数理解析研究所講究録
1097 巻 1999 年 82-88
82
ON THE VANISHING OF IWASAWA INVARIANTS OF
ABSOLUTELY ABELIAN p-EXTENSIONS
GEN YAMAMOTO (山本現)
ABSTRACT. Let be any odd prime. We determine all absolutely abelian p-extension
and -invariants of the cyclotomic -extension are
fields such that Iwasawa
zero, in terms of congruent conditions, pth power residues, and genus fields.
$p$
$\lambda_{p},$
$\mathbb{Z}_{p}$
$\nu_{p}$
$\mu_{p}$
1. INTRODUCTION
the ring of -adic integers. Let be a finite extension of
Let be a prime and
, and
the n-th layer of
a -extension of
the rational number field
the -Sylow subgroup of the ideal class group of . Iwasawa proved the well-known
that there exist integers
of
$\mu=$
theorem about the order
such that
, and
,
$\mathbb{Q},$
$k$
$p$
$\mathbb{Z}_{p}$
$p$
$k_{\infty}$
$k,$
$\mathbb{Z}_{p}$
$A_{n}$
$k_{\infty}/k$
$k_{n}$
$k_{n}$
$p$
$\neq A_{n}$
$\mu(k_{\infty}/k)\geq 0$
$l\text{ノ}=\nu(k_{\infty}/k)$
$\lambda=\lambda(k_{\infty}/k)\geq 0,$
$A_{n}$
$n_{0}\geq 0$
$\neq A_{n}=p^{\lambda n}+\mu p^{n}+\nu$
$\mu=\mu(k_{\infty}/k)$ and $\nu=\nu(k_{\infty}/k)$ are called
for all $n\geq n_{0}$ . These integers
is the cyclotomic -extension of , we write
for . If
Iwasawa invariants of
invariants,
respectively.
the
for
above
and
In [7], Greenberg conjectured that if is a totally real, $\lambda_{p}(k)=\mu_{p}(k)=0$ . We call this
-invariants of abelian p-extension
conjecture Greenberg conjecture. For Iwasawa
fields of , there are results by Greenberg ([7], V), Iwasawa $([9])$ , Fukuda, Komatsu,
, and the author $([12])$ , etc. On the other hand, Ferrero
Ozaki and Taya $([6]),$
and Washington have shown that $\mu_{p}(k)=0$ for any abelian extension field of .
In this paper, we will consider a stronger condition than Greenberg conjecture that
$\lambda_{p}(k)=\mu_{p}(k)=\nu_{p}(k)=0$ and determine all absolutely abelian
-extensions , i.e.
$\lambda_{p}(k)=\mu_{p}(k)=\nu_{p}(k)=0$
is an abelian extension of the rational number field , with
.
for an odd prime , using the results of G. Cornell and M.
$\lambda=\lambda(k_{\infty}/k),$
$k_{\infty}/k$
$\lambda_{p}(k),$
$k_{\infty}$
$p$
$k$
$\mathbb{Z}_{p}$
$\nu_{p}(k)$
$\mu_{p}(k)$
$k$
$\lambda_{p},$
$\mu_{p}$
$\mathbb{Q}$
$\mathrm{F}\mathrm{u}\mathrm{k}\mathrm{u}\mathrm{d}\mathrm{a}([4])$
$k$
$\mathbb{Q}$
$k$
$p$
$k$
$\mathbb{Q}$
$\mathrm{R}_{\mathrm{o}\mathrm{S}\mathrm{e}\mathrm{n}}([1])$
$p$
2. MAIN THEOREM
Throughout this section, we fix an odd prime . For an absolutely abelian p-extension
.
is the minimum positive integer with
be its conductor, i.e.
field , let
and
integer
, where is a non-negative
Then, it follows easily that
by the
are distinct primes which are congruent to 1 modulo . We denote
is the maximal unramified abelian extension of such that
genus field of . So
is an abelian extension of odd degree, then
is an abelian extension. In general, if
$p$
$k$
$k\subseteq \mathbb{Q}(\zeta_{f_{k}})$
$f_{k}$
$f_{k}$
$f_{k}=p^{a}p_{1}\cdots p_{i}$
$p_{1},$
$a$
$k_{G}$
$p$
$\cdots,p_{t}$
$k$
$k$
$k_{G}$
$k/\mathbb{Q}$
$k_{G}/\mathbb{Q}$
83
it has shown by Leopoldt that
$[k_{G} :
k]= \frac{e_{1}e_{2}.\cdots e_{t}}{[k.\mathbb{Q}]}$
,
. Hence in our
where
are ramification indices of primes which ramify in
case,
is also an abelian -extension of . For instance we denote by
the p-th
power residue symbol, i.e., for integers
if and only if is the p-th power
modulo .
