AN ABSTRACT OF THE THESIS OF
WILLIAM EUGENE MILLER for the
(Name)
in
MATHEMATICS
(Major)
MASTER OF SCIENCE
(Degree)
presented on Cx\x_\--Date)
Title: THE QUADRATIC INTEGRAL DOMAINS Ra[47171] AND Ra[]
Abstract approved:
Redacted for Privacy
William H. Simons
This paper records a study of two quadratic number fields. In
the first field, denoted by Ra[1],
the unique factorization theo-
rem holds. In the second field, denoted by Ra[ii.-0],
it is demon-
strated that the unique factorization theorem does not hold and there-
fore ideals are introduced to restore this property.
The Quadratic Integral Domains R [ -.1] and Ra[Nr7)]
by
William Eugene Miller
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Master of Science
June 1969
APPROVED:
Redacted for Privacy
Professor of Mathematics
in charge of major
Redacted for Privacy
Acting Chairman of Department of M t matics
_Redacted for Privacy
Dear)! of Graduate School
Date thesis is presented
Typed by Clover Redfern for
William Eugene Miller
TABLE OF CONTENTS
Page
Chapter
I.
II.
THE QUADRATIC INTEGRAL DOMAIN Ra[471-.1]
THE QUADRATIC INTEGRAL DOMAIN Ra[
BIBLIOGRAPHY
IT)]
1
23
52
THE QUADRATIC INTEGRAL DOMAINS Ra[] AND Ra[]
I.
THE QUADRATIC INTEGRAL DOMAIN Ra[
]
Consider a quadratic equation, irreducible over the rational
field, of the form ax 2 + bx + c = 0,
where the coefficients are ra-
We can assume that a, b, c
tional and a k 0.
are rational integers
since this does not alter the roots of the equation. We will consider
the case: b 2 - 4ac = -11d 2 where
Let
p
s
is a rational integer not zero.
be one of the irrational roots of the above quadratic
the set of numbers
equation. Denote by Ra(p)
and
d
r + sp where
range over the rational field Ra.
Theorem 1. 1: There exists a rational integer m without a
repeated factor such that Ra(p) = Ra(J).
P1
From the above, b 2 - 4ac = -11d 2, d 0.
-b+tr1-112
-b-t./-11d 2 . We will let
2a
p2
2a
The roots are
1)
p-
-b+NI-11d 2
2a
as the argument is similar for the other root. Then 2)
Z.712
p + qp
= b + 2ap.
Equation 1) shows that every number of the form
is of the form k + it./2-11d
number of the form p + qt\I-11d 2
Ra(p) = Ra(r/-11d2).
Now
Equation 2) shows that every
is of the form k + ip.
p + q Nr-T-1-712 = p +
Ra(p) = Ra(4-11012) = Ra(47-1.1),
and m = -11.
so
Hence
2
Theorem 1. 2: The set
Since
Ra(47171)
Ra(
is a field.
1)
is a subset of the complex number field, both
the associative and commutative properties for the operations of addition and multiplication carry over to the elements of Ra(
Also the distributive property carries over to the elements of
Ra(T.1). We observe that: i) the operations of addition and multiplication are closed in Ra(471-.1) since the sum of two rational numbers is rational and the product of two rational numbers is rational;
ii) the additive identity is
identity is
is
0 + 0 NFT1 = 0,
iii) the additive inverse of
1 + 0471-.1 = 1;
-(p+wr-1:1) = -p + (-q)47T.1
p+q
(
p
-Nr171,
for p
and
2) + ( 2-q
p +11q
while the multiplicative
q
2-47-F1),
p + q1
and the multiplicative inverse of
not both zero, is
an element of Ra(4:7-.1)
since the
p +1 1 q
reciprocal of a non zero rational number is rational.
Theorem 1. 3: Every number of
Ra(471-.1)
satisfies a quad-
ratic equation with rational coefficients.
Let a = p + q1 be any number of Ra(Nr-T.1).
Then
a
satisfies the equation 1) (x-p) 2 - q (-11)= x2 - 2px + p 2 + llq 2 = 0.
i. e.
,
substituting
2
a for
= q (-11) - q2( -11) = 0.
of
a = p + q471.1.
x we have (p+q471.1-p) 2 - q 2 (-11)
Equation 1) is called the principal equation
Its constant term N(p+q(4=7.1)) = p 2 + llq 2 is
3
called the norm of p + qt\r:T.1
x,
is called the trace of
1) = 2p,
T(p+qt\
and the negative of the coefficient of
are both rational since
and trace of a = p + qt\F-T1
The norm
p + qt\r-T.1.
and
p
rational. Also, it is seen that N(a) = as where a
q
are
is the complex
conjugate of a
Integers of Ra(n): The set of all numbers of
Ra(47T.1)
that satisfy equations of the form x2 + bx + c = 0 where
are rational integers constitute the integral domain,
and c
of
-47T.1],
Numbers of Ra[471-1] will be called integral numbers
Ra(Nr-T.1).
We need to find such numbers.
Ra(
of
Ra[
b
Theorem 1. 4: Every rational integer is in Ra[r-T.1].
Every
number of Ra[r---1.1] which is rational is a rational integer.
If
is a rational integer, its principal equation is
b
2
x2 - 2bx + b = 0.
Therefore, from the definition,
is a member
Ra[r:T.1].
of
If,
conversely,
a+b
cipal equation becomes
a2
b
=
M2/4.
must have
Since
2
2a
x2
-1471:1
is rational, then b = 0. The prin-
- tax + a 2
and
a2
= 0.
Let
then
are both rational integers,
M
as a factor, hence a = M/ 2 is a rational integer.
Theorem 1. 5: The conjugate of a number of
Ra[r--1:1].
2a = M,
Ra[471-1]
is in
4
be in Ra[
Let a = a + bt4:1:1
1].
2
tion for a is x2 - tax + a + l lb2 = 0.
Then the principal equa-
But the principal equation
is also x2 - 2ax + a 2 + llb 2 = 0.
for a = a - 134..1
Hence
aE Ra[Nr-TI ].
Theorem 1. 6: The numbers of Ra[
a +13
where
.471.1,
and
a
1]
are given by
are both integers or both halves of
b
odd integers.
The principal equation for a number a = a + b
x2 - 2ax + a
2
+ llb 2
= 0.
If
a is in
an equation of the form x2 + px + q = 0,
Ra[
1]
is
-NFT.1
then
where p and
satisfies
a
q
are
rational integers. Hence 2a = p and a 2 + llb 2 = q are rational
integers. There are two cases to consider: either a = p/ 2 is a
rational integer or half an odd rational integer.
i) Suppose a = p/ 2
is a rational integer. If b is not a ra-
tional integer, then since
llb 2 = q - a 2 is a rational inte-
ger and b
is rational, we have the square of the denomi-
nator of
divides
b
we conclude that b
ii) Suppose
a = p/ 2
a = (n+1)/ 2,
11
which is impossible. Therefore,
is a rational integer.
is half an odd rational integer. Then
where n is a rational integer. We have
5
llb 2 = q - a 2 = q - (4n 2 +4n+1)/4
2
= (4q-4n -4n-1)/ 4,
so
4 1 lb
2
= 4q - 4n 2 - 4n - 1.
Since the right hand member of this equation is an odd ra-
tional integer, it is necessary that
be half an odd ration-
b
al integer.
