for cyclic fields of prime degree l

LXXIV.4 (1996)

Congruence of Ankeny–Artin–Chowla type modulo p

for cyclic fields of prime degree l

by

Stanislav Jakubec (Bratislava)

Introduction. Let p ≡ 1 (mod 4), let T + U √

p > 1 be the fundamental unit, and let h be the class number of Q( √

p). The following congruence (Ankeny–Artin–Chowla congruence) holds:

h U

T ≡ B

(mod p).

For a cubic field K ⊂ Q(ζ

+ ζ

), p ≡ 1 (mod 3) the analogous congruence was proved by Feng Ke Qin in [1]. In general, for fields of the lth degree, the analogous congruence is proved in [5].

The aim of this paper is to prove a congruence of Ankeny–Artin–Chowla type modulo p

for real Abelian fields of a prime degree l and a prime conductor p. The importance of such a congruence can be demonstrated by the following example. Let K be a cubic field. In [5], the following congruence is proved:

(1) h

S

S

≡ −

B

B

(mod p).

As is well known, h

< p for a cubic field. If B

B

≡ 0 (mod p), then S

S

≡ 0 (mod p), hence the congruence (1) does not provide any information about h

. Note that such a prime exists, e.g. p = 5479. We have

B

= B

≡ 0 (mod 5479).

In this paper two applications of the main theorem (Theorem 1), for a quadratic and for a cubic field, will be given.

Let l and p be primes such that p ≡ 1 (mod l) and let K ⊂ Q(ζ

+ ζ

) with [K : Q] = l. Let a be a primitive root modulo p. As is well known, the conjugates of the unit

1991 Mathematics Subject Classification: Primary 11R29.

[293]

η

= N

p+ζ⁻¹_p )/K



ζ

1 − ζ

1 − ζ

 ,

generate the group of cyclotomic units C(K) of K. Consider the unit η

= N

p+ζ_p⁻¹)/K

(ζ

+ ζ

).

It is easy to prove that if 2 is not an lth power modulo p, then the conjugates of the unit η

generate the group C(K). Let hεi be the group generated by all conjugates of the unit ε.

According to ([3], Lemma 1, p. 69) for a cyclic field K with [K : Q] = l, there is a unit δ such that [U

: hδi] = f , where (p, f ) = 1.

The following is taken from [5]. According to [8] and [9] (see also [10], p. 284), we have h

= [U

: C(K)], where C(K) = hη

i is the group of cyclotomic units of K. From [U

: hδi] = f we have η

∈ hδi. Let [hδi : hη

i] = e. Clearly [hη

i : hη

i] = f

.

Consider two towers of groups

hη

i ⊂ hδi ⊂ U

, hη

i ⊂ hη

i ⊂ U

. This implies ef = h

f

and hence e = h

f

. Let

η

= δ

σ(δ)

. . . σ

(δ)

. It is easy to prove that

e = N

(c

+ c

ζ

+ . . . + c

ζ

).

Note that a unit δ for which f = 1 is called a strong Minkowski unit. As is well known, a strong Minkowski unit exists for a cyclic fields K with l < 23.

The problem of existence of a strong Minkowski unit for cyclic fields of small non-prime degree is solved in [7].

Let a be a fixed primitive root modulo p, let χ be the Dirichlet character of order n, n | p − 1, χ(x) = ζ

. Let g be such that g ≡ a

(mod p) and g

≡ 1 (mod p

). Denote by p a prime divisor of Q(ζ

) such that p | p and 1/g ≡ ζ

(mod p

).

Define the rational numbers A

(n), A

(n), . . . , A

(n) in the following way:

(2)

A

(n) = −1/n,

τ (χ

)

≡ n

A

(n)

(−p)

(mod p

), A

(n) ≡ (p − 1)/n

(i(p − 1)/n)! (mod p), where τ (χ) is the Gauss sum.

Put m = (p − 1)/2, and

G

(X) = A

(m)X

+ A

(m)X

+ . . . + A

(m), F

(X) = 1

(p − 1)! X

+ 1

(p + 1)! X

+ 1

(p + 3)! X

+ . . . + 1

(p + 2j − 1)! .

Define

E

= E

(2n)! for n = 1, 2, 3, . . . ,

where E

are the Euler numbers, i.e. E

= 1, E

= −1, E

= 5, E

= −61, E

= 1385, E

= −50521, E

= 2702765, E

= −199360981, . . .

