dividing l, and b K the λ-adic completion of K. For α, β in K

LXXIII.2 (1995)

Norm residue symbol and cyclotomic units

by

Charles Helou (Media, Penn.)

Introduction. Let l be a prime number ≥ 5, ζ a primitive lth root of unity in C, K = Q(ζ), O

= Z[ζ] the ring of integers of K, λ = 1 − ζ, prime element of O

dividing l, and b K the λ-adic completion of K. For α, β in K

= K − {0}, denote by (α, β)

the norm residue symbol (i.e. Hilbert symbol) as defined in [5] where it is written

, or in [2] where the symbol is the inverse of the one in [5] used here. Let [α, β] denote the element of the finite field F

= Z/lZ defined by

(α, β)

= ζ

.

The symbol [α, β] is a bilinear skew-symmetric function of α, β, with respect to the multiplicative and additive structures of K

and F

respectively; and if [α, β] = 0, then α and β are said to be orthogonal.

The cyclotomic units, in O

, are u

= 1 − ζ

1 − ζ ,

for n ∈ Z not divisible by l (or 1 ≤ n ≤ l − 1). They generate a subgroup C of the multiplicative group of all units of O

.

Let a ∈ Z and α

= a − ζ. Terjanian conjectured ([9]) that every prime number l ≥ 5 has the property (called LC) that the following two conditions are equivalent:

(i) α

is orthogonal to C, (ii) a ≡ 0, 1, or −1 (mod l

).

He showed that (ii) implies (i), and if (i) holds then a is divisible by l

or a

≡ 1 (mod l

). But the latter condition, in conjunction with the congruence a ≡ 0, 1, or −1 (mod l), implies a ≡ 0, 1, or −1 (mod l

). Thus, Terjanian’s conjecture amounts to the assertion that if α

is orthogonal to C then a ≡ 0, 1, or −1 (mod l).

In this paper, we use the Artin–Hasse explicit reciprocity law to give an

[147]

expression for [α

, u

], when a 6≡ 1 (mod l), involving the Fermat quotient q(x) = x

− 1

l (for x ∈ Z, l - x) and the polynomial sums in F

S

(a) = X

1≤k≤l−1 k≡r (mod n)

a

k (for n, r ∈ Z, l - n).

Some properties of these sums are established, in order to simplify the ex- pression for [α

, u

]. This yields necessary and sufficient conditions for α

to be orthogonal to C, in the form of equations over F

.

Furthermore, we assume l ≡ 1 (mod 4) and consider the quadratic sub- field E = Q( √

l) of K. When n is not a quadratic residue modulo l, the norm of u

in K|E is an element of C which, by Dirichlet’s class number formula, is equal to ε

, where ε is the fundamental unit and h the class number of E. Also, by a result on real quadratic fields, h < √

l, so that l - h.

Thus, if α

is orthogonal to C then it is orthogonal to ε. Upon using again the Artin–Hasse law, an expression is obtained for [α

, ε] in terms of ε and the polynomial

F (X) = X

 k l



2

X

, where

2

is the Legendre symbol. It follows that, assuming the validity of a conjecture by Ankeny, Artin and Chowla ([1]) concerning ε, if α

is orthogonal to C, then a is a root of F in F

. This polynomial F has in F

the trivial roots 0, 1, −1 (with 1 of multiplicity (l − 1)/2) and takes in K as values for X = ζ

(1 ≤ m ≤ l − 1) the quadratic Gauss sums; it can also be written in terms of the polynomial

A(X) = Y

1≤k≤(l−1)/2

(X − ζ

),

whose coefficients lie in O

. The polynomials F and A were studied by Jacobi ([6]) and Dirichlet ([4]) respectively.

1. Application of the Artin–Hasse law. Let a ∈ Z, a 6≡ 1 (mod l), α

= a−ζ and c ∈ Z such that (a−1)c ≡ 1 (mod l). For any n ∈ Z with l - n, let σ

be the element of the Galois group of K|Q defined by σ

(ζ) = ζ

. Denote by log α the λ-adic logarithm of any unit α in b K; in particular, for α = 1 + x with x ≡ 0 (mod λ),

log(1 + x) = X

(−1)

k x

.

