The 2-Sylow subgroups of the tame kernel of imaginary quadratic fields

Transcription

1 ACTA ARITHMETICA LXIX The 2-Sylow subgous of the tame kenel of imaginay uadatic fields by Houong Qin Nanjing To Pofesso Zhou Boxun Cheo Peh-hsuin on his 75th bithday 1. Intoduction. Let F be a numbe field and O F the ing of its integes. Many esults ae known about the gou K 2 O F, the tame kenel of F. In aticula, many authos hae inestigated the 2-Sylow subgou of K 2 O F. As comaed with eal uadatic fields, the 2-Sylow subgous of K 2 O F fo imaginay uadatic fields F ae moe difficult to deal with. The objectie of this ae is to oe a few theoems on the stuctue of the 2-Sylow subgous of K 2 O F fo imaginay uadatic fields F. In ou Ph.D. thesis see [11], we deelo a method to detemine the stuctue of the 2-Sylow subgous of K 2 O F fo eal uadatic fields F. The esent ae is motiated by some ideas in the aboe thesis. 2. Notations and eliminaies. Let F be a numbe field and O F the ing of integes in F. Let Ω be the set of all laces of F. Fo any finite lace P, denote by P the discete aluation on F coesonding to P. Fo any {x, y} K 2 F, the tame symbol is defined by τ P {x, y} 1 P x P y x P y y P x mod P. Fo any P Ω, the Hilbet symbol P of ode 2 on FP, the comletion of F at P, is defined as follows: Gien non-zeo elements α, β F P, α,β P 1 if αξ 2 + βη 2 1 has a solution ξ, η F P, othewise the symbol is defined to be 1. In aticula, suose P is a non-dyadic lace in Ω. By a fomula in Theoem 5.4 of [9], if {x, y} K 2 F and τ P {x, y} 1 then x, y P [153]

2 154 H. Qin And if F Q, the ational numbes field, and x, y ae the units in Q 2, then x, y x 1/2 y 1/2 2 see Theoem 5.6 in [9]. The following Poduct Fomula fo the Hilbet symbol is well known: α, β P P Ω Fo any odd ime, let P denote the Legende symbol. We hae Lemma 2.1 Legende. Suose a, b, c ae suae fee, a, b b, c c, a 1, and a, b, c do not hae the same sign. Then the Diohantine euation ax 2 + by 2 + cz 2 0 has non-tiial intege solutions if and only if fo eey odd ime abc, say a, bc 1. P o o f. See Theoem 4.1 and its Coollay 2 in [3]. In this ae, we use K 2 O F 2 to denote the 2-Sylow subgou of K 2 O F. Let F be an imaginay uadatic field. By [13], we hae [ : F 2 ] 4, whee {z F { 1, z} 1}. In Section 5, we will detemine fo some imaginay uadatic fields. 3. Geneal esults Lemma 3.1. Let F Q d d a ositie suae-fee intege. Fo any α x + y d F, ut S {P 1,..., P n } {P τ P { 1, α} 1}. Without loss of geneality, we can assume that i P i Z is not inet fo 1 i n. Then x 2 + dy 2 ε 1... n z 2, whee ε {1, 2} and z Q. Conesely, suose that 1,..., n ae distinct imes in Z and P 1,..., P n ae ime ideals of O F such that P i Z i fo 1 i n. If thee is an ε {1, 2} such that the euation x 2 + dy 2 ε 1... n z 2 is solable in Q euialently in Z, then thee is an α F such that S {P τ P { 1, α} 1} {P 1,..., P n }. P o o f. Suose α x + y d F and S {P 1,..., P n } {P τ P { 1, α} 1}. Then x + y d σ P 1... P n a 2, whee 2 and σ 0 o 1 and a is a factional ideal of O F in F. Hence, x 2 + dy 2 ε 1... n z 2. Conesely, if x 2 + dy 2 ε 1... n z 2 has a solution x, y, z Z, then fo any 1 i n, eithe x+y d P i o x y d P i. So suitably choosing δ Q, we can assume that δx+y d e P 1... P n a 2, whee 2, e 0

