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A Note on a result of Iwasawa
Bruno Angles, Thomas Herreng
To cite this version:
Bruno Angles, Thomas Herreng. A Note on a result of Iwasawa. 2005. �hal-00014967�
ccsd-00014967, version 1 - 1 Dec 2005
On a result of Iwasawa
Bruno Angl` es, Thomas Herreng
Universit´ e de Caen
Laboratoire de Math´ ematiques Nicolas Oresme CNRS UMR 6139
Boulevard Mar´ echal Juin B.P. 5186
14032 Caen Cedex FRANCE
E-mail: bruno.angles@math.unicaen.fr, thomas.herreng@math.unicaen.fr
Abstract
We recover a result of Iwasawa on the p-adic logarithm of principal units of Q
p(ζ
pn+1by studying the value at s = 1 of p-adic L-functions.
Key words: p-adic L-functions, Iwasawa theory, cyclotomic fields.
Math Subject classification: 11R23, 11R18.
Since the ninteenth century, it is known that values of L-functions at s = 1 contain deep arithmetic information. This result has much importance because it links analytic formulas with arithmetic invariants.
Kubota and Leopoldt have defined an analogous L-function on the p- adic fields with analytic techniques. Iwasawa has shown how to construct this p-adic L-function algebraically. Our aim in this paper is to give some algebraic interpretations to the analytic formulas giving the value at s = 1 of p-adic L-functions. It leads us to study some properties of the p-adic logarithm which enable us to establish the Galois module structure of the logarithm of principal units. The result obtained (theorem 1.10 below) had been discovered first by Iwasawa ([6]) in 1968 using explicit reciprocity laws. The first corollary we state to this theorem is an important result also due to Iwasawa ([5]) which gives the structure of the plus-part of the principal units modulo cyclotomic units.
In the second section we use the theorem 1.10 to study the minus part of the projective limit of these units. The aim is to obtain a theorem which looks like the one of Iwasawa ([5]). But in the minus-part there are no more cyclotomic units. Yet by considering the p-adic logarithm of principal units we are able to obtain an Iwasawa-like result (theorem 2.3) and its global counterpart.
The authors would like to thank Maja Volkov for her careful reading
of the paper and great improvements in the presentation.
For all the paper, fix an odd prime number p and denote by v
pthe normalized valuation at p. We are interested in cyclotomic p-extensions.
We start with θ an even character of Gal(Q(µ
pd)/Q) of conductor f
θ= d or pd with d ≥ 1 and p ∤ d. For n ≥ 0, let q
n= p
n+1d. For all integer m, we identify both groups
Gal(Q(µ
m)/Q) ≃ (Z/mZ)
×σ
a↔ a where σ
a(ζ
m) = ζ
ma.
Fix a primitive dth root of unity α. We note ζ
qn= αζ
pn+1with ζ
pn+1defined such that ζ
ppn+1= ζ
pnand ζ
p∈ µ
p\ { 1 } . Let F be the Frobenius endomorphism of Q
p(α)/Q
p(i.e. F (α) = α
p), and ∆ = Gal(Q(µ
pd)/Q).
Then F corresponds to σ
p∈ ∆. As θ ∈ ∆ we write b θ(F ) for θ(σ
p).
Let K
n= Q
p(µ
qn) and K
∞= S
n≥0
K
n. We know that Γ
n= Gal(K
n/K
0) ≃ Z/p
nZ and Γ = Gal(K
∞/K
0) ≃ Z
p. Moreover γ
0= σ
1+q0is a topological generator of Γ. We have the decomposition Gal(K
n/Q) ≃ ∆ × Γ
σ
a7→ (δ(a), γ
n(a)).
Finally, we fix some notations from Iwasawa theory. Let O be the ring of integers of Q
p(θ) and Λ = O [[T ]]. By a fundamental results of Iwasawa there exists a power series f (T, θ) ∈ Λ such that for all n ≥ 0, χ ∈ b Γ
n, and s ∈ Z
pL
p(s, θχ) = f (χ(γ
0)(1 + q
0)
s− 1, θ).
1 Value at s = 1 of p-adic L-functions
1.1 A Lemma on the Iwasawa algebra
Let Λ = Z
p[[T ]] be the Iwasawa algebra, g ∈ Λ a formal power series and ω
nthe element (1 + T )
pn− 1. It is known that the Iwasawa algebra can be described as follows. Let Γ be a multiplicative topological group iso- morphic to Z
p, Γ
nits quotient Γ/Γ
pn≃ Z/p
nZ, and Z
p[[Γ]] = lim
←−
Z
p[Γ
n].
Let γ
0be a topological generator of Γ and γ
0its image in Z
p[Γ
n].We thus have an isomorphism
Z
p[Γ
n] ≃ Z
p[T ]/((1 + T )
pn− 1) γ
07→ 1 + T
which, passing to the projective limit, gives an isomorphism Z
p[[Γ]] ≃ Z
p[[T ]]
γ
07→ 1 + T
(see for example [7]). Via this isomorphism the power series g can be writ- ten as a sequence of elements ǫ
n∈ Z
p[Γ
n] compatible with the restriction morphisms. The aim of this paragraph is to express ǫ
naccording to g.
We can write g as g(T ) =
p
X
n−1b=0
a
n(b)(1 + T )
b+ ω
n(T )Q
n(T )
with Q
n(T ) ∈ Λ. The canonical isomorphism described above implies that ǫ
n= P
pn−1b=0
a
n(b)γ
0b. Pick a character χ ∈ b Γ
n= Hom(Γ
n, Q
×p), and let e
χ=
p1nP
γ∈Γn
χ(γ)γ ∈ Q
p[Γ
n] be the associate idempotent. Thus
e
χǫ
n= e
χ(
p
X
n−1b=0
a
n(b)γ
0b)
= (
p
X
n−1b=0
a
n(b)χ(γ
0)
b).e
χ= g(χ(γ
0) − 1)e
χ.
Summing over all characters of Γ we deduce the following result.
Lemma 1.1. Let g be a power series in Z
p[[T ]] and ǫ
nits image in Z
p[T ]/ω
n(T )Z
p[T ]. Via the canonical isomorphism between Z
p[[T ]] and Z
p[[Γ]] the element ǫ
nin Z
p[Γ
n] writes as
ǫ
n= X
γ∈Γn
1 p
nX
χ∈bΓn
g(χ(γ
0) − 1)χ(γ)
γ
where γ
0is a topological generator of Γ.
1.2 Value of the p-adic L-function at s = 1 : alge- braic interpretation
Our aim is to give an algebraic interpretation of the following formula (which can be found in [7], theorem 5.18) :
Proposition 1.2. Let χ be an even nontrivial character of conductor f and ζ be a primitive fth root of unity. We have
L
p(1, χ) = − (1 − χ(p) p ) τ (χ)
f X
fa=1
χ(a) log
p(1 − ζ
a) where τ (χ) is the Gauss sum τ (χ) = P
fa=1
χ(a)ζ
a.
