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



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.
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
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.
The p-adic logarithm
logp(1 + x) = X
converges for all xmK. Following Iwasawa [14], the
p-adic logarithm extends to pZ(1 + mCp)C×
that logp(p) = 0. It is known that the p-adic logarithm
defines an isomorphism from 1 + mr
Kif r > e
where eis the ramification index. However, the case
r= 1 is different, and logp: 1 + mKmKis 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
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
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.
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:QpZ{∞}. We extend the valuation
vpon Qpto the finite extension Kdenoted by vthat
makes v(z) = vp(z)whenever zQpby the following
1 Formal group law and p-adic logarithm 70
v(λ) = 1
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:FGover Ris
a power series f(T)R[[T]], having no constant term,
f(F(X, Y )) = G(f(X), f(Y)).
If further there exists another homomorphism ψR[[T]],
ψ:GFsuch 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:
FFsatisfies 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-
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=RZQ. 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
3T3+··· R0[[T]],
satisfying logF(F(X, Y )) = logF(X) + logF(Y).
Proof. Define the map RR0=RZQby r7→
r1for rR. 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 RZQ. 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 +
Integrating, we get
logF(T) = ZT
0ω(τ) = ZT
FX(0, T )dT
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=RZQsuch 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) = tt2
3 · ··
and converges in mK.
Proof. If π OKis a uniformizer, we can choose either
por π=ζp1, for some pth root of unity.
It can be checked that the differential form ω(x) = dx
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
1 + t=xx2
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) = xx2
3 ··· OK[[x]].
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) = Pn0anxnK[[x]], by the
Cauchy-Hadamard formula, the radius of convergence
r[0,]is given by
r= lim
For the p-adic logarithm (1.5), we have therefore an=
n, and hence
r= lim
= lim
= lim
n→∞ pvp(n)/n.
Since 1pvp(n)n, we get 1pvp(n)/n n
Also n
n1as n . In particular, n
as n . Hence the radius of convergence is r= 1,
i.e., 1< x 1. At the end point x= 1, we have
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 {xK:|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-
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 aZp, the binomial expension
(1 + b)a=
a(a1) ···(an+ 1)
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
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)
(i) [0]F= 0 and for n0,
[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) = 22 + 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
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, log1
M=x+x2+··· Z2[[x]]. We,
of course, know that this inverse function is 1
1). This inverse function is also convergent for xmK.
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 m2, 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
all coefficients in Zpand all but first-degree coefficients
are in pZp.
Proof. Writing out the series
plogM(px) = xpx2
3 · ·· ,
and looking at the denominator, the n-th denomina-
tor in the coefficient satisfy vp(n)logp(n), while the
numerator has vp(pn1) = n1. 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 m1, 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 C2C2. 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
2F, 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
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
the second isomorphism follows from a map f:Z2
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
f(a, b) = (0 + 2Z2,0+2Z2)
(a, b) = (2u, 2v), u, v Z2
(a, b) = 2(u, v)2(Z2×Z2)=2Z2
Also C2
2= (Z2/2Z2)2, and hence
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.,
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,
K=π2OK, and m3
K=π·π2OK=π·2OK, since
π2OK= 2OK. Then [πiOK:πi+1OK] = 2,for i0,
because OKOK=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+ 4ω2 + 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+Mz0 (mod 4OK), and hence the
factor group OK/2OKhas the structure C2C2.
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 z2Z2, then logM(z)
4Z2; and that if w4Z2, then there is z2Z2with
logM(z) = w.
We start with z2Z2. We know that if z4Z2,
then logM(z)4Z2, by Lemma 2.4, so that we need
to consider only the case where z0 (mod 2) but z6≡
0 (mod 4): in other words, z2 (mod 4). Now consider
w=z+M(2) = z22zz20 (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 w4Z2, then w=
logM(z)for zZ2, and since 4Z22Z2, 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 z2 (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:
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 zπmK, 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/21 (mod mK), where we
use the fact that the residue-class field is F2=Z/2Z.
And of course, 2/2 = 11 (mod mK). Thus,
(z2)/20 (mod mK), and z20 (mod 2mK).
2 Computation of basis of 2-adic logarithm 74
The last congruence says v(z2) 3
2. Now we get
z+M(2) = z22z, in which V2(z2) 3
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
which says that logM(z)2mK. So logM(πmK) =
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 + 2z, with a unit
of Z2and zZ2. 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 n0.
