Commun. Math. Stat. (2017) 5:175–197 DOI 10.1007/s40304-017-0107-8

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).

B

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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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, − 2 or − p with p ≡ 3 mod 4. In this paper, we shall investigate the Hilbert genus field E of K under the conditions that K 0 is real and K is imaginary. Our results are stated in Theorems 3.5, 4.2, 5.8 and 6.2. From now on, we list some notations: √ √ 1. K = Q( δ, −d) is a biquadratic field, where either δ = p with p a prime or δ = 2 p, p1 p2 with p ≡ p1 ≡ p2 ≡ 3 mod 4 primes and d a squarefree positive integer; √ 2. K 0 = Q(√ δ) is a real quadratic field with odd class number; 3. E = K ( ) is the Hilbert genus field of K with K ∗2 ⊂  ⊂ K ∗ ; 4. N K = N K /K 0 (K ∗ ) is the norm set of K /K 0 ; 5. s is the number of finite primes of K 0 ramified in K ; q−1 6. q ∗ = (−1) 2 q is the discriminant of odd prime q. Define a subset of K ∗ as follows: D K = {x ∈ K ∗ | vp (x) ≡ 0 mod 2 for all finite primes p of K }. We have the following result.

123

(1.2)

Hilbert Genus Fields of Imaginary Biquadratic Fields

177

Proposition 1.1 Let the notations be as above.

√ 1. For any x ∈ D K , all nondyadic primes of K are unramified in K ( x). Moreover,  ⊂ DK . 2. The sequence φ

1 −→ U K /U K2 −→ D K /K ∗2 −→ C(K )[2] −→ 1 is exact, where φ([x]) = [I ] if xO K = I 2 . Hence, we have       r2 (C(K )) = r2 /K ∗2 = r2 D K /K ∗2 − 2 = s + 1 −r2 U K 0 /U K 0 ∩ N K , (1.3) where for a finite abelian group A, r2 (A) is the 2-rank of A. Proof 1. The proof is similar to that of [9, Lemma 2.1]. 2. Since D K = {x ∈ K ∗ | (x) = I 2 for some fractional ideal I of K }, it is clear that φ is epimorphic. If φ([x]) = [I ] = 1, then I = aO K ,a ∈ O∗K and xO K = I 2 = a 2 O K , so x = a 2 u,u ∈ U K and [x] = [u] in D K /K ∗ 2 . We have proved the exact sequence. Since r2 (U K /U K2 ) = 2 and r2 (/K ∗ 2 ) = r2 (C(K )), the second equality of Eq. (1.3) follows. About the third equality, we have the following results: (i) C(K 0 ) has odd order, (ii) C(K )Gal(K /K 0 ) has no 4-torsion, (iii) r2 (C(K )) = r2 (C(K )Gal(K /K 0 ) ), (iv) there is the class number formula ( see [4, Lemma 4.1, p. 307]) for cyclic extensions |C(K )Gal(K /K 0 ) | = |C(K 0 )| · We prove the third equality.

[U K 0

2s+1 . : UK0 ∩ N K ] 

By Proposition 1.1, we first study the group U K 0 /U K 0 ∩ N K to obtain the 2-ranks of /K ∗2 . Then we find a set of representatives of /K ∗2 .

2 Local and Global Computation For a number field or local field F, we let O F be the ring of integers of F and U F the unit group of O F . If F is a number field and p is a prime of F, we let Fp be the (n) completion of F at p. If F is a local field, let U F = 1+π n O F where π is a uniformizer of F. An integral solution (x0 ,y0 ,z 0 ) of a Diophantine equation ax 2 + by 2 + cz 2 = 0 is called primitive if the great common divisor of the components is 1. 2.1 Local Computation We list several properties of quadratic extensions of local field Q2 . The proofs of these results are routine, which we omit here.

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√ √ Lemma 2.1 ([1, Lemma 2.4]) Suppose F = Q2 ( −3) and ω = (−1+ −3)/2 √ ∈ F. If a ∈ U F and a ≡ x or ω · x or ω2 · x mod 4 for some odd integer x, then F( a)/F is unramified if and only if x ≡ 1 mod 4. √ √ Lemma 2.2 Suppose F = Q2 ( −1). Then π = −1 + −1 is a uniformizer of F and (5) (2) (5)  (5) 1. U F = (U F )2 , U F2 = U F (−1) · U F . √ √ 2. F( 3) = F( −3) is unramified over F. √ √ Lemma 2.3 Suppose F = Q2 ( 3). Then π = −1 + 3 is a uniformizer of F and (5) (2) (5)  (5) 1. U F = (U F )2 and U F2 = U F (1 + π 2 + π 3 + π 4 )U F . √ √ 2. F( −1) = F( −3) is unramified over F. √ √ Lemma 2.4 Suppose F = Q2 ( 2n) where n is an odd integer. Then π = 2n is a uniformizer of F and (5) (2) (5)  (5) (1 + π 2 + π 3 )U F . 1. U F = (U F )2 and U F2 = U F √ √ √ 2. F( 1 + π 2 + π 3 + π 4 ) = F( 1 + π 4 ) = F( 5) is unramified over F. Lemma 2.5 Let n be an odd integer, then, √ √ √ √ 4 = −1+ 3. 1. If n ≡ 3 mod 8, then in the field Q2 ( 3), √ n ≡√ 3 mod √ π , where π 2. If n ≡√7 mod 8, then in the field Q2 ( −1), n ≡ −1 mod π 4 , where π = −1 + −1. 2.2 Solutions of Quadratic Diophantine Equations Lemma 2.6 Suppose that p ≡ 1 mod 4 is a prime.   1. If q is an odd prime such that qp = 1, then the equations ±qz 2 = x 2 − py 2 have non-trivial solutions in Z. 2. If furthermore p ≡ 1 mod 8, then there exists a primitive solution (x0 ,z 0 ) of 2z 2 = x 2 − p such that 4 | z 0 . 3. Moreover, in the case p ≡ 1 mod 8, −2z 2 = x 2 − py 2 has a primitive solution (x,y,z) = (u 0 ,v0 ,w0 ) such that 4 | w0 . Proof 1. It suffices to compute the Hilbert symbols associated to the equations. We just prove the result for the equation qz 2 = x 2 − py 2 . The proof for the equation −qz 2 = x 2 − py 2 is similar. First, the equation has real solutions. It is clear that qz 2 = x 2 − py 2 ⇔ py 2 +qz 2 = x 2 . Let  be a prime number, then for  = 2, p,q, we have that ( p,q) = 1. For  = p or q, we have that       p p q = 1, ( p,q) p = = = 1. ( p,q)q = q p q Finally, for  = 2, we have that ( p,q)2 = (−1)

123

p−1 q−1 2 · 2

=1

Hilbert Genus Fields of Imaginary Biquadratic Fields

179

Hence, by Hasse’s local–global principle, qz 2 = x 2 − py 2 has non-trivial rational solutions, hence non-trivial integer solutions. 2. Any integral solution is clearly primitive, and moreover, x0 is odd and z 0 even. Replace (x0 ,z 0 ) by (3x0 + 4z 0 ,2x0 + 3z 0 ) if necessary, we can get z 0 such that 4 | z 0 . The last assertion follows from this immediately. 3. Let (x0 ,z 0 ) be as given in (2) and (x1 ,y1 ) be a primitive solution of −1 = x 2 − py 2 . Then (x,y,z) = (x0 x1 + py1 ,x0 y1 + x1 ,z 0 ) is a solution of −2z 2 = x 2 − py 2 . Since p ≡ 1 mod 8, (x1 ,y1 ) ≡ (0,1) mod 2, thus x0 x1 + py1 and x0 y1 + x1 are odd integers. Since 4 | z 0 , (u 0 ,v0 ,w0 ) = (x0 x1 + py1 ,x0 y1 + x1 ,z 0 ) gives a 

primitive solution with 4 | w0 . Lemma 2.7 Suppose that p,q are odd primes with p ≡ 3 mod 4.   1. If qp = 1, then the equation q ∗ z 2 = x 2 − py 2 has a primitive solution (x0 ,y0 ,z 0 ) q−1

