Let $p$ be an odd prime and let $F/\mathbb{Q}$ be a Galois extension with Galois group $D_{2p}$, let $K$ be the intermediate quadratic extension of $\mathbb{Q}$, and $L$ an intermediate degree $p$ extension:
$\;\;\;K$ $L$
$\;\;\;\;|{\tiny 2}\;\;/{\tiny p}$
Then $$ \frac{h(F)}{h(K)h(L)^2}\in\left\{1,\frac{1}{p},\frac{1}{p^2}\right\}.\;\;\;\;\;\;\;\;\;\;(1) $$ One can also prove a similar statement over any given base field, but let's stick with this simple special case for now. The only way I know of proving this is rather roundabout:

  1. One first replaces the class numbers by regulators, using the analytic class number formula.

  2. One interprets the resulting quotient of regulators as an invariant of the isomorphism class of the Galois module $\mathcal{O}_F^\times$.

  3. One uses a classification of all $\mathbb{Z}$-free $\mathbb{Z}[G_{2p}]$-modules due to M. Lee from the 60s and computes the corresponding regulator quotient for all the modules that can occur in the above situation. The fact that this module is not $\mathbb{Z}$-free is a subtlety that one has to take care of separately.

It seems to me that there should be a much easier way, or at least a more direct one, one that actually gives insight into the reasons for the dependence of class numbers on each other. My question is:

Does anyone know of a direct way of proving (1), one that only uses properties of class groups, but not the analytic class number formula, nor any integral representation theory?

Here are the few things I know:

  • The fact that this class number quotient will have trivial $q$-adic valuation for any $q\neq p$ follows from the formalism of cohomological Mackey functors and is due to R. Boltje, Class group relations from Burnside ring idempotents, J. of Number Theory, 66, (1997). Essentially, the two ingredients are:
    (a) class groups form a cohomological Mackey functor, and
    (b) for any $q\neq p$, there exists an isomorphism $$ \mathbb{Z}_q[G/1]\oplus\mathbb{Z}_q[G/G]^{\oplus 2}\cong\mathbb{Z}_q[G/C_p]\oplus\mathbb{Z}_q[G/C_2]^{\oplus 2}. $$ I am happy with that part, so the remaining bit is the p-primary part.
  • I can immediately see from class field theory that the coprime-to-2 part of the class number of $L$ divides the class number of $F$, because the compositum of $F$ with the coprime-to-2 bit of the Hilbert class field of $L$ gives an abelian unramified extension of $F$ of the same degree. Similarly, if $3^r|h(K)$, then $3^{r-1}|h(F)$.

But already the simple statement that

if $p|h(F)$, then either $p|h(K)$ or $p|h(L)$

is not clear to me. E.g. the Hilbert class field of $F$ needn't be abelian over $L$, nor even Galois. Am I missing something elementary? A direct proof of this would already be nice. Or in the opposite direction, if the class number of $L$ is divible by $p^{100}$, why must the class number of $F$ compensate that? (Note that in the quotient, $h(L)$ is squared, while $h(F)$ isn't).

The article of Franz Lemmermeyer, Class groups of dihedral extensions gives a pretty extensive overview of the known variants of Spiegelungssätze for dihedral extensions, but as far as I can see, (1) does not follow from any of them (dear Franz, I call upon thee to confirm or to correct my assessment). Some very special cases do follow, but it seems to me that there should be a general direct proof.

  • $\begingroup$ 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
  • $\begingroup$ @Kevin: of course, you are right. Thanks! I was getting carried away in my pessimism. But it needn't be abelian over $L$, e.g. it could be dihedral. If we just look at the odd degree bit of the Hilbert class field (which we may wlog), then the Galois group over $L$ is a semi-direct product of an abelian group and $C_2$. I guess, if it's not abelian, then one should be able to say something about the Galois group of $Hilb(F)/K$. I may be beginning to see an argument for the last boxed statement... $\endgroup$ – Alex B. Aug 17 '11 at 6:29
  • $\begingroup$ @Alex: Here's the beginning of an argument: Consider only the p-torsion Hilbert class field over $F$, Hilb$_p$, which is an $\mathbb{F}_p$ vector space. Consider the action of the involution $\sigma\in$Gal$(F/L)$. If there is a +1 eigenvector, then the corresponding $p-$extension of $F$ is abelian over $L$, and so there is a $p-$unramified extension of $L$. Otherwise, $\sigma$ acts by -1 everywhere, hence lies in the center of the action of Gal$(F/Q)$ on Hilb$_p$. Then Gal$(F/K)$ lies in the kernel, and Hilb$_p$ is abelian over $K$. I'm not sure how to proceed from here... $\endgroup$ – Kevin Ventullo Aug 17 '11 at 7:15
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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