Our main theorem gives a necessary and sufficient condition for $\lambda_{p}(k)=\mu_{p}(k)=$
$\nu_{p}(k)=0$ in terms of p-th power residue symbol, congruent conditions and genus fields:
$e_{1},$
$k/\mathbb{Q}$
$\cdots e_{t}$
$k_{G}$
$(-.)_{p}$
$\mathbb{Q}$
$p$
$x,$ $y,$
$x$
$( \frac{x}{y})_{p}=1$
$y$
Theorem 1. Let
be an abelian -extension
decomposition of its conductor, where primes
$k$
of
$p$
$p_{1},$
$\mathbb{Q}$
)
$\cdots,p_{t}$
and $f_{k}=p^{a}p_{1}\cdots p_{t}$ the prime
are distinct. If
$\lambda_{p}(k)=\mu_{p}(k)=\nu_{p}(k)=0$
,
(1)
then $t\leq 2$ . Conversely, in each case of $t=0$ or 1 or the
and sufficient condition of (1):
In case of
: (1) holds.
In case of
: (1) is equivalent to $k_{1}=k_{1,G}and_{f}$
$2_{f}$
$following_{\mathit{8}}$
are a necessary
$\mathrm{t}=0$
$\mathrm{t}=1$
(mod
$( \frac{p}{p_{1}})_{p}\neq 1orp_{1}\not\equiv 1$
In case
of
$\mathrm{t}=2$
: (1) is equivalent to
$( \frac{p}{p_{i}})_{p}\neq 1,$
$and_{f}$
there exist
$x,$ $y,$
$z\in \mathrm{F}_{p}$
$( \frac{p_{j}p^{x}}{p_{i}})_{p}=1,$
$k_{1}=k_{1,Gf}$
$( \frac{p_{i}}{p_{j}})_{p}\neq 1,p_{j}\not\equiv 1$
$p^{2}$
(2)
).
and for $(i,j)=(1,2)$ or
(mod
$p^{2}$
$(2, 1)_{f}$
),
(3)
such that
$( \frac{pp_{i}^{y}}{p_{j}})_{p}=1,p_{i}p_{\mathrm{j}}^{z}\equiv 1$
(mod
$p^{2}$
), and
$xyz\neq-1$ in
(4)
$\mathrm{F}_{p}$
In case of $t=2$ , the conditions in Theorem 1 are complicated. So we will give an
example. We consider the case $p=3,p_{1}=7$ and $p_{2}=19$ . We denote
by
the subfield of
with degree 3 ver . As for the condition $k_{1}=k_{1,G}$ ,
there exists a field $F$ such that
and $F\neq k7k19,$
, where
is the
first layer of cyclotomic -extension of . Then
is a nontrivial unramified
extension and
. But, for
is abelian, hence
, it
follows easily that $F_{1}=F_{1,G}$ . If we restrict
case is unramified in , i.e. $a=0$ , then
the statement $k_{1}=k_{1,G}$ can be simplified to $k=k_{G}$ because
. This restriction
is not so strong: In general, for an absolutely abelian -extension
, there exists an
absolutely abelian extension field such that is unramified in and
. Note
that if is the maximal subfield of
( $m=p^{a}p_{1}\cdots p_{t}$ as above) which is abelian
-extension of , then $k=k_{G}$ .
We continue to examine the above example. If we put $(i,j)=(1,2)$ , then $p_{j}=19\equiv 1$
(mod ), so the condition (3) is not satisfied. But if we put $(i,j)=(2,1)$ , then we can
$k_{7}(\mathrm{r}\mathrm{e}\mathrm{s}_{\mathrm{P}}.
