On the representation of integers by binary cubic forms ... - Springer Link

Inventiones mathematicae

Invent. math. 73, 117-138 (1983)

9 Springer-Verlag 1983

On the Representation of Integers by Binary Cubic Forms of Positive Discriminant J.-H. Evertse University of Leiden, Department of Mathematics, Wassenaarseweg 80, Postbus 9512, NL-2300 RA Leiden, The Netherlands

w 1. Introduction Let F(x,y) be an irreducible binary cubic form with integral coefficients and negative discriminant. Delone [-3] and Nagell [-8] showed that the equation ~(x, y) = 1

(1)

has at most five solutions in integers x, y. This can not be improved for if F(x, y) equals x 3 - x y2 +y3 then F has discriminant - 2 3 and (1) is satisfied by (1, 0), (0, 1), ( - 1 , 1), (1, 1), ( 4 , - 3 ) . Delone and Nagell proved their result by considering units in the algebraic number field L = ~ ( e ) , where e is the real root of F(x, 1)=0. In both the proofs of Delone and Nagell the fact that the group of units in the ring of integers of K is generated by one fundamental unit is essential. Now suppose that F(x, y) satisfies the same conditions as above except that its discriminant is positive. Let L = ~ ( e ) where e is some root of F(x, 1)=0. We cannot apply the methods of Delone and Nagell since L has a system of two fundamental units. However, it is possible to reduce (1) to a set of exponential equations to which a local method of Skolem can be applied (cf. [-12], [13], [14], [6], [1], [-7] Ch.23). In this way, (1) was solved for the forms F(x,y)=x 3 - 3 x y 2 + y 3 of discriminant 81 and F ( x , y ) = x 3 + x Z y - 2 x y 2 - y 3 of discriminant 49 by respectively Ljunggren [6] and Baulin [2]. In the first case, the six solutions of (1) are (1, 0), (0, 1), ( - 1 , - 1), (1, 3), ( - 3 , - 2 ) , ( 2 , - 1) whereas in the second case (1) is only satisfied by the nine pairs (1,0), ( 0 , - 1), ( - 1 , 1), ( - 1 , - 1 ) , (2,-1), (-1,2), (5,4), (4,-9), (-9,5). Note that in the first case L = ~ ( e 2~r//9 + e -2'ri/9) and in the second case, L = ff~(e2'~i/v +e-2'ri/7). In the proofs of Ljunggren and Baulin, use is made of the explicit values of some fundamental units in a quadratic extension of L. So it does not seem easy to derive general results on (1) by their method. It is possible to derive more general, though ineffective, results on (1) by means of an approximation method in which hypergeometric functions are

118

J.-H. Evertse

used. In this way, Siegel ([91, [i1]) showed that the number of solutions of the inequality IF(x, Y)I _- 0 or (x, y) --- (1, 0), where k is a positive integer, is at most 18 if the discriminant of F is sufficiently large compared with k. In a student paper from 1949, A.E. Gel'man showed, by refining Siegel's estimates, that 18 can be replaced by 10. (For a proof of this, we refer to [4], Chap. 5). In particular this implies that (1) has at most ten solutions if the discriminant of F is large enough. We shall give a uniform upper bound for the number of solutions of (1). Theorem 1. Let F be a binary cubic form with integral coefficients and non-zero

discriminant. 7hen the equation F(x, y) = 1

(1)

has at most twelve solutions in integers x, y. As far as I know, no cubic forms F are known for which (1) has more than nine solutions and it is likely that our result can be improved. The method which we shall apply in the proof of Theorem 1 allows us also to prove a result on the number of solutions of (2). Before we can state it we have to introduce the notion of a reduced cubic form. For any cubic form F(x, y) its quadratic covariant H(x, y) is defined by

l {02F 02F H(x,y)= --~ \Ox2 0y 2

{ 0 2 F ]2] \OxOy! ] = A x 2 + B x y + C y 2

say, where A, B, C are certain polynomials in the coefficients of F. It is wellknown that if F has discriminant D, then H has discriminant B 2 - 4 A C = - 3 D and that H is positive definite if F has positive discriminant. Furthermore, H satisfies the covariance property, i.e. if a11, a12, a21, a22 are real (or complex) numbers with a l l a 2 2 - a 1 2 a 2 1 = l , then H(allx+a~2y, a21x+az2y) is the quadratic covariant of F(a 1~x + a 12 Y, a2 t x + a 22 Y). Definition. A binary cubic form of positive discriminant is called reduced if its quadratic covariant H is reduced, i.e. if C ~ A > [B[. Since every positive definite quadratic form is equivalent to a reduced quadratic form, the covariance property yields that every cubic form of positive discriminant is equivalent to a reduced cubic form. Theorem2. Let F be a reduced, irreducible, binary cubic form of positive dis-

