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All content in this area was uploaded by Mabud Ali Sarkar on Aug 09, 2023

Content may be subject to copyright.

Graduate J. Math. 888 (2023), 69 – 77

On the bases of the image of 2-adic logarithm

on the group of principal units

Mabud Ali Sarkar and Absos Ali Shaikh

Abstract

This paper computes the bases of the image of the 2-adic logarithm

on the group of the principal units in all 7 quadratic extensions of

Q2. This helps one understand the free module structure of the 2-

adic logarithm at arbitrary points on its domain. We discuss some

applications at the end.

Keywords: p-adic numbers, p-adic logarithm, formal group, Iwa-

sawa theory.

MSC 2020. Primary: 11F85; Secondary: 12F05, 14L05.

Introduction

The p-adic logarithm

logp(1 + x) = X

n≥1

(−1)n−1xn

n∈K[[x]]

converges for all x∈mK. Following Iwasawa [14], the

p-adic logarithm extends to pZ(1 + mCp)⊂C×

psuch

that logp(p) = 0. It is known that the p-adic logarithm

deﬁnes an isomorphism from 1 + mr

K→mr

Kif r > e

p−1,

where eis the ramiﬁcation index. However, the case

r= 1 is diﬀerent, and logp: 1 + mK→mKis not an

isomorphism. The image logp(1 + mK)is unknown for

an arbitrary extension Kof Qp.

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 crucial in Iwasawa theory. More specif-

ically, in [12], Iwasawa explicitly computed the formu-

las for the norm residue symbol by using the image of

the p-adic logarithm on the group of principal units of

certain cyclotomic extensions of Qp, which is our main

motivation.

There are plenty of other potential applications of

the image or value of the p-adic logarithm. In [15],

the formula for normalized p-adic regulator contains the

image of the p-adic logarithm on the principal units. In

the study of p-adic L-functions, there appears a p-adic

logarithm. To compute the p-adic L-function L(1, χ),

one of the key challenges is the computation of the p-

adic logarithm of arbitrary elements in its domain, see

[9].

The value of the p-adic logarithm has an important

role in the formula of the Sen operator ([4]) in the the-

ory of p-adic representation as well as in the formula of

p-adic heights [16] in the theory of arithmetic dynam-

ics. Therefore, it is of novelty to understand the image

and value of the p-adic logarithm on its domain. Since

by Proposition 1.4, logp(1 + mK)is a Zp-module, once

we know the bases of logp(1 + mK), we can understand

the value of the p-adic logarithm at an arbitrary point

in its domain by forming a linear combination of those

basis elements with scalars from Zp.

The current paper undertakes an initial challenge to-

wards computing the image of p-adic logarithm on the

group of principal units, by computing the bases of the

p-adic logarithm. The basis of a free module is as im-

portant as the basis of a vector space. In Theorem 2.13

and Theorem 2.14 of Section 2, we compute the bases

of the 2-adic logarithm for all 7quadratic extensions of

Q2. Moreover, we ﬁnd the exact images of the 2-adic

logarithm in some cases. In Section 3, we discuss some

more applications.

Notations

Qpis the p-adic number ﬁeld with the ring of integers

Zp, unique maximal ideal pZpand a residue ﬁeld Fp:=

Zp/pZp. Let ¯

Qpbe the algebraic closure and take the

p-adic completion Cp:= b

¯

Qpwith the maximal ideal

mCpand the units C×

p. Let Kbe a ﬁnite extension of

Qpwith a ring of integers OK, unique maximal ideal

mK. The quotient ﬁeld κK:= OK/mKis the residue

ﬁeld of OK. We also have the standard p-adic additive

valuation vp:Qp→Z∪{∞}. We extend the valuation

vpon Qpto the ﬁnite extension Kdenoted by vthat

makes v(z) = vp(z)whenever z∈Qpby the following

69

1 Formal group law and p-adic logarithm 70

formula

v(λ) = 1

[K:Qp]vpNK

Qp(λ),

where NK

Qpis the ﬁeld-theoretic norm, the multiplica-

tive mapping from K×to Qp. We use the common

notation vthroughout the paper.

Acknowledgements: The authors are deeply grate-

ful to Professor Jonathan Lubin for helpful correspon-

dence and suggestions. The authors would like to thank

the anonymous referee for careful reading and sugges-

tions that improved the article. The ﬁrst author is

grateful to CSIR, Government of India, for the grant

with File no.-09/025(0249)/2018-EMR-I.

1 Formal group law and p-adic logarithm

In this section, we construct the p-adic logarithm from

one-dimensional multiplicative formal group law.

Deﬁnition 1.1. [5] Let Rbe a commutative ring. A

formal group law over Ris a formal power series F(X, Y )

∈R[[X, Y ]] with the following properties

(i) F(X, Y ) = X+Y+higher degree terms,

(ii) F(X, F (Y , Z)) = F(F(X, Y ), Z),

(iii) F(X, 0) = Xand F(0, Y ) = Y,

(iv) There exists a unique power series ι(T)∈R[[T]]

such that

F(X, ι(X)) = F(i(Y), Y ) = 0,

which is the inverse power series in the formal

group law.

