Available at: http: //www. ictp. trieste . it/~pub_of f 10/98/51

United Nations Educational Scientific and Cultural Organizationand

International Atomic Energy Agency

THE ABDUS SALAM INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS

TAME KERNELS AND TATE KERNELSOF QUADRATIC NUMBER FIELDS

Qin Hourong*Department of Mathematics, Nanjing University,

Nanjing 210093, People's Republic of Chinaand

The Abdus Salam International Centre for Theoretical Physics, Trieste, Italy.

Abstract

Let F be a quadratic field. We obtain the necessary and sufficient conditions for an element oforder two in the tame kernel of F to be a fourth power in the tame kernel of F. This enables usto compute the 8-rank of the tame kernel of F. In the case when F is an imaginary quadratic fieldwith the 8-rank of A ^ O F = 0, the explicit structure of the Tate kernel of F can be obtained byour method.

MIRAMARE - TRIESTE

June 1998

E-mail: hrqin@ictp.trieste.it

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1.INTRODUCTION

Let F = Q(\/d), d E 7L square-free, be a quadratic field, OF the ring of integers of F. We knowthat any element of order 2 in A'2 OF can be written as the form { — 1, x], where x has been givenexplicitly by A.Schinzel and J. Browkin [1]. Therefore, we can determine the 2n-rank of A ^ O F ifwe can give the necessary and sufficient conditions for { — 1, x] = a2 with a E A ^ O F - We havedone this for n = 2 in [16] and [17]. The purpose of this paper is to consider the same problem forthe case n = 3. The paper is organized as follows.

Section 2 contains a review of known material, some of which is stated in the form needed here.In section 3, we study the fourth power problem in A'2 OF, some results are valid for the A'2

group of a general field. We present the main theorem of the paper, which gives the necessary andsufficient conditions for an element { — 1, 2;} 6 A'2 OF being a fourth power in A'2 OF, where F is aquadratic field.

Finally, in sections 4 and 5, we apply the results in section 3 to the imaginary quadratic fields andthe real quadratic fields respectively. The 8-rank of A'2 OF has been determined for any quadraticfield whose discriminant has only one odd prime divisor. At the same time the Tate kernel of F isgiven if F is an imaginary quadratic field with the 8-rank of A'2 OF = 0.

2.PRELIMINARIES

In this section, we will introduce some notations, recall some known facts, and give some im-mediate corollaries to known results.

We begin with the following lemma, which is very useful in this paper.

Lemma 2.1(Legendre's Theorem). Suppose that a, b, c are square-free, (a, b) = (a, c) = (b, c) =1 and a, b, c do not have the same sign. Then the Diophantine equation

(2.1) aX2 + bY2 + cZ2 = 0

has nontrivial solutions if and only if for every odd prime p abc, say, p I a, I I = 1.\ P J

Proof. See [11].

We have alsoLemma 2.2(Holzer's Theorem). / / (2.1) is solvable, then it has a nontrivial integer solutionwith

\X\ < y/\bc], \Y\ < y/m, \Z\ < y/\ab\

Proof. See [11].

Corollary 2.3. Suppose that the Diophantine equation X2 + dY2 + mZ2 = 0 has a nontrivial

solution and w EM with (w,m) = 1. Then it has a solution X\,Y\,Z\ £ Z satisfying (Z\,w) = 1.

Here (Z\, w) = 1 means that (Z\, w) has no any odd prime divisor.

Proof. Let X0,Y0, Zo E 7L with (Xo, Yo, Zo) = 1 be a solution of X2 + dY2 + mZ2 = 0. Then forany x ETL, one can verify that

X = X0(x2 + m) - 2x(Xox + mZ0)

Y = Y0(x2 + m)

Z = Zo(x2 + m) — 2(Xox + mZo)

is also a solution. (See [11] for the result on the form of the general solutions of aX2-\-bY2-\-cZ2 = 0.)For any odd prime p, if p \ (Zo, w), set x = 1 (mod p); if p \ w,p\ Zo, then

Z0(x2 + m) - 2(Xox + mZo) = Z0(x + s)2 +t (mod p),

where s,t £ Z . Choose r ^ 0 (mod p) such that Zo(x + s)2 + t = r (mod p) is solvable. Then by

the Chinese Remainder Theorem, there are X\ ,Y\,Z\ ETL with (Z\, w) = 1 satisfying X2 + dY2 +mZ2 = 0. This completes the proof.

Remark 2.3.1. In view of Lemma 2.2 and the above proof, we see that Z\ can be chosen dependingonly on (Zo, w) and d.

From now on, when we say a (quadratic) Diophantine equation is solvable, we always mean thatit has nontrivial solutions.

Let d ^ 0 be an integer. It is convenient for us to use the following notation:

{±1,±2}, if d > 0,

{1,2}, if d < 0.

For any abelian group A, let A^ denote the 2-Sylow subgroup of A, and let 2̂ 4 = {x E A | x2 =1}. Let F be a number field, OF the ring of integers of F, denote by £1 the set of all places of F.

For any p £ fi and a,b E Fp, I —-— I is the Hilbert symbol with order 2 on Fp. In particular, on

Q2, w e have

' « , n = ( _ 1 ) i ^ x ^ i a n d ^ ^ = ( _ 1 ) ^ ij

where u,v are units in (Q>2- For more about the Hilbert symbols, we refer to [12] or [13]. For anyfinite place p of F, we use vp(-) to denote the discrete valuation on F with respect to p and rp forthe tame symbol at p. For any integer n, put y " = {a £ A ^ O F \ a = /3n for some /3 £ I

L e m m a 2 . 4 [1 ] . Let F = Q(\/d), d £ Z square-free. Then ^K^OF can be generated by

{ — 1, m}, m | d;

together with

if { — l,±2}f)NF ^ 0, where Ui £ Z S«C/J iftai uj — d = Ciwf for some u>i £ 7L and c

Lemma 2.5 [16],[17]. Let F = Q(\/d),d £ 7L square-free. Suppose that m \ d and write d =u2 — 2w2 with u,w E TL if 2 £ NF. Then { — 1, m} £ y if and only if one can find an e £ S(d)such that

(i) I m~ = I — I for any odd prime pi - i i p j

(ii) ( — J = | - J , for any odd prime p

and { — 1, m(u + \fd)} £ y 2 ' / and on^V tf

, for any odd prime p \ m;

,for any odd prime p \ —.

For a real quadratic number field F, we have the following map

It is clear that for any a £ A'2OF, if a = /32 with /3 £ K2F, then the image of a is (1,1). Onecan see that if there are m £ N and e < 0 such that both (i) and (ii) in Lemma 2.5 are satisfied,then the image of a is either (—1, 1) or (1, —1), where a E K'ZOF with or = { — 1, g}. Hence, if— 1, —2 ^ NF, then { — 1, m) (£ XJ . This fact has been used to determine the structure of (K^OF)^for some real quadratic fields in [15]. On the other hand, if —1 or —2 6 NF and there are m £ Nand e < 0 such that both (i) and (ii) are satisfied, then one can show that both (i) and (ii) are alsovalid for the same m and — e. Therefore, we have

Lemma 2.6. Let F = Q(y/d) be a quadratic field and a G 2-̂ 2 Of- If a £ \/4, then there is ane £ {1,2} such that

emz'2 = x2 + dy2 is solvable if a = { — l,m),or

em(u + w)z2 = x2 + rfj/2 «s solvable if a = { — 1, m(w + v rf)}.

