Timeline for k-th powers in the field of p-adics
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 11, 2011 at 14:12 | vote | accept | Silvain Rideau | ||
| Jul 9, 2011 at 6:05 | comment | added | LSpice |
@Auguste, I'm not sure what the typos are; I guess you meant that $\mathbb Z_p^\times = \mathbb F_p^\times \times (1 + p\mathbb Z_p)$? I'm not sure that I can see it directly from there. Anyway, thanks for the pointer to the refined version of Hensel's lemma (which Dave Marker calls Hensel–Rychlik); that certainly does it, and to call it a use of series would be stretching a point, so I won't. :-)
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| Jul 6, 2011 at 19:45 | comment | added | Auguste Hoang Duc | I am sorry, there are typos in my previous comment. I wanted to say that $(O_K^\times)^n$ is open. And a $\times$ is missing. | |
| Jul 6, 2011 at 19:14 | comment | added | Auguste Hoang Duc | @L Spice : Because $Z_p = F_p^\times \times Z_p \times Z$ (as topological groups). It is aslo true that for $K$ a p-adic field $O_K^\times$ is open. Milne proves it with Newton's Lemma in his notes on Class Field Theory (look at page 22, (1.7) after corollary 1.5) jmilne.org/math/CourseNotes/CFT.pdf | |
| Jul 6, 2011 at 19:05 | comment | added | LSpice | @Jérôme, or, again, one can avoid series and say that integers are dense, so that there exists one in $x + \wp^r$. | |
| Jul 6, 2011 at 19:03 | comment | added | LSpice | @Auguste, how do you see the open-ness? Perhaps one wishes to argue based on the fact that the derivative is a surjection, but I suspect that a proof based on this observation will still involve series somewhere. | |
| Jul 6, 2011 at 13:20 | comment | added | Jérôme Poineau | @Auguste: Indeed! That's a nice argument. @Silvain: Write x as a series $\sum_{i\ge 0} a_i p^i$ and let y be $\sum_{i\ge r} a_i p^{i-r}$. | |
| Jul 6, 2011 at 12:37 | comment | added | Silvain Rideau | Writing x=n+yp^r makes absolute sense if x is in Q (it is a simple euclidian division) but how does it work when x is not? | |
| Jul 6, 2011 at 12:26 | comment | added | Auguste Hoang Duc | You can notice that $(Z_p^\times)^k$ is open. It avoids the use of series. | |
| Jul 6, 2011 at 11:27 | history | answered | Jérôme Poineau | CC BY-SA 3.0 |