Timeline for Does $0\to I\to\mathrm{Gal}_K\to\mathrm{Gal}_k\to 0$ always split?
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8 events
| when toggle format | what | by | license | comment | |
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| Jan 21 at 22:57 | comment | added | user267839 | Furthermore, why in last part the assumed maximality of $K$ (wrt those extensions having purely insep'ble residue extension) should imply that the sequence written down would be nonsplit? | |
| Jan 21 at 22:57 | comment | added | user267839 | Could you explain why to show splitting as above for $K$ it is wlog allowed to pass to alg extension $K'/K$ with purely insep'ble residue extension? Do I understand it correctly that this only works due to the purely insep assumption for residue field $k'$ forcing $\text{Gal}_k= \text{Gal}_{k'}$ and assumping we have the split for $\text{Gal}_{K'}$ it just prolonges via natural inclusion $\text{Gal}_{K'} \subset \text{Gal}_{K} $ to split over $K$? Is this the "trick" or is there more involved? | |
| Mar 14, 2021 at 20:19 | comment | added | ali | @Asvin I think enlarging $K$ is also essential for the last part of the argument. If $K$ is not maximal but only have divisible value group then the last exact sequence could spilt | |
| Mar 11, 2021 at 21:24 | comment | added | Peter Scholze | The book of Gabber-Ramero on Almost ring theory has a nice Section 6.2 on ramification theory. See in particular 6.2.17 (and 6.2.12 for the assertion that the non-tame part is pro-$p$). | |
| Mar 11, 2021 at 16:46 | comment | added | user39380 | One thing I am a bit confused about: given that $K^{ur}$ has a divisible value group, how do we know that its tame extensions are trivial? Would you explain a bit more? Thanks! | |
| Mar 11, 2021 at 12:30 | comment | added | Asvin | Is the only reason to enlarge K to ensure that I is pro-p? That's a really interesting technique! | |
| Mar 11, 2021 at 12:02 | vote | accept | CommunityBot | ||
| Mar 11, 2021 at 10:35 | history | answered | Peter Scholze | CC BY-SA 4.0 |