On the linear independence of p-adic L-functions modulo p

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On the linear independence of p-adic L-functions modulo p

Bruno Angles, Gabriele Ranieri

To cite this version:

Bruno Angles, Gabriele Ranieri. On the linear independence of p-adic L-functions modulo p. 2009.

�hal-00370802�

On the linear independence of p-adic L-functions modulo p

BRUNO ANGLES` , GABRIELE RANIERI

Laboratoire de math´ematiques Nicolas Oresme, CNRS UMR 6139, Universit´e de Caen BP 5186,

14032 Caen cedex, France E-mail: bruno.angles@math.unicaen.fr

E-mail: ranieri@math.unicaen.fr

ABSTRACT

Let p ≥ 3be a prime. Let n ∈ N such thatn ≥ 1, let χ₁, . . . , χ_n be characters of conductord not divided bypand let ω be the Teichm¨uller character. For alli between1andn, for allj such that0≤j ≤(p−3)/2, set

θi,j =

(χiω^2j+1 ifχi is odd;

χiω^2j ifχi is even.

LetKbe the least extension ofQ_p which contains all the values of the characters χi and let π be a prime of the valuation ring O_K ofK. For alli, j let f(T, θi,j) be the Iwasawa series associated to θ_i,j and f(T, θ_i,j)its reduction modulo (π).

Finally letF_pbe an algebraic closure ofF_p. Our main result is that if the characters χi are all distinct modulo(π), then1and the series f(T, θi,j), for1 ≤i ≤ nand 0≤j ≤(p−3)/2, are linearly independent over a certain fieldΩwhich contains F_p(T).

1 Introduction

Letp be an odd prime. Letn ∈ N such thatn ≥ 1, letχ₁, . . . , χ_nbe characters of conductord not divided bypand let ω be the Teichm¨uller character. For alli between1andnandjsuch that0≤j ≤(p−3)/2, set

θi,j =

(χiω^2j+1 ifχi is odd;

χiω^2j ifχi is even.

Observe that, by definition,θi,j is an even character for alli, j.

Set κ0 = 1 +dp and K = Q_p(χ1, . . . , χn) (i.e. the least extension of Q_p generated by all the images ofχi for alli). Letπ be a prime of the valuation ring O_K ofK and letF_q beO_K/πO_K. For allibetween1andnandj between0and (p−3)/2, setf(T, θi,j)the Iwasawa power series attached to thep-adicL-function L_p(s, θ_i,j)(see [4, Theorem 7.10]) andf(T, θ_i,j)its reduction modulo(π).

LetF_p be an algebraic closure ofF_p. ForF(T) ∈ F_p[[T]], we say thatF(T) is a pseudo-polynomial if and only if there exist r ∈ N, a1, . . . , ar ∈ Z_p and c₁, . . . , c_r ∈F_p such that

F(T) = Xr

i=1

ci(T + 1)^ai.

Then the set of the pseudo-polynomials is a ring which we denote byA. Moreover we denote by Ωthe quotient field of A. The elements of Ω are called pseudo- rational functions. Angl`es (see [1, Theorem 4.5]) shows that for all non-trivial even character of the first kind θ, f(T, θ)is not a pseudo-rational function. We shall prove the following generalisation of this result:

Theorem 1. Suppose that the charactersχi are all distinct modulo(π)(i.e. for all integer i 6= j there exist a ∈ (Z/dZ)^∗ such thatχi(a) 6≡ χj(a) mod (π)).

Then the elements of the set

{1, f(T, θ_i,j), 1≤i≤n, 0≤j ≤(p−3)/2}

are linearly independent overΩ.

Observe that in the statement of Theorem 1 it is necessary to suppose that the characters χi are all distinct modulo(π). Indeed suppose that there existi 6= k between 1and n such that χi and χk are congruent modulo(π). Since p is odd

this implies thatχi andχk have the same parity. Then for all0 ≤j ≤ (p−3)/2 we have that θi,j is congruent to θk,j modulo (π), which implies f(T, θi,j) = f(T, θ_k,j). Thus in this case the seriesf(T, θ_i,j)are dependent.

Observe also that if the charactersχi are distinct modulo(π), then for alli, i^′ between1andn, j, j^′ between0and(p−3)/2, θi,j is congruent toθi^′,j^′ modulo (π)if and only ifi = i^′, j = j^′. It is clear thati = i^′ impliesj = j^′ andj = j^′ implies i = i^′. Then suppose that i 6= i^′, j 6= j^′ and that θi,j is congruent to θi^′,j^′ modulo (π). Moreover suppose that χi and χi^′ are even (the other case is identical). Hence there exists an integerasuch thatω^2j(a) 6≡ ω^2j′(a) mod (π).

Sincepdoes not dividedthere exists an integercsuch that1 +cd≡a mod (p).

Then θi,j(1 + cd) ≡ ω^2j(a) mod (π) and θi^′,j^′(1 + cd) ≡ ω^2j′(a) mod (π).

Sinceθ_i,jandθ_i^′_,j^′ are equivalent modulo(π), we getω^2j(a)≡ω^2j′(a) mod (π) which is a contradiction.

As in the proof of [1, Theorem 4.5], the main ingredient in the proof of The- orem 1 is a remarkable result due to Sinnott. Before the statement of that result we must define the following equivalence relation: leta, b∈Z^∗_p. We say thatais equivalent tob mod (Q^∗)(a ≡ b mod (Q^∗)) if and only if there existsc ∈ Q^∗ such thatab⁻¹ =c.

Proposition 1. ([3, Proposition 1]) LetF be a finite field of characteristicpand letr1(T), . . . , rs(T)∈F(T)∩F[[T]]. Letc1, . . . , cs ∈Z_p− {0}and suppose that

Xs i=1

ri((T + 1)^ci −1) = 0.

Then for alla ∈Z_p,

X

ci≡a mod (Q^∗)

ri((T + 1)^ci−1)∈F.

Let’s describe briefly the strategy of the proof of Theorem 1. First (section 2) we recall some properties of thep-adic Leopoldt transform (most of them already proved in [1]) which we will often use.

