On the bases of the image of $2$-adic logarithm on the group of principal units

TL;DR

This paper determines the bases of the 2-adic logarithm's image on principal units across all quadratic extensions of Q_2, enhancing understanding of its algebraic structure and potential applications.

Contribution

It provides explicit bases for the 2-adic logarithm's image in all quadratic extensions of Q_2, revealing the module structure of the logarithm.

Findings

Explicit bases for the 2-adic logarithm in quadratic extensions

Insights into the free module structure of the logarithm

Applications discussed for algebraic and number theoretic contexts

Abstract

This paper computes the bases of the image of $2$-adic logarithm on the group of the principal units in all 7 quadratic extensions of $\mathbb{Q}_2$. This helps one to understand the free module structure of $2$-adic logarithm at arbitrary points on its domain. We discuss some applications at the end.

Equations

$$F(X,\iota (X))=F(i(Y),Y)=0,$$

$$f(F(X,Y))=G(f(X),f(Y)).$$

$$f(\psi (T))=\psi (f(T))=T,$$

$$\omega (T)=F_X(0,T{)}^{-1}dT.$$

$${\log }_F(T)={\int }_0^T\omega (\tau )=T+\frac{a_2}{2}{T}^2+\frac{a_3}{3}{T}^3+\cdots \in R^{\prime }[[T]],$$

$$p(T)=F_X(0,T{)}^{-1}=1+\sum _{i=1}^{\infty }{a}_i{T}^i\in R^{\prime }[[T]].$$

$${\log }_F(T)={\int }_0^T\omega (\tau )={\int }_0^T\frac{1}{F_X(0,T)}dT=T+\frac{a_1}{2}{T}^2+\frac{a_2}{3}{T}^3+\cdots \in R^{\prime }[[T]].$$

$$F_X(X,F(Y,Z))=F_X(F(X,Y),Z)\cdot F_X(X,Y).$$

$$p(X)\cdot F_X(0,X)=p(0)=1.$$

$$\frac{\partial }{\partial X}{\log }_F(X)=p(X).$$

$$\frac{\partial }{\partial X}{\log }_F(F(X,Y))=\frac{\partial }{\partial X}{\log }_F(X).$$

$${\log }_F(F(X,Y))={\log }_F(X)+C(Y),\text{where C(Y) is an arbitrary functions in Y},$$

$${\log }_F(F(X,Y))-{\log }_F(X)=C(Y).$$

$${\log }_F(F(0,Y))-{\log }_F(0)=C(Y)$$

$$\Rightarrow$$

$${\log }_{\mathcal{M}}(t)={\log }_p(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}-\cdots$$

$${\log }_{\mathcal{M}}(x)={\int }_0^x\omega (t)dt={\int }_0^x\frac{dt}{1+t}=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots ,$$

$${\log }_{\mathcal{M}}(x)={\log }_p(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots \in {\mathcal{O}}_K[[x]].$$

$$\frac{1}{r}=\underset{n\to \infty }{\overline{\lim }}\sqrt[n]{|a_n{|}_p}.$$

$$\frac{1}{r}=\underset{n\to \infty }{\overline{\lim }}\sqrt[n]{{\left|\frac{(-1{)}^{n-1}}{n}\right|}_p}=\underset{n\to \infty }{\overline{\lim }}\frac{1}{\sqrt[n]{|n{|}_p}}=\underset{n\to \infty }{\overline{\lim }}{p}^{v_p(n)/n}.$$

$$(1+b{)}^a=\sum _{n=0}^{\infty }\frac{a(a-1)\cdots (a-n+1)}{n!}{b}^n\text{converges}.$$

$${\log }_{\mathcal{M}}(x)=\frac{1}{2}{\log }_{\mathcal{M}}(2x)=x-x^2+\frac{4x^3}{3}-2x^4+\frac{16x^5}{5}-\frac{16x^6}{3}+\cdots ,$$

$$2{+}_{\mathcal{M}}2\omega =2+2\omega +4\omega \equiv 2+2\omega (\text{mod}4{\mathcal{O}}_K).$$

$$\{\begin{array}{rl}For,& z=2,2{+}_{\mathcal{M}}2=\mathcal{M}(2,2)\equiv 0(\text{mod}4{\mathcal{O}}_K),\\ For,& z=2\omega ,2\omega {+}_{\mathcal{M}}2\omega =\mathcal{M}(2\omega ,\omega )\equiv 0(\text{mod}4{\mathcal{O}}_K),\\ For,& z=2+2\omega ,(2+2\omega ){+}_{\mathcal{M}}(2+2\omega )=\mathcal{M}(2+2\omega ,2+2\omega )\equiv 0(\text{mod}4{\mathcal{O}}_K).\end{array}$$

$$\cdots \subset 1+2{\mathfrak{m}}_K=1+{\pi }^2{\mathfrak{m}}_K=1+{\pi }^3{\mathcal{O}}_K\subset 1+\pi {\mathfrak{m}}_K=1+{\pi }^2{\mathcal{O}}_K\subset 1+{\mathfrak{m}}_K=1+\pi {\mathcal{O}}_K,$$

$$2{\mathfrak{m}}_K={\pi }^2{\mathfrak{m}}_K={\pi }^3{\mathcal{O}}_K\subset \pi {\mathfrak{m}}_K={\pi }^2{\mathcal{O}}_K\subset {\mathfrak{m}}_K=\pi {\mathcal{O}}_K.$$

$${\log }_{\mathcal{M}}(z)={\log }_{\mathcal{M}}(z{+}_{\mathcal{M}}(-2))-{\log }_{\mathcal{M}}(-2)={\log }_{\mathcal{M}}(z{+}_{\mathcal{M}}(-2))\ge \frac{3}{2},$$

$$\pi {+}_{\mathcal{M}}(-\pi )=\mathcal{M}(\pi ,-\pi )=\pi +(-\pi )+\pi (-\pi )=-{\pi }^2=-2.$$

$${\log }_{\mathcal{M}}(z)={\log }_{\mathcal{M}}(z{+}_{\mathcal{M}}\pi )-{\log }_{\mathcal{M}}(\pi )={\log }_{\mathcal{M}}(z{+}_{\mathcal{M}}\pi )\in {\log }_{\mathcal{M}}(\pi {\mathfrak{m}}_K)=2{\mathfrak{m}}_K.$$

$${\log }_{\mathcal{M}}(z)$$

$$={\log }_{\mathcal{M}}(z{+}_{\mathcal{M}}\pi )\in {\log }_{\mathcal{M}}(\pi {\mathfrak{m}}_K)=2{\mathfrak{m}}_K,({\log }_{\mathcal{M}}(\pi )=0)$$

Full text

Introduction

The $p$-adic logarithm $\log_{p}(1+x)=\sum_{n\geq 1}(-1)^{n-1}\frac{x^{n}}{n}\in K[[x]]$ converges for all $x\in\mathfrak{m}_{K}$. Following Iwasawa [14], the $p$-adic logarithm extends to $p^{\mathbb{Z}}(1+\mathfrak{m}_{\mathbb{C}_{p}})\subset\mathbb{C}_{p}^{\times}$ such that $\log_{p}(p)=0$. It is 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, the case $r=1$ is different i.e, $\log_{p}:1+\mathfrak{m}_{K}\to\mathfrak{m}_{K}$ is not an isomorphism. The image $\log_{p}(1+\mathfrak{m}_{K})$ is not known for arbitrary extension $K$ of $\mathbb{Q}_{p}$.