We use the notations of the introduction. Recall the well-known rela- tion for all n ≥ 0 and i ≤ n
N
Kn/Ki(ζ
Fqni−n− 1) = ζ
qi− 1. (1) Assume θ 6 = 1. By the results of the previous paragraph we know there exists (ǫ
n(θ))
n≥0∈ lim
←−
O [Γ
n] corresponding to f(
1+q1+T0− 1, θ) and that for all n ≥ 0 and χ ∈ b Γ
ne
χǫ
n= L
p(1, θχ)e
χ.
Remark. Assume θ 6 = 1. We have for all χ ∈ b Γ
nand n ≥ 0 L
p(1, θχ) 6 = 0. Thus for all n ≥ 0 ǫ
n∈ Frac( O )[Γ
n]
×.
Let χ ∈ Γ b
n. We assume that χ 6 = 1 if f
θ= d. Then the conductor of
the product θχ is f
θχ= q
kfor an integer 0 ≤ k ≤ n.
Lemma 1.3. We have 1
p
nθ(F
n−k)τ (θχ)L
p(1, θχ) = − X
δ∈∆
θ(δ)e
χlog
p(1 − ζ
qδn).
Proof. Start from the formula in proposition 1.2. Using the fact that τ (χ)τ (χ) = f for a character χ of conductor f we find
τ (θχ)L
p(1, θχ) = − X
δ∈∆,γ∈Γk
θ(δ)χ(γ) log
p(1 − ζ
qδγk).
Then using the equality (1) τ (θχ)L
p(1, θχ) = − X
δ∈∆,γ∈Γn
θ(δ)χ(γ) log
p(1 − ζ
δγFqk k−n)
= − X
δ∈∆,γ∈Γn
θ(δ)θ(F
n−k)χ(γ) log
p(1 − ζ
qδγn)
= − p
nθ(F
n−k) X
δ∈∆
θ(δ)e
χlog
p(1 − ζ
qδn) which proves the lemma.
Assume now that f
θ= pd and let ˙ T
n= P
ni=0
F
n−ip
i−nζ
qi. Proposition 1.4. For all n ≥ 0 we have
X
δ∈∆
θ(δ)ǫ
n(θ)( ˙ T
nδ) = − X
δ∈∆
θ(δ) log
p(1 − ζ
qδn).
Proof. Recall that for a character χ of second kind (i.e. whose conductor is f
χ= p
k+1) and an integer l 0 ≤ l ≤ n the χ-part e
χζ
pl+1vanishes if and only if l 6 = k (see for example [2]). Let χ be a character of Γ
n. We have al- ready shown that P
δ∈∆
θ(δ)e
χǫ
n(θ)( ˙ T
nδ) = P
δ∈∆
θ(δ)L
p(1, θχ)e
χ( ˙ T
nδ).
Moreover
e
χ( ˙ T
nδ) = 1 p
nX
γ∈Γn
p
k−nχ(γ)ζ
qδγFk n−kwhere k is such that f
χ= p
k+1when χ 6 = 1 and k = 0 when χ = 1. Thus we have 0 ≤ k ≤ n and f
θχ= q
k. Then
X
δ∈∆
e
χθ(δ)ǫ
n(θ)( ˙ T
nδ) = L
p(1, θχ) p
nX
δ∈∆
γ∈Γn
p
k−nθ(δ)χ(γ)ζ
qδγFk n−k= θ(F
n−k)L
p(1, θχ) p
nX
δ∈∆
γ∈Γn
p
k−nθ(δ)χ(γ)ζ
qδγk= θ(F
n−k)L
p(1, θχ) p
nX
δ∈∆
γ∈Γk
θ(δ)χ(γ)ζ
qδγk= 1
p
nθ(F
n−k)L
p(1, θχ)τ(θχ)
= − X
δ∈∆
θ(δ)e
χlog
p(1 − ζ
qδn).
The last equality follows from the previous lemma. Summing over all characters χ ∈ Γ b
nwe obtain the required equality.
Of course there is a similar equality in the case where f
θ= d instead of pd. However the result is slightly more complicated to state. We need the following lemma.
Lemma 1.5. If f
θ= d then
• P
δ∈∆
θ(δ) log
p(1 − ζ
qδ0) = (θ(F) − 1) P
y∈(Z/dZ)×
θ(y) log
p(1 − α
y) and
• τ (θ) = − P
δ∈∆
θ(δ)ζ
qδ0It follows that for all n ≥ 0, 1
p
nτ (θ)L
p(1, θ) = − (1 − θ(F) p ) X
δ∈∆
θ(δ)e
χ0log
p(1 − ζ
qδn)
where χ
0is the trivial character of Γ
n. We can now state the theorem which includes all the results of this section.
Theorem 1.6. Let θ be an even non-trivial character of Gal(Q(µ
pd)/Q) and (ǫ
n(θ))
n≥0∈ lim
←−
O [Γ
n] the element corresponding to f(
1+q1+T0− 1, θ).
We define T ˙
n=
P
ni=0
F
n−ip
i−nζ
qiwhen f
θ= pd
P
ni=1
F
n−ip
i−n(1 −
Fp)ζ
qi− p
−nF
n(f − 1)ζ
q0when f
θ= d.
We thus have for all n ≥ 0, X
δ∈∆
θ(δ)ǫ
n(θ)( ˙ T
nδ) = − E(θ) X
δ∈∆
θ(δ) log
p(1 − ζ
qδn)
where E(θ) is a kind of Euler factor : E(θ) = (1 −
θ(F)p) when f
θ= d and E(θ) = 1 when f
θ= pd.
1.3 An application
We now apply theorem 1 to the case where f
θ= pd. We slightly change our notations. For the rest of the paper we denote by ∆ the group (Z/pZ)
×, θ
1a character of ∆ and θ
2a character of conductor d with d dividing p − 1 such that θ = θ
1θ
2is even. We assume first that θ
16 = 1, ω. Both cases will be treated separately in the sequel. Let W = Z
p[θ
2] and (ǫ
n(θ))
n≥0∈ W [[T ]] the element corresponding to f(
1+pd1+T− 1, θ). Recall that, as usual, α is a primitive dth root of unity. In this situation we have the following result.
Lemma 1.7. Let T
n= P
ni=0
p
i−nζ
pi+1. Then τ (θ
2)ǫ
n(θ)e
θ1T
n= − X
y∈(Z/dZ)×
θ
2(y)e
θ1log
p(α
y− ζ
pn+1).
Proof. Theorem 1 can be restated as X
δ∈(Z/pdZ)×
θ(δ)ǫ
n(θ)( ˙ T
nδ) = − X
δ∈(Z/pdZ)×
θ(δ) log
p(1 − ζ
qδn)
where ˙ T
n= P
ni=0
p
i−nζ
qiand q
i= dp
i+1. A straightforward calculation gives
X
δ∈(Z/pdZ)×
θ(δ)ǫ
n(θ)( ˙ T
nδ) = X
δ∈(Z/pZ)× y∈(Z/dZ)×
θ
2(y)θ
1(δ)ǫ
n(θ)(
X
ni=0
p
i−nα
yζ
pδd+1)
= (p − 1) X
y∈(Z/dZ)×
θ
2(y)α
yǫ
n(θ)e
θ1( X
ni=0
p
i−nζ
pi+1)
= (p − 1)τ (θ
2)ǫ
ne
θ1T
n.