If in addition F(X, Y ) = F(Y, X )holds then Fis called

commutative. The commutativity property comes triv-

ially as long as Rhas no elements that are both torsion

and nilpotent.

Example 1.1. The one-dimensional formal additive

group law is deﬁned by F(X, Y ) = X+Y, and the

one-dimensional formal multiplicative group law is de-

ﬁned by M(X, Y ) = X+Y+X Y.

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→Gover Ris

a power series f(T)∈R[[T]], having no constant term,

satisfying

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

If further there exists another homomorphism ψ∈R[[T]],

ψ:G→Fsuch that

f(ψ(T)) = ψ(f(T)) = T,

then we say that Fand Gare isomorphic over Rwhile

fis an isomorphism over R. An endomorphism f:

F→Fsatisﬁes f(F(X, Y )) = F(f(X), f(Y)).

Deﬁnition 1.3. [6] An invariant diﬀerential of a formal

group law Fover an arbitrary ring Ris a diﬀerential

form ω(T) = p(T)dT ∈R[[T]]dT that satisﬁes

p(F(T, S))FX(T, S) = p(T),

where FX(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 Fgiven by the for-

mula

ω(T) = FX(0, T )−1dT.

We now construct the formal logarithm, of a formal

group law, obtained by integrating the invariant diﬀer-

ential. But integrating an invariant diﬀerential form

ω(T)produces denominators in each coeﬃcient. So we

assume Rto be a commutative ring of characteristic 0

so that we don’t have further diﬃculty with 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 Ris a torsion-free ring

and let R0=R⊗ZQ. If ω(T) = p(T)dT is a normalized

invariant diﬀerential of a formal group law F, then the

formal group law Finduces a formal logarithmic series

logFon R0given by

logF(T) = ZT

0ω(τ) = T+a2

2T2+a3

3T3+··· ∈ R0[[T]],

satisfying logF(F(X, Y )) = logF(X) + logF(Y).

Proof. Deﬁne the map R→R0=R⊗ZQby r7→

r⊗1for r∈R. The kernel of this map is the set of

torsion elements of R. But as Ris the torsion-free ring,

the kernel is trivial. Therefore, there is a one-to-one

correspondence between Rand R⊗ZQ. This allows us

to take coeﬃcients of the formal logarithm in Q.

As p(T)dT is a normalised invariant diﬀerential, we

have p(0) = 1 and so we can assume

p(T) = FX(0, T )−1= 1 + ∞

X

i=1

aiTi∈R0[[T]].

Integrating, we get

logF(T) = ZT

0ω(τ) = ZT

0

1

FX(0, T )dT

=T+a1

2T2+a2

3T3+··· ∈ R0[[T]].

It remains to be shown that logF(F(X, Y )) = logF(X)+

logF(Y). Considering the partial derivatives of the as-

sociativity relation F(X, F (Y , Z)) = F(F(X, Y ), Z)with

respect to X, we obtain

FX(X, F (Y , Z)) = FX(F(X, Y ), Z)·FX(X, Y ).(1.1)

Let p(X)be the formal power series with coeﬃcients in

Rdeﬁned by

p(X)·FX(0, X) = p(0) = 1.(1.2)

1 Formal group law and p-adic logarithm 71

Now logF(X)is the unique formal power series with

coeﬃcients in R0=R⊗ZQsuch that ∂

∂X logF(X) =

p(X). Then, using the equation (1.2) and making the

changes X= 0, Y →X, Z →Yin the equation (1.1),

we get

∂

∂X logF(F(X, Y )) = ∂

∂X logF(X).(1.3)

Integrating (1.3), we get logF(F(X, Y )) = logF(X) +

C(Y), where C(Y)is an arbitrary function of Y. There-

fore, we have the relation

logF(F(X, Y )) −logF(X) = C(Y).(1.4)

Putting X= 0 in (1.4), we have

logF(F(0, Y )) −logF(0) = C(Y)

⇒logF(Y) = C(Y),since logF(0) = 0 and F(0, Y ) = Y.

Hence from (1.4), we conclude logF(F(X, Y )) = logF(X)

+ logF(Y).This completes the proof.

Proposition 1.2. Let Kbe a ﬁnite extension of Qp

with a ring of integers OKand a unique maximal ideal

mK. Then the one-dimensional multiplicaive formal

group law M(x, y) = x+y+xy deﬁned over OKinduces

the p-adic logarithm given by

logM(t) = logp(1 + t) = t−t2

2+t3

3− · ··

and converges in mK.

Proof. If π∈ OKis a uniformizer, we can choose either

π=p−1

√−por π=ζp−1, for some pth root of unity.

It can be checked that the diﬀerential form ω(x) = dx

1+x

is the invariant diﬀerential for M(x, y) = x+y+xy,

and hence the corresponding logarithmic series logMis

obtained by

logM(x) = Zx

0ω(t)dt =Zx

0

dt

1 + t=x−x2

2+x3

3−··· ,

which satisﬁes logM(M(x, y)) = logM(x) + logM(y).

This logarithmic series is called the p-adic logarithm

and is denoted by

logM(x) = logp(1 + x) = x−x2

2+x3

3− ··· ∈ OK[[x]].

