On some p-adic power series attached to the arithmetic of $\mathbb Q(\zeta_p).$

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On some p-adic power series attached to the arithmetic of Q(ζ _p).

Bruno Angles

To cite this version:

Bruno Angles. On some p-adic power series attached to the arithmetic of Q(ζ_p).. 2005. �hal-

00011266�

ccsd-00011266, version 1 - 14 Oct 2005

On some p-adic power series attached to the arithmetic of Q (ζ _p )

Bruno Angl`es Universit´e de Caen,

Laboratoire Nicolas Oresme, CNRS UMR 6139, Campus II, Boulevard Mar´echal Juin,

BP 5186, 14032 Caen Cedex, France.

E-mail: angles@math.unicaen.fr October 11 2005

Let p be a prime number, p ≥ 5. Let θ be an even and non-trivial character of Gal(Q(µ

)/Q). Let f (T, θ) ∈ Z

[[T ]] be the Iwasawa power series associated to the p-adic L-function L

(s, θ) (see [12]), i.e.:

∀n ≥ 1, n ≡ 0 (mod p − 1), f ((1 + p)

− 1, θ) = L(1 − n, θ), where L(s, θ) is the usual Dirichlet L-series. In 1979, in their celebrated article [4], Ferrero and Washington have proved that (see also [1]):

f (T, θ) 6≡ 0 (mod p).

Thus we can write:

f(T, θ) ≡ T

U ¯ (T ) (mod p),

where ¯ U(T ) ∈ F

[[T ]]

. The Iwasawa lambda-invariant λ(θ) is not well- understood. By a heuristic argument due to Ferrero and Washington, one could expect that for sufficiently large p (see [5]):

X

θeven,θ6=1

λ(θ) ≤ Log(p)

Log(Log(p)) .

Furthermore, if p < 4000000, we have λ(θ) ≤ 1, and one would reasonably expect (see [8] ):

λ(θ) < p.

The only known result is due to Ferrero and Washington ([4]): for sufficiently large p we have:

λ(θ) ≤ p

. Now observe that:

f

(T, θ) ≡ λ(θ)T

U ¯ (T ) + T

U ¯

(T ) (mod p).

Thus if λ(θ) ≥ 1 and λ(θ) 6≡ 0 (mod p), we have that f

(T, θ) 6≡ 0 (mod p).

The aim of this paper is to prove that for all θ even, θ 6= 1, we have (Corollary 4.4):

f

(T, θ) 6≡ 0 (mod p).

This fact comes from properties of some power series that are connected to polynomials introduced by Mirimanoff at the beginning of the XXth century.

1 Notations

Let p be a prime number, p ≥ 5. Let Q

be an algebraic closure of Q

. All the extensions of Q

considered in this paper are contained in Q

. Let v

be the p-adic valuation on Q

such that v

(p) = 1. We denote the Iwasawa p-adic logarithm on Q

by Log

.

If A is a commutative ring, we denote the set of invertible elements of A by A

.

For every integer d, d ≥ 1, set µ

= {z ∈ Q

| z

= 1}. If ρ ∈ ∪

µ

, we denote the order of ρ by o(ρ). Set µ

= ∪

µ

. For all n ≥ 0, let ζ

∈ µ

such that: ζ

6= 1 and ∀n ≥ 0, ζ

= ζ

.

Let:

- K

= Q

(µ

),

- O

= Z

[ζ

],

- π

= ζ

− 1, - U

= 1 + π

O

, - Γ

= Gal(K

/K

), - ∆ = Gal(K

/Q

), - K

= Q

(µ

), - Γ = Gal(K

/K

).

Let γ

∈ Γ be such that ∀ζ ∈ µ

, ζ

= ζ

. For a ∈ Z

, we write:

a = ω(a) < a >,

where ω is the Teichm¨ uller character (i.e. ω(a) = lim

a

∈ µ

) and

< a >≡ 1 (mod p).

Let n ≥ 0. For a ∈ Z

, let σ

(a) ∈ Gal(K

/Q

) be such that:

ζ

= ζ

.

We have σ

(a) = δ(a)γ

(a), where δ(a) ∈ ∆ and γ

(a) ∈ Γ

. Observe that:

ζ

= ζ

. For a ∈ Z

, let σ(a) ∈ Γ × ∆ be such that:

∀ζ ∈ µ

, ζ

= ζ

.

Note that ∆ = {σ(ω(a)) | a = 1, · · · , p − 1}. For θ ∈ ∆ = Hom(∆, µ b

), we set:

e

= 1 p − 1

X

δ∈∆

θ

(δ) δ ∈ Z

[∆].