Then (section 3) we consider the cased ≥ 2. Some results about the p-adic Leopoldt transform of [1] and Proposition 1 will allow us to reduce the proof of Theorem 1 to the computation of the rank of a certain matrix whose entries depend on the values of the charactersχ_i (Lemma 4). After such computation the proof of this case of the theorem will follow by some simple remarks of linear algebra.

In section 4 we study the case d = 1. In that case we have to consider a

¡¡perturbation¿¿ of the functionsf(T, θi,j)to be able to apply Proposition 1. Then the proof is not very different from the proof of the previous case (actually it is simpler since it does not request a result similar to Lemma 4) and some remarks of linear algebra will imply the assumption.

Finally we give a possible link between Theorem 1 and Ferrero-Washington’s heuristic (see [2]). Letibe an integer between1and(p−3)/2. Write

f(T, ω²ⁱ) = X+∞

k=0

a_k(ω²ⁱ)T^k.

Theλ-invariant off(T, ω²ⁱ), denoted byλ(ω²ⁱ), is the leastksuch thatak(θ)6≡0 mod (p). We set

λ⁻=

(p−3)/2X

i=1

λ(ω²ⁱ).

Ferrero and Washington make the following hypothesis to define a heuristic to make previsions about possible bounds forλ⁻.

Ferrero-Washington’s hypothesis: Every coefficient of f(T, ω²ⁱ) is random mod (p)and independent from the other coefficients.

Theorem 1 implies that

1, f(T, ω⁰), f(T, ω²), . . . , f(T, ω^p−3)

are linearly independent over Ω. Thus our result seems to confirm Washington’s hypothesis.

2 Preliminaries

In this section we shall list some properties of thep-adic Leopoldt transform which will be very important in the proof of Theorem 1. LetL be a finite extension of Q_p, O_L its valuation ring and F_q^′ its residue field. Let κ a topological generator of1 +pZp and, for alla ∈ Z^∗_p, setω(a)the unique(p−1)th root of unity inZ_p congruent toa mod (p). Following [1] for allδ ∈Z/(p−1)Zwe define p-adic Leopoldt transformΓδthe unique continuousO_L-linear endomorphism ofO_L[[T]]

such that for alla∈Z_p,

Γδ((T + 1)^a) = (

ω^δ(a)(T + 1)

logp(a)

logp(κ) ifa ∈Z^∗_p;

0 otherwise

(see [1, Sections 2., 3.] for the proof of the fact thatΓδis well-defined and unique).

In an obvious way we can define a similarF_q′-linear continuous endomorphism of F_q′[[T]]that we denote byΓδ. Observe that ifa∈Z^∗_pwe havea≡ω(a) mod (p).

Thus, for alla∈Z^∗_p, we have

Γ_δ((T + 1)^a) =a^δ(T + 1)

logp(a) logp(κ).

In the proof of Theorem 1 we use otherO_L-linear endomorphisms ofO_L[[T]]

already introduced by Angl`es in [1]. Let us recall their definition. Letµp−1 ⊆Z^∗_p be the group of (p−1)th roots of unity. For all δ ∈ Z/(p−1)Z andF(T) ∈ O_L[[T]], we set

γδ(F(T)) = 1 p−1

X

η∈µp−1

η^δF((T + 1)^η−1).

Observe thatγδγδ^′ = 0forδ6=δ^′,γ_δ² =γδandP

δ∈Z/(p−1)Zγδ =Id^O_L[[T]]. ForF(T)∈O_L[[T]]set

D(F(T)) = (T + 1) d dTF(T) U(F(T)) = F(T)− 1

p X

ζ∈µp

F(ζ(T + 1)−1)∈O_L[[T]].

In an obvious way we can define the F_q′-linear endomorphism ofF_q′[[T]]γ_δ, D andU. Observe that:

• U² =U;

• DU =UD;

• γδU =Uγδ for allδ∈Z/(p−1)Z;

• Dγδ=γδ+1Dfor allδ ∈Z/(p−1)Z;

• U =D^p−1.

In the following lemma we shall list some properties ofΓδ whose we need to prove Theorem 1.

Lemma 1. Letδ∈Z/(p−1)ZandF(T)∈O_L[[T]]. Then

1. Γδ(F(T)) = Γδγ−δ(F(T)) = Γδγ−δUF(T).

2. Suppose that Γδ(F(T)) is a pseudo-polynomial. Then γ−δU(F(T)) is a pseudo-polynomial.

3. Γδ+1(F(T)) = ΓδD(F(T)).

Proof. 1. See [1, Proposition 3.2(2)].

2. The assumption immediately follows from [1, Proposition 3.1].

3. SinceΓδ is aF_q′-linear continuous endomorphism ofF_q′[[T]], it suffices to prove that the assumption is true forF(T) = (T + 1)^awitha ∈ Z_p. Ifpdivides awe have

0 = Γδ+1((T + 1)^a) = Γδ(D((T + 1)^a)) and the assumption is trivial.

Suppose thatpdoes not dividea. We have

Γ_δ(D((T + 1)^a)) = Γ_δ(a(T + 1)^a))

=aΓδ((T + 1)^a)

=a^δ+1(T + 1)

logp(a) logp(κ)

= Γδ+1((T + 1)^a).

Letθbe a Dirichlet character of the first kind such that

θ =χω^δ+1,

withδ∈Z/(p−1)Zandχa character of conductordnot divided byp. Observe thatκ0 = 1+dpis a topological generator of1+pZpand from now on setκ=κ0. Suppose thatQ_p(χ)⊆L. Set

Fχ(T) = Xd

a=1

χ(a)(T + 1)^a 1−(T + 1)^d

andFχ(T)∈F_q′(T)its reduction modulo the maximal ideal ofO_L. In the follow- ing lemma we list some properties of Fχ(T) and we recall the relation between Fχ(T)andf(T, θ).

