
$\;\;\;K$ $L$






Here are the few things I know:

    (a) class groups form a cohomological Mackey functor, and

But already the simple statement that




  • $\begingroup$ Actually the Hilbert class field of $F$ is always Galois over $L$. In fact it is Galois over $\mathbb{Q}$: if $\sigma\in$Gal$(\overline{\mathbb{Q}}/\mathbb{Q})$, then $\sigma(Hilb(F))$ is unramified abelian over $\sigma(F)=F$. $\endgroup$ – Kevin Ventullo Aug 17 '11 at 6:16
  • $\begingroup$ let $K$ ..., and $L$ an intermediate cubic. Do you mean an intermediate degree-$p$ extensions? Also, presumably $p$ is a prime. $\endgroup$ – Chandan Singh Dalawat Aug 17 '11 at 11:42
  • $\begingroup$ Notation: $h(F)$ is the order of the class group of $F$ (it took me some time to find this...) $\endgroup$ – YCor Mar 13 '18 at 9:19

I normally don't like to cite my own work on MO, but this time the preprint arXiv:1803.04064 was written, together with L. Caputo, having the OP's question in mind; and so, first of all, let me thank Alex for having asked it.

The main result is a purely algebro-arithmetic proof that for any base field $k$ and for every dihedral extension $F/k$ of degree $2q$ with $q$ odd, it holds $$ \frac{h(k)^2h(F)}{h(L)^2h(K)}=\frac{\lvert\widehat{H}^0(D_{2q},\mathcal{O}_F^\times\otimes\mathbb{Z}[1/2])\rvert}{\lvert\widehat{H}^{-1}(D_{2q},\mathcal{O}_F^\times\otimes\mathbb{Z}[1/2])\rvert} $$ in the OP's notations (which are different from the ones used in the preprint). When I say that the proof is ``algebro-arithmetic'' I mean that the main ingredient is class field theory and some group cohomology: no $\zeta$- or $L$-functions are involved, neither is the classification of integral representation of dihedral groups. The key point is that, if we call $G_q=\operatorname{Gal}(F/K)\cong\mathbb{Z}/q\subseteq D_{2q}$, then $G_q$-cohomology of a $D_{2q}$-module (typically: units, ad`eles, local units, etc.) has an action of $\operatorname{Gal}(K/k)$ and when the module is uniquely $2$-divisible, this induces identifications $\widehat{H}^i(G_q,-)^+=\widehat{H}^i(D_{2q},-)$ as well as $\widehat{H}^i(G_q,-)^-=\widehat{H}^{i+2}(D_{2q},-)$, where $\pm$ are the eigenspaces with respect to the action of $\operatorname{Gal}(K/k)$. Therefore we can use class field theory for the abelian extension $F/K$ to deduce information about cohomology groups for $D_{2q}$.

As corollary of the above formula we show that, for every prime $\ell$, the following bounds hold $$ -av_\ell(q)\leq v_\ell\left(\frac{h(k)^2h(F)}{h(L)^2h(K)}\right)\leq bv_\ell(q) $$ where $a=\operatorname{rank}_\mathbb{Z}\mathcal{O}_K^\times + \beta_K(q) + 1$, $b=\operatorname{rank}_\mathbb{Z}\mathcal{O}^\times_{k}+\beta_k(q)$ and $v_\ell$ denotes the $\ell$-adic valuation; the "defect" $\beta_M(q)\in\{0,1\}$ (for $M=K,k$) is defined to be $1$ if $\mu_M(q)$ is non-trivial, and $0$ otherwise. From this, we deduce even sharper bounds in case $K$ is either CM (with totally real subfield equal to $k$) or if it is totally real. Since when $k=\mathbb{Q}$ this is always the case, we prove as a special result that in every dihedral extension of $\mathbb{Q}$ of degree $2p$ the formula required by the OP holds, again without resorting to any analytic or ``hard'' representation-theoretic result. Actually, restricting to the prime case $q=p$ has no utility whatsoever, and when $F/\mathbb{Q}$ is any dihedral extension of degree $2q$ ($q$ odd!) we deduce that the ratio of class numbers verifies $$ 0 \geq v_\ell\left(\frac{h(F)}{h(K)h(L)^2}\right)\geq \begin{cases} -2&\text{if $K$ is real quadratic}\\ -1&\text{if $K$ is imaginary quadratic} \end{cases} $$ where, again, $v_\ell$ is the $\ell$-adic valuation. It is perhaps interesting to observe that ramification plays little to no role in our proof (ramification indexes only appear as well-controlled orders of cohomology groups of local units) and so assuming particular ramification behaviours wouldn't simplify it.

