relations between class numbers of quadratic extensions

Question

Let $h_m$ is the class number of $\mathbb{Q}[\sqrt m]$ and let $p>2$ a prime number. Is there a known connections between $h_p$ and $h_{-p}$? e.g. if $q^i$ divides $h_p$ then it also divides $h_{-p}$, or the other way around?

The only relevant result I've found is the following (ex' 10.6 in Washington's book "Intro to cyclotomic fields"): if $0 < d$ is an even, square-free integer, $r = $ 2-rank of $\mathbb{Q}[\sqrt d]$, $d = $ 2-rank of $\mathbb{Q}[\sqrt{-d}]$, then $r \leq s \leq r+1$.

Is there a similar result for any $d$?

Thanks

Answer 1

Let $m$ be a squarefree number, and let $d$ run through the discriminants of quadratic number fields coprime to $m$. Then the $2$-rank of the class group of ${\mathbb Q}(\sqrt{dm})$ is, up to a small term depending on the residue class of $m$ modulo $4$ and the sign of the fundamental unit involved, essentially the number of prime factors of $md$ minus $1$. This follows from Dedekind's version of Gauss's genus theory in quadratic number fields. Thus the answer to your question is yes for any $d$ (I've just seen that I switched your $d$ and $m$).

You will get slightly cleaner formulas is you replace the usual class group by the class group in the strict sense, since the dependence on the fundamental unit will disappear.

Answer 2

The short answer to your question is basically no, there's essentially no connection between the prime powers $q^i$ dividing $h_p$ and $h_{-p}$.

It's true that there's a general relationship between the 2-ranks of the class groups of $\mathbb{Q}(\sqrt{m})$ and $\mathbb{Q}(\sqrt{-m})$ as per Franz's answer and the result from Washington you cite (which, incidentally, can be pushed further to 4-ranks, 8-ranks, etc., getting pretty close to a full comparison of the 2-parts of the class numbers given knowledge of the fundamental unit of the real quadratic), but for your case this is almost contentless by genus theory. For other primes $q$, or the class number in whole, it's hard to give a conclusive justification for the "no relationship" claim, though two points bear mentioning:

1. Larges tables of these class numbers are available or easily generated, a quick survey of which is pretty compellingly against any correlation; and
2. There are all sorts of heuristics out there about how the two classes of class numbers should behave, some of which great imply a lack of relationship. For example, it's very unlikely that $q^i$ dividing $h_{-p}$ could tell you anything about $q^i$ dividing $h_p$ since $h_{-p}$ is non-trivial for all but nine examples, whereas (probably) $h_p=1$ infinitely often.