Theorem 1. 7: The numbers
1
and 0 = 2 + 21
1
form a
-471:1
basis for Ra[i---1.1].
Let (y131471:1) E Ra[
-477.].
We need to show that this numb2
ber can be put in the form a2 1 + b20 = a2 +
+
2
471.1,
where
b2
are rational integers. We put al = a2 + 7 so
a2
and
2a
= 2a 2 + b2,
b2
b
b2
If
1
al
and
b
1
a
2
1
so
2b1 = b2,
are halves of odd integers,
Also, if a 2 1 + b 2
gers,
=
1+b
then
a
2
= al - b 1.
are rational integers, so are a2 and b2.
and b1
al and
b
2
0
0
a
2
and
b
If
are integers.
2
is such that a 2 and b 2 are rational inte-
is in Ra[ i:T.1].
Because, if a 2 and b 2
are rational integers and b 2 is even, then al and
al integers; if b 2 is odd,
al
and
b
1
b
1
are ration-
are both halves of odd
rational integers.
That the numbers are given without repetition follows from the
6
fact that
1
and
1
(-2+-2
r--- 1 1) are linearly independent.
is a basis for Ra[1],
Theorem 1. 8: If 01, 02
basis of Ra[ r:71]
every
is given by
01' = a 1101 + a 1202
a2202
a2191
where the al s
are rational integers and
all
a
Conversely, every such pair
Ra[
= f 1.
a 22
011' 021
constitutes a basis for
71].
Let
02
21
alt
01'
and
02
are in Ra[ r:T.1]
1)
011
a1181
be a basis for Ra[r:T.1].
and
a1202,
1]
2) 01 = b1101
and
0
11'
b128 2
82 + a2101
02
a2202
1
0
1
where the al s
and
82
are in
form a basis,
02 = b 2101 + b 2202
where the bl s
are rational integers. Substituting the values of
2) we have
and
0
01, 02 form a basis,
are rational integers. Also, since
Ra[
Since
011,
2
in
7
(b11a1 1+b 1 2a21
1
)01 +
(311 a1 2+b 1 2a22)02
02 = (b21a11 +b 22a 21 )01 + (b21a12 +b22a22)02.
Since
01
and
form a basis they are linearly independent and
02
we can equate coefficients as follows:
b
b
11
11
a11
1
b
al + b12a22 = 0
b
+ b12a21 =
21
a11
+ b22a21 =
a
+b
21 12
a
22 22
0
= 1,
or
b
11
b
12
1
12
all 11
0
=1
b
21
b
a
22
21
a
0
22
Hence each determinant of the coefficients is
Now, if 01, 02 is a basis for
a1181
bination of
a1202,
821
a2101
Solving these two equations for
(a
02 = (a
but by hypothesis
Ra[ r-T.1]
1
or
-1.
then define
Clearly every linear com-
a2202.
with rational coefficients is in Ra[.71-.1].
Olt, 021
el1
+1
.
0
1
and
02
we have
0 1-a 12 0 2 ) a a 1
all
12 21-a 22
22 1
2
01 1-a
1
1102) a12a21- a22a11
8
all
a
12
=a
a
a
21
11
a
22
- a 21 a 12
= ±1.
22
b's such
Therefore we have shown the existence of rational integral
that
Al = b 11011 + b 12021
02 = b21011 + b 2202'.
Thus, since every number of Ra[1]
combination of
Ra[
-NTT.1]
Al
and
can be written as a linear
it is seen that every number of
02'
can also be written as a linear combination of
with rational integral coefficients. Hence
1' 02'
01" 02
constitute a basis
for Ra[Ni7T1].
Example: Since
we have
1
1
r---
1, 2 + 2 N1-11
form a basis for Ra[
1
Al' = 3.1 + 2 (7+-2-1)=
4+
1
1
1
5
Nr-T.1)=192+-N/71-1
02' = 7 1 + 5 (-2+2
2
,
form a basis for Ra[ r-T.1].
Note that
all
a
a21
a22
12
3
2
7
5
= 1.
In general, all
-Nnn.]
01.1, AZ
may be written as
9
I
1
11)
all + a12(1+7
el'
1
02' = a 21 + a 22(-2+-2
.1)
where
all
a
a
a
12
= ±1
21
22
Theorem 1. 9: The norm of the product of two numbers in
Ra[ 17-171]
is equal to the product of the norms.
Let a = p + qT.1
=r+s
and
N(a(i) = N(a)N((3).
be in
-Nil
Ra[471-.1],
N(aP) = [(p+q,\T-171)(r+sq71-.1)]
= N[(pr-Ilqs) + (ps+qr)
= p 2r
=
=
2
+
11q2r
2+11q2)r 2
2
+
+
11p2s
11s2(p
.Nr:T1]
2
+
121q2s
2
2+11q2)
2+11q2)(r 2+11s2)
= N(a)N(P)
Theorem 1. 10: The norm of a quotient is equal to the quotient
of the norms.
N(41)
-
Rf31N
,
p
0.
Let a, p be as in Theorem 1. 9. Then
10
N(a)
N(p+q4.T.11
r+s\r"--11'
=
N[
=
N(
Nf7.1) ,
(p+q .147.1.)(r-sT
r 2+11s 2
pr+11qs
clr-Ps
2+11s2+r2+11s2
r
p2 r
2+
22pqrs + 121q2s
2+
11(q2r
2-
2pqrs+p2s 2)
(r 2+ 11s2)2
p2 r
2+
11q2r
2
+
11p2s
2+
121q2s
2
(r 2+11s2)2
(p2+ 11q2)r
2
+
11s2(p
2+
11q2)
(r2+11s2)2
2+
(p2+11q2)(r 11s2)
2+11s2)2
(r
p2+11q 2
r 2+11s2
N(a)
N(13)
Units of Ra[47171]: If a and
a 13= y,
Ra[
then
a and
A number of Ra[
1.
are in Ra[ r-7.1]
are divisors of
iff. there exists a
i=171]
divides
3
13
-1471-1]
3
in
Y.
Rak -r.1]
a divides
and
y
in
such that a p. = -y.
is called a unit of Rak1 if it
11
Theorem 1. 11: A number, E, of Ra[ r-1.1]
The units of
N(E) = 1.
is a unit, there exists an
el
If
are
Ra[Nr.1]
1,
E
2
is a unit iff.
-1.
such that
Since
Theorem 1. 9 we have N(eiE2) = N(E1)N(Ez) = 1.
e
E
1
2
= 1.
N(a)
By
is a
positive rational integer for all a E Ra[NFT.1], MEd = N(E2) = 1.
If
N(E) = 1
E = a + btr-T.1.
Hence
2
2
then we have N(E) = a + 11b = 1,
The only solutions to this equation are a = ±1, b = 0.
= ±1.
E
Prime Numbers of Ra[ 171-.1]: A number
neither
if
where
0
Tr = a1i
Tr
is a prime if it is
nor a unit, and is divisible only by a unit and itself. (i.e.,
implies that a or
is a unit. ) A number that is non-
(3
zero, not a unit, and not a prime, is composite.
Example: It can be seen that the number
1 + Nr:T.1
posite because
1
= 2(-2+-21)
1
1+
and neither of the factors,
2,
+
1
1
2
2
- ii,
is a unit.
