On D 5-polynomials with integer coefficients

We give a family of D5-polynomials with integer coefficients whose splitting fields over Q are unramified cyclic quintic extensions of quadratic fields. Our polynomials are constructed by using Fibonacci, Lucas numbers and units of certain cyclic quartic fields.


Introduction
The following is a fundamental problem in the theory of quadratic fields: For a given positive integer N , find quadratic fields whose class number is divisible by N .Several authors (for example, T. Nagell [6], N. C. Ankeny and S. Chowla [1], Y. Yamamoto [12], P. J. Weinberger [11] and H. Ichimura [3]) gave an infinite family of quadratic fields whose class number is divisible by arbitrary given integer N .If limited to the case N = 5, C. J. Parry [8], J.-F.Mestre [5], M. Sase [10] and D. Byeon [2] gave a family of quadratic fields whose class number is divisible by 5.In particular, Sase [10] gave a family of polynomials whose splitting field is a D 5 -extension of Q and an unramified C 5 -extension of containing the quadratic field.In the present paper, we will give other such polynomials by the use of the result of our previous result [4].Then we get a new family of quadratic fields whose class number is divisible by 5.As a consequence, the following conjecture arises: Conjecture 1.1.Let F n denote the n-th number in the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, . . .

Then the class number of the quadratic field
The organization of this paper is as follows.In Section 1, we state the main theorem that the above conjecture is true under some conditions.In Section 2, we give a parametric D 5 -polynomial with integer coefficients.It is a review of [4].In Section 3, we study the Fibonacci and the Luca sequences.In Section 4, we give a proof of the main theorem.We give a numerical example in the last Section 5.
We list here those symbols which will be used throughout this article.
Let Q denote the field of rational numbers and Z denote the ring of rational integers.
For an integer n, let C n and D n denote the cyclic group of order n and the dihedral group of order 2n, respectively.
For an extension L/K, denote the norm map and the trace map of L/K by N L/K and by Tr L/K , respectively.For simplicity, we denote N L and Tr L if the base field K = Q.For a Galois extension L/K, we denote the Galois group of L/K by Gal(L/K).
For a polynomial f (X) and a field K, we denote the minimal splitting field of f (X) over K by Spl K (f ).

The main Theorem
To state the main theorem, we prepare some notations.
Let (F n ) and (L n ) be the Fibonacci and the Luca sequences, respectively, defined as follows: Let ζ := e 2πi/5 be a primitive fifth root of unity.For a non-negative integer m, we set k m := Q( √ −F 2m+1 ).Moreover we define a cyclic quartic field M m as follows: ).As we will see in Section 4, δ m is a unit in M m .Let τ be a generator of On D 5 -polynomials For a non-negative integer m, we define a polynomial g m (X) of degree 5 with integer coefficients by Under the above notations and assumptions, we have the following.
Remarks 2.2.The author think the assumption "δ 2+3τ +τ 2 m is not a fifth power in M m " can be excluded (cf.Remark 6.2).

Construction of D 5 -polynomials
This section is a review of [4] For an element γ ∈ M , we define a polynomial f γ (X) by Applying [4, Theorem 2.1, Corollary 2.6] to the case p = 5, we get the following proposition.

Proposition 3.1. Let the notation be as above. Then for
Remarks 3.2.In [4], we can see that every

Properties of Fibonacci and Lucas numbers
There are many relations between Fibonacci and Lucas numbers.(See for example [9] and [7].)In this section, we show the following four properties which we need in the next section.(A) The power of ε = (1 + √ 5)/2 is expressed by (B) For positive integer m, we have We easily get the property (A) by mathematical induction on m, using and the property (B).We will prove the property (B) by mathematical induction on m.The equations (4.1) and (4.2) hold clearly for m = 1.Assume that (4.1) and (4.2) hold for m.Then we have and Hence (4.1) and (4.2) hold for m + 1.
Before proving the property (C), we show that the relation holds for positive integers n, m.For any n, we have (For convenience we define F 0 = 0.) Then (4.3) holds for m = 1, 2. Assume that (4.3) holds for m = k, k − 1: Then we have Next we prove namely, Then by (4.3), we have Hence (4.4) is proved.
For positive integers n and m, express n = qm + r, 0 ≤ r ≤ m.Then we claim Using (4.3) and (4.4), we have Here we have From this together with F m | F qm , we have gcd(F qm−1 , F m ) = 1.Then we get (4.5).Hence by using the Euclidean algorithm, the property (D) follows.
Remarks 4.1.The inverse of the property (C) also holds true.