∗ with  2 | y0 , where q = (−1) 2 q. 2. If 2qp = 1, then the equation qz 2 = x 2 − 2 py 2 has non-trivial solution in Z if

and only if q ≡ 1,3 mod 8. −qz 2 = x 2 − 2 py 2 has non-trivial solution in Z if and only if −q ≡ 1,3 mod 8. For any primitive solution (u 0 ,v0 ,w0 ) of the equations, we must have 2  w0 . Proof 1. For the solvability of the equations, it suffices to compute the Hilbert symbols associated to the equations, which is clear. √ √ Let  p = x1 + y1 p be the fundamental unit of Q( p), then by [10, Theorem 1.1], 2 | x1 . Let (x2 ,y2 ,z 2 ) be a primitive solution. Obviously, 2  z 2 . If 2 | y2 , there is nothing to prove. If 2  y2 , then 2 | x2 and (x0 ,y0 ,z 0 ) = (x1 x2 + py1 y2 ,x1 y2 + x2 y1 ,z 2 ) gives a primitive solution with 2 | y0 . 2. For the existence of integral solutions, it suffices to compute the Hilbert symbols associated to the equation, which is clear. Now 2  w0 is also clear, for if 2 | w0 , 

then 2 | u 0 and v0 , which is impossible.   Lemma 2.8 Suppose that p1 , p2 ,q are odd primes with p1 , p2 ≡ 3 mod 4, p1qp2 = q−1

1 and q ∗ = (−1) 2 q.   1. If pq1 = 1, then the equation q ∗ z 2 = x 2 − p1 p2 y 2 has non-trivial integral   solutions. If pq1 = −1, then the equation −q ∗ z 2 = x 2 − p1 p2 y 2 has non-trivial integral solutions. 2 = x 2 − p p has non-trivial 2. If p1 ≡ p2 ≡ 3 mod 8, then the equation 1 2   −2z p1 solution in Z. If p1 ≡ p2 ≡ 3 mod 8 and q = −1, then the equation 2q ∗ z 2 = x 2 − p1 p2 y 2 has non-trivial integral solution. 2 3. If p1 ≡ p2 ≡ 7 mod 8, then there exists a primitive solution  (x0 ,z 0 ) of 2z = x 2 − p1 p2 such that 4 | z 0 . If p1 ≡ p2 ≡ 7 mod 8 and pq1 = −1, then the equation −2q ∗ z 2 = x 2 − p1 p2 y 2 has a primitive solution (x,y,z) = (u 0 ,v0 ,w0 ) such that 4 | w0 .

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Proof The proof of (1), (2) is trivial. Now we prove (3). Since p1 ≡ p2 ≡ 7 mod 8, similar to the proof of Lemma 2.6 (2), − pi = x 2 − 2z 2 (i = 1,2) has primitive solution (xi ,z i ) such that 4 | z i . Thus p1 p2 = x 2 − 2z 2 has primitive solution (x0 ,z 0 ) such that 4 | z 0 . Similarly to the proof of Lemma 2.6 (3), The last assertion of (3) can be obtained. 

2.3 Decomposition and Congruence q−1 √ Lemma 2.9 Assume that p ≡ 1 mod 4 is a prime, F = Q( p) and q ∗ = (−1) 2 q, where q is an odd prime satisfying Lemma 2.6 (1). Let (x0 ,y0 ,z 0 ) be a primitive √ √ x + py solution of q ∗ z 2 = x 2 − py 2 . Take α = x0 + py0 if 2  z 0 and α = 0 2 0 if 2 | z 0 . Let α be the conjugate of α in F. Then

1. The element α ∈ O F and the ideal αO F is relatively prime to αO F . 2. If 2  z 0 , then α ≡ x0 + y0 mod 4O F . √ √ field Q2 ( p) = Q2 ( −3), α ≡ 3. If p ≡ 5 mod 8 and 2 | z 0 , then in the local √ ω(−x0 ) or ω2 (−x0 ) mod 4, where ω = −1+2 −3 . 4. If p ≡ 1 mod 8 and 2 | z 0 , then D1 = (2,α) = D2 = (2,α) are the two dyadic primes of F, and α ≡ x0 mod D22 and α/2e ≡ x0 mod D12 O FD1 for an even integer e.

Proof The proof of (2)–(4) is similar to that of [1, Lemma 2.6]. Now we prove (1). One can check that αα and α + α ∈ Z, so α ∈ O F . Let p be a prime of O F and p divide both αO F and αO F . Hence, α,α ∈ p. Thus α + α ∈ p. If p is an odd prime, we have x0 or 2x0 = α + α ∈ p ∩ Z = (), then  | x0 . Since  | z 0 , we must have  | y0 , which is impossible. If p is a dyadic prime, then 2 | z 0 and x0 = α + α ∈ p ∩ Z = (2), 

i.e., 2 | x0 , hence 2 | y0 , which is also impossible. Remark 1 Note that replacing p with p1 p2 , Lemma 2.9 is also true. √ Lemma 2.10 Suppose that p ≡ 1 mod 8 is a prime and F = Q( p). 1. Suppose (x0 ,z 0 ) is a√solution of 2z 2 = x 2 − p as given in Lemma 2.6 (2). Let α = √ x0 + p x − p and α = 0 2 its conjugate in F. Then D1 = (2,α) and D2 = (2,α) are 2 the two dyadic primes of F, and α ≡ x0 mod D23 and α/2e1 ≡ x0 mod D13 O FD1 for an odd integer e1 . 2. Suppose (u 0 ,v0 ,w0 ) is a primitive solution of −2z 2 = x 2 − py 2 as√given in u +v p Lemma 2.6 (3) such that (u 0 ,v0 ,w0 ) ≡ (x0 ,1,0) mod 4. Let β = 0 20 and β the conjugate of β in F. Then (2,β) = D1 , (2,β) = D2 and β ≡ u 0 mod D23 and β/2e2 ≡ −u 0 mod D13 O FD1 for an odd integer e2 . 2z 2 Proof 1. We have αα = 0 ≡ 0 mod 8 and α + α = x0 ∈ Z, hence α ∈ O F . 4 By the same technique of Lemma 2.9 (1), we can show that αO F is relatively prime to αO F . Moreover, by the fact that αα ∈ 8Z, we know D1 = (2,α) and D2 = (2,α) are the two dyadic primes of F, and α ∈ D13 and α ∈ D23 . Then

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α = x0 − α ≡ x0 mod D23 and α ≡ x0 mod D13 . If 2k z 0 , k ≥ 2, then by z 02 α · α · 2−2(k−1)−1 = 2k ≡ 1 mod D13 O FD1 ( because of the square of an odd 2 integer congruence to 1 modulo 8), α ≡ α −1 ≡ x0 mod D13 O FD1 . 22(k−1)+1 2. Since u 0 ≡ x0 mod 4 and v0 ≡ 1 mod 4, (2,β) = (2,α) = D1 and (2,β) = (2,α) = D2 . The rest of (2) follows from the same technique as in the proof of (1). 

Lemma 2.11 Suppose that p1 ≡ p2 ≡ 3 mod 8 are different primes and F = √ Q( p1 p2 ). 2 2 1. Suppose that√(x0 ,z 0 ) is a solution √ of −2z = x − p1 p2 as given in Lemma 2.8 (2). x 0 + p1 p2 x 0 − p1 p2 Let α = and α = , which is the conjugate of α, so D1 = (2,α) 2 2 and D2 = (2,α) are the two dyadic ideals of F. If 2z 0 , then α ≡ x0 + 2 mod D22 and α/2 ≡ −(x0 + 2) mod D12 O FD1 . If 4 | z 0 , then α ≡ x0 mod D23 and α/2e1 ≡ −x0 mod D13 O FD1 for an odd integer e1 . 2. Suppose that (u 0 ,v0 ,w0 ) is a primitive solution of 2q ∗ z 2 = x 2 − p1 p2 y√2 as given u +v p p in Lemma 2.8 (2) such that (u 0 ,v0 ) ≡ (x0 ,1) mod 4. Let β = 0 02 1 2 and u −v

