Unramified extensions of some cyclic quartic fields

Let K be a cyclic quartic field such that its 2-class group CK,2 is isomorphic to Z/2Z × Z/2Z × Z/2Z. In this paper we give the generators of CK,2 and we determine the fourteen unramified extensions of K.


Introduction
Let K be an imaginary cyclic quartic extension of the rational field Q, K (1) 2 be the Hilbert 2-class field of K, K (2) 2 be the Hilbert 2-class field of K (1) 2 , K ( * ) be the genus field of K, that is the maximal absolute abelian subfield of K (1) 2 /K and let C K,2 be the 2-class group of K.If C K,2 is isomorphic to Z/2Z × Z/2Z, then A. Azizi and M. Talbi have studied this situation and answered concretely to the capitulation problem of C K,2 in the three subfields of K (1) 2 /K (see [1,2,3,4]. . .).Let K = Q( −2pε √ ℓ) where ℓ ≡ 5 (mod 8) and p ≡ 1 (mod 4) are different primes and ε is the fundamental unit of k = Q( √ ℓ).If p l = −1, then C K,2 is isomorphic to Z/2Z × Z/2Z and the capitulation problem has been studied in [3].But if p l = 1, then K ( * ) K (1) 2 and there exist two prime ideals B 1 , B 2 of k such that B 1 B 2 = (p).If h 0 denotes the class number of k, then B h0 1 = (a + b √ ℓ) and B h0 2 = (a − b √ ℓ); one can show that four prime ideals of k ramify in K which are: the prime ideal of k above 2, ( √ ℓ), B 1 and B 2 (see [8]), so [6,Theorme 4,p. 68]).By class field theory, there are seven unramified quadratic fields over K and seven unramified biquadratic fields over K which are contained in Hilbert 2-class field K (1) 2 .The following diagram illustrates the situation.
In the case where we give, in this paper, the generators of C K,2 and we will build the fourteen unramified abelian extensions within K (1) 2 using [7] and [2,3] .

The generators of C K,2
Lemma 2.1.Let p ≡ 1 (mod 4) and ℓ ≡ 5 (mod 8) be different primes.Put Proof: As ℓ and p are the unique primes of Q different from 2 which ramify in K, of ramification index e ℓ = 4 and e p = 2 respectively, since the relative discriminant of K/k is ∆ K/k = (8p √ ℓ); then, according to [9, Theorem 4, p.48 − 49], we have where ℓ is a prime number such that ℓ ≡ 5 (mod 8), h 0 be the class number of k, and p be a prime number such that p ≡ 1 (mod 4).
, where P 1 and P 2 are the prime ideals in K above p and H is the prime ideal in K above 2.

The class [P h0
1 P h0 2 ] is of order 2. In fact, if P h0 1 P h0 2 = (α) for any α in K, this is equivalent to (p h0 ) = (α 2 ) in K. Then there exists ε ′ a unit of K such that ℓ with c and d in k, and as {ε} is a fundamental system of units of K and √ −1 / ∈ K, then p h0 ε ′ ∈ k therefore c = 0 or d = 0.If d = 0, then p h0 ε ′ = c 2 , thus ±p h0 = c 2 or p h0 ε = c 2 in k, which gives that √ ±p ∈ k in the first case and √ −1 ∈ Q in the second case, which is impossible, and similarly, if c = 0 we find that ±ℓ is a square in Q, which is not the case.
The class [P h0 i ] is of order 2. In fact, suppose that P h0 i = (α) for any α in K, then multiplying by π j for j = i, we get p h0 ε ′ = π j c 2 .So by applying the norm in k/Q, we find that ±p h0 = N k/Q (c) 2 , this means that ±p is a square in Q, which is impossible.In a similar way if c = 0, we get ±ℓp is a square in Q, which is impossible.
The class [H] is of order 2. In fact, suppose that H = (α) for any α in K, then , so by applying the norm in k/Q, we find that ±l is a square in Q, which is impossible.
The class [HP h0 i ] is of order 2. In fact, suppose that HP h0 i = (α) for any α in K, then so by applying the norm in k/Q, we find that ±p is a square in Q, which is impossible.If c = 0, we get ±ℓp is a square in Q, which is impossible.

The unramified extensions of K
Let M = N ( √ α) be an extension of a number field N contains the 2-roots of unity, where α is a square free element of N coprime to 2, it is well known that M is unramified extension of N if and only if the principal ideal generated by α is the square of an ideal of N and the equation α ≡ x 2 (mod 4) admits solution in N (see [7]).Lemma 2.1 and Lemma 2.2 allow us to deduce the following Theorem: where ε is the fundamental unit of Q( √ ℓ), ℓ ≡ 5 (mod 8) and p ≡ 1 (mod 4) are different primes such that p ℓ = 1 and 2 p = − p ℓ 4 .Then the fourteen unramified extensions of the imaginary cyclic quartic field K are given by: , then • The unramified quadratic extensions of K are: • The unramified biquadratic extensions of K are: Unramified extensions of some cyclic quartic fields = p ℓ 4 , then • The unramified quadratic extensions of K are: 2 ).
• The unramified biquadratic extensions of K are: Note that π i is defined in the Lemma 2.2.
Proof: According to Lemma 2.1, = p ℓ 4 , then, from Lemma 2.2, the equation −π i ≡ x 2 (mod 4) admits solution in k i.e. admits solution in K. Since B h0 i = (π i ) and B i ramifies in K, then (−π i ) = (P h0 i ) 2 with P i an ideal of K. Therefore K( √ −π i ) is an unramified quadratic extension of K, this implies that 2 ).On the other hand, the extensions F i are pairwise different for i = 1, 2, 6, 7, because for example if F 1 = F 2 , then there exists t ∈ K such that π 1 = t 2 π 2 , this yields that p h0 = t 2 π 2 2 , which is not the case, since √ p / ∈ K. Similarly, we show the other cases.Also F i = F j for (i, j) ∈ {1, 2, 6, 7}×{3, 4, 5} (see the following Remark).It is easy to see that F 1 ≃ F 2 and F 6 ≃ F 7 .
2. We proceed as in 1.We conclude easily that ✷ Remark 3.1.The base field K admits tree unramified quadratic extensions absolutely abelian of type (2,4), which are intermediate fields between K and its genus field K ( * ) , and four unramified quadratic extensions absolutely non-Galois which are F 1 , F 2 , F 6 and F 7 .Moreover, the field K admits tree unramified biquadratic extensions absolutely Galois which are L 1 , L 4 and L 6 , and four unramified biquadratic extensions absolutely non-Galois which are L 2 , L 3 , L 5 and L 7 .