On the image of $p$-adic logarithm on principal units

TL;DR

This paper investigates the image of the $p$-adic logarithm on principal units in specific ramified extensions of $Q_p$, providing explicit descriptions in cases where the logarithm is not a bijection.

Contribution

It computes the image of the $p$-adic logarithm on principal units for certain ramified extensions where the logarithm is not bijective, extending understanding beyond the well-known cases.

Findings

Computed $ ext{log}_p(1+rak{m}_K)$ for $K=Q_p(zeta_p)$ with ramification index $p-1$.

Determined the image of $ ext{log}_p$ for quadratic extensions of $Q_2$ with ramification indices 1 and 2.

Extended the description of the $p$-adic logarithm's image in ramified cases where bijectivity fails.

Abstract

The $p$-adic logarithm appears in many places in number theory. Hence having a good description of the image of the $p$-adic logarithm could be useful, and in particular, to figure out the image of $1 + \mathfrak{m}_K$, where $K$ is an algebraic extension of $\mathbb{Q}_p$ and $\mathfrak{m}_K$ its maximal ideal. If the ramification index of $K$ is strictly less than $p-1$ then it is well known that the $p$-adic logarithm is a bijection of $1+\mathfrak{m}_K$ onto $\mathfrak{m}_K$. If the ramification index is equal or greater than $p-1$ than the $p$-adic logarithm is no more a bijection and the situation is more complicated. Our main result is the computation of $\log_p(1+\mathfrak{m}_K)$ in two cases: \begin{enumerate} \item[$\bullet$] for $K=\mathbb{Q}_p(\zeta_p)$, with $\zeta_p^p=1$, totally ramified $p$-cyclotomic extension of $\mathbb{Q}_p$ (ramification index equal $p-1$) \item[$\bullet$] for $K$ a quadratic extension of $\mathbb{Q}_2$ (ramification index equal 1, 2). \end{enumerate}

Equations

$${\mathfrak{m}}_K\supset {\mathfrak{m}}_K^2\supset {\mathfrak{m}}_K^3\supset \cdots ,$$

$$1+{\mathfrak{m}}_K\supset 1+{\mathfrak{m}}_K^2\supset 1+{\mathfrak{m}}_K^3\cdots$$

$$\{y\in {\mathbb{Z}}_p[{\zeta }_p]|y={\pi }^{2}z,y\equiv -\frac{b^2}{2}\text{mod}\pi {\mathbb{Z}}_p[{\zeta }_p],b\in ({\mathbb{Z}}_p[{\zeta }_p]{)}^{\times }\}.$$

$$b=\sum _{n\ge n_0}{\alpha }_n{\pi }^n,$$

$$L(a)={\log }_p(1+\pi b)=\sum _{n=1}^{\infty }(-1{)}^{n-1}\frac{{\pi }^n{b}^n}{n}.$$

$$y=\sum _{n=1}^{\infty }(-1{)}^{n-1}\frac{{\pi }^n{b}^n}{n},b\in ({\mathbb{Z}}_p[{\zeta }_p]{)}^{\times }.$$

$$b^p\pi =-\frac{b^p{\pi }^p}{p}.$$

$$\sum _{n=1}^{\infty }(-1{)}^{n-1}\frac{{\pi }^n{b}^n}{n}$$

$$=\pi b-\frac{{\pi }^2{b}^2}{2}+\frac{{\pi }^3{b}^3}{3}-\frac{{\pi }^4{b}^4}{4}+\cdots -\frac{{\pi }^{p-1}{b}^{p-1}}{p-1}-b^p\pi -\sum _{n\ge p+1}\frac{(-b\pi {)}^n}{n}+\cdots \cdots$$

$$=(b-b^p)\pi -\frac{{\pi }^2{b}^2}{2}+\frac{{\pi }^3{b}^3}{3}-\frac{{\pi }^4{b}^4}{4}+\cdots -\frac{{\pi }^{p-1}{b}^{p-1}}{p-1}-\sum _{n\ge p+1}\frac{(-b\pi {)}^n}{n}+\cdots \cdots$$

$$y=-\frac{{\pi }^2{b}^2}{2}+\frac{{\pi }^3{b}^3}{3}-\frac{{\pi }^4{b}^4}{4}+\cdots -\frac{{\pi }^{p-1}{b}^{p-1}}{p-1}+p\pi {\mathbb{Z}}_p[{\zeta }_p].$$

$$y=-\frac{b^2}{2}+\frac{\pi b^3}{3}-\cdots -\frac{{\pi }^{p-3}{b}^{p-1}}{p-1}-\sum _{n\ge p+1}\frac{(-b{)}^n{\pi }^{n-2}}{n},b\in ({\mathbb{Z}}_p[{\zeta }_p]{)}^{\times }$$

$$f(t)=-z-\frac{t^2}{2}+\frac{\pi t^3}{3}-\cdots -\frac{{\pi }^{p-3}{t}^{p-1}}{p-1}-\sum _{n\ge p+1}\frac{(-t{)}^n{\pi }^{n-2}}{n}$$

$$\{\begin{array}{rl}y=& {\pi }^{2}z\\ z\equiv & -\frac{b^2}{2}\text{mod}\pi {\mathbb{Z}}_p[{\zeta }_p],b\in {\mathfrak{R}}_p\end{array}$$

$$f(t)=-z-\frac{t^2}{2}+\frac{\pi t^3}{3}-\cdots -\frac{{\pi }^{p-3}{t}^{p-1}}{p-1}-\sum _{n\ge p+1}\frac{(-t{)}^n{\pi }^{n-2}}{n}\in {\mathbb{Z}}_p[{\zeta }_p][[t]],$$

$$\begin{array}{rl}{f}^{\prime }(t)\equiv & t\text{mod}\pi {\mathbb{Z}}_p[{\zeta }_p][[t]]\\ f^{\prime }(t)\equiv & b\text{mod}\pi {\mathbb{Z}}_p[{\zeta }_p]\end{array}\}\Rightarrow f^{\prime }(b)\not\equiv 0\text{mod}\pi {\mathbb{Z}}_p[{\zeta }_p].$$

$$\{\begin{array}{rl}{f}^{\prime }(b)\equiv & 0\text{mod}\pi {\mathbb{Z}}_p[{\zeta }_p],\\ f^{\prime }(b)\not\equiv & b\text{mod}\pi {\mathbb{Z}}_p[[{\zeta }_p]].\end{array}$$

$$\left({\log }_p(1+\pi ({\mathbb{Z}}_p[\zeta ]{)}^{\ast })\right)=\{y\in {\mathbb{Z}}_p[{\zeta }_p]|y={\pi }^{2}z,z\equiv -\frac{b^2}{2}\text{mod}\pi {\mathbb{Z}}_p[{\zeta }_p],b\in ({\mathbb{Z}}_p[{\zeta }_p]{)}^{\times }\}.$$

$$(1+{\mathfrak{m}}_K)-(1+{\mathfrak{m}}_K^2)\underset{\text{p-adic logarithm}}{\overset{L(x)}{\to }}\{y\in {\mathbb{Z}}_p[{\zeta }_p]|y={\pi }^{2}z,z\equiv -\frac{b^2}{2}\text{mod}\pi {\mathbb{Z}}_p[{\zeta }_p],b\in ({\mathbb{Z}}_p[{\zeta }_p]{)}^{\times }\}.$$

$${\pi }^2({\mathbb{Z}}_p[{\zeta }_p]{)}^2=({\mathbb{Z}}_p[{\zeta }_p]{)}^2=\{{\pi }^2{\zeta }^2(1+\pi u),u\in {\mathbb{Z}}_p[{\zeta }_p]\}\cup \{0\}.$$

$${\pi }^2({\mathbb{Z}}_p[{\zeta }_p{]}^{\times }{)}^2={\pi }^2\cdot ⟨{\zeta }^2⟩\cdot (1+\pi {\mathbb{Z}}_p[{\zeta }_p]).$$

$${\log }_p((1+{\mathfrak{m}}_K)-(1+{\mathfrak{m}}_K^2))=\{y\in {\mathbb{Z}}_p[{\zeta }_p]|y={\pi }^{2}z,z\equiv -\frac{b^2}{2}\text{mod}\pi {\mathbb{Z}}_p[{\zeta }_p],b\in ({\mathbb{Z}}_p[{\zeta }_p]{)}^{\times }\},$$

$$\{y\in {\mathbb{Z}}_p[{\zeta }_p]|y={\pi }^{2}z,z\equiv -\frac{b^2}{2}\text{mod}\pi {\mathbb{Z}}_p[{\zeta }_p],b\in ({\mathbb{Z}}_p[{\zeta }_p]{)}^{\times }\}={\pi }^2\cdot \left(-\frac{1}{2}\right)\cdot ⟨{\zeta }_p^2⟩\cdot (1+\pi {\mathbb{Z}}_p[{\zeta }_p]).$$

$$\{y\in {\mathbb{Z}}_p[{\zeta }_p]|y={\pi }^{2}z,z\equiv -\frac{b^2}{2}\text{mod}\pi {\mathbb{Z}}_p[{\zeta }_p],b\in ({\mathbb{Z}}_p[{\zeta }_p]{)}^{\times }\}={\pi }^2\cdot ⟨{\zeta }^2⟩\cdot (1+\pi {\mathbb{Z}}_p[{\zeta }_p])={\mathfrak{m}}_K^2.$$

$${\pi }^2=-3,{\pi }^4=3^2,{\pi }^6=-3^3,{\pi }^8=3^4,{\pi }^{10}=-3^5,{\pi }^{12}=3^6,{\pi }^{14}=-3^7,{\pi }^{16}=3^8,\cdots$$