criminant with integral coefficients and let k be a positive integer. 7hen the inequality IF(x, Y)I < k (2) has at most nine solutions in integers x,y with (x,y)= 1, y>_ 121/4 k a/2. In fact, b o t h Theorem 1 and Theorem 2 are consequences of Theorem 3. Let F be an irreducible, binary cubic form with integral coefficients,

discriminant D with D > 0 and quadratic covariant H. Let k be a positive integer.

Representation of Integers by Cubic Forms

119

Then the number of solutions of the inequality. IF(x, Y)I 5 k

(2)

in integers x, y with H(x, y)>=~(a D) ~/2 k a,

(x,y)=l,

y>0

or

(x, y) = (1, 0)

(3)

is at most 9. In the proof of Theorem3 we have used ideas from [4], Chap. 5, [9] and

[lo]. w2. Proofs of Theorem I and Theorem 2 In this section we shall derive Theorem 1 and Theorem 2 from Theorem 3. In the sequel, let F be a binary cubic form with integral coefficients, discriminant D and quadratic covariant H. Theorem 2 is an immediate consequence of Theorem 3 and the lemma below.

Lemma 1. I f D > 0 and if F is reduced and irreducible then

H(x,y)>=88

2

H(x,y)>=~(3D)~/2y 2

for x, yeT~, for x, y e Z with Ixl>12yi.

(4) (5)

Proof. We may assume that y ~ 0 . Put H ( x , y ) = A x 2 + B x y + C y E = y E f ( z ) , where z = x/y and f (z) = A z 2 + B z + C. Note that for z = - B/2 A, f (z) assumes a minimum on the reals which is equal to 3D/4A. Furthermore F, whence H, is reduced, hence A22,

f ( z ) > rain(f(2), f ( - 2)) = 4A - 2 [BI + C = 4 A + C - 2(4AC - 3 D) t/2 = :g(A, C). We shall minimalize g(A, C) on the (A, C)-plane. Note that (A, C) belongs to the area {(A, C)EN2:1 < A < C, A2>=4AC-3D>__O}, since H is reduced. On this area, g(A, C) is minimal in the point ((3D)1/2/3, 5(3D)1/2/6). This minimum equals 3(3D)1/2/2, which proves our lemma completely (cf. Fig. 1). [] We shall now prove Theorem 1. Suppose D + 0. Firstly we assume that F is reducible. Then there are integers a 1,a2,bl, b2,b 3 such that

F(x,y) = ( a 1 x + a 2 y)(b 1 x 2 + b 2 x y + b 3 y2). Hence solutions of (1) must satisfy

alx+a2y=+l ,

blx2-t-b2xy-bb3y2=-t-1.

120

J.-H. Evertse I

/

(1,z(3D*l)~I,"

//

A2=4.AC-3D

/ ofif, 4-~ i

AC=,3-DW

/

//

//

p

/

/

/

.

/

/

A=I

I (1.0)

A ---,'-

Fig. 1 Therefore, a~ b 1 x 2 -t-a 2 b 2 x(a 1 x -T-1)+ b 3(a 1 x-T- 1)2 = _ a~.

Since D 4=0, the latter equation has at most two solutions in x for each choice of the sign of a 1 x + a 2 Y . Hence (1) has at most four solutions. We shall now assume that F is irreducible. We shall also assume that D > 0, that F is reduced and that its coefficients of x 3 is positive. By the result of Delone and Nagell and by the fact that the number of solutions of (1) does not change when F is replaced by an equivalent form, these are no restrictions. It follows easily from Theorem3, that (1) has at most nine solutions in integers x, y for which H(x, y) ~ 3(3D)1/2/2. Note that by L e m m a 1, H ( x , y ) > 3 ( 3 D ) l / 2 / 2 if [xyl>2. Hence it suffices to show that (1) has at most three solutions with Ix y[ < 1. First of all, there can be at most three solutions with lyl= 1, Ixl < 1. F o r such solutions belong to the set of six pairs (x,y) with x e { - 1 , 0 , 1}, y e { - 1 , 1} and for each solution (x,y) belonging to this set, the pair ( - x , - y ) , which is clearly no solution of (1), also belongs to it. Hence we m a y restrict ourselves to the case that (1, 0) is a solution of (1). We may assume without restriction that the number of solutions with y = 1, txl < 1 is not less than the number of solutions with y = - 1, Ixl < 1 and moreover, that there are at least two solutions with y = 1, [xl < 1. If there are three such solutions, we have F(x, y ) = x ( x + y ) ( x - y) + ya = xa - x y2 + y a,

