Proceedings of the 44th Symposium on Ring Theory and ...

Theorem. Assume r is a common prime divisor <strong>of</strong> f and t, and n is a divisor <strong>of</strong> q ∗ ,
where q ∗ q = r − 1. Then p splits completely in O n and if <strong>the</strong>re exists µ ∈ O n such that p
does not divide I(µ), <strong>the</strong>n n ≦ p. In particular, for n > p, p is a common index divisor <strong>of</strong>
O n namely, p divides I(γ) for all γ ∈ O n .
Let c be a primitive root for r, let χ be a character <strong>of</strong> order n defined by χ(c) = ω
where ω = e 2πi
n and let g(χ) = ∑ a∈F r
χ(a)ζ a be <strong>the</strong> Gauss sum <strong>of</strong> χ where F r is a finite
field <strong>of</strong> order r. Let σ(ζ) = ζ c be a generator <strong>of</strong> <strong>the</strong> Galois group G <strong>of</strong> K over Q and set
T n := 〈σ n 〉.
For simplicity, we set g 0 = −1, g k = g(χ k ) for n > k > 0 and θ k = θ σk for n > k ≧ 0
where θ = ∑ τ∈T n
ζ τ is a trace <strong>of</strong> ζ.
It is known that L n = Q(θ) and θ is a normal basis element <strong>of</strong> O n over Z (see [9, p.61,
p.74])
The next Lemma is useful to our object. It only needs to assume r is prime and n is a
divisor <strong>of</strong> r − 1 in this Lemma. This pro<strong>of</strong> is essentially in <strong>the</strong> first equation <strong>of</strong> (1) due
to [9, p.62]. This idea <strong>of</strong> classifying primitive roots goes back to Gauss; <strong>the</strong> regular 17
polygon construction by ruler and compass.
Lemma.
(1) g k = ∑ n−1
s=0 ωks θ s for 0 ≦ k < n and nθ k = ∑ n−1
s=0 ¯ωks g s for 0 ≦ k < n where ¯ω is
<strong>the</strong> complex conjugate <strong>of</strong> ω.
(2) Using (1), determinants <strong>of</strong> cyclic matrices A n , B n are given by
∣ ∣ θ 0 θ 1 . . . θ n−1 ∣∣∣∣∣∣∣ g
n−1
0 g 1 . . . g n−1 ∣∣∣∣∣∣∣ θ
|A n | :=
n−1 θ 0 . . . θ n−2
∏
n−1
g . . . ..
= g k and |B n | :=
n−1 g 0 . . . g n−2
∏
.
.
k=0
. . ..
= n n θ k .
.
k=0
∣
θ 1 θ 2 . . . θ
∣
0 g 1 g 2 . . . g 0
(3) We have
{
r n−1 if n is odd,
d(L n ) =
(−1) r−1
2 r n−1 if n is even.
Some results in [7, 8] are proved again in <strong>the</strong> next
Corollary. Let r be a common prime divisor <strong>of</strong> f and t. Then we have
(1) p ≡ 1 or r ≡ 1 mod 4 (see [7, Lemma, (4) ]).
(2) q ≡ −1 mod 9 in case p = 3 and f divides t (see [8, Corollary, (a)]).
–122–