√

p p

β = 0 02 1 2 , so (2,β) = D1 and (2,β) = D2 are two dyadic ideals of F. If 2w0 , then β ≡ u 0 + 2 mod D22 and β/2 ≡ u 0 + 2 mod D12 O FD1 . If 4 | w0 , then β ≡ u 0 mod D23 and β/2e2 ≡ u 0 or 5u 0 mod D13 O FD1 for an odd integer e2 . 2z 2 Proof 1. We have αα = − 0 ≡ 0 mod 2 and α + α = x0 ∈ Z, hence α ∈ O F . 4 By the same technique of Lemma 2.9 (1), we can show that αO F is relatively prime to αO F . Moreover, by the fact that αα ∈ 2Z, we know D1 = (2,α) and D2 = (2,α) are the two dyadic ideals of F. If 2z 0 , then α ∈ D1 and α ∈ D2 . Thus α = x0 − α ≡ x0 mod D2 ≡ x0 + 2 mod D22 and α ≡ x0 + 2 mod D12 . Then z2 α α ·α ·2−1 = − 02 ≡ −1 mod D12 O FD1 and ≡ α −1 ≡ −(x0 +2) mod D12 O FD1 . 2 2 If 4 | z 0 , then αα ∈ 8Z, thus α ∈ D13 , α ∈ D23 . Then α = x0 − α ≡ x0 mod D23 z 02 and α ≡ x0 mod D13 . If 2k z 0 , k ≥ 2, then by α · α · 2−2(k−1)−1 = − 2k ≡ 2 −1 mod D13 O FD1 , α 22(k−1)+1

≡ α −1 ≡ −x0 mod D13 O FD1 .

2. Since u 0 ≡ x0 mod 4 and v0 ≡ 1 mod 4, (2,β) = (2,α) = D1 and (2,β) = (2,α) = D2 . If 2w0 , then β ∈ D1 and β ∈ D2 . Thus β = u 0 − β ≡ u 0 mod w2 D2 ≡ u 0 + 2 mod D22 and β ≡ u 0 + 2 mod D12 . Then β · β · 2−1 = q ∗ 20 ≡ 2

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β −1 ≡ β ≡ u 0 + 2 mod D12 O FD1 . 2 If 4 | w0 , then ββ ∈ 8Z, thus β ∈ D13 , β ∈ D23 . Then β = u 0 − β ≡ u 0 mod D23 w2 and β ≡ u 0 mod D13 . If 2k w0 , k ≥ 2, then by β · β · 2−2(k−1)−1 = q ∗ 2k0 ≡ 2 1 or 5 mod D13 O FD1 , 1 mod D12 O FD1 and thus

β 22(k−1)+1

≡β

−1

≡ u 0 or 5u 0 mod D13 O FD1 . 

√ Lemma 2.12 Suppose p1 ≡ p2 ≡ 7 mod 8 are different primes and F = Q( p1 p2 ). 1. Suppose √ (x0 ,z 0 ) is a solution√of 2z 2 = x 2 − p1 p2 as given in Lemma 2.8 (3). Let x 0 + p1 p2 x − p p α= and α = 0 2 1 2 be its conjugate in F. Then D1 = (2,α) and 2 D2 = (2,α) are the two dyadic primes of F and α ≡ x0 mod D23 and α/2e1 ≡ x0 mod D13 O FD1 for an odd integer e1 . 2. Suppose (u 0 ,v0 ,w0 ) is a primitive solution of −2q ∗ z 2 = x 2 − p1 p2√y 2 as given u +v p p in Lemma 2.8 (3) such that (u 0 ,v0 ) ≡ (x0 ,1) mod 4. Let β = 0 02 1 2 , β the conjugate of β in F. Then (2,β) = D1 , (2,β) = D2 and β ≡ u 0 mod D23 and β/2e2 ≡ −u 0 or 3u 0 mod D13 O FD1 for an odd integer e2 . Proof The proof of (1) is similar to that of Lemma 2.10 (1). The proof of (2) is similar to that of Lemma 2.11, just recall that −q ∗ ≡ −1 or 3 mod 8. 

3 The Case K = Q(

√

√ p, −d) with p ≡ 1 mod 4

√ √ √ In this section, p ≡ 1 mod 4, K 0 = Q( p) and K = Q( p, −d). Let  p > 1 be the fundamental unit of K 0 . We always write d=

n j=1

q j or 2

n

qj

(3.1)

j=1

with p,q1 , . . . ,qn distinct odd primes such that the Legendre symbols 

p qj



 =

1 −1,

if 1 ≤ j ≤ m, if m + 1 ≤ j ≤ n.

(3.2)

Choices of α j , α0 , β0 . Suppose m ≥ 1, by Lemma 2.6 each equation q ∗j z 2 = x 2 − py 2 for 1 ≤ j ≤ m has a primitive solution (x j ,y j ,z j ) satisfying the following rules: • if there exists z j odd, then choose x j and y j such that x j + y j ≡ 1 mod 4; • if for every primitive solution z j is even, then choose x j ≡ 1 mod 4 if p ≡ 1 mod 8 and x j ≡ 3 mod 4 if p ≡ 5 mod 8.

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Then set αj = xj +

√

py j if 2  z j , and α j =

xj +

√

py j

2

if 2 | z j .

(3.3)

Now assume p ≡ 1 mod 8. Set α0 =

x0 + 2

√

p

, with (x0 ,z 0 ) ≡ (1,0) mod 4 as given in Lemma 2.6 (2).

(3.4)

Let (u 0 ,v0 ,w0 ) be any primitive solution of −2z 2 = x 2 − py 2 such that 4 | w0 and u 0 ≡ 1 mod 4 as given in Lemma 2.6 (3), then set β0 =

u0 +

√

pv0

2

.

(3.5)

Lemma 3.1 The elements −1, ±q j (1 ≤ j ≤ n), α j , α0 and β0 defined as above belong to D K . If d ≡ 2,3 mod 4, then ±2 ∈ D K . Proof −1 ∈ D K is trivial. Since q j is ramified in K , we see that ±q j ∈ D K for 1 ≤ j ≤ n. q j z2

For α j , we know that α j α j = ±q j z 2j or ± 4 j and that q j are ramified in K . By Lemma 2.9, α j O K 0 is relatively prime to α j O K 0 , hence α j O K is relatively prime to α j O K . We see that in O K , α j α j O K is a square of an ideal, thus α ∈ D K . The proof of α0 and β0 is similar. 

If d ≡ 2,3 mod 4, 2 is ramified in K , thus ±2 ∈ D K . Lemma 3.2 Suppose p ≡ 1 mod 8. Then √ 1. If −d ≡ 3 mod 4, then K ( −1)/K is unramified at the dyadic primes and so is √ K (  p )/K . √ at the dyadic primes and so 2. If −d ≡ 2 mod 8, then K ( 2)/K is unramified √ √ −2)/K is unramified at the dyadic ≡ 6 mod 8, then K ( is K ( α0 )/K . If −d √ primes and so is K ( β0 )/K . Proof √ (1), (2) In two cases, −d ≡ 2 or 3 mod 4, we know that 2 is ramified in Q( −d)/Q, so 2O K = D12 D22 , where D 1 = D1 ∩ O K 0 ,

D 2 = D2 ∩ O K 0

are the dyadic primes of K 0 as given in Lemma 2.10. For any dyadic√prime D of K , √ let D = D ∩ O K 0 , then K 0,D = Q( p) D  Q2 . Hence, K D  Q2 ( −d). √ √ If −d ≡ 3 mod √ 4, then K D  Q2 ( 3) or Q2 ( −1) which is ramified over Q2 , hence K D ( −1)/K D is unramified. Since  p is a 2-adic unit in K 0,D  Q2 , √ K D (  p )/K D is also unramified. √ √ √ ( 10). ( Thus K If −d ≡ 2 mod 8, then K D  Q2 ( 2) or Q2√ D √2)/K √ √D is unram√ ified. By Lemma 2.10 (1), K D1 ( α0 )  Q2 ( 2, 2x0 ) or Q2 ( 10, 2x0 ). Since

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√ √ x0 ≡ 1 mod 4, K D1 ( α0 )/K D1 is unramified. Similarly, K D2 ( α0 )/K D2 is also √ at the dyadic primes of K . unramified. Therefore, K ( α0 )/K is unramified √ √ √ If −d ≡ 6 mod 8, then K D  Q2 ( 6) or Q2 ( −2). Hence, K D ( −2)/K D is unramified. By Lemma 2.10 (2), we have β0 ≡ −u 0 mod D13 O FD1 and β0 ≡ u 0 mod D23 , 2e where e is an odd √ √ √ integer. √ √ √ β0 ) √ Q2 ( −d, √ u 0 ). Since Then K D1 ( β0 )  Q2 ( −d, −2u 0 ), K D2 ( √ ≡ 1 mod 4 and√ −d ≡ 6 mod 8, both Q2 ( −d, −2u 0 )/Q2 ( −d) and u0 √ √ 