$\mathbb{Q}(\zeta_{7})(\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{p}.\mathbb{Q}(\zeta_{1}9))$
$k7\mathbb{Q}1$
$k_{7}\underline{\subseteq}F\subset\sim k_{7}k_{19}\mathbb{Q}_{1}$
$F\subseteq k_{\tau}k_{19}\mathbb{Q}1\subseteq F_{G}$
$k_{7}k_{19}\mathbb{Q}_{1}$
$\mathbb{Q}\mathrm{l}$
$k_{7}k_{19}\mathbb{Q}_{1}/F$
$\mathbb{Q}$
$\mathbb{Z}_{3}$
$F_{1}=k_{7}k_{19}\mathbb{Q}_{1}$
’
$k$
$\mathrm{t}\mathrm{h}\mathrm{e}$
$p$
$k_{1}=k\mathbb{Q}_{1}$
$\mathrm{f}\mathrm{i}\mathrm{e}\mathrm{l}\dot{\mathrm{d}}k$
$p$
$k’$
$k$
$3^{2}$
$p$
$\mathbb{Q}(\zeta_{m})$
$\mathbb{Q}$
$p$
k_{19})$
$\mathbb{Q}$
$\mathit{0}$
$k’$
$k_{\infty}=k_{\infty}’$
84
verify that $p_{i}=19$ and $p_{j}=7$ satisfy the conditions (3) and (4). Hence, for example,
which is 3-extension of , then If satisfies the
if $K$ is the maximal subfield of
conditions of Theorem 1. Therefore we get
$\mathbb{Q}$
$\mathbb{Q}(\zeta_{7\cdot 19})$
$\lambda_{p}(K)=\mu_{p}(K)=\nu p(K)=0$
.
As for Greenberg conjecture, we can also get the following: In general, it is known that
and $\mu_{p}(L)\leq\mu_{p}(M)$ for number fields $L,$ $M$ . Hence for
then
if
theJn $\lambda(pk)=\mu_{p}(k)=0$ .
which is 3-extension of , i.e.
any subfield of
This consideration is generalized as follows:
$L\subseteq M$
$\lambda_{p}(L)\leq\lambda_{p}(M)$
$k$
Corollary 2. Let
subfield of
$k$
$\mathbb{Q}(\zeta_{m})$
$k\underline{\subseteq}K,$
$\mathbb{Q}$
$\mathbb{Q}(\zeta_{7\cdot 19})$
satisfy the condition (2) or (3), (4). Then for any
which is -extension of , Greenberg conjecture for and is valid.
$m=p^{a}p_{1}\cdots p_{t}$
3. THE
$k$
$\mathbb{Q}$
$p$
RESULTS OF
$p$
G. CORNELL AND M. ROSEN
be an abelian -extension, a
In this section, we review briefly part of [1]. Let
, A. Fr\"ohlich determined all such fields with class number prime to
prime. In the
. Cornell and M. Rosen reconsidered this problem in the case where
(cf. [2]). In
is an elementary
is an odd prime, and reduced the problem to the case when
$m$
.
for some integer
abelian -group, i.e.
is an abelian -group. Then the
We suppose that is an odd prime and
of $K$ is also abelian -extension. If does not divide the class number
genus field
of $K$ , then $K$ does not have any non-trivial unramified abelian -extension by class
. Further, we
. In the following we will assume
field theory, hence
of $K$ such
If,
is
-extension
of
the
maximal
-class
field
i.e.
the
central
consider
is in the center
is Galois and
is abelian and unramified,
that
that
terminates
in the
central
series
a
lower
a
must
-group
.
