Timeline for Proof of a Simple Converse in Algebraic Number Theory
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 7, 2012 at 20:39 | vote | accept | Jon Cohen | ||
| May 1, 2012 at 2:07 | comment | added | KConrad | Jon: I changed the primes over $\mathfrak p$ in your question to ${\mathfrak P}_i$ instead of $\beta_i$: $\mathfrak P$ is a capital $P$, not anything like $B$. Don't $\mathfrak P$ and $\mathfrak B$ look completely different? :) | |
| May 1, 2012 at 2:05 | comment | added | KConrad | Zev's question is more reasonable since all but finitely many primes in $K$ are unramified in $L$, hence most $e$'s are 1 anyway. A Galois extension would have all residue field degrees over each $\mathfrak p$ equal, not just all ramification indices, so the converse is natural to ask with the residue field degrees taken into account also. | |
| May 1, 2012 at 2:01 | history | edited | KConrad | CC BY-SA 3.0 |
added 17 characters in body
|
| Apr 30, 2012 at 23:34 | answer | added | Alex B. | timeline score: 3 | |
| Apr 30, 2012 at 23:33 | comment | added | Kevin Buzzard | Just demanding that all the exponents are equal isn't enough. Just take any non-Galois extension unramified everywhere -- e.g. take a number field $A$ whose Hilbert class field has a non-trivial Hilbert class field $B$ (these exist), and then take a non-normal subgroup of $Gal(B/A)$ giving an extension $C/A$, non-Galois but with every prime unramified. | |
| Apr 30, 2012 at 23:08 | comment | added | Zev Chonoles | You might be interested in my question here: mathoverflow.net/questions/34180, though I was requiring that the residue degrees also agreed for all primes above $\mathfrak{p}$. | |
| Apr 30, 2012 at 22:49 | history | asked | Jon Cohen | CC BY-SA 3.0 |