Hilbert Genus Fields of Imaginary Biquadratic Fields
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Hilbert Genus Fields of Imaginary Biquadratic Fields Zhe Zhang1 · Qin Yue2
Received: 23 May 2016 / Revised: 17 April 2017 / Accepted: 26 April 2017 / Published online: 15 June 2017 © School of Mathematical Sciences, University of Science and Technology of China and Springer-Verlag Berlin Heidelberg 2017
√ Abstract Let K 0 = Q δ be a quadratic field. For those K 0 with odd class number, much work has been done on the explicit construction of the Hilbert genus field of a √ √ δ, d over Q. When δ = 2 or p with p ≡ 1 mod 4 biquadratic extension K = Q a prime and K is real, it was described in Yue (Ramanujan J 21:17–25, 2010) and Bae and Yue (Ramanujan J 24:161–181, 2011). In this paper, we describe the Hilbert genus field of K explicitly when K 0 is real and K is imaginary. In fact, we give the explicit construction of the Hilbert genus field of any imaginary biquadratic field which contains a real quadratic subfield of odd class number. Keywords Class group · Hilbert symbol · Hilbert genus fields Mathematics Subject Classification 11R65 · 11R37
Research partially supported by National Key Basic Research Program of China (Grant No. 2013CB834202) and National Natural Science Foundation of China (Nos. 11501429, 11171150 and 11171317) and Fundamental Research Funds for the Central Universities (Grant No. JB150706).
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Zhe Zhang [email protected] Qin Yue [email protected]
1
School of Mathematics and Statistics, Xidian University, Xi’an 710126, People’s Republic of China
2
Department of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, People’s Republic of China
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Z. Zhang, Q. Yue
1 Introduction For a number field K , the Hilbert 2-elementary class field E of K is called the Hilbert genus field of K (cf. [1]). Let H be the Hilbert class field of K and G = Gal(H/K ) the Galois group of H/K . Then G is isomorphic to the ideal class group C(K ) of K and Gal(E/K ) G/G 2 C(K )/C(K )2 . Hence, there exists a unique multiplicative group , K ∗2 ⊂ ⊂ K ∗ such that E=H∩K
√
√ K∗ = K .
(1.1)
It is natural to ask how to construct the Hilbert √ genus field E of K explicitly. Suppose that quadratic field K 0 = Q( δ) has odd class number, then (i) either δ = p for p a prime or δ = 2 p, p1 p2 for p, p1 and p2 primes congruence to 3 modulo 4; (ii) δ = −1, − 2 or − p for p ≡ 3 mod 4 a prime (cf. [2]). There has been long √a √ history of study on the Hilbert genus field of a quadratic extension K = Q( δ, d) over K 0 . When δ = p with p ≡ 1 mod 8 and d ≡ 3 mod 4, Sime [7] used Herglotz’s results [3] to give the Hilbert genus field of K , under √ the condition that√2-Sylow √ subgroups of the class groups of K 0 = Q( p), K 1 = Q( d) and K 2 = Q( pd) are elementary. Later, Yue [9] and Bae–Yue [1] improved Sime’s result to p ≡ 1 mod 4 or p = 2 and d a squarefree positive integer without the condition on the class groups. More recently, Ouyang and Zhang ([6]) worked out the case that K is biquadratic and √ K 0 = Q( δ) is imaginary with odd ideal class number, i.e., δ = −1, −
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