Denote by Tr the trace from b K to the field Q

of l-adic numbers (whose restriction to K coincides with the trace from K to Q). Further notations will be introduced as needed.

Proposition 1. For a 6≡ 1 (mod l) and for any 1 ≤ n ≤ l − 1, [α

, u

] = n(a

− a)

l(a − 1) − n + 1

2 q(a − 1) + cq(c) − n l Tr

 log(1 + cλ) σ

(λ)

 . P r o o f. We have α

= a − 1 + λ ≡ (a − 1)(1 + cλ) (mod λ

), so that ([9], (R10) or [5], p. 54)

[α

, u

] = [a − 1, u

] + [1 + cλ, u

].

For the first bracket, since u

=

n−1

X

k=0

ζ

≡

n−1

X

k=0

(1 − kλ) ≡ n − n(n − 1)

2 λ (mod λ

), we have ([9], (R22) or [5], p. 110)

[a − 1, u

] = n − 1

2 q(a − 1).

For the second bracket, since u

= σ

(λ)/λ, we have

[1 + cλ, u

] = [1 + cλ, σ

(λ)] − [1 + cλ, cλ] + [1 + cλ, c].

Moreover ([9], (R19) or [5], p. 54),

[1 + cλ, σ

(λ)] = n[1 + cσ

(λ), λ], and ([9], (R18) or [5], p. 55)

[1 + cλ, cλ] = [1 + cλ, 1] + [1, cλ] = 0, and ([9], (R23) or [5], p. 110)

[1 + cλ, c] = cq(c).

Therefore

(1) [α

, u

] = n − 1

2 q(a − 1) + n[1 + cσ

(λ), λ] + cq(c).

By the Artin–Hasse reciprocity law ([2], Ch. 12, Th. 10, whose third part is missing a factor 1/λ of ζ log α; cf. [5], p. 94),

(2) [1 + cσ

(λ), λ] = − 1 l Tr

 ζ

λ log(1 + cσ

(λ))

 .

Let b σ

be the Q

-automorphism of b K extending σ

. For any α ∈ b K,

Tr(b σ

(α)) = Tr(α), and if α ≡ 1 (mod λ), then, by continuity of b σ

in

the λ-adic topology, b σ

(log α) = log(b σ

(α)). Therefore, in the right-hand

side of (2), the argument of Tr may be replaced by

n(λ)

log(1 + cλ), which gives, since ζ

= 1 − σ

(λ),

(3) Tr

 ζ

λ log(1 + cσ

(λ))



= Tr

 log(1 + cλ) σ

(λ)



− Tr(log(1 + cλ)).

Similarly, letting N be the norm in b K|Q

, and using the λ-adic continuity of the b σ

’s and the multiplicative-additive homomorphism property of log, we have, for α ≡ 1 (mod λ),

(4) Tr(log(α)) = log(N (α)) ≡ N (α) − 1 (mod l

).

In particular, for α = 1 + cλ = 1 + c − cζ, the norm is (1 + c)

− c

, so that

(5) 1

l Tr(log(1 + cλ)) ≡ (c + 1)

− c

− 1

l (mod l).

Moreover, in F

, we have c = 1/(a − 1) and (a − 1)

= a − 1, so that (6) (c + 1)

− c

− 1

l = 1

a − 1

 a

− a

l − (a − 1)q(a − 1)



, in F

. The result now follows from (1), (2), (3), (5) and (6).

Corollary. We have [α

, u

] = n(a

− a)

l(a − 1) − n + 1

2 q(a − 1) + cq(c) + n l

X

(−c)

k Tr

 λ

σ

(λ)

 . P r o o f. Let D be the different of the extension b K|Q

([8], Ch. 3). For x in K, if x ∈ (l b

)D

then Tr(x) lies in the ideal (l

). Since the ring of integers of b K is Z

[ζ] and the irreducible polynomial of ζ over Q

is Φ(X) =

, we have

D = (Φ

(ζ)) = (lλ

) = (λ

).