hence F has discriminant - 23 < 0. If we have two such solutions, then F(x, y) = (x - p y)2 (x - q y) + y3


121

for some p, q e { - 1,0, 1}, since by our a s s u m p t i o n that F ( 1 , 0 ) = 1, F(x, 1)= 1 can not have any non-integral, rational solution. If three are also solutions with y = - 1, then (x + p)2 (x + q) = 2, hence x + q = 2 , x + p = l , q = p + l . But then F is equivalent to the form x2(x _ y) + y3 = x 3 _ x 2 y + y3 of negative discriminant - 23. This implies that (1) has at most two solutions with [ y l = l , I x l ~ l if F ( 1 , 0 ) = I which completes the p r o o f of T h e o r e m 1. [] Note that there are infinitely m a n y reduced forms F for which (1) has three solutions with H(x, y) < 3(3 D)1/2/2. F o r take

F(x,y)=x3 + a x 2 y - ( a + 3)x y2 + y 3, where a is some fixed integer. Then F has quadratic covariant ( a 2 + 3 a + 9 ) ( x 2 - x y + y2) and discriminant (a2+ 3 a + 9) 2. It is easy to check that F is irreducible and that (1, 0), (0,1), ( - 1 , - 1) are solutions of (1). F o r these solutions one has H(x, y) = D 1/2. Note that (1) has at m o s t one solution with H(x,y)< 89 1/2 if F is irreducible and of positive discriminant. F o r by the covariance p r o p e r t y of H we m a y assume that F is reduced. Hence, by L e m m a l , H(x,y)>3(3D)l/2/2 if Ix Yi >--2. F u r t h e r m o r e , H ( _ 1, 0) = A, H(0, +_ 1) = C, H ( + 1 ; + 1) > A - {B[ + C and A - [ B I + C >_C>_(AC) 1/2 = 89

+ Ba) ~/2 _->89 D) 1/2.

If D exceeds some absolute constant, D o say, one can prove, similar to Theor e m 3 , that T h e o r e m 3 with k = l holds even if the lower b o u n d (3/2)(30) 1/2 in (3) is replaced by 89 ~/2. Hence for D>Do, (1) has at most ten solutions. In fact, one can show that D o can be taken as 5 x 10 l~ but we shall not work this out here.

w 3. Preliminaries to the P r o o f of Theorem 3

In the sequel we shall assume that F(x, y) is an irreducible, binary cubic form of positive discriminant D, F(x, y) = a x a + b x 2 y + c x y2 + dy3 say, where a, b, c, are rational integers. T h e quadratic covariant and the cubic covariant of F are defined by

H(x,y)= - ~

1 ,--'~ 02F \0X2 Oy 2

t 02F \O•

oF ou ]

G(x'Y)=Ox Oy

oF Oy Ox

respectively. Here

H(x, y) = A x 2 + B x y + C y 2, G ( x , y ) = a ' x 3 + 3 b ' x 2 y + 3 c ' x y Z + d ' y 3, where A, B, C, a', b', c', d' are rational integers given by

(6)

122

J.-H. Evertse

A=b2-3ac, B=bc-9ad, C=c2-3bd, a'=9abc-2ba-27a2d, b'=6ac2-b2c-9abd, c'=9acd+bc2-6b2d, d'=27ad2+2ca-9bcd. We have the following identity: (cf. [7], Chap. 24)

4 H(x, y) 3= G(x, y)2 + 27 DF(x, y)2.

(7)

Let V~-3D be a fixed choice of the square-root of - 3 D . Let M =Q(]f-S~ and let (9 be the ring of integers in M. We define a subset (90 of (9 by ( m + n l~-/ - 3 D e(9: m, neZ}. (9o=~ It is easy to check that (90 is a subring of (9, on noting that D=B2(mod4), i.e. D = 0 , 1 ( m o d 4 ) and that (m+nl/-Z~))/2e(9 if and only if m=-nD(mod2). For any complex number 2 we denote its complex conjugate by 2. We have clearly 2e(9 0 if and only if 2e(9, 2 - 2 e Z . l / - 3 D .