Let T be an indeterminate over Q

. Let L be a finite extension of Q

. Let O

be the valuation ring of L. The restriction maps Res

: Γ

→ Γ

induce maps Res

: O

[Γ

] → O

[Γ

]. Thus we can take the inverse limit of the group rings O

[Γ

] with respect to these maps, and by [12], Theorem 7.1,we have:

O

[[T ]] ≃ lim

←

O

[Γ

],

where T corresponds to γ

− 1. We set Λ

= O

[[T ]] and Λ = Λ

. For all n ≥ 0, we set ω

(T ) = (1 + T )

− 1. Recall that:

∀n ≥ 0, Λ

ω

(T )Λ

≃ O

[Γ

].

Note that forl all n ≥ 0, N

(U

) ⊂ U

, where N

is the norm map from K

to K

. We denote the inverse limit of the principal units U

with respect to the norm maps by U

. Note that U

is a Λ[∆]- module.

2 Logarithmic derivatives

Let u = (u

)

be an element in U

, recall that:

∀n ≥ 0, N

(u

) = u

.

There exists an unique element f

(T ) ∈ Λ such that ([12], Theorem 13.38, see also [3]):

∀n ≥ 0, f

(π

) = u

. Furthermore, if x = P

b

σ(a) ∈ Z

[Γ × ∆] and if u ∈ U

, then we have ([12], Lemma 13.48):

f

(T ) = Y

f

((1 + T )

− 1)

. For u ∈ U

and for n ≥ 0, we set:

D

(u) = ζ

f

(π

)

f

(π

) ∈ O

.

We call the map D

: U

→ O

the nth logarithmic derivative of Coleman.

Lemma 2.1

i) Let u ∈ U

, ∀n ≥ 0, D

(u

) ≡ 0 (mod p

).

ii) ∀θ ∈ ∆, b ∀u ∈ U

, ∀n ≥ 0, D

(u

) = e

D

(u).

Proof i) We have:

f

(T ) = f

((1 + T )

− 1) f

(T ) . Thus:

(1+T ) f

(T )

f

(T ) = (1+p)

(1+T )

f

((1 + T )

− 1)

f

((1 + T )

− 1) −(1+T ) f

(T )

f

(T ) .

Note that:

(1 + π

)

− 1 = π

. Therefore:

D

(u

) = ζ

((1 + p)

− 1) f

(π

)

f

(π

) ≡ 0 (mod p

).

ii) Let a ∈ Z

and let u ∈ U

. Observe that:

D

(u

) = aσ

(a)(D

(u)).

Let u ∈ U

, we get:

D

(u

) = 1 p − 1

X

a=1

θ

(a)D

(u

).

We have:

D

(u

) = 1 p − 1

X

a=1

θ

(a) ω(a) σ

(ω(a))(D

(u)).

The Lemma follows. ♦

Proposition 2.2 Let θ ∈ ∆, θ b 6= 1, ω. The map D

gives rise to a mor- phism of Z

[Γ

]-modules:

φ

: U

U

→ e

O

p

e

O

,

where T acts on the right via (1 + p)(1 + T ) − 1.

Proof We have ([12], Theorem 13.54):

U

≃ U

U

.

Let v ∈ U

, then there exists u = (u

)

∈ U

such that v = u

. We set:

φ

(v) ≡ D

(u) (mod p

).

By Lemma 2.1, we get a morphism of Z

-modules:

φ

: U

U

→ e

O

p

e

O

.

Let v ∈ U

, and let u = (u

)

∈ U

such that v = u

. We have:

φ

(u

) ≡ D

(u

) (mod p

).

But:

D

(u

) ≡ ((1 + p)γ

− 1)(D

(u)) (mod p

).

The Proposition follows. ♦

Theorem 2.3 For all n ≥ 0, set T

= P

d=0

ζ

. Let θ ∈ ∆, θ b 6= 1, ω.

Then:

∀n ≥ 0, Im(φ

) = Z

[Γ

]e

T

p

e

O

.

Proof Let Λ

be the maximal order of Q

[Γ

]. Then ([6], see also [7]):

O

= Λ

[∆] T

. Now observe that p

Λ

⊂ pZ

[Γ

]. Thus:

p

e

O

⊂ p Z

[Γ

]e

T

.

Recall that there exists α ∈ µ

\ {1} such that ([12], Corollary 13.37):

∀n ≥ 0, U

= ( α − ζ

ω(α − 1) )

. For α ∈ µ

\ {1}, set:

ρ

(α) = ( α − ζ

ω(α − 1) )

∈ U

. We have:

D

(ρ

(α)) = −ζ

α − ζ

.

Thus Im(φ

) is the Z

[Γ

]-module generated by the e

pn+1

, α ∈ µ

\ {1}. For α ∈ µ

\ {1}, for all n ≥ 0, set:

u

(α) =

pⁿ⁺¹

X

k=1, k6≡0 (modp)

α

σ

(k) ∈ Z

[Γ

× ∆].

