AN ABSTRACT OF THE THESIS OF
Adelbert Frederick Hackert for the
(Name)
Date thesis is presented
Title
G2erf-,LY-/,
QUADRATIC INTEGRAL DOMAINS IN
Abstract approved
Mathematics
(Major)
M.S. in
(Degree)
/ ?6
5
Ra(q5) AND Ra(J -13)
Redacted
ed for Privacy
(Major professor)
Some of the properties of the numbers of two quadratic num-
ber fields are explored. Among these properties is the existence
of
unique prime factorization of the integers of the field and the impor-
tance
of the
concept of ideal numbers in restoring unique factorization
when it does not exist.
Some consideration is given to the relationship
between the nature of the ideals of an integral domain and the existence
of unique
factorization in that domain.
QUADRATIC INTEGRAL DOMAINS IN
Ra (^/5)
AND
by
ADELBERT FREDERICK HACKERT
A THESIS
submitted to
OREGON STATE UNIVERSITY
in
partial fulfillment
of
the requirements for the
degree of
MASTER OF SCIENCE
June 1966
Ra N-13)
APPROVED:
Redacted
Professor
of
for Privacy
Mathematics
In Charge of Major
Redacted
for Privacy
Chaiíriíän of Mathematics Departments
Redacted
for Privacy
Dean of Graduate School
Date thesis is presented
Typed by Carol Baker
August 9, 1965
TABLE OF CONTENTS
Page
Chapter
I.
INTRODUCTION
II.
THE NUMBERS
Ra (45)
III.
THE NUMBERS
Ra
IV.
THE IDEALS OF
V.
1
(i -13)
Ra [ 4-13]
5
35
42
A NECESSARY AND SUFFICIENT CONDITION FOR
UNIQUE FACTORIZATION
66
BIBLIOGRAPHY
71
QUADRATIC INTEGRAL DOMAINS IN
I.
Ra ('J5)
AND
Ra ('J --13)
INTRODUCTION
An algebraic number is defined to be a complex number which
satisfies
a polynomial equation with
rational coefficients. Every alge-
braic number satisfies many such polynomial equations, but among
these is one
of
least degree
determines the degree
(3, p.
of the
1
The degree of this equation
-2).
number. A number which satisfies an
irreducible quadratic equation is therefore called a quadratic number.
Suppose
p
is a quadratic number. We are first interested
are rational
in the set of numbers
al
+
numbers. For every
p
there exists some rational integer m,
b1p,
where
al
and b1
without a repeated prime factor, such that the set
identical to the set
al
+
a +
b'Im is
b1p (3, p. 280 -283).
The purpose of this paper is to consider two such sets, one in
which
m
=
5,
the other in which
m
=
-13.
The paper will show
that these sets are fields, and that in each field there is a particular
subset which is an integral domain, and whose elements will be called
integers.
1
In the
first set unique factorization
of
these integers into
prime factors will be demonstrated. In the second set this property
is absent, so ideal numbers will be introduced, which will
restore
To avoid confusion, the term integer will be used in reference to
1
a number of quadratic integral domain, while the ordinary integers
of arithmetic will be called rational integers.
2
the unique factorization.
In the development, the following definitions will be used:
1.
1
A group is a mathematical system composed of a set of
elements with a well defined binary operation and:
1.
The system is closed under the operation.
2.
The operation is associative.
(a
+
b)
+ c =
a
+ (b +
for every element
3.
a, b and
+
0=
0+ a
=
Every element
a+ a=
of the
a
a
of the
such that
set.
has an inverse
a
+
0
set.
=
a
such that
0
An abelian group is one in which the operation is
commutative. That is
a
+
b
=
b + a
for every element
1. 3.
c
a
for every element a
1. 2.
c)
There exists an identity element
a
4.
That is
a,b
of the group.
ring is a mathematical system consisting of a set of
elements closed under two well defined binary operations,
addition ( +) and multiplication (X) and subject to the
following:
A
3
The elements form an abelian group relative to
1.
addition.
2.
Multiplication is associative.
3.
Multiplication is distributive over addition. That
is, for every element a,b and c in the system
aX
1. 4.
1. 5.
(b
+c) =aX
b
+aX c.
An integral domain is a ring in which:
1.
The operation multiplication is commutative.
2.
There exists an identity element for multiplication.
3.
There are no proper divisors of zero.
integral domain in which every element
except the additive identity has an inverse under multiplication.
A field is an
It will be
assumed that the complex number system has been
developed and shown to be a field and that sufficient background to
establish the following specific properties from number theory and
algebra have been developed.
1.
6a.
1. 6b.
1.
6c.
The natural numbers are well ordered.
Every composite rational integer has a unique factorization into a finite number of prime factors.
If
or
a
1
is a rational integer, a2 is congruent to
mod 4, and in particular for
0
4
a=
1
a=0
mod 2, a2 =
1
mod 4
mod 2, a2
0
mod 4
FL-
and conversely.
1.
6d.
In an integral domain ab = ac
b =c for all b and c.
and
a #
0
implies
5
II.
Definition
and
Ra
THE NUMBERS
(N/5)
The set of numbers
2. 1.
a
+
b'ß/5,
where
a
range independently over the field of rational numbers, will
b
be called
Ra ('J5).
Theorem
is a field.
Ra (45)
2. 1.
Proof: a. The set is closed under addition and multiplication,
for if we take a = a + b'J 5 and ß = c + dN/ 5
we have
+c)+
a
+ß
=
(a
a
ß
=
(ac
+
(b
5bd)
+
+d)'ß/5
(ad
+
bc)
N/5
and a, b, c and d are rational numbers. Then
so are the coefficients a + c, b + d, and so on.
b.
Both operations are associative and commutative,
and multiplication distributes over addition since
Ra (q5) is a subset of the complex field.
c.
0 = 0 + ON/ 5
d.
If
e.
Each element a + b'./ 5 with not both a and b
equal to 0 has a multiplicative inverse in Ra(^/5).
a
+
b
N/
5
and
1
is in
=
1
are in
+ 0'x/ 5
Ra(N/ 5)
so is
For
1
a - bN/5
a + bN/5 - a2 5b2
R a (N/ 5) .
-a
- b^/ 5.
6
If a2 - 5b2 = 0 either a2 = b2 = 0, which by
b2
hypothesis is impossible, or
¡ O. Then
and ^/5 = a/b which is impossible
5 = a2 /b2
since that would mean N/5 is rational. Therefore
a2 - 5b2 ¡ 0 and we have
a
1
a
b\/5
+
conjugate of
the norm of
a
=
á
=
The product
and is denoted
a
'
Ra(I5).
The number
a +b'/5.
a2 - 5b2
a2 - 5b2
an element of
Definition 2. 2.
b^/5
is called the
a -b^/5
a a
=
a2 - 5b2
is called
N(a).
From this definition the following properties are clearly true.
a.
b.
á
a
=
.
if
=
a
c. N(a)
=
Theorem
2. 2.
Ti
a
is a rational number.
N(a).
a
á are the two roots
and
monic quadratic equation with rational coefficients.
Proof of existence:
Then
a
and
Let
b^/5
=
a
+
a=
a
-b^/5.
a
á satisfy the equation
(x - a)2 - 5b2
=
x2 - 2ax
+
a2 - 5b2
= 0
of a unique
7
and the coefficients
are rational since a and
a2 - 5b2
are.
b
Proof
á
and
2a
=
of
Case I,
uniqueness:
b
= O.
Then
a
=
a
and
The equation must have equal rational roots and so it is of
a.
the form
(x -
with
(a -
r)
=
0
a rational integer. Since
r
2
r)2
=
0, r
=
a
satisfies this equation
and the equation must be
a
(x - a)2
=
0
so the equation is unique.
Case II,
b #
Lemma.
a
linear equation
=
of the
O.
a
+
b^/5, b #
0
form
x
-r
=0,
Suppose the contrary. Then
a +bN/5
=
N/5
=
r
,
r
- a
b
does not satisfy a rational
8
which cannot be since
Suppose
Then
a
+
Hence the lemma.
satisfies two monic rational quadratic equations
a
x2
is irrational.
^/5
p lx
+ q
and
= 0
1
x2
+
p2x
+
q2
= O.
satisfies the equation formed by subtracting the second of
these from the first
;
(Pl
-
p2)x
+
ql - q2
=
0
.
This equation must be identically zero, otherwise it contradicts the
above lemma.
Therefore pl
=
p2
and
q1
=
and the two
q2
equations are identical.
Definition
2. 3. The
equation of theorem 2.
2
is called the
principal equation of a.
Corollary
of
a
is
The constant term of the principal equation
2. 2.
N(a).
Theorem
The conjugate of the product (sum) of two
2. 3.
is equal to the product (sum) of the conjugates.
numbers of Ra(^/5)
Proof:
Let
a
=
al
+ b1N/5
and
ß
=
a2
+
b2N/5.
Then
9
aß
And a +ß
=
ala2
+
=
ala2
- alb2N/5 +
=
a1(a2 - b2q5)
_
(a1 - b1,45) (a2 - b2q5)
=
á
=
al
+
=
al
- b1'./5
=
a +
ß
5b1b2 - (alb2
5bib2 - a2b1'J5
- b1'./5 (a2 -
a2 - (bl
b2)N/5
+
+a2 -b2q5
ß.
N(aß)
the product of two numbers of
be two numbers of
Let
a
:_.
aß
artF
=
aß
aß
by theorem 2.
3
=
aá
ß ß
by theorem 2.
1
=
N(a)nT((3)
Corollary
# 0,
of
is equal to the product of their norms.