Thus x ≡ 0 (mod λ

) implies Tr(x) ≡ 0 (mod l

). Since log(1 + cλ)

σ

(λ) ≡ X

(−1)

k c

λ

σ

(λ) (mod λ

), it follows that

Tr

 log(1 + cλ) σ

(λ)



≡ X

(−1)

k c

Tr

 λ

σ

(λ)



(mod l

).

Hence the result follows by substitution into Proposition 1.

Notations. For k ∈ Z such that l - k, the element 1/k is well defined in

F

; by abuse of notation, it will also be used in congruences modulo l. By

contrast, the appearance of l in a denominator means that the factor of 1/l

is divisible by l and that we are considering the quotient. For an integer

1 ≤ n ≤ l − 1, we denote by n

the specific representative of (1/n) (mod l) contained in the same interval, i.e. 1 ≤ n

≤ l − 1 and nn

≡ 1 (mod l).

For a real number x, the greatest integer ≤ x is written [x]. For a positive integer n and any m ∈ Z, we denote by res

(m) the least residue ≥ 0 of m modulo n, i.e.

res

(m) = m − n[m/n].

Lemma 1. For any integers 1 ≤ n ≤ l − 1 and 2 ≤ k ≤ l, Tr

 λ

σ

(λ)



= l X

0≤s<kn⁰/l

 k − 1 [sl/n

]



(−1)

, while for k = 1,

Tr

 λ

σ

(λ)



= l − n

. P r o o f. For 1 ≤ n ≤ l − 1 and 1 ≤ k ≤ l,

Tr

 λ

σ

(λ)



= X

(1 − ζ

)

1 − ζ

.

The general term in the latter sum can be written in the form 1 − ζ

1 − ζ

(1 − ζ

)

= X

i=1

(1 − ζ

)

n

X

ζ

. Moreover, for any m ∈ Z,

(7)

X

ζ

=

 l − 1 if l | m,

−1 if l - m.

The case k = 1 now follows easily, and so we assume 2 ≤ k ≤ l. In view of what precedes, and after replacing (1 − ζ

)

by its binomial expansion, we get

(8) Tr

 λ

σ

(λ)



=

n

X

k−1

X

t=0

 k − 1 t

 (−1)

X

ζ

.

The inner sum in (8) is equal to l − 1 or to −1 according as t ≡ −jn or t 6≡ −jn (mod l). Therefore the right-hand side of (8) splits into two sums obtained by replacing the inner sum respectively by −1 for all terms, and by l for those terms such that t ≡ −jn (mod l). The first sum is

(−1)

n

X

k−1

X

t=0

 k − 1 t



(−1)

= −

n

X

(1 − 1)

= 0,

and (8) is thus reduced to the second sum, i.e.

(9) Tr

 λ

σ

(λ)



= l X

t∈T

 k − 1 t

 (−1)

,

where T is the set of integers t satisfying the two conditions: 0 ≤ t ≤ k − 1, and t ≡ −jn (mod l) for some 0 ≤ j ≤ n

− 1. The latter condition is equivalent to : tn

+ j = sl for some s ∈ N and 0 ≤ j ≤ n

− 1, i.e.

j = res

(sl) and t = [sl/n

] for some s ∈ N. The former condition on t is then equivalent to 0 ≤ sl/n

< k. The result now follows by substitution for t in terms of s in (9).

Lemma 2. (a) For any integer 1 ≤ k ≤ l − 1, the binomial coefficient

is divisible by l and we have

1 k ≡ 1

l

 l k



(−1)

(mod l).

(b) For any algebraic integer α, we have the congruence (in any ring of algebraic integers containing α)

X

α

k ≡ (α − 1)

− α

+ 1

l (mod l).

(c) For any x, y ∈ Z not divisible by l, we have q(xy) ≡ q(x) + q(y) (mod l).

