Timeline for Is there a finite set of primes such that if K over Q is completely split at all those primes then it is Q?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Jan 8, 2011 at 23:59 | comment | added | Qiaochu Yuan | @Guillermo: no. Take {2, 3}. | |
| Jan 8, 2011 at 23:27 | answer | added | George Lowther | timeline score: 7 | |
| Jan 8, 2011 at 23:04 | comment | added | Kevin Buzzard | However there will be an infinite, density zero set of primes with this property. Because you can just enumerate the non-trivial extensions of the rationals as $K_1$, $K_2,\ldots$ and then for each $n$ choose a prime $p>10^{10^n}$ which isn't completely split in $K_n$ and add it to $S$. The ridiculous growth rate I assumed forces the set to be very sparse. | |
| Jan 8, 2011 at 22:49 | comment | added | Guillermo Mantilla | I don't think such S exists. If you can find an odd prime q such that all the primes in S are congruent to 1 mod q then every prime in S is totally split in q-cyclotomic field. So, my example boils down to: Given a finite set S of primes is it always possible to find a prime q as above? maybe some analytic number theorist knows the answer of this of the top of his/ger head. | |
| Jan 8, 2011 at 22:38 | vote | accept | Makhalan Duff | ||
| Jan 8, 2011 at 22:37 | answer | added | Felipe Voloch | timeline score: 5 | |
| Jan 8, 2011 at 22:35 | answer | added | Qiaochu Yuan | timeline score: 15 | |
| Jan 8, 2011 at 22:25 | history | asked | Makhalan Duff | CC BY-SA 2.5 |