Timeline for Prime splitting in cubic field, congruence
Current License: CC BY-SA 3.0
8 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Oct 10, 2013 at 21:26 | comment | added | Faisal | @i707107: That's because if you can characterize the totally split primes in a number field via congruence conditions alone, then said number field will necessarily be abelian/Q. | |
| Sep 24, 2013 at 8:56 | comment | added | Sungjin Kim | In Faisal's example, $p$ splits completely in $\mathbb{Q}(2^{1/3})$ iff $p$ has the form $p=x^2+27y^2$. It doesn't seem to be expressed as a single congruence. | |
| Sep 22, 2013 at 18:04 | comment | added | Noam D. Elkies | I thought this was well-known: an unramified prime $p$ splits as $\wp_1 \wp_2$ in a cubic field iff the discriminant is not a square mod $p$; so for non-cyclic fields there's a necessary congruence condition for $p$ to split completely. But finding a criterion that's both necessary and sufficient is harder. | |
| Sep 22, 2013 at 9:00 | comment | added | Felipe Voloch | That's a nice answer. My negative answer in the non-galois case was for a full characterization of the non-split primes but I didn't know you could characterize the primes that factor into two prime ideals this way. | |
| Sep 22, 2013 at 6:13 | vote | accept | Sungjin Kim | ||
| Sep 22, 2013 at 6:13 | comment | added | Sungjin Kim | Thank you, so it is achievable no matter $K$ is Galois or not. Very nice! | |
| Sep 22, 2013 at 5:43 | history | answered | Faisal | CC BY-SA 3.0 |