$${\pi }^3=-3\pi ,{\pi }^5=3^2\pi ,{\pi }^7=-3^3\pi ,{\pi }^9=3^4\pi ,{\pi }^{11}=-3^5\pi ,{\pi }^{13}=3^6\pi ,{\pi }^{15}=-3^7\pi ,\cdots$$

$${\log }_3(1+\pi b)$$

$$=\pi b-\frac{{\pi }^2{b}^2}{2}+\frac{{\pi }^3{b}^3}{3}-\frac{{\pi }^4{b}^4}{4}+\frac{{\pi }^5{b}^5}{5}-\frac{{\pi }^6{b}^6}{6}+\cdots ,b\in ({\mathbb{Z}}_3[{\zeta }_3]{)}^{\times },$$

$$=b^2(\frac{3}{2}-\frac{3^2{b}^2}{4}+\frac{3^3{b}^4}{6}-\frac{3^4{b}^6}{8}+\frac{3^5{b}^8}{10}-\frac{3^6{b}^{10}}{12}+\cdots )+\pi b(1-\frac{3b^2}{3}+\frac{3^2{b}^4}{5}-\frac{3^3{b}^6}{7}+\frac{3^4{b}^8}{9}-\frac{3^5{b}^{10}}{11}+\cdots ),$$

$$=b^2{S}_1(b)+\pi b{S}_2(b),\text{where}$$

$$S_1(b):=\frac{3}{2}-\frac{3^2{b}^2}{4}+\frac{3^3{b}^4}{6}-\frac{3^4{b}^6}{8}+\frac{3^5{b}^8}{10}-\frac{3^6{b}^{10}}{12}+\cdots ,$$

$$S_2(b):=1-\frac{3b^2}{3}+\frac{3^2{b}^4}{5}-\frac{3^3{b}^6}{7}+\frac{3^4{b}^8}{9}-\frac{3^5{b}^{10}}{11}+\cdots .$$

$$\underset{n\to \infty }{\lim }{S}_1(b)$$

$$\underset{n\to \infty }{\lim }{S}_2(b)$$

$${\log }_3(1+\pi b)=b^2{z}_1+\pi b{z}_2.$$

Full text

1. $\bullet$

for $K=\mathbb{Q}_{p}(\zeta_{p})$, with $\zeta_{p}^{p}=1$, totally ramified $p$-cyclotomic extension of $\mathbb{Q}_{p}$ (ramification index equal $p-1$) 2. $\bullet$

for $K$ a quadratic extension of $\mathbb{Q}_{2}$ (ramification index equal 1, 2).

Introduction

Motivations

This paper is a continuation of the authors’ previous paper [20], which computes the bases of the image of $p$-adic logarithm on principal local units $1+\mathfrak{m}_{K}$ as $\mathbb{Z}_{p}$-module. In $p$-adic number theory, the $p$-adic logarithm plays an essential role. From [3, 8, 13, 14, 22], we know the image of $p$-adic logarithm on the group of principal units is very crucial in the theory of Iwasawa. Iwasawa explicitly computes the formulas for the norm residue symbol in [13] by utilising the image of $p$-adic logarithm on the group of principal units of a particular cyclotomic extension of $\mathbb{Q}_{p}$, which is one of the primary motivations. The image/value of the $p$-adic logarithm has a wide range of other potential applications. The key ingredient in the definition of normalised $p$-adic regulator of a number field in [12] is the image of $p$-adic logarithm on the principal units. In [7], the authors use the $p$-adic logarithm in their study to compute the set of rational points on smooth, projective, geometrically integral curve from LMFDB [23]. In [18], the p-adic logarithm is used in the investigation of the $p$-adic counterpart of the classical Birch-Swinnerton-Dyer conjecture, where the $L$-function of an elliptic curve is considered. In [5], the $p$-adic logarithm plays crucial role in the study of $p$-adic $L$-functions of modular elliptic curves. To compute the $p$-adic $L$-function $L(1,\chi)$, one of the key challenges is the computation of $p$-adic logarithm of arbitrary elements in its domain, e.g., [11]. The $p$-adic logarithm has an important role in the formula of Sen operator ([4]) in the theory of $p$-adic representation as well as in the formula of $p$-adic heights [19] in the theory of arithmetic dynamics. In [6], the author defines cyclotomic and anticyclotomic characters of a $\mathbb{Q}_{p}$-vector space by $p$-adic logarithm. Consequently, it is novel to comprehend the image of the $p$-adic logarithm on its domain. It is well-known that the $p$-adic logarithm defines an isomorphism from $1+\mathfrak{m}_{K}^{r}\to\mathfrak{m}_{K}^{r}$ if $r>\frac{e}{p-1}$, where $e$ is the ramification index. However, things are not the same when $r=1$; in fact, $\log_{p}:1+\mathfrak{m}_{K}\to\mathfrak{m}_{K}$ is not an isomorphism. The image $\log_{p}(1+\mathfrak{m}_{K})$ is unknown for arbitrary extension $K$ of $\mathbb{Q}_{p}$. The present paper takes the initiative to compute the image of $p$-adic logarithm on the group of principal units. In Subsection 1.1, we explicitly compute the images of the $p$-adic logarithm of maximum ideals in the cyclotomic extension of $\mathbb{Q}_{p}$, and related results. In Subsection 2.2, we compute image of $p$-adic logarithm on principal units in the ramified quadratic extensions of $\mathbb{Q}_{2}$. In Subsection 2.3, we apply our result to compute normalized $p$-adic regulator of a number field (see Theorem 2.21).

Notations

Consider the $p$-adic number field $\mathbb{Q}_{p}$, where $p$ is a prime, and where $\mathbb{Z}_{p}$ is the ring of integers, where $p$ is the unique maximum ideal, and where $\mathbb{F}_{p}:=\mathbb{Z}_{p}/p\mathbb{Z}_{p}$ is the residue field. Let $\bar{\mathbb{Q}}_{p}$ be the algebraic closure, and take the $p$-adic completion $\mathbb{C}_{p}:=\widehat{\bar{Q}}_{p}$, where $\mathfrak{m}_{\mathbb{C}_{p}}$ is the maximal idea, and $\mathbb{C}_{p}^{\times}$ are the units. Let $K$ be a finite extension of $\mathbb{Q}_{p}$, where $\mathcal{O}_{K}$ is the ring of integers and $\mathfrak{m}_{K}$ is the unique maximal ideal. The residue field of $\mathcal{O}_{K}$ is $\kappa_{K}:=\mathcal{O}_{K}/\mathfrak{m}_{K}$. We also have the standard $p$-adic additive valuation $v_{p}:\mathbb{Q}_{p}\to\mathbb{Z}\cup\{\infty\}$, which extends uniquely to the valuation $v_{p}$ on $\mathbb{Q}_{p}$ on a finite extension $K$ of $\mathbb{Q}_{p}$ denoted by $v$ that makes $v(z)=v_{p}(z)$ whenever $z\in\mathbb{Q}_{p}$ by the following formula $v(\lambda)=\frac{1}{[K:\mathbb{Q}_{p}]}v_{p}\left(N_{\mathbb{Q}_{p}}^{K}(\lambda)\right),$ where $N_{\mathbb{Q}_{p}}^{K}$ is the field norm. Throughout this study, we will refer to $v$ as the standard notation.

1. Image of $p$-adic logarithm in the cyclotomic extension

In this section, we calculate the images of the p-adic logarithm, $\log_{p}(1+\mathfrak{m}_{K})$, where $\mathfrak{m}_{K}$ is the maximum ideal in the cyclotomic extension $\mathbb{Q}_{p}(\zeta_{p})$, $\zeta_{p}$ being the $p^{th}$ root of unity. Our goal is to determine the result of the p-adic logarithmic series $L(x)=\log_{p}(1+x)$ when the maximal ideal $\mathfrak{m}_{K}$ is plugged in. Since any finite abelian extension of $\mathbb{Q}_{p}$ is a subfield of a cyclotomic field according to the local application of the Kronecker-Weber theorem, knowing the answer also applies to the case of any finite abelian extension.

The p-adic logarithm is an isometry between the additive group of elements with $|x|<(\frac{1}{p})^{\frac{1}{p-1}}$ and the multiplicative group of elements of the form $1+y$ with $|y|<(\frac{1}{p})^{\frac{1}{p-1}}$. Given the maximal ideal $\mathfrak{m}_{K}$ of $\mathbb{Q}_{p}(\zeta_{p})$, the following chains

[TABLE]

holds. The power series for the p-adic logarithm is defined only on $1+\mathfrak{m}_{K}\subset\mathbb{Q}_{p}(\zeta_{p})$ and $\mathfrak{m}_{K}$ contains elements of precise size $(\frac{1}{p})^{\frac{1}{p-1}}$. If we restrict ourselves on $1+\mathfrak{m}_{K}^{2}$, then it can be demonstrated that $\log_{p}(1+\mathfrak{m}_{K}^{2})=\mathfrak{m}_{K}^{2}$ using homomorphism and isometry property of p-adic logarithm $L(x)$. In Theorem 1.4, we further demonstrate that $L(\mathfrak{m}_{K})=\log_{p}(1+\mathfrak{m}_{K})=\mathfrak{m}_{K}^{2}$. However, the situation is more interesting for the annular region $(1+\mathfrak{m}_{K})-(1+\mathfrak{m}_{K}^{2})$ i.e., finding the image of p-adic logarithm on those elements $u$ belonging to $1+\mathfrak{m}_{K}$ but not to $1+\mathfrak{m}_{K}^{2}$. In Theorem 1.1, we therefore address the following interesting question:

How does the p-adic logarithm treat the elements of $1+\mathfrak{m}_{K}$ that are not in $1+\mathfrak{m}_{K}^{2}$ ?

