Timeline for Covolumes of unit groups of division algebras
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 2, 2021 at 15:28 | comment | added | Radu T | Great! Thank you! I'll just add here to complete the answer to my question: using the argument above, the approximation in the question reduces to showing that $(\mathcal{O}/J)^1$ is of the same order of magnitude as $\mathcal{O}/J$, which is true by checking it on matrix algebras. | |
| Nov 2, 2021 at 15:23 | vote | accept | Radu T | ||
| Nov 2, 2021 at 14:59 | comment | added | John Voight | To your second question about $J^r = p\mathcal{O}$: I think for quaternion orders you can ensure equality? But you shouldn't need this, there is still an isomorphism $(1+p\mathcal{O})/(1+J^r) \simeq p\mathcal{O}/J^r$ so everything past the first part of the filtration is computed by the index. | |
| Nov 2, 2021 at 14:49 | history | edited | John Voight | CC BY-SA 4.0 |
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| Nov 2, 2021 at 14:44 | comment | added | John Voight | You're quite right: the filtration is only on $\mathcal{O}^\times$. If you want norm 1 units, you need to consider the ratio $[\mathrm{nrd}(\mathcal{O}_m^\times):\mathrm{nrd}(\mathcal{O}^\times)]=[\mathbb{Z}_p^\times:\mathrm{nrd}(\mathcal{O}^\times)]$ (the reduced norm is surjective on the units of a maximal order, Theorem (33.4) of Reiner's "Maximal Orders"). This factor can be a bit delicate, but it is bounded: we have $\mathbb{Z}_p \subseteq \mathcal{O}$ so $\mathbb{Z}_p^{\times n} \leq \mathcal{O}^\times$ if the central simple algebra has degree $n$, so the index is bounded by $n$. | |
| Nov 2, 2021 at 14:33 | history | edited | John Voight | CC BY-SA 4.0 |
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| Nov 1, 2021 at 16:22 | comment | added | Radu T | Also, I have only found references for $J^r \subset p\mathcal{O}$ but you write $J^r = p\mathcal{O}$. Why can we assume that? | |
| Nov 1, 2021 at 13:11 | comment | added | Radu T | Thank you very much for the great answer! I am still trying to work out the details. One thing that I noticed is that, as far as I understand it, $1 + p \mathcal{O}$ need not lie in the norm 1 units. I computed that $[\mathcal{O}_m^\times : \mathcal{O}^\times] = [\mathcal{O}_m^1 : \mathcal{O}^1] \cdot [N(\mathcal{O}_m^\times) : N(\mathcal{O}^\times)]$. How can we also control the size of the quotient of norms? Have I made a mistake somewhere? | |
| Oct 29, 2021 at 17:15 | history | answered | John Voight | CC BY-SA 4.0 |