Example: Assume
2 +31 = ar4
where
a = p + q47-171,
13 = r + sr\M-1.
is corn-
12
Then
r-T.1
1) = N(p+qtr-1.1 ).N(r+sN)
N(2 +3
103 = (p2+11q2)(r
2+11s2).
are all rational integers then one of a, p is a
Now, if p, q, r, s
unit. Otherwise we have the two cases:
i) p, q
are rational integers and r, s are both halves of odd
rational integers. Let r = (2m+1)/ 2, s = (2n+1)/ 2, m and
n
are rational integers. Then
4 103 = (p 2 +11q 2 )[(2m+1)
2+
11(2n+1)
2].
Now, we have the following possibilities:
2
2
a) p + llq
2
b) p + llq
2
2
2
2
2
c) p + llq
d) p + llq
2
2
2
2
2
2
= 2
(2m+1) + 11(2n+1)
=
4
(2m+1) + 11(2n+1)
=
103
(2m+1) + 11(2n+1)
=
206 (2m+1)
2
+ 11(2n+1)
2
=
206
=
103
=
4
= 2
a) has no solution because no rational integers exist such
that p
2
+
llq2
= 2.
b) has no solution because the left hand
member of the equation
2
2
(2m+1) + 11(2n+1) = 103
is al-
ways an even rational integer. c) and d) have no solutions
because (2m +1)2 + 11 (2n+1 )2>
11
for m, n rational integers.
13
ii) p, q, r, s are all halves of odd rational integers. Let
P=
(2m+1)
2
q=
(2n+1)
'
2
r=
(2k+1)
2
'
s-
(2j+1)
2
where m, n, k, j are all rational integers. Then
[(2n+1)2+11(2m+1)21[(2k+1)2+11(2j+1)2] = 103.16
and we consider the following possibilities:
a) [(2n+1)2+11(2m+1)2]
=
103
[(2k+1)2+11(2j+1)2}
=
b) [(2n+1)2+11(2m+1)2]
=
206
[(2k+1)2+11(2j+1)2]
= 8
c) [(2n-1-1)2+11(2m+1)2]
=
412
[(2k+1)2+11(2j+1)2]
=
d) [(2n+1)2+11(2m+1)2]
=
824
[(2k+1)2+11(2j+1)2]
= 2
e) k2n+1)2+11(2m+1)2]
=
1648
[(2k+1)2+11(2j+1)2]
= 1
16
4
a) has no solution because the left side of the first equation is always
even for n, m rational integers. The other cases have no solutions
because in each set of equations,
[(2k+1)2+11(2j+1)2] > 11
for k, j
rational integers.
Hence it is concluded that 2 + 34:T.1
the product of a unit and itself, therefore
can only be written as
2 + 34--T.1
is prime.
Furthermore it can be proved in a manner similar to the above
example, that if
a. e
RaH\FT1]
and
N(a)
is a prime rational
14
integer, then a is prime in Ra[4::Th.
But it can be seen, from
the following example, that a prime rational integer is not necessarily
a prime of Ra[Nr:T.1].
Let
then
a = 11,
a = (47T.1)(-N :F.1)
and since neither of
these factors is a unit of Ra[i2-11.] but both factors are in Ra[ r:TI],
therefore
11
is not a prime of Ra[
is any integer of Ra[1] and
Theorem 1.12: If a
any non-zero integer of Ra[
Ra[ r:I.-1]
.1471-.1].
there exists an integer
1],
is
of
such that N(a-y13) < N((3).
By Theorem 1.8,
-
r-
form a basis for Ra[[7.1].
- 2 + 2N- 11
1,
Let
a
where
p = r + r1
integers nearest
r---=p+q(--2+-21 N-11)
and
q = s + s 1,
and
p
r and
1
If
=r+
1
1
2
2
-Nr7-.1)
then
and
being the rational
respectively. Therefore
q
kit
a
s
1
= r + s 1 (--12 +7,4711)
1
15
2
=
- r1s1 + 3s1.
Now if
Ir 1<
I si I <
and
47
1
,
we have
r 12 - r s + 3s 12 < 1.
1)
1
Also, if either
Ir
or
I
choose the signs of r 1
1
1 slI
sl
and
or both equal
1/2,
then we can
so that they are alike and hence
r 12 - r s + 3s 12 < 1.
1
1
If
1
N5
and
< 1r 1 1 <
1
1
2
p;r5
1
L.
r1, si have the same sign equation 1) holds.
have opposite signs, for
depending whether
r1
we put
r2
N13\
rs-
is of the same sign as sl,
r
Hence
r 2 = r1 + 1
rl is negative or positive.
Irzi <
and
< 1s1 I <
2
If
or
Then
1
in which case
- r2s1 + 3s 12 < 1.
r1
and
r2 = r 1 -
s
1,
1
16
N(2*- -11) < 1
SO
N(a-p.P)< N(13)
and
Y=
Let a =
Example:
3
+
a
so
r = 1 , s = -2
7
I=-
5
+
1
tf-.1,
then
2
19
,--=-3
-(--2+-2 N/-11)
9
1
1
and
= 1 - 2(-12-4-12-) = 2 -
N(a-03) =N[(4+-P7T.1)
- (2-471-1)(-;+2 47T-.1)]
= N(1) = 1 < N((3) = 9.
Theorem 1. 13: If
a and p be any two integers of Ra[r-T.1]
prime to each other, there exists two integers,
Ra[
1]
such that
If either
integers,
and
a or
and
11,
of
a +1311 = 1.
13
is a unit, then the existence of the required
is evident. i. e. If a is a unit then so is
1/a
= 1/a, ii = O.
In case neither a nor p be a unit, assume N(() < N(a)
17
which does not limit the generality of the proof.
By Theorem 1.12 there exists an integer
Thus
N(a-µR) < N(13).
(3.
such that
al,
a-1-0 are a pair of integers,
and
prime to each other and
Pi,
II
N(a-11.13)
is less than both N(a) and
N(P)
If, now, two integers
ti,
exist such that
all
that is Pti +
= 1,
then we have
= tl
=
The determination of
al
nor
112
13111 =1
ti, Ili for al, Pi may, if neither
is a unit, be made to depend similarly upon that of
131
for a pair of integers a 2,
132
prime to each other and such
that the norm of one of them is less than both N(a 1) and
N(131).
By continuing this process, we are always able to make the determination of
t
and
n
for a pair of integers an,
Since the existence of
of
t
and
Ti
depend eventually upon that of
13n.'
tn,
one of which is a unit.
and lin
is evident,
the existence
is proved.
The following example will serve as a method by which
may be found.
t
and
18
Example: Let
number of Ra[t\FT.1].
Suppose
N(p)
N(a)
25
is a prime
a.
Then a
is a
But the norm of each number of Rakr-7.11
47
rational integer. Therefore
3
is divisible by
p
a
This means that N(a)N(y) = N(P) or
would exist such that ay = p.
N(y)
r-p = 17 +7N-ii.
a= 6 + 47711,
p
is not divisible by
a,
a and
so
are relatively prime.
N(P) = 25 < N(a) = 47.
By Theorem 1. 12 there exists
such that N(a-
µ
Using the method previously exampled we find a p. -=
a- p.P = -
So
nor
find a p.i
and
1
-
1
ra
= pi, p =
1
+
3
= al.
such that N(ar p.ipi) < N((3i)< N(ai).
1
= p..
Then
al
1
-1
P1
a2 = pl.
