For each prime number $p$ and number field $k$, there exists at least one extension $k_{\infty}/k$ with Galois group isomorphic to $\mathbb{Z}_p$, the *cyclotomic $\mathbb{Z}_p$-extension*. If $k_p/k$ is the maximal abelian $p$-unramified pro-$p$ extension of $k$, and if $G_p = \mathrm{Gal}(k_p/k)$, then $\mathrm{Gal}(k_{\infty}/k)$ is a quotient of $G_p$. The group $\mathbb{Z}_p$ is a free object within the theory of pro-p groups, so in fact there is a (non-unique) decomposition $G_p = \mathbb{Z}_p \times \Gamma$ for some pro-p group $\Gamma$.

When $k$ is totally real, it is known that the $p$-adic regulator of $k$ is nonzero if and only if $\Gamma$ is finite. My question is, when the $p$-adic regulator is nonzero, what is known about the relationship between its p-valuation and the group $\Gamma$? In particular, is there a relationship between this valuation and the exponent of $\Gamma$?

** Edit**: this question arose from the following: suppose $p$ is odd, $\zeta_p$ is a primitive $p$th root of unity, $L/k$ is a degree $2p$ cyclic extension, $\zeta_p \in L$, and $F = k(\zeta_p)/k$ is the quadratic subextension of $L/k$. (It follows that both $F$ and $L$ are CM). Suppose further that:

- all prime ideals dividing $p$ ramify in $L/k$,
- none of these primes splits in $K/k$,
- the component of the p-primary part of the class group of $F$ on which complex conjugation acts by inversion is nontrivial,
- the norm map from the corresponding component of $L$ down to this component of $F$ has trivial kernel.

I can show in this case, contingent on the Brumer-Stark conjecture, that the values of the S-imprimitive partial zeta functions for $L/k$ at $s=0$, which are rational, are either $p$-integral or have valuation $-1$, and that the former occurs precisely when there exists a larger extension $L'/k$ containing $L/k$, cyclic of degree $2p^2$ and unramified away from $p$. (Here $S$ is set of Archimedean places of k combined with the finite places that ramify in $L/k$, which in this case can be shown to be precisely the prime ideals dividing $p$.) On the other hand, Gras showed in "Sur les denominateurs des fonctions zeta partielles" that the valuations of these partial zeta values can be computed in terms of the valuation of the $p$-adic regulator of $k$. I haven't written down the exact inequality that you'll get, because it looks like it may be rather messy in general, but there definitely is one.

When $k$ is real quadratic, $p = 3$, there exists an extension $L/k$ as above, and, say $3$ is inert in $k$, then this simplifies to the following: the $3$-adic regulator of $k$ has $3$-valuation equal to either $2$ or $3$, and it is $2$ precisely when $\Gamma$ has exponent $3$.

Examples of both types of real quadratic fields:

$k=\mathbb{Q}(\sqrt{29})$ has an extension $L$ meeting the criteria above, with $L$ generated over $k$ by $\zeta_3$ and a root of $x^3 - 6x - \sqrt{29}$. If $u$ is the fundamental unit of $k$, then $u^8-1$ is divisible exactly by $3^3$, so the p-adic regulator has 3-valuation 3. PARI/GP shows that $\Gamma$ can be taken to be cyclic of order $9$.

$k=\mathbb{Q}(\sqrt{113})$ -- here we may let $L$ be generated over $k$ by $\zeta_3$ and a root of $x^3 - 15x - 2\sqrt{113}$. In this case, $u^8-1$ is divisibl exactly by $3^2$, so the 3-adic regulator has 3-valuation equal to $2$. In this case, PARI/GP shows that $\Gamma$ has order $3$.