ARKIV FOR MATEMATIK Band 4 nr 37

ARKIV FOR MATEMATIK Band 4 nr 37

R e a d 8 N o v e m b e r 1961

O n t h e n u m b e r o f r e p r e s e n t a t i o n s o f a n A - n u m b e r i n a n a l g e b r a i c f i e l d

By

w 1. Introduction

1. Let a be an integer =~ 0 in the algebraic field ~ . If :r is representable as the sum of two integral squares in ~ , we say, for the sake of brevity, t h a t is an A-number in g'~. We s a y t h a t

where ~ a n d ~] are integers in ~ , is a primitive representation if the ideal (~,~) is the unit ideal, a n d otherwise an imprimitive representation. I n the following we shall use the terms A.prime a n d A-unit. The representations ~ = x 2 + y 2 with x = • y = - 4 - 7 and x = - 4 - ~ , y = • are considered to be one a n d the same.

W h e n the degree of ~ is ~> 2 the integer zt is said to be a prime when (~r) is a prime ideal. The relation 1 = 12 + 02 is called the trivial representation of the n u m b e r 1.

Design b y G an infinite (abelian) group of units belonging to ~ (composition = multiplication). B y the rank of G we u n d e r s t a n d the m a x i m a l n u m b e r of inde- pendent units (of infinite order) in G. T h e rank of t h e group consisting of all the units in ~ is r = r l + r ~ - 1, where r 1 is the n u m b e r of real conjugated fields and 2r~ the n u m b e r of imaginary conjugated fields.

Design b y It a ring of integers contained in ~ b u t not in a n y sub-field of ~ . If R contains the n u m b e r 1, it contains an infinity of units a n d it is well-known t h a t the unit-group of R has the r a n k r.

w 2. The representations of A-units and A-primes 2. We first prove

Theorem 1. When there are more representations o/ the number 1 than the trivial one, then there are in/initely many representations.

Proo/. Suppose t h a t

1 =~2+~2,

where ~ a n d ~ are integers in ~ and ~ 4 0 . P u t for n = 1 , 2 , 3 , . . . ,

~n + ~ni = (~ + ~i) n,

467

T. r(A(;ELL, Number of representations of an A-number where

a n d

T h e n we clearly h a v e a n d

Hence

(& + v.i) (~. - ~.i) = (~ + ~i) ~ (~ - ~i)" = (~2 + v~)..

T h u s the Diophantine equation

x 2 + y2 = 1 has the integral solutions

x = ~ , , y = ~ .

I t is easy to prove t h a t these solutions are all different.

I n fact, if we have (for n # m ) ,

we get

(~ + i~) m = (~ + i @ ,

(U

(2)

(3)

Hence ~ + i~ is a root of unity. Suppose t h a t

~ + i ~ = ~ is a primitive N t h root of unity. Since

we get

1 1

~ = ~ ( e + -1), ~ = ~ ( ~ _ ~ - 1 ) .

I t is easy to show t h a t these n u m b e r s are n o t integers if N # 4 , 4=2 a n d # 1.

Suppose first t h a t N is a powder of 2 a n d >=8. If ~ ( ~ - 1 ) were an integer, t h e n u m b e r

N

should also be an integer. B u t this is n o t the case.

ARKIV F()R MATEMATIK. B d 4 nr 37 Suppose n e x t t h a t N is divisible b y the o d d prime p. If -~(o 2 - 1) were an integer, the n u m b e r

2N

~(e ~ - 1)

should also be an integer. Hence, if x is an a r b i t r a r y primitive p t h root of unity, the n u m b e r y = ~ (x - 1) should be an integer. B u t the n u m b e r s y clearly are the roots of the irreducible algebraic equation

1

~ y

with integral coefficients. Hence t h e y are n o t integers.

Since the values N = 4, 2 or 1 i m p l y either ~ = 0 or ~ =: 0, theorem 1 is proved.

3. We next prove

T h e o r e m 2. There is exactly one representation o~ every A-prime, i/ the number 1 has only the trivial representation. Otherwise there is an infinity o/representations.

This result also holds /or every A-unit.

Proo/. Suppose t h a t the n u m b e r 1 has only t h e trivial representation. L e t be an A-prime with the two representati6ns

a n d

= ~ + ~ ,

where ~t, fl, al a n d fll are integers in the field. F r o m these representations we get / f - t~, ~) = ~ t ~ * - ~ * ~ .