have
Since
of
. So we are interested in which
if and only if
identity, one sees that
$K_{C}=K$
is an elementary abelian
case
when
the
. This can be reduced
case
-group by the following result:
$K/\mathbb{Q}$
$p$
$p$
$1950’ \mathrm{s}$
$[1],\mathrm{G}$
$p$
$p$
$\mathrm{G}\mathrm{a}1(K/\mathbb{Q})$
$\mathrm{G}\mathrm{a}1(K/\mathbb{Q})\simeq(\mathbb{Z}/p\mathbb{Z})^{m}$
$p$
$\mathrm{G}\mathrm{a}1(K/\mathbb{Q})$
$p$
$IC_{\mathrm{G}}$
$p$
$p$
$p$
$h_{K}$
$p$
$I\mathrm{f}_{G}=K$
$I\mathrm{f}_{G}=I\acute{\mathrm{t}}$
$I\zeta_{C}$
$I\mathrm{f}_{C}$
$p$
$p$
$\mathrm{G}\mathrm{a}1(Kc/K)$
$K_{C}/\mathbb{Q}$
$I\mathrm{f}_{C}/K$
$\mathrm{G}\mathrm{a}1(I\mathrm{f}c/\mathbb{Q})$
$p$
$I\mathrm{f}_{C}=I\acute{\mathrm{t}}$
$p\wedge h_{I\prec_{\mathrm{L}}’}$
$\mathrm{G}\mathrm{a}1(K/\mathbb{Q})$
$p$
Let
be an abelian -extension with
Lemma 3 ([1] Theorem 1). Let
$K$
is
such that
and
be the maximal intermediate extension between
to
the
-rank
equal
is
an elementary abelian -group. Then -rank of
of
$K/\mathbb{Q}$
$I\mathrm{f}_{G}=K.$
$p$
$k$
$\mathrm{G}\mathrm{a}1(k/\mathbb{Q})$
$\mathbb{Q}$
$\mathrm{G}\mathrm{a}1(kc/k)$
$\mathrm{G}\mathrm{a}1(I\zeta_{C}/K)$
$p$
$p$
$p$
.
is an elementary abelian -group, by the results of Furuta and
In the case
Tate, we have the following lemma:
$\mathrm{G}\mathrm{a}1(K/\mathbb{Q})$
$p$
Lemma 4 ([1] Section 1). Let $K$ be an absolutely abelian -extension such that
. Then, we have
is an elementary abelian -group and
$p$
$p$
$I\{\mathrm{i}_{G}=K$
$\mathrm{G}\mathrm{a}1(Kc/I\mathrm{f})\simeq \mathrm{C}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}(\oplus^{n}i=1\wedge^{2}(G_{i})arrow\wedge^{2}(G))$
where
’
$G_{i^{S}}$
$\mathrm{G}\mathrm{a}1(K/\mathbb{Q})$
are the decomposition groups
of primes ramified in
$K/\mathbb{Q}$
,
and
$G=\mathrm{G}\mathrm{a}1(K/\mathbb{Q})$
.
85
We will assume
. Let
be the primes ramified in and
the class number of If. From genus theory, it follows that if
is not divisible by
, then $t=m$ . Also it follows that if $m\geq 4$ then divides
by Lemma 4. So, we
assume $t=m$ and $m=2$ or 3. (In case of $t=m=1,$
. cf. [8].)
$\mathrm{G}\mathrm{a}1(K/\mathbb{Q})\simeq(\mathbb{Z}/p\mathbb{Z})^{m}$
$p_{1},$
$I\iota’$
$\cdots p_{t}$
$h_{K}$
$h_{I\mathrm{t}^{r}}$
$p$
$h_{K}$
$p$
$p\parallel h_{K}$
Lemma 5 ([1] Proposition 2). Suppose
and only if $p=1$ and
.
and
$m=2$
$p_{i}\neq p$
for $i=1,2$ . Then
$p|h_{K}$
if
$( \frac{p_{2}}{p_{1}})_{p}=1$
Next, we consider the case where one of the ramified primes is . Suppose $m=2$ and
and
are the only primes ramified in $K$ . Then we can get easily
and
(mod ), where
is the unique subfield of
which is cyclic over of
degree
is a primitive -th root of unity, and
is the first layer of the cyclotomic
-extension of .
$p$
$p$
$K=k(p_{1})\mathbb{Q}_{1}$
$p_{1}$
$p_{1}\equiv 1$
$k(p_{1})$
$p$
$p,$
$\zeta_{\mathrm{P}1}$
$\mathbb{Q}(\zeta_{p_{1}})$
$\mathbb{Q}$
$\mathbb{Q}_{1}$
$p_{1}$
$\mathbb{Q}$
$\mathbb{Z}_{p}$
Lemma 6 ([1] Proposition 3). Suppose $m=2$ and
,
and
ramified in K. Then
if and only if
$p|h_{I\mathrm{t}}$
Suppose $t=m=3$ and
and
decomposition field of $p_{i}(i=1,2,3)$ in
$p_{1},p_{2}$
$p_{3}$
$p$
and
$p_{1}\equiv 1$
$( \frac{p}{p_{1}})_{p}=1$
are the only primes
(mod ).
