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



relations between class numbers of quadratic extensions
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.