1.1. Main Results

Theorem 1.1**.**

Assume $p\neq 2$, and consider the cyclotomic extension $K:=\mathbb{Q}_{p}(\zeta_{p})$ of $\mathbb{Q}_{p}$, where $\zeta_{p}$ is the primitive $p^{th}$ root of unity, and $\mathfrak{m}_{K}$ be its maximal ideal. Then the image of $(1+\mathfrak{m}_{K})-(1+\mathfrak{m}_{K}^{2})$ under p-adic logarithm is

[TABLE]

Proof.

Assume that $p>2$. We know that the elements of $\mathbb{Q}_{p}(\zeta_{p})$ can be represented by their Hensel expansion. Then for $a\in\mathbb{Q}_{p}(\zeta_{p})$, let

[TABLE]

where $\pi$ is an uniformizer of the local field field $\mathbb{Q}_{p}(\zeta_{p})$ and $\alpha_{n}$ are taken from a complete set of representative $\mathfrak{R}_{p}$ of the residue field $\kappa_{K}\simeq\mathbb{F}_{p}$ of $\mathbb{Q}_{p}(\zeta_{p})$, for example, $\mathfrak{R}_{p}=\{0,1,2,\cdots,p-1\}$ or $\mathfrak{R}_{p}=\{\omega(0),\omega(1),\omega(2),\cdots,\omega(p-1)\}$, $\omega$ is the Teichm$\ddot{u}$ller character, and $\pi$ is a uniformizing parameter of the local field $\mathbb{Q}_{p}(\zeta_{p})$. The uniformizing parameter of the local field $\mathbb{Q}_{p}(\zeta_{p})$ is usually taken as $\pi=(\zeta_{p}-1)$ or $\pi=\sqrt[p-1]{-p}$. Also $v(\pi)=\frac{1}{p-1}$, where $v$ is the p-adic valuation on $\mathbb{Q}_{p}(\zeta_{p})$. We eventually settle on the notation $\pi=\sqrt[p-1]{-p}$ for our computing needs. We know that $\mathbb{Z}_{p}[\pi]=\mathbb{Z}_{p}[\zeta_{p}]$. Thus the unique maximal ideal in $\mathbb{Q}_{p}(\zeta_{p})$ is $\mathfrak{m}_{K}=\pi\mathbb{Z}_{p}[\zeta_{p}].$

Since $a\in(1+\mathfrak{m}_{K})-(1+\mathfrak{m}_{K}^{2})$, in the Hensel expansion of $u$, we have $n_{0}=0,\ \alpha_{0}=1,\ \alpha_{1}\neq 0$. Therefore, from equation (1), we have $a=1+\pi b,\ b\in(\mathbb{Z}_{p}[\zeta_{p}])^{\times},$ where $(\mathbb{Z}_{p}[\zeta_{p}])^{\times}$ is the multiplicative group of invertible elements in $\mathbb{Z}_{p}[\zeta_{p}]$. Therefore, we can derive the following from the definition of the p-adic logarithm:

[TABLE]

Now we want to characterise those $y\in\mathbb{Q}_{p}(\zeta_{p})$ such that

[TABLE]

We will use the uniformizer $\pi=\sqrt[p-1]{-p}$ of the local field $\mathbb{Q}_{p}(\zeta_{p})$, which is well suited for our problem because it is a root of the polynomial $x+\frac{x^{p}}{p}$. The polynomial is thus satisfied by $\pi$, and we have $\pi+\frac{\pi^{p}}{p}=0$ i.e., $\pi=-\frac{\pi^{p}}{p}$, consequently,

[TABLE]

So by the above choice of the uniformizer $\pi$ and by using (3), we get the following:

[TABLE]

By Fermat’s little theorem $b^{p}\equiv b\ (\text{mod}\ p)$ i.e., $\frac{b-b^{p}}{p}\in\mathbb{Z}_{p}[\zeta_{p}]$. This implies $b-b^{p}\in p\mathbb{Z}_{p}[\zeta_{p}]\subset\mathbb{Z}_{p}[\zeta_{p}]$. Hence from equation (2),

[TABLE]

It is interesting to note that power $p$ disappeared. So the first necessary condition is $y=\pi^{2}z$ with $z\in(\mathbb{Z}_{p}[\zeta_{p}])^{\times}$ and the second necessary condition is $z\equiv-\frac{b^{2}}{2}\ \text{mod}\ \pi\mathbb{Z}_{p}[\zeta_{p}]$.

Therefore our problem is now reduced to find the $y\in(\mathbb{Z}_{p}[\zeta_{p}])^{*}$ such that

[TABLE]

or equivalently: For which $z\in(\mathbb{Z}_{p}[\zeta_{p}])^{\times}$ the Taylor series

[TABLE]

does have a zero at $x_{0}\in(\mathbb{Z}_{p}[\zeta_{p}])^{\times}$. We now demonstrate that the necessary conditions

[TABLE]

are also sufficient. The problem is now further reduced to a classical Hensel’s lemma.

Take $y\equiv-\frac{b^{2}}{2}\ \text{mod}\ (\pi\mathbb{Z}_{p}[\zeta_{p}])$, and consider the Taylor series:

[TABLE]

its radius of convergence is $\rho=|\frac{1}{\pi}|_{p}=p^{\frac{1}{p-1}}>1.$ So it follows that

[TABLE]

So we have

[TABLE]

By Hensel lemma, there exists $x_{0}\in\mathbb{Z}_{p}[\zeta_{p}]$ such that $f(x_{0})=0$ and $x_{0}\equiv b\ \text{mod}\ \pi\mathbb{Z}_{p}[\zeta_{p}]$. Hence:

[TABLE]

Therefore, we conclude

[TABLE]

∎

Corollary 1.1.1**.**

For $p\equiv 1,3~{}(\text{mod}~{}8)$, the image of $(1+\mathfrak{m}_{K})-(1+\mathfrak{m}_{K}^{2})$ under $p$-adic logarithm $\log_{p}(1+x)$ is the subgroup $\mathfrak{m}_{K}^{2}$.

Proof.

For $p\neq 2$, every element of the group of principal units $1+\pi\mathbb{Z}_{p}[\zeta_{p}]$ is square. Further $\mathbb{Z}_{p}/\pi\simeq\mathbb{F}_{p}$ and $(\mathbb{Z}_{p}[\zeta_{p}]/\pi)^{\times}$ has representatives $\{\zeta\}$, where $\zeta$ is the primitive $(p-1)^{th}$ root of unity in $\mathbb{Q}_{p}(\zeta_{p})$. The binomial series for $(1+u)^{1/2}$ converges for $|u|_{p}<1$ when $p\neq 2$. Thus ,

[TABLE]

Taking into account only the unit elements, we get the following decomposition:

[TABLE]

By Theorem 1.1, we obtain

[TABLE]

which we can express in the following way:

[TABLE]

We now have to address the factor $\left(-\frac{1}{2}\right)$. By quadratic reciprocity law, $\left(-\frac{1}{2}\right)$ is a square in $(\mathbb{Z}_{p}[\zeta_{p}]/\pi)^{\times}\simeq\mathbb{F}_{p}^{\times}$ if $-2$ is a quadratic residue modulo $p$. Moreover, $\left(\frac{-2}{p}\right)=1\Leftrightarrow p\equiv 1,3~{}(\text{mod}~{}8)$. Hence if $p\equiv 1,3~{}(\text{mod}~{}8)$, taking into account the equations (4) and (5), we have

[TABLE]

This proves the result. ∎

The following lemma is useful for our purposes:

Lemma 1.2**.**

[17]** Each $f(X)\in\mathbb{Z}_{p}[[X]]$ converges in $D(1^{-})=\{x\in\mathbb{C}_{p}||x|_{p}<1\}$, where $.\mathbb{Z}_{p}[[X]]$ is a ring of formal power series.

We will use the following method to justify the preceding Corollary 1.1.1 by testing at $p=3$, specifically. This case is interesting due to the following behaviour of the uniformizer $\pi=\sqrt{-3}$ or $\pi^{2}=-3$:

Theorem 1.3**.**

For $p=3$, the image of $(1+\mathfrak{m}_{K})-(1+\mathfrak{m}_{K}^{2})$ is $\mathfrak{m}_{K}^{2}$ under $3$-adic logarithm.

Proof.

For $p=3$ case, the uniformizer $\pi=\sqrt[(p-1)]{-p}$ reduces to $\pi=\sqrt{-3}$, we have

[TABLE]

That is, $\pi^{2n}=(-3)^{n}$ and $\pi^{2n+1}=(-3)^{n}\pi$. Thus, similar to Theorem 1.1, we observe

[TABLE]

[TABLE]

Clearly, the series $S_{1}$ and $S_{2}$ converges $3$-adically as the exponents of $3$ are increasing as $n\to\infty$. Because all the coefficients of $S_{1}$ and $S_{2}$ belong to $\mathbb{Z}_{3}$, so by applying Lemma 1.2, we deduce that

[TABLE]

Hence, we have

[TABLE]

We will now establish a connection between the R.H.S. quantity and $m_{K}^{2}=1+\pi^{2}\mathbb{Z}_{3}[\zeta_{3}]=\pi^{2}\cdot\left\langle\zeta_{3}\right\rangle\cdot(1+\pi\mathbb{Z}_{3}[\zeta_{3}])$. Since $z_{1},z_{2}\in D(1^{-1})$, we have $|z_{1}|_{3}\leq 1/3$ and $|z_{2}|_{3}\leq 1/3$. That is, $z_{1},~{}z_{2}\in 3\mathbb{Z}_{3}[\zeta_{3}]=\mathfrak{m}_{K}^{2}$, and so $b^{2}z_{1}+\pi bz_{2}\in\mathfrak{m}_{K}^{2}$. More precisely, since $z_{1},z_{2}\in 3\mathbb{Z}_{3}[\zeta_{3}]$, we get

[TABLE]

∎

We now prove the following result:

Theorem 1.4**.**

The image of $1+\mathfrak{m}_{K}$ under p-adic logarithm $\log_{p}(1+x)$ is $\mathfrak{m}_{K}^{2}$. Further, the image of $(1+\mathfrak{m}_{K})-(1+\mathfrak{m}_{K}^{2})$ under p-adic logarithm is $\mathfrak{m}_{K}^{2}.$

Proof.

