Timeline for Is diagonalizability a local property?
Current License: CC BY-SA 4.0
15 events
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| Jun 9, 2023 at 2:07 | history | edited | Yikun Qiao | CC BY-SA 4.0 |
added 43 characters in body
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| Jun 8, 2023 at 16:12 | history | became hot network question | |||
| Jun 8, 2023 at 15:37 | comment | added | YCor | To make sense of items 2, 3, 4, would you introduce $\mathfrak{p}$ somewhere? and specify "for all $\mathfrak{p}$" or "for the given $\mathfrak{p}$, or something. | |
| Jun 8, 2023 at 15:11 | comment | added | Gro-Tsen | (I mean, I thought the question was about the openness or constructibility of the set of diagonalizable matrices, with various conditions on diagonalizability, inside all matrices. So, yes, I misread, but I still think this is very confusing.) | |
| Jun 8, 2023 at 15:08 | comment | added | Gro-Tsen | @R.vanDobbendeBruyn Aaaaaaah! The matrix is fixed in the whole question, the thing that is made to vary is where it is considered! OK, I misread the question, but, considering the upvotes my comment got, I'm not the only one who was confused (and the title of the question is especially confusing). This deserves at least a warning or clarification. | |
| Jun 8, 2023 at 14:49 | vote | accept | Yikun Qiao | ||
| Jun 8, 2023 at 14:36 | comment | added | R. van Dobben de Bruyn | @Gro-Tsen if you believe this locus is constructible, then it is actually open since it is also stable under generalisation [Tag 0903]: if $\mathfrak p \subseteq \mathfrak q$ and $A_{\mathfrak q}$ is diagonalisable, then $A_{\mathfrak p}$ is diagonalisable as well. I think you might be confusing stalks with fibres, but it was already noted that (4) is only constructible. | |
| Jun 8, 2023 at 14:18 | answer | added | R. van Dobben de Bruyn | timeline score: 19 | |
| Jun 8, 2023 at 14:13 | comment | added | Yikun Qiao | @Gro-Tsen I do not think this is a counterexample. Your example satisfies (1) since $A$ is the identity matrix (is this what in your mind?), and then (1) is true for any $R$. Then (2) holds for every prime of $\mathrm{Spec}(R)$. | |
| Jun 8, 2023 at 14:08 | comment | added | Yikun Qiao | @DaveBenson That is Spec(R), just edited. | |
| Jun 8, 2023 at 14:07 | history | edited | Yikun Qiao | CC BY-SA 4.0 |
Change to Spec(R)
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| Jun 8, 2023 at 11:31 | comment | added | Dave Benson | I'm confused as to what you mean by Spec($A$). | |
| Jun 8, 2023 at 9:15 | comment | added | Gro-Tsen | I'm confused: if $A=1$ is the identity and $R=\mathbb{C}$, there's no Euclidean neighborhood of $1$ consisting only of diagonal matrices, because each neighborhood will contain a matrix equal to $1$ plus a nonzero off-diagonal term, which is not diagonalizable. So in particular, there's no such neighborhood in the Zariski topology, and I believe (2) is not open, merely constructible. | |
| Jun 8, 2023 at 8:25 | history | edited | Yikun Qiao | CC BY-SA 4.0 |
added 744 characters in body
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| Jun 8, 2023 at 8:12 | history | asked | Yikun Qiao | CC BY-SA 4.0 |