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Sep 7, 2019 at 3:29 comment added Venkataramana Yes. I meant to write $I\subset {\mathbb Z}[\alpha]$ contains $J=NO-k$.
Sep 6, 2019 at 21:17 comment added Ami you meant $I$ contains $J=NO_k$ ?
Sep 6, 2019 at 14:29 comment added Venkataramana $I\subset {\mathbb Z}[\alpha]$ 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
Sep 6, 2019 at 10:12 comment added Ami Thank you very much, can you please refer to a source regarding the ideal property you mentions?
Sep 6, 2019 at 4:20 comment added Venkataramana 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.
Sep 6, 2019 at 3:36 comment added Venkataramana 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 $\mathbb Z$.
Sep 5, 2019 at 23:41 history edited Ami CC BY-SA 4.0
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Sep 5, 2019 at 18:31 comment added Ami 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 $Λ=\mathbb Z$.
Sep 5, 2019 at 18:19 comment added Venkataramana where in Raghunathan's paper, do you see the statement for $G({\mathbb Z})$?
Sep 5, 2019 at 18:15 history edited YCor CC BY-SA 4.0
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Sep 5, 2019 at 17:42 history asked Ami CC BY-SA 4.0