\input amstex
\define\Q{{\bold Q}}
\define\GL{{\operatorname{GL}}}
\define\Cl{{\operatorname{Cl}}}
\define\Gal{{\operatorname{Gal}}}
\define\la{\langle}
\define\ra{\rangle}
\documentstyle{amsppt}

\mag=\magstep1

\document
\nopagenumbers

Imaginary quadratic number fields $K=\Q(\sqrt{d})$ with $|d|<1000$,
where $d$ is the discriminant of $K$,
can be classified simply as follows:

\eightpoint
\NoBlackBoxes
$$
\matrix
	\format \l & \l & \l \\
	\matrix
		\format \l \\
		\noalign{\vskip18pt}
		\left\{
		\matrix
		\\
		\noalign{\vskip27pt}
		\\
		\endmatrix
		\right. 
	\endmatrix &
	\hskip-5pt
	\matrix
		\format \l \\
		\noalign{\vskip18pt}
		d_E\nmid d\Bigg\{ \\
		\\
		\noalign{\vskip9pt}
		\\
		d_E\mid d\Bigg\{
	\endmatrix &
	\hskip-3pt
	\matrix
		\format \l & \l & \l \\
		\matrix
			\format \l & \l \\
			\matrix
				\format \l \\
				\noalign{\vskip6pt}
				\hskip-1pt
				d\ne-856,-996\biggm\{ \\
			\endmatrix &
			\hskip-1pt
			\matrix
				\format \l & \l \\
				h(K_g)=h(K)/[K_g:K] \biggm\{ &
				\matrix
					\format \l \\
					h(K)=1 \\
					\noalign{\smallskip}
					h(K)>1
				\endmatrix \\
				h(K_g)>h(K)/[K_g:K] &
			\endmatrix
		\endmatrix &
		\hskip-5pt
		\left.
		\matrix
			\format \l & \c \\
			\cdots & K_{ur}=K \\
			\noalign{\smallskip}
			\cdots & K_{ur}=K_1 \\
			\noalign{\smallskip}
			\cdots & K_{ur}=K_2 \\
		\endmatrix
		\right\} &
		K_{ur}=(K_g)_1 \\
		\noalign{\smallskip}
		\matrix
			\format \l \\
			d=-856,-996 \\
			\noalign{\smallskip}
			\matrix
				\format \l & \l \\
				\hskip-1pt
				d=d_E \Bigm\{ &
				\matrix
					\format \l \\
					d_E=-p\text{\,:\,prime}\\
					\noalign{\smallskip}
					d_E\text{\,:\,composite}
				\endmatrix
			\endmatrix \\
			\noalign{\smallskip}
			d=d'd_E
			(d'\text{\,:\,fundamental quadratic discriminant})
		\endmatrix &
		\hskip-5pt
		\left.
		\matrix
			\format \l & \c \\
			\cdots & l\geqq 3 \\
			\noalign{\smallskip}
			\cdots & K_{ur}=K_3 \\
			\noalign{\smallskip}
			\cdots & l\geqq 2 \\
			\noalign{\smallskip}
			\cdots & l\geqq 3 \\
		\endmatrix
		\right\} &
		K_{ur}\supsetneqq (K_g)_1
	\endmatrix
\endmatrix
$$
\tenpoint
\noindent
Here, we denote by $K_g$ the genus field of $K$.
We mean by $d_E\mid d$ that the discriminant $d$ of $K$ is
divisible by the discriminant of some quartic number field $E$.
In this classification, there are some possible exceptions.
More precisely, for $18$ fields $K$ with $d_E\nmid d\ne -856,-996$,
we have not succeeded in showing $K_{ur}=(K_g)_1$.
(Among them, for $15$ fields, we have $l=1$ and
in order to conclude $K_{ur}=K_1$, we must show that
$K_1$ does not have an unramified $A_5$-extension.
For the other $3$ fields we have $K_{ur}=K_l, 2\leqq l\leqq 3$,
and we expect $l=2$.)
For ($12$ fields) $K$ with $K_{ur}\supsetneqq (K_g)_1$,
there exists an $S_4$-extension $M$ of $\Q$
such that the compositum $K_gM$ is an unramified $A_4$-extension of $K_g$.

%The first known imaginary quadratic number field with $l\geqq 4$ is
%$K=\Q(\sqrt{-1415})$.
%It is thought likely that this field is
%really the first such field.
\medpagebreak

\proclaim{Question} 
What is the first imaginary quadratic number field
having an unramified nonsolvable Galois extension ?
{\rm(}What is the first $K$ with $K_{ur}\ne K_l$ ?{\rm)}
\endproclaim

\noindent
Recent data for quintic number fields enable us to
give a partial answer:

\proclaim{Proposition 1}
The field $\Q(\sqrt{-1507})$ is the first imaginary quadratic number field
having an unramified $A_5$-extension which is normal over $\Q$
in the sense that none of $\Q(\sqrt{d})$ of discriminant $d$
with $0>d>-1507$ has such an extension.
Moreover, such an extension of $\Q(\sqrt{-1507})$ is
given by the composite field of it with the splitting field of
the quintic polynomial $X^5-5X^3+5X^2+24X+4$,
which is an $A_5$-extension of $\Q$.
\endproclaim

\noindent
It is reasonable to expect that the field $\Q(\sqrt{-1507})$ gives the answer
to the question above.

we may expect that $[K_{ur}:(K_g)_1]$ is small for fields we consider
on the ground of the following.

