Fundamental systems of units in bicubic parametric number fields

Arch. Math., Vol. 36, 524--536 ( 1 9 8 1 ) 0003-889X/81/3606-0013 $ 01.50 + 0.20/0 �9 1981 Birlda~user Verlag, Basel

Fundamental systems of units in bicubic parametric number fields

By

Gi~H~.R F~v.~ *)

0. Let D and d be integers with D > 0, d dividing D and d square f~ee and suppose tha t M = D 2 ~- 4d > 1. We introduce ~ and /~ as the two distinct roots of the quadratic equation

x 2 - D x - d = ( x - = ) ( x - 8 ) ; tha t is

+ f l = D , ~ f l = - - d, (~--fl)~-----D 2 + 4d = M .

We now set

= - = + =

e~ ,~- ~ , u ~ , ~ - =~ , ~ - f l

Then we proved in [3] tha t the set

is an independent system of units in the real algebraic number field Z~ n = K n (~) = Q(o~, ~) of degree 2n over the rationals Q.

Note tha t our ~] is the reciprocal of the ~ in [3] and tha t K= ---- Q(w). S (L2n) is a full independent system if n = 1, 2, 3, 4, 6. When S is a fundamental

system of units has already been investigated for the cases n = 1, 2 in [2]. We shaI] now s tudy the same problem for n = 3.

1. We suppose tha t M ~ D ~ ~ 4d > 1 be square free, Ma ~ D 3 ~ 3 D d > 1 be cube free and not a square, d i d , D > 0, d, D e E and (D, d) ~= (5, - - 5). I f we write D ---- A d , then M3 ----- })3 ~_ 3 D d = d2A (A~d ~ 3) -----/g2 with (], g) ~ 1, where / contains all linear prime factors and g contains all square prime factors of Ms. Hence dig . We now require in addition tha t A (A 2d -~ 3) be square free, tha t is d ----- g and / = A (A2d -Jr- 3). These conditions shall prevail throughout, except for the initial propositions 1 and 2 which we prefer to formulate in ~ more general form.

*) Prof. B. :L. van der Waerden zum 77. Geburtstag gewidmet.

Vol. 36, 1981 Fundamental systems of units 525

Then the conditions Ms > 1 cube free and no t a square can be replaced b y the condition / > 1.

2. P u t

Then

(~ - ~) (o~ - ~) s ~ where

- ~ f l - ~ + ~ ( ~ + ~ ) ~ - ~ + f12- 3 ~ + 3~(~ + ~ ) - (~ + ~)~ - ~ - 3 ~

ojs - - 3 w 2 D + 3 c o d ~ - - Da (co - - D)a

3Dd 3Dd

and we can conclude f rom Satz (I) in [6].

Proposition 1. Let M3 = D 3 --~ 3Dd > 1 be cube/tee and not a square and

M = D 2 - k 4 d > l , d iD, D > 0 .

Then

8 ~

is a/undamental unit in the ]/ek~ Ka = Q (~Ma). (Note tha t e > 0 (see [3])). I f we take into consideration t h a t / = A(A2d -k 3) is no t square free ff (D, d) -----

(3, - - 1) we get f rom theorem 2 in [2] the following

-rid \~[ D e ) Proposition 2. Let M ---- D2 + 4d > 1 and l--Y- -~ 3 both be square/tee, diD and D > O. Then

= - ~ i I d # 4-1 and (D,d) ~ (5, --5),

o=ly ,0

is a ]undamental unit in L2 = Q (~-M).

We pu t ( D - - ~ r U

e = - - = , u - - - - and ~-- // ~ e

Then we know from [3] t ha t S = {e, u, 0} is a full independent system of

units in L 6 = Q(}/MM, ] / ~ ) ---- Q(~, r Note tha t e > 0, u > 0 and hence ~ > 0 (see [3], 3.3).

"We are now going to show, b y using a method of t t . -J . Stender, t h a t {e, ~, O} and hence S is a fundamenta l system of units in L~, i.e. <-- 1, e, ~, 0} = E(L~) --- unit group of L 6.

526 G. FI~EI AECH. MATH.

3. We introduce the field isomorphisms a and ~ determined by

a: (0),c$)~->(9m,~) where 9 2 ~ - 1 = - 0 ,

~ : (0), ~) ~ (0), - ~ ) .

