On Jacobi Sums in Q(zeta_p)

HAL Id: hal-00947148

https://hal.archives-ouvertes.fr/hal-00947148

Submitted on 21 Feb 2014

HAL is a multi-disciplinary open access archive for the deposit and dissemination of sci- entific research documents, whether they are pub- lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.

L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

Bruno Angles, Filippo Alberto Edoardo Nuccio Mortarino Majno Di Capriglio

To cite this version:

Bruno Angles, Filippo Alberto Edoardo Nuccio Mortarino Majno Di Capriglio. On Jacobi Sums in

Q(zeta_p). Acta Arithmetica, Instytut Matematyczny PAN, 2010, 142 (3), pp.199-218. �hal-00947148�

142.3 (2010)

On Jacobi sums in Q(ζ

)

by

Bruno Angl` es (Caen) and Filippo A. E. Nuccio (Roma) Let p be a prime number, p ≥ 5. Iwasawa has shown that the p-adic properties of Jacobi sums for Q(ζ

) are linked to Vandiver’s Conjecture (see [5]). In this paper, we follow Iwasawa’s ideas and study the p-adic properties of the subgroup J of Q(ζ

)

generated by Jacobi sums.

Let A be the p-Sylow subgroup of the class group of Q(ζ

). If E denotes the group of units of Q(ζ

), then if Vandiver’s Conjecture is true for p, by Kummer theory and class field theory, there is a canonical surjective map

Gal(Q(ζ

)( √

E)/Q(ζ

)) → A

/pA

.

Note that J is, for the “minus” part, the analogue of the group of cyclotomic units. We introduce a submodule W of Q(ζ

)

which was already considered by Iwasawa [6]. This module can be thought of, for the minus part, as the analogue of the group of units. We observe that J ⊂ W and if the Iwasawa–

Leopoldt Conjecture is true for p then W (Q(ζ

)

)

= J(Q(ζ

)

)

. We prove that if pA

= { 0 } then (Corollary 4.8) there is a canonical surjective map

Gal(Q(ζ

)( √

W )/Q(ζ

)) → A

/pA

.

The last part of our paper is devoted to the study of the jacobian of the Fermat curve X

+ Y

= 1 over F

where ℓ is a prime number, ℓ 6 = p.

It is well-known that Jacobi sums play an important role in the study of that jacobian. Following ideas developed by Greenberg [4], we prove that Vandiver’s Conjecture is equivalent to some properties of that jacobian (for a precise statement see Corollary 5.3).

1. Notations. Let p be a prime number, p ≥ 5. Let ζ

∈ µ

\ { 1 } , and let L = Q(ζ

). Set O = Z[ζ

] and E = O

. Let ∆ = Gal(L/Q) and let

∆ b = Hom(∆, Z

). Let I be the group of fractional ideals of L which are

2010 Mathematics Subject Classification: 11R18, 11R29, 11R23.

Key words and phrases: Jacobi sums, ideal class group, Iwasawa theory, Vandiver’s con- jecture.

DOI: 10.4064/aa142-3-1 [199] c Instytut Matematyczny PAN, 2010

prime to p, and let P be the group of principal ideals in I . Let A be the p-Sylow subgroup of the ideal class group of L.

Set π = ζ

− 1, K = Q

(ζ

), U = 1 + π

Z

[ζ

]. Observe that if A ∈ P , then there exists α ∈ L

∩ U such that A = α O . If H is a subgroup of U, we will denote the closure of H in U by H. Let ω ∈ ∆ b be the Teichm¨ uller character, i.e.

∀ σ ∈ ∆, σ(ζ

) = ζ

. For ρ ∈ ∆, b we set

e

= 1 p − 1

X

δ∈∆

ρ

(δ)δ ∈ Z

[∆].

If M is a Z

[∆]-module, for ρ ∈ ∆, b we set M (ρ) = e

M.

For ψ ∈ ∆, ψ b odd, recall that B

= 1

p

p−1

X

a=1

aψ(a).

Set

θ = 1 p

p−1

X

a=1

aσ

∈ Q[∆],

where σ

∈ ∆ is such that σ

(ζ

) = ζ

. Observe that we have the following equality in C[∆]:

θ = N

2 + X

ψ∈∆, ψb odd

B

e

, where N = P

δ∈∆

δ.

If M is a Z[∆]-module, we set

M

= { m ∈ M : σ

(m) = − m } , M

= { m ∈ M : σ

(m) = m } . If M is an abelian group of finite type, we set

M [p] = { m ∈ M : pm = 0 } , d

M = dim

M/pM .

2. Background on Jacobi sums. Let Cl(L) be the ideal class group of L. Then Cl(L) ≃ I / P . Note that we have a natural Z[∆]-morphism (see [6, pp. 102–103])

φ : (Ann

Cl(L))

→ Hom

(Cl(L), E

/(E

)

).

For the convenience of the reader, we recall the construction of φ. Let x ∈ (Ann

Cl(L))

and A ∈ I . We have A

= γ

O , where γ

∈ L

∩ U. Now,

γ

= ε

γ

for some ε

∈ E

∩ U. One can prove that we obtain a well-defined morphism of Z[∆]-modules φ(x) : Cl(L) → E

/(E

)

, class of A 7→ class of ε

. In this section, we will study the kernel of the morphism φ.