Q2 ( −d, u 0 )/Q2 ( −d) are unramified extensions. √ Lemma 3.3 Suppose that p ≡ 1 mod 4 and α j is defined as above. Let K 0 = Q( p)  √ and let D be a dyadic prime of K 0 . Then K 0 ( qi∗ )/K 0 and K 0 ( α j )/K 0 are always qi −1

unramified at D, where qi∗ = (−1) 2 qi .  √  Proof Since p,qi∗ ≡ 1 mod 4, Q( p, qi∗ )/Q is unramified at 2. Thus K 0 ( qi∗ )/K 0 is unramified at the dyadic prime(s) of K 0 . √ If 2  z j , then by Lemma 2.9 (2), α j ≡ x j + y j ≡ 1 mod 4 in K 0,D = Q2 ( p). √ √ √ Thus Q2 ( p)( α j )/Q2 ( p) is unramified. √ If 2 | z j and p ≡ 1 mod 8, then K 0,D  Q2 . By Lemma 2.9 (4), K 0,D ( α j )  √  √ K 0,D ( x j ) or K 0,D ( x j + 4)  K 0,D or K 0,D ( −3), because x j ≡ 1 mod 4. Thus √ K 0,D ( α j )/K 0,D is unramified. √ If 2 | z j and p ≡ 5 mod 8, then K 0,D  Q2 ( −3). By Lemma 2.9 (3), α j ≡ ω(−x j ) or ω2 (−x j ) mod 4. Since now x j ≡ 3 mod 4, by Lemma 2.4, √ 

K 0,D ( α j )/K 0,D is unramified. Lemma 3.4 Let conventions on d be as above. Then we have the following table: p

−d

s

r2 (/K ∗2 )

1 mod 4 1 mod 8 5 mod 8

1 mod 4 2,3 mod 4 2,3 mod 4

m+n m+n+2 m+n+1

m+n−1 m+n+1 m+n

Proof For −d ≡ 1 mod 4, there are m + n finite primes ramified in K /K 0 , and for −d ≡ 2,3 mod 4, there are m + n + 1 (resp. m + n + 2) finite primes ramified in K /K 0 if 2 is inert (resp. split) in K 0 , i.e., p ≡ 5 mod 8 (resp. 1 mod 8). We thus get the values of s in the table. To know r2 (/K ∗2 ), by Proposition 1.1, it suffices to know r2 (U K 0 /U K 0 ∩ N K ). / N K . If  p ∈ N K , then −1 ∈ N K /Q (K ∗ ), Since K 0 is real, we have −1, −  p ∈ which is also impossible, because the image of the norm map N K /Q (K ∗ ) = NQ(√−d)/Q (N K /Q(√−d) (K ∗ )) is positive. Hence, r2 (U K 0 /U K 0 ∩ N K ) = 2. We thus get the values of r2 (/K ∗2 ) in the table. 

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Theorem 3.5 Let p and d be as above, then the Hilbert genus field E of K = √ √ Q( p, −d) is given by the following table. Case

p

−d

Hilbert genus field E

I

1 mod 4

1 mod 4

Q

II

1 mod 8 5 mod 8

3 mod 4

III

1 mod 8

2 mod 8

√  √ √  p, q1∗ , . . . , qn∗ , α1 , . . . , αm √ √ √  √ √ √ √ Q  p,√−1, q1 , . . . , qn ,  p , α1 , . . . , αm √ √ √ √ √ Q  p, −1, q 1 , . . . , q n , α1 , . . . , αm  √ √ √  √ √ ∗ Q p, 2, q1 , . . . , qn∗ , α0 , α1 , . . . , αm √ √   √ √ Q p, 2, q1∗ , . . . , qn∗ , α1 , . . . , αm √ √  √ √ √  Q p, −2, q1∗ , . . . , qn∗ , β0 , α1 , . . . , αm √ √  √ √  Q p, −2, q1∗ , . . . , qn∗ , α1 , . . . , αm

5 mod 8 IV

1 mod 8

6 mod 8

5 mod 8

Here q ∗ = (−1)

q−1 2

q, α j ,α0 ,β0 are given by (3.3)–(3.5).

Proof of Case I By Lemma 3.4, we have r2 (/K ∗2 ) = m +n −1 and r2 (D K /K ∗2 ) = m + n + 1. We first show the set 

∗ (3.6) −1,q1∗ , . . . ,qn−1 ,α1 , . . . ,αm , p is a set of representatives of D K /K ∗2 . It suffices to show its elements are independent modulo K ∗2 .  ∗bi  c j ∈ {0,1}, qi∗ Consider ξ =  ap · i qi j α j , where a,bi ,c j √ ∗ ∗ ∈ {−1,q1 , . . . ,qn−1 }, α j ∈ {α1 , . . . ,αm }. Let K 2 = Q( − pd), then N K /K 2 (ξ ) = (−1)a ·



qi2bi

i



c

q j j · λ2 ,

λ ∈ K2.

j

 Suppose ξ ∈ K ∗2 , then N K /K 2 (ξ ) ∈ K 2∗2 , thus a = c j = 0. Now ξ = i qi∗bi ∈ K ∗2 , √ √ √ since K has only three quadratic subfields: Q( p), Q( −d), Q( − pd), we must have bi = 0. Therefore, the set (3.6) is a representative set of D K /K ∗2 . ∗ ,α , . . . ,α }. It suffices We now show that /K ∗2 is generated by {q1∗ , . . . ,qn−1 1 m  ∗ √ to show K ( qi ), K ( α j )/K are unramified extensions. By Proposition 1.1 (1), we only need to show they are unramified at the dyadic primes of K . LetD be a dyadic prime of K and D = D ∩ O K 0 . By Lemma  3.3, both √ K 0,D ( qi∗ )/K 0,D and K 0,D ( α j )/K 0,D are unramified. Hence, K D ( qi∗ )/K D and √ 

K D ( α j )/K D are also unramified. ∗2 ) = Proof of Case II 1. If p ≡ 1 mod 8, then by Lemma 3.4, √ we have r2 (/K √ m + n + 1. By Lemma 3.2 (1) and Proposition 1.1, K ( −1)/K and K (  p ) are unramified extensions of K . Similar to the proof of Case I, we see that the set

 −1,q1 , . . . ,qn−1 ,α1 , . . . ,αm , p

is a set of representatives of /K ∗2 .

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2. If p ≡ 5 mod 8, then r2 (/K ∗2 ) = m + n and {−1,q1 , . . . ,qn−1 ,α1 , . . . ,αm } is a set of representatives of /K ∗2 .



Proof of Case III and IV If p ≡ 1 mod 8 and −d ≡ 2 mod 8, we see that by √ Lemma 3.2 (2) and Proposition√1.1, K ( α0 )/K is unramified. Similarly, if p ≡ 1 mod 8 and −d ≡ 6 mod 8, K ( β0 )/K is unramified. Then by the same method of the proof of Case I and II, the results follow. 