Lemma 2. We have:

1. Ifd≥2, thenFχ(T)∈O_L[[T]].

2. Ifd= 1, thenγα(Fχ(T))∈O_L[[T]]for allα∈Z/(p−1)Zandα6= 1.

3. Ifd ≥ 2, thenF_χ((T + 1)⁻¹ −1) = εF_χ(T)where ε = 1ifχ is odd and ε=−1ifχis even.

4. Ifd= 1, thenFχ((T + 1)⁻¹−1) =−Fχ(T)−1.

5. Ifddivides the positive integergwe have Fχ(T) =

Pg

a=1χ(a)(T + 1)^a 1−(T + 1)^g . 6. Γδγ−δU(Fχ(T)) =f((T + 1)⁻¹−1, θ).

Proof. For 1., 2., 3., 4. see [1, Lemma 4.1].

5. We have:

Pg

a=1χ(a)(T + 1)^a 1−(T + 1)^g =

Xd a=1

χ(a) X

b≡a mod (d)

χ(b)(T + 1)^b 1−(T + 1)^g

= Xd

a=1

χ(a)(T + 1)^a+ (T + 1)^a+d+. . .+ (T + 1)^a+g−d 1−(T + 1)^g

= Pd

a=1χ(a)(T + 1)^a(1−(T + 1)^g) (1−(T + 1)^g)(1−(T + 1)^d)

=Fχ(T).

6. Remember thatUγ−δ =γ−δU. Then apply [1, Lemma 4.4].

Consider the ringO_L[[t]]. It acts overO_L[[T]]via:

(t+ 1)F(T) = F((T + 1)^κ0 −1)∈O_L[[T]], for allF(T)∈O_L[[T]].

Lemma 3. LetH(T)∈O_L[[T]]. Then for allF(T)∈O_L[[T]]we have H(T)Γδ(F(T)) = Γδ(H(t)F(T)).

Proof. SinceΓδisO_L-linear and continuous it suffices to prove the assumption in the case P(T) = (T + 1)^a, where a ∈ Z_p. By [1, Proposition 3.2(3)] for all b ∈Z^∗_pwe have

Γδ(F(T + 1)^b−1) =ω^δ(b)(T + 1)

logp(b)

logp(κ)Γδ(F(T)).

Then we get

Γδ((t+ 1)^aF(T)) = Γδ(F((T + 1)^κa0 −1)

=ω^δ(a)(T + 1)

logp(κa0 )

logp(κ0)Γδ(F(T))

= (T + 1)^aΓδ(F(T)).

3 The case d ≥ 2

The aim of this section is to prove Theorem 1 in the case where the conductord of the charactersχi is≥2.

Proof of Theorem 1 in the case d ≥ 2. Let χ1, χ2, . . . , χn characters of conductor d ≥ 2 distinct modulo (π). Without loss of generality we can sup- poseχ1, . . . , χr odd andχr+1, . . . , χneven for a certain integer r ≤ n. For alli between1andnandjbetween0and(p−3)/2set

θi,j =

(χiω^2j+1 if1≤i≤r;

χiω^2j otherwise. (1)

Suppose that for all1 ≤ i≤ n, 0 ≤j ≤ (p−3)/2, there existgi,j(T)∈ Ωsuch that

Xn i=1

(p−3)/2X

j=0

gi,j(T)f(T, θi,j)∈Ω

andgi,j(T)6= 0for somei, j. Sethi,j(T) =gi,j(1/(T + 1)−1)for alli, j. Then we have

Xn i=1

(p−3)/2

X

j=0

hi,j(T)f(1/(T + 1)−1, θi,j)∈Ω

andhi,j(T)6= 0for certaini, j. Observe that we can suppose thathi,j ∈Afor all i, j and that

Xn i=1

(p−3)/2

X

j=0

hi,j(T)f(1/(T + 1)−1, θi,j)∈A. (2) By Lemma 2(6.) and (1) for allibetween1andnandjbetween0and(p−3)/2, we have:

f 1

T + 1 −1, θi,j

=

(Γ2jγ−2j(Fχi(T)) if1≤i≤r;

Γ_2j−1γ_−2j+1(F_χi(T)) otherwise. (3) Moreover by Lemma 1(3.), for allδ∈Z/(p−1)Zwe have

Γδ+1γ_−δ−1 = Γδγ_δD.

Hence we can rewrite relation (2) in the following way:

Xr i=1

(p−3)/2X

j=0

hi,j(T)Γ0γ₀(D^2jFχi(T))+

Xn i=r+1

(p−3)/2X

j=0

hi,j(T)Γ0γ₀(D^2j−1Fχi(T))∈A.

By Lemma 1(1.) this last relation implies that Xr

i=1

(p−3)/2

X

j=0

hi,j(T)Γ0γ₀U(D^2jFχi(T))+

Xn i=r+1

(p−3)/2

X

j=0

hi,j(T)Γ0γ₀U(D^2j−1Fχi(T))∈A.

(4) Applying Lemma 3 and 1(2.) to (4) we get

Xr i=1

(p−3)/2X

j=0

h_i,j(t)γ₀U(D^2jF_χi(T)) + Xn i=r+1

(p−3)/2X

j=0

h_i,j(t)γ₀U(D^2j−1F_χi(T))∈A.

(5)

Set for all1≤i≤n,0≤j ≤(p−3)/2, Fi,j(T) =

(U D^2j(Fχi(T)) if1≤i≤r;

U D^2j−1(F_χi(T)) otherwise. (6) Then we can rewrite relation (5) in the following way:

Xn i=1

(p−3)/2

X

j=0

hi,j(t)γ₀(Fi,j(T))∈A. (7)

Now recall that by Lemma 2(3.), we have Fχi((T + 1)⁻¹−1) =

(Fχi(T) if1≤i≤r;

−F_χi(T) ifr+ 1 ≤i≤n.

Moreover observe that for all1≤i≤n,0≤k ≤p−2we have

(U D^kF_χi)((T + 1)⁻¹−1) = (−1)^kU D^k(F_χi((T + 1)⁻¹−1)).

