Timeline for Image of the norm map for Artin-Schreier extensions
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 9, 2022 at 9:04 | vote | accept | Daniel Loughran | ||
| Dec 7, 2022 at 19:17 | answer | added | Philip | timeline score: 4 | |
| Dec 3, 2022 at 19:56 | answer | added | Lubin | timeline score: 4 | |
| Dec 2, 2022 at 21:55 | comment | added | Lubin | I’m working computationally, which is getting information; but I think the approach is wrong-headed. Keep tuned — | |
| Dec 2, 2022 at 14:25 | comment | added | Daniel Loughran | @Lubin: That would be excellent! Feel free to make some simplifying assumptions, e.g. $n=-1$ or $p=2$. | |
| Dec 2, 2022 at 13:29 | comment | added | Lubin | Both comments above get right to the point. The action is all in the principal units, clearly. I’ll see whether I can stir my brain up to understand the situation. | |
| Dec 2, 2022 at 11:05 | comment | added | Daniel Loughran | @ArnoFehm; Yes this is exactly what I'm asking for. In fact I think that even the image of the units has index $p$, so the problem is about determining which units are norms. I was hoping given that the situation is so explicit this might be possible! Even if a complete description is not possible, it would be nice to know even some units which are not norms. | |
| Dec 2, 2022 at 7:08 | comment | added | Arno Fehm | It certainly is wildly ramified if $n<0$ and $p\nmid n$. For the wildly ramified prime cyclic case $L/k$, Serre's Local Fields, Remark at the end of V.§3 seems to say that $(k^\times:NL^\times)=p$. Your question is which subgroup $NL_n^\times$ is precisely? | |
| Dec 1, 2022 at 18:42 | history | edited | Daniel Loughran | CC BY-SA 4.0 |
deleted 197 characters in body
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| Dec 1, 2022 at 13:19 | history | asked | Daniel Loughran | CC BY-SA 4.0 |