(1.5)

This series converges in mK. For, we note that the

Cauchy-Hadamard convergence formula holds in the p-

adic ﬁeld as well. If f(x) = Pn≥0anxn∈K[[x]], by the

Cauchy-Hadamard formula, the radius of convergence

r∈[0,∞]is given by

1

r= lim

n→∞

n

q|an|p.

For the p-adic logarithm (1.5), we have therefore an=

(−1)n−11

n, and hence

1

r= lim

n→∞

n

v

u

u

t

(−1)n−1

np

= lim

n→∞

1

n

q|n|p

= lim

n→∞ pvp(n)/n.

Since 1≤pvp(n)≤n, we get 1≤pvp(n)/n ≤n

√n.

Also n

√n→1as n→ ∞. In particular, n

q|n|p→1

as n→ ∞. Hence the radius of convergence is r= 1,

i.e., −1< x ≤1. At the end point x= 1, we have

an=±1

nand its p-adic norm 1

|n|pdoesn’t tends to 0

because it is one when p-n. Therefore, the disc of

convergence of the p-adic logarithm in Kis the open

unit disc {x∈K:|x|p<1}=mK.

Thus, the important thing is to note that the one-

dimensional multiplicative formal group law Mgives a

group structure on mKas follows:

Lemma 1.3. Let Fbe a formal group law over OK.

Then the set mKbecomes a group under the group law

F, via the law of combination α+Fβ=F(α, β).

Proof. The proof is obvious using the previous deﬁni-

tions and basic properties of formal group law. The

only thing that needs to be ensured is the convergence

of the inﬁnite series F(α, β), which follows from the

completeness of mKunder mK-adic topology.

The following standard proposition describes the mod-

ule structure of the p-adic logarithm on principal units.

The proof is based on the basic concept of p-adic loga-

rithm, e.g., [2, Chap. IV], [3, Sec. 4, Chap. 5].

Proposition 1.4. logp(1 + mK) = logM(mK)is a Zp-

module.

Proof. The action of the ring Zpon logp(1 + mK)can

be described as Zp×logp(1 + mK)→logp(1 + mK)

deﬁned by a·logp(b) = logp(ba). This is valid because

for a∈Zp, the binomial expension

(1 + b)a=∞

X

n=0

a(a−1) ···(a−n+ 1)

n!bnconverges.

Remark. The notation logp(1 + x)of the p-adic loga-

rithm indicates the considered ﬁeld Qpwhile the no-

tation logM(x)indicates that the p-adic logarithm is

induced by the one-dimensional multiplicative formal

group law M(X, Y ) = X+Y+X Y . The use of the for-

mal group Mmakes notation and concept much clearer

here, because, the original multiplicative group 1 + mK

gets described, using M, as just mK. Most of the time,

in this article, we use the notation logMif no confusion

arises.

2 Computation of basis of 2-adic logarithm 72

2 Computation of basis of 2-adic logarithm

In this section, we compute the bases of the images of

the 2-adic logarithm log2(1 + mK), where mKare max-

imal ideals in the 7quadratic extensions K=Q2(√d)

of Q2,d=−1,±2,±3,±6. Throughout this section,

we use the notation logM(x)instead of the notation

log2(1 + x)to use the properties of the one-dimensional

multiplicative formal group law M.

The multiplication-by-nmap n→[n]Fis an endo-

morphism of a formal group law Ffrom the ordinary

integers Zinto the ring of endomorphisms EndOK(F)

satisfying:

(i) [0]F= 0 and for n≥0,

[n+ 1]F(X) = F([n]F(X), X).

(ii) [n]F=ιF◦[−n]F, if n < 0.

We have the following relations for the one-dimensional

multiplicative formal group law M:

Lemma 2.1. We have the following consequences,

(i) M(−2,−2) = [2]M(−2) = 0.

(ii) logM(−2) = 0.

(iii) logM(2) = 0 (mod 4OK).

(iv) M(2,2) = [2]M(2) = 0 (mod 4OK).

Proof. (i) We have M(x, y) = x+y+xy. Then,

M(−2,−2) = −2−2 + 4 = 0.

(ii) It follows from (i).

(iii) logM(2) = M(2,2) = 2+2+4 = 8 ≡0 (mod 4OK).

(iv) It follows from (iii).

We note the following result:

Theorem 2.2. [7] There are 7quadratic extensions of

Q2given by Q2(√d)for d=−1,±2,±3,±6. All these

extensions except Q2(√−3) are ramiﬁed.

Each of these 7extensions has multiple equally good

descriptions, for instance, Q2(√−3) = Q2(√5). Of

these seven, Q2(√−3) is the only unramiﬁed extension,

and Q2(√−1) is the extension with additional 2-power

roots of unity.

Lemma 2.3. Let Kbe a ﬁnite extension of Q2. Then

the logMseries is convergent on the maximal ideal mK

and maps the set 2mKonto 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

logM(x) = 1

2logM(2x)

=x−x2+4x3

3−2x4+16x5

5−16x6

3+··· ,

whose coeﬃcients are all in Z2. Since logMhas its

ﬁrst-degree coeﬃcient in Z×

2, the units of Z2, it has an

inverse power series, log−1

M=x+x2+··· ∈ Z2[[x]]. We,

of course, know that this inverse function is 1

2(exp(2x)−

1). This inverse function is also convergent for x∈mK.