Letibe between1andrandjbe between0and(p−3)/2. Then Fi,j((T + 1)⁻¹−1) = (U D^2jFχi)((T + 1)⁻¹−1))

= (−1)^2jU D^2j(Fχi((T + 1)⁻¹−1))

=U D^2j(Fχi(T))

=Fi,j(T).

With exactly the same computation we can prove that Fi,j((T + 1)⁻¹−1) =Fi,j(T) for alli, j, also in the case wherer+ 1 ≤i≤n. Then

Fi,j((T + 1)⁻¹−1) =Fi,j(T), ∀i, j. (8) We recall thatF_q is, by definition, the residue field of the least extension of Q_p which contains all the images of the characters χi. Consider the least field which contains F_q and all the coefficients of hi,j(T)for all i, j. Since hi,j(T) ∈ A ⊆ Fp[[T]], such field is a finite extension ofF_q. Call itF_q1 and writehi,j(t) =

P

b∈Zpci,j,b(t+ 1)^b withci,j,b ∈F_q1. Moreover observe that sincehi,j(T)6= 0for certain integeri, j, there existi, j, bsuch thatci,j,b 6= 0. Let

Gb(T) = Xn

i=1

(p−3)/2X

j=0

ci,j,bFi,j(T).

By Lemma 2(1.),Gb(T)∈F_q1[[T]]∩F_q1(T). Since

(t+ 1)^b(Gb(T)) = Gb((T + 1)^κb0 −1), by (7) we have

γ₀(X

b∈Zp

Gb((T + 1)^κb0 −1))∈A. (9) Choose a subset ofµp−1whose elements represent all the classes ofµp−1/{−1,1}

and call itS. From (8) it follows that

Gb(T) = Gb((T + 1)⁻¹−1) (10) for allb. Thus by (9) we get

X

η∈S

X

b∈Zp

G_b((T + 1)^ηκb0 −1)∈A.

SinceGb(T) = 0for all but finitely manyb∈Z_p, there exists a positive integeru such that

X

η∈S

Xu k=1

Gbk((T + 1)^ηκbk0 −1)∈A.

Moreover we setGk(T) =Gbk(T). By Proposition 1 there exist an integerl ≤u, b₁, b₂, . . . b_l ∈ Z_p,b_i 6=b_j fori6=j, η₁, η₂, . . . η_l ∈ µ_p−1withη_iκ^b₀ⁱ ∼Q^∗ η_jκ^b₀^j for alli,jandηiκ^b₀ⁱ 6=ηjκ^b₀^j fori6=j such that

Xl k=1

Gk((T + 1)^ηkκbk0 −1)∈A. (11) For all1≤k ≤l write

η_kκ^b₀^k =η₁κ^b₀¹x_k,

wherexj ∈Q^∗ ∩Z^∗_pandxi 6=xkifi6=k. Recall that by (10), we haveGk(T) = Gk((T + 1)⁻¹−1). Hence we can suppose thatxk>0for allk. By (11) we get

Xl k=i

G_k((T + 1)^xk−1)∈A.

Therefore there exist some positive integersN1, N2, . . . Nl not divided bypsuch that1≤N1 < N2 < . . . < Nland

Xl k=i

Gk((T + 1)^Nk −1)∈A.

If we rewriteG_k(T)as a combination ofF_i,j(T), we get Xl

k=1

Xn i=1

(p−3)/2

X

j=0

ci,j,kFi,j((T + 1)^Nk−1)∈A.

By [1, Lemma 3.5] if H(T) ∈ F_q1(T), then H(T) ∈ A if and only if there exists m ∈ N such that (T + 1)^mH(T) ∈ F_q1[T]. Since the denominator of Fi,j((T + 1)^Nk−1)is relatively prime with(T + 1)for alli, j, k, we get

Xl k=1

Xn i=1

(p−3)/2X

j=0

ci,j,kFi,j((T + 1)^Nk−1)∈F_q1[T]. (12) Observe that, by Lemma 2(5.), we have

F_χi(T) = Xdp a=1

χi(a)(T + 1)^a 1−(T + 1)^dp. Then for all1≤i≤nand0≤j ≤(p−3)/2, we have

Fi,j(T) =

(Pdp−1 a=1,p6|a

χi(a)a^2j(T+1)^a

1−(T+1)^dp if1≤i≤r;

Pdp−1 a=1,p6|a

χi(a)a^2j−1(T+1)^a

1−(T+1)^dp otherwise. (13) Then replacing in (12) using (13), we get

Xl k=1

Xr i=0

(p−3)/2X

j=0

ci,j,k

Xdp a=1,p6|a

a^2jχi(a)(T + 1)^aNk 1−(T + 1)^dpNk

+ (14)

+ Xl k=1

Xn i=r+1

(p−3)/2

X

j=0

ci,j,k

Xdp a=1,p6|a

a^2j−1χi(a)(T + 1)^aNk 1−(T + 1)^dpNk

∈ F_q1[T].

To finish the proof we shall prove that (14) is satisfied only if ci,j,k = 0 for all i, j, k, obtaining a contradiction.

We need the following remark:

Remark 1. Let V be a vectorial space over a fieldF andW a sub-space ofV. Moreover let φ be an endomorphism of V such that φ(W) ⊆ W. We remark that ifmis a positive integer and v1, v2, . . . , vm ∈V are eigen-vectors ofφwith non-zero eigen-valuesλ1, λ2, . . . , λmsuch thatλi 6=λj ifi6=j and if

v1+v2+. . .+vm ∈W, thenv_i ∈W for alli.

If we apply Remark 1 in the particular case where F = F_q1, V = F_q1(T), W =F_q1[T],m=p−1,φ =D,λ_b =b for all1≤b≤p−1and

vb =Vb(T) = Xl k=1

Xr i=0

(p−3)/2

X

j=0

ci,j,k

X

aNk≡b mod (p)

a^2jχi(a)(T + 1)^aNk 1−(T + 1)^dpNk

+

+ Xl k=1

Xn i=r+1

(p−3)/2

X

j=0

c_i,j,k

Xdp aNk≡b mod (p)

a^2j−1χ_i(a)(T + 1)^aNk 1−(T + 1)^dpNk

,

by (14) we getV_b(T)∈F_q1[T]. MultiplyV_b(T)by1−(T + 1)^dpNl. Then (1−(T + 1)^dpNl)Vb(T)∈(1−(T + 1)^dpNl)Fq1[T].