Since z is a prime, either of the n u m b e r s alfl + aft1 a n d : q f l - aft1 m u s t be di- visible b y ~. W e m a y choose the sign of fll such t h a t we obtain

o~lfl~ocfll (rood zO.

Multiplying together the two representations of ~t, we get

~ = ( ~ + P ~ ? + ( ~ - ~ ) ~ .

Since ~lfl-~tfll is divisible b y zt, the n u m b e r ~t~ 1 +tiff1 is so. I f we p u t

*t~ + tiff1 = zt~l a n d ~d ~ - ~tfll = ~ 1 , where ~ a n d ~h are integers, we get

1 = rj 2 + ~ .

469

T. NAGELL, Number o f representations of an A-number

B y hypothesis this equation is possible only for ~ = 0 or ~ h = 0 . F o r ~ = 0 and 7 1 = - - - l w e get

~ 1 = - 8 8 1 a n d ~ l f l - ~ f l l = • whence b y elimination of 81,

~ 1 _ ~ ( ~ + 8 2 ) = ~ = + ~ "

0fif~ ~- ~ ~ __

Hence g l = +--8 a n d 81 = • a.

F o r ~h = 0 a n d ~7 = - 1 we get

gift = ~81 a n d a a l + 881 = +--:~, whence b y elimination of ~1

a~i + ~ a ~ = a ~ (a~ + f ) = a ~ = + ~ .

Hence (~1 = -~ g a n d 81 = • fl" Thus there is only a single representation of the prime. The proof also holds when g is a unit.

Suppose next t h a t the equation (3) has an infinity of solutions x = ~ , y = ~ given b y (1) a n d (2). L e t eo be an A-number with the representation

CO = ~ 2 - ~ 8 2 ,

:r a n d 8 being integers in ~ . P u t for n = 1, 2, 3 ...

where

Then we have a n d

Hence

~n = ~ - 8~- a n d 8~ = ~ n + 8 ~ "

(~n + 8hi) (an - 8~i) = ( ~ + V~)" ( as + 8 ~) = ~"

I t is easy to see that, in this way, we get an infinity of representations of m.

I n fact, supposing we get

B u t in the proof of theorem 1 we showed t h a t this relation is possible only for m = n . Thus we have p r o v e d t h e o r e m 2. Moreover we have p r o v e d the more general result: If the n u m b e r 1 has an infinity of representations, there is an infinity of representations of every A-number.

ARKIV FOR MATEMATIK. Bd 4 nr 37 w 3. The representations o f an arbitrary A - n u m b e r

4. Owing to the a b o v e proof we h a v e a l r e a d y established the result expressed in the second p a r t of

Theorem 3. I / the number 1 has only the trivial representation, the number o/

representations o/ every A-number is finite. Otherwise there is an infinity o/ rep- resentations.

Proo/. Suppose t h a t t h e n u m b e r 1 has only the trivial representation. L e t 0) be an A - n u m b e r having an infinity of different representations

0) = a ~ +fl~, (n = 1 , 2 , 3 .... )

an and fin being integers, with anfl~ + 0. Then we h a v e for all indices m and n ( m ~ n ) : a , ~ • fl~:t: • ang = • and f l ~ _+am.

A m o n g these representations of 0) there m u s t exist a t least two different rep- resentations

2 2

am+tim and a ~ + f l ~ , (4)

which satisfy t h e congruence conditions

a m m a n (mod 0)) and flm~fl~ (mod 0)). (5) I n fact, the n u m b e r of residue classes modulo co is [No) I , a n d thus the remainders of the four n u m b e r s an, tim, a~ a n d fl~ m a y be combined in a t m o s t IN0)[ 4 ways.

Multiplying the two representations

0) = a~ + fl~ a n d 0) = a~ +

fl~,

we get

0) 2 = ( a m f l n - - ~ m a n ) 2 + a r e a n -~- f l m f l n ) 2.

I t follows from (5) t h a t the two n u m b e r s

amZ.-- ~ a . a n d area. + fire/3.

are divisible b y 0). H e n c e we m a y p u t

a,~Sn -- flman = 0)~ a n d aman + flmfln = C0~1, (6) where ~/ a n d ~h are integers. Then we get

1 = ~/2 + ~/12.

