Timeline for On the relative class number of a cyclotomic extension
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 9, 2022 at 21:58 | comment | added | user164898 | @JohnKlein I don't know anywhere where it's given explicitly in the literature, but I'm not the right person to ask, since I don't know the literature on class numbers as well as I'd like. But I am positive that the argument is not new at all, since it only uses ideas which were known to Kummer 130 years ago. | |
| Oct 9, 2022 at 21:54 | comment | added | user164898 | @OscarLanzi Yes, that was indeed my typo! Sorry for the confusing typo. | |
| Oct 9, 2022 at 21:42 | comment | added | Oscar Lanzi | "$h_-$ is divisible by $p$, hence fails to be a power of $p$. Probably mistyped "...fails to be a power of $2$". | |
| Oct 9, 2022 at 21:39 | comment | added | John Klein | @A.S. I feel like an idiot: you wrote, "$h_-$ is divisible by $p$, hence fails to be a power of $p$." I don't see why this conclusion follows. Can you please explain? Thanks in advance. | |
| Oct 9, 2022 at 21:21 | comment | added | John Klein | @A.S., thanks. Do you know if this argument appears in the literature? I am not a number theorist, so I don't know the literature (I asked my colleague who is one, and he was unable to tell me). | |
| Oct 9, 2022 at 20:42 | comment | added | user164898 | Hmm, try this: suppose $p$ is irregular. Then $p$ divides the class number $h(p) = h^+h^-$, so $p$ divides either $h^-$ or $h^+$. A thm. of Kummer establishes that if $p$ divides $h^+$, then it also divides $h^-$ (see Thm.9 in Lozano-Robledo's "Bernoulli numbers, Hurwitz[...]" for a nice exposition). So if $p$ is irr., then $p$ divides $h^-$. There are infinitely many irrregular primes, so for infinitely many primes $p$, $h^-$ is divisible by $p$, hence fails to be a power of $p$. (Vandiver's conj. is that $p$ never divides $h^+$ when $p$ is irr., but you don't need it, due to Kummer's thm.) | |
| Oct 9, 2022 at 19:25 | comment | added | John Klein | Yes, I am using Milnor's notation for the "first factor" in $h(p)$--it's the one given by the order of the kernel of the norm map. Cf. Milnor's, Algebraic K-theory, p. 30. | |
| Oct 9, 2022 at 19:01 | comment | added | user164898 | By $h_1$ you mean the "first factor" in $h(p)$, the negative part $h^-$, the factor not coming from the max'l real subfield of the cyclotomic field, right? If so, then I think this ought to be doable: I suspect you can prove from the formula for $h_1$ in terms of Bernoulli numbers (Iwasawa's "Lectures on p-adic L-functions" pg. 90) that if $p$ is irregular, then $p$ divides $h_1$. It works out for $p=37$ and $59$ but I haven't had a moment to check in general. Infinitude of the irregular primes then gives you that $h_1(p)$ is divisible by $p$ (hence not a power of $2$) for infinitely many $p$. | |
| Oct 9, 2022 at 17:17 | history | asked | John Klein | CC BY-SA 4.0 |