If x ≡ y (mod l), then

q(x) ≡ q(y) − 1

x · x − y

l (mod l).

If xy ≡ 1 (mod l), then

q(y) ≡ −q(x) − xy − 1

l (mod l).

(d) For any x ∈ Z not divisible by l, we have X

k=1

 kx l

 1

k ≡ xq(x) (mod l).

(e) For any integer 0 ≤ r ≤ l − 1, we have

 l − 1 r



≡ (−1)

 1 − l

X

1 j



(mod l

).

P r o o f. (a) The divisibility of the binomial coefficient by l is clear, and we have

1 l

 l k



= (l − 1)(l − 2) . . . (l − (k − 1))

k! ≡ 1

k! (−1)

(k − 1)! (mod l).

Hence the result.

(b) It follows from (a) and the binomial formula that X

k=1

α

k ≡ − 1

l X

 l k



(−α)

≡ − 1

l ((1 − α)

+ α

− 1) (mod l).

(c) First,

q(xy) = x

− 1

l y

+ y

− 1

l ≡ q(x) + q(y) (mod l).

Next, if x ≡ y (mod l), let x = y + hl, with h in Z, then, by the binomial formula,

q(x) = (y + hl)

− 1

l ≡ y

+ (l − 1)hly

− 1

l ≡ q(y) − hy

(mod l).

The conclusion follows, since y

≡ (1/y) (mod l).

Now if xy ≡ 1 (mod l), then, by what precedes, q(x) + q(y) ≡ q(xy) ≡ q(1) − 1

xy · xy − 1

l (mod l), and the result follows.

(d) For every 1 ≤ k ≤ l − 1, we have

kx ≡ res

(kx) (mod l) and kx − res

(kx)

l =

 kx l

 , so that, by (c) above,

q(k) + q(x) ≡ q(kx) ≡ q(res

(kx)) − 1 kx

 kx l



(mod l).

Hence, by summation, X

k=1

q(k) + (l − 1)q(x) ≡ X

q(res

(kx)) − 1 x

X

 kx l

 1

k (mod l).

Since the map k 7→ res

(kx) induces a permutation of the set {1, . . . , l − 1}, then P

k=1

q(res

(kx)) = P

k=1

q(k), and the result follows.

(e) The property holds trivially for r = 0, so we assume 1 ≤ r ≤ l − 1.

In Z[X], we have the congruence

(X − 1)(X − 2) . . . (X − r) ≡ (−1)

r! + (−1)

 X

r!

j



X (mod X

).

Hence we get the congruence in Z (l − 1)(l − 2) . . . (l − r) ≡ (−1)

r!

 1 −

 X

j=1

1 j

 l



(mod l

).

The result follows upon division of both sides by r!.

Proposition 2. For a 6≡ 1 (mod l) and for any 1 ≤ n ≤ l − 1, [α

, u

] = n − 1

2 q(a − 1) − q(n

) a − 1 + n

X

(−1)

k(a − 1)

X

1≤s<kn⁰/l

 k − 1 [sl/n

]



(−1)

.

P r o o f. Substituting the expressions for Tr(λ

/σ

(λ)), obtained in Lemma 1, into the sum that occurs in the expression for [α

, u

], established in the Corollary to Proposition 1, and separating the terms corresponding to k = 1 and to k = l from the others, we get, in F

,

(10) n l

X

(−c)

k Tr

 λ

σ

(λ)



= − nc + n X

(−c)

k

X

0≤s<kn⁰/l

 k − 1 [sl/n

]



(−1)

+ n l



cn

− c

n

X

 l − 1 [sl/n

]



(−1)

 . By Lemma 2(e),

n

X

 l − 1 [sl/n

]



(−1)

≡

n

X

 1 − l

[sl/n

X

1 j

 (11)

≡ n

− l

n

X

X

1≤j≤sl/n⁰

1

j (mod l

).

In the last double sum, every term 1/j (for 1 ≤ j ≤ l − 1) occurs as many times as there are integers jn

/l ≤ s ≤ n

− 1, i.e. n

− 1 − [jn

/l] times.