Consider the formal expressions G

(E

) and F

(E

), where (E

)

= E

.

Let β

, β

, . . . , β

be the integral basis of the field K formed by the Gauss periods. Let

δ = x

β

+ x

β

+ . . . + x

β

. Associate with the unit δ the polynomial f (X) as follows:

f (X) = X

+ d

X

+ d

X

+ . . . + d

, where

d

= −lA

(l) x

+ x

g

+ x

g

+ . . . + x

g

x

+ x

+ . . . + x

for i = 1, . . . , l − 1. Put

S

= S

(d

, . . . , d

) = sum of the jth powers of the roots of f (X) for j = 1, . . . , 2l − 1. Hence

S

= −d

, S

= d

− 2d

, S

= −d

+ 3d

d

− 3d

, . . . Define the numbers T

, . . . , T

as follows:

T

= − 1

(i(p − 1)/l)! 2

(2

−1)B

−i p − 1

4l G

(E

) for i = 1, . . . , l − 1, and

T

= 1 − q

2 , where q

= 2

− 1

p ,

T

= − 1

(p − 1 + i(p − 1)/l)! 2

(2

− 1)

× B

+

 p − 1

2 + i p − 1 2l



F

(E

) for i = 1, . . . , l − 1.

Define

α

= c

+ c

g

+ c

g

+ . . . + c

g

for i = 1, . . . , 2l − 1.

Let X

, . . . , X

∈ Q and let

g(X) = X

+ Y

X

+ . . . + Y

be a polynomial such that

X

= sum of the jth powers of the roots of g(X).

Define the mapping Φ : Q

→ Q

as follows:

Φ(X

, . . . , X

) = (1 − pY

, Y

− pY

, . . . , Y

− pY

).

Now the main theorem of this paper be formulated.

Theorem 1. Let l and p be primes with p ≡ 1 (mod l) and let K ⊂ Q(ζ

+ ζ

) with [K : Q] = l. Suppose that 2 is not an l-th power mod- ulo p. Let δ be a unit of K such that [U

: hδi] = f , (f, p) = 1. Let η

= δ

σ(δ)

. . . σ

(δ)

and α

= c

+ c

g

+ . . . + c

g

for i = 1, . . . , 2l − 1. The following congruence holds:

(3) ε

 x

+ x

+ . . . + x

−l



Φ(α

S

, . . . , α

S

)

≡ (2 + 2p)

Φ(f T

, . . . , f T

) (mod p

), where ε = ±1.

R e m a r k. The class number h

appears in the preceding congruence implicitly, via the congruence

h

f

≡ α

. . . α

(mod p

).

This congruence can be proved in the following way:

We have the congruence 1/g ≡ ζ

(mod p

) and hence σ

(c

+ c

ζ

+ . . . + c

ζ

) ≡ α

(mod p

);

this yields

h

f

= e = N

(c

+ c

ζ

+ . . . + c

ζ

) ≡ α

. . . α

(mod p

).

The congruence (3) gives l congruences (one in each component). If B

B

. . . B

6≡ 0 (mod p),

then from the congruence (3), the numbers α

, . . . , α

modulo p

can be calculated. Using the congruence

h

f

≡ α

. . . α

(mod p

), also h

can be calculated modulo p

. If

B

B

. . . B

≡ 0 (mod p),

then the numbers α

, . . . , α

and hence also h

can be calculated at most

modulo p.

Before proving Theorem 1, we show its applications to quadratic and cubic fields.

The quadratic case K = Q( √

p), p ≡ 5 (mod 8). Let δ = x

β

+ x

β

= x

−1 + √ p 2 + x

−1 − √ p 2 > 1 be a fundamental unit of Q( √

p). Then

d

= −2A

(2) x

− x

x

+ x

. Hence

S

= 2A

(2) x

− x

x

+ x

, S

= 4A

(2)

(x

− x

)

(x

+ x

)

, S

= 8A

(2)

(x

− x

)

(x

+ x

)

. For A

(2) we have

τ (χ)

≡ 4A

(2)

(−p) (mod p

), A

(2) ≡ (p − 1)/2

((p − 1)/2)! (mod p).

Hence

A

(2)

≡ − 1

4 (mod p

), A

(2) ≡ (p − 1)/2

((p − 1)/2)! (mod p).