So
O. a + (-1)p
2
(- 1 )
( -1 )[ (3-
2
11 1 Pl)
) (a
=1
1
P)] = 1
p.a + (-1-p.2)p
Then
1
.7
Now neither
be a unit so we use the method of Theorem 1.12 again to
P1
=
5
-
< N((3).
P 2'
We find a
which is a unit,
al
19
=21
=
Corollary 1. 13: If
1
(-1-112) =4+-1
and
a
- 21
are any two integers of
3
there exists a common divisor,
Ra[4717.1],
that every common divisor of a and
t
exist two integers,
a
and
p
divides
Ra[r-T.1]
of
a and p such
of
5,
and there
8,
such that
= 5-
+
a and
If
p
are relatively prime then 6 = 1 because by
Theorem 1.13 there exist t and
a and
If
f3= (31y
where
p
al
such that at +
Ti
= 1.
are not relatively prime then let a = a 1 y and
and
are relatively prime. Then by Theo-
p
1
rem 1.13 there exist integers t and
Ti
such that
1.
We multiply both sides of this equation by y to obtain
a.lyt + Pon = y
3
divides
y
or
Every common divisor of a and
at + Ri = y.
and hence
5 = y.
The number
greatest common divisor of a and
5
p.
Theorem 1. 14: If the product of two integers,
Ra[tr-T.1]
is divisible by a prime number,
integers is divisible by
Let up= yrr,
is called the
Tr,
a and
p,
at least one of the
Tr.
where
y
is a number of Ra[471:1],
and
of
20
assume
is not divisible by v.
a,
a and
Then
are prime to
it
each other and by Theorem 1.13 there exist two integers,
such that
71,
ag +
g
Multiplying this equation by
= 1.
p,
and
we
have
Rat + RTr = R,
and therefore
Tr(Yt +PrI) = 13.
Since
(-y g+ Pi)
is in Ra[t
we have
1]
is divisible by
p
Tr.
Corollary 1. 14; If the product of any number of integers of
Rab17.-1]
is divisible by a prime number,
integers is divisible by
and such that al. a2 a3
If
an = Trp.
-Nr-
1}
Hence
this manner if
it
proved. Otherwise
it
it
y1 = az a3... an,
Let
divides either
a
2-
y2 = Tr13
1
divides either a 2 or
divides
it
so
are in Ra[Nr-7.i.]
an,v,13
the corollary is proved. If
al
then let y2 = a3 a 4... an
Ra[
al, a2, a3,
and by Theorem 1.14
divides
it
at least one of the
Tr.
Suppose the numbers
aiyi = TrP
Tr,
it
al and
yi.
divides
yi
for some
Y2.
then
(3
1
in
Continuing in
ak, k = 1,2, ... , n-2 the corollary is
divides y. = a.j+1 a.3+2 ... an, j = 1,2, ..., n-2
and the number of factors of
Y.
is reduced one at a time until there
are only two left. Then by Theorem 1.14 one of these two is divisible
by
v.
21
Theorem 1. 15: (Unique Factorization Theorem).
ber of Ra[
Every num-
can be represented in one and only one way as the
-NFT-1]
product of prime numbers.
a be in Ra[kr, ]
Let
.
a is not prime, we have
If
where neither of y, p is a unit. Then N(a) = N(R)N(y),
a = (3y,
and since
N((3) # 1, N(y)
we have
1,
N((3) < N(a)
and
N(y) < N(a).
If
B = 16
'
l yr
where neither of
is a unit and as before N((31)< N((3) and N(yi)< N(p).
131' YI
(31
is not a prime number then
p
If
is not a prime, we proceed in the same manner, and, since
Na3), N((3
N(13 ).
1),
form a decreasing series of positive rational
.
2
integers, we must after a finite number of such factorizations reach
in the series
factor
a
and we have
Tri,
al
If
1T=
1
2
CI
2
,
13' P1' P2'
a prime
Tr
1.
Thus
a has a prime
a = Trial.
is not a prime number we proceed similarly to obtain
where
Tr
2
is a prime, and hence
a = Tr1Tr2a2.
form a decreasing
Since the series N(a), N(a1), N(a2),
series of positive rational integers, if we continue this process we
must reach in the series a, a 1 , a 2,
we have
Hence
a = 1T
a
I
TT
2
Tr3... 11-n ,
...
a prime number
where the
Tr 's
Trn.
are all prime.
can be represented as a product of a finite number of
Thus
22
factors all of which are prime.
Now suppose
of
a
a = Pi P2 P3-
is another representation
Pm
as a product of prime factors, then
1)
"1 "2 "3.
P1 P2 P3.
"n
Pm'
say pl,
By Corollary 1.14, then, at least one of the p's,
is divisible by
Dividing 1) by
;
Tr
1
,
Tr
1
that is
pl = ±1
we have
Tr
1
1
is a prime.
= (±1)p2 p
3...
Tr2
since
Prn
Again, by Corollary 1.14 at least one of the remaining p's, say
P2,
i s divisible by
Tr3
"4*
p2 = ±1
Thus
Tr2.
1)
'icrn =
1)P3
P4.
Tr2
and
Pm = ±
p3 134 Pm'
Proceeding in this manner, we see that with each
associated at least one
and, if two or more
p,
Tr
there is
are associat-
Tr's
ed with each other, at least as many p's are associated with these
Tr's,
and hence with one another.
In the same manner we can show that with each
sociated at least one
Tr,
there is as-
and, if two or more p's are associated
with one another, at least as many
p's,
p
Tr's
are associated with these
and hence with one another.
Since in all questions of divisiblity we consider two associated
factors as the same, (i. e. ,
a + bc\ir.T.1
is considered the same as
-a - b471-1), the two representations are the same. Hence the repre-
sentation of a number of Ra[
-77.1]
in prime factors is unique.
23
THE QUADRATIC INTEGRAL DOMAIN Rakr1-6
II.
Consider a quadratic equation with rational coefficients, irreducible over the rational field, of the form ax2+ bx + c = 0, a
0.
We can assume that a, b, c are rational integers since this assump-
tion does not alter the roots of the equation. We will consider the
case b 2 -4ac = 10d 2,
Let
p
where
d
is a rational integer not zero.
be one of the roots of the above quadratic.
p
is not
a rational number since the equation is irreducible. Denote by Ra(p)
the set of numbers
where r
r + sp
and
range over the ra-
s
tional field Ra.
Theorem 2. 1: There exists a rational integer,
repeated factor such that
m,
without a
Ra(p) = Ra(Nir71).
For the above considered case m = 10.
this theorem is similar to that for m = -11,
Since the proof of
we omit the proof.
When the proof of a theorem of Ra() is similar to the
proof of the corresponding theorem of Ra(NFT.1),
the theorem will
be stated without proof.
Example: Let
2x2 + 8x + 3 = 0,
p
then
b2
- 4ac = 10.4 and
-8+47175
-8+24T.0
-4+41-0
4
4
2
24
So
+ qp = p + q(
Also, since p =
-4+4-171)
1r
2+
Ni 10'
2
- (p-2q)+
we have
q
4-1-0 = 2p + 4
versely. Hence
p + q147.0
and
And it can be seen that
p + q47.0 = p + q(2p+4) = (p+4) + 2qp = r + sp.
numbers of the form
10=r+s
are of the form r + sp and con-
Ra(p) = Ra(NFIT)).
Theorem 2. 2: The set
is a field.