Let W be the set of elements f ∈ Hom

( I , L

) such that:

• f ( I ) ⊂ U,

• there exists β (f ) ∈ Z [∆] such that f (α O ) = α

for all α ∈ L

∩ U . One can prove that if f ∈ W then β(f ) is unique, the map β : W → Z [∆]

is an injective Z[∆]-morphism and β( W ) ⊂ Ann

(Cl(L)) (see [2]). If B denotes the group of Hecke characters of type (A

) that have values in Q(ζ

) (see [6]), then one can prove that B is isomorphic to W .

Lemma 2.1 . Ker φ = β( W

).

Proof. We just prove the inclusion Ker φ ⊂ β( W

). Let x ∈ Ker φ. Let A ∈ I . Then there exists a unique γ

∈ L

∩ U such that γ

γ

= 1 and

A

= γ

O .

Let f : I → L

, A 7→ γ

. It is not difficult to see that f ∈ Hom

( I , L

) and f ( I ) ⊂ U. Now, if α ∈ L

∩ U, we have

f(α O ) = α

u

for some u ∈ E. Since x ∈ Z[∆]

and α, f (α O ) ∈ U, we must have u = 1.

Therefore f ∈ W

and x = β(f ).

Now, we recall some basic properties of Gauss and Jacobi sums (we refer the reader to [12, Sec. 6.1]).

Let P be a prime ideal in I and let ℓ be the prime number such that ℓ ∈ P. We fix ζ

∈ µ

\ { 1 } . Set F

= O /P . Let χ

: F

→ µ

be such that

∀ α ∈ F

, χ

(α) ≡ α

(mod P ), where N P = |O /P | . For a ∈ Z/pZ, we set

τ

(P) = − X

α∈FP

χ

(α)ζ

. We also set τ (P ) = τ

(P). For a, b ∈ Z/pZ, we set

j

(P ) = − X

α∈FP

χ

(α)χ

(1 − α).

Then:

• if a + b ≡ 0 (mod p), we have:

(i) if a 6≡ 0 (mod p), then j

(P ) = 1,

(ii) if a ≡ 0 (mod p), then j

(P ) = 2 − N P,

• if a + b 6≡ 0 (mod p), we have

j

(P ) = τ

(P )τ

(P ) τ

(P ) .

Observe that τ (P ) ≡ 1 (mod π), and therefore (see [5, Theorem 1])

∀ a, b ∈ Z /p Z , j

(P ) ∈ U.

Let Ω be the compositum of the fields Q(ζ

) where ℓ runs through the prime numbers distinct from p. The map P 7→ τ (P ) induces by linearity a Z[∆]-morphism

τ : I → Ω(ζ

)

.

Let G be the Z[∆]-submodule of Hom

( I , Ω(ζ

)

) generated by τ. We set J = G ∩ Hom

( I , L

).

Let S be the Stickelberger ideal of L, i.e. S = Z[∆]θ ∩ Z[∆]. Then one can prove the following facts (see [2]):

• J ⊂ W ,

• the map β : W → Z[∆] induces an isomorphism J ≃ S of Z[∆]- modules.

Lemma 2.2 . Let N ∈ Hom

(I

, L

) be the ideal norm map. Then, as a Z-module,

J = N Z ⊕

(p−1)/2

M

n=1

j

Z.

Proof. Recall that, for 1 ≤ n ≤ p − 2 and a prime P in I , we have j

(P ) = − X

α∈FP

χ

(α)χ

(1 − α) = τ (P )τ

(P ) τ

(P ) . Thus, for 1 ≤ n ≤ p − 2,

j

= τ

= τ τ

τ

, where τ

= τ

for a ∈ F

. Observe that

∀ a ∈ F

, τ

τ

= N .

Thus N ∈ J . Since J ≃ S , J is a Z -module of rank (p + 1)/2. It is not difficult to show that (see [5, Lemma 2])

J = τ

Z ⊕

(p−1)/2

M

a=1

τ

τ

Z . Observe also that, for 2 ≤ n ≤ p − 2, we have

j

= j

.

Let V be the Z -submodule of J generated by N and the j

, 1 ≤ n ≤ (p − 1)/2. Then j

∈ V for 1 ≤ n ≤ p − 2. Furthermore,

p−2

Y

n=1

j

= τ

N .

Therefore τ

∈ V. Since τ

τ

= N , τ

τ

∈ V. Now, let 2 ≤ r ≤ (p − 1)/2 and assume that we have proved that τ

τ

∈ V. We have

j

= τ τ

τ

= N τ τ

N τ

. Thus

τ

= j

τ

τ

and τ

τ

= j

τ

τ

. Hence τ

τ

∈ V and the lemma follows.

Lemma 2.3 . Let ℓ be a prime number, ℓ 6 = p. Let P be a prime ideal of O above ℓ and let a ∈ { 1, . . . , p − 2 } . Then Q(j

(P )) = L if and only if ℓ ≡ 1 (mod p) and a

+ a + 1 6≡ 0 (mod p) if p ≡ 1 (mod 3).

Proof. Since j

(P) ≡ 1 (mod π

) and j

(P )j

(P )

= ℓ

where f is the order of ℓ in ( Z /p Z )

, we have

∀ σ ∈ ∆, j

(P)

= j

(P) ⇔ j

(P )

O = j

(P ) O . Recall that

∀ σ ∈ ∆, j

(P )

O = j

(P ) O ⇔ P

= O . Since j

(P )

= j

(P ), we can assume ℓ ≡ 1 (mod p). Let σ ∈ ∆. We have to consider the following equation in C[∆]:

(σ − 1)(1 + σ

− σ

)θ = 0.

