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(A) It is an old, outstanding problem to show that there are infinitely many real quadratic fields with class number one.

(B) On the other hand, Weber's class number problem (for $p=2$) asks to show that the degree $2^n$ cyclotomic extension $F_n=\mathbb Q(\cos(2\pi/2^{n+2}))$ of $\mathbb Q$ has class number one, for all $n$.

According to p.41 of this thesis, showing (B) for infinitely many $n$ implies (A), but no idea of the proof is given. This does not seem to follow immediately for me, but perhaps I am missing something simple here. Certainly $F_n$ contains a real quadratic subfield, but why should its class number also be 1?

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  • $\begingroup$ From the second page of the thesis: "Perhaps the subject goes back to Fermat, who stated in 1964 that for an odd prime $p$..." $\endgroup$
    – KConrad
    Commented Jan 3, 2017 at 18:36
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    $\begingroup$ A more serious comment: on p. 41 of the thesis, the author says Heilbronn showed GRH implies for fundamental discriminants $D < 0$ that $h(\mathbf Q(\sqrt{D})) \rightarrow \infty$ as $|D| \rightarrow \infty$. This is a misleading formulation, because failure of GRH is too broad as a hypothesis. For example, if GRH is false for the zeta-function of $\mathbf Q(\sqrt[3]{2})$, Heilbronn can't deduce anything about class numbers of imaginary quadratic fields. The correct hypothesis is failure of a restricted version of GRH: just for $L$-functions of imaginary quadratic (Dirichlet) characters. $\endgroup$
    – KConrad
    Commented Jan 3, 2017 at 18:53

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This statement is most certainly nonsense; I guess what she meant to write was that in order to prove the existence of infinitely many number fields with class number $1$ it is sufficient to prove $h(F_n) = 1$ for infinitely many (and therefore for all) $n$.

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  • $\begingroup$ The context is very much consistent with this interpretation -- this weaker analogue of (A) is mentioned a few lines higher up in the thesis. $\endgroup$
    – David Loeffler
    Commented Jan 4, 2017 at 17:01
  • $\begingroup$ Yep - and Japanese names ending with "ko" are feminine -) $\endgroup$
    – Franz Lemmermeyer
    Commented Jan 4, 2017 at 17:12
  • $\begingroup$ Thanks! I thought this was the case (that it proves the existence of infinitely many number fields..), but wanted to be sure. $\endgroup$
    – Tian An
    Commented Jan 4, 2017 at 17:54
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Let $K$ be any subextension of $F_n$. Then the natural map of class groups $\text{Cl}(K) \rightarrow \text{Cl}(F_n)$ is an injection. Indeed, you can see it in terms of unramified abelian extension via global class field theory. If $L$ is an unramified abelian extension of $K$, since $F_n$ is totally ramified at primes over $2$ in $K$, we have $L \cap F_n = K$. Therefore the compositum $L \cdot F_n$ has Galois group $\text{Gal}(L/K)$ over $F_n$.

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    $\begingroup$ I seem to be missing something. There is only one quadratic field in $F_{\infty}$ , namely $\mathbb{Q}(\sqrt{2})$. $\endgroup$
    – Chris Wuthrich
    Commented Jan 3, 2017 at 17:09
  • $\begingroup$ I just answered to the question : "... but why should its class number also be 1 ?" $\endgroup$
    – Emmanuel Lecouturier
    Commented Jan 3, 2017 at 17:11

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