Lemma 2.7 [16],[17]. Let F = Q(\/d),d £ 7L square-free, be a quadratic field. For any a =

x + y\/d £ F, put S = {pi, ..., pn} = {p, l rp{~l, a} = ~1}- Without loss of generality, we can

assume that pi = pj f lZ «s noi inert for 1 < i < n. Then x2 — dy2 = ep\ • • -pnz2, where e G S(d)

and z G Q. Conversely, suppose that p\, ...,pn ore distinct primes in 7L and p\, ...,pn are prime

ideals of OF such that p{C\7L = pi for 1 < i < n. If there is an e G S(d) such that the equation

x2 + dy2 = epi • • • pnz2 is solvable in Q (equivalently in 7L), then there is an a G F* such that

S= {pkp{-l,a} = -1} = {pi,-,Pn}-

L e m m a 2 . 8 [ 1 4 ] . Let F be a field, 7 G K2F with -f4 = 1 . Then there are x,y G - F * , S « C / J i / j a i

7 = {x,x2 + l}{-l , j /} .

3. FOURTH POWERS IN K2OF

Although we will focus on quadratic fields, we first give some results on symbols in the generalcase.

L e m m a 3 . 1 . Let F be a field. Suppose that x,y,c G F* with x + y = cn. Then

Proof, {x, y) = {x, c" - x} = [x, c»(l - J-)} = {x, c"} [x, 1 - ^

This completes the proof.

The identities in the following lemma will be used repeatedly.

L e m m a 3 . 2 . Let F be a field. If a £ F* and a = x 2 + y 2 with x,y,x — y £ F*. Then

If a = s2 ± 2t2 with s,t £ F*, then

Proof. {-l,a} = {-l,x2

2j/ x2 + y2

2 ' a;2

« T 0 1 / 2 y *2 + y 2 \ ( 2 * y * y \ „By L e m m a 6.1, < , ^— > = < -^ TT, > . Hence,

I x xz J ^ x + V x )

Now suppose that a = s2 + 2t2. Then

{-2, a} = {-2, s2 + 2t2} = {-2t2, s2 + 2t2} | i , s2 + 2

r 2 t 2 1 2

By Lemma 3.1, {-2t2, s2 + 2t2} = I — - , s } , hence,[sz + 2tz J

{-2,a} = {2t2,s}2{-,s2 + 2t2}2.r

Similarly, we can verify that {2, a} = {2t2 ,s}2 < —, s2 — 2t2 > if a = s2 — 2t2. The proof is

complete.

Corol lary 3 .3 . Let F be afield, and let a £ F*. If a = x2 + y2 = (c2 - 2d2)(e2 + 2 / 2 ) (especially,c2 — 2d2 or e2 + 2f2) for some x,y,c,d,e, f 6 F*, then there is an element 7 6 K^F with7 4 = { — l , a } . In particular, if —1 = x'2 + y2, then { — 1 , - 1 } = -f4 holds for some 7 6 K2F, inother words, { — 1, — 1} = a 2 if and only if { — 1, — 1} = /34 , where a , / ? £ K2K.

Proof. The result is just a consequence of Lemmas 3.1 and 3.2.

Lemma 3.4. Let F = Q(\/d) be a quadratic field.(1) Suppose m \ d. Assume that m > 0 if d > 0 and m = 1 (mod 4) if d = 1 (mod 8). T/jen

i/jere «s a prime p = 1 (mod 4) SMC/J iftai

epmZ2 = X2 + dY2

«s solvable for e = 1 or 2.fny Suppose 2 £ 7V_F, d = u2 — 2w2, where u,w £ Z anc? m | rf. Assume that mu > 0 if d > 0

and m(u + w) = 1 (mod 4) if d = 1 (mod 8). T/jen i/jere «s a prime p = 1 (mod 4) S«C/J

«s solvable for e = 1 or 2.

Proof, (i). We may rewrite the equation

2 = X 2

as

2(3.1) epZ2 = m l 2 + —Ym

By Lemma 2.1, (3.1) is solvable if and only if there is a prime p = 1 (mod 4) such that

-y- = ( — j , for any odd prime / | m,

m\ (ep\ . d— = — , lor any odd prime / —.I / \ I / m

~\I f 2 d, t h e n w e c h o o s e p = 1 ( m o d 4 ) s u c h t h a t b o t h I — I = ± 1 a n d ( — 1 = 1 h o l d .

P \PJ

If d > 0 is odd, then we want to find a prime p = 1 (mod p) such that both I -^- I I —=- I =1 TO I I a '

TO

— ) a n d ( — 1 = 1 ho ld for s o m e e = 1 or 2.

Suppose that m = 1 (mod 4) or — = 1 (mod 4), then we can take e = 1.m

Suppose that d = 5 (mod 8) and m = 3 (mod 4), then we can take e = 2.Suppose that rf = 1 (mod 8), then we must have m = 1 (mod 4), hence, we can take e = 1.If d < 0 is odd, then we require

(3.2)- \m Im 1

/

/I m

-\ TO

\\ _ e p

\\d1

Suppose that \d\ = 3 or 5 (mod 8), then it is easy to see that there is a prime p = 1 (mod 4)

such that both I — I = 1 and (3.2) hold for an e = 1 or 2.\PJ

Suppose that \d\ = 1 (mod 8), one can verify that

= 1.

Suppose that \d\ = 7 (mod 8), then the only possibility is that m = 1 (mod 4). Hence, by theassumption, we have

d

Therefore, in both cases, taking e = 1, we see that there is a prime p = 1 (mod 4) such that both

- ) = 1 and (3.2) hold.

The proof of (ii) is similar, so we complete the proof.

Lemma 3.5. Let F = Q(\/d),d £ 7L square-free. Suppose that prime p \ d. Then there is a7 G K2F such that

f 2 if o v(i) {2,p] = 7

2 with r p 7 = * y\ ifp = 1, 7 (mod 8);1̂ 1 otherwise

(ii) {-2,p} = 72 withrp~f = \ ' «/p = 3(mod8).1̂ 1 otherwise

Proof. If p = 1 or 7 (mod 8), then p = a2 — 2b2 with a, b £ 7L. By Lemma 3.2, we have

Put 7 = {2&2, a} < — ,p >. Then it is easy to see that 7 has the desired property.

In the case where p = 3 (mod 8),p = a2 + 2b2 with a,b £ Z . We have

{-2,p} = {2b2,a}2{a-,p}2.

