Timeline for Any reason why K_23(Z) has order 65520?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Jan 1, 2010 at 3:18 | comment | added | Jonah Sinick | I believe that Quillen proved that the K-groups of rings of integers are finitely generated and that Borel computed the ranks | |
| Dec 31, 2009 at 23:59 | history | edited | Simon | CC BY-SA 2.5 |
Made changes as suggested by Bjorn
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| Dec 31, 2009 at 23:52 | comment | added | Simon | Fair enough. Thanks for keeping me honest! | |
| Dec 31, 2009 at 21:59 | comment | added | Bjorn Poonen | OK, but if you really intended to interpret the formula as 0=0 in the odd i case, then I think there was no need to include the regulator! | |
| Dec 31, 2009 at 19:33 | comment | added | Simon |
In the totally real case, at least, what I said seems to be true (if the size of $K_{2i-1}(\mathfrak{o})$ is infinite when $i$ is odd, and $\zeta_F(1-i)$ is also zero in this case). But as Bjorn pointed out, you can also get the leading coefficient by restricting to the torsion subgroups. Moreover, this works even if $F$ is not totally real. I have no idea what the actual $K$ groups (rather than just the sizes of the torsion subgroups) tell us. I'd love to know though!
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| Dec 31, 2009 at 19:13 | comment | added | Bjorn Poonen | Finally, there is more than one abelian group of order 65520, so the formula by itself does not fully answer Ilya's question about why it is Z/65520Z. | |
| Dec 31, 2009 at 19:06 | comment | added | Bjorn Poonen | And on the right hand side, shouldn't the K-groups be replaced by their torsion subgroups? | |
| Dec 31, 2009 at 19:04 | comment | added | Bjorn Poonen | Shouldn't the left hand side of the formula be the absolute value of the leading coefficient of the Taylor series centered at 1-i? | |
| Dec 31, 2009 at 15:14 | history | edited | Simon | CC BY-SA 2.5 |
added 16 characters in body
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| Dec 31, 2009 at 6:04 | comment | added | Reid Barton | Shouldn't there be a regulator factor in the Lichtenbaum conjecture formula (even when F is totally real)? | |
| Dec 31, 2009 at 5:48 | history | answered | Simon | CC BY-SA 2.5 |