# $U$ unipotent $\mathbb{Q}(\alpha)$-group then no infinite index subgroup is Zariski-dense

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 $$n\geq 3$$

• No. Take $U={\Bbb G}_a$. Then $$\Lambda={\Bbb Z}\subset\mathcal{O}_k=U(\mathcal{O}_k)$$ is a counter-example. – Mikhail Borovoi Sep 12 at 2:45
• 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. – LSpice Sep 12 at 2:45
• @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. – Ami Sep 12 at 3:37
• @MikhailBorovoi I forgot to add that $k-rank(U)\geq 2$ – Ami Sep 12 at 3:50
• No. Take $U=\Bbb G_a\times \Bbb G_a$, $\Lambda=\Bbb Z\times\Bbb Z$. – Mikhail Borovoi Sep 12 at 4:35