Hence ~(~,~)--- - ( : r "r(~r (~,:r a s = id, ~2 = id and ~ ~ ~a. F r o m L~ ~ = L2 = Q(5) and L~ ~> = K3 = Q(~o) we get the following norm relations (see also [3], corollary 3.1.2)

NLs/1,:s (e) = NLs/K3 (u) = ~,

NLs/L~(O) = 03,

- - 1 if d = + l NLs/K3(O)= + 1 otherwise ,

N~/L2 (~) -~- 1 = NLs/Ks (~), N~.~/K2(e) = N~:3/Q(e), NLs/Ks(O) = N ~ / Q ( O ) .

or (D,d) ---- (5, - - 5 ) ,

4. We shall first prove t h a t ~ is not the power of a uni t in Ls (proposition 7). To t h a t end we consider the integers a in K s ---- Q (co), where

ooS= M z = D S + 3 D d = / d 2 with / = A ( A 2 d ~ 3 )

square free, (/, d) = 1 and D = Ad. According to Dedekind (see [1], w 3 and w 4), t hey can be expressed in the form

O = l . ( x § 2 4 7 x,y, zeTZ,

where _ _ 0 ) 2 s _ _

0 ) = y ~//~d,

and x -- y ~ z ~- 0 modulo 3, i f ]2 ~ d 2 modulo 9, i.e. if M3 �9 i 1 modulo 9. Since, in fact , M3 ~ 0 modulo 9 if D -= 0 modulo 3 and M3 ~ 4- (1 -b 3d)

modulo 9 if D -= • 1 modulo 3, and since D �9 0 modulo 3 implies d m 0 modulo 3. we have Ms �9 -j= 1 modulo 9 and hence the integers in K s are of the form

a = x + y 0 ) + z ~ with x,y, z e F ,

t h a t is 1, 0), ~ is an integral basis of K s .

5. Suppose now t h a t ~ e o(L6) , i.e. t ha t 2 is an integer in Ls. We can write

2 = al ~- a2 (~, where a l , ~ ~ K8 and ($ = ~/]1 = ~/~-2 ~- 4 d.

I f ~ ' = 2 3 = 01-- ~26

is the conjugate in L6 with respect to K s , we get

2 § 2.' = 2 ~1 e o (Ls) n K 3 = D (Ks )


and (~ (2 - - 4 ' ) ~-- 2 ~2 M e {) (L6) (5 K 3 = 0 (K3).

F rom 4. we infer

2 0~ = y0 ~- Y~ w -b 0)~ with Y0, y~, Y2 e Z and

therefore

X2 0) 2 2 0~ M ---- xo + x~ 0) ~- -d - wi th xo, x~, x2 e 7] and

x2 y2 ) ~)' = ~61 + M 62 ---- �89 xo + xl o + ~ - ~2 § Yo ~ + Yl ~0) + ~ - ~0)2 .

F r o m the images of 5)` under G -~ (a , T}, i .e. f rom the conjugates of ~), we ge t

3 x~ --- - - (2 -~- ~ ) ' z + ~ 2 ~ _ ) / _ ~) ,~6 _ ~ 4 ~ ) , 0)

1

d~

d

and hence

where

A ' A I ~ t ~

= I)'t + 1261 + 14~'l + L)"l + 14~~ + 12~~

6. Suppose fur ther t h a t 4, ~ e E (Le) wi th

NL,/L2(~) ---- I L t K 3 (D = 1

and t h a t = 2 n wi th n ~ l .

Then ~ = (2x) n and hence 12 ~- ] ---- ~ V ~ for any ~ e G = ; a , T}. I f fu r the rmore > 1, then we have ] ~6 ] = ] ~o~[ < 1 and I ~" ] < 1 and therefore 146 ] ---- ] 2~ < 1

and 12" I < 1. Since I ~r61 = ]~ '~ I we obta in f rom 5.

< (~/~ + 2 ~ + 31.

528 G. F~.~I ARCH. MATH.

Altogether we have

Lemma 3. Let 2 and ~ be units in Le with ~ = ~n, n >= 1, ~ > 1 and

NL~/L2(}) = lVZ6/K3(}) = 1, and

~ = ~- x0 + xl Co + --~ o~ 2 + Y0 ~ + Yi do~ + ~o~ 2 , Xo, xi , x2, Yo, Yi, Y~ ~ 77.