In $p$-adic number theory, the $p$-adic logarithm plays an essential role. From ([8], [11], [12], [13], [14]), we know the image of $p$-adic logarithm on the groups of principal units is very crucial in the theory of Iwasawa. More specifically in [12], Iwasawa explicitly computed the formulas for the norm residue symbol by using the image of $p$-adic logarithm on the group of principal units of certain cyclotomic extension of $\mathbb{Q}_{p}$, which is our main motivation. There are plenty of other potential applications of the image/value of $p$-adic logarithm. In [15], the formula for normalized $p$-adic regulator contains the image of $p$-adic logarithm on the principal units. In the study of $p$-adic $L$-functions there appears $p$-adic logarithm. 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., [9]. The value of $p$-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 [16] in the theory of arithmetic dynamics. More applications about the image of $p$-adic logarithm are discussed in Section 3. Therefore, it is of novelty to understand the image/value of $p$-adic logarithm on its domain. Since by Proposition 1.4, $\log_{p}(1+\mathfrak{m}_{K})$ is a $\mathbb{Z}_{p}$-module, once we know the bases of $\log_{p}(1+\mathfrak{m}_{K})$, we can understand the value of the $p$-adic logarithm at an arbitrary point in its domain by forming a linear combination of that basis elements with scalars from $\mathbb{Z}_{p}$. Thus, this current paper undertakes an initial challenge towards computing the image of $p$-adic logarithm on the group of principal units, by computing the bases of $p$-adic logarithm. A basis of a free module is as important as a bases of a vector space. In Theorem 2.13 and Theorem 2.14 of Section 2, we compute the basis of $2$-adic logarithm for all $7$ quadratic extensions of $\mathbb{Q}_{2}$. Moreover, we find the exact images of $2$-adic logarithm in some cases. In Section 3, we discuss some more applications.

Notations

$\mathbb{Q}_{p}$ is the p-adic number field with ring of integers $\mathbb{Z}_{p}$, unique maximal ideal $p\mathbb{Z}_{p}$ and residue field $\mathbb{F}_{p}:=\mathbb{Z}_{p}/p\mathbb{Z}_{p}$. Let $\bar{\mathbb{Q}}_{p}$ be the algebraic closure and take the $p$-adic completion $\mathbb{C}_{p}:=\widehat{\bar{Q}}_{p}$ with maximal ideal $\mathfrak{m}_{\mathbb{C}_{p}}$ and the units $\mathbb{C}_{p}^{\times}$. Let $K$ be a finite extension of $\mathbb{Q}_{p}$ with ring of integers $\mathcal{O}_{K}$, unique maximal ideal $\mathfrak{m}_{K}$. The quotient field $\kappa_{K}:=\mathcal{O}_{K}/\mathfrak{m}_{K}$ is the residue field of $\mathcal{O}_{K}$. We also have the standard $p$-adic additive valuation $v_{p}:\mathbb{Q}_{p}\to\mathbb{Z}\cup\{\infty\}$. We extend the valuation $v_{p}$ on $\mathbb{Q}_{p}$ to the finite extension $K$ denoted by $v$ that makes $v(z)=v_{p}(z)$ whenever $z\in\mathbb{Q}_{p}$ by the following formula $v(\lambda)=\frac{1}{[K:\mathbb{Q}_{p}]}v_{p}\left(\mathcal{N}_{\mathbb{Q}_{p}}^{K}(\lambda)\right),$ where $\mathcal{N}_{\mathbb{Q}_{p}}^{K}$ is the field-theoretic norm, the multiplicative mapping from $K^{\times}$ to $\mathbb{Q}_{p}$. We use the common notation $v$ throughout the paper.

1. Formal group law and $p$-adic logarithm

In this section, we construct the $p$-adic logarithm from $1$-dimensional multiplicative formal group law.

Definition 1.1**.**

[5] Let $R$ be a commutative ring. A formal group law over $R$ is a formal power series $F(X,Y)\in R[[X,Y]]$ with the following properties

1. (i)

$F(X,Y)=X+Y+\text{higher degree terms}$, 2. (ii)

$F(X,F(Y,Z))=F(F(X,Y),Z),$ 3. (iii)

$F(X,0)=X$ and $F(0,Y)=Y$, 4. (iv)

There exists a unique power series $\iota(T)\in R[[T]]$ such that

[TABLE]

which is the inverse power series in the formal group law.

If in addition $F(X,Y)=F(Y,X)$ holds then $F$ is called commutative. The commutativity property comes trivially as long as $R$ has no elements that are both torsion and nilpotent.

Example 1.1**.**

The $1$-dimensional formal additive group law is defined by $F(X,Y)=X+Y$ and the $1$-dimensional formal multiplicative group law is defined by $\mathcal{M}(X,Y)=X+Y+XY.$

Now, we discuss maps between formal group laws.

Definition 1.2**.**

[6] Let $F,~{}G$ be two formal group laws over a ring $R$. A homomorphism $f:F\to G$ over $R$ is a power series $f(T)\in R[[T]]$, having no constant term, satisfying

[TABLE]

If further there exists another homomorphism $\psi\in R[[T]]$, $\psi:G\to F$ such that

[TABLE]

then we say that $F$ and $G$ are isomorphic over $R$ while $f$ is an isomorphism over $R$. An endomorphism $f:F\to F$ satisfies $f(F(X,Y))=F(f(X),f(Y)).$

Definition 1.3**.**

[6] An invariant differential of a formal group law $F$ over an arbitrary ring $R$ is a differential form $\omega(T)=p(T)dT\in R[[T]]dT$ that satisfies $p(F(T,~{}S))F_{X}(T,~{}S)=p(T)$, where $F_{X}(X,~{}Y)$ is the partial derivative with respect to its first variable. An invariant differential is said to be normalized if $p(0)=1$. Then there exists a unique normalized invariant differential of $F$ given by the formula

[TABLE]

We now construct the formal logarithm, of a formal group law, obtained by integrating the invariant differential. But integrating an invariant differential form $\omega(T)$ produces denominators in each coefficient. So we assume $R$ to be a commutative ring of characteristic [math] so that we don’t have further difficulty of division. The following construction of the $p$-adic logarithm can be found in [5, Section 5.8], but for the convenience of readers we add a proof here, providing minor details:

Proposition 1.1**.**

Assume that $R$ be a torsion-free ring and let $R^{\prime}=R\otimes_{\mathbb{Z}}\mathbb{Q}$. If $\omega(T)=p(T)dT$ is a normalized invariant differential of a formal group law $F$. Then the formal group law $F$ induces a formal logarithmic series $\log_{F}$ on $R^{\prime}$ given by

[TABLE]

satisfying $\log_{F}(F(X,Y))=\log_{F}(X)+\log_{F}(Y)$.

Proof.

Define the map $R\to R^{\prime}=R\otimes_{\mathbb{Z}}\mathbb{Q}$ by $r\mapsto r\otimes 1$ for $r\in R$. The kernel of this map is the set of torsion elements of $R$. But as $R$ is the torsion-free ring, the kernel is trivial. Therefore there is a one-to-one crorrespondence between $R$ and $R\otimes_{\mathbb{Z}}\mathbb{Q}$. This allows us to take coefficients of the formal logarithm in $\mathbb{Q}$.

Now as $p(T)dT$ is normalised invariant differential, we have $p(0)=1$ and so we can assume

[TABLE]

Integrating, we get

[TABLE]

Now it is remains to show that $\log_{F}(F(X,Y))=\log_{F}(X)+\log_{F}(Y).$ To do this we take partial derivatives with respect to $X$ in the associativity relation $F(X,F(Y,Z))=F(F(X,Y),Z)$, we get

[TABLE]

Let $p(X)$ be the formal power series with coefficients in $R$ defined by

[TABLE]

Now $\log_{F}(X)$ is the unique formal power series with coefficients in $R^{\prime}=R\otimes_{\mathbb{Z}}\mathbb{Q}$ such that

[TABLE]

Then using the changes $X=0,\ Y\to X,\ Z\to Y$ in (1) and using (2), we get

[TABLE]

Integrating (4), we get

[TABLE]

which can be written as

[TABLE]

Putting $X=0$ in (6), we have

[TABLE]

Hence from (6), we conclude $\log_{F}(F(X,Y))=\log_{F}(X)+\log_{F}(Y).$ This completes the proof.

∎

Proposition 1.2**.**

Let $K$ be a finite extension of $\mathbb{Q}_{p}$ with ring of integers $\mathcal{O}_{K}$ and unique maximal ideal $\mathfrak{m}_{K}$. Then the 1-dimensional multiplicaive formal group law $\mathcal{M}(x,y)=x+y+xy$ defined over $\mathcal{O}_{K}$ induces the $p$-adic logarithm (analogue to classical logarithm) given by

[TABLE]

, and converges in $\mathfrak{m}_{K}$.