Proof:
ß
b2q5)
.
Theorem 2. 4. The norm
Ra(J5)
a2b1)^/5
+
2. 4.
and
If
ß
a, ß
are two numbers
then
N( a )
ß
N(a)
NW)
of
Ra('./5).
Then
Ra('./5)
and
10
By definition 2. 2,
if
#
ß
0,
N(ß) #
and
N(a)
then
N( ß
N(R)
N(f3)
f3)
Definition
2. 4.
A
2. 2, so
N(a)
# N(a)
N(a)
N(a)
.
number of Ra('./5) is an integer of
if its principle equation has rational integral coefficients.
Ra(N/5)
The set of integers of
Theorem
Every number
2. 5.
Ra[
of
Proof:
a
If
O.
are rational integers by definition
N(ß)
N()
of
N(ß) #
If
a
Ra('./5) will be denoted
Ra[N/5]
.
Every rational integer is in Ra[ Ñ5]
N/5]
.
which is rational is a rational integer.
is a rational integer, the principal equation
is
x2 - 2ax
+
a2
=
0
and its coefficients are rational integers.
If
a =
a
+
b'ß/5
is rational,
b
= 0
+
a2
and
implies the principal equation
x2 - 2ax
=
0
a
in
Ra[N/5]
11
of
has rational integral coefficients.
a
integer so is
a
Theorem
=
But if
a
2
is a rational
a.
2. 6.
If
is in Ra[ 45]
a
then so is
,
á.
This is so since both have the same principal equation.
Theorem 2.7. A number
only if it is of the form
Ra(J5)
of
where
a +b'ß/5
integers, or where both a and
a
is in Ra[45]
and
are halves
b
b
if and
are rational
of odd
rational
integers.
Proof:
Let
a
be a number of
a
-
a1 +
where
al,
and
#
c
b1
b
0
1
111
Ra(g5). Then
^/5
cl
are rational integers with no common factor
and c1
to avoid the previous case where
may be considered positive without loss of generality. The prin-
cipal equation
of
a
is
2a1
x
x+
-
If in addition
a
Ra['45]
is in
al2 -5b12
c1
1
2
-
,
2a1
(1)
is rational. Now
a
c1
is a rational integer;
U
.
12
a 2 - 5b1
(2)
2
is a rational integer
cl2
.
Then one of the following is true:
(i) c 1
If
#
c
factor and by
(2)
c
or
1
or 2, then by (1)
(ii)
2
=
c
=
2,
c1
and from
4
=
w 0
mod 2,
b12
al =
mod 2,
mod 4
:I- 0
of
1
c
b1
.
have a common
contrary to the
(2)
4,
5b 12 mod
a 12
b
al and
=
are relatively prime.
and c
a12 - 5b12 = 0 mod
If
(iii) c
2
this factor is also a factor
al, bl
hypothesis that
If
1
4,
by property
a12 = 0 mod 4
and
1
6c.
1.
So
0
which makes
contradiction to hypothesis.
a12 =
a
+
mod 4.
1
b ./5
If
al =
Then
to be an integer
1
a
b1 =
1
al,
mod 2,
even in
b1 and c1
b12 =
1
mod 4,
so
mod 2. Thus, for this case, for
and
b
must be halves of odd rational
integers.
If
alt
- 5b1
al
and
c
=
1,
a
=
al
+
b1 .i5
is an integer since
2a1
and
are rational integers for all rational integral values
bl.
of
13
Thus it follows that any number
have
and
a
a +1345
rational integers or both a and
b
Ra['/5]
of
must
halves
b
of
rational integers.
Conversely any number
this form is in Ra['/5]
of
for the
equation
x2 - 2ax
+
2
2
a2 - 5b2
has rational integral coefficients if
and if
and
a
are halves
b
0
and b
a
of odd
=
integers
are rational integers,
is a rational
2a
integer and
a2-5b 2- n2-5m2
4
where
n= m
n2 - 5m2
°0
°
1
mod 2.
mod 4. Thus
Definition 2. 5.
02
So
form a basis for
given in the form
n2
m2 = 5m2 _
1
mod 4, and
is a rational integer.
a2 - 5b2
Two linearly independent numbers
if every member of
Ra[ J/5]
a01
°
+
be2
where
a
and
b
1
and
0 =
01
Ra['./5]
and
is
range independ-
ently over the rational integers.
Theorem
a
basis for
Ra[ 45]
Proof:
The numbers
2. 8.
2+ + 2
.
Consider the sets
S1
=
al
+ b1N/5
and
N/5
form
14
S2
a2
=
+
or halves
b2(
2
1
-1 +
2
integers. By theorem 2.7,
If a number of
al
+
+
is
S1
are rational integers
bl
and
rational integers and a2
of odd
tal
al
where
'/5)
and
Ra[N/5]
.
equals a number of
S1
b
b
bl/5
=
a2
+
2
+
2y5
=
2a2
+
b2
+
2a1
=
2a2
+
b2
=
2b
,
b2
are rational
b2
that is
52,
2,45
,
2
b2,45
;
(i)
,
(ii)
1
a2
If
al
and
=
al
- b1
are rational integers or halves
b
1
integers, then by (ii) and (iii),
So
a2
and
b2
of odd
rational
are rational integers.
51 S2.
If
al
(iii)
.
and
2a1
and
a2
b1
and
b2
are rational integers and b2
are rational integers by
2b1
(i) and (ii).
are odd rational integers so al
halves of odd integers and S2C Si. Therefore
(1, 0)
is a basis for
Ra[ q5]
Theorem
Ra[q5]
2. 9.
tion and multiplication.
b2
If
and
S1
is even,
=
b1
S2
is odd,
are
and
.
is closed under addition, subtrac-
15
Proof:
numbers of
Let
Ra[ ^/5]
a
al
+
±ß
=
aß
=
a1a2
82
=
(2+ 2^/5)2 = z+
aß
=
a la2 + b lb2 + (a lb2 + a2b
=
and
b10,
ß =
a2
+
be two
b20
.
.
}a2)+
(al
+
(b1
f b2)e
b1b202
+
(alb2
+
a2b1)0
.
And
2
= 1
+2 +2N5 =0 +1.
So
From theorems
2. 1, 2.
5
and 2.
9
+
b lb2)0.
and the fact that we are
using complex number multiplication so can have no proper divisors
of
is an integral domain.
zero it follows that Ra[ 45]
Theorem
2. 10.
is a basis of
01, 02
If
Ra[ 45]
,
the
necessary and sufficient condition that
with
a11,
a12, a21
01
=
a1101
+
a1202
02
=
a2101
+
a2202
and
a22
rational integers be a basis also is
all
a12
a21
a22
=
f1
16
Proof:
b..'s
'J
where the
So, since
If
01
=
b1101
+
b1202
02
=
b21 01 +
b02
22
are rational integers. Then
- (alibi].
+
a21b1201
+
02
=
(allb21
+
a21b22)01
+
and
(a
b
(a12b11
(a12b21
+
a22b1202'
+
a22b22)02.
are linearly independent,
02
+
a21b12
=
11 21 +
a21b22
=
b
is a basis,
01
01
allbll
a
01, 02
1
a12b11
0
a
+
b
12 21 +
a22b12
=
0
1.
a22b22 -
From these four equations, it follows
(bul
b21
b12
b
all
22
a21
b12
all
a12
1
0
a
0
1
22
)
Henc e
b11
a12
=
b21
b22
a21
1
a22
and the determinant of each matrix on the left divides
1
so is
17
either
+1
Thus
-1.
or
all
a12
t
=
a22
a21
is a necessary condition for
Since
el,
to be a basis.
02
is a basis, if
01'02
1
and
e
we have
ei
=
all
l
+
e2
=
a21e1
+
a
a
1202
2202
are rational integers.
where the a..'s
'3
Then
el
a12
all
el
e2
a22
a21
e2
a12
all
a12
a
a21
a22
02 -
e
1
all
a21
a22
all
a12
=
a21
a22
f
1
,
e2
are in Ra[ ^/5]
,
18
both
01
and
and thus all numbers of
e2
0i and
sed as linear combinations of
Ra[ 45]
can be expres-
with rational integral
02
coefficients. This establishes the sufficiency condition of the theorem.
Definition 2.6.
and
a
and
el
and
nant of the numbers
A(a,ß)
Theorem
2. 11.
invariant under change
and
a
form a basis for
02
are any two numbers
ß
Proof:
If
of the
Ra[45]
domain, the discrimi-
is
ß
a101
+
b102
a201
+
b202
a
+
b
a201
+
b202
2
=
1-U 1
102
0(01, 02)
of
where
01, 02
is a basis is
basis.
From definition
2.
6
2
01
02
81
02
2
al
a2
b1
b2
0(a,ß)=
and
2
G(01
By
theorem
2. 10,
0)
if
So
01
02
el
e2
a, ß
0i, 02,
is a basis
2
o(ei,e2)=
_
.(3s 102)
then
a1b2 -a2b1 = f1.
19
Definition 2. 7.
and denoted
Ra[ ^/5]
will be called the discriminant of
A(01,02)
L1[ N/5]
Theorem
2. 12.
Proof:
Take (1,
.
A[ N/5]
1
+
2
=
5
1
.
as a basis.
2 .Í5)
Then
2
L1 [ Nl
5]
2
=
1
Theorem
61, 82
2. 13.
be a basis for
Proof:
2,45
+
1
(2 -2í5
2
-2-
2
2^/5) =( -45)
22
-
A
necessary and sufficient condition that
=
5.
2^/5
I
is that
Ra[ '45]
A(01,02)
=
5.