Taking 7 = {2&2, a} | - , | )> yields the result. The lemma is proved.

Lemma 3.6. Let F = Q(Vrf), d £ 7L square-free. Suppose that p \ d and p = 5 (mod 8) is a prime.

Then there is a prime q with I — I = 1 such that

(1) pqZ2 = X2 + 2dY2 is solvable if d is even or d = 3 (mod 8);

(ii) 2pqZ2 = X2 + 2dY2 is solvable if d = 5 or 7 (mod 8).Moreover, in each case, there is a 7 £ K2F with

ifp\P,

ifp = q2, where qiq2 = qOF,

such that 72 = {2,p}.

Proof, (i). Suppose d is even. Let d = 2d'. Rewrite

pqZ2 = X2 + 2dY2

as

(3.3) qZ2=pX2 + —Y2

P

By Lemma 2.1, we require that there is a prime q with I — I = 1 and I -—- I = 1 if we want (3.3)\iJ \\d'\J

to be solvable. Hence we may choose q = 1 (mod 8), so I — I = 1.

Now let X0,Y0,Z0 <EZ with (X0,Y0) = (X0,Z0) = (Y0,Z0) = 1 be a solution of (3.3). It follows

from I — I = 1 that ( 1 = 1. Hence, q = a2 + 2b2 for some a, b £ 7L.

W \ 9 JSince X2 + 2dY0

2 = X2 + 2(VdY0)2, by Lemma 3.2,

{-2,X2 + 2dY2} = {2dY2,X0}2 { ^ T ^ o + ̂ Y2^ and

Therefore, {-2,p} = {-2,pqZ2}{qZ2,-2}= {-2, X2 + 2dY2}{q, -2}{Z0, - 2 } ^

2= {2dY2,X0}2 { ^ T ^ o +2rfY0

2} {a,2b2}2 {q^}" {Z0,-2}2.Put

7 = {2rfY02, Xo} j - ^ , X2 + 2rfY0

2| {a, 2b2} {q, ^ } {Zo, - 2 }

Since I — I = 1, we have qC>F = ^1^2, where qi, q2 are the prime ideals of OF-

Clearly, for i = 1,2, Tqt{2dY2, X2} = Tqt{2b2, a} = 1. We may assume further that for i =l , 2 , r q , { Z 0 , 2 } = l .

On the other hand, for i = 1,2,

) ^ (modq . )dY0

and

x j X2b2

We have ra -f2 = 1 (mod a,-), since I —p= I = ^—- = 1 (mod </).

Note that it is impossible that both

Tqi7 = c (mod qi)

andrq27 = c (mod q2)

hold at the same time, where c = 1 or — 1. In fact, if it is the case, then

Xo b _= c I mod Q;) lor i = 1,2,

Y aVdY0 a

hence, —= = c (mod q), therefore, X§b — cayrfYo = 0 (mod q), this is a contradiction sinceV dY0 a

(q,Xob) = l.

A computation shows thatrp7 = 2 (mod p),for p \p

and

rp7 = 1 (mod p),for any p\pq.

The similar discussion works for the other cases and our lemma is proved.

One can use the same method to show the following

Lemma 3.7. Let F = Q(y/d), d £ 7L square-free, and let d — 1 (mod 8). Suppose that m \ d withm > 0.

(i) If m = 1 (mod 8), then we can find a prime q with ( — 1 = 1 such that qmZ2 = X2 — 2dY2

is solvable. Furthermore, there is a 7 £ K2F with

2, if p\ m;

- 1 , ifp = q1;

1, if p = q2(q1q2 = qOF);

1, otherwise

such that {2,m} = 72.

A « ) / / d < 0 a n c ? m = 5 ( m o d 8 ) , t h e n w e c a n f i n d a p r i m e q w i t h ( — 1 = 1 s u c h t h a tW

—2qmZ2 = X2 — 2dY2 is solvable. Furthermore, there is a 7 £ K2F with

2, if p\ m;

, - 1 , «/p = qi;

1, «/p = q2(qiq2 = gOf);

1, otherwise

such that {2, m} = 7 .(m,) / / d > 0 anc? m = 5 (mod 8), then we can find a prime q with q = 3 (mod 8) SMC/J inai

qmZ = X — 2dY is solvable. Furthermore, there is a 7 6 K2F with

{ 2, if p\ m;

i(i2 = - 1 (mod p)), if p = qOF;

1, otherwisesuch that {2, m}{-\, q} = ~f2•Lemma 3.8. Let d £ 7L square-free. Suppose that there are u,w £ 2 such that d = u — 2w .Then there is a prime q = 1 or 3 (mod 8) such that(3.4) X2 + 2dY2 = uqZ2

is solvable if one of the following conditions is satisfied(1) d^l (mod 8) or d < 0;

( u \— 1 = 1 .

For the case where d = 1 (mod 8), d > 0 and ( — j = —1, there is a prime q = 7 (mod 8) SMC/J

(3.5) X2

«s solvable.

Proof. This is proved by explicit calculations of Jacobi symbols. We will not go into details.

/ u \ f-u\Remark 3.8.1. If d = 1 (mod 8), d < 0 and — = - 1 , then — - = 1.

\\d\J \ \ d \ JWith the same notations as above, suppose that X2 + 2dY2 = uqZ2 is solvable and XO,YQ,ZQ

N a solution. PutXo- 2wY0

n = Yo, m = .u

By the choice of Xo, Yo and Zo, we can assume t h a t n,m £ Z wi th (n, m) = 1. Let

g = m2 - 2n2, e = m2 + 2n2

eu + 4wmn = qZ0 and g — e = —8m n .

and

x = —2wng + mqZ0 , y = me,

a = 2wmg + 2nqZ2, b = 2ne.

Then(a + bVd)2 + 2(x + y\fd)2 = (u + \fd~){2eZ0)

2.In fact, we have

Hence, a2 + b2d + 2(x2 + y2d)= a2 + 2x2 + (b2 + 2y2)d= (2wmg + 2nqZ2)2 + 2(-2wng + mqZ2)2 + (4n2e2 + 2m2e2)(«2 - 2w2)= 4w2m2g2 + An2q2Z?i + 8w2n2g2 + 2m2q2Z?i + 2e3(u2 - 2w2)= 4w2g2e + 2q2Zte + 2e3(u2 - 2w2)

2Z^ + e2u2)+4ew2(g2-e2)2ZQ + e2u2) — 32ew2 m2 n2

2ZQ + qZ2(eu — Awmn))= 2eqZ2(qZ2 + eu — 4wmn)= qu(2eZ0)

2

and

4xy + 2ab = Am qeZ0 + 8n qeZ0 = q(2eZo) .

WriteH = x + yvd, G = a + b\fd.

The above computation shows

G2 + 2H2 = q{u + Vd){2eZ0)2.

