Timeline for Why is $(\mathbb{Z}/3\mathbb{Z})^3$ not a class group of an imaginary quadratic number field ?
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Jun 7, 2013 at 12:12 | history | edited | Cam McLeman | CC BY-SA 3.0 |
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| Jun 7, 2013 at 12:10 | comment | added | Cam McLeman | Good point, will do. | |
| Jun 7, 2013 at 9:06 | comment | added | Dietrich Burde | It seems that the orders of groups not occuring are $27,32,54,64,64,81,81$. Perhaps this should be edited above (and at stackexchange). Anyway, thanks for all answers ! | |
| Jun 6, 2013 at 7:35 | vote | accept | Dietrich Burde | ||
| Jun 6, 2013 at 4:26 | comment | added | v08ltu | The work of Watkins (or Arno et.al.) can be minutely simplified for $(Z/3)^3$. This is by the "structure of minima" they use. You cannot have a form of order $>3$, and forms of odd order pair conjugately, so every nonprincipal form $(a,b,c)$ must have $a^{(3+1)/2}\ge \sqrt{d/4}$ rather than $a^{(27+1)/2}\ge\sqrt{d/4}$ as with general order 27. I don't know how much this eases the situation, but the papers use similar facts to reduce the sieving problems in their cases. I would not be surprised if handling $(Z/3)^k$ for $k=5$ or even more is feasible. Even order case is much more difficult. | |
| Jun 5, 2013 at 21:42 | comment | added | Cam McLeman | Thanks, Noam. I had meant to mention Cohen-Lenstra at some point, but hadn't thought to do the actual automorphism count to make it explicit. That's pretty enlightening. | |
| Jun 5, 2013 at 16:52 | comment | added | Noam D. Elkies | P.S. The 2-part of the class group must be treated specially because of genus theory. | |
| Jun 5, 2013 at 16:50 | comment | added | Noam D. Elkies | The other ingredient is the Cohen-Lenstra heuristics, which suggest that among the groups of given order those with many automorphisms should be rarer. Since $({\bf Z}/3{\bf Z})^3$ has $26 \cdot 24 \cdot 18 = 11232$ automorphisms, and ${\bf Z}/27{\bf Z}$ only $18$, we expect the cyclic group to arise about $624$ times more often in the real quadratic case, and are not surprised that it does not arise at all among the few imaginary quadratic fields (IQFs) with $h=27$. It may be reasonable to conjecture that no odd elementary abelian group of rank $3$ or more is a class group of an IQF. | |
| Jun 5, 2013 at 15:30 | comment | added | Dietrich Burde | Thank you, Cam. Which groups are these, of orders $32$, $64$ and $81$ ? | |
| Jun 5, 2013 at 14:19 | history | edited | Cam McLeman | CC BY-SA 3.0 |
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| Jun 5, 2013 at 14:13 | history | answered | Cam McLeman | CC BY-SA 3.0 |