Characterization of a generalized Shanks sequence - Department of ...

CHARACTERIZATION OF A GENERALIZED SHANKS SEQUENCE 203
It is clear that δ(1/α) ≥ δ(α)−2, δ(−α) ≥ δ(α)−2 and δ(α +n) = δ(α) for any
n ∈ . Furthermore, Mendès France [1973] has shown that for any α ∈ , n ∈ + ,
δ(nα) ≥
δ(α) − 1
− 1,
κ(n) + 2
where κ(n) is a positive valued function whose values depend only on n. Consequently,
for any sequence αi ∈ satisfying limi→∞ δ(αi) = ∞ and any n ∈ + we
have limi→∞ δ(αi/n) = ∞. It follows that, if T > 1, then
(6-2) lim
i→∞ δ
A2i
= lim
i→∞ δ
1
= ∞.
B2i
(Eξ
F2i
i+1 + f2i)
Period length of ωn. By our criteria for a kreeper we must have for some a, b ∈ ,
for all n ∈ I . Hence we must have
(6-3)
(n−V )/k
an + b = lp(ωn) >
(n−V )/k
i=W
δ
A2i
B2i
j=W
p2 j
< an + b.
By (6-2) there exists a γ > W such that δ (A2i/B2i) > k(a +1) for all i ≥ γ . Then,
(6-4)
(n−V )/k
i=W
δ
A2i
B2i
> k(a + 1) (n − V )/k − γ .
When n > b + (a + 1) V + kγ we have (6-4) is greater than an + b. And since
all the terms on the right side of this inequality are bounded, there must exist an
infinitude of values of n ∈ I such that lp(ωn) > an + b for any fixed a, b. In
conclusion Dn can not be a kreeper if T > 1. In other words, we must have T = 1
and U = x.
Our only remaining objective is to show that necessarily N = 1. Clearly, there
is no loss in generality in supposing that N is not a power of x. In Section 5 we
established the existence of the following complete quotient in all kreepers,
θh2i = ωn + S1/d − Nsr2il2iu2iu ′ 2i y2i y ′ 2ic2i xn−ki /d
r2im2il2i(su2i y2i z2i) 2N xn−ki .
By taking i = ⌊n/k⌋ we find an element, η ∈ n, whose norm can be written as
RN xν , where R Dn, (N, En) = 1 and 0 ≤ ν < k. Hence the norm of η is bounded
independently of n, and coprime to the conductor of the order ∗ n . First, recall the
ideal bn ∈ ∗ n from page 195, which has norm x g , with g fixed independently of