Timeline for Have we ever proved any non-solvable case of reciprocity without the Langlands program ?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Nov 7, 2014 at 2:08 | answer | added | Tommaso Centeleghe | timeline score: 6 | |
| Mar 29, 2013 at 7:25 | comment | added | user30035 | [just to tidy up: of course Chandan is right. According to Cox, Euler conjectured that if p=1 mod 3 then 2 was a cube mod p iff p=x^2+27y^2; it was Gauss who proved it.] | |
| Mar 28, 2013 at 9:26 | answer | added | Chandan Singh Dalawat | timeline score: 21 | |
| Mar 23, 2013 at 2:47 | comment | added | Chandan Singh Dalawat | Dear Joël, look at Philippe Satgé's paper (numdam.org/numdam-bin/fitem?id=AIF_1977__27_4_1_0) about which I asked a question here some time ago (mathoverflow.net/questions/119305/…). | |
| Mar 22, 2013 at 13:41 | comment | added | Chandan Singh Dalawat | @wccanard : The fact that a finite galoisian extension $K$ of $\mathbf{Q}$ is determined by the set of prime numbers which split completely in $K$ is due to M. Bauer, Über einen Satz von Kronecker, Arch. der Math. u. Phys. (3) 6, 218--219 (1903). | |
| Mar 22, 2013 at 13:33 | comment | added | Chandan Singh Dalawat | @wccanard : The cubic character of $2$ was first determined by Gauß in his Disquisitiones, as far as I know. | |
| Mar 22, 2013 at 12:35 | comment | added | Joël | David, yes I am looking primarily for not solvable example, but I realized that I don't exactly know how people were feeling with solvable non abelian example before Langlands, so that I should be able to learn something already from this basic example. | |
| Mar 22, 2013 at 11:44 | comment | added | David E Speyer | $S_3$ is solvable. I thought you wanted nonsolvable examples. | |
| Mar 22, 2013 at 7:39 | comment | added | user30035 | PS Keith isn't it just something like $p=x^2+27y^2$? My source for all of this was my memory of the first few sections of Cox's book. When I get into the office my new source will be Cox's book and perhaps I'll have to retract some assertions :-/ | |
| Mar 22, 2013 at 7:34 | comment | added | user30035 | Joel -- I'm not in the office right now (and I wasn't when I wrote that last comment) but my memory is that some historical references are in Cox' "primes of the form $x^2+ny^2$". I'll try and remember to look today. | |
| Mar 21, 2013 at 23:16 | comment | added | KConrad | A rule for when $2 \bmod p$ is a cube, for $p \equiv 1 \bmod 3$, is in Ireland & Rosen's book. Just google "cubic character of 2" and look at the hit it gives you on their book. Prop. 9.6.2 on p. 119 is a result in that direction, although it's of course not formulated as a simple congruence condition on $p$. | |
| Mar 21, 2013 at 21:39 | comment | added | Joël | @wccanard. It's funny, I think I know you in real life, though I can't be sure :-) I like your Euler's example: but did he know the prime where $x^3-2$ splits completely with a proof ? That would be surprising, since the hadn't a proof of the simpler quadratic reciprocity. Then, who proved it? regarding your second comment, I completely agree, this question is much older than Langlands, in a sense it dates back to Euler, and in the nineteen's century I think it was a well-known problem and motivation in some form. | |
| Mar 21, 2013 at 21:11 | comment | added | paul garrett | There is Shimura's 1966 Crelle paper "A reciprocity law in non-solvable extensions". | |
| Mar 21, 2013 at 20:05 | comment | added | user30035 | Joel -- an afterthought -- I think that this question was raised explicitly way way before Langlands. There are exercises in Cassels-Froehlich about the primes that split completely in an (arbitrary) finite extension of number fields (proving e.g. that these primes determine the extension up to iso etc etc), plus some sort of assertion that beyond abelian situations [so perhaps implicitly allowing solvable cases] no-one has a clue how to classify these primes. I know C-F was after Langlands but one can easily imagine that the question was much older... | |
| Mar 21, 2013 at 19:47 | comment | added | user30035 | I think that Euler already knew the primes $p$ for which $x^3-2$ splits completely mod $p$; one can dress the result up as some weight 1 modular form defined by theta series, and of course the Galois group is soluble, but I am pretty sure that Euler didn't know about the Langlands program. On the other hand of course this doesn't answer the question, because the weight 1 forms that you can access using theta series are precisely those ones giving dihedral Galois reps. | |
| Mar 21, 2013 at 17:10 | history | asked | Joël | CC BY-SA 3.0 |