4 The Case K = Q(

√

√ √ √ p, −d) or Q( 2 p, −d) with p ≡ 3 mod 4

In this section, let p be odd primes such that p ≡ 3 mod √ 4. Let δ = p or 2 p, we can verify easily that the fundamental unit δ of K 0 = Q( δ) can be write as δ = 2u 2δ √ √ where u δ ∈ K 0 (see [5, p. 91] or [10, Lemma 3.2]). Let K = Q( δ, −d) and let Q = {q1 ,q2 , . . . ,qn } = the set of odd prime divisors of d,

(4.1)

and inside Q, the subsets     δ Q + = q1 , . . . ,qm | q j satisfies =1 , qj     δ Q − = qm+1 , . . . ,qn | q j satisfies = −1 . qj

(4.2) (4.3)

We set r2 (Q + ) = the 2-rank of the subgroup of μ2 generated by σ (q) =

  2 for q ∈ Q + , q (4.4)

and if Q + = ∅, we set r2 (Q + ) = 0. We denote by the above subgroup Q + . If r2 (Q + ) = 1, choose q1 ∈ Q + such that σ (q1 ) is a generator of Q + , i.e., q1∗ ≡ 5 mod 8. Choices of α j . Suppose Q + = ∅. By Lemma 2.7 (1), for any 1 ≤ j ≤ m, the equation q ∗j z 2 = x 2 − py 2 has an integral solution with the y-component even. If r2 (Q + ) = 0, we know that each q ∗j ≡ 1 mod 8 for 1 ≤ j ≤ m, so we choose a primitive solution (x j ,y j ,z j ) of q ∗j z 2 = x 2 − py 2 with 4|y j . If r2 (Q + ) = 1, for each 2 ≤ j ≤ m and q ∗j ≡ 1 mod 8(resp. q ∗j ≡ 5 mod 8), we choose a primitive solution (x j ,y j ,z j ) of q ∗j z 2 = x 2 − py 2 (resp. q1∗ q ∗j z 2 = x 2 − py 2 ) with 4|y j . Then set αj = xj +

123

√

py j .

(4.5)

Hilbert Genus Fields of Imaginary Biquadratic Fields

187

Choices of β j . By Lemma 2.7 (2), if r2 (Q + ) = 0, for any 1 ≤ j ≤ m, the equation q ∗j z 2 = x 2 − 2 py 2 has a non-trivial integral solution. If r2 (Q + ) = 1, the equation q ∗j z 2 = x 2 − 2 py 2 (resp. q1∗ q ∗j z 2 = x 2 − 2 py 2 ) has a non-trivial integral solution if q ∗j ≡ 1 mod 8 (resp. q ∗j ≡ 5 mod 8). For each j, choose a primitive solution (x j ,y j ,z j ) of q ∗j z 2 = x 2 − 2 py 2 (resp. q1∗ q ∗j z 2 = x 2 − 2 py 2 ) if q ∗j ≡ 1 mod 8 (resp. j > 1 and q1∗ ≡ q ∗j ≡ 5 mod 8). Then set βj = xj +



2 py j with x j + y j ≡ 1 mod 4.

(4.6)

By the same method of Lemma 3.1, we can show that α j , β j ∈ D K . Lemma 4.1 Suppose conventions on d are the same as above. Then we have the following table: δ

−d

r2 (Q + )

s

r2 (/K ∗2 )

p

1,3 mod 4

0 1 0 1 0 1 0 1

m+n m+n m+n+1 m+n+1 m+n m+n m+n+1 m+n+1

m+n m+n−1 m+n+1 m+n m+n m+n−1 m+n+1 m+n

2 mod 4 2p

1 mod 4 or 6 mod 8 3 mod 4 or 2 mod 8

Proof If δ = p and −d ≡ 1 mod 4, there are m + n finite primes ramified in K /K 0 , and if −d ≡ 2,3 mod 4, there are m + n + 1 finite primes ramified in K /K 0 . If δ = 2 p and −d ≡ 1 mod 4 or 6 mod 8, there are m + n finite primes ramified in K /K 0 , and if −d ≡ 3 mod 4 or 2 mod 8, there are m + n + 1 finite primes ramified in K /K 0 . We thus get the values of s in the table. To know r2 (/K ∗2 ), by Proposition 1.1, it suffices to know r2 (U K 0 /U K 0 ∩ N K ). / N K . Thus we just have to check whether Since K 0 is real, we have −1, − δ ∈ δ ∈ N K or not, equivalently, if (2, − d)p = 1 for every ramified prime p of K 0 . For the real prime of K 0 , (2, − d)p = 1. For q ∈ Q + , q splits in K 0 . For every prime q above q, we have (2, − d)q =

  2 . q

For q ∈ Q − , q is inert in K 0 . Let q be the prime above q. By Lemma 3.3 of [8], we have (2, − d)q = (N K 0 /Q (2),d)q = (4,d)q = 1.

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Z. Zhang, Q. Yue

Since p ≡ 3 mod 4, 2 is ramified in K 0 , the product formula gives (2, − d) D = 1 where D is the dyadic prime of K 0 . We thus get the values of r2 (/K ∗2 ) in the table. 

Theorem 4.2 Let p and d be as above, then the Hilbert genus field E of K = √ √ Q( p, −d) is given by the following table. Case

δ

−d

r2 (Q + )

I

p

1,3 mod 4

0 1

II

p

2 mod 4

0

Hilbert genus field √ √ √  √ √ √ Q  p,√−1, q1 , . . . , qn , α1 , . . . , αm  √ √ √ √ √ Q  p, −1, q1 , . . . , qn , α2 , . . . , αm √ √ √  √ √ √ √ p, −1, 2, q1 , . . . , qn , α1 , . . . , αm Q √ √ √ √ √  √ √ Q p, −1, 2, q1 , . . . , qn , α2 , . . . , αm √ √  √ √  2 p, −2, q1∗ , . . . , qn∗ , β1 , . . . , βm Q √ √  √ √  Q 2 p, −2, q1∗ , . . . , qn∗ , β2 , . . . , βm √ √ √ √ √  √ √ 2 p, −1, 2, q1 , . . . , qn , β1 , . . . , βm Q √ √ √ √ √  √ √ Q 2 p, −1, 2, q1 , . . . , qn , β2 , . . . , βm

1 III

2p

1 mod 4 or 6 mod 8

IV

2p

3 mod 4 or 2 mod 8

0 1 0 1

Here α j ,β j are given by (4.5) and (4.6).  √ We note the fact that K ( qi∗ )/K is always unramified. Hence, if K ( −1)/K is √ unramified, then K ( qi )/K is unramified. Proof of Case I 1. If r2 (Q + ) = 0, then by Lemma 4.1, r2 (/K ∗2 ) = m + n. We now show that /K ∗2 is generated by {−1,q1 , . . . ,qn−1 ,α1 , . . . ,α √m }. The linear independence can be proved similarly to Theorem 3.5. Since K ( −1)/K √ is unramified, it suffices to show K ( α j )/K is unramified. By Proposition 1.1, it √ suffices to show K ( α j )/K is unramified at the dyadic primes of K . Let D be a dyadic prime of K and let D = D ∩ O K 0 . If p ≡ 3 mod 8, then √ K 0,D  Q2 ( 3). Since q ∗j ≡ 1 mod 8, y j ≡ 0 mod 4. By Lemma 2.4 (1), we have √ √ √ α j = x j + y j p = x j + y j + (−1 + p)y j ≡ x j + y j + (−1 + 3)y j mod π 5 , √ √ where π = −1 + 3 is a uniformizer of Q2 ( 3). Thus α j ≡ x j + y j mod π 5 . √ Since x j + y j ≡ ±1, ± 3 mod 8, by Lemma 2.3, K 0,D ( α j )/K 0,D is unramified, √ thus K D ( α j )/K D is also unramified. √ If p ≡ 7 mod 8, then K 0,D  Q2 ( −1). Since q ∗j ≡ 1 mod 8, y j ≡ 0 mod 4. By Lemma 2.4 (2), we have αj = x j + yj

√

p = x j + y j + (−1 +

√

p)y j ≡ x j + y j + (−1 +

√

−1)y j mod π 5 ,

√ √ where π = −1+ −1 is a uniformizer of Q2 ( −1). Thus α j ≡ x j + y j mod π 5 . √ Since x j + y j ≡ ±1, ± 3 mod 8, by Lemma 2.2, K 0,D ( α j )/K 0,D is unramified, √ thus K D ( α j )/K D is also unramified.

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Hilbert Genus Fields of Imaginary Biquadratic Fields

189

2. If r2 (Q + ) = 1, then by Lemma 4.1, r2 (/K ∗2 ) = m + n − 1. If q ∗j ≡ 5 mod 8, √ then q1∗ q ∗j ≡ 1 mod 8. Then by the construction of α j (2 ≤ j ≤ m), K ( α j )/K is also unramified. Thus {−1,q1 , . . . ,qn−1 ,α2 , . . . ,αm } is a representative set of /K ∗2 .