That this is a necessary condition follows immedi-
ately from the last two theorems.
To prove it also sufficient, let
independent numbers of
where
0
62
Ra[
a
a
a
=
ß
=b161+b262
+
202;
is a basis and
al
± 1.
1,l
b2
all
and
3
be two linearly
which do not form a basis. Then
X15 ]
101
a
20
Since
al, a2, b1 and
are rational integers and
b2
and
a
ß
are linearly independent it follows that
a2
a
1
>
1
,
b2
bl
2
al
bl
b2
01
02
a2
>
1
2
2
A(a,(3)
Example:
a
+
aI
a2
>
_
0l
bß/5, c
+
bl
02
.
b2
is a basis for
dß/5
5.1
Ra[
N/5]
if and only
if
2
a+bN/5 c+dq5
2
= [
ac-5bd+(bc-ad)^/5-ac+5bd-(ad-bc)/5]
= [
2(bc-ad)q5]
a-bN/5 c-d^/5
=
20(bc-ad)2
=
2
5
or
(bc - ad)2
The pair
a and b
=
4
must be rational integers or both halves of odd
rational integers and likewise for the pair
infinite number
of
c
and
d.
So any of an
values will do, in particular the values
,
21
5
2+
Definition
(3,
denoted
For
2. 8.
2 +ZN/5.
ZN/5
a,
when there exists
131a,
a
and
y
ß
and
Theorem
2. 14.
If
Ra[ ./5] such that
and
a
is called
ß
is a multiple of
y,
a
each integer of a sequence al, a2,
, an
of integers
is a multiple of the one that succeeds it, each integer is
a multiple of every integer which follows it for any
If
rational integer n
c.
in
y.
is a multiple of ß and
is a multiple of y.
a
then
b.
y
is divisible by
a
,
are called divisors or factors of a,
multiple of
a.
Ra[N/5]
=ßY
a
13
in
ß
> 2.
two integers a and (3, are multiples of a third
integer y, then a + ßrß is a multiple of y where
and ri are any integers of Ra['./5]
If
.
Proof:
and YIP implies a = f ß and ß = r1y where
and r) are integers. So a = r)Y and the integers
being closed under multiplication, a is a multiple of y.
a.
(31a
b.
For n = 2 the theorem is obviously true Assume the
theorem is true for sequences with k terms, k> 2.
.
Let
a i+ l
al, a2,
l
ai'
,
i
ai
=
1
=
ak, ak +l be a sequence where
.k. Then
iak
i
=
1...k-1
(1)
22
by the induction hypothesis. Also by hypothesis
X
ak
So
from
ai
=
(1) and (2)
Xi
(Xkak+1)
or ai - Fliak +1
and taking
ak
=
ai
=
(2)
kak+ 1
µk
µkak
=
for
i =
1
k-1,
for
i
1
k -1
=
in (2)
Xk
+l
So
for i=
k+1
1k.
(3)
Also by induction hypothesis, each a. is a multiple
of each ai +j where i = 1. k- 1 and i +j < k so
from this and (3) each aj is a multiple of ai+j
where i = 1 k and i+ j < k +l .
Since the theorem is true for n = 2 and is true for
n = k+ 1 whenever it is true for n = k, then it is true
for all n > 2 .
c.
a = P1Y,
ß
a
=
P1'P2
P2Y,
+ ß r)
=
p a
+
integers
p2ar
1
=
and
a
+ 131
(Pl
+
is a multiple of
P2r1)Y
y.
;
23
Theorem
If
2. 15.
is divisible by
a
is divisible
NW),
by
Proof:
and
unit of
Ra[ N/5]
a
=
V!,
N(a)
=
N(I3)N(y)
An integer which divides
2. 9.
A
2. 16.
necessary and sufficient condition that an
integer be a unit is that its norm be
Proof:
If
divides
N(1)
or
EE
In the second
and
Therefore
= 1.
if
E
case
E
(
-1)
N(E)
1
=
is an integer and
-1. In the first case
=
±1.
is a unit it divides
E
Conversely,
= 1
is called a
1
.
Theorem
N(E)
by theorem 2. 4.
being rational integers it follows that N(ß )I N(c).
N(a), N(ß)N(y)
Definition
Er
N(a)
13,
and
-111
E
so by theorem 2.15,
± 1.
N(E)
=
±1,
then
is a unit by definition.
so by theorem 2. 14a
E
1
is a unit.
E
Corollary
2. 16.
The product of two units and the quotient of
two units are units.
Theorem
RaP./5]
I
.
2. 17.
There are an infinite number of units
of
24
Proof:
N(E)
=
-4 - 4
Consider
so
-1
=
=
E
21
+
1
Every positive power
is a unit.
E
which is an integer of Ra[45].
-2,45
of
is a
E
unit for
N(En)
according as
Now,
we have
since
n
E
=
n
is even or odd.
E , E
2, E 3,
E
m.
(-1)n
=
or -1
+1
=
are all different or for some
'
latter case
In the
E
n> m
which is impossible
n -m =1
Therefore every positive integral power
> 1.
E
[N(E )] n
_
of
is a
E
unique unit.
Theorem
if it is of the form
±E
number
of
where
E
A
2. 18.
n
is a unit if and only
Ra[^/5]
1/5)
= -1 (1 +
n is any
and
rational integer.
Theorem
Proof:
unit for
n >
O.
If n
=
2. 17
establishes the proof that
0, En= 1,
a unit.
If
then
n < 0,
is a
En
n
E
1
_
E
where
m
_
-n
m
and is thus a unit since the quotient of two units
> 0
is a unit.
To show that all units
unit. Then
where
a
integers,
-
and
E
1
and
b
are
-T1
of the
form
are units.
±E
If
n,
E
are rational integers or halves
let
=
E
be a
a +b^/5
of odd
rational
25
-
E
=
-a-bq5,
=
a-b^I5,
=
--a+bq 5.
1
T
-E
1
One of the above four units has both coefficients positive and so is
positive and greater than
1.
we suppose that one to be
three units will be
Either
E
-E
=
E
1,
of
and designate it
E
a +bß/5
-
and
E
or
n
There will be no lack
E
The other
respectively.
1
n< 1<
E
generality if
E
n +1
where
n
is a non
negative rational integer. If the latter case is true, then
E1
1
--n
<
<
(1)
.
E
E
Since the quotient of two units is a unit we may write
E1
n
=x +yq5
-
E
where
x and y
are rational integers or halves
of odd
rational
integers. Then
(x+ yg5)(x- y"./5)
Since by (1)
or
x
+
yq5
>
t
-
1
1
by theorem 2. 16.
,
Ix-yÑSI<1
-1< x-yq5 <1.
(2)
26
From
(1) and (2)
<2x<2+
0
To satisfy this, since
25
is a rational integer or half an odd rational
x
integer, its value must be
or
L
1.
But from (1),
if x
=
2,
y
must be positive and half an odd rational integer. No such value will
satisfy
(1).
x
If
=
y must be
1,
positive and a rational integer and
again no such value will satisfy (1).
Thus it is impossible that
n
E
<
E
E
=
E
<
En+1
E
holds. So
Then
E
E
.
= -E
-E
And since
n
n
1_± 1,
_:
1
E1
-:
1
E
-
f
E
-n
Finally
-
-E
_ -E
-n
Hence the theorem.
Definition
a
2. 10.
An integer of
Ra[45]
by only a unit factor is called an associate of
which differs from
a.
If an
integer is
not a unit nor zero and has no factors except units or its associates,
27
it is called a prime. An integer is composite if it has factors other
than units and its associates.
Theorem
2. 19.
there exist integers
Proof:
=
r+
r1,
b
a and b so
=
Let
s +
µ
-r
+
se,
of
a
=
where
2 (1 +b5) and
being rational integers nearest to
+ IA
r and s
Isil
µ-
N(á
0 =
<
r1
+
s10,
=-
Ir2 +r
s
- s 21 <
1
Then multiplying by
INT(ß)IIN(
a
._.
2
1
- 11)1
I
N(f3
- µ)I
µ
a
)
1.
1<
2
which is not zero since
I
P µI <
N(
N(ß)
we have
= p
=
Ra[J5]
that
then
ß
and setting
are integers
0
that
a
I
#
IN(P)I < IN(P)I
Ir11
Set
ß
of the domain such
p
P
ß
sl,
and
a
and
µ
=pµ+
a
a
If
RN+
P
IN(P)I < IN(R)I
3
#
0
28
Definition 2.11.
and
5
I
a,
then
51 ß
a,
If
and
ß
is a common divisor of
5
addition every common divisor of
a
ß
a
and
ß
ß
If in
.
5,
is
5
and denoted
).
ß
Theroem
2. 20.
If
and
a
are any two integers
ß
not both zero there exists a greatest common divisor
5
Ra[45]
of
of
and
a
such that
P
aµ +
where
5 =
of
Ra[g5]
of
and
a
divides
and
called the greatest common divisor of
(a,
are integers
6
a
p.
=
and
are integers.
ri
Proof:
If
P.
a L ß
If
ßr) = 6
a
=
0,
#
ß
0
is unique up to associates.
5
then
5 = ß
.
If
a
=
ß, then
and neither one is zero we may, without loss
generality, assume
I
there exists integers
and
p
a
N(a)
=ßp
I
>
I
o
N(ß
)
I
Then by theorem 2.
.