We need a relation between primitive characters and unprimitive ones which is given by the following lemma.
Lemma 1.8. Let θ
2be a character whose conductor is d
2with d
2| d. Then X
y∈(Z/dZ)×
θ
2(y)e
θ1log
p(α
y− ζ
pn+1) =
X
y∈(Z/d2Z)×
θ
2(y)e
θ1log
p(α
y2− ζ
pn+1)
x(θ)
where α
2= α
d/d2and x(θ) ∈ Z
p[θ].
Proof. It is sufficient to consider the case where d = ld
2with l a prime number. Let S be the left-hand side sum in lemma 1.8.
• First case : l 6 | d
2. For y ∈ (Z/d
2Z)
×we want to write the elements z of (Z/dZ)
×as z = y + kd
2for 0 ≤ k ≤ l − 1. Howerver for all y there exists a k
y, 0 ≤ k
y≤ l − 1, such that y + k
yd
2≡ 0 mod l (i.e.
y + k
yd
26∈ (Z/dZ)
×). We can write S as
S = X
y∈(Z/d2Z)×
θ
2(y)e
θ1log
pl−1
Y
k=0 k6=ky
(α
kd2+y− ζ
pn+1).
Note that α
d2is a lth root of unity. Since Q
ζ∈µl
(X − ζY ) = (X
l− Y
l) we get S
S = X
y∈(Z/d2Z)×
θ
2(y)e
θ1log
p
α
y2− ζ
pln+1α
y−ky d2 l
2
− ζ
pn+1
= (θ
1(l) − θ
2(σ
y+ky d2 l)) X
y∈(Z/d2Z)×
θ
2(y)e
θ1log
p(α
y2− ζ
pn+1).
• Second case : l | d
2. Then all z = y + kd
2with y ∈ (Z/d
2Z)
×and 0 ≤ k ≤ l − 1 are invertible modulo d. Thus
S = X
y∈(Z/d2Z)×
θ
2(y)e
θ1log
p(α
y2− ζ
pln+1)
= θ
1(l) X
y∈(Z/d2Z)×
θ
2(y)e
θ1log
p(α
y2− ζ
pn+1).
For a given d dividing p − 1 the two previous lemmas yield the equality τ (θ
2)x(θ
1θ
2)ǫ
n(θ
1θ
2)e
θ1T
n= − X
y∈(Z/dZ)×
θ
2(y)e
θ1log
p(α
y− ζ
pn+1). (2) Lemma 1.9. Let d be an integer dividing p − 1 and α a primitive dth root of unity. Fix θ
1∈ ∆ b different from 1, ω. Then there exists u
n∈ Z
p[Γ
n] such that
u
ne
θ1T
n= e
θ1log
p(α − ζ
pn+1).
Proof. Suppose first that θ
1is even. Summing the equality (2) over all the θ
2such that the product θ
1θ
2is even we obtain
X
fθ2|d θ2even
τ (θ
2)x(θ
1θ
2)ǫ
n(θ
1θ
2)
e
θ1T
n= − X
y∈(Z/dZ)×
X
fθ2|d θ2even
θ
2(y)
e
θ1log
p(α
y− ζ
pn+1).
We have
X
fθ2|d θ2even
θ
2(y) =
0 when y 6≡ ± 1 mod d
ϕ(d)
2
when y ≡ ± 1 mod d.
Hence the right-hand side of the above is equal to −
ϕ(d)2e
θ1log
p(α − ζ
pn+1)(α
−1− ζ
pn+1) . Since θ
1is even we have e
θ1log
p(α
−1− ζ
pn+1) = e
θ1log
p(α − ζ
p−1n+1) =
e
θ1log
p(α − ζ
pn+1). The right-hand side thus equals − ϕ(d)e
θ1log
p(α − ζ
pn+1). Set
u
n= − 1 ϕ(d)
X
fθ2|d θ2even
τ (θ
2)x(θ
1θ
2)ǫ
n(θ
1θ
2)
.
As u
nis Galois-invariant we have u
n∈ Z
p[Γ
n]. The result follows in this case.
When θ
1is odd, the proof runs the same except that X
fθ2|d θ2odd
θ
2(y) =
0 if y 6≡ ± 1 mod d
±
ϕ(d)2if y ≡ ± 1 mod d
and e
θ1log
p(α
−1− ζ
pn+1) = − e
θ1log
p(α − ζ
p−1n+1).
Theorem 1.10. Let θ ∈ ∆, b θ 6 = 1, ω. Let U
nbe the group of principal units of Q
p(ζ
pn+1) defined by U
n= 1 + (ζ
pn+1− 1)Z
p[ζ
pn+1] and T
n= P
ni=0
p
i−nζ
pi+1. Then for all n ≥ 0,
e
θlog
pU
n= Z
p[Γ
n]e
θT
n.
Proof. By lemma 1.9 there exists u
n∈ Z
p[Γ
n] such that u
ne
θT
n= e
θlog
p(α − ζ
pn+1). By the results of [7] section 13.8 there exists an in- teger d dividing p − 1 and a primitive dth root of unity α such that e
θlog
pU
n= Z
p[Γ
n]e
θlog
p(α − ζ
pn+1). Hence e
θlog
pU
n⊆ Z
p[Γ
n]e
θT
n. The converse inclusion will be proved in the following section (lemma 1.11 and proposition 1.12). Just note it implies u
n∈ Z
p[Γ
n]
×for such an α.
We now derive some corollaries from theorem 1.10, the first of which is a well-known result of Iwasawa.
Corollary 1 (Iwasawa). ([5] or [7] theorem 13.56) Let θ be an even nontrivial character in ∆. Let b (ǫ
n(θ))
n≥0∈ lim
←−
Z
p[Γ
n] be the element corresponding to the power series f(
1+T1+p− 1 , θ). Let C
nthe closure of the cyclotomic units in U
n. For all n ≥ 0 we have an isomorphism
e
θU
n/C
n≃ Z
p[Γ
n]/ǫ
n(θ)Z
p[Γ
n].
Proof. By basic results on cyclotomic units we have e
θlog
pC
n= Z
p[Γ
n]e
θlog
p(1 − ζ
pn+1).
Lemma 1.9 allpied to d = 1 shows that there exists u
n∈ Z
p[Γ
n] such that u
ne
θT
n= e
θlog
p(1 − ζ
pn+1). The definition of u
ngives u
n= − ǫ
n(θ). We deduce that log
pC
n= Z
p[Γ
n]ǫ
n(θ)e
θT
n. Moreover theorem 2 shows that log
pU
n= Z
p[Γ
n]e
θT
n. As θ is even and nontrivial we have isomorphisms
e
θU
n/C
n≃ e
θlog
pU
n/ log
pC
n≃ Z
p[Γ
n]/ǫ
n(θ)Z
p[Γ
n].