Proof.

If $\pi\in\mathcal{O}_{K}$ is a uniformizer, we can choose either $\pi=\sqrt[p-1]{-p}$ or $\pi=\zeta_{p}-1$, for some $p^{th}$ root of unity. It can be checked that the differential form $\omega(x)=\frac{dx}{1+x}$ is the invariant differential for $\mathcal{M}(x,y)=x+y+xy$, and hence the corresponding logarithmic series $\log_{\mathcal{M}}$ is obtained by

[TABLE]

which satisfies $\log_{\mathcal{M}}(\mathcal{M}(x,y))=\log_{\mathcal{M}}(x)+\log_{\mathcal{M}}(y).$ This logarithmic series is called the p-adic logarithm and is denoted by

[TABLE]

This series converges in $\mathfrak{m}_{K}$. For, we note that the Cauchy-Hadamard convergence formula holds in the $p$-adic field as well. If $f(x)=\sum_{n\geq 0}a_{n}x^{n}\in K[[x]]$, by Cauchy-Hadamard formula, the radius of convergence $r\in[0,\infty]$ is given by

[TABLE]

Therefore, for $p$-adic logarithm (7), we have $a_{n}=(-1)^{n-1}\frac{1}{n}$. Then, we have

[TABLE]

Since $1\leq p^{v_{p}(n)}\leq n$, we get $1\leq p^{v_{p}(n)/n}\leq\sqrt[n]{n}$. Also $\sqrt[n]{n}\to 1$ as $n\to\infty$. In particular $\sqrt[n]{|n|_{p}}\to 1$ as $n\to\infty$. Hence the radius of convergence $r=1$ i.e., $-1<x\leq 1$. At the end point $x=1$, we have $a_{n}=\pm\frac{1}{n}$ and its $p$-adic norm $\frac{1}{|n|_{p}}$ doesn’t tends to [math] because it is $1$ when $p\nmid n$. Therefore the disc of convergence of the $p$-adic logarithm in $K$ is the open unit disc $\{x\in K:|x|_{p}<1\}=\mathfrak{m}_{K}$. ∎

Thus the important thing is to note that the $1$-dimensional multiplicative formal group law $\mathcal{M}$ gives a group structure on $\mathfrak{m}_{K}$ as follows:

Lemma 1.3**.**

Let $F$ be a formal group law over $\mathcal{O}_{K}$. Then the set $\mathfrak{m}_{K}$ becomes a group under the group law $F$, via the law of combination $\alpha+_{F}\beta=F(\alpha,\beta)$.

Proof.

The proof is obvious using the previous definitions and basic properties of formal group law. The only thing that needs to ensure is the convergence of the infinite series $F(\alpha,\beta)$, which follows from the completeness of $\mathfrak{m}_{K}$ under $\mathfrak{m}_{K}$-adic topology. ∎

The following standard proposition describes the module structure of $p$-adic logarithm on principal units. The proof is based on basic concept of $p$-adic logarithm, e.g., [2, Chap. IV], [3, Sec. 4, Chap. 5].

Proposition 1.4**.**

$\log_{p}(1+\mathfrak{m}_{K})=\log_{\mathcal{M}}(\mathfrak{m}_{K})$* is a $\mathbb{Z}_{p}$-module.*

Proof.

The action of the ring $\mathbb{Z}_{p}$ on $\log_{p}(1+\mathfrak{m}_{K})$ can be described as $\mathbb{Z}_{p}\times\log_{p}(1+\mathfrak{m}_{K})\to\log_{p}(1+\mathfrak{m}_{K})$ defined by $a\cdot\log_{p}(b)=\log_{p}(b^{a})$. This is valid because for $a\in\mathbb{Z}_{p}$, the binomial expension

[TABLE]

∎

Remark*.*

The notation $\log_{p}(1+x)$ of $p$-adic logarithm indicates the field $\mathbb{Q}_{p}$, we are considering. While the notation $\log_{\mathcal{M}}(x)$ indicates that the $p$-adic logarithm is induced by the $1$-dimensional multiplicative formal group law $\mathcal{M}(X,Y)=X+Y+XY$. The use of the formal group $\mathcal{M}$ makes notation and concept much clearer here, because, the original multiplicative group $1+\mathfrak{m}_{K}$ gets described, using $\mathcal{M}$, as just $\mathfrak{m}_{K}$. Most of the time, in this article we use the notation $\log_{\mathcal{M}}$ if no confusion arises.

2. Compuation of basis of $2$-adic logarithm

In this section, we compute the bases of the images of the $2$-adic logarithm $\log_{2}(1+\mathfrak{m}_{K})$, where $\mathfrak{m}_{K}$ are maximal ideals in the $7$ quadratic extensions $K=\mathbb{Q}_{2}(\sqrt{d})$ of $\mathbb{Q}_{2}$, $d=-1,\pm 2,\pm 3,\pm 6$. Throughout this section, we use the notation $\log_{\mathcal{M}}(x)$ instead of the notation $\log_{2}(1+x)$ to use the properties of the $1$-dimensional multiplicative formal group law $\mathcal{M}$.

The multiplication-by-$n$ map $n\to[n]_{F}$ is an endomorphism of a formal group law $F$ from the ordinary integers $\mathbb{Z}$ into the ring of endomorphisms $\text{End}_{\mathcal{O}_{K}}(F)$ satisfying:

1. (i)

$[0]_{F}=0$ and for $n\geq 0$, $[n+1]_{F}(X)=F([n]_{F}(X),X),$ 2. (ii)

$[n]_{F}=\iota_{F}\circ[-n]_{F}$, if $n<0$.

We have the following relations for the $1$-dimensional multiplicative formal group law $\mathcal{M}$:

Lemma 2.1**.**

We have the following consequences,

1. (i)

$\mathcal{M}(-2,-2)=[2]_{\mathcal{M}}(-2)=0$. 2. (ii)

$\log_{\mathcal{M}}(-2)=0$. 3. (iii)

$\log_{\mathcal{M}}(2)=0\ (\text{mod}\ 4\mathcal{O}_{K})$. 4. (iv)

$\mathcal{M}(2,2)=[2]_{\mathcal{M}}(2)=0\ (\text{mod}\ 4\mathcal{O}_{K})$.

Proof.

1. (i)

We have $\mathcal{M}(x,y)=x+y+xy$. Then, $\mathcal{M}(-2,-2)=-2-2+4=0.$ 2. (ii)

It follows from $(i)$. 3. (iii)

$\log_{\mathcal{M}}(2)=\mathcal{M}(2,2)=2+2+4=8\equiv 0\ (\text{mod}\ 4\mathcal{O}_{K})$. 4. (iv)

It follows from $(iii)$.

∎

We note the following result:

Theorem 2.2**.**

[7]** There are $7$ quadratic extensions of $\mathbb{Q}_{2}$ given by $\mathbb{Q}_{2}(\sqrt{d})$ for $d=-1,\pm 2,\pm 3,\pm 6$. All these extensions except $\mathbb{Q}_{2}(\sqrt{-3})$ are ramified.

Each of these $7$ extensions has multiple equally-good descriptions, for instance $\mathbb{Q}_{2}(\sqrt{-3})=\mathbb{Q}_{2}(\sqrt{5})$. Of these seven, $\mathbb{Q}_{2}(\sqrt{-3})$ is the only unramified extension and $\mathbb{Q}_{2}(\sqrt{-1})$ is the extension with additional $2$-power roots of unity.

Lemma 2.3**.**

*Let $K$ be a finite extension of $\mathbb{Q}_{2}$. Then the $\log_{\mathcal{M}}$ series is convergent on the maximal ideal $\mathfrak{m}_{K}$ and maps the set $2\mathfrak{m}_{K}$ onto itself in a one-to-one way. *

Proof.