T h u s b y our hypothesis we m u s t h a v e either ~ = 0 or ~/1 = 0 . I f ~/=0, it fol- lows from (6)

a m f l n = f l m a n a n d O~mOf. n .qt_ ~ m f l n = "q- tO,

whence b y elimination of fin,

471

T. NAGELL, Number of

representations

~ ~ + 8 ~ ~ (~+/3~): ~

~m ~m ~m

Hence ~ = + ~ and ~ = • For ~ 1 = + - 1 we get from (6):

whence by elimination of 8m,

~ m ^{~m ~rn}

Hence ~ = + ~ and fi~= + ~ .

From this we conclude t h a t the representations (4) cannot be different. Con- sequently, the number of representations must be finite.

w 4. The totally real fields and the imaginary quadratic fields 5. We next prove

Theorem 4. I n the totally real field ~ there is only a finite number o/ represen.

ta, tions of a given A-number. There is exactly on representation of the number 1 and likewise of every A - p r i m e and of every A-unit. A unit is an A-number only when it is a square.

Proof. A real field is called totally real when all the conjugate fields are real.

L e t s be an A-number in ~ with the representation

=~+~,

where ~ and fl are integers in ff~. Then the conjugate equations

also hold, Since the conjugater arc al! real, we ge~

I~,k,I < IV_Z,k,i

for every value of k. Hence there is only a finite number of Possibilities for ~ when is given.

Consider in particular the ease ~ = 1 . If we suppose 8 = 0 , w e get I~ik)[<l, hence ~ = 0 .

When ~ is a prime or a unit, we m a y apply theorem 2.

Finally, suppose t h a t e is a unit with the representation

E=~2+f,

and fl being integers in ~ . Then we get b y squaring 472

ARKIV F6R MATRMATIK. Bd 4 nr 37

whence

e~ = (a~ _ ~ ) 2 + (2afl)~,

/~2

Since the number 1 has only the trivial representation, this implies either a2 ~2 = 0 or ~fl = 0 ; but it is clear t h a t ~z_fl2 = 0 is impossible when ~ is a unit.

6. We add the following result:

Theorem 5. I n the [ield K ( ~ - ~ ) there is only a / i n i t e number o/representations o / a given A-number. There is exactly one representation o / t h e number 1 and likewise o/ every A-prime.

Proo[. B y theorems 2 and 3 it is sufficient to show t h a t the n u m b e r i has only the trivial representation. The equation

1 = :r +/32,

where ~ and /3 are integers in K (]/~ 1) leads to either of the following systems:

o r

~ + f l i = l , o~- fl i = l o~ + fl i = i, ~ -- fl i = - i.

B u t the first system implies t h a t f l = 0 and the second t h a t ~ = 0 . This proves theorem 5.

I t is easy to prove

Theorem 6. I n the imaginary quadratic/ield K ( ~ - D ) there is an in/inity o/ rep- resentations o/every A-number, except when the/ield is K (V---l).

Proo/. According to theorem 3 it suffices to show t h a t the number 1 has a non trivial representation, I n fact, since the equation

x 2 - D y 2 = 1

has solutions in rational integers x and y, y # 0, the number 1 has the non ~riv- ial representation

1 = x ~ +

(yV-~) 2.

w 5. T h e m a i n result o n t h e r e p r e s e n t a t i o n s

7. Theorems 4, 5, and 6 are contained in the following general result:

Theorem 7. There is an in/inity o/ representations o/ every A-number in a~ : al- gebraic field ~ except in the /oUowing cases:

~73

T. NAGELL, Number of representations of an A-number 1 ~ ~'~ is the Gaussian field K ( V - l ) .

2 ~ ~ is totally real.

Proo/. I n v i r t u e of t h e o r e m 3 i t is sufficient to p r o v e t h a t t h e r e is a n i n f i n i t y of r e p r e s e n t a t i o n s of t h e n u m b e r 1, p r o v i d e d t h a t ~ is n o t one of t h e excep- t i o n a l fields in t h e o r e m 7. B y t h e o r e m 1 i t suffices to show t h a t t h e r e is a non- t r i v i a l r e p r e s e n t a t i o n of t h e n u m b e r 1.