We have:

Res

(u

(α)) =

pⁿ⁺²

X

k=1, k6≡0 (modp)

α

σ

(k).

Thus :

Res

(u

(α)) =

pⁿ⁺¹

X

k=1, k6≡0 (modp)

σ

(k) ( X

ℓ=0

α

).

Finally, we get:

Res

(u

(α)) = u

(α).

Now, we have:

α − 1

αζ

− 1 − 1 = (αζ

)

− 1 αζ

− 1 − 1.

Thus:

α − 1

αζ

− 1 − 1 =

pⁿ⁺¹−1

X

k=1

α

ζ

.

Therefore:

α − 1

αζ

− 1 − 1 =

pⁿ⁺¹

X

k=1,k6≡0 (modp)

α

σ

(k)(ζ

) +

p

X

k=1

α

ζ

. Finally, we get:

α − 1

αζ

− 1 − 1 = u

(α)(T

).

Now set:

u

(θ, α) =

pⁿ⁺¹

X

k=1, k6≡0 (modp)

θ(k)ω

(k)α

γ

(k).

We have:

(u

(θ, α))

∈ Λ.

Furthermore, we have:

e

−ζ

α − ζ

= e

1 α

ζ

− 1 . Thus:

e

−ζ

α − ζ

= 1

α

− 1 u

(θ, α

)(e

T

).

Now: X

α∈µp−1\{1}

u

(θ, α) = (p − 1)θ(−1)ω

(−1) ∈ Z

.

Thus there exists α ∈ µ

\ {1} such that for all n ≥ 0, u

(θ, α) ∈ Z

[Γ

]

. The theorem follows. ♦

We will also need the following Lemma:

Lemma 2.4 Let Λ

be the maximal order of Q

[Γ

]. Under the isomorphism Λ/ω

(T )Λ ≃ Z

[Γ

], p

Λ

corresponds to an ideal U

of Λ such that ω

(T )Λ ⊂ U

⊂ (p, ω

(T )) and lim

U

= {0}.

Proof Set:

e

= 1 p

X

γ∈Γn

γ, and for 1 ≤ d ≤ n, set:

e

= X

χ∈Γcn, fχ=p^d+1

e

,

where the sum is over all the characaters of Γ

of conductors p

and : e

= 1

p

X

γ∈Γn

χ

(γ)γ.

We have:

p

Λ

= ⊕

Z

[Γ

]p

e

⊂ pZ

[Γ

].

Thus it is clear that ω

(T )Λ ⊂ U

⊂ (p, ω

(T )). Now:

p

e

≡ p(

pⁿ−1

X

ℓ=0

(1 + T )

) (mod ω

(T )).

Thus:

p

e

≡ p ω

(T )

T (mod ω

(T )).

Let d be an integer, 1 ≤ d ≤ n. We have:

p

e

≡ p(

pⁿ−1

X

ℓ=0

(1 + T )

( X

fχ=p^d+1

χ(1 + p)

)) (mod ω

(T )).

Thus:

p

e

≡ p(

p

X

ℓ=0

T r

(ζ

)(1 + T )

) (mod ω

(T )).

Now recall that T r

(ζ

) = 0 if v

(ℓ) < d − 1. Therefore:

p

e

≡ p

(

p^n−d+1

X

ℓ=0

(1 + T )

T r

(ζ

)) (mod ω

(T )).

But:

p^n−d+1

X

ℓ=0,ℓ≡0 (modp)

(1 + T )

T r

(ζ

) = (p − 1) ω

(T ) ω

(T ) , and

p^n−d+1−1

X

ℓ=0,ℓ6≡0 (modp)

(1 + T )

T r

(ζ

) = ω

(T )

ω

(T ) − ω

(T ) ω

(T ) . Thus:

p

e

≡ p

ω

(T )

ω

(T ) − p

ω

(T )

ω

(T ) (mod ω

(T )).

The Lemma follows. ♦

3 Mirimanoff ’s power series

Recall that Mirimanoff has introduced the following polynomials in F

[T ] :

∀j, 1 ≤ j ≤ p − 1, ϕ

(T ) = X

a=1

a

T

.

These polynomials have many beautiful properties and we refer the inter- ested reader to [9], [11] and [2]. In this section, we will introduce some power series that are related to these polynomials.

Let L be a finite extension of Q

. Let π

be a prime of L. Let θ ∈ ∆. b Let a ∈ O

such that a(a − 1) 6≡ 0 (mod π

). For all n ≥ 0, we set:

M

(θ, a) =

pⁿ⁺¹

X

k=1,k6≡0 (modp)

a

a

− 1 θ(k)ω

(k)γ

(k) ∈ O

[Γ

].

