Timeline for Are the abelian absolute Galois groups of these local fields isomorphic?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 12, 2016 at 23:24 | answer | added | Lubin | timeline score: 8 | |
| Mar 11, 2016 at 23:03 | answer | added | znt | timeline score: 9 | |
| Mar 10, 2016 at 17:22 | comment | added | Pablo | @GHfromMO It is well known that these absolute (abelian) Galois groups are finitely generated (as profinite groups). Finitely generated profinite groups are isomorphic if and only if they have the same continuous finite images (you can see this in the book of Ribes-Zalesskii on Profinite groups). | |
| Mar 10, 2016 at 16:22 | comment | added | GH from MO | I don't see that two statements are equivalent. If $G_1$ and $G_2$ are the two Galois groups in the first statement, then the second statement merely says that for any open subgroup $H_1\leq G_1$ there is an open subgroup $H_2\leq G_2$ such that $G_1/H_1\cong G_2/H_2$ and vice versa. For example, the latter property is true if $G_1$ and $G_2$ are factors of each other, but I don't see how this implies that $G_1$ and $G_2$ are isomorphic. | |
| Mar 9, 2016 at 10:16 | history | edited | Pablo | CC BY-SA 3.0 |
added 241 characters in body
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| Mar 9, 2016 at 9:53 | history | asked | Pablo | CC BY-SA 3.0 |