(8)

This implies that

121>__1Im21___89

if 2 e ( 9 o - 7/..

(9)

Put

G(x, y) + 3 ~ U(x, y ) =

F(x, y) 2

'

(10)

V (x, y) =

G(x, y) - 3 ~

2

F(x, y)

Then U(x, y), V(x, y) are cubic forms with coefficients belonging to M such that corresponding coefficients of U and V are complex conjugates. Since F must be also irreducible over M, U(x,y) and V(x,y) do not have factors in common. Moreover, by (7), U(x,y)V(x,y)=H(x,y) 3, hence U(x,y), V(x,y) must be cubes of linear forms, ~(x,y), q(x, y) respectively say, of which the corresponding coefficients are complex conjugates. On noting that r must be a quadratic form which is a cube root of H(x, y)3 and for which the coefficient of x 3 is a positive, real number, we have that

~(x,y)3-q(x,y)3= 3l,/- 3O F(x,y), (x, y)a + q (x, y)3 = G(x, y), ~(x, y),7(x, y) = U(x, y).

(11)

These identities imply that for any pair of rational integers Xo, Yo, the numbers ~(Xo,Yo),rl(Xo,Yo) are complex conjugate algebraic integers and, since H has


123

discriminant - 3 D, 4(x,y)

e. r . 4(l, -~ MIx, yJ,

n(x,y)

~em[x,

.

r

(12)

yl.

A pair of linear forms 4, t/with properties (11), (12) is called a pair of resolvent

forms 1. L e m m a 2. (i) 4 (x, y) 3 e (9o Ix, y], t/(x, y) 3 e (9 o Ix, y]. (ii) If (Xl, Yl), (x2, Y2) are two pairs of rational integers then

4(x~,yl) tl(xz, Y2)G(9O, 4(x 1, y~)2 4(x2 ' y2)G(90 ' ~/(Xl, yl)2 ~/(x2, y2) ff(9o"

Proof. Let 4(x,y)=elx+g2y. Clearly, el,e 2 are algebraic integers and so are e~, e2 ~2,el e2,~3. Note that by (6) and (10),

4(x,y)3=e3x3 + 3e~e2x2y+ 3el e~xyZ +e~y3=U(x,y) - a ' + 3 a l / - a ~ - D 2 x3+3b'+bl/-3Dxzy+3C'+Cl/-3D2

2

xY2

_~d' + 3d l~-3Dy3 2 Hence e3~,e2 e2, el ~, e3 belong to (9o. This clearly implies that 4(x, y)3e(9o Ix, y] and since the coefficients of t/(x,y) 3 are complex conjugate to the corresponding coefficients of 4(x, y)3 that tl(x, y)3~(9 o [x, y]. Furthermore this implies that 4(xl,yt) 2 4(x2,Y2) belongs to (90 since this number can be written as a linear combination with rational integral coefficients in e3~,e2%,e~e~,e 3. But then, t/(x ~, y a)2 t/(x z, Y2) = ~(x 1, Y~)z ~(x 2, Y2) also belongs to (9o. Finally we note that the linear transformation (x, y) ~ (4, t/) has determinant + If--- 3 D, since the product of 4 and t/is a quadratic form of discriminant - 3 D. Hence

4(x~, Yl) q(x2, Yz)- ~(xl, Y~)g/(x2, Y2) =4(Xl,Yl)rl(x2,Y2)--4(x2,Y2)tl(Xl,yl) = ~ ] f ~ ( x t

(13)

y2--X2 yl).