Let us define

[TABLE]

and in general $U^{(n)}=1+\mathfrak{m}_{K}^{n}$, the $n^{th}$ unit group for $n\geq 1$. We also denote the group of units by $U^{(0)}=\mathbb{Z}_{p}[\zeta_{p}]^{\times}$.

Since we are working on not only p-adic metric space but also on a group structure, instead of considering $U^{(1)}-U^{(2)}$, rather we consider the quotient group $U^{(1)}/U^{(2)}$. Therefore, $\log_{p}U^{(2)}=\mathfrak{m}_{K}^{2}$, since $\log_{p}(1+\mathfrak{m}_{K}^{2})=\mathfrak{m}_{K}^{2}$ follows from previous discussions. Consider the ring of integers $\mathcal{O}_{K}$, and the residual field $\kappa_{K}=\mathcal{O}_{K}/\mathfrak{m}_{K}$ of $K=\mathbb{Q}_{p}(\zeta_{p})$. We can express $U^{(n)}=\{u\in U^{(0)}=\mathcal{O}_{K}^{\times}:~{}u-1\in\mathfrak{m}_{K}^{n}\}$. Then $U^{(0)}$ and $U^{(n)}$ are open and closed and compact in $K^{\times}=(\mathbb{Q}_{p}(\zeta_{p}))^{\times}$ with induced topology. Hence we have the following isomorphisms of topological groups:

[TABLE]

Because $\mathbb{Q}_{p}(\zeta_{p})$ is totally ramified extension of $\mathbb{Q}_{p}$ and $\mathbb{F}_{p}\cong\mathbb{Z}_{p}/p\mathbb{Z}_{p}\cong\kappa_{K}$, we infer that the quotient group $U^{(1)}/U^{(2)}$ has order $p$. We now find a system of representatives of the quotient group $U^{(1)}/U^{(2)}$. We take $\pi=\zeta_{p}-1$ as an uniformizer. Then $\zeta_{p}=1+\pi\in U^{(1)}-U^{(2)}$ and hence the cyclic group $\left\langle\zeta_{p}\right\rangle$ is a subgroup of $U^{(1)}$, with $\left\langle\zeta_{p}\right\rangle\cap U^{(2)}=(1)$. This implies that $U^{(1)}=\left\langle\zeta_{p}\right\rangle\times U^{(2)}$. Therefore,

[TABLE]

Note the kernel of the p-adic logarithm is the set of $n^{th}$ roots of unity $\zeta_{p}$ in $\mathbb{C}_{p}^{\times}$ and hence $\log_{p}(\zeta_{p})=0$. Thus from equation (6), we get

[TABLE]

[TABLE]

[TABLE]

Direct product decomposition $U^{(1)}=\left\langle\zeta_{p}\right\rangle\times U^{(2)}$ implies that every $u_{1}\in U^{(1)}-U^{(2)}$ can be uniquely written as $u_{1}=\omega u_{2}$, where $\omega$ is the non-trivial $p$-th root of unity. But the non-triviality of $\omega$ plays no role as $\log_{p}$ of a root of unity is null, so that $\log_{p}(u_{1})=\log_{p}(u_{2})$. We have already seen that $\log_{p}(U^{(2)})=\mathfrak{m}_{K}^{2}$. Thus $\log_{p}(U^{(1)}-U^{(2)})=\log_{p}(U^{(2)})=\mathfrak{m}_{K}^{2}$.

∎

For a visual representation of the results of this subsection, refer to Figure 1.1:

2. Image of $2$-adic logarithm in $\mathbb{Q}_{2}(\sqrt{d})$

This section is divided into two subsections, 2.1 and 2.2, focusing, respectively, on tool development and main results.

2.1. Newton polygon, valuation function $and$ some developments

The goal of this subsection is to explore the Newton polygon and the valuation function/copolygon of the $p$-adic logarithm and to develop some concepts that will be useful in the subsequent subsection. Newton polygon is an important tool to study the roots of a $p$-adic power series.

Definition 2.1**.**

[2] Let $K$ be a finite extension of $\mathbb{Q}_{p}$, with extended $p$-adic additive valuation $v$, and let $0\neq f(x)=\sum_{m=0}^{\infty}a_{m}x^{m}\in K[[x]]$, then the Newton polygon of $f(x)$, denoted $\mathcal{N}_{f}$, is the lower convex hull of the points $(m,~{}v(a_{m}))$.

There is another geometric tool, namely, the valuation function having equal information as Newton polygon. The valuation function is the boundary of the unbounded convex polygon $\cap_{m\geq 0}\{(\xi,\eta):\eta\leq v(a_{m})+m\xi\}$ if the polynomial is $f(x)=\sum_{m\geq 0}a_{m}x^{m}$. That is, there is a convex body described as the intersection of half-planes, and the valuation function’s graph is the boundary of this body. Let us, then, make a formal definition:

Definition 2.2**.**

[15] Let $0\neq f(x)=\sum_{m=0}^{\infty}a_{m}x^{m}\in K[[x]]$. For each $m\geq 0$, let $H_{m}$ be the closed half-plane $H_{m}=\{(\xi,\eta)\in\mathbb{R}^{2}:\eta\leq m\xi+v(a_{m})\}$. Let $\psi$ be the closed convex set $\cap_{m}H_{m}$, which is known as Newton copolygon. Finally, if $\psi$ is not empty, we define $\psi_{f}$ to be the real-valued function whose graph is the boundary of $\psi$. This $\psi_{f}$ is the valuation function of $f$.

The domain of $\psi_{f}$ is a closed subset of $\mathbb{R}$. A few examples are given. If $f$ is a polynomial, then the domain of $\psi_{f}$ is the whole of $\mathbb{R}$. But if $f(x)=\sum_{0}^{\infty}x^{m}$, so that every $v(a_{m})=0$, then we see that $\psi$ is the whole $4^{th}$ quadrant of $(\xi,\eta)$-space, and the domain of $\psi_{f}$ is the closed half-line $[0,\infty)$. Consider, however, the series $f(x)=\sum_{m\geq 1}p^{-m^{2}}x^{m}$. then $H_{m}$ is the half-plane to the right of the line $\eta=m\xi-m^{2}$, whose $\xi$-intercept is the point $(m,0)$. We see in this case that the intersection of the half-planes $H_{m}$ is empty. There is a connection between the domain of $\psi_{f}$ and the set of $\alpha$ in an algebraic closure of $K$ for which the series $f(\alpha)$ is convergent.

Proposition 2.1**.**

[15]** The vertices of the Newton polygon $\mathcal{N}_{f}$ are in one-to-one correspondence with the segments of the valuation function $\psi_{f}$. That is, if $V$ is a vertex of $\mathcal{N}_{f}$ and $l$ is a segment of $\psi_{f}$, then the $x$-coordinate of $V$ is the negative sole of $l$ and the $y$-coordinate of $V$ is the $y$-intercept of $l$.

Example 2.1**.**

The vertex $V=(a,b)$ of $\mathcal{N}_{f}$ corresponds to a segment $l:y=-ax+b$ of $\psi_{f}$.

Proposition 2.2**.**

[15]** There is one-to-one correspondence between the non-vertical segments of the Newton polygon $\mathcal{N}_{L}$ and the vertices of the valuation function $\psi_{L}$. That is, if $l$ is a non-vertical segment of $\mathcal{N}_{L}$ and $V$ is the corresponding vertex of $\psi_{L}$, then the $x$-coordinate of $V$ is the slope of $l$ and the $y$-coordinate of $V$ is the $y$-intercept of $l$.

Example 2.2**.**

The segment $l:y=ax+b$ of $\mathcal{N}_{f}$ corresponds to the vertex $V=(a,b)$ of $\psi_{f}$.

Proposition 2.3**.**

Every monomial of a polynomial does not have an effect on the valuation function.

Proof.

Let’s look at some polynomials over $\mathbb{Q}_{p}$. First, consider the monomial $f(x)=px^{2}$, as a function into which we may make substitutions for the variable $x$. When $\beta$ is an element of $\mathbb{Q}_{p}$ or some algebraic extension of this, with $v(\beta)=b$, then valuation is $v(f(\beta))=2b+1$, without exception. Now look at the binomial expression, say $f(x)=p^{2}-px^{2}$, and substitute $\beta$ for $x$ here. Then the $v$-value of the first monomial is $2$, while the $v$-value of the second is again $2b+1$, assuming $b=v(\beta)$. When one of the numbers $2$ and $2b+1$ is smaller than the other, that is the $v$-value of $f(\beta)$. Since $2=2b+1$ exactly when $b=\frac{1}{2}$, when $v(\beta)<\frac{1}{2}$, $2b+1<2$ and $v(f(\beta))=2b+1=2v(\beta)+1$, while when $v(\beta)>\frac{1}{2}$, then $2<2b+1$ and $v(f(\beta))=2$. Indeed, for $\beta=\sqrt{p}$, we know that the $f(\beta)=0$, so it has an infinite $v$-value. Let us look at the graphs of valuation functions.

