Q)$.) Jun 25, 2022 at 19:50
• @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. Jun 27, 2022 at 11:20
• @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.
– YCor
Jun 27, 2022 at 15:22

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 $$G$$.
• 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. Jun 27, 2022 at 12:31
• @WillSawin That is a nice argument. Your proof works for any algebraic variety defined over any field $K$, and any field extension $L\supset K$. Jun 27, 2022 at 19:11