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
wmK. 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+β2with α, β Q2, when ¯w=wso that
αβ2 = (α+β2), we conclude α= 0, since
{1,2}is a Q2-basis of K. Or, using logM(π) = +2z
with both uand zin Z2, since the conjugate of logM(π)
is logM(π)in this case, the 2-component has to be
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 ππ2
in the evaluation of the logM, since all others have v2-
value 3
2. Using the relation π2+ 2π2=0, we have
π4= 4 8π+ 4π2= 12 16π. Thus, ππ2
2π44 =
π1 + π3+4π=4+6π, which has a v2-value 3
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 πM¯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
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 zmKnow, then as before, v(z)
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+π0 (mod πmK). Again, we calculate
z+Mπ=z+π+πz in which both z+πand πz
are in πmK, so that z+MππmK. 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 zZ2.
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 zZ2.
6. For d= 6,logM(mK)has a basis {π+2z , 4}, with
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+MππmK. 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) = 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) = , there is also an element of mK,
namely w= [u1]M()whose logM(w) = π=
2. Recall that [u1]Mis the multiplication-by-
u1map, 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=31. 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 uZ2, 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+ 2,
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 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
= 2ω2ω2+8ω3
4+··· ,
where the missing terms all have v2-value greater than
one. Indeed, the only terms that interest us are 2ω
2ω2. But ω2=ω1, 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
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 ¯
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, biZ2.
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, β2Q2¯
β1log2(α1) + β2log2(α2)=0
β1(2πa1+ 4b1) + β2(2πa2+b2) = 0
ai, biZ2and β1, β2Q2¯
b+πa = 0,where b= 4(b1β1+b2β2),
and a= 2(a1β1+a2β2)Q2
b=a= 0,since π /Q2π /¯
(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:XYis
continuous and is a measure on X, then we obtain
a measure on Yby defining
ZYf =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 +2Z24Z2is continuous
in Lemma 2.8. Thus, from (3.2), we have
f =Z1+2Z2
f(log2(1 + 2Z2))dφ,
f =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, p25
(mod 8), the equation p1=pz
2has a solution in Z2.
Proof. For, p15(mod 8), we have
v2(log2(p1)) = v2(log2(p1)log2(1)),since log2(1) = 0
=v2(p11),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)
We know log2(xn) = nlog2(x)holds for nZ,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)×Z21+4Z2. Com-
posing with the continuous map log2, we get log2(xz) =
zlog2(x)for all zZ2. So taking the 2-adic logarithm
on both sides of p1=pz
2, we obtain
log2(p1) = zlog2(p2)z=log2(p1)
Thus, the equation p1=pz
2has a solution in Z2i.e., z
is a 2-adic integer satisfying p1=pz
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.
It would be interesting to compute the bases of the
p-adic logarithm on the principal units of any abelian
extensions of Qp.
[1] M. A. Sarkar, A. A. Shaikh,On the
image of p-adic logarithm on principal units,
[2] N. Koblitz,p-adic Numbers, p-adic Analysis,
and Zeta-functions, Springer, New York, 1984.
[3] A. M. Robert, A course in p-adic Analysis,
Springer, New York, 2000.
[4] L. Berger, An introduction to the the-
ory of p-adic representations, Lecture se-
ries in “Dwork trimester”(2001), Padova,
[5] M. Hazewinkel,Formal Groups and Applica-
tions, AMS Chelsea Publishing, 1978.
[6] J. H. Silverman,The Arithmetic of Elliptic
curves, Springer Publication, 1992.
[7] G. Shimura, Arithmetic of Quadratic Forms,
Springer, New York, 2009.
[8] B. Angles and T. Herreng, On a result of
Iwasawa, International Journal of Number Theory,
[9] C. Fieker and Y. Zhang, An application of the-
adic analytic class number formula, LMS Journal
of Computation and Mathematics 19(1), 217-228,
[10] A. Brumer, On the units of algebraic number
fields, Mathematika 14(2), 121-124, 1967.
[11] K. Iwasawa, On some modules in the theory of
cyclotomic fields, J. Math. Soc. Japan 16, 42-82,
[12] K. Iwasawa, On explicit formulas for the norm
residue symbol, J. Math. Soc. Japan 20, 151-165,
[13] J. Coates,p-adic L-functions and Iwasawa’s the-
ory, in Algebraic number fields (Durham sympo-
sium, 1975, ed. by A. Fr¨ohlich), 269-353, Academic
Press, London 1977.
[14] L.C. Washington, Introduction to cyclotomic
fields, Sprinjer Verlag, 1997.
[15] G. Georges, The p-adic Kummer-Leopoldt
constant-Normalized p-adic regulator, Interna-
tional Journal of Number Theory, 14(2), 329-337,
[16] B. Mazur, W. Stein and J. Tate, Compu-
tation of p-adic Heights and Log Convergence,
Mabud Ali Sarkar and Absos Ali Shaikh
Department of Mathematics,
The University of Burdwan
Burdwan-713104, India.
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Full-text available
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