Therefore, by the fact that 4(x~,yt)rl(x2,y2) is an algebraic integer, that 4(Xl,Yx)/4(x2,Y2) belongs to M by (12), that 4(x2,y2)rl(x~,y2) is a rational integer and by (8), 4(xl,yl)rl(x2,y2) belongs to (90. [] Note that there are exactly three pairs of resolvent forms, for if (4, t/) is one pair, then (p~, p~ t/) and (p2 4, pt/) are the other two pairs, where p is a primitive third root of unity. Let (4:, ~), (~2, th), (43, t/a) be the three pairs of reOne can show that for xo,Yo~Z, ~(xo,Yo), r/(xo,Yo) are Lagrange resolvents, i.e. they can be written as co+pco'+p2co'', where co,co',co" are conjugate cubic numbers and where p is a third root of unity (cf. [4], p. 167). We shall not use this fact

124

J.-H. Evertse

solvent forms. We say that a pair of rational integers (x, y) is related to a pair of resolvent forms (r r/) if 1 - r / ( x ' Y ) = rain

~(x,y)

1

~__ 0

\

u

+m)...(r+ 1/2-g/2+m) (m+l)(2r+2-g+m) = m+l

gin+ 1 (r+ 1 - g + m ) ( r +2/3 gm

go=(-r-l~2+g/21(-1)~

( _ 1 ) . + * (-r-1/2+g/2~m+lI

, ,.(

!

128

J.-H. Evertse

We shall deduce (22) from (19) and (27). Note that by (19) with g = 0 and by (19) with instead of r and g = 1,

r+h

(30)

A,, o(Z) Br+h, l(z)--Ar+h, l(z) B,, o(Z) = z 2,+h P~,h(z)

Pr,h(Z). 2r+h.

for some power-series However, the left-hand side of (30) is a polynomial of degree at most Hence P,,h(z) is a constant. By inserting z = l in (30) and using (27), it follows that this constant is non-zero. Hence the lefthand side of (30) equals 0 if and only if z = 0 which proves (22). Finally we shall prove (24). It clearly suffices to prove that -''(a/3)(3~)nEOo

for each pair of integers a, n with n>0. This is obvious

(a/3)3'3"-~'/2 is an

for n = 0 and for n > l it suffices to show that

algebraic

integer. For this implies that for even n with n > 2

is a rational integer, whence belongs to (9o and that for odd n with n > 1

(a/n3)(3]/~-3D)"=(a/n3)3(3"-lv2(-D)~"-l'/zI/~-3D also belongs to C o. We shall now show that

u(n):=(a/n3)33"-x is

a p-adic

integer for each prime p and for each integer n with n > 1. First of all, if p ~e 3, then whence is a p-adic integer. Moreover, since

a/3,

u(n)

u(n)= (3 ( 3 - 1 ) . . . ( 3 - n + 1 ) / n [ ) 2 33"-l=(a(a -

3)(a-6)...(a-3(n-1))Z~(n0

and the number of times that 3 divides (n !)2 is equal to 2 j9~ = 1 the number

.=

~

--/'/,

u(n) is a 3-adic integer. This completes the proof of Lemma 3.

[]

w5. Proof of Theorem 3

F(x,y)

Let be an irreducible, binary cubic form with ficients and positive discriminant. Let (~, 7) be a pair constructed in w3. We have to show that at most three ing (3) are related to (~, r/) i.e. that there are at most

rational integral coefof resolvent forms as solutions of (2) satisfythree pairs of rational


129

integers (x, y) related to (4, r/) with

[4(x,y)3--tl(x,y)3[O

or

[r189

A1/4

(31)

(x, y) = (1, 0),

where A = 3D. It is k n o w n that 49, 81, 148 are the three smallest positive integers which are discriminant of an irreducible cubic form and that for these values of D only one equivalence class of cubic forms exists. 2 Hence by the results of Ljunggren and Baulin we m a y assume that D > 1 4 8 if k = l and D > 4 9 if k>=2, i.e. A_>_444 if k = l and A=>147 if k > 2 and we shall do so in the sequel. F o r each solution of (31) we have, by the fact that A1/4 kT/2> 4441/4> 23/2 3-1/2, 1

t/(x'y)3 < 3kA1/2

23/z

- < < 1. 4 (x, y)3 =14 (x, y)[3 = 31/2 A 1/4 k7/2

(32)

This fact will be used in the p r o o f of the lemma below. L e m m a 4. (i) I f (x, y) is a solution of (31) related to (4, rl) then 1-r/(x'Y) T l ~ ( x , y ) l

,

, z

9

(34)

Proof (i) I shall use an a r g u m e n t which was suggested to me by F. Beukers. By (32) and by the fact that I~/(x, y)3/4(x, y)31 = 1 we have [arg(rl(x,y)3/4(x,y)3)l444s/2,

3kA1/2

0 l , ge{0, 1}, and A,,g(z), B,.g(z) are the polynomials constructed in Lemma 3. Lemma 5.