Let us recall that pdoes not divide Nk for allk. Observe that if ζ is a primitive dNl-th root of unity, thenζ−1is a root of(1−(T+ 1)^dpNl)Vb(T). SinceNl > Nk

for allk < l,ζ−1is a zero of (1−(T + 1)^dpNl)

Xl−1 k=1

Xr i=0

(p−3)/2X

j=0

c_i,j,k X

aNk≡b mod (p)

a^2jχi(a)(T + 1)^aNk 1−(T + 1)^dpNk

+

+(1−(T+ 1)^dpNl) Xl−1 k=1

Xn i=r+1

(p−3)/2X

j=0

ci,j,k

Xdp aNk≡b mod (p)

a^2j−1χi(a)(T + 1)^aNk 1−(T + 1)^dpNk

.

Then we get

Xr i=1

(p−3)/2X

j=0

ci,j,l

X

aNl≡b mod (p)

a^2jχi(a)ζ^aNl+ (15)

+ Xn i=r+1

(p−3)/2X

j=0

ci,j,l

X

aNl≡b mod (p)

a^2j−1χi(a)ζ^aNl = 0.

Observe that since pdoes not divide d, {a, p+a, . . . , p(d−1) +a} is a set of representants of all the classes modulo d. Moreover observe that, since ζ is a primitive dNl-th root of unity, ζ^′ = ζ^Nl is a primitive d-th root of unity. Let k ∈Z/pZsuch thatkNl ≡b mod (p). We can rewrite (15) as

Xr i=1

(p−3)/2X

j=0

ci,j,lk^2j Xd−1 h=0

χi(h)ζ^′h+ Xn i=r+1

(p−3)/2X

j=0

ci,j,lk^2j−1 Xd−1 h=0

χi(h)ζ^′h = 0. (16) Then for all primitived-th root of unity, (16) must be satisfied.

Set

xi,k =

(P(p−3)/2

j=0 c_i,j,lk^2j, if1≤i≤r;

P(p−3)/2

j=0 ci,j,lk^2j−1 otherwise and let

{ζ₁, ζ₂, . . . , ζ_φ(d)}

be the set of primitive d-th roots of unity inF_p (recall thatp does not divided).

Then by (16) we have

Xn i=1

xi,k

Xd−1 h=0

χi(h)ζ_c^h = 0. (17)

for all1≤c ≤φ(d). Hence we have a system ofn unknows (x_1,k, . . . , x_n,k) and φ(d)equations (one for all primitived-th root of unity). Observe thatn < φ(d).

Indeed by definitionnis less than the number of the characters of conductord≥2 distinct mod (π). Since we have onlyφ(d)characters whose conductor divides dand since the trivial character has conductor16=d, we haven ≤φ(d)−1. Thus the number of equations of the system is greater then the number of its unknowns.

We shall prove that the system has the unique solution(0,0, . . . ,0).

LetB be the matrix associated to the system (17). Then

B =







Pd−1

h=0χ1(h)ζ₁^h Pd−1

h=0χ2(h)ζ₁^h · · · Pd−1

h=0χn(h)ζ₁^h Pd−1

h=0χ1(h)ζ₂^h Pd−1

h=0χ2(h)ζ₂^h · · · Pd−1

h=0χn(h)ζ₂^h

... ... . .. ...

Pd−1

h=0χ₁(h)ζ_φ(d)^h Pd−1

h=0χ₂(h)ζ_φ(d)^h · · · Pd−1

h=0χ_n(h)ζ_φ(d)^h





.

Observe thatB =CE, where

C =







1 ζ1 · · · ζ₁^d−1 1 ζ₂ · · · ζ₂^d−1

... ... . .. ... 1 ζφ(d) · · · ζ_φ(d)^d−1





, E =







χ₁(0) χ₂(0) · · · χ_n(0) χ1(1) χ2(1) · · · χn(1)

... ... . .. ... χ1(d−1) χ2(d−1) · · · χn(d−1)





.

In the following lemma will prove thatker(B)is equals to{(0,0, . . . ,0)}, which impliesx1,k =x2,k =. . .=xn,k = 0.

Lemma 4. We have:

1. The rank ofCisφ(d).

2. The set

B={(1, ζ, . . . , ζ^d−1), ζ ∈µd, ζ not primitive}

is a basis ofker(C).

3. ker(B) ={(0,0, . . . ,0)}.

Proof. 1. Let C^′ be the matrix whose columns coincide with the first φ(d) columns of the matrixC. ThenC^′is a square matrix equals to

C^′ =









1 ζ1 · · · ζ₁^φ(d)−1 1 ζ2 · · · ζ₂^φ(d)−1

... ... . .. ... 1 ζφ(d) · · · ζ_φ(d)^φ(d)−1







.

Observe thatC^′ is a Vandermonde matrix. Thus its determinant is equals to Y

1≤r<s≤φ(d)

(ζr−ζs).

Sincepdoes not divided, we have

ζr 6=ζs.

Thus the determinant of C^′ is non-zero. Therefore C has rank φ(d), since its squareφ(d)×φ(d)sub-matrixC^′ has non-zero determinant.

2. SinceC has rankφ(d), the dimension ofker(C)isd−φ(d). Observe that Bhasd−φ(d)elements. HenceBis a basis ofker(C)if and only ifB⊆ker(C) and the elements ofBare linearly independent.