Hence

n

X

X

1≤j≤sl/n⁰

1 j =

X



n

− 1 −

 jn

l

 1 (12) j

= (n

− 1) X

1 j −

X

 jn

l

 1 j .

Applying Lemma 2(b) and (d) to evaluate the last two sums in (12) modulo l,

then substituting the result into (11), we get

(13)

n

X

 l − 1 [sl/n

]



(−1)

≡ n

(1 + lq(n

)) (mod l

).

Substituting (13) into (10), we obtain, in F

, (14) n

l X

(−c)

k Tr

 λ

σ

(λ)



= − cn − cq(c) − cq(n

) + n

X

(−c)

k

X

0≤s<kn⁰/l

 k − 1 [sl/n

]



(−1)

. In the last double sum, we isolate the terms corresponding to s = 0 and we use Lemma 2(b) to evaluate the resulting sum, which is (in F

)

X

(−c)

k = c − (c + 1)

− c

− 1

l = c − 1

a − 1

 a

− a

l − (a − 1)q(a − 1)

 . Rewriting (14) accordingly and then substituting it into the expression of [α

, u

] in the Corollary to Proposition 1 yields the desired result.

2. A formula for [α

, u

]

Lemma 3. For any integers 1 ≤ n, n

≤ l − 1 such that nn

≡ 1 (mod l) and 2 ≤ k ≤ l − 1, if d = (nn

− 1)/l, then

X

1≤s<kn⁰/l

 k − 1 [sl/n

]



(−1)

= X

X

1≤j≤k−1 j≡[(rn−1)/d] (mod n)

 k − 1 j

 (−1)

.

P r o o f. If n = n

= 1 then d = 0 and the equality holds trivially. We may thus assume n, n

> 1, so that d ≥ 1. Let s = hd + r, with h, r ∈ N and 0 ≤ r ≤ d − 1 (euclidean division of s by d). We first prove that, for 1 ≤ s < kn

/l,

(15)

 sl n



= hn +

 rn − 1 d

 .

If r = 0 then one can easily see that both sides in (15) are equal to hn − 1, and (15) holds. Assume then 1 ≤ r ≤ d − 1. Since

sl

n

= (hd + r)ln

dl + 1 < hn + rn d , and rn/d is not an integer (d being prime to n), we have

 sl n



≤ hn +

 rn − 1 d



.

The inverse inequality is obtained as follows: since s < kn

/l, we have (hd + r)n < ld; hence

hn +

 rn d



< l ≤

 rn − d

 rn d



l, so that

hn +

 rn d



≤ (hd + r)ln dl + 1 = sl

n

,

which implies the desired inequality and ends the proof of (15).

Now set j = [sl/n

], and note that 1 ≤ s < kn

/l if and only if 1 ≤ j ≤ k − 1. Moreover, since n

< l and d < n, the maps s 7→ [sl/n

] and r 7→ [(rn − 1)/d] are strictly increasing and therefore injective. Furthermore, any integer 1 ≤ j ≤ k − 1 satisfying j ≡ [(rn − 1)/d] (mod n), for some 0 ≤ r ≤ d − 1, is of the type j = [sl/n

] for a convenient 1 ≤ s < kn

/l (namely, if j = hn + [(rn − 1)/d], then s = hd + r). Therefore, in view of (15), the integers j = [sl/n

], for 1 ≤ s < kn

/l, are partitioned into the congruence classes j ≡ [(rn−1)/d] (mod n) (1 ≤ j ≤ k −1), for 0 ≤ r ≤ d−1 or, what is the same, for 1 ≤ r ≤ d. Hence the identity of the statement.

Lemma 4. Let m, n be positive integers, r ∈ Z and ζ

a primitive n-th root of unity in C. Then

X

0≤j≤m j≡r (mod n)

 m j



(−1)

= 1 n

n−1

X

k=1

ζ

(1 − ζ

)

.