By Lemma 3.2,

. 2

{2, q(u + Vd)(2eZ0)2} = {2, G2 + 2H2} = {2H2, G}2 \ —, G2 + 2H2

If q = 1 or 3 (mod 8), then there are r,s £ Z such that q = r2 + 2s 2 . Hence,

So, we get

{2, u + Vd} = {2, q(u + Vd){2eZ0)2}{q, 2}{(2eZ0)2, 2}

= {2H2, G}2 1 1 , G2 + 2 F 2 | 2 {r, 2S2}2 {q, ^ } ' {2eZ0, 2}2.

Similarly, if q = 7 (mod 8), then q = r2 — 2s2 with r,s £ Z . In this case,

Hence,

f G I 2 r r->2

{2,u + Vd}{-l,q} = {2H2,G}2 l—,G2 + 2H2\ {r, 2s2}2 |g, - | {2eZ0,2}2.

Put

7 = {2H2, G] \^,G2 + 2H2 \ {r, 2s2} \q, -} {2eZ0,2}.

10

Lemma 3.9. Let everything be the same as above. Then by the choice of solutions of (3.4) or(3.5), we can assume that for any non-dyadic finite place p,

= qOF);

otherwise

if d ^ 1 (mod 8) or d < 0 or d = 1 (mod 8) together with d > 0, ( - ] = 1; and

, (i2 = - 1 (mod p)), if p = qOF;

other

—) = - i .

Proof. It follows from (m, n) = 1 that (e, m) = (e, n) = 1, since e = m 2 + 2n2 . We can assume that

(g,w) = 1, by Lemma 2.2, (Zo,w) = 1, hence, (e,w) = 1 and (e,q) = 1. Clearly, we can assumethat (n, u) = 1, so (n, ZQ) = 1.

We have

e = m 2 + 2n2 = - £ • + 2Y02 = — ( X 2 + 2rfY0

2) = ^ - (mod w).£Hence,

1 + yy d = mqZ0 + meyd (mod w)

'o H —yd) (mod w)u

i + yd) (mod w).a

Hence, for any non-dyadic finite place p | (u + \fd),vp(x + y\fd) > —vp(u + \fd). Similarly,I

vp(a + b\fd) > —vp(u + Vd). Let g = vp(G) = vp(a + b\fd) and h = vp(H) = vp(x + yVd).

Then min(g,h) = -Vp^ + yQ), since G2 + 2_ff2 = q(u + yrd){2eZ0)'2, (e,w) = 1 and Z0,w) = 1.

Therefore, if g > h, then

TP7 = —Fr

= 2h

p)

H2

= 2h (mod p)

A computation shows that r p 7 = 22vf(u+^d) (mod p) holds also for g < h.One can verify that

mx + myvd + no + nbvd = e yd.

Therefore we have vp(G) = vp(H) = 0 for any pOn the other hand, for any prime I \ e,

+ yv d = x ^ 0 (mod /) and a + by d = a ^ 0 (mod /).

11

It is easy to see that

Tp7 = 1 if p j q(u + yd).

Now suppose p | q.

If I — 1 = 1 , then qOp = qq. We have

W— G s _ G s

rq7 = — • - (mod q), rq7 = — • - (mod q).

G2 s2

For p = q or q, since — - • — = 1 (mod p), we see that

— • ^ = ±1 (modp).

We claim that if — • — = c (mod q), then — • — = —c (mod q), where c = ±1.H r H r

In fact, if both — • - = c (mod q) and — • - = c (mod q) hold, thenH r H r

G s / j \ , 2wmg + 2ne\/d s— • - — c (mod q), hence, -= • - — c (mod q).H r -2wng + meyd r

It follows that

—2cwngr = 2wmgs (mod q), cmer = 2nes (mod g).

This is the same as to say that

—cnr = ms (mod g), cmr = 2ns (mod g).

Hence, we have m + 2n = 0 (mod g). This is impossible since e = m + 2n and (e, g) = 1. So,we can assume that rq i7 = —1 and rq27 = 1, where qiq2 = qOp.

'd\ , „ „ „ , . , . G2

If g = 7 (mod 8), then I — I = — 1. Hence, gOi? is a prime ideal of OF- In this case, —r = —2\qj Hl

2 1 r<

(mod qOp) and — = — (mod qOp). So, if we write i = — • —, then TqoF~i = i (mod qOp) with

i2 = —1 (mod qOp). The lemma is proved.

In view of Lemma 2.5, we assume that d £ 7L square-free, m \ d and emZ2 = X2 + dY2 issolvable for e £ {1, 2}. Let Xo, Yo, Zo £ Z with (Xo, Yo) = 1 be a solution. Put

emZ2

By Lemma 3.2, p4{m, 2}2 = { — 1, m}. We need to compute the tame symbol of p.

Lemma 3.10. With the same notations and assumptions, we can assume

TPP =

-4;— ( m o d P), tf p \ Zo;

V Xo J2 (mod p), tfp\m;1, otherwise.

12

Proof. cmZ2 = X2 + dY02 and m \ d imply that m \ Xo, let XQ = mX'o. We can assume that

(d, Zo) = 1, hence, (m,Yo) = 1. If p \ m, then vp[Xo) = 0 (mod 2), and

TnP = \ — J-f 7~77—T * Z I I I 1 O Q» ' v ' / \ — l-\-V p(A.o)

o) (mod p)

o) (mod p)

' o ) (mod p)

= 2 (mod p).

We have used the fact that eZl = mX0 -\ Y02.

For p | Zo,

(mod p)

(

Tf rf

+ u (xo-YoVd\If p —, then vp — = 0, hence, rpp = 1.

m \ XQ Ili p \ d,p \ Zo, then it is easy to see that Tpp = 1. This completes the proof.Now we are going to deal with the case when 2 6 NF. In this case, d = u'2 — 2w'2, u, w 6 Z and

for any odd prime divisor p of d,p = ±1 (mod 8).Suppose that m \ d and em(u+w)Z'2 = X'2+dY'2 is solvable, where e = 1 or 2. Let Xo, Yo, Zo £ ^

be a nontrivial solution. Put, v Xp - wYph = Y0, g = .

u + w

By t h e choice of Xo, Yo a n d Zo, we can a s s u m e t h a t g,h £ Z w i t h (g, h) = 1. As in [16] a n d [17],wr i t e

a = g2 + h2, 6 = (g2 - h2 + 2gh)w, A = (g2 - h2 - 2gh)w.

We have the following identities

au + e = mZ2, X2 + 92 = 2a2w2.

Letx = mZ2(g + h) + X(g - h), y = a(g + h);

a = mZ2(g - h) + X(g + h), b = a(g - h).

13

Then

And

Let

Then

Put

Then

x2 + a2 + (y2 +b2)d

= m2Z%(g + h)2 + X2(g - h)2 + m2Z\{g - h)2 + X2(g + h)2

+ a2{g + h)2d + a2{g-h)2d

= 2m2Z^a + 2X2a + 2a3(u2 - 2w2)

= 2a(rn2Zt + A2 + 2a2u2 - 2a2w2)

= 2a{m2Z%-92 +a2u2)

= 2a(au + 6)(2au)

= mu(2aZ0)2.