The Proposition 1.2 shows the $p$-adic logarithm is convergent in the maximal ideal. For the second claim, looking at the power series

[TABLE]

whose coefficients are all in $\mathbb{Z}_{2}$. Since $\log_{\mathcal{M}}$ has its first-degree coefficient in $\mathbb{Z}_{2}^{\times}$, the units of $\mathbb{Z}_{2}$, it has an inverse power series, $\log_{\mathcal{M}}^{-1}=x+x^{2}+\cdots\in\mathbb{Z}_{2}[[x]]$, of course we know that this inverse function is $\frac{1}{2}(exp(2x)-1)$. This inverse function is also convergent for $x\in\mathfrak{m}_{K}$. Thus we have two series in $\mathbb{Z}_{2}[[x]]$, inverse to each other and thus $\log_{\mathcal{M}}$ is one-to-one and onto $2\mathfrak{m}_{K}$. ∎

Lemma 2.4**.**

The image of $4\mathbb{Z}_{2}$ under 2-adic logarithm $\log_{\mathcal{M}}$ is $4\mathbb{Z}_{2}$. Furthermore, if $z=2^{m}u$ for a 2-adic unit $u$ and with $m\geq 2$, then $\log_{\mathcal{M}}(z)=2^{m}u^{\prime}$ for a 2-adic unit $u^{\prime}$. In other words, if $v_{2}(z)\geq 2$, then $v_{2}(\log_{\mathcal{M}}(z))=v_{2}(z)$.

Proof.

Applying Lemma 2.3 the proof follows. Therefore $\log_{\mathcal{M}}$ maps $2\cdot 2\mathbb{Z}_{2}$ onto itself in a one-to-one way. ∎

Lemma 2.5**.**

For $p>2$, the $\mathbb{Q}_{p}$-series $\frac{1}{p}\log_{\mathcal{M}}(px)$ has all coefficients in $\mathbb{Z}_{p}$ and all but first degree coefficients are in $p\mathbb{Z}_{p}$.

Proof.

Writing out the series $\frac{1}{p}\log_{\mathcal{M}}(px)=x-\frac{px^{2}}{2}+\frac{p^{2}x^{3}}{3}-\cdots,$ and looking at the denominator, the $n$-th denominator in the coefficient satisfy $v_{p}(n)\leq\log_{p}(n)$, while the numerator has $v_{p}(p^{n-1})=n-1$. Since for $p>2$ and $n>1$, we have the inequality $n>\log_{p}(n)+1$, the result follows immediately. ∎

Lemma 2.6**.**

Let $p>2$, the image of $p\mathbb{Z}_{p}$ under the $p$-adic logarithm is $p\mathbb{Z}_{p}$. Furthermore, if $z=p^{m}u$ for a $p$-adic unit $u$ with $m\geq 1$, then $\log_{\mathcal{M}}(z)=p^{m}u^{\prime}$ for a $p$-adic unit $u^{\prime}$. In other words, if $v_{p}(z)\geq 1$, then $v_{p}(\log_{\mathcal{M}}(z))=v_{p}(z)$.

Proof.

Applying previous Lemma 2.5, the result follows immediately. ∎

Let $C_{n}$ refer to the cyclic group of order $n$. We can think of it as the additive group $\mathbb{Z}/n\mathbb{Z}$, but in many cases, the group will be written multiplicatively.

Lemma 2.7**.**

Let $K$ be a quadratic extension of $\mathbb{Q}_{2}$, its ring of integers $\mathcal{O}_{K}$, and the maximal ideal $\mathfrak{m}_{K}=\pi\mathcal{O}_{K}$ for uniformizer $\pi\in\mathcal{O}_{K}$. Then, in the additive group $\mathcal{O}_{K}$, the subgroup $\mathcal{O}_{K}/2\mathcal{O}_{K}$ has order $4$, and the factor group $\mathcal{O}_{K}/2\mathcal{O}_{K}$ has the structure $C_{2}\oplus C_{2}$. The same is true for the quotient $\mathfrak{m}_{K}/2\mathfrak{m}_{K}$. Under the formal group addition $+_{\mathcal{M}}$, the subgroup $2\mathfrak{m}_{K}$ of $\mathfrak{m}_{K}$ still has index $4$.

Proof.

Using the structure theorem of a finitely generated module over a principal ideal domain, we see that a finitely generated module over $\mathbb{Z}_{2}$ will have the decomposition $\mathbb{Z}_{2}^{m}\oplus\mathcal{F}$, where the first part $\mathbb{Z}_{2}^{m}$ is the free part, which is torsion-free and $\mathcal{F}$ is a direct sum of modules of the form $\mathbb{Z}_{2}/2^{m}\mathbb{Z}_{2}$, which is torsion group. Thus $\mathcal{O}_{K}\cong\mathbb{Z}_{2}^{m}$ for some $m$, the maximal cardinality of a $\mathbb{Z}_{2}$- linearly independent subset of $\mathcal{O}_{K}$. But such a set is also a basis of the quadratic extension $K$ over $\mathbb{Q}_{2}$.

Thus we have $\mathcal{O}_{K}/2\mathcal{O}_{K}\cong\mathbb{Z}_{2}^{2}/2\mathbb{Z}_{2}^{2}\cong(\mathbb{Z}_{2}/2\mathbb{Z}_{2})^{2}$, the second isomorphism follows from a map $f:\mathbb{Z}_{2}^{2}=\mathbb{Z}_{2}\times\mathbb{Z}_{2}\to(\mathbb{Z}_{2}/2\mathbb{Z}_{2})^{2}=\mathbb{Z}_{2}/2\mathbb{Z}_{2}\times\mathbb{Z}_{2}/2\mathbb{Z}_{2}$ defined by $f(a,b)=(a+2\mathbb{Z}_{2},b+2\mathbb{Z}_{2})$ for $(a,b)\in\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. This map is a surjective homomorphism. The kernel is $2\mathbb{Z}_{2}^{2}$ because $f(a,b)=(0+2\mathbb{Z}_{2},0+2\mathbb{Z}_{2})\Leftrightarrow(a,b)=(2u,2v),\ u,v\in\mathbb{Z}_{2}\Leftrightarrow(a,b)=2(u,v)\in 2(\mathbb{Z}_{2}\times\mathbb{Z}_{2})=2\mathbb{Z}_{2}^{2}$. Also $C_{2}^{2}=(\mathbb{Z}_{2}/2\mathbb{Z}_{2})^{2}$ and hence $\mathcal{O}_{K}/2\mathcal{O}_{K}\cong C_{2}^{2}\cong C_{2}\oplus C_{2}$. This proves the first assertion. Since $\mathfrak{m}_{K}=\pi\mathcal{O}_{K}$, the multiplication-by-$\pi$ map is a $\mathbb{Z}_{2}$-homomorphism from $\mathcal{O}_{K}$ onto $\pi\mathcal{O}_{K}$, and the next assertion follows i.e., $\mathfrak{m}_{K}/2\mathfrak{m}_{K}\cong C_{2}\oplus C_{2}$.

The structure of $\mathfrak{m}_{K}$ under the addition $+_{\mathcal{M}}$ may be different from the familiar additive structure, but at least $2\mathfrak{m}_{K}$ is still a subgroup of $\mathfrak{m}_{K}$ of index four. For, if $K$ is ramified extension over $\mathbb{Q}_{2}$, then $\mathfrak{m}_{K}=\pi\mathcal{O}_{K}$, $\mathfrak{m}_{K}^{2}=\pi^{2}\mathcal{O}_{K}$, and $\mathfrak{m}_{K}^{3}=\pi\cdot\pi^{2}\mathcal{O}_{K}=\pi\cdot 2\mathcal{O}_{K}$, since $\pi^{2}\mathcal{O}_{K}=2\mathcal{O}_{K}$. Then $[\pi^{i}\mathcal{O}_{K}:\pi^{i+1}\mathcal{O}_{K}]=2,$ for $i\geq 0$, because $\mathcal{O}_{K}/\pi\mathcal{O}_{K}=\mathcal{O}_{K}/\mathfrak{m}_{K}=\kappa_{K}=\mathbb{F}_{2}$.