19
such that
where
+ a-,
IN(o-)I < IN(ß)I
and by continuing the process
ß
=c-p1+c1
0
=
0-1p2+0
(rk-3
=
c'k
6k-2
I
(rk -1pk + Gk '
IN(ß)I
IN(a-2)I < IN(cr1)I,
,
2pk-1+`rk-1
NO- 1)I <
'
I
N(crk-1) II N(°rk-2)
IN(0-k)I`IN(crk-1)I'
I'
29
IN(a)
is a non negative integer when
I
is an integer, so
a
in a finite number of steps the process will result in a
N(6k)1
= O.
Then 6k
tions successively
0
=
crk -2
'
6k-1
=
6k-2
crk = 0,
If
k-3
_ b
=
so
b
=
'
,
=
bpk,
pkpk-1
+
to obtain
6
aµ +
so
8(PkPk-1
6k
divides
and we may eliminate from these equak -3
b
ßrl
b
divides
ó'
+ 1)
,
Continuing, we see that the left member of
is a common divisor of
Since
divides
b,
That
by assuming
aµ
b =
so
6
bl
b
and
6k -2'
each of the above series of equations is a multiple of
b
such that
Crk
+
VI,
a
S.
Therefore
a
and
ß
a
and
P.
P.
and
any common divisor of
is a greatest common divisor of
is the only greatest common divisor may be seen
is also a greatest common divisor of
T hen
b
=K
ló1
by definition 2. 11, and
ó = K1
K2ö.
51
=
K2ó
a
and
P.
30
Then
hl
and
K2
N(5)
=
N(kl)N(K2)N(6)
1
="
N( kl)N(K2),
are units and
Example: To find the
and
6
g. c. d.
and
-2
of
N(S)
so
0
are associates.
81
and
+ 2N/5
13 -
7^/5,
note
that
IN(-2
2N/5)
+
and
13 - 7N/5
-2 + 2N/5
or
13 - 7,45
and
IN(-3
so
IN(-3+ ^/5)
16,
=
-11
IN(13 - 7N/51
+ 3N/5
=
(-2
-3
_
4
(-3 +N/5)
+ 2N/5)
1
+ N/5 +
+
76
=
(-3
- ^/ 5
4
+ N/5)
4-N/5)1 = 4
I
<
IN(-2
+ 2N/5)
I
In the same manner
-2+
So
-3
+ N/5
= (
-1 -./5)( -3
+ N/5)
is the g. c. d..
We may
write
2N/5
(1) (13 - 7N/5) + (3 - N/5)( -2 + 2N/5)
Definition 2. 12.
Two integers
=
-3
+
N/5.
are said to be relatively prime
if every common divisor is a unit.
Corollary
2. 20.
If
a
and
ß
are relatively prime, there
31
exist integers
µ
and
+TIP
[La
Proof:
integers
µl
a unit.
1.
=
By definition 2. 12 and
and
theorem
2. 20
there exist
such that
rll
µ1a
E
such that
T1
+
TIlß
=
E
Then
fl
la
1
+
Tlß
E
E
E
E
and the units being closed under division and the integers closed under
multiplication it follows that
µ¢
+rlß
=
1.
Theorem 2.21. If a prime
uct
of
aß
of two
integers
rr
of
Ra [45]
of the domain, then
divides a prod-
Tr
divides at least one
a.
Then by corollary
the integers.
Proof:
2. 20
Suppose
there exist integers
Tr
µ
does not divide
and
+Tri=
Multiplying by
ß
we have
1.
ri
such that
32
ßaN-
or since
and
Tr
I
Tr
ß
aß
I
Tr(X
integers
a2.
2. 21.
a1a2
a
Proof:
I
By
µ+ßTl)=ß
n
=
If a
prime
divides a product of several
Tr
it divides some one of them.
n
theorem
2. 21, if
Tr
I
a1a2,
corollary is true for the case
So the
true for
=ß
.
Corollary
Tr
+ßTr
Then if
k.
n
then
=
2.
Trial
Suppose it is
divides the product of k
Tr
or
+
integers
1
we may write without loss of generality
Tr
and either
Tr
divide
ak+ l'
of the
integers
I
(ala2
I
(ala2
or
ak)
ak)
Tr
i
ak+1
ak +l
or both.
If
Tr
does not
then by the induction hypothesis it divides some one
al, a2,
product of any
n
Theorem
,
ak.
k.
integers,
2. 22.
Thus the theorem is true for the
n >
2
.
Every composite number of
factored into a finite number
of
Ra[N/5]
primes, and this factorization is
unique up to associates.
Lemma
1.
can be
Every composite number of Ra [4-5]
factored into a finite number of primes.
can be
33
Proof:
#
a
a unit or a
be the proposition that every integer
P(n)
Ra[.5]
of
0
Let
where
I
=
1,
is a unit and
a
If
a
is a prime, then
If
a
is composite, then
is a unit nor an associate of
IN(a)I
I
N(ß)
=
I
=
I
and
I
N(ß)
I
I
N(y)
I
N(y)
I
I
P(1)
a.
a
So
=
ßy
I
N(K)I <
IMO
=
n
primes.
N(ß)
are less than
n.
where neither
11,
I
N(a)
I
nor
ß
N(y)
I
#
1
y
and
since
I
and the norms are rational integers.
Now suppose that every composite integer
I
of
is true.
is true for all
P(n)
is either
natural number)
n (a
prime or can be factored into a finite number
If n
both
N(a)
K
with
has a finite prime decomposition.
ß
and
Y
would then be
ß
=
ß1ß2...ßr
Y=lY2...Ys'
'
products of finite numbers of primes and
a =
ß1ß2...
a
product of a finite number
of
mathematical induction,
Lemma
2.
of
ßrViY2...
Ys
primes. Thus by the second principle
P(n)
is true for all
n > 1.
The decomposition of a composite integer into
primes is unique.
Proof:
Suppose there are two prime decompositions of
a,
34
say
a
So
er
=
Tr
and by corollary 2. 21,
X1
-
E
1711
1
where
ir2
r
<
s,
divides some
after
divides some
X
say
Then
X1.
is a unit and
E
7317
If
_
)
rrl
Tr2Tr3.
Then
r
.
.
Xs
X1X2.
=
Tr
1(Tr 2.
X1X2...Xs
=
zrr
=
X.,
say
TTr
4.
=
E
E
.
.
Xs
1X2X3.
and
X2,
1 2X 3X 4. .
Xs
.
steps we have
r
1=
..
E
E
rs-r
Xs
This implies
1
=
N(X
s-r )
and this is impossible as each
a
rational integer not equal to
impossible. Then r
=
s
1
=
N(X
s
X.
±1.
is a prime and hence
Similarly the case
r
> s
is
and
1E2...ES
and the prime factorization of a composite integer of
unique up to associates.
is
N(X.)
Ra[45]
is
35
THE NUMBERS
III.
Theorem
b
Ra(.- 13)
The set
3. 1.
Ra(N/ -13)
range independently over the field
of
Theorem 3.2. The numbers
a
=
+
where
b'J -13
a
and
rational numbers is a field.
and
a
of
a
Ra(N/-13)
1
satis-
fy a unique monic quadratic equation with rational coefficients.
Proof:
as does
a
=
a
+
satisfies
b^/ -13
(x-a)2
+
13b2
N(a)
=
a a
=
x2 - 2ax
a2
+
13b2
= 0
d.
Definition
3. 1.
.
Theorem 3.3. For every number
N(a)
+
a
#
0
Ra('J- 13),
of
is a positive rational number.
Proof:
Let
a
=
a
+
b'J -13
=
a
where
a
and
b
are rational
numbers. Then
N(a)
d
=
a2
+
13b2
The proof of this theorem as well as those of a number of others
1
in this chapter are essentially no different from the proofs of the
corresponding theorems of Ra(^/5) and will be omitted for sake of
brevity. For the same reason, only major theorems and definitions
will be restated in this chapter.
36
which is a rational number since
since
a2
and
are squares
b2
Theorem
3. 4.
Theorem
3. 5.
aß
N(aß)
Definition 3.2.
for
.
is an integer of
a
Ra ['J- 13]
Theorem
3
Every number of
.
.
Every rational integer is in Ra[g-13]
6.
Ra[g-13]
is in
a
A number of
3. 8.
.
which is rational is a rational integer.
Ra[g-13]
Theorem 3.7. If
Theorem
if its principal
Ra(^/ -13)
The set of integers will
equation has rational integral coefficients.
be denoted
numbers of Ra("/- 13).
a, ß
N(a)N(ß)
=
are, and positive
b
real numbers not both zero.
of
ß
a
=
and
a
and only if it is of the form
Ra('J- 13)
bq-13
a
+
+
b1'J -13
so is
,
where
is in
a
a
.
Ra[ \/-13]
and
b
if
are
rational integers.
a1
Proof:
with
al, bl
Let
and
a
-
Then the principal equation of
2
-
a
is in
a
is
2a1
-
1
and if
Ra('J- 13)
relatively prime rational integers and
c1
x
be a number of
c
Ra[ N/-13]
a12
x
+
13b12
= 0
+
cl
b1
O.
..
37
2a1
(1)
cl
a12
13b12
+
(2)
cl 2
are rational integers. One of the following cases must hold.
(i) c
1
or 2,
(ii) c 1= 2,
(iii)
c 1= 1.
Case (i) may be eliminated by exactly the reasoning which
disposed of the similar case of theorem
If
case (ii) should hold, then
a12 + 13b12
s0
mod 4. Then
b2
a12
.... O
1
and we have
al,
hypothesis.
If
b1
b1 E
mod
4
°
O
mod
2
1
2
b 12
°
a 12
°3
1
4
all with factor
c
mod
=
and from (2)
If
b1
- 0 mod 2
mod 4,
0
and
c 12
- -13b12 mod 4.
a 12 =
al
2. 7.
mod
4
,
mod 4
which is impossible by principle
1.
