Class Groups in Dihedral Extensions – Spiegelungssatz?

algebraic-number-theoryclass-field-theorynt.number-theoryrt.representation-theory

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:
$\;\;\;F$
$\;\;\;\;|\;\;\;\backslash$
$\;\;\;K$ $L$
$\;\;\;\;|{\tiny 2}\;\;/{\tiny p}$
$\;\;\;\mathbb{Q}$
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:

One first replaces the class numbers by regulators, using the analytic class number formula.
One interprets the resulting quotient of regulators as an invariant of the isomorphism class of the Galois module $\mathcal{O}_F^\times$.
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.

Best Answer

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.

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[Math] Hilbert Class Field Galois over Q

I would first look at the corresponding local question. So $\mathbf{Q}$ gets replaced by $\mathbf{Q}_p$ (where $p$ is a prime), $K$ becomes a finite extension of $\mathbf{Q}_p$, and $H$ is the maximal abelian unramified extension of $K$ (here the adjective "abelian" is superfluous).

Is it possible for $H$ to be galoisian over $\mathbf{Q}_p$ while $K$ itself fails to be so ? If such is the case, the galoisian closure $\tilde K$ of $K$ over $\mathbf{Q}_p$ will be unramified over $K$, so let's look for a $K$ for which $\tilde K$ over $K$ is ramified (of degree $>1$).

To begin somewhere, let's look for $K$ which have degree $p$ over $\mathbf{Q}_p$. If $p=2$, there are no examples because every quadratic extension is galoisian. But if $p\neq2$, by looking at the parametrisation of degree-$p$ extensions of $\mathbf{Q}_p$ as in arXiv:1005.2016, it is easy to see that such $K$ do indeed exist.

Degree 17 Number Fields Ramified Only at 2

Thank you for calling this problem to my attention. I computed $K$ en route to AWS (though this year's topics are a rather different flavor of number theory...). After some simplification (gp's $\rm polredabs$), it turns out that the field $K$ is generated by a root of $$ f(x) = x^{17} - 2x^{16} + 8x^{13} + 16x^{12} - 16x^{11} + 64x^9 - 32x^8 - 80x^7 $$ $$ \qquad\qquad\qquad {} + 32x^6 + 40x^5 + 80x^4 + 16x^3 - 128x^2 - 2x + 68. $$ gp quickly confirms:

p = Pol([1,-2,0,0,8,16,-16,0,64,-32,-80,32,40,80,16,-128,-2,68])
poldegree(p)
F = nfinit(p);
factor(F.disc)

returns

[2 79]

So the field has discriminant $2^{79}$.

Let $F = {\bf Q}(\zeta_{64}^{\phantom{0}} - \zeta_{64}^{-1})$, with $F/\bf Q$ cyclic of degree $16$ of $\bf Q$. It was known that $K$ is contained in an unramified cyclic extension $L/F$ of degree $17$. Hence $L(\zeta_{17})$ is a Kummer extension of $F_{17} := F(\zeta_{17})$. Now $F_{17}$ is a $({\bf Z} / 16 {\bf Z})^2$ extension of $\bf Q$. Fortunately it was not necessary to work in the unit group of this degree $256$ field: the Kummer extension is $F_{17}(u^{1/17})$ with $u \in F_{17}^* / (F_{17}^*)^{17}$ in an eigenspace for the action of ${\rm Gal}(F_{17}/{\bf Q})$, which makes it fixed by some index-$16$ subgroup of this Galois group. Thus $u$ could be found in one of the cyclic degree-$16$ extension of $\bf Q$ intermediate between $\bf Q$ and $F_{17}$, and eventually a generator of $K$ turned up whose minimal polynomial, though large, was tractable enough for gp's number-field routines to do the rest.

Best Answer

Related Solutions

[Math] Hilbert Class Field Galois over Q

Degree 17 Number Fields Ramified Only at 2

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