If $K$ is an unramified extension of $\mathbb{Q}_{2}$, we take the basis $\{1,\omega\}$ of $\mathcal{O}_{K}$ over $\mathbb{Z}_{2}$, where $\omega$ is the primitive cube root of unity, i.e., $1+\omega+\omega^{2}=0$. Then,

[TABLE]

We note,

[TABLE]

Thus we get $z+_{\mathcal{M}}z\equiv 0\ (\text{mod}\ 4\mathcal{O}_{K})$ and hence the factor group $\mathcal{O}_{K}/2\mathcal{O}_{K}$ has the structure $C_{2}\oplus C_{2}$. ∎

2.1. Main results

Lemma 2.8**.**

If we consider $K=\mathbb{Q}_{2},~{}\mathcal{O}_{K}=\mathbb{Z}_{2},~{}\mathfrak{m}_{K}=2\mathbb{Z}_{2},~{}\pi=2$, then $\log_{\mathcal{M}}(\mathfrak{m}_{K})=\log_{\mathcal{M}}(2\mathfrak{m}_{K})=2\mathfrak{m}_{K}$.

Proof.

We must show that if $z\in 2\mathbb{Z}_{2}$, then $\log_{\mathcal{M}}(z)\in 4\mathbb{Z}_{2}$; and that if $w\in 4\mathbb{Z}_{2}$, then there is $z\in 2\mathbb{Z}_{2}$ with $\log_{\mathcal{M}}(z)=w$.

We start with $z\in 2\mathbb{Z}_{2}$. We know that if $z\in 4\mathbb{Z}_{2}$, then $\log_{\mathcal{M}}(z)\in 4\mathbb{Z}_{2}$, by Lemma 2.4, so that we need to consider only the case where $z\equiv 0~{}(\text{mod}~{}2)$ but $z\not\equiv 0~{}(\text{mod}~{}4)$: in other words, $z\equiv 2~{}(\text{mod}~{}4)$. Now consider $w=z+_{\mathcal{M}}(-2)=z-2-2z\equiv z-2\equiv 0~{}(\text{mod}\ 4)$. We have $\log_{\mathcal{M}}(w)\in 4\mathbb{Z}_{2}$ by Lemma 2.1, and at the same time $\log_{\mathcal{M}}(w)=\log_{\mathcal{M}}(z)+\log_{\mathcal{M}}(-2)=\log_{\mathcal{M}}(z)$ because $\log_{\mathcal{M}}(-2)=0$. So $\log_{\mathcal{M}}(z)\in 4\mathbb{Z}_{2}$.

The other direction is trivial: if $w\in 4\mathbb{Z}_{2}$, then $w=\log_{\mathcal{M}}(z)$ for $z\in\mathbb{Z}_{2}$, and since $4\mathbb{Z}_{2}\subset 2\mathbb{Z}_{2}$, we are done. ∎

Alternative proof of Lemma 2.8.

What is to be proved is that $\log_{\mathcal{M}}(2\mathbb{Z}_{2})=\log_{\mathcal{M}}(4\mathbb{Z}_{2})$. What the proof depends on is the fact that $\log_{\mathcal{M}}(2)=4u$ for $u$ a $\mathbb{Z}_{2}$-unit, in other words, that $v_{2}(\log_{\mathcal{M}}(2))=2$. For this, we do a direct computation using the series expansion for $\log_{\mathcal{M}}$: the first two terms, $2$ and $-2^{2}/2$, cancel each other, and the most interesting one is $-2^{4}/4=-4$. All other terms have $v_{2}$-value at least $3$, so the claim about $v_{2}(\log_{\mathcal{M}}(2))$ is verified.

Now suppose $v(z)=1$, then $z+_{\mathcal{M}}2=z+2+2z$, in which $z\equiv 2~{}(\text{mod}\ 4)$ so that $v(z+2)\equiv 0~{}(\text{mod}\ 4)$, and hence it follows that $v(z+_{\mathcal{M}}2)\geq 2$. Now, we have $\log_{\mathcal{M}}(z)=\log_{\mathcal{M}}(z+_{\mathcal{M}}2)-\log_{\mathcal{M}}(2)$, in which both elements on the right-hand side have $v$-value $2$, so that their difference on the left has $v$-value $\geq 2$. ∎

Now, in case $K$ is quadratic ramified over $\mathbb{Q}_{2}$, with ring of integers $\mathcal{O}_{K}$ and maximal ideal $\mathfrak{m}_{K}=\pi\mathcal{O}_{K}$, we have the chain of multiplicative subgroups of $1+\mathfrak{m}_{K}$:

[TABLE]

where the inclusions are strict, and each index $[\pi^{m}\mathcal{O}_{K}:\pi^{m+1}\mathcal{O}_{K}]=2$. Translating to the group structure of $\mathfrak{m}_{K}$ furnished by $+_{\mathcal{M}}$, we see the same inclusions:

[TABLE]

Now we will associate here $\log_{\mathcal{M}}(\pi)$ with $\log_{\mathcal{M}}(\mathfrak{m}_{K})$. We know that $\log_{\mathcal{M}}(2\mathfrak{m}_{K})=2\mathfrak{m}_{K}$, and since in the $+_{\mathcal{M}}$-group structure a representative of the nonzero element of $\pi\mathfrak{m}_{K}/2\mathfrak{m}_{K}$ is $-2$, which has the property that $\log_{\mathcal{M}}(-2)=0$, we can say that $\log_{\mathcal{M}}(\pi\mathfrak{m}_{K})=2\mathfrak{m}_{K}$. The following lemma involves more efficient method to show $\log_{\mathcal{M}}(\pi\mathfrak{m}_{K})=2\mathfrak{m}_{K}$:

Lemma 2.9**.**

For the uniformizer $\pi$ in the quadratic extension $K$ of $\mathbb{Q}_{2}$, we have $\log_{\mathcal{M}}(\pi\mathfrak{m}_{K})=2\mathfrak{m}_{K}$.

Proof.

We do know that $\log_{\mathcal{M}}(2\mathfrak{m}_{K})=2\mathfrak{m}_{K}$, by Lemma 2.8. An element of $\pi\mathfrak{m}_{K}$ either has $v$-value $1$ or greater; in the latter case, it will be in $2\mathfrak{m}_{K}$. So, let’s take $z\in\pi\mathfrak{m}_{K}$, in other words, $v(z)\geq 1$. We need only consider the case of equality, $v(z)=1$. Now, as above, $v(z/2)=0$, so that $z/2\equiv 1~{}(\text{mod}\ \mathfrak{m}_{K})$. (Here we use the fact that the residue-class field is $\mathbb{F}_{2}=\mathbb{Z}/2\mathbb{Z}$.) And of course $-2/2=-1\equiv 1~{}(\text{mod}\ \mathfrak{m}_{K})$. Thus $(z-2)/2\equiv 0~{}(\text{mod}\ \mathfrak{m}_{K})$, and $z-2\equiv 0~{}(\text{mod}\ 2\mathfrak{m}_{K})$. The last congruence says $v(z-2)\geq\frac{3}{2}$. Now we get $z+_{\mathcal{M}}(-2)=z-2-2z$, in which $V_{2}(z-2)\geq\frac{3}{2}$ and $v(-2z)\geq 2$, so that $v(z+_{\mathcal{M}}(-2))\geq\frac{3}{2}$. So we observe

[TABLE]

which says that $\log_{\mathcal{M}}(z)\in 2\mathfrak{m}_{K}$. So $\log_{\mathcal{M}}(\pi\mathfrak{m}_{K})=2\mathfrak{m}_{K}$. ∎

In these 6 ramified cases, thus, we need only to determine $\log_{\mathcal{M}}(\pi)$ in order to identify $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ because we already know $\log_{\mathcal{M}}(\pi\mathfrak{m}_{K})=2\mathfrak{m}_{K}$ by Lemma 2.9 and since the group index $[\mathfrak{m}_{K}:\pi\mathfrak{m}_{K}]$ is equal to $2$ with respect to the group structure furnished by $+_{\mathcal{M}}$, we can look at this as saying that the cosets of $\mathfrak{m}_{K}$ modulo $\pi\mathfrak{m}_{K}$ are two in number, namely $\pi\mathfrak{m}_{K}$ and $\pi+_{\mathcal{M}}\pi\mathfrak{m}_{K}$ and when we apply $\log_{\mathcal{M}}$, these two cosets go over to $\log_{\mathcal{M}}(\pi\mathfrak{m}_{K})=2\mathfrak{m}_{K}$ and $\log_{\mathcal{M}}(\pi+_{\mathcal{M}}\pi\mathfrak{m}_{K})=\log_{\mathcal{M}}(\pi)+\log_{\mathcal{M}}(\pi\mathfrak{m}_{K})=\log_{\mathcal{M}}(\pi)+2\mathfrak{m}_{K}$.

