Timeline for The normalizer of $\mathrm{GL}(n,\mathbf Z)$ in $\mathrm{GL}(n,\mathbf Q)$
Current License: CC BY-SA 3.0
11 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Nov 13, 2011 at 11:58 | comment | added | Homology | @KConrad However the UFD case follows from the localization (I edited my answer below). | |
| Nov 12, 2011 at 17:31 | comment | added | Henri | @KConrad No, the very same matrix also conjugates $M_2(\mathbb{Z}[\frac{1+\sqrt{-15}}{2}])$. Homology's answer shows that $A$ conjugates $\mathrm{GL}_n(R)$ iff $I^n =(det(A))$ where $I$ is the (fractional) ideal generated by the coefficients of $A$, so that one can always take a bigger $R' \subset \mathrm{Frac}(R)$. | |
| Nov 12, 2011 at 17:21 | comment | added | KConrad | In Henri's example, ${\mathbf Z}[\sqrt{-15}]$ is not the full ring of integers of ${\mathbf Q}(\sqrt{-15})$. Perhaps that "explains" the example. Emerton's argument goes through for PIDs. Does it go through more generally for any Dedekind domain, perhaps by a localization argument to reduce to the case of a PID? | |
| Nov 12, 2011 at 17:02 | comment | added | Henri | It is not true in general, as $\begin{pmatrix} 2 \sqrt{-15} & 5 \\ 5 & - \sqrt{-15} \end{pmatrix} \in \mathrm{GL}_2(\mathbb{Q}(\sqrt{-15}))$ conjugates $M_2(\mathbb{Z}[\sqrt{-15}])$ but is not a scalar multiple of an element of $\mathrm{GL}_2(\mathbb{Z}[\sqrt{-15}])$ (its determinant, $5$, is not a square times a unit). | |
| Nov 12, 2011 at 12:43 | answer | added | Olod | timeline score: 4 | |
| Nov 12, 2011 at 11:53 | answer | added | Homology | timeline score: 5 | |
| Nov 11, 2011 at 16:05 | vote | accept | Olod | ||
| Nov 11, 2011 at 13:58 | answer | added | Emerton | timeline score: 24 | |
| Nov 11, 2011 at 9:55 | history | edited | Olod | CC BY-SA 3.0 |
Link to the question at mathunderflow
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| Nov 11, 2011 at 9:48 | history | asked | Olod | CC BY-SA 3.0 |