Leopoldt's Conjecture and Congruence Subgroups of Prime Ideal Powers

Question

Let $K$ be a number field and let $\mathfrak{p} \subset \mathcal{O}_K$ be a nonzero prime ideal in its ring of integers. I read here that the following question is equivalent to (a version of) Leopoldt's conjecture:

Does every finite-index subgroup $U \subset \mathcal{O}_K^\times$ contain a congruence subgroup of the form $$ \mathcal{O}_K^\times(\mathfrak{p}^m) := \{ x \in \mathcal{O}_K^\times : x \equiv 1 \text{ mod } \mathfrak{p}^m \} $$ for some $m \in \mathbb{N}$ ?

As far as I know, the Leopold conjecture for a number field $K$ and a prime number $p \in \mathbb{Z}$ states that the $p$-adic regulator $R_p$ of $K$ vanishes.

Question: Is the above problem open? If so, is it equivalent to Leopoldt's conjecture?

Answer 1

Probably better to consider the following related problem. Fix a prime $p$ and a number field $K$:

Does every finite-index subgroup $U \subset \mathcal{O}_K^\times$ of $p$-power index contain a congruence subgroup of the form $$ \mathcal{O}_K^\times(p^m) := \{ x \in \mathcal{O}_K^\times : x \equiv 1 \text{ mod } p^m \} $$ for some $m \in \mathbb{N}$ ?

This is literally the same as Leopoldt's conjecture as formulated by Iwasawa. (I don't know what the previous answer is talking about when it says the problems are different because Leopoldt is arithmetic.)

Consider this with $U = \mathcal{O}^{\times p}_K$. Then this question becomes:

if $m$ is big enough, is any unit $u \equiv 1 \bmod p^m$ a $p$th power?

Equivalently, let $U_v$ be the local units for $v | p$, and $E$ the group of global units ($\mathcal{O}^{\times}_K$). I claim the question above is equivalent to:

is the map $E \otimes \mathbf{Z}_p \rightarrow \prod_{v} (U_v \otimes \mathbf{Z}_p)$ injective?

if this map is not injective, then there is a saturated element $v$ in the kernel. Since the image of $E$ in $E \otimes \mathbf{Z}_p$ is dense, it means that, for any $m$, there is a unit $u_m \in E$ which is equal to $v$ up to an element of $p^m (E \otimes \mathbf{Z}_p) = E^{p^m} \otimes \mathbf{Z}_p$. It follows that $u_m$ is a global unit which is $1 \bmod p^m$ but not a $p$th power.

Conversely, if $u_m$ is a sequence of global units which are $1 \bmod p^m$ but not $p$th powers, then their limit $v \in E \otimes \mathbf{Z}_p$ (which is compact, since it is just a finitely generated $\mathbf{Z}_p$-module) will lie in the kernel.

Answer 2

The formulation of Leopoldt's conjecture as the non vanishing of the $p$-adic regulator is limited to totally real field. The general formulation, for any number field, is due to Iwasawa (I think) in the algebraic part of Iwasawa theory, see e.g. Washington's "Introduction to Cyclotomic Fields", §13.1. In short :

Let $E$ be the group of units of a number field $K$, $p$ a fixed prime, $U_v$ the groups of local units at all primes $v$ of $K$ above $p$. The natural injection of $E$ into $\prod U_v$ induces a map $E \otimes \mathbf Z_p \to \prod U_v \otimes \mathbf Z_p$. Leopoldt's conjecture: this map is injective. For abelian $K$, this has been proved by Brumer, using $p$-adic transcendental methods. The general case is still a conjecture. In spite of the apparent parenthood with the congruence subgroup theorem given in your link, the two problems are different, at least because Leopoldt's singles out a prime $p$, hence is arithmetic in nature.