Lemma 2.10**.**

Let $d$ be one of $\{\pm 6,\pm 2\}$, and $\pi=\sqrt{d}$. Then $\log_{\mathcal{M}}(\pi)$ is one of the form $u\pi+2z$, with a unit of $\mathbb{Z}_{2}$ and $z\in\mathbb{Z}_{2}$. Further, in the case $d=2$, $z=0$.

Proof.

In most cases, we can tell $v_{p}(\log_{\mathcal{M}}(w))$ from the knowledge of $v_{p}(w)$, unless $v_{p}(w)$ is the $\xi$-coordinate of a vertex of the Newton copolygon of $\log_{\mathcal{M}}(x)$. The fact that the $\xi$-coordinates of the vertices of the copolygon, in the case $p=2$, are all at the numbers $\frac{1}{2^{n}}$ for $n\geq 0$. In each case under consideration here, we see that the terms $\frac{\pi^{2}}{2}$ and $\frac{\pi^{4}}{4}$ have $v$ value equal to [math]. This means that we can not ascertain the value of $\log_{\mathcal{M}}(\pi)$ without some direct calculation. We see, however, that the first term in the series $\log_{\mathcal{M}}(\pi)$, namely, $\pi$, dominates. This demonstrates the first claim.

For the special situation $d=2$, we use Galois theory, looking at the $\mathbb{Q}_{2}$-automorphism of $k=\mathbb{Q}_{2}(\sqrt{2})$, which sends $\pi=\sqrt{2}$ to $-\pi$. We denote this automorphism by $w\to\bar{w}$. Because the coefficients of $\log_{\mathcal{M}}$ all are in $\mathbb{Q}_{2}$, we have the relation $\log_{\mathcal{M}}(\bar{w})=\overline{\log_{\mathcal{M}}(w)}$ whenever $w\in\mathfrak{m}_{K}$. In particular, $\log_{\mathcal{M}}(-\pi)=\overline{\log_{\mathcal{M}}(\pi)}$. But we claim that $\pi+_{\mathcal{M}}(-\pi)=-2$: indeed

[TABLE]

By the homomorphic property of $\log_{\mathcal{M}}$, we get $\log_{\mathcal{M}}(\pi)+\log_{\mathcal{M}}(-\pi)=\log_{\mathcal{M}}(-2)=0$, leading to the conclusion that $\log_{\mathcal{M}}(-\pi)=-\log_{\mathcal{M}}(\pi)$. For an element $\alpha+\beta\sqrt{2}$ with $\alpha,\beta\in\mathbb{Q}_{2}$, when $\bar{w}=-w$ so that $\alpha-\beta\sqrt{2}=-(\alpha+\beta\sqrt{2})$, we conclude $\alpha=0$, since $\{1,\sqrt{2}\}$ is a $\mathbb{Q}_{2}$-basis of $K$. Or, using $\log_{\mathcal{M}}(\pi)=u\pi+2z$ with both $u$ and $z$ in $\mathbb{Z}_{2}$, since the conjugate of $\log_{\mathcal{M}}(\pi)$ is $-\log_{\mathcal{M}}(\pi)$ in this case, the $2$-component has to be zero. ∎

Lemma 2.11**.**

Let $d=3$, $\pi=-1+\sqrt{3}$. Then $v_{2}(\log_{\mathcal{M}}(\pi))=1$, and $\log_{\mathcal{M}}(\pi)\in\mathbb{Z}_{2}$.

Proof.

The evaluation of $v_{2}(\log_{\mathcal{M}}(\pi))$ is computational, and depends only on the polynomial $\pi-\frac{\pi^{2}}{2}-\frac{\pi^{4}}{4}-\frac{\pi^{8}}{8}$ in the evaluation of the $\log_{\mathcal{M}}$, since all others have $v_{2}$-value $\geq\frac{3}{2}$.

Using the relation $\pi^{2}+2\pi-2=0$, we have $\pi^{4}=4-8\pi+4\pi^{2}=12-16\pi$. Thus $\pi-\frac{\pi^{2}}{2}-\pi^{4}-4=\pi-1+\pi-3+4\pi=-4+6\pi$, which has $v_{2}$-value $\frac{3}{2}$. Since $v_{2}(\frac{\pi^{8}}{8})=1$, this term dominates, and the $v$-value of $\log_{\mathcal{M}}(\pi)$ is $1$.

The second claim follows by looking at the elements of $1+\mathfrak{m}_{K}$ corresponding to $\pi$ and $\bar{\pi}$. These are $1+\pi=\sqrt{3}$ and $-\sqrt{3}$, whose quotient is $-1$, a number whose $p$-adic logarithm is [math]. Thus $\pi-_{\mathcal{M}}\bar{\pi}=-2$, and when we take logarithms, we get $\log_{\mathcal{M}}(\pi)+\log_{\mathcal{M}}(\bar{\pi})=\log_{\mathcal{M}}(\pi)+\overline{\log_{\mathcal{M}}(\pi)}=0,$ so that $\log_{\mathcal{M}}(\pi)$ is its own conjugate and hence in $\mathbb{Q}_{2}$, the fixed field of conjugation. Since $v_{2}(\log_{\mathcal{M}}(\pi))=1$, and $\log_{\mathcal{M}}(\pi)\in\mathbb{Q}_{2}$, we conclude $\log_{\mathcal{M}}(\pi)\in\mathbb{Z}_{2}$. ∎

Lemma 2.12**.**

Let $d=-1$, $\pi=-1+i$. Then $\log_{\mathcal{M}}(\pi)=0$. For $d=-1$, $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ has basis equal to the basis of $2\mathfrak{m}_{K}$.

Proof.

The element of $1+\mathfrak{m}_{K}$ corresponding to $\pi$ is $1+\pi=i$, multiplicatively of order $4$, so that its $\log_{\mathcal{M}}(\pi)=\log_{\mathcal{M}}(-1+i)=\log_{\mathcal{M}}(-1)+\log_{\mathcal{M}}(i)=0$. In another way, since $\pi=-1+i$, we have $\pi^{2}+2\pi=2$. Now, $2\log_{\mathcal{M}}(\pi)=\log_{\mathcal{M}}(\pi+_{\mathcal{M}}\pi)=\log_{\mathcal{M}}(\pi+\pi+\pi^{2})=\log_{\mathcal{M}}(2\pi+\pi^{2})=\log_{\mathcal{M}}(-2)=0$, by Lemma 2.1. Next, the conclusion is just another way of saying that $\log_{\mathcal{M}}(\mathfrak{m}_{K})=2\mathfrak{m}_{K}$. we know that $\log_{\mathcal{M}}(\pi\mathfrak{m}_{K})=2\mathfrak{m}_{K}$. Consider $z\in\mathfrak{m}_{K}$ now, then as before, $v(z/\pi)\equiv 1~{}(\text{mod}~{}\mathfrak{m}_{K})$, and thus $z/\pi+1\equiv 0~{}(\text{mod}~{}\mathfrak{m}_{K})$, again using the fact that the residue field is $\mathbb{F}_{2}$. As a result, $z+\pi\equiv 0~{}(\text{mod}~{}\pi\mathfrak{m}_{K})$. Again we calculate $z+_{\mathcal{M}}\pi=z+\pi+\pi z$ in which both $z+\pi$ and $\pi z$ are in $\pi\mathfrak{m}_{K}$, so that $z+_{\mathcal{M}}\pi\in\pi\mathfrak{m}_{K}$.Now we see that:

[TABLE]

Hence $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ has basis equal to the basis of $2\mathfrak{m}_{K}$ for $d=-1$. ∎

We are now ready to describe $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ in the $6$ ramified cases $d=-1,\pm 2,3,\pm 6$. We would describe the set $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ by giving a basis $\{\alpha,\beta\}$, because it is a free $\mathbb{Z}_{2}$-module of rank $2$. We know that $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ is a group containing $\log_{\mathcal{M}}(\pi\mathfrak{m}_{K})=\log_{\mathcal{M}}(2\mathfrak{m}_{K})=2\mathfrak{m}_{K},$ with the associated index being $2$. Here we note that in ramified extensions, we have a $\mathbb{Z}_{2}$-basis $\{1,\pi\}$ for $\mathcal{O}_{K}$ and to get a $\mathbb{Z}_{2}$-basis of $\pi^{m}\mathcal{O}_{K}$, we just take $\{\pi^{m},\pi^{m+1}\}$. In case $m=2r$ is even, an equally good basis will be $\{2^{r},2^{r}\pi\}$, and in case $m=2r+1$ is odd, an equally good basis will be $\{2^{r}\pi,2^{r+1}\}$. We will use the basis $\{2\pi,4\}$ for $2\mathfrak{m}_{K}$ to see how it extends when we adjoin $\log_{\mathcal{M}}(\pi)$.

