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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:

  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.

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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 $\sigma(F)=F$. –  Kevin Ventullo Aug 17 '11 at 6:16
    
@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... –  Alex B. Aug 17 '11 at 6:29
    
@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... –  Kevin Ventullo Aug 17 '11 at 7:15
    
let $K$ ..., and $L$ an intermediate cubic. Do you mean an intermediate degree-$p$ extensions? Also, presumably $p$ is a prime. –  Chandan Singh Dalawat Aug 17 '11 at 11:42
    
Thank you, corrected. –  Alex B. Aug 17 '11 at 11:51
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