Ideal Class Number If $R$ and $S$ are both Dedekind domain and $R$ is a subring of $S$ then can we
show $\mathfrak{cl}(S)=1$ implies $\mathfrak{cl}(R)=1$ ? Is it true?
Also, is there any relation or theorem for us to determine the class number
between those rings and subrings. 
Thank you very much! 
 A: First of all, you can always embed a domain in its field of fractions. So take some Dedekind domain with nontrivial class group, something like $\mathbb{Z} [\sqrt{-5}]$, and then embed it in $\mathbb{Q} (\sqrt{-5})$. But this is probably not what you have in mind, since a field is a very uninteresting example of a Dedekind domain.
(As usually $\operatorname{Cl} (K)$ denotes $\operatorname{Pic} (\mathcal{O}_K)$, this might be confusing. What I mean is that $\operatorname{Pic} (K)$ is trivial for any field $K$, while for $\mathcal{O}_K \subset K$ the group $\operatorname{Pic} (\mathcal{O}_K)$ is very often nontrivial. In particular, for $K = \mathbb{Q} (\sqrt{-5})$.)
For a nontrivial example, one can look at quadratic subfields of cyclotomic fields. For instance, $\mathbb{Q} (\zeta_{20})$ contains $\mathbb{Q} (\sqrt{-5})$, and while $h_{\mathbb{Q} (\zeta_{20})} = 1$, we have $h_{\mathbb{Q} (\sqrt{-5})} = 2$.
A: Ken Yamamura, The determination of the imaginary abelian number fields with class number one, Math. Comp. 62 (1994) 899-921, has a table of the 172 imaginary abelian number fields with class number one. Looking through this table, it's easy to find examples that have subfields that don't have class number one. For example, ${\bf Q}(\sqrt{-3},\sqrt5)$ has class number one, and contains ${\bf Q}(\sqrt{-15})$, which doesn't; similarly, ${\bf Q}(\sqrt{-1},\sqrt5)$ versus ${\bf Q}(\sqrt{-5})$; ${\bf Q}(\sqrt{-3},\sqrt2)$ versus ${\bf Q}(\sqrt{-6})$; ${\bf Q}(\sqrt{-7},\sqrt5)$ versus ${\bf Q}(\sqrt{-35})$; and so on. 
I believe the paper is freely available here. 