Lemma 3.1

(M

(θ, a))

∈ lim

←

O

[Γ

].

Proof We must prove that:

∀n ≥ 0, Res

(M

(θ, a)) = M

(θ, a).

Now observe that:

Res

(M

(θ, a)) =

pⁿ⁺¹−1

X

k=1,k6≡0 (modp)

θ(k)ω

(k)γ

(k) 1 a

− 1 (

X

ℓ=0

a

).

The Lemma follows.♦

By the above Lemma, (M

(θ, a))

corresponds to a power series M (T, θ, a) ∈ Λ

. If θ = ω

, 1 ≤ j ≤ p − 1, observe that:

M(0, θ, a) ≡ ϕ

(a)

a

− 1 (mod π

).

Therefore we call M (T, θ, a) the Mirimanoff’s power series attached to θ and a.

Lemma 3.2

M (T, θ, a) = −θ(−1)ω

(−1)M (T, θ, a

).

Proof

We have:

M

(θ, a) =

pⁿ⁺¹

X

k=1,k6≡0 (modp)

a

a

− 1 θ(p

− k)ω

(p

− k)γ

(p

− k).

Thus:

M

(θ, a) = −

pⁿ⁺¹

X

k=1,k6≡0 (modp)

(a

)

(a

)

− 1 θ(p

−k)ω

(p

−k)γ

(p

−k).

The Lemma follows. ♦ Theorem 3.3

M

(T, θ, a) ≡ 0 (mod π

) if and only if a ≡ −1 (mod π

) and θ is odd.

Proof By Lemma 3.2, M (T, θ, −1) = 0 if θ is odd. Thus, if a ≡ −1 (mod π

) and if θ is odd, we have M

(T, θ, a) ≡ 0 (mod π

).

The proof of this Theorem is based on Sinnott’s proof that the Iwasawa µ-invariant vanishes for cyclotomic Z

-extensions of abelian number fields ([10]) as exposed in Washington’s book([12], paragraph 16.2).

Now, let’s suppose that M

(T, θ, a) ≡ 0 (mod π

). We have:

∀n ≥ 0, M (T, θ, a) ≡

pⁿ⁺¹

X

k=1,k6≡0 (modp)

a

a

− 1 θ(k)ω

(k)(1+T )

(mod ω

(T )), where i(k) = Log

(k)/Log

(1 + p). Thus, for all n ≥ 1, we have:

(1+T )M

(T, θ, a) ≡

pⁿ⁺¹

X

k=1,k6≡0 (modp)

i(k) a

a

− 1 θ(k)ω

(k)(1+T )

(mod (p

, ω

(T ))).

Therefore, for all n ≥ 1, we get:

pⁿ⁺¹

X

k=1,k6≡0 (modp)

i(k) a

a

− 1 θ(k)ω

(k)(1 + T )

≡ 0 (mod (π

, ω

(T ))).

Recall that i(k) ≡ i(k

) (mod p

) if and only if

p

≡

(mod p

).

Therefore changing i(k) to

p

permutes exponents modulo p

and do not affect divisibility by π

. Thus:

∀n ≥ 1,

pⁿ⁺¹

X

k=1,k6≡0 (modp)

< k > −1 p

a

a

− 1 θ(k)ω

(k)(1+T )

≡ 0 (mod (π

, ω

(T ))).

Let α ∈ µ

. For n ≥ 1, set:

h

(t) ≡

pⁿ⁺¹

X

k=1,k≡α (modp)

α

k − 1 p

a

a

− 1 θ(k)ω

(k)(1+T )

−1k−1

p

(mod (p

, ω

(T ))).

Note that:

h

(T ) ≡ h

(T ) (mod (p

, ω

(T ))).

Now recall that:

Λ

≃ lim

←

Λ

(p

, ω

(T )) . Therefore, there exists h

(T ) ∈ Λ

such that:

∀n ≥ 1, h

(T ) ≡ h

(T ) (mod (p

, ω

(T ))).

Thus, we have: X

α∈µp−1

h

(T ) ≡ 0 (mod π

).

And also: X

α∈µp−1

(1 + T )h

((1 + T )

− 1) ≡ 0 (mod π

).

Now, note that:

(1 + T )h

((1 + T )

− 1) ≡ f

((1 + T )

− 1) (mod (p

, ω

(T ))), where

f

(T ) ≡

pⁿ⁺¹

X

k=1,k≡α (modp)

α

k − 1 p

a

a

− 1 θ(k)ω

(k)(1+T )

(mod (p

, ω

(T ))).