This is equivalent to

∀ ψ ∈ ∆, ψ b odd, (ψ(σ) − 1)(1 + ψ(a) − ψ(1 + a)) = 0.

Assume that ω

(σ) 6 = 1. Then

1 + ω

(a) − ω

(1 + a) = 0.

This implies a

+ a ≡ 0 (mod p), which is a contradiction. Thus ω

(σ) = 1.

Suppose that σ 6 = 1. We get 1 + ω(a) = ω(1 + a), which is equivalent to a

+ a + 1 ≡ 0 (mod p).

Conversely, one can see that if p ≡ 1 (mod 3), a

+ a + 1 ≡ 0 (mod p), and ω

(σ) = 1, then

∀ ψ ∈ ∆, ψ b odd, (ψ(σ) − 1)(1 + ψ(a) − ψ(1 + a)) = 0.

The lemma follows.

For x ∈ Z

, let [x] ∈ { 0, . . . , p − 1 } be such that x ≡ [x] (mod p). We set η =

Y

n=1

j

∈ J

. Lemma 2.4 .

(a) Let ψ ∈ ∆, ψ b 6 = ω, ψ odd. Then e

X

n=1

(1 + σ

− σ

)[n

]

∈ Z

e

. (b) We have

1

p e

X

n=1

(1 + σ

− σ

)[n

]

∈ Z

e

. Proof. (a) Write ψ = ω

, k odd, k ∈ { 3, . . . , p − 2 } . We have X

n=2

(1 + ψ(n) − ψ(1 + n))[n

] ≡

p−1

X

n=1

1 + n

− (1 + n)

n ≡ k (mod p).

This implies (a).

(b) We have

∀ a ∈ F

, ω(a) ≡ a

(mod p

).

Thus 1 p

X

n=1

(1 + ω(n) − ω(1 + n))[n

] ≡ −

p−1

X

n=1

X

k=1

p!

(p − k)!k!p n

(mod p), and we get

1 p

X

n=1

(1 + ω(n) − ω(1 + n))[n

] ≡ − 1 (mod p).

This implies (b).

Lemma 2.5 . Let ℓ be a prime number, ℓ 6 = p. Let V

be the Z[∆]- submodule of L

/(L

)

generated by { f(P ) : f ∈ J } where P is some prime of I above ℓ. Let ψ ∈ ∆, ψ b odd and ψ 6 = ω. Then

V

(ψ) = F

e

η(P ).

Proof. Let E = L(ζ

). Then L

(L

)

(ψ) ֒ → E

(E

)

(ψ).

Now, in

(ψ), we have V

(ψ) = F

e

τ (P). It remains to apply Lemma

2.4.

Finally, we record the following lemma:

Lemma 2.6 . We have ( J

: Z[∆]η) = 2

1

p Y

ψ∈∆, ψb odd

X

n=1

(1 + ψ(n) − ψ(1 + n))[n

] .

Furthermore ( J

: Z [∆]η) 6≡ 0 (mod p).

Proof. Set J e

= (1 − σ

) J ⊂ J

. Then (see [12, Sec. 6.4]):

( J

: J e

) = 2

.

Now, by the same kind of argument as in [12, Sec. 6.4], we get ( J e

: Z[∆]η) = 1

p Y

ψ∈∆, ψb odd

X

n=1

(1 + ψ(n) − ψ(1 + n))[n

] .

It remains to apply Lemma 2.4 to conclude the proof.

3. Jacobi sums and the ideal class group of Q (ζ

). Recall that the Iwasawa–Leopoldt Conjecture ([9, p. 258]) asserts that A is a cyclic Z

[∆]-module. This conjecture is equivalent to:

∀ ψ ∈ ∆, ψ b odd, ψ 6 = ω, A(ψ) ≃ Z

/B

Z

. It is well-known (see [12, Theorem 10.9]) that

∀ ψ ∈ ∆, ψ b odd, ψ 6 = ω, A(ωψ

) = { 0 } ⇒ A(ψ) ≃ Z

/B

Z

. In this section, we will study the links between Jacobi sums and the structure of A

.

We fix ψ ∈ ∆, ψ b odd and ψ 6 = ω. We set m(ψ) = v

(B

).

Recall that, by [12, Sec. 13.6], we have | A(ψ) | = p

. Let p

be the exponent of the group A(ψ). Then

B

≡ 0 (mod p

).

Lemma 3.1 . Let P be a prime ideal in I above a prime number ℓ. Then e

η(P) O = 0 in I / I

⇔ ψ(ℓ) 6 = 1 or B

≡ 0 (mod p).

Proof. First note that, if ρ ∈ ∆, b then e

P = 0 in I / I

if and only if ρ(ℓ) 6 = 1. By the Stickelberger Theorem, we have

η(P) O =

X

n=1

(1 + σ

− σ

)[n

]

(1 − σ

)θP.

Recall that e

θ = B

e

. The lemma follows.

Lemma 3.2 . Let f ∈ W

. Then f lies in W

if and only if f (P ) ∈ (L

)

for all prime ideals P ∈ I .

Proof. Let f ∈ W

be such that f (P ) ∈ (L

)

for all prime ideals P ∈ I . Let A ∈ I . Then there exists γ

∈ L

∩ U such that γ

γ

= 1 and f ( A ) = γ

. Observe that β(f ) ∈ p(Z[∆])

. Let g : I → L

, A 7→ γ

. Then one can verify that f = g

and g ∈ W

.

