On a minimal algebraic number field which satisfies the principal ideal theorem

Question

By an algebraic number field, we mean a finite extension field of the field of rational numbers. Let $k$ be an algebraic number field, we denote by $\mathcal{O}_k$ the ring of algebraic integers in $k$. Let $K$ be a finite extension field of an algebraic number field $k$. Suppose for every ideal $I$ of $\mathcal{O}_k$, $I\mathcal{O}_K$ is principal. Then $K$ is called a PIT(Principal Ideal Theorem) field over $k$. Let $K$ be a PIT field over $k$. We say $K$ is a minimal PIT field over $k$ if $L/k$ is not PIT for every proper subextension $L/k$ of $K/k$.

(1) Let $k$ be an algebraic number field and $K/k$ be a finite extension. Is $K/k$ a minimal PIT if and only if $K/k$ is the Hilbert class field?

(2) Let $K/k$ and $L/k$ be minimal PITs. Are $K/k$ and $L/k$ isomorphic?

Answer 1

The answer to your first question is "no". In general, if $K/k$ is a cyclic unramified Galois extension of odd order, then the order of the capitulation kernel (the subgroup of the class group of $k$ that dies when base-changing to $K$), is $[K:k]\cdot [\mathcal{O}_k^\times: N_{K/k}\mathcal{O}_K^\times]$. The second factor is the index in the integral units of $k$ of the subgroup generated by norms of units of $K$, and it can be non-trivial. See Iwasawa, A Note on Capitulation Problem for Number Fields for a concrete example of a field $k$ whose class group capitulates in a proper subfield of the Hilbert class field, using the above observation.

Answer 2

The answer to the second question is also "no". Take $k= \mathbb{Q}(\sqrt{-5})$. Then the only non-trivial class capitulates in $H=k(i)$ and it also does in $K=k(\sqrt{-3})$, yet $H$ and $K$ are not isomorphic.