For α ∈ µ

, let s

(α) ∈ {1, · · · , p − 1} such that α ≡ s

(α) (mod p). We have:

f

(T ) ≡ θ(α)ω

(α)

p

X

ℓ=0

α

(s

(α) + pℓ) − 1 p

a

a

− 1 (1+T )

(mod (p

, ω

(T ))).

But:

pⁿ−1

X

ℓ=0

a

a

− 1 (1 + T )

≡ a

(1 + T )

a

(1 + T )

− 1 (mod (p

, ω

(T ))),

and

p

X

ℓ=0

ℓ a

a

− 1 (1+T )

≡ −a

(1+T )

a

(1 + T )

(a

(1 + T )

− 1)

(mod (p

, ω

(T ))).

Set:

f

(T ) = θ(α)ω

(α) a

(1 + T )

a

(1 + T )

− 1 ( α

s

(α) − 1

p − α

a

(1 + T )

a

(1 + T )

− 1 ).

Then:

∀n ≥ 1, f

(T ) ≡ f

(T ) (mod (p

, ω

(T ))).

We have obtained:

X

α∈µp−1

f

((1 + T )

− 1) ≡ 0 (mod π

).

Observe that for all α ∈ µ

, f

(T ) ∈ Λ

∩L(T ). Thus, by Sinnott’s Lemma ([12], Lemma 16.9), for all α ∈ µ

there exists c

∈ O

such that:

f

(T ) + f

((1 + T )

− 1) ≡ c

(mod π

).

Observe that:

f

(T ) ≡ θ(α)ω

(α) a

a

− 1 ( α

s

(α) − 1

p −α

a

a

− 1 )(1+T )

(mod (π

, T

)), and

f

((1 + T )

− 1) ≡ θ(−α)ω

(−α)

(

+ α

)(1 + T )

(mod (π

, T

)).

Thus, for all α ∈ µ

, we must have:

a

(

−α

) ≡ θ(−1)a

(

+α

) (mod π

).

For α = 1, we get:

a a

a

− 1 ≡ −θ(−1)a

( a

a

− 1 − 1) (mod π

).

Thus:

a

≡ −θ(−1) (mod π

).

We obtain a ≡ −1 (mod π

) and θ is odd or θ is even and a

≡ −1

(mod π

). For the second case, if we consider all the equations obtained

when α runs through µ

, we obtain that for all b ∈ {1, · · · , p − 1}, b even, we must have:

b ≡ ω(b) + p (mod p

), and for all b ∈ {1, · · · , p − 1}, b odd, we must have:

b ≡ ω(b) (mod p

),

This leads to a contradiction and the Theorem is proved. ♦ We will need the following Lemma:

lemma 3.4 There exists α ∈ µ

such that for all prime numbers ℓ, ℓ ≡ α (mod p) and ℓ ≥ p

, we have:

T r

( ζ

+ ζ

(ζ

− 1)

) 6≡ 0 (mod p), where ζ

is a primitive ℓth root of unity.

Proof For a ∈ Z, let [a]

∈ {0, · · · , ℓ − 1}, such that a ≡ [a]

(mod ℓ). Let Φ

(X) be the ℓth cyclotomic polynomial, then:

Φ

(1) = ℓ, Φ

(1) = ℓ(ℓ − 1)

2 , and

Φ

(1) = ℓ(ℓ − 1)(ℓ − 2)

3 .

Now: Φ

(X)

Φ

(X) = X

ρ∈µℓ\{1}

1 X − ρ ,

and Φ

(X)

Φ

(X) − ( Φ

(X)

Φ

(X) )

= − X

ρ∈µℓ\{1}

1 (X − ρ)

. Therefore:

T r

( 1

ζ

− 1 ) = 1 − ℓ 2 , and

T r

( 1

(ζ

− 1)

) = (ℓ − 1)

4 − (ℓ − 1)(ℓ − 2)

3 .

Furthermore, if a ∈ Z, a 6≡ 0 (mod ℓ), we have:

T r

( ζ

ζ

− 1 ) = ℓ + 1

2 − [a]

. Set:

S = T r

( ζ

+ ζ

(ζ

− 1)

).

Let m be the order of p modulo ℓ, set:

a = [1 + p

]

and b = [1 − p

]

. We have:

S = T r

( ζ

+ ζ

(ζ

− 1)

).

Since ℓ ≥ p

, we have b ≥ 3, and thus:

a = ℓ + 2 − b.

Now:

S = T r

( ζ

− 1

(ζ

− 1)

) + T r

( ζ

− 1

(ζ

− 1)

) + 2T r

( 1 (ζ

− 1)

).

We obtain:

S = −b

+ b(ℓ + 2) − ℓ

+ 6ℓ + 5

6 .

