TITLE: How is class of composition of two quadratic fields is related class numbers of quadratic field?
QUESTION [5 upvotes]: Let  $K_1=\Bbb Q(\sqrt{d_1})$ ,  $K_2=\Bbb Q(\sqrt{d_2})$ and  $K=\Bbb Q(\sqrt{d_1},\sqrt{d_2})$.Suppose $h_1,h_2,h$ be class number of $K_1,K_2,K$ respectively.
(i) Can we express $h$ in terms of $h_1,h_2$?
(ii) Knowing the divisibility properties  of  $h_1,h_2$, I want help with   concluding about the divisibility of $h_1,h_2.$

REPLY [10 votes]: I assume you mean $K_1=\mathbb Q(\sqrt{d_1})$.
1) If by $K$ you mean $\mathbb Q(\sqrt{d_1d_2})$, then there is no simple relation
between $h$ and $h_1$ and $h_2$.
2) If by $K$ you mean the quartic biquadratic field $\mathbb Q(\sqrt{d_1},\sqrt{d_2})$, a theorem of Herglotz says that $h=h_1h_2h_3/2^j$, where
$h_3$ is the class number of $\mathbb Q(\sqrt{d_1d_2})$ and $j=0,1,2$ which can
be computed in terms of the units of $K$.