3 Imaginay uadatic fields 155 o 1 and a is a factional ideal of O F in F. Then taking α δx + y d yields the esult. Theoem 3.2. Let F Q d d a ositie suae-fee intege, and m d, a ositie intege. If any ime facto of m satisfies i 1 mod 4, fo 1 i n, then thee is an α K 2 O F with α 2 { 1, m} if and only if thee is an ε {1, 2} such that the Diohantine euation εmz 2 X 2 +dy 2 is solable in Z. P o o f. Fom the assumtion, we hae m x 2 + y 2, whee x, y Z. Let { x α y, x2 + y 2 } { x y 2 y, m } y 2. Then α 2 { 1, m}. It is easy to check that {P τ P α 1} {P 1,......, P n }, whee P i Z i 1 i n. Theefoe, the esult follows fom Lemma 3.1. Lemma 3.3. Let F Q d d a ositie suae-fee intege. ŁThen m x 2 + y 2 fo x, y F, whee m is an intege satisfying: m d if d 1 mod 8 o m d togethe with m 3 mod 4 if d 1 mod 8. P o o f. Let Ω denote the set of all laces of F. In iew of the Hasse Minkowski Theoem see [10], it is enough to oe that 1,m P 1 fo any P Ω. Clealy, if P is the uniue Achimedean lace in Ω, then 1,m P 1. In the non-dyadic cases, 1,m P 1 follows fom 2.1. If d 1 mod 8, then thee is a uniue dyadic lace P in Ω. Then the Poduct Fomula yields 1,m P 1. Now suose d 1 mod 8. Hee m 3 mod 4. Let P 1, P 2 denote the two dyadic laces in Ω. Then F P1 FP2 Q2. Hence, by 2.2, we hae 1 fo i 1, 2. This comletes the oof. 1,m P i R e m a k. By a theoem due to Bass and Tate see [8], a necessay and sufficient condition fo { 1, m} α 2 with α K 2 F is that m x 2 + y 2 fo x, y F. On the othe hand, if d 1 mod 8, m d and m 3 mod 4, then by 2.2, 1,m P i 1, whee i 1, 2, and P1, P 2 ae the two dyadic laces of F. So { 1, m} α 2 fo any α K 2 F. Now, let F Q d be an imaginay uadatic field. Suose m 1 o m δ 1..., whee δ 1 o 1 fo 1 i, i 3 mod 4 is a ime, and suose m X 2 + Y 2 fo X, Y F. Wite X x + y d, Y x + y d, whee x, y, x, y, z Z. z z Clealy m X 2 + Y 2 imlies that xy x y and 3.1 mz 2 x 2 dy 2 + x 2 dy 2.

4 156 H. Qin Without loss of geneality, we can assume that x, y, x, y 1 below. Lemma 3.4. Notations being as aboe, suose k 0 is an intege and is a ime with 2k+1 x 2 + x 2. Then thee is a ime ideal P of F such that P x+y d > 0, P x +y d > 0. Conesely, if P x+y d > 0 and P x + y d > 0, then P x 2 + x 2 > 0. P o o f. It follows fom 3.1 that mx 2 z 2 x 2 + x 2 x 2 dy 2. Then 2k+1 x 2 + x 2 imlies x 2 dy 2 note that 1 mod 4. But x + y dx y d x 2 + dy 2 x 2 + dy 2 0 mod P. So we may assume that P x + y d > 0. Similaly, P x + y d > 0. Conesely, suose P x + y d > 0 and P x + y d > 0. Let P Z. Then x 2 + dy 2, x 2 + dy 2. If x, then x o y since xy x y. But y also imlies x. Hence x 2 + x 2. Now we assume that x and x. It is easy to eify that 3.2 x + y d x x So, P x 2 + x 2 > 0. x + x 2 x x + y d. x 2 Lemma 3.5. With notations being as aboe, suose P is a non-dyadic ime ideal of O F, P Z and m. If P x 2 + x 2 > 0 and P x + y d > 0, then x 2 + x 2 P x + y d mod 2. P o o f. Fist suose that P x 0. Then is unamified in O F, hence x 2 + x 2 P x 2 + x 2 and P x + y d P x 2 + dy 2. It can be deduced fom the identity 2 x + y d x + y 2 d m + z z that P x + y d P x + y d. If P x + y d P x 2 + x 2, then 3.2 yields P x 2 + x 2 > P x + y d P x 2 + dy 2. Hence, P x 2 + dy 2 P x 2 + dy 2 x 2 + x 2 P x 2 + dy 2. Now, P x 2 +x 2 P x 2 +x 2 mod 2 is a conseuence of the obseation that x 3.3 x 2 dy x 2 mz 2. Then suose that P x > 0. In this case, the only ossibility is that d, y and y. Then x x 2 dy x 2 mz 2 imlies x 2 P x + y d P x 2 + x 2 1 mod 2. x 2