First let us prove thatB⊆ker(C). Letζ ∈µdbe ad-th root of unity which is not primitive. Observe that(1, ζ, . . . , ζ^d−1)∈ker(C)if and only if

Xd−1 h=0

ζ_i^hζ^h = 0

for alli. Since

ζiζ Xd−1 h=0

ζ_i^hζ^h = Xd−1 h=0

ζ_i^hζ^h

andζiζ 6= 1becausepdoes not divided, it follows that Xd−1

h=0

ζ_i^hζ^h = 0.

Thus (1, ζ, . . . , ζ^d−1) ∈ ker(C)for all non primitived-th root of unityζ. Then B⊆ker(C).

Denote still by ζ an element of µd whose order is different from d. Set βζ: Z/dZ → µd the character which sends i ∈ Z/dZ to ζⁱ. Observe that if ζ^′ ∈µ_dandζ 6=ζ^′,

βζ(1) =ζ 6=ζ^′ =βζ^′(1),

since p does not divide d. Thus the characters βζ are all distinct. Hence the theorem of the linear independence of characters imply that βζ are linearly in- dependent overF_p. From this fact it follows that the vectors (1, ζ, . . . , ζ^d−1)are linearly independent for all non primitived-th root of unityζ. HenceBis a basis ofker(C).

3. Using the previous notation for all non primitived-th root of unityζ, letβζ

be the function which sendsj ∈Z/dZtoζ^j. Observe that every non trivial linear combination of the vectors(χi(0), χi(1), . . . , χi(d−1))for 1 ≤ i ≤ n is not in

ker(C) if and only if the functions χi andβζ for 1 ≤ i ≤ n and non primitive d-th root of unity ζ are linealry independent over F_p (here χi is considered as a function of Z/dZ over F_q ⊆ F_p). Suppose that such fonctions are dependent.

Then we can choose a minimalr, non-zeroλi andµζinF_psuch that Xr

i=1

λiχi+ X

ζnot primitive

µζβζ = 0. (18)

First observe thatr ≥2. Indeed ifr= 0then (18) would imply the linear depen- dence of the elements ofBagainst 2.. Moreover ifr = 1then there would exist a character χ_i of conductor dsuch that (χ_i(0), χ_i(1), . . . , χ_i(d−1)) ∈ ker(C).

Then we would have

Xd−1 h=0

ζ_j^hχ_i(h) = 0

for all primitived-th root of unityζ_j, which contradicts [4, Lemma 4.8].

Letb∈(Z/dZ)^∗ such thatχ1(b)6=χ2(b)(suchbexists because recall that, by hypothesis,χ1is different fromχ2modulo(π)). Hence by (18), for allz ∈Z/dZ we have

Xr i=1

λiχi(bz) + X

ζnot primitive

µζβζ(bz) = 0. (19) Observe that the function which sends z ∈ Z/dZ to βζ(bz) coincides with the functionβ_ζ^b. Hence we can rewrite (19) as

Xr i=1

λ_iχ₁(b)χ_i(z) + X

ζnot primitive

χ₁(b)µ_ζβ_ζ^b(z) = 0. (20) If we multiply (18) byχ1(b)and we subtract it to (20),we get a non trivial relation with less thanrcharactersχi and it is impossible by the minimality ofr.

Finally consider the matrix B. Let v = (λ1, λ2, . . . , λn) ∈ ker(B). Since B =CE, thenE(v)∈ker(C). This fact implies that

Xn i=1

λi(χi(0), χi(1), . . . , χi(d−1))∈ker(C).

But we have previously proved that this relation is possible only ifλi = 0for all i. Thusv = (0,0, . . . ,0).

By the previous lemma we immediately getxi,k = 0for all1≤i≤nand for all1≤k ≤p−1. Remember that, by definition,

xi,k =

(P(p−3)/2

j=0 ci,j,lk^2j, if1≤i≤r;

P(p−3)/2

j=0 ci,j,lk^2j−1 otherwise.

We shall prove that the relation xi,k = 0for all i, k impliesci,j,l = 0 for all i, j.

We just consider the casei≤r (the proof in the other case is very similar). Leti be an integer between1andrand setci,j,l=yj. Since xi,k = 0for allkbetween 1andp−1, we have the following relations:

(p−3)/2

X

j=0

yjk^2j = 0. (21)

The matrix M associated to the first (p−1)/2 equations of the system (21) is given by:

M =









1 1 · · · 1 1 4 · · · 4^(p−3)/2

... ... . .. ... 1 ^(p−1)₄ ² · · ·

(p−1)² 4

(p−3)/2









=









1 α1 · · · α^(p−3)/2₁ 1 α2 · · · α^(p−3)/2₂

... ... . .. ... 1 α_(p−1)/2 · · · α^(p−3)/2_(p−1)/2









SoM is a Vandermonde matrix and its determinant is equals to:

det(M) = Y

1≤r<s≤(p−1)/2

(α_r−α_s)6= 0.

It follows that the only solution of the system (21) isyj = 0for allj.

Then we have proved that the coefficients c_i,j,l = 0 for all i and j. Since Nl−1 > Nkfor allk < l−1, if we replacel withl−1with the same procedure we can prove that the coefficientsci,j,l−1 = 0for alli, j and so on. Thusci,j,k = 0 for alli, j, k, which is a contradiction.

4 The case d = 1

The aim of this section is to prove Theorem 1 in the case where d = 1. In other words, letχthe trivial character. We shall prove that

dimΩ(Ω + Ωf(T, χω⁰) + Ωf(T, χω²) +. . .+ Ωf(T, χω^p−3)) = p+ 1 2 . As we have already remarked is Section 1 we shall modify the proof of the case d≥2. Let us give a preliminar reason for this. Letδ∈Z/(p−1)Zbe odd. Then by Lemma 2(6.) we have

f((T + 1)⁻¹−1, χω^δ+1) = Γδγ−δU((Fχ(T)).

Sinceχis the trivial character we have Fχ(T) =

Xp−1 a=0

(T + 1)^a

1−(T + 1)^p −1 =−1

T −1. (22)

Thus Fχ(T) 6∈ F_p[[T]]and we shall see that this fact does not allow us to apply Proposition 1 (observe that ifχ^′ is not the trivial character thenF_χ^′(T)∈ F_q[[T]]

and see also Remark 2). The following lemma explains how we can solve this problem.