$p_{1}$
$p^{2}$
all the primes ramified in If. We put
the
. In [1], the following simple result is given:
$D_{pi}$
$K$
Lemma 7 ([1] Theorem 2). Suppose $t=m=3$ . Following statements (a) and (b)
are equivalent:
(a)
is not divisible by
(b)
and
.
$h_{K}$
$p_{f}$
$[D_{p_{1}} :
\mathbb{Q}]=[D_{p_{2}} :
\mathbb{Q}]=[D_{p_{3}} :
\mathbb{Q}]=p$
$D_{p_{1}}D_{p2}D_{p\mathrm{s}}=K$
In the next section, we shall prove Theorem l,using these results.
4. PROOF
OF
THEOREM 1
Notations are as in previous section.
Firstly, we suppose $\lambda_{p}(k)=\mu_{p}(k)=\nu_{p}(k)=0$ . Clearly, this condition is equivalent
to $A(k_{n})=0$ for any sufficiently large . Then,
satisfies $k_{n}=k_{n,G}$ and $k_{1}=k_{1,G}$ ,
because all ramified primes are totally ramified in
. Since is also an abelian
-extension of , we can apply the results of ornell-Rosen:
Let
be the maximal subfield of
such that
is an elementary abelian
extension of . Since $k_{n}=k_{n,G},$
is the direct sum of the inertia groups
of primes ramified in
, hence it follows that $L=k(p1)\cdots k(Pt)\mathbb{Q}1$ . By Lemma 3,
$A(k_{n})=0$ is equivalent to
. Note that if $t\geq 3$ then we always have
as in the
previous section. Hence we may examine in each case of $t=0$ or 1 or 2.
If $t=0$ then
, hence it is well known that $A(L)=A(\mathbb{Q}_{1})=0$ (cf. [8]).
If $t=1$ then
. By lemma 6, we get the statement in Theorem 1.
In the following we assume that $t=2$ . In this case, $L=k(p_{1})k(p2)\mathbb{Q}_{1}$ . Let
$G_{pi}(i=$
$1,2)$ be the decomposition groups for
in
and let
be the fixed
, respectively. We note that $D_{p}\subset k(p_{1})k(p2),$
field of
and
$k_{n}$
$n$
$k_{n}/k_{1}$
$p$
$k_{n}$
$\mathrm{C}$
$\mathbb{Q}$
$L$
$k_{n}$
$\mathbb{Q}$
$\mathrm{G}\mathrm{a}1(L/\mathbb{Q})$
$\mathrm{G}\mathrm{a}1(k_{n}/\mathbb{Q})$
$k_{n}/\mathbb{Q}$
$p\parallel h_{L}$
$p|h_{L}$
$L=\mathbb{Q}_{1}$
$L=k(p_{1})\mathbb{Q}_{1}$
$G_{p},$
$p,p_{i}$
$G_{p},$
$k(p_{1})\mathbb{Q}_{1}$
$\mathrm{G}\mathrm{a}1(L/\mathbb{Q})$
$G_{p_{i}}$
$D_{p},$
$D_{p_{i}}$
$Dp1\subset k(p_{2})\mathbb{Q}_{1}$
$D_{p_{2}}\subset$
.
Now, from our assumption
, we have
$D_{p}D_{p1}D_{p2}=L$ by Lemma 7. Here, we assume that either
$p\parallel h_{L}$
$[D_{p} :
\mathbb{Q}]=[D_{p1} :
and
$p=1$ or
\mathbb{Q}]=[D_{p_{2}}2^{\cdot}\mathbb{Q}]=p$
$p=1$
or
86
$p_{2}\equiv 1$
(mod
$p^{2}$
$p=1$
) holds, and either
equivalent to
$D_{p}=k(p_{i})$
or
$D_{pi}=k(pj)$
or
or
or
$( \frac{p_{2}}{p_{1}})_{p}=1$
$p_{1}\equiv 1(\mathrm{m}o\mathrm{d}p^{2})$
for $(i,j)=(1,2)$ and
$D_{p_{J}}=\mathbb{Q}_{1}$
$(2, 1)$
. This is
(5)
,
.