  • 1
    $\begingroup$ Very nice, Filippo! I am very happy that this question persuaded you to polish that paper and make it public; that effect alone made this question worthwhile! $\endgroup$ – Alex B. Apr 12 '18 at 11:24

Let $L$ be a dihedral extension of the rationals with degree $2\ell$ (where $\ell$ is an odd prime number), $k$ its quadratic subextension and $K$ one of the conjugate fields of degree $\ell$. I will only consider the case where $k$ is complex, as it already shows that the whole thing is far from being trivial.

Let $\tau$ be complex conjugation and $\sigma$ an automorphism of order $\ell$; then $\tau\sigma\tau = \sigma^{-1}$. It is known that the conductor of the cyclic extension $L/k$ is an integer $f$ [Martinet, Sur l'arithmétique des extensions galoisiennes à groupe de Galois diédral d'ordre $2p$, Ann. Inst. Fourier 19 (1969), 1-80; Sect. IV] and that prime number $p$ dividing $f$ satisfy $p \equiv (\Delta/p) \bmod \ell$, where $\Delta$ is the discriminant of $k$.

We now distinguish three cases.

  1. $f = 1$. In this case, $L/k$ is unramified and $\ell \mid h(K)$ since $L/k$ is an abelian unramified extension-

  2. If $f$ is divisible by some prime number $p \equiv 1 \bmod \ell$, let $F$ denote the subfield of degree $\ell$ in the field of $p$-th roots of unity. By Abhyankar's Lemma, the extensions $KF/K$ and $LF/L$ are unramified and abelian, so $\ell \mid h(K)$.

  3. $f$ is only divisible by primes $p \equiv -1 \bmod \ell$ that remain inert in $k/{\mathbb Q}$. We assume that $\ell \nmid h(k)$ and have to show that $\ell \mid h(K)$.

    The ambiguous class number formula tells us that $\# Am(L/k) = \ell^{t-1}/(E_k:E_k \cap NL^\times)$, where $t$ is the number of primes dividing $f$, and where $E_k$ is the unit group of $k$. The number of strongly ambiguous ideals (those generated by ideals fixed by $\sigma$) is given by $\# Am_{st}(L/k) = \ell^{t-1}/(E_k: NE_L)$.

    If $\ell > 3$, or if $\ell = 3$ and $k \ne {\mathbb Q}(\sqrt{-3}\,)$, then the index $(E_k:E_k \cap NL^\times) = 1$ because $E_k = \{-1, +1\}$ and these are norms of units. Thus there must be at least two primes dividing $f$. But then the number of strongly ambiguous ideal classes is $\# Am_{st}(L/k) = \ell^{t-1}$ since the index $(E_k:NE_L) = 1$. Thus there is a strongly ambiguous ideal class of order $\ell$ generated by a prime ideal above one of the primes dividing $f$. But then the prime ideal in $K$ above this prime $p$ cannot be principal, hence has order $\ell$, and we conclude that $\ell \mid h(K)$.

The case $\ell = 3$ and $k = {\mathbb Q}(\sqrt{-3})$ requires studying the plus and minus parts of the class group of $L$ (the plus part is the piece on which $\tau$ acts trivially). The key to the proof is the exact sequence $$ 1 \longrightarrow Am_{st}(L/k)^- \longrightarrow Am(L/k)^- \longrightarrow E_F \cap NK^\times / NE_K \longrightarrow 1 $$ for dihedral extensions $L/F$. The idea then is to show that if $\ell \nmid h(K)$, then the plus part of the ambiguous ideal class group is trivial, hence everything is in the minus part. There all ambiguous ideal classes are strongly ambiguous, and this will lead to a contradiction similarly as above.

The proof given here is extracted from Bölling [Zur Klassenzahl nicht galoisscher Körper in Diedererweiterungen, Math. Nachr. 118 (1984), 271--284).

  • $\begingroup$ Thank you! Given that this proof also uses a class number formula and the passage via the units, maybe my hunch that this should be avoidable is simply false. This proof certainly has the merit that it avoids Lee's classification, although I cannot see straightaway whether one can go all the way to the full strength of (1) without it, and also whether one can avoid saying anything about the Galois module structure of the units in the real case. $\endgroup$ – Alex B. Aug 23 '17 at 10:35
  • $\begingroup$ The ambiguous class number formula is a purely arithmetic result, and it does not even admit an analytic proof. $\endgroup$ – Franz Lemmermeyer Aug 23 '17 at 10:44

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