6e.
2
contrary to
38
Therefore case (iii),
c
=
holds and
1
and
a
b
are
rational integers.
Conversely,
if
a
=
a
+
rational integers, then 2a
and
and the principal equation of
a
Theorem
for
Ra
[
N/-131
3. 9.
and
bel- 13,
a2
13b2
+
and
a
b
are
are rational integers
has rational coefficients.
The numbers
and
1
J-13
forma basis
.
This theorem is an immediate consequence of theorem 3. 8.
Clearly Ra P4-13]
is closed under addition, subtraction
and multiplication and this, with theorems 3.
1
and 3. 6, and the
fact that multiplication is complex number multiplication and allows
no proper divisors of zero, show that
Ra [ q-13]
is an integral
domain.
Theorem
If
3. 10.
01
and
form a basis
02
of
the necessary and sufficient condition that
where the
01
=
a1101
+
a1202
02
=
a2101
+
a2202
are rational integers,
a..'s
ij
all
a12
=
a21
a22
±1
is also a basis is
Ra[ N/ -13],
39
Theorem
and
which
02
The discriminant of a pair of integers
3. 11.
forma basis
0
1
Ra [ N/-13]
of
2
01
02
e1
e2
° (A1, 62) =
is invariant under change of basis.
Theorem
Proof:
AN-13]
3. 12.
By
theorems
_ -
52
.
3. 9 and 3. 11,
2
1
AN-13]
A(1, N/-13)
=
01, 92
aIß
3. 14.
-13
=
- 52
-N/-13
necessary and sufficient condition that
is that
Ra [N/-13]
If
a
and
ß
A(01,02)
are members
=
of
- 52.
Ra[ "./-13]
then N(a)IN((3).
Definition 3.3.
only if it divides
units.
A
form a basis for
Theorem
and
3. 13.
N/
=
1
Theorem
-
A
number
of
Ra[N/ -13]
is a unit if and
1.
Theorem
3. 15.
a
Theorem
3. 16.
The product and quotient of two units are
is a unit if and only if
N(a)
=
1.
40
Theorem
are
Ra[ 4-13]
The only units of
3. 17.
+1
and
-1.
Proof:
and
x
and
Let
+
y q-13
N(E)
=
x2 + 13y2
=
1
are rational integers so x
y
Then
be a unit.
x
=
E
±1,
=
It was at this point in the development of
theorems leading to the proof
of unique
y
=
0.
Ra[ ^/5]
that
factorization were introduced.
It can be shown that the analogous theorems are not true in
Ra[ Nl-13],
but it will suffice to show that there is at least one composite integer
in
which does not have a unique prime factorization.
Ra[ .J -13]
First let
prime. For
us observe that
an integer of the domain, is
if not we would have
2 = (x +
with
2,
+
v4-13)
rational integers, and since
x, y, u and v
4
y'J -13) (u
(x2
=
+
13y2) (u2
+
13v2)
and either
x2
x2
+ 13y2 = 4
13y2
= 2
u2 + 13v2
= 2
+
or
u2
+
13v2
=
1
N(0)=N(a)N(13),
41
But the case on the right is impossible and that on the left has rational
integer solutions only if x
=
±2, y
=
0,
u
=
±1, y
= O.
So
2
is
prime.
In the same manner it can be shown that
1
-
q-13
7,
1+ NI -13,
and
are prime.
Now
consider
14
By theorem 3. 17 it is
=
2-7
=
(1+N/-13)(1-4-13).
clear that neither factor in the first pair is an
associate of a number in the second pair. Thus
prime factorizations.
14
has two distinct
42
Ra [4- 13]
THE IDEALS OF
IV.
In Chapter III, it was shown that the unique
Ra[J -13]
does not apply to the composite integers of
ter the concept
of
factorization law
.
In this chap-
ideal numbers will be introduced, and it will be
shown that unique factorization of an ideal into prime ideals exists.
Definition 4.
1.
An ideal of
is an additive sub-
Ra[ i- 13]
group of integers, which is closed under multiplication by all the
integers
of the domain.
al, a2,
If
,an is
then the set of integers
X
a
a set of
+
X2a2
n
+
integers
+
consists
of
all multiplicities
of a
=
(al' a2,
,
single integer
domain is called a principal ideal and denoted
Definition
write A
=
B,
4. 2.
Two ideals
A
and
,X
n
An ideal which
by integers of the
(a).
B
A
=
B
if and only if
are equal, and we
every integer defining
is a linear combination of the integers defining
B
2,
is clearly an
,
a).
a
X ,X
,
when every number of one is a number of the other.
It follows that
defining
Ra[Nl- 13]
where
Xnan,
range independently over the integers of Ra [N/-13]
ideal. We denote such an ideal A
of
B
A
and every integer
is a linear combination of the integers defining A using
integers of Ra[4-13]
as coefficients in both cases.
43
Example:
(2,3
(2,3
- ^/-13) =
- ^Í-13,
1
-
'/ -13). Obviously both
numbers in the left hand ideal are expressible as linear combinations
of the
1
-
N/
integers in the ideal on the right. And since
-13
2X-
=
1
+ 3 -
q-13, it is equally clear that all three integers
in the right ideal are linear combinations of those defining the left
hand ideal.
(al, a2,
Theorem 4.1. If
es
,
is an ideal, any one of the
an)
may be eliminated from the symbol of the ideal provided it is a
linear combination
remaining integers in the symbol. Likewise
of the
an integer may be placed in the symbol for the ideal if it is any num-
ber
of the
ideal.
Proof:
of
RaN/-13]
,
Let
al
and
a
a
where
X
1,
X
a
,
2,
= X
_
=
X
X
µ2a2
=
= X
+
lal
X
2a2
are integers
n
+
+
1µ2a2
+
+
(X1µ2
+µnap,
+
x
2)a2+
Ra[4-13]
+
+ X
lµnan
+
X
2a2 +
+
X
...
Then
Xnan
Ra[4-13]
of
µnan)
X
+
+
µ3,
N.2,
be any number of the ideal.
1(µ2a2
Since the integers of
µ3a3
+
.
.
Then
+
X
+
flan,
Xnan,
n)an
are closed under addition and
µn
44
multiplication, it follows that
(al'a2'...,an)
=
'
Similarly, any number
. .
(a2
,an)
ideal is a linear combination of
of the
the integers in the symbol for the ideal and including it among them
will not change the ideal.
Theorem
there exist two integers wl' w2
In every ideal
4. 2.
such that the numbers of the ideal are given by kiwi
kl
and
k2
c
a
is in A,
c2
A 0
a
=
al
Every number of an ideal
with
+ c2'ß/-13
-a
and
c
is in A.
Let
w2
A
of
be a number of
c2
is positive and minimal.
a2^/ -13
of
A
a - k2w2
and
r2
= 0
with
Then for any number
0
<r2 <c2.
=
al +a2^/ -13 - k2(cl +c2'./ -13),
=
al +(k2c2
+
al
+
=
is in A
A
If
we can write
a2= k2c2 +r2
Then
is of
Ra['4-13]
rational integers.
c2
in which
+
where
k2w2
are rational integers.
Proof:
the form
+
-
k2c
r2)ß/-13
-
k2(c1
2(c
1
+ c N/- 13)
2
r2^/ -13
otherwise the definition of
w2
would be
45
violated. So
is a rational integer.
a - k2w2 = b
Since for any
in
A,
Let
col
a
tive rational integers.
is in A,
a-a
contains posi-
A
be one of these which is least.
Then
So
b
=
w1k1+r1,
a- k2w2 - k10)1
=
b -
is in
A
r
and
integer in A.
0 <
=
k1w1
or else
= 0
r1
<
wl.
r1
is not the least positive rational
wl
Hence
a
Definition
4. 3.
=klw1 +k2w2.
A
pair of numbers
col, w2
theorem
4. 2
form a minimal basis for the ideal.
Example
1:
To show that
(7 -
i -13, -10
1
+ ^/ -13
is a basis for
one observes that any number of the ideal
2,4-13)
+
2,
derived as in
is of the form
(7
X
-,4-13)
+
A2(-10
+
2,4-13)
1
where
X
X
2
are integers
of
Ra['/-13]
,
and if
2,1+,4-13
is to be a basis for the ideal it must be possible to find rational
integers
k1
and
k2
which satisfy the equation
k1(2)+k2(1+^/-13) =(a+b,4-13)(2)+(c+d,4-13)(-10+24-13)
46
for all rational integral values
a, b, c, and d. Expanding and
of
equating coefficients of powers of
2k1
+
k2
7a
+
k2 - -a
+
=
one obtains the system
'.i -13
13b - lOc - 26d
7b
2c - 10d
+
and this is equivalent to the system
k
=
4a
+
3b - 6c - 8d
k2
=
-a
+
7b
+
2c -10d
which satisfies the requirments.
Example 2:
3,
there would be
1
rational integral value for
a
rational integer
3k1 + (1
Then
is not a basis for
+4-13
a, b, c, and
d
k1
1
+
í -13).
and
k2
If it was
for every
in
Nf-13)k2
=
(a +b'\1- 13)(3) + (c
3k1 + k2
=
3a
+
(3,
+ c -
13d
k2=3b+c+
d
+d./- 13)(1
+'.1 -13).
which is equivalent to
-3b-
14d
k2=3b+ c+
d
3k1= 3a
and it is impossible to find a
k1
which is a rational integer for
47
every rational integral value
Corollary
divisible by
If not
col
a, b, c, and d.
of
Every rational integer of ideal A
4. 2a.