Corollary 2. Assume d ≥ 3. There exists an odd character θ
2whose conductor divides d such that there are at least √ p − 2 odd character θ
1∈ ∆, b θ
16 = ω satisfying
f(T, θ
1θ
2) ∈ W [[T ]]
×.
Proof. By a result of Angl`es ([1], theorem 5.4) there are at least √ p − 2 odd characters θ
1∈ ∆, such that b θ 6 = ω and for all n ≥ 0,
e
θ1log
pU
n= e
θ1log
p(α − ζ
pn+1).
For such a character we know that u
n= X
fθ2|d θ2odd
τ(θ
2)x(θ
1θ
2)ǫ
n(θ
1θ
2) ∈ Z
p[Γ
n]
×.
Therefore there exists at least one θ
2such that ǫ
0(θ
1θ
2) ∈ W
×. Since θ
1θ
2is a character of the first kind it follows that f(T, θ
1θ
2) ∈ W [[T ]]
×.
A similar proof yields the following result.
Corollary 3. Let θ
1∈ ∆, b θ
16 = 1, ω. There exists θ
2whose conductor divides p − 1 such that θ
1θ
2is even and the generalized Bernoulli number B
1,θ1θ2ω−1is rpime to p
1.4 Some index computations
Let us recall some notations. As usual p is an odd prime number and O
nis the ring of integers of Q
p(ζ
n+1p). Let T
nbe P
ni=0
p
i−nζ
pi+1, e
0=
1 pn
P
γ∈Γn
γ and for 1 ≤ d ≤ n
e
d= X
χ∈bΓn χof conductorpd+1
e
χ.
We first want to compute the index [
p1ne
θO
n: Z
p[Γ
n]e
θT
n] with θ ∈ ∆. b Let Λ
nthe maximal order of Q
p[Γ
n]. Then Λ
n[∆] is the maximal order of Q
p[∆ × Γ
n] and the Leopoldt theorem (see [2]) states that O
nis a free Λ
n[∆]-module generated by T
n= P
ni=0
ζ
pi+1. We can therefore substitute
p1ne
θO
nby Λ
ne
θP
ni=0 ζpi+1
pn
in our calcu- lations. The index [
p1ne
θO
n: Z
p[Γ
n]e
θT
n] with θ ∈ ∆ is the product b
[Λ
ne
θX
ni=0
ζ
pi+1p
n: Λ
ne
θX
ni=0
ζ
pi+1p
n−i].[Λ
ne
θT
n: Z
p[Γ
n]e
θT
n].
By a standard calculation of discriminants we have [Λ
ne
θT
n: Z
p[Γ
n]e
θT
n] = [Λ
n: Z
p[Γ
n]] = p
pn−1p−1. In order to find [Λ
ne
θP
ni=0 ζpi+1
pn
: Λ
ne
θP
n i=0ζpi+1 pn−i
] we notice that Λ
n= L
nd=0
Z
p[γ
n]e
d. Thus [Λ
ne
θX
ni=0
ζ
pi+1p
n: Λ
ne
θX
ni=0
ζ
pi+1p
n−i] = Y
ni=0
[Z
p[Γ
n]e
iζ
pi+1p
n: Z
p[Γ
n]e
iζ
pi+1p
n−i]
= Y
i
p
i(pi−pi−1),
because the Z
p-rank of Z
p[Γ
n]ζ
pi+1is p
i− p
i−1. Moreover we have the equality
P
ni=0
d(p
i− p
i−1) = np
n−
pp−1n−1.
We have thus shown the following result.
Lemma 1.11. Let θ ∈ ∆. We have b [
p1ne
θO
n: Z
p[Γ
n]e
θT
n] = p
npn. The next index we want to calculate is [
p1ne
θO
n: e
θlog
pU
n].
Proposition 1.12. Let θ ∈ ∆. We have b
[ 1
p
ne
θO
n: e
θlog
p(U
n)] =
p
npnwhen θ 6 = 1, ω
p
npn+1when θ = 1
p
npn+n+1when θ = ω.
Proof. The proof is based on a result of John Coates, see [3]. We first show the inclusion log
pU
n⊆
p1nO
n. Define π
n= ζ
pn+1− 1 and let u ∈ U
n. Then there exists a ∈ Z such that
uζ
pn+1= 1 mod π
n2.
Therefore log
pU
n= log
p(1 + π
2nO
n). Moreover we have u ≡ 1 mod π
n2implies v
p(u
pn− 1) ≥
p−12for u ∈ U
n. Then by lemma 5.5 of [7] we have log
pU
n⊆
p1nO
n.
We now calculte the index. We introduce the group V = 1 + p O
n. By standard properties of the p-adic logarithm we know that log
pV = p O
n. Then
[ 1
p
ne
θO
n: e
θlog
pV ] = [ 1
p
ne
θO
n: pe
θO
n] = p
pn(n+1). It remains to compute [e
θlog
pU
n: e
θlog
pV ]. Consider the morphism
(1 + π
2nO
n)/V −→
logplog
pU
n/ log
pV,
the kernel of which consists of the roots of unity. Then for θ 6 = ω we have e
θlog
pU
n/ log
pV ≃ e
θ(1 + π
2nO
n)/V =
pn(p−1)−1
Y
i=2
e
θ1 + π
inO
n1 + π
ni+1O
n. For all integers i ≥ 1 we have an isomorphism of Z
p-modules
1+πinOn
1+πi+1n On
−→ O
n/π
nO
n≃ F
p1 + xπ
ni7→ x mod π
n. Hence the θ-part of
1+πinOn1+πni+1On
is F
por { 0 } . Moreover for all δ ∈ ∆ we have x
δ≡ x mod π
n. Thus for θ ∈ Delta \ we have
(1 + xπ
ni)
eθ≡ Y
δ∈∆
(1 + 1
p − 1 θ(δ)x(π
in)
δ) mod (π
ni+1).
However (π
ni)
δ≡ ω
i(δ)π
inmod (π
ni+1) which yields (1 + xπ
ni)
eθ≡ 1 − ( X
δ∈∆
θ(δ)ω
i(δ))xπ
inmod (π
ni+1).
Thus for θ = ω
kwe get e
θ1 + π
inO
n1 + π
ni+1O
n=
0 when i 6≡ k mod (p − 1) F
pwhen i ≡ k mod (p − 1),
and [e
θ(1 + π
2nO
n) : e
θ(1 + p O
n)] = p
C(k)where C(k) is the number of integers i such that 2 ≤ i ≤ p
n(p − 1) − 1 and i ≡ k mod (p − 1). When k 6 = 0, 1 (i.e. θ 6 = ω, 1) we have C(k) = p
nand when k = 0, 1 we have C(k) = p
n− 1. Hence the result for θ 6 = ω.
The case θ = ω is similar except that ζ
p, · · · , ζ
pnare in the kernel of
the morphism log
p: (1 + π
n2O
n)/V −→ log
pU
n/ log
pV .