Therefore

S

= 2A

(2) x

− x

x

+ x

, S

= − (x

− x

)

(x

+ x

)

, S

= −2A

(2) (x

− x

)

(x

+ x

)

. Let

x

+ x

−2 + x

− x

2

√ p = T + U √ p > 1.

It can be proved that |η

| = |N

p+ζ_p⁻¹)/Q(√

p)

(ζ

+ ζ

)| < 1, hence it is necessary to start from the unit T − U √

p. We have S

= 2A

(2) U

T , S

= − U

T

, S

= −2A

(2) U

T

. For the numbers T

, T

, T

we have

T

= − 1

((p − 1)/2)! 2

(2

− 1)B

− p − 1

8 G

(E

), T

=

(1 − q

),

T

= − 1

(3(p − 1)/2)! 2

(2

− 1)B

+ (3(p − 1)/4)F

(E

).

It is easy to see that Φ(X

, X

, X

) =



1 − p X

− X

2 , −X

− p



− 1

6 X

+ 1

2 X

X

− 1 3 X



. Hence

ε

 x

+ x

−2



Φ(α

S

, α

S

, α

S

) ≡ (2+2p)

Φ(T

, T

, T

) (mod p

).

Since (x

+ x

)/(−2) = T and α

= α

= α

= h, we get

εT

Φ(hS

, hS

, hS

) ≡ (2 + 2p)

Φ(T

, T

, T

) (mod p

), where ε = ±1. It can be proved that ε = (−1)

, where r is the number of quadratic residues modulo p in the interval (p/4, p/2).

The cubic case. Let p be a prime such that p ≡ 1 (mod 3), p 6= a

+ 27b

and let

δ = x

β

+ x

β

+ x

β

. Then

d

= −3A

(3) x

+ x

g + x

g

x

+ x

+ x

, d

= −3A

(3) x

+ x

g

+ x

g x

+ x

+ x

. For the numbers A

(3), A

(3) we have

τ (χ)

≡ 27A

(3)

(−p) (mod p

), A

(3) ≡ (p − 1)/3

((p − 1)/3)! (mod p), τ (χ

)

≡ 27A

(3)

(−p)

(mod p

), A

(3) ≡ (p − 1)/3

(2(p − 1)/3)! (mod p).

As is well known, τ (χ)

= pJ(χ, χ), where J(χ, χ) is the Jacobi sum.

Let J(χ, χ) = a + bζ

, a ≡ −1, b ≡ 0 (mod 3) and p = a

− ab + b

. Hence

−

 a + b 1

g



≡ 27A

(3)

(mod p

), A

(3) ≡ (p − 1)/3

((p − 1)/3)! (mod p).

The number A

(3) is determined by the congruence

−1 ≡ 27

A

(3)

A

(3)

(mod p

), A

(3) ≡ (p − 1)/3

(2(p − 1)/3)! (mod p).

For the numbers T

, . . . , T

we have T

= − 1

(i(p − 1)/3)! 2

(2

−1)B

−i p − 1

12 G

(E

) for i = 1, 2, and

T

=

(1 − q

),

T

= − 1

(p − 1 + i(p − 1)/3)! 2

(2

− 1)

× B

+

 p − 1

2 + i p − 1 6



F

(E

) for i = 1, 2.

It is easy to prove that for Φ(X

, . . . , X

) we have Φ(X

, . . . , X

)

=

 1 − p



− 1

6 X

+ 1

2 X

X

− 1 3 X

 ,

− X

− p

 1

24 X

+ 1

3 X

X

+ 1

8 X

− 1

4 X

X

− 1 4 X

 , 1

2 (X

− X

) − p



− 1

120 X

+ 1

4 X

X

+ 1 6 X

X

+ 1

12 X

X

− 1

6 X

X

− 1

8 X

X

− 1 5 X



. Hence

±

 −1 3



(x

+ x

+ x

)

Φ(α

S

, . . . , α

S

)

≡ (2 + 2p)

Φ(T

, . . . , T

) (mod p

).

P r o o f o f T h e o r e m 1. Let g

(X), . . . , g

(X) be polynomials such that g

(X) 6≡ 0 (mod X). Let

g(X) ≡ g

(X)

. . . g

(X)

(mod X

).