Thus we have two series in Z2[[x]], inverse to each other,

and thus logMis one-to-one and onto 2mK.

Lemma 2.4. The image of 4Z2under 2-adic logarithm

logMis 4Z2. Furthermore, if z= 2mufor a 2-adic

unit uand with m≥2, then logM(z)=2mu0for

a 2-adic unit u0. In other words, if v2(z)≥2, then

v2(logM(z)) = v2(z).

Proof. Applying Lemma 2.3, the proof follows. There-

fore, logMmaps 2·2Z2onto itself in a one-to-one way.

Lemma 2.5. For p > 2, the Qp-series 1

plogM(px)has

all coeﬃcients in Zpand all but ﬁrst-degree coeﬃcients

are in pZp.

Proof. Writing out the series

1

plogM(px) = x−px2

2+p2x3

3− · ·· ,

and looking at the denominator, the n-th denomina-

tor in the coeﬃcient satisfy vp(n)≤logp(n), while the

numerator has vp(pn−1) = n−1. Since for p > 2and

n > 1, we have the inequality n > logp(n)+1, the result

follows immediately.

Lemma 2.6. Let p > 2, the image of pZpunder the

p-adic logarithm is pZp. Furthermore, if z=pmufor

ap-adic unit uwith m≥1, then logM(z) = pmu0for

ap-adic unit u0. In other words, if vp(z)≥1, then

vp(logM(z)) = vp(z).

Proof. Applying the previous Lemma 2.5, the result fol-

lows immediately.

Let Cnrefer to the cyclic group of order n. We can

think of it as the additive group Z/nZ, but in many

cases, the group will be written multiplicatively.

Lemma 2.7. Let Kbe a quadratic extension of Q2, its

ring of integers OK, and the maximal ideal mK=πOK

for the uniformizer π∈ OK. Then, in the additive

group OK, the subgroup OK/2OKhas order 4, and the

factor group OK/2OKhas the structure C2⊕C2. The

same is true for the quotient mK/2mK. Under the for-

mal group addition +M, the subgroup 2mKof mKstill

has index 4.

Proof. Using the structure theorem of a ﬁnitely gener-

ated module over a principal ideal domain, we see that

a ﬁnitely generated module over Z2will have the de-

composition Zm

2⊕F, where the ﬁrst part Zm

2is the free

part, which is torsion-free, and Fis a direct sum of

modules of the form Z2/2mZ2, which is torsion group.

Thus OK∼

=Zm

2for some m, the maximal cardinality

2 Computation of basis of 2-adic logarithm 73

of a Z2- linearly independent subset of OK. But such a

set is also the basis of the quadratic extension Kover

Q2. Thus we have OK/2OK∼

=Z2

2/2Z2

2∼

=(Z2/2Z2)2,

the second isomorphism follows from a map f:Z2

2=

Z2×Z2→(Z2/2Z2)2=Z2/2Z2×Z2/2Z2deﬁned by

f(a, b)=(a+ 2Z2, b + 2Z2)for (a, b)∈Z2×Z2. This

map is a surjective homomorphism. The kernel is 2Z2

2

because

f(a, b) = (0 + 2Z2,0+2Z2)

⇔(a, b) = (2u, 2v), u, v ∈Z2

⇔(a, b) = 2(u, v)∈2(Z2×Z2)=2Z2

2.

Also C2

2= (Z2/2Z2)2, and hence

OK/2OK∼

=C2

2∼

=C2⊕C2

This proves the ﬁrst assertion. Since mK=πOK, the

multiplication-by-πmap is a Z2-homomorphism from

OKonto πOK, and the next assertion follows, i.e.,

mK/2mK∼

=C2⊕C2.

The structure of mKunder the addition +Mmay

be diﬀerent from the familiar additive structure, but at

least 2mKis still a subgroup of mKof index four. For,

if Kis ramiﬁed extension over Q2, then mK=πOK,

m2

K=π2OK, and m3

K=π·π2OK=π·2OK, since

π2OK= 2OK. Then [πiOK:πi+1OK] = 2,for i≥0,

because OK/πOK=OK/mK=κK=F2.

If Kis an unramiﬁed extension of Q2, we take the

basis {1, ω}of OKover Z2, where ωis the primitive

cube root of unity, i.e., 1 + ω+ω2= 0. Therefore,

2 +M2ω= 2 + 2ω+ 4ω≡2 + 2ω(mod 4OK). We

ﬁnally note,

•For z= 2,2 +M2 = M(2,2) ≡0 (mod 4OK),

•For z= 2ω, 2ω+M2ω=M(2ω, ω)≡0 (mod 4OK),

and •For z= 2 + 2ω,

(2+2ω)+M(2+2ω) = M(2+2ω, 2+2ω)≡0 (mod 4OK).

Thus, we get z+Mz≡0 (mod 4OK), and hence the

factor group OK/2OKhas the structure C2⊕C2.

2.1 Main results

Lemma 2.8. If we consider K=Q2,OK=Z2,mK=

2Z2, π = 2, then

logM(mK) = logM(2mK) = 2mK.

Proof. We must show that if z∈2Z2, then logM(z)∈

4Z2; and that if w∈4Z2, then there is z∈2Z2with

logM(z) = w.