5 Imaginay uadatic fields 157 Theoem 3.6. Let F Q d d a ositie suae-fee intege, and let m be an intege with m d if d 1 mod 8, and with m d and m 3 mod 4 if d 1 mod 8. Moeoe, if m 1, then fo any ime facto of m, 3 mod 4. Then thee is an α K 2 O F with α 2 { 1, m} if and only if thee is an ε {1, 2} such that d/m ε fo any ime m; m ε fo any ime d, m. P o o f. We know that thee ae x, y, x, y, z Z with x, y, x, y 1 such that mz 2 x + y d 2 + x + y d 2. Wite { x + y d β x + y d, mz 2 } x + y. d 2 Then β 2 { 1, m} and it is not had to check that 1 P z P x+y d τ P β if P Z m and P x + y d P x + y d, 1 othewise. In iew of Lemma 3.5 and elacing β by β{ 1, δ} fo a suitable δ Z if necessay allows us to assume that τ P β 1 if and only if 2k+1 x 2 +x 2, whee P Z and k is a non-negatie intege. Hence, by Lemma 3.1 we conclude that thee is a β K 2 O F with β 2 { 1, m} if and only if thee is an ε {1, 2} such that the Diohantine euation 3.4 εx 2 + x 2 Z 2 X 2 + dy 2 is solable in Z. Obiously, we can assume that x 2 + x 2 is suae-fee. Let us assume that c is the geatest common diiso of εx 2 + x 2 and d. Then 3.4 can be witten as εx 2 + x 2 Z 2 cx 2 + d c c Y 2. By Lemma 2.1, it is solable in Z if and only if c d/c fo any ime x 2 + x 2 3.5, c εx 2 + x 2 d/c 3.6 fo any ime c,

6 158 H. Qin and 3.7 c εx 2 + x 2 /c Note that the identity 3.3 can be witten as fo any ime d c. 3.8 mcx 2 d mc y2 x2 + x 2 z 2, c so that mc d/mc fo any ime x 2 + x 2. c This is euialent to 3.5, because x2 +x 2 c imlies that 1 mod 4. In othe wods, 3.5 is tiial. If c, then d/mc x 2 + x 2 /c. So 3.6 is euialent to d/mc εd/c, i.e., m So does the case d, c and m. If m, then d/mc x 2 + x 2 /c. In this case, 3.7 is euialent to d/mc εc, i.e., This concludes the oof. d/m ε. ε. Coollay 3.7. Let the assumtions and the notations be as in Theoem 3.6, and assume that n is a ositie intege satisfying n d and fo any ime facto of n, 1 mod 4. Then thee is a β K 2 O F such that β 2 { 1, mn} if and only if i fo any ime mn, d/mn ε, ii fo any ime d, mn, mn ε, whee ε 1 o 2. P o o f. Conside εx 2 + x 2 nz 2 X 2 + dy 2 in lace of 3.4. Coollay 3.8. Let F Q d be an imaginay uadatic field. Then { 1, 1} α 2 with α K 2 O F if eithe d 1 o d 2 o fo any odd ime d, 1 mod 4 o fo any odd ime d, 1 o 3 mod 8. Othewise, { 1, 1} α 2 fo any α K 2 O F, in aticula, { 1, 1} 1. Lemma 3.9. Let m 3 mod 4 and d 1 mod 8. Then fo ε 1 o 2, the Diohantine euation εmz 2 X 2 dy 2 has no solutions in Z.