Let s

be the homomorphism defined in [4]. We have

s

(g(X)) = b

s

(g

(X)) + b

s

(g

(X)) + . . . + b

s

(g

(X)), for j = 1, . . . , M − 1. Define

X

= b

s

(g

(X)) + b

s

(g

(X)) + . . . + b

s

(g

(X)) for j = 1, . . . , M − 1. Let

g(X) ≡ C

+ C

X + C

X

+ . . . + C

X

(mod X

).

Consider the reciprocal polynomial F (X) = X

+ C

C

X

+ . . . + C

C

.

By the definition of the homomorphism s

(see [4]) we have X

= sum of the jth powers of the roots of F (X).

The numbers C

/C

, C

/C

, . . . , C

/C

can be calculated by the Newton recurrence formula.

According to [2] and [4] we have:

Proposition 1. There is a number π ∈ Q(ζ

+ ζ

), π | p such that (i) N

p )/Q

(π) = (−1)

p,

(ii) σ(π) ≡ gπ (mod π

), g ≡ a

(mod p

), (iii) ζ

+ ζ

≡ P

i=0

a

π

(mod π

), where 0 ≤ a

< p and a

≡ (2/(2i)!) (mod p) for i = 1, . . . , m.

The numbers a

for i = 1, . . . , m, are defined by a

= 2 p − 1 − p(p + 1)B

p .

If 2 is a primitive root modulo p then the coefficients a

, a

, . . . , a

are given by the recurrence formula a

≡ 1

4

− 4

 4

a

− b

p + b



(mod p), where b

and b

are the coefficients of X

and X

, respec- tively, in the polynomial

 2 + 2

2! X + . . . + 2

(p − 1)! X

+ a

X

+ . . . + a

X



. Proposition 2. Let K ⊂ Q(ζ

+ ζ

) with [K : Q] = n. There is a number π ∈ K with π | p such that

(i) N

(π) = (−1)

p,

(ii) σ(π) ≡ gπ (mod π

), g ≡ a

(mod p

), (iii) β

≡ P

i=0

a

π

(mod π

), where 0 ≤ a

< p and where a

≡ ((p − 1)/n)/(i(p − 1)/n)! (mod p) for i = 1, . . . , n.

From now on we will use the following transformation. Let γ ≡ c

+ c

π + c

π

+ . . . + c

π

(mod π

).

From π

≡ −p (mod π

) we have

γ ≡ (c

− pc

) + (c

− pc

)π + . . . + (c

− pc

)π

(mod π

).

Lemma 1. Let α, β ∈ K with αβ 6≡ 0 (mod π). Let

α ≡ e

+ e

π + . . . + e

π

(mod π

),

β ≡ b

+ b

π + . . . + b

π

(mod π

).

Relate to the conjugation σ

(α) the polynomial

f

(X) = e

+ e

g

X + e

g

X

+ . . . + e

g

X

. Let

F (X) ≡ f

(X)

f

(X)

. . . f

(X)

≡ d

+ d

X + . . . + d

X

(mod X

).

Define X

= s

(F (X)). Then

α

σ(α)

. . . σ

(α)

≡ β (mod p

) if and only if

d

Φ(X

, . . . , X

) ≡ (b

, b

, . . . , b

) (mod p

).

P r o o f. Clearly

α

σ(α)

. . . σ

(α)

≡ d

+ d

π + . . . + d

π

(mod π

).

The coefficients d

/d

, d

/d

, . . . , d

/d

can be expressed using the num- bers X

, . . . , X

by the Newton recurrence formula. Clearly

d

+ d

π + . . . + d

π

≡ (d

− pd

) + (d

− pd

)π + . . . + (d

− pd

)π

(mod π

).

From the definition of the mapping Φ it follows that

(d

−pd

, d

−pd

, . . . , d

−pd

) ≡ d

Φ(X

, X

, . . . , X

) (mod p

).

Lemma 1 is proved.

Proposition 2 gives

β

≡ −1/l + (a

− pa

)π + . . . + (a

− pa

)π

(mod π

), where 0 ≤ a

< p and a

≡ ((p − 1)/n)/(i(p − 1)/n)! (mod p) for i = 1, . . . , l − 1. According to [4] we have

a

− pa

≡ A

(l) (mod p

).

Let δ = x

β

+ x

β

+ . . . + x

β

. Then δ ≡ − 1

l (x

+ x

+ . . . + x

) +

X

A

(l)(x

+ x

g

+ x

g

+ . . . + x

g

)π

(mod π

).