\proclaim{Proposition 2} 
Let $L$ be the Hilbert class field of the gunus field $K_g$ of
an imaginary abelian number field $K$.
Then for any prime number $p$ with $p\nmid [L:\Q]$,
the $p$-class group $\Cl^{(p)}(L)$ of $L$ is trivial or noncyclic.
\endproclaim

For all $K$ with $|d|<1000$
such that $h(K_g)>h(K)/[K_g:K]$,
which is equivalent to $(K_g)_1\supsetneqq K_1$,
we have $K_2=(K_g)_1$. %, that is,
%the second Hilbert class field of $K$ coincides with 
%the Hilbert class field of the genus field $K_g$ of $K$.
For $h(K_g)>h(K)/[K_g:K]$, $h(K)$ must necessarily be even.
$h(K)$ is even if and only if $d$ has (at least) two distinct prime factors,
however, for most $K$, this inequality holds.
In fact, if a quadratic subfield$\ne K$ of $K_g$
has class number divisible by an odd prime $p$,
then we have $h(K_g)\geqq ph(K)/[K_g:K]$.
Thus, the following problem arises naturally:

\proclaim{Problem 1} 
Characterize the imaginary quadratic number fields $K$
with $K_2=(K_g)_1\ne K_1$.
\endproclaim

\noindent
This is very difficult.
(But we can easily find $K$ with $K_2\supsetneqq(K_g)_1\supsetneqq K_1$.
In fact, if the $p$-rank of the $p$-class group of $K$
is larger than one for some odd prime $p$,
then $K_2\supseteqq (K_g)_1K_2^{(p)}\supsetneqq (K_g)_1$,
and therefore for such $K$ with $(K_g)_1\ne K_1$,
we have $K_2\ne(K_g)_1\ne K_1$.
For example, let $K=\Q(\sqrt{-3896})$.
Then we have $\Cl(K)\cong C_{12}\times C_3,
K_g=\Q(\sqrt{2},\sqrt{-487})$, and $\Cl(K_g)\cong C_{42}\times C_3$.
Therefore $K_2\supsetneqq K_1^{(3)}(K_g)_1=(K_g)_1\supsetneqq K_1$.)
Thus, it is reasonable to consider the following
before treating Problem 2.

\proclaim{Problem 2} 
Characterize the imaginary quadratic number fields $K$ with
$$
K_2^{(2)}=(K_g)_1^{(2)}\ne K_1^{(2)}.
$$
\endproclaim

\noindent
By known results on the $2$-class field tower,
if the $2$-class group of $K_1^{(2)}$ is nontrivial and cyclic,
$K_2^{(2)}=(K_g)_1^{(2)}\ne K_1^{(2)}$ holds.
No example of $K$
with $K_2^{(2)}\supsetneqq (K_g)_1^{(2)}\supsetneqq K_1^{(2)}$ is known.
\medpagebreak

Note that $h(K_g)=h(K)/[K_g:K]$ is equivalent to $(K_g)_1=K_1$
and that all fields with odd class number satisfy this.
For some $K$ with $(K_g)_1=K_1$,
we can estimate $h(K_1)$ from below under some assumptions
by considering the action of Galois groups on class groups.

\proclaim{Proposition 3} 
Let $K$ be an imaginary quadratic number field
with cyclic class group of odd order $n\geqq 7$.
Assume that for all intermediate fields $E$ of $K_1/K$, we have $E_1=K_1$.
Then if $h(K_1)>1$ {\rm(}this is equivalent to $K_2\ne K_1${\rm)},
we have $h(K_1)\geqq 2^6=64$.
{\rm(}This is valid also for real quadratic number fields.
In the case where $K$ is imaginary and $h(K)$ is a prime {\rm(}$\geqq 7${\rm)},
$h(K_1) \geqq 64$ can be improved to $h(K_1)\geqq 13^2=169$.
Moreover, in the case where $h(K)\geqq 11$ and $h(K)\ne 21, 63$,
$h(K_1)\geqq 64$ can be improved to $h(K_1)\geqq 2^8=256$
{\rm(}without assuming the imaginarity of $K$.{\rm)}
{\rm)}
\endproclaim

\noindent
The first known imaginary quadratic number field
with $h(K)=q\geqq 7$ ($q$\,:\, an odd prime) and $l\geqq 2$
is $\Q(\sqrt{-14947})$.
(For $K=\Q(\sqrt{-14947})$, we have $h(K)=17$ and $256\mid h(K_1)$.)
It is thought likely that this field is
really the first such field.

\proclaim{Proposition 4} 
Let $K$ be an imaginary quadratic number field
with nontrivial cyclic $2$-class group.
\newline
{\rm (i)} Assume $\Cl(K)\cong C_{2^m} (m\geqq 3)$ and
that the unique unramified cyclic quartic extension $E$ of $K$
has class number $2^{m-2}${\rm(}, which implies $h(K_g)=h(K)/2$.{\rm)}
If $h(K_1)>1$, then we have $h(K_1)\geqq 7^2=49$.
\newline
{\rm (ii)} Assume $h(K)=2q$ {\rm(}$q$\,:\, an odd prime $\geqq 7${\rm)} and $h(K_g)=q(=h(K)/2)$.
If $h(K_1)>1$, then we have $h(K_1)\geqq 2^6=64$.
\newline
{\rm(}These are valid also for real quadratic number fields
if $K_g$ is the unique unramified quadratic extension of $K$.{\rm)}
\endproclaim

\enddocument