Proof of the main theorem
The goal of this section is to give a proof of our main theorem.By the definition, M m is the associated field with k m .Hence we can apply Proposition 3.1.Now let us calculate f δm (X); We note that τ satisfies the following: Put ε := ε τ ; then we have by using Here, by (4.1) we have Therefore we get N M (δ m ) = 1.By similar calculations, we have

Y. Kishi
Substituting them into f δm (X), we have Assume that δ 2+3τ +τ 2 m is not a fifth power in M m .Then we have 5 , Assume in addition that m ≡ 12 (mod 5 2 ).We will show that the cyclic quintic extension Spl Q (g m )/k m is unramified.Let θ be a root of g m (X).Let q be a prime number in general.
and [k m : Q] are relatively prime.Hence we have only to verify that no primes are totally ramified in Q(θ).This can be proved by using the following Sase's result.For a prime number p and for an integer m, we denote the greatest exponent µ of p such that p µ | m by v p (m). Proposition 5.1.[10, Proposition 2] Let p ( = 2) and q be prime numbers.Suppose that the polynomial ϕ(X) = X p + p−2 j=0 a j X j , a j ∈ Z is irreducible over Q and satisfies the condition v q (a j ) < p − j for some j, 0 ≤ j ≤ p − 2. (5.1) Let θ be a root of ϕ(X).
(1) If q is different from p, then q is totally ramified in Q(θ)/Q if and only if (2) The prime p is totally ramified in Q(θ)/Q if and only if one of the following conditions (S-i), (S-ii) holds: and where ϕ (j) (X) is the j-th differential of ϕ(X).Now let us apply Proposition 4.1 to our polynomial g m (X).First, we easily verify that g m (X) satisfies (4.1) for each prime.Next, we see from Proposition 4.1 (1) that no primes except for 5 are totally ramified in Q(θ)/Q because the greatest common divisor of the coefficient of X 3 and that of X is equal to 5. We will show, therefore, that 5 is not totally ramified.Denote the constant term of g m (X) by c 0 ; Since c 0 is not divisible by 5, the condition (S-i) does not hold.By the assumption m ≡ 12 (mod 5 2 ), we have 2m + 1 ≡ 0 (mod 5 2 ).Then by the property (C), in Section 3, we have F 2m+1 ≡ 0 (mod 5 2 ), and hence −c 0 ≡ 4 (mod 5 5 ).Therefore we have m (−c 0 ) ≡ 60 • 4 2 − 60 ≡ 0 (mod 5 2 ).
Then the condition (S-ii-4) does not hold.(We can easily check that (S-ii-1), (S-ii-2) and (S-ii-3) hold.)Hence 5 is not totally ramified in Q(θ).This completes the proof of the main theorem.Next, we consider when δ 2+3τ +τ 2 m is not a fifth power in M m .and hence by the property (D), From this together with (5.2), we can express for some A q ∈ Z. Then we have This implies that (A q , L 25q ) is an integer solution of the equation (5. 3) The values of L 25q (q is prime), of course, are different from each other.However by Siegel's theorem, there are only finitely many integer solutions (X, Y ) of Eq. ( 5.3), This is a contradiction.and g 12 (X) = f δ 12 (X) is given by X 5 − 10X 3 − 20X 2 + 562875062485X + 37771618494049996.

Numerical examples
Since g 12 (X) is irreducible over Q, it follows from Proposition 5.2 that δ 12 is not a fifth power in M m .Then by the main theorem, the splitting field of g 12 (X) is an unramified cyclic quintic extension of Q( √ −F 25 ).
In the following table, we list the prime decompositions of −F 2m+1 and the structure of the ideal class groups of k m = Q( √ −F 2m+1 ) for m ≤ 87 with m ≡ 12 (mod 5 2 ).Remarks 6.2.For this table we use GP/PARI (Version 2.1.5).By using the same calculator, the author verified that g m (X) is irreducible over Q for every m, 0 ≤ m ≤ 2000.

√ 5 )
in the case p = 5 and the base field Q.Let ζ be a primitive fifth root of unity, and let k = Q( √ D) be a quadratic field which does not coincide with Q( √ 5).Then there exists a unique proper subextension of the bicyclic biquadratic extension k(ζ)/Q( √ 5) other than k( and Q(ζ).We denote it by M .Then M is a cyclic quartic field.Let us call M the associated field with k.Fix the generator τ of Gal(k(ζ)/k) with ζ τ = ζ 2 , and define a subset M(k) of k(ζ) × as follows: .2) (C) Let m be a positive integer.If m is divisible by 5 2 , then so is F m .(D) Let n and m be positive integers.If d = gcd(n, m), then we have