Let

δ

σ(δ)

. . . σ

(δ)

= η

.

Then X

corresponding to the product on the left-hand side is equal to

c

s

(δ) + c

g

s

(δ) + . . . + c

g

s

(δ) = α

S

for j = 1, . . . , 2l − 1. Hence the number corresponding to the left-hand side is 

x

+ x

+ . . . + x

−l



Φ(α

S

, . . . , α

S

).

Now we prove that the right-hand side is equal to (2 + 2p)

Φ(f T

, . . . , f T

).

By Proposition 1 we have ζ

+ ζ

≡

X

a

π

(mod π

),

where 0 ≤ a

< p and a

≡ (2/(2i)!) (mod p) for i = 1, . . . , m, and hence ζ

+ ζ

≡ A

(m) + A

(m)π + . . . + A

(m)π

(mod π

).

Consider the polynomial

g(X) = A

(m)X

+ A

(m)X

+ . . . + A

(m).

Now we shall calculate the numbers

s

= sum of the ith powers of the roots of g(X)

for i = 1, . . . , 2m − 1 modulo p

. It is easy to see that for i > m − 1 it is enough to determine s

modulo p. Let W

, W

, . . . be a linearly recurrent sequence modulo p of order m − 1 defined by

W

= −1

(2i)! 2

(2

− 1)B

for i = 1, . . . , m − 1.

For i > m − 1 we have W

= −

 1

2! W

+ 1

4! W

+ . . . + 1

(2m − 2)! W

 . Lemma 2. The following congruence holds:

s

≡ −1

(2n + 2)! 2

(2

− 1)B

− n + 1

2 G

(E

) (mod p

) for n + 1 < m.

P r o o f. It is easy to see that A

(m) ≡ 2 + 2p (mod p

). Clearly (2 + 2p) 1 − p

2 ≡ 1 (mod p

).

Consider the polynomial

g

(X) = X

+ C

X

+ . . . + C

, where

C

≡ 1 − p

2 A

(m) (mod p

).

Obviously s

is equal to the sum of the nth powers of the roots of g

(X).

Write

C

= c

+ b

p, C

= c

+ b

p, . . . , C

= c

+ b

p.

It is known that there exists a polynomial f

(x

, . . . , x

) such that s

= f

(c

+ b

p, . . . , c

+ b

p).

Write s

= r

+ t

p. Clearly r

+ t

p = −(c

+ b

p). Put t

= −b

. According to the Newton recurrent formula we have

r

+ t

p + (c

+ b

p)(r

+ t

p) + 2(c

+ pb

) = 0.

The last equation can be rewritten in the form

r

+ c

r

+ 2c

+ p(t

+ c

t

+ r

b

+ 2b

) = 0.

Define F

= r

b

+ 2b

. Hence r

+ c

r

+ 2c

+ p(t

+ c

t

+ F

) = 0. Put t

= −(c

t

+ F

).

Further we have

r

+ t

p + (c

+ b

p)(r

+ t

p) + (c

+ b

p)(r

+ t

p) + 3(c

+ b

p) = 0, hence

r

+ c

r

+ c

r

+ 3c

+ p(t

+ c

t

+ c

t

+ b

r

+ b

r

+ 3b

) = 0.

Define F

= b

r

+ b

r

+ 3b

. Then we put t

= t

(c

− c

) + c

F

− F

,

t

= t

(−c

+ 2c

c

− c

) − (c

− c

)F

+ c

F

− F

,

t

= t

(c

− 3c

c

+ 2c

c

+ c

− c

) − (−c

+ 2c

c

− c

)F

− (c

− c

)F

+ c

F

− F

. Consider the numbers

K

= −c

, K

= c

− c

, K

= −c

+ 2c

c

− c

, . . .

The numbers c

, c

, c

, . . . are equal to 1/2!, 1/4!, 1/6!, . . . modulo p. Define K

= K

/(2n)!.

Then

K

(2n)! + 1

(2)! · K

(2n − 2)! + . . . + 1

(2n − 2)! · K

2! + 1

(2n)! = 0, hence

1 (2n)!

