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

CHARACTERIZATION OF A GENERALIZED SHANKS SEQUENCE 211
This shows that the length of the expansion up to h2nt−1 is linear in n. It remains
to show that there exists some h2nt−1, where Qh2nt−1 = 1, and if Q j = 1 then
j = h2nt−1 for some t independent of n.
11. Finding an element of norm 1
We now examine the product of the elements in the expansion. By Theorem 9.1
and (9-2),
So,
Hence,
Qh2nt−1
as well as
h2nt−1+1 =
d 2nt−1
2nt−2
j
h2i−1+1 = (αβ)i α i−1
j=1 ε2 j−1
(αβ) nt α kt
d 2nt−1 2nt−2
j=0 j
j=0
Qh2n−1
2
d 2n−1
2n−2
j
d 2i−1 2i−2
j=0 j
.
and |N(h2nt−1+1)| = Qh2nt−1 .
= N (αβ) nt N α kt = N (αβ) n N(α k ) t
j=0
2
= N (αβ) n N(α k ) .
Thus, Qh2nt−1 / t Qh2n−1 ∈ . Since λ2 j−1 ≡ jk (mod n) we have that λ2ni−1 = 0
for positive i. Hence, λ2n−1 = λ2nt−1 = 0 and so
Thus,
Qh2n−1 = l2n−1m2n−1r2n−1 (su2n−1z2n−1y2n−1) 2 ,
Qh2nt−1 = l2nt−1m2nt−1r2nt−1 (su2nt−1z2nt−1y2nt−1) 2 .
l2nt−1m2nt−1r2nt−1
∈ .
(l2n−1m2n−1r2n−1) t
Since l2n−1, m2n−1, r2n−1 are each squarefree and relatively prime, if 2 t then
√ l2nt−1m2nt−1r2nt−1 ∈ , which implies l2nt−1m2nt−1r2nt−1 = 1. Conversely, if
2 t, then l2nt−1m2nt−1r2nt−1 = l2n−1m2n−1r2n−1.
Now, we construct an element of norm 1. Put
1 if l2n−1m2n−1r2n−1 = 1,
ε =
0 if l2n−1m2n−1r2n−1 = 1,