because
If $D_{p}=k(p_{1})$ , then $D_{p_{2}}\neq k(p_{1})$ because $D_{p}D_{p_{1}}D_{p_{2}}=L$ . Hence by (5) (put
, which contra. Then
$(i,j)=(2,1))$ , we have
and we have
dicts $D_{p}D_{p1}D_{p2}=L$ . In the same way, if $D_{p}=k(p_{2})$ , then
(5) cause
assumption
the
follows
Thus,
it
that
by (5), which contradicts.
contradiction. Therefore, for $(i,j)=(1,2)$ or $(2, 1)$ ,
$p\neq 1$ , and
(mod ).
is inert
Without loss of generality, we may assume $(i,j)=(1,2)$ . Since
, where
is the Artin symbol, and generates
. Hence
in
. We often regard $<\sigma>=\mathrm{G}\mathrm{a}1(k(p1)k(p_{2})/k(p_{2}))$
,
and
in the natural way. Similarly, we put
or
.
and
then
. Then
such that
, there exists
Since
$[D_{p} :
\mathbb{Q}]=[D_{p_{1}} :
\mathbb{Q}]=[D_{p_{2}} :
\mathbb{Q}]=p$
$D_{p_{2}}\subseteq k(p_{1})\mathbb{Q}_{1}=D_{p}D_{\mathrm{P}1}$
$D_{p_{1}}=\mathbb{Q}_{1}$
$D_{p_{1}}\neq k(p_{2})$
$D_{p_{2}}=\mathbb{Q}_{1}$
$p_{j}\not\equiv 1$
$( \frac{p}{pi})_{p}\neq 1,$
$p^{2}$
$( \frac{p}{p_{1}})_{p}\neq 1,$
$k(p_{1})$
$\sigma=(\frac{k(p_{1})/\mathbb{Q}}{p})\neq 1$
$( \frac{k(p1)/\mathbb{Q}}{p})$
$p$
$\sigma$
$\mathrm{G}\mathrm{a}1(k(p_{1})/\mathbb{Q}):<\sigma>=\mathrm{G}\mathrm{a}1(k(p_{1})/\mathbb{Q})$
$\tau=(\frac{k(p2)/\mathbb{Q}}{p_{1}})$
$\mathrm{G}\mathrm{a}1(L/k(p2)\mathbb{Q}1)$
$<\tau>=\mathrm{G}\mathrm{a}1(k(p_{2})/\mathbb{Q})$
$<\eta>=\mathrm{G}\mathrm{a}1(\mathbb{Q}_{1}/\mathbb{Q})$
$( \frac{p_{2}p^{x}}{p_{1}})_{p}=1$
$x\in \mathrm{F}_{p}$
$( \frac{p}{p_{1}})_{p}\neq 1$
$\eta=(\frac{\mathbb{Q}_{1}/\mathbb{Q}}{p_{2}})$
$( \frac{p_{2}p^{x}}{p_{1}})_{p}=1\Leftrightarrow(\frac{k(p_{1})/\mathbb{Q}}{p_{2}p^{x}})=(\frac{k(p_{1})/\mathbb{Q}}{p_{2}})(\frac{k(p_{1})/\mathbb{Q}}{p})^{x}=1$
Therefore
$p_{1}P_{2}^{\chi}\equiv 1$
Since
(mod
$p^{2}$
.
Similarly, we obtain
and
), and hence
$( \frac{k(p_{1})/\mathbb{Q}}{p_{2}})=\sigma^{-x}$
$( \frac{k(p2)/\mathbb{Q}}{p})=\tau^{-y}$
$y,$
Therefore, when we
$\mathrm{c}o$
nsider
in
$G_{p}$
$G_{p}=<\eta,$
such that
$( \frac{pp_{1}^{y}}{p_{2}})_{p}=1$
.
is the fix field of
and
$( \frac{\mathbb{Q}_{1}/\mathbb{Q}}{P1})=\eta^{-z}$
$( \frac{k(p_{1})k(p2)/\mathbb{Q}}{p})=(\frac{k(p1)/\mathbb{Q}}{p})(\frac{k(p2)/\mathbb{Q}}{p})=\sigma\tau^{-y},$
$k(p1)k(P2)$ .
