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\topmatter
\title
construction of tables of biquadratic bicyclic number fields
\endtitle
\author By Ken Yamamura \endauthor
\affil Department of Mathematics,\\
National Defense Academy \endaffil
\rightheadtext{construction of table}
\leftheadtext{construction of table}
%\address Hashirimizu Yokosuka 239-8686, Japan \endaddress
\endtopmatter

\document
Let $K$ be a biquadratic bicyclic number field
and $k_1$, $k_2$, $k_3$ its quadratic subfields.
If an odd prime $p$ divides $d(K)$, the discriminant of $K$,
then $p$ divides exactly two of $d(k_1)$, $d(k_2)$, $d(k_3)$.
On the other hand, if $d(K)$ is even, then there are two cases:
\roster
\item"" Exactly two of $d(k_1)$, $d(k_2)$, $d(k_3)$ are even;
\item"" All $d(k_1)$, $d(k_2)$, $d(k_3)$ are even.
\endroster
We call $K$ of type B in the latter case,
and call $K$ of type A otherwise (including the case $d(K)$ is odd).
Since $d(K)=d(k_1)d(k_2)d(k_3)$,
if $K$ is of type A, then $d(K)=D^2$,
where $D$ is a fundamental (quadratic) discriminant that is not prime,
and if $K$ is of type B, then $d(K)=(16D_0)^2$,
where $D_0$ is an odd fundamental discriminant, or $1$.
Thus, one way to construct a table is
to list up all $K$ with $d(K)=D^2$ for each $D$
and all $K$ with $d(K)=(16D_0)^2$ for each $D_0$.

Now we describe how to list.
First we treat $K$ with $d(K)=D^2$.
Such $K$ are divided into two types:
\roster
\item"(A1)" $(d(k_1), d(k_2))=1$ if we renumber suitably.
(In this case $d(k_3)=d(k_1)d(k_2)$.)
\item"(A2)" $(d(k_i), d(k_j))\ne 1$ for any disctinct $i$, $j$.
\endroster
$K$ of type A1 is obtained from a factorization $D=D_1D_2$
into a product of fundamental discriminants:
$$
d(k_1)=D_1,\ d(k_2)=D_2,\ d(k_3)=D.
$$
(Thus, the number of $K$ of type A1 with $d(K)=D^2$ for fixed $D$ is
$(2^{\omega(D)}-2)/2=2^{\omega(D)-1}-1$,
where $\omega(D)$ is the number of the prime divisors of $D$.
This is valid also for prime discriminants.)
$K$ of type A2 is obtained from a factorization $D=D_1'D_2'D_3'$
into a product of fundamental discriminants:
$$
d(k_1)=D_1'D_3',\ d(k_2)=D_2'D_3',\ d(k_3)=D_1'D_2'.
$$
We note that if $8\mid D$, then $-D$ is also a fundamental discriminant,
however, $-D$ is not if $8\nmid D$.
Thus, the number of $K$ of type A with $d(K)=D^2$ for fixed $D$ is
$(3^{\omega(D)-1}-1)/2$ if $8\nmid D$,
and $3^{\omega(D)-1}-1)$ if $8\mid D$.

Now we treat $K$ with $d(K)=(16D_0)^2$.
For such $K$, we may assume $-4$, $-8$, $8$ are
factors (as fundamental prime discriminant) of $d(k_1)$, $d(k_2)$, $d(k_3)$,
respectively.
Such $K$ are divided into two types:
\roster
\item"(B1)" $(d(k_1), d(k_2))=4$, or $(d(k_2), d(k_3))=8$,
or $(d(k_1), d(k_3))=4$.
\item"(B2)" $(d(k_i), d(k_j))$ has an odd prime factor
for any disctinct $i$, $j$.
\endroster
$K$ of type B1 is obtained from a factorization $D_0=D_1D_2$:
\roster
\item"" $d(k_1)=(-4)D_1,\ d(k_2)=(-8)D_2,\ d(k_3)=8D_0$;
\item"" $d(k_1)=(-4)D_0,\ d(k_2)=(-8)D_2,\ d(k_3)=8D_1$;
\item"" $d(k_1)=(-4)D_1,\ d(k_2)=(-8)D_0,\ d(k_3)=8D_2$.
\endroster
Note that the factorization $D_0=D_1D_2$ may be trivial, that is,
it may be $D_1=1$, or $D_2=1$.
$K$ of type B2 is obtained from a factorization $D_0=D_1'D_2'D_3'$
into a product of fundamental discriminants:
$$
d(k_1)=(-4)D_1'D_3',\ d(k_2)=(-8)D_2'D_3',\ d(k_3)=8D_1'D_3'.
$$
Thus, the number of $K$ of type B with $d(K)=(16D_0)^2$ for fixed $D_0$ is
$3^{\omega(D_0)}$.
(This is valid also for $D_0=1$.)

Finally we touch a formula for $N_4(V_4,X)$,
the number of biquadratic bicyclic number fields
whose discriminants not exceeding $X$.
Form the above, we have
\NoBlackBoxes
$$
\aligned
&N_4(V_4,X)\\
=&\sum_{n\leq\sqrt{X}/16}3^{\omega(n)}
+\sum_{n\leq\sqrt{X}/8}(3^{\omega(n)}-1)
+\frac{\,1\,}{\,2\,}\sum_{n\leq\sqrt{X}/4}(3^{\omega(n)}-1)
+\frac{\,1\,}{\,6\,}
\sideset\and{}^{\prime}\to\sum_{n\leq\sqrt{X}}(3^{\omega(n)}-3).
\endaligned
$$
Here, all the summations are taken over squarefree positive odd integer $n$.
In the last summation, $n=1$ is excluded.
Hence we have the following formula.
$$
N_4(V_4,X)=
\frac{\,1\,}{\,6\,}
\sum_{\scs n\leq\sqrt{X}\atop\scs\text{$n$ odd}}f_X(n)|\mu(n)|3^{\omega(n)}
-\frac{\,1\,}{\,2\,}N_2(C_2,\sqrt{X})-\frac{\,1\,}{\,6\,},
$$
where $N_2(C_2,X)$ denotes the number of quadratic number fields
whose discriminants not exceeding $X$ and
$$
f_X(n)=
\cases
16 & \quad\text{if $1\leq n\leq\sqrt{X}/16$} \\
10 & \quad\text{if $\sqrt{X}/16<n\leq\sqrt{X}/8$} \\
4 & \quad\text{if $\sqrt{X}/8<n\leq\sqrt{X}/4$} \\
1 & \quad\text{if $\sqrt{X}/4<n\leq\sqrt{X}$}.
\endcases
$$
\enddocument