 K

+

 2n 2



K

+ . . . + 1



= 0,

and therefore K

= E

, the Euler number. Using induction by n we put (4) t

= t

E

(2n)! − F

E

(2n − 2)! − . . . − F

,

where

F

= b

r

+ 2b

, F

= b

r

+ b

r

+ 3b

, F

= b

r

+ b

r

+ b

r

+ 4b

, . . . Since t

= −b

, we have

t

= − b

E

(2n)! − E

(2n − 2)! (b

r

+ 2b

)

− . . . − (b

r

+ b

r

+ . . . + (n + 1)b

).

In the last equation we first cancel the brackets by multiplication and then make new brackets by factoring out b

, . . . , b

. The summand obtained by factoring out b

is

(5) b

 E

(2n)! + r

E

(2n − 2)! + . . . + r

 . According to [4, Lemma 3],

r

≡ − 1

(2i)! 2

(2

− 1)B

(mod p).

By substitution into (5) we get the sum E

(2n)! − E

(2n − 2)! · 1

2! 2(2

− 1)B

− E

(2n − 4)! · 1

4! 2

(2

− 1)B

− . . . − 1

(2n)! 2

(2

− 1)B

. The following identity holds:

(6) E

(2n)! − E

(2n − 2)! · 1

2! 2(2

− 1)B

− E

(2n − 4)! · 1

4! 2

(2

− 1)B

− . . . − 1

(2n)! 2

(2

− 1)B

= (n + 1) E

(2n)! . This identity can be proved in the following way. For the functions sec x and tan x, we have

sec x = 1 − E

2! x

+ E

4! x

− E

6! x

+ . . . , tan x = 2

(2

− 1)B

x

2! − 2

(2

− 1)B

x

4! + . . . Then

tan x · sec x = 2

(2

− 1) 2! B

x −

 E

B

2

(2

− 1)

2!2! + 2

(2

− 1)B

4!



x

+ . . . The identity (6) follows from the equation

d(sec x)

dx = tan x · sec x.

Using the identity (6), by induction (n + 1 → 1) we have (7) t

= −(n + 1)

 b

E

(2n)! + b

E

(2n − 2)! + . . . + b

 . The numbers a

, . . . , a

from Proposition 1 are

a

≡ 2

(2i)! (mod p), 0 < a

< p.

Define the numbers D

as follows:

1 − p

2 a

+ pD

≡ 1

(2i)! (mod p

), 0 ≤ D

< p, hence

D

≡ 1 p

 1

(2i)! − 1 − p 2 a



(mod p).

Let v

be the sum of the (n+1)th powers of the roots of the polynomial X

+ 1 − p

2 a

X

+ . . . + 1 − p 2 a

.

According to [4, Lemma 3], the sum of the (n + 1)th powers of the roots of the polynomial

X

+ 1

2! X

+ 1

4! X

+ . . . + 1 (2m − 2)! , is equal to

−1

(2n + 2)! 2

(2

− 1)B

. Hence

v

= −1

(2n + 2)! 2

(2

− 1)B

+ p(n + 1)



D

E

(2n)! + D

E

(2n − 2)! + . . . + D

 . Therefore

s

= −1

(2n + 2)! 2

(2

− 1)B

(8)

+ p(n + 1)



D

E

(2n)! + D

E

(2n − 2)! + . . . + D



− p(n + 1)

 b

E

(2n)! + b

E

(2n − 2)! + . . . + b

 . Since

D

≡ 1 p

 1

(2i)! − 1 − p 2 a



,

from (8) we get s

= −1

(2n + 2)! 2

(2

− 1)B

− (n + 1)E

(2n + 2)!

(9)

+ (n + 1) p − 1 2

 a

E

(2n)! + a

E

(2n − 2)! + . . . + a



+ p n + 1 2



a

E

(2n)! + a

E

(2n − 2)! + . . . + a

 , where a

, . . . , a

, a

, a

, . . . are the numbers from Proposition 1. Since A

(m) = a

− pa

, from the equality (9) we get

s

= −1

(2n + 2)! 2

(2

− 1)B

− n + 1

2 G

(E

).

Lemma 3. The following congruence holds:

W

≡ − 1

(p − 1 + 2j)! 2

(2

− 1)B

+ (m + j)F

(E

) (mod p) for j = 1, . . . , m − 1.

P r o o f. Let n > 2p − 3. Consider the polynomial h(X) = X

+ 1

2! X

+ 1

4! X

+ . . . + 1 (2n)! .