Let m ≥ 1 be such that p

> | A | . Set n = | Cl(L) | / | A | . Let e

(ψ) ∈ Z[∆]

be such that

e

(ψ) ≡ e

(mod p

).

Set

β

= 2np

e

(ψ) ∈ Z [∆]

.

Since np

e

(ψ) ∈ (Ann

Cl(L))

, by Lemma 2.1 there exists a unique element f

∈ W

such that β(f

) = β

. Recall that

(Ann

A)(ψ) = p

Z

e

. Therefore, for 0 ≤ k ≤ m,

(W⁻)^pk

(ψ) is cyclic of order p

generated by the image of f

. We set

W = { f( A ) : A ∈ I , f ∈ W} , J = { f ( A ) : A ∈ I , f ∈ J } .

Observe that J is a Z[∆]-submodule of W, and it is called the module of Jacobi sums of Q (ζ

). Note that, by Lemma 3.2 and the fact that

(ψ) 6 = { 0 } (recall that ψ is odd and ψ 6 = ω), we have

W (L

)

(L

)

(ψ) 6 = { 0 } .

Theorem 3.3 . The map f

induces an isomorphism of groups A(ψ) ≃ W (L

)

(L

)

(ψ).

Proof. First observe that m ≥ k(ψ) + 1. Let P be a prime in I . Then f

(P) O = P

.

Let ρ ∈ ∆, ρ b 6 = ψ. Then

e

(ρ)e

(ψ) ≡ 0 (mod p

).

Therefore, there exists γ ∈ L

∩ U such that:

P

= γ

σ

(γ )

O .

But (1 − σ

)e

(ψ) = 2e

(ψ). Thus, there exists α ∈ L

∩ U, ασ

(α) = 1, and

f

(P )

= α

.

Therefore, e

f

( I ) = 0 in L

/(L

)

. It is clear that f

induces a mor- phism

I

( I )

P (ψ) → L

(L

)

(ψ).

Now, let P be a prime in I such that e

f

(P) = 0 in

(L^∗)^pk(ψ)

(ψ). Then, by the above remark, we get f

(P ) = 0 in L

/(L

)

. Thus, there exists γ ∈ L

∩ U such that

P

= γ

O . Thus P

= γ O . This implies

e

P = 0 in I

( I )

P (ψ).

Thus our map is injective. Now, observe that the image of the map induced by f

is

(ψ)

(L^∗)^pk(ψ)

(ψ) and that A(ψ) ≃

(ψ). The theorem follows.

Recall that

η =

Y

n=1

j

∈ J

. Set

z = (1 − σ

)

p−2

X

n=1

(1 + σ

− σ

)[n

] ∈ Z[∆]

. We have β(η) = zθ.

Corollary 3.4 .

(1) The map η induces an isomorphism of groups A(ψ) ≃ J(L

)

(L

)

(ψ).

(2)

(ψ) 6 = { 0 } if and only if A(ψ) is Z

-cyclic.

Proof. (1) Let P be a prime in I . Then one can show that f

(P )

= η(P )

.

The first assertion follows from Theorem 3.3.

(2) Note that A(ψ) is Z

-cyclic ⇔ m(ψ) = k(ψ). Thus, if A(ψ) is Z

- cyclic, then

J (L

)

(L

)

(ψ) = W (L

)

(L

)

(ψ) 6 = { 0 } .

By the proof of (1), if k(ψ) < m(ψ) and if P is a prime in I , then η(P )

∈ (L

)

. Therefore, we get (2).

4. The p-adic behavior of Jacobi sums. Let M be a subgroup of L

/(L

)

. We say that M is unramified if L( √

M)/L is an unramified ex- tension. Note that Kummer’s Lemma asserts that ([12, Theorem 5.36])

∀ ρ ∈ ∆, ρ b even, ρ 6 = 1, E

E

(ρ) is unramified ⇒ B

≡ 0 (mod p).

It is natural to ask if this implication is in fact an equivalence (see [1], [3]).

We will say that the converse of Kummer’s Lemma is true for the character ρ if

E

E

(ρ) is unramified ⇔ B

≡ 0 (mod p).

In this section, we will study this question with the help of Jacobi sums.

Let F/L be the maximal abelian p-extension of L which is unramified outside p. Set X = Gal(F/L). We have an exact sequence of Z

[∆]-modules ([12, Corollary 13.6])

0 → U/E → X → A → 0.

Let ρ ∈ ∆ b and observe that:

• if ρ = 1, ω then X (ρ) ≃ Z

,

• if ρ is even, ρ 6 = 1, then X (ρ) ≃ Tor

X (ρ),

• if ρ is odd, ρ 6 = ω, then X (ρ) ≃ Z

⊕ Tor

X (ρ).

Lemma 4.1 . Let ψ ∈ ∆, ψ b odd, ψ 6 = ω. Then d

Tor

X (ψ) = d

A(ωψ

).

Proof. This is a consequence of the proof of Leopoldt’s reflection theorem ([12, Theorem 10.9]). For the convenience of the reader, we give the proof.

Let H be the Galois group of the maximal abelian extension of L which is unramified outside p and of exponent p. Then H is a Z

[∆]-module and we have:

• H(1) ≃ F

and corresponds to L(ζ

)/L,

• H(ω) ≃ F

and corresponds to L( √

p)/L,

• if ρ is even, ρ 6 = 1, then d

H(ρ) = d

Tor

X (ρ),

• if ρ is odd, ρ 6 = ω, then d

H(ρ) = 1 + d

Tor

X (ρ).