Lemma 5. Let

Ffχ(T) = (p+ 1)Fχ((T + 1)^p+1−1)−Fχ(T).

ThenFf_χ(T)∈Z_p[[T]].

Moreover

T f((T + 1)⁻¹−1, χω^δ+1) = Γ_δγ_−δU(fF_χ(T)) for all oddδ ∈Z/(p−1)Z.

Finally1, T f((T + 1)⁻¹−1, χω⁰), . . . , T f((T + 1)⁻¹−1, χω^p−3)are inde- pendent overΩif and only if1, f(T, χω⁰), . . . , f(T, χω^p−3)are independent over Ω.

Proof. Observe that

Ffχ(T) = (p+ 1)Fχ((T + 1)^p+1−1)−Fχ(T)

=− (p+ 1)

(T + 1)^p+1−1+ 1 T −p

= (T + 1)^p+1−(p+ 1)T −1 T((T + 1)^p+1−1) −p.

We remark that T² exactly divides T((T + 1)^p+1 − 1). Moreover T² divides (T + 1)^p+1−(p+ 1)T −1. Then we immediately getFfχ(T)∈Z_p[[T]].

LetL be a finite extension of Q_p and let O_L be its valuation ring. Let d be an integer not divided by p and set κ0 = 1 + dp. Remember that in Section 2 we have defined an action of O_L[[t]] overO_L[[T]]such that, for all a ∈ Z_p and F(T)∈O_L[[T]],

(t+ 1)^aF(T) =F((T + 1)^κa0 −1).

SetL=Q_p,d= 1andκ0 = 1 +p. Then, since

Feχ(T) =Fχ((T + 1)^p+1−1)−Fχ(T) +pFχ((T + 1)^p+1−1), we have

Ffχ(T)≡tFχ(T) mod (p).

By Lemma 3(2.) we get

Γδγ−δU((Ffχ(T)) = Γδγ−δU(tFχ(T)) =TΓδγ−δU(Fχ(T)).

Then, by Lemma 2(6.), we immediately get

T f((T + 1)⁻¹−1, χω^δ+1) = Γδγ_−δU(Ffχ(T)).

Finally1, T f((T + 1)⁻¹−1, χω⁰), . . . , T f((T + 1)⁻¹−1, χω^p−3) are inde- pendent overΩif and only if1, f(T, χω⁰), . . . , f(T, χω^p−3)are independent over Ω, sinceT ∈Ωand the image ofΩvia the endomorphism ofF_p((T))which sends F(T)∈F_p((T))inF((T + 1)⁻¹−1)isΩ.

By the previous lemma to finish the proof of Theorem 1 it suffices to show that 1, T f((T + 1)⁻¹−1, χω⁰), . . . , T f((T + 1)⁻¹−1, χω^p−3)are linearly indepen- dent overΩ. Since, always by the previous lemma, for all oddδ∈Z/(p−1)Zwe have

T f((T + 1)⁻¹−1, χω^δ+1) = Γδγ_−δU(Ffχ(T))

and Ffχ(T) ∈ F_p[[T]], to show the linear independence we can easily adapt our proof of Theorem 1 in the cased≥2.

Proof of Theorem 1 in the case d = 1. First observe that by Lemma 5 to prove the assumption it suffices to show that

1, T f((T + 1)⁻¹−1, χω⁰), . . . , T f((T + 1)⁻¹−1, χω^p−3)

are linearly independent overΩ. Suppose that this is not the case. Then there exist h₀(T), . . . , h_(p−3)/2(T)∈Ωsuch that

(p−3)/2X

j=0

hj(T)T f 1

T + 1 −1, ω^2j

∈Ω, (23)

with hj(T) 6= 0 for a certain j. Observe that without loss of generality we can suppose thathj(T)∈Afor alliand that

(p−3)/2X

j=0

hj(T)T f 1

T + 1 −1, ω^2j

∈A.

By Lemma 5 we get

(p−3)/2X

j=0

hj(T)Γ2j−1γ_−2j+1U(Ffχ(T))∈A.

By Lemma 1(3.) and sinceDγδ =γδ+1Dfor allδ ∈Z/(p−1)Z, we have

(p−3)/2X

j=0

hj(T)Γ−1γ₁D^2jU(fFχ(T))∈A.

Moreover, applying Lemma 3, we get

(p−3)/2X

j=0

Γ−1γ₁D^2jU(hj(t)fFχ(T))∈A.

From Lemma 1(2.) it follows

(p−3)/2X

j=0

γ₁D^2jU(hj(t)fFχ(T))∈A. (24) For allj such that0≤j ≤(p−3)/2set

F_j(T) =D^2jU(fF_χ(T)).

Then we can rewrite (24) in the following way:

(p−3)/2

X

j=0

γ₁(hj(t)Fj(T))∈A. (25) By Lemma 2(4.) we have

Fχ((T + 1)⁻¹−1) = −Fχ(T)−1.

Then

Ffχ((T + 1)⁻¹−1) =Fχ((T + 1)^−(p+1)−1)−Fχ((T + 1)⁻¹−1)

=−Fχ((T + 1)^p+1−1)−1 +Fχ(T) + 1

=−Ffχ(T).

Moreover observe that for allj between0and(p−3)/2,

(D^2jU(Ffχ))((T + 1)⁻¹−1) =D^2jU(Ffχ((T + 1)⁻¹−1).

Then for allj

(D^2jU(fFχ))((T + 1)⁻¹−1) =−D^2jU(fFχ(T)).

It follows that

Fj(T) = −Fj((T + 1)⁻¹−1) (26) for allj.