P r o o f. For any k ∈ Z, writing the binomial expansion of (1−ζ

)

, then partitioning the resulting sum according to the congruence classes modulo n of the exponent, we get

(1 − ζ

)

=

n−1

X

t=0

 X

0≤j≤m j≡t (mod n)

 m j

 (−1)

 ζ

.

Multiplying this identity by ζ

, for k = 0, 1, . . . , n − 1 respectively, then adding the resulting equalities, we get

(16)

n−1

X

k=0

ζ

(1 − ζ

)

=

n−1

X

t=0

 X

0≤j≤m j≡t (mod n)

 m j

 (−1)



X

k=0

ζ

.

Since P

k=0

ζ

is equal to n if t ≡ r (mod n) and equal to 0 if t 6≡ r (mod n), the right-hand side of (16) is equal to n P

0≤j≤m j≡r (mod n)

m j

(−1)

.

Hence the result.

Proposition 3. For a 6≡ 1 (mod l) and 1 ≤ n ≤ l − 1, letting d = (nn

− 1)/l (with 1 ≤ n

≤ l − 1 such that nn

≡ 1 (mod l)), we have

[α

, u

] = n − 1

2 q(a − 1) + q(n) a − 1 − 1

a − 1

n−1

X

t=1

f

(ζ

),

where f

(X) = g

(X)h

(X) is the product of the following two polynomials in Z[X]

g

(X) = X

X

, h

(X) = (X − a)

− (X − 1)

+ (a − 1)

l(X − 1) .

Also, ζ

is a primitive n-th root of unity in C, and P

t=1

f

(ζ

) is an element of Z independent of the choice of ζ

.

P r o o f. It follows from Proposition 2 and Lemmas 3 and 4 that [α

, u

] = n − 1

2 q(a − 1) − cq(n

) (17)

+

n−1

X

t=1

X

ζ

n

X

(−c)

k (1 − ζ

)

. By Lemma 2(b), for 1 ≤ t ≤ n − 1, we have in Z[ζ

] the congruence (18) (1 − ζ

)

X

(−c)

k (1 − ζ

)

≡ (c(ζ

− 1) − 1)

− c

(ζ

− 1)

+ 1

l − c(ζ

− 1) (mod l).

Since (a − 1)c ≡ 1 (mod l) and c(ζ

− 1) − 1 ≡ c(ζ

− a) (mod l), so that (a − 1)

c

≡ 1 (mod l

) and (c(ζ

− 1) − 1)

≡ c

(ζ

− a)

(mod l

), the fractional term in the right-hand side of (18) is

≡ c

(ζ

− a)

− (ζ

− 1)

+ (a − 1)

l ≡ c(ζ

− 1)h

(ζ

) (mod l).

Also, the right-hand side of (18) is an integral multiple of 1 − ζ

, which is relatively prime to l. Therefore, we can divide (18) through by 1 − ζ

to get (19)

X

(−c)

k (1 − ζ

)

≡ −ch

(ζ

) + c (mod l).

Substituting (19) into (17), and taking into account that, by Lemma 2(c),

q(n

) ≡ −q(n) − d (mod l), we obtain

[α

, u

] = n − 1

2 q(a − 1) + cq(n) + cd (20)

− c

n−1

X

t=1

 X

r=1

ζ

n

 h

(ζ

)

+ c X

n−1

X

t=1

ζ

n

.

In the last double sum, P

t=1

ζ

n

= −1, for 1 ≤ r ≤ d, since [(rn − 1)/d] is not divisible by n. Moreover, P

r=1

ζ

n

= g

(ζ

), for any t. Hence the result by substitution into (20).

Note that the polynomial (X−a)

−(X−1)

+(a−1)

has all its coefficients divisible by l (by the binomial expansion), and admits 1 as a root, so that h

∈ Z[X]. Moreover, any element of the Galois group of Q(ζ

)|Q permutes the nth roots of unity ζ

in the sum θ = P

t=1

f

(ζ

), and therefore it leaves θ invariant. Thus θ lies in Q, and since it is an integral element, it lies in Z.

Lemma 5. Let n be a positive integer.

