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 | |
|---|---|---|---|---|---|
| Jul 11, 2011 at 15:17 | comment | added | Geoff Robinson | Sure, it's OK if you allow rational $n$. | |
| Jul 11, 2011 at 14:03 | comment | added | Silvain Rideau | Well I guess then you have to allow n to be rationnal (or at least of the form 1/n where n is an interger). | |
| Jul 11, 2011 at 13:54 | vote | accept | Silvain Rideau | ||
| Jul 11, 2011 at 14:12 | |||||
| Jul 11, 2011 at 11:04 | comment | added | Geoff Robinson | Thanks. Well, it seems to be fine if $\nu(x) >0$, since we can just choose replace $x$ by $\frac{x}{m}$ for some integer $m$ with $\nu(m) = \nu(x)$. Since the conclusion is just about $\frac{x}{n}$ for some integer $n$, this doesn't change anything. But if $\nu(x) <0$, I don't see what to do, if you're only allowed to divide by inetgers. | |
| Jul 11, 2011 at 10:03 | comment | added | Silvain Rideau | I am under the impression this generalises nicely if v(x) is not 0. Indeed, at some point I began to wonder if Qp*/(Qp*)^k was finite and the explicit bound here seems to point that way. | |
| Jul 11, 2011 at 9:57 | vote | accept | Silvain Rideau | ||
| Jul 11, 2011 at 13:53 | |||||
| Jul 9, 2011 at 21:44 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
added more necessary details for case $p=2$.
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| Jul 7, 2011 at 0:09 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
typos
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| Jul 6, 2011 at 12:19 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Typos
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| Jul 6, 2011 at 11:00 | history | answered | Geoff Robinson | CC BY-SA 3.0 |