Then

where

ly~l < D~ (]/~ -~ 2 j /~-~ I --~ 3)

(i = o, l , 2 ) ,

(i = o, l , 2 ) ,

i ~ Idl C~----(~D~ and D o - - 3 ' D i - 3 ( o ' D2 = 3m 2 �9

7. Fur thermore we shall need the

Lemma 4. Let ,~ e E (Le) with I Ls/Ks(~) = 4- 1 and

Then

(i) (x~, y2) ~ (0, 0),

M (ii) Ix2 [ -->--~- or x2 = 0,

(iii) i / (M, 3 ) = 1 , then ]x21>__M or x 2 = O .

xo, x l , x2, Yo, Yl, Y2 e "/7.

P r o o f . F r o m the norm relation

we get

(1)

(2)

(3)

x~ -+- 2 x i x 2 / d -- y ~ M -- 2 y i y 2 / d M ---- 4. 4 M ,

2xox l ~ ~ f -- 2yoyx M -- ~ M /---- O,

2 2 +-y~o~2 --y~ M-- y y o y ~ = O.


I f @2, Y~) = (0, 0), then we conclude from (3) tha t M ---- \~-1] which contradicts

the hypothesis tha t M be square free. This yields (i). To get (ii) and (lii) we first derive from (1), (2) and (3) respectively the divisibility

conditions

(4) M{ (X~o + 2xlx2/d),

(5) M I (2 XoXl + ~ f), (6) M l(dx ~ + 2x0x2).

From (4) and (5) we conclude

(7) M I ( ~ 2 / a - ~" ~ox2/). The conditions (6) and (7) yield

(8) M I 9 ~ x 2 / d

and

(9) M{ 9~o~/ ,

and from the conditions (4) and (8) we get

(I0) M{O~xz. By squaring (5) we obtain

M ] (36~x21 H 3 6 x o x z ~ / H 9x~] 2)

and hence with the help of (9) and (10)

Since M is square-free we have

F r o m M = d d H 4 a n d / = d H 3 we see that (M, /) = l or 3,

since D must be odd and hence , 4 = 1 ; therefore

M] 3x2.

M Hence Ix2{ > ~ - or x2 = 0; and in case (M, 3) = 1 we have {x21 > M or x2 = 0.

q.e.d.

8. In order to show that

<~> = Ez (Lr ---- {~ e E (L6) ] NL61Ka (~) =- "4- 1, NLs/L2 (~) "= 1}

we begin with proving the

Ard~iv der Mathcmatik 36 34

530 G. FREI ARCH. MATH.

Lemma 5. Ve, 1/$ ~ L6.

P r o o f . a) Since

U q~e 8

e e 2 - - e 2 '

we have t h a t y ~ e Le if and only if I/~ e Le.

b) I f $ .-= 1/~ e Le, then

Ls = K3 ($) = K3 (O),

t h a t is

~=~q-fl$ and ~'=~--fl$ with ~ , f l e K 3 ,

where 6' is the conjugate in K s (y ~) wi th respect to K s . Since 6 ,2 = 62 = M e K3 we get t h a t 8' = • 6.

I f 6' - - 8, t hen 6 = ~ and hence Ka (6) = K3 which cannot be.

I f 6' = - - 6, t hen 6----- fl$ and h e n c e - ( - - f l e K s . As $ is a uni t in Ls we ge t

(NKs/Q(~))2 _--__ NL6/Q (~) = .~- NL6/Q(6) = -~- (NL2/Q (6))8 = Z~ .M 3 '

and since

-NKs/Q (fl) = N e Q

we conclude f rom/V2 = M 3 t h a t M is a square, which contradicts the hypothesis . q .e .d .

I n the nex t s tep we establish some est imates .

Lemma 6. a) I / d < O, then

169 T < T ] 2 < ~ < 913,

13 , �9 2

b) I / d > O, then

144 < ( - - ~/)2 < ~ <: ( _ r/)5/2, except when (D, d) = (7, 7),

17 -~-< (--7)< l~r~l < (" ~)2.

P r o o f . Since = ~7 3 q- 1 we have

60

7 - 1 1


a) I f d < 0, we know from [3], lemma 2.3.1 that 2 / = -~ > 2. Since 2/s + 1 <

(2/+ 1) 3, we have

2/-- l < i / U 3 + 1 - - 1 < 2/.

As 2 / > 2 we have ~9 < 32/s(2/_ 1) + 3~4(2/-- 1) 2 + (~ -- 1) 3 from which we conclude ~9(2/s + 1) < (~4 + (~ _ 1))a and hence

i / ~ + 1 - 2 / < 2 / - 1 2/3

I f ~ > ~ we have 2/9 > 1 + 32/4 + 3~ s and therefore 2/9(2/8 + 1) > (1 + 2/4)3, from which we get

1 s _ 2/3 < ~/~]3 + 1 --2/ .

