Units become powers when lifted to unramified extensions?

Question

Suppose $k$ is an algebraic number field, and $K$ is an unramified extension. I know:

• non-units $p\in k$ cannot become a power in $K$, or else the ideal they generate would become ramified in ${\cal O}_K$.

• units of finite order can become powers, as demonstrated here.

• What about units of infinite order? These are guaranteed to exist by Dirichlet's Unit Theorem. Is anything at all known?

Answer 1

I have the following attempt:

If $u\in\cal O$$_K^\times$, then the discriminant of $k(\sqrt[p]{u})/k$ is a power of $p$ times a power of $u$. An ideal is ramified iff it divides the discriminant. So the only ideals that can be ramified in this extension are those dividing $\langle pu\rangle$. Since $u$ is a unit, none will divide $\langle u\rangle$; thus $u$ becomes an $p$-th power in some unramified extension if and only if $k(\sqrt[p]{u})/k$ is unramified, if and only if $p\cal O$ is unramified in $k(\sqrt[p]{u})$.

Answer 2

If $p$ is properly irregular (which is equivalent to $p$ irregular if Vandiver's conjecture holds), then the unramified $p$-extensions of $\mathbf{Q}(\zeta_p)$ constructed by Ribet in his proof of the converse of Herbrand's theorem are generated by $p$th roots of cyclotomic units.