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Kirk
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Let $\mathcal{O}$ be the ring of integers in an algebraic number field and let $R \subset \mathcal{O}$ be an order. For instance, we might have $\mathcal{O} = \{\text{$x+i y$ $|$ $x,y \in \mathbb{Z}$}\}$ and $R = \{\text{$x+i y \in \mathcal{O}$ $|$ $y$ even}\}$. Question : Is $\text{SL}_n(R)$ a finite-index subgroup in $\text{SL}_n(\mathcal{O})$? I might be missing something obvious here, but I'm having trouble proving this (maybe because it isn't true?).

Let $\mathcal{O}$ be the ring of integers in an algebraic number field and let $R \subset \mathcal{O}$ be an order. For instance, we might have $\mathcal{O} = \{\text{$x+i y$ $|$ $x,y \in \mathbb{Z}$}\}$ and $R = \{\text{$x+i y \in \mathcal{O}$ $|$ $y$ even}\}$. Question : Is $\text{SL}_n(R)$ a finite-index subgroup in $\text{SL}_n(\mathcal{O})$? I might be missing something obvious here, but I'm having trouble proving this.

Let $\mathcal{O}$ be the ring of integers in an algebraic number field and let $R \subset \mathcal{O}$ be an order. For instance, we might have $\mathcal{O} = \{\text{$x+i y$ $|$ $x,y \in \mathbb{Z}$}\}$ and $R = \{\text{$x+i y \in \mathcal{O}$ $|$ $y$ even}\}$. Question : Is $\text{SL}_n(R)$ a finite-index subgroup in $\text{SL}_n(\mathcal{O})$? I might be missing something obvious here, but I'm having trouble proving this (maybe because it isn't true?).

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Kirk
  • 53
  • 3

Is SL_n of an order in a number ring finite-index in SL_n of the number ring?

Let $\mathcal{O}$ be the ring of integers in an algebraic number field and let $R \subset \mathcal{O}$ be an order. For instance, we might have $\mathcal{O} = \{\text{$x+i y$ $|$ $x,y \in \mathbb{Z}$}\}$ and $R = \{\text{$x+i y \in \mathcal{O}$ $|$ $y$ even}\}$. Question : Is $\text{SL}_n(R)$ a finite-index subgroup in $\text{SL}_n(\mathcal{O})$? I might be missing something obvious here, but I'm having trouble proving this.