.
there is a rational integer
and
in
a
rational integers. But then
r
A
a - w1k
and this contradicts the hypothesis that
in A
such that
0<r<col
a=co1k+r
with k
is
col
=
r
would be
is the smallest
positive rational integer in A.
Corollary
4. 2. b.
x
If
1w1 +
gives a number of A
for
Let
2w2,
A'
=
{X
A
=
{kiwi
Then AG A'
and
wl
X
+
1w1
So
and
X
2w2
A' GA.
lwl
+ X
+
k2w2
is a minimal basis for
col' w2
2w2
x
and
X
any integers of Ra['J-13]
X
1
,
X
X
2
integers of Ra[ .J -13]
are integers
of
A,
so
are integers of A for all
Then A
},
X
X
lwl,
X
X
Ra[s/ -13]
2w2,
and
from Ra[s/ -13]
=A'.
In exactly the same way as was done
can prove
.
rational integers}
k1,k2
since any rational integer is an integer of
w2
A, then
for theorem
2. 10, one
.
48
Theorem
two numbers of
A
4. 3.
necessary and sufficient condition that any
A
=
ul
w2
=
a
a
11
0.)
1
a12w2
+
21(01 + a22w2
a..'s rational integers and
with the
be also a basis of
A
a12
=
of
ideal A,
f1
.
a22
a21
4. 4.
basis
is
all
Theorem
a
col, w2
has a minimal basis of the
Every ideal A
form
k,p
where
0
k
+
r^/-13
is the smallest positive rational integer in
A
and
<p<k.
Proof:
Let
col =
k,
w2 =
m
+
r^/-13
mined by theorem 4. 2. Then
and
wl
= col =
are numbers
of
m =qk
+p,
k,
w2
- q col
A
and
= w2
0 <
=
p < k
p + r'./ -13
be a basis as deter-
49
1
0
-q
1
=1
w,
So by theorem 4. 3,
Theorem
basis
ra,
co* =
r
and
Proof:
p+
r'/-13
and
Every ideal A
4. 5.
0 < p <
w2 =
a
By
k.
r(b
theorem
=
ar
+
t,
0 <
t
r
<
=
kN/-13 - ap -
=
-ap
+ (k -
=
-ap
+
which by theorem
k
wl,
A,
4.
2
arq-13,
arN-13
,
tN/-13
is impossible unless
I
w1 =
+
=
.
r k. Then
p
a,
Set
kN/-13 - aw2
which case
and since
has a basis
A
4. 4,
the smallest positive rational integer in
k
k
Then
0<b<a, b2+ 13-0 mod
N/-13),
+
are positive rational integers.
with
= w2
is in A,
has a minimal
Ra[s/-13]
of
form
of the
where
is a basis of A
w2
r^/ -13
ra
is in A,
w2 =
p
so is
+
r^/-13
p^/ -13 - 13r.
t
=
0,
in
50
Set
tl
and
b
0<tl<r,
p=br+t1,
rational integers. Then
-13r
is in
+
pq-13 -bw2
and
A
-13r
=
_
-13r
=
(-13r
+
pq-13
- bp +
-bp-brq-13,
(p-br)^/-13,
- pb) +
tl,4-13
for the same reason t=
t1 =0
0
above.
r (b
+
Then
r p
i
and
w '1 =
is a basis for
Since
A.
wl
ra,
w*
and
wl
=
r
= w2 =
/-13 )
are positive by theorem
4. 2,
is positive and of course a rational integer. Since by theorem 4. 4
a
0<p <k,
0<rb <ra
Since
and
w24-13 - bw2
is a rational integer in
A,
0
=
<b <a.
r(b
+ ^/-
13)( -b
+
it is divisible by
v-13)
wl
=
=
rb2
ra,
- 13
r
according
to corollary 4. 2a. So
b2
Definition
4. 4.
+ 13
=
0
mod a.
The basis defined in theorem 4.
5
is called a
canonical basis.
Example: 2,
A
=
(2,
1
1
+^/-13
+ ^/ -13)
is clearly a canonical basis for the ideal
since the coefficient of
is the least positive coefficient of
^/ -13
s/ -13
is one and so surely
in any number
a
+
b i-13
51
of
A,
if not
and
then
is the least positive rational integer of
2
1
is in A
1
where
For
and so
=
2(x+y'./ -13)
+ (1
+^/ -13)(u
x, y, u and v are rational integers.
1
A.
=
0 =
2x
+
-13)
So
u - 13v
v.
2y+ u+
When the second equation is subtracted from the
-2y-
1= 2x
+ vN/
first
14v
is obtained, a relation which has no integral solutions since the left
number is odd and the right is even.
Definition
and
A
If
4. 5.
B
are ideals, the product AB
is the set formed by multiplying every number of A by every number of
B
and then taking all possible linear combinations of these products,
using as coefficients integers of
If
A =(wl,
w),
B
Ra [ N/-13]
(4J1,142),
=
.
then AB
is the set of all
numbers given by
klwll
where, by corollary
integers or integers
+ k2w1412 + k3w2th. + k4w2142
4. 2b,
of
kl, k2, k3 and k4 may
Ra[
.i
-13]
.
be either rational
52
It is evident that the product of ideals is an ideal of the same
domain, and that ideal multiplication is both commutative and associ-
ative.
Example:
(1 +N/ -13, 2
14,4
16 + 24-13,
Theorem
-
-4-13)
-i -13,
(2,3
1- N/-13)
=
(2 +2q -13,
24-13, -7, - 5'./ -13, -11 -34-13).
If
4. 6.
every number
of an
ideal A
Ra["./ -13]
of
is replaced by its conjugate, the resulting set is an ideal of Ra[4-13].
Proof:
If
(al, a2,
,
-
an)
is an ideal, then any number
of the ideal is given by
a
with the
A
=X1a1 +X a2
integers
.'s
+
Ra[4-13]
of
+Xnan
and
J
a
= X
1
and since the
X
.'s
1
+'2a2
range over
+
.
. .
+
Xnan
so do the
Ra['./ -13]
-X
.'s
and
(1,2,...,an)
is an ideal of
Ra[ q-13]
Definition 4.
6.
.
The ideal defined by theorem 4.
the conjugate ideal. The conjugate of
A
6
is denoted A.
is called
a
53
Corollary
is a basis for
á
Let
since
A
=
If
4. 6.
be any number of
k2
A
=
=
then
w1, w2
Then
7i..
is in A
a
and
=
klwl
+
k2w2
=
k1w1 + k2w2
(wl, w2).
Theorem 4.7.
AB
A,
some rational integers. Then
á
and
is a basis for
(col, w2)
and
k1
w2
A.
a
for
w
and
B
(col, w2),
B
A
If
are ideals
of
A B.
Proof:
Let
A
=
=
(q1,
Then
442).
AB
=
(w141, w142, w241, w2w2)
AB
=
(w1,1, w142,
_
(wl`P1'w14 ,w2`í1'w2
_
(w1w2)(4j1442)
and
=A B.
w2441, w242)
)
Ra[' -13]
,
54
Theorem
Ra[ ^/ -13]
If
4. 8.
A
(ar, r(b
=
Any number of
Proof:
ka2r2
b2
+ 13
.k , X
=
0
N/
is an ideal of
-13))
then
,
A A
where
+
,
+
and
µ
mod a.
=
+
range over
Set
b2 + 13
x
=
.
is given by
AÁ.
Aar2(b+\/-13)
v
k1,
(r2a)
=
µar2(b-q-13)+
Ra[ \/ -13]
ac,
c a
=x1+v1, µ=x
.
v
By
r2(b2+
13)
theorem
4. 5,
rational integer, and let
v=µ1
1,
Then the set of numbers AA is given by
k1
a2r2+X 1ar2(b+q-13)+v1 ar2(b+N1-13)+X lar2(b-^/-13)+µlr2ac
=
I(1a2r2
+ 2X
1abr2+µ1ar2c+v1ar2(b+q-13).
Conversely every number in this second set is a number of AA.
3W
so
b2 =
0
or
b2
1
mod
+ 13
=
and
4
or
1
but
b2+
so it is impossible for both
a
13
and
2
13
-
1
mod
4
mod 4
=ac
c
to be even.
Let
d
be the
55
Then
of a, 2b, and c.
g. c. d.
both even, and
b2
So
since
1,
=
13
2
+ 13
°
ac =
- 13
°
0
K1r2a2
by some integer of
r2a
Since
a, 2b and
xa
xr2a2
+
2.
.
1(2r2ab)
+
µ1r2ac
Then
1.
is a multiple
.
are relatively prime rational integers,
c
+
+ X
x,y and
2yb
+ zc
2ybr2a +zcr2a
such that
z
=
1.
=
r2a,
So every number which is a multiple of
Ra[/ -13]
mod d
mod d2
Ra[NI -13]
there exist rational integers
Then
0
has no square factors other than
any number of the set
of
are not
c
di b.
Then
and d
is odd since a and
d
r2a by an integer of
2 2
is a number in the set
k1r2a2
+ X
2
12r ab
+µ1r 2 ac and
we have
(r2a2, 2r2ab, r2ac, r2a(b+ q-13))
=
(r2a, r2a ( b+ q-13))
.
But every number of the ideal on the right is clearly a number of
(r2a)
and conversely. So
Definition 4. 7.
the norm of
A
AA
=
(r2a).
The number r2a of theorem 4.
and written
N(A).
8
is called
56
Theorem
The norm of the product of two ideals is equal
4. 9.
to the product of their norms.