Put t/2

r/1

r,,,g = 4~ A,,g(z 1) - 4-]- B,,g(z 1) a(r,g)= 2-zg/3 (2;) n, z(r,g)=2g/3 ~(3.1)2r+ l-g / r - g + l / 3 ~

r-

= a(r, g) kA g/314113"+1 -g 142]- 2, H,,, = z(r, g)k 1'+ 1-gAr-g/6[421 [411 - 3 r -

)/(

+l-

~~" r,g

2(1 - g )

If Z,,g #: O, then 1 --A l/2.

(41)

If g = 1 then we have, by Lemma 2 and (24), that A~,~O o. Furthermore, A3r For suppose otherwise. Then we have for some third root of unity p, that A~,I = p A,, 1. Hence by Lemma 2,

Zrl=4-(3~4~lpA~,l=~-(3~4~lpr13~rl (42 A t/3r (At, 1 (1

=P~;~

~13]

(1_~]_41B 41t/2 B

,~1-421t--~ ,,1

(1_413] )

(1__~]]

.~ IIeM(P) 9

Hence Z,. 1 has degree at most 2 over M. However, by (12) and Lemma 2, there exists a non-zero element 0 of M such that Z,.~=(rl(1,O)/4(1,O))O. Choose such that 4(a, 1)=r/(e, 1). Then, by (12), a~M(t/(1, 0)/4(1, 0)) = M(2;,, 1)


133

and by (11), F(~t, 1)=0. Therefore, by the irreducibility of F, ~ , 1 has degree 3 over M. Thus obtain a contradiction. Hence A ~. 3 1 can not belong to Z. This implies by (9) that }At, 11~(1A 1/2)113 Together with (41) this yields that for positive integers r and for g~{0, i},

IA,, g[> 2 -g/3 ,d 1/2 -g/3.

(42)

We shall now estimate IA~,gl from above. By (38), (39), (19), I1-z~[---1, (21), (32), (20), (38), (33) we have 'Ar, g' ~- ]r l l3r + l -g

1~211(~22- i ) A~,g(z,) + z~ ~+~-g F~,g(z,)

[r-g+1/g](r-1/3) -~ \ r + l - g !

3kA1/2]2~+a-g)

,, 3. , x ,

-~

. lr .

.

.

],

= 2 - g / 3 A1/Z-g/3( lg~r,g ~' +2/-/r,g)"

Together with (42) this proves our lemma.

[]

Lemma6. S,g+Ofor (r,g)=(1, 1), (1,0), (2,0), (3,0).

Proof First of all we shall show that Z~, 1~0. Suppose Z1,l =0. Then ?/2~ t Al,l(zl)=?/t ~{ ~Bl,l(Zl), i.e. /11 Therefore, 1

/'/t 2 /2

?/221 lr/3\ /2

1(1 1r

r/13~12 2 5 2(r

17/32 ,r/~

This implies that {,/~h 3 3 +?/a~/~3=2, where {~/n~=l, which is impossible by (11) and the irreducibility of F.

134

J.-H. Evertse

We now assume that Z , , o = 0 for some rs{1,2,3}. Let q*(r) be the smallest positive rational integer such that both q*(r)Ar.o(Z) and q*(r) B,,o(Z ) have rational integral coefficients. Put u = ~ 3, v=~a-r/3,1 A,(z)=q*(r)Ar, o(Z), B,(z) = q* (r) B~, o(z), and similarly for A* (x, y),/~* (x, y). Then, by our assumption,

~z~ _ (u- v)(/~* (u, v)) 3

~

u(/i* (u, v)) 3

Let ar be the integral ideal in M generated by u.4*(u,v) 3 and ( u - v ) B * ( u , v)3. Then the ideal generated by u(A*(u, v))3 - ( u - v ) ( B * ( u , v)) 3 divides (4 3-q23).ar. Hence by (31)

lul 3,+~ I~i,(z i) 3 - (1 - z0/~,(z 0 31 = Iu (~i* (u, v)? - (u - v) (/~* (u, v))31 < N(a,) 1/21~ 3 - t13[< N(Or) 1/2" 3 k A 1/2

