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Let $G$ be a simply connected complex Chevalley group and let $T$ be some maximal torus in $G$ with $\dim T\geq 2$.

From "A note on generators for arithmetic subgroups of algebraic groups" by Raghunathan, we have that for every ideal $I\lhd\mathbb Z$, $$\Gamma(I):=\langle U_\phi(I)|\phi\in\Phi \rangle$$ is of finite index in $G(\mathbb Z)$.

Is it also true for every ideal $I\lhd\mathbb Z[\alpha]$ that $Γ(I)$ of finite index in $G(\mathbb Z[\alpha])$?

Specifically, I'm looking at $G=SL_n$

Where:

  • $\Phi$ denote the root system of $G$ with respect to $T$;
  • $U_\phi$ is the root subgroup corresponding to $\phi\in\Phi$;
  • $\alpha$ is an element that is integral over $\mathbb Z$;
  • if $I$ is an ideal of the ring $A$ and $H\leq G$ we mark $H(I):=\{x\in H(A)|x\equiv1(\mathrm{mod}\,I)\}$ where $H(A):=H\cap \mathrm{GL}_n(A)$.
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    $\begingroup$ where in Raghunathan's paper, do you see the statement for $G({\mathbb Z})$? $\endgroup$
    – Venkataramana
    Commented Sep 5, 2019 at 18:19
  • $\begingroup$ I believe in page 365 (Theorem 1.2). He uses in the paper the ring of S-integers in $k$, I don't think I understand it enough but from what I gather for $k=\mathbb Q$ we can get $Λ=\mathbb Z$. $\endgroup$
    – Ami
    Commented Sep 5, 2019 at 18:31
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    $\begingroup$ If he uses the ring of S integers in the number field k, then that answers your question; in any case, the answer to your question is indeed yes; that is what is proved in Raghunathan's paper. Not just the special case of $\mathbb Z$. $\endgroup$
    – Venkataramana
    Commented Sep 6, 2019 at 3:36
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    $\begingroup$ The reason is that $G({\mathbb Z}[\alpha ])$ is commensurate to $G(O_k)$ where $k$ is the field over $\mathbb Q$ generated by $\alpha$; every non-zero ideal $I$ of ${\mathbb Z}[\alpha]$ contains a nonzero ideal $J$ of $O_k$; thus Raghunathan's theorem applied to $O_k$ and $J$ gives what you want. $\endgroup$
    – Venkataramana
    Commented Sep 6, 2019 at 4:20
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    $\begingroup$ $I\subset {\mathbb Z}[\alpha]$ has finite index in $O_k$. Hence $J$ contains $I=NO_k$ for some non-zero integer $N$ since $O_k$ is a finitely generated torsion free abelian group. Now $I$ is a nonzero ideal $\endgroup$
    – Venkataramana
    Commented Sep 6, 2019 at 14:29

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