Let $K$ be a number field with ring of integers $R$ such that the class group of $R$ is non trivial. Then, is it possible that for all $k \in K$, we can write $k = a/b$ with $(a,b) = 1$ and $a,b \in R$?
Certainly if $R$ has trivial class group, this is possible but what about the general case?
It's not possible: if we write $k=a/b$ with $a$, $b\in R$ then the ideal class of $\left<a,b\right>$ is an invariant of $k$, and this can be non-trivial, e.g., $k=\sqrt{-6}/2$.