D e n o t e b y n t h e degree of t h e field ~ ; b y r 1 t h e n u m b e r of real c o n j u g a t e fields ~(h~, b y r 2 t h e n u m b e r of pairs of i m a g i n a r y c o n j u g a t e fields a n d b y r = r 1 + r~ - 1 t h e n u m b e r of u n i t s in a f u n d a m e n t a l s y s t e m of units in t h e field ~ . W e first consider t h e case t h a t ~'~ c o n t a i n s t h e n u m b e r ~ / - ~ . I n t h i s case we h a v e n >~ 4. Since r >/1, t h e r e exists in ~ a u n i t E which is n o t a root, of u n i t y .

T h e n t h e e q u a t i o n

1 = ~2 + f12

is satisfied b y t h e following numbers:

a = l ( E m + E - ~ ) a n d

1

fl:= ~i ( E m - E-"),

where m is a n a r b i t r a r y r a t i o n a l :nteger. L e t us choose t h e n u m b e r m as a m u l - t i p l e of q (2), where fi~ (2) d e n o t e s t h e n u m b e r of r e s i d u e classes m o d u l o 2 in which are p r i m e to 2. T h e n we h t v e for a n y integer 7 in ~'~ which is p r i m e t o 2,

7 ~ 1 (mod 2).

H e n c e t h e n u m b e r s a a n d /~ are i ltegers in ~ ; for m # 0 we h a v e a f t # 0 . Consider n e x t t h e case t h a t ~ does n o t c o n t a i n t h e n u m b e r ~ / ~ 1. A d j o i n i n g t h i s n u m b e r t o ~ we g e t t h e field ~ ( l / - 1) = ~1. This field has t h e degree 2n.

D e n o t e b y R 1 t h e n u m b e r of real c o n j u g a t e fields ~ ) , b y R~ t h e n u m b e r of p a i r s of i m a g i n a r y c o n j u g a t e fields a n d b y R = R 1 + R 2 - 1 t h e n u m b e r of u n i t s in a f u n d a m e n t a l s y s t e m of units il. t h e field s

I f ~ is a g e n e r a t i n g n u m b e r of ~ , one m a y f i n d a r a t i o n a l u such t h a t t h e 2n c o n j u g a t e fields ~'~(1 k) (k = 1, 2, 3 . . . 2n) a r e g e n e r a t e d b y t h e 2n n u m b e r s

co = ~(h) __ u ~/-- 1,

where ~(h) r u n s t h r o u g h t h e s y s t e m of n n u m b e r s c o n j u g a t e t o ~ (see f. ex.

H e c k e [4], p. 67). I f ~(hl is real, i t is e v i d e n t t h a t co is i m a g i n a r y , since u # 0 . I f ~(n) is i m a g i n a r y , i t is e v i d e n t t h a t co m a y be real for a t m o s t t w o special v a l u e s of u, fcr all o t h e r v a l u e s of u t h e n u m b e r eo is i m a g i n a r y . Hence, all t h e 2n c o n j u g a t e fields ~(1 k) are i m a g i n a r y . T h u s we h a v e RI='O, R ~ = r l + 2 r ~ a n d

R = R I + R 2 - 1 = r x + 2 r 2 - 1 =r +r~.

Since ~ is n o t t o t a l l y real, we h a v e r ~ > 1 a n d t h u s

ARKIV FOR MATEMATIK, Bd 4 nr 37 R > r .

R is t h e r a n k of t h e g r o u p of all t h e units in ~1, a n d r is t h e r a n k of t h e g r o u p of all t h e u n i t s ~ . L e t us consider t h e r i n g consisting of the n u m b e r s in

~ 1 h a v i n g t h e form c + di, where c a n d d are integers in ~ . The u n i t - g r o u p G of t h i s r i n g has t h e r a n k R. T h e s u b - g r o u p G 1 consisting of t h e squares of t h e u n i t s in G c l e a r l y has t h e s a m e r a n k R. The u n i t s in G 1 c a n n o t all be e q u a l to t h e p r o d u c t of a u n i t in ~ a n d a r o o t of u n i t y since r < R . Hence we con- elude t h a t t h e r e e x i s t s a u n i t E = a + b i in t h e ring, a a n d b integers in ~'~, such t h a t ab ~= O, a n d such t h a t E 2 is n o t e q u a l to t h e p r o d u c t of a u n i t in s a n d a r o o t of u n i t y . T h e n t h e n u m b e r E 1 = a - bi is also a u n i t in ~1- H e n c e a 2 + b ~ is a u n i t in ~ . T h e n t h e e q u a t i o n

1 ~ a2 ~_ ~2 is satisfied b y t h e following n u m b e r s :

a n d

E ~z + E~ a

= 2 ( a ~ + b2) a

E ~ _ E~, = fi 2 i ( a 2+b2) m'

where m is a n a t u r a l n u m b e r . I t is e v i d e n t t h a t :r a n d fl are integers in ~ , since a a n d b are so. The h y p o t h e s i s ~ = 0 leads to

E4m = E~ m.