(a) If ζ

is a primitive n-th root of unity in C, and m ∈ Z, then

n−1

X

t=1

ζ

ζ

− 1 = n − 1

2 − res

(m − 1).

(b) If d is a positive integer relatively prime to n, then

d−1

X

j=1

 jn d



= (d − 1)(n − 1)

2 .

P r o o f. Both statements are trivially true if n = 1, since the sums in- volved, as well as the right-hand sides, are then equal to 0. We thus assume n ≥ 2.

(a) Let r = res

(m − 1), so that m ≡ r + 1 (mod n) and 0 ≤ r ≤ n − 1.

Then (21)

n−1

X

t=1

ζ

ζ

− 1 =

n−1

X

t=1

ζ

− 1 ζ

− 1 +

n−1

X

t=1

1 ζ

− 1 . For the first sum in the right-hand side of (21), we have

n−1

X

t=1

ζ

− 1 ζ

− 1 =

X

n−1

X

t=1

ζ

= n − 1 − r.

For the second sum, consider the polynomial f (X) =

n−1

Y

t=1

(X − ζ

) = X

− 1 X − 1 =

n−1

X

j=0

X

. Then

f

(X) f (X) =

n−1

X

t=1

1 X − ζ

, and thus

n−1

X

t=1

1

1 − ζ

= f

(1) f (1) =

P

j=1

j

n = n − 1 2 . Hence the result, by substitution into (21).

(b) Note that, for a real number x 6∈ Z, we have [n − x] = n − [x] − 1.

Since d is prime to n, it follows that, for 1 ≤ j ≤ d − 1, jn/d 6∈ Z. Hence, for 1 ≤ j ≤ d − 1, we have

 (d − j)n d

 +

 jn d



= n − 1.

Therefore 2

d−1

X

j=1

 jn d



=

d−1

X

j=1

 jn d

 +

 (d − j)n d



= (d − 1)(n − 1).

Hence the result.

Lemma 6. In the notations and under the conditions of Proposition 3, we have

(a) For any integer 1 ≤ e ≤ n − 1,

n−1

X

t=1

ζ

h

(ζ

) ≡ (e − 1)

 a

− a

l − (a − 1)q(a − 1)



+

n−1

X

r=1

r X

1≤k≤l−1 k≡l+e−r−1 (mod n)

a

− 1

k (mod l).

(b) Also,

n−1

X

t=1

f

(ζ

) ≡ (d − 1)(n − 1) 2

 a

− a

l − (a − 1)q(a − 1)



+

d−1

X

j=0 n−1

X

r=1

r X

1≤k≤l−1 k≡l+[−jn/d]−r (mod n)

a

− 1

k (mod l).

P r o o f. (a) For any 1 ≤ t ≤ n − 1, using the binomial expansions of (ζ

− a)

and (ζ

− 1)

in the defining expression of h

, we have

(ζ

− 1)h

(ζ

) = (a − 1)

− a

+ 1

l −

X

1 l

 l k



(−1)

(a

− 1)ζ

. Using Lemma 2(a), we convert this into a congruence modulo l in Z[ζ

];

then dividing by ζ

− 1, which is relatively prime to l, we get, in the ring obtained by localization of Z[ζ

] at a prime ideal above l,

h

(ζ

) ≡ (a − 1)

− a

+ 1

l · 1

ζ

− 1 − X

a

− 1

k · ζ

ζ

− 1 (mod l).

Hence, using Lemma 5(a), we deduce

n−1

X

t=1

ζ

h

(ζ

) ≡ (a − 1)

− a

+ 1 l

 n − 1

2 − (e − 1)



− X

a

− 1 k

 n − 1

2 − res

(l − k + e − 1)



(mod l).

In view of Lemma 2(b), X

a

− 1

k ≡ (a − 1)

− a

+ 1

l (mod l),

so that (22)

n−1

X

t=1

ζ

h

(ζ

)

≡ (e − 1) a

− (a − 1)

− 1

l +

X

res

(l + e − k − 1) a

− 1

k (mod l).