(43)

where N(a,) denotes the absolute norm of the ideal a r. Here we used, that the norm of an element of an imaginary quadratic field is equal to the square of its absolute value. From (19) we infer that there exists a polynomial K,(z), with rational integral coefficients, such that A,(z) 3 - ( 1 - z ) B , ( z ) 3 = z z'+ l K~(z). Hence, by (43), IZllZr + l lg,(zl)l < N(a~)X/z lu[ -3~-a 3 k A1/Z=lzl]3, + x N(a~) uz Iv[ -3r-1 3 k A l/z. Therefore, by (31), 1 < N(ar)l/2 Ivl- 3r- 1 iztl r" 3 k A 1/2 < N(a,)U2 [v[- 3r-- 1 Ig,(z01

(3k A1/2) r+ l I~11-3r.

lgr(z01

This implies, by the fact that k 2 A >444,

{N(a,) 1/=Ivl-

I~1< I-

~

3,-1 )1/3,

3(r+t)/ar(444)(1-Sr)/6r'k3A"

A straightforward computation shows that, by (18), Al(z)=3-2z, .42(z) = 54 - 63 z + 14 z z, .43(z) = 81 - 135 z + 63 z 2 - 7 z 3,

Bl(z)=3-z, Bz(z ) = 5 4 - 45 z + 5 z z, /~3(z) = 81 - 108 z + 36z z - 2z 3, KI(z)=2-z, K2(z ) = 756 - 756 z + 125 z 2, K3(z) = 1 6 2 - 2 4 3 z + 97 z Z - 8 z 3.

(44)


135

Hence, by (38) [Kl(z0l > 1,

lKz(zt)l > 755,

IK3(zl) [ > 161.

(45)

We shall now estimate N(a,) from above. We choose a finite extension M' of M in which the ideal generated by u and v is principal, with generator w say. We put u'= u/w, v'= v/w and we denote the extension of a, to M' by a',. The ideal in M' generated by cq,c~2,..., % is denoted by (cq,e2,...,%). Put r, --, t --, = (A, (u,t v),B, (u,t vt )). N o t e that A,* (x, x - y) =/3" (y, y - x). This can be verified for r e { I , 2 , 3 } by straightforward computation but follows also from the proof of (23). Hence a; = w 3' +~ (u' A * (u', v ' p , (u' - v')/~* (u', v') ~) ~ W3r + l (ut, ut ~w3r+

r v3 - v )t( u ,tB ,- , (u,v) )(u P- v , pA ,--~g(u t,v')S)r~

(46)

I(u,,~*r (O,v,)3)(U , _ I ) , A ) r - *, ( v , v , 3 ) r r3

2DW3r+ 1 ( u ,t B , ( 0, ,

v')3)(u ' - v , ,B ,-(, O , v ), 3 ) r ,3= w 3 , + ~ /~*(0, 1)3 r,.3

We have the following identities: A* (x, y) - / ~ (x, y) = - y, (3 x - y) A* (x, y) - (3 x - 2 y)/~* (x, y) = - 4 y3, (27 x - 5 y) A~ (x, y ) - (27 x - 14 y)/3" (x, y) = - 72 x yZ,

A~(x,y)-;B~(x,y)=(-27x2 + 2 7 x y - 5 y 2 ) y , (3x-y)A](x,y)-(3x-2y)B](x,y)=(-6x

+ 3y)y 3.

These imply

rl ~(3u'-v',v')~(v'), r z ~ (72 u' v'Z,4v'3)~(72u',4v')v 'z ~72(v') 2, r3 ~ ( 6 u ' - 3v')v'3,(27 u'2- 27 u' v' + 5 v'2)v ') ~3v'3(2u'-v',27u'Z-27u'v' + 5v '2) = 3 v'3(2u'-v',27u'a-27u'(2u')+

5(2u') 2)

= 3 v'3(2 u ' - v', 7 u'Z) ~ 21(v') 3. Hence by (46),

a'l~(w4v'3)D(v) 4,

a~(3603wVv'6)~3603(v) 7,

a~ ~(423 w l~ v'9) ~ 423 (v) 1~ Therefore,

N(%)l/21vl-4~l ,

N(a2)l/21vl-74441/6, H1, l =3(1, 1) k 2 A 5/6 [42114l[-- 3 ~'L'(1, 1)(2 n) 1/2 k 5/2 A 5/61~11-1 _1. We shall now show that ft