Therefore, S ≡ 0 (mod p) implies that:

ℓ

+ 2

3 ∈ (F

)

. Now observe that the function field F

(T,

q

3

) has genus zero and thus has exactly p + 1 places of degree one. The Lemma follows. ♦

Theorem 3.5 Let θ ∈ ∆, θ b even. There exists α ∈ µ

such that for all prime numbers ℓ, ℓ ≡ α (mod p) and ℓ ≥ p

, we have:

X

ρ∈µℓ\{1}

M

(T, θ, ρ) 6≡ 0 (mod p).

Proof

Let ℓ be a prime number, ℓ 6= p. Suppose that we have:

X

ρ∈µℓ\{1}

M

(T, θ, ρ) ≡ 0 (mod p).

For α ∈ µ

, set:

f

(T ) = X

ρ∈µℓ\{1}

θ(α)ω

(α) ρ

(1 + T )

ρ

(1 + T )

− 1 ( α

s

(α) − 1

p −α

ρ

(1 + T )

ρ

(1 + T )

− 1 ).

By the proof of Theorem 3.3, we get:

X

α∈µp−1

f

((1 + T )

− 1) ≡ 0 (mod p).

Therefore by [12], Lemma 16.9, for all α ∈ µ

there exists c

∈ Z

such that:

f

(T ) + f

((1 + T )

− 1) ≡ c

(mod p).

But:

f

(T ) ≡ −(1 + T )T r

( ζ

(ζ

− 1)

) (mod (p, T

)), and

f

((1 + T )

− 1) ≡ −(1 + T )T r

( ζ

(ζ

− 1)

) (mod (p, T

)).

Thus we get:

T r

( ζ

+ ζ

(ζ

− 1)

) ≡ 0 (mod p).

It remains to apply Lemma 3.4. ♦

4 p-adic L-functions

Let θ ∈ ∆. b We set:

- f (T, θ) = 0 if θ is odd,

- if θ is even, f (T, θ) is the Iwasawa power series associated to the p-adic L-function L

(s, θ) (see [12], paragraph 7.2.).

Recall that if θ 6= 1, f (T, θ) ∈ Λ, anf if θ is the trivial character then:

(1 − 1 + p

1 + T )f (T, θ) ∈ Λ

. For α ∈ µ

, we set:

ρ

(α) = α − ζ

ω(α − 1) ∈ U

, and

ρ

(α) = (ρ

(α))

∈ U

. We set:

η

= Y

α∈µp−1\{1}

ρ

(α)

∈ U

, and

η

= (η

)

∈ U

. Lemma 4.1 Let θ ∈ ∆, θ b 6= 1, ω. Then:

U

= (η

)

. Proof By the proof Theorem 2.3:

φ

(ρ

(α)

) ≡ M

(θ, α

)e

T

(mod p

).

Thus:

φ

(η

) ≡ ( X

α∈µp−1\{1}

(α − 1)M

(θ, α))e

T

(mod p

).

But: X

α∈µp−1\{1}

(α − 1)M

(θ, α) = (p − 1)θ(−1)ω

(−1) ∈ Z

. The Lemma follows. ♦

Let θ ∈ ∆, θ b 6= 1, ω. We denote by G(T, θ) ∈ Λ

the power series that corresponds to (( P

α∈µp−1\{1}

(α − 1)M

(θ, α)))

. Observe that:

X

α∈µp−1\{1}

(α−1)M

(θ, α) = (p−1)θ(−1)ω

(−1)γ

(p−1)

pⁿ

X

ℓ=1,ℓ6≡0 (modp)

θ(ℓ)ω

(ℓ)γ

(ℓ).

Let θ ∈ ∆, θ b 6= 1, θ even. By π

we mean the unique (p − 1)

th root in U

which is congruent to 1 modulo π

of:

p−1

Y

a=1

( ζ

− 1

ζ

− 1 )

. We set:

π

= (π

)

∈ U

. Theorem 4.2

i) Let θ ∈ ∆, θ b even and non-trivial. Then:

(π

)

= (η

)

. ii) Let θ ∈ ∆, θ b 6= 1, ω. Then:

∀α ∈ µ

\ {1}, (ρ

(α)

)

= (η

)

. Proof Recall that:

φ

(ρ

(α)

) ≡ M

(θ, α

)e

T

(mod p

), and

φ

(η

) ≡ ( X

α∈µp−1\{1}

(α − 1)M

(θ, α))e

T

(mod p

).

Thus ii) follows from Lemma 4.1, Lemma 2.4, and Proposition 2.2. Note that:

φ

(π

) ≡ e

1

ζ

− 1 (mod p

).

Let f(X) =

, then:

(X − 1)Xf

(X) + Xf (X) = p

X

. Therefore:

1

ζ

− 1 = 1 p

pⁿ⁺¹

X

k=1

kζ

.