7 P o o f. Conside the conguence Imaginay uadatic fields 159 εmz 2 X 2 dy 2 mod 8, i.e., εmz 2 X 2 + Y 2 mod 8. Note that fo any a Z, a 2 0 o 4 o 1 mod 8. Then the esult follows. As a conseuence of Lemma 2.1, Coollay 3.7 and Lemma 3.9, we hae: Theoem Let F Q d d a ositie suae-fee intege, and m d an intege. Then thee is an α K 2 O F with α 2 { 1, m} if and only if thee is an ε {1, 2} such that the Diohantine euation εmz 2 X 2 dy 2 is solable in Z. Next, we conside the case when 2 NF. Just as befoe, we always discuss imaginay uadatic fields. Let F Q d d a ositie suae-fee intege. Then 2 NF if and only if d u 2 2w 2 with u, w Z see [2]. When d is not a ime, the symbol d denotes the Jacobi symbol. Note that u+w d simlicity of notation, denote by ψ the Jacobi symbol u+w d u w d. Fo. Lemma Let d be a ositie suae-fee intege with d u 2 2w 2, whee u, w Z. Then thee is a ime 1 mod 4 with d, u + w and uw such that the Diohantine euation 3.9 X 2 dy 2 u + wz 2 is solable in Z if d 1 mod 8, and 3.10 X 2 dy 2 ψu + wz 2 is solable in Z if d 1 mod 8. P o o f. Clealy, d u+w 1. Hence, if d 7 mod 8, then by the oeties of the Jacobi symbol see [6], we hae u+w d 1. Fo any ime l d, we choose a ime 1 mod 4 with u + w and uw such that l u+w l. Put d 2d if 2 d. Fo any ime l d, we choose a ime with u + w, uw such that l u+w l and 1 mod 8 if d 1 o 5 mod 8 if d 1. In both cases, d 1. If d 1 mod 8, we choose a ime 1 mod 4 such that u + w, uw and fo any ime l d, l ψu+w l. We also hae d 1. Then by Lemma 2.1, the esult follows. R e m a k. In the oof of the aboe theoem, we used the emakable fact that any aithmetic ogession contains infinitely many imes. By choice of X, Y, Z, a solution of euation 3.9 o 3.10, we can find g, h Z such that h Y, u + wg + wh X and g, h 1. Put 3.11 α g 2 + h 2, θ g 2 h 2 + 2ghw.

8 160 H. Qin Clealy, if d 1 mod 8 and u+w d 1, then u+w d 1. Without loss of geneality, we can assume that ψ 1. Then αu+θu+w u+wg+wh 2 +u 2 2w 2 h 2 X 2 dy 2 u+wz 2, hence, 3.12 αu + θ Z 2. Theefoe, u + θ α 2ααu + θ α 2 ξ 2 + η 2, whee ξ, η Q with αξ, αη Z. It follows fom uw and g, h 1 that, α, θ 1. Moeoe, we can assume that αξ, αη, 1. Let x αξz 2 + αηλ, y α 2 ξ, a αηz 2 αξλ, b α 2 η, whee λ g 2 h 2 2ghw. Note that λ 2 + θ 2 2α 2 w 2. Then 3.16 x + y d 2 + a + b d 2 u + d2αz 2 2. and On the othe hand, α 2 ξ 2 + η 2 0 mod Z 2, hence, x 2 + dy 2 αηλ 2 + dα 4 η 2 αηλ 2 + u 2 2w 2 α 4 η 2 αη 2 λ 2 + α 2 u 2 2α 2 w 2 αη 2 θ 2 + α 2 u 2 0 mod Z 2 x 2 + dy 2 αξz 2 2 u 2 α 2 ξ 2 α 2 ξ 2 Z 2 2 α 2 u 2 0 mod w. Similaly, a 2 + db 2 0 mod Z 2 and a 2 + db 2 0 mod w. Lemma With the notations as aboe, set E x + y d, F a + b d and { E β F, E2 + F 2 } F 2. Then β 2 { 1, u + d} K 2 O F and thee is a β K 2 O F with β 2 { 1, u + d} if and only if the Diohantine euation 3.17 u + wn 2 S 2 dt 2 is solable in Z. P o o f. We only need to conside non-dyadic laces of F. It is easy to see that fo any lace P, if P E P F, then τ P β 1 and if P E P F,

