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An MSc student asked me if I knew an example of a prime $p$ and some finite layer $K_n$ in the cyclotomic $\mathbf{Z}_p$-extension of $\mathbf{Q}$ (so $[K_n:\mathbf{Q}]=p^n$) which had non-trivial class group. My gut feeling was that fixing a $p$ and then going up the tower was a bad idea in the sense that going along might find a counterexample quicker, so I tried going along instead, but my computer is having trouble looking at the bottom level when $p$ starts getting bigger than about 30. So now I'm just confused. Presumably this question has been raised before? Can anyone enlighten me with a counterexample or reassurance that this is a standard open question?

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Not being an expert can you tell me in more down-to-earth terms what $K_n$ is? Class field theory is fine, but more concrete descriptions are even better. Thanks. – GH from MO Dec 2 '11 at 16:33
For $p>2$ it's this: if you adjoin all the $p^{n+1}$st roots of unity to $\mathbf{Q}$ then you get an extension of degree $p^n(p-1)$ with cyclic Galois group. Then $K_n$ is the subfield of this with degree $p^n$ over $\mathbf{Q}$, so it's like a complement to $\mathbf{Q}(\zeta_p)$. – Kevin Buzzard Dec 2 '11 at 16:55
GH: Pick $a \in (\mathbf{Z}/p^{n+1}\mathbf{Z})^{\times}$ which generates a subgroup isomorphic to $(\mathbf{Z}/p\mathbf{Z})^{\times}$; then $K_n=\mathbf{Q}(\zeta+\zeta^a+\zeta^{a^2}+\dots+\zeta^{a^{p-2}})$ where $\zeta$ is a primitive $p^{n+1}$st root of unity. – David Hansen Dec 2 '11 at 16:55
John Coates gave a talk in Warwick not long back where he said that it was a long-standing open conjecture (of Kronecker I think?) that every field in the cyclotomic $\mathbb{Z}_2$ extension of $\mathbb{Q}$ had trivial class group. – David Loeffler Dec 2 '11 at 16:56
Kevin and David: Thanks to both of you, your responses were very useful. – GH from MO Dec 2 '11 at 17:01
up vote 9 down vote accepted

I found the notes of Coates' seminar I alluded to in my comment above. He said the following:

For $n \ge 1$, let $h(n)$ denote the class number of the unique cyclic degree $n$ extension contained in the compositum of all the cyclotomic $\mathbb{Z}_p$-extensions for $p \mid n$. It is apparently not too difficult to show that if $n \mid m$, then $h(n) \mid h(m)$, and that if $n$ is a power of $p$, then $h(n)$ is prime to $p$.

Then Weber (not Kronecker, but close!) has conjectured that $h(2^n) = 1$ for all $n$, and this is known for $n \le 5$ and for $n=6$ conditional on GRH. More generally, it is a folklore conjecture that $h(n) = 1$ for all prime power values of $n$.

Horie, Fukuda and Komatsu have recently shown that $h(62)$ is divisible by 31, the first known example of an integer $n$ where $h(n) > 1$.

So the verdict seems to be that your student's question is a well-known open problem.

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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. – GH from MO Dec 2 '11 at 17:18
@GH: You misunderstand. Coates' $h(p^n)$ is the class number of the $n$th layer in the cyclotomic $\mathbb{Z}_p$-extension. – David Loeffler Dec 2 '11 at 17:47
Coates also spoken about this in his Part III course this term. In recent work ( ) Fukuda and Komatsu show that if a prime p divides $h(2^n)$ for some $n$ then $p > 1.2 \times 10^8$ and $p \not \equiv \pm 1 \mod 16$. – Henri Johnston Dec 2 '11 at 18:01
Folks, please look at the comments to Cam's answer at…. – KConrad Dec 3 '11 at 3:31
Newsflash: at 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 $n = 7$. – KConrad May 7 '14 at 3:06

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