6
$\begingroup$

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?

$\endgroup$
12
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
10
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.