9 then 3.18 τ P β 1 P αz 2 w/f. Imaginay uadatic fields 161 Obiously, if P E P F, then P Z dz 2 w. We deduce fom α y, α b and, α 1 that fo any lace P, if P Z α, then τ P β τ Pβ, whee P P is the conjugation of P. Thus, multilying β by { 1, c} fo a suitable c Z if necessay allows us to assume that τ P β 1 fo any P Z α. On the othe hand, if P Z w, P E P F, then P F P w, since x 2 + dy 2 a 2 + db 2 0 mod w. Hence τ P β 1 fo any P Z w. Finally, since d 1, PP. It follows fom x 2 + dy 2 a 2 + db 2 0 mod Z 2,, α 1 and 3.16 that if P E + P F 0 then eithe P E o P F 1 mod 2. Hence, τ P β 1 and τ Pβ 1. If P E P F 0, then τ P β 1 and τ Pβ 1. By Lemma 3.1, we see that thee is a β K 2 O F with β 2 { 1, u + d} if and only if the Diohantine euation 3.19 εn 2 S 2 + dt 2 is solable in Z fo ε 1 o 2. This is euialent to saying that the Diohantine euation 3.17 is solable in Z. This oes ou theoem. The following theoem is a conseuence of the aboe lemma and Theoem 3.6. Theoem Let F Q d d a ositie suae-fee intege with d u 2 2w 2 fo u, w Z, and let m d. Then thee is a β K 2 O F with β 2 { 1, mu + d} if and only if the Diohantine euation 3.20 mu + wn 2 S 2 dt 2 is solable in Z. P o o f. Fist, we obsee that if d 1 mod 8 and u+w d 1 togethe with m 1 mod 4, then 3.20 has no solutions in Z. In fact, conside mu + wn 2 S 2 dt 2 mod 4, i.e., 3N 2 S 2 + T 2 mod 4; then the esult follows. Next, if d 1 mod 8 and u+w d 1 togethe with m 1 mod 4, then thee is no β K 2 F with β 2 { 1, mu + d}. Then, by Lemma 3.12 and Theoem 3.6, the assetion follows ank K 2 O F. Fo any numbe fields F a 4-ank K 2 O F fomula is oed in [7] comae also [5]. Fo uadatic field, we efe to [1], [11]. Hee, we aly Theoems 3.10 and 3.13 to detemine the 4-ank K 2 O F fo any imaginay uadatic field F. Let F Q d d a ositie suae-fee intege. Put d 1 2d o d accoding as 2 d o not. Wite K {m m d, m 1, d, 2 m} and V {u + dm d u 2 2w 2 with u, w Z,

10 162 H. Qin w > 0, m K {1, d }} and ut K {k K εkz 2 X 2 dy 2 is solable in Z fo ε 1 o 2}, V 0 {mu + d mu + d V, V {mu + w mu + d V 0 }. mu + wz 2 X 2 dy 2 is solable in Z}, Theoem 4.1. With the aboe notations, let F Q d be an imaginay uadatic field. Then 4 4 ank K 2 O F log 2 4, whee +2 #K V. P o o f. Fo any ositie intege n, let n K 2 O F denote the subgou geneated by all elements of ode n. By [2], 2 K 2 O F can be geneated by the following elements: { 1, k} k K, { 1, mu + d} mu + d V if d u 2 2w 2 with u, w Z. Since [ : F 2 ] 4, thee ae the only two elements δ, d /δ K o δ, d /δu + d 2 V 0 satisfying δ, d/δ. Suose that a 1,..., a 4 geneate 4 K 2 O F. Then a 2 i { 1, b i} 2 K 2 O F 1 i 4. Set b 0 δ. Then by Theoems 3.10 and 3.13, #{b i1... b ik, d/b i1... b ik i 1,..., i k {0, 1,..., 4 }} #K V 0 #K V. It is easy to eify that #K V So 4 log as desied. Coollay if and only if #K V 2. Coollay if and only if K K and #V #V. 5. The stuctue of K 2 O F 2. In this section, we aly Theoems 3.10, 3.13 and 4.1 to detemine the stuctue of K 2 O F 2 fo imaginay uadatic fields F. Theoem 5.1. Let F Q d be an imaginay uadatic field with d eithe o 2 o o 2, whee,, ae distinct odd imes. If 2 NF, ut u + w, whee u, w Z ae such that d u 2 2w 2. Let δ be an element such that F 2 2F 2 δf 2 2δF 2. Then we hae the tables gien below. If F is a field as in Table III, then 2 2, 4 0, othewise excet fo the case d 2 with,, 7, 5, 3 mod 8 2 2, 4 1. N o t e s. 1. Only when 2 d and d/2 1 mod 8, the altenatie can occu in Table II.