Let V be the Z[∆]-submodule of L

/(L

)

which corresponds to H, i.e.

H = Gal(L( √

V )/L). Let M be the Z [∆]-submodule of L

/(L

)

generated by E and 1 − ζ

. We have an exact sequence

0 → M → V → A[p] → 0.

Let ψ ∈ ∆, ψ b odd, ψ 6 = ω. By Kummer theory we have

1 + d

Tor

X (ψ) = d

V (ωψ

),

and, by the above exact sequence,

d

V (ωψ

) = 1 + d

A(ωψ

).

The lemma follows.

Lemma 4.2 . Let ρ ∈ ∆, ρ b even and ρ 6 = 1. If

(ρ) is ramified then d

A(ρ) = d

A(ωρ

).

Proof. We keep the notations of the proof of Lemma 4.1. Let V

⊂ V correspond via Kummer theory to A/pA. Then

V

(ρ) ≃ A

pA (ωρ

).

But

(ρ) is ramified if and only if V

(ρ) ֒ → A[p](ρ). Now recall that d

A(ρ) ≤ d

A(ωρ

). The lemma follows.

Lemma 4.3 . There exists a unique Z [∆]-morphism ϕ : K

→ Z

[∆] such that

∀ x ∈ K

, ϕ(x)ζ

= Log

(x).

Furthermore,

Im ϕ = M

ρ=1,ω

p Z

e

⊕ M

ρ6=1,ω

Z

e

.

Proof. Let λ ∈ K

be such that λ

= − p. Then K

= λ

× µ

× µ

× U.

Recall that:

• the kernel of Log

on K

is equal to λ

× µ

× µ

,

• Log

(U ) = π

Z

[ζ

].

For ρ ∈ ∆, b set

τ (ρ) = X

a=1

ρ(a)ζ

∈ Z

[ζ

].

Then e

ζ

= τ (ρ

). But recall that Z

[ζ

] = Z

[∆]ζ

. Thus e

Z

[ζ

] = Z

τ (ρ

).

If ρ = ω

, k ∈ { 0, . . . , p − 2 } , we have

v

(τ (ρ

)) = k p − 1 . Therefore

π

Z

[ζ

] = M

ρ=1,ω

pZ

τ (ρ

) ⊕ M

ρ6=1,ω

Z

τ (ρ

).

The lemma follows.

Let P be a prime in I . We fix a generator r

∈ F

such that χ

(r

) = ζ

.

For x ∈ F

, let Ind(P, x) ∈ { 0, . . . , N P − 2 } be such that x = r

.

We recall the following theorem (see also [10] for a statement similar but weaker than part (2) below):

Theorem 4.4 . (1) ϕ(1 − ζ

) = P

ρ∈∆, ρ6=1, ρb even

− (p − 1)

L

(1, ρ)e

.

(2) Let ψ ∈ ∆, ψ b odd, ψ 6 = ω. Write ψ = ω

, k ∈ { 2, . . . , p − 2 } . Then e

ϕ(η(P )) ≡ 2k Ind

P,

p−1

Y

a=1

1 − ζ

1 − ζ

e

(mod p).

Proof. (1) Let ρ ∈ ∆, b ρ even, ρ 6 = 1. By [12, Theorem 5.18], we have L

(1, ρ)τ (ρ

) = − (p − 1)e

Log

(1 − ζ

).

Thus the first assertion follows.

(2) Let ψ ∈ ∆, b ψ odd, ψ 6 = ω. By a beautiful result of Uehara ([11, Theorem 1]), we have

e

Log

(η(P )) ≡ 2k Ind

P,

p−1

Y

a=1

1 − ζ

1 − ζ

τ (ψ

) (mod p).

This implies the second assertion.

Theorem 4.5 . Let ψ ∈ ∆, b ψ 6 = ω, ψ odd. We have exact sequences 0 → Tor

X (ψ) → A(ψ) → W (ψ)/U

(ψ) → 0,

0 → Tor

X (ψ) → A(ψ) → J (ψ)/U

(ψ) → 0.

Proof. This is a consequence of the method developed by Iwasawa [5].

We briefly recall it.

Let f ∈ W . For n ≥ 2, set P

= { α O : α ≡ 1 (mod π

) } . Observe that f ( P

) ⊂ 1 + π

Z

[ζ

].

Let

X e = lim ←− I / P

.

If F e is the maximal abelian extension of L which is unramified outside p, then, by class field theory,

X ≃ e Gal( F /L). e

By [12, Theorem 13.4], the natural surjective map X → X e has a finite kernel of order prime to p. Thus f induces a map

f ¯ : X → U.

Furthermore,

f ¯ (U ) = U

⊂ f ¯ ( X ).

Now let ψ ∈ ∆, b ψ odd, ψ 6 = ω. We have a map f ¯ : X (ψ) → U (ψ).

But

X (ψ) ≃ Z

⊕ Tor

X (ψ) and U (ψ) ≃ Z

. Thus, if e

β(f) 6 = 0, we get

Ker( ¯ f : X (ψ) → U (ψ)) = Tor

X (ψ).

Therefore, if e

β(f) 6 = 0, we get the following exact sequence induced by f : 0 → Tor

X (ψ) → A(ψ) → f ¯ ( X )(ψ)/U

(ψ) → 0.

It remains to apply this construction to f

and η to get the desired exact sequences.