In Figure 2.2, we draw exactly the behavior we describe in the preceding discussion; in Figure 2.2, we add the monomial to give the polynomial $f(x)=p^{2}+p^{2}x-px^{2}$. In these graphs, we label the horizontal axis $\xi$ instead of $x$ and vertical axis $\eta$ instead of $y$. In terms of these, both the valuation functions are the same, we call them as $\psi_{f}$, and we see that $\psi_{f}(\xi)=2\xi+1$ if $\xi\leq\frac{1}{2}$, while $\psi_{f}(\xi)=2$ if $\xi\geq\frac{1}{2}$. Our point in looking at the polynomial $p^{2}+p^{2}x-px^{2}$ is to emphasize the proof of the theorem. ∎

In order to accomplish our goals, we must develop a Newton polygon and valuation function for the most significant series, the $p$-logarithmic $L(x)$. This entire Subsection 2.2 often makes use of the following proposition.

Proposition 2.4**.**

The vertices of the Newton polygon $\mathcal{N}_{L}$ of the $p$-adic logarithm $L(x)=-\sum_{n\geq 1}\frac{(-x)^{n}}{n}$ are given by $(p^{m},-m),\ n=p^{m}$.

Proof.

We have $L(x)=-\sum_{n\geq 1}\frac{(-x)^{n}}{n}=\sum_{n\geq 1}a_{n}x^{n},$ say. Now look at the following:

[TABLE]

Thus the lower convex hull of all points $(n,v(a_{n}))$ are formed of the points

[TABLE]

all of which lie below the horizontal axis except the point $(1,0)$, and all other points lie on the horizontal axis. Therefore the vertices of the Newton polygon of $p$-adic logarithm in Figure 2.3 are given by { $(p^{m},-m),\ n=p^{m}$}.

∎

Proposition 2.5**.**

The vertices of the valuation function $\psi_{L}$ of the $p$-adic logarithm $L(x)=-\sum_{n\geq 1}\frac{(-x)^{n}}{n}$ are given by $\left(\frac{1}{p^{m-1}(p-1)},\frac{1}{p-1}+1-m\right),\ m\geq 1.$

Proof.

The rightmost segment, extending infinitely to the right, is given by $\eta=\xi$, while to its left, the second segment is given by $\eta=p\xi-1$, so that their intersection point is $\left(\frac{1}{p-1},\frac{1}{p-1}\right).$ This verifies our claim for $m=1$.

For general $m$, we need the intersection of $\eta=p^{m}\xi-m$ and $\eta=p^{m+1}\xi-m-1$. The first coordinate is $\frac{1}{p^{m-1}(p-1)}$ and the second coordinate directly comes from this. ∎

Proposition 2.6**.**

Consider the $p$-adic logarithm $L(x)=-\sum_{n\geq 1}\frac{(-x)^{n}}{n}$ of the formal group law $\mathcal{M}(x,y)=x+y+xy$, defined over $\mathbb{Q}_{p}$. Assume $\psi_{L}$ and $\mathcal{N}_{L}$ be its valuation function and Newton polygon respectively. Then the following two are equivalent:

1. (i)

the vertices of the Newton polygon $\mathcal{N}_{L}$ are $(p^{m},-m),\ n=p^{m}$. 2. (ii)

the graph of $\psi_{L}$ has the only vertices $\left(\frac{1}{p^{m-1}(p-1)},\frac{1}{p-1}+1-m\right),\ m\geq 1.$

Proof.

$(i)\Rightarrow(ii)$:  

By Proposition 2.1, the vertex $(p^{m},-m)$ correspond to the segment $l_{1}:\eta=p^{m}\xi-m$    and that of $(p^{m+1},-(m+1))$ correspond to the segment $l_{2}:\eta=p^{m+1}\xi-m-1$ of valuation function $\psi_{L}$. The vertex $\left(\frac{1}{p^{m-1}(p-1)},\frac{1}{p-1}+1-m\right)$ of $\psi_{L}$ is the intersection of $l_{1}$ and $l_{2}$.

$(ii)\Rightarrow(i)$:

By Proposition 2.2, the vertex $\left(\frac{1}{p^{m-1}(p-1)},\frac{1}{p-1}+1-m\right)$ correspond to the segment $l_{1}:y=\frac{1}{p^{m-1}(p-1)}x+\frac{1}{p-1}+1-m$ and that of $\left(\frac{1}{p^{m}(p-1)},\frac{1}{p-1}-m\right)$ correspond to the segment $l_{2}:y=\frac{1}{p^{m}(p-1)}x+\frac{1}{p-1}-m$ of the Newton polygon $\mathcal{N}_{L}$. The vertex $(p^{m},-m)$ of the Newton polygon $\mathcal{N}_{L}$ is the intersection of $l_{1}$ and $l_{2}$.  This completes the proof. ∎

Proposition 2.7**.**

The only monomials that affect the graph of $\psi_{L}$ are those of degree $p^{s}$, for all $s\geq 0$.

Proof.

We instantly see that the lines $\eta=p^{s}\xi-s$ have segments on the boundary of $\psi$ and others don’t have. Lets go into details:

Notice that the only vertices of the polygon of $\log_{p}(1+x)=-\sum_{n=1}^{\infty}\frac{(-x)^{n}}{n}$ are $\{(p^{m},-m)\}$. We know by Proposition 2.1 that the vertices of polygon are in direct correspondence with the segments of the valuation funtion/copolygon. In general, there is a vertex coming from $a_{n}x^{n}$ at $(n,v(a_{n}))$ if and only if there is segment with equation $\eta=n\xi+v(a_{n})$. We have an equivalent condition for both phenomena above (appearance of a vertex of the polygon and appearance of a segment of the copolygon) is that for possible $n_{0}<n<n_{1}$, we get the inequality $v(a_{n})(n_{1}-n_{0})<v(a_{0})(n_{1}-n)+v(a_{1})(n-n_{0})$. For, in the polygon case, it’s necessary and sufficient to have a vertex at $(n,v(a_{n}))$ that for every pair $(n_{0},n_{1})$ with $n_{0}<n<n_{1}$, is

[TABLE]

[TABLE]

In the same way, we can see that there’s a segment of the copolygon along the line in copolygon $(\xi,\eta)$-space given by the equation $l:\eta=n\xi+v(a_{n})$ if and only if a similar inequality is satisfied. What does mean it by saying that there is a line segment $l:\eta=n\xi+v(a_{n})$ of the copolygon ? It just means for every pair $n_{0},n_{1}$ with $n_{0}<n<n_{1}$, the lines $\eta=n_{1}\xi+v(a_{n_{1}})$ and $\eta=n_{0}\xi+v(a_{n_{0}})$ intersect above the line $l$. We see that the point of intersection is

[TABLE]

i.e., the $\eta$-value is bigger than the $\xi$-value of the point on the line segment $l$. That is,

[TABLE]

which is the required inequality in this case of Newton copolygon. Now the only vertices of the polygon of $\log_{p}(1+x)=-\sum_{n=1}^{\infty}\frac{(-x)^{n}}{n}$ are $\{(p^{m},-m)\}$. It follows that the inequalitites (7) and (8) are satified for every $n=p^{m}$ in the case of $\log_{p}(1+x)$ and nowhere else.

∎

2.2. Main Results

This subsection is dedicated to the images of the $2$-adic logarithm on the principal units of quadratic extensions of $\mathbb{Q}_{2}$. We denote the $2$-adic absolute value $|.|$ normalised by $|2|=\frac{1}{2}$. We also consider the $2$-adic valuation $v=v_{2}$ on $\mathbb{C}_{2}$, normalised by $v_{2}(2)=v(2)=1$. We want to compute the image of the maximal ideal $\mathfrak{m}_{K}$ of the local field $K=\mathbb{Q}_{2}(\sqrt{d})\subset\mathbb{C}_{2}$ for $d=-1,\pm 2,\pm 3,\pm 6$ by the $2$-adic logarithm.

[TABLE]

This amounts to finding

[TABLE]

and then to solve the equation

[TABLE]

To solve equation (10) we will use the theory of Newton polygons ([1], [9], [10], [16]), for a Taylor series $f(x)=\sum_{n\geq 0}a_{n}x^{n}\in K[[x]]$, or what amounts to the same to use the $p$-adic Weierstrass Preparation Theorem for Taylor series with coefficients in a $p$-adic field. We will study the zeroes of $\log_{2}(1+x)-\alpha$ with $\alpha\in K$ and $|\alpha|\leq r_{K}$, and we are only interested in the zeroes $x\in\mathfrak{m}_{K}$.

We know that the zeros of $\log_{p}(1+x)$ are the $(\gamma-1)$ where $\gamma$ is a $p^{h}$ roots of unity for $h\geq 0$, and that $\log_{p}(1+x)=\log_{p}(1+y)\Rightarrow(1+x)=\gamma(1+y)$.

We recall without proof two forms of Hensel’s lemma which will be useful later on. First the so-called Master Factorisation Theorem as stated by Kedlaya in [16] or Christol [9].

Theorem 2.8**.**

*(Master Factorization Theorem, [16]).

Let $R$ be a non-Archimedean, not necessarily commutative ring. Suppose that the nonzero elements $a,b,c\in R$ and additive subgroups $U,V,W$ satisfy the following conditions:*

1. (a)

The spaces $U$ and $V$ are complete under the $p$-adic norm and $UV\subset W$, 2. (b)

The map $f(u_{0},v_{0})=av_{0}+bu_{0}$ is a surjection of $U\times V$ onto $W$, 3. (c)

There exists $\lambda>0$ such that

[TABLE] 4. (d)

We have $ab-c\in W$ and $|ab-c|<\lambda^{2}|c|.$

Then there exists a unique pair $(x,y)\in U\times V$ such that

[TABLE]

Next, the classical Hensel’s lemma for polynomials:

Lemma 2.9**.**

*(Classical Hensel’s Lemma [10]).