This completes the proof of theorem 1.10. We might want to refor- mulate it according to Leopoldt’s element T
n= P
ni=0
ζ
pi+1. Let l
n= P
ni=0
p
n−ie
d∈ Q
p[Γ
n]. This element is constructed to satisfy the identity l
nT
n= T
n. Notice that l
n◦ ( P
ni=0
p
i−ne
d) = 1 so that l
n∈ Q
p[Γ
n]
×. Let L
n= l
n◦ log
p.
Corollary 4. Let θ ∈ ∆with b θ 6 = 1, ω. Then, e
θL
nU
n= Z
p[Γ
n]e
θT
n.
1.5 The case of the Teichm¨ uller character
For technical reasons the results in this section are only valid when p ≥ 5.
Let us begin with the following proposition.
Proposition 1.13. There exists α ∈ µ
p−1\ {± 1 } such that for all n ≥ 0, e
ωlog
pU
n= Z
p[Γ
n]e
ωlog
p(α − ζ
pn+1).
Proof. A careful reading of the proof of theorem 13.54 in [7] shows that lim
←−n≥0e
ωlog
pU
nis a free Λ-module of rank 1 and that when (ǫ
n)
n≥0∈ lim
←−n≥0e
ωlog
pU
nis such that e
ωlog
pU
0= Z
p[ǫ
0] then for all n ≥ 0,
e
ωlog
pU
n= Z
p[Γ
n]ǫ
n.
Therefore it is sufficient to show there exists α ∈ µ
p−1, α 6 = ± 1 such that e
ωlog
pU
0= Z
pe
ωlog
p(α − ζ
p). (3) Let us prove (3). Recall that e
ωlog
pU
0= e
ωπ
2nO
0= p τ (ω
−1)Z
p. Let π denote the element ζ
p− 1.
Lemma 1.14. Let x ∈ π
2Z
p[ζ
p]. Then log
p(1+x) ≡
(1+x)pp−1mod (pπ
2).
Proof. We have the congruences
• for all n ≥ p, v
π(
xnn) ≥ p + 1 and
• for 1 ≤ k ≤ p − 1, (
pk)
p
≡
(−1)kk+1mod p.
We then have
log
p(1 + x) = P
n≥1 (−1)n+1
n
x
n≡ P
p−1 n=1(
np)
p
x
nmod (pπ
2)
≡
(1+x)pp−1−xpmod (pπ
2)
≡
(1+x)pp−1mod (pπ
2), and the lemma is proved.
Lemma 1.15. Let α ∈ µ
p−1\ { 1 } and θ ∈ ∆, b θ 6 = 1. We have e
θlog
p(α − ζ
p) ≡ − θ( − 1)
ω(α − 1) τ(θ)
p−1
X
k=1
( − 1)
kp k
p θ(k)α
k!
mod (pπ
2).
Proof. Let γ =
ω(α−1)α−ζpζ
pω(α−1). Notice that log
p(γ) = log
p(α − ζ
p) and γ ≡ 1 mod (π
2), which allows us to apply lemma 1.14. We get
log
p(α − ζ
p) ≡
(α−ζp)p ω(α−1)
− 1
p mod (pπ
2).
Taking the θ-part and expanding the sum yields the required result.
Since e
ωlog
pU
0= p τ (ω
−1)Z
pand v
π(τ(ω
−1)) = 1 we deduce from lemma 1.15 the following result.
Lemma 1.16. Let α ∈ µ
p−1\ { 1 } . The element e
ωlog
p(α − ζ
p) is a generator of e
ωlog
pU
0if and only if
p−1
X
k=1
( − 1)
kp k
p ω(k)α
k6≡ 0 mod (p
2).
Lemma 1.17. There exists α ∈ µ
p−1\{± 1 } such that P
p−1k=1
( − 1)
k(
pk)
p
ω(k)α
k6≡
0 mod (p
2).
This is precisly what we need to complete the proof of proposition 1.13.
Proof. Let us consider the polynomial P(X) = P
p−1k=1
( − 1)
k(
pk)
p
ω(k)X
k∈ Z
p[X ]. We know that
pk/p ≡ ( − 1)
k+1/k mod p and thus P (X) ≡ − X Y
α∈µp−1\{1}
(X − α) mod p.
We can therefore use Hensel’s lemma which shows that P (X) = − X Q
α∈µp−1\{1}
(X − a(α)) where a(α) ∈ Z
×pand a(α) ≡ α mod p. Notice that P ( − 1) ≡ 0
mod (p
2).
Let us assume the lemma is false. Then for all α ∈ µ
p−1\ { 1 } , we have a(α) ≡ α mod p
2. Thus P (X) ≡ − X
Xp−1X−1−1mod p
2. Comparing with the expression of P (X ) we deduce that for all k ∈ { 1, · · · , p − 1 } , we have
( − 1)
kp k
p ω(k) ≡ − 1 mod p
2.
Let us apply this congruence with k = 2 and k = 4. The assump- tion p ≥ 5 is essenial in what follows. We obtain on the one hand (p − 1)ω(2)/2 ≡ − 1 mod p
2, that is to say ω(2)/2 ≡ p + 1 mod p
2. On the other hand we have ω(4)/4 ≡ 1 + (11/6)p mod p
2. But ω(4)/4 = (ω(2)/2)
2≡ 1 + 2p mod p
2. This implies 11/6 ≡ 2 mod p
2which is not possible.
This finishes the proof of proposition 1.13.
In order to apply some of the results of section 1.2, we let θ
2be an odd character of conductor d with d dividing p − 1. Recall that
τ (θ
2)ǫ
n(θ
2ω)e
ωT
n= − X
y∈(Z/dZ)×
θ
2(y)e
ωlog
p(α
y− ζ
pn+1) where α is a primitive dth root of unity and T
n= P
ni=0
p
i−nζ
pi+1. Since θ
2(p) = 1 we have for all characters χ and all integers n ≥ 1,
L
p(1 − n, χ) = − (1 − χω
−n(p)p
n−1) B
n,χω−nn
from which we deduce immediatly that L
p(0, θ
2ω) = 0. Thus if f(T, θ
2ω) denotes the power series associated to L
p(s, θ
2ω) we have the factorization f(T, θ
2ω) = T h(T, θ
2ω). Then there exists H(T, θ
2ω) ∈ W [[T ]] such that
f( 1 + pd
1 + T − 1, θ
2ω) = (T − pd)H(T, θ
2ω).
As usual T corresponds to σ
1+pd− 1 ∈ W [[ Gal(K
∞/K
0)]]. Let ǫ e
nbe the element of lim
←−
W [Γ
n] associated to H via the isomorphism in lemma 1.1.