We start with z∈2Z2. We know that if z∈4Z2,

then logM(z)∈4Z2, by Lemma 2.4, so that we need

to consider only the case where z≡0 (mod 2) but z6≡

0 (mod 4): in other words, z≡2 (mod 4). Now consider

w=z+M(−2) = z−2−2z≡z−2≡0 (mod 4).

We have logM(w)∈4Z2by Lemma 2.1, and at the

same time logM(w) = logM(z) + logM(−2) = logM(z)

because logM(−2) = 0. So logM(z)∈4Z2.

The other direction is trivial: if w∈4Z2, then w=

logM(z)for z∈Z2, and since 4Z2⊂2Z2, we are done.

Alternative proof of Lemma 2.8. What is to be proved

is that logM(2Z2) = logM(4Z2). What the proof de-

pends on is the fact that logM(2) = 4ufor uaZ2-unit,

in other words, that v2(logM(2)) = 2. For this, we

do a direct computation using the series expansion for

logM: the ﬁrst two terms, 2and −22/2, cancel each

other, and the most interesting one is −24/4 = −4. All

other terms have v2-value at least 3, so the claim about

v2(logM(2)) is veriﬁed.

Now suppose v(z) = 1, then z+M2 = z+ 2 + 2z,

in which z≡2 (mod 4) so that v(z+ 2) ≡0 (mod 4),

and hence it follows that v(z+M2) ≥2. Now, we have

logM(z) = logM(z+M2) −logM(2), in which both

elements on the right-hand side have v-value 2, so that

their diﬀerence on the left has v-value ≥2.

Now, in case Kis quadratic ramiﬁed over Q2, with a

ring of integers OKand maximal ideal mK=πOK, we

have the chain of multiplicative subgroups of 1 + mK:

··· ⊂ 1+2mK= 1 + π2mK

= 1 + π3OK

⊂1 + πmK= 1 + π2OK

⊂1 + mK= 1 + πOK,

where the inclusions are strict, and each index [πmOK:

πm+1OK] = 2. Translating to the group structure of

mKfurnished by +M, we see the same inclusions:

2mK=π2mK=π3OK⊂πmK=π2OK⊂mK=πOK.

Now we will associate here logM(π)with logM(mK).

We know that logM(2mK) = 2mK, and since in the

+M-group structure a representative of the nonzero el-

ement of πmK/2mKis −2, which has the property that

logM(−2) = 0, we can say that logM(πmK)=2mK.

The following lemma involves more eﬃcient method to

show logM(πmK)=2mK:

Lemma 2.9. If πis an uniformizer in the ring of in-

tegers of the quadratic extension Kof Q2, then

logM(πmK) = 2mK

Proof. We do know that logM(2mK) = 2mK, by Lemma

2.8. An element of πmKeither has v-value one or

greater; in the latter case, it will be in 2mK. So, let’s

take z∈πmK, in other words, v(z)≥1. We need only

consider the case of equality, v(z)=1. Now, as above,

v(z/2) = 0, so that z/2≡1 (mod mK), where we

use the fact that the residue-class ﬁeld is F2=Z/2Z.

And of course, −2/2 = −1≡1 (mod mK). Thus,

(z−2)/2≡0 (mod mK), and z−2≡0 (mod 2mK).

2 Computation of basis of 2-adic logarithm 74

The last congruence says v(z−2) ≥3

2. Now we get

z+M(−2) = z−2−2z, in which V2(z−2) ≥3

2and

v(−2z)≥2, so that v(z+M(−2)) ≥3

2. So we observe

logM(z) = logM(z+M(−2)) −logM(−2)

= logM(z+M(−2)) ≥3

2,

which says that logM(z)∈2mK. So logM(πmK) =

2mK.

In these 6 ramiﬁed cases, we need only to deter-

mine logM(π)in order to identify logM(mK)because

we already know logM(πmK) = 2mKby Lemma 2.9

and since the group index [mK:πmK]is equal to 2

with respect to the group structure furnished by +M,

we can look at this as saying that the cosets of mK

modulo πmKare two in number, namely πmKand

π+MπmKand when we apply logM, these two cosets

go over to logM(πmK) = 2mKand logM(π+MπmK) =

logM(π) + logM(πmK) = logM(π)+2mK.

Lemma 2.10. Let dbe one of {±6,±2}, and π=√d.

Then logM(π)is one of the form uπ + 2z, with a unit

of Z2and z∈Z2. Further, in the case d= 2,z= 0.

Proof. In most cases, we can tell vp(logM(w)) from the

knowledge of vp(w), unless vp(w)is the ξ-coordinate of

a vertex of the Newton copolygon of logM(x). The fact

that the ξ-coordinates of the vertices of the copolygon,

in the case of p= 2, are all at the numbers 1

2nfor n≥0.

In each case under consideration here, we see that the

terms π2

2and π4

4have vvalue equal to 0. This means

that we can not ascertain the value of logM(π)without

some direct calculation. We see, however, that the ﬁrst

term in the series logM(π), namely, π, dominates. This

demonstrates the ﬁrst claim.

