Timeline for Class Numbers and 163
Current License: CC BY-SA 2.5
19 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Nov 22, 2016 at 12:55 | answer | added | Reimer Brüchmann | timeline score: 4 | |
| Oct 7, 2014 at 3:58 | comment | added | NAME_IN_CAPS | @user9167 You might as well just call it the Gelfond theorem, as ex post facto Stark showed you get away with linear forms in 2 logs, with he had previously solved.ams.org/journals/proc/1969-021-01/S0002-9939-1969-0237461-X/… | |
| Sep 15, 2010 at 17:35 | comment | added | Franz Lemmermeyer | At the end of their paper, Cohen and Lenstra remark that the prime-to-3 part of the class groups of cyclic cubic fields seems to be trivial for about 85 % of all fields. Thus we may expect that the class number is 1 for fields with prime conductor with about the same probability. And for small primes there is a bias towards class number 1 coming from the Odlyzko bounds. | |
| Sep 15, 2010 at 16:00 | history | edited | Cam McLeman | CC BY-SA 2.5 |
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| Sep 12, 2010 at 0:52 | comment | added | Cam McLeman | Agreed with David on all fronts. Nomenclature aside, I feel like I need to comment (the word "admitted" bothers me) that Stark wrote a defense of Heegner's approach (ams.org/mathscinet-getitem?mr=241384, the title alone proves the point), arguing that any serious gap in Heegner's argument is inherited from corresponding gaps in Weber's work. | |
| Sep 11, 2010 at 22:49 | comment | added | David E Speyer | I started a response on the merits of your question but decided I didn't want to get drawn into this argument. Instead, I'll note: You surely mean aimath.org, not aim.org. (3) There is no paper by that title at aimath.org/preprints.html , nor can Google find any paper with that title at all. | |
| Sep 11, 2010 at 22:24 | comment | added | user9167 | Why is there so much resistance to state it and call it as it is: Heegner Theorem for quadratic field. The other guys did it after Heegner. One of them, stark admmited it that he saw Heegner paper before he did his work, see his paper on "Class number of quadartic fields" at AIM.org. | |
| Sep 11, 2010 at 10:47 | answer | added | Franz Lemmermeyer | timeline score: 12 | |
| Sep 11, 2010 at 0:04 | answer | added | Richard Borcherds | timeline score: 5 | |
| Sep 10, 2010 at 21:14 | comment | added | Will Jagy | I overheard the test in the 1980's while at UCSD, I can see that some things may have changed. When you have gotten all the prizes for this and write a book about it, please sign a copy for me. Just as a side note, I can show that your insight implies $P \neq NP,$ it is just a little long to fit into this comment box. | |
| Sep 10, 2010 at 20:52 | comment | added | Cam McLeman | or c) Maybe I'm really on to something here, and this brilliant insight could've overcome the Stark Heuristic and proved one from the other. Let's not forget that option. :) | |
| Sep 10, 2010 at 20:50 | comment | added | Cam McLeman | Okay, now I see what you're saying. Though a) Is it clear that BHS is easier than Schoof? Certainly BHS was more momentous...but Schoof was later and takes a substantial amount of computing power, and b) I think this would rule out only a very direct statement of the form you suggested in your second comment. | |
| Sep 10, 2010 at 20:32 | comment | added | Will Jagy | See if I can get Stark's test correct. Use apostrophe for negation, S for Baker-Heegner-Stark, C for some conjecture. If we have the implication S' --> C, then Stark believes C pending further investigation. Why, you ask? Worry not, I shall tell you. The contrapositive is C' --> S. If C actually turned out to be false, this would provide an easy proof of S | |
| Sep 10, 2010 at 20:10 | comment | added | Will Jagy | No, I take it back. Stark has a very accurate test for conjectures based on the first result being the most difficult proof in that area of mathematics. As the number 163 was surely known for both problems, an easy answer to my problem, or your original, gives an easy proof of Baker-Heegner-Stark, therefore there is no such easy relationship. The relationship is a coincidence. Quod Erat Demonstarkum. | |
| Sep 10, 2010 at 19:58 | comment | added | Will Jagy | Cam, a possibly easier question is whether the Baker-Heegner-Stark answer had to be less than or equal to the Schoof answer, once the former was known to be finite. | |
| Sep 10, 2010 at 19:25 | comment | added | Cam McLeman | That is indeed reassuring. | |
| Sep 10, 2010 at 19:24 | comment | added | Will Jagy | My computer screen smells the same with this displayed as it usually does. | |
| Sep 10, 2010 at 19:16 | history | asked | Cam McLeman | CC BY-SA 2.5 |