(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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

2 Answers 2


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) $n$.

  • $\begingroup$ The context is very much consistent with this interpretation -- this weaker analogue of (A) is mentioned a few lines higher up in the thesis. $\endgroup$ Commented Jan 4, 2017 at 17:01
  • $\begingroup$ Yep - and Japanese names ending with "ko" are feminine -) $\endgroup$ 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

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 $F_n$.

  • 5
    $\begingroup$ I seem to be missing something. There is only one quadratic field in $F_{\infty}$ , namely $\mathbb{Q}(\sqrt{2})$. $\endgroup$ Commented Jan 3, 2017 at 17:09
  • $\begingroup$ I just answered to the question : "... but why should its class number also be 1 ?" $\endgroup$ Commented Jan 3, 2017 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.