Let $\alpha$ be an integral element and set $k=\mathbb Q(\alpha)$, and $U$ is a $k$-defined unipotent subgroup of $\operatorname{GL}_n(k)$ with $k-rank\geq 2$.
Let $Λ$ be a subgroup of $U(\mathcal{O}_k)$ which is Zariski-dense in $U$, I'm trying to prove there is an ideal $I\lhd\mathcal{O}_k$ such that $U(I)\subset Λ$.

The congruence subgroup property applies for unipotent groups under the $\mathcal{O}_S$ ring so I think my question comes down to:

Is $Λ$ of finite index in $U(\mathcal{O}_k)$?
Specifically, I'm looking at $U$ to be all the upper triangular matrices with ones on the diagonal in $\mathrm{SL}_n(k)$ for $n\geq 3$

  • $\begingroup$ No. Take $U={\Bbb G}_a$. Then $$\Lambda={\Bbb Z}\subset\mathcal{O}_k=U(\mathcal{O}_k)$$ is a counter-example. $\endgroup$ – Mikhail Borovoi Sep 12 at 2:45
  • $\begingroup$ You seem to be asking many questions (e.g., 1 and 2) that are variants of the same underlying question, which seems to indicate that you're not formulating the question you really mean. I think appropriate etiquette is to figure out fewer questions—ideally just one—rather than asking a lot of variants on the same theme. $\endgroup$ – LSpice Sep 12 at 2:45
  • $\begingroup$ @LSpice indeed I'm working on a big research question, I am trying to find the big path to the answer myself but sometimes I get stuck on the small stuff in unfamiliar math. I will try to figure out fewer questions. $\endgroup$ – Ami Sep 12 at 3:37
  • $\begingroup$ @MikhailBorovoi I forgot to add that $k-rank(U)\geq 2$ $\endgroup$ – Ami Sep 12 at 3:50
  • $\begingroup$ No. Take $U=\Bbb G_a\times \Bbb G_a$, $\Lambda=\Bbb Z\times\Bbb Z$. $\endgroup$ – Mikhail Borovoi Sep 12 at 4:35

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