$z\in \mathrm{F}_{p}$
.
$D_{p}$
$\mathrm{G}\mathrm{a}1(L/\mathbb{Q})$
$\sigma\tau-y>$
$<\sigma\tau^{-y}>$
in
,
.
And similarly,
$G_{p1}=<\sigma,\tau\eta^{-}z>$
,
$G_{p_{2}}=<\tau,$
,
and
in
$\mathrm{G}\mathrm{a}1(L/\mathbb{Q})$
$\eta\sigma^{-}>x$
.
By a direct computation, we have,
.
$G_{p}\cap G_{p}1=<\sigma\tau^{-y}\eta^{yz}>$
Hence,
$\eta\sigma^{-}x>>$
$G_{p^{\cap G}p1^{\cap G_{p}}}2=<\sigma\tau^{-}\eta yyz\mathrm{n}<\mathcal{T},$
$=\{$
But , our assumption
$D_{p}D_{p1}D_{p2}=L$
$<\sigma\tau^{-}\eta>\{1\}yyz$
implies
,’
.
$\mathrm{i}\mathrm{f}xyz=\mathrm{i}\mathrm{f}xyz\neq-1-1’$
$G_{p}\cap G_{p_{1}}\cap G_{p_{2}}=\{1\}$
. Hence
$xyz\neq-1$
.
87
Conversely, we assume satisfies the conditions of Theorem 1 in case of $t=2$ . Since
$k_{1}=k_{1,G}$ , it follows easily that $L=k(p_{1})k(p2)\mathbb{Q}_{1}$ is the maximal intermediate ext\’ension
between and $k_{n}(n\geq 1)$ such that
is an elementary abelian -group. Without loss of generality, we may assume $(i,j)=(1,2)$ . Since
and is unramified in $k(p_{1})k(p2),$ must decompose in $k(p_{1})k(p_{2})$ . But the condition
is inert in $k(p_{1})\subset k(p_{1})k(p_{2})$ , hence we obtain
. Simi$p\neq 1$ implies
larly,
and
(mod ) imply
. Therefore, as in
, by $xyz\neq-1$ .
the above computation of
, we have
$k$
$\mathrm{G}\mathrm{a}1(L/\mathbb{Q})$
$\mathbb{Q}$
$p$
$\mathrm{G}\mathrm{a}1(k(p1)k(p_{2})/\mathbb{Q})\simeq(\mathbb{Z}/p\mathbb{Z})^{2}$
$p$
$p$
$[D_{p} :
$p$
$( \frac{p_{1}}{p_{2}})_{p}\neq 1$
$p^{2}$
$p_{2}\not\equiv 1$
$G_{p},$
$[D_{p_{1}} :
\mathbb{Q}]=[D_{\mathrm{P}2} :
\mathbb{Q}]=p$
\mathbb{Q}]=p$
$D_{p}D_{p1}D_{\mathrm{P}2}=L$
$G_{p_{i}}$
$\square$
5. REMARKS
The condition of Theorem 1 in [12] means $xyz=0$ which is a special case of $xyz\neq-1$ .
Hence, our Corollary 2 conta\’ins some known results and there exist infinitely many fields
satisfying the conditions of Theorem 1 (cf. [12]).
If $K=k(p1)k(P2)$ satisfies the conditions of Theorem 1, then $\lambda_{p}(k)=\mu_{p}(k)=0$
for any field
with $[k : \mathbb{Q}]=p$ . This is a result of Fukuda [4]. He has shown
this result using a technic of capitulation of ideal class group. The case $xyz=-1$ is a
difficult case. But we can get some results:
$k\subseteq K$
Proposition 8. Notations are as in section 3. Assume that $p\neq 1,$
. Then $\lambda_{p}(k)=\mu_{p}(k)=0$ for the decomposition field
and
$p\neq 1_{\text{ノ}}$
$p_{2}\not\equiv 1(\mathrm{m}o\mathrm{d}p^{2})$
$k(p_{1})k(F2)$
$k$
.
of
$p$
in
Proof, We apply a result of [6]:
Lemma 9 ([6] Corollary 3.6). Let
following conditions are equivalent:
(a) $\lambda_{p}(k)=\mu_{p}(k)=0$ ,
(b) For any prime ideal of
the order of the ideal class
$w$
$k$
be a cyclic extension
which is prime to
of is prime to .