Let W

be the sum of the ith powers of the roots of h(X). By [4, Lemma 3], W

= − 1

(2i)! 2

(2

− 1)B

. The characteristic polynomial of the sequence W

is

X

+ 1

2! X

+ 1

4! X

+ . . . + 1 (2m − 2)! .

Hence W

is the sum of the ith powers of the roots of this polynomial.

Let

c

= 1

2! , c

= 1 4! , . . . Obviously

W

= W

+ mc

,

W

= W

+ c

W

+ (m + 1)c

− c

mc

. We can write

G

= mc

, G

= (m + 1)c

− c

G

.

By induction we can prove that

W

= W

+ c

W

+ (c

+ c

E

)W

(10)

+ (c

+ c

E

+ c

E

)W

+ . . .

+ (c

+ c

E

+ . . . + c

E

)W

+ G

. Now we rewrite the right side of (10). We cancel the brackets by multi- plication and make new brackets by factoring out c

, c

, . . .

E.g., the summand obtained by factoring out c

is c

(W

+ E

W

+ E

W

+ . . . + E

W

).

From (6) we get

c

(W

+ E

W

+ E

W

+ . . . + E

W

) = c

jE

. By repeating this procedure with each summand we have

(11) W

= W

+ jc

E

+ (j − 1)c

E

+ . . . + c

E

+ G

. By induction it is easy to prove

(12) G

= (m + 1)c

+ (m + j − 1)c

E

+ . . . + mc

E

. Substituting (12) into (11) we get

W

= W

+ (m + j)(c

+ c

E

+ . . . + c

E

).

Since η

= N

p+ζ⁻¹_p )/K

(ζ

+ ζ

) we have η

≡ r

+

X

r

π

(mod π

),

where r

= (2 + 2p)

. From Lemmas 1–3 we deduce that the right- hand side is equal to

(2 + 2p)

Φ(f T

, . . . , f T

).

It remains to prove that T

= (1 − q

)/2. From the proof of Lemma 3 we have

T

≡ −1

(p − 1)! 2

(2

− 1)B

+ p − 1

2 · 1

(p − 1)! , hence

T

≡ −1

(p − 1)! 2

2

− 1

p pB

+ p − 1

2 · 1

(p − 1)! .

From pB

≡ −1 (mod p) the required congruence follows. Theorem 1 is

proved.

Example 1 (p = 13, l = 3, K ⊂ Q(ζ

+ ζ

), [K : Q] = 3). By Proposition 1 we have

ζ

+ ζ

≡ 2 + π + 12π

+ 3π

+ 4π

+ 9π

+ 11π

+ 2π

+ 12π

+ 8π

+ 10π

+ 4π

(mod π

), hence

ζ

+ ζ

≡ A

(m) + A

(m)π + . . . + A

(m)π

≡ 28 + 144π + 25π

+ 68π

+ 43π

+ 126π

(mod π

).

It is easy to see that η

= N

p+ζ_p⁻¹)/K

(ζ

+ ζ

)

≡ (28 + 144π + 25π

+ 68π

+ 43π

+ 126π

)

× (28 − 144π + 25π

− 68π

+ 43π

− 126π

)

≡ 56 + 86π

+ 50π

(mod π

).

The number β

= ζ

+ ζ

+ ζ

+ ζ

is the Gauss period. By Proposition 2 we have

β

≡ −

+ A

(3)π + A

(3)π

(mod π

)

(π is a prime divisor of the field K, [K : Q] = 3, π | p). For A

(3), A

(3) we have

τ (χ)

≡ 27A

(3)

(−13) (mod p

), A

(3) ≡ 4

4! (mod 13),

−1 ≡ 27

A

(3)

A

(3)

(mod 169), A

(3) ≡ 4

8! (mod 13).

The number g satisfies g ≡ 2

≡ 146 (mod 169), and τ (χ)

= pJ(χ, χ), J(χ, χ) = −4 − 3ζ

. Hence A

(3) ≡ 4/4! ≡ 11 (mod 13). Thus

−



− 4 − 3 1 146



≡ 27(11 + 13k)

(mod 169).

It follows that A

(3) = 50, A

(3) = 86.

The fundamental unit of the field K is δ = β

. Hence

d

= −3 · 50 · 146

≡ 80 (mod 169), d

= −3 · 86 · 146 ≡ 19 (mod 169).

Therefore

S

= −80, S

= 109, S

= 2, S

= 5, S

= 4.