Ra(T.0)
The proof of this theorem is similar to that of Theorem 1. 2.
We will show that each non-zero number of
rocal in
Ra(41-70)
where not both p, q
Let (p+qt\TO) e Ra(Nr17)),
Ra(NITO).
has its recip-
are zero.
p-qq1-6
p+qq-1-0 -p2-10q 2
-q
2
p2-10q
1
This is in RaNTO) provided that
Suppose
= NriO
p2-
But both
this implies that
1 Oq
2
p, q
Nr1-0
p
2
p -10q2
2
- 10q 2
0.
p, q
then
2
= 0.
Since
0
12- = 1 0
2
are rational numbers and so is
P
so
and
is rational, a contradiction, therefore the re-
ciprocal of a non-zero number of RaNT.0) is in
Theorem 2. 3:
NIT°
Every number of Ra(47.0)
ratic equation with rational coefficients.
Ra(4-1-0).
satisfies a quad-
25
Let a = p + q41-0 be in
Ra(41-0),
then
2
2.
a satisfies the
equation
x2 - 2px + p - 1 Oq
This equation is called the principal equation of
stant term N(p+q0) = p 2 - 10q 2 is called the norm of
can be seen that N(a) = a a,
of the coefficient of
p + q41-0.
where
a=p-
The con-
a.
a.
and it
The negative
q4170.
x, T(p+q) = 2a is called the trace of
are rational,
Since p, q
N(p+q4-170)
and
T(p+q,\TO)
are also rational.
Integers of Ra(41-6): The set of all numbers of
Ra(N/To)
satisfy equations of the form x2 + bx + c = 0 where b and
are rational integers constitute the integral domain Ra[TO]
that
c
of
Ra(N5.7)).
Theorem 2. 4: Every rational integer is in
number of Ra[iro]
Ra[N5.15].
which is rational is a rational integer.
Theorem 2. 5: The conjugate of a number of Ra[ITO]
Ra[Nrib] .
Theorem 2. 6: The numbers of Rakr7)]
a + birro,
Every
where a, b
are given by
range over all rational integers.
The principal equation for a number a = a + 13,./0 is
is in
26
2
x2 - tax + a If
Ra[rio]
aE
x2 + px + q = 0
and
a
2
then
a
2
consider: either
2
= 0.
satisfies an equation of the form
where p, q
- 10b = q
1 Ob
are rational integers. Hence 2a = p
are rational integers. There are two cases to
a = p/2 is a rational integer or half an odd ra-
tional integer.
a = p/2 is half an odd rational integer.
Suppose
So
2
2
a = (2n+1)/2 where n is a rational integer. From a - 10b = q
we have
4 10b2 = a
1)
Now, if
2b
2
- q = 4(n 2 +n-q) + 1
were not a rational integer then the square of its de-
nominator would divide
10
which is impossible. Then the left side
of 1) is an even rational integer while the right side of 1) is an odd
rational integer, a contradiction. Thus a = p/2 must be a rational
integer.
We have shown that
2
a - 10b
From
2
=q
a = p/2 must be a rational integer.
we have
10b
2
2
a rational integer. If
=a -q
b were not a rational integer, the square of its denominator would
divide
10
which is impossible. Therefore
b
must be a rational
integer.
Moreover it can be seen from the principal equation that if a, b
are rational integers then (a+b,\FF)) e RaNT0].
27
Ra[tsin]
Basis for Ra[r.5.75]: Every number of
out repetition in the form a
1 + bN/T.0
where
a
and
is given withb
range in-
dependently over all rational integers and every such number is in
Ra[r1.-0].
We call two such numbers,
1, NI0,
a basis for
Ra[r1.15].
Theorem 2. 7: If
every basis of
1
Ra[NnT)]
and
47.0
is a basis for Ra[5.71],
is given by el'
all + a12NIT6'
02 = a 21 + a 22 ,r10 where
all
a12
= ±1 .
a
21
a22
Theorem 2. 8: The norm of a product of two numbers in Ra[4.7]
is equal to the product of the norms.
N(a13) = N(a) N(I3)
Theorem 2. 9: The norm of a quotient is the quotient of the
norms.
N(R)
N(P)
P # O.
It was shown in Theorem 2. 2 that if p # 0 then N((3) # 0.
The rest of the proof of this theorem is similar to the proof of Theorem 1. 10.
28
Units of Ra[rr.0]: If a and
a
then
y,
Ra[T.0]
a and
are in
are divisors of
13
iff. there exists a
in
Ra[ ri.-0]
and
a divides
y.
y
in
such that a. (3 = y.
Ra[Nil3]
is called a unit of Ra[iro]
A number of Ra[t.n7)]
vides
(3
if it di-
1.
Theorem 2. 10: A number
E
of Ra[]
is a unit iff.
N(E ) = ±1.
Theorem 2.11: All units of Ra[n.0] have the form
(3+4) n,
where n is a positive or negative rational integer or
zero, and all numbers of this form are units of Ra[n--.0].
Let E = 3 + \M.
So
N(En)
[N(E)]n
(-1)n = ±1.
=
Hence
En
is a unit.
Also, since
-n
En
E
= 1,
-n
E
is a unit.
Moreover, since E = (3+470) > 1,
are all greater than
1
the positive powers of
E
and continually increase. Hence no two are
equal.
Since
-n
1
n
-1
it is clear that
E
are all less than
1
and hence that the
<1
E
negative powers of
e
and continually decrease.
Therefore no two negative powers are equal, and no negative power is
equal to any positive power. So every power of
Ra[477],
E
is a unit of
and two different powers give different units.
29
Now we need to show that the powers of
multiplied by ±1
E
are all the units of Ra[ir0].
-a -
-a + bt4T-0,
are also units of Ra[41-0]
134170
remaining three will be
1
>1
_iv
and
11
n
E
<E
<
Denote that one
11 1,
so the
-7.1
it follows that either
1
.
a > 0 and b > 0 by
of these four units which has
Since
Then a - brrr.0,
be a unit of Ra[r1-0].
Let a + bt.rro
=E
1
or
n+1
1
where n is a positive rational integer or zero.
Dividing 1) by en,
we have
11
<
2)
n
<E
;
E
where
1
/en
is a unit, since the quotient of two units is a unit.
Let
= x + yArib
.
Then
(x+y4T.0)(x-y\TO) = ±1,
and hence, since x +
yNri.7) > 1,
it follows that
<1
30
or
-1 < x - yqr.0 < 1
This combined with 2), gives
0
1
< x < 2+
2
r--
Ni 10 ,
and since x must be a rational integer,
x = 1, 2, or
If
RaNT.0].
nor
by 2)
x = 1,
If
2 + Nrr.0
but
y = 1,
then
x = 2,
y=0
3.
1+417-0
or
1,
is not a unit of
but neither
And if x = 3, y = 0,
is a unit.
but
2 + 0NFF.0
3 + ONTO
is
not a unit.
Hence 1) is impossible, and we have
n
111 =
and therefore
= -E
11
t
1
n
and since 1101' = ±1,
= ±n1= ±E-n
and
-TV = TE
E
Therefore, if
we have
= ±E
.1-1
is any one of the four units
n where
n
is a rational integer.
ri it, -ill,
31
Prime numbers of Ra[ r1-0]: A number
prime if it is neither
0
it
is
Rakn.-0]
of
nor a unit, and is divisible only by a unit
and itself.