We have
ǫ
n(θ
2ω)e
ωT
n= ǫ e
n(θ
2ω)(σ
1+pd− 1 − pd)e
ωT
n(4)
= ǫ e
n(θ
2ω)(σ
1+ps− (1 + p)
s)e
ωT
n, (5) where s = log
p(1 + pd)/ log
p(1 + p) ∈ Z
×p. So we have
σ
s1+p− (1 + p)
s= (σ
1+p− 1 − p).u
with u ∈ Z
p[Γ
n]
×. The same kind of calculation than in the general case shows that there exists a unit u
n∈ Z
p[Γ
n] such that
u
n(γ
0− 1 − p)e
ωT
n= e
ωlog
p(α − ζ
pn+1) where γ
0= σ
1+pand α is a primitive dth root of unity.
Lemma 1.18. We have
[Z
p[Γ
n]e
ωT
n: Z
p[Γ
n](γ
0− 1 − p)e
ωT
n] = p
npn+n+1. Proof. By lemma 1.11 we have [
p1ne
ωO
n: Z
p[Γ
n]e
ωT
n] = p
npn. Also
[Z
p[Γ
n]e
ωT
n: Z
p[Γ
n](γ
0− 1 − p)e
ωT
n] = [Λ/ω
nΛ : (T − p)Λ/ω
nΛ]
= [Λ : (ω
n, T − p)]
where Λ ≃ Z
p[[T ]] is the Iwasawa algebra. The Λ-module M = Λ/(T − p) has no Z
p-torsion. Its characteristic polynomial is T − p and it is a standard result that | M/ω
nM | = Q
ζ∈µpn
(ζ − p − 1). Then [Λ : (ω
n, T − p)] ∼ p Y
ζ∈µpn ζ6=1
(ζ − 1) ∼ p
n+1where ∼ means ’has same p-adic valuation as’.
We have thus established the following result.
Theorem 1.19. Let p ≥ 5 be a prime number. We have e
ωlog
pU
n= Z
p[Γ
n](γ
0− 1 − p)e
ωT
nwhere T
n= P
ni=0
p
i−nζ
pi+1. With the notations ofsection 1.4, this writes as
e
ωL
nU
n= Z
p[Γ
n](γ
0− 1 − p)e
ωT
nwhere T
n= P
n i=0ζ
pi+1.
1.6 The case of the trivial character
The main difference between the trivial character and the other ones is that the power series associated to L
phas not integral coefficients. Thus We have to work with the power series g(T ) defined by g(T ) = (1 −
1+q0
1+T
)f(T, 1) ∈ Λ. See [7], chapter 7 for more details on this particularly proposition 7.9. In our situation we have q
0= p. Let h(T ) = g(
1+T1+p− 1) =
− T f(
1+T1+p− 1, 1) and η
n∈ lim
←−
Z
p[Γ
n] be the element corresponding to h.
This means that we have, as in section 1.1, for all χ ∈ b Γ
ndifferent from 1,
e
χη
n= (1 − χ(γ
0))L
p(1, χ)e
χ. Let T
∆= P
δ∈∆
δ the norm element of Z
p[Gal(K
0)/Q].
Proposition 1.20. Let E
nbe the group of units in K
n, E
nits closure in U
nand T f
n= P
ni=1
p
i−nζ
pi+1(note that the sum begins at i = 1). We have
T
∆log
pE
n= Z
p[Γ
n]T
∆T f
n.
Proof. Let the field B
n∈ Q(ζ
pn+1) be such that [B
n: Q] = p
n. Iwasawa has showed [4] that p does not divide the class number of B
n. Thus [7], theorem 8.2 shows that
T
∆log
pE
n= T
∆log
pC
nwhere C
nis the closure of C
nin U
n.
Moreover we have T
∆log
pC
n= Z
p[Γ
n]T
∆log
p((1 − ζ
pn+1)
γ0−1). Now fix a character χ ∈ Γ b
ndifferent from 1 whose conductor is f
χ= p
k+1with 1 ≤ k ≤ n. We have
1
p
nτ (χ) = e
χT
∆T f
nand then
1
p
n(1 − χ(γ
0))τ (χ)L
p(1, χ) = e
χη
nT
∆T f
n. A little calculation gives
τ (χ)L
p(1, χ) = − X
δ∈∆
γ∈Γd
χ(δ) log
p(1 − ζ
pδγd+1)
= − p
ne
χT
∆log
p(1 − ζ
pn+1).
This proves that for all χ ∈ Γ b
n, χ 6 = 1, we have the following equality e
χT
∆log
p((1 − ζ
pn+1)
γ0−1) = − e
χT
∆T f
n.
We check that this equality also holds when χ = 1 and summing over all the characters gives the required result.
Lemma 1.21. We have [T
∆1
p
nO
n: Z
p[Γ
n]T
∆(pζ
p+ T f
n)] = p
npn+n+1.
Proof. We already know that [Λ
n: Z
p[Γ
n]] = p
pn−1p−1where Λ
nis the maximal order of Q
p[Γ
n]. It remains to calculate [
p1nT
∆O
n: Λ
nT
∆(pζ
p+
f
T
n)]. Decompose into characters yields [ 1
p
nT
∆O
n: Λ
nT
∆(pζ
p+ T f
n)] = Y
nd=0
[e
d1
p
nT
∆O
n: e
dΛ
nT
∆(pζ
p+ T f
n)].
For 1 ≤ d ≤ n we have e
d 1pn
T
∆O
n= Z
p[Γ
n]T
∆(p
−nζ
pd+1) and e
dΛ
nT
∆(pζ
p+ f
T
n) = Z
p[Γ
n]T
∆(p
d−nζ
pd+1). Then when d 6 = 0 we have [e
d 1pn
T
∆O
n: e
dΛ
nT
∆(pζ
p+ T f
n)] = p
d(pd−pd−1). For e
0we notice that e
0T
∆ 1pn
O
n=
1
pn
Z
pand e
0T f
n= 0. Then we have [e
0 1pn
T
∆O
n: e
0Λ
nT
∆(pζ
p+ T f
n)] = [
p1nO
n: pZ
p] = p
n+1. The lemma is now proved.
Recall that proposition 1.12 gives the index [ 1
p
nT
∆O
n: T
∆log
pU
n] = p
npn+1.
Moreover note that pζ
p∈ log
pU
nso that T
∆(pζ
p+ T f
n) ∈ T
∆log
pU
n. However the value of the index furnished by proposition 1.12 implies that the Z
p[Γ
n]-module generated by T
∆(pζ
p+ T f
n) cannot be equal to T
∆log
pU
n. Let us define M = Z
p[Γ
n]T
∆pζ
p+ Z
p[Γ
n]T
∆T f
n. We check that M ⊆ T
∆log
pU
n.
Now we want to calculate the index [ M : Z
p[Γ
n]T
∆(pζ
p+ T f
n)]. Note that T
∆pζ
p6∈ Z
p[Γ
n]T
∆(pζ
p+ T f
n). From
p
nT
∆pζ
p= p
ne
0(T
∆(pζ
p+ T f
n)) ∈ Z
p[Γ
n]T
∆(pζ
p+ T f
n)
we deduce that the index is less or equal to p
n. Let p
hthe order of T
∆pζ
p. Then p
hT
∆pζ
p= u
nT
∆(pζ
p+ T f
n) with u
n∈ Z
p[Γ
n]. This implies that p
he
0∈ Z
p[Γ
n] and h ≥ n. Thus we have [ M : Z
p[Γ
n]T
∆(pζ
p+ T f
n)] = p
nwhich proves the following result.