Let $K$ be a complete non-Archimedian valued field with valuation $|\cdot|$ and valuation ring $\mathcal{O}_{K}$. Let $f\in\mathcal{O}_{K}[x]$ and $\beta\in\mathcal{O}_{K}$. Suppose that*

[TABLE]

Then there is an $\bar{\beta}\in\mathcal{O}_{K}$ such that

[TABLE]

Notations

For $\rho>0$ a real number we denote by $B(0,\rho)^{+}\ \left(\text{resp.}\ B(0,\rho)^{-}\right)$

[TABLE]

Let $\mathscr{A}(K,\rho^{+})$ be the space of the Taylor series $f\in K[[t]]$ converging on $B(0,\rho)^{+}$.

Theorem 2.10**.**

(Weierstrass Preparation Theorem, Proposition 8.3.2 in [16]). Let $K$ be a complete non-Archimedean-valued field with an absolute value of $|.|$. For $f\in\mathscr{A}(K,\beta^{+})$, there exists a monic polynomial $P\in K[t]$ and a unit $g(t)\in\mathscr{A}(K,\beta^{+})$ such that $f=Pg.$

2.2.1. First case (unramified): $K_{-3}=\mathbb{Q}_{2}(\sqrt{-3})$

Let $\mathcal{O}_{-3}$ represent its ring of integers, $\mathcal{O}_{-3}^{\times}$ represent its invertible elements, and $\mathfrak{m}_{-3}$ represent its maximal ideal.

The field $K_{-3}$ is unramified, and complete, having only two roots of unity of order a power of $2$, $\pm 1$. It contains also the cubic roots $j$ and $j^{2}$ of $1$,

[TABLE]

The residue field of $K_{-3}$ is $\mathbb{F}_{4}$ the finite field with $4$ elements. A set of representatives of $\mathbb{F}_{4}$ in $\mathcal{O}_{-3}$ is the set $\{0,1,j,j^{2}\}$. The group of valuation of $K_{-3}$ is $\mathbb{Z}$ if we normalize it by $v_{2}(2)=1.$ Every element $x\in K_{-3}$ has an Hensel expansion

[TABLE]

We have $\text{Image}(\log_{2}(1+\mathfrak{m}_{K}))\subset K\cap B(0,2^{-1})^{+}$. But we can be more precise.

Lemma 2.11**.**

Let $x_{1}=2+4z_{1}$, $x_{j}=2j+4z_{j}$, $x_{j^{2}}=2j^{2}+4z_{j^{2}}$ with $z_{1},z_{j},z_{j^{2}}\in\mathcal{O}_{-3}$ and let $x_{2}=4z_{2}$ with $z_{2}\in\mathcal{O}_{-3}$. Then

[TABLE]

or equivalently $\text{Image}(\log_{2}(1+\mathfrak{m}_{-3}))\subset\{2+4\mathcal{O}_{-3}\}\cup 4\mathcal{O}_{-3}.$

Proof.

We have

[TABLE]

Similarly, we have

[TABLE]

because $1+j+j^{2}=0$. Similarly, we have

[TABLE]

because $1+j+j^{2}=0$. Let us now consider the situation when $x_{2}=4z_{2}$, and $z_{2}\in\mathcal{O}_{-3}$, where we have classically

[TABLE]

The proof is complete. ∎

Corollary 2.11.1**.**

Let $K=\mathbb{Q}[\sqrt{-3}]$ then, we have

[TABLE]

Proof.

We just rephrase the relations (11), (12), (13), (14). ∎

Corollary 2.11.2**.**

Let $K=\mathbb{Q}[\sqrt{-3}]$, then the image of $\mathfrak{m}_{-3}$ by the function $\log_{2}(1+x)$ doesn’t contain the sets $2j+4\mathcal{O}_{-3}$ and $2j^{2}+\mathcal{O}_{-3}$.

Proof.

It follows from Lemma 2.11. ∎

We shall now demonstrate that $\text{Image}(\log_{2}(1+\mathfrak{m}_{-3}))=(2+4\mathcal{O}_{-3})\cup(4\mathcal{O}_{-3})$. To accomplish this, we will use the theory of the Newton polygon and the Weierstrass Preparation Theorem or equivalently Hensel’s lemma.

Lemma 2.12**.**

Let $\alpha\in 4\mathcal{O}_{-3}$ then there exists $x_{i}\in 2+4\mathcal{O}_{-3}$ and $x_{2}\in 4\mathcal{O}_{-3}$ such that

[TABLE]

Proof.

We are looking for zeros of the function $x\mapsto\log_{2}(1+x)-\alpha$ contained in $\mathfrak{m}_{-3}$. The Newton polygon of $\log_{2}(1+x)-\alpha$ (Figure 2.5) shows that it has infinitely many zeros, $(x_{i})_{i\geq 1}$, in $\mathbb{C}_{2}$ with respect to $2$-adic valuation corresponding to the slopes of the Newton polygon:

[TABLE]

We are only interested in zeros with an integral $2$-valuation; hence only $x_{1}$ and $x_{2}$ are good candidates as a pre-image of $\alpha\in 4\mathcal{O}_{-3}$. We just have to check that $x_{1},x_{2}\in\mathfrak{m}_{-3}$. This is obvious because the length of the first two edges of the Newton polygon on Figure 2.5 is $1$. ∎

To deal with the next case $\alpha\in 2+4\mathcal{O}_{-3}$, we need to look at the proof of Hensel’s lemma.

Lemma 2.13**.**

Let $\alpha\in 2+4\mathcal{O}_{-3}$ then there exists $x_{j}\in 2j+4\mathcal{O}_{-3}$ and $x_{j^{2}}\in 2j^{2}+4\mathcal{O}_{-3}$ such that $\log_{2}(1+x_{1})=\log_{2}(1+x_{2})=\alpha.$

Proof.

For $\alpha\in 2+4\mathcal{O}_{-3}$, the Newton polygon of the function $\log_{2}(1+x)-\alpha$ corresponds to Figure 2.7. The only edge which has a slope in $\mathbb{N}$ is the first one, which is of length $2$. Hence, there are at most two solutions in $\mathfrak{m}_{-3}$.

We want to prove that the two roots of the Weierstrass polynomial corresponding to this edge are in $\mathfrak{m}_{-3}$. We will go back to the proof of Hensel’s lemma. We have

[TABLE]

The hypothesis on $\alpha$ guaranties that $\left|\frac{\alpha}{2}\right|=1$. We have, with $b(t)=1$

[TABLE]

Take $R$, the space of analytic functions on the closed ball of radius $1$ with coefficients in $\mathcal{O}_{-3}$, hence $R=\mathcal{O}_{-3}[[t]]$ equipped with the Gauss-norm. Hence $c(t)\in R$, take for $U$ the space of polynomials of $4\mathcal{O}_{-3}[t]$ of degree $1$ at most and take for $V$ the space $4\mathcal{O}_{-3}[[t]]$ and take $V=W$.

We can apply the Master Factorisation Theorem with $a,b,c$, and $\lambda=1$. It is easy to check that all the conditions are fulfilled:

1. (a)

$U$ and $V$ are obviously complete for the Gauss-norm, and $UV\subset W$ 2. (b)

The map $f(u_{0},v_{0})=av_{0}+bu_{0}$ is a surjection of $U\times V$ onto $W$ 3. (c)

$||f(u_{0},v_{0})||=\max\{||a||||v_{0}||,||b||||u_{0}||\}$ because the degree of $u_{0}$ is less or equal to $1$ and hence there can be no compensations between monomials of the same degree in $av_{0}$ and $bu_{0}$. 4. (d)

$||a(t)b(t)-c(t)||<\frac{1}{4}||c(t)||$.

Hence there exists $x(t)\in U$ and $y(t)\in V$ such that

[TABLE]

The conditions on $\alpha$ and on $a(t)$ ensure that

[TABLE]

Hence, by Hensel’s lemma, the two solutions $x_{1}$ and $x_{2}$ of $\log_{2}(1+x)-\alpha=0$ such that $|x_{1}|=|x_{2}|=2^{-1}$ are solutions of the polynomial $a(t)+x(t)=0$. By the Classical Hensel Lemma 2.9 for polynomials, $\frac{x_{1}}{2}$ and $\frac{x_{2}}{2}$ can be obtained from the roots of the polynomial $a(t)=1+t+t^{2}$ by Hensel’s procedure. But we notice that

[TABLE]

which means that the roots of $a(t)$, $j$ and $j^{2}$ are approximate roots of $P(t)=a(t)+x(t)$. It is clear that

[TABLE]

We can apply the Classical Hensel’s Lemma for polynomials, and so, if $|\alpha|=\frac{1}{2}$, then $c(t)=\frac{1}{2}\left(\log_{2}(1+2t)-\alpha\right)$ has two roots in $\mathcal{O}_{-3}$ close to $j$ and $j^{2}$. This means that $\log_{2}(1+t)-\alpha$ has two zeros: $x_{1}=x_{j}\in 2j+4\mathcal{O}_{-3}$ and $x_{2}=x_{j^{2}}\in 2j^{2}+4\mathcal{O}_{-3}$. This proof is complete. ∎

Hence we have proved that:

Theorem 2.14**.**

Let $\mathcal{O}_{-3}=\mathbb{Q}_{2}[\sqrt{-3}]$, the image of $\mathfrak{m}_{-3}$ by the function $x\mapsto\log_{2}(1+x)$ is exactly:

[TABLE]

and the pre-image is two-to-one.

Proof.

It is just a restatement of Lemma 2.11, Lemma 2.12 and Lemma 2.13. ∎

We do not have, for the moment, an explicit formula for the pre-image of $\alpha\in\text{Image}(\log_{2}(1+\mathfrak{m}_{K}))$ except in one case; however, the theory of the Newton polygon gives an easy algorithm to compute the pre-images. If $\alpha\in 4\mathcal{O}_{K}$ the pre-image in $4\mathcal{O}_{K}$ is given by $x=e^{\alpha}-1$.