Proof:
(N(AB))
Then by definition
N(A)N(B).
the norms in
=
AA
=
=
(N(A))(N(B))
are mult' pies
(N(AB))
and
BB
(N(A))(N(B)) is a multiple of
any number of
4. 5
N(AB)
Then
N(AB).
of
divide each other, but since
N(A)N(B)
are rational integers it follows that
Ra[ N/ -13]
N(A)N(B).
Theorem
Proof:
and
4. 10. If A, B
Let
are ideals
S
SA
=
SB
A
=
B
then
A
AB
AB
The numbers of
it must be that
N(AB)
=
Raki-13]
and
,
.
basis for
col' w2 be a
of
Then any number of
A.
is given by
X,
where
X
(s)
are given by
Any number of
µsX
where
/I
of the
form
l+
,1
c,2
are integers of Ra[^/ -13]
X
the numbers of
Ra[^/- 13] ,
w
112
s a
lwl
where
a
is in
N(S)
µ
_=
s
then
ranges over
is given by
µsX 2cä2
are in Ra['/ -13]
Let
where
µs,
(s)A
+
.
.
So
A.
=
r)saal
+ 112sw2
every number of
If
(s)A
is
57
=SB,
SA
then
SSA
=
SS B
(s)A
=
(s)B
and for every number
in
a
,
there is
A
a
number
in
ß
B
such that
and
sa
=
sß
a
=
ß
and conversely. Thus every
so
A
A
=
denoted
A, B, and
If
4. 8.
and every
B
CIA.
C
and
If
A
and
and
B IA.
4. 11.
if and only if
Proof:
If
are ideals
C
divides
BC, we say that
Theorem
AI C
B
is in
ß
A,
B.
=
Definition
and
is in
a
every number
A and
divides
C
are called factors
B
are ideals
C
of
C
is in
of
A,
A.
Ra['J -13]
of
,
A.
then there exists an ideal
AI C,
Ra[ ^/ -13]
of
B
of R.a['/ -13]
such that
AB=C.
Let A
=
(w, w),
B
=
(,
Xlwlgl +X
where the
's
42).
Then any number of
2w11142+X3w2g,1
are in Ra[ N/ -13]
.
C
is given by
+4w24/2
But this can be written
58
(X
or
+
111'1
2Vw1
X
(x lwl + x 3w2)
+ (X
34'1
+ (A
+
+
2wl
'P1
4442)w2
X
x 4w2N2
first arrangement every number
so in the
of
the second arrangement every number of
every number
Now suppose
number
of
is in
CA
AA
integer. Then all numbers
ranges over
Ra[ ./-13]
every two numbers
133a,
ß4a
and
ßla +ß2a
for every
X
and so the set
=
and
131a
of
ß3a,
ß1
+ß2 =ß3'
of
all the
AC
and by theorem 4.
Theorem
=
10
4. 12.
.
there are numbers
ß2a
=
la= ß5a
ß4a,
XP
ß1
ß's
(a)B
ß2
-ß4'
i`ß1 =ß5
is an ideal.
=
AAB
=
AB
Then
.
positive rational integer
a finite number of ideals of
so for
So
CG
A
,
P
such that
131a -
Ra[ ^/ -13]
where
13a,
Ra[ s/ -13]
CA
of
B.
is some positive rational
are given by
ß 2a
CA
in
B
CA
a
and in
Then every
is in A.
C
is an ideal of
CA
.
ß 5a
of
is also in
C
where
(a)
=
of
is in A,
C
Ra[ i -13]
.
t
occurs in only
59
Proof:
Let
ra, r(b
t and let
corollary
4.
Ra[ ^/-13]
and by theorem 4. 5,
and
Theorem
An ideal
4. 13.
rem
By
4. 8.
divides
(a)
=
theorem
where
4. 11,
a
A
ideals which contain
Definition
of
a, b and r,
of
Ra['./ -13]
Ra[N/ -13]
a
is divisible by
.
is a positive number, by theo-
is in A
But by theorem 4. 12,
A.
a.
ideals which can contain t.
t3
AA.
are positive
is positive or zero but less than
b
only a finite number of ideals of
Proof:
Then by
r and a
there are no more than t possibilities for each
thus no more than
which contains
be a canonical basis of A.
'J-13)
+
rait
2a,
rational integers
So
be an ideal cf
A
and in every ideal which
there are but a finite number
of
a. Hence the theorem.
4. 9.
An ideal which divides every ideal of the
domain is called a unit ideal.
Theorem
Proof:
4. 14.
The only unit ideal in
Ra['./ -13]
is (1).
The ideal (1) is the set of all mutliples of
numbers of Ra[N/ -13]
ideal of Ra[q-13]
and thus is the set
consists
of
Ra[ q-13]
.
1
by
Since any
numbers from Ra['.J -13] only,
every ideal is divisible by (1).
Let
A
be any ideal of
Ra[ N/-13]
which divides all the
60
ideals
of the domain.
every number
Then
of (1) is in
But then, by theorem 4. 11,
Al (1).
and
A
A
(1).
=
An ideal, different from the unit ideal, and
Definition 4. 10.
divisible only by itself and the unit ideal is called a prime ideal.
Every ideal not prime is said to be composite.
Example: (2,
ideals
A
=
1
+
N/
a2,
(a
is a prime ideal.
-13)
,
(2,
But then A
Let
ai
=
a
and
B
+ bN/ -13
- b
+
N/
-13)
=
(ß
=
AB
l'
13
2'
,
so we may write
A
=
(al,
...
,
an, 2,
1
+
q-13),
B
=
(ß1,ß2,
...
'Pm, 2,
1
+
NI
al, a2,
be any integers
=
b(1 +N -13)
such that
)
.
+ ^/-13)
1
ß
+a
'
,
-13).
Then
an.
- b
is a rational integer so
or
in the
1
B
both divide (2,
a.
and a
and
an)
there would be
If not
first case
a.
q-13)
a.
=
b(1
+
+
2c
a.
=
b(1
+ N/-13) +
2c
+
1;
is a linear combination of
2,
1
+
N/
-13
and
61
so may be dropped from the symbol for
In the second case we
A.
have
a. - b(1
and
A
=
= 1
may be inserted in the symbol for
1
(1).
Since
A
(1).
or
Ni-13) - 2c
+
=
in which case
A,
was arbitrary we find that either A
ai
Similarly for
(2,
1+ .i -13) =
(1)(1)
=
=
(2,
1
+ s/ -13)2
or
=
(2,
1
+ N1-13) (1)
or
=
(1)(2,
(2,
1
definition
(2,
1
+
+
NI
4. 4,
-13) # (1)
(2,
and (2,
1
1
1
+I -13)2
+q-13)2
+q-13) # (2)
=
since
two equations can be true and
An ideal
and B if
G IA
divides
G.
G
is not in
1
(2,
1
+
q-13). Also
since
(4,
1
2 +
+
(2,
2v -13)
q-13
1
+
- 12 +
2v -13)
is prime.
N/-13)
=
(2)
So only the
last
is a prime ideal.
is the greatest common divisor of the ideals
4. 15.
Every pair of ideals A and
have a unique greatest common divisor.
a + 3
+ ^/ -13).
A
and GIB and if every common divisor of A and B
Theorem
bers
1
for it was shown in the example following
that the integer
i -13) # (2,
+ ,4- 13)
(1)
or
But
1
either
So
B.
(2,
=
where
a
ranges over
A
B of
Ra[4- 13]
It is composed of all num-
and
Proof: Consider any two numbers yt
ß
and
ranges over
y2
of the
B
62
set
G
B.
Let
numbers of the form
of
=a1 +ß1
Y1
Then
1
and
a +
y2
a2
=
with
3
X
y2
of
a + ß
A
So
G
is closed under
is in
G.
ß 2).
=
RaN-13] and
of
in
and
in
ß
+ß2.
(al ± a2) + (13 11
addition and subtraction, and of course
Y1
a
0
Also for any
G
ñ.(a+ß)=Xa+Xß
is in
G.
So
is an ideal of
G
Every number
in
G
is a common divisor of
Let
C
be any common divisor of
and
CI G
K'G'
and
domain. So
and
=
K'KG
K(1),
G
and
G
G'
=
=
where
KG
K'KG.
By
so
K
=
K'
4. 12.
=
theorem
(1)
G
and
Two ideals
greatest common divisor is
1:
K
G
d's of A and
K'
4. 14,
G
=
(3
)'s
C
of
B.
of
B.
=
(1)G
so
But then
K'(1)
.
are relatively prime if their
(1).
the g. c. d of (2,
1
- ^/-13)
G.
Then
are ideals of the
K'K.
G'
(a + ß
Then
B.
and
=
B.
all the numbers
g. c.
and, by theorem 4. 10, (1)
Definition
Example
are two
G'
is
B
B.
and
A
is a g. c. d. of A and
G
Suppose
(1)G
A and
and
A
is closed under addition and so contains all the
Thus
G =
of
a
.
and every number of
G
so
G,
contains all the numbers
C
is in
A
of
Ra[ ^/-13]
and
(4,
3 + ^/-13)
is
63
the set
xi(2) +X2(1- './-13) +X3(4) +X4(3+ J-13)
where the
X
's
range over
G
(2,
=
=
since
1
Example
-
N/
-13
1-q-13, 4,
(2,
1
(2,
3 +
-
q-13,
So
.
+'4-13),
3
3 +
q-13),
J -13)
2 (2) - (3 + ^/ -13).
=
(2,
2;
Ra[ -13]
1
+q-13)
and
(3, 4
+
N/
are relatively prime
-13)
since
G
Corollary
N/-13, 3, 4 + N/-13),
=
(2,
1
+
=
(2,
1
+N/-13, 3, 4 +
=
(1).