Theorem 1.22. The Z
p[Γ
n]-module T
∆log
pU
nis generated by T
∆T f
nand p.
Corollary 5. The Z
p[Γ
n]-module T
∆U
nis generated by T
∆E
nand (1+p).
In particular, let U
n′= { u ∈ U
n| N
Kn/Qp(u) = 1 } . Then T
∆log
pU
n′=
T
∆log
pE
n= T
∆log
pC
n.
2 A result ` a la Iwasawa in the odd part
We recall that K
n= Q(ζ
pn+1). Let π
n= ζ
pn+1− 1. In this section ǫ
n(θ) is the element of lim
←−
Z
p[Γ
n] corresponding to the power series f(
1+T1− 1, ωθ) where θ is an odd character of ∆. We assume that θ 6 = ω
−1.
2.1 The main result
Lemma 2.1. Let χ be an odd character in ∆, with b χ 6 = ω
−1. We have e
χ 1πn
∈ Z
p[ζ
pn+1].
Proof. It is enough to show that π
ne
χ 1πn
≡ 0 mod π
n. A straightforward calculation gives
(p − 1)π
ne
χ1 π
n= X
σ∈∆
χ(σ)π
nσ(π
n)
≡ X
σ∈∆
ω
−1(σ)χ(σ) mod π
n≡ 0 mod π
nbecause χ 6 = ω
−1, which proves the lemma.
Recall that
ǫ
n(χ) = 1 p
n+1X
1 ≤ a ≤ p
n+1p 6 | a
aθω
−1(a)γ
n(a)
where γ
n(a) is as defined in the introduction.
Lemma 2.2. Let T
n∈ K
nbe the sum T
n= P
nd=0
ζ
pd+1. We have e
χ1
π
n= ǫ
n(χ)e
χT
n. Proof. Let f(X) =
Xpn+1−1
X−1
. It is easily checked that Xf
′(X) =
pn+1
X
−1k=1
kX
kand (X − 1)Xf
′(X ) + Xf (X ) = p
n+1X
pn+1. Letting X = ζ
pn+1we obtain
1 π
n= 1
p
n+1pn+1
X
−1k=1
kζ
pkn+1= 1
p
n+1p
X
n+1a=1 p6|a
aζ
pan+1+ 1 p
np
X
n−1k=1
kζ
pkn.
By induction we get e
χ 1πn
= P
nk=1
ǫ
k(χ)e
χ(ζ
pk+1). For m ≤ n the re-
striction of ǫ
n(χ) to K
mis ǫ
m(χ). The lemma follows.
We now deal with the case where χ = ω. Consider the power series g(T ) = (1 −
1+T1+p)f(T, 1). We have
g( 1
1 + T − 1) = (1 − (1 + p)(1 + T ))f( 1
1 + T − 1, 1).
Let (ǫ
n)
n≥0correspond to g(
1+T1− 1). The same calculation as in lemma 2.3 shows that for all n ≥ 0
(1 − (1 + p)γ
0)e
ω−11 π
n= ǫ
ne
ω−1T
n. Computing the sum P
χodd
ǫ
ne
χ(T
n− T
n) and using theorem 1.10 we easily obtain the following result.
Theorem 2.3. There exists ν
∞= (ν
n)
n≥0∈ lim
←−
U
n−= U
∞−such that each ν
n∈ U
n−is unique modulo µ
pn+1and
log
pν
n= X
ni=0
p
i−ne
d( 1 π
n− 1
π
n− 2(1 + p)γ
0e
ω−11 π
n).
For all odd characters χ ∈ ∆ b \ { ω } we have e
χU
∞−/Λe
χν
∞≃ Λ/ f( e 1
1 + T − 1, ωχ)Λ
where f e (
1+T1− 1, ωχ) = f(
1+T1− 1, ωχ) when χ 6 = ω
−1and f( e
1+T1− 1, ωχ) = (1 − (1 + p)(1 + T ))f(
1+T1− 1, 1) when χ = ω
−1.
For χ = ω we have
e
χlog
pU
∞−/Λe
χlog
pν
∞≃ Λ/f( 1
1 + T − 1, ω
2)Λ.
2.2 A result ` a la Stickelberger
The aim of this section is to obtain a global result from theorem 2.3 which is of local nature. In order to achieve this, set
ǫ
n= 1 p
n+1X
1 ≤ a ≤ p
n+1p 6 | a
a σ
a∈ Q[G
n]
where G
n= ∆ × Γ
n. This element looks like the Stickelberger one and our computation are inspired by this analogy. Notice that the restriction of ǫ
n+1to K
nis not ǫ
nbut ǫ
n+ (p − 1)/2N
nwhere N
nis the norm element of the group algebra Z[G
n]. We should thus consider the element (j − 1)ǫ
nwhere j is the complex conjugation which is compatible with the canonical morphisms Z[G
n+1] → Z[G
n]. By lemma 2.2 we have
1 πn
= P
nk=0
ǫ
k(ζ
pn+1). We deduce that (j − 1)
π1n= (j − 1)ǫ
nT
n. Let θ
n=
π1n
−
π1n. Define the ideal I of Z[G
n] by I = Z[G
n](σ
c− c
∗) where c
∗is the inverse of c modulo p
n+1. However contrary to the standard case (the one of the Stickelberger element), I is not the order associated to
π1n
. We also define the ideal I = Z[G
n]ǫ
n∩ Z[G
n].
We need two more ideals.
E
n= Z[G
n]T
nand C
n= Iθ
nNotice that C
n= C
−n. The aim of this section is to show the following result.
Theorem 2.4. We have [ E
−n: C
n] = 2
|Gn2|−1· h
−pn+1where h
−pn+1is the quotient of the class number of K
nby the class number of the maximal real subfield of K
n.
Proof. We need several lemmas.
Lemma 2.5. We have I = Iǫ
n.
Proof. We want to show that for all β ∈ Z[G
n], βǫ
n∈ Z[G
n] is equivalent to β ∈ I. Let β ∈ I. We can assume that β = (σ
c− c
∗). Then
(σ
c− c
∗) = X
a
{ a
p
n+1} σ
ac− c
∗{ a p
n+1} σ
a= X
a
( { ac
∗p
n+1} − c
∗{ a p
n+1} )σ
a.
As cc
∗≡ 1 mod p
n+1we obtain (σ
c− c
∗)ǫ
n∈ Z[G
n].Conversely, notice first that p
n+1= p
n+1− 1 + σ
1+pn+1∈ I. Let β = P
a
x
aσ
awith x
a∈ Z and assume that βǫ
n∈ Z[G
n]. Then
β · ǫ
n= X
a
X
c
x
a{ c p
n+1} σ
ac= X
b
X
a
x
a{ a
∗b p
n+1} σ
b. When b = 1 our assumption implies that p
n+1| P
a
x
aa
∗so that P
a
x
aa
∗∈ I. Finally we get
β = X
x
aa
∗= X
x
a(σ
a− a
∗) + X
x
aa
∗∈ I.