The condition ~ > ~ is now always satisfied, since (D, d) ~= (5, --5) . Indeed, from

D + DV-D-~ 4 d 1 D ~2 D ~ = D -- ]/D2 + 4d -- 2d (D2 + 2d + D V-D~ + 4g) >= 2 - I + >-2-

we get tha t D => 7 implies 2 / > 7. Since M = D 2 + 4d > 1 has to be square free we get D = 1 (rood 2) and from

-----d + 3 > l square free we infer ~ + 3 ~ 0 (mod4) and h e n c e d - - - 3

(rood 4). One verifies now quickly that D < 13 implies (D, d) = (5, -- 1) in which case 2 /= {- (23 + 5 ~/~) > 22.

13 We therefore get, in fact, that in the admissible cases always ~ > -~- . Hence

we have

169 - i - < ~2< ~< 2/3.

From

~50\2 ~ = ~ ~ = ~ + ~ + a 2

~ / ~ + 112 + 2/i/~ + 1 + 2/2

= r i / ~ + 1)~ + i/2/~ + 1 + 1

we see immediately that

7] 2 < l~va]2 < 2/4.

34*


b) I f d > 0, we p u t

I v = - ~ - _ ~ .

T h e n we k n o w f r o m [3], l e m m a 2.3.1 t h a t Iv > 2. S i m i l a r l y as before we can w r i t e

I + ~(--~)3-- I i i + ~IV3-- i i

_ ~ _ ~ / ( _ ~)3 _ 1 ( - ~) Iv - ~ / ~ _ ~ Iv

Since 3 I = D 2 -~ 4 d > 1 a n d J = -d- + 3 > 1 a r e b o t h s q u a r e f ree we h a v e

as before d - 3 ( rood 4) a n d h e n e e D ~ 7, a n d s ince (D, d) = (7, 7) was e x c l u d e d , we ge t D => 11 a n d hence

1 D IV=-~(D2 + 2d + DI/i)~ + 4e) > E + I + } I / ~ + 4g > D + I > 1 2 .

F r o m IV3 __ 1 > (IV -- 1) 3 we n o w in fe r

IV < 1 . ~ i/iv 3 - - 1 < 1 ~- IV,

a n d f r o m

3 IVg(IV + 1) - - 3IVg/2(IV + 1)2 + (IV § 1)3 < IV~/~

we ge t (IV9/2 _ (IV ~ 1))3 > IV21/2(IV3 _ 1) a n d hence

~_ I + I V

F r o m 3IV s - - 3IV 3 ~- 1 > IV6 we d e d u c e (IV3 _ 1)3 < IVs(IV3 _ 1) a n d the re fo re

3. 1 IV _ I/IV3 _ 1 < IV~ .

A l t o g e t h e r we h a v e

144 < IV~ < ~ < IV~/~.

F o r

we h a v e

where t h e s econd i n e q u a l i t y fol lows f r o m 1 < ~ i - - IV-3(IV _ 1), i .e .

i r i _ Iv-3( iv + 1 ) < ~ / ~ - Iv-~)~ (iv~ - 1 ) ,

which im p l i e s

i/(1 - I v - ~ + ~/1 - i v - - g < Iv~ ~ - Iv-3)~ - I v h - Iv-3,


and therefore

Pro= ~-3)2 + i / r - ~-3 + 1 < i / r 1)2 _ ~ r 1 + 1. q.e.d.

Proposition 7. ~ is not the power o] a unit in Ls , i.e. i]

= ~n with ~ E ( L 6 ) , A > I

then n ---- 1.

P r o o f . a) B y lemma 5 we have t h a t n mus t be odd. h) Suppose now tha t ~ = A n ~%h A ~ E (L6), ~ > 1 and n > 3. Wri te

&t = �89 o + xl co + 0) 2 + yo ~ + Yl ~o~ + &o 2

with z~, y~ e Z for i = 0, 1, 2. 3 _ _ .

We note t h a t ]71 § 2 4 7 , as soon as ]71 >~- . Then we infer from lemma 2.3.1 in [3] and from lemma 3 and lemma 6 for the ease d < 0

o- 1~1 (i/~: + 2~/T~Wl + a) lull < ~

< FLT~ (171 + 21/i7 P + 3) < (31~1)- ~2 = ~ < 1 .