The sum in the right-hand side of (22) can be rewritten in terms of r = res

(l + e − k − 1) which takes the values 0 ≤ r ≤ n − 1, each of which corresponds to those 1 ≤ k ≤ l − 1 such that k ≡ l + e − r − 1 (mod n). The result then follows immediately from (22).

(b) For every 1 ≤ j ≤ d, let e(j) = n − [(jn − 1)/d]. From the definitions, we have

n−1

X

t=1

f

(ζ

) = X

n−1

X

t=1

ζ

h

(ζ

).

Thus, in view of (a) above,

n−1

X

t=1

f

(ζ

) ≡

 a

− a

l − (a − 1)q(a − 1)

 X

j=1

e(j) − d

 (23)

+ X

n−1

X

r=1

r X

1≤k≤l−1 k≡l+e(j)−r−1 (mod n)

a

− 1

k (mod l).

Moreover, for 1 ≤ j ≤ d − 1, e(j) = n − [jn/d], while for j = d, e(d) = 1, so that, in view of Lemma 5(b), P

j=1

e(j) = (d − 1)(n − 1)/2 + d. Also, for 1 ≤ j ≤ d, −[(jn − 1)/d] − 1 = [−jn/d], so that in the triple sum in (23), the inner summation is for k ≡ l + [−jn/d] − r (mod n); furthermore, the latter congruence class (mod n) is the same for j = d as for j = 0, so that the summation may take place for 0 ≤ j ≤ d − 1. Hence the result.

Definition. For a, n, r ∈ Z, with n > 0, we introduce the polynomial sums in F

S

(a) = X

1≤k≤l−1 k≡r (mod n)

a

k .

Proposition 4. For a 6≡ 1 (mod l) and 1 ≤ n ≤ l − 1, let d = (nn

− 1)/l (with 1 ≤ n

≤ l − 1 such that nn

≡ 1 (mod l)). Then

[α

, u

] = d(n − 1)

2 q(a − 1) − (d − 1)(n − 1)

2(a − 1) · a

− a

l + q(n) a − 1

− 1

a − 1

d−1

X

j=0 n−1

X

r=1

r(S

n

(a) − S

n

(1)).

P r o o f. The result is derived from Proposition 3, by substituting for P

t=1

f

(ζ

) its expression from Lemma 6(b), and writing the latter in terms of the above defined sums S

(a) and S

(1) in F

.

3. The sums S

(a) and the expression for [α

, u

]

Lemma 7. The sums S

(a) have the following properties, for any a, n, r ∈ Z, with n > 0:

(a) If a 6≡ 0 (mod l), then

S

(a) = −aS

(a

),

where a

is an inverse of a modulo l in Z.

(b) For any a ∈ Z,

n−1

X

r=0

S

(a) = (a − 1)

− a

+ 1 l

(the summation could be, instead, over any residue system modulo n in Z).

(c) For any integer e ≥ 1, S

(a) =

e−1

X

h=0

S

(a).

(d) If l - n, then

a

S

(a) = X

rl/n<j<(r+1)l/n

a

nj , provided a 6≡ 0 (mod l) if r ≤ 0.

(e) If l - n and a

≡ 1 (mod l), then

n−1

X

r=1

ra

S

(a) = q(n).

(f) If l - n and m ∈ Z, then

n−1

X

r=1

rS

(a) = X

res

(m − k) a

k ,

n−1

X

r=1

rS

(a) = X

res

(k) a

k . (g) For any a ∈ Z,

S

(a) = 1

2 · (a − 1)

− (a + 1)

+ 2

l , S

(a) = 1

2 · (a − 1)

− 2a

+ (a + 1)

l .

(h) If l - n and a 6≡ 1 (mod l), then

n−1

X

r=1

rS

(a) = X

q(j)(a

− a

) = −n X

 n

k l

 a

k

= X

res

(dk) a

k ,

where 1 ≤ n

≤ l − 1 satisfies nn

≡ 1 (mod l) and d = (nn

− 1)/l.