Corollary 4.6 .

(1) Let ψ ∈ ∆, b ψ odd, ψ 6 = ω. Then

d

A(ψ) = 1 + d

A(ωψ

) ⇔ B

≡ 0 (mod p) and W (ψ) = U (ψ).

(2) Let ρ ∈ ∆, b ρ even and ρ 6 = 1. Assume that B

≡ 0 (mod p) and that W (ωρ

) = U (ωρ

). Then the converse of Kummer’s Lemma is true for the character ρ.

Proof. (1) We apply Theorem 4.5. We identify Tor

X (ψ) with its image in A(ψ). We can write A(ψ) = B ⊕ C, where C is cyclic of order p

and B ⊂ Tor

X (ψ). Now,

(C : C ∩ Tor

X (ψ)) = (W (ψ) : U

(ψ)).

It remains to apply Lemma 4.1 to get the desired result.

(2) We apply the first assertion and Lemma 4.1 to deduce that d

A(ρ) = d

A(ωρ

) − 1. It remains to apply Lemma 4.2.

We set

W

= { α ∈ W : α ∈ U

} .

Let ψ ∈ ∆, b ψ odd, ψ 6 = ω. We assume that B

≡ 0 (mod p). Write A(ψ) = Z /p

Z ⊕ · · · ⊕ Z /p

Z ,

where t = d

A(ψ) and 1 ≤ e

≤ · · · ≤ e

= k(ψ). Set

n(ψ) = |{ i ∈ { 1, . . . , t } : e

= k(ψ) }| .

Corollary 4.7 . We have

n(ψ) − 1 ≤ dim

W

(L

)

/(L

)

≤ n(ψ).

Furthermore,

dim

W

(L

)

/(L

)

= n(ψ) ⇔ W (ψ) 6 = U (ψ).

Proof. By Theorems 4.5 and 3.3, we have

W

(L

)

/(L

)

≃ Ker(A(ψ) → W (ψ)/U

(ψ)).

The corollary follows.

Corollary 4.8 . Assume that pA

= { 0 } . Then we have an isomor- phism of groups

Gal(L( √

W

)/L) ≃ A

/pA

.

Proof. This is a consequence of Kummer theory, Corollary 4.7 and Corol- lary 4.6.

Note that the above results lead to the following problem (which is a restatement of the converse of Kummer’s Lemma): do we have ϕ(W

) = (Im ϕ)

? Observe that e

ϕ(W

) = e

(Im ϕ)

, and since K

(Z) = { 0 } , we have A(ω

) = { 0 } (see [7]) and therefore e

ϕ(W

) = e

(Im ϕ)

.

5. Remarks on the jacobian of the Fermat curve over a finite field. First we fix some notations and recall some basic facts about global function fields.

Let F

be a finite field having q elements. Let ℓ be the characteris- tic of F

, ℓ 6 = p. Let F

be a fixed algebraic closure of F

and let f F

= S

n≥1, n6≡0 (modp)

F

⊂ F

. Let k/F

be a global function field such that F

is algebraically closed in k. We set:

• D

: the group of divisors of k,

• D

: the group of divisors of degree zero of k,

• P

: the group of principal divisors of k,

• J

: the jacobian of k; note that

∀ n ≥ 1, J

(F

) ≃ D

/P

,

• g

: the genus of k,

• L

(Z) ∈ Z [Z ]: the numerator of the zeta function of k; we recall that L

(Z)

(1 − Z )(1 − qZ ) = Y

vplace ofk

(1 − Z

)

, furthermore deg

L

(Z ) = 2g

and L

(1) = | J

( F

) | ,

• C

(F

) = J

(F

) ⊗

Z

,

• d e

J

= d

C

( f F

); observe that there exists an integer m 6≡ 0 (mod p) such that C

(f F

) = C

(F

).

Write

L

(Z) =

2g_k

Y

i=1

(1 − α

Z ).

For simplicity, we assume that v

(α

− 1) > 0 for i = 1, . . . , 2g

. In this case, C

(f F

) = C

(F

).

Set P

(Z) = Q

i=1

(Z − (α

− 1)). Let γ be the Frobenius of F

, and set C

(k) = C

(F

).

Let C

(k) = S

n≥0

C

(k), and set

M

= Hom( Q

/ Z

, C

(k)).

Then M

is isomorphic to the p-adic Tate module of J

. Set Λ = Z

[[Z]]

where Z corresponds to γ − 1. Then it is well-known that:

• M

is a Λ-module of finite type and of torsion,

• as a Z

-module, M

is isomorphic to Z

,

• M

/ω

M

≃ C

(k), where ω

= (1 + Z)

− 1,

• Char

M

= P

(Z)Λ,

• the action of Z on M

is semisimple, i.e. the minimal polynomial of the action of Z on M

has only simple roots.

Now, let ℓ be a prime number, ℓ 6 = p. We fix a prime P of O above ℓ and we view O /P as a subfield of F

, thus F

= O /P ⊂ F f

. We identify ζ

with its image in F

. Let X be an indeterminate over F

. We set k = F

(X, Y ) where X

+ Y

= 1, and we set T = X

. For a, b ∈ Z, let τ

∈ Gal(F

k/F

(T )) be such that

τ

(X) = ζ

X and τ

(Y ) = ζ

Y.

Let a ∈ { 1, . . . , p − 2 } . Let H

be the subgroup of Gal(F

k/F

(T )) generated by τ

. Set

E

= F

(T, XY

).