Ifd > 0 we apply the inequalities 17 ]518 A- 2 1712/~ + 3 < 3 I~1 ~/~ and ~ - - < 1.

15 The first inequali ty holds, as soon as [ 7 ] > ~ - and the second holds since -- 2 fl < ~,

which implies

_ ~11/7 < (~8 + / ~ 3 ) 4 = ~ 1 2

Again by lemma 3 and lemma 6 we now get for (D, d) # (7, 7)

lul l< Idl d/~+2P]~---i + 3)< : - - = i I l l d l p 3~2 - ~o~ (,7,~/~ + 2 , 7 +3)

< Ij_JL i , p / . _ I : ~ I " ~z0 = V ] / = ~ . I~l <1 ( 0 2 ( ,012 �9

I n the case (D, d) = (7, 7) we compute directly

I~l < ~gj tvr + 2 1 7 p + s) = o.o~... < 1.

Hence always yz = 0. c) For x2 we get from lemma 3 and lemma 6 and the preceding computat ions

1~1< 1/~1~1 (~/~+2~/]~,~1 + 3 ) < V ~ < M 3o) 2

and hence by lemma 4 (iii) x2 ~- 0, if (M, 3) = 1.


In the remaining case 3[ M one has 3 ~ VM, since M = D 2 + 4d =~ 3, 6, and therefore

M

L e m m a 4 (ii) now gives x~ = 0 also in this ease. As (x~, Y2) = (0, 0) contradicts l emma 4 (i) the proposi t ion is proved, q .e .d .

L e m m a 8. ~0, ]/6~, ~0~-----i6L6.

P r o o f . a) Suppose ] / ~ L 6 . Then L6 ----- L2(co) = L2(?)) where co ~ ~ / ~ and

?) = ~6. We can therefore write

? ) = : r wi th a , / ~ , ? ~ L 2 , a n d s o

?), = ?)a ~__ (x + fl0eo + y02CO2.

As L<6 *> = L2 we get ?)3 = 0 = ?),3. Hence

(1) ?)'-----?) or ? ) ' = Q ? ) or ? ) ' - - 0 2 ? ) .

One now has (see [1], w 2)

f ? ) ' - ?) = f l~(~ - ~) + ~ * ~ ( o ~ - ~) = (o - 1 ) ( ~ o - ~ o ~ ) 0 (2) ' / ~ , - - Q?)-_- ~(1 - o) + r ~ 2 ( o 2 - o) = (o - l ) (r co2e - ~ ) ,

[?) - - 02?) ---~ ~(1 - - 02 ) + /~co(0 - - 09) --~ (0 - - 1)(~02 - - /~~

As g, fl, 7 ~ L2 but 0, co 6 L2 tho three possible equalities (1) toge ther wi th (2) lead to the following contradictions.

a l) ? ) ' = ?) implies fl = ?----0, hence ? ) = :r ~ L2 which contradicts the hypothesis, t h a t 0 is a fundamenta l uni t in L2.

a2) ?)' ---- 0?) implies cr = ? ----- 0 and hence ?) =/Sco, t h a t is - - = - - ~/52. The

norm now yields, since 0 is a uni t in L2

NLc;/Q ~ - N L 2 / Q ~ N L 2 / Q = N L 2 / Q = -~- M 3 .

NLe/Q(~ , )= .NL~/Q(1)=N~Q we get N 3 = -VM~ and hence M 2 m u s t be As k'i" l \ r " /

a cube, which contradicts the nature of M3. a3) ?)'-----0e?) implies in a similar manner :r = 0 hence ?) - - - -yw 2 t h a t is

?)co ~- ? M3 e L2. The norm relat ion

2Vr6/Q (?)co) - - (2VL2/q (?)~))3 _ 2VL~/Q (?)~ w 3) = NL2/Q (M3 0) = • M~

again implies t ha t M3 mus t be a cube, therefore ~0----- ?) 6 L~.

(b) Suppose ?) = ~ e i 6 . Since ~ 6 L2, also 0~ 6 L~ and therefore ?) = ] / ~ 6 L2. Hence we have as before L~ = L2 (co) = L2 (?)). As before, we conclude (with the


6O role o f co and ~0 interchanged) either - - e L2 or ~0r e L~. Therefore

~0

~o3 M 3 - - e L 2 or v2~o)~=O~M~L~ ,

�9 W ~ 0r

t h a t is ~ e L2, which is a contradiction. Similarly ]/0--~ ~ ~ L6. q.e .d.

We shall apply several times the following useful criterion (see [4], p. 167 or [5], Hilfssatz 4).