Let θ ∈ ∆, b for θ 6= 1, set:

v

(θ) = − 1 p

pⁿ⁺¹−1

X

k=1,k6≡0 (modp)

kθ(k)ω

(k)γ

(k),

and if θ = 1, set:

v

(θ) = −(1 − (1 + p)γ

(1 + p)) 1 p

pⁿ⁺¹−1

X

k=1,k6≡0 (modp)

kω

(k)γ

(k).

Then by [12], paragraph 7.2, (v

(θ))

corresponds to f (

−1, θ) if θ 6= 1, and to (1 − (1 + p)(1 + T ))f (

− 1, θ) if θ is trivial. Observe that if θ is even and non-trivial:

e

1

ζ

− 1 = −v

(θ)e

T

, and if θ = 1 :

(1 − (1 + p)γ

(1 + p))e

1

ζ

− 1 = −v

(θ)e

T

. The Theorem follows. ♦

Theorem 4.3 Let d be an integer, d ≥ 2, d 6≡ 0 (mod p). Let θ ∈ ∆. b Then:

X

ρ∈µd, o(ρ)=d

M (T, θ, ρ) = −f ( 1

1 + T −1, θ)( X

ℓdivides d

ℓµ( d

ℓ )θ(ℓ)ω

(ℓ)(1+T )

Logp(ℓ) Logp(1+p)

),

wher µ(.) is the M¨obius function.

Proof If θ is odd, the result is clear by Lemma 3.2. Thus we assume that θ is even. Let Φ

(X) be the dth cyclotomic polynomial, i.e.

Φ

(X) = Y

ρ∈µd, o(ρ)=d

(X − ρ) ∈ Z[X].

Then:

Φ

(X) = Y

ℓdivides d

(X

− 1)

.

If we take logarithmic derivatives, we get:

Φ

(X)

Φ

(X) = X

ℓdivides d

ℓµ( d

ℓ ) X

X

− 1 . It follows that:

X

ρ∈µd, o(ρ)=d

ζ

ζ

− ρ = X

ℓdivides d

ℓµ( d

ℓ ) ζ

ζ

− 1 . Thus:

X

ρ∈µd, o(ρ)=d

1

ρζ

− 1 = ( X

ℓdivides d

ℓµ( d

ℓ )σ

(ℓ)) 1 ζ

− 1 . Now:

ρ

− 1

ρζ

− 1 − 1 =

pⁿ⁺¹−1

X

k=1

ρ

ζ

. Thus:

1

ρζ

− 1 − 1 ρ

− 1 =

pⁿ⁺¹−1

X

k=1

ρ

ρ

− 1 ζ

.

We are working in the extension Q

(µ

, ζ

)/ Q

(µ

) and we identify Gal( Q

(µ

, ζ

)/ Q

(µ

)) with Γ

× ∆. Therefore:

e

1

ρζ

− 1 = M

(θ, ρ)e

ζ

+ e

(

p

X

k=1

(ρ

)

(ρ

)

− 1 ζ

).

Now observe that the map µ

→ µ

, ρ 7→ ρ

, is an isomorphism. Thus:

e

( X

ρ∈µd, o(ρ)=d

1

ρζ

− 1 ) = ( X

ρ∈µd, o(ρ)=d

M

(θ, ρ))e

T

.

But T

generates a normal basis for the field extension K

/Q

, thus:

X

ρ∈µd, o(ρ)=d

M

(θ, ρ) = −( X

ℓdivides d

ℓµ( d

ℓ )θ(ℓ)ω

(ℓ)γ

(ℓ))v

(θ),

where v

(θ) is as in the proof of Theorem 4.2. The Theorem follows. ♦

Corollary 4.4 Let θ ∈ ∆, θ b even and non-trivial. Then:

f

(T, θ) 6≡ 0 (mod p).

Proof Let α ∈ µ

as in Theorem 3.5. Let ℓ be a prime number such that ℓ ≥ p

and ℓ ≡ α (mod p

) (note that there exist infinitely many such primes). We know that:

X

ρ∈µℓ\{1}

M

(T, θ, ρ) 6≡ 0 (mod p).

But by Theorem 4.3:

X

ρ∈µℓ\{1}

M (T, θ, ρ) = −f( 1

1 + T − 1, θ)(ℓθ(ℓ)ω

(ℓ)(1 + T )

Logp(ℓ)

Logp(1+p)

− 1).

But since ℓ

≡ 1 (mod p

), we have:

Log

(ℓ)

Log

(1 + p) ≡ 0 (mod p).

Thus, if we take derivatives and reduce modulo p, we get:

X

ρ∈µℓ\{1}

M

(T, θ, ρ) ≡ 1

(1 + T )

f

( 1

1 + T −1, θ)(ℓθ(ℓ)ω

(ℓ)(1+T )

Logp(ℓ)

Logp(1+p)

−1) (mod p).