For the special situation d= 2, we use Galois theory,

looking at the Q2-automorphism of k=Q2(√2), which

sends π=√2to −π. We denote this automorphism

by w→¯w. Because the coeﬃcients of logMall are in

Q2, we have the relation logM( ¯w) = logM(w)whenever

w∈mK. In particular, logM(−π) = logM(π). But we

claim that π+M(−π) = −2: indeed

π+M(−π) = M(π, −π)

=π+ (−π) + π(−π) = −π2=−2.

By the homomorphic property of logM, we get logM(π)+

logM(−π) = logM(−2) = 0, leading to the conclu-

sion that logM(−π) = −logM(π). For an element

α+β√2with α, β ∈Q2, when ¯w=−wso that

α−β√2 = −(α+β√2), we conclude α= 0, since

{1,√2}is a Q2-basis of K. Or, using logM(π) = uπ+2z

with both uand zin Z2, since the conjugate of logM(π)

is −logM(π)in this case, the 2-component has to be

zero.

Lemma 2.11. Let d= 3,π=−1 + √3. Then

v2(logM(π)) = 1, and logM(π)∈Z2.

Proof. The evaluation of v2(logM(π)) is computational

and depends only on the polynomial π−π2

2−π4

4−π8

8

in the evaluation of the logM, since all others have v2-

value ≥3

2. Using the relation π2+ 2π−2=0, we have

π4= 4 −8π+ 4π2= 12 −16π. Thus, π−π2

2−π4−4 =

π−1 + π−3+4π=−4+6π, which has a v2-value 3

2.

Since v2(π8

8) = 1, this term dominates, and the v-value

of logM(π)is one.

The second claim follows by looking at the elements

of 1 + mKcorresponding to πand ¯π. These are 1 +

π=√3and −√3, whose quotient is −1, a number

whose p-adic logarithm is 0. Thus π−M¯π=−2, and

when we take logarithms, we get logM(π) + logM(¯π) =

logM(π)+ logM(π) = 0,so that logM(π)is its own con-

jugate and hence in Q2, the ﬁxed ﬁeld of conjugation.

Since v2(logM(π)) = 1, and logM(π)∈Q2, we conclude

logM(π)∈Z2.

Lemma 2.12. Let d=−1,π=−1+i. Then logM(π) =

0. For d=−1,logM(mK)has a basis equal to the basis

of 2mK.

Proof. The element of 1+ mKcorresponding to πis 1 +

π=i, multiplicatively of order 4, so that its logM(π) =

logM(−1 + i) = logM(−1) + logM(i) = 0. In an-

other way, since π=−1 + i, we have π2+ 2π= 2.

Now, 2 logM(π) = logM(π+Mπ) = logM(π+π+

π2) = logM(2π+π2) = logM(−2) = 0, by Lemma

2.1. Next, the conclusion is just another way of saying

that logM(mK) = 2mK. We know that logM(πmK) =

2mK. Consider z∈mKnow, then as before, v(z/π)≡

1 (mod mK), and thus z/π + 1 ≡0 (mod mK), again

using the fact that the residue ﬁeld is F2. As a re-

sult, z+π≡0 (mod πmK). Again, we calculate

z+Mπ=z+π+πz in which both z+πand πz

are in πmK, so that z+Mπ∈πmK. Now we see that:

logM(z) = logM(z+Mπ)−logM(π)

= logM(z+Mπ)∈logM(πmK)=2mK.

Hence logM(mK)has a basis equal to the basis of 2mK

for d=−1.

We are now ready to describe logM(mK)in the 6

ramiﬁed cases d=−1,±2,3,±6. We would describe

the set logM(mK)by giving a basis {α, β}, because it

is a free Z2-module of rank 2. We know that logM(mK)

is a group containing logM(πmK) = logM(2mK) =

2mK,with the associated index being 2. Here we note

that in ramiﬁed extensions, we have a Z2-basis {1, π}

for OKand to get a Z2-basis of πmOK, we just take

{πm, πm+1}. In case m= 2ris even, an equally good

basis will be {2r,2rπ}, and in case m= 2r+ 1 is odd,

an equally good basis will be {2rπ, 2r+1 }. We will use

the basis {2π, 4}for 2mKto see how it extends when

we adjoin logM(π).

3 Applications 75

Theorem 2.13. If K=Q2(√d)for d=−1,±2,3,±6,

where π=√din case of even dand π=−1 + √dfor

odd d, then:

1. For d=−1,logM(mK)has a basis {2π, 4}.

2. For d=−2,logM(mK)has a basis {πu + 2z, 4}

with z∈Z2.

3. For d= 2,logM(mK)has a basis {π, 4}.

4. For d= 3,logM(mK)has a basis {2π, 2}.

5. For d=−6,logM(mK)has a basis {π+ 2z, 4}

with z∈Z2.

6. For d= 6,logM(mK)has a basis {π+2z , 4}, with

z∈Z2.

