Examples
If the ring of integers of K is a unique factorization domain, in particular, if then K is its own Hilbert class field.
By contrast, let . By analyzing ramification degrees over, one can show that is an everywhere unramified extension of K, and it is certainly abelian. Hence the Hilbert class field of K is a nontrivial extension and the ring of integers of K cannot be a unique factorization domain. (In fact, using the Minkowski bound, one can show that K has class number exactly 2.) Hence, the Hilbert class field is .
To see why ramification at the archimedean primes must be taken into account, consider the real quadratic field K obtained by adjoining the square root of 3 to Q. This field has class number 1, but the extension K(i)/K is unramified at all prime ideals in K, so K admits finite abelian extensions of degree greater than 1 in which all primes of K are unramified. This doesn't contradict the Hilbert class field of K being K itself: every proper finite abelian extension of K must ramify at some place, and in the extension K(i)/K there is ramification at the archimedean places: the real embeddings of K extend to complex (rather than real) embeddings of K(i).
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