2.2.2. Second case (ramified): $K_{-1}=\mathbb{Q}_{2}(\sqrt{-1})$.

The field $K_{-1}:=\mathbb{Q}_{2}(\sqrt{-1})$ is ramified over $\mathbb{Q}_{2}$, because if $i$ is one of the roots of the polynomial $x^{2}+1$ in an algebraic closure of $\mathbb{Q}_{2}$ we have $(1+i)^{2}=2i$, which means that $|1+i|=\frac{1}{\sqrt{2}}$. It contains four roots of unity: $\pm 1,~{}\pm i$. We denote by $\pi_{-1}=\pi=(1+i)$ a uniformizer parameter. Any element $x\in K_{-1}$ has an Hensel expansion

[TABLE]

As before, let $\mathcal{O}_{-1}$ be its ring of integers, $\mathcal{O}_{-1}^{\times}$ be the invertible elements of $\mathcal{O}_{-1}$, and let $\mathfrak{m}_{-1}$ be its maximal ideal. We have

[TABLE]

which shows that we can write $\pm 1,\ \pm i$ as $1+x$ with $x\in\mathfrak{m}_{-1}$. The image $\text{Image}(\log_{2}(1+\mathfrak{m}_{-1}))\subset K_{-1}\cap B(0,1)^{+}$. But we can be more precise.

Lemma 2.15**.**

Let

[TABLE]

then

[TABLE]

Proof.

The identity $\pi+\pi^{2}z_{1}^{\prime}=\pi+2z_{1},\ z_{1}^{\prime},z_{1}\in\mathcal{O}_{-1}$ comes from the equality

[TABLE]

and the same for $x_{2}$. We have

[TABLE]

because $\frac{(\pi+2z_{1})^{3}}{3}\equiv 0\ (\text{mod}\ \pi^{3}\mathcal{O}_{-1})$ and $z_{1}-z_{1}^{2}\equiv 0\ (\text{mod}\ 2\mathcal{O}_{-1})$. Similarly,

[TABLE]

because $\frac{(2+2\pi z_{2})^{3}}{3}\equiv 0\ (\text{mod}\ 4\pi\mathcal{O}_{-1}).$

Similarly, $\log_{2}(1+x_{3})\equiv 0\ (\text{mod}\ 2\pi\mathcal{O}_{-1}).$ The proof is complete. ∎

Lemma 2.16**.**

Let $\alpha\in 2\pi\mathcal{O}_{-1}$, then there exist only two $x_{1},x_{2}\in\pi\mathcal{O}_{-1}=\mathfrak{m}_{-1}$ such that

[TABLE]

Proof.

We are looking for the zeros of the function $x\mapsto\log_{2}(1+x)-\alpha$ with $v_{2}(\alpha)\geq\frac{3}{2}>1$ continued in $\mathfrak{m}_{-1}$. The Newton polygon of $\log_{2}(1+x)-\alpha$ corresponds to Figure 2.5. It shows that $\log_{2}(1+x)-\alpha$ has infinitely many zeros, $(x_{i})_{i\geq 1}$, in $\mathbb{C}_{2}$ with $2$-adic valuation corresponding to the slopes of the Newton polygon:

[TABLE]

We are only interested in zeros with an integral or half-integral $2$-valuation; hence, only $x_{1},x_{2},x_{3}$ and $x_{4}$ are good candidates as pre-image of $\alpha\in 2\pi\mathcal{O}_{-1}$. They correspond to the green and orange edges on Figure 2.5 according to the theory of Newton polygon. We just have to check that $x_{1},x_{2},x_{3},x_{4}\in\mathfrak{m}_{-1}$ under the hypothesis $\alpha\in 2\pi\mathfrak{m}_{-1}$. The first two edges in green of the Newton polygon on Figure 2.5 correspond to solutions $x_{1},x_{2}\in\mathfrak{m}_{-1}$, because the length of the first two edges is $1$. Moreover, we have $v_{2}(x_{1})=v_{2}(\alpha)$, $v_{2}(x_{2})=1.$ It remains to be checked whether or not $x_{3},x_{4}\in\mathfrak{m}_{-1}$ where $x_{3}$ and $x_{4}$ correspond to the organe edge in Figure 2.5. To do this, we will use the Master Factorisation Theorem again. Consider

[TABLE]

The Newton polygon of $c(t)$ is given by Figure 2.8.

The roots $x_{3}$ and $x_{4}$ correspond, in the $t$-variable, to the horizontal orange edge of the Newton polygon given by Figure 2.8. Consider with the notations of Theorem 2.8, and with $\alpha=2\pi z$ and $z\in\mathcal{O}_{-1}$

[TABLE]

Let $R=\mathcal{O}_{-1}[[t]]$ be equipped with the Gauss-norm

[TABLE]

Let $U$ be the set of polynomials of degree $1$ at most, in $\mathcal{O}_{-1}[t]$ and $V$ be the module $\mathcal{O}_{-1}[[t]]$ and let $W=\mathcal{O}_{-1}[[t]].$ It is clear that $U,V$ are complete topological spaces for the Gauss norm. It is also clear that $UV\subset W$ and $l(u_{0},v_{0})=av_{0}+bu_{0}$ is a surjection of $U\times V$ onto $W$. We have

[TABLE]

because $b=t^{2}$, $||a||=1$, and degree of $u_{0}$ is less or equal to $1$. It is clear that

[TABLE]

Hence we can apply the Theorem 2.8 and there exists $x(t)\in U$ and $y(t)\in V$ such that

[TABLE]

The polynomial $a(t)+x(t)$ contains all the zeros of $c(t)$ inside the unit closed ball. According to the classical Hensel Lemma 2.9, the roots of $a(t)+x(t)$ are close to the two primitive fourth-roots of $-1$ and so do not belong to $K_{-1}=\mathbb{Q}_{2}(\sqrt{-1})$. The proof of the lemma is complete. ∎

Theorem 2.17**.**

Let $\mathcal{O}_{-1}=\mathbb{Q}_{2}[\sqrt{-1}]$, the image of $\mathfrak{m}_{-1}$ by the function $x\mapsto\log_{p}(1+x)$ is exactly:

[TABLE]

and the pre-image is two-to-one.

Proof.

It is just a restatement of Lemma 2.16. ∎

2.2.3. Third case (ramified) : $K_{3}=\mathbb{Q}_{2}(\sqrt{3})$.

The field $K_{3}=\mathbb{Q}_{2}(\sqrt{3})$ is ramified over $\mathbb{Q}_{2}$, because if $\alpha=\sqrt{3}$ is one of the roots (the other being $-\alpha$) of the polynomial $x^{2}-3$ in an algebraic closure of $\mathbb{Q}_{2}$ then $\beta=\alpha+1$ is a root of the polynomial

[TABLE]

which is an Eisenstein polynomial. Hence $K_{3}$ is ramified containing two roots of unity $\pm 1$. We denote by $\pi_{3}=\pi=(\alpha+1)$ an uniformizing parameter. We deduce from $\eqref{28}$ that

[TABLE]

Any element $x\in K_{3}$ has Hensel expansion

[TABLE]

Let $\mathcal{O}_{3}$ be its ring of integers, $\mathcal{O}_{3}^{\times}$ be the units of $\mathcal{O}_{3}$ and $\mathfrak{m}_{3}$ be its maximal ideal. We have

[TABLE]

The image $\text{Image}\left(\log_{2}(1+\mathfrak{m}_{3})\right)\subset K_{3}\cap B(0,1)^{+}$. But we can be more precise.

Lemma 2.18**.**

Any element of $\mathfrak{m}_{3}$ is either of type $x_{1}$ or $x_{2}$ or $x_{3}$ with:

[TABLE]

then

[TABLE]

Proof.

The identity $\pi_{3}+\pi_{3}^{2}z_{1}^{\prime}=\pi_{3}+2z_{1},\ z_{1}^{\prime},z_{1}\in\mathcal{O}_{3}$ coming from (19) and the same for $x_{2}$. It is easy to see that

[TABLE]

Using (23), we have

[TABLE]

because $z_{1}-z_{1}^{2}\equiv 0\ (\text{mod}\ 2\mathcal{O}_{3})$. Similarly, it is to see that:

[TABLE]

So using (24),

[TABLE]

Similarly, it is easy to see that,

[TABLE]

Using (25), we have

[TABLE]

The proof is complete.

∎

Lemma 2.19**.**

Let $\alpha\in 2\pi_{3}\mathcal{O}_{3}$, then there exists $x_{1},x_{2},x_{3},x_{4}\in\pi_{3}\mathcal{O}_{3}=\mathfrak{m}_{3}$ such that

[TABLE]

Let $\alpha\in(2+2\pi_{3}\mathcal{O}_{3})$ then there exist $x_{1},x_{2}\in\pi_{3}\mathcal{O}_{3}=\mathfrak{m}_{3}$ such that

[TABLE]

Proof.

We are looking for the zeros of the function $x\mapsto\log_{2}(1+x)-\alpha$ with

[TABLE]

contained in $\mathfrak{m}_{3}$.