Lemma 9. Let K be an algebraic number/ield with unit group E (K) o/ranIr r ~ 1.

(1) ~ e E ( K ) is a/ irs t /undamental unit in K i/ and only i/ ~ is not o/ the ]orm

/or n e 7/, n > 1, v e E (K) and e a root o/ unity.

(2) I] $1, $2, . . . , ~ - 1 are elements o] a basis o/ E (K), 2 <_ k <-- r, and ~ e E (K), thez~ ~1, $2 . . . . , ~ - 1 , $ belong to a system o/]undamental units in K i] and only i] is not o~ the/orm

/or mi, . . . , m~-i , n e 7/, n ~= -4- 1, ~, e E(K) and ~ a root o] unity.

Lemma 10. There is no 2 e L6 such that NL~/L~ (2) = O.

P r o o f . Suppose 2 e L8 with Nz,8/L~(1) = O. Put Nr.~/K~(2) = e r, then we pre tend t h a t el ---- ~, e2 = 240-1e -2r and e~ = e is a fundamenta l system of units in L~.

Indeed, el is a first fundamenta l uni t in L6 according to lemma 9. e~ is a second fundamenta l unit, as f rom e2 = 240 -1 e -2r --= 4- ~m ~n for ~ e E (L6) after applying the norm from 156 to L2 we get t h a t n = 4- 1. Similarly, e3 is a th i rd fundamenta l unit, as the norm from L6 to Ks applied to a relat ion

83 ~--- e = 4- ~ n , (2 4 O- 18-2r)m~ ~

implies n = =k 1. Since ez, e2, e3 is a system of fundamenta l units we have

O ~--- 81m* 8m~2 8m33 f o r m l ~ m 2 ~ m 3 ~ ~ �9

After applying the norms from L6 to L2 and to Ka we obtain 0 = e m~e s Hence 1 2" L0 or or d e p e n a m g on hether mz = 0, or 2 mod o 3.

I n each case we get a contradict ion to lemma 8. q.e.d.

We can now easily deduce the main

Theorem 11. Let M = D z + 4d > 1 and / : --~ ---d- ~- 3 > 1 both be square

Ree, D, d e ~ , d[D, D > 0 , (D,d) ~= ( 5 , - - 5 ) and M s = D a + 3 D d = f d % Then the unit group in L6 is

_E (L6) - - < - 1, e, u, 0>,

536 G. FR~z ~RCH. ~ATH.

where

and

e= , = , with

if d = 4 - I .

Proof . By proposition 7 and ]emma 9, ~ is a first f~damenta! unit in Le. By ]emma 9, 0 is a second fundamenta! u~it m L6, as the norm from L6 to L~ applied to 0 = • ~ ~e]ds 03 ___ 4- (NLs/Z2~P and hence ~ ---- i 1 because of ]emma I0. Similarly e is a third fundamental ~ t since the norm from L6 to K3 for the re]ation e = 4- ~ 0 m' ~ implies n = 4- 1. Therefore

Ee - - ( - - 1, ~, 0, ey = <- - 1, e, u, 0y. q .e .d.

This paper was suppor ted by ~he Nat ional Research Council of Canada.

References

[1] R. DEDEKIND, ~ber die Anzahl der Idealklassen in reinen kubisehen Zahlk6rpern. J. Reine Angew. Math. 121, 40--123 (1900).

[2] G. FREI, Fundamental systems of units in biquadratic parametric number fields. To appear in J. Number Theory.

[3] G. ~ E I and C. LEVESQUE, Independent systems of units in certain algebraic number fields. J. Reine Angew. Math. $11/312, 116--144 (1979),

[4] E. LA_~AU, Vorlesungen iiber Zahlentheorie I I I . Leipzig 1927. Reprint Chelsea 1969. [5] H.-J. STE~-DF~, ~ber die Einheitengruppe der reinen algebraischen Zahlk6rper seehsten

Grades. J. Reine Angew. Math. 268/269, 78--93 (1974). [6] H.-J. STE~r~E~, Eine Formel ffir Grundeinheiten in reinen algebraischen Zahlk6rpern dritten,

vierten und sechsten Grades. J. Number Theory 7, 235--250 (1975).

Eingegangen am16.9.1980

Anschrift des Autors:

Gfmther Frei Ddpartement de mathdmatiques Universit6 Laval Ste-Foy, Qu& Canada, G1K 71:'4.

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Fundamental systems of units in bicubic parametric number fields