2{xy + ab) + AmZ2a2 = m(2aZ0)2.

E = x + yVd, F = a + bVd.

E2 + F2 = m{u + Vd){2aZ0)2.

2EF F-E'ib =

F{2,F}{aZ0,-2}.

tp4{m(u + \fd),2}2 = {-1, (u + Vd)m}.

Continuing to use the notations as above, we have

Lemma 3.11. By a suitable choice of Xo,Yo, Zo, a solution of the Diophantine equation

e(u + w)mZ2 = X2 + dY2,

we can assume thatE~F

2,1,

ifp\m;otherwise.

Proof. We can take e = 1, since 2 6 NF. Rewrite m(u-\-w)Z2 = X2-\-dY2 in the form (u-\-w)Z2 =

mX' -\ Y2, where X' = —X. Applying Corollary 2.3, we can assume that (dw,Zo) = 1. Them m

assumption (g,h) = 1 implies that (a,Zo) = 1. In fact, if there is an odd prime / | (a,Zo), then/ | 6, since au + 6 = mZ2. But / j w, so we must have / | (g2 — h2 — 2gh). On the other hand,g2 - h2 - 2gh = a - 2h(g + h). Hence, / | 2h(g + h), therefore, / | h or / | (g + h). If / | h, thenI \ g, contradicting the assumption (g, h) = 1. If / | (g + h), then we would have / \ g — h, since(g + h)2 + (g — h)2 = 2a2. Then / | (g, h), is also a contradiction. The above discussion also showsthat (a,w) = 1. The identity x2 + a2 + (y2 + b2)d = m(2aZo)2 implies that massume that for any p \ m,

mm(vp(E),vp(F)) = vp(F - E) = 1.

Hence, in any case, for any p \ m, we can show

X. So we can

= 2.

14

Note that x2 - y2 d = a2 -b2d (mod w) and E2 + F2 = m(u + yrd){2aZ0)2. Hence, we obtain that

for any p | (u + vd),

im-n(vp(E),vp(F)) = -vp(u + Vd).

Moreover, we can assume that vp(F — E) = —vp(u + \/~d). Hence, we can show that if p

then

As an illustration, we suppose vp(F) < vp(E) = vp(F — E) = —vp(u + yd), then

/ 2EF

F

E2 + F2

2E(F - E)

• 2U^F) (mod p)

U^F ) (mod p)

Since (a, ZQ) = 1, a | y and a \ b, we do not need to consider any p | a.Finally, for any p | ZQ , we have

1 — _F\ "(mod p)

F

-2F2 )

(mod p).

(mod p)

Remark 3.11.1. With the same notations as above, we discuss more about the tame symbol of ip.

Clearly, ( — j = - 1 (mod p). It follows from a2d + X2 = a2u2 - 62 - 2a2w2 + X2 + 62 = 0

(mod ZQ) and (a, ZQ) = 1 that (A, ZQ) = 1. Suppose that p | ZQ and i £ Z with i2 = —1 (mod p).If J>O_F = pp, one can verify that it is impossible that both Tpip = i and Tpip = i hold at the sametime. This fact will be used later.

Lemma 3.12. Let F = Q(\/d),d £ 7L square-free and let a £ K2F with a4 = 1. If for anyp £ il, Tpa = ±1 , then a = (3*f with /32 £ A ^ O F and j 2 = 1.

Proof. By the assumption, we see that for any p £ i l , r p a 2 = 1, hence, a2 £ A ^ O F - Since( a 2 ) 2 = 1, we get a2 = { — 1, m } , where m | rf or m = n(u + \fd) with n \ d. Let f32 = { — 1, m } .Then a 2 = /32 . Hence, a = (3*f with 7 2 = 1 as desired. This completes the proof.

Let { — l,g} G A ^ O F with { — 1,^} = j 4 . Then for any non-dyadic place p £ il, ( r p 7 ) 4 = 1(mod p) . When we consider the problem if { — l,g} £ \ / , i n most cases, we can choose 7 withTp7 = ± 1 . In this case, we can give a criterion for { — l,g} to be in V or not. So we give thefollowing

Def in i t ion 3 .13 . Let { — l,g} £ A ^ O F with { — l,g} = 7 4 . We call a case the normal case if forany non-dyadic place p £ Q, TPJ = ± 1 .

For convenience, we introduce the following notations. For any square-free integer d and i =1, 3, 5, 7, denote by rf8- the product of all prime divisors of d which = i (mod 8). Note that rf8- = 1if and only if d has no prime divisor = i pmodH. So if d is odd, then \d\ = did^d^dj, if d is even,then \d\ = 2didsd$d7.

With several preparatory results, we are now in a position to give our main theorem.

15

Theorem 3.14. Let d be a square-free integer and F = Q(y/d), and let m \ d. Write m =±i7iii7i3ni5i7i7 with mi | rf8- for i = 1, 3, 5, 7. Assume there is an e G {1, 2} such that

(3.6) emZ2 = X2 + dY2

is solvable, and let Xo, Yo,Zo £ H with {Zo, d) = 1 be a solution of (3.6).(A) Suppose that 2 (jz NF. Then { — 1, m] £ y if and only if for i = 1,3,5,7, there are h{

in particular, hi = 1 is permitted, and e G {±1, i 2 } such that for any odd prime I \ d,

d \eh3h7m5Z0

\, if I I m3,1 \h3 or I \ hih$, I \ m$;

h3h3h7 , if I I n i 3 , l I h 3 ;

2eh3h7 Zom5 | h5,l | m5

/m 3 hih 5 \ _ (eh3h7m5Z0\

" V i Je——m5Zo

h3h7

, ifl\ hih3h5h7m3m5;

, if I I h3h7,l\m3;

2eh3h7 Zof^ I ,ifl | m5,l\h5.

(B) Suppose that 2 e NF.

(1) Then { —l,m} £ y 4 tf and only if for i = 1,7, there are hi \e G {±1} such that for any odd prime I \ d,

h7Zotflihlhr>

t d \

,ifl\h

(hi = 1 is permitted) and

= (— ^ifllh,

ehi{u + w)\ _ I'h7Zp

)

,ifl\h7

16

= (h^\ if

(ii) Suppose that there is an e 6 {1,2} such that

(3.7) em(u + w)Z2 = X2 + dY2

is solvable and let Xo, Yo, Zo £ H with (Zo, dw) = 1 be a solution. Then { — 1, m(u + Vd)} £ y 4 ' /and only if for i = 1, 7, there are hi | rf8- (hi = 1 is permitted) and e £ {±1} such that for any oddprime I \ d,

h7uZ0

d 7

V7uZo

, tfl\hih7;

,ifl\h7

andd\

= ( —j )

ehy (u •

h7

7uZ0h7

andId. .

e—- (u + w)hi

h7uZo

\

Proof. (A) We divide the proof into the following cases:Case 1. d£l (mod 8).By Lemmas 3.2 and 3.10, we know that

where p 6 K2F with

TPP= S

hi.