Case 1: $v_{2}(\alpha)\geq\frac{3}{2}$. Then the Newton polygon of $\log_{2}(1+x)-\alpha$ corresponds to Figure 2.5. It shows that $\log_{2}(1+x)-\alpha$ has infinitely many zeros, $(x_{i})_{i\geq 1}$, in $\mathbb{C}_{2}$ with $2$-adic valuation corresponding to the slopes of the Newton polygon:

[TABLE]

We are only interested in zeros with an integral or half-integral $2$-valuation, hence only $x_{1},x_{2},x_{3}$ and $x_{4}$ are good candidates are pre-image of $\alpha\in 2\pi_{3}\mathcal{O}_{3}$. They correspond to the green and organe edges on Figure 2.8 by the theory of the Newton polygon. We just have to check whether or not $x_{1},x_{2},x_{3},x_{4}\in\mathfrak{m}_{3}$ under the hypothesis $\alpha\in 2\pi_{3}\mathcal{O}_{3}$. The first two edges in green of the Newton polygon on Figure 2.5 correspond to solutions $x_{1},x_{2}\in\mathfrak{m}_{3}$, because the length of the first two edges is $1$. Moreover we have $v_{2}(x_{1})=v_{2}(\alpha)$, $v_{2}(x_{2})=1$. Concerning $x_{3}$ and $x_{4}$, we prove exactly as in the case $K_{-1}$, that they do not belong to $K_{3}$, because they are close to

[TABLE]

Case 2: $v_{2}(\alpha)=1$. The Newton polygon corresponding to $-\alpha+\log_{2}(1+x)$ looks like Figure 2.7. We must demonstrate that that each of the zeros of $-\alpha+\log_{2}(1+x)$ corresponding to the first two edges of the Newton polygon (the green and orange ones) has two roots in $K_{3}$. Because each of these edges has a length $2$, it suffices to show that $-\alpha+\log_{2}(1+x)$ contains at least one zero with a valuation of $1$ and one zero with a valuation of $\frac{1}{2}$. We are looking for solutions of the following form:

[TABLE]

Consider first $x=2+2\pi_{3}t$ with $t\in\mathcal{O}_{3}$. The equation $-\alpha+\log_{2}(1+x)$ becomes

[TABLE]

It immediately gives us

[TABLE]

By the same argument as above, we see that there exists one $t\in\mathfrak{m}_{3}$ such that

[TABLE]

The theory of the Weierstrass Preparation Theorem is rational on the field of coefficients which means that Weierstrass polynomial corresponding to the slope $-1$ of the Newton polygon on Figure 2.8 is in $K_{3}[t]$ and of degree $2$. Hence if one root in $K_{3}$ it implies that the second one is also in $K_{3}$. This proves that $x_{1},x_{2}\in 2+2\pi_{3}\mathcal{O}_{3}$.

Consider now the roots corresponding to the slope $-\frac{1}{2}$. We are looking for roots of the form $\pi_{3}+\pi_{3}^{2}t$ with $t\in\mathcal{O}_{3}$. The equation $-\alpha+\log_{2}(1+x)$ then becomes

[TABLE]

From this, we immediately get

[TABLE]

with $\beta_{0}\in\mathcal{O}_{3}$. It is easy to see that the polynomial

[TABLE]

has no roots in $\mathcal{O}_{3}$. The proof of the lemma is complete. ∎

Theorem 2.20**.**

Let $K_{3}=\mathbb{Q}_{2}[\sqrt{3}]$, the image of $\mathfrak{m}_{3}$ by the function $x\mapsto\log_{2}(1+x)$ is exactly:

[TABLE]

and the pre-image is two-to-one for $2\pi\mathcal{O}_{3}$ and also two-to-one for $(2+2\pi_{3}\mathcal{O}_{3})$.

Proof.

It is just restatement of Lemma 2.18 and Lemma 2.19. ∎

Remark*.*

For $p=2$, we can treat in parallel the remaining ramified cases. For $p\geq 3$, we expect the above developed techniques of Newton polygon to compute the image of $p$-adic logarithm on the principal units in any quadratic extensions $\mathbb{Q}_{p}(\sqrt{p}),~{}\mathbb{Q}_{p}(\sqrt{n}),~{}\mathbb{Q}_{p}(\sqrt{np})$, where $n$ is quadratic non-residue. The computation of the image of the $p$-adic logarithm on the principal units of any abelian extension of $\mathbb{Q}_{p}$ remains an exercise.

The key findings of Section 1 and Section 2 are summerised as follows:

1. (i)

$\log_{p}(1+\mathfrak{m}_{K})=\mathfrak{m}_{K}^{2}$, where $\mathfrak{m}_{K}=\pi\mathbb{Z}_{p}[\zeta_{p}]$. (Cyclotomic extension $\mathbb{Q}_{p}(\zeta_{p})$) 2. (ii)

$\log_{p}((1+\mathfrak{m}_{K})-(1+\mathfrak{m}_{K}^{2}))=\mathfrak{m}_{K}^{2}$. (Cyclotomic extension $\mathbb{Q}_{p}(\zeta_{p})$) 3. (iii)

$\log_{2}(1+\mathfrak{m}_{-3})=(2+4\mathcal{O}_{-3})\cup(4\mathcal{O}_{-3})$. (Quadratic extension $\mathbb{Q}_{2}(\sqrt{-3})$) 4. (iv)

$\log_{2}(1+\mathfrak{m}_{-1})=2\pi\mathcal{O}_{-1}$. (Quadratic extension $\mathbb{Q}_{2}(\sqrt{-1})$) 5. (v)

$\log_{2}(1+\mathfrak{m}_{3})=2\pi\mathcal{O}_{3}\cup(2+2\pi\mathcal{O}_{3})$. (Quadratic extension $\mathbb{Q}_{2}(\sqrt{3})$)

Remark*.*

It is known from [20] that the images of $p$-adic logarithm are $\mathbb{Z}_{p}$-modules. In addition, the preceding results demonstrate that the images are compact and open subsets of the respective ring of integers. As studied by Iwasawa in [14], such compact modules serve a crucial role in the theory of cyclotomic fields.

2.3. Direct application to class field theory

We conclude the paper by applying one of our result to compute normalized $p$-adic regulator of the global field $K:=\mathbb{Q}(\zeta_{p})$, where $\zeta_{p}^{p}=1$. Note that Leopold conjecture states that $p$-adic regulator of a number field doesn’t vanish. Using $p$-adic logarithm, Leopold propsed a definition of $p$-adic regulator $\mathcal{R}_{p}$ attached to a number field for a prime number $p$. The conjecture is known for abelian extension of $\mathbb{Q}$.

Assume $p$ is a regular prime number. The ring of integers of $K$ is $\mathcal{O}_{K}:=\mathbb{Z}[\zeta_{p}]$. The notation $\mathfrak{p}\mid p$ means prime ideals of $K$ dividing $p$. Consider the group of $p$-principal global-units of $K$ denoted by

[TABLE]

Now consider, for each $\mathfrak{p}\mid p$, let $K_{\mathfrak{p}}$ be the $\mathfrak{p}$-adic completion of $K$ and $\bar{\mathfrak{p}}$ be the corresponding prime ideal of the of integers of $K_{\mathfrak{p}}$, let $U_{K}$ be the group of principal local-units denoted by

[TABLE]

which is a $\mathbb{Z}_{p}$-module.Then there is a diagonal embedding $E_{K}\hookrightarrow U_{K}$ and its scalar extension $E_{K}\otimes_{\mathcal{O}_{K}}\mathbb{Z}_{p}\hookrightarrow U_{K}$ whose image is $\bar{E}_{K}$, the topological closure of $E_{K}$ in $U_{K}$.

Theorem 2.21**.**

Consider the global field $K=\mathbb{Q}(\zeta_{p})$, $p$ is regular prime. Then the normalized $p$-adic regulator of the global field $K$ is 1.

Proof.

As it is well known, the Leopold conjecture is true for abelian extension (and hence for cyclotomic extension) of $\mathbb{Q}$, thus one can apply the results of Gras [12], which defines the normalized $p$-adic regulator

[TABLE]

Since $\bar{E}_{K}$ is the topological closure of $E_{K}$ inside principal local units $U_{K}$, we have $\log_{p}(\bar{E}_{K})=\mathbb{Z}_{p}\log_{p}(E_{K})$. Therefore,

[TABLE]

Note that the $p$-adic logarithm behaves same way on principal global-units $E_{K}$ as it behaves on principal local-units $U_{K_{\mathfrak{p}}}$ so that we also have $\log_{p}(1+\mathfrak{p})=\mathfrak{p}^{2}$.

Note that we have obtained $\mathbb{Q}_{p}(\zeta_{p})$ by completing $F=\mathbb{Q}(\zeta_{p})$ at $\mathfrak{p}$. Since $\mathbb{Q}_{p}(\zeta_{p})$ is totally ramified extension of $\mathbb{Q}_{p}$, its ring of integers is monogenic over $\mathbb{Z}_{p}$. Therefore $\mathfrak{p}\mathbb{Z}_{p}=\bar{\mathfrak{p}}$, which implies the quotient $\bar{\mathfrak{p}}^{2}/\mathfrak{p}^{2}\mathbb{Z}_{p}\simeq\bar{\mathfrak{p}}^{2}/\bar{\mathfrak{p}}^{2}$ is trivial. Thus the quotient $\frac{\oplus_{\mathfrak{p}\mid p}\bar{\mathfrak{p}}^{2}}{\oplus_{\mathfrak{p}\mid p}\mathfrak{p}^{2}\mathbb{Z}_{p}}$ is trvial, and hence its torsion subgroup is trivial. So $\#\text{tor}\left(\frac{\oplus_{\mathfrak{p}\mid p}\bar{\mathfrak{p}}^{2}}{\oplus_{\mathfrak{p}\mid p}\mathfrak{p}^{2}\mathbb{Z}_{p}}\right)=1$. Therefore the normalized $p$-adic regulator of the global field $\mathbb{Q}(\zeta_{p})$ is 1. ∎

Remark*.*

From [20], it is known that the images of $p$-adic logarithm are $\mathbb{Z}_{p}$-module. Further the above results shows that the images are compact and open subsets as well. These types of compact modules plays important role in the theory of cyclotomic fields as studied by Iwasawa in [14].

Acknowledgement

The authors are indebted to Professors Daniel Barsky and Jonathan Lubin for their insightful correspondences and suggestions. We appreciate the helpful comments made by the anonymous referee. The first author is supported by The Council Of Scientific and Industrial Research (CSIR), Government of India, with the award of Senior Research Fellowship.