For the number fields K whose ring of integer is UFD,
why the Hilbert class field of K is K itself???
thanks
it's actually an if and only if statement. the reason is quite clear: $\displaystyle \mathcal{O}_K,$ the ring of integers of $\displaystyle K$, is a UFD if and only if it's a PID because $\displaystyle \mathcal{O}_K$ is always a Dedekind domain.
on the other hand by definition $\displaystyle H(K),$ the ideal class group of $\displaystyle K,$ is trivial if and only if every ideal of $\displaystyle \mathcal{O}_K$ is principal, i.e. if and only if $\displaystyle \mathcal{O}_K$ is a PID. so $\displaystyle |H(K)|,$ the class
number of $\displaystyle K,$ is 1 if and only if $\displaystyle \mathcal{O}_K$ is a PID. finally the dimension of the Hilbert class field of $\displaystyle K,$ as a vector space over $\displaystyle K,$ is exactly $\displaystyle |H(K)|$ and the result follows.