The Corollary follows. ♦

The case of the trivial character is treated in the last section.

5 Other results

Let θ be an even Dirichlet character of conductor d or pd, where d ≥ 1, d 6≡ 0 (mod p). For all n ≥ 0, set: q

= p

d. Let g(T, θ) be the power series introduced in [12], paragraph 7.2, i.e.

g(T, θ) = T − q

1 + T f (T, θ),

where f (T, θ) is the power series associated to the p-adic L-function L

(s, θ) (see [12], Theorem 7.10). Set L = Q

(θ), and let π

be a prime of L. Then, the Ferrero-Washington Theorem states (see [12], paragraph 16.2):

g(T, θ) 6≡ 0 (mod π

).

We have:

Theorem 5.1

g

(T, θ) 6≡ 0 (mod π

).

Proof For y ∈ Q , set:

B(y) = (1 + q

){y} − {(1 + q

)y} − q

2 ∈ Z

,

where {y} is the fractional part of y. recall that ([12], proof of Theorem 7.10):

∀n ≥ 0, g(T, θ) ≡

qn

X

a=1,(a,q0)=1

B( a

q

) θω

(a) (1 + T )

(mod ω

(T )), where i(a) =

p(1+q0)

. Let’s suppose that g

(T, θ) ≡ 0 (mod p). We have, for all n ≥ 1 :

(1+T )

g

(T, θ) ≡ − P

a=1,(a,q0)=1

(i(a)+1) B(

n

) θω

(a) (1+T )

(mod (p

, ω

(T ))).

Therefore, we get:

∀n ≥ 1,

qn

X

a=1,(a,q0)=1

(i(a)+1) B( a q

) θω

(a) (1+T )

≡ 0 (mod (π

, ω

(T ))).

Now, changing i(a) to (1 + q

)

p

permutes exponents modulo p

and does not affect divisibility by π

. Thus, for all n ≥ 1 :

P

a=1,(a,q0)=1

((1+q

)

p

+1) B(

n

) θω

(a) (1+T )

≡ 0 (mod (π

, ω

(T ))).

For n ≥ 1 and for α ∈ µ

, set:

H

(T ) ≡ P

a≡α (modp)

((1 + q

)

+ 1) B(

) θω

(a) (1 + T )

−1a−1 p

(mod (p

, ω

(T ))).

Note that:

H

(T ) ≡ H

(T ) (mod (p

, ω

(T ))).

Thus, there exists H

(T ) ∈ Λ

such that:

H

(T ) ≡ H

(T ) (mod (p

, ω

(T ))).

We get: X

α∈µp−1

H

(T ) ≡ 0 (mod π

).

Therefore:

X

α∈µp−1

(1 + T )

H

((1 + T )

− 1) ≡ 0 (mod π

).

Let f

(T ) ∈ Λ

as in [12], Lemma 16.8. Set:

F

(T ) = α

p (1 + T )f

(T ) + (1 − 1 + q

p )f

(T ).

It is not difficult to see that F

(T ) ∈ Λ

∩ L(T ). By [12], Lemma 16.8, we have:

F

(T ) ≡ P

a≡α (modp)

((1+q

)

+1) B(

) θω

(a) (1+T )

(mod (p

, ω

(T ))).

Therefore, we get:

X

α∈µp−1

F

((1 + T )

− 1) ≡ 0 (mod π

).

Now apply [12], Lemma 16.9, and we get that for all α ∈ µ

, there exists b

∈ O

such that:

F

(T ) + F

((1 + T )

− 1) ≡ b

(mod π

).

But observe that:

F

(T ) = F

((1 + T )

− 1).

Thus, for all α ∈ µ

, there exists c

∈ O

such that:

α

p (1 + T )f

(T ) + (1 − 1 + q

p )f

(T ) ≡ c

(mod π

).

Now we take α = 1, and we set:

G

(T ) = (1+q

) P

0<a<q0,a≡1 (modp)

θω

(a) (1+T )

− P

0<a<q²₀+q0,a≡1 (modp)

θω

(a) (1+

T )

.

We get:

(1 −

)((1 + T )

− 1)G

(T ) +

(1 + T )(((1 + T )

− 1)G

(T ) − q

(1 + q

)(1 + T )

G

(T )) ≡ c

((1 + T )

− 1)

(mod π

).

Note that:

G

(T ) ≡ −(1 + T ) (mod (1 + T )

), ans

G

(T ) ≡ −1 (mod (1 + T )

).

Thus we must have:

c

≡ 0 (mod π

).

The coefficient of 1 + T in the left-side is: 1 − d and the coefficient of (1 + T )

in the left side is 2d − 1 +

. The Theorem follows. ♦

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