Theorem 2.13**.**

If $K=\mathbb{Q}_{2}(\sqrt{d})$ for $d=-1,\pm 2,3,\pm 6$, where $\pi=\sqrt{d}$ in case of even $d$ and $\pi=-1+\sqrt{d}$ for odd $d$, then:

1. (1)

For $d=-1$, $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ has basis $\{2\pi,4\}$. 2. (2)

For $d=-2$, $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ has basis $\{\pi u+2z,4\}$ with $z\in\mathbb{Z}_{2}$. 3. (3)

For $d=2$, $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ has basis $\{\pi,4\}$. 4. (4)

For $d=3$, $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ has basis $\{2\pi,2\}$. 5. (5)

For $d=-6$, $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ has basis $\{\pi+2z,4\}$ with $z\in\mathbb{Z}_{2}$. 6. (6)

For $d=6$, $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ has basis $\{\pi+2z,4\}$, with $z\in\mathbb{Z}_{2}$.

Proof.

1. (1)

The basis of $2\mathfrak{m}_{K}$ is $\{2\pi,4\}$. By Lemma 2.12, the basis of $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ is equal to the basis of $2\mathfrak{m}_{K}$. Therefore, the basis of $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ is $\{2\pi,4\}$.

Alternatively, we can directly show that $\log_{\mathcal{M}}(\mathfrak{m}_{K})=2\mathfrak{m}_{K}$. For, we know that from Lemma 2.9, $\log_{\mathcal{M}}(\pi\mathfrak{m}_{K})=2\mathfrak{m}_{K}$. Consider $z\in\mathfrak{m}_{K}$, then $v(z/\pi)\geq 0$, and as before the only one case that concerns us is when $v(z/\pi)=0$ so that $z/\pi\equiv 1$ (mod $\mathfrak{m}_{K}$), and thus $z/\pi+1\equiv 0$ (mod $\mathfrak{m}_{K}$), by using the fact that $1=-1$ (mod 2) because $\mathbb{Q}_{2}(\sqrt{-1})$ is ramified extension of degree 2 and its residue field is $\mathbb{F}_{2}$. As a result, $z+\pi\equiv 0$ (mod $\pi\mathfrak{m}_{K}$). Again we calculate $z+_{\mathcal{M}}\pi=\mathcal{M}(z,\pi)=z+\pi+\pi z$, in which both $z+\pi$ and $\pi z$ are in $\pi\mathfrak{m}_{K}$, so that $z+_{\mathcal{M}}\pi\in\pi\mathfrak{m}_{K}$. Finally,

[TABLE] 2. (2)

This follows from Lemma 2.10. 3. (3)

We have already seen, in case $K=\mathbb{Q}_{2}(\sqrt{2})$, that $\log_{\mathcal{M}}(\pi)=u\pi$ where $u$ is a $\mathbb{Z}_{2}$-unit. We also know that $\log_{\mathcal{M}}(\pi\mathfrak{m}_{K})=2\mathfrak{m}_{K}$, which has basis $\{2\pi,4\}$. Since the “new” element of $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ is $\log_{\mathcal{M}}(\pi)=u\pi$, there is also an element of $\mathfrak{m}_{K}$, namely $w=[u^{-1}]_{\mathcal{M}}(u\pi)$ whose $\log_{\mathcal{M}}(w)=\pi=\sqrt{2}$. Recall that $[u^{-1}]_{\mathcal{M}}$ is the multiplication-by-$u^{-1}$ map, which is an endomorphism of the 1-dimensional multiplicative formal group $\mathcal{M}$. Also $\log_{\mathcal{M}}([2]_{\mathcal{M}}(w))=2\pi$, which belong to $\log_{\mathcal{M}}(\mathfrak{m}_{K})$. This demonstrate the basis $\{\pi,4\}$. 4. (4)

For the case $d=3$, $\pi=\sqrt{3}-1$. We know $\log_{\mathcal{M}}(\pi\mathfrak{m}_{K})=2\mathfrak{m}_{K}$ by Lemma 2.9, which has basis $\{2\pi,4\}$. The image of $\log_{\mathcal{M}}$ contains at least $2\pi\mathcal{O}_{K}=2\mathfrak{m}_{K}$, and the important thing to note that on this set, the 2-adic logarithm and 2-adic exponential are perfect inverses of each other. Now, the index of the quotient group $\mathfrak{m}_{K}/2\mathfrak{m}_{K}$ is $4$ by Lemma 2.7. Note that the quotient group $\mathfrak{m}_{K}/2\mathfrak{m}_{K}$ is $\mathbb{F}_{2}$-vector space with basis $\{\pi,2\}$. Let us now look at what the logarithm $\log_{\mathcal{M}}$ does to the elements of $\mathfrak{m}_{K}/2\mathfrak{m}_{K}$. We have already seen that $\log_{\mathcal{M}}(\pi)=2u$ for a unit $u\in\mathbb{Z}_{2}$, and thus $\log_{\mathcal{M}}([1/u](\pi)=2$. So $2$ is in the image of $\log_{\mathcal{M}}$. By direct calculation, one sees that $\log_{\mathcal{M}}(-\pi)$ has the form $2z+2w\pi$, where $z$ is something in $\mathbb{Z}_{2}$ and $w$ is a unit in $\mathbb{Z}_{2}$. And that gives us the “new” element of the image of the logarithm $\log_{\mathcal{M}}$, giving as basis of $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ the pair $\{2\pi,2\}$. 5. (5)

The proof follows from Lemma 2.10. 6. (6)

The proof follows from Lemma 2.10.

∎

It only remains to dispose of the single unramified case, $K=\mathbb{Q}_{2}(\sqrt{-3})$ which is proved in the following theorem:

Theorem 2.14**.**

A $\mathbb{Z}_{2}$-basis of $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ is $\{2,4\omega\}$ in the unramified case $K=\mathbb{Q}_{2}(\omega)$, where $\omega^{2}+\omega+1=0$.

Proof.

Here, $2$ is a prime, but the residue field is $\mathbb{F}_{4}=\mathbb{F}_{2}(\omega)$ with $\omega^{2}+\omega+1=0$. Back here in characteristic zero, there is still an $\omega\in\mathcal{O}_{K}$, $\omega^{2}+\omega+1=0$, and we may take $\omega=\frac{-1+\sqrt{-3}}{2}$, a cube root of unity.