If we set U = T and V = XY

, then V

− U (1 − U )

= 0 and of course E

= F

(U, V ). We set

E = F

E

, F = F

k,

and observe that F f

= f F

. It is clear that F

= E. Finally, we set G = Gal(E/ F

(T)).

Note that g

= (p − 1)/2.

Lemma 5.1 . We have

L

(Z ) = Y

σ∈∆

(1 − j

(P)

Z).

Proof. Let χ ∈ G b be such that χ(g) = ζ

, where g ∈ G is such that g(XY

) = ζ

XY

. Note that

L

(Z ) = Y

σ∈∆

L(Z, χ

), where L(Z, χ) = Y

vplace ofFq(T)

(1 − χ(v)Z

)

. Since 2g

= p − 1, we get deg

L(Z, χ) = 1.

For b ∈ F

\ { 0, 1 } , we denote the Frobenius of T − b in E/F

(T ) by Frob

. We have

Frob

(XY

) = (b(1 − b)

)

XY

. But

L(Z, χ) ≡ 1 + X

b∈Fq\{0,1}

χ(Frob

)

X (mod X

).

Thus

L(Z, χ) = 1 + X

b∈Fq\{0,1}

χ(Frob

) X.

But we can write

j

(P) = −

p−1

X

i=0

N

ζ

,

where N

= |{ α ∈ F

\{ 0, 1 } : (α(1 − α)

)

≡ ζ

(mod P) }| . Therefore j

(P) = − X

b∈Fq\{0,1}

χ(Frob

).

The lemma follows.

Theorem 5.2 . Let n be the smallest integer (if it exists) such that 3 ≤ n ≤ p − 2, n is odd and e

j

(P ) 6∈ U

. Then

J

( F f

)

⊗

Z

≃ ( Z /p Z )

. If such an integer does not exist then:

(1) d e

J

= p − 1, (2) we have

J

( F f

)

⊗

Z

≃ (Z/pZ)

⇔ ℓ

6≡ 1 (mod p

).

Proof. The proof is based on ideas developed by Greenberg [4]. Write

H = H

. Let P

be the prime of E above T , P

the prime of E above T − 1

and P

the prime of E above 1/T . Recall that in D

we have p(P

− P

) = (T),

p(P

− P

) = (T − 1), P

− P

+ a(P

− P

) = (XY

).

Thus, by [4, Sec. 2],

J

(F

)

≃ Z/pZ,

and J

(F

)

is generated by the class of P

− P

. Observe also that F/E is unramified and cyclic of order p. Let us start with the exact sequence

0 → F

→ F

→ P

→ 0.

We get

P

/P

≃ Z/pZ,

and P

/P

is generated by the image of P

− P

in D

. In particular, P

/P

≃ J

( F

)

.

Note that we also have

0 → H

(H, P

) → H

(H, F

) → H

(H, F

).

But F/E is unramified and cyclic, therefore every element of F

is a norm in the extension F/E. Thus

H

(H, P

) ≃ Z/pZ.

Now, we look at the exact sequence

0 → P

→ D

→ J

(F

) → 0.

Since F/E is unramified,

H

(H, D

) = { 0 } .

Therefore, we have obtained the following exact sequence:

0 → J

( F

)

→ J

( F

) → J

( F

)

→ Z /p Z → 0.

Now, it is not difficult to deduce that, for all n ≥ 1, we have the exact sequence

0 → Z/pZ → J

(F

) → J

(F

)

→ Z/pZ → 0.

From this, we get the following exact sequence of Z

[G]-modules and Λ- modules:

0 → M

→ M

→ Z/pZ → 0.

Recall that in our situation, by Lemma 5.1, P

(Z) = Y

σ∈∆

(Z − (j

(P )

− 1)).

Furthermore the actions of G and Z commute on M

. Now, we have:

• Char

M

= Char

M

= P

(Z)Λ,

• M

≃ Z

as Z

-modules,

• M

/ω

≃ C

(F )

. Observe that

C

(F)

= J

( F f

)

⊗

Z

.

Note also that the minimal polynomial of the action of Z on M

is Irr(j

(P ) − 1, Q

; Z) := G(Z).

Set N = P

δ∈G

δ. Then one can see that

N M

= N M

= { 0 } .

Thus M

is a Z

[G]/N Z

[G]-module. Now, we identify Z

[G]/N Z

[G] with Z

[ζ

]. Since M

≃ Z

, there exists m ∈ M

such that

M

≃ Z

[ζ

].m,

i.e. M

is a free Z

[ζ

]-module of rank one. Therefore there exists an element x ∈ Z

[ζ

] such that Zm = xm. Now set

D(Z) = Y

σ∈∆

(Z − x

) ∈ Λ.

Then D(Z)M

= { 0 } . Therefore G(Z) divides D(Z ) in Λ. Thus there exists σ ∈ ∆ such that

x

= j

(P ) − 1.

But

C

(F )

≃ M

/ZM

≃ Z

[ζ

]/x Z

[ζ

].

Therefore, we get

J

( F f

)

⊗

Z

≃ Z

[ζ

]/(j

(P ) − 1) Z

[ζ

].

Recall that j

(P ) ≡ 1 (mod π

). Thus

v

(j

(P) − 1) = v

(Log

(j

(P ))).

