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Nov 25, 2020 at 18:53 comment added David Corwin Does anyone have a reference for these notes?
Jan 6, 2015 at 17:24 comment added grad student @RobertPollack: You can find background on this question and some reasoning based on Cohen-Lenstra heursistics in this recent paper: arxiv.org/pdf/1410.2921v4.pdf
May 7, 2014 at 3:06 comment added KConrad Newsflash: at arxiv.org/pdf/1405.1094v1.pdf John Miller has announced a proof that $h(2^n) = 1$ for $n = 6$ (that is, the real cyclotomic field ${\mathbf Q}(\zeta_{128})^+$ has class number 1) and under GRH it's true for $n = 7$.
Dec 3, 2011 at 3:31 comment added KConrad Folks, please look at the comments to Cam's answer at mathoverflow.net/questions/41219/….
Dec 2, 2011 at 23:16 comment added GH from MO @David: It seems that you use the symbol $n$ in two different meanings in your original definition of $h(n)$. If I am not mistaken $h(n)$ is the class number of the compositum of the following fields: for each $p^m\parallel n$ we consider the $m$-th layer of the cyclotomic $\mathbb{Z}_p$-extension of $\mathbb{Q}$. And my apologies, I forgot to read the sentence "it is a folklore conjecture that $h(n)=1$ for prime power values of $n$". I was too busy with other MO posts and some private math email exchanges.
Dec 2, 2011 at 22:34 comment added sibilant Do you know if there is any theoretical reason behind the conjecture that $h(n)=1$? Or even just $h(p^n)=1$ for $p$ prime?
Dec 2, 2011 at 18:01 comment added Henri Johnston Coates also spoken about this in his Part III course this term. In recent work (jtnb.cedram.org/item?id=JTNB_2010__22_2_359_0 ) Fukuda and Komatsu show that if a prime p divides $h(2^n)$ for some $n$ then $p > 1.2 \times 10^8$ and $p \not \equiv \pm 1 \mod 16$.
Dec 2, 2011 at 17:56 vote accept Kevin Buzzard
Dec 2, 2011 at 17:47 comment added David Loeffler @GH: You misunderstand. Coates' $h(p^n)$ is the class number of the $n$th layer in the cyclotomic $\mathbb{Z}_p$-extension.
Dec 2, 2011 at 17:18 comment added GH from MO This is very interesting, but I don't see the conclusion that "your student's question is a well-known open problem". You only talk about $K_1$'s (for various primes $p$) and their compositums. Sorry if I am missing something, I am no expert as I said.
Dec 2, 2011 at 17:07 history answered David Loeffler CC BY-SA 3.0