We still have the result that $\log_{\mathcal{M}}(2\mathfrak{m}_{K})=2\mathfrak{m}_{K}$, but since $\mathcal{O}_{K}$ has basis $\{1,\omega\}$ and $\mathfrak{m}_{K}=2\omega$, the module $2\mathfrak{m}_{K}$ has $\mathbb{Z}_{2}$-basis $\{4,4\omega\}$. Since $\mathfrak{m}_{K}$ has basis $\{2,2\omega\}$, we have $[\mathfrak{m}_{K}:2\mathfrak{m}_{K}]=4$. Let us use our knowledge that $\log_{\mathcal{M}}(-2)=0$ to say that $[\log_{\mathcal{M}}(\mathfrak{m}_{K}):\log_{\mathcal{M}}(2\mathfrak{m}_{K})]\leq 2$. The only question is what $\log_{\mathcal{M}}(2\omega)$ might be. We get

[TABLE]

where the missing terms all have $v_{2}$-value greater than $1$. Indeed, the only terms that interest us are $2\omega-2\omega^{2}$. But $\omega^{2}=-\omega-1$, the $v$-value of $2+4\omega$ is $1$, so that $\log_{\mathcal{M}}(\mathfrak{m}_{K})$ in this case has basis $\{2,4\omega\}$. ∎

3. Applications

The basis of vector space is as important as the basis of the free module. So Theorems 2.13 and 2.14 give important directions. We write a few applications of our results as follows:

Let $p$ be a prime number, $\mathbb{Q}_{p}$ the field of $p$-adic numbers, and $\mathbb{C}_{p}$ the completion of the algebraic closure of $\mathbb{Q}_{p}$. Let $U_{p}$ be the units $(1+\mathfrak{m}_{\mathbb{C}_{p}})$ of $\mathbb{C}_{p}$, the $p$-adic version of Baker theorem is:

Theorem 3.1**.**

(Brumer) Let $\alpha_{1},\cdots,\alpha_{n}$ be elements of principal units $U_{p}$ which are algebraic over the rationals $\mathbb{Q}$ and their $p$-adic logarithms $\log_{p}(\alpha_{1}),\cdots,\log_{p}(\alpha_{n})$ are linearly independent over $\mathbb{Q}$. These logarithms $\log_{p}(\alpha_{1}),\cdots,\log_{p}(\alpha_{n})$ are then linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{C}_{p}$.

We prove a thin result similar to the above result for $p=2,~{}n=2$.

Theorem 3.2**.**

If $\alpha_{1},\alpha_{2}$ are two elements of the group of principal units $U_{1}=1+\mathfrak{m}_{K}$ of $K=\mathbb{Q}_{2}(\sqrt{-1})$ such that their 2-adic logarithms $\log_{2}(\alpha_{1}),\log_{2}(\alpha_{2})$ are linearly independent over $\mathbb{Q}$. Then their logarithms $\log_{2}(\alpha_{1}),\log_{2}(\alpha_{2})$ are linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_{2}$.

Proof.

Note the uniformizer of $K=\mathbb{Q}_{2}(\sqrt{-1})$ was $\pi=-1+\sqrt{-1}$. Since $\{2\pi,4\}$ is a $\mathbb{Z}_{2}$-basis of $\log_{2}(U_{1})$ by Theorem 2.13, we can write

[TABLE]

But given that $\log_{2}(\alpha_{1}),\log_{2}(\alpha_{2})$ are linearly independent over $\mathbb{Q}$, therefore

[TABLE]

We claim $\log_{2}(\alpha_{1}),\log_{2}(\alpha_{2})$ are also linearly independent only over $\mathbb{Q}_{2}\cap\bar{\mathbb{Q}}$. For, if $\beta_{1},\beta_{2}\in\mathbb{Q}_{2}\cap\bar{\mathbb{Q}}$ such that

[TABLE]

By relation (8), this system has a trivial solution $\beta_{1}=\beta_{2}=0$. Therefore $\log_{2}(\alpha_{1}),\log_{2}(\alpha_{2})$ are linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_{2}$. This proves the theorem. ∎

Remark*.*

In the above theorem, we can replace $\mathbb{Q}_{2}(\sqrt{-1})$ by other 6 quadratic extensions of $\mathbb{Q}_{2}$.

Proposition 3.3**.**

The $2$-adic logarithm induces a measure on $4\mathbb{Z}_{2}$ from a measure on $1+2\mathbb{Z}_{2}$.

Proof.

By [14, Sec. 12.2, Chap. 12], if $h:X\to Y$ is continuous and $d\phi$ is a measure on $X$, then we obtain a measure $d\psi$ on $Y$ by defining

[TABLE]

for some measurable function $f$.

Take $X=1+2\mathbb{Z}_{2},~{}Y=4\mathbb{Z}_{2}$, and $h=\log_{2}$. Recall that $p$-adic logarithm is a continuous function. Infact, the 2-adic logarithm $\log_{2}:1+2\mathbb{Z}_{2}\to 4\mathbb{Z}_{2}$ is continuous in Lemma 2.8. Thus from (9), we have

[TABLE]

Therefore, we obtain a measure on $4\mathbb{Z}_{2}$ induced from a 2-adic measure on $1+2\mathbb{Z}_{2}$. ∎

Proposition 3.4**.**

For two arbitrary primes $p_{1},p_{2}\equiv 5$ (mod 8), there is a 2-adic integer $z$ such that $p_{1}=p_{2}^{z}$.

Proof.

For, $p_{1}\equiv 5$ (mod 8), we have

[TABLE]

Similary, $v_{2}(\log_{2}(p_{2}))\equiv 2~{}\text{mod 8}$. Thus $v_{2}\left(\frac{\log_{2}(p_{1})}{\log_{2}(p_{2})}\right)=v_{2}(1)=0$, and so $\frac{\log_{2}(p_{1})}{\log_{2}(p_{2})}\in\mathbb{Z}_{2}$.

We know $\log_{2}(x^{n})=n\log_{2}(x)$ holds for $n\in\mathbb{Z}$, $x\in 1+4\mathbb{Z}_{2}$. Now the map $(x,n)\mapsto x^{n}$ from $(1+4\mathbb{Z}_{2})\times\mathbb{Z}\to 1+4\mathbb{Z}_{2}$ is continuous with respect to $2$-adic topology, and thus extends to $(1+4\mathbb{Z}_{2})\times\mathbb{Z}_{2}\to 1+4\mathbb{Z}_{2}$. Composing with continuous map $\log_{2}$, we get $\log_{2}(x^{z})=z\log_{2}(x)$ for all $z\in\mathbb{Z}_{2}$. So taking the 2-adic logarithm both sides of $p_{1}=p_{2}^{z}$, we obtain

[TABLE]

Thus the equation $p_{1}=p_{2}^{z}$ has a solution in $\mathbb{Z}_{2}$ i.e., $z$ is a 2-adic integer satisfying $p_{1}=p_{2}^{z}$. ∎

$p$-adic logarithm in Iwasawa theory

We repeat from the introduction section that, the reference to Iwasawa is the most important application of $p$-adic logarithm. In [12], Iwasawa has extensively studied certain types of modules made up of the images of $p$-adic logarithm on the principal units $U_{n}$ of the field $\mathbb{Q}_{p}(\zeta_{p^{n+1}})$ and used those special modules in [11] to prove explicit formulas for the norm residue symbol, e.g., [12, Lemma 1], [11, Theorem 1]. There are other similar results along the lines of Iwasawa in [8, Theorem 1.10], [14, Section 13.8] and in [13], where the image of $p$-adic logarithm on the principal units $U_{n}$ of the cyclotomic extension of $\mathbb{Q}_{p}$ has played a crucial role.

Conclusion

The paper takes an initial effort to compute the image of $p$-adic logarithm on the principal units of quadratic extensions of $\mathbb{Q}_{2}$. We find the $\mathbb{Z}_{2}$-bases of $\log_{p}(1+\mathfrak{m}_{K})$. In the later work [1], the authors explicitly compute the images of $\log_{p}(1+\mathfrak{m}_{K})$ on the principal units in all 7 quadratic extensions of $\mathbb{Q}_{2}$ as well as for cyclotomic extensions of $\mathbb{Q}_{p}$ for all $p$. Iwasawa’s work is our main motivaton to study the image of $p$-adic logarithm on the group of principal units.

Question

It would be interesting to compute the bases of $p$-adic logarithm on the principal units of any finite extensions of $\mathbb{Q}_{p}$.

Acknowledgement

The authors are deeply grateful to Professor Jonathan Lubin for helpful correspondence and suggestions. The first author is grateful to CSIR, Govt. of India, for the grant with File no.-09/025(0249)/2018-EMR-I.