Cyclotomic Zp-extensions 109
2. Some lemmas. In this section, we shall describe some properties of
n(r)
0and n(r)
2. The following lemma, which is a basic fact, is an immediate
consequence of the definitions of n(r)
0and n(r)
2.
Lemma 1. For each integer r≥0,
(1) r+ 1 ≤n(r)
0≤n(r)
2,
(2) n(r)
0≤n(r+1)
0≤n(r)
0+ 1,
(3) n(r)
2≤n(r+1)
2≤n(r)
2+ 1.
In particular ,if n(r)
0=r+ 1 (resp. n(r)
2=r+ 1) for some integer r≥0,
then n(s)
0=s+ 1 (resp. n(s)
2=s+ 1) for all integers s≥r.
Let kpbe the completion of a real quadratic field kat pand fix a prime
element of kpthroughout this section. Let Ωpdenote the completion of
the algebraic closure of kpand e
Dthe subgroup of the multiplicative group
Ω×
pconsisting of elements u∈Ωpsuch that vp(u−1) >0, vpbeing the
p-adic normalized valuation on Ωp. Moreover, we denote by logpthe p-adic
logarithm extended to Ωpso that logp(v) = 0 for all v∈Ωp\e
Dand that
logp(uv) = logp(u) + logp(v) for all u, v ∈Ωp(cf. [10] or [13]). Since psplits
in k,kpis isomorphic to the field Qpof p-adic numbers. Therefore, vpand
logpcan be essentially identified with the p-adic valuation vpand the p-adic
logarithm logp, respectively. However, we will use the former notation to
specify the fixed prime.
Let E∗
nbe the group of p-units in kn, i.e., the group of elements εnof kn
with vl(εn) = 0 for all prime ideals lof knoutside p. The following lemma
is now obvious, but its proof will be given for the sake of completeness.
Lemma 2. For each integer r≥0,
(1) n(r)
0= min{vp(logp(Nr,0(ε∗
r))) |ε∗
r∈E∗
r},
(2) n(r)
2= min{vp(logp(Nr,0(εr))) |εr∈Er}.
P r o o f. We first prove (2). Let εr∈Er. Then Nr,0(εr) = ±εa(E0:Nr,0(Er)),
where εdenotes the fundamental unit of kand a∈Z. If a= 0, then
vp(logp(Nr,0(εr))) = ∞. Thus we may assume that a6= 0. From the defi-
nition of n2and Lemma 5.5 of [13], it follows that vp(logp(ε)) = n2, which
implies that
vp(logp(Nr,0(εr))) = vp(a(E0:Nr,0(Er)) logp(ε))
=vp(a) + vp((E0:Nr,0(Er))) + n2
=vp(a) + n(r)
2≥n(r)
2.
On the other hand, there exists an element εrof Ersuch that Nr,0(εr) =
±ε(E0:Nr,0(Er)), so that a= 1. Hence the assertion holds.