11 Imaginay uadatic fields 163 Table I F, mod δ Q 5, Q 2 3, Q 3, 5 [2] 1 0 5, Q 2 3, , Table II F, mod 8 The Legende symbols 4 δ , 1 1 o δ 1 1, Q 1 0 δ 1 Q 2 1, Q 1, , 7 1 3, 3 1 5, If mod 8 o mod 8 o mod 8, then the condition on the Legende symbols, say, should be undestood as: if thee is a choice of,, with the Legende symbols satisfying. Fo examle, in the case 7, 7, 5 in Table III, the condition on the Legende symbol is

12 164 H. Qin Table III F,, mod 8 The Legende symbols δ Q Q 2 7, 7, 5 7, 7, 3 7, 5, 1 1 if 1; if 1 1 if 1; if 1 1 if 1; if 1 1 if 1 1 if 1; if 1 7, 3, 1 1 if 1 3, 3, 3 1 Q 7, 5, 5 7, 3, 3 5, 5, 3 5, 3, 3 1 if 1; if 1 1 if 1; if 1 1 if 1; if 1 1 if 1; if 1 1 if 1; if 1 5, 3, 1 1 if 1 7, 5, 5 1 if 1; if 1 7, 5, 3 7, 3, 3 1 if 1; if 1 5, 5, 3 Q 2 5, 5, 1 5, 3, 3 1 if 1; if if 1; if 1 1 if 1; if 1 5, 3, 1 1 if 1 3, 3, In actice, we identify 1 with 1. Hence, if 1 then we also hae 2 0. P o o f o f T h e o e m 5.1. We will eeatedly use the notations K, K, V and V which ae defined in Section 4. It is not had to eify the coectness of the statement 4 0 when δ

13 Imaginay uadatic fields 165 Table IV F,, mod 8 The Legende symbols 4 δ 7, 5, , 5, othewise 1 Q 1 2 5, 5, 1 othewise , 3, 1 othewise Q 2 5, 5, othewise , 1, Q othewise 1 Q , 1, othewise 1 has been listed. In fact, one can easily check that K {δ, d/δ} o K {δ, d/2δ} and V. Then the esult follows fom Theoem 4.1. On the othe hand, 4 2 if and only if K K and V V. Hence, it is also easy to eify the coectness of the statement 4 2. Now, fo Tables I, II we only need to conside the following cases: 1,1; 1,7; 7,7. T h e c a s e 1, 1. Clealy, 2 2 and 1 K. Suose 1. Then Z 2 X 2 dy 2 has no solutions in Z, hence ± K, so 4 1. If 1, then V, hence 4 1, theefoe 4 1. If 1, then V, hence 4 1. If, then V, hence 4 0. Suose 1. Then ± K, hence 4 1. If 1, o 1,

14 166 H. Qin Table V F,, mod 8 The Legende symbols , 7, othewise 0, othewise 1 7, 7, othewise 1 0 Q Q 2 othewise , 1, 1 1, , 1 1 othewise othewise 2 1, 1, 1, 1 2 1, 1 othewise othewise 1 0 then V, hence 4 1, so 4 1. If 1, then K K and V V, hence 4 2. T h e c a s e 1, 7. We hae 2 2 and 1 K, hence 4 1. Suose 1. Then ± K, hence 4 0. Suose 1. If 1, then K {, }, V, hence 4 0. If 1, then K, hence 4 1, so 4 1. T h e c a s e 7, 7. We hae 2 2 and 1 K, hence 4 1. Suose 1. Then K. If 1, then V ; if 1, then V, hence 4 1, so 4 1; if, then V, hence 4 0. The oof of Table III is diect.

15 Imaginay uadatic fields 167 Fo Table IV, we only need to conside the thee cases: 7, 5, 3, 5, 5, 1 and 3, 3, 1 d. Fo the case 7, 5, 3, we hae 1 K, hence 4 1. But it is easy to see that, K, V V, hence 4 1, so 4 1. Fo the case 5, 5, 1 o 3, 3, 1, we hae 1 K, V V. If 1, then K K, hence 4 2. Othewise, we hae 1 o 1. If 1, 1, then ± K, ± K; if 1, then ± K, hence, we hae 4 1. Finally, we conside Table V. Clealy, in any case, 2 3. Without loss of geneality, when 2 d, o 7 mod 8, we can assume 1. On the othe hand, when d 2 with 1 mod 8, if 1, then it is easy to see that V. Hence, we always assume 1. T h e c a s e 7, 7, 7. We hae 1,,, K, hence 4 1. Suose, then ±, ±, ± K, hence 4 0. Since 1, thee ae the following ossibilities: 1, 1, 1. Suose 1. Then K. If 1, then V, and if 1, 1, then V, hence 4 1, so 4 1. If 1, 1, o 1, 1, then V, hence 4 0. Similaly, suose 1. Then K. If 1, then V, and if 1, 1, then V, hence 4 1. Othewise, 4 0. Suose 1. Then K. If 1, then V, and if 1, 1, then V, hence 4 1. Othewise 4 0. T h e c a s e 7, 7, 1. We hae 1 K. Suose 1. Then ± K. If 1, then K and ±, K, and if 1, then K and ±, K. Hence 4 1. If 1, then V ; if 1, and 1, then V ; if 1 and 1, then V ; if 1 and 1, then V. Hence 4 1. So 4 1. This discussion also woks fo 1. Suose 1. Then, K. If 1, then eithe 1 and 1, o 1 and 1. In both cases, ±, ±, ± V. If 1, then eithe 1 o 1, theefoe eithe V o V, hence 4 2, so 4 2.