Example 1: To determine whether 5 - 2\70
is prime or com-
posite. Put
5 - 20 = (a+bNIT0)(c+d4170),
then
-15 = (a
2-
1
Ob2)(c
2-1
Od2)
There are only four distinct cases to be considered:
a2-
2
= -5
a2-
10b
a2-
= -3
1 Ob
2
=
1
iii) and iv)
ii)
i)
c2 - 10d
a - 10b
a2-
=3
1 Ob
2
c
=5
2
- 10d 2 = *15.
Both iii) and iv) have a + bt\r1-0 a unit and therfore need not be con-
sidered. From i) we have
b
2
=
a
2-3
10
and this makes it necessary that a 2
7- 3
mod. 10.
rational integer exists. Hence i) has no solutions.
From ii) we have
b2 -
a2+3
10
But no such
32
2
and it is necessary that a 7, 7 mod. 10.
But no such rational in-
teger exists. Hence ii) has no solutions. Therefore 5 - 241.-0 is
prime.
Example 2: It can be seen that the number
since
Also
26 = 13- 2.
26
is not prime
and neither of these
26 = (6-EN/T.0)(6-40)
factors is a unit. It remains to be seen whether they are prime factors.
where neither factor is
Assume (6+4i-0) = (a+b1.5-5)(c+d,\70),
2)(c2
-10d
a unit, then 26 = (a2 -10b
2).
There are two cases to consider:
a2- 10b2
= 13
i)
10b
c2-
10d = -2
Since b
2
± 13
- a 10
requires that a 2
such rational integer,
(6-NiT.-0)
a,
= -13
ii)
c2- 10d2 = 2
2
2
a2-
3
2
or 7 mod. 10 and no
exists, we conclude that
(6-I-Nri-0)
and
are prime factors.
Therefore it is clear that
26 7-- 13
2 = (6+4T.0)(6-1\570)
cannot be factored uniquely into prime factors.
The introduction of so-called ideal numbers restore this property of unique factorization in terms of these ideal numbers.
33
not both zero, had a
If every pair of numbers of RaNT.0],
greatest common divisor (g. c. d. ) expressed linearly in terms of the
numbers we could prove unique factorization. (See Theorems 1. 12
through 1.15. )
Example: Consider the positive rational integers congruent to
1 modulo 4. This set is closed under multiplication. A number of
this set is considered prime if it cannot be written as the product of
two numbers E 1 mod. 4,
where neither of the two numbers be a
unit. Then
2541 = 33-77 = 121-21
33, 77, 121, 21 are all primes.
where
The cause of the failure of the unique factorization law is because of the absence of the remaining positive integers. Let
denote the g. c. d. of a and b
11 = (33, 121) = (77, 121),
Then
(a, b)
so
7 = (77, 21),
2541 = (33, 121)(77, 121)(77, 21)(33, 21)
3 = (33, 21).
is uniquely factored in-
to prime "ideal" numbers.
Ideals of Ra[47.0]: If al, a2,
Ra[41-0]
, an
are any numbers of
not all zero, then the set
{'1` ial+X2a2+ ..+Xn anIX1, X2,
,X
n
range over RaNT.01}
34
constitutes an ideal
A = (ai, a2,
,
of Ra[0].
A
We will denote A by
an).
Theorem 2. 12: In every ideal there exists two numbers w1 ,w 2
such that the numbers of the ideal are given by
kiwi + k2w2
where k1, k2
range over the rational integers.
Two such numbers are called a minimal basis for the ideal.
Let
of the ideal
1, Nr1-0
be a basis for Rakr1-01.
then A contains *a a = ±N(a),
A,
contains positive integers. Let
Of all numbers
as
one such in which
a. = al + a 2 \TO
w
1
is a number
and so
A
be the smallest positive integer
/ 2 is positive and minimal. Let
be any number of A.
a2
0
/1 + / 2 Nri.-0 in A having / 1 , 0, choose
in A.
(02
a,
If
=
k +r
2 2
2
Write
0 < r2 < Q
Then
a - k2w2 = (al-k2/1) +
is in
A,
and if
be violated. Thus
r2Nr1:0
r2 were not zero, the definition of
a - k 2w2 = a 1 -k 212 =
b = wlkl + r1,
b.
0<r1
Now write
<w
,
1
w
2
would
35
so that a - k2w2 - kiwi = r1.
was minimal,
Since
r1 = 0,
and
a = klwl + k2w2.
Corollary 2.12: Every rational integer in A is divisible by
Because if a is a rational integer then we can let k 2 = 0,
so
a = k1
Theorem 2. 13: If
in
is a minimal basis for an ideal A
w1, w2
every minimal basis is given by
RakrT.-01,
i,11 1=a
w +a12
w2'
11 1
where the a's
col=a
2
21
0.)
+a22w2
,
are rational integers such that
all
a12
a21
a22
and every such pair wi, w2
= t1,
is a minimal basis.
The proof is similar to that of Theorem 1. 8.
Theorem 2. 14: Every ideal A has a minimal basis
k, f + r40,
0< Q < k.
where k is the smallest positive integer in A and
36
In the proof of Theorem 2.12 we saw that we could choose a
basis wl = k,
integer in
A.
co
2
=m+
where k was the smallest positive
rNiTO,
Set
m = qk +
In the transformation
= k,
= (A)
(.0
0<
0j12 =
1
1
<k.
2
- qco
=
1
+ rNr1-0
we
have the determinant of coefficients
1
0
-q
1
= 1,
so the result follows from Theorem 2.13.
Theorem 2. 15: Every ideal A has a minimal basis of the
form
1
where r
and
a
= ra,
co2 = r(b+4-1:0),
are positive integers, and
b2
0 < b < a. Moreover,
- 10 a 0 modulo a.
Such a basis is called a canonical basis.
Using the notation of Theorem 2.14, since k
is in
A.
Set
k = ar + t,
0 < t < r.
is in A, 1c4T0
37
Then
- a(,) ' =
kNi-f-0
()
1
Since
Q + rNri-0
= ra,
(02 =
is in
A,
= br + t
and
0 - bo)
lOr +
Then
r
2
r
and a
0 < rb < ra,
Since
+ rNiTO .
so is
/4-1-.0 + 10r.
Set
0 < t < r.
1
1
= 10r - b./ + t 1 NT: 0 is in
A,
so
t =0
1
= ra,
w2 = r(b+Nr1-0),
are positive. Since, by Theorem 2. 14,
0 < b < a.
we have
(,)2NI-T0
Hence
Hence there is a basis
divides
1
where
This is impossible unless
A.
r divides k.
in which case
t = 0,
is in
+ tNr1-0
- b(.02 =
lOr - rb2
is a rational integer in
A,
it is divisible by ra by Corollary 2.12.
i. e.
b2 - 10 :S. 0 modulo a.
The product AB of two ideals A and
B
is defined to be
the set of all numbers obtained by multiplying every number of A
by every number of
and then adding and subracting these num-
B,
bers until no new numbers are obtained. This set of numbers satisfies the definition of ideal.
If
A = (w1,0,2),
and
B = (xi, x2),
then AB consists of the
38
numbers
k
co
1
1
x + k 2w1 x2 + k 3 co 2 x 1 + k 4'42 x2 ,
1
range over all numbers of
where k1, k2, k3, k4
Ra[ NiT.0]
.
It is evident that ideal multiplication is associative and commutative.
