Timeline for Set of all primes $p$ that split in $\mathbb{Q}\left(\sqrt{-k}\right)$
Current License: CC BY-SA 4.0
12 events
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| Sep 12, 2022 at 10:55 | history | edited | GH from MO | CC BY-SA 4.0 |
added 125 characters in body
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| Sep 12, 2022 at 10:16 | comment | added | GH from MO | @user491084 OK, so by "parametrized $k$" you meant "$k$ lying in a given arithmetic progression". Please update your question to reflect this precise meaning, and then accept my answer officially (so that it turns green). Thanks in advance! | |
| Sep 12, 2022 at 10:14 | history | undeleted | GH from MO | ||
| Sep 12, 2022 at 10:13 | history | deleted | GH from MO | via Vote | |
| Sep 12, 2022 at 10:13 | comment | added | user491084 | Okay, that makes sense. Thanks!!! | |
| Sep 12, 2022 at 10:11 | comment | added | GH from MO | @user491084 My answer should make it clear that the answer to your general question is also yes. The prime $p$ must divide the modulus of the arithmetic progression of the $k$'s, and for such a prime either every $-k$ is a quadratic residue modulo $p$ or every $-k$ is a quadratic non-residue modulo $p$ (which can be checked by plugging in the relevant residue into the quadratic residue symbol modulo $p$). | |
| Sep 12, 2022 at 10:07 | comment | added | user491084 | Sure, I have now edited the question and asked in a more general context :) | |
| Sep 12, 2022 at 10:06 | comment | added | GH from MO | @user491084 Yes, $p=3$ is the only prime that splits simultaneously in every $\mathbb{Q}(\sqrt{-k})$ for $k\equiv 11\pmod{24}$. If you like my answer, please accept it officially (so that it turns green). Thanks in advance! | |
| Sep 12, 2022 at 10:04 | comment | added | user491084 | Oh, so $3$ is the only prime that splits in $K$ for $k\equiv11\pmod{24}$? | |
| Sep 12, 2022 at 10:01 | history | undeleted | GH from MO | ||
| Sep 12, 2022 at 10:00 | history | deleted | GH from MO | via Vote | |
| Sep 12, 2022 at 10:00 | history | answered | GH from MO | CC BY-SA 4.0 |