Proof. (1) The basis of 2mKis {2π, 4}. By Lemma

2.12, the basis of logM(mK)is equal to the ba-

sis of 2mK. Therefore, the basis of logM(mK)

is {2π, 4}. Alternatively, we can directly show

that logM(mK) = 2mK. For, we know that from

Lemma 2.9, logM(πmK) = 2mK. Consider z∈

mK, then v(z/π)≥0, and as before the only one

case that concerns us is when v(z/π)=0so that

z/π ≡1(mod mK), and thus z/π + 1 ≡0(mod

mK), by using the fact that 1 = −1(mod 2) be-

cause Q2(√−1) is ramiﬁed extension of degree 2

and its residue ﬁeld is F2. As a result, z+π≡

0(mod πmK). Again, we calculate z+Mπ=

M(z, π) = z+π+πz, in which both z+πand πz

are in πmK, so that z+Mπ∈πmK. Finally,

logM(z) = logM(z+Mπ)−logM(π)

= logM(z+Mπ)∈logM(πmK) = 2mK,

since logM(π)=0.

(2) This follows from Lemma 2.10.

(3) We have already seen, in case K=Q2(√2), that

logM(π) = uπ where uis a Z2-unit. We also

know that logM(πmK)=2mK, which has a ba-

sis {2π, 4}. Since the “new” element of logM(mK)

is logM(π) = uπ, there is also an element of mK,

namely w= [u−1]M(uπ)whose logM(w) = π=

√2. Recall that [u−1]Mis the multiplication-by-

u−1map, which is an endomorphism of the one-

dimensional multiplicative formal group M. Also,

logM([2]M(w)) = 2π, which belongs to logM(mK).

This demonstrate the basis {π, 4}.

(4) For the case d= 3,π=√3−1. We know

logM(πmK)=2mKby Lemma 2.9, which has ba-

sis {2π, 4}. The image of logMcontains at least

2πOK= 2mK, and it is important to note that on

this set, the 2-adic logarithm and 2-adic exponen-

tial are perfect inverses of each other. Now, the in-

dex of the quotient group mK/2mKis 4by Lemma

2.7. Note that the quotient group mK/2mKis F2-

vector space with basis {π, 2}. Let us now look at

what the logarithm logMdoes to the elements of

mK/2mK. We have already seen that logM(π) =

2ufor a unit u∈Z2, and thus logM([1/u](π)=2.

So 2is in the image of logM. By direct calculation,

one sees that logM(−π)has the form 2z+ 2wπ,

where zis something in Z2and wis a unit in Z2.

And that gives us the “new” element of the im-

age of the logarithm logM, giving as the basis of

logM(mK)the pair {2π, 2}.

(5) The proof follows from Lemma 2.10.

(6) The proof follows from Lemma 2.10.

It only remains to dispose of the single unramiﬁed

case, K=Q2(√−3) which is proved in the following

theorem:

Theorem 2.14. AZ2-basis of logM(mK)is {2,4ω}in

the unramiﬁed case K=Q2(ω), where ω2+ω+ 1 = 0.

Proof. Here, 2is a prime, but the residue ﬁeld is F4=

F2(ω)with ω2+ω+ 1 = 0. Back here in characteristic

zero, there is still an ω∈ OK,ω2+ω+ 1 = 0, and we

may take ω=−1+√−3

2, a cube root of unity.

We still have the result that logM(2mK)=2mK,

but since OKhas basis {1, ω}and mK= 2ω, the mod-

ule 2mKhas Z2-basis {4,4ω}. Since mKhas a basis

{2,2ω}, we have [mK: 2mK]=4. Let us use our

knowledge that logM(−2) = 0 to say that [logM(mK) :

logM(2mK)] ≤2. The only question is what logM(2ω)

might be. We get

logM(2ω)=2ω−4ω2

2+8ω3

3−16ω4

4+···

= 2ω−2ω2+8ω3

3−16ω4

4+··· ,

where the missing terms all have v2-value greater than

one. Indeed, the only terms that interest us are 2ω−

2ω2. But ω2=−ω−1, the v-value of 2+4ωis one, so

that logM(mK)in this case has basis {2,4ω}.

3 Applications

The basis of vector space is as important as the basis

of the free module. So the Theorems 2.13 and 2.14 give

important directions. We make a few applications of

our results as follows:

Let pbe a prime number, Qpthe ﬁeld of p-adic num-

bers, and Cpthe completion of the algebraic closure of

Qp. Let Upbe the units (1 +mCp)of Cp, then the p-adic

version of the Baker theorem is as follows:

Theorem 3.1. (Brumer) Let α1,·· · , αnbe elements

of principal units Up= 1 + mCpof Cpwhich are alge-

braic over the rationals Qand their p-adic logarithms

3 Applications 76

logp(α1),··· ,logp(αn)are linearly independent over Q.

These logarithms logp(α1),·· · ,logp(αn)are then lin-

early independent over the algebraic closure Aof Qin

Cp.

We make no claim to justify Brumer’s result in any

way, rather we prove the following independent result

along the same spirit of Brumer for p= 2, n = 2, but

on the restricted domain U= 1+ mKof K=Q2(√−1)

rather than the big domain U2= 1 + mC2.

Theorem 3.2. If α1, α2are two elements of the group

of principal units U= 1 + mKof K=Q2(√−1) such

that their 2-adic logarithms log2(α1),log2(α2)are lin-

early independent over Q. Then their logarithms log2(α1),

log2(α2)are linearly independent over the algebraic clo-

sure Aof Qin Q2i.e., on ¯

Q∩Q2.