Proof √ of Case II The proof of Case II is similar to that of Case I, just recall that 

K ( 2)/K is an unramified extension. Proof of Case III 1. If r2 (Q + ) = 0, then by Lemma 4.1, r2 (/K ∗2 ) = m + n. We claim that

∗ ,β1 , . . . ,βm −2,q1∗ , . . . ,qn−1



√ is a representative set of /K ∗2 . It suffices to show K ( α j )/K is unramified. By √ Proposition 1.1, it suffices to show K ( α j )/K is unramified at the dyadic primes √ of K . Let √ D be a dyadic prime of K and let D = D∩O K 0 . Then K 0,D  Q2 ( 2 p). Let π = 2 p be a uniformizer of K 0,D . Since (x j ,y j ,z j ) is a primitive solution of q ∗j z 2 = x 2 − 2 py 2 and q ∗j ≡ 1 mod 8, we must have x j ,z j odd and 2 | y j . Recall that we choose x j , y j such that x j + y j ≡ 1 mod 4. If x j ≡ 1 mod 4, y j ≡ 0 mod 4, we have  β j = x j + y j 2 p ≡ 1,5 mod π 5 . If x j ≡ 3 mod 4, y j ≡ 2 mod 4, we have  β j = x j + y j 2 p ≡ 1 + π 2 + π 3 or 1 + π 2 + π 3 + π 4 mod π 5 .  By Lemma 2.4, in both cases, K 0,D ( β j )/K 0,D is unramified. Therefore,  K D ( β j )/K D is also unramified. 2. If r2 (Q + ) = 1, then r2 (/K ∗2 ) = m +n. If q ∗j ≡ 5 mod 8, then q1∗ q ∗j ≡ 1 mod 8.  Then by the construction of β j (2 ≤ j ≤ m), K ( β j )/K is also unramified. Thus

 ∗ −2,q1∗ , . . . ,qn−1 ,β2 , . . . ,βm is a representative set of /K ∗2 .



Proof √ of Case IV The proof of Case IV is similar to that of Case III, just recall that 

K ( −1)/K is an unramified extension.

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190

5 K = Q(

Z. Zhang, Q. Yue

√

√ p1 p2 , −d) with p1 ≡ p2 ≡ 3 mod 4

√ In this section, let K 0 = Q( p1 p2 ) with p1 ≡ p2 ≡ 3 mod 4 odd primes, √ √ √ d >√0 squarefree and prime to p p . Let K = K ( −d) = Q( p p , −d) or 1 2 0 1 2 √ √ K 0 ( p1 d) = Q( p1 p2 , p1 d). Let  p1 p2 > 1 of K 0 be the fundamental integral unit of K 0 , then  p1 p2 = p1 u 2p1 p2 where u p1 p2 ∈ K 0 (see [5] or [10, Lemma 3.2]). √ √ Let K = Q( p1 p2 , −d). Q = {q1 ,q2 , . . . ,qn } = the set of odd prime divisors of d,

(5.1)

and inside Q, the subsets     p1 p2 Q + = q1 , . . . ,qm | q j satisfies =1 , qj     p1 p2 Q − = qm+1 , . . . ,qn | q j satisfies = −1 . qj

(5.2) (5.3)

We set  r2 (Q + ) = the 2-rank of the subgroup of μ2 generated by σ (q) =

p1 q

 for q ∈ Q + ,

(5.4) and if Q + = ∅, we set r2 (Q + ) = 0. We denote by the above subgroup Q + . If r2 (Q + ) = 1, choose q1 ∈ Q + such that σ (q1 ) is a generator of Q + . Proposition 5.1 Let√p1 , p2 ,d and K 0 be as above. (1) If K = K 0 ( d), then prime q ∈ Q + splits in K 0 and every prime q of K 0 above q is ramified in K and  ( p1 p2 , − d)q =

p1 q

 .

Prime q ∈ Q − is inert in K 0 , and the prime q above q in K 0 is ramified in K and ( p1 p2 , − d)q = 1. If p1 p2 ≡ 1 mod 8, then 2 splits in K 0 and for D a dyadic prime of K 0 , we have ( p1 p2 , − d) D =

⎧ ⎨ (−1) d+1 2

if 2  d

⎩ (−1)

if 2 | d.

p12 −1 d/2+1 8 + 2

If p1 p2 ≡ 5 mod 8, then 2 is inert in K 0 , the dyadic prime D of K 0 is ramified in K if and only if d ≡ 2 or 3 mod 4, and ( p1 p2 , − d) D = 1.

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Hilbert Genus Fields of Imaginary Biquadratic Fields

191

√ (2) If K = K 0 ( p1 d), then all the assertions in (1) hold if replacing d by p1 d. 

Proof Similar to the calculation in Lemma 4.1. 5.1 p1 p2 ≡ 5 mod 8 Proposition 5.1 tells us that

√ m +n if d ≡ 1 mod 4 and m +n +1 if d ≡ 2 Lemma 5.2 If K = K 0 ( −d), then s = √ or 3 mod 4, and t = r2 (Q + ). If K = K 0 ( − p1 d), then s = m + n if p1 d ≡ 1 mod 4 and m + n + 1 if p1 d ≡ 2 or 3 mod 4, and t = r2 (Q + ). q j is chosen as follows: Suppose Q + = ∅. For any j such that r2 (Q + ) + 1 ≤ j ≤ m,  q j = q ∗j . • If r2 (Q + ) = 0, then for all 1 ≤ j ≤ m, let  • If r2 (Q + ) = 1, then σ (q j ) = σ (q1 )a for a ∈ {0,1}. Let  q j = q1∗a q ∗j for 2 ≤ j ≤ m. By checking the corresponding Hilbert symbols and then by Lemma 2.8 (1), we have Lemma 5.3 The equation  q j z 2 = x 2 − p1 p2 y 2 is solvable in Z and has a primitive integer solution (x j ,y j ,z j ) satisfying either (i) 2  z j and x j + y j ≡ 1 mod 4 or (ii) (x j ,z j ) ≡ (3,2) mod 4. For such a solution, we set αj = xj +

√

p1 p2 y j if 2  z j , and α j =

xj +

√

p1 p2 y j if 2 | z j . 2

(5.5)

√ By the√ same method of Lemma 3.1, we can show that α j ∈ D K for K = K 0 ( −d) or K 0 ( − p1 d).  √ Lemma 5.4 Suppose p1 ≡ p2 ≡ 3 mod 4, then both K ( qi∗ )/K and K ( α j )/K are always unramified. Proof By Proposition 1.1 (1), it suffices to show they are all unramified at the dyadic primes of K . Let D be a dyadic prime of K  and let D = D ∩ O K 0 . Similar to the √ proof of Lemma 3.3. We can show that K 0,D ( qi∗ )/K 0,D and K 0,D ( α j )/K 0,D are √ always unramified. Hence, K D ( q ∗j )/K D and K D ( α j )/K D are also unramified.

 Now we have the following theorem. Theorem 5.5 1. Let √ p1 , p2 and d be as above, then the Hilbert genus field E of √ K = Q( p1 p2 , −d) is given by the following √ table. √ 2. The Hilbert genus field E of K = Q( p1 p2 , p1 d) is obtained by replacing d by p1 d in (1).

