Timeline for Why is the regulator of the degree 11 extension of $\mathbb{Q}$ so large?
Current License: CC BY-SA 4.0
15 events
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| Mar 23, 2021 at 20:51 | vote | accept | Thomas Browning | ||
| Mar 23, 2021 at 19:52 | history | edited | GH from MO |
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| Mar 23, 2021 at 19:02 | answer | added | Will Sawin | timeline score: 19 | |
| Mar 22, 2021 at 21:18 | history | edited | Thomas Browning | CC BY-SA 4.0 |
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| Mar 22, 2021 at 21:14 | history | rollback | Thomas Browning |
Rollback to Revision 3
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| Mar 22, 2021 at 21:05 | history | edited | Thomas Browning | CC BY-SA 4.0 |
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| Mar 22, 2021 at 21:04 | comment | added | YCor | (I made an unsuccessful attempt to use the spoiler mode to hide the links and make them click visible: it replaces the whole thing with a huge grey area so this is essentially useless) | |
| Mar 22, 2021 at 20:59 | history | edited | YCor | CC BY-SA 4.0 |
[Edit removed during grace period]
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| Mar 22, 2021 at 20:57 | comment | added | Will Sawin | I guess when you go from $K_11$ to $K_{33}$, the primes lying over $3$ ramify, so the local contribution ends up being exactly the same, while going from $K_{11}$ to $K_{22}$ renders the primes lying over $3$ inert which drops the local contribution from $(3/(3-1))^11$ to $(9/(9-1))^11$, a loss of a factor of about $24$, which tracks with the explicit values... | |
| Mar 22, 2021 at 20:47 | history | edited | Thomas Browning | CC BY-SA 4.0 |
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| Mar 22, 2021 at 20:34 | comment | added | Thomas Browning | Also, the fact that $3$ splits in $K_{11}$ is equivalent to $3^{10}\equiv1\pmod{11^2}$. So maybe this is a "consequence" of the coincidence $3^5=2\cdot11^2+1$. | |
| Mar 22, 2021 at 20:30 | comment | added | Thomas Browning | I graphed a few of the partial Euler products and this seems right to me. Do you think that this also explains why this doesn't happen for $n=22$ and does for $n=33$? | |
| Mar 22, 2021 at 20:17 | comment | added | Will Sawin | If we view the zeta function as counting primes of the number field, the small split prime $3$ gives a large contribution to $\zeta_{K_{11}}$. This is quite exceptional - the next $p$ for which $3$ is split in $K_p$ is $1006003$, and the first $p$ for which $2$ is split in $K_p$ is $1093$. Maybe this could provide an explanation? | |
| Mar 22, 2021 at 20:07 | history | edited | Thomas Browning | CC BY-SA 4.0 |
Compute n=29
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| Mar 22, 2021 at 19:51 | history | asked | Thomas Browning | CC BY-SA 4.0 |