Let us return to the proof of theorem 2.4. On the one hand, by Leopoldt’s theorem we know that E
nis a free Z[G
n]-module of rank one.
We thus have E
−n= Z[G
n]
−· T
n. A straightforward calculation shows that Z[G
n]
−= (j − 1)Z[G
n]. On the other hand, we have θ − n = (j − 1)ǫ
nT
n(see beginning os this section). Then we get
C
n= (j − 1) I ǫ
nT
n= (j − 1)(I)T
nIn order to complete the proof of the theorem we have to express I
−according to (j − 1) I and to compute the index of I
−in Z[G
n]
−. By definition we have
I
−= I ∩ Z[G
n]
−= Z[G
n] · ǫ
n∩ Z[G
n]
−.
The following result resembles a theorem by Iwasawa (see [7] theorem
6.19).
Proposition 2.6. We have [ Z[G
n]
−: I
−] = h
−pn+1.
Proof. The proof is the same as the one of Iwasawa’s theorem. Let us recall the main steps : first complete and then worke at each prime. We define the ideal I
q= Z
q[G
n] I for a prime q. We have the following results.
Lemma 2.7. 1. Z
q[G
n]
−= (1 − j)Z
q[G
n] 2. I
q= Z
q[G
n] · ǫ
n∩ Z
q[G
n]
3. I
q−= I
q∩ Z
q[G
n] = Z
q[G
n] · ǫ
n∩ Z
q[G
n]
−4. I
q−= I
−· Z
q5. When p 6 = q, I
q= Z
q[G
n]ǫ
nThe proof runs just as in [7], lemma 6.20. By lemma 2.7 we have an isomorphism Z
q[G
n]
−/ I
q−≃ (Z[G
n]
−/ I
−) ⊗ Z
q. It is enough to show the following result.
Proposition 2.8. The index [ Z
q[G
n]
−: I
q−] is the q-part of h
−pn+1for all primes q.
Proof. Assume to begin with that q 6 = 2, p. Then (1 ± j)/2 ∈ Z
q[G
n] so we can separate the plus-part and the minus-part. We obtain
I
q−= 1 − j
2 I = Z
q[G
n]
−· ǫ
nWe have to calculate the index [ Z
q[G
n]
−: Z
q[G
n]
−· ǫ
n] which equals to the q-part of the determinant of the map
ϕ : Z
q[G
n]
−→ Z
q[G
n]
−x 7→ x ǫ
n.
Compute this determinant in Q
q[G
n]
−= ⊕
χodde
χQ
q[G
n]. However we have e
χσ = χ(σ)e
χfor all σ ∈ G from which we deduce that
e
χǫ
n= B
1,χe
χ. Then we have
[ Z
q[G
n]
−: I
q−] = q − part of det(ϕ) = q − part of Y
χodd
B
1,χ= q − part of h
−pn+1.
Let us now deal with the case p = 2. The trick is to modify ǫ
nand to define ˜ ǫ
n= ǫ
n− 1/2N where N is the norm element of the group algebra Z[G
n]. We easily check that
1−j2e ǫ
n= ˜ ǫ
nso that ˜ ǫ
n∈ Q
2[G
n]
−.
Lemma 2.9. We have 1. I
−2
⊆ Z
2[G
n]˜ ǫ
n2. [ Z
2[G
n]˜ ǫ
n: I
−2
] = 2
Proof. The first part is obvious. For the second one, notice that if x ∈ Z
2[G
n] then either x ˜ ǫ
n∈ Z
2[G
n] or x ˜ ǫ
n− ǫ ˜
n∈ Z
2[G
n], and that Z
2[G
n]˜ ǫ
n∩ Z
2[G
n] = I
2.
The end of the proof for p = 2 runs the same as previously except that the map ϕ is now the multiplication by ˜ ǫ
n. Moreover we have e
χ˜ ǫ
n= e
χǫ
nwhen χ is odd. Then we get
[ Z
2[G
n]
−: Z
2[G
n]
−˜ ǫ
n] = 2 − part of det ϕ = 2
(1/2)|Gn|−1(2 − part of h
−pn+1).
Note that
Z
2[G
n]
−˜ ǫ
n= (1 − j)Z
2[G
n]˜ ǫ
n= Z
2[G
n](2˜ ǫ
n) = 2Z
2[G
n]˜ ǫ
n. It follows that [ Z
2[G
n]˜ ǫ
n: Z
−2[G
n]˜ ǫ
n] = 2
Z2−rank ofZ2[Gn]−= 2
(1/2)|Gn|. Together with lemma 2.6 this yields the required result
[ Z
2[G
n]
−: I
2−] = 2 − part of h
−pn+1.
It remains to deal with the case where q = p. Let us consider again the element ˜ ǫ
n= ǫ
n−
12N. Notice that as in [7] we have the equivalence x˜ ǫ
n∈ Z
p[G
n]
−⇐⇒ xǫ
n∈ Z
p[G
n] for x ∈ Z
p[G
n]. The equality [ Z
p[G
n]˜ ǫ
n: Z
p[G
n]˜ ǫ
n∩ Z
p[G
n]
−] = p
nfollows. However we also have
Z
p[G
n]˜ ǫ
n∩ Z
p[G
n]
−= I
p−and Z
p[G
n]˜ ǫ
n= Z
p[G
n]
−ǫ
n, thus
[ Z
p[G
n]
−ǫ
n: I
p−] = p
n. We define the map
ϕ : Z
p[G
n]
−→ Z
p[G
n]
−, x 7→ p
nǫ
nx.
Then [ Z
p[G
n]
−: p
nZ
p[G
n]
−ǫ
n] = p
(n/2)|Gn|(
p1n)(p − part of h
−pn+1). We obtain
[ Z
p[G
n]
−: I
p−] = p − part of h
−pn+1which concludes the proof of proposition 2.8.
Proposition 2.6 is now proved.
Lemma 2.10. The index [ I
−: (1 − j) I ] equals [ I
2−: (1 − j) I
2] = 2
|Gn2 |−1.
Proof. Localize at each prime by applying part 4 of lemma 2.7. Then [ I
−: (1 − j) I ] = Y
p
[ I
p−: (1 − j) I
p]
= [ I
2−: (1 − j) I
2] · [ I
p−: (1 − j) I
p] · Y
q6=2,p
[ I
q−: (1 − j) I
q].
When q 6 = 2, p lemma 2.7 implies [ I
q−: (1 − j) I
q] = 1. Moreover
1−j
2
∈ Z
p[G
n] so I
p−=
(1−j)2I
pand [ I
p−: (1 − j) I
p] = 1. It remains to calculate the index at q = 2. Notice that (1 − j) I
2= (1 − j)Z
2[G
n]ǫ
n= 2Z[G
n]˜ ǫ. We have [Z
2[G
n]˜ ǫ
n: I
−2