2 (modp),

1

(mod p), if p I Zo;

if p I m;

otherwise.

By Lemmas 3.5, 3.6 and 3.7, we see that

{m,2} = /34,

17

where /3 £ K2F. Let UI5 = pi-.-Pj be the primes factorization. Then

2'if p I mim5m7 ;

- 1 , if p = q,-;

. 1, otherwise.

Here, q8q8 = qiC>F and g8 are primes corresponding to p8 as in Lemma 3.7. Let q = q\...qj. One cancheck that there is a prime p = 1 (mod 4) and eo = 1 or 2 such that copZoH2 = S2 + dT2 is solvable.

I T //-/ *?2 I rl^^ 1Let 50 , To, i?0 with (50 , To) = 1 be a solution. And let £ = <̂ ~^—, o2 ° > { - l , - f f 0 } - It is

So 'immediate to verify that ^4 = 1 and

-to1

So, if

So. 1,

if p I p;

otherwise.

Since p = 1 (mod 4), p = a2 + b2, where a,b £ 7L. Note that for any p

2

(T0Vd\= —1

(mod p), and for p \ p,TpVd

So ,

V 5o 7= - 1 (mod p), ( - J = - 1 (mod p). Let 77 = j - , — | and let

7 = /?/?£ 1?y. Then 74 = { — 1, m}. We may assume that

Now put the above result and Lemma 3.12 together, we see that { — 1, m} £ y if and only if fori = 1,3,5,7 there are hi | rf8 and e\ 6 S(d) such that

(3.8) elPqm3rZ2 = X2 - dY2

is solvable, where r = 1 (mod 4) is a prime such that

(3.9) e2h1h3h5h7rZ2 = X2 + dY2

is solvable for some e2 G S(—d).

But (3.8) can be rewritten as cipqrZ2 = m3X2 Y2. Since for any odd prime / | d,

UI3

if/ /11/13/15/17,

/ d \

\

/ 7 N

and f — J = I —-— ), it is no problem now to check that (3.8) is solvable if and only if the conditions

in (A) hold.

18

Case 2. d= 1 (mod 8).One can easily see that the above method works also for the case d < 0 (d = 1 (mod 8)), since

there is a prime q with ( — 1 = 1 such that coqm^Z2 = X2 — 2dY2 (CQ = 1 or —2) is solvable and\PJ

there is a prime r equivl (mod 8) such that e\Z$rZ2 = X2 + dY2 is solvable for e\ = 1 or — 1.So we assume d = 1 (mod 8) with d > 0 below. There are six possibilities altogether. We need

to consider the cases one by one.( Z \

(PI) m5 = 5 (mod 8), ( -j- ] = - 1 .

= l,d3d7= 1.

= 1.

= -l,d3d7^l.

(P2) m5 = 5 (mod 8), ( -j

"'Zo(P3) m5 = 5 (mod 8),

(P4) m5 = 1 (mod 8),

(P5) m5 = 1 (mod 8),

(P6) m5 = 1 (mod 8), ( -j- = - 1 , rf3rf7 = 1.

Remember that we have an element p 6 A'2-F with

TpP =if P

otherwise

such that { — 1, m) = p4{m, 2} 2 .

If ni5 — 5 (mod 8), then by Lemma 3.7 we have a prime q with I — I = —1 such that qm^Z2 =

X2 — 2dY2 is solvable, hence we can find an element /3 £ K2F with

1

i{i2 = - 1 (mod p)), if p = qOF;

1, otherwise,

such that /32 = {ms, 2}.

If TO5 = 1 (mod 8), then also by Lemma 3.7 we have a prime q with ( — 1 = 1 such that

qm^Z2 = X2 — 2dY2 is solvable, and we can find an element j3 6 K^F with

if p I m5;12'- 1 , ifp = q;

1, ifp = q(qq =1, otherwise,

such that /32 = {m^, 2}.For (PI), there is a prime r = 1 (mod 4) such that rqZoH2 = S2 + rf?12 is solvable. Let

So,To, Ho £ Z with (So, To) = 1 be a solution and let

19

where a, b £ 7L such that a2 + b2 = r. We have £4 = 1 and

\ So

T0VdSo 5

- 1 ,

1,

1,

if P I Zo;

if p = qOF;

if p = t;if p = t, where ttotherwise.

It is clear that for p

withSo

rp7 =

= — 1 (mod p). Hence, we can find an element 7 6 K2F

- 1 , i fp = t;

1, if p = t(tt =

1, otherwise

such that 74 = { — 1, m}. This is the normal case and the same argument just as in the case d ^ 1gives the desired result.

For (IP2), there is a prime r = 1 (mod 4) such that rqZop'H = S + dT is solvable, wherep' I ^3^7 is a prime. Constructing £ as in (PI), one sees that the £ has the same property. Sowe have returned to the similar setting. We point out that in this case, /J3/J7 in the theorem willbe hsh~/p', if we use the same notation as before. But everything remains valid, since /J3/J7 in thetheorem is arbitrary.

For (P3), there is a prime r = 1 (mod 4) such that

rZ0H2 = S2 + dT2

is solvable. Then a similar process shows that there is an element 7 £ K2F with

i{i2 = - 1 (mod p)), if p = qOF\T P 7 =

±1, otherwise.

such that 74 = { — l ,m} . Hence, { — l ,m} £ y if and only if we can find an element r\ 6 K^Fsatisfying that for any p 6 £1, r p 7 = rp?y and ryt = 1. By Lemma 2.8, there are x,y E F* such that77 = {a;, x2 + !}{- ! , y} since rf = 1. Note that

Tvf] =- 1 , if p = qOF;

1, otherwise.

Therefore, rp(?y2{ —1, g}) = 1 holds for any p 6 £1. In other words, ?y2{ —1, g} 6that there is an

. This impliesd such that ?y2{ —1, q) = { — 1, m}. But, r/2 = {x2, x2 + 1} = { — 1, a;2 + 1}.

Hence, {-l,q(x2 + 1)} = { - l , m } . It follows that q(x2 + 1) = m/ 2 or 2m/ 2 , where / £ F*. So gis the sum of two squares in F. Since d = 1 (mod 8), there are two dyadic places in £1, say, q, q.

) ( ^ ) = — 1, since Fq = Fq = Q2- Hence, it is impossible that q is theWe have

\J4.sum of two squares in F. So, { — 1, m)On the other hand, one can see that the conditions in (A) read as the following in this case.

(Al) if/ I hih5,l \ m5.

20

(A2)2e Zo

if / | /15, / I UI5

(A3)f

if/ \ hih5,l \ m5

(A4) if / \ h5,l I m 5 .

Let m5 = TiT-2 and = S1S2, where Ti = (m5,h5),T2 = -=-,Sim5 Ti

We get from Al, A2, A3 and A4 the following

( , hih5) andm5

(Bl)

\

Si

_ /em5Z0\

~ \ Si J

(B2)

d \

m5

(B3)em5Z0

S-2

(B4)

respectively.

