CONSTRUCTION OF TABLES OF

BIQUADRATIC BICYCLIC NUMBER FIELDS

By Ken Yamamura

Department of Mathematics, National Defense Academy

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:

Exactly two of d(k_{1}), d(k_{2}), d(k_{3}) are even;

All d(k_{1}), d(k_{2}), d(k_{3}) are even.

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: (A1) (d(k_{1}), d(k_{2})) = 1 if we renumber suitably. (In this case d(k_{3}) = d(k_{1})d(k_{2}).) (A2) (d(k_{i}), d(k_{j})) ≠ 1 for any disctinct i, j. K of type A1 is obtained from a factorization D = D_{1}D_{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^{ω(D)} − 2)/2 = 2^{ω(D)−1} − 1, where ω(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 | D, then −D is also a fundamental discriminant, however, −D is not if 8 ∤ D. Thus, the number of K of type A with d(K) = D^{2} for fixed D is (3^{ω(D)−1} − 1)/2 if 8 ∤ D, and 3^{ω(D)−1} − 1) if 8 | 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:

(B1) (d(k_{1}), d(k_{2})) = 4, or (d(k_{2}), d(k_{3})) = 8, or (d(k_{1}), d(k_{3})) = 4.

(B2) (d(k_{i}), d(k_{j})) has an odd prime factor for any disctinct i, j.

Typeset by AMS-TEX

K of type B1 is obtained from a factorization D_{0} = D_{1}D_{2}:

d(k_{1}) = (−4)D_{1}, d(k_{2}) = (−8)D_{2}, d(k_{3}) = 8D_{0};

d(k_{1}) = (−4)D_{0}, d(k_{2}) = (−8)D_{2}, d(k_{3}) = 8D_{1};
d(k_{1}) = (−4)D_{1}, d(k_{2}) = (−8)D_{0}, d(k_{3}) = 8D_{2}.

Note that the factorization D_{0} = D_{1}D_{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^{ω(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

\begin{matrix}N_{4}(V_{4}, X) \\ = & ∑_{n≤\sqrt{X}/16} 3^{ω(n)} + ∑_{n≤\sqrt{X}/8}(3^{ω(n)} − 1) + \frac{1}{2} ∑_{n≤\sqrt{X}/4}(3^{ω(n)} − 1) + \frac{1}{6} \begin{matrix}∑ & ′ \\ n≤\sqrt{X} \end{matrix}(3^{ω(n)} − 3). \end{matrix}

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} ∑_{{n≤\sqrt{X} \atop n odd}} f_{X}(n)|μ(n)|3^{ω(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) = \{ \begin{matrix}16 & if 1 ≤ n ≤ \sqrt{X}/16 \\ 10 & if \sqrt{X}/16 < n ≤ \sqrt{X}/8 \\ 4 & if \sqrt{X}/8 < n ≤ \sqrt{X}/4 \\ 1 & if \sqrt{X}/4 < n ≤ \sqrt{X}. \end{matrix}