H e n c e E E s 1 s h o u l d be a r o o t of u n i t y = E 2 , a n d we s h o u l d h a v e E 2 = (a S + b 2) E~.

B u t this is c o n t r a r y t o our a s s u m p t i o n on E. Thus, for m # 0, we h a v e aft # 0, a n d t h e p r o o f of t h e o r e m 7 is complete.

R e m a r k s on previous p a p e r s on A - n u m b e r s .

I n two p r e v i o u s p a p e r s , [1] a n d [2], we h a v e a l r e a d y e s t a b l i s h e d a n u m b e r of t h e o r e m s on A - n u m b e r s . The p r o o f of t h e o r e m 21 in p a p e r [1] was n o t c o m p l e t e as we d i d n o t show t h a t m m a y be chosen such t h a t a f t # 0 . This l a c u n a was r e p a i r e d in t h e a b o v e p r o o f of t h e o r e m 7. T h e o r e m s 2 a n d 3 in t h i s p a p e r cor- r e s p o n d t o t h e o r e m s 16 a n d 17 in p a p e r [1] w i t h a c e r t a i n c o r r e c t i o n in t h e proof.

I n t h e o r e m 2 in [1] i t is n e c e s s a r y to a d d t h e following condition: The ideal (a, fl) is e i t h e r t h e u n i t i d e a l or t h e p o w e r of a p r i m e ideal p which does n o t d i v i d e 2. T h u s t h e t h e o r e m o u g h t t o be p r o n o u n c e d as follows:

Let o~ and fl be A-numbers in the field ~ with the primitive representations in 475

T. NAGELL, N u m b e r o f r e p r e s e n t a t i o n s o f a n A . n u m b e r o~ = a 2 A- b 2

and

fl = c 2 + d 2.

I / (~,fl)=pro, m>~O, where the prime ideal p is prime to (2), then the product aft has a 2,~'imitive representation o / t h e / o r m

aft = (ac • bd) s + (ad ~ bc) 2, either/or the upper o r / o r the lower sign.

This restriction in the theorem does not make necessary any alterations in the proofs of theorems 29-31 in [1].

The following misprints in paper [1] ought to be noticed: Page 24, in line 14 replace e by ~1 in the right-hand side of the equation. Page 33, in line 7 the first equation shall be ( ~ ) = + 1 . Page 41, i n l i n e 11 from betow add, after tile word even, >~ 2. Page 46, in the last line replace db 1 by r 1. Page 50, in line 5 from below replace ~ by ft. Page 58, in line 11 from below add, after E, the square of which. Page 68, in line 9 the first factor shall be (~/2+ 1).

The last 11 lines on page 34 in [1] ought to be replaced by: This congruence is possible only when one of the numbers b'and c is divisible by 4 and the other one is ----=2 (rood 4). Since 2 v = a c + b d , where v is even, we get ac =- - b d (rood 4).

Thus, a and d being odd, both b and c should be divisible b y 4. Since this is impossible we conclude that, the numbers a, b, c and d are all even.

In paper [2] on pape 279, line 12, read q instead of 5.

w 6. The complete solution o f ~s + ~i2 = 1 in a quadratic field

8. According to theorems 4 and 5 it suffices to consider the imaginary quad- ratic fields K (~/-D), where D is a square-free natural number > 1.

First case. - D -- 2 or =--- 3 (rood 4).

The equation in question is

(a + c V ~ ) 2 + (b + d V---D) s = 1, (7)

where a, b, c and d are rational i~tegers. Hence we get the system aS + b S - D(cS + d 2) =1, a c = - b d .

If c = 0 we must have b = 0 (d = 0 gives the trivial solution). Hence

a s - Dd s = 1. (8)

Suppose next c d # 0 . By elimination of a we obtain

ARKIV F6R MATEMATIK. Bd 4 n r 37 l = b2d2c -2 % b 2 - D (c 2 §

d2).

T h e n we get

c 2 = (c 2 -~- d 2) (b 2 - Dc2), which is impossible since d~=0.