If all the numbers of an ideal A are multiples by numbers of
Rakri-01
of one number
a,
the ideal A is called principal and
is written (a).
The conjugate of
A,
denoted by A,
is an ideal formed by
replacing every number of A by its conjugate.
Theorem 2. 16: AB = A B .
Let A = (0)1, (02), B = (x1, X2),
AB =
=
then
w1X2, w2X1, w2X2)
1
X
1, w 1 X2, w2x1, w2x2)
=AB
Theorem 2. 17: If A = (ra, r(b+14-1-0)),
The number
r 2a
X
is called the norm of
= (ra, r (b-0))
then A A = (r2a).
A,
written N(A).
39
so
A A = (r2a2, r2a(b-4T-0), r2a(b+4-1-0), r2(b2-10)).
Then A A consists of all numbers
2
2
kr 2 a 2+ Xr a(b+NIT0) + p.r a(b-47.0) + vr (b
1)
where
k,
p.,v
2-10)),
range over all numbers of Ra[riT)]. By Theorem
2.15
c_
2-10
b
a
is an integer. The transformation
X =Xi + v1,
k = k1,
p. =
,
v = µ1
takes the set of numbers 1) into the set
2)
2
2
2
2 2
k1r a + X12r ab +p.ir ac +vir a(b+4171)
.
Hence every number of 1) is in 2). The converse is true, since
k1 = k,
Suppose that
b
2
X
1
a
- 10 = ac E 0 mod 4.
=
and
But
=v,
11
v
1
=X - p..
were both even. Then
c
b
2
0
or 1 mod 4
according as
is even or odd, and neither of these agrees with the fact that
b
40
Hence
10 m 2 mod 4.
Let g = (a, 2b, c).
is odd and so
g
c
are not both even.
and c
Since
a
b.
Then
divides
b2implies
and
a
10 = ac
10 s 0 mod g 2.
Since
are not both even,
0 mod g 2
10
has no square factor > 1,
g = 1.
We can now see that the set of numbers
3)
{klr2a2 +X12r2ab
is the same as the set
{pr
2
+11
r2ac}
alp ranges over RakriOn. Clearly,
every number of 3) is in {pr2a}.
Since
a, 2b, c
are relatively
prime, there exist rational integers p, q, t such that
1 = pa + 2qb + tc.
2a.
Multiply through by r
Then
r 2 a = pr 2 a 2 + 2qr 2 ab + tr2ac
2a}
is in 3).
so that every number of {pr
The set 2) is now seen to be equal to the set
4)
{p r 2a + v r 2a. (bi-Nrr.0 )}
g
41
r2a,
Obviously every number of 4) is a multiple of
every multiple of r2a
is in the set; i. e.
v1 = 0.
and, conversely,
Thus
A A = (r2a).
where A and
Theorem 2.18: N(AB) = N(A)N(B),
B
are
ideals.
By Theorem 2.16 N(AB) = AB AB
A. E
= AB
=AA BB
= N(A) N(B)
where
Theorem 2. 19: If SA = SB,
and B are
S, A,
ideals, then A = B.
The numbers of A are given by
kiwi + k2w2,
where wi,w2 form a basis for
Let
s = N(S)
X e Ra[Jr170].
1
1
and
k
1,
k
2
The numbers of (s) are given by
are in Ra[ T.0].
Xs,
where
Thus the numbers of (s)A consist of the numbers
Xki swi + Xk2sw2 =
where
A,
r1lswl
range over Ra[ri.0].
+ 1-25,,2 = s(Ti iwi
Thus every number of (s)A
42
is of the form
where a is in A.
Sa
or
then SSA = SSB,
SA = SB,
If
(s)A = (s)B,
where
s
is a rational integer. That is, for every number a in
A there is a number
in
p
so. = sp,
such that
B
0. = p,
and conversely. Hence every a is in
and every
B
p
is in
A,
so that A = B.
Divisors of Ideals: If three ideals of
AB = C,
B
we say that
A
divides
and
C
B
are such that
divides
C.
A and
are called factors of C.
Theorem 2. 20: A divides
C
Ra[4-170]
is in
C
if an only if every number of
A.
Let A = (GT (02), B =
x2),
then AB = C consists of all
numbers
kcol X1 + )1/4u X2 +1-1402X1 + vc.02 X2
1
where
k,X,}1,v
range over
following two ways,
Ra[r.5.70]
.
Putting this form in the
43
(kw1 +µw.2)X1
(Xco
+vxdco 2
(kxi + Xx2)(01 +
+v(.0
1
it can be seen that every number of
A and
is in both
C
Conversely, suppose that every number of
every number of CA is in AA = (a),
where
is in
C
Then
where
(3a,
of numbers of Ra[47.0 ] .
B
Since CA is an ideal, for every two numbers
132a
A.
is a positive in-
a
teger. That is, all numbers of CA are given by
ranges over a certain set
B.
of CA there are numbers
133a,
13
and
4a,
(31a
p5a
and
of CA
such that
(31 a - 132a = (34a,
(31a + 32a = 133a,
for every k
in
131
so that
B
Ra[0].
2
133)
kr3
la = p5a
Hence
P1
p
4,
= (35
kr3
,
1
is an ideal.
It follows from Theorem 2. 19 and
AC = (a) B = AAB
that C = AB.
Theorem 2. 21: A positive integer t occurs in but a finite
number of ideals.
p
44
Let the ideal
ra divides
t.
By Corollary 2. 12,
r > 0, a > 0, 0 < b < a.
where
(ra, rb+rt..70),
have a canonical basis
containing t
A
there are not more than
For a given t,
choices for each of the positive integers r, a,
3
fore not more than t
such ideals
Theorem 2. 22: An ideal
and
t
and there-
b,
A.
is divisible by only a finite num-
C
ber of ideals.
We have
orem 2. 20,
c
C C = (c) where
is in
is a positive integer. By The-
c
and also in every ideal which divides C.
C
By Theorem 2. 21 there is but a finite number of such ideals.
different from the unit ideal,
Prime Ideals: If an ideal P,
(1),
is divisible by no ideal other than itself and
are composite.
a prime ideal. All other ideals, except (1),
Greatest Common Divisor: An ideal
common divisor of A and
B
if
Theorem 2. 23: Every pair of ideals,
a
G.
G
is called a greatest
G divides both A and
and if every common divisor of A and
a unique g. c. d. ,
it is called
(1),
B
divides
A and
B,
G.
B,
possesses
It is composed of all numbers a. + 3 where
ranges over A and
p
over
B.
The set G of all numbers a + p satisfies the definition of
45
ideal. Since every number of
B
is in
A
G
and every number of
is in G, G is a common divisor of A and
Let
E
B.
be any common ideal divisor of A and
B;
that is,
any ideal containing all the numbers of A and all the numbers of
B.
of
Since it is closed under addition, it contains all numbers a. + p
G and hence divides
G.
G and
Suppose that
are two g. c. d. 's of A and
G1
Then G = K 1 G 1, G1 = KG,
B.
so that
(1)G = K KG.
1
Hence, by Theorem 2.19
Since
K1K = (1).
N(K K) = N(K1)N(K) = N(1) = 1,
1
K = K = (1).
1
So
G = GI
Two ideals are called relatively prime if their g. c. d. is
Theorem 2. 24: If A and
exists an a in A and a
Since A and
that is G =
But since
tion of wl, w 2,
B
13
X
such that a + p = 1.
where A =
is a number of
1, X2;
B
are relatively prime their g. c. d. is
X1' X2) = ( 1 )
1
are relatively prime, there
B
in
(1).