123

192

Z. Zhang, Q. Yue

−d

r2 (Q + )

Hilbert genus field E

1 mod 4

0

Q

3 mod 4

0 1

2 mod 8

0

√  √ √ √  − p1 , − p2 , q1∗ , . . . , qn∗ , α1 , . . . , αm √  √ √ √  Q − p1 , − p2 , q1∗ , . . . , qn∗ , α2 , . . . , αm √ √  √ √ √ √ √ Q √−1, p1 , p2 , q1 , . . . , qn , α1 , . . . , αm  √ √ √ √ √ √ Q  −1, p1 , p2 , q 1 , . . . , qn , α2 , . . . , αm   √ √ √ √ √ Q 2, − p1 , − p2 , q1∗ , . . . , qn∗ , α1 , . . . , αm √ √  √ √ √  Q 2, − p1 , − p2 , q1∗ , . . . , qn∗ , α2 , . . . , αm √  √ √ √ √  Q −2, − p1 , − p2 , q1∗ , . . . , qn∗ , α1 , . . . , αm √  √ √ √ √  Q −2, − p1 , − p2 , q1∗ , . . . , qn∗ , α2 , . . . , αm

1

1 6 mod 8

0 1

5.2 p1 p2 ≡ 1 mod 8 5.2.1 The Cases −d ≡ 1 mod 4 or (−d, √ p1 ) ≡ (2,7) mod 8 or (−d, p1 ) ≡ (6,3) mod 8 for K 0 ( −d) and − p1 d ≡ 1 mod 4 or √ (− p1 d, p1 ) ≡ (2,7) mod 8 or (− p1 d, p1 ) ≡ (6,3) mod 8 for K 0 ( − p1 d) We note that − p1 d ≡ 1 mod 4 is nothing but −d ≡ 3 mod 4. The form we adopt here is to illustrate the symmetry between −d and − p1 d. Proposition 5.1 gives the following lemma: √ Lemma 5.6 √ If −d ≡ 1 mod 4 (resp. − p1 d ≡ 1 mod 4 ) for K = K 0 ( −d) (resp. K = K 0 ( − p1 d) ), then s = m + n and t = r2 (Q + ). If (−d, p1 ) ≡ (2,7) mod 8 or (−d, p1 ) ≡ (6,3) mod√8 (resp. (− p1 d, p1 ) ≡√(2,7) mod 8 or (− p1 d, p1 ) ≡ (6,3) mod 8 ) for K = K 0 ( −d) (resp. K = K 0 ( − p1 d) ), then s = m + n + 2 and t = r2 (Q + ). Suppose Q + = ∅. For any j such that r2 (Q + ) + 1 ≤ j ≤ m,  q j is chosen as follows: • If r2 (Q + ) = 0, then for all 1 ≤ j ≤ m, let  q j = q ∗j . a • If r2 (Q + ) = 1, then σ (q j ) = σ (q1 ) for a ∈ {0,1}. Let  q j = q1∗a q ∗j for 2 ≤ j ≤ m. By checking the corresponding Hilbert symbols and then by Lemma 2.8 (1), we have Lemma 5.7 The equation  q j z 2 = x 2 − p1 p2 y 2 is solvable in Z and has a primitive integer solution (x j ,y j ,z j ) satisfying either (i) 2  z j and x j + y j ≡ 1 mod 4 or (ii) (x j ,z j ) ≡ (1,0) mod 4. For such a solution, we set αj = xj +

√

p1 p2 y j if 2  z j , and α j =

xj +

√

p1 p2 y j if 2 | z j . 2

(5.6)

If p1 ≡ p2 ≡ 3 mod 8, let (x0 ,z 0 ) be a primitive solution of −2z 2 = x 2 − p1 p2 as given in Lemma 2.8 (2). Then set α0 =

x0 +

123

√ 2

p1 p2

with x0 ≡ 3 mod 4 if 2z 0 and x0 ≡ 1 mod 4 if 4 | z 0 . (5.7)

Hilbert Genus Fields of Imaginary Biquadratic Fields

193

If p1 ≡ p2 ≡ 7 mod 8, set β0 =

u0 +

√

p1 p2

2

with (x0 ,z 0 ) ≡ (1,0) mod 4 as given in Lemma 2.8 (3). (5.8)

By of Lemma 3.1, we can show that α0 ,β0 ,α j ∈ D K for K = √ the same method √ K 0 ( −d) or K 0 ( − p1 d). Theorem 5.8 1. Let √ p1 , p2 and d be as above, then the Hilbert genus field E of √ K = Q( p1 p2 , −d) is given by the following table. −d

p1

1 mod 4

r2 (Q + )

Hilbert genus field E

0

Q

1 2 mod 8

7 mod 8

6 mod 8

3 mod 8

1 0 0 1

√  √ √ √  − p1 , − p2 , q1∗ , . . . , qn∗ , α1 , . . . , αm √  √ √ √  Q − p1 , − p2 , q1∗ , . . . , qn∗ , α2 , . . . , αm √ √  √ √ √ √  Q 2, − p1 , − p2 , q1∗ , . . . , qn∗ , β0 , α1 , . . . , αm √ √  √ √ √ √  2, − p1 , − p2 , q1∗ , . . . , qn∗ , β0 , α2 , . . . , αm Q √  √ √ √ √ √  Q −2, − p1 , − p2 , q1∗ , . . . , qn∗ , α0 , α1 , . . . , αm √  √ √ √ √ √  Q −2, − p1 , − p2 , q1∗ , . . . , qn∗ , α0 , α2 , . . . , αm

α j ,α0 ,β0 are given by (5.6), (5.7) and (5.8), √ respectively. 2. The Hilbert genus fields E of K = K 0 ( − p1 d) for the cases − p1 d ≡ 1 mod 4 or (− p1 d, p1 ) ≡ (2,7) mod 8 or (− p1 d, p1 ) ≡ (6,3) mod 8 are obtained by replacing −d by − p1 d in (1). √ √ Proof We only show the case that K = K 0 ( −d). √ The case K = K 0 ( − p1 d) is √ similar. It suffices to prove that K ( α0 )/K and K ( β0 )/K is unramified at the dyadic primes of K . Let Di (i = 1,2) be the two dyadic primes of K and let Di = D ∩ O K 0 . √ First, we show that K Di ( α0 )/K Di is unramified. By Lemma 2.11 (1), if 2z 0 , then we have α0 ≡ −(x0 + 2) mod D12 O FD1 and α0 ≡ x0 + 2 mod D22 . 2 √ √ √ √ √ Then K D1 ( α0 )  Q2 ( −d, −2(x0 + 2)) or Q2 ( −d, −2(x0 + 6)). Since √ x0 ≡ 3 mod 4 and −d ≡ 6 mod 8, K D1 ( α0 )/K D1 is unramified. √ Similarly, K D2 ( α0 )/K D2 is also unramified. If 4 | z 0 , then by Lemma 2.11 (1), α0 ≡ −x0 mod D13 O FD1 and α0 ≡ x0 mod D23 , 2 e2 √ √ √ √ where an odd integer. Then K D1 ( α0 )  Q2 ( −d, −2x0 ),√K D2√ ( α0 )  √ e2 is √ 4 and −d ≡ 6 mod 8, both Q2 ( −d, −2x0 )/ Q2 (√−d, x0 ). Since √ x0√≡ 1 mod √ Q2 ( −d) and Q2 ( −d, x0 )/Q2 ( −d) are unramified.

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By√Lemma 2.12 (1) and similar to the proof of Lemma 3.2 (2), we can show that 

K Di ( β0 )/K Di is also unramified. √ 8 and −d ≡ 2 mod 8 for K ( −d) or 5.2.2 The Cases p1 ≡ p2 ≡ 3 mod 0 √ − p1 d ≡ 2 mod 8 for K 0 ( − p1 d) Proposition 5.1 gives the following lemma:

√ 8 ) for K = K 0 ( −d) (resp. Lemma 5.9 √ If −d ≡ 2 mod 8 (resp. − p1 d ≡ 2 mod ∗2 K = K 0 ( − p1 d) ), then s = m + n + 2 and r2 (/K ) = m + n + 1. Suppose Q + = ∅. For any 1 ≤ j ≤ m,  q j is chosen as follows: • Let  q j = q ∗j if σ (q j ) = 1 and let  q j = 2q ∗j if σ (q j ) = −1. By checking the corresponding Hilbert symbols and then by Lemma 2.8 (1), we have Lemma 5.10 There exists a primitive solution (x j ,y j ,z j ) for  q j z 2 = x 2 − p1 p2 y 2 satisfying 1. If  q j is odd, then either z j odd and x j + y j ≡ 1 mod 4, or z j even and x j ≡ 1 mod 4. 2. If  q j is even, then either 2 || z j and x j ≡ 3 mod 4, or 4 | z j even and x j ≡ 1 mod 4. For such a solution, we set αj = xj +

√

p1 p2 y j if 2  z j , and α j =

xj +

√

p1 p2 y j if 2 | z j . 2

(5.9)

Theorem 5.11 1. Assume p1 ≡ p2 ≡ 3√mod 8 and√−d ≡ 2 mod 8  as above, √ √ then Hilbert genus field E of K = K 0 ( −d) is Q( 2, − p1 , − p2 , q1∗ , . . . ,  ∗√ √ qn , α1 , . . . , αm ). 2 mod 8 as  above, then 2. Assume p1 ≡ p2 ≡ 3 mod 1d ≡  Hilbert √ 8 and − p√ √ √ √ genus field E of K = K 0 ( − p1 d) is Q( 2, − p1 , − p2 , q1∗ , . . . , qn∗ , α1 , √ . . . , αm ). √ Proof In all cases, it suffices to show that K ( α j )/K is unramified at every dyadic prime D of K . The proof is similar to that of Theorem 5.8. For the case  q j even, one needs Lemma 2.11 (2). 