Proof. Note the uniformizer of K=Q2(√−1) is π=

−1 + √−1. Since {2π, 4}is a Z2-basis of log2(U)by

Theorem 2.13, we can write

log2(α1)=2πa1+4b1,log2(α2) = 2πa2+4b2, ai, bi∈Z2.

But given that log2(α1),log2(α2)are linearly indepen-

dent over Q, therefore

det a1b1

a2b26= 0 (3.1)

We claim log2(α1),log2(α2)are also linearly inde-

pendent only over Q2∩¯

Q. For, if β1, β2∈Q2∩¯

Qsuch

that

β1log2(α1) + β2log2(α2)=0

⇒β1(2πa1+ 4b1) + β2(2πa2+b2) = 0

⇒4(b1β1+b2β2)+2π(a1β1+a2β2)=0,

ai, bi∈Z2and β1, β2∈Q2∩¯

Q

⇒b+πa = 0,where b= 4(b1β1+b2β2),

and a= 2(a1β1+a2β2)∈Q2

⇒b=a= 0,since π /∈Q2⇒π /∈¯

Q∩Q2.

⇒(a1β1+a2β2= 0

b1β1+b2β2= 0

By relation (3.1), this system has a trivial solution

β1=β2= 0. Therefore, log2(α1),log2(α2)are linearly

independent over the algebraic closure Aof Qin Q2.

This proves the theorem.

Remark. In the above theorem, we can replace Q2(√−1)

by other 6 quadratic extensions of Q2.

Proposition 3.3. The 2-adic logarithm induces a mea-

sure on 4Z2from a measure on 1+2Z2.

Proof. By [14, Sec. 12.2, Chap. 12], if h:X→Yis

continuous and dφ is a measure on X, then we obtain

a measure dψ on Yby deﬁning

ZYfdψ =ZXf(h(X))dφ, (3.2)

for some measurable function f.

Take X= 1 + 2Z2, Y = 4Z2, and h= log2. Recall

that p-adic logarithm is a continuous function. In fact,

the 2-adic logarithm log2: 1 +2Z2→4Z2is continuous

in Lemma 2.8. Thus, from (3.2), we have

Z4Z2

fdψ =Z1+2Z2

f(log2(1 + 2Z2))dφ,

⇒Z4Z2

fdψ =Z1+2Z2

f(4Z2)dφ, (by Lemma 2.8).

Therefore, we obtain a measure on 4Z2induced from a

2-adic measure on 1+2Z2.

Proposition 3.4. For two arbitrary primes p1, p2≡5

(mod 8), the equation p1=pz

2has a solution in Z2.

Proof. For, p1≡5(mod 8), we have

v2(log2(p1)) = v2(log2(p1)−log2(1)),since log2(1) = 0

=v2(p1−1),since log2is isometry from

1+4Z2to 4Z2by Lemma 2.8

≡v2(5 −1) mod 8 ≡2mod 8.

Similary, v2(log2(p2)) ≡2mod 8. Thus

v2 log2(p1)

log2(p2)!=v2(1) = 0

and so log2(p1)

log2(p2)∈Z2.

We know log2(xn) = nlog2(x)holds for n∈Z,x∈

1+4Z2. Now the map (x, n)7→ xnfrom (1+4Z2)×Z→

1+4Z2is continuous with respect to 2-adic topology,

and thus extends to (1 + 4Z2)×Z2→1+4Z2. Com-

posing with the continuous map log2, we get log2(xz) =

zlog2(x)for all z∈Z2. So taking the 2-adic logarithm

on both sides of p1=pz

2, we obtain

log2(p1) = zlog2(p2)⇒z=log2(p1)

log2(p2)∈Z2.

Thus, the equation p1=pz

2has a solution in Z2i.e., z

is a 2-adic integer satisfying p1=pz

2.

p-adic logarithm in Iwasawa theory

We reiterate that the reference to Iwasawa is the most

important application of the p-adic logarithm. In [12],

Iwasawa has extensively studied certain types of mod-

ules made up of the images of p-adic logarithm on the

principal units Unof the ﬁeld Qp(ζpn+1 )and used those

special modules in [11] to prove explicit formulas for the

norm residue symbol, e.g., [12, Lemma 1], [11, Theo-

rem 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 the p-adic logarithm on the

principal units Unof the cyclotomic extension of Qphas

played a crucial role.

4 Conclusion 77

4 Conclusion

The paper takes an initial eﬀort to compute the im-

age of the p-adic logarithm on the principal units of

quadratic extensions of Q2. We ﬁnd the Z2-bases of

logp(1 + mK). We encourage the readers to explore the

next sequel [1], where the authors explicitly compute

the images of logp(1 + mK)on the principal units in all

7 quadratic extensions of Q2as well as for cyclotomic

extensions of Qpfor all p. Iwasawa’s work is our main

motivation to study the image of the p-adic logarithm

on the group of principal units.

Question

It would be interesting to compute the bases of the

p-adic logarithm on the principal units of any abelian

extensions of Qp.

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Mabud Ali Sarkar and Absos Ali Shaikh

Department of Mathematics,

The University of Burdwan

Burdwan-713104, India.

E-mail address:mabudji@gmail.com

aashaikh@math.buruniv.ac.in