G,
(1);
w 2), B = (Xl, X2).
it must be a linear combina-
46
that is,
klwl + k2(.02 + /1x1 + 2X2 = 1,
where k1, k
But
k
1'
1(01
2
+ k2w2
are in Ra[ TO]
is in A and /1 x 1 + I 2x2
is in
B,
and
and is prime to
B,
then
we have
a + P = 1.
Theorem 2. 25: If A divides
A
divides C.
If
a
BC
A
is prime to
in A and a
p
in
then by Theorem 2. 24 there exists an
B,
such that
B
a + P = 1.
Multiplying through by y we have
ay + 13y =
For every
of
BC
y
is in
Theorem 2. 20,
C.
in
A.
Since
So is
A
BC,
the number
and so therefore is
ay,
A divides
divides
y.
yP
Then, by
C.
Theorem 2. 26: (Unique factorization of composite ideals). Every
composite ideal can be factored into prime ideals in one and, except
47
for order of the factors, in only one way.
That every ideal can be factored into a finite number of prime
ideals follows from Theorem 2. 22.
If C
is a composite ideal then
C = Al A A3... Am
2
where the A's are prime ideals.
Suppose that
C = B1B 2 B3... Bn
is another representation of
C,
where the B's are prime ideals.
Then
Al A A3... Am = B1B2 B3... Bn.
2
Since
Al
Theorem 2. 25,
is a prime ideal dividing B1B2B3... Bn then by
Al
divides some Bk.
rem 1.15. ) Also, since Bk
(As in the proof of Theo-
is prime we must have Al = Bk.
Since we can rearrange the order of the B's,
Al = B 1.
Now, since Al
we may assume
(0)
A2 A3... Am = B2 B3... Bn.
As before A2
since A2
and
divides one of the remaining B's,
B2
are both prime,
say
B2,
and
A2 = B2. We continue in this
48
or all the
manner until all the A's
ly m = n,
to
B's
are exhausted. Evident-
or otherwise we should have a product of primes equal
(1).
Hence every ideal can be written uniquely as a product of prime
ideals.
Example: To factor
into prime ideals.
(26)
(26) = (2)(13) = (6+Nri-.0)(6-4T.0)
= (4, 2,47.0, 2N/T-.0, 10)
(2) = (2, NiT0)(2,
but
(13) = (13, 6+14-170)(13, 6-Nr1-.0)
and
Now, suppose
(2, NIT.0)
is not a prime ideal. Then
(2, 4-170) = AB,
where neither A nor
B = (1). Let
A = (a1, a2, a3, .
. .
B = (b1, b2, b3,
,
am),
,
am)(b 1, b 2, b 3 ,
. ,
bn)
then
(2, Nri--0) = (al, a
By Theorem 2. 20,
2
2,
a
3
,
and
Nr1-0
,
bn).
are numbers of both A and
So
(2, \ITO) = (al, a2, ..., am, 2, 4170)(bl, b 2 ,
,b n,
2, N70).
B.
49
Let
a
k
=
be any one of the
p+c4-1-0
2r, 2r + 1,
or
2r-1.
a's.
Then p
is of the form
We have
+ 2r
1)
=
qa.
2)
=
qa.
NrITO + 2r + 1
3)
=
qa.
Nr170 + 2r
- 1.
If 1) is the case we may omit ai from the symbol
A.
If 2) is the
case, we have
-
and
1
qa.
NT° - Zr =
may be introduced in the symbol A,
so
A = (1).
If 3) is
the case, we have
=1
cr\ITO + 2r -
and again A = (1).
Continuing in this manner we find that either all numbers
al, a2,
, am are linear combinations of
A = (2, 14170),
A = (1).
or
1
2, Nrio,
in which case
may be introduced in the symbol A and so
Similar conclusions hold for
sible factorizations for
(2, t4-1:0)
B.
Therefore the only pos-
are as follows:
4)
(2, moo) = (1)(1) = (1)
5)
(2, Nr1-0) = (2,1\riso)(2, \r0) = (2)
6)
(2,
= (2, Nri70)(1)
= (1)(2,
Nri.-0).
50
If 4) be the case then
that is
must be in (2, trio),
1
2(x+rrr.0) +
0(u+vt4T.0) = 1,
2x + 10v = 1 ,
2y + u = O.
or
Since no two rational integers satisfy the equation
2x + 10v = 2(x+5v) = 1,
4) is impossible.
5) is impossible because
is not a multiple of 2.
41-0
Hence we have shown the ideal,
Suppose
(13, 6+4-1-.0)
(2, 4-170),
is composite. Then, as before,
(13,6 +41-0) = AB = (al, a2,
and by Theorem 2. 20
and
B.
13
and
for the numbers of Ra[4.5],
. . .
and
r,
am)(bi, b2,
1
,
bn),
are numbers of both A
6 + Niro form a basis
and
we can express any one of the num-
in the form a. = p + q(6+41-.0). Since
, am
a rational integer,
,
6 + NITo
Since, by Theorem 2.7,
bers al, a 2,
to be a prime ideal.
p
is
p
is of the form 13r ±n, n = 0, 1, 2, 3, 4, 5, 6,
a rational integer. Then we have,
- q(6+NaT)) + 13r] = ±n.
[a.
If
n = 0 we may omit ai
from the symbol A.
If
n
is any
other of its permissible values then we may introduce that number in
the symbol A and since
13
is also in
A,
we will have two ra-
tional integers in A which are relatively prime. Therefore, rational integers,
a, (3,
exist such that na + 13P = 1.
So
1
may
51
be introduced in the symbol
hence A = (1).
A,
Continuing in this manner we find that either all numbers
a1
,a
m
or
A = (13, 6+14-17j),
A = (1).
in which case
are linear combination of 13, 6 +41-5,
may be introduced in the symbol
1
Similar conclusions hold for
sible factorizations for
(13, 6+NiT.0)
B.
are as follows:
(13, 6+,41-6)
=
(1)(1) = (1)
8)
(13, 6+Nr1:0)
=
(13, 6+4-170)(13, 6+4-1-0)
9)
(13, 6+41-0)
=
(13, 6+470)(1)
= (1)(13,
is not a member of
1
so
Therefore the only pos-
7)
Since
A,
6+4-1-.0)
(13, 6+Nr1:0),
7) is impossible.
$) is impossible, for if not, then Theorem 2. 19 would imply that
but
(13, 6+Nrr.0) = (1)
1
is not in (13, 6+4T.0).
Hence we have shown the ideal
(13, 6-1-Nrr.0)
to be a prime
ideal.
Since
1, 6-N70
also form a basis for
Ra[J r-To]
a similar
(13, 6_) to be a prime ideal.
argument shows
Therefore
(26) = (2, 41-0)(2, Nr1-0)(13, 6+4-1-.0)(13, 6-Nr1-0)
and by Theorem 2. 26, this representation of
prime ideals is unique.
(26)
as a product of
52
BIBLIOGRAPHY
1.
MacDuffee, Cyrus Colton. An introduction to abstract algebra.
New York, Wiley, 1940. 303 p.
2.
Reid, Legh Wilber. The elements of the theory of algebraic
numbers. New York, MacMillan, 1910. 454 p.