4. 15.
and
A
If
B
which are relatively prime, there is an
q-13,
1
) ,
are two ideals
a
in
A
of
and a
Ra[^/-13]
R
in B
such that
a + ß
By theorem 4. 15,
numbers
tion 4.
12
a + ß
this
where
g. c. d is (1),
=
1
and
B
have a g. c. d composed of all
is in
A
and
A
a
so
1
ß
is in
B.
By defini-
is a number of the g. c. d.
64
Theorem
B
or
AI BC,
If
4. 16.
then A divides at least one of
C.
Proof:
for some
Suppose
in
a
A
is prime to
and some
A
in
ß
Then
B.
for every
So
B.
a + ß
=
Y
1
in
C
Ya+Yß =Y.
Since
Ya
the numbers
A BC,
I
is in A
so
ya
Corollary
+ Yß
4. 16.
If a
of
Yß
is in A.
BC
are in
Then
A
is in
y
and of course
and A IC.
A
prime ideal divides a product of ideals,
then it divides at least one of the ideals making up the product.
The proof is similar to that of corollary 2. 21.
Theorem
factored into
a
4. 17.
Every composite ideal
of
can be
Ra[ 4-13]
finite number of prime ideals, and the factorization
is unique except for the arrangement of the factors.
Proof:
are ideals
A
If
and
is any composite ideal of
C
B
of the domain,
C
Either
A2.
A
=
Ra[ NI- 13]
neither equal to (1), such that
AB.
is prime or it can be decomposed into factors
Then each of these is prime or it can be decomposed.
process is finite by theorem
there
,
4. 13.
The factor
B
Al
and
The
can be treated
65
similarly. Thus the ideal
C
has a finite prime factorization.
The proof that there is a unique factorization into primes
on
corollary
4. 16, and is
similar to that
of
theorem 2.22.
Example: In Chapter III it was shown that, in Ra[ './ -13]
14
factors into two sets
(1 +
N/
prime integers,
of
(2) is not
2.7 and
=
=
(1 +
i-13)(1
The ideal
q-13).
-
(2)
=
(2,
1
+
N/-13)2.
(7)
=
(7,
1
+
N/-13)(7,
1
- N/-13),
=
(7,
1
+
N/-13)(2,
1
+
^/-13),
=
(7,
1
-
^/-13)(2,
1
+
4-13),
(1 + N/-13
)
(1 - N/-13)
It was shown in the example
is a prime ideal.
-
(2)(7)
prime for
Similarly
1
the integer
,
-13)(1 - J-13).
New consider (14)
(7,
rests
after definition
In the same manner
(7,
'/ -13) can be shown to be primes.
(2)(7)
=
(2,
1
+
N/-13)2 (7,
1
4. 10
1
+
that
(2,
I -13)
and
1
+
i -13)
So
+
ti/-13)(7,
1
- N/-13)
and
(1 +
N/
-13)(1
- n/-13) = (7,
1
+./- 13)(2,
1
+
./- 13)(7,
1
4- 13)(2,
1
+
'/ -13).
Thus (14) has this decomposition into prime ideal factors which by
Theorem
4. 17 is
unique.
66
V.
A NECESSARY AND SUFFICIENT CONDITION
FOR UNIQUE FACTORIZATION
It has been shown that the introduction of ideal
numbers into
of unique
prime factor-
In much the same way, the ideals of the domain
Ra['.f5] may
the domain Ra[',/ -13]
restored the property
1
ization of the integers of the domain.
be discussed and theorems analagous to theorems 4.
1
through 4.
17
derived. In addition we have
Theorem 5.1. Any ideal
Proof:
Then
wl
2. 20,
and
w1
71
of
and
and
Let
w2
w2
Ra[45]
of
is principal.
RaN5]
be a basis for any ideal of
(wl, 02)
are integers
6
Ra[45]
.
and, by theorem
Ra[45]
of
have a g. c. d
and there are integers
µ
such that
p.wl +
Then
A
w2 =
(w1, w2)
=
6'
(w1, w2, s
by theorem 4. 1. Thus every ideal of
)
=
(s
Ra[45]
)
is principal.
That a similar theorem does not hold in
shown by considering the ideal
(2,
1
+ ^/ -13).
Ra['J/ -13]
If
can be
this ideal is
Actually, it is the principal ideal generated by the integer which
1
has this property.
67
principal, there must be an integer
any
X1,
X1(2)
and in particular a
a
divides both
so their g. c. d
(2,
+ 1/-13)
1
X2(1
2
+
pla,
p
of the
^/-13)
=
pa
+
1
./ -13.
But each of these is prime
and it must be that
*1
This contradicts the fact that it has already been
shown in a previous example the ideal
tain the integer
domain such that
+^/- 13 =p2a.
1
and
a
such that, for
such that
p2
can only be
b
=(1).
+
and
pl
2 =
Then
there is
Ra[A./-13] ,
of
X
Ra[./ -13]
of
a
(2,
1
+ ^/ -13)
does not con-
1.
The foregoing theorem and example suggest a possible con-
nection between the existence of a unique factorization law and the
form
of the
ideals
of an
integral domain. It can be shown, in fact,
that any quadratic domain
has a unique prime factoriza-
Ra['/m]
tion law if and only if every ideal
of
the domain is principal.
Lemma. 1. A necessary and sufficient condition that two num-
bers
6
a
and
ß
of
Ra[ qm]
such that
6 =
X , p.
have a greatest common divisor
in
Ra[ ^/m]
Aa
is that the ideal
+
µß
(a,
3 )
be principal.
68
Proof:
theorem
5. 1,
If
and
a
that
(a, ß)
Conversely, if
domain there is a
a
and a
X
Xa
p6
=
Then in particular there exist
Also there is a
1
is a g. c.
6
2. 22
of
of the
that
p
such that
=p26.
ß
=
d of
and
a
so that
µ1
a+µ1ß
It can be shown in a manner
theorem
and
1
and a
X
X
Therefore
p
µ
6Iß.
and
a
6
X ,
.
p
a=p16,
So
and for any
of the domain such
µ
µß
+
then for any given
(6),
=
of the domain;
p
it follows as in
6,
is principal.
(6)
=
(a, ß)
there exists
Ra[ s/m]
have a g. c. d.
p
6.
ß
.
similar to that used for lemma
1
that any integer of any quadratic integral domain has a
finite decomposition into prime factors.
Lemma
up to
2.
Prime factorization in Rai ^/m]
associates, if and only
prime,
TT
I
ßy
Proof:
implies
If
Tr
I
ßy
Tr
I
P
if for
or
implies
TT
Tr
ß
,
,y
is unique,
Ra[ qm]
in
,
Tr
a
k.y.
Tr
I
ß
or
Tr
I
y
then uniqueness
69
of
prime factorization follows as in Chapter II as demonstrated by
corollary
theorem
2. 21 and
2. 22.
factorization into primes is unique and
If
13
where the
y
Y's
and
ßß's
3 1
1
° 2.
So
Tr
I
or
[3
Theorem
Tr
5. 2.
torization into primes
is one of the
Tr
I
then
ß y,
Ym
are primes. Since
prime factorization is unique,
v .' s.
.. RnY12. ..
Tr
is a prime and
Tr
or one of the
ßß's
k.
A
necessary and sufficient condition that fac-
of
integers
of
RaN/m]
be unique is that
every ideal shall be principal.
Proof:
be any pair of relatively prime integers
a, ß
and suppose every ideal of the domain is principal.
Ra[qm]
of
Let
Then by lemma
1,
there exist
Xa
Then
+
µß
yX.a +Yµ13
for any
y
in
Ra['/m]
If
.
and
X
=
p.
in
Ra['./m]
such that
1 .
= Y
al ß y
then
a ly,
and there is
unique factorization by lemma 2.
If
Ra[\/m]
By lemma 2,
TrI
Y
has unique factorization:
if
Tr
is a prime and
Tr
I
RY,
TO or
70
If
is a prime
Tr
where neither A nor
any
ß
But if
is
B
Tr
I
Let
which means
Tr
every number of A is a multiple
a
B.
P
or
A
So one of
=AB
(Tr)
Then for any a in A and
(Tr).
in B a (3 is a multiple of
Similarly for
a
is a prime ideal, otherwise
(Tr)
is
B
be any prime ideal of
Tr
Tr
of
a
Tr
or
Tr
so
A
I
ß
=
.
(Tr).
a contradiction.
,
Ra[gm]
I
Then any number
.
P may be written
of
e
a
=
en
e
Tr2
Tr
. .
.
Trn
1
the
Tr'
s
being primes of
and the e's natural numbers.
Ra['/m]
Then
e
(a)
and each
number of
(Tr.)
P
= (Tr
)
2
1
(Tr
PI (a).
...
Then
P
A
the
=
of
P1P2
n) n
is one of the
for ideals there is unique factorization, and
Also, any ideal A
(Tr
But every number of
is a prime ideal.
so
2)
Ra NIm
.
P
(Tr
is a
(a)
.)'s
since
is principal.
factors
Pn
P's being prime ideals and thus principal ideals. Clearly the
product of principal ideals is itself a principal ideal, so A
principal.
is
71
BIBLIOGRAPHY
1.
Birkhoff, Garrett and Saunders Mac Lane. A survey of modern
algebra. Rev. ed. New York, MacMillan, 1963. 472 p.
2.
MacDuffee, Cyrus Colton. An introduction to abstract algebra.
New York, John Wiley and Sons, 1940. 303 p.
3.
Reid, Legh Wilber. The elements of the theory of algebraic
numbers. New York, MacMillan, 1910. 454 p.