Conclusion: We o b t a i n all the solutions of (7) w h e n b = c = 0 a n d a a n d d satisfy e q u a t i o n (8).

Second case. - D ~ 1 (rood 4).

T h e n t h e e q u a t i o n is

(a + c V - D) ~ + (b + d V - D ) ~ = 4 , (9) where a, b, c a n d d are r a t i o n a l integers, a a n d c are of t h e same p a r i t y , a n d so are b a n d d. H e n c e we get t h e system

a 2 + b e - D (c ~ + d 2) = 4, ac = - bd.

I f c = 0 we m u s t h a v e b = 0 . T h u s we get

a ~ - D d 2 = 4. (10)

Suppose n e x t c d * O . B y e l i m i n a t i o n of a we o b t a i n

4c ~ = (c ~ + d ~) (b ~ - Dc~). ( l I )

P u t ( c , d ) = g , c = g c l , d = g d 1 a n d (cl, d l ) = I , where g, c 1 a n d d 1 are r a t i o n a l inte- gers. T h e n we get from (11)

4Cl ~ = (Cl 2 + d~) (b 2 - Dg~c~).

H e n c e b is divisible b y c r P u t t i n g b = c l ] we get 4 = (c~ + d b (/~ - Dg~).

This is possible o n l y for c~ = ~ = 1. H e n c e

/~ - D a ~ = 2 . ( 1 2 )

I n this r e l a t i o n / a n d g are clearly o d d n u m b e r s . H e n c e we m u s t have D - - ~ - 1 (mod 8).

Conclusion: W e o b t a i n all t h e solutions of (9) from t h e f o r m u l a

a ' + (a V:--D)' = 4,

a n d , if e q u a t i o n (12) is solvable, from t h e f o r m u l a (/ + g V - - D ) S + ( / - g V - D ) 2 = 4 .

E q u a t i o n (12) is n o t always solvable for D - - - 1 (mod 8). T h u s it is solvable for D = 7 b u t n o t for D ~ 1 5 .

477

T, NAGELL, Number of representations of an A-number

O u r r e s u l t s in t h i s s e c t i o n m a y b e i n t e r p r e t e d in t h e D i r i c h l e t - f i e l d K (i, I / - D )

in t h e f o l l o w i n g m a n n e r . D e s i g n b y s t h e f u n d a m e n t a l u n i t in K (VD), s > 1, a n d b y E t h e f u n d a m e n t a l u n i t in K (i, V - D ) , ]El > 1 a n d S > 1, if E is real. T h e n w e h a v e , for D > 3, e i t h e r E = ~ o r E = l/ei. T h e n e c e s s a r y a n d s u f f i c i e n t c o n d i t i o n for t h e l a t t e r case is t h a t t h e i d e a l (2) is t h e s q u a r e of a principal i d e a l in K (VD). F o r t h e p r o o f see [3], p. 11-15. H e n c e we m a y c o n c l u d e : T h e s o l u t i o n s of ~ 2 + ~ 2 = 1 a r e g i v e n b y • + ~ M a c c o r d i n g a s N ( ~ ) i s = + l o r = - 1 . I n t h i s w a y we g e t all t h e s o l u t i o n s e x c e p t w h e n D ~ - 1 (rood 8) a n d t h e i d e a l (2) is t h e s q u a r e of a p r i n c i p a l i d e a l in K 0 / D ) in w h i c h case we h a v e t h e f u r t h e r s o l u t i o n s • E~ M. T h e e x p o n e n t M is a n a r b i t r a r y r a t i o n a l i n t e g e r .

R E F E R E N C E S

1. NAGELL, T., On the representations of integers as the sum of two integral squares in algebraic, mainly quadratic fields. Nova Acta Soc. Sci. upsal., Ser. IV, Vol. 15, No. 11. Upp- sala 1953.

2. NAGELL, T., On the sum of two integral squares in certain quadratic fields. Arkiv fSr matem., Bd. 4, nr. 20. Uppsala 1961.

3. ~AGELL, T., Sur quelques questions dans la th6orie des corps biquadratiques. Arkiv f6r matem., Bd. 4, nr 26. Uppsala 1961.

4. HI, eKE, E., Theorie der algebraischen Zahlen, Leipzig 1923.

Tryckt den 29 januari 1962

Uppsala 1962. Almqvist & Wiksells Boktryckeri AB