$k_{\infty}$
$w$
$p$
of
and
$\mathbb{Q}$
of degree . Then
$p$
ramified in
$k_{\infty}/\mathbb{Q}_{\infty}$
the
,
$p$
If $xyz\neq-1$ then we have $\lambda_{p}(k)=\mu_{p}(k)=0$ by Corollary 2. So we only consider
the case $xyz=-1$ . In this case we have $k\neq k(p_{i})$ $(i=1,2)$ . It follows easily that
$A(k)$ , the -part of the ideal class group of
, is cyclic of order , and it is generated
by products of primes of above . On the other hand, for $i=1,2$ , the prime of
above generates $A(k)$ , and is inert in
. Since the primes of above is principal
by the natural mapping $A(k)arrow A(k_{n})$ (cf. [7]), is principal in
for some
.
Since the primes ramified in
are
and , which is principal in
, we can
apply Lemma 9 and obtain $\lambda_{p}(k)=\mu_{p}(k)=0$ .
$k$
$p$
$k$
$p$
$p$
$\mathfrak{p}_{i}$
$k$
$k_{\infty}/k$
$p_{\mathrm{i}}$
$k_{n}$
$\mathfrak{p}_{i}$
$k_{\infty}/\mathbb{Q}_{\infty}$
$\mathfrak{p}_{1}$
$k$
$p$
$k_{\infty}$
$k_{\infty}$
$\mathfrak{p}_{2}$
$\square$
Recently, Fukuda verified Greenberg conjecture for various cubic cyclic fields with
$f_{k}=p_{1}p_{2}$ and $p=3$ .
He gives an example, which is the case $p_{1}=7$ and $p_{2}=223$ .
Note that there exist two such fields, and these and do not satisfy condition (3) in
Theorem 1. He verified
for one of such fields by using his result concerning.
with the unit group of (cf. [5]).
$k$
$p_{1}$
$\lambda_{3}=\mu_{3}=0$
$k$
$p_{2}$
88
, there are a few results for
When $t\geq 3$ , i.e. at least 3 primes are ramified in
$A(k)$
is greater than 2. Greenberg $([7])$
Greenberg conjecture. In this case, the -rank of
gave the following example, but the proof are omitted in his paper: $p=3$ and is an
. He mentioned that
and 3 is inert in
cubic cyclic field with conductor 7
by ”delicate” arguments one can show $\lambda_{3}(k)=\mu_{3}(k)=0$ . The author had a chance
to contact Prof. Greenberg, and asked him about this example. He kindly taught the
author the ”delicate” arguments, which is a system to examine relations of the ideal
. Applying his idea, we can show the following
class group of intermediate fields of
result:
Theorem 10 ([13]). Let be any odd prime. For any integer $0\leq m\leq p-1_{f}$ there exist
infinitely many cyclic extension fields of with $[k:\mathbb{Q}]=p$ such that p-rankA $(k)=m$
and $\lambda_{p}(k)=\mu_{p}(k)=0$ .
$k/\mathbb{Q}$
$p$
$k$
$\cdot 13\cdot 19$
$k/\mathbb{Q}$
$k\mathbb{Q}_{1}$
$p$
$k$
$\mathbb{Q}$
REFERENCES
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(1988), 453-461.
, Central extensions, Galois groups, and ideal class groups of number fields, Contemp.
2.
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degree , Proc. Japan Acad. 73A (1997), 108-110.
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of
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$l$
$\mathrm{A}.\mathrm{F}\mathrm{r}\ddot{\mathrm{o}}\mathrm{h}\mathrm{l}\mathrm{i}_{\mathrm{C}}\mathrm{h}$
$\mathbb{Z}_{p}$
$\mathbb{Q}$
$\mathbb{Q}$
$\lambda_{p}$
$\mathbb{Q}$
$p$
$\mathbb{Q}$
DEPARTMENT OF MATHEMATICAL SCIENCE, SCOOL OF SCIENCE AND ENGINEERING., WASEDA
UNIVERSITY, 3-4-1, OKUBO SHINJUKU-KU, TOKYO 169-8555, JAPAN
-mail address: [email protected]
$E$