Timeline for "Understanding" $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Jun 21 at 16:25 | comment | added | LSpice | CharlesRezk's remark to JSE's answer. | |
| Jun 21 at 13:42 | comment | added | Alon Amit | I hope I’m not violating some MO etiquette by saying this years later: this is a beautiful and revealing point of view. Amazing. | |
| Dec 1, 2014 at 3:44 | history | edited | Peter Humphries | CC BY-SA 3.0 |
Fixed more LaTeX
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| Dec 1, 2014 at 2:51 | history | rollback | François G. Dorais |
Rollback to Revision 3
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| Sep 5, 2013 at 5:46 | history | undeleted | Kim Morrison | ||
| Jun 21, 2013 at 6:20 | history | deleted | user631 | ||
| Jun 21, 2013 at 5:54 | history | rollback | user631 |
Rollback to Revision 1
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| Nov 6, 2012 at 23:06 | comment | added | Filippo Alberto Edoardo | Very very nice!!! | |
| Nov 6, 2012 at 17:24 | history | edited | Daniel Miller | CC BY-SA 3.0 |
deleted 29 characters in body
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| Nov 6, 2012 at 17:18 | history | edited | Daniel Miller | CC BY-SA 3.0 |
cleaned up the LaTeX
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| Nov 1, 2009 at 23:29 | comment | added | user631 | Charles, yes, that's correct, although literally to have a Frobenius at $p$ one should work with extensions unramified at $p$, which amounts to working with $\pi_1(\mathbb{Z}[1/N])$ for some set $N$, and taking $p$ not dividing $N$. (For $p \mid n$, one has to work with the still complicated group $Gal(\bar{\mathbb{Q}}_p/\mathbb{Q}_p)$, rather than $Gal(\bar{F}_p/F_p)$, which is topologically cyclic and generated by Frobenius. | |
| Nov 1, 2009 at 20:13 | comment | added | Charles Rezk | I hate the fact that you can't even use html characters in comments. | |
| Nov 1, 2009 at 20:12 | comment | added | Charles Rezk | Thanks! So what I take from this is that the only canonical constructions are the completions $\mathbb{Q}_p$ (with $\mathbb{Q}_\infty = \mathbb{R}$). So if $X$ is the algebraic space associated to $\mathbb{Q}$, then for each $p$ we get a covering space $X_p \to X$, associated to the field of algebraic numbers in $Q_p$. If $X$ had a distinguished base point, we could name the $X_p$'s using subgroups of $\pi_1X$. Since we can't, all we know is the conjugacy class of $\pi_1X_p$ inside $\pi_1X$. For rational primes $p$, this is the congugacy class of the subgroup "generated" by $Frob_p$. | |
| Nov 1, 2009 at 18:44 | comment | added | Jonah Sinick | Thanks, this is a point that I have long wondered about and your remark clarifies the matter completely. | |
| Nov 1, 2009 at 17:25 | history | answered | user631 | CC BY-SA 2.5 |