Putting the four identities together and using S\Ti =

\ / d \

\

^ and S2T2 =

t \d

\m5 )

, we obtain

/ v \Clearly, the left hand side is 1. But the right-hand side is —1, since d = 1 (mod 8), ( —— ) = 1

\ a jand I I = — 1. This gives the desired result.

\m5jClearly, (P4) is the same as the case d ^ 1 (mod 8).(P5) is the same as (P2). In fact, there is a prime r = 1 (mod 4) such that rqZop'Z = X -\-dY

is solvable, the rest is the same as before.(P6) is the same as (P3). More precisely, in this case, { — 1, m) £ XJ (but there is 7 6 Kp such

that 74 = { — 1, m}) and at least one of the conditions in (A) fails to be true.

21

(B) 2We mention that if d = u2 — 2w2, then the fact (3A = 1 with /32 £ will imply that

f32 = { — 1, TO} with TO | d or f32 = { — l,n(u + \fd)} with n \ d. We have dealt with the case wherej32 = { — 1, TO}. If/32 = { — 1, n(u-\- vrf)}, then using Lemma 3.12, we see that there is an a 6 K^OFwith a 2 = /32 and

1, if p = p(pp = POF)',

1, otherwise,

where p = 1 (mod 4) with ( — 1 = 1 being a prime such that pn(u + w)Z2 = X2 + dY2 is solvable.\PJ

It is obvious that rf3 = rf5 = 1, since 2 £ 7V_F. Let TO | d with TO > 0. Then we always have{TO, 2} = /32 , where j3 e A'2-F with

r 13= { 2' if P ' m'I 1, otherwise.

We also have {u + \fd, 2} = 72 . For the tame symbol of 7, see Lemma 3.9 .Let us see each case below.Case 1. d < 0 or d ^ 1 (mod 8).In this case, there is a prime r = 1 (mod 8) such that erZoZ2 = X2 + dY2 is solvable for

e = 1 or — 1. On the other hand, we have {u + \fd, 2} = 72 and for any non-dyadic place p, if

p \ (u + \fd), then r p 7 = - , otherwise ±1 . Hence, the discussion in the case where d ^ 1 (mod 8)

with 2 ^ 7V_F is also valid here.Case 2. d> 0,rf= 1 (mod 8).First, we consider if { — 1, n(u + v d) 6 V .We have six possibilities:

«) ( f ) - G)=-'•

(P'l) is the normal case.For (IP'2), there is a prime r = 1 (mod 4) such that rqZ^Z2 = X2 + dY2 is solvable, this is also

the normal case.For (P'3), there is a prime r = 1 (mod 4) such that rqZ0d7Z

2 = X2 + dY2 if d7 ^ 1 orrqZo(u + w)Z2 = X2 + dY2 if d? = 1, is solvable, the former is the normal case. For the latter,constructing E, F as was done just before L emma 3.11, we have E2 + F2 = rqZo(u + \/d)(2aZ)2.

Let 6 = I j , E t,2F I {-1, a 2 } {v , 7 j } , where a, & £ Z such that r = a2 + b2. Then 6>4 = 1.t,2

We may assume that

7 j

p

- 1 ,

P\ZO;

if p = i (xx = rOF)

otherwise.

22

f E\Since I — I = — 1 (mod qZo), we get an element 7 6 K2F with Tp-f = — 1 only for p = t,

\FJ

E—F

otherwise 1 such that -f4 = { — 1, n(u + vrf)}- This is again the normal case.(P'4) is the same as (P'3).(P'5) and (P'6) are the same as (P3).Finally, we study if { — 1, n) £ y 4 - There are three cases:

(P''2). Uf\ = -1, either d7 ± 1 or (^f) = -1-

Clearly, (P"l) is the normal case. Regarding u + \/d disappears, we easily see that (P"2) and(P"3) are the same as (P'4) and (P'5) respectively.

Now, we complete the proof of our theorem.

Corollary 3.15. Let F = Q(\/d),d £ 7L square-free be a real quadratic field, and let e be thefundamental unit of F. If Ne = —1, then in Q(\/—d), { — 1,-1} £ y 4 ; hence, if the 8-rank ofI<2Omv^1) = 0, then {-1, - 1 } = 1.

Proof. It follows from Ne = — 1 that X2 — dY2 = —4 has a nontrivial solution in Z, hence, we cantake ZQ = 2. In view of Theorem 3.14, one sees that our result follows.

Remark 3.15.1. The converse of the above corollary is also true if d is odd, more precisely, wehave shown that if the 8-rank of K^O^i^zrd) = 0> then { — 1, —1} = 1 in K2Q(V—d) if and only ifNe = — 1. See [18] for more in details.

4. THE 8-RANK OF K2OF AND THE TATE KERNELS OF IMAGINARY QUADRATIC FIELDS

In this section, we apply Theorem 3.14 to the imaginary quadratic fields case. We computethe 8-rank of A ^ O F in some cases. For a given number field (not totally real), it would be aninteresting problem to give an explicit structure of the Tate kernel. In [16], we have done this forsome imaginary quadratic fields. Here we will also deal with this problem in some new cases.

Recall that for a number field F, the Tate kernel of F is defined to be A = {a £ F* \ {-1, a} =1}. For an imaginary quadratic field F, we know from [20] that [A : F*2] = 4. Hence, A =F*2U2F*2USF*2U2SF*2. So it is enough for us to find such Se F*.

It is convenient for us to fix some necessary notations.

u, w : u,w e Z , u ! — 2w2 = d.

v : v = u + w.

r2» : 2 n - rank of K2OF(n £ N.)

Z_i : -Z-x2 = X2 +dY2, where X,Y,Z^ £ N with (X,Y) = 1.

Zv : vZ2v = X2 + dY2, where X, Y,ZV £ H with {X, Y) = 1.

Theorem 4.1. Let F = Q(\/—d) be an imaginary quadratic field. Then we have the followingTable 1.

Table 1

23

d

—p (p = 1 (mod 16))

-p (p = 9 (mod 16))

- p (p = 7 (mod 8))

-2p (p = 1 (mod 16))

-2p (p = 9 (mod 16))

—2p (p = 7 (mod 8))

1

1

1

1

1

1

r4

1

1

0

1

0

0

?*8

1

0

0

0

if(zI7) = - l

0l f ( ^ ) = -1>(zT7) = 1

1i f(£) = (zir) = 1

0

0

Tate Kernels (S)

- 1

u + v dwith u -\- w = 1 (mod 4)

(f)u + Vd

- 1

- 1

Remark 4.1.1. Let p be a prime, F = Q(i/—p). The following result in the above Theorem

1, i f p = 1 (mod 16);

0, ifp = 9 (mod 16)

was first conjectured by J.Hurrelbrink.

Remark 4-1-2. For the convenience of readers, we list the known results for r2 and r4 ( see [1] and[4]). It is an easy consequence of [1] that r2 = 0 for d = —p or —2p with p ^ ±1 (mod 8).