√ 5.2.3 The Cases p1 ≡ p2 ≡ 7 mod √ 8 and −d ≡ 6 mod 8 for K 0 ( −d) or − p1 d ≡ 6 mod 8 for K 0 ( − p1 d) Proposition 5.1 gives the following lemma:

√ Lemma 5.12 √ If −d ≡ 6 mod 8 (resp. − p1 d ≡ 6 mod∗28 ) for K = K 0 ( −d) (resp. K = K 0 ( − p1 d) ), then s = m + n + 2 and r2 (/K ) = m + n + 1. Suppose Q + = ∅. For any 1 ≤ j ≤ m,  q j is chosen as follows:

123

Hilbert Genus Fields of Imaginary Biquadratic Fields

195

• Let  q j = q ∗j if σ (q j ) = 1 and let  q j = −2q ∗j if σ (q j ) = −1. By checking the corresponding Hilbert symbols and then by Lemma 2.8 (1), we have Lemma 5.13 There exists a primitive solution (x j ,y j ,z j ) for  q j z 2 = x 2 − p1 p2 y 2 satisfying 1. If  q j is odd, then either z j odd and x j + y j ≡ 1 mod 4, or z j even and x j ≡ 1 mod 4. 2. If  q j is even, then either 2 || z j and x j ≡ 3 mod 4, or 4 | z j even and x j ≡ 1 mod 4. For such a solution, we set αj = xj +

√

p1 p2 y j if 2  z j , and α j =

xj +

√

p1 p2 y j if 2 | z j . 2

(5.10)

Theorem 5.14 1. Assume p1 ≡ p2 ≡ 7√mod 8 and √ −d ≡ 6 mod 8 as above,  then √ √ Hilbert genus field E of K = K 0 ( −d) is Q( −2, − p1 , − p2 , q1∗ , . . . ,  ∗√ √ qn , α1 , . . . , αm ). mod 8 as above, 2. Assume p1 ≡ p2 ≡ 7 mod  then Hilbert  genus √ 8 and − p1 d√≡ 6 √ √ √ field E of K = K 0 ( − p1 d) is Q( −2, − p1 , − p2 , q1∗ , . . . , qn∗ , α1 , √ . . . , αm ). √ Proof In all cases, it suffices to show that K ( α j )/K is unramified at every dyadic prime D of K . The proof is similar to that of Theorem 5.8. For the case  q j even, one needs Lemma 2.12 (2). 

√ 5.2.4 The √ Cases −d ≡ 3 mod 4 for K 0 ( −d) and − p1 d ≡ 3 mod 4 for K 0 ( − p1 d) Proposition 5.1 gives the following lemma:

√ Lemma 5.15 √ If −d ≡ 3 mod 4 (resp. − p1 d ≡ 3 mod∗24 ) for K = K 0 ( −d) (resp. K = K 0 ( − p1 d) ), then s = m + n + 2 and r2 (/K ) = m + n + 1. Suppose Q + = ∅. For any 1 ≤ j ≤ m,  q j is chosen as follows: q j = −q ∗j if σ (q j ) = −1. • Let  q j = q ∗j if σ (q j ) = 1 and let  By checking the corresponding Hilbert symbols and then by Lemma 2.8 (1) and (2), we have Lemma 5.16 The equation  q j z 2 = x 2 − p1 p2 y 2 is solvable in Z. Let (x j ,y j ,z j ) be any primitive solution of above equation, we set αj = xj +

√

p1 p2 y j if 2  z j , and α j =

xj +

√

p1 p2 y j if 2 | z j . 2

(5.11)

By the same method of Lemma 3.1, we can show that α j ∈ D K .

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Theorem 5.17 1. Assume p1 p2 ≡ 1 √ mod 8 and √ −d ≡ 3 mod 4 as above, then √ √ √ √ Hilbert genus field E of K = K 0 ( −d) is Q( −1, p1 , p2 , q1 , . . . , qn , √ √ α1 , . . . , αm ). − p1 d ≡ 3 mod 4 as above, then Hilbert genus field 2. Assume p1 p2 ≡ √1 mod 8 and √ √ √ √ √ √ √ E of K = K 0 ( − p1 d) is Q( −1, p1 , p2 , q1 , . . . , qn , α1 , . . . , αm ). √ Proof (1) It suffices to show that K ( α j )/K is unramified at every dyadic prime D of K . √ Since p1 p2 ≡ 1 mod 8, K 0,D  Q2 and K D  Q2 ( d). If 2  z j , then α j is a √ 2-adic unit in Q2 . Since d ≡ 3 mod 4, K D ( α j ) is unramified over K D . If 2 | z j , then by the same method of Lemma 2.11, we can show that αj ≡ x j or − x j mod D12 O K 0,D1 according to  q j ≡ 1 or − 1 mod 4 and α j ≡ x j mod D22 , 2e

√ where e is an even integer. Thus there exists odd integers u j ,v j such that K D1 ( α j )  √ √ √ √ √ √ Q2 ( d, u j ) and K D2 ( α j )  Q2 ( d, v j ). Since d ≡ 3 mod 4, K Di ( α j )/K Di (i = 1,2) is unramified. The proof of (2) is similar to that of (1). 

√ √ 6 The Case K = Q( 2, −d) √ √ √ √ In this section, K 0 = √ Q( √2), K = Q(√ 2,√ −d). Let  = 1 + 2 be the fundamental unit of K 0 . Since Q( 2, −d) = Q( 2, −2d), without loss of generality, we can assume −d ≡ 1 or 3 mod 4. We write d=

n

qj

(6.1)

j=1

withq1 , . . . ,qn distinct odd primes such that q j ≡ ±1 mod 8 if 1 ≤ j ≤ m (i.e.,  2 = 1) and q j ≡ ±3 mod 8 if m + 1 ≤ j ≤ n. For 1 ≤ j ≤ m, choose qj (x j ,y j ) to be a primitive solution of q j = x 2 − 2y 2 (resp. −q j = x 2 − 2y 2 ) such that x j + y j ≡ 1 mod 4 if q j ≡ 1 mod 8 (resp. q j ≡ −1 mod 8). Set √ α j = x j + y j 2.

(6.2)

Lemma 6.1 Let the notations be as above. Then we have the following table: −d

s

r2 (/K ∗2 )

1 mod 4 3 mod 4

m+n m+n+1

m+n−1 m+n

123

Hilbert Genus Fields of Imaginary Biquadratic Fields

197

Proof For −d ≡ 1 mod 4, there are m + n finite primes ramified in K /K 0 , and for −d ≡ 3 mod 4, there are m + n + 1 finite primes ramified in K /K 0 . We thus get the values of s in the table. To know r2 (/K ∗2 ), by Proposition 1.1, it suffices to know r2 (U K 0 /U K 0 ∩ N K ). / N K . If  ∈ N K , then −1 ∈ N K /Q (K ∗ ), Since K 0 is real, we have −1, −  ∈ which is also impossible, because the image of the norm map N K /Q (K ∗ ) = NQ(√−d)/Q (N K /Q(√−d) (K ∗ )) is positive. Hence, r2 (U K 0 /U K 0 ∩ N K ) = 2. We thus 

get the values of r2 (/K ∗2 ) in the table. √ √ Theorem 6.2 Let d be as above, then the Hilbert genus field of K = Q( 2, −d) is given by the following table. Case

−d

Hilbert genus field E

I

1 mod 4

Q

II

3 mod 4

√  √ √  2. q1∗ , . . . , qn∗ , α1 , . . . , αm √ √ √ √  √ √ −1, 2, q1 , . . . , qn , α1 , . . . , αm Q

Proof The proof is similar to the proof of Theorem 4.2, Case III.



Acknowledgements The first author would like to thank his advisor, Prof. Yi Ouyang for his constant support and encouragement.

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