Timeline for Probability of an extension being normal
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Jun 29, 2010 at 1:46 | answer | added | JSE | timeline score: 5 | |
| Jun 18, 2010 at 7:20 | comment | added | Dror Speiser | If you fix the degree and count by coefficients of polynomials you get 0. Most of the polynomials have full galois group, preventing their corresponding extension to be normal (van der Waerden). On the other hand, if you count all number fields by their discriminant (possible by Hermite), then the first thing to note is that "most" number fields are in fact quadratic or cubic. This can be seen from the degree of the discriminant. Hence, in this case, the density will actually be positive. Finally, if you fix the degree and count by discriminant, then you can read the comments above. | |
| Jun 18, 2010 at 2:36 | comment | added | KConrad | Gerry: I knew about Davenport--Heilbronn for the cubic case, but off the top of my head I remembered Bhargava's result for the quartic case so I just went with that. | |
| Jun 18, 2010 at 1:19 | comment | added | Gerry Myerson | Before Bhargava, there was Andrew Baily, On the density of discriminants of quartic fields, J Reine Angew Math 315 (1980) 190-210, MR 81c:12006. Andy proved (among other things) that the number of Galois quartic number fields with discriminant $d$, $|d|\lt X$, is asymptotic to a constant times $\sqrt X\log^2X$. Before Baily, there were results on cubic fields by H Cohn, and by Davenport and Heilbronn. | |
| Jun 18, 2010 at 1:06 | comment | added | KConrad | Bhargava has theorems on the number of number fields with fixed degree (4 or 5) and increasing discriminant, and I believe that among quartic fields the ones with Galois group S_4 and D_4 contribute positive proportions and the rest (in particular, Galois quartics) contribute density 0. | |
| Jun 18, 2010 at 1:04 | comment | added | KConrad | Why don't you expect it to be 0? Quadratic fields are misleading. Think about the cubic case. To be Galois requires the discriminant is a perfect square and already that is not so easy to stumble across by chance. Now think about the chance that a random quartic will be Galois. Wouldn't you kind of surprised if you wrote one down and it was actually normal? The book "Field Arithmetic" has a treatment of probabilistic questions about random elements of the absolute Galois group of Q. Maybe something in there addresses your question with precision. | |
| Jun 18, 2010 at 0:42 | history | edited | Daniel Barter |
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| Jun 18, 2010 at 0:35 | history | asked | Daniel Barter | CC BY-SA 2.5 |