Proof. We know that in all our cases K^OF can be generated by { — 1,-1} and { — 1, u + \/d).If d = —p or —2p with p = 7 (mod 8), then — Z2 = X2 + dY2 is not solvable in 7L, hence,

{-1, - 1 } £ v 2 , so r4 = 0.If d = —p or —2p with p = 1 (mod 8), then we have X, Y, Z_\ G Z< with (X, Y) = 1 and

(4.1) X2 +dY2 = -Z_i2.

Let us investigate the two cases individually below.The case d = —p, p = 1 (mod 16). It follows from — p = u2 — 2w2 that — p = (u-\-w)(u — w) — w2 .

Hence, we have f - 1 = 1 . So X2-pY2 = vZ2 is solvable, and we let X, Y, Zv <E Z with (X, Y) = 1

be a solution. By Corollary 3.15, we have { — 1, —1} G V • O n the other hand, — p = u2 — 2w2

24

implies that u = ±3 (mod 8) if p = 9 (mod 16) and u = ±1 (mod 8) if p = 1 (mod 16). Also from

-p = u2 - 2w2, we see that ( - ) = ( ^- ). Hence, ( - ) = - 1 if p = 9 (mod 16) and ( - ) = 1 if\PJ \\U\J \PJ \PJ

p = 1 (mod 16). Note that we always have I — I = I — 1 = 1. So we obtain from Theorem 3.14\Pj \P J

the desired result for this case.

The case d = —2p,p = 1 (mod 8). It is clear that { — 1,—1} 6 V • It follows from —2p = u2 —2w2

tha t 2 I u. We have p = 1 (mod 16) if and only if I I = 1, equivalently, { — l , u + vrf} £ V\ | M + w J

and p = 9 (mod 16) if and only if I I = —1, equivalently, { — l , u + vrf} ^ V . Let\\u + w J

= 2M'. Then -p = 2M'2 - w2. So, (—) = 1. Furthermore, ( — ) = ( — ) = 1. IfVM'/ \p J \ p J

— ) = —1, then vZ2 = X2 — 2pY2 has no nontrivial solution in 7L. Hence, { — 1, u + \fd) (£ \J4.v\J

If (^—] = 1, then vZ2 = X2 - 2pY2 is solvable in Z. Hence, if ( — ) = ( — ) = 1, then\\v J \ZV) \Z-iJ

4 a n d { l u + V d } e V 4 ; i f ( | ) = 1 \ ) = X t h e n i 1 1 } t V 4{-1,-1} E V4 and {-l,u + Vd} e V4; if ( | " ) = 1\Y~) = " X ' t h e n i " 1 ' " 1 } t V

but {-1, M + Vd} £ V4, thus u + Vd is in the Tate kernel; if ( — ) = - 1 , ( —— ) = 1, then\ZV J \Z-iJ

/ o \ / 2 \{-1, -1} G V4 but {-1, M + Vrf} £ V4, thus - 1 is in the Tate kernel; if — = = - 1 ,

\ZV J \Z-iJthen {-1, -1} , { - 1 , M + \fd] g XJ4 but {-1, -u - \fd] £ y 4 ; thus -u - \fd is in the Tate kernel.

This completes the proof.

25

5. THE 2-SYLOW SUBGROUPS OF K2OF FOR REAL QUADRATIC FIELDS

Theorem 5.1. With the same notations as in section 4. Let F = Q(\/d) be a real quadratic field.Assume u > 0. Then we have the following Table 2.

Table 2

d

p (p = 1 (mod 8))

p {p = 7 (mod 16))*

p (p = 15 (mod 16))

2p (p = 1 (mod 8))

2p (p = 7 (mod 8))

r2

3

2

2

3

2

1if M + w = 1 (mod 4)

0otherwise

1

1

1

0otherwise

1

1otherwise

0otherwise

0

1

1

1

0otherwise

0

1

0

?*16

0

0

0

1

0

0

1

0otherwise

0

Remark 5.1.1. The results for r2 and r4 are known. In particular, r2 = 2 and r4 = 0 if d = p or2p with p ^ ±1 (mod 8). For (*), see also [3], [22].

Proof. Clearly, { — 1, —1}, { — 1, U{ + v d} generate 2K2OF, where uf — d = C{w'f, v.{, W{, C{ G Z and

26

Ci G NF fl { — 1, 2, —2}. Of course, we only need to consider whether u + y/d £ y 2 o r V4; where9 79 t\ 9

u — a = 2w .The case rf = p = 1 (mod 8).

If I — I = —1, equivalently, i> = 3 (mod 4)(v > 0), then { — 1, u + \fd\ 4 y 2 , hence, r^ = 0.

Suppose that ( - ) = 1. If ( — ) = ( — ), then ( - ) = ( — ), hence, {-l,u + \/d} G y 4 ,\PJ \ u ) \Zv ) \Pj \ P J

so rg = 1, otherwise, { — 1, u + \fd} 4_ y 4 , hence, rg = 0.The case d = p = 7 (mod 8).

It is obvious that vZ2 = X2 + pY2 is solvable. Moreover, we have ( — 1 = 1 . It follows

V P Jfrom p = u2 — 2w2 that I — I = I I. Hence, if p = 7 (mod 16) then u = ±3 (mod 8), hence,

\uJ \ u Ju\ . .— I = —1, taking e = —1, we obtain from Theorem 3.14 that rg = 1 and V\Q = 0 since e < 0. Ifpj

fu\p = 15 (mod 16), then u = ± 1 (mod 8), hence, I — I = 1 , taking e = 1, we see that rg = 1 and

Wr16 = 1 since e > 0.

The case rf = 2p,p = 1 (mod 8).If (—) = - 1 , then ( - ] = - 1 , hence, {-l,u + Vd} 4_ y 2 , so r4 = 0.

\ v J \pj

Suppose ( f ) = 1. If (Zl) = ( ^ ( u , = lu), then ( H ) = ( ^ ) , hence, { - l , u + Vd} G

y 4 , hence, rg = 1, otherwise, rg = 0.The case rf = 2p,p = 7 (mod 8)

If I — I = 1, then v = ±1 (mod 8), hence, I — I = 1, since I I = 1. On the other hand,W \PJ \ v J

w e h a v e I — I = 1 . O n e c a n v e r i f y t h a t I —— I = I — I . H e n c e , i f I — 1 = 1 t a k i n g e = 1 , w e\pj \pj \ZV) \ZV)

( 2 \see at once that rg = r\§ = 1; if I — I = —1, taking e = —1, we get rg = 1 and r\§ = 0.

V Zv JThis proves the theorem.

ACKNOWLEDGMENTS

The author would like to thank the Abdus Salam International Centre for Theoretical Physics,Trieste, Italy, for hospitality, where this work was done. He is also grateful to Professors J.Browkin,J.Hurrelbrink, F.Keune, M.Kolster and A.O.Kuku for their valuable help.

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