Now

Log

(j

(P )) = 1

2 f Log

(ℓ) + X

ψ∈∆, ψb odd

e

Log

(j

(P )),

where f is the order of ℓ in (Z/pZ)

. Let ψ ∈ ∆, b ψ = ω

, n odd. If e

Log

(j

(P)) 6 = 0, then

v

(e

Log

(j

(P))) ≡ n

p − 1 (mod Z ), and furthermore

v

(e

Log

(j

(P ))) > n

p − 1 ⇔ e

j

(P) ∈ U

.

Note also that

v

(e

Log

(j

(P ))) > 1 p − 1 . The theorem follows.

Observe that the proof of the above theorem implies that we have an isomorphism of Z[G]-modules

J

( F f

) ≃ J

( F f

)

.

Corollary 5.3 . Let n ∈ { 3, . . . , p − 2 } , n odd. Let a ∈ { 1, . . . , p − 2 } be such that 1 + a

− (1 + a)

6≡ 0 (mod p). The following assertions are equivalent:

(1) A(ω

) = { 0 } ,

(2) there exists a prime number ℓ 6 = p such that d e

J

= n, where E

= F

(U, V ) and V

− U (1 − U )

= 0.

Proof. Observe that (2) implies (1) by Theorems 5.2 and 4.5. Write ψ = ω

. Let ℓ be a prime number, ℓ 6 = p. Write

F

= O /ℓ O and D

= F

/( F

)

.

Observe that D

is a Z

[∆]-module. Let Cyc be the group of cyclotomic units of L. We denote the image of Cyc in D

by Cyc

. Then Theorem 4.4 asserts that e

Cyc

= { 1 } in D

if and only if e

j

(P) ∈ U

, where P is a prime of O above ℓ. Let

B = L( p

Cyc).

We assume that (1) holds. By the Chebotarev density theorem applied to the extension B/L, there exist infinitely many primes ℓ such that:

• e

Cyc

= { 1 } for ρ 6 = ψ,

• e

Cyc

6 = { 1 } .

It remains to apply Theorem 5.2 and the above remarks to get (2).

Now, let ℓ be a prime number. Let p be an odd prime number, p 6 = ℓ. Let T be an indeterminate over F

and let E

/F

(T ) be the imaginary quadratic extension defined by

E

= F

(T, X) where X

− X + T

= 0.

Let n be an odd integer, n ≥ 3. Let S

(ℓ) denote the set of primes p such that d e

J

= n. By our results above, if p ∈ S

(ℓ) then A(ω

) = { 0 } . Observe that if ℓ

6≡ 1 (mod p) then p 6∈ S

(ℓ), and therefore S

(ℓ) is a finite set. Set S(ℓ) = S

n

S

(ℓ), where n runs through the odd integers. Observe

that if the order of ℓ modulo p is even then p 6∈ S(ℓ). Therefore, by a classical

result of Hasse (see [8]) there exist infinitely many primes p not in S(ℓ) (in

fact at least “2/3 of the prime numbers” are not in S(ℓ)). Thus, we ask the following question: is S(ℓ) infinite?

Acknowledgments. The authors thank Cornelius Greither for inter- esting discussions on the converse of Kummer’s Lemma which led us to the study of the analogous statement for the odd part. The second author thanks the mathematicians of the Laboratoire de Math´ematiques Nicolas Oresme for their hospitality during his stay at Caen.

References

[1] B. Angl`es, Units and norm residue symbol, Acta Arith. 98 (2001), 33–51.

[2] B. Angl`es and T. Beliaeva, On Weil numbers in cyclotomic fields, Int. J. Number Theory 5 (2009), 871–884.

[3] J. Assim et T. Nguyen Quang Do, Sur la constante de Kummer–Leopoldt d’un corps de nombres, Manuscripta Math. 115 (2004), 55–72.

[4] R. Greenberg, On the jacobian variety of some algebraic curves, Compos. Math. 42 (1980), 345–359.

[5] K. Iwasawa, A note on Jacobi sums, in: Symposia Math. XV, Academic Press, London, 1975, 447–459.

[6] —, Some remarks on Hecke characters, in: Algebraic Number Theory (Kyoto, 1976), S. Iyanaga (ed.), Japan Soc. Promotion Sci., Tokyo, 1977, 99–108.

[7] M. Kurihara, Some remarks on conjectures about cyclotomic fields and K-groups of Z, Compos. Math. 81 (1992), 223–236.

[8] J. C. Lagarias, The set of primes dividing the Lucas numbers has density 2/3, Pacific J. Math. 118 (1985), 449–461.

[9] S. Lang, Units and class groups in number theory and algebraic geometry, Bull.

Amer. Math. Soc. 6 (1982), 253–316.

[10] F. Thaine, On the coefficients of Jacobi sums in prime cyclotomic fields, Trans.

Amer. Math. Soc. 351 (1999), 4769–4790.

[11] T. Uehara, On a congruence relation between Jacobi sums and cyclotomic units, J. Reine Angew. Math. 382 (1987), 199–214.

[12] L. C. Washington, Introduction to Cyclotomic Fields, 2nd ed., Springer, 1997.

Bruno Angl`es

Universit´e de Caen, CNRS UMR 6139

Campus II, Boulevard Mar´echal Juin, B.P. 5186 14032 Caen Cedex, France

E-mail: bruno.angles@math.unicaen.fr

Filippo A. E. Nuccio Istituto Guido Castelnuovo Universit` a “La Sapienza”

5, Piazzale Aldo Moro 00186 Roma, Italy E-mail: nuccio@mat.uniroma1.it Received on 14.2.2008

and in revised form on 12.5.2009 (5644)