16 168 H. Qin T h e c a s e 7, 1, 1. We hae 1 K, hence 4 2. Suose 1. Then,, K. If 1, then V, hence 4 2, so 4 2. Othewise, ± V, hence ±, ±, ± V, since,, K. Hence 4 1. Suose 1 o 1 o 1. Then ±, ±, ± K, hence 4 0. Suose 1 and 1. Then K. If 1, then V ; if 1 and 1, then V, hence 4 1. Othewise, V, hence 4 0. Suose 1 and 1. Then K. If 1, then V, and if 1 and 1, then V, hence 4 1. Othewise, V, hence 4 0. T h e c a s e 1, 1, 1. We hae 1 K. Suose 1. Then K K, hence 4 2. If 1, then V V, hence 4 3. Othewise, K, hence 4 2. Suose 1, 1. Then ±, ±, ± K, hence 4 1. If 1, then V ; if 1 and 1, then V ; if 1 and 1, then V ; if 1 and 1, then V. In any case, 4 1, so Then 4 1. If 1, Suose then V ; if 1 and 1, then V ; if 1 and 1, then V ; if 1 and 1, then V. In any case, 4 1, so 4 1. Suose 1 and 1. Then ± K, ±, ± K. If 1, then V ; if 1 and 1, then V. In both cases, 4 2. Othewise V, hence 4 1. This concludes the oof of the theoem. R e m a k s. 1. Ou method can be alied to any imaginay uadatic field. 2. Simila esult fo eal uadatic fields hae been obtained by the autho see [12]. Acknowledgements. I would like to thank the efeee fo the aluable comments and sending me the ae [4] fom which I know that most esults of Tables I and II hae been obtained by P. E. Conne and J. Huelbink by a diffeent method. I would also like to thank Pof. J. Bowkin fo the helful suggestions which hae been incooated heein. Finally, I would like to thank Pof. Tong Wenting fo his hel.

17 Imaginay uadatic fields 169 Refeences [1] B. B a u c k m a n n, The 2-Sylow subgou of the tame kenel of numbe fields, Canad. J. Math , [2] J. Bowkin and A. Schinzel, On Sylow 2-subgous of K 2 O F fo uadatic fields F, J. Reine Angew. Math , [3] J. W. S. Cassels, Rational Quadatic Foms, Academic Pess, London, [4] P. E. Conne and J. Huelbink, Examles of uadatic numbe fields with K 2 O containing no elements of ode fou, eint. [5],, The 4-ank of K 2 O, Canad. J. Math , [6] K. Ieland and M. Rosen, A Classical Intoduction to Moden Numbe Theoy, Singe, New Yok, [7] M. Kolste, The stuctue of the 2-Sylow subgou of K 2 O, I, Comment. Math. Hel , [8] J. Milno, Intoduction to Algebaic K-theoy, Ann. of Math. Stud. 72, Pinceton Uniesity Pess, [9] J. Neukich, Class Field Theoy, Singe, Belin, [10] O. T. O Meaa, Intoduction to Quadatic Foms, Singe, Belin, [11] H. Qin, K 2 and algebaic numbe theoy, Ph.D. Thesis, Nanjing Uniesity, [12], The 2-Sylow subgous of K 2 O F fo eal uadatic fields F, Science in China Se. A , [13] J. Tate, Relations between K 2 and Galois cohomology, Inent. Math , DEPARTMENT OF MATHEMATICS NANJING UNIVERSITY NANJING, , P.R. CHINA Receied on and in eised fom on