Consider the least field which containsF_pand all the coefficients ofhj(T)for allj. Sincehj(T)∈A⊆F_p[[T]], such field is a finite extension ofF_p. Call itF_q1

and writehj(t) =P

b∈Zpcj,b(t+ 1)^bwithcj,b ∈F_q1. Moreover observe that since hj(T)6= 0for certain integerj, there existj, bsuch thatcj,b 6= 0. Set

Gb(T) =

(p−3)/2X

j=0

cj,bFj(T)

and observe thatGb(T)∈F_q1[[T]]∩F_q1(T)for allb. Since (t+ 1)^b(Gb(T)) = Gb((T + 1)^κb0 −1), by (25) we have

γ₁(X

b∈Zp

Gb((T + 1)^κb0 −1))∈A. (27) Choose a subset ofµ_p−1whose elements represent all the classes of the group µp−1/{−1,1}and call itS. From (26) it follows that

G_b(T) = −G_b((T + 1)⁻¹−1) (28) for allb. Thus by (27) we get

X

η∈S

X

b∈Zp

ηG_b((T + 1)^ηκb0 −1)∈A.

SinceGb(T) = 0for all but finitely manyb ∈ Z_p, there exists an integerusuch

that X

η∈S

Xu k=1

ηGbk((T + 1)^ηκbk0 −1)∈A.

Moreover setG_k(T) =G_bk(T). SinceG_k(T)∈F_q1[[T]]∩F_q1(T)for allk, we can apply Proposition 1, obtaining that there exist an integerl ≤u,b1, b2, . . . bl∈ Z_p, bi 6= bj for i 6= j, η1, η2, . . . ηl ∈ µp−1 with ηiκ^b₀ⁱ ∼Q^∗ ηjκ^b₀^j for all i, j and ηiκ^b₀ⁱ 6=ηjκ^b₀^j fori6=j such that

Xl k=1

ηkGk((T + 1)^ηkκbk0 −1)∈A. (29) Remark 2. We remark that that the fact that for allk,Gk(T)∈F_q1[[T]]∩F_q1(T) is necessary to apply Proposition 1. This relation is verified since for all j, we

haveFj(T)∈F_p[[T]]∩F_p(T)which is an immediate consequence of the fact that Ffχ(T)∈F_p[[T]]∩F_p(T).

Finally observe that in the cased≥ 2, for all characterχ^′ of conductordwe have Fχ^′(T) ∈ F_q[[T]]∩F_q(T). For this reason it is not necessary ¡¡to pertur- bate¿¿Fχ^′(T)to apply Proposition 1.

For all1≤k ≤lwrite

ηkκ^b₀^k =η1κ^b₀¹xk,

wherexk ∈ Q^∗ ∩Z^∗_p andxi 6=xk ifi6=k. Recall that by (28), we haveGk(T) =

−Gk((T + 1)⁻¹−1). Hence we can suppose thatxk >0for allk. By (27) we get Xl

k=1

ηkGk((T + 1)^xk −1)∈A.

Therefore there exist some integersN₁, N₂, . . . N_lnot divided byp, such that1≤ N1 < N2 < . . . < Nland

Xl k=1

ηkGk((T + 1)^Nk −1)∈A.

By definition ofGk(T)this last relation becomes:

Xl k=1

ηk (p−3)/2

X

j=0

cj,kFj((T + 1)^Nk−1)∈A. (30) Now we want to computeFj(T)for allj. First remember that

Fχ(T) = Xp−1

a=0

(T + 1)^a

1−(T + 1)^p −1.

Then

D^2j(F_χ(T)) = Xp−1 a=1

a²ⁱ(T + 1)^a 1−(T + 1)^p for allj between0and(p−1)/2. SinceU =D^p−1we have

D²ⁱU(F_χ(T)) = Xp−1 a=1

a²ⁱ(T + 1)^a 1−(T + 1)^p.

Remember that

Ff_χ(T) = F_χ((T + 1)^p+1−1)−F_χ(T) =tF_χ(T).

Then

D^2jU(fF_χ(T)) =tD^2jU(F_χ(T))

=t X^p−1

a=1

a²ⁱ(T + 1)^a 1−(T + 1)^p

= Xp−1 a=1

a²ⁱ(T + 1)^a(p+1) 1−(T + 1)^p(p+1) −

Xp−1 a=1

a²ⁱ(T + 1)^a 1−(T + 1)^p. Replacing in (30) we get

Xl k=1

ηk (p−3)/2

X

j=0

cj,k

Xp−1 a=1

a^2j(T + 1)^aNk(p+1)

1−(T + 1)^pNk(p+1) − a^2j(T + 1)^aNk 1−(T + 1)^pNk

∈A.

Since by [1, Lemma 3.5] a rational functionH(T)∈ Aif and only if there exists an integernsuch that(T + 1)ⁿH(T)is a polynomial, we get

Xl k=1

ηk (p−3)/2

X

j=0

cj,k

Xp−1 a=1

a^2j(T + 1)^aNk(p+1)

1−(T + 1)^pNk(p+1) − a^2j(T + 1)^aNk 1−(T + 1)^pNk

∈F_q1[T] (31) (recall that, by definition,F_q1 is the extension ofF_pgenerated bycj,kfor allj, k).

We apply Remark 1 in the particular case where F = Fq1, V = F_q1(T), W =F_q1[T],m=p−1,φ =D,λb =b for all1≤b≤p−1and

vb =Vb(T) :=

Xl k=1

ηk (p−3)/2X

j=0

cj,k

a^2j_b,k(T + 1)^ab,kNk(p+1)

1−(T + 1)^pNk(p+1) − a^2j_b,k(T + 1)^ab,kNk 1−(T + 1)^pNk

, whereab,ksatisfies the relationab,kNk ≡b mod (p). Then by Remark 1,Vb(T)∈ F_q1[T]. Let us recall thatpdoes not divideN_k for allk between1andl. Sincep does not divideb, we haveab,k∈F^∗_p for allb, k.

Letζbe a primitive(p+ 1)Nlth root of unity and multiplyVb(T)by the poly- nomial1−(T + 1)^p(p+1)Nl. We get

Q(T) +ηl (